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DANSO, DANIEL BOAKYEi

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Table of conTenT PREFACE iii ACKNOWLEDGEMENT iv PROJECT viii TABLE OF GREEK ALPHABETS ix TABLE OF SYMBOLS x SETS 1 ARITHMETIC OF REAL NUMBERS 33 SURDS 38 BINARY OPERATIONS 61 INDICES 76 LOGARITHMS 85 FUNCTIONS 97 POLYNOMIAL FUNCTIONS 123 QUADRATIC 145 RATIONAL FUNCTIONS 173 SEQUENCES AND SERIES 198 TRIGONOMETRY 226 COORDINATE GEOMETRY ( THE STRAIGHT LINE) 284 COORDINATE GEOMETRY (THE CIRCLE) 315 CALCULUS (DIFFERENTIATION) 347 CALCULUS (INTEGRATION) 393 VECTORS 426 MATRICES 468 FUNDAMENTAL PRINCIPLES OF COUNTING, PERMUTATIONS AND COMBINATIONS 495 BINOMIAL THEOREM 509 PROBABILITY 513 BINOMIAL PROBABILITY DISTRIBUTION 529 MULTIPLE CHOICE QUESTIONS 534

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PREFACE

This book has been primarily written for students of Senior High Schools and Colleges to appreciate the importance of studying mathematics in everyday life. The book is also recommended for students who are studying for their West African Senior Schools Certificate Examination and Diploma Examination in Mathematics, undergraduates can also make use of most of the topics covered in the book. The author believes that errors are committed by humans and when corrected, lead to perfection. Therefore, all errors that come to the forefront of the author or project editors will be welcome and dealt with.

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acknowledgement My heartfelt gratitude goes to the First Author and Creator of the universe. I am thankful to the following people for their contribution towards the coming out of this book: Mr. David Danso-Abeam, Mrs. Becky Anaquah, Lydia Gaisei, Mr. Yaw Danso, Mr. Clifford Arthur, Madam Rosemond Sowah, Mr. and Mrs. Ayensu-Adu of Tema, community 11, all lecturers, especially Dr. Henry Amankwah, Dr. Ernest Yankson and Prof. K. Essel, at the Department of Mathematics and Statistics of the University of Cape Coast (UCC), The Pastor and members of Present Faith Ministry, Tema, current and past teachers of Aburaman Senior High School, Mr. Edward Mensah, Mr. Issah Usman, Edward Marfo, Timothy Abban, Bernard Nyame Annor Yeboah and all who finance the printing of this book.

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PROJECT: Advisor: Mr. David Danso-Abeam Editors: Kataboh Kingsley Pascal, Richmond Essieku, Margaret Hayford and the late Alberta Asare Typesetters: Audrey Mac-Canaan Ayensu-Adu, Mrs. Mabel Mireku, Abigail Opoku Secretary: Sanahu Abdul-Karim Printed by: Abundance of Grace printing press Cape Coast Tel: +233547204084 Cover design: Isaac Oduro of Abundance of Grace Printing Press

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Regional Correspondence: Region Name Mobile Number Greater Accra: Kataboh Kingsley Pascal +233246360068 Eric Oppong +233276168644 Madam. Rosemond Sowah +233242178888 Mrs. Doris Ayensu-Adu +233576862724 Hariet Owusu Ansah +233243124066 Ashanti: Opoku Danquah +233540998199 Opoku Abigail +233247232919 Appiah Kojo Ennim +233243913095 Sarah Agyapomaa +233248052905 Central: Richmond Essieku +233246762009 Afred Nana Kyei +233548934030 Western: Theresah Ayebeng +233266741035 Patrick Agyei Yeboah +233546418187 Shadrack Forson +233267032337 Eastern: Vincent Agyenim Boateng +233545344675 Volta: Samuel Gbeti Doe +233248890717

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For bulk purchase, comments, enquiries and tuition contact the above correspondence or The author: Danso, Daniel Boakye e-mail: dboaks@rocketmail.com Post: P. O. Box Ce 11015 Tema. Tel: +233246627439/+233236106765 © Copyright 2015. All rights reserved. Seek the permission of the author before the reproduction of this work in any format.

ISNB 978− 9988− 2 − 1884 − 3

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To

The late

Stephen Danso-Abeam and Ruben Arthur

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Table of Greek lower case alphabets 휶 − alpha 휷 − beta 휸 − gamma 휹 − delta 휺 − epsilon 휻 − zeta 휼 − eta 휽 − theta

휾 − iota 휿 − keppa 흀 − lambda 흁 − mu 흂 −nu 흃 − xi 흄 −omicron 흅− pi

흆 − rho 흈 − sigma 흉 − tau 흊 −upsilon 흓− phi 흌 −chi 흍−psi 흎−omega

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Table of symbols

Symbol Meaning Symbol Meaning ℕ ℕ ℕ ℤ ℝ ℚ ℚ′ 핊 ℙ 피 핆 ∈ ∉ 퐴′ ∪ ∩ ⊂ ⊆ ∅ :/∣ ∀

Set of natural numbers Set of positive natural numbers Set of negative natural numbers Set of integers Set of real numbers Set of rational numbers Set of irrational numbers Set of square numbers Set of prime numbers Set of positive even numbers Set of positive odd number An element of/a member of Not an element of/not a member of Compliment of set A Union Intersection Subset Proper subset Null/empty set Such that For all

= ≠ ≡ ≢ ≈ ≉ + − × ÷ ! ∎

푥 ≤ 푎 푎 ≤ 푥 푥 < 푎 푎 < 푥 ∴

퐿퐻푆 푅퐻푆

⟺ ⟹ ⊥

w. r. t

Equal to Not equal to Identical to Not identical to Equivalent to Not equivalent to Plus/addition Minus/subtraction Times/multiplication Division Factorial End of proof 푥 is less than or equal to 푎 푥 is greater than or equal to 푎 푥 is less than 푎. 푥 is greater than 푎 There fore Left Hand Side Right hand side With respect to If and only if Implies/implication Perpendicular to

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SETS A set is a well-defined collection of objects or elements of similar characteristics. Sets are usually denoted by upper case letters such as 퐴,퐵,퐶 e.t.c. The elements/objects also called 푚푒푚푏푒푟푠 of a set are listed in a pair of curly braces separated by commas. For instance, if we say that a set 퐴 is the set of the vowels in the English alphabets, we write

퐴 = {푎, 푒, 푖,표,푢} We will choose the members of a set from the 푟푒푎푙푛푢푚푏푒푟푠푦푠푡푒푚 Real numbers 1. Natural/Counting numbers Natural numbers are numbers starting with 1 to infinity. The set of natural numbers is denoted by:

ℕ = {1, 2, 3, … } 2. Integers Integers are positive and negative natural numbers including the element 0. The set of integers is denoted by:

ℤ = {… ,−3,−2,−1, 0, 1, 2, 3, … } A careful study of the set reveals the following set of real numbers.

Even numbers Even numbers are integers that are exactly divisible by 2. We will denote the set of positive even numbers by:

피 = {2, 4, 6, … } Odd numbers

Odd numbers are integers that leaves a remainder of 1 when divided by 2. We will denote the set of positive odd numbers by:

핆 = {1, 3, 5, … } Square numbers

If 푥 is an integer, then the square of 푥, written as 푥 , refers to the number that is formed by multiplying 푥 by itself. We will denote square numbers by:

핊 = {1, 4, 9, 16, 25, 49, 64, … } Prime numbers

A Prime number is an integer with only two distinct factors; 1 and the number.

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Definition Factors of a number,푥, refers to numbers that exactly divides 푥. For instance, the positive factors of the number 6 are 1, 2, 3 and 6 because 1, 2, 3 and 6 exactly divides 6. Definition Multiples of a number 푥 are 푥, 2푥, 3푥, 4푥,…. For instance the multi- ples of 2 are 2, 4, 6, 8… We will denote the set of positive prime numbers by:

핡 = {2, 3, 5, 7, 11, … } Method for finding finite number of prime numbers

This method is called풕풉풆풔풊풆풗풆풐풇푬풓풂풔풕풐풕풉풆풏풆풔. Let assume that we want to find the first 10 prime numbers. First, the sieve suggests that we list numbers between 1 and 51 as follow: 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 Secondly, with the knowledge that 2 is the first prime number, we put an asterisk (*) on 2 and cancel out all multiples of 2. The next number not cancelled is prime and we put an asterisk (*) on the number. This is followed by cancelling all multiples of the number. We will do the same to the subsequent numbers until we get required number of prime numbers. The whole process is shown below:

2 3 4 5 6 7 8 9 10

11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

31 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50

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∴, the first ten positive prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29

3. Composite numbers Composite numbers are integers having more than two factors. 4. Rational numbers Rational numbers are numbers that can be written in the form , provided

푏 ≠ 0 and 푎 and 푏 Integers. For example, , ,− ,− e.t.c. Since every integer is divisible by 1, we conclude that every integer is a rational number. We denote the set of rational numbers by ℚ. 5. Irrational numbers Irrational numbers are numbers that cannot be written in the form , 푏 ≠ 0.We can describe them as real numbers that fall outside the domain of rational numbers. They include real numbers such as √2,√3, 2√3,휋, 풆 e.t.c.

Real numbers include rational numbers and irrational numbers. We will denote real numbers by ℝ.

Description of sets 1. Listing of members A set can be described by listing the members/elements of the set. For example, 퐴 = {2, 4, 6} or 퐵 = {1, 3, 5, 7, … }. The three dots after the 7 in 퐵 is read as ‘and so on’. They represent the missing members of 퐵. 2. Statements A set can also be described by a single statement. The statement gives the characteristics of the elements of the set. For example, the sets 퐴 and 퐵 above can be written as 퐴 = {Evennumbersfrom2to6} and 퐵 = {Positiveoddnumbers} 3. Set builders In set theory, the inequality signs <, >,≤, and ≥ are used with variables to describe the members of sets. We write푥 > 푎 (read ‘푥 is greater than 푎’), where 푎 is a constant

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and 푥 is the variable describing the elements greater than 푎. We also write 푎 ≤ 푥 ≤ 푏 (read ‘푥 is greater than or equal to 푎 or 푥 is less than or equal to 푏), where 푎 and 푏 are constant and 푥 is the variable describing the elements from 푎 to 푏. For example, we can write sets 퐴 and 퐵 as 퐴 = {푥: 2 ≤ 푥 ≤ 6} or 퐴 = {푥|2 ≤ 푥 ≤ 6}, where 푥is an integerand 퐵 = {푥: 푥 ≥ 1}, where 푥 is an odd integer. Set 퐴 is read as ‘푥 is such that 푥 is greater than or equal to 2 or 푥 is less than or equal to 6’. The symbols : and ⃓ mean 푠푢푐ℎ푡ℎ푎푡.

Example 1.1

List the members of the following sets i. 퐴 = {Primenumberslessthan20} ii. 퐵 = {Integersbetween1and12inclusive} iii. 푃 = {푥: 푥 ≥ 5,푥isaninteger} iv. 푄 = {푥: 3 < 푥 < 11,푥isaninteger}

Solution i. 퐴 = {2, 3, 5, 7, 11, 13, 17, 19} ii. 퐵 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} iii. 푃 = {5, 6, 7, … } iv. 푄 = {4, 5, 6, 7, 8, 9, 10} Try: 1. Write down the first 30 prime numbers. 2. List the members of the following sets i. 푋 = {Factorsof24} ii. 푌 = {Multiplesof3lessthan20} iii. 푍 = {푥:−1 ≤ 푥 < 10,푥isaninteger}

Definition of terms 1. Members of a set The members of a set refer to the distinct elements that belong to the set. If 푎 is a member of set 푃, we write 푎 ∈ 푃 (read ‘ 푎 is a member of 푃’). For example, given that 퐴 = {2, 3, 5,7}, we write 2 ∈ 퐴. Since 4 does not belongs to 퐴, we write 4 ∉ 퐴. The symbols ∈ and ∉ mean is a member/element of and is not a member of /is not an element of respectively. 2. Number of elements of a set.

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The number of elements of a set refers to the total amount of elements that belong to the set. For instance, the set 퐴 = {2, 3, 5, 7} has 4 elements and we write 푛(퐴) = 4 (read ‘number of A’) 3. Finite set A set is said to be finite if it has a known number of elements. In other words, a finite set is a set which last member is known. For instance the set 퐴 = {2, 3, 5, 7} is a finite set. It has 4 number of elements. 4. Infinite set A set is said to be infinite if it has unknown number of elements. An example of an infinite set is the set of integers

ℤ = {… ,−3,−2,−1, 0, 1, 2, 3, … } because it has an unknown number of elements. 5. Null or empty set. A null set is a set which has no element. The null set is denoted by {} or ∅. Note that {0} is 푛표푡 a null set and {∅} is not well defined. 6. Unit set or singleton A unit set is a set with only one member. For example 퐴 = {0} is a unit set. 7. Subsets and Proper subsets Let 퐴 and 퐵 be two sets. If all the elements of 퐴 can be found in 퐵 then we conclude that set 퐴 is a subset of set 퐵. We then write 퐴 ⊂ 퐵(read ‘A is a subset of B’). For example, given that 퐴 = {1, 2, 3} and 퐵 = {1, 2, 3, 4, 5}, we conclude that

퐴 ⊂ 퐵 If all the members of 퐴 are the members of 퐵, but some members of 퐵 cannot be found in 퐴, then we say that 퐴 is a proper subset of 퐵. We then write 퐴 ⊆ 퐵 (read ‘ is a proper subset of 퐵’). The symbols ⊂ and ⊆ mean 푠푢푏푠푒푡 and 푝푟표푝푒푟푠푢푏푠푒푡 respectively.

Determining the subsets of a set In set theory, if 퐴 is a set, then the first and last subsets of 퐴 are the null set and the itself. The other subsets of 퐴 are formed by writing the members of 퐴 as unit, pairs, triple and so on depending the number of elements 퐴 has. In general, if 퐴 has 푥 elements, then the number of subset 퐴 has is given by

푛(⊂) = 2

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Example 1.2 Given that 퐴 = {1, 2, 3, 4}, find the number of subsets of 퐴 and write down these sets.

Solution

We have 퐴 = {1, 2, 3, 4}. We see that 퐴 has 4 members so we write 푛(퐴) = 4 The number of subsets 퐴 has is given by: 푛(⊂) = 2 , where 푥 the number of members of 퐴

∴ 푛(⊂) = 2 = 16 The subsets of 퐴 are {}, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1,4}, {2, 3}, {2, 4}, {3, 4}, {1, 2, 3}, {1,24}, {1, 3, 4}, {2, 3, 4} and {1, 2, 3, 4} Try: Find and write down the subsets of the following sets 푃 = {1, 2, 3} and 푄 = {2, 3, 5, 7, 11} 8. Equal and equivalent sets Let 퐴 and 퐵 be two sets. We say that 퐴 = 퐵 if the two sets have the same members and number of elements. We say that 퐴 ≈ 퐵 (read ‘퐴 is equivalent to 퐵′) if the two sets have the same number of elements. For instance, given that 퐴 = {2, 3, 5} and 퐵 = {6, 7, 9}, we conclude that 퐴 ≈ 퐵 because 푛(퐴) = 푛(퐵) = 3

Example 1.3 Given that 퐴 = {2,9, 11, 13} and 퐵 = {2,푎 , 11, 13}, find 푎 if 퐴 = 퐵

Solution We have퐴 = {2,9, 11, 13} and 퐵 = {2,푎 , 11, 13}. If 퐴 = 퐵, then 푎 = 9 ⟹ 푎 = 3 (By taking square roots of both sides) 9. Universal set The universal set also called the universe of discourse refers to the set in which all other sets are studied from. We denote the universal set by 푈 or 휉. The universal set for the set of odd numbers is the set of integers.

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Operations on sets Definition An operator is a symbol used to operate/combine or perform calculations on two or more expressions. Operators enable us to get new forms of expressions. Operators used on sets include: ′ (compliment), ⋃(union) and ⋂(intersection). a. Compliment of a set The compliment of a set 퐴 refers to the set which contains elements that are not in set퐴 but are in the universal set.The compliment of a set is denoted by 퐴 .

Example 1.4 If 퐴 = {1, 2, 4,7, 9}is a subset of 푈 = {푥: 1 ≤ 푥 < 11,푥 ∈ ℤ}, find 퐴 .

Solution 퐴 = {1, 2, 4,7, 9}. By listing the members of 푈, we have 푈 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} 퐴 = {3, 5, 6, 8, 10} since 3, 5, 6, 8, 10 are not members of 퐴. 푥 ∈ ℤ means that 푥 is a member of integers. Try: If 푄 = {1, 3, 4, 6} is a subset of 푈 = {푥: 1 ≤ 푥 ≤ 10}, find 푄 . b. Intersection of sets. Let 퐴 and 퐵 be two sets. The intersection of 퐴 and 퐵, written as 퐴⋂퐵, refers to the set that contains the elements that can be found in 퐴 and 퐵. For instance if 퐴 = {2, 4, 6, 8} and 퐵 = {3, 4, 6,9, 10}, then 퐴⋂퐵 = {4, 6} because 4, 6 belong to 퐴 and 퐵. Definition: Two sets are said to be disjoint if there is푛표 member that belongs to both sets. If 퐴 and 퐵 are disjoint we write 퐴 ∩ 퐵 = ∅

Example 1.5 Given that 푈 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, 퐴 = {2, 4, 5, 7},퐵 = {5, 8, 9, 10} and 퐶 = {2, 3}, where 퐴,퐵 and 퐶 are subsets of 푈, find 퐴 ∩ 퐵,퐴 ∩ 퐶 and 퐵 .

Solution We have 푈 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} 퐴 = {2, 4, 5, 7} 퐵 = {5, 8, 9, 10} 퐶 = {2, 3} ∴ 퐴 ∩ 퐵 = {5} 퐴 ∩ 퐶 = {3} 퐵 = {0, 1, 2, 3, 4, 6, 7}

Example 1.6

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If 푃 = {5, 6,7, 9}푄 = {푥: 푥 > 2, 푥 ∈ ℤ} and 푅 = {푥: 3 ≤ 푥 ≤ 7,푥 ∈ ℤ} are subsets of 푈 = {−1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, find 푄 ∩ R, 푃′ ∩ Q′ and 푅 .

Solution By listing the elements of Q and R, we have 푈 = {−1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} 푃 = {5, 6, 7, 9} 푄 = {3, 4, 5, 6, 7, 8, 9, 10} 푅 = {3, 4, 5, 6, 7} 푄 ∩ 푅 = {3, 4, 5, 6, 7}. To find푃′ ∩ 푄′, first find 푃′ and 푄 . 푃 = {−1, 0, 1, 2, 3, 4, 8, 10} 푄 = {−1, 0, 1, 2} ∴,푃 ∩ 푄 = {−1, 0, 1, 2} 푅 = {−1, 0, 1, 2, 8, 9, 10} Try: The sets 푋 = {2, 5, 6, 9, 11} and 푌 = {Primenumberslessthan20} are subsets of 휉 = {푥: 1 ≤ 푥 ≤ 20,푥 ∈ ℕ}. Find i. 푋 ∩ 푌 ii. 푋′ iii. 푌′. c. Union of sets. Let 퐴 and 퐵 be two sets. The union of 퐴 and , written as 퐴⋃퐵, refers to the set that contain all the element that are in both sets. For example if 퐴 = {2, 3, 4, 6} and 퐵 = {3, 6, 9,12} then

퐴 ∪ 퐵 = {2, 3, 4, 6, 9, 12} Note: If a member appears in both sets it is listed once.

Example 1.7 Given that 푈 = {Naturalnumberslessthan16},푃 = {푥:−2 > 2푥 − 12},and 푄 = {푥: 푥 > 5}, where, 푃 and 푄 are the subsets of 푈, find 푃 ∪ 푄 and 푃 ∩ 푄′.

Solution 푈 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} 푃 = {푥:−2 > 2푥 − 12} 푄 = {6, 7, 8, 9, 10, 11, 12, 13, 14, 15} = {푥:−2 + 12 > 2푥} 푄 = {1, 2, 3, 4} ={푥: 10 > 2푥} ={푥: 5 > 푥} ⟹ 푃 = {1, 2, 3, 4} 푃 = {6, 7, 8, 9, 10, 11, 12, 13, 14, 15} 푃 ∪ 푄 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} 푃 ∩ 푄 = {}

Example 1.8 Given that 퐴 = {Oddnumbersgreaterthan6}, 퐵 = {Primenumberslessthan15} and 퐶 = {1, 4, 9, 16} are subsets of 푈 = {Naturalnumberslessthan18},find

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i. 퐴 ∪ 퐵 ii. 퐴 ∩ 퐵 iii. 퐴 ∪ 퐵′ iv. (퐴 ∩ 퐵)′ v. 퐶′ What conclusion can you draw about your answers in iii and iv?

Solution By listing the elements of 푈,퐴 and C, we have 푈 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17} 퐴 = {7, 9, 11,13,15} 퐵 = {2, 3, 5, 7, 11, 13} 퐶 = {1, 4, 9, 16} i. 퐴 ∪ 퐵 = {2, 3, 5, 7, 9, 11, 13, 15} ii. 퐴 ∩ 퐵 = {7, 11, 13} iii. To find 퐴′ ∪ 퐵′, first find 퐴′ and 퐵′. 퐴 = {1, 2, 3, 4, 5, 6, 8, 10, 12, 14, 16, 17} 퐵 = {1, 4, 6, 8, 9, 10, 12, 14, 15, 16, 17}

∴,퐴 ∪ 퐵 = {1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 16, 17} iv. To find (퐴 ∩ 퐵)′, first find (퐴 ∩ 퐵). 퐴 ∩ 퐵 = {7, 11, 13}

∴, (퐴 ∩ 퐵) = {1, 2, 3, 4, 5, 6, 8, 9, 10,12, 14, 15, 16, 17} v. 퐶′ = {2, 3,5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17}

퐴 ∪퐵 = (퐴 ∩ 퐵)′

Properties of sets operations Let 퐴,퐵 and 퐶 be three intersecting sets of the universal set 푈. The following results are true: 1.퐴 ∩ 퐴 = 퐴 ∪ 퐴 = 퐴 (Idempotent property) 2a. 퐴 ∪ 퐵 = 퐵 ∪ 퐴 b. 퐴 ∩ 퐵 = 퐵 ∩ 퐴 (Commutative property) 3a. (퐴 ∪ 퐵) ∪ 퐶 = 퐴 ∪ (퐵 ∪ 퐶) b. (퐴 ∩ 퐵) ∩ 퐶 = 퐴 ∩ (퐵 ∩ 퐶) (Associative property) 4a. 퐴 ∪ (퐵 ∩ 퐶) = (퐴 ∪ 퐵) ∩ (퐴 ∪ 퐶) b. 퐴 ∩ (퐵 ∪ 퐶) = (퐴 ∩ 퐵) ∪ (퐴 ∩ 퐶) (Distributive property) In 4a, union is distributed over intersection while in 4b intersection is distributed over union. The following are also valid: i. 퐴 ∪ 푈 = 푈 ii. 퐴 ∩ 푈 = 퐴 iii. 퐴⋂∅ = ∅ iv. 퐴 ∪ ∅ = 퐴 v. 퐴 ∩ 퐴′ = ∅ vi. 퐴 ∪ 퐴 = 푈 vii. (퐴 ) = 퐴

Example 1.9 Given that 퐴 = {1, 3, 4, 5},퐵 = {2,4,5,7,9},퐶 = {3, 5,7,10} are subsets of the universal set 푈 = {1, 2, 3, … . , 10}, show that i. 퐴 ∩ 퐴 = 퐴 ∪ 퐴 = 퐴 ii. 퐴 ∪ 퐵 = 퐵 ∪ 퐴 iii. 퐴 ∩ 퐵 = 퐵 ∩ 퐴 iv. (퐴 ∪ 퐵) ∪ 퐶 = 퐴 ∪ (퐵 ∪ 퐶) v. (퐴 ∩ 퐵) ∩ 퐶 = 퐴 ∩ (퐵 ∩ 퐶)

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vi.퐴 ∪ (퐵 ∩ 퐶) = (퐴 ∪ 퐵) ∩ (퐴 ∪ 퐶) vii. 퐴 ∩ (퐵 ∪ 퐶) = (퐴 ∩ 퐵) ∪ (퐴 ∩ 퐶) viii. 퐴 ∪ 푈 = 푈 ix. 퐴 ∩ 푈 = 퐴 x. 퐴 ∩= ∅ xi. 퐴 ∪ ∅ = 퐴 xii. 퐴 ∩ 퐴′ = ∅ xiii. 퐴 ∪ 퐴 = 푈 xiv. (퐴 ) = 퐴

Solution

We have 퐴 = {1, 3,4, 5}퐵 = {2,4,5,7,9}퐶 = {3, 5,7,10}푈 = {1, 2, 3, 4, 5, 6, 7, 8, 910} i. 퐴 ∩ 퐴 = {1, 3,4, 5} = 퐴퐴 ∪ 퐴 = {1, 3, 4, 5} = 퐴 ⟹ 퐴 ∩ 퐴 = 퐴 ∪ 퐴 = 퐴 ii. 퐴 ∪ 퐵 = {1, 2, 3, 4, 5, 7, 9}퐵 ∪ 퐴 = {1, 2, 3, 4, 5, 7, 9} ⟹ 퐴 ∪ 퐵 = 퐵 ∪ 퐴 iii. 퐴 ∩ 퐵 = {4, 5}퐵 ∩ 퐴 = {4, 5} ⟹퐴 ∩ 퐵 = 퐵 ∩ 퐴 iv.퐵 ∪ 퐶 = {2, 3,4, 5, 7, 9, 10} (퐴 ∪ 퐵) ∪ 퐶 = {1, 2, 3, 4, 5, 7, 9} ∪ {3, 5,7,10} = {1, 2, 3, 4, 5, 7,9, 10} 퐴 ∪ (퐵 ∪ 퐶) ={1, 3,4, 5} ∪ {2, 3, 4,5, 7, 9, 10} = {1, 2, 3, 4, 5, 7,9, 10}

∴ (퐴 ∪ 퐵) ∪ 퐶 = 퐴 ∪ (퐵 ∪ 퐶) v. (퐴 ∩ 퐵) ∩ 퐶 = {4, 5} ∩ {3, 5,7,10} = {5}퐵 ∩ 퐶 = {5, 7} 퐴 ∩ (퐵 ∩ 퐶) = {1, 3,4, 5} ∩ {5, 7} = {5} ∴ (퐴 ∩ 퐵) ∩ 퐶 = 퐴 ∩ (퐵 ∩ 퐶) vi. 퐴 ∪ (퐵 ∩ 퐶) = {1, 3,4, 5} ∪ {5, 7} = {1, 3, 4, 5, 7}퐴 ∪ 퐶 = {1, 3, 4, 5, 7, 10} (퐴 ∪ 퐵) ∩ (퐴 ∪ 퐶) = {1, 2, 3, 4, 5, 7, 9} ∩ {1, 3, 4, 5, 7, 10} = {1, 3, 4, 5,7}

∴ 퐴 ∪ (퐵 ∩ 퐶) = (퐴 ∪ 퐵) ∩ (퐴 ∪ 퐶) vii. 퐴 ∩ (퐵 ∪ 퐶) = {1, 3, 4, 5} ∩ {2, 3, 4, 5, 7, 9, 10} = {3, 4, 5} (퐴 ∩ 퐵) ∪ (퐴 ∩ 퐶) = {4, 5} ∪ {3, 5, } = {3,4, 5}

∴ 퐴 ∩ (퐵 ∪ 퐶) = (퐴 ∩ 퐵) ∪ (퐴 ∩ 퐶) viii. 퐴 ∪ 푈 = {1, 2, 3, 4, 5, 6, 7, 8, 910} = 푈 ix. 퐴 ∩ 푈 = {1, 3,4, 5} = 퐴 x. 퐴 ∩ ∅ = {1, 3,4, 5} ∩ { } = ∅ xi. 퐴 ∪ ∅ = {1, 3,4, 5} ∪ { } = {1, 3,4, 5} = 퐴 xii. 퐴 = {2, 6, 7, 8, 9, 10}퐴 ∩ 퐴 = ∅ xiii. 퐴 ∪ 퐴 = {1, 2, 3, 4, 5, 6, 7, 8, 910} = 푈 xiv. (퐴 ) = {1, 3,4, 5} = 퐴 ∎

De Morgan’s law of sets operations

Let set 퐴 and 퐵 be the subsets of the universal set 푈. Then 1. (퐴 ∪ 퐵) = 퐴′ ∩ 퐵′ 2. (퐴 ∩ 퐵) = 퐴′ ∪ 퐵′

Example 1.10 Given that 퐴 = {푥:푥 ≥ −8 − 3푥}퐵 = {푥:−3 ≤ 5 − 2푥} and 푈 = {Integersbetween2and16}, where 퐴 and 퐵 are subsets of 푈, show that i. (퐴 ∪ 퐵) = 퐴′ ∩ 퐵′ii.(퐴 ∩ 퐵) = 퐴′ ∪ 퐵′

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Solution 푈 = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} 퐴 = {푥: 푥 ≥ −8 − 3푥} 퐵 = {푥: − 3 ≤ 5 − 2푥} ={푥: 푥 − 3푥 ≥ −8} = {푥:−3 − 5 ≤ −2푥} = {푥: − 2푥 ≥ −8} = {푥: − 8 ≤ −2푥} ={푥: 푥 ≤ 4} ={푥: 푥 ≥ 4} 퐴 = {3, 4} 퐵 = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} i. 퐴 ∪ 퐵 = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} (퐴 ∪ 퐵) ={ } 퐴 = {5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} 퐵 = {3} (퐴′ ∩ 퐵′) = {} ∴ (퐴 ∪ 퐵) = 퐴′ ∩ 퐵′ ∎

ii. (퐴 ∩ 퐵) = {4} (퐴 ∩ 퐵) = {3,5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} 퐴 ∪ 퐵 = {3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} (퐴 ∩ 퐵) = 퐴′ ∪ 퐵′ ∎ Try: If 퐴 = {푥: 푥 > 6},퐵 = {푥: 푥 ≤ 10} and 퐶 = {Factorsof16} are subsets of the universal set 푈 = {푥: 1 ≤ 푥 ≤ 17}, find i.퐴 ∪ 퐵 ii. 퐴 ∩ 퐵 iii. (퐴 ∪ 퐵) ∪ 퐶 iv. (퐴 ∩ 퐵) ∩ 퐶 v.퐴 ∪ (퐵 ∩ 퐶) vi. (퐴 ∩ 퐵) ∪ (퐴 ∩ 퐶) vii. (퐴 ∩ 퐵) viii. 퐴′ ∩ 퐵′

Representation of sets in a Venn diagram

A good way of representing set diagrammatically is by using a Venn diagram. The Venn diagram consists of a rectangle with circles drawn in it. The rectangle represents the universal set and the each circle represents a set which is a subset of the universal set. The part of the rectangle which is not covered by circle(s) represents the compliment of a set. The Venn diagram helps to solve problems associated with sets.The Venn diagrams for intersection, compliment, and subsets are shown below:

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Example 1.11

Given that 퐴 = {Evennumberslessthan12} and 퐵 = {Primenumberslessthan16} are subsets of 푈 = {푥: 2 ≤ 푥 ≤ 15,푥 ∈ ℤ}. a. Find i. 퐴 ∩ 퐵 ii. 퐴 ∪ 퐵iii. 퐴′ iv. (퐴 ∩ 퐵)′ v. (퐴 ∪ 퐵)′ v. 퐴′ ∩ 퐵 vi. 퐴 ∩ 퐵′ vii. 퐴′ ∩ 퐵′ b. List only the results for each of the set formed in (a) above in a Venn diagram. c. shade the space for each of the set formed in (a) above in a Venn diagram.

Solution Listing the members of 퐴,퐵 and 푈, we have 퐴 = {2, 4, 6, 8, 10} 퐵 = {2, 3, 5, 7, 11, 13} 푈 = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} a. i. 퐴 ∩ 퐵 = {2} ii. 퐴 ∪ 퐵 = {2, 3, 4, 5, 6, 7, 8, 9} iii. 퐴 = {3, 5, 7, 9, 11, 12, 13, 14, 15} iv. (퐴 ∩ 퐵) = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} v. 퐴 ∩ 퐵 = {3, 5, 7, 11, 13} vi. Let’s first find 퐵 :퐵′ = {4, 6, 8, 9, 10, 12, 14, 15} ⟹ 퐴 ∩ 퐵 = {4, 6, 8, 10} vii. 퐴 ∩ 퐵 = {9, 12, 14, 15} b.

퐴 ∩ 퐵 퐴 ∪ 퐵 퐴′ (퐴 ∩ 퐵)′

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퐴′ ∩ 퐵 퐴 ∩ 퐵′ 퐴′ ∩ 퐵 c.

퐴 ∩ 퐵 퐴 ∪ 퐵 퐴 (퐴 ∩ 퐵)′

퐴 ∩ 퐵 퐴 ∩ 퐵′ 퐴′ ∩ 퐵′ Try: Using example 1.10 a. Find i. 퐴 ∪ 퐵′ ii. 퐴′ ∪ 퐵′ iii.(퐴 ∪ 퐵)′ b. List only the results for each of the set formed in (a) above in a Venn diagram. c. shade the space for each of the set formed in (a) above in a Venn diagram.

Two sets problems

Let 퐴 and 퐵 be two intersecting subsets of the universal set 푈. Consider the Venn diagram below.

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푎= only 퐴 = 푛(퐴 ∩ 퐵 ) 푏 =only B=푛(퐴 ∩ 퐵) 푐 = 푛(퐴′ ∩ 퐵) 푑 = 푛(퐴 ∪ 퐵)′ When solving two sets problems, we make use of the following three equations: 푛(퐴) = 푎 + 푏… … …(1) 푛(퐵) = 푏 + 푐… … … (2) 푛(푈) = 푎 + 푏 + 푐 + (푑) … … …(3) Also 푛(퐴 ∪ 퐵) = 푎 + 푏 + 푐 = (푎 + 푏) + (푏 + 푐) − 푏 = 푛(퐴) + 푛(퐵) − 푛(퐴 ∩ 퐵) ∴ 푛(퐴 ∪ 퐵) = 푛(퐴) + 푛(퐵)− 푛(퐴 ∩ 퐵) … … …(4) From (4) 푛(퐴 ∩ 퐵) = 푛(퐴) + 푛(퐵) − 푛(퐴 ∪ 퐵) … … …(5) 푛(퐴) = 푛(퐴 ∩ 퐵′) + 푛(퐴 ∩ 퐵) ⟹푛(퐴) − 푛(퐴 ∩ 퐵) = 푛(퐴 ∩ 퐵 ) … … … (6) Similarly, 푛(퐵) = 푛(퐴 ∩ 퐵) + 푛(퐴 ∩ 퐵) ⟹ 푛(퐵) − 푛(퐴 ∩ 퐵) = 푛(퐴 ∩ 퐵) … … …(7) ⟹ 푛(퐴 ∪ 퐵) = 푛(퐴 ∩ 퐵′) + 푛(퐴 ∩ 퐵) + 푛(퐴 ∩ 퐵) … … …(8) Statements such as ‘each student read one of the two subjects’ means푑 = 0. If 푑 = 0, 푛(퐴 ∪ 퐵) = 푛(푈)

Example 1.12 a. If 퐴 and 퐵 are two intersecting subsets of the universal set 푈, find 푛(퐴 ∪ 퐵) if 푛(퐴) = 12,푛(퐵) = 15 and 푛(퐴 ∩ 퐵) = 10. b. 푃 and 푄 are two intersecting subsets of the universal set 푈. Find 푛(푃 ∩ 푄)if 푛(푃 ∪ 푄) = 50, 푛(푃) = 35 and 푛(푄) = 18.

Solution a. From (4) 푛(퐴 ∪ 퐵) = 푛(퐴) + 푛(퐵) − 푛(퐴 ∩ 퐵)

⟹ 푛(퐴 ∪ 퐵) = 12 + 15 − 10 = 17 ∴,푛(퐴 ∪ 퐵) = 17 b. From (5) 푛(푃 ∩ 퐵) = 푛(푃) + 푛(푄) − 푛(푃 ∩ 푄) ⟹ 푛(푃 ∩ 퐵) = 35 + 18 − 50 = 3 ∴ 푛(푃 ∩ 퐵) = 3 Try: Find 푛(퐴),if 푛(퐴 ∪ 퐵) = 45,푛(퐴 ∩ 퐵) = 16 and 푛(퐵) = 27.

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Example 1.13 In a class of 35 students, 20 read Mathematics and 18 read Economics. Each student reads at least one of the two subjects. Represent this information in a Venn diagram and find the number of students who read both Mathematics and Economics.

Solution Let 푈 represents student in the class, 푀 represents Mathematics student and 퐸 represent Economics student. Let 푛(푀) only= 푎, 푛(푀 ∩ 퐸) = 푥 푛(퐸) only=푏 ⟹ 푛(푈) = 푛(푀 ∪ 퐸) = 35 푛(푀) = 20 푛(퐸) = 18 Using equations (1) (2) and (3) we have: 푛(푀) = 푎 + 푥 ⟹ 20 = 푎 + 푥 ⟹ 20 − 푥 = 푎 푛(퐸) = 푏 + 푥 ⟹ 18 = 푏 + 푥 ⟹ 18 − 푥 = 푏

푛(푈) = 푎 + 푥 + 푏 ⟹ 35 = 20 − 푥 + 푥 + 18 − 푥 ⟹ 35 = 20 + 18 − 푥 ⟹ 35 = 38 − 푥 ⟹ 푥 = 38 − 35 = 3 ∴ 3 students read both Mathematics and Economics.

Example 1.14

In a sports team of 22 players, 13 players play Football, 9 players play Tennis and 4 play neither Football nor Tennis. Represent this information in a Venn diagram and find the number of players who play i. both Football and Tennis. ii. only one game.

Solution Let 푈 be the players in the team, 퐹 be Football players and 푇 be Tennis players. Let 푛(퐹) only= 푎 푛(퐹 ∩ 푇) = 푥 and 푛(푇)only=푏 ⟹ 푛(푈) = 22 푛(퐹) = 13 푛(푇) = 9 푛(퐹 ∩ 푇) = 4 i. Using equations (1) (2) and (3) we have:

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푛(퐹) = 푎 + 푥 ⟹ 13 = 푎 + 푥 ⟹ 13 − 푥 = 푎 푛(푇) = 푏 + 푥 ⟹ 9 = 푏 + 푥 ⟹ 9− 푥 = 푏 푛(푈) = 푎 + 푥 + 푏 + 4 ⟹ 22 = 13 − 푥 + 푥 + 9 − 푥 + 4 ⟹ 22 = 13 + 9 + 4 − 푥 ⟹ 22 = 26 − 푥 ⟹ 푥 = 26 − 22 = 4 ∴ 4 players play both Football and Tennis ii. Only one game= 푎 + 푏 = 13 − 푥 + 9 − 푥 = 13− 4 + 9 − 4 = 14 ∴ 14 players play only one game.

Example 1.15 In a class of students, 23 students like reading Fiction books, 17 like reading Non-Fiction books and 7 students like reading only Fiction. Each student likes reading at least one of the two types of books. Represent this information in a Venn diagram. Find the number of students i. in the class ii. who read only Non-Fiction.

Solution Let 푈 = {Studentintheclass}, 퐹 represent Fiction students and N represent Non-Fiction students. Let 푛(푁) only=푎 푛(퐹 ∩ 푁) = 푥 푛(푈) = 푦 ⟹ 푛(퐹) = 23 푛(퐹) ony=7 푛(푁) = 17 i. Using equations (1), (2) and (3) we have 푛(퐹) = 7 + 푥 ⟹ 23 = 7 + 푥 ⟹ 푥 = 23 − 7 = 16 푛(N) = 푎 + 푥 ⟹ 17 = 푎 + 16 ⟹ 푎 = 17 − 16 = 1 푛(푈) = 7 + 푥 + 푎 ⟹ y = 7 + 16 + 1 = 24 ∴ there are 24 students in the class. ii. Students who like reading only Non-Fiction books= 푎 = 1 ∴ only one student read Non-Fiction books.

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Example 1.16

In a class of 54students, 12 of them study both Economics and Computer Science and the number of students who study Economics is half the number of students who study Computer Science. If a student studies at least one of the two courses: a. i. Illustrate the above information in a Venn diagram. ii. Find the number of students who study Economics and Computer Science. iii. Find the number of students who study Economics but not Computer Science. b. Provide an alternative solution for this problem without Venn diagram

Solution Let 푈 = {푆푡푢푑푒푛푡푠푖푛푡ℎ푒푐푙푎푠푠}, E be students who study Economics and C be students who study Computer Science. Let 푛(퐶) = 푥 푛(퐶) only= 푎 푛(퐸) only= 푏 ⟹ 푛(푈) = 54 푛(퐸) = 푥 = 0.5푥 푛(퐸 ∩ 퐶) = 12 Using equations (1), (2) and (3) we have: a. 푛(퐸) = 푎 + 12 ⟹ 0. 5푥 = 푎 + 12 ⟹ 0. 5푥 − 12 = 푎 푛(퐶) = 푏 + 12 ⟹ 푥 = 푏 + 12 ⟹ 푥 − 12 = 푏 푛(푈) = 푎 + 12 + 푏 ⟹ 54= 0.5푥 − 12 + 12 + 푥 − 12 ⟹ 54 = 0.5푥 + 푥 − 12 ⟹ 54 = 1.5푥 − 푥 ⟹ 54 + 12 = 1.5푥

⟹ 66 = 1.5푥 ⟹ 푥 =661.5 = 44

ii. ∴ 44 students study Computer Science and (44) = 22 study Economics. iii. Students who study Economics but not Computer Science=푛(퐸) only= 푏 ⟹ b = 푥 − 12 = 44 − 12 = 32 ∴ 32 students study only Economics but not Computer Science. b. We have 푛(퐸 ∪ 퐶) = 54 푛(퐸 ∩ 퐶) = 12 But From (6) and (7) 푛(퐸) = 푎 + 12 ⟹ 푎 = 0.5푥 − 12and 푛(퐶) = 푏 + 12 ⟹ 푏 = 푥 − 12

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From (8) we have 푛(퐸 ∪ 퐶) = 푛(퐸 ∩ 퐶′) + 푛(퐸 ∩ 퐶) + 푛(퐸 ∩ 퐶) ⟹ 54 = 0.5푥 − 12 + 12 + 푥 − 12 ⟹ 54 = 1.5푥 − 12푥 =

.= 44

(Compute the answers for the other questions.) Note: If a question is given in percentages we take 푛(푈) = 100% Try: In a class of 200 students, 45% read Classical and 75% read Philosophy. Each student read at least one of the two courses. Find the number of students who study both courses.

Example 1.17 Given that 푃 = {2, 4, 6, 7, 8},푄 = {1, 3, 4, 7,9, 10} and 푅 = {2, 3, 5, 7, 11} are subsets of the universal set 푈 = {푥: 1 ≤ 푥 ≤ 12,푥 ∈ ℤ}, a. find i. 푃 ∩ 푄 ∩ 푅 ii. 푃 ∪ (푄 ∪ 푅) iii. 푃 ∩ (푄 ∩ 푅) iv. 푃 ∩ (푄 ∩ 푅) v. 푃 ∩ (푄 ∩ 푅 ) vi. 푃 ∩ (푄 ∩ 푅 ) vii. 푃 ∩ (푄 ∩ 푅 ) viii. 푃 ∩ (푄 ∩ 푅) ix. (푃 ∩ 푄 ∩ 푅)′ x. (푃 ∪ 푄 ∪ 푅)′ b. list only the results for each of the set formed in (a) above in a Venn diagram. c. shade the space for each of the set formed in (a) above in a Venn diagram.

Solution

a. We have 푃 = {2, 4, 6, 7, 8},푄 = {1, 3, 4, 7,9, 10}, 푅 = {2, 3, 5, 7, 11} and 푈 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} i. 푃 ∩ 푄 ∩ 푅 = {7} ii. 푃 ∪ (푄 ∪ 푅) = {2, 4, 6, 7, 8} ∪ {1, 2, 3, 4, 5, 7, 9, 10, 11} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} iii. Let first find푃 :푃 = {1, 3, 5, 9, 10, 11, 12} and 푄 ∩ 푅:푄 ∩ 푅 = {3, 7} ⟹ 푃 ∩ (푄 ∩ 푅) = {3} iv. Let’s first find 푄 :푄 = {2, 5, 6, 8, 11, 12} and 푄 ∩ 푅:푄 ∩ 푅 = {2, 5, 11} ⟹ 푃∩ (푄 ∩ 푅)={2} v. Let’s find 푅 :푅 = {1, 4, 6, 8,9, 10, 12} and 푄 ∩ 푅 = {1, 4, 9, 10} ⟹ 푃∩ (푄 ∩ 푅 ) = {4} vi. 푄 ∩ 푅 = {2, 5, 6, 8, 11, 12} ∩ {1, 4, 6, 8,9, 10, 12} = {6, 8, 12}

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⟹ 푃∩ (푄 ∩ 푅 ) = {6, 8} vii.푄 ∩ 푅 = {1, 3, 4, 7,9, 10} ∩ {1, 4, 6, 8,9, 10, 12} = {1, 4, 9, 10} ⟹ 푃 ∩ (푄 ∩ 푅 ) = {1, 3, 5, 9, 10, 11, 12} ∩ {1, 4, 9, 10} = {1, 9, 10} viii. 푃 ∩ (푄 ∩ 푅) = {1, 3, 5, 9, 10, 11, 12} ∩ {2, 5, 11} = {5, 11} ix. (푃 ∩ 푄 ∩ 푅) = {1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12} x. (푃 ∪ 푄 ∪ 푅) = {12} b.

i. 푃 ∩ 푄 ∩ 푅 ii. 푃 ∪ (푄 ∪ 푅) iii. 푃 ∩ (푄 ∩ 푅)

iv.푃 ∩ (푄 ∩ 푅) v. 푃 ∩ (푄 ∩ 푅 ) vi. 푃 ∩ (푄 ∩ 푅 )

vii.푃 ∩ (푄 ∩ 푅 ) viii. 푃 ∩ (푄 ∩ 푅) ix. (푃 ∩ 푄 ∩ 푅)′

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x. (푃 ∪ 푄 ∪ 푅)′ c.

i. 푃 ∩ 푄 ∩ 푅 ii. 푃 ∪ (푄 ∪ 푅) iii. 푃 ∩ (푄 ∩ 푅)

iv. 푃 ∩ (푄 ∩ 푅) v. 푃 ∩ (푄 ∩ 푅 ) vi. 푃 ∩ (푄 ∩ 푅 )

vii. 푃 ∩ (푄 ∩ 푅 ) viii. 푃 ∩ (푄 ∩ 푅) ix. (푃 ∩ 푄 ∩ 푅)′

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x. (푃 ∪ 푄 ∪ 푅)′ Try: Using example 1.15 a. find i. 푃 ∩ (푄 ∪ 푅) ii. 푃 ∩ (푄 ∪ 푅) iii. 푃 ∪ (푄 ∩ 푅 ) iv. (푃 ∩ 푄 ∩ 푅)′ v. (푃 ∩ 푄 ∪ 푅)′ b. list only the results for each of the set formed in (a) above in a Venn diagram. c. shade the space for each of the set formed in (a) above in a Venn diagram.

Three sets problems

Let 퐴,B and 퐶 be three intersecting subsets of the universal set 푈. Consider the Venn diagrambelow. 푎 =only 퐴 = 푛(퐴 ∩ 퐵 ∩ 퐶 ) = 퐴butno푡퐵and퐶. 푏 =only 퐵 = 푛(퐴 ∩ 퐵 ∩ 퐶 ) = 퐵butnot퐴and퐶. 푐 =only 퐶 = 푛(퐴 ∩ 퐵 ∩ 퐶) = 퐶butno푡퐴and퐵. 푑 =only A and C= 푛(퐴 ∩ 퐵 ∩ 퐶) = 퐴and퐶butnot퐵 푒 = only A and B= 푛(퐴 ∩ 퐵 ∩ 퐶 ) = 퐴and퐵butno푡퐶. 푓 = only B and C= 푛(퐴 ∩ 퐵 ∩ 퐶) = 퐵and퐶butnot퐴 푔 = 퐴and 퐵 and 퐶 = 푛(퐴 ∩ 퐵 ∩ 퐶) ℎ = neither퐴nor퐵nor퐶 = 푛(퐴 ∪ 퐵 ∪ 퐶)′

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When solving three sets problems, we make use of the following six equations: 푛(퐴 ∩ 퐵) =e +푔… … …(1) 푛(퐴 ∩ 퐶) = 푑 + 푔… … …(2) 푛(퐵 ∩ 퐶 = 푓 + 푔… … …(3) 푛(퐴) = 푎 + 푑 + 푒 + 푔… … …(4) 푛(퐵) = 푏 + 푒 + 푓 + 푔… … … (5) 푛(퐶) = 푐 + 푑 + 푓 + 푔… … …(6) 푛(푈) = 푎 + 푏 + 푐 + 푑 + 푒 + 푔 + ℎ… … …(7) Also 푛(퐴 ∪ 퐵 ∪ 퐶) = 푎 + 푏 + 푐 + 푑 + 푒 + 푓 + 푔 = (푎 + 푑 + 푒 + 푔)− (푒 + 푔) + (푏 + 푒 + 푓 + 푔)− (푓 + 푔) +(푐 + 푑 + 푓 + 푔)− (푑 + 푔) + 푔 = 푛(퐴) − 푛(퐴 ∩ 퐵) + 푛(퐵) − 푛(퐵 ∩ 퐶) + 푛(퐶)− 푛(퐴 ∩ 퐶) + 푛(퐴 ∩ 퐵 ∩ 퐶) ∴ 푛(퐴 ∪ 퐵 ∪ 퐶)

= 푛(퐴) + 푛(퐵) + 푛(퐶) − 푛(퐴 ∩ 퐵) − 푛(퐴 ∩ 퐶)− 푛(퐵 ∩ 퐶) + 푛(퐴 ∩ 퐵 ∩ 퐶) If 푛(퐴 ∩ 퐵 ∩ 퐶) = ℎ = 0,then 푛(퐴 ∪ 퐵 ∪ 퐶) = 푛(푈)

Example 1.18 In a class of 33 students, 20 read Mathematics, 18 read Economics and 15 read Computer Science. 8 read Mathematics and Economics, 10 read Mathematics and Computer Science and 7 read Economics and Computer Science. Each student read at least one of the three subjects and 푥 student read all the three subjects. a. Illustrate the information above in a Venn diagram. b. Find, i. the value of 푥. ii. the number of students who read only one subject. iii. the number of students who read only two subjects.

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Solution Let 푈 = {studentsintheclass}, 푀 = {studentswhoreadMathematics}, 퐸 = {studentswhoreadEconomisc} and 퐶 = {studentswhoreadComputerScience} Let 푛(푀) only= 푎 푛(퐸) only= 푏 푛(퐶) only= 푐 푛(푀∩ 퐸) only= 푑 푛(푀∩ 퐶) only= 푒 푛(퐸 ∩ 퐶) only= 푓 ⟹ 푛(푈) = 50푛(푀) = 20푛(퐸) = 18푛(퐶) = 15푛(푀 ∩ 퐸) = 8 푛(푀 ∩ 퐶) = 10and 푛(퐸 ∩ 퐶) = 7 a. b. i. Using equations (1) through we have:

푛(푀 ∩ 퐸) = 푑 + 푥 ⟹ 8 = 푑 + 푥 ⟹ 8 − 푥 = 푑 푛(푀 ∩ 퐶) = 푒 + 푥 ⟹ 10 = 푒 + 푥 ⟹ 10 − 푥 = 푒

푛(퐸 ∩ 퐶) = 푓 + 푥 ⟹ 7 = 푓 + 푥 ⟹ 7 − 푥 = 푓 푛(푀) = 푎 + 푑 + 푒 + 푥 ⟹ 20 = 푎 + 8 − 푥 + 10 − 푥 + 푥 푛(퐸) = 푏 + 푑 + 푓 + 푥 = 푎 + 8 + 10 − 푥 ⟹ 18 = 푏 + 8 − 푥 + 7 − 푥 + 푥 = 푎 + 18 − 푥 = 푏 + 8 + 7 − 푥 ⟹ 20 − 18 + 푥 = 푎 = 푏 + 15 − 푥 ⟹ 2 + 푥 = 푎 ⟹ 18 − 15 + 푥 = 푏 ⟹ 3 + 푥 = 푏 푛(퐶) = 푐 + 푒 + 푓 + 푥 ⟹ 15 = c + 10 − 푥 + 7 − 푥 + 푥 ⟹ 15 = c + 7 + 10− 푥 ⟹ 15 − 17 + 푥 = 푐 ⟹ −2 + 푥 = 푐 푛(푈) = 푎 + 푏 + 푐 + 푑 + 푒 + 푓 + 푥 ⟹ 33 = 2 + 푥 + 3 + 푥 + (−2 + 푥) + 8 − 푥 + 10 − 푥 + 7 − 푥+푥 ⟹ 33 = 2 + 3− 2 + 8 + 10 + 7 + 푥 + 푥 + 푥 + 푥 − 푥 − 푥 − 푥 ⟹ 33 = 28 + 푥 ⟹ 푥 = 33 − 28 = 5 ∴ 푥 = 5 ii. Students who read only one book= 푎 + 푏 + 푐 = 2 + 푥 + 3 + 푥 + (−2 + 푥) = 2 + 5 + 3 + 5 − 2 + 5 = 11

∴ 11 students read only one subject.

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iii. Students who read only two books = 푑 + 푒 + 푓 = 8 − 푥 + 10 − 푥 + 7 − 푥 = 8 − 5 + 10− 5 + 7 − 5 = 10

∴ 10 students read only two subjects.

Example 1.19 In a class of 76 students, 8 read only French, 22 read English but not Fine Art, 34 read English and French, 37 read French and Fine Art, 48 read English and 57 read French. Every student read at least one of the three Programmes. a. Illustrate the information above in a Venn diagram b. Use your Venn diagram to find the number of students who read i. Fine Art. ii. Fine Art and English but not French. iii. At least two programs.

Solution Let:푈 = {studentsintheclass} 퐸 = {studentswhoofferEnglish} 퐹 = {studentswhoofferFrench} 퐾 = {studentswhoofferFineArt} 푛(퐸) only= 푎 푛(퐾) only= 푏 푛(퐸 ∩ 퐹)only= 푐 푛(퐸 ∩ 퐾)only= 푑 푛(퐹 ∩ 퐾)only= 푒 푛(퐸 ∩ 퐹 ∩ 퐾) = 푥 푛(퐾) = 푌 ⟹ 푛(푈) = 76 푛(퐹) only= 8 푛(퐸 ∩ 퐹) = 34 푛(퐹 ∩ 퐾) = 37 푛(퐸) = 48 푛(퐹) = 57 a. b. 푛(EnglishbutnotFineArt) = 푎 + 푐 ⟹ 22 = 푎 + 푐 i. Using equation (1) through (7)we have: 푛(퐸 ∩ 퐹) = 푐 + 푥 ⟹ 34 = 푐 + 푥 ⟹ 34 − 푥 = 푐 푛(퐹 ∩ 퐾) = 푒 + 푥 ⟹ 37 = 푒 + 푥 ⟹ 37− 푥 = 푒 푛(퐸) = 푎 + 푐 + 푑 + 푥 푛(퐹) = 8 + 푐 + 푒 + 푥 ⟹ 48 = 22 + 푑 + 푥 ⟹ 57 = 8 + 34 − 푥 + 37 − 푥 + 푥

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⟹ 48 − 22 − 푥 = 푑 ⟹ 57 = 8 + 34 + 37− 푥 ⟹ 26 − 푥 = 푑 ⟹ 57 = 79− 푥 ⟹ 푥 = 79 − 57 = 22 푛(퐾) = 푏 + 푑 + 푒 + 푥 푛(푈) = 8 + 푎 + 푏 + 푐 + 푑 + 푒 + 푥 ⟹ 푌 = 푏 + 26 − 푥 + 37 − 푥 + 푥 ⟹ 76 = 8 + 푎 + 푐 + 푌 − 41 + 26− 푥 + 37 − 푥 + 푥

⟹ 푌 = 푏 + 63 − 푥 ⟹76=8+22+26+37 − 41− 푥 + 푌 ⟹ Y = 푏 + 63 − 22 ⟹ 76 = 52− 22+푌 ⟹ 푌− 41 = 푏 ⟹ 76 = 32 + Y ⟹ 푌 = 76 − 32 = 44 ∴,44 student offer Fine Art. ii. Fine Art and English but not French= 푐 = 34− 22 = 12 ∴, 12 students offer Fine Art and English but nor French. iii. At least two programs= 푐 + 푑 + 푒 + 푥 = 12 + 26 − 22 + 37 − 22 + 22 = 53 ∴,53 students offer at least two programmes

Example 1.20 In a class of 80 students, 5 offer only Biology, 7 offer only Chemistry and 16 offer Biology and Chemistry. 36 offer Biology, 39 offer Chemistry and 44 offer Physics. If 3 students offer all the three courses, find i. the number of students who offer only Physics. ii. the number of student who did not offer any of the three courses. iii. the number of students who offer Chemistry and Physics.

Solution Let: 푈 = {studentsintheclass} 퐵 = {studentswhoofferBiology} 퐶 = {studentswhoofferChemistry}

푃 = {studentswhoofferPhysics} 푛(퐵 ∩ 푃)표푛푙푦 = 푎 푛(퐶)표푛푙푦 = 푏

푛(퐵 ∩ 퐶)표푛푙푦 = 푐 푛(퐶 ∩ 푃)표푛푙푦 = 푑 푛(퐵 ∪ 퐶 ∪ 푃) = 푦 푛(퐵 ∩ 푃)표푛푙푦 = 푥 푛(퐵 ∩ 퐶) = 푐 + 3 ⟹ 16 = 푐 + 3 ⟹ 16− 3 = 푐 ⟹ 13 = 푐 푛(퐵 ∩ 푃) = 푎 + 3 ⟹ 푥 = 푎 + 3 ⟹ 푥 − 3 = 푎 푛(퐶 ∩ 푃) = 3 + 푑

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푛(퐵) = 5 + 3 + 푐 + 푎 푛(퐶) = 7 + 3 + 푐 + 푑 ⟹ 36 = 8 + 푎 + 푐 ⟹ 39 = 10 + 푐 + 푑 ⟹ 36 − 8 = 푎 + 푐 ⟹ 39 − 10 = 푐 + 푑 ⟹ 28 = 푎 + 13 ⟹ 29 = 13 + 푑 ⟹ 28 − 13 = 푎 ⟹ 29 − 13 = 푑 ∴ 푎 = 15 ∴ 16 = 푑 푛(푃) = 3 + 푎 + 푏 + 푑 푛(푈) = 5 + 7 + 3 + 푎 + 푏 + 푐 + 푑 + 푦 ⟹ 44 − 3 − 푎 − 푑 = 푏 ⟹ 80 = 15 +15+10+13+16+푦 ⟹ 41− (15 + 16) = 푏 ⟹ 80 = 69 + 푦 ⟹ 41− 31 = 푏 ⟹ 80 − 69 = 푦 ∴ 푏 = 10 ∴ 푦 = 11 i. The number of students who offer only Physics= 푏 = 10

∴ 10 students offer only 푃ℎ푦푠푖푐푠 ii. The number of students who did not offer any of the three courses= 푦 = 11

∴ 11 students did not offer any of the three courses. iii. The number of students who offer Chemistry and Physics= 푧 = 3 + 푑 = 3 + 16 = 19

∴, 19 students offer Chemistry and Physics

Example 1.21 In a survey of 75 students who went to a school garden reveals that: 50 used cutlasses, 25 used hoes and 40 used rakes. 14 students used exactly two of the garden tools. Every student used at least one of the tools. How many students used i. all the three tools? ii. only one tool?

Solution Let:

푈 = {studentswhowenttothegarden}, 퐶 = {studentswhousedcutlasses}, 퐻 = {studentswhousedhoes} and 푅 = {studentswhousedrakes} Let 푛(퐶) only= 푎 푛(퐻) only= 푏 푛(푅) only= 푐 푛(퐶 ∩ 푅) only= 푑 푛(퐶 ∩ 퐻)only= 푒 푛(퐻 ∩ 푅) = 푓 ⟹ 푛(푈) = 75푛(퐶) = 50푛(퐻)25푛(푅) = 40 푑 + 푒 + 푓 = 14

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Using equation (1) through (7) we have 푛(퐶 ∩ 퐻) = 푒 + 푥 푛(퐶 ∩ 푅) = 푑 + 푥 푛(퐻 ∩ 푅) = 푓 + 푥 푛(퐶) = 푎 + 푒 + 푑 + 푥 ⟹ 50 = 푎 + 푒 + 푑 + 푥 ⟹ 50 − 푑 − 푒 − 푥 = 푎 푛(퐻) = 푏 + 푒 + 푓 + 푥 ⟹ 25 = 푏 + 푒 + 푓 + 푥 ⟹ 25 − 푒 − 푓 − 푥 = 푏 푛(푅) = 푐 + 푑 + 푓 + 푥 ⟹ 40 = 푐 + 푑 + 푓 + 푥 ⟹ 40 − 푑 − 푓 − 푥 = 푐 푛(푈) = 푎 + 푏 + 푐 + 푑 + 푒 + 푓 + 푥 ⟹ 75 = 50 − 푑 − 푒 − 푥+25 − 푒 − 푓 − 푥 + 40 − 푑 − 푓 − 푥 + 푑 + 푒 + 푓 + 푥 = 50 + 40 + 25 − 푑 − 푒 − 푓 − 2푥 = 115 − 2(푑 + 푒 + 푓) − 2푥 = 115 − 2(14)− 2푥 = 115 − 28 − 2푥 ⟹ 2푥 = 87 − 75 ⟹ 푥 = = 6 i. Therefore,6 students used all the three tools. ii. Only one tool= 푎 + 푏 + 푐 = 50 − 푑 − 푒 − 푥 + 25 − 푒 − 푓 − 푥 + 40 − 푑 − 푓 − 푥 = 50 + 25 + 40 − 2푑 − 2푒 − 2푓 − 3푥 = 115− 2(푑 + 푒 + 푓) − 3푥 = 115− 2(14)− 3(6) = 69 Therefore, 69 used only one of the tools. Try: 1. There are 40 pupils in a class. Thirty of them study Biology, 22 study Physics and 21

study Chemistry. 15 study Physics and Biology, 10 study Physics and Chemistry and 13 study Biology and Chemistry. 2 study Physics only, 3 study Chemistry only and 7 study Biology only. i. Represent this information in a Venn diagram. ii. How many pupils study all the three subjects? iii. If a pupil is selected at random, what is the probability that he studies either

Physics or Chemistry? (SSSCE) 2. In a survey of the 100 out-patient who reported at a hospital one day, it was found

that 70 complained of fever, 50 had stomach ache and 30 were injured. All the 100 patients had at least one of the complaints and 44 had exactly two of the complaints. How many patients had all the three complaints? (SSSCE)

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Final Exercises

1. Write down the members of the following sets: 퐴 = {푥: 푥 > 3,푥 ∈ ℤ} 퐵 = {푥: 1 ≤ 푥 ≤ 10,푥 ∈ ℤ} 퐶 = {푥:−1 < 푥 ≤ 6,푥 ∈ ℤ} 2. Given that 퐴 = {1, 2, 3, 7, 9},퐵 = {푥: 2 ≤ 푥 ≤ 10} and 퐶 = {factorsof24} are Subsets of the universal set 푈 = {1, 2, 3, … , 15} . Find i. 퐴 ∩ 퐵 ii. 퐴 ∩ 퐶 iii. 퐴 ∩ 퐵 ∩ 퐶 iv. 퐴 ∪ 퐵 ∪ 퐶 v. 퐴 ∪ 퐵′ 3. Given that 퐴 = {2, 3, 5,7, 11, 13},퐵 = {multiplesof3lessthan18} and 퐶 = {푥: 3 ≤ 푥 ≤ 19} are subsets of the universal set 푈 = {1, 2, 3, … ,20}. Find i. 퐴 ∪ 퐵 ii. 퐵 ∪ 퐶 iii. (퐴 ∪ 퐵) ∩ 퐶 iv. 퐴 ∪ 퐵 ∪ 퐶′ v. 퐴′ ∪ 퐵′ 4. Given that 푈 = {Naturalnumberslessthan21},퐴 = {푥: 푥 > 3}, 퐵 = {multiplesof2} and 퐶 = {푓푎푐푡표푟푠표푓48}, where 퐴,퐵 and 퐶 are subsets of 푈. Find: i. 퐴 ∩ 퐵 ii. 퐴 ∩ 퐵 ∩ 퐶 iii. 퐴′ ∪ 퐵′ iv. (퐴 ∩ 퐵)′ v. (퐴 ∪ 퐵)′ ∩ 퐵 5. Given that 푈 = {integersbetween2and16},푃 = {factorsof50}, 푄 = {푥: 2 < 2푥 < 20},푅 = {Multiplesof3}, where 푃,푄and 푅 are subsets of 푈. a. Find i. (푃 ∩ 푄) ∩ 푅 ii. (푃 ∩ 푄)′ ∩ 푅 iii. 푃′ ∩ 푄′ iv. (푃 ∪ 푄) ∪ 푅′ b. Represent (iv) in a Venn diagram. 6. The sets 퐴 = {푥: 푥isafactorofof72}, 퐵 = 푥: 푥 + 3 < 2푥 − 3 and 퐶 = {푥: 푥 < 20} are subsets are subsets of ℤ. a. list the elements of i. (퐴 ∩ 퐵)∩ 퐶 ii. 퐴 ∩ (퐵 ∩ 퐶) iii. 퐴 ∪ 퐵 iv. 퐵 ∪ 퐴 b. What conclusion can you draw about your answers in (i) and (ii) and (iii) and (iv). 7. Given that 푈 = {1,2,3, … ,12}, 퐴 = {2,4, 6,7, 9, 11}, 퐵 = {1,2, 5, 11} and 퐶 = {10,11, 12}. Fin d i. 퐴 ∩ 퐵 ii. (퐴 ∩ 퐵) ∩ 퐶 iii. 퐵 ∩ 퐶 iv. 퐴′ ∩ 퐶′ v. 퐴′ ∩ 퐵′ ∩ 퐶 8. Let 푈 = {1, 2, 3, 4, 5, … , 10} 퐴 = {2, 4, 6, 8} 퐵{3, 4, 5} 퐶{1, 2, 3, 9}. Find i. 퐴 ∪ 퐵 ∪ 퐶 ii. (퐵 ∪ 퐶 ) ∩ (퐴 ∪ 퐵′) iii. (퐴 ∩ 퐵)′ ∪ 퐶 9. Given that 퐴 = {1, 4, 9, 16}. i. Find the number of subsets of 퐴. ii. Write down all the subsets of 퐴. 10. Given that 퐴 = {−1, 0, 1, 2, 3}. i. Find the number of subsets of 퐴. ii. Write down all the subsets of 퐴. 11. In a class of 300 students, 200 of them are Day students and 150 of then are Boarders. If 30 of them are both Day students and Boarders; a. Illustrate the above information on a Venn diagram.

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b. Find the number of students who are not days students nor Boarders. 12. In a class of 42 students, 16 offer only Chemistry and 22 offer only Biology. If a student study at least one of the two subjects; find i. Those who offer Chemistry. ii. Those who offer Biology. 13. There are 80 students in a class. 50 of them read Physics and 40 of them read Computer Science. If a student read a least one of the two programmes; find i. Those who read all the three programmes. ii. Those who read only computer Science. 14. In a call of 33 students, the number of students who study Geography is twice the number of students who study history. If a student studies at least one of the two course; find i. The number of students who study Geography. ii. The number of students who study History. 15. In a class, the number of students who read Communicative Skills is the same as the number of students who read only English. If 30 students read read only English and 5 students read both programmes; find The number of students in the class. 16. In a class of 52 students, 30 students offer Geography and 9 students offer only History. If a student study at least one of the two subjects. Find; a. The number of students who offer History. b. The number of students who offer both subjects 17. In a class of 74 students, 푥 of them study Mathematics and 2푥 of them study Music. 10 of them study both subjects. i. Illustrate the information in a Venn diagram. ii. Find the number of students who study Mathematics but not Music. 18. In a class of 16 students, 4 read only History and 7 read only Government. If 2 students did not read any of the two subjects, find the number of students who study government. 19. Given that 푛(퐴 ∪ 퐵) = 30, 푛(퐴) = 18, 푛(퐵) = 16 and 푛(퐴 ∪ 퐵) = 2. Find 푛(퐴 ∩ 퐵). 20. In a class of 50 students, 23 study Geography and 34 study History. If 푥 of them study both subjects, find the value of 푥. 21. In a class of 70 students, 30 of them offer Cost Accounting, 25 0f them offer Further Mathematics and 36 offer Geography. The number of students who offer Further Mathematics is 2 times the number of students who of Geography only. 4 students

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offer only Cost Accounting and Further Mathematics, 2 students off only Further Mathematics and Geography and 3 students offer only Geography and Cost Accounting. 4 of them do not offer any of the three subjects and 푥 students offer all the three subjects. a. Illustrate the above information on a Venn diagram b. Find; i. the number of students who offer Further Mathematics. ii. the number of students who offer Geography. iii. the number of students who offer Costing but not Further Mathematic. 22. In a class, every student study at least one of the following subjects: English, Economics and Geography. 17 students study English, 21 students study Economics and 25 students study Geography. 10 students study Economics and English, 7 students study Economics and Geography, 3 students study only English and Geography. Find i. the number of students who study all the three subjects. ii. the number of students in the class. iii. the number of students who study only English. 23. In a class of 53 students, 31 read Accounting, 32 read Social studies and 23 read Statistics. 14 students read Accounting and Social Studies, 8 students read Accounting and Statistics, 3 students read only Accounting. If 5 students did not read any of the three programmes. a. Illustrate the above information on a Venn diagram. b. Find the number of students who read; i. Social studies and Statistics ii. Only Statistic. 24. In a class of 36 students, 19 read GKA, 16 read Food and Nutrition and 15 read Clothing and Textiles. 6 read only GKA, 4 read only Clothing and Textiles and 12 read GKA and Clothing and Textiles. Find; i. the number of students who read only Food and Nutrition. i. the number of students who read GKA but not Food and Nutrition. 25. In a class of 46 students, 20 read Biology, 30 read Physics and 35 read Chemistry. 12 read Biology and Physics, 10 read Biology and Chemistry and 15 read Physics and Chemistry. Each student read at least one of the three subjects. a. Find the number of students who study all the three subjects.

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b. Find the number of students who study only two subjects. 26. There are 45 students in a class. 25 of them play Tennis, 23 of them play Football and 18 of them play Handball. 8 students play all the three games, 16 play Tennis and Football, 12 play Tennis and Handball while 6 Football and Handball. i. Illustrate the information on a Venn diagram. ii. Find the number of students who play at least two games. 27. In a survey of eating habit of 110 participants, 50 of them like eating in the Morning, 60 of them like eating in the Afternoon and 40 of them like eating in the Evening. Exactly 22 participants like eating in the Morning and Afternoon or Morning and Evening or Afternoon and Evening. Find the number of students who eat in the morning and Afternoon and Evening. 28. In a small town of 300 people, 120 of them speak Twi, 110 of them speak Ga and 130 of them speak Fante. 43 of them speak only Twi and Ga, 32 of them speak only Twi and Fante and 41 of them speak only Ga and Fante and 푥 speak all the three languages. Find the value of 푥. 29. In a class of 48 students, 22 read French, 30 read English and 푥 students read both French and English. Each student read at least one of the two subject. Find, i. the value of 푥 ii.The number of students who read only French. 30. In a class of 65 students, 30 read Mathematics, 22 read Economics and 26 read Geography. 15 students read Mathematics and Economics, 13 students read Mathematics and Geography and 11 read Economics and Geography. Each student read at least one of the three courses. a. Represent this information in a Venn diagram. b. Find, the number of students who read, i. all the three courses. ii. only Geography iii. Only two subjects. c. If the probability of selecting a student at random from the class is estimated as 푃(푁) =

, estimate the probability of selecting a student who

read only one subject. 31. In a class of 60 students, of them offer Government and of them offer Geography. Find the fraction of students who offer both subjects if every student offer at least one of the two subjects. 32. In a mathematics class, 20 liked Calculus and only 6 liked Trigonometry. 5 students did not like any of the two topics.

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(a) Illustrate the above information in a Venn diagram. (b) Find the total number of students in the class. 33. List all the possible subsets of 퐴 = {1, 2, 3}, and state the compliment of each. (CCE) 34. Given that 퐴 = {푎, 푏, 푐,푑},퐵 = {푏,푑,푓, ℎ, 푗} and 퐶 = {푎, 푐, 푒,푔, 푖};verify the associative property of intersection of sets. (CCE) 35. A number of students were asked whether they liked football, athletics or boxing. Twelve students said they liked football, 16 liked athletics and 21 liked boxing. Only three students said they liked all three sports. Five students liked football and athletics (this includes those who liked all three sports), 8 liked football and boxing and 12 students like athletics and boxing. How many students liked (i) football only? (ii) athletics only? (iii) boxing only? (iv) football or athletics or boxing? (CCE)

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ARITHMETICS OF REAL NUMBERS The set of real numbers,ℝ, is defined by:

ℝ = ℚ∪ℚ

The four basic operations on real numbers are +(퐴푑푑푖푡푖표푛), −(푆푢푏푡푟푎푐푡푖표푛), ×/∙(푀푢푙푡푖푝푙푖푐푎푡푖표푛)1 and ÷ (퐷푖푣푖푠푖표푛). Natural Numbers The set of natural numbers is defined by:

ℕ = {1, 2, 3, … } We can classify ℕ into the following groups: Set of even numbers= {2, 4, 6, … } Set of odd numbers = {1, 3, 5, … } Set of prime numbers= {2, 3, 5, … } Set of square numbers= {1, 4, 9, … } The set of natural numbers is well-defined (closed) under the operations + and ×. This means that if푎 and푏 are natural numbers, then 푎 + 푏 and 푎 × 푏 are also natural numbers. For instance, 2 + 3 = 5 and 2 × 3 = 6. It is clearly seen that 5 = (2 + 3) and 6 = (2 × 3) are all members of ℕ. However, the set of odd numbers described above is closed under the operation × but not +. That is if 푎 and 푏 are odd, it is not true that 푎 + 푏 is odd. For example, 3 and 5are odd numbers but 3 + 5 = 8 is not odd. But 3× 5 = 15; which is odd. The set of even numbers is closed under both + and ×. That is, if 푎 and 푏 are even numbers, then 푎 + 푏 and 푎 × 푏 are also even numbers. For example, 2 + 4 = 6 and 2 × 4 = 8 The set of prime numbers is not closed under both + and ×. This means that, if 푎 and 푏 are prime numbers, it is not always true that 푎 + 푏 or 푎 × 푏 is prime. For example, 2+3=5 is prime but 2 + 7 = 9 is not prime. Also 2 × 3 = 6 is not prime 1 We will sometimes use ∙ for multiplication. Note that 푎. 푏 ≠ 푎 ∙ 푏

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We conclude by saying that: ∀푎, 푏 ∈ ℕ,푎 + 푏 ∈ ℕ and 푎 × 푏 ∈ ℕ. The symbol, ∀, means ‘for all’. Some properties of natural numbers 1. i. 푎 + 푏 = 푏 + 푎 ii. 푎 ∙ 푏 = 푏 ∙ 푎∀푎,푏 ∈ ℕ (Commutative property) 2. i.(푎 + 푏) + 푐 = 푎 + (푏 + 푐) ii. (푎 ∙ 푏) ∙ 푐 = 푎 ∙ (푏 ∙ 푐)∀푎,푏, 푐 ∈ ℕ (Associative property) 3. i. 푎 ∙ (푏 + 푐) = 푎 ∙ 푏 + 푎 ∙ 푐∀푎,푏, 푐 ∈ ℕ (Distributive property)

Example 2.1

Given that 푎 = 2푏 = 3 and 푐 = 4,∀푎, 푏, 푐 ∈ ℕ, show that 푎 + (푏 × 푐) ≠ (푎 + 푏) × (푎 + 푐)

Proof We have 푎 = 2푏 = 3 and 푐 = 4

푎 + (푏 × 푐) = 2 + (3 × 4) But 2 + 12 ≠ (2 + 3) × (2 + 4) ∎ The set of natural number is not closed under − because, if 푎,푏 ∈ ℕ,it is not always true that 푎 − 푏 ∈ ℕ. For instance, the result of 2 − 3 cannot be found in the set of natural numbers. This will lead us to define another set which is closed under +,× and −. Integers The set of integers is defined by:

ℤ = ℕ ∪ {0} ∪ ℕ , where ℕ and ℕ denote the set of negative and positive natural numbers respectively. The presence of negative numbers helps broaden the real number system. We can clearly see that the set of natural numbers is a subset of the set of integers. So we write

ℕ ⊂ ℤ The set of integers is closed under the operations +,− and ×. That is, ∀푎,푏 ∈ ℤ,then 푎 + 푏 ∈ ℤ,푎 − 푏 or 푏 − 푎 ∈ ℤ and 푎 × 푏 ∈ ℤ. For example, −2 + (−4) = −6 is an integer, −2 − (−4) = 2 is an integer, and −2 × −4 = 8 is also an integer. Note that, If 푎 and 푏 are integers, then

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i. –푎 − 푏 =– (푎 + 푏) ii. 푎 − (−푏) = 푎 + 푏 iii. –푎 ×–푏 = 푎푏 iv. – 푎 − (−푏) =– 푎 + 푏 For example – 2 − 5 =– (2 + 5) = −7 Try: Simplify the following: 1. 2 − 3 + 6 × −3 + 4 − 11 2. – 7 − 5 + 36— 6 × 3 3. – 5 × 3 + 4 × 5 − 22 + – 3 (Use the BODMAS principle)

Some properties of integers ∀푎,푏, 푐 ∈ ℤ

1. 푎 + 0 = 0 + 푎 = 푎 (Identity property) 2. 푎 + –푎 = –푎 + 푎 = 0 (Additive inverse property) 3. i. 푎 + 푏 = 푏 + 푎 ii. 푎 ∙ 푏 = 푏 ∙ 푎 (Commutative property) 4. i. 푎 ∙ (푏 + 푐) = 푎 ∙ 푏 + 푎 ∙ 푐 (Distributive property) 5. 푎 ∙ 푏 = 0,then 푎 = 0or 푏 = 0

Rational numbers

Rational numbers are numbers of the form ,where 푎 and 푏 are integers and 푏 ≠ 0. We call 푎 푛푢푚푒푟푎푡표푟and 푏 푑푒푛표푚푖푛푎푡표푟. Definition: A rational number whose numerator is 푙푒푠푠푒푟 than the denominator is called proper fraction. For example, is a proper fraction because 1 < 2. Definition: A rational number whose numerator is 푔푟푒푎푡푒푟than the denominator is called improper fraction. For instance, is an improper fraction because 9 > 2. Integers are rational number because every integer can be written as a ratio of two integers. For example 2 = = . We therefore conclude that ℤ ⊂ ℚ. A fraction can be a 푡푒푟푚푖푛푎푡푖푛푔 푑푒푐푖푚푎푙 or 푟푒푐푢푟푟푖푛푔 푑푒푐푖푚푎푙. For instance, = 0.5 is a terminating decimal but = 0. 3̇ is a recurring decimal.

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If a decimal number recur, we will put a dot on the number(s) which is recurring. The set of rational numbers is closed under the operations +,− and ×. It is not closed under ÷ because of division of by 0. That is not defined in the set of rational numbers, where 푎 is an integer. Note the following: 1. + = 2. 푎 + = 3. − = 4. 푎 − =

5. × = 6. 푎 × = 7. ÷ = × = In all the above, the denominators are not 0.

Example 2.2 Simplify the following: i. + ii. 6 + iii. − iv. × v. ÷ vi. 7 ×

Solution i. + = ( )( ) ( )( )

( )( )= = ii. 6 + = × = =

iii. − = ∙ ∙∙

= = iv. × = ××

=

v . ÷ = × = ( )( )( )( )

= ( )( ) = v. 7 × = × =

Try: Simplify the following:

1. + − 2. ÷ × ÷ 3. ÷ × + − 4. ÷

(Use the BODMAS principle)

The absolute value of a real number The absolute value of e real number 푎, wriiten as |푎| is defined by:

|푎| = 푎if푎 ≥ 0– 푎if푎 < 0

That is, the absolute value of positive real number is the same number but the absolute value of a negative real number is positive. For example |2| = 2 and |−2| = 2.

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Irrational numbers

In the set of real numbers, if a number is not rational, then it is irrational. Examples of irrational numbers are √3, 2√3, 3 − √2,휋, 풆and so on. In general, if √푎, cannot the written as a ratio of two integers, then it is irrational, where 푎 is positive if 푛 is even and negative if 푛 is odd. Therefore we can conclude that the nth root of all prime numbers is irrational. Traditionally, irrational numbers of the form √2, √3, 3√2, √7, and so on are called surds. For our level we will consider surds of the square root group.

Final exercises 1. Using the terms 푐푙표푠푒푑, 푐표푚푚푢푡푎푡푖푣푒,푎푠푠표푐푖푎푡푖푣푒,푑푖푠푡푟푖푏푢푡푖푣푒, 푎푑푑푖푡푖표푛, 푠푢푏푡푟푎푐푡푖표푛, 푎푛푑푚푢푙푡푖푝푙푖푐푎푡푖표푛,complete the statement of the property of real numbers illustrated below. (a) 2 × (3− 4) = (2 × 3)− (2 × 4). This means that “for all real numbers,…………….. is …………… over ……………. ”. (b)(2 × 3) × 4 = 2 × (3 × 4). This means that “all real numbers are ……………… under…………… multiplication”. (c) 2+3=5.This means that “real numbers are …………….. closed under …………….”. (CCE)

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SURDS Surds of the square root group include: √2, 3√3, 2√5,√7, 7√11, and so on. Reduction of a surd into its basic form The square root of prime numbers such as √2, √3,√5 … are examples of surds in their basic form. In general, if √푎 cannot be simplified further, then it is said to be in its basic form. For our level if√푎can be simplified further, then we need to find a square number say 푏 and a number say 푐 whose product 푏 × 푐 = 푎. For example, for us to simplify √8,we only need to find the square number 4 and the number 2 whose product4 × 2 = 8. Then we carry the simplification on as follow: √8 = √4 × 2 = √4 × √2 = 2√2 The table below will enable us to identify some square numbers.

풃 √풃 풃 √풃 풃 √풃 4 2 100 10 324 18 9 3 121 11 361 19 16 4 144 12 400 20 25 5 169 13 441 21 36 6 196 14 484 22 49 7 225 15 529 23 64 8 256 16 576 24 81 9 289 17 625 25

Example 3.1 Simplify the following surds: i. √18 ii. √72 iii. √125 iv. √128 v. √180 vi. √486

Solution i. √18 = √9 × 2 = √9 × √2 = 3√2 ii. √72 = √36 × 2 = √36 × √2 = 6√2 iii. √25 × 3 = √25 × 5 = √25 × √5 = 5√5 iv. √128 = √64 × 2 = √64 × √2 = 8√2 v. √180 = √36 × 5 = √36 × √5 = 6√5 vi. √486 = √81 × 6 = √81 × √6 = 9√6

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Try: Simplify: i. √27 ii. √98 iii. √120 iv. √50 vi. √80 v. √300 vi. √48 vii.√147

Addition and subtraction of surds Let 푚√푎 and 푛√푎 be two surds. The addition and subtraction of the two surds is defined by: 푚√푎 + 푛√푎 = (푚 + 푛)√푎 and 푚√푎 − 푛√푎 = (푚 − 푛)√푎, where 푎,푚and 푛 are rational numbers and 푎 > 0. Note that we can only add and subtract surds of the same kind. That is, the number in the radical sign must be the same.

Example 3.2

Simplify: i. √18 + √72 + √50 ii. √48 − √27 − √243 iii. 3√12 − 2√75 + √27− √192 iv. 3√24 − √6 + √216 − √54 v. √5 + √125 − √45 − √8 + √50

Solution i. √18 + √72 + √50 = √9 × 2 + √36 × 2 + √25 × 2 = 3√2 + 6√2 + 5√2 = (3 + 6 + 5)√2 = 14√2 ii. √48 − √27− √243 = √16 × 3 − √9 × 3 − √81 × 3 = 4√3 − 3√3 − 9√3 = (4 − 3 − 9)√3 = −8√2 iii. 3√12 − 2√75 + √27− √192 = 3 √4 × 3 − 2 √25 × 3 + √9 × 3 − √64 × 3 = 3 2√3 − 2 5√3 + 3√3 − 8√3 =6√3− 10√3 + 3√3 − 8√3 =(6 − 10 + 3 − 8)√3 = −9√3 iii. 3√24 − √6 + √216 − √54 = 3 √4 × 6 − √6 + √36 × 6 − √9 × 6

= 3 2√6 − √6 + 6√6 − 3√6 = 6√6 − √6 + 6√6 − 3√6

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= (6 − 1 + 6 − 3)√6 = 8√6 iv. √5 + √125 − √45 − √8 + √50

= √5 + √25 × 5 − √9 × 5 − √4 × 2 + √25 × 2

= √5 + × 5√5− × 3√5 − 2√2 + × 5√2

= √5 + √5 − √5 − 2√2 + √2

= + − 1 √5+ −2 + √2

= √5 + ( )√2

= 13√5 + √2

Example 3.3 For what value of 푎 is 3√45 − √180 − 6√5 = 푎√5.

Solution

We have 3√45 − √180 − 6√5 = 푎√5 ⟹ 3√9 × 5 − √36 × 5 − 6√5 = 푎√5 ⟹ 9√5 − 6√5 − 6√5 = 푎√5 ⟹ (9− 6 − 6)√5 = 푎√5 ⟹ −3√5 = 푎√5 ∴ 푎 = −3 (By dividing both sides by √5) Example 3.4 If 3√6 − 4√27− 4√216− √75 = 푎√3 + 푏√6,find the value of 푎 and 푏.

Solution We have 3√6− 4√27 − 4√216 − √75 = 푎√3 + 푏√6 ⟹ 3√6 − 4(√9 × 3) − 4√36 × 6 − √25 × 3 = 푎√3 + 푏√6 ⟹ 3√6 − 4(3)√3 − 4(6)√6 − 5√3 = 푎√3 + 푏√6 ⟹ 3√6 − 24√6 − 12√3 − 5√3 = 푎√3 + 푏√6 ⟹ (3− 24)√6 −(12 + 5)√3 = 푎√3 + 푏√6 ⟹ −21√6 − 17√3 = 푎√3 + 푏√6 ⟹ −17√3 − 21√6 = 푎√3 + 푏√6 By comparing co-efficients 푎 = −17 and 푏 = −21 ∴ 푎 = −17,푏 = −21 Try: 1. Simplify the following without using calculator.

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i. √7 + √63 + √252 ii. 2√11 − √44 − √99 iii. √98 − √200 + √32

iv. √76 + √475− 3√19 − √931 v. √117 + √13 − √52 + √208

2. a. For what value of 푥 is √1000− √40 + √250 − √160 = 푥√10. b. Find the values of 푥 and 푦 for which

−2√2 − 3√75 + 2√12 − √8 − √32 = 푥√2 + 푦√3.

Multiplication of surds Let 푠 and 푠 be two surds. The product of 푠 and 푠 is the number푠 ∙ 푠 ∈ ℝ. Let’s note the following properties of surds multiplication: 1. √푎 × √푏 = √푎푏 2. √푎 × √푎 = √푎 ∙ 푎 = √푎 = 푎 3. 푎 × 푏√푐 = 푎푏√푐 4. 푎√푏 × 푐√푑 = 푎푐√푏푑 Example 3.5 Simplify the following: i. √3 × √3 ii. √5 × √2 iii. 3√3 × √2 iv. 7√7 × 10√5 v. 8√6 × √5 vi. 2√12 × √12 × √2 Solution i. √3 × √3 = √3 = 3 ii. √5 × √2 = √5 × 2 = √10 iii. 3√3 × √2 = 3 √3 × 2 = 3√6 iv. 7√7 × 10√5 = (7 × 10) √7 × 10 = 70√70 v. 8√6 × √5 = 8( 6 × 5) = 8√30 vi. 2√12 × √12 × √2 = 2(√12 × 12 × 2) = 2 √12 × 2 = 24√2

Example 3.6 Expand and simplify the following: i. 2 − √3 1 + 2√3 ii. √6 − 5 √2 − 2 iii. √5 3 − 2√5 iv. 2 − √5 2 + √5 v. (푚√푛 − 푚)2

Solution

i. 2 − √3 1 + 2√3 ii. √6 − 5 √2 − 2 = √6 √2 − 2 − 5(√2 − 2) = 2 1 + 2√3 − √3 1 + 2√3 = √12 − 2√6 − 5√2 + 10

42

= 2 + 4√3 − √3− 2√3 = 4√3 − 5√2 − 2√6+10 = 2 − 6 + (4 − 1)√3 = −4 + 3√3 iii. √5 3 − 2√5 = 2√5 − 2√5 iv. 2 − √5 2 + √5 = 2 2 + √5 − √5 2 + √5 = 2√5 − 10 = 4 + 2√5 − 2√2 − √5 = 2(√5 − 5) = 4 − 5 = −1 v.(푚√푛 − 푚)2 = 푚√푛 −푚 푚√푛 − 푚 = 푚√푛 푚√푛 −푚 −푚(푚√푛 −푚) = 푚 √푛 −푚 √푛 −푚 √푛 + 푚 = 푚 푛 − 2푚 √푛 + 푚 Try: Expand and simplify the following: 1. (9 − 2√6)(6 − √6) 2. 11 − √7 (√5 − 1) 3. 2√3 − √2 (√2 − 2√3)

4. √13( √13 − 4) 5. √3 − 8 1 − √3 6. 3√2 − 2 2

7. √18 − √12 √12 + √2 8. 3 − √7 3 + √7

Conjugate of a surd Let 푠 denote the sum/difference of two different a surds. The conjugate of 푠, denoted by 푠̅, is the surd chosen so that 푠 ∙ 푠̅ ∈ ℚ. In general, if √푎 + √푏 is the sum of two different surds, then its conjugate is the surd

√푎 − √푏 chosen such that (√푎 + √푏 )(√푎 − √푏) = √푎 −√푏 = 푎 − 푏 ∈ ℚ. Similarly, if √푎 − √푏 is the difference of two surd different surds, then its conjugate is

the surd √푎 + √푏chosen such that(√푎 − √푏 )(√푎 + √푏) = √푎 −√푏 = 푎 − 푏 ∈ ℚ.

Example 3.7 Write down the conjugate of the following surds and find the product of the surd and its conjugate. i. √3 + √2 ii. √2 − 3 iii. √5 + 4 iv. 3 − √7

Solution

i. The conjugate of 푠 = √3 + √2 is 푠̅ = √3 − √2

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푠 ∙ 푠̅ = (√3 + √2)(√3− √2) = √3 √3 − √2 + √2 √3 − √2 = √3 − √6 + √6 − √2 = 3 − 2 = 1 ii. The conjugate of 푠 = √2 − 3 is 푠̅ = √2 + 3

푠 ∙ 푠̅ = √2 − 3 √2 + 3 = 2 − 3 = 2 − 9 = −7 iii. The conjugate of 푠 = √5 + 4 is 푠̅ = √5 − 4. 푠 ∙ 푠̅ = √5 + 4 √5 − 3 = √5 − 4 = 5− 16 = −11. iv. The conjugate of 푠 = 3 − √7 is 푠̅ = 3 + √7

푠 ∙ 푠̅ = 3 − √7 3 + √7 = 3 − 7 = 9 − 7 = 2 Try: Write down the conjugate of the following surds and find the product of the surd and its conjugate. i. 3+√2 ii. 2 − √5 iii. √3 − √7

Division of surds (Rationalising the denominator)2

In our quest to divide a surd by another surd we get 푓푟푎푐푡푖표푛of two surds. We are not interested in leaving the denominator as a surd so we carry on the process of rationalisation. The process of making fraction whose denominator is irrational rational is called rationalising the denominator.

Example 3.8 Rationalise the denominator of the following:

i. √√

ii. √√

iii. √√

√ √ iv. √

√√ √

Solution

i. To rationalise √√

, multiply the numerator and the denominator by √푐.

⟹ √√

= √√∙ √√

= √√

= √

ii. To rationalise √√

, multiply the numerator and the denominator by √푏.

2 We can also rationalise the numerator

44

⟹ √√

= √√∙ √√

= √√

= √

iii. To rationalise √√

√ √, multiply the numerator and the denominator by √푎 − √푏 (the

conjugate) of √푎 + √푏 ) .

⟹ √ √√ √

= √ √√ √

√ √√ √

= √ (√ √ ) √ (√ √ )√ √

= √ √ √ √

= √

iv. To rationalise √ √√ √

, multiply the numerator and the denominator by √푎 − √푏

(the conjugate of √푎 + √푏 ).

⟹ √√

√ √= √ √

(√ √ )√ √√ √

= √ (√ √ ) √ (√ √ )(√ )

= √ √ √ √

= √

Note: = √√

Example 3.9 a. Rationalise the denominator of the following fractions:

i. √

ii. √

iii. √

iv. √√

v. √√

vi. √√

Solution

i.√

=√∙ √√

= √√

= √ = √5

ii. √

=√∙ √√

= √√

= √ = √3

iii. √

=√

√√

= √√

= √ = + √2

iv. √√

= √√

√√

= √ √ ( √ )√

= √ √ √ = √

= − √3

v. √√

= √√

√√

= √ √ ( √ )√

= √ √ √

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= + √ − √7 − √

vi. √√

= √( √ )

√√

= √ √ ( √ )( ( √ ) )

= √ √ √( )

= √

= + √3

Example 3.10

Express √√

in the form 푎 + 푏√3,where 푎and푏 are rational numbers.

Solution

By simplifying √√

we have

√√

= √√

√√

= √ √ √√

= √ = − √3

Comparing − √3with 푎 + 푏√3푎 = , 푏 = −

∴, 4−√34+√3 = 19

13−8

13√3

Alternatively,

Let √√

= 푎 + 푏√3 ⟹ 4 − √3 = 푎 + 푏√3 4 + √3

= 4푎 + 푎√3 + 4푏√3 + 3푏 = 4푎 + 3푏 + (푎 + 4푏)√3 ⟹ 4푎 + 3푏 = 4 … … …(1) and 푎 + 4푏 = −1 … … … (2) From (2) 푎 = −1 − 4푏 put 푎 = −1 − 4푏 into (1) ⟹ 4(−1 − 4푏) + 3푏 = 4 ⟹ −4 − 16푏 + 3푏 = 4 ⟹ −13푏 = 4 + 4 ⟹ 푏 = − put 푏 = − into (2)⟹ 푎 = −1 − 4 − =

∴ 4−√34+√3 = 19

13 −8

13√3

Try:

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1. Evaluate √ √√ √

(SSSCE)

2. Simplify the expression √ √ √√

(SSSCE)

3. Express √√

in the form 푝 + 푞√5, where 푝 and 푞 are rational numbers (SSSCE )

4. Express √ √√ √

in the form 푎 + 푏√푐,where 푎,푏 and 푐 are rational numbers. (SSSCE)

5. Express √√ √

in the form 푚√3 + 푛√2, where 푚 and 푛 are rational numbers.

(SSSCE)

6. Express √ √√ √

in the form 푝+ 푞√푟,where 푝,푞 and 푟 are constant. (WASSCE)

Equality of surds Two surds 푚√푎 and 푛√푏 are equal if and on if 푚 = 푛 and 푎 = 푏.

Equations Two or more mathematical expressions joined by the sign = (equal to) forms an equation. For example, 푎푥 + 푏푦 = 푐and 푎푥 + 푏푥 + 푐 = 0, where 푎,푏 and 푐 are constants and 푥 and 푦 are variables. A variable is anything whose magnitude can change. That is, a variable can take different values. A constant is anything whose magnitude cannot change. In the equation 푎푥 + 푏푦 = 푐, the expression 푎푥 + 푏푦 forms the left hand side(퐿퐻푆) of the equation whereas 푐forms the right hand side (푅퐻푆)of the equation. If a constant is attached to a variable, it is called the coefficient of the variable.

Linear equations in one variable A linear equation in one variable is an equation defined in terms of only one variable whose power is 1. For example, 푎푥 + 푏 = 푐, where 푎, 푏and 푐 are constant and 푎 ≠ 0, is a linear equation in one variable 푥whose power is 1. For us to solve a linear equation in one variable we only need to follow the procedure below: 1. Multiply both sides of the equation by the least common multiple(퐿퐶푀)if the equations involve fractions.

47

2. Expand braces if any. 3. Group like terms at one side of the equation. 4. Divide both side of the equation by the coefficient of the variable. First, let’s try to solve the value of 푥 in the equation 푎푥 + 푏 = 푐. Since the equation does not involve fractions and brackets, we move on to procedure. In doing so, we will try to group 푏 and 푐 at one side of the equation. This carried is on as follows:

푎푥 + 푏 − 푏 = 푐 − 푏 ⟹ 푎푥 = 푐 − 푏 We then divide both side by the coefficient of 푥 as follows:

= ⟹ 푥 = ∴ the value of 푥 is (To make things simple for us, we will now agree that if a term is positive, we can move it to one side of the equation by negating it and vice versa.)

Example 3.11 Find the values of 푥 and 푦 in the following equations: i.푥 + 4 = 12 − 3푥 ii. 2(푥 + 5) − 3(1− 푥) = 17 iii. + = 6 −

iv. √2푥 + 4 = −√3푥 − 4 v. ( ) + ( ) = − (푦 + 2)

Solution i. We have 푥 + 4 = 12 − 3푥 Grouping like terms gives 푥 + 3푥 = 12 − 4 ⟹ 4푥 = 8 ⟹ 푥 = = 2 ∴ 푥 = 2. ii. We have 2(푥 + 5) − 3(1 − 푥) = 17 Expanding braces gives 2푥 + 10 − 3 + 3푥 = 17 Grouping like terms gives 2푥 + 3푥 = 17 − 10 + 3 ⟹ 5푥 = 10 ⟹ 푥 = = 2 ∴ 푥 = 2 iii. We have + = 6 − . The LCM of the denominators is 15. Multiply both sides by the denominator gives 15 + = 15 6 − ⟹ 15 × + 15 × = 15 × 6 − 15 ×

⟹ 5 × 2푦+ 3 × 4 = 90 − 3 × 3푦 ⟹ 10푦 + 12 = 90 − 9푦 Grouping like terms gives 10푦 + 9푦 = 90 − 12 ⟹ 19푦 = 78 ⟹ 푦 = ∴ 푦 =

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iv. Given √2푥 + 4 = −√3푥 − 4 ⟹ √2푥 + √3푥 = −8 ⟹ √2 + √3 푥 = −8

⟹ 푥 = −8√2+√3 = −8

√2+√3√2−√3√2−√3 = −8√2+8√3

22− 32= −8√2+8√3

2−3 = 8√2 − 8√3

∴ 푥 = 8(√2 − √3) v. given ( ) + ( ) = − (푦 + 2) multiplying both sides by the LCM gives.

12 ×(6푦 + 2)

4+ 12 ×

3(푦 − 1)6

= 12 ×32− 12 × (푦 + 2)

⟹ 3(6푦 + 2) + 2 × 3(푦 − 1) = 6 × 3 − 12푦 − 24 ⟹ 18푦 + 6 + 6푦 − 6 = 18− 12푦 − 24 Grouping like terms gives 18푦 + 6푦 + 12푦 = 18 − 24 + 6 − 6 ⟹ 36푦 = −6 ⟹ 푦 = −

∴ 푦 = −16

Try: Find the values of 푥 and 푧 in the following equations: i .푥 − 4 = 12 + 6푥 ii. 2(푧 − 35)− (4 + 푧) = 17 iii. + = −6 −

iv. √6푥 + 4√2 = √2푥 − 4 v. ( ) + ( ) = (푧 − 2)

Linear equations in two and three variables A linear equation in two variables is an equation of the form 푎푥 + 푏푦 = 푐,where 푎, 푏,and 푐 are constants, both 푎 and 푏 are not zero and 푥 and 푦 are variables. Consider the following set of equations:

푎 푥 + 푏 푦 = 푐 … … …(1) 푎 푥 + 푏 푦 = 푐 … … …(2)

These are two linear equations in two unknown variables 푥 and 푦,where 푎 , 푎 , 푏 ,푏 ,푐 and 푐 constants. A solution this system of equations means a pair of values for 푥 and 푦 which satisfy these equations simultaneously. System of equations can be inconsistent, dependent or independent. In this book we will work only on equations which are consistent and independent.

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A set of linear equations in equations in three variables 푥,푦 and 푧 can be written in form:

푎 푥 + 푏 푦 + 푐 푧 = 푑 … … …(1) 푎 푥 + 푏 푦 + 푐 푧 = 푑 … … …(2) 푎 푥 + 푏 푦 + 푐 푧 = 푑 … … … (3)

where 푎 , 푎 , 푎 , 푏 , 푏 , 푏 , 푐 , 푐 , 푐 , 푑 , 푑 and푑 are constants.

Methods for solving system of linear equations For now we will discuss two exact methods for solving system of linear equations.

1. Elimination method This method involves elimination of variable(s) of our choice in other to get the value of other variable(s). For two linear equations in two unknown variables, we try to eliminate one variable of our choice in other to get the other variable. We then repeat the same approach to find the variable we first eliminated. For three linear equations in three unknown variables, we try to eliminate a variable of our choice in other to get two linear equations in two unknown variables. We then carry on the steps for eliminating two linear equations in two unknown variables. For us to understand the elimination process well, we will treat coefficients of variables and the sign in front of the coefficients in a given system of equations separately. Let’s note down the following steps: 1. Choose a variable you want to eliminate in the equations given. 2. Check the coefficients and the signs in front of that variable. 3. If the coefficients and the sign are the same, subtract one equation from the other. If the signs differ add one equation to the other. 4. If the coefficients are not the same, then choose a number that when you multiply by the equations, will make the coefficients the same. After this; repeat step 3. 5. Start from step 1 through 4 and eliminate the other variable.

Example 3.12

Solve for the values of the variables in the following equations using the elimination method.

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Solution i. We have 푥 + 푦 = 4 … … … (1) and 푥 − 푦 = 6 … … … (2) Let’s first eliminate 푥. We can easily see that 푥 has the same coefficient in both equations. We can also see that the arithmetic sign in front of 푥 in both equations is the same. Then, by applying the rule: Subtract (1) from (2) ⟹ 푥 − 푥 − 푦 − 푦 = 6 − 4 ⟹ −2푦 = 2 ⟹ 푦 = − = −1 Now, starting from (1) and (2) and applying the rules we can eliminate 푦 by adding (1) to (2). ⟹ 푥 + 푥 + 푦 − 푦 = 4 + 6 ⟹ 2푥 = 10 ⟹ 푥 = = 5.

∴ 푥 = 5,푦 = −1. ii. We have 푥 + 2푦 = 17 … … … (1) and 2푥 − 푦 = 9 … … … (2) Let’s first eliminate 푥. We can easily that the coefficients of 푥 in both equations are not the same. So we proceed by multiplying (1) by 2 and (2) by 1. ⟹ 2(푥 + 2푦) = 2 × 17 ⟹ 2푥 + 4푦 = 34⋯⋯ (1푎) and 2푥 − 푦 = 9⋯⋯ (2) Now, we can eliminate 푥 by subtracting (1푎) from (2) ⟹ 2푥 − 2푥 − 푦 − 4푦 = 9 − 34 ⟹ −5푦 = −25 ⟹ 푦 = 5 Now, starting from (1) and (2) and applying the rules we can eliminate 푦 by first multiplying (1) by 1 and (2) by 2. ⟹ 푥 + 2푦 = 17⋯⋯ (1) and 2(2푥 − 푦) = 2 × 9 ⟹ 4푥 − 2푦 = 18⋯⋯ (2푎) Now, add (1) to (2푎) ⟹ 4푥 + 푥 − 2푦 + 2푦 = 18 + 17 ⟹ 5푥 = 35 ⟹ 푥 = = 7

∴ 푥 = 5, 푦 = 7 iii. We have 2푥 + 3푦 − 푧 − 8 = 0 … … … (1) −푥 + 푦 + 2푧 + 4 = 0 … … …(2) and

3푥 + 푦 − 2푧 − 2 = 0 … … … (3) Let’s eliminate 푥 from (1) and (2) ⟹ 2푥 + 3푦 − 푧 − 8 = 0 … … … (1) and 2(−푥 + 푦 + 2푧 + 4) = 2 × 0 ⟹ −2푥 + 2푦 + 4푧 + 8 = 0 … … … (2푎) adding (1) to (2푎) gives −2푥 + 2푥 + 2푦 + 3푦 + 4푧 − 푧 + 8 − 8 = 0 ⟹ 5푦 + 3푧 = 0 ⟹ 푦 = 0 … … … (4)

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Let us eliminate 푥 from (2) and (3) ⟹ 3(−푥 + 푦 + 2푧 + 4) = 3 × 0 ⟹ −3푥 + 3푦 + 6푧 + 12 = 0 … … … (2푏) and 3푥 + 푦 − 2푧 − 2 = 0 … … … (3) Adding (2푏) to (3) gives 3푥 − 3푥 + 푦 + 3푦 − 2푧 + 6푧 − 2 + 12 = 0 ⟹ 4푦 + 4푧 + 10 = 0 … … …(5) Let us eliminate 푦 from (4) and (5) ⟹ 4(5푦+ 3푧) = 4 × 0 ⟹ 20푦 + 12푧 = 0 … … … (4푎) and

5(4푦 + 4푧 + 10) = 5 × 0 ⟹ 20푦+ 20푧 + 50 = 0 … … … (5푎) Subtract (4푎) from (5푎) ⟹ 20푧 − 12푧 + 50 = 0 ⟹ 8푧 = −50 ⟹ 푧 = − = − Let’s eliminate 푧 from (4) and (5) ⟹ 4(5푦+ 3푧) = 4 × 0 ⟹ 20푦 + 12푧 = 0 … … … (4푎) and

3(4푦 + 4푧 + 10) = 5 × 0 ⟹ 12푦 + 12푧 + 30 = 0 … … … (5푏) Subtract (4푎) from (5푏) ⟹ 12푦 − 20푦 + 30 = 0 ⟹ −8푦 = −30 ⟹ 푦 = = Let’s eliminate 푧 from (1) and (2)

⟹ 2(2푥 + 3푦 − 푧 − 8) = 2 × 0 ⟹ 4푥 + 6푦 − 2푧 − 16 = 0 … … … (1푎) Adding (1푎) to (2) gives

−푥 + 4푥 + 푦 + 6푦 + 2푧 − 2푧 + 4− 16 = 0 ⟹ 3푥 + 7푦 − 12 = 0 … … … (6) Let’s eliminate 푧 from (2) and (3) Adding (2) to (3) gives

3푥 − 푥 + 푦 + 푦 − 2푧 + 2푧 − 2 + 4 = 0 ⟹ 2푥 + 2푦 + 2 = 0 … … … (7) Let’s eliminate 푦 from (6) and (7) ⟹ 2(3푥 + 7푦 − 12) = 2 × 0 ⟹ 6푥 + 14푦 − 24 = 0 … … … (6푎) and 7(2푥 + 2푦 + 2) = 5 × 0 ⟹ 14푥 + 14푦 + 14 = 0 … … … (7푎) Subtract (6푎) from (7푎) ⟹ 14푥 − 6푥 + 10 − (−24) = 0 ⟹ 8푥 = −38 ⟹ 푥 = − = −

∴ 푥 = − ,푦 = , 푧 = −

2. Substitution method To substitute means to replace. This method works in the following way: 1. Make a variable in any of the equations the subject.

52

2. Substitute the value of that variable in terms of the other variable(s) in the other equation(s). 3. Solve for the value of the other value variable and replace the value in any of the equation. This gives the value of the variable you made the subject. Note: For linear equations in three unknown variables, it may sometimes be convenient to combine the two methods.

Example 3.13 Solve the equations in example 3.12 using the substitution method.

Solution i. We have 푥 + 푦 = 4 … … … (1) and 푥 − 푦 = 6 … … … (2) From (1)푥 = 4 − 푦 Substitute 푥 = 4 − 푦 into (2) ⟹ 4 − 푦 − 푦 = 6 ⟹ 4− 2푦 = 6 ⟹ −2푦 = 6 − 4 ⟹ 푦 = − = −1 Put 푦 = −1 into (1) ⟹ 푥 = 4 − (−1) = 5 ∴ 푥 = 5,푦 = −1 ii. We have 푥 + 2푦 = 17 … … … (1) and 2푥 − 푦 = 9 … … … (2) From (1) 푥 = 17 − 2푦Substitute 푥 = 17 − 2푦 into (2) ⟹ 2(17 − 2푦) − 푦 = 9 ⟹ 34 − 4푦 − 푦 = 9⟹ 34 − 9 = 4푦 + 푦 ⟹푦 = = 5 Put 푦 = 5 into (1) ⟹ 푥 = 17 − 2(5) = 7 ∴ 푥 = 7,푦 = 5 iii. We have 2푥 + 3푦 − 푧 − 8 = 0 … … … (1) −푥 + 푦 + 2푧 + 4 = 0 … … … (2) and 3푥 + 푦 − 2푧 − 2 = 0 … … … (3) From (2) 푦 = 푥 − 2푧 − 4 By substituting 푦 = 푥 − 2푧 − 4 into (1) gives 2푥 + 3(푥 − 2푧 − 4)− 푧 − 8 = 0 ⟹ 2푥 + 3푥 − 6푧 − 12 − 푧 − 8 = 0 ⟹ 5푥 − 7푧 − 20 = 0 … … … (4) By substituting 푦 = 푥 − 2푧 − 4 into (3) gives 3푥 + 푥 − 2푧 − 4 − 2푧 − 2 = 0 ⟹ 4푥 − 4푧 − 6 = 0. Dividing both sides by 2 gives

2푥 − 2푧 − 3 = 0 … … … (5) From (5) 푥 = . Put 푥 = into (4)

⟹ 5 − 7푧 − 20 = 0 ⟹ − 7푧 − 20 = 0

⟹ 2 − 2 × 7푧 − 2 × 20 = 0 ⟹ 15+10푧 − 14푧 − 40 = 0

⟹ −4푧 = 40 − 15 ⟹ 푧 = −

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Put 푧 = − into (5) ⟹ 푥 =( )

= ÷ 4 = = −

Put 푥 = − , 푧 = − into (2) ⟹ 푦 = − − 2 − − 4 = =

∴ 푥 = − ,푦 = ,푧 = − Try: Solve for the value of the variables in the following equations. i. 푥 − 3푦 − 1 = 0 and ii. 2푥 − 3푦 = 8 and iii. 푥 + 푦 = 3 −2푦+ 3푥 = 7 3푥 + 2푦 = 42 푥 − 3푦+ 푧 = 10 and 푦 − 3푥 − 2푧 = 17 vi. 3푥 + 2푦 = 15 and ii. 3푥 + 5푦 = 14 and iii. 푥 + 푦 − 3푧 = 3 푦 + 2푥 = 65 푦 + 3푥 = 4 2푥 − 3푦 − 푧 = −11 and 푦 + 3푥 − 2푧 = 15

Quadratic equations3

A quadratic equation is an equation of the form 푎푥 + 푏푥 + 푐 = 0, where 푎, 푏 and 푐 are constants and 푎 ≠ 0. For now, we are interested in finding the values of the variable in a given quadratic equation. The values of 푥 in the quadratic equation 푎푥 + 푏푥 + 푐 = 0 are given by the formula:

푥 =−푏± 푏2−4푎푐

2푎 Where 푎isthe coefficient of is 푥 ,푏is the coefficient of 푥 and 푐 is a constant.

Example 3.14 Determine the values of 푥 and 푦in the following equations i.2푥 − 3푥 − 10 = 0 ii. 푥 − 2푥 + 1 = 0 iii. 5푦 − 6푦 − 3 = 0.

Solution i. We have 2푥 − 3푥 − 10 = 0. Comparing our given equation with the general equation 푎푥 + 푏푥 + 푐 = 0 ⟹ 푎 = 2,푏 = −3 and푐 = −10 The values of 푥 are obtained by the formula:

3 We will discuss quadratic equations in later topics

54

푥 =−푏± 푏2−4푎푐

2푎 ⟹ 푥 =−(−3)± (−3)2−4(2)(−10)

2(2) = 3± 9+804 = 3±√89

4

∴ 푥 = √ ≈ 3.1085 or 푥 = √ ≈ −1.6084 ii. We have 푥 − 2푥 + 1 = 0. Comparing our given equation with the general equation 푎푥 + 푏푥 + 푐 = 0 ⟹ 푎 = 1,푏 = −2 and푐 = 1 The values of 푥 are obtained by the formula:

푥 =−푏± 푏2−4푎푐

2푎 ⟹ 푥 = ( )± ( ) ( )( )( )

= ±√ = = 1

∴ 푥 = 1 iii. . We have 5y − 6푥 − 3 = 0. Comparing our given equation with the general equation 푎푦 + 푏푦 + 푐 = 0 ⟹ 푎 = 5, 푏 = −6 and푐 = −3 The values of 푦 are obtained by the formula:

푦 =−푏± 푏2−4푎푐

2푎 ⟹ 푦 =−(−6)± (−6)2−4(5)(−3)

2(5) = 6± 36+6010 = 6±√96

10

∴ 푦 = √ ≈ 1.5798 or 푥 = √ ≈ −0.3798 Try: Solve for the values of 푥 and 푦 in the following equations: i. 2푥 + 2푥 − 3 ii. 푥 − 5푥 + 6 iii. 3푦 − 3푦 − 16 = 0 iv. 2푥 − 푥 = 6

Example 3.15 Find the positive square root of the following: i. 3 − 2√2 ii. 12.5 + 5√6

Solution

i. The 3 − 2√2 = √푎 − √푏⟹ 3 − 2√2 = √푎 − √푏

⟹ 3 − 2√2 = 푎 + 푏 − 2√푎푏. By comparing 푎 + 푏 = 3 … … … (1) and −2√2 = −2√푎푏

(Dividing both sides by −2gives) √2 = √푎푏 ⟹ √2 = (√푎푏) ⟹ 2 = 푎푏… … … (2) From (1) 푏 = 3 − 푎 Put 푏 = 3− 푎 into (2) ⟹ 푎(3 − 푎) = 2 ⟹ 3푎 − 푎 = 2 ⟹ 푎 − 3푎 + 2 = 0 The values of 푎 can be obtained from the quadratic formula such that

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푎 =−(−3) ± (−3) − 4(1)(2)

2(1) =3 ± √9 − 8

2 =3 ± 1

2

⟹ 푎 = = 2 or 푎 = = 1 Put the values of 푎 into (1) That is, when 푎 = 2,푏 = 3 − 2 = 1 and when 푎 = 1, 푏 = 3 − 1 = 2

퐵푢푡푎 = 2and 푏 = 1 since 푥 − 푦√푧 ≥ 0 ∴ 3− 2√2 = √2 − 1

ii. The 12.5 + 5√6 = √푎 + √푏 ⟹ 12.5 + 5√6 = √푎 − √푏

⟹ 12.5 + 5√6 = 푎 + 푏 + 2√푎푏. By comparing 푎 + 푏 = 12.5 … … … (1) and 5√6 = 2√푎푏

⟹ 5√6 = (2√푎푏) ⟹ 150 = 4푎푏… … … (2) From (1) 푏 = 12.5 − 푎 Put 푏 = 12.5 − 푎 into (2) ⟹ 4푎(12.5− 푎) = 150 ⟹ 50푎 − 4푎 = 150 ⟹ 4푎 − 50푎 + 150 = 0 (Dividing both sides by 2gives) 2푎 − 25푎 + 75 = 0 The values of 푎 can be obtained from the quadratic formula such that

푎 = ( )± ( ) ( )( )( )

= ±√ = ±√

⟹ 푎 = = 15 or 푎 = = 10 Put the values of 푎 into (1) That is, when 푎 = 15,푏 = 12.5 − 15 = −3.5 and when 푎 = 10, 푏 = 12.5− 10 = 2.5

퐵푢푡푎 = 10and 푏 = 2.5 since 푏 ≥ 0 and 푥 − 푦√푧 ≥ 0 ∴, 12.5 − 5√6 = √10 − 2.5 Try: Find the positive square root of the following: i.16 − √2 ii. 25 + √3

Radical equations

These are equations involving the radical sign √. Note the following: 1. (√푎 + √푏) = √푎 + √푏 √푎 + √푏 = 푎 + 2√푎푏+ 푏 2.(푎 + 푏) = (푎 + 푏)(푎 + 푏) = 푎 + 2푎푏 + 푏 3. (√푎 − √푏) = √푎 − √푏 √푎 − √푏 = 푎 − 2√푎푏+ 푏

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4. (푎 − 푏) = (푎 − 푏)(푎 − 푏) = 푎 − 2푎푏 + 푏 Example 3.15

Solve the following equations: i. √푥 + 3 = 4 ii. (푥 − 6) = 64 iii. √2푥 − √푥 + 2 = 0 iv. √푥 + 2 − √푥 + 1 = 1 v. √푥 − 3 + 푥 = 8 vi. √3푥 − 2 − √푥 − 5 = 3

Solution

i. Given √푥 + 3 = 4 (square both sides) ⟹ √푥 + 3 = 4 ⟹ 푥 + 3 = 16 ⟹ 푥 = 16 − 3 = 13 Since it is true that√13 + 3 = 4, ∴, 푥 = 16

ii. Given (푥 − 6) = 64 (square both sides) ⟹ (푥 − 6) = 64

⟹ (푥 − 6) = 4096 (take cube root of both side) ⟹ (푥 − 6) = √4096 ⟹ 푥 − 6 = 16 ⟹ 푥 = 16 + 6 = 22 Since it is true that (22− 6) = 64, ∴, 푥 = 16 iii. Given √2푥 − √푥 + 2 = 0 (rearrange the equation) ⟹ √2푥 = √푥 + 2

(Square both sides) ⟹ √2푥 = √푥 + 2 ⟹ 2푥 = 푥 + 2 ⟹ 2푥 − 푥 = 2 ⟹ 푥 = 2 Since it is true that 2(2)− √2 + 2 = 0, ∴ 푥 = 2 iv. Given √푥 + 2 − √푥 + 1 = 1 (rearrange the equation) ⟹ √푥 + 2 = 1 + √푥 + 1

(Square both sides) ⟹ √푥 + 2 = 1 + √푥 + 1 ⟹ 푥 + 2 = 1 + 2√푥 + 1 + 푥 + 1 ⟹ 푥 − 푥 + 2 − 2 = 2√푥 + 1 ⟹ 0 = 2√푥 + 1

(Dividing both sides by 2 gives)√푥 + 1 = 0 (Square both sides) ⟹ √푥 + 1 = 0 ⟹ 푥 + 1 = 0 ⟹ 푥 = −1. Since it is true that (−1) + 2 − (−1) + 1 = 0, ∴ ,푥 = −1 v. Given √푥 − 3 + 푥 = 8 (rearrange the equation) ⟹ √푥 − 3 = 8 − 푥

(Square both sides) ⟹ √푥 − 3 = (8 − 푥) ⟹ 푥 − 3 = 64 − 16푥 + 푥 ⟹ 푥 − 16푥 − 푥 + 64 + 3 = 0 ⟹ 푥 − 17푥 + 67 = 0 The values of 푥 of the equation can be obtained by the formula

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푥 =−푏± 푏2−4푎푐

2푎 ⟹ 푥 = ( )± ( ) ( )( )( )

= ±√

푥 = √ ≈ 10.7923 or 푥 = √ ≈ 6.2087

Now, we see that √10.7923− 3 + 10.7923 ≉ 8 but √6.2087 − 3 + 6.2087 ≈ 8. ∴ 푥 = 6.2087

vi. Given √3푥 − 2− √푥 − 5 = 3 (rearrange the equation) ⟹ √3푥 − 2 = 3 + √푥 − 5

(Square both sides) ⟹ √3푥 − 2 = 3 + √푥 − 5 ⟹ 3푥 − 2 = 9 + 6√푥 − 5 + 푥 − 5 ⟹ 3푥 − 푥 − 2 − 9 + 5 = 6√푥 − 5 ⟹ 2푥 − 6 = 6√푥 − 5 (Divide both sides by 2) ⟹ 푥 − 3 = 3√푥 − 5

(Square both sides) ⟹ (푥 − 3) = 3√푥 − 5 ⟹ 푥 − 6푥 + 9 = 9(푥 − 5)⟹ 푥 − 6푥 + 9 = 9푥 − 45 ⟹ 푥 − 6푥 − 9푥 + 9 + 45 = 0 ⟹ 푥 − 15푥 + 54 = 0 The values of 푥 of the equation can be obtained by the formula

푥 =−푏± 푏2−4푎푐

2푎 ⟹ 푥 = ( )± ( ) ( )( )( )

= ±√

푥 = = 9 or 푥 = = 6

Now, we see that 3(9) − 2 − √9 − 5 = 3 and 3(6)− 2 − √6 − 5 = 3. ∴ 푥 = 6or 푥 = 9 Try: Solve the following equations: 1. √푥 − 1 = 6 2. √푥 + 1 + 푥 = 3 3. √4푥 − 3 − √푥 − 1 = 0 4. √2푥 − 3 + √1 − 2푥 = 1 5. (푥 − 3) − 64 = 0 (SSSCE) 6. √3푥 + 4 − √푥 − 3 = 3 (SSSCE) 7. √푥 + 7 − √3푥 − 2 = 1 (SSSCE) 8. √3푥 + 4 − √푥 + 5 = 1 (WASSCE) 9. 푦 + 7 − 3푦 − 3 = 1 (WASSCE)

58

Final Exercises 1. Simplify the following. (i) √18 + √2 (ii) √27 − 13 (iii) √18 + √50 + √2− √3 (iv)√6 + 2√6 − √16 + √72 (v). √200 + √500 − √10 2. Simplify the following: i. √27 + √6 + √57 ii. √2 + √32 − √8 iii. √18 + √54− √108 iv. √18 + √12 v. √20 − √45 vi. √40 3. Expand and simplify the following: i. 2 + √5 √5− 2 ii. 3 − √6 2√5 − 2√3 iii. √2 + 3 4 + √2

iv. √3 − 1 √3 + 8 v. √3(√3 − 5) iv. 2 + √3

vi. 2 − √3 vii. √ √

4. Expand and simplify the following: i. √푏 푛 + √푘 √푘 − 푛 ii. 푚 − √푛 푚√푛 − 푛√푚 iii. 푚 √푚 + 3 4 + √푛

iv. √27 √3 + 8 v. √18(√3− 5) iv. 1 − √푛

vi. √2 √44− √242 vi. √

3√2 −√

viii. 3√2 − 2√2 2√2 + 3

5.Write down the conjugate of the following surds and find their product. i. 4√3 − 2 ii. 3 − √3 iii. 3 + 4√3 iv. √7 − √3 v. √3 − √17 6. Rationalise the denominator of the following:

i. √

ii. √ √

iii. √ √

iv. √ √

v. √√

vi. √ √√ √

vii. √ √√ √

viii. √

ix. √ √

7. Show that √ √

√= ( ) √ and hence simplify the following:

(i) √ √

√ (ii) √ √

√ (iii)√ √

√ √ (iv). √ √

√ √

8. If + = , evaluate the following:

(i). √√

+ √√

(ii).√√

+ √√

(iii). √ √√ √

− √ √√ √

(iv).√ √√ √

− √ √√ √

(v). √ √√

− √√ √

(vi).√ √√

+ √√

(vii). √ √√

+ + √ √√ √

(viii). √√

− √ √√

(x). √ √√ √

− √ √√ √

59

(xi). √√

− √√

9. Solve the following equations: i .2푥 − 3 = 12 + 5푥 ii. 2(푧 − 5) − (4 − 푧) = 11 iii. + = −6 −

iv. √3푥 + √2 = 5푥 + 3 v. ( ) + ( ) = (푥 + 1) vi. 4(2푥 − 5) + 3(2푥 − 3) = 18

vii. 푦 − = 8 − 푦 viii. + = 10. Show that the general solutions for the equations 푎 푥 + 푏 푦 = 푐 and 푏 푥 + 푎 푦 = 푑 is 푥 = , 푦 = and hence Solve the equations:

i.3푥 + 2푦 = 4 and 2푥 + 3푦 = 5 ii.√3푥 + √2푦 = 1 and √2푥 + √3푥 = 4 11. Solve the equations: 푥 − 푦 = 8and 푥 + 푦 = 4

푥 + 푦 = 16and 푥 − 푦 = 2 12. Solve the following quadratic equations: i.푥 − 3 = 0 ii. 2푥 − 푥 − 3 = 0 iii. 푥 + 푥 = 0 iv. 3푥 − 푥 + 6 = 0 v. 10푥 + 11푥 + 1 = 0 vi.1 − 3푥 − 2푥 = 0 vii. 3푥 − 13푥 + 10 = 0 viii. 16 − 15푥 − 푥 = 0 13. Express the following in this form 푎 + 푏√푐

i. √√

ii.√ √√ √

iii. √√

iv. √√

v. √√

14. Solve for the following: i. 푥 − 푦 = 8 and 푥 + 푦 = 4 ii. 2푥 − 푦 = 8 and 4푥 + 3푦 = 12 iii. −4푦 + 3푧 = 16 and 3푦 − 4푧 = 17 iv. 6푎 + 7푏 = 24 and 4푎 + 8푏 = 20 v. −7푓 + 4푔 = 10 and 4푓 − 3푔 = 1 15. By writing = 푎 and = 푏, find the values of 푥 and 푦 in the following systems of

equations: i. − = 3 and + = 4 ii. − = + 6 and − = −1

iii. − = and + =

16. Solve for the values of 푥,푦 and 푧 in each of the following:

i. 푥 + 푦 + 푧 = 3

3푥 + 2푦 − 푧 = 11푥 + 푦 + 2푧 = −9

ii. 푥 + 2푦 − 3푧 = −3푥 − 4푦 − 푧 = 13−푥 − 4푧 = 23

iii. 푥 + 푧 = 3

3푥 − 2푦 − 푧 = −1푥 + 푦 − 푧 = 0

17. Find the following

60

i. 10 − 2√5 ii. 17 − 5√3 iii. 19 − 2√15 iv. 11 + 8√3 18. Solve for the following: i. √푥 + 2 = 10 ii. √푥 + 2 + √푥 + 4 = 7 iii. 푦 − 1 + 4푦 + 2 = 8

iv. (푥 + 5) = 82 v. 푦 − 4 + 2푦 + 1 = 8 vi. √푥 + 2 + 푥 = 24 19. Solve for the following: i. √푥 + 5 − √푥 + 3 = 2 ii. √4 − 푥 + √4 + 푥 = 6 iii. 2√푥 = 푥 − 3 iv. √푥 − 1 = 4 − 푥 v. √푥 − 1 + 2 = √푥

20. Express √√

in the form 푎 + 푏√5.

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BINARY OPERATIONS A binary operation ∗ on a non-empty set 푆 is a rule for combining two elements of 푆to produce another unique element of 푆. Binary operations are usually denoted by symbols such as

+,−,×,÷,∗,∪,∩,∙, ∘, ⋀, ⦁,⧍,⨀,⨁,⨂,… Notation

If ∗ is a binary operation on a set 푆, we write 푎 ∗ 푏,∀푎,푏 ∈ 푆. This is called infix notation.

Closure If ∀푎,푏 ∈ 푆,푎 ∗ 푏 ∈ 푆, then we say that the set 푆 is closed under 푆. For example, if ∗ is defined on the set ℤ of integers by 푎 ∗ 푏 = 푎 − 푎푏,∀푎, 푏 ∈ ℤ,then 1 ∗ 2 = 1 − (1)2 = −1 which is also an integer. Similarly, −1 ∗ −2 = −1 − (−1)(−2) = −1 + 2 = 1 ∈ ℤ. That is, a set is closed under an operation if we combine two elements of the set the third it produces by the operation is also an element of the set.

Example 4.1 a. An operation ∗ is defined on the set ℕ = {1,2, 3, … } by 푎 ∗ 푏 = 푎 + 푏,for all 푎, 푏 ∈ ℕ. Determine whether or not ℕ is closed under∗.

Solution If ℕis closed under ∗ then the result of 푎 ∗ 푏 = 푎 + 푏 must be a member of ℕ. The set ℕ is closed under ∗ since, for example, 1 ∗ 2 = 1 + 2 = 3,2 ∗ 3 = 2 + 3 = 5, 3 ∗ 4 = 3 + 4 = 7and so on are all members of ℕ.

Example 4.2

A binary operation ⦁ is defined on the set 푆 = {1, 2, 3, 4} by the table beside. Determine whether or not 푆 is closed under ⦁.

Solution If 푆 is closed under ⦁then the result of two elements of 푆 must also be a member of 푆. From the table, we can see that the results of combining two elements of 푆 under ⦁are all members of 푆. For example, 1⦁1 = 1, 2⦁1 = 2 and so on are all elements of 푆.

⦁ 1 2 3 4 1 1 2 3 4 2 2 3 4 1 3 3 4 1 2 4 4 1 2 3

62

∴, set 푆 is closed under ⦁. Example 4.3

An operation ∘ is defined on the set 푆 = {1, 3, 4, 9} by 푥 ∘ 푦 = 푥 + 푦. Show that 푆 is not closed under ∘.

Solution If the set 푆 = {1, 3, 4, 9} is closed under ∘,then the results of 푥 ∘ 푦 = 푥 + 푦 are elements of 푆. Although, for example, 1 ∘ 4 = 1 + √4 = 3 is a member of 푆,3 ∘ 4 = 3 +√4 = 5, 4 ∘ 9 = 4 + √9 = 7 and 1 ∘ 3 = 1 + √3 are not members of 푆.Therefore, 푆 is not closed under ∘.

Example 4.4 A binary operation ∗ is defined on the set of 푟푒푎푙푛푢푚푏푒푟푠 by 푎 ∗ 푏 = 푎

2+ 푏, for all

푎, 푏 ∈ ℝ and 푏 ≥ 0. i. Calculate 2 ∗ 25 ii. Find 2 ∗ 3 leaving your answer in the form 푟 + 푠√3,where 푟 and 푠 are real numbers. iii. Find the value of 푥 such that 4 ∗ 푥 = −1.

Solution

i. Given 푎 ∗ 푏 = 푎2+ 푏

, 2 ∗ 25 = 22+ 25

= 22+5 = 2

7

ii. 2 ∗ 3 = 22+√3 = 2

2+√32−√32−√3 = 4−2√3

(2)2−(√3)2 = 4−2√34−3 = 4 − 2√3.

Comparing 4 − 2√3with 푟 + 푠√3, 푟 = 4푎푛푑푠 = −2. ∴ 2 ∗ 3 = 4 − 2√3 Alternatively: Let 2 ∗ 3 = 푟 + 푠√3 But 2 ∗ 3 =

√ ⟹

√= 푟 + 푠√3

⟹ 2 = 2 + √3 (푟 + 푠√3) ⟹ 2 = 2푟 + 2푠√3 + 푟√3 + 3푠 ⟹ 2 = 2푟+ 3푠 + (푟 + 2푠)√3 By comparing the 퐿퐻푆푤푖푡ℎ푅퐻푆, 2푟 + 3푠 = 2 … … … (1) and 푟 + 2푠 = 0 … … … (2) (solve (1) and (2) simultaneously) From (2) 푟 = −2푠. Put 푟 = −2푠 into (1). ⟹ 2(−2푠) + 3푠 = 3 ⟹ −4푠 + 3푠 = 2 ⟹ 푠 = −2. Put 푠 = −2 into (2). ⟹ 푟 = −2(−2) = 4.

∴ 2 ∗ 3 = 4 − 2√3

63

Try: 1. A binary operation ∘ is defined on the set 푆 = {푎,푏, 푐} by the table beside. Show that 푆 is closed under ∘. 2. A binary operation ∇ is defined on the set 푅 of real numbers by: 푎∇푏 = 푎 − 2푎푏+ 푏 , 푎, 푏 ∈ 푅. Find; a. 푟 such that 2∇ − 5 = √푟

b.( )∇

, 푥 ≠ 0 simplifying your answer as far as possible. (SSSCE)

3. A binary operation ∗ is defined on the set, 푅, of real numbers by 푎 ∗ 푏 = , 푎 ≠ −푏.

(i) Evaluate 3 ∗ √2,leaving your answer in the form 푝 + 푟√2,where 푝 and 푟 are real numbers. (ii) Find 푥 such that 2 ∗ 푥 = . (SSSCE)

Commutative property

A binary operation ∗ defined on the set 푆 is said to be commutative if it satisfies the condition that

푎 ∗ 푏 = 푏 ∗ 푎,for all 푎, 푏 ∈ 푆. For instance, the binary operation + defined on the set of 푟푒푎푙푛푢푚푏푒푟푠 is commutative because 푎 + 푏 = 푏 + 푎,∀푎, 푏 ∈ ℝ. Similarly, × defined on the set of 푟푒푎푙푛푢푚푏푒푟푠 is commutative because 푎 × 푏 = 푏 × 푎. That is 2 + 3 = 3 + 2 = 5 ; 2 × 3 = 3 × 2 = 6,∀2,3 ∈ ℝ. The binary operations – and ÷ defined on the set of 푟푒푎푙푛푢푚푏푒푟푠 are not commutative since 푎 − 푏 ≠ 푏 − 푎 and 푎 ÷ 푏 = 푏 ÷ 푎,∀푎, 푏 ∈ ℝ. That is, 2 − 3 ≠ 3 − 3; 2 ÷ 3 ≠ 3 ÷ 2,∀2, 3 ∈ ℝ.

Example 4.5 A binary operation ⨂ is defined on the set, ℚ, of rational numbers by 푎⨂푏 = 푎 + 2푎푏, for all푎, 푏 ∈ ℚ. i. Find 2⨂− 2 , 3⨂ and ⨂ . ii. Show that ⨂ is not commutative.

Solution

∘ 푎 푏 푐 푎 푎 푏 푐 푏 푏 푐 푎 푐 푐 푎 푏

64

i. Given 푎⨂푏 = 푎 + 2푎푏, 2⨂−2 = 2 + 2(2)(−2) = 2 − 8 = −6.

3⨂ = 3 + 2(3) = 3 + 3 = 6 ⨂ = + 2 = + = ii. If ⨂ is commutative, then푎⨂푏 = 푏⨂푎. But given ℚ푎 + 2푎푏 ≠ 푏 + 2푏푎

∴ ⨂ is not commutative. Check: 2⨂3 ≠ 3⨂ 2 since 2 + 2(2)(3) ≠ 3 + 2(3)(2) ∎

Example 4.6

A binary operation∗ is defined on the set 푆 = {2, 3,4, … } by 푥 ∗ 푦 = 푥 , for all 푥, 푦 ∈ 푆. i. Find, 2 ∗ 5 ii. 6 ∗ 7. iii. Show that ∗ is not commutative.

Solution i. Given 푥 ∗ 푦 = 푥 , 2 ∗ 5 = 2 = 32 ii. 6 ∗ 7 = 6 = 279936 ii. If ∗ is commutative, then 푥 ∗ 푦 = 푦 ∗ 푥. But given set 푆 = {2, 3,4, … },푥 ≠ 푦

∴∗ is not commutative. Check: 2 ∗ 3 ≠ 3 ∗ 2 since 2 ≠ 3 ∎

Example 4.7 A binary operation ⦁ is defined on the set of real numbers by 푎⦁푏 = 푎 − 푏 + 푎푏. Show whether or not ⦁ is commutative.

Solution If⦁ is commutative then, 푎⦁푏 = 푏⦁푎. But given the set of real numbers

푎 − 푏 + 푎푏 ≠ 푏 − 푎 + 푏푎. ∴, ⦁is not commutative Check: 2⦁3 ≠ 3⦁2 since 2 − 3 + 2(3) ≠ 3 − 2 + 3(2) ∎

Example 4.8

A binary operation ∘ is defined by 푢 ∘ 푣 = 푢 + 푣 + 푢푣. 훼. Form a table of the operation ∘ on the set 푆 = {1,3, 5} 훽. State with reasons whether or not i. 푆 closed under ∘; ii.∘ is Commutative under 푆.

Solution

∘ 1 3 5 1 3 13 31 3 13 27 49

65

훼. Given 푢 ∘ 푣 = 푢 + 푣 + 푢푣 1 ∘ 3 = 1 + 3 + 1(3) = 13

1 ∘ 5 = 1 + 5 + 1(5) = 31 3 ∘ 5 = 3 + 5 + 3(5) = 49

훽. i. From the table above, some results of 푢 ∘ 푣 are not members of 푆. For example, 1 ∘ 3 = 13 is not a member of 푆. Therefore, the set 푆 = {1,3, 5} is not closed under ∘. ii. If ∘ is commutative, then 푢 ∘ 푣 = 푣 ∘ 푢. 푢 ∘ 푣 = 푢 + 푣 + 푢푣 푣 ∘ 푢 = 푣 + 푢 + 푣푢 = 푢 ∘ 푣 ⟹ 푢 ∘ 푣 = 푣 ∘ 푢

∴, ∘ is commutative. Try: 1. An operation ∗ is defined on the set 푇 = {1, 2, 3, 4} by 푎 ∗ 푏 = 푎 + 푏 − 푎푏,where 푎, 푏 ∈ 푇. a. i. Copy and complete the table below. (SSSCE) ii. is 푇closed with respect to ∗? 2. Given that 푚∇푝 = 푚 + 푝 −푚푝, evaluate 4∇3. (SSSCE) 3. Given that 푚∆푝 = 푚 + 푝 −푚푝,evaluate 4∆3. (SSSCE) 4. A binary operation ⦁ is defined by 푚⦁푛 = (푚− 푛). a. Form a table of the operation ⦁ on the set 푇 = {2, 4, 6}. b. State with reason whether or not ⦁ is i. closed under 푇; ii. Commutative under 푇. (SSSCE) 5. A binary operation ∗ is defined on the set 푅, of real numbers by 푝 ∗ 푞 = 푝 + 푞 − 푝푞. a. Determine whether or not ∗ is commutative. b. Evaluate −2 ∗ 7. (SSSCE)

Associative property A binary operation ∗ defined on a set 푆 is said to be associative if

5 31 49 75

∗ 1 2 3 4 1 1 1 1 1 2 1 −1 −2 3 1 −1 4 1 −2 8

66

(푎 ∗ 푏) ∗ 푐 = 푎 ∗ (푏 ∗ 푐), ∀푎,푏 ∈ 푆. The binary operations + and × defined on the set of 푟푒푎푙푛푢푚푏푒푟푠 are associative because (푎 + 푏) + 푐 = 푎 + (푏 + 푐) and (푎 ∙ 푏)∙ 푐 = 푎 ∙ (푏 ∙ 푐),for all푎, 푏, 푐 ∈ ℝ. For instance , (2 + 3) + 4 = 5 + 4 = 2 + (3 + 4) = 2 + 7 = 9and (2 × 3) × 4 = 6 × 4 = 2 × (3 × 4) = 2 × 12 = 24, ∀2, 3,4 ∈ ℝ. The binary operations – and ÷are not associative because, for instance, (2 − 3) − 4 ≠2 − (3 − 4) and (2 ÷ 3) ÷ 4 ≠ 2 ÷ (3 ÷ 4),∀2, 3, 4 ∈ ℝ.

Example 4.9

A binary operation ∗ is defined on the set of real numbers by 푥 ∗ 푦 = 푥 + 2푦, ∀푥,푦 ∈ ℝ. i. Evaluate −3 ∗ (2 ∗ 4) ii. Show whether or not∗ is associative under ℝ.

Solution i. Given 푥 ∗ 푦 = 푥 + 2푦, −3 ∗ (2 ∗ 4) = −3 ∗ 2 + 2(4) = −3 ∗ 10 = −3 + 3(10) = 27 (BODMAS) ii. If ∗ associative, then (푥 ∗ 푦) ∗ 푧 = 푥 ∗ (푦 ∗ 푧),∀푥, 푦, 푧 ∈ ℝ Now, (푥 ∗ 푦) ∗ 푧 = (푥 + 2푦) ∗ 푧 = 푥 + 2푦 + 2푧

푥 ∗ (푦 ∗ 푧) = 푥 ∗ (푦+ 2푧) = 푥 + 2(푦 + 2푧) = 푥 + 2푦 + 4푧 Since (푥 ∗ 푦) ∗ 푧 ≠ 푥 ∗ (푦 ∗ 푧), ∗is not associative. ∎

Example 4.10 An operation ∇ is defined on the set of푛푎푡푢푟푎푙푛푢푚푏푒푟푠by 푎∇푏 = 푎 − 2푏 + 1, ∀푎,푏 ∈ ℕ. 훼. Evaluate (3∇5)∇7. 훽. Show that i. ℕ is not closed under ∇. ii. ∇ is not associative under ℕ.

Solution 훼.Given 푎∇푏 = 푎 − 2푏 + 1 , (3∇5)∇7 = (3 − 2 × 5 + 1)∇7 = −6∇7 = −6 − 2 × 7 + 1 = −19 (BODMAS) 훽.i. If ℕ is closed under ∇, then any result of 푎∇b must be a member of ℕ. But, for example, 1∇2 = 1 − 2(2) + 1 = −2 is not a member of ℕ. Therefore, ℕis not closed under ∇ ∎ ii. If ∇ is associative then (푎∇푏)∇푐 = 푎∇(푏∇푐),∀푎,푏, 푐 ∈ ℕ

67

Now, (푎∇푏)∇푐 = (푎 − 2푏 + 1)∇c = 푎 − 2푏 + 1 − 2푐 + 1 = 푎 − 2푏 − 2푐 + 2 푎∇(푏∇푐) = 푎∇(푏 − 2푐 + 1) = 푎 − 2(푏 − 2푐 + 1) + 1 = 푎 − 2푏 + 4푐 − 2 + 1

= 푎 − 2푏 + 4푐 − 1 Since (푎∇푏)∇푐 ≠ 푎∇(푏∇푐), ∇ is not associative ∎

Example 4.11

A binary operation ⨀ is defined on the set of 푟푒푎푙푛푢푚푏푒푟푠by 푎⨀푏 = 푎 + 푏 + 2푎푏, ∀푎,푏 ∈ ℝ. i. Evaluate (−2⨀−3)⨀5 ii. Determine whether or not the operation ⨀ is (훼) commutative (훽) associative. iii. Find the value of 푥 such that 2⨀ ⨀푥 = −1,푥 ∈ℝ.

Solution i. Given 푎⨀푏 = 푎 + 푏 + 2푎푏, (−2 ⨀−3) ⨀ 5 = (−2 + (−3) + 2 × −2 × −3)⨀ 5 = 7 ⨀ 5 = 7 + 5 + 2 × 7 × 5 = 82 (BODMAS) ii. (훼) If ⨀ is commutative, then 푎⨀푏 = 푏⨀푎 Now, 푎⨀푏 = 푎 + 푏 + 2푎푏 , 푏⨀푎 = 푏 + 푎 + 2푏푎 ⟹ 푏⨀푎 = 푏 + 푎 + 2푏푎 = 푎⨀푏 Therefore, ⨀ is commutative. (훽) If ⨀ is associative, then (푎⨀푏)⨀푐 = 푎⨀(푏⨀푐) ,∀푎, 푏,푐 ∈ ℝ Now, (푎⨀푏)⨀푐 = (푎 + 푏 + 2푎푏)⨀푐 = 푎 + 푏 + 2푎푏 + 푐 + 2(푎 + 푏 + 2푎푏)푐

= 푎 + 푏 + 2푎푏 + 푐 + 2푎푐 + 2푏푐 + 4푎푏푐 푎⨀(푏⨀푐) = 푎⨀(푏 + 푐 + 2푏푐) = 푎 + 푏 + 푐 + 2푏푐 + 2푎(푏 + 푐 + 2푏푐)

= 푎 + 푏 + 푐 + 2푏푐 + 2푎푏 + 2푎푐 + 4푎푏푐 Since (푎⨀푏)⨀푐 = 푎⨀(푏⨀ 푐) ,⨀ is associative. ∎

iii. 2⨀ ⨀푥 = −1 ⟹ 2 + + 2(2) ⨀푥 = −1 ⟹ ⨀푥 = −1

⟹ + 푥 + 2 (푥) = 1 ⟹ + 20푥 = −1 ⟹ 19 + 40푥 = −1

⟹ 푥 = − = − ∴ 푥 = − Try: 1. A binary operation ∆ is defined on the set of a non-zero rational numbers by 푎∇푏 = . Determine whether ∆ is; a. commutative b. associative. (SSSCE)

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2. a. A binary operation∗ is defined on the set of real numbers by 푎 ∗ 푏 = 푎 + 푏√2. Calculate each of the following leaving your answer in surd form. i. −2 ∗ 1 ii. 3 ∗ (2 ∗ 3) b. Find the set of real values of 푥 such that (2푥 − 1)(푥 + 2) ≥ 0. (SSSCE) 3. A binary operation ∗ is defined on the set 푆 = {2, 3, 5, 7} by 푎 ∗ 푏 = 푎 + 푏 − 2,where

푎, 푏 ∈ 푆. a. Copy and complete the table below. b. i. Show whether or not 푆is closed under ∗.ii. Evaluate 3 ∗ (2 ∗ 5). (SSSCE)

Identity/neutral element Let ∗ be a binary operation defined on a set 푆.If for all 푎, 푏 ∈ 푆푎 ∗ 푏 = 푏 ∗ 푎 = 푎, then 푏 is called an identity element. We will now use the letter 푒 instead of 푏 to denote the identity element.

푎 ∗ 푒 = 푒 ∗ 푎 = 푎 That is, either 푎 ∗ 푒 = 푎 (right identity) or 푒 ∗ 푎 = 푎 (left identity). The identity element is unique, i.e. there is one and only one identity element under a particular binary operation. For instance, 0 and 1 are the identity elements for addition (+) and multiplication (×) of real numbers respectively. That is if 푎 ∈ ℝ, then 푎 + 0 = 0 + 푎 = 푎 and 푎 × 1 = 1 × 푎 = 푎.

Example 4.12 A binary operation ⦁ is defined on the set of 푟푒푎푙푛푢푚푏푒푟푠by 푥⦁푦 = 푥 + 푥푦, for all 푥, 푦 ∈ ℝ. i. Evaluate 2⦁ − 3 ii. Find the identity element.

Solution i. Given 푥⦁푦 = 푥 + 푥푦, 2⦁ − 3 = 2 + 2(−3) = 2 − 6 = −4. ii. If an identity element exist then 푥⦁푒 = 푒⦁푥 = 푥, where 푒 is the identity element.

∗ 2 3 5 7 2 2 3 5 7 3 3 4 8 5 5 8 7 7 8 10

69

Take 푥⦁푒 = 푥 ⟹ 푥 + 푥푒 = 푥 ⟹ 푥푒 = 푥 − 푥 ⟹ 푥푒 = 0⟹ 푥 = = 푒 ∴ , the identity element 푒 = 0

Example 4.13

An operation ∗ is defined on the set 푆 = {1, 2, 3} by the table below. Use it to answer the questions that follow. i. Evaluate (2 ∗ 3) ∗ 1 ii. Find the identity element.

Solution i. From the table, (2 ∗ 3) ∗ 1 = 1 ∗ 1 = 1 ii. From the table, we can easily see that 1 ∗ 1 = 1, 2 ∗ 1 = 2and3 ∗ 1 = 3. Therefore, the identity element 푒 = 1

Example 4.14 Show that the operation푎 ∗ 푏 = 1 + 푎푏 the set of 푖푛푡푒푔푒푟푠 has no identity element.

Proof If an identity element exist, then 푎 ∗ 푒 = 푒 ∗ 푎 = 푎, where 푒 is the identity element. Take 푎 ∗ 푒 = 푎 ⟹1 + 푎푒 = 푎 ⟹ 푎푒 = 푎 − 1 ⟹ 푒 = .

Assume,푎 = 1, 푡ℎ푒푛푒 = = 0. If 푎 = −1, then푒 = = 2. So we conclude that if 푎 ≠ 0, every integer 푎 has its own identity element. But the identity element is unique (one and only one) for a particular operation. Therefore, the operation 푎 ∗ 푏 = 1 + 푎푏 defined on the set of 푖푛푡푒푔푒푟푠 has no identity element.

∎

Inverse of an element Let ∗ be a binary operation on a set 푆 in which there is an identity element 푒.If for all 푎, 푏 ∈ 푆, 푎 ∗ 푏 = 푏 ∗ 푎 = 푒,then 푏 is called an inverse element of 푎. We will now use the 푎 to denote the inverse of 푎.

푎 ∗ 푎 = 푎 ∗ 푎 = 푒 That is either 푎 ∗ 푎 = 푒 (right inverse) or 푎 ∗ 푎 = 푒. The inverse of an element is unique.

∗ 1 2 3 1 1 2 3 2 2 3 1 3 3 1 2

70

Example 4.15

A binary operation ∘ is defined on the set ℤby 푎 ∘ 푏 = 푎 + 푏 − 1,∀푎, 푏 ∈ ℤ. Find i. the identity element; ii. the inverse of 6 under ℤ.

Solution i. We have 푎 ∘ 푏 = 푎 + 푏 − 1,∀푎,푏 ∈ ℤ. If there exist an identity element, then 푎 ∘ 푒 = 푒 ∘ 푎 = 푎. Let’s take 푎 ∘ 푒 = 푎 . ⟹ 푎 + 푒 − 1 = 푎 ⟹ 푒 = 푎 − 푎 + 1

∴ 푒 = 1 ii. If an inverse of an element exist, then 푎 ∘ 푎 = 푒, where 푎 is the inverse of 푎. ⟹ 푎 + 푎 − 1 = 1, since 푒 =1. ⟹ 푎 = 1 + 1 − 푎 ⟹ 푎 = 2 − 푎 푎 (6) = 2 − 6 = −4. Therefore, the inverse of 6 under ℤ is −4.

Example 4.16 A binary operation ∗ is defined on the set of 푟푒푎푙푛푢푚푏푒푟푠 by 푎 ∗ 푏 = 푎 + 푏 − 2푎푏, where 푎,푏 ∈ ℝ. Find i. the identity element; ii the inverse of 2 and 8; iii. the inverse of 푟, 푟 ∈ ℝ.State the value of 푟 for which no inverse exist.

Solution i. We have 푎 ∗ 푏 = 푎 + 푏 − 2푎푏. If there exist an identity element, then 푎 ∗ 푒 = 푎. where 푒 is the identity element, ⟹ 푎 + 푒 − 2푎푒 = 푎 ⟹ 푒 − 2푎푒 = 푎 − 푎 ⟹ 푒(1− 2푎) = 0 ⟹ 푒 = = 0. ∴, 푒 =0 ii. If an inverse of an element exist, then 푎 ∗ 푎 = 푒, where 푎 is the inverse of 푎. ⟹ 푎 + 푎 − 2푎푎 = 0 since 푒 = 0. ⟹ 푎 − 2푎푎 = −푎 ⟹ 푎 (1− 2푎) = −푎 ⟹ 푎 =

∴ 푎 (2) =( )

= and 푎 (8) =( )

= .

iii. From 푎 = , the inverse of a real number 푟, is 푟 = .To find the value of

푟 for which no inverse exist put 1 − 2푟 to zero. ⟹ 1 − 2푟 = 0 ⟹ 푟 =

∴ 푟 = , 푟 ≠ Try: 1. A binary operation∆ is defined on the set 푅 of real numbers by: 푚∆푛 = 푚 + 푛 + 10. Find a. The identity element

71

b. The inverse of 2 and −5 under 푅. (SSSCE) 2. A binary operation ∆ is defined on the set of real numbers 푅 by 푎∆푏 = 푎 + 푏 − 5푎푏, where푎, 푏 ∈ 푅and푅 is closed under ∆. Find under ∆,the (a) Identity element. (b) Inverse of − .

3. A binary operation ∗ is defined on the set, 푅, of real numbers by 푎 ∗ 푏 = 푎 + 푏 + , 푎,푏 ∈ 푅.Find; i. The identity element of the operation; ii The inverse of a real number 푥; iii. The inverse of . (SSSCE) 4. The table below shows the operation ∆ on the set {푎, 푏, 푐,푑} Use it to answer the following questions: a. State the identity element. b. Find the inverse of the element 푎and 푑. c. State whether or not the operation is commutative.

Distributive property Let ∗ and ∘ be two binary operations on a set 푆.If for all 푎, 푏, 푐 ∈ 푆 푎 ∗ (푏 ∘ 푐) = (푎 ∗ 푏) ∘ (푎 ∗ 푐),we say that ∗ is left distributive over ∘. Similarly, if (푏 ∘ 푐) ∗ 푎 = (푏 ∗ 푎) ∘∗ (푐 ∗ 푎), then we say that ∗ is right distributive over ∘. Note that the binary operation + is not distributive over × on the set of 푟푒푎푙푛푢푚푏푒푟푠.

Example 4.17 The binary operations ∗ and ∘are defined on the set of 푟푒푎푙푛푢푚푏푒푟푠 by 푎 ∗ 푏 = 2푎+ 푏 and 푥 ∘ 푦 = 푥푦 − 푥respectively, where 푎,푏, 푥, 푦 ∈ ℝ. Evaluate, i. 3 ∗ (4 ∘ −1) ii.(4 ∘ 3) ∗ (−3 ∗ 2)iii.(7 ∗ 0) ∘ 8

Solution i. We have 푎 ∗ 푏 = 2푎 + 푏 and 푥 ∘ 푦 = 푥푦 − 푥

3 ∗ (4 ∘ −1) = 3 ∗ (4 × (−1)− 4) = 3 ∗ −8 = 2(3) − 8 = 6 − 8 = −2 ii. (4 ∘ 3) ∗ (−3 ∗ 2) = (4(3)− 4) ∗ (2(−3) + 2) = 8 ∗ −4 = 2(8)− 4 = 16 − 4 = 12 iii. (7 ∗ 0) ∘ 8 = (2(7) + 0) ∘ 8 = 14 ∘ 8 = 14(8)− 14 = 112 − 14 = 98.

∆ 푎 푏 푐 푑 푎 푏 푐 푎 푑 푏 푐 푑 푏 푎 푐 푎 푏 푐 푑 푑 푑 푎 푑 푐

72

(BODMAS) Try:

The binary operations ∗ and ∆ are defined on the set 푅 of real numbers by 푎 ∗ 푏 = 2푎 + 푎푏 and 푥∆푦 = 푥 + 푦 − 푥푦 respectively, where 푎,푏, 푥,푦 ∈ 푅.

a. Determine whether or not ∆ is commutative. b. Evaluate (3 ∗ 2)∆(5∆3). (WASSCE)

Example 4.18 A binary operation ∘ is defined on the set of 푟푒푎푙푛푢푚푏푒푟푠 by 푠 ∘ 푡 = 푠 + + 푠푡, ∀푠, 푡 ∈ ℝ. a. Show whether or not ∘ is commutative. b. Find the identity element. c. Find the inverse of a real number 푎, and hence find the inverse of 3. d. Find the possible values of 푥such that ( )∘ = , 푥 ≠ 0.

Solution a. Given 푠 ∘ 푡 = 푠 + + 푠푡. If ∘ is commutative, then 푠 ∘ 푡 = 푡 ∘ 푠.

Now, 푠 ∘ 푡 = 푠 + + 푠푡 푡 ∘ 푠 = 푡 + + 푡푠 ⟹ 푠 ∘ 푡 ≠ 푡 ∘ 푠 since ≠ . Therefore ∘ is not commutative. Check: 2 ∘ 4 = 2 + + 2(4) = 10 and 4 ∘ 2 = 4 + + 4(2) = 13 ⟹ 2 ∘ 4 ≠ 4 ∘ 2 b. If there exist an identity element, then 푠 ∘ 푒 = 푠, where 푒 is the identity element. ⟹ 푠 + + 푠푒 = 푠 (Multiply both sides by 2) ⟹2푠 + 푒 + 2푒푠 = 2푠

⟹ 푒 + 2푒푠 = 2푠 − 2푠⟹푒 + 2푒푠 = 0 ⟹ 푒(1 + 2푠) = = 0. ∴ 푒 = 0 c. If an inverse exist, then 푠 ∘ 푠 = 푒, where 푠 is the inverse of 푠.

⟹ 푠 + + 푠푠 = 0 since 푒 = 0. ⟹ 2푠 + 푠 + 2푠푠 = 0

⟹ 푠 + 2푠푠 = −2푠 ⟹ 푠 (1 + 2푠) = −2푠 ⟹ 푠 = − .

Therefore, the inverse of a real number 푎 is 푎 = − . 푎 (3) = − ( )( )

= −

d. Using 푠 ∘ 푡 = 푠 + + 푠푡 ⟹(푥 − 1) ∘ 푥 = (푥 − 1) + + (푥 − 1)푥.

⟹ ( )∘ =( ) ( )

⟹ ( ) ( )

=

73

⟹ 2 (푥 − 1) + + (푥 − 1)푥 = 푥 ⟹ 2(푥 − 1) + 푥 + 2푥(푥 − 1) = 푥 ⟹ 2푥 − 2 + 푥 + 2푥 − 2푥 − 푥 = 0 ⟹ 2푥 − 2 = 0 (Divide both sides by 2) ⟹ 푥 − 1 = 0 ⟹ (푥 − 1)(푥 + 1) = 0 ⟹ 푥 − 1 = 0 or 푥 + 1 = 0\ ⟹ 푥 = 1 or 푥 = −1 ∴ 푥 = −1or 푥 = 1

Final exercises

1. A binary operation ∗ is defined on the set of real numbers by 푎 ∗ 푏 = √√

,푎 > 0.

Evaluate i. 9 ∗ 2 ii. 3 ∗ 2 iii. 1 ∗ 5 iv. 100 ∗ 3 v. −2 ∗ 8 2. A binary operation ∗ is defined on the set ℝ, of real numbers by 푥 ∗ 푦 = 푥푦 + 2푥 a. Find i. −1 ∗ −2 ii. 3 ∗ −4 iii. −1 ∗ −3 b. Show that ∗ is not commutative c. Find the identity element if any. 3. The binary operation ∘is defined on the set ℕ = {1, 2, 3, … } by 푥 ∘ 푦 = 푥 + 푦 − 푥푦 i. Determine whether or not ℕ is closed under ∘. ii. Solve, 2 ∘ √푥 = 3 iii. Determine whether or not 표 is one to one. 4. The binary operations ∗ and 표are defined on the set of real numbers by 푎 ∗ 푏 = 푎 + 푏 + 푎푏 and 푥 ∘ 푦 = 푥푦 − 푥 a. Evaluate i. 2 ∗ 1 ii. 2 ∘ −1 iii. (4 ∘ 1) ∗ (2 ∘ −1) b. Show that ∗ is commutative. c. Find the identity element under ∗ Solve the equations 푚 ∗ 2 = 푛 − 1 and 푚 ∗ −1 = 2푛 − 1 5. The binary operations ∆ is defined on the set of real numbers by 푎∆푏 =

√ , 푎 ≥ 0

i. Evaluate 5∆2, leaving your answer in the form 푝+ 푞√푟 ii. Solve the equation ( ∆ ) = 푥 ∆1. 6. An operation ⊗is defined on the set ℝ,ofrealnumbersby:푎 ⊗ 푏 = 푎푏 − 3. i. Show that ⊗ is commutative. ii. Determine whether or not ⊗ is associative. 7. The binary operation ∗ is defined on the set of real numbers by 푥 ∗ 푦 = 2(푥 + 푦 + 푥푦) a. Evaluate 5 ∗ −7 b. Show that ∗ is commutative. c. Find the identity element. d. Find the inverse of −2 and 6under ∗. 8. A binary operation ∘ is defined by 푢 ∘ 푣 = 2푣 + 푢.

74

훼. i. Form a table of the operation ∘ on the set 푆 = {1,3, 5} 훽. State with reasons whether or not i.푆 is closed under ∘; ii.∘ Commutative under 푆 9. A binary operation ∗ is defined on the set 푅, of real numbers by 푥 ∗ 푦 = 푥 + 푦 − 1, where 푥, 푦 ∈ 푅. i. Calculate −3 ∗ (1 ∗ 2) ii. Determine whether or not ∗ (훼) is commutative; (훽) is associative 10. A binary operation ∆ is defined on the set of real numbers 푅 by 푎∆푏 = 푎 + 푏 + 12푎푏, where 푎,푏 ∈ ℝ and ℝ is closed under ∆. Find under ∆,the (a) Identity element. (b) Inverse of . 11. A binary operation is defined on the set of real numbers by 푎∇푏 = 푎 + 푎푏 − 푏. Find. i. 4∇6 ii. 6∇3 iii. −3∇12 iv. −2∇ − 5 12. A binary operation ∗ is defined on the set of real numbers by 푥 ∗ 푦 = 푥 + 푦 − 2푥푦. i. Determine whether or not ∗ is commutative and associative. ii. Find the identity element under the operation ∗ iii. Determine the inverse of an element 푥 under the operation ∗ and find the inverse

of 4. 13. A binary operation ⋄ is defined on the set of real numbers by 푎 ⋄ 푏 = 푎 + 푏 + 2푎푏. a. Calculate i. 6 ⋄ (−2) ii. (−4) ⋄ 6. b. Find the unique identity element under the operation ⋄ c. Determine the inverse of a real number 푎 under ⋄. 14. A binary operation is defined on the set of real numbers by 푎 ∗ 푏 = 푎 + , 푧 ≠ 0.

Calculate. i. 6 − 2√3 ∗ 8 + √3 ii. (3 − 7√4 ∗ 1 + 3√4 15. The operation • is defined on the set of non –zero numbers by 푎 • 푏 = 푎 − 2푎푏

determine whether • is i. commutative ii. associative iii. Calculate a. −1 • (−2)푏. 3 • −23 16. A binary operation • on the set ℝ, of real numbers is defined by푟 • 푠 = 2푟 + 2푠 + 푠푟 Find i. the identity (neutral) element. ii. The inverse 푟 of푟 iii. The inverse of 1 show that • is associative

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17. The operation ⋄ is defined on the set ℝ of positive real numbers by푥 ⋄ 푦 = .

Determine whether or not the operation ⋄ is i. commutative ii. associative. 18. An operation ∗ is defined on the set ℤ of integers by 푎 ⋄ 푏 = 푎 + 푎푏, i. Construct a table for this operation on the subset S={−3,−2,−1, 0, 1, 2, 3}. ii. Find, from your table, a number 푏 ∈ 푆, such that 푦 ∗ 푧 = 푦 for all 푦 ∈ 푆. 19. An operation ∗ is defined on the set of real numbers by 푚 ∗ 푛 = 푚 + 푛 + 5푚푛. a. Determine whether or not ∗ is associative. b. Find the identity element 푒 of R under the operation ∗. c. Determine the inverse under ∗ of an element 푚 ∈ ℤ, stating the element of 푚 for

which no inverse exists. 20. A binary operation ∿ is defined on the set ℝ, of real numbers by 푝∿푞 = 푝 +푞 + 푝푞

, where 푎, 푏 ∈ 푅. i. Determine whether or not the operation ∿ is associative. ii. Calculate √2∿

√ and

√∿√3.

iii. Use your result in i. above to calculate √2∿√

∿√

∿√3

21. The binary operation ∗ defined on the set 푆 = {푎,푏, 푐} by the table below ∗ 푎 푏 푐 푎 푎 푏 푐 푏 푏 푐 푎 푐 푐 푎 푏

Determine whether or not a. S is closed under ∗ b. ∗ is commutative c. Determine the identity element for the operation∗ d. Determine the inverse (under ∗) of each element of S.

22. An operation ∗ is defined on the set 푇 = {1, 2, 3, 4} by푎 ∗ 푏 = 푎 + 푏 − 푎푏 a. i. Copy and complete the table. ii. Is T closed with respect to∗

b. Evaluate (2 ∗ 3) ∗4

∗ 1 2 3 4 1 1 2 4 3 9 4 16

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INDICES/ POWERS OF NUMBERS A good way of appreciating large numbers is by writing them as a simple 푏푎푠푒and an 푒푥푝표푛푒푛푡/푖푛푑푒푥4. If 푏 = 푎 , then 푏 = 푎 × 푎 × … × 푎푛 factors, 푛 ∈ ℕ. Here 푎 is known as the base and 푛 is called the index/power/exponent. For instance,2 = 2 × 2 × 2. Question: What if 푛 ∈ ℤor ∈ ℝ? This calls for us to generalise our idea of indices into laws.

Laws of indices 1. 푎 × 푎 = 푎 For example, 푎 × 푎 = 푎 = 푎 . 2. 푎 ÷ 푎 = 푎 For example, 푎 ÷ 푎 = 푎 = 푎 3. (푎 ) = 푎 For example, (푎 ) = 푎 × = 푎

Example 5.1

Evaluate the following: i. 7 × 7 ii. 3 ÷ 3 iii. 푥 × 푥 ÷ 푥 iv. (2푥 푦)(2푦푥) ÷ (2푥 )

v. ÷

Solution i. 7 × 7 = 7 = 7 = 823543. ii. 3 ÷ 3 = 5 = 5 = 25 ii. 푥 × 푥 ÷ 푥 = 푥 ( ) ÷ 푥 = 푥 ÷ 푥 = 푥 = 푥 iii. (2푥 푦)(2푦푥) ÷ (2푥 ) = (2푥 푦)(2 푦 푥 ) ÷ (2푥 ) =2 × 2 × 푥 × 푥 × 푦 × 푦 ÷ 2푥 = 2 ( ) × 푥 × 푦 ( ) = 2 푥 푦

iv. ÷ = ÷ = ÷ = =

4 The plural of index is indices.

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Let us observe the table of index of some numbers below:

Properties of indices

1. 푎 = 1 (This is true because 푎 ÷ 푎 = 푎 = 푎 푎푛푑 = 푎 ÷ 푎 = 1) 2. 1 = 1 (1 exponent any number is 1) 3.i. (푎 ∙ 푏) = 푎 ∙ 푏 For instance,(4 × 3) = 4 × 3 = 64 × 27 = 1728.

ii. = For instance, = =

4.i.푎 = , 푎 ≠ 0. (This is true because 푎 ÷ 푎 = 푎 = 푎 and푎 ÷ 푎 = ).

For instance, 2 = =

ii. = , 푎 ≠ 0and푏 ≠ 0. For instance, = = =

5. 푎 = √푎

From this property we have 11 132 43 4; ; ;...a a a a a a

For example, 11 132 43 44 4 2;27 27;625 625 5

In general, 푎 = √푎 .

That is, 3416 = √16 = √4096 = 8

Index of 2

Index of 3

Index of 4

Index of 5

Index of 6

Index of 7

Index of 11

2 = 2 3 = 1 4 = 4 5 = 5 6 = 6 7 = 7 11 = 11 2 = 4 3 = 9 4 = 16 5 = 25 6 = 36 7 = 49 11 = 121 2 = 8 3 = 27 4 = 64 5 = 125 6 = 216 7 = 343 11 = 1331

2 = 16 3 = 81 4 = 256 5 = 625 6 = 1296 7 = 2401 2 = 32 3 = 243 4 = 1024 5 = 3125 6 = 7776 2 = 64 3 = 729 4 = 4096

2 = 128 3 = 2187 2 = 256 3 = 6561 2 = 512

78

Example 5.2 Simplify the following without using calculator.

i. ii. iii. (36) √ iv. v. vi. (0.125)

Solution

i. = =×

×=

ii. = =× /

× / = =

iii. (36) √ = √ =×√

=×√

= √

iv. = = =×

/ =√

=√∙ √√

= √ = 3√3

v. = = =×

×=

vi. . (0.125) =. 1251000

−23 = 1000

125

23 = 103×2

3

53×23

= 102

52 = 10025 = 4

Try: Simplify the following without using calculator.

a. 푥 ÷ 푥 b. 2 × ÷ 2 c. 8 ÷ 32 × 2 d. e. (1 ) Example 5.3

Simplify the following:

i. /

× /

( ) / ii. / × / × /

iii. ×

√ iv. ×

( )

Solution

i. /

× /

( ) / =/

× × /

( ) / =× /

= 2 × 2 / ÷ 2 = 2 = 2 / = /

ii. / × / × /

/ = ( × ) / × / × /

× / =× / × / × /

/ = 2 / / × 3 / /

= 2 / × 3 /

79

= √ ∙ √

iii.

1 1 23 3 3 2 2

3 312

33 1 3 13 2

2 (2 )

nn nn n n

n nn

x x x x x xxxx

2 2 2 2 2 23 3 3 3 3 3 2 2 1 1

3 3 4 21 1 1 1 18 8 8 4 2

2 1 2 13 4 3 2

2 42 4iv. ( ) ( )

( )( ) ( )

x y x y x y x y x yx yx y x y

x y

3 8 6 812 12

5 1612

x y

x y

Try: Solve the following:

a. / × /

/ b. ×( )

c. ÷

×

Indicial/exponential equations

If 푎 = 푎 , then 푥 = 푦, where 푥 and 푦 are variables and 푎 ≠ −1, 1or 0.

훼. Indicial equations leading to linear equations Example 5.4

a. Solve for the values of 푥 and 푦 in the following equations:

i. 2 = 2 ii. 5 = 5 iii. 3 = iv. 16 = 4 v. 9√ =

Solution i. 2 = 2 ⟹ 푥 = 2 ∴ 푥 = 2 ii. 5 = 5 ⟹ 푥 + 1 = 2푥 − 3 ⟹ 1 + 3 = 2푥 − 푥 ⟹ 푥 = = 2 ∴ 푥 = 2

iii. 3 = ⟹ 3 = (3 ) ( ) ⟹ 2푦 − 1 = 푦 − 1 ⟹ 4푦 − 2 = 푦 − 2 ⟹ 4푦 − 푦 = −2 + 2 ⟹ 3푦 = 0 ∴ 푦 = 0

80

iv. 16 = 4 ⟹ 4 ( )=4 ⟹ 2푦 = 푦 + 15 ⟹ 2푦 − 푦 = 15 ∴ 푦 = 15

v. 9√ = ⟹ 3 √ = 3 ⟹ 2√푥 = −1 ⟹ √푥 = − ⟹ (√푥) = −

⟹ 푥 = ∴ 푥 =

Example 5.5 i. If 2 ∙ 3 = 12, find the value of 푦. ii. Find the truth set of the equation √2 = √32 .

Solution i. 2 ∙ 3 = 12⟹ 2 ∙ 3 = (2 ∙ 3) ⟹ 푦 = 2 or 푦 − 1 = 1 ⟹ 푦 = 2. ∴ 푦 = 2

ii. √2 = √32 ⟹ (2 ) = (32 ) ⟹ (2 ) × = (32 ) × ⟹ 2 = 2 ( ) ⟹ 2푥 − 3 = 5(2 − 푥) ⟹ 2푥 − 3 = 10 − 5푥 ⟹ 2푥 + 5푥 = 10 + 3 ∴,푥 = Try: 1. Find the truth set of the equation 2√ = 8 . (SSSCE) 2. If 125 = √5 , find the value of 푥. (WASSCE)

훽.Indicial equations leading to simultaneous equations Example 5.6

Find the values of 푥 and 푦 in the following systems of equations.

i. 2 = 2 and 3 = ii. 4 = 8 and = ( ) iii. 5 = 1 and 3푥 + 2푦 = 2 iv. 343 = 7 and2 = 4

Solution i. We have 2 = 2 ⟹ 푥 − 푦 = 4 … … …(1) and 3 = ⟹ 3 = 3 ⟹ 2푥 − 푦 = −1 … … … (2). From (1) 푥 = 푦 + 4 Put 푥 = 푦 + 4 into (1) ⟹ 2(푦+ 4) − 푦 = −1⟹ 2푦 + 8 − 푦 = −1 ⟹ 2푦 − 푦 = −1 − 8 ⟹ 푦 = −9 Put 푦 = −9 into (1). ⟹ 푥 = −9 + 4 = −5 ∴,푥 = −5,푦 = −9 ii. We have 4 = 8 ⟹ 2 = 2 ⟹ 2 푥 + 푦 = 3 ⟹ 2푥 + 푦 = 3 … … … (1)

and = ( ) ⟹ (9 ) = (3 ) ⟹ (3 ) = (3 ) ⟹ −2(푦 − 2푥) = −1(3푦 + 1) ⟹ −2푦 + 4푥 = −3푦 − 1 ⟹ 4푥 + 푦 = −1 … … … (2)

81

From (1) 푦 = 3 − 2푥 Put 푦 = 3 − 2푥 into (2) ⟹ 4푥 + 3 − 2푥 = −1 ⟹2푥 = −4푥 = − = −2 Put 푥 = −2 into (1) ⟹ 푦 = 3 − 2(−2) = 7 ∴,푥 = −2,푦 = 7 iii. We have 5 = 1 ⟹ 5 = 5 ⟹ 2푥 + 3푦 = 0 … … … (1) and 3푥 + 2푦 = 2 … … … (2) From (1) 푥 = − 푦 Put 푥 = − 푦 into (2)

⟹ 3(− 푦) + 2푦 = 2 ⟹ −9푦+ 4푦 = 4 ⟹ 푦 = − Put 푦 = − into (2)

⟹ 푥 = − = − ∴ 푥 = − , 푦 = −

iv. We have 343 = 7 ⟹ 7 ( ) = 7 ⟹ 3(4 − 푥) = 1 − 푦 ⟹ 12 − 3푥 = 1 − 푦 ⟹ 3푥 − 푦 = 11⋯⋯⋯ (1) and2 = 4 ⟹ 2 = 2 ( ) ⟹ 푦 + 2푥 = −2푦 ⟹ 2푥 + 3푦 = 0⋯⋯⋯ (2) From (1) 푦 = 3푥 − 11 Put 푦 = 3푥 − 11 into (2). ⟹ 2푥 + 3(3푥 − 11) = 0 ⟹ 2푥 + 9푥 − 33 = 0 ⟹ 푥 = = 3 Put 푥 = 3 into (1) ⟹ y = 3(3)− 11 = −2

∴ 푥 = 3, 푦 = −2 Try: 1. Find the values of 푎 and 푏 in the following equations.

i. = and = ii. 3푎 + 푏 = 4 and 4 = 0.25

2. Solve the simultaneous equations: 3 = 243 and 3푥 + 푦 = 9 (SSSCE) 3. Solve the simultaneous equations: 4 = ; 2 ∙ 4 = 8. (SSSCE)

Indicial equations leading to quadratic equations Example 5.7

Find the truth set of the following equations: i. 2 − 8 ∙ 2 = 0 ii. 3 − 3 − 3 = −9 iii. 2 + 2 − 2 = 28 iv. 3 = 10 ∙ 3 − 1 v. 5 − 30 + 125 ∙ 5 = 0 vi. 7 =

Solution i. We have 2 − 8 ∙ 2 = 0⟹ 2 ( ) − 8 ∙ 2 = 0Let 2 = 푦 ⟹ 푦 − 8푦 = 0. Comparing 푦 − 14푦 + 32 = 0with 푎푦 + 푏푦 + 푐 = 0 , 푎 = 1, 푏 = −8and푐 = 0. The values of 푦 can be determined by the formula

82

푦 =−푏± 푏2−4푎푐

2푎 ⟹ 푦 = ( )± ( ) ( )( )( )

= ±√ = ±

⟹ 푦 = = 8 or 푦 = = 0 But 2 = 푦 ⟹ 2 = 8 = 2 ⟹ 푥 = 3 or 2 = 0 (not defined) ∴ 푥 = 3 ii. We have 3 − 3 − 3 = −9 ⟹ 3 × 3 − 9 ∙ 3 − 3 × 3 = −9 ⟹ 3( ) × 3− 9 × 3 − 3 × 3 = −9 let 3 = 푦 ⟹ 3푦 − 푦 − 11푦 = −9⟹ 3푦 − 12푦 + 9 = 0 Comparing 3푦 − 4푦+ 9 = 0 With 푎푦 + 푏푦 + 푐 = 0 , 푎 = 3,푏 = −4and푐 = 9. The values of 푦 can be determine by the formula

푦 =−푏± 푏2−4푎푐

2푎 ⟹ 푦 = ( )± ( ) ( )( )( )

= ±√ = ±

⟹ 푦 = = 3 or 푦 = = 1 But 3 = 푦 ⟹ 3 = 3 ⟹ 푥 = 1 or 3 = 1 = 3 ⟹ 푥 = 0 ∴ 푥 = 1, 0 iii. 2 + 2 − 2 = 28 ⟹ 2 × 2 + 2 + 2 × 2 = 28 ⟹ 2 ( ) × 2 + 3 − 2 × 2 = 28 Let 2 = 푥 ⟹ 2푥 + 푥 − 2푥 = 282푥 − 푥 − 28 = 0Comparing 2푥 − 푥 − 28 = 0 With 푎푥 + 푏푥 + 푐 = 0 , 푎 = 2,푏 = −1and푐 = −28. The values of 푥 can be determined by the formula

푥 =−푏± 푏2−4푎푐

2푎 ⟹ 푥 = ( )± ( ) ( )( )( )

= ±√ = ±

⟹ 푥 = = 4 or 푥 = = − But 2 = 푥

⟹ 2 = 4 = 2 ⟹ 푦 = 2 or2 = − (not defined)∴ 푥 = 2

iv. We have 3 = 10 ∙ 3 − 1 ⟹3 = 10 ∙ 3 ÷ 3 − 13 ( ) = 10 ∙ − 1

let 3 = 푥푥 = 10 ∙ − 13푥 = 10푥 − 33푥 − 10푥 + 3 = 0 Comparing 3푥 − 10푥 + 3 = 0 With 푎푥 + 푏푥 + 푐 = 0 , 푎 = 3,푏 = −10and푐 = 3. The values of 푥 can be determine by the formula

푥 =−푏± 푏2−4푎푐

2푎 ⟹ 푥 = ( )± ( ) ( )( )( )

= ±√ = ±

⟹ 푥 = = 3 or 푥 = = But 3 = 푥

83

⟹ 3 = 3 ⟹ 푦 = 1 or 3 = = 3 ⟹ 푦 = −1 ∴ 푥 = −1, 1 v.5 − 30 + 125 ∙ 5 = 0 ⟹ 5 − 30 + 125 ∙ = 0 let 5 = 푦

⟹푦 − 30 + 625 ∙ = 0 ⟹푦 − 30푦 + 125 = 0⟹푦 − 25푦 − 5푦+ 125 = 0

⟹ 푦(푦 − 25)− 5(푦 − 25) = 0 ⟹(푦 − 25)(푦 − 5) = 0 ⟹ 푦 − 25 = 0 or 푦 − 5 = 0 ⟹ 푦 = 25 or 푦 = 5 But 5 = 푦 ⟹ 5 = 25 = 5 ⟹ 푥 = 2 or 5 = 5푥 = 1 ∴,푥 = 1, 2 vi. We have 7 = ⟹ 7 = 7 ⟹ 푥 − 3푥 = −2 ⟹ 푥 − 3푥 + 2 = 0 ⟹ 푥 − 2푥 − 푥 + 2 = 0⟹푥(푥 − 2) − (푥 − 2) = 0 ⟹ (푥 − 2)(푥 − 1) = 0 ⟹ (푥 − 2) = 0 or (푥 − 1) = 0 ∴ 푥 = 2 or 푥 = 1

Example 5.8

Find the values of 푥 if 4 + 4 = 17 Solution

Given 4 + 4 = 17 ⟹ 2 + 2 = 17

⟹ 2( ) + 2( ) = 17 ⟹ 2 × 2 + = 17 let 2 = 푦

⟹ 2푦 + = 17 ⟹ 2y + 8 = 17

⟹ 2푦 − 17푦 + 8 = 0⟹ 2푦 − 16푦 − 푦 + 8 = 0 ⟹ 2푦(푦 − 8)− (푦 − 8) = 0 ⟹ (푦 − 8)(2푦 − 1) = 0

∴ 푦 = 8or푦 = But 2 = 푦

⟹ 2 = 8 = 2 ⟹ 푥 = or 2 = = 2 ⟹ 푥 = −

∴ 푥 = − , Try: 1. Find the truth set of: 2 − 5(2 ) + 1 = 0,푥 ∈ ℝ. (SSSCE) 2. Solve the equation 2 − 9(2 ) + 2 = 0 (WASSCE) 3. Solve 3 ( ) − 3 = 3 − 3 (WASSCE) 4. Solve 3 − 5(3 ) + 6 = 0 (WASSCE)

Final exercises 1. i. 3 × 3 ii. 6 ÷ 6 iii. 푦 × 푦 ÷ 3푦 iv. (8푚 푛)(2푚푛) ÷ (16푚 )

84

v. ÷

2. i. /

× /

( ) / ii. / × / × /

/ iii. × /

√ iv. ×

( × )

3. Solve: i. 2 − 2 + 2 = 1 ii. 2 − 2 + 2 = 28 iii. 3 − 3 + 3 = 1 iv. 3 − 3 + 3 = −1 v. 5 = 3 vi. 푎 = 푎 vii. 3 − 6 ∙ 3 − 7 = 0 viii. 3 + 3 + 3 + 3 = 112 ix. 25 = 5 x. 3 = 1

4. i. 2 = 2 and 3 = ii. 5 = 25 and = iii. 2 = 2 and 3푥 + 2푦 = 2 iv. 343 = 7 and6 = 36

5. i. 11 = 121 ii. 144 = 12 iii. 2 . =( . )

iv. 16 = 4 v. 9 √ = 6. Find the truth set of i. 2 − 5(2 ) + 4 = 0 ii. 3 − 3 −3 + 1 = 0 (CCE)

85

LOGARITHMS The logarithm of a number 푏 to the base 푎, (written as log 푏)has a value 푥 if and only if the base 푎 exponent 푥 gives 푏. That is, log 푏 ⇔ 푎 = 푏. log 푏 is read ‘ 푏 base 푎’.

Example 6.1 Evaluate the following: i. log 27 ii. log 2 iii. log√ 128 iv. log √ 5√5 v. log √64

Solution i. Let log 27 = 푥 ⟹ 3 = 27 (By definition) ⟹ 3 = 3 ⟹ 푥 = 3 ∴ log 27 = 3 ii. Let log 2 = 푥 ⟹ 8 = 2 (By definition) ⟹ 2 = 2 ⟹ 3푥 = 1 ⟹ 푥 =

∴ log 2 =

iii. Let log√ 128 = 푥 ⟹ √2 = 128 ⟹ 2 / ( ) = 2 ⟹ 푥 = 7 ⟹ 푥 = 2 × 7 = 14 ∴ log√ 128 = 14

iv. Let log √ 5√5 = 푥 ⟹ √5 = 5√5 ⟹ 5 / ( ) = 5 ∙ 5 / ⟹ 푥 = 1 +

⟹ 푥 = ⟹ 2푥 = 9 ⟹ 푥 = ∴ log √ 5√5 =

86

v. log √64 = 푥 ⟹ = √64 ⟹ = 64 / ⟹ 2 ( ) = 2

⟹ −푥 = 3 ∴ log √64 = −3

Example 6.2 If log 5 = 5√5, find the value of 푥.

Solution

log 5 = 5√5 ⟹푥 √ = 5 ⟹ 푥 √ ×√ = 5 √ ∴ 푥 = 5 √ = 5

√= 5√

Try: Evaluate the following i. log 343 ii. log 81 iii. log√ 36 iv. log √3 v. log 729 vi. log 0.001

Properties of Logarithms 1. log 푎 = 1

Proof Let log 푎 = 푥 ⟹푎 = 푎 ⟹ 푥 = 1 ∴ log 푎 = 1

∎ 2. log 푎 = 푘

Proof Let log 푎 = 푥 ⟹ 푎 = 푎 ⟹ 푥 = 푘 ∴ log 푎 = 푘.

∎ 3. log 1 = 0

Proof Let log 1 = 푥 ⟹푎 = 1 ⟹푎 = 푎 ⟹ 푥 = 0 ∴ 푥 = 0

∎ 4. 푎 = 푏

Proof Let 푎 = 푥 … … …(1) and let log 푏 = 푦 ⟹ 푎 = 푏. Put 푏 = 푎 into (1) Now 푎 = 푎 ∴ 푎 = 푏 ∎ 5. log 푏 is not defined if 푎 ≤ 1.

Example 6.3 Given that log 4 = √2 , find log 푥.

Solution

87

log 4 = √2 ⟹ 푥√ = 4 ⟹ 푥√ ×√ = 4√ ⟹푥 = 2 ×

√ = 2 ×√= 2√

∴ log 푥 = log 2√ = √2 log 2 = √2

Laws of Logarithms 1. Addition/Multiplication rule

log 푥 + log 푦 = log 푥푦 It states that, the sum of the logarithm of two numbers of the same base is equal to the logarithm of the product of the two numbers to the base and vice versa. For instance, log 3 + log 5 = log (3 × 5) = log 15 log 7 + log 2 = log (7 × 2) = log 14 log 35 = log 5 × 7 = log 5 + log 7

Proof

To prove that log 푥 + log 푦 = log 푥푦 Let log 푥 = 푏 and log 푦 = 푐 ⟹ 푎 = 푥… … …(1) and 푎 = 푦… … …(2). Multiply (1) and (2) ⟹푎 × 푎 = 푥 × 푦 ⟹ 푎 = 푥푦. Take log of base 푎 of both sides.

⟹ log 푎 = log 푥푦 ⟹ 푏 + 푐 = log 푥푦 ∴ log 푥 + log 푦 = log 푥푦 since log 푥 = 푏 and log 푦 = 푐 ∎ 2. Subtraction/Division rule

log 푥 − log 푦 = log

It states that, the difference of the logarithm of two numbers of the same base is equal to the logarithm of the ratio of the numbers to the base and vice versa. For instance, log 3 − log 5 = log log 14 − log 2 = log ( ) = log 7

log = log 8 − log 3 Proof

To prove that log 푥 − log 푦 = log

Let log 푥 = 푏 and log 푦 = 푐 ⟹ 푎 = 푥… … …(1)and 푎 = 푦… … …(2). (1) ÷ (2) ⟹푎 ÷ 푎 = 푥 ÷ 푦 ⟹ 푎 = . Take log of base 푎 of both sides.

⟹ log 푎 = log 푥푦 ⟹ 푏 − 푐 = log 푥푦

88

∴ log 푥 − log 푦 = log since log 푥 = 푏 and log 푦 = 푐 ∎

3. Rule of exponent

log 푥 = 푛 log 푥 It states that, the logarithm of a number with an exponent is equal to the product of the exponent and the logarithm of the number and vice versa. For instance, log 7 = 2 log 7 log 8 = log 2 = 3 log 2 3log 6 = log 6 = log 216

Proof To prove that log 푥 = 푛 log 푥

Let 푛 log 푥 = 푏 ⟹ log 푥 = ⟹ 푎 = 푥. Take exponent 푛 of both sides.

⟹ 푎 = 푥 ⟹ 푎 = 푥 . By taking logarithm of base 푎 of both side we get log 푎 = log 푥 ⟹ 푏 = log 푥 ∴ log 푥 = 푛 log 푥 since 푛log 푥 = 푏 ∎ Note: (log 푥) ≠ 푛 log 푥 4. Change of base

log 푥 = . If 푥 = 푏, then log 푥 =

It states that, the logarithm of a number to a given base is equal to the ratio of the logarithm of the number and the logarithm of the base to a different base. For example, log 9 =

Proof To prove that log 푥 =

Let log 푥 = 푐 ⟹ 푎 = 푥. Take logarithm of base 푏 of base sides. ⟹ log 푎 = log 푥 ⟹ 푐 log 푎 = log 푥 ⟹ 푐 =

∴ log 푥 = since log 푥 = 푐 ∎

Note: Logarithms written without bases are in base 10. Logarithms that has base of 10 are called 푐표푚푚표푛푙표푔푎푟푖푡ℎ푚.

89

Example 6.4 Simplify the following: i. log 3 + log 8 − log 6 ii. log + log − log − log ( 6)

iii. 2log 6 + log 16 − log 4 − log 72 Solution

i. log 3 + log 8 − log 6 = log (3 × 8) − log 6 = log = log 4 = log 2 = 2 ∴ log 3 + log 8 − log 6 = 2

ii. log + log − log − log ( 6)

= log + log − log + log (6)

= log × − log × 6

= log − log

= log ÷

= log ×

= log

∴ log ( ) + log − log − log ( 6) = log iii. 2log 6 + log 16 − log 4 − log 72 = log 6 + log 16 − log 4 / − log 72 = log(36 × 16)− (log 2 + log 72) = log ×

×

= log(4)

∴ 2log 6 + log 16 −12 log 4 − log 72 = log 4 = log 2 = 2 log 2

Example 6.5

Simplify,2log 푥 + log 푦 + log 푧 − log 2. Evaluate the expression if 푥 = 5, 푦 = 3 and 푧 = 6.

Solution

2log 푥 +12 log 푦 +

13 log 푧 − log 2 = log 푥 + log 푦 × / + log 푧 × / − log 2

90

= log × ×

∴ 2log 푥 + log 푦 + log 푧 − log 2 = log

If 푥 = 5, 푦 = 3 and 푧 = 6, then log = log × × = log 675

Try:

1. Given that log 푎 = 3, log 푏 = −2 and log 푐 = 5,evaluate log 8 (SSSCE)

2. Without using tables simplify (SSSCE)

3. Find the value of log 8 + log 625− log 256 (SSSCE)

Equations involving logarithms Recall that log 푏 ⇔ 푎 = 푏

Example 6.6 Solve for the following: i. log (푥 + 2) − log (2푥 − 3) = 2 ii. log (2푥 − 1)− log 3 = 0 iii. log (푥 + 1)− log (3푥 − 1) = 1 iv. 3log (푥 − 1) = 3

Solution i.log (푥 + 2)− log (2푥 − 3) = 2 ⟹ log = 2 ⟹ = 2 (By definition)

⟹ 푥 + 2 = 4(2푥 − 3) ⟹ 푥 + 2 = 8푥 − 12 ⟹ 2 + 12 = 8푥 − 푥 ∴ 푥 = = 2

ii. log (2푥 − 1)− log 3 = 0 ⟹ log = 0 ⟹ = 2 (By definition)

⟹ 2푥 − 1 = 3(1)⟹2푥 = 3 + 1 ∴ 푥 = = 2

iii. log (푥 + 1)− log (3푥 − 1) = 1 ⟹ log = 1 ⟹ = 5 (By definition)

⟹ 푥 + 1 = 5(3푥 − 1) ⟹ 푥 + 1 = 15푥 − 5 ⟹ 1 + 5 = 15푥 − 푥 ∴ 푥 = = iv. 3log (푥 − 1) = 3 ⟹ log (푥 − 1) = 3 ⟹ (푥 − 1) = 3 ⟹ 푥 − 1 = 3

∴ 푥 = 3 + 1 = 4

Example 6.7 Solve for the following: i. log (푥 − 푥) = 1 ii. log (푥 + 1) + log (푥 − 1) = 1 iii. log (푥 − 6푥 + 2) = 2 iv. (log 푥 + 4) + log (푥 + 4) = 8

91

v. log(푥 + 3) + log(2푥 − 3) = log(1− 푥) Solution

i. log (푥 − 푥) = 1 ⟹ (푥 − 푥) = 2 (By definition) ⟹ 푥 − 푥 − 1 = 0 Comparing 푥 − 푥 − 1 = 0 with 푎푥 + 푏푥 + 푐 = 0, 푎 = 1, 푏 = −1 and 푐 = −2. The values of 푥 can be determined by the formula:

푥 =−푏± 푏2−4푎푐

2푎 ⟹ 푥 =−(−1)± (−1)2−4(1)(−2)

2(1) = 1±√92

⟹ 푥 = = 2 or 푥 = = −1 ∴ 푥 = 2 or 푥 = −1. ii. log (푥 + 1) + log (푥 − 1) = 1 ⟹ log (푥 + 1)(푥 − 1) = 1 ⟹ (푥 + 1)(푥 − 1) = 5 ⟹ 푥 − 1 = 5 ⟹ 푥 = 5 + 1 = 6 ∴ 푥 = √6 (Take square root of both sides) iii. log (푥 − 6푥 + 2) = 2 ⟹ (푥 − 6푥 + 2) = 3 ⟹ 푥 − 6푥 + 2 − 9 = 0 ⟹ 푥 − 6푥 − 7 = 0 Comparing 푥 − 6푥 − 7 = 0 with 푎푥 + 푏푥 + 푐 = 0, 푎 = 1,푏 = −6 and 푐 = −7. The values of 푥 can be determined by the formula:

푥 =−푏± 푏2−4푎푐

2푎 ⟹ 푥 =−(−6)± (−6)2−4(1)(−7)

2(1) = 6±√642

⟹ 푥 = = 7 or 푥 = = −1 푥 = 7 since 푥 = −1 is a false solution. iv. (log 푥 + 4) + log (푥 + 4) = 8 ⟹ (log 푥 + 4) + 2log (푥 + 4) = 8 Let log (푥 + 4) = 푦 ⟹ 푦 + 2푦 = 8 ⟹ 푦 + 2푦 − 8 = 0 ⟹ 푦 − 2푦 + 4푦 − 8 = 0 ⟹ 푦(푦 − 2) + 4(푦 − 2) = 0 ⟹ (푦 − 2)(푦 + 4) = 0 ⟹ 푦 − 2 = 0 ⟹ 푦 = 2 or 푦 + 4 = 0 ⟹ 푦 = −4 But log (푥 + 4) = 푦 ⟹ log (푥 + 4) = 2 ⟹ 푥 + 4 = 3 ⟹푥 = 9− 4 = 5 Or log (푥 + 4) = −4 ⟹ 푥 + 4 = 3 ⟹ 푥 = − 4 = −

∴ 푥 = 5 since 푥 = − is a false solution.

v. log(푥 + 3) + log(2푥 − 3) = log(1− 푥) ⟹ log(푥 + 3)(2푥 − 3) = log(1 − 푥) × /

log(푥 + 3)(2푥 − 3) = log(1 − 푥) ⟹ (푥 + 3)(2푥 − 3) = 10 ( ) ⟹ (푥 + 3)(2푥 − 3) = (1 − 푥) ⟹ 2푥 + 3푥 − 9 = 1 − 푥 ⟹ 2푥 + 4푥 − 10 = 0 (Dividing both sides by 2 gives) 푥 + 2푥 − 5 = 0 Comparing 푥 + 2푥 − 5 = 0 with 푎푥 + 푏푥 + 푐 = 0, 푎 = 1, 푏 = 2 and 푐 = −5.

92

The values of 푥 can be determined by the formula:

푥 =−푏± 푏2−4푎푐

2푎 ⟹ 푥 =−(2)± (−2)2−4(1)(−5)

2(1) = −2±√242

⟹ 푥 = √ = −1 + √6 or 푥 = √ = −1 − √6

∴ 푥 = −1 + √6

Example 6.8 Find the truth set of the following equations: i. log 푥 + log 푦 = 3 and 2log 푥 − log 푦 = 6 ii. log 푦 = 2 and 푥 + 푦 = 6 iii. log 푥푦 = 3 and log (푥 + 1) + log 푦 = 4

Solution i. log 푥 + log 푦 = 3 ⟹ log 푥푦 = 3 ⟹ 푥푦 = 2 ⟹ 푥푦 = 8 … … … (1)

and 2log 푥 − log 푦 = 6 ⟹ log = 6 ⟹ = 2 ⟹ = 64 … … … (2)

From (2) 푦 = Put 푦 = into (1)

⟹ 푥 = 8 ⟹ 푥 = 64 × 8 ⟹ 푥 = √513 = 8 Put 푥 = 8 into (2)

⟹푦 = = 1 ∴ 푥 = 8, 푦 = 1 ii. log 푦 = 2 ⟹ 푦 = 푥 … … … (1) and 푥 + 푦 = 6 … … . . . (2) Put 푦 = 푥 into (2) ⟹ 푥 + 푥 = 3 ⟹ 푥 + 푥 − 6 = 0 푥 − 2푥 + 3푥 − 6 = 0 ⟹ 푥(푥 − 2) + 3(푥 − 6) = 0 ⟹ (푥 − 2)(푥 + 3) = 0 ⟹ (푥 − 2) = 0 ⟹ 푥 = 2 or (푥 + 3) = 0 ⟹ 푥 = −3 Put the values of 푥 into (1).When 푥 = 2,푦 = 2 = 4 When 푥 = −3,푦 = (−3) = 9 But 푥 = −3,푦 = 9 is a pair of false solution. ∴ 푥 = 2, 푦 = 4 iii. log 푥푦 = 3 ⟹ 푥푦 = 3 ⟹ 푥푦 = 27 … … … (1) and log (푥 + 1) + log 푦 = 4 log (푥 + 1)(푦) = 4 ⟹ 푦(푥 + 1) = 2 ⟹ 푦(푥 + 1) = 16 … … … (2)

From (2) 푦 = Put 푦 = into (1) ⟹ 푥 = 27

⟹ 16푥 = 27(푥 + 1) ⟹ 16푥 = 27푥 + 27 ⟹ 16푥 − 27푥 = 27 ⟹ 푥 = −

Put 푥 = − into (1) ⟹ 푦 − = 27 ⟹ 푦 = − × = −11

∴ 푥 = −11, 푦 = −

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Example 6.9 Solve for the following: i. log 푥 + log 푥 = 3 ii. 5 = 8 iii. Log (푥 − 2푥) + log (푥 − 1) = log 6푥 iv. 5 + 5 = 6

Solution i. log 푥 + log 푥 = 3 ⟹ log 푥 = 3 ⟹ 푥 = 5 ∴ 푥 = 3 ii. 5 = 8 (Take log of both side) ⟹log 5 = log 8 ⟹ 푥 log 5 = log 8

∴ 푥 = ≈ 1.2920

iii. Log (푥 − 2푥) + log (푥 − 1) = log 6푥 ⟹ Log (푥 − 2푥)(푥 − 1) = log 6푥 ⟹ (푥 − 2푥)(푥 − 1) = 7 ⟹ 푥 − 푥 − 2푥 + 2푥 = 6푥 ⟹ 푥 − 3푥 + 2푥 − 6푥 = 0 ⟹ 푥 − 3푥 − 4푥 = 0 ⟹ 푥(푥 − 3푥 − 4) = 0 ⟹ 푥(푥 − 4)(푥 + 1) = 0 ∴ 푥 = 0 or 푥 = 4 or 푥 = −1 iv. 5 + 5 = 6 (Factorise out5 ) ⟹ 5 (1 + 5) = 6 ⟹ 5 (6) = 6 5 = 1(Take log of both sides) ⟹ log 5 = log 1⟹ (푥 − 1) log 5 = 0 ⟹ (푥 − 1) = = 0 ⟹ 푥 − 1 = 0 ∴ 푥 = 1

Example 6.10 Solve for the following equation: i. log 푥 − log 8 = 2 ii. log (푥 + 1) + log( ) 5 = 2

Solution i. log 푥 − log 8 = 2 (Change to a common base 2)⟹ log 푥 − = 2

⟹(log 푥)(log 2) − log 2 = 2log 푥 ⟹ (log 푥) − 3 = 2 log 푥 Let log 푥 = 푦 ⟹푦 − 3 = 2푦 ⟹푦 − 2푦 − 3 = 0 Comparing 푦 − 2푦 − 3 = 0 With 푎푦 + 푏푦 + 푐 = 0 , 푎 = 1, 푏 = −2and푐 = −3. The values of 푦 can be determine by the formula

푦 =−푏± 푏2−4푎푐

2푎 ⟹ 푦 = ( )± ( ) ( )( )( )

= ±√

⟹ 푦 = = 3 or 푦 = = −1 But log 푥 = 푦

⟹ log 푥 = 3 ⟹ 푥 = 2 = 8 or log 푥 = −1 ⟹ 푥 = 2 = ∴ 푥 = 8, ii. log (푥 + 1) + log( ) 5 = 2 (Change to common base 5)

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⟹ log (푥 + 1) +( )

= 2 ⟹ log (푥 + 1)(log (푥 + 1)) + log 5 = 2 log (푥 + 1)

⟹ (log (푥 + 1)) + 1 = 2 log (푥 + 1) Let log (푥 + 1) = 푦 ⟹ 푦 + 1 = 2푦 ⟹ 푦 − 2푦+ 1 = 0⟹(푦 − 1)(푦 − 1) = 0 ⟹푦 − 1 = 0⟹푦 = 1 But log (푥 + 1) = 푦 ⟹ log (푥 + 1) = 1 ⟹푥 + 1 = 5 ∴ 푥 = 5 − 1 = 4

Example 6.11 If 푎 + 푏 − 23푎푏 = 0, a. Prove that log = (log푎 + log푏) b. Show that log 푎 × log 푏 × log 푐 = 24

Proof i. If 푎 + 푏 − 23푎푏 = 0, then 푎 + 푏 + 2푎푏 − 2푎푏 − 23푎푏 = 0 ⟹(푎 + 푏) − 25푎푏 = 0 ⟹ (푎 + 푏) = 25푎푏 By taking log of both sides we have log(푎 + 푏) = log 25푎푏 ⟹ 2 log(푎 + 푏) = log 25 + log푎 + log 푏 ⟹ 2 log(푎 + 푏) − log 25 = log푎 + log 푏 ⟹ 2 log(푎 + 푏) − log 5 = log푎 + log푏 ⟹ 2 log(푎 + 푏)− 2 log 5 = log푎 + log푏 ⟹ 2[log(푎 + 푏) − log 5] = log푎 + log 푏 ⟹ 2 log = log푎 + log 푏

∴ log = (log푎 + log푏) ∎ ii. By taking the 푙푒푓푡ℎ푎푛푑푠푖푑푒 of the equation a changing the bases to a common base 푎 we have:

log 푎 × log 푏 × log 푐 = × × log 푐

= × × 4 log 푐 = 2 × 3 × 4 = 24 ∎

Example 6.12

Given that log (푝+ 2) + log 푞 = 푟 − and log (푝 − 2) − log 푞 = 2푟 + 1, show that 푝 = 32 + 4. If 푟 = 1,find the possible values of 푝and 푞.

Proof We have log (푝 + 2) + log 푞 = 푟 − and log (푝 − 2)− log 푞 = 2푟 + 1

⟹ log (푝 + 2) 푞 = 푟 − and log = 2푟 + 1

By definition (푝 + 2)푞 = 8 … … …(1) and = 2 … … …(2)

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From (2) 푞 = (2 ∗) put (2 ∗) into (1)

⟹ (푝 + 2) = 8 ⟹ (푝+ 2)(푝 − 푞) = 8 × 2

⟹ 푝 − 4 = 2 × 2 = 2 ⟹ 푝 − 4 = 2 = 32 ∴ 푝 = 32 + 4 ∎ If 푟 = 1 then 푝 = 32 + 4 = 36 ⟹ 푝 = √36 = ±6 Put 푟 = 1 and 푝 = 6 or 푝 = −6 into (2 ∗) ⟹ 푞 = = or 푞 = = −1

But 푝 = −6,푞 = −1 give a false solution.∴ 푝 = 6, 푞 =

Example 6.13 If (푥 − 3) ( ) = 27(푥 − 3) , find the value of 푥.

Solution We have (푥 − 3) ( ) = 27(푥 − 3) (Take log of both sides) ⟹ log (푥 − 3) ( ) = log[27(푥 − 3) ] ⟹ log(푥 − 3) ∙ log(푥 − 3) = log 27 + log(푥 − 3) ⟹ (log(푥 − 3)) = 3 + 2 log(푥 − 3) Let log(푥 − 3) = 푦 ⟹ 푦 = 3 + 2푦 ⟹ 푦 − 2푦 − 3 = 0 ⟹ 푦 − 3푦 + 푦 − 3 = 0 ⟹ 푦(푦 − 3) + (푦 − 3) = 0 ⟹ (푦 − 3)(푦 + 1) = 0 ⟹ 푦 = 3 or 푦 = −1 But log(푥 − 3) = 푦 ⟹ log(푥 − 3) = 3 ⟹ (푥 − 3) = 10 = 1000 ⟹ 푥 = 1000 + 3 = 1003 or log(푥 − 3) = 1 ⟹ (푥 − 3) = 10푥 = 13

∴ 푥 = 13, 1003 Try: 1. Find the truth set of the equation log√ 8 = 푥 (SSSCE)

2. Evaluate 2log 5 − log 16 + 4 log 2 (SSSCE) 3. If log (5푥 + 2)− log (푥 − 1) = 0.7782 , find the value of 푥. (SSSCE) 4. Find the values of 푥 for which a. log (푥 − 2) + log (푥 − 6) = 1 b. log√ 9 = 푥 (WASSCE) 5. Given that log 4 = −2, evaluate log 푥 (WASSCE) 6. Given that log 2푥 + 2 log(푦 + 1) = log(푥 + 1), a. express 푦 in terms of 푥. b. determine the possible value of 푦, if 푥 = 8. (WASSCE)

Final Exercises 1. Simplify the following

96

i. log 256 ii. log 4096 iii. log iv. log 625v. log 3125 vi. log √ √343

vi. log vii. log 3 2. Solve the equations: i. log (푥 − 2) + log (푥 + 1) = 2 ii. log (푥 + 1)− 2log 3 = 0 iii. log (푥 − 4) + log (푥 + 2) = 0.84510 iv. log (3푥 − 1) + log (푥 + 1) = v. 2(log 푥) + log 푥 = 9 vi. (log 푥) + log 푥 − 8 = 0 vii. log(log (푥 − 1) = 0.47713 viii. 2 = 0 3. If log 푥푦 = 4 and log 푥 푦 = 5, find the values of 푥. 4. Find the values of 푦 if log (푦 − 10푦 + 28) = 2. 5. Find 푥 if i. log 27 = 2. ii. log 푥 = 2 iii. log 12 = 3 iv. log 128 = 푥 6.Evaluate the following:

i. 3 ii. 3 ( ) ( ) iii. (푥 + 3)10 ( ) ( ) 7. Given that log 2 = 푛, express log in terms of 푥. 8. Given that 푥 = 2 and 푦 = 2 , find log 푥 + log 3 in terms of 푥. 9. Solve the equations:

i. 3 = 8 ii. 2 ∙ 3 = 5 iii. 2 = 100 ivv. 6 ∙ 3 = 5 v. = 4 10. Find the truth set of the equation log (푎 − 4) + log (푎 + 3) = 2 11. If log 8 − log푦 = 2 log푥, express 푥 in terms of 푦.

12. Prove that (푥 + 푦) log( )

= (푥 + 푦)(푥 − 푦)

13. Simplify the following:

i. 2log 5(2푥 + 1) − 5 log 푥 − 2 log (2푥 − 푥 − 1) ii. log iii. log (푥 + 4)− 4 log (푥 − 3) + 2 log (푥 − 9) 14. Solve for 푥 if log(푥 + 3) + log푥 = 1 15. Evaluate i. log 9 ii. log 16 iii. log (3푥 + 8) = 1 + log + 1 (CCE) 16. If log 푎 = log 푏 + 1, find a relationship between 푎 and 푏 without any logarithm. (CCE) 17. Find the value of 푥 in the following: i. (푥 − 2) ( ) = (푥 − 2) ii. (푥 − 1) ( ) = 100(푥 − 1)

18. If (푥 − 푎) ( ) = (푥 − 푎), show that 푥 = 1 + 푎. 19. Write down the following as single logarithms:

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i. log 2 + 2 log 2 ii. log 14 − log 9 − log 2 iii. log(푥 − 1)− log(푥 + 1)

iv. log + log − log v. (6 log푥 − 3 log푦 − 2 log 푧) 20. Solve the following equations:

i. 2 = ii. 3 = iii. log (3푥 + 2) = 4iv. log (푥 − 6) = 21. Solve for the values of 푎 in the following equations: i. 2 log푎 = log 4 + log(푎 + 2) ii. log 3 + log푎 = log 6 + log(푎 − 1) iii. log푎 = log 3 − log(3푎 − 7) iv. 2 log 푎 − log(푎 + 4) = log푎 v. log 푎 + log 푎 = 6 vi. (log 푎) + 11 = 6 log 푎 + 6 log 3 22. Another interesting form of logarithms are logarithms with base 푒. They are called 푛푎푡푢푟푎푙logarithms. The natural log of 푥 is written as log 푥 = ln 푥 . Natural logarithms obey all the properties and law of the logarithms studied so far. For instance, log e = ln푒 = 1. Use natural logarithms to solve the following equations: i. 푒 = 2 ii. 3 ∙ 5 = 4 iii. 200(1 + 푟) = 5000 23. Find the truth set of the following equations: i. 10 ( ) = 푥 ii. 3 ( ) = 11푥 iii. 2 = 1 iv. 5 ( ) = 3푥

RELATIONS AND FUNCTIONS

Definition: A relation of the form 푦 = 푎푥 + 푏, where 푎푎푛푑푏are constants, is a rule uniting the two variables (quantities) 푥 and 푦. in the given relation 푥 is called independent variable while 푦 is called the dependent variable. The reason is that in other to know the value of 푦 the value of 푥 must be given to us. That is, the value of 푦 depends on the value of 푥. Definition: A function (mapping) is a special relation where for every member of 푥 (independent variable) there exist 표푛푒푎푛푑표푛푙푦표푛푒 value of 푦 (dependent variable) associated to it. In functional notation, the independent variable 푥 is called 푑표푚푎푖푛표푟푎푟푔푢푚푒푛푡 whereas the dependent variable푦 is called 푐표-푑표푚푎푖푛 We use the term 푟푎푛푔푒 to describe the set of possible values of co-domain associated to a given domain.

98

Consider the diagram below:

In the relation above, the set of domain is {푥 ,푥 ,푥 }. The set of co-domain is {1, 2, 3} and the range is {1, 2}. Therefore, we can conclude that the range is the subset of the co-domain.

Functional notation Functions are denoted by their names followed by a pair of braces with the arguments of the functions written in the braces. For example, 푓(푥) read ‘f of 푥’, is a function 푓 which has an argument 푥.sometime we put the sign : (푠푢푐ℎ푡ℎ푎푡) in front of the name of the function followed by the argument of the function. For example, 푓: 푥 read ′푓 such that 푥′, denotefunction 푓 which has an argument 푥. Let 푓(푥) be a function of 푥and let 푥 ,푥 ,푥 ,. , … , 푥 be the domain of 푓(푥). then we can find the range of values for the domain {푥 ,푥 ,푥 , … , 푥 } by substituting each element into the function. That is, 푥 → 푓(푥 ),푥 → 푓(푥 ),푥 → 푓(푥 ), … ,푥 → 푓(푥 ). The elements of the domain of 푓 domain are sometimes called 푝푟푒 − 푖푚푎푔푒푠 of 푓(푥). The values of 푓(푥 ),푓(푥 ),푓(푥 ), … , 푓(푥 ) are sometimes called the 푖푚푎푔푒푠 of their respective pre-images.

Example 7.1 A function ℎ with domain {−1,0, 2, 5} is defined as ℎ(푥) = . Find the images of each element in the domain.

Solution We have the domain {−1,0, 2, 5} and ℎ(푥) =

ℎ(−1) = ( )( )

= − ∴ −1 → − ℎ(0) =( )

= − ∴ 0 → −

ℎ(2) = ( )( )

= − ∴ 2 →− ℎ(5) = ( )( )

= −2 ∴ 5 → −2

Example 7.2 Find the value of 푥 for which 푓(푥) = −4 given that 푓(푥) = 3푥 − 5.

Solution If 푓(푥)=4 and 푓(푥) = 3푥 − 5 then 3푥 − 5 = −43푥 = −4 + 5 ∴ 푥 =

99

Example 7.3 A function 푓 defined by 푓(푥) = 푟푥 − 푠, where 푟 and 푠 are constants. Find the values of 푟 and 푠 if 푓(2) = −1 and 푓(−1) = 4.

Solution 푓(푥) = 푟푥 − 푠 푓(2) = 푟(2)− 푠 푓(−1) = 푟(−1)− 푠 but 푓(2) = −1 and 푓(−1) = 4 ⟹ 2푟 − 푠 = −1 … … … (1) and −푟 − 푠 = 4 … … … (2) From (2) 푠 = −푟 − 4 Put 푠 = −푟 − 4 into (2) ⟹ 2푟 − (−푟 − 4) = −1 ⟹ 2푟 + 푟 + 4 = −1 ⟹ 3푟 = −1 − 4 ⟹ 푟 = −

Put 푟 = − into (2) ⟹ 푠 = − − − 4 = = − ∴ 푟 = − ,푠 = −

푓(푥) = − 푥 +

Types of relations 푀푎푛푦푡표푀푎푛푦푟푒푙푎푡푖표푛 This is the relation where several elements of the domain correspond (map) to several elements of the co-domain. Consider the diagram below:

푂푛푒푡표푀푎푛푦푟푒푙푎푡푖표푛 This is the relation where one element of the domain corresponds (maps) to corresponds to several elements of the co-domain. Consider the following diagram below:

푀푎푛푦푡표푂푛푒푟푒푙푎푡푖표푛 This is the relation where several elements of the domain correspond(map) to one element of the co-domain. This type of relation is considered a function. Consider the diagram below:

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푂푛푒푡표푂푛푒푟푒푙푎푡푖표푛 This is the relation where each element of the domain corresponds (maps) to exactly one element of the co-domain. This type of relation is also considered a function. Consider the diagram below:

Vertical line test

If the vertical line cuts the graph of a function푓 at one point along the curve, the 푓 is considered a function.

Example 7.4 Show that the graph of 푦 = 푥 is a not function.

Proof Let 푥 = 9, then 푦 = 9 ⟹ 푦 = ±√9 = ±3. Now we have the ordered pair (9, 3) and (9,−3). This follows that for a given value of 푥 the exist two values of 푦. Hence, the curve 푦 = 푥 is not a function. See the vertical line test below.

푦 3 ⦁ 푦 = 푥 Vertical line

9 푥 -3 ⦁

Example 7.2

101

Determine with a vertical line the curve which is a function. i. ii. iii. iv. v.

Solution

i. If a vertical line is drawn to cut the curve, it will cut more than one point. Hence the curve (i) is not a function. ii. If a vertical line is drawn to cut the curve, it will cut at only one point. Hence the curve (ii) is a function. iii. If a vertical line is drawn to cut the curve, it will cut at only one point. Hence the curve (ii) is a function. Try iv and v (퐴푛푠푤푒푟푠: iv. A function v. a function)

Even and Odd functions

102

Let 푓(푥) be a function in 푥. If for all values of 푥,푓(−푥) = 푓(푥), then 푓(푥) is an 푒푣푒푛푓푢푛푐푡푖표푛. If for all values of 푥, 푓(−푥) = −푓(푥), then 푓(푥) is an 표푑푑 function.

Example 7.5 Given that 푓(푥) = 푥 − 2푥 + 3, show that 푓(푥) is an even function.

Proof If 푓(푥) is an even function, then 푓(−푥) = 푓(푥). Now, 푓(−푥) = (−푥) − 2(−푥) + 3 = 푥 − 2푥 + 3 = 푓(푥). Hence, 푓(푥) is even. Try: Show that the curve 푓(푥) = 푥 − 4 is an even function. Show that the function 푓(푥) = 푥 − 3푥 is not an even function.

Example 7.6 Show that the 푓(푥) = 푥 − 푥 is an odd function.

푷풓풐풐풇 If 푓(푥) is an odd function, then 푓(−푥) = −푓(푥). Now, 푓(−푥) = (−푥) − (−푥) = −푥 + 푥 + 1 = −(푥 − 푥) = −푓(푥). Hence, 푓(푥) is odd.

Periodic function Let 푓(푥) be a function and let 푡 be a parameter called period. If 푓(푥)is a periodic function,then 푓(푥 + 푡) = 푓(푥). The function푦 = sin 푥 is periodic with period 2휋 such that sin(푥 + 2휋) = sin 푥 ,where 휋 = 180°.

One to One function A function is said to be one to one if for each element of its domain there exist one and only one element of the range associated to it. In functional notation, a function 푓(푥) is one to one if

푓(푚) = 푓(푛) ⟺ 푚 = 푛 or 푓(푚) ≠ 푓(푛) ⟺ 푚 ≠ 푛 Example 7.7

A function 푓 defined on the domain{−1, 1, 2, 3} is given as 푓(푥) → 푥 − 2. i. Show pictorially whether 푓(푥) is one to one. ii. Show algebraically whether 푓(푥) is one to one. iii. Show whether or not 푓(푥) is an even function.

Solution i. We have 푓(푥) → 푥 − 2 and domain {−1, 1, 2, 3}.

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푓(−1) = (−1) − 2 = −1 푓(1) → 1 − 2 = −1 푓(2) → 2 − 2 = 2 푓(3) → 3 − 2 = 7 푥 푓(푥)

−112−3

−127

.

From the diagram, we see 푓(푥) is not one to one since the elements −1 and 1 of the domain maps one element in the range. ii. If 푓(푥) is one to one, then 푓(푚) = 푓(푛) ⟺ 푚 = 푛. Now, 푓(푚) = 푓(푛) ⟹ (푚) − 2 = (푛) − 2 ⟹ 푚 − 2 = 푛 − 2 ⟹ 푚 = 푛 − 2 + 2 ⟹ 푚 = √푛 = ±푛. Since 푚 = −푛 or 푚 = 푛, the function 푓 is not one to one. iii. If 푓(푥) is an even function, then푓(−푥) = 푓(푥). Now, 푓(−푥) = (−푥) − 2 = 푥 − 2 = 푓(푥). Hence 푓(푥) is even.

Example 7.8 Show that the function 푓(푥) = 2푥 − 1 is one to one.

Solution 푓(푥) = 2푥 − 1 is one to one if for 푓(푚) = 푓(푛) ⟹푚 = 푛 Now, 푓(푚) = 푓(푛) ⟹ 2푚 − 1 = 2푛 − 1 ⟹ 2푚 = 2푛 − 1 + 1 ⟹2푚 = 2푛 ⟹ 푚 = 푛.Therefore, 푓(푥) is one to one.

Onto function A function 푓 is said to be onto if for each element in the co-domain there exist an element in the domain associated to it. To show that a function 푓(푥) is onto: 1. Set 푓(푥) = 푦. 2. Find an expression for 푥 in terms of 푦. 3. Choose some arbitrary values for 푦and determine their corresponding values of 푥. 4. If for each value of 푦 there exist a value of 푥 associated to it in the domain of 푓, the 푓(푥) is onto. (The values of 푦 should be taken from the taken from the domain of 푓(푥))

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Example 7.9 Show that the function 푓(푥) = 푥 − 4 defined on the set ℤ = {… ,−2,−1, 0, 1, 2, 3, … } is onto.

Proof We have 푓(푥) = 푥 − 4 and ℤ = {… ,−2,−1, 0, 1, 2, 3, … }. Let 푓(푥) = 푦 ⟹ 푥 − 4 = 푦 ⟹ 푥 = 푦 + 4 ⟹ 푥 = 푦 + 4. When 푦 = 1, 푥 = 1 + 4 = 5.When 푦 = 0,푥 = 4. We can easily see that for every given element of the co-domain,푦, there exists an element in the domain ℤ = {… ,−2,−1, 0, 1, 2, 3, … }associated to it. Therefore, 푓(푥) is onto

Example 7.10 Show that the function 푓(푥) = 2푥 + 3 defined on the set of 푟푎푡푖표푛푎푙 numbers is one to one and onto.

Solution We have 푓(푥) = 2푥 + 3. If 푓(푥) is one to one the 푓(푚) = 푓(푛) ⟺ 푚 = 푛 Now, 푓(푚) = 푓(푛) ⟹ 2푚 + 3 = 2푛 + 3 ⟹ 2푚 = 2푛 + 3 − 3 ⟹ 2푚 = 2푛 ⟹ 푚 = 푛 Therefore, 푓(푥) is one to one. Let 푓(푥) = 푦 ⟹ 2푥 + 3 = 푦 ⟹ 2푥 = 푦 − 3 ⟹ 푥 =

When 푦 = 1,푥 = = −1 Given any value of 푦 from the set of 푟푎푡푖표푛푎푙 numbers there exists an element associated to it. Therefore, 푓(푥) is onto. Try: Show that the function 푓(푥) = 3푥 − 4 defined on the set ℤ = {… ,−2,−1, 0, 1, 2, 3, … } is not onto. Show that the function푔(푥) = 2푥 defined on the set of 푛푎푡푢푟푎푙푛푢푚푏푒푟is one to one but not onto.

Domain of a function The domain of a function is refers to the values of the argument of the function which make the function well-defined. That is the domain excludes any value of the argument of the function which makes the function not defined in real terms. The domain of functions of the form ( )

( ),excludes any value of 푔(푥) which set 푔(푥) to

zero.

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Example 7.11 State the largest possible domain of the following functions defined on the set of real numbers: i. 푓(푥) = 3 − 2푥 ii. 푓(푥) = iii. 푔(푥) = 푥 + 2

Solution i. We have 푓(푥) = 3 − 2푥. We see that the domain of this function exist for all real numbers. Therefore, domain 퐷 = {푥: 푥 ∈ ℝ} ii. The domain of the function 푓(푥) = exists for all real numbers except when 푥 − 2 = 0 ⟹ 푥 = 2. Therefore, domain 퐷 = {푥: 푥 ∈ ℝ,푥 ≠ 2} iii. The domain of the function 푔(푥) = 푥 + 2 exist for all real numbers. Therefore, domain 퐷 = {푥: 푥 ∈ ℝ}

Example 7.12 State the domain of the following functions. i. 푓(푥) = √3 − 3푥 ii. 푓(푥) = √푥 − 푥 iii. ℎ:푥 =

√

Solution i. The domain of the function 푓(푥) = √3 − 3푥 exists for all real numbers except when 3− 3푥 < 0 ⟹ −3푥 < −3 ⟹ 푥 > 3. This is true because √푎 ≥ 0,푎 ∈ ℝ. Therefore, domain 퐷 = {푥: 푥 ∈ ℝ, except, 푥 > 3} ii. The domain of the function 푓(푥) = √푥 − 푥 exists for all real numbers except when 푥 − 푥 < 0 ⟹ 푥(1 − 푥) < 0 ⟹ 푥 < 0 or1 − 푥 < 0 − 푥 < −1 ⟹ 푥 > 1. This is true because √푎 ≥ 0,푎 ∈ ℝ. Therefore, domain 퐷 = {푥: 푥 ∈ ℝ, except, 0 > 푥 > 1} iii. The domain of the function ℎ: 푥 =

√exists except when 2푥 − 1 ≤ 0 ⟹ 푥 ≤

This is true because √푎 ≥ 0,푎 ∈ ℝ and domain of a rational function exist except when the function at the denominator is equal to 0. Therefore, domain 퐷 = 푥: 푥 ∈ ℝ, except,푥 ≤

Range of a function

106

Let 푓(푥) = 푦. The range of the function 푓 is the possible value(s) of 푦 which makes the domain푥,well defined. The range excludes all values of 푦 which makes a given domain not defined in real terms. To find the range of a function 푓(푥). 1. set 푓(푥) = 푦2. Express 푥 in terms of 푦. 3. State the value(s) of 푦 for which makes 푥 is well defined.

Example 7.13 State the range of the following functions. i. 푓(푥) = ,푥 ≠ −1 ii. ℎ: 푥 = 3푥 − 1 iii. 푘(푥) = iv. 푓: 푥 = √4 − 푥

Solution i. We have 푓(푥) = . Let푓(푥) = 푦 ⟹ = 푦 ⟹ 3푥 − 4 = 푦(2푥 + 2) ⟹3푥 − 4 = 2푥푦 + 2푦⟹ 3푥 − 2푥푦 = 2푦 + 4 ⟹ 푥(3− 2푦) = 2푦 + 4 ⟹ 푥 = . Therefore, range 푅 = 푦:푦 ∈ ℝ, 푦 ≠ ii. We haveℎ: 푥 = 3푥 − 1. Let ℎ:푥 = 푦 ⟹3푥 − 1 = 푦⟹3푥 = 푦 + 1 ⟹ 푥 = Therefore, range 푅 = {푦:푦 ∈ ℝ} iii. We have푘(푥) = . Let 푘(푥) = 푦⟹ = 푦⟹2푥 + 1 = 3푦 ⟹푥 = Therefore, range 푅 = {푦:푦 ∈ ℝ} iv. We have 푓: 푥 = √4 − 푥 . Let 푓: 푥 = 푦 ⟹√4 − 푥 = 푦⟹4 − 푥 = 푦 ⟹푥 = 4 − 푦 ⟹ 푥 = ± 4 − 푦 Therefore, range 푅 = {푦: 푦 ∈ ℝ, except, 푦 > ±2} Try: Find the range of the following functions. i. 푓(푥) = ii.푓(푥) = 2푥 + 5 iii. 푔(푥) = Inverse of one to one functions Let 푓 be a function. If the inverse of 푓, denoted by 푓 , exist, then 푓 is one to one and onto. That is, a function is invertible if it is one to one and onto and vice versa. In functional notation

푓(푥) = 푦 ⟺푓 (푦) = 푥. Example 7.14

107

A function 푓 is defined by 푓(푥) = {(1, 2), (3, 2), (−1, 4)}, find 푓 (푥), the inverse of 푓(푥). Solution

Given 푓(푥) = {(1, 2), (3, 2),−1, 4)}, to find 푓 (푥) interchange the elements of 푥 for 푦. ∴ 푓 (푥) = {(2, 1), (2, 3), (−1, 4)}

Example 7.15

Show that the function 푓(푥) = , 푥 ≠ , defined on set of real numbers is one to one. Hence, find the inverse of 푓(푥).

Solution i. If 푓(푥) is one to one, then 푓(푚) = 푓(푛) ⟺ 푚 = 푛 Now, 푓(푚) = 푓(푛) ⟹ = ⟹ (푚 − 4)(2푛 − 1) = (푛 − 4)(2푚− 1) ⟹ 2푚푛 −푚 − 8푛 + 4 = 2푚푛 − 푛 − 8푚 + 4 ⟹ 8푚 −푚 = 8푛 − 푛 + 2푚푛 − 2푚푛 + 4 − 4 ⟹ 7푚 = 7푚 ⟹ 푚 = 푛 ∴,푓(푥) is one to one. Inverse of 푓(푥) Let 푓(푥) = 푦 ⟹ = 푦 ⟹ 푥 − 4 = (2푥 − 1)푦 ⟹ 푥 − 4 = 2푥푦 − 푦

⟹ 푦 − 4 = 2푥푦 − 푥 ⟹ 푦 − 4 = 푥(2푦 − 1) ⟹ 푥 =

Interchange 푥 for 푦.that is, 푦 = ∴ 푓 (푥) =

Example 7.16 Find the inverses of the following functions defined on the set of real numbers. i. 푓(푥) = ,푥 ≠ − ii. 푔(푥) = , 푥 ≠ 2

Solution i. Given 푓(푥) = . To find the inverse of 푓(푥) let 푓(푥) = 푦

⟹ = 푦(Find 푥 in terns of 푦) ⟹ 푥 = 푦(3푥 + 1) ⟹ 푥 = 3푥푦 + 푦

⟹ 푥 − 3푥푦 = 푦 ⟹ 푥(1 − 3푦) = 푦 ⟹ 푥 = .

Interchange 푥 for 푦. ⟹ 푦 = ∴ 푓 (푥) = ii. given 푔(푥) = . Let 푔(푥) = 푦 ⟹ = 푦 ⟹ 5푥 + 2 = 푦(푥 − 2) ⟹ 5푥 + 2 = 푥푦 − 2푦 ⟹ 2푦 + 2 = 푥푦 − 5푥 ⟹ 2푦+ 2 = 푥(푦 − 5)

108

⟹ 푥 = . Interchange푥 for 푦. ⟹ 푦 = ∴ 푔 = ( ) ,푥 ≠ 5

Try: Find the inverses of the following functions defined on the set of real numbers. 1. ℎ: 푥 = 2. 푝(푥) = , 푥 ≠ −

Example 7.17 If 푓 (푥) = , 푥 ≠ −4, find 푓(푥).

Solution Given 푓 (푥) = , to find 푓(푥) let 푓 (푥) = 푦 ⟹ = 푦 ⟹ 3푥 − 1 = 푦(4 + 푥) ⟹ 3푥 − 1 = 4푦 + 푥푦 ⟹ 3푥 − 푥푦 = 4푦 + 1 ⟹ 푥(3 − 푦) = 4푦 + 1 ⟹ 푥 = . Interchanging푥 for 푦gives푦 = ∴ 푓(푥) = , 푥 ≠ 3.

Example 7.18 A function 푓 is defined on the set of real numbers by 푓(푥) = ,푥 ≠ − .

i. Find 푓(푥), 푓 and the domain of 푓 . ii. Show that 푓(푥) is one to one. iii. Find the images of the domain {1, 2, 4} under 푓. iv. Find the values of 푥 for which 푓(푥) = 푓 (푥)

Solution i. 푓(푥) = ,푥 ≠ − . 푓(푥) = ∙ = ∴ . 푓(푥) =

To find 푓 let 푓(푥) = 푦 ⟹ = 푦 ⟹ 5푥 = 3푥푦 + 2푦 ⟹ 5푥 − 3푥푦 = 2푦

⟹푥(5 − 3푦) = 2푦 ⟹ 푥 = Interchange 푥 for 푦. ∴ 푓 (푥) = , 푥 ≠

The domain of 푓 exists for all real number except when 5 − 3푥 = 0푥 =

Therefore, domain 퐷 = 푥: 푥 ∈ ℝ, 푥 ≠ ii. 푓(푥) is one to one if for 푓(푚) = 푓(푛) ⟹ 푚 = 푛 Now, 푓(푚) = 푓(푛) ⟹ = ⟹ 5푚(3푛 + 2) = 5푛(3푚 + 2) ⟹ 15푚푛 + 10푚 = 15푚푛 + 10푛⟹10푚 = 15푚푛 − 15푚 + 10푛⟹ 10푛 = 10푚 ⟹ 푚 = 푛. Hence, 푓(푥) is one to one. iii. We have the domain {1,−1, 2,−2}

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푓(1) = ( )( )

= 1 ∴ 1 → 1 푓(−1) = ( )( )

= ∴ −1 →

푓(4) = ( )( )

= ∴ 4 → iv. 푓(푥) = 푓 (푥) ⟹ = ⟹ 5푥(5 − 3푥) = 2푥(3푥 + 2) ⟹ 25푥 − 15푥 = 6푥 + 4푥⟹ 21푥 − 21푥 = 0 ⟹ 21푥(푥 − 1) = 0 ⟹ 21푥 = 0 ⟹ 푥 = 0 or ⟹ 푥 − 1 = 0 ⟹ 푥 = 1 ∴,푥 = 0 or 푥 = 1

Example 17.19 A function 푔 is defined on the set of 푟푒푎푙푛푢푚푏푒푟푠 as 푔(푥)=푎푥 + 푏푥, where 푎 and 푏are constants. If 푔(3) = 9 and 푔(−3) = 14,find the values of 푎 and 푏.

Solution We have 푔(푥) = 푎푥 + 푏푥. 푔(3) = 푎(3) + 3푏 = 9 ⟹ 9푎 + 3푏 = 9 (Dividing both sides by 3 gives) 3푎 + 푏 = 3 … … … (1) ⟹ 푔(−3) = 푎(−3) + 푏(−3) = 14 ⟹ 9푎 − 3푏 = 14 … … … (2) From (1) 푏 = 3 − 3푎 Put 푏 = 3 − 3푎 into (2) ⟹ 9푎 − 3(3 − 3푎) = 14 ⟹ 9푎 − 9 + 9푎 = 14 ⟹ 18푎 = 14 + 9 ⟹ 푎 = Put 푎 = into (1)

⟹ 푏 = 3 − 3 = − ∴ 푎 = ,푏 = −

Graphs and region values of functions The graphs of functions we shall describe in this book include 1. The graph of straight line function of the form 푦 = 푚푥 + 푐, where 푚 is the gradient of the function and 푐 is the 푦 intercept. We can sketch the graph of this function if we know the values of the 푥 and 푦 intercepts. The 푥 intercept is a value on the 푥 for which 푦 = 0 while the 푦 intercept is the value on the 푦 axis for which 푥 = 0 2. The graphs of quadratic and cubic function of the forms 푦 = 푎푥 + 푏푥 + 푥 and 푦 = 푎푥 + 푏푥 + 푐푥 + 푑 3. The graphs of parabolic function of the form푦 = 4푎푥 and 푥 = 4푎푦. 4. The graph of a circle with centre (푎,푏) and radius 푟. We shall know that the equation of the circle is given by (푥 − 푎) +(푦 − 푏) = 푟 5. The graphs of logarithmic functions of the y = log 푏 Definition: The region values (boundary values) of a function are values that satisfies a

110

given function or set of functions. Composite functions

Let 푓(푥) and 푔(푥) be two functions. The function 푓(푔(푥)) is the composite function formed when we assign 푔(푥) into 푓(푥). Composition of functions involves putting the range of one function into the domain of another function.

Notation The composition of functions involves two notations 1. Infix notation This notation involves the use of the ring operator ‘∘ ′ between functions. Under this notation we write 푓 ∘ 푔 to mean the composition of 푔 followed by 푓. This literally mean put the domain of푔 into 푓. 2. Generic notation This notation does not involve the use of any operator. Under this notation we write 푓푔 to mean the composition of 푔 followed by 푓. Note: 1.푓푔 = 푓 ∘ 푔 but 푓 ∙ 푔 ≠ 푓 ∘ 푔 2. 푓 ∘ 푔 ≠ 푔 ∘ 푓 3. (푓 ∘ 푔) = 푔 ∘ 푓 4. 푓 ∘ 푔 is well defined if the co-domain of 푔 is contained in the domain of 푓.

Example 7.20 Two functions 푓 and 푔 are defined on the set of 푟푒푎푙푛푢푚푏푒푟푠 as 푓(푥) = 3푥 + 2 and 푔(푥) = ,푥 ≠ 2. a. Find i. (푓 ∘ 푔)(푥) ii.(푔 ∘ 푓)(푥) iii. 푓 (푥) b. Show that (푓 ∘ 푓 )(푥) = 푥

solution a. We have 푓(푥) = 3푥 + 2 and 푔(푥) = i.

(푓 ∘ 푔)(푥) = 푓 푔(푥) = 3 + 2 = 3( )

+ 2 = ( )( )

=

=

ii. (푔 ∘ 푓)(푥) = 푔 푓(푥) = ( )( )

= ( ) = = + = 1 +

iii. For 푓 (푥) let 푓(푥) = 푦 ⟹ 3푥 + 2 = 푦 ⟹ 푥 = ⟹ 푥 = ±

111

Interchanging 푥for 푦 gives 푦 = ± ∴ 푓 (푥) = ±

b. (푓 ∘ 푓 )(푥) = 푓 푓 (푥) = 3 ± + 2 = 3 × + 2 = 푥 − 2 + 2 = 푥 ∎

Example 7.21 The functions 푓,푔 and ℎ are defined on the set of real numbers as 푓: 푥 = 2푥 − 1, 푔: 푥 = 푥 − 푥 and ℎ: 푥 = 푥 − 3. a. Find i. (푓 ∘ 푔)(푥) ii. (푓 ∘ 푔 ∘ ℎ)(푥) b. Find the truth set of the equations: (푓 ∘ 푔)(푥) = (푓 ∘ 푔 ∘ ℎ)(푥) and (푔 ∘ ℎ)(푥) = (푓 ∘ 푔 ∘ ℎ)(푥)

Solution a. We are given 푓: 푥 = 2푥 − 1,푔:푥 = 푥 − 푥 andℎ: 푥 = 푥 − 3

i. (푓 ∘ 푔)(푥) = 푓 푔(푥) = 2(푥 − 푥)− 1 = 2푥 − 2푥 − 1

ii. (푓 ∘ 푔 ∘ ℎ)(푥) = 푓 푔 ℎ(푥) so let’s first find 푔(ℎ(푥)).

푔 ℎ(푥) = 푥 − 3 − 푥 − 3 = 푥 − 3 푥 − 3 − 1 = 푥 − 3 푥 − 4

= 푥 − 푥 + 12

Now, (푓 ∘ 푔 ∘ ℎ)(푥) = 2 푥 − 푥 + 12 − 1 = 푥 − 7푥 + 23

b. The equation (푓 ∘ 푔)(푥) = (푓 ∘ 푔 ∘ ℎ)(푥) ⟹2푥 − 2푥 − 1 = 푥 − 7푥 + 23 (Multiplying both sides by 2 gives) 4푥 − 4푥 − 2 = 푥 − 14푥 + 46

(rearranging like terms gives) 3푥 + 10푥 − 48 = 0 Comparing 7푥 + 6푥 − 52 = 0 with 푎푥 + 푏푥 + 푐 = 0 gives 푎 = 3, 푏 = 10and 푐 = −48. The values of 푥 can be determined by the formula:

푥 =−푏± 푏2−4푎푐

2푎 푥 = ± ( ) ( )( )( )

= ±√ = ±√

∴ 푥 = = or 푥 = = −

The equation (푔 ∘ ℎ)(푥) = (푓 ∘ 푔 ∘ ℎ)(푥) ⟹ 푥 − 푥 + 12 = 푥 − 7푥 + 23 (Multiplying both sides by 4 gives) 푥 − 14푥 + 48 = 2푥 − 28푥 + 92

112

(rearranging like terms gives) 푥 − 14푥 + 44 = 0 Comparing 푥 − 14푥 + 44 = 0 with 푎푥 + 푏푥 + 푐 = 0 gives 푎 = 1, 푏 = −14and 푐 = 44. The values of 푥 can be determined by the formula:

푥 =−푏± 푏2−4푎푐

2푎 푥 = ( )± ( ) ( )( )( )

= ±√ = ±√ = ± √

∴ 푥 = 7 + √5 or 푥 = 7 − √5

Example 7.22 If 푓(푥) = 푥 − 1 and 푔(푥) = 3푥 − 5,푥 ∈ ℝ,find 푔(푥).

Solution We have 푓(푥) = 푥 − 1 and (푓 ∘ 푔)(푥) = 3푥 − 5. Let 푔(푥) = 푎. Now, (푓 ∘ 푔)(푥) = 푓 푔(푥) = 푎 − 1 ⟹ 푎 − 1 = 3푥 − 5 ⟹ 푎 = 3푥 − 5 + 1 = 3푥 − 4 ∴ 푔(푥) = 3푥 − 4. Alternatively, If (푓 ∘ 푔)(푥) = 3푥 − 5 and 푓(푥) = 푥 − 1, then 푔(푥) must be a function of the form 푔(푥) = 푎푥 + 푏, where 푎 and 푏 are constants. Now, (푓 ∘ 푔)(푥) = 3푥 − 5 ⟹ 푎푥 + 푏 − 1 = 3푥 − 5. Comparing coefficients gives 푎 = 3 and 푏 − 1 = −5 ⟹ 푏 = −5 + 1 = −4 ∴ 푔(푥) = 3푥 − 4

Example 7.23 Given that 푝(푥) = 3푥 − 4, find 푞(푥) if (푝 ∘ 푞)(푥) = 푥 − 2.

Solution We have 푝(푥) = 3푥 − 4 and (푝 ∘ 푞)(푥) = 푥 − 2. Let 푞(푥) = 푎. Now, (푝 ∘ 푞)(푥) = 푝 푞(푥) = 3푎 − 4

⟹3푎 − 4 = 푥 − 2⟹3푎 = 푥 − 2 + 4 ⟹푎 = ∴ 푞(푥) = Alternatively, If (푝 ∘ 푞)(푥) = 푥 − 2 and 푝(푥) = 3푥 − 4, then 푞(푥) is a function of the form 푞(푥) = 푎푥 + 푏푥 + 푐,where 푎,푏 and 푐 are constants. Now, (푝 ∘ 푞)(푥) = 푥 − 2 ⟹3(푎푥 + 푏푥 + 푐) − 4 = 푥 − 2 ⟹ 3푎푥 + 3푏푥 + 3푐 − 4 = 푥 − 2 Comparing coefficients gives 3푎 = 1 ⟹ 푎 = , 3푏 = 0 ⟹ 푏 = 0 and 3푐 − 4 = −2 ⟹ 푐 =

113

∴ 푞(푥) = 푥 + =

Example 7.24 Given that 푓(푥) = 2푥 + 3 and (푔 ∘ 푓)(푥) = 4푥 − 1, 푥 ∈ ℝ, find 푔(푥).

Solution If 푓(푥) = 2푥 + 3 and (푔 ∘ 푓)(푥) = 4푥 − 1,then 푔(푥) must be a function of the form 푔(푥) = 푎푥 + 푏푥 + 푐, where 푎, 푏 and 푐 are constants. Now, (푔 ∘ 푓)(푥) = 4푥 − 1 ⟹푎(2푥 + 3) + 푏(2푥 + 3) + 푐 = 4푥 − 1 ⟹ 푎(4푥 + 12푥 + 9) + 2푏푥 + 3푏 + 푐 = 4푥 − 1 ⟹ 4푎푥 + 12푎푥 + 9푎 + 2푏푥 + 3푏 + 푐 = 4푥 − 1 ⟹ 4푎푥 + (12푎 + 2푏)푥 + 9푎 + 3푏+ 푐 = 4푥 − 1 Comparing coefficients gives 4푎 = 4 ⟹ 푎 = 1,12푎+ 2푏 = 0 ⟹ 12(1) + 2푏 = 0 ⟹ 2푏 = −12 ⟹ 푏 = −6 and 9푎 + 3푏 + 푐 = −1 ⟹ 9(1) + 3(−6) + 푐 = −1⟹ 푐 = −1− 9 + 18 = 8

∴ 푔(푥) = 푥 − 6푥 + 8 Try: 1. Given that 푓(푥) = 푥 − 1 and (푓 ∘ 푔)(푥) = 3푥 − 2, 푥 ∈ ℝ, find 푔(푥). 2. If 푓(푥) = 푥 − 1 and (푔 ∘ 푓)(푥) = 2푥, 푥 ∈ ℝ,find 푔(푥). 3. If 푠(푥) = 푥 + 4 and (푟 ∘ 푠)(푥) = 푥 − 푥 + 3,푥 ∈ ℝ,find 푟(푥) 4. Given that 푓(푥) = 2푥 + 3 and (푔 ∘ 푓)(푥) = 푥 + 4, find 푔(푥).

Example 7.25 Given that 푓: 푥 = 푎푥 + 푏 and 푔: 푥 = 푐푥 + 푑, where 푎,푏, 푐and 푑 are constants. a. If 푓 and 푔 are defined on the set of 푟푒푎푙푛푢푚푏푒푟푠, show that (푓 ∘ 푔)(푥) = (푔 ∘ 푓)(푥) if and only if 푑(푎 − 1) = 푏(푐 − 1). b. Given that (푓 ∘ 푔)(−1) = 4 and (푔 ∘ 푓)(2) = 5, find the values of 푎 and 푐 if 푏 = 3 and

푑 = −2

Solution a. We have 푓: 푥 = 푎푥 + 푏 and 푔(푥) = 푐푥 + 푑 i. (푓 ∘ 푔)(푥) = 푓 푔(푥) = 푎(푐푥 + 푑) + 푏 = 푎푐푥 + 푎푑 + 푏

(푔 ∘ 푓)(푥) = 푔 푓(푥) = 푐(푎푥 + 푏) + 푑 = 푎푐푥 + 푏푐 + 푑 (푓 ∘ 푔)(푥) = (푔 ∘ 푓)(푥) ⟹ 푎푐푥 + 푎푑 + 푏 = 푎푐푥 + 푏푐 + 푑 ⟹ 푎푐푥 − 푎푐푥 + 푎푑 − 푑 = 푏푐 − 푏⟹푎푑 − 푑 = 푏푐 − 푏⟹푑(푎 − 1) = 푏(푐 − 1)

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∎ b. (푓 ∘ 푔)(−1) = 푎푐(−1) + 푎푑 + 푏 = 4 ⟹ −푎푐 + 푎푑 + 푏 = 4 But 푏 = 3, 푑 = −2 ⟹ −푎푐 + 푎(−2) + 3 = 4 ⟹−푎푐 − 2푎 = 4 − 3 ⟹ 푎(−푐 − 2) = 1 … … … (1) and (푔 ∘ 푓)(2) = 푎푐(2) + (3)푐 − 2 = 5 ⟹ 2푎푐 + 3푐 = 5 + 2 ⟹ 2푎푐 + 3푐 = 7 … … … (2) From (1) 푎 = = −

Put 푎 = − into (2) ⟹ 2푐 − + 3푐 = 7 ⟹− + 3푐 = 7 (Multiplying both sides by 푐 + 2gives −2푐 + 3푐(푐 + 2) = 7(푐 + 2) ⟹ −2푐 + 3푐 + 6푐 = 7푐 + 14

(Rearranging like terms gives) 3푐 − 3푐 − 14 = 0 Comparing 3푐 − 3푐 − 14 = 0 with 푎푥 + 푏푥 + 푑 = 0 gives 푎 = 3,푏 = −3and 푐 = −14. The values of 푐 can be determined by the formula:

푐 =−푏± 푏2−4푎푑

2푎 ⟹ 푐 = ( )± ( ) ( )( )( )

= ±√ = ±√

⟹ 푐 = √ ≈ 2.7174 or 푐 = √ ≈ −1.7174

Put the values of 푐 into (1). That is, when 푐 ≈ 2.7174푎 ≈ −.

≈ −0.2120

And when 푐 = −1.7174,푎 ≈ −.

=≈ −3.5386 ∴ 푎 ≈ −0.2120, 푐 ≈ 2.7174 or 푎 ≈ −3.5386, 푐 ≈ −1.7174

Example 7.26

The functions 푓 and 푔 are defined on the set of 푟푒푎푙푛푢푚푏푒푟푠 as 푓(푥) = 푥 − 2푥 and 푔(푥) = ,푥 ≠ 3. i. State the domain of the two functions. ii. State the range of 푔 iii. Show that 푔(푥) is one to one and onto. iv. Find (푔 ∘ 푓)(푥) and state its domain.

Solution i. We have 푓(푥) = 푥 − 2푥 and 푔(푥) = , 푥 ≠ 3. The domain of 푓(푥) exists for all real numbers. Therefore, domain 퐷 = {푥: 푥 ∈ ℝ} The domain of 푔(푥) exists for all real numbers except when 푥 − 3 = 0 or 푥 = 3 Therefore, domain 퐷 = {푥: 푥 ∈ ℝ,푥 ≠ 3} ii. For the range of 푔(푥) let 푓(푥) = 푦 ⟹ = 푦 ⟹ 푥 = 푦(푥 − 3) ⟹ 푥 = 푥푦 − 3푦

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⟹ 푥푦 − 푥 = 3푦 ⟹ 푥(푦 − 1) = 3 ⟹ 푥 = .∴ range 푅 = {푦:푦 ∈ ℝ,푦 ≠ 1}

iii. 푔(푥) is one to one if 푔(푚) = 푔(푛) ⟹ 푚 = 푛. Now, 푔(푚) = 푔(푛) ⟹ = ⟹ 3푚(푛 − 3) = 3푛(푚 − 3) ⟹ 3푚푏 − 9푚 = 3푚푛 − 9푛 ⟹ −9푚 = 3푚푛 − 3푚푛 − 9푛 ⟹ −9푚 = −9푛 ⟹ 푚 = 푛 Hence, 푔(푥) is one to one.

iv. (푔 ∘ 푓)(푥) = 푔 푓(푥) = ( )( )

= ( ) = ( )( ) ( )

= ( )( )( )

The domain of 푔 푓(푥) exist except when (푥 − 3)(푥 + 1) = 0 푥 − 3 = 0 or 푥 + 1 = 0 ⟹ 푥 = 3 or 푥 = −1 ∴, domain 퐷 = {푥:푥 ∈ ℝ,푥 ≠ −1, 3}

Example 7.27

Given that 푓(푥) = 2푥 + 5, ℎ(푥) = and (푓 ∘ 푔)(푥) = 2푥 − 2,푥 ∈ ℝ. a. Find i. 푔(푥) ii. (푔 ∘ 푓)(푥) iii. (푓 ∘ 푔 ∘ ℎ)(푥) b. If ℎ(푥) ≡ 퐴 + , where 퐴and 퐵 are constants, find the values of 퐴and 퐵. c. Solve the equation (푓 ∘ 푔 ∘ ℎ)(푥) = 0, leaving your answers in surd form.

solution a. We have 푓(푥) = 2푥 + 5, ℎ(푥) = and (푓 ∘ 푔)(푥) = 2푥 − 2 i. If 푓(푥) = 2푥 + 5 and (푓 ∘ 푔)(푥) = 2푥 − 2, then 푔(푥) is a function of the form 푔(푥) = 푎푥 + 푏푥 + 푐, where 푎, 푏 and 푐 are constants.

Now, (푓 ∘ 푔)(푥) = 2푥 − 2 ⟹ 2(푎푥 + 푏푥 + 푐) + 5 = 2푥 − 2 ⟹2푎푥 + 2푏푥 + 2푐 + 5 = 2푥 − 2 Comparing the 푅퐻푆 with 퐿퐻푆 gives 2푎 = 2 ⟹ 푎 = 1,2푏 = 0 ⟹ 푏 = 0 and 2푐 + 5 = −22푐 = −2 − 5푐 = −

∴ 푔(푥) = 푥 −

ii. (푔 ∘ 푓)(푥) = 푔 푓(푥) = (2푥 + 5) − = 4푥 + 20푥 + 25 − = 4푥 + 20푥 +

iii. (푓 ∘ 푔 ∘ ℎ)(푥) = 푓 푔 ℎ(푥) so let us first find 푔 ∘ ℎ(푥).

푔 ∘ ℎ(푥) = − = − =( )

=( )

=

116

b. ℎ(푥) ≡ 퐴 + ⟹ ≡ 퐴 + ≡ ( ) ⟹ 1 − 2푥 ≡ 퐴(3− 푥) + 퐵 1 − 2푥 ≡ 3퐴 − 퐴푥 + 퐵 ⟹ 1 − 2푥 ≡ 3퐴 + 퐵 − 퐴푥 Comparing the 푅퐻푆 with the 퐿퐻푆 gives –퐴 = −2 ⟹ 퐴 = 2 and 3퐴 + 퐵 = 1 ⟹ 3(2) + 퐵 = 1 ⟹ 퐵 = 1 − 6 = −5 ∴ ℎ(푥) ≡ 2 −

c. The equation (푓 ∘ 푔 ∘ ℎ)(푥) = 0 ⟹ = 0푥 + 34푥 − 61 = 0 Comparing 푥 + 34푥 − 61 = 0 with 푎푥 + 푏푥 + 푐 = 0 gives 푎 = 1,푏 = 34and 푐 = −61. The values of 푥 can be determined by the formula:

푥 =−푏± 푏2−4푎푐

2푎 ⟹ 푥 = ( )± ( ) ( )( )( )

= ±√ = √

∴ 푥 = √ = −17 + 5√14 or 푥 = √ = −17 − 5√14

Example 7.28 The functions 푝 and 푞 are defined on the set of 푟푒푎푙푛푢푚푏푒푟푠as 푝(푥) = ,푥 ≠ and

푞(푥) = 푥 ≠ 2. a. Find i. (푝 ∘ 푞) (푥) ii. (푞 ∘ 푝 )(푥) iii. The image of 3 under (푞 ∘ 푝 )(푥) b. Show that 푝 is one to one and onto. c. Solve the equation (푝 ∘ 푞) (푥) + (푞 ∘ 푝 )(푥) = 0

solution a. We have 푝(푥) = and 푞(푥) = .

i.To find (푝 ∘ 푞) (푥) first find (푝 ∘ 푞)(푥).

(푝 ∘ 푞)(푥) = 푝 푞(푥) =3 3푥−1

푥−2 −1

2 3푥−1푥−2 −7

=9푥−3−(푥−2)

푥−26푥−2−7(푥−2)

푥−2=

9푥−3−푥+2푥−2

6푥−2−7푥+14푥−2

= 8푥−112−푥

Now , for (푝 ∘ 푞) (푥), let (푝 ∘ 푞)(푥) = 푦 ⟹ = 푦 ⟹ 8푥 − 1 = 푦(12 − 푥) ⟹ 8푥 − 1 = 12푦 − 푦푥 ⟹ 8푥 + 푥푦 = 12푦 + 1 ⟹ 푥(8 + 푦) = 12푦+ 1 ⟹ 푥 = interchanging 푥 for 푦 gives 푦 = ∴ (푝 ∘ 푞) (푥) = ,푥 ≠ 8

ii. For 푞 (푥) let 푞(푥) = 푦 ⟹ = 푦 ⟹ 3푥 − 1 = 푦(푥 − 2)

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⟹ 3푥 − 1 = 푥푦 − 2푦 ⟹ 3푥 − 푥푦 = 1 − 2푦 ⟹ 푥(3 − 푦) = 1 − 2푦 ⟹ 푥 =

Interchanging 푥 for y gives 푞 (푥) =

For 푝 (푥) let 푝(푥) = 푦 ⟹ = 푦 ⟹ 3푥 − 1 = 푦(2푥 − 7) ⟹ 3푥 − 1 = 2푥푦 − 7푦 ⟹ 3푥 − 2푥푦 = 1 − 7푦 ⟹ 푥(3 − 2푦) = 1 − 7푦

⟹ 푥 = Interchanging 푥 for y gives 푝 (푥) =

Now ,

(푞 ∘ 푝 )(푥) = 푞 푝 (푥) =1−2 1−7푥

3−2푥3− 1−7푥

3−2푥=

(3−2푥)−2(1−7푥)3−2푥

3(3−2푥)−(1−7푥)3−2푥

=12푥+13−2푥푥+8

3−2푥= 12푥+1

푥+8

b. 푝 (푥) is one to one if 푝 (푚) = 푝 (푛) ⟹ 푚 = 푛 Now, 푝 (푚) = 푝 (푛) ⟹ = ⟹ (1 − 7푚)(3 − 2푛) = (1 − 7푛)(3 − 2푚) ⟹ 3 − 2푛 − 21푚 + 14푚푛 = 3 − 2푚− 21푛 + 14푚푛

⟹−21푚+ 2푚 = −21푛 + 2푛 + 3 − 3 + 14푚푛 − 14푚푛 ⟹ −19푚 = −19푛 ⟹ 푚 = 푛 Hence, 푝 is one to one.

c. The equation (푝 ∘ 푞) (푥) + (푞 ∘ 푝 )(푥) = 0 ⟹ 12푥+1

푥+8 + 12푥+1푥+8 = 0

⟹ 12푥 + 1 + 12푥 + 1 = 0 ⟹ 24푥 = −2 ∴ 푥 = − = − Try: 1. Two functions 푓 and 푔 are defined on the set of real numbers by, 푓: 푥 → ,푥 ≠ −4 and 푔(푥) → , 푥 ≠ 1 Find: i. 푔 , the inverse of 푔. ii. the image of −1 under 푓 ∘ 푔. (SSSCE) 2. Two functions 푓 and 푔 are defined on the set of real numbers by 푓: 푥 → 7푥 + 4, 푔(푥) = , 푥 ≠ 2. a. Find i. 푔 ∘ 푓 ii. 푔 b. i. Express 푔(푥) in the for 푔(푥) = 퐴 + , 푥 ≠ 2, where 퐴 and 퐵 are constants. ii. Hence find (SSSCE)

[퐹표푟푙푖푚푖푡푠푟푒푓푒푟푡표퐶푎푙푐푢푙푢푠] 3. Given that 푓:푥 → 푥 + 1 and 푔(푥) → , 푥 ≠ 1.

118

Find; i. 푔 ∘ 푓 ii. 푔 ∘ 푓(2) (SSSCE) 4. The functions 푔 and ℎ are defined on the set of real numbers by 푓: 푥 → 푥 and ℎ: 푥 → 푥 − 5

Find : i. 푔 ∘ ℎ(2) ii. ℎ ∘ 푔(푥) (SSSCE) 5. A function is defined on the set, 푅, of real numbers by 푓: 푥 → 5푥 − 9. Find the inverse of 푓 (SSSCE) 6. Two functions 푓 and 푔 are defined on the set of real numbers by 푓: 푥 → 2푥 + 3, 푔(푥) → (푥 − 2). Find i. 푔 ii. 푓 ∘ 푔 (SSSCE) 7. The functions 푓 and 푔 are defined by 푓: 푥 → √푥 − 1,and 푔: 푥 → ,푥 ≠ 0 a. State the largest possible domain of f. b. Determine whether or not g is one to one mapping. (SSSCE) 8. Two functions f and g are defined on the set R of real numbers by f: x → , x ≠ − and g: x → ,푥 ≠ −1 Find; a. 푔 , the inverse of 푔 b. 푔 ∘ 푓. (SSSCE ) 9. Two functions 푓 and 푔are defined by 푓: 푥 → 2푥 − 푥 + 3 and 푔(푥) = 3 − 푥. Find i. 푓 ∘ 푔 ii. the values of 푥 such that 푓 ∘ 푔(푥 + 1) = 푓 ∘ 푔(푥)(SSSCE) 10.Two functions 푓 and 푔 are defined on the set 푅 of real numbers by 푓(푥) → 푥 + 1 and 푔:푥 → 푥 − 2. Find a. 푓 ∘ 푔 b. the values of 푥 for which 푓 ∘ 푔(푥) = 5. (SSSCE) 11. Two functions 푓 and 푔 are defined on the set 푅, of real numbers by 푓: 푥 → 푥 − 1 and 푔(푥) = . Find a. 푔 (푥)b. 푓 ∘ 푔 (2) (SSSCE ) 12. Two functions 푓 and 푔 are defined by 푓(푥) = 푥 + 3 and 푔(푥) = √푥 − 3, where 푥 ∈ 푅 the set of real numbers. a. State the domain of 푔. b. Find 푓 ∘ 푔(푥) c. Determine whether or not 푓 is one to one. (WASSCE) 13. Two functions 푔 and ℎ are defined on the set 푅, of real numbers by 푔: 푥 → 푥 − 2 and ℎ: 푥 → , 푥 ≠ −2. Find

119

i. ℎ , the inverse of ; ii. 푔 ∘ ℎ(− ) (SSSCE )

14. a. Given that 푝(푥) = 2푥 and 푞(푥) = 푥 − 1, find the range of 푝 푞(푥) for the domain {−2,−1, 0, 1, 2}

b. Two functions 푓 and 푔 are defined on 푅,the set of real numbers by 푓: 푥 → 푥 + 3 and 푔: 푥 → 푥 − 2. Find i. 푔 푓 (3) , ii. the value of 푥 for which 푔 푓(푥) = 푓 푔(푥) . (WASSCE) 15. Two functions 푔 and ℎ are defined on the set 푅, of real numbers by 푔: 푥 → 푥 + 1 and ℎ: 푥 → , find a. ℎ , the inverse of ℎ.b. ℎ ∘ 푔 (WASSCE) 16. 푓 and 푔 are defined on the set푅, of real numbers by 푓: 푥 → 푥 + 1 and 푔(푥) → 5− 3푥. Find a. 푔 (푥) b. the values of 푥 such that 푔 (푥) = 푓(푥).(WASSCE)

Final Exercises 1. A function 푓 is defined on the domain {−4,−3,−1, 0, 1, 2, 3, 4} by 푓(푥) = 푥 + 2푥 − 1

i. Find the image of each element in the domain. ii. Show that 푓(푥) is not one to one.

2. A function is defined on the set of 푟푒푎푙 numbers as 푓(푥) = .

i. Find 푓(1),푓 , 푓(0),푓 − and 푓 −

ii. Evaluate 푓(√2), leaving your answer in surd form. iii. Show that 푓(푥) is onto.

3. State the domain of the following functions defined on the set of real numbers: i. 푓(푥) = ii. 푓(푥) = iii. 푓(푥) = iv. 푔(푥) =

v. 푔(푥) = vi. ℎ(푥) = 푥 − vii. ℎ(푥) = 푥 + 1 viii. ℎ(푥) = 푥 + 2푥 4. Find the range of the following functions defined on the set of real numbers:

i. 푓(푥) = 푥 − 1 ii. 푓(푥) = 6푥 − 4 iii. 푓(푥) = iv. 푔(푥) =

v. 푔(푥) = vi. 푔(푥) = 푥 vii. ℎ(푥) = viii. ℎ(푥) = 5. Find the inverses of the following functions defined on the set of real numbers:

120

i. 푓(푥) = ii. 푓(푥) = iii. 푔(푥) = iv. 푔(푥) =

v. ℎ(푥) = 푥 − 1 vi. ℎ(푥) = 1− 푥 vii. ℎ(푥) = 푥 + 3 viii. ℎ(푥) = 푥 − 5 6. If

i. 푓 ∘ 푔(푥) = 푥 and 푔(푥) = 푥 + 1, find 푔(푥). ii. 푔 ∘ 푓(푥) = 2푥 and 푔(푥) = 4푥 − 1, find 푓(푥).

iii. ℎ ∘ 푔(푥) = 푥 − 2푥 and ℎ(푥) = 푥 − 1, find 푔(푥). iv. 푝 ∘ 푞(푥) = 푥 − 8 and 푝(푥) = , find 푞(푥).

7. Given that 푓(푥) = 푎푥 + 푏, 푥 ∈ ℝ, find the values of 푎 and 푏 if i. 푓(−1) = 2 and 푓(−2) = 6. ii. 푓(4) = 2 and 푓(−3) = 4. iii. 푓(1) = 6 and 푓 = 1. iv. 푓 = −3 and 푓 − = −2.

8. The following functions are defined on the set of real numbers. Determine whether or not the functions are one to one and onto. i. 푓(푥) = 푥 − 3 ii. 푓(푥) = , 푥 ≠ 0 iii. 푔(푥) = iv. 푔(푥) =

v. ℎ(푥) = vi. ℎ(푥) = 푥 − 1

9. Given that 푓(푥) = 푥 − 3,푔(푥) = 푥 − 5푥 and ℎ(푥) = 6 − 푥, 푥 ∈ ℝ, find a. i. 푓 ∘ 푔(푥) ii. 푔 ∘ ℎ(푥) iii. 푔 ∘ ℎ ∘ 푓(푥) b. the values of 푥 for which 푔 ∘ ℎ ∘ 푓(푥) = 푓 ∘ 푔 ∘ ℎ.

10. Given that 푓(푥) = , 푥 ≠ − . i. Determine whether or not 푓(푥) is one to one. ii. Find 푓 , the inverse of 푓 iii. Find the values of 퐴 and 퐵 for which 푓(푥) ≡ 퐴 +

11. Two functions 푓 and 푔 are defined on the set of 푟푒푎푙푛푢푚푏푒푟푠 as 푓:푥 → ,푥 ≠ 5

and 푔: 푥 → ,푥 ≠ 6 i. Find the range of 푔 ii. Show that 푓(푥) is one to one. iii. Find 푓 ∘ 푔(푥) and 푔 ∘ 푓(푥) and hence solve the equation 푓 ∘ 푔(푥) + 푔 ∘ 푓(푥) = 1

12. Two functions 푔 and ℎ are defined on the set of real numbers by 푔: 푥 → 푥 + 1 and ℎ(푥) → ,푥 ≠ 0.

i. State the domain of 푓(푥). ii. Show that 푔(푥) is an even function. iii. Write ℎ(푥) in the form 퐴 + ,where 퐴 and 퐵 are constants. Iv. Find ℎ , the

121

inverse of ℎ. 13. Two functions 푓 and 푔 are defined on the set of real numbers by

푓(푥) = ,푥 ≠ 2 and 푔(푥) = , 푥 ≠ , where 푎,푏 and 푐are constants. i. Show that (푓 ∘ 푔) = 푔 ∘ 푓 (푥). ii. Find the values of 푎 and 푏 if 푓(−1) = 6 and 푓(−3) =

14. A function ℎ is defined on the set of real numbers as ℎ: 푥 → . i. State the domain ofℎ. ii. Find the range ofℎ. iii. Find ℎ , the inverse ofℎ. iv. Show that ℎ is one to one. 15. The functions 푓,푔and ℎdefined on the set of real numbers as: 푓: 푥 → 푥 − 3, 푔(푥) → and ℎ(푥) → a. Find i. (푓 ∘ 푔 ∘ ℎ)(푥) ii. (푔 ∘ ℎ ∘ 푔)(푥) iii. (푔 ∘ 푓) (푥) b. Solve the equation (푓 ∘ 푔 ∘ ℎ)(푥) = 1 − (푔 ∘ ℎ ∘ 푔)(푥) 16. The functions 푝 and 푞 are defined on the set of real numbers by

푝: 푥 → 3 and 푞: 푥 → ,푥 ≠ −2 a. Find 푞 ,the inverse of 푞 b. Solve the equation (푝 ∘ 푞)(푥) = 27. 17. The functions 푟 and 푠 are defined on the set ℝ,by 푟(푥) = 4푥 + 푘 and 푠(푥) = 3푥 − 1 i. Find 푠 , the inverse of푠. ii. Find the values of 푘 for which (푟 ∘ 푠)(푥) = (푠 ∘ 푟)(푥). 18. Two functions 푓 and 푔 are defined on the set ℝ,ofrealnumbersas 푓(푥) = 푥 + 2푥 + 1 and 푔(푥) = √푥 i. Show that (푔 ∘ 푓)(푥) = 푥 + 1 ii. Evaluate 푓 ∘ (푔 ∘ 푓) (푥). iii. Show that 푔(푥) is not onto but one to one. 19. The functions 푝,푞and 푟 are defined on the set of real numbers as

푝(푥) = , 푞(푥) = and 푟(푥) = 3푥 − 1 a. Simplify, 푝(푥) ∙ 푞(푥) b. Find i. (푝 ∘ 푞)(푥) ii. (푝 ∘ 푟)(푥) iii. (푞 ∘ 푟)(푥) c. Find the solution set of the equation (푝 ∘ 푟)(푥) + (푞 ∘ 푟)(푥) = 3. 20. The function 푓 are defined on the set of real numbers 푓(푥) = .

122

i. State the largest possible domain of 푓. ii. Show that 푓(푥) is one to one. iii. Find the range of 푓. 21. Two functions 푓 and 푔 are defined as 푓(푥) = 푥 − 1 and 푔(푥) = √푥 − 1,푥 ∈ ℝ. i. State the domain of 푓(푥) ii. Find (푔 ∘ 푓)(푥). iii. Find the values of 푥 for which(푔 ∘ 푓)(푥) = 푔(푥). iv. Find the values of 푥 for which (푓 ∘ 푔)(푥) = 푥 − 2 22. Given that (푔 ∘ 푓)(푥) = , 푥 ≠ 2 and 푔(푥) = 푥 − 1, a. Find 푓(푥) and hence state the domain of 푓. b. Show that 푔표푓(푥) is one to one. c. Evaluate (푔 ∘ 푓 )(푥) 23. Given that 푔(푥) = 푎푥 + 푏푥, 푥 ∈ ℝ, find a. The values of 푎 and 푏 if 푓(−1) = 6 and 푓(1) = −3 Another function 푓 is defined on the domain of 푔 by 푓(푥) = . b. Find 푔표푓(푥) and state the domain of (푔 ∘ 푓)(푥) c. Find the values of 푥 for which (푔 ∘ 푓)(푥). 24. Show that the function 푓(푥) = 푥 defined on the set of real numbers is one to one. Find the range and inverse of 푓. 25. Two functions 푓 and 푔 are defined on the as: 푓(푥) = log (5푥 + 3) and 푔(푥) = log (2푥 − 4), 푥 is a member of positive real numbers. i. State the domain and range of 푓. ii. Determine the values of 푥 for which 푓(푥) = 푔(푥). 26. Three functions are defined on the set of real numbers as: 푓: 푥 → 푥 − 2푥 + 3,푔: 푥 → 2푥 + 3푥 − 5 and ℎ: 푥 → 6 − 2푥 + 푥 . Evaluate 푓 ∙ (푔 − ℎ) b. Determine the element whose image under the mapping 푝: 푥 → , 푥 ≠ 2 is −5. c. Find 푝 (−4). (CCE) 27. Three functions are defined on real numbers by 푓(푥) = 3푥 + 1,푔(푥) = 2푥 + 3 and ℎ(푥) = 푥 − 2. Show that (a) (푓 ∘ 푔) ∘ ℎ = 푓 ∘(푔 ∘ ℎ) (b) (푓 + 푔) ∘ ℎ = (푓 ∘ ℎ) + (푔 ∘ ℎ) (CCE)

123

POLYNOMIAL FUNCTIONS A Polynomial in 푥 is any expression of the form 푓(푥) = 푎 푥 + 푎 푥 + 푎 푥 + ⋯+ 푎 , where 푎 is a constant term and 푛 is the highest degree of 푥. Therefore the equation푓(푥) = 푎 푥 + 푎 푥 + 푎 푥 + ⋯+ 푎 = 0 is a polynomial equation of degree 푛. For example 푓(푥) = 4푥 + 8푥 − 6푥 + 3 is a polynomial function of degree 3 with a constant term3.

124

A polynomial function with degree 1 is called a linear or monomial function. For example 푓(푥) = 3푥 + 4.

A polynomial function of degree 3 is termed quadratic function. For example 푝(푥) = 푥 + 6푥 − 9

A polynomial function of degree 3 is called cubic function. For example 푓(푥) = 푥 + 푥 + 4푥 + 16

A polynomial function of degree 4 is called 푞푢푎푟푡푖푐function. for instance푓(푥) =푥 + 1

A polynomial of degree 5 is called 푞푢푖푛푡푖푐 function. Addition of polynomials

Addition of polynomial functions is done by adding the corresponding powers of the 푢푛푘푛표푤푛.

Example 8.1 Add the following polynomial functions i. 푓(푥) = 푥 −2푥 − 3푥 and 푃(푥) = 푥 − 2푥 − 15 ii. ℎ(푥) = 푥 − 2푥 − 5푥, 푔(푥) = 3푥 − 푥 + 1 and 푘(푥) = 2푥 − 2푥 + 1

solution

i.푓(푥) + 푝(푥) =푥3−2푥2−3푥

+푥2−2푥−15푥3−푥2−5푥−15

ii. ℎ(푥) + 푔(푥) + 푘(푥) =푥3−2푥2−5푥

+3푥2−푥+1푥3+푥2−6푥+1+2푥2−2푥+1푥3+3푥2−8푥+2

Subtraction of polynomials Subtraction of polynomial functions is done by subtracting the corresponding powers of the 푢푛푘푛표푤푛.

Example 8.2 i. If푓(푥) = 6푥 + 3푥 + 2푥 + 10 and 푔(푥) = 2푥 + 4푥 + 8, find 푓(푥)− 푔(푥). ii. If 푔(푦) = 8푦 + 푦 − 3푦 − 9 and 푘(푦) = 6푦 − 6푦+ 4, find 푔(푦)− 푘(푦).

Solution

푓(푥)− 푔(푥) =6푥3+3푥2+2푥+10−2푥3+4푥+84푥3+3푥2−2푥−2

ii. 푔(푦)− 푘(푦) =8푦3+푦2−3푦−9

−6푦3−6푦+42푦3+푦2−9푥−13

Multiplication of polynomials

Given the polynomials 푓(푥) and 푔(푥), the product of the two functions is done by distributing on function over the other.

125

Note that after the multiplication process, we add the terms according to the powers of the unknown.

Example 8.3 i. If 푓(푥) = 푥 + 2 and 푔(푥) = 푥 + 푥 + 5,find 푓(푥) ∙ 푔(푥). ii. Find 푝(푥) ∙ 푞(푥) if 푃(푥) = 3푥 + 푥 and ℎ(푥) = 6푥 + 3.

Solution i. 푓(푥) ∙ 푔(푥) = (푥 + 2)(푥 + 푥 + 5) = 푥(푥 + 푥 + 5) + 2(푥 + 푥 + 5) = 푥 + 푥 + 5푥 + 2푥 + 2푥 + 10 = 푥 + 3푥 + 7푥 + 10 ii. 푝(푥) ∙ 푞(푥) = (3푥 + 푥)(6푥 + 3) = 3푥 (6푥 + 3) + 푥(6푥 + 3) = 18푥 + 9푥 + 6푥 + 3푥 = 18푥 + 15푥 + 3푥

Division of Polynomials This is done by the long division approach. If we write ( )

( ) and the degree of 푝(푥) is greater than that of 푓(푥), then푝(푥) can be

written as 푝(푥) = 푓(푥) ∙ 푞(푥) + 푟(푥)

푝(푥),푓(푥),푞(푥) and 푟(푥) are called 푑푖푣푖푑푒푛푑,푑푖푣푖푠표푟,푞푢표푡푖푒푛푡and 푟푒푚푎푖푛푑푒푟respectively. In the other way round, if we write ( )

( ) and the degree of 푓(푥) is greater or equal to that

of 푝(푥), then 푝(푥) can be written as

푝(푥) = 푞(푥) +푟(푥)푓(푥)

That is 푑푖푣푖푑푒푛푑 = 푞푢표푡푖푒푛푡 + If 푓(푥) exactly divides 푝(푥), then 푟(푥) = 0 The 푞푢표푡푖푒푛푡is found at the at top of the long division sign.

Exercise 8.4 Find the quotient and the remainder in each of the following i. (푥 + 푥 − 3푥 + 6) ÷ (푥 − 1) ii. (푥 − 5푥 + 2) ÷ (푥 + 1) iii. (2푥 − 4푥 + 3푥 − 1) ÷ (푥 + 1) iv. (4푥 − 9푥 + 1) ÷ (2푥 − 1)

Solution

126

i. We have 12

)(

63)1(

2

23

23

xx

xx

xxxx Therefore, the quotient is 푥 + 2푥 − 1

)22(

322

2

xx

xx

)1(6

xx

5 and the remainder is 5

ii. We have 777

)(

2005)1(

23

34

234

xxx

xx

xxxxx

)77(

0723

23

xx

xx

)77(

072

2

xx

xx

)77(27

xx

9 Therefore, the quotient is 푥 − 7푥 + 7푥 − 7 and the remainder is 9

iii. We have 962

)22(

1342)1(

2

23

23

xx

xx

xxxx Therefore, the quotient is 2푥 − 6푥 + 9

)66(

362

2

xx

xx

127

)99(19

xx

8 and the remainder is −3

iv. We have 42

)24(

1904)12(

2

23

23

xx

xx

xxxx Therefore, the quotient is 2푥 + 푥 − 4

)2(

922

2

xx

xx

)48(18

xx

3 and the remainder is −3 Try: Divide 푝(푥) = 2푥 − 7푥 + 5푥 − 9 by (푥 − 3)

The Remainder Theorem If a polynomial function or expression is divided by the monomial (푥 − 푎) or (푥 + 푎), then the remainder is the value of 푝(푎) or 푝(−푎) evaluated by putting 푥 = 푎 or 푥 = −푎 into the function or expression.

Example 8.5 Find the remainder when i. 푓(푥) = 푥 + 푥 − 6푥 + 4 is divided by (푥 − 2)

Solution We have 푓(푥) = 푥 + 푥 − 6푥 + 4 let 푥 − 2 = 0 ⟹ 푥 = 2 Now, 푓(2) = (2) + (2) − 6(2) + 4 = 8 + 4 − 12 + 4 = 4 Therefore, the remainder of 푓(푥) = 푥 + 푥 − 6푥 + 4when divided by (푥 − 2) is 4. ii. 푝(푥) = 2푥 − 푥 + 2푥 − 1 is divided by (푥 + 1)

Solution We have 푝(푥) = 2푥 − 푥 + 2푥 − 1 let 푥 + 1 = 0 ⟹ 푥 = −1 Now, 푝(−1) = 2(−1) − (−1) + 2(−1)− 1 = −2 − 1 − 2 − 1 = −6

128

Therefore, the remainder of 푝(푥) = 2푥 − 푥 + 2푥 − 1 when divided by (푥 + 1) is −6. iii. ℎ(푦) = 4푦 − 3푦 − 2푦 is divided by (푦 + 3)

Solution We have ℎ(푦) = 4푦 − 3푦 − 2푦 let 푦 + 3 = 0 ⟹ 푦 = −3 Now, ℎ(−3) = 4(−3) − 3(−3) − 2(−3) = 324 + 81 + 6 = 411 Therefore, the remainder of ℎ(푦) = 4푦 − 3푦 − 2푦 when divided by (푦 + 3) is 411. iv. 푓(푥) = 푥 + 9푥 + 26푥 + 24 is divided by (푥 − 4)

Solution We have 푓(푥) = 푥 + 9푥 + 26푥 + 24 let 푥 − 4 = 0 ⟹ 푥 = 4 Now, 푓(4) = (4) + 9(2) + 26(2) + 24 = 64 + 36 + 56 + 24 = 180 Therefore, the remainder of 푓(푥) = 푥 + 9푥 + 26푥 + 24 when divided by (푥 + 4) is 180.

The Factor Theorem Let 푝(푥) be a polynomial and let 푓(푥) = (푥 ± 푎).If 푓(푥) exactly divides 푝(푥),such that 푝(푥) leaves a remainder of 0 when divide by 푓(푥), then 푓(푥) = (푥 ± 푎) is a factor of the polynomial.

Example 8.6 Factorise the following i. 푥 + 2푥 − 푥 − 2 ii. 푥 − 푥 − 푥 − 2 iii. 푥 − 1 iv. 2푥 − 푥 + 2푥 − 1

Solution i. Let 푓(푥) = 푥 + 2푥 − 푥 − 2 consider the factors of −2: {±1, ±2} Let’s find the remainders of the factors. Now, 푓(1) = (1) + 2(1) − 1 − 2 = 1 + 2 − 1 − 2 = 0 Therefore (푥 − 1) is the factor of the polynomial. We can write: 푓(푥) = (푥 + 1)Quotient+remainder So we carry on the division of 푓(푥) by (푥 + 1) as follows:

23

)(

22)1(

2

23

23

xx

xx

xxxx Therefore, the quotient is 푥 + 3푥 + 2

129

)33(

32

2

xx

xx

)22(22

xx

0 and the remainder is 0 Therefore, 푓(푥) = (푥 − 1)(푥 + 3푥 + 2) = (푥 − 1)(푥 + 푥 + 2푥 + 2) = (푥 − 1)(푥 + 1)(푥 + 2)

Solution ii. Let 푓(푥) = 푥 − 푥 − 푥 − 2 Consider the factors of −2: {±1, ±2} Let us find the remainders of the factors. Now, 푓(1) = (1) − (1) − 1 − 2 = 1 − 1 − 1 − 2 ≠ 0 푓(−1) = (−1) − (−1) − 1 − 2 = −1 − 1 − 1 − 2 ≠ 0 푓(2) = (2) − (2) − 2 − 2 = 8 − 4 − 2 − 2 = 0 Therefore (푥 − 2) is the factor of the polynomial. We can write: 푓(푥) = (푥 − 2)Quotient+remainder So we carry on the division of 푓(푥) by (푥 − 2) as follows:

1

)2(

2)2(

2

23

23

xx

xx

xxxx Therefore, the quotient is 푥 + 푥 + 1

)2( 2

2

xx

xx

)2(2

xx

0 and the remainder is 0 Therefore, 푓(푥) = (푥 − 2)(푥 + 푥 + 1)

Solution iii. Let 푓(푥) = 푥 − 1 Consider the factors of −1: {−1, 1} Let’s find the remainders of the factors.

130

Now, 푓(−1) = (−1) − 1 = −1 − 1 ≠ 0 푓(1) = (1) − 1 = 1 − 1 = 0 Therefore (푥 − 1) is the factor of the polynomial. We can write: 푓(푥) = (푥 − 1)Quotient+remainder So we carry on the division of 푓(푥) by (푥 − 1) as follows:

1

)(

100)1(

2

23

23

xx

xx

xxxx Therefore, the quotient is 푥 + 1

)(

02

2

xx

xx

)1(1

xx

0 and the remainder is 0 Therefore, 푓(푥) = (푥 − 1)(푥 + 푥 + 1)

Solution iv. Let 푓(푥) = 2푥 − 푥 + 2푥 − 1 Consider the factors of −1: {−1, 1} Let’s find the remainders of the factors. Now, 푓(−1) = 2(−1) − (−1) + 2(−1)− 1 = −2 − 1 − 2 − 1 ≠ 0 푓(1) = 2(1) − (1) + 2(1)− 1 = 2 − 1 + 2 − 1 ≠ 0

Let’s try 푓 = 2 − + 2 − 1 = − + 1 − 1 = 0

Therefore (2푥 − 1) is the factor of the polynomial. We can write: 푓(푥) = (2푥 − 1)Quotient+remainder So we carry on the division of 푓(푥) by (2푥 − 1) as follows:

10

)2(

122)12(

2

23

23

xx

xx

xxxx Therefore, the quotient is 푥 + 푥 + 1

)00(

202

2

xx

xx

131

)12(12

xx

0 and the remainder is 0 Therefore, 푓(푥) = (2푥 − 1)(푥 + 1) Try: Factorise 푥 + 1

Curve Sketching I To sketch the graph of a polynomial function 푓(푥) 1. find the zeros of 푝(푥) or the 푥 −intercept(s) 2. find the 푦 −intercept(s) and the make appropriate sketch. For instance, if 푓(푥) = 푎푥 + 푏푥 + 푐, 푎 > 0, we have a minimum curve below. 푦 푓(푥) = 푎푥 + 푏푥 + 푐, 푎 > 0 푥

Also, if 푓(푥) = 푎푥 + 푏푥 + 푐, 푎 < 0, we have a maximum curve below. 푦 푓(푥) = 푎푥 + 푏푥 + 푐, 푎 < 0 푥

132

Example 8.7

Sketch the following i. 푓(푥) = 6 − 푥 − 푥 ii. 푔(푥) = 푥 + 5푥 + 6

Solution i. We have 푓(푥) = 6 − 푥 − 푥 For the zeros of 푓(푥), let 푓(푥) = 0 ⟹ 6 − 푥 − 푥 = 0 or 푥 + 푥 − 6 = 0 or 푥 − 2푥 + 3푥 − 6 = 0 6 or 푥(푥 − 2) + 3(푥 − 2) = 0 푓(푥) = 6 − 푥 − 푥 or (푥 − 2)(푥 + 3) = 0 ⟹ 푥 = 2 or 푥 = −3 -4 -2 2 4 For the 푓(푥)−intercept, let 푥 = 0. 푓(푥) = 6 − 0 − (0) = 6 -6

Solution ii. We have 푔(푥) = 푥 + 5푥 + 6 For the zeros of 푔(푥), let 푔(푥) = 0 ⟹ 푥 + 5푥 + 6 = 0 or 푥 + 2푥 + 3푥 + 6 = 0 6 or 푥(푥 + 2) + 3(푥 + 2) = 0 푓(푥) = 6 − 푥 − 푥 or (푥 + 2)(푥 + 3) = 0 ⟹푥 = −2 or 푥 = −3 -3 -2 2 For the 푔(푥)−intercept, let 푥 = 0. 푔(푥) = (0) + 5(0) + 6 = 6 -6 Try Sketch the following functions i. 푝(푥) = 푥 − 9 ii. 푓(푥) = 5 − 2푥 − 3푥

Example 8.8 The polynomial equation 푥 +3푥 −7푥 − 15푥 + 18 = 0 has a double root at 푥 = −3. Find all the possible zeros of the polynomial.

133

Solution Let 푓(푥) = 푥 +3푥 −7푥 − 15푥 + 18.If 푥 = −3 is a double root, then (푥 + 3)(푥 + 3) =푥 + 6푥 + 9 is a factor of the polynomial 푓(푥) = 푥 +3푥 −7푥 − 15푥 + 18. So we can write 푓(푥) = (푥 + 6푥 + 9)푄푢표푡푖푒푛푡 + 푟푒푚푎푖푛푑푒푟 Since 푥 + 6푥 + 9 is a factor of the polynomial we expect the remainder to be zero when we divide 푓(푥) by 푥 + 6푥 + 9 .

2

2 4 3 2

4 3 2

3 2( 6 9) 3 7 15 18

( 6 9 )

x xx x x x x x

x x x

3 2

3 2

3 16 15

( 3 18 27 )

x x x

x x x

2

2

2 12 18

(2 12 18)0

x x

x x

So, we can write 푓(푥) = (푥 + 6푥 + 9)(푥 − 3푥 + 2) For the zeroes of 푓(푥), we set 푓(푥) = 0. ⟹ (푥 + 6푥 + 9)(푥 − 3푥 + 2) = 0 ⟹ (푥 + 3) (푥 − 2푥 − 푥 + 2) = 0 (푥 + 3) (푥 − 2)(푥 − 1) = 0 ⟹ (푥 + 3) = 0 or 푥 − 2 = 0 or 푥 − 1 = 0 ∴, 푥 = −3 (double root) or 푥 = 2 or 푥 = 1

Example 8.9 When the polynomial 푓(푥) = 2푥 + 푝푥 + 푞푥 − 6 where 푝 and 푞 are constants is divided by (푥 + 1) the remainder is −4. When divided by (푥 − 1) the remainder is 8. a. Find the values of 푝 and 푞 b. Show that 푓(−3) = 0 and hence write 푓(푥) as a product of three linear factors.

Solution We have, 푓(푥) = 2푥 + 푝푥 + 푞푥 − 6 . let 푥 + 1 = 0 ⟹ 푥 = −1. and let 푥 − 1 = 0 ⟹ 푥 = 1. a. We want to find the values of 푝 and 푞 such that 푓(−1) = −4 and 푓(1) = 8. 푓(−1) = 2(−1) + 푝(−1) + 푞(−1) − 6 = −2 + 푝 − 푞 − 6

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⟹ −2 + 푝 − 푞 − 6 = −4 ⟹ 푝 − 푞 = −4 + 2 + 6 ⟹ 푝 − 푞 = 4 … … …(1) 푓(1) = 2(1) + 푝(1) + 푞(1)− 6 = 2 + 푝 + 푞 − 6 ⟹ 2 + 푝 + 푞 − 6 = 8 ⟹ 푝 + 푞 = 8 + 6 − 2 ⟹ 푝 + 푞 = 12 … … …(2) Add (1) to (2) ⟹ 푝 + 푝 + 푞 − 푞 = 12 + 4 ⟹ 2푝 = 16 ∴ 푝 = 8 Put 푝 = 8 into (2) ⟹ 8 + 푞 = 12 ∴ 푞 = 12 − 8 = 4. b. Now, 푓(푥) = 2푥 + 8푥 + 4푥 − 6 푓(−3) = 2(−3) + 8(−3) + 4(−3) − 6 = −54 + 72 − 12 − 6 = 0 ∎

Example 8.10 If 푝(푥) = 푥 + 푎푥 + 푏. When 푝(푥) is divided by (푥 − 2) the remainder is 8. When 푝(푥) is divided by (푥 + 3) the remainder is 18. Find the values of 푎 and 푏.

Solution We have, 푝(푥) = 푥 + 푎푥 + 푏. let 푥 − 2 = 0 ⟹ 푥 = 2. and let 푥 + 3 = 0 ⟹ 푥 = −3. a. We want to find the values of 푎 and 푏 such that 푓(2) = 8 and 푓(−3) = 18. 푓(2) = (2) + 푎(2) + 푏 = 4 + 2푎 + 푏 ⟹ 4 + 2푎 + 푏 = 8 ⟹ 2푎 + 푏 = 8 − 4 ⟹ 2푎 + 푏 = 4 … … …(1) 푓(−3) = (−3) + 푎(−3) + 푏 = 9 − 3푎 + 푏 ⟹ 9 − 3푎 + 푏 = 18 ⟹ −3푎 + 푏 = 18 − 9 ⟹ −3푎 + 푏 = 9 … … …(2) Subtract (1) from (2) ⟹ −3푎 − 2푎 + 푏 − 푏 = 9 − 4 ⟹ −5푎 = 5 ∴, 푎 = −1

Put 푎 = −1 into (1) ⟹ 2(−1) + 푏 = 4 ∴, 푏 = 4 + 2 = 6. 푝(푥) = 푥 − 푥 + 6

Example 8.11

When the polynomial 푓(푥) = 푎푥 + 푏푥 + 푐 is divided by (푥 + 1) and (푥 − 2) they leave remainders are −2 and 4 respectively. If 푥 + 2 is a factor of the polynomial, a. find the values of 푎,푏 and 푐. b. factorise the polynomial completely.

Solution We have, 푓(푥) = 푎푥 + 푏푥 + 푐 let: 푥 + 1 = 0 ⟹ 푥 = −1, 푥 − 2 = 0 ⟹ 푥 = 2 푥 + 2 = 0 ⟹ 푥 = −2. a. We want to find 푎,푏 and 푐 such that 푓(−1) = −2, 푓(2) = 4 and 푓(−2) = 0 푓(−1) = 푎(−1) + 푏(−1) + 푐 = 푎 − 푏 + 푐 ⟹ 푎 − 푏 + 푐 = −2 … … … (1)

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푓(2) = 푎(2) + 푏(2) + 푐 = 4푎 + 2푏 + 푐 ⟹ 4푎+ 2푏 + 푐 = 4 … … … (2) 푓(−2) = 푎(−2) + 푏(−2) + 푐 = 4푎 − 2푏 + 푐 ⟹ 4푎 − 2푏 + 푐 = 0 … … … (3)

Now, solve the system: 푎 − 푏 + 푐 = −2 … … … (1)4푎 + 2푏 + 푐 = 4 … … … (2)4푎 − 2푏 + 푐 = 0 … … … (3)

Subtract (3) from (2) ⟹ 4푎 − 4푎 + 2푏 + 2푏 = 4 − 0 ⟹ 4푏 = 4 ∴, 푏 = 1 Subtract (1) from (2) ⟹ 4푎 − 푎 + 2푏 + 푏 + 푐 − 푐 = 4 + 2 ⟹ 3푎 + 3푏 = 6 (Divide both sides by 3) ⟹ 푎 + 푏 = 2 ∴, 푎 = 2 − 푏 = 2 − 1 = 1 Since 푏 = 1. Put 푎 = 1 and 푏 = 1into (1) ⟹ 1− 1 + 푐 = −2 ∴, 푐 = −2 Hence, 푓(푥) = 푥 + 푥 − 2 b. Let (푥 + 푘) be the second factor of 푓(푥). So we can write 푥 + 푥 − 2 ≡ (푥 + 2)(푥 + 푘) ⟹ 푥 + 푥 − 2 ≡ 푥 + 푥푘 + 2푥 + 2푘 ⟹ 푥 + 푥 − 2 ≡ 푥 + (푘 + 2)푥 + 2푘 By comparing the LHS with the RHS we get −2 = 2푘 ∴,푘 = −1 ∴, 푓(푥) = (푥 + 2)(푥 − 1)

Example 8.12 When 푓(푥) = 푥 +푎푥 − 3푥 + 4 is divided by (푥 − 3) the remainder obtained is twice the remainder when the polynomial is divided by (푥 − 2). Find the value of 푎.

Solution We have, 푓(푥) = 푥 + 푎푥 − 3푥 + 4 let: 푥 − 3 = 0 ⟹ 푥 = 3, 푥 − 2 = 0 ⟹ 푥 = 2 We want to find 푎 such that 푓(3) = 2푓(2) 푓(3) = (3) + 푎(3) − 3(3) + 4 = 27 + 9푎 − 9 + 4 = 22 + 9푎 푓(2) = (2) + 푎(2) − 3(2) + 4 = 8 + 4푎 − 6 + 4 = 6 + 4푎 Now, 푓(3) = 2푓(2) ⟹ 22 + 9푎 = 2(6 + 4푎) ⟹ 22 + 9푎 = 12 + 8푎 ⟹ 9푎 − 8푎 = 12 − 22 ∴, 푎 = −10

Example 8.13

(푥 − 2) is a factor of the polynomial 푓(푥) = 푥 + 푎푥 + 푏 where 푎 and 푏 are constants. The remainder when 푓(푥) is divided by (푥 + 3) is 25. Find a. the values of 푎 and 푏. b. the remainder when 푓(푥) is divided by (푥 − 3)

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c. the truth set of 푓(푥) = 0 Solution

We have, 푓(푥) = 푥 + 푎푥 + 푏 let: 푥 − 2 = 0 ⟹ 푥 = 2, 푥 + 3 = 0 ⟹ 푥 = −3 a. We want to find 푎 and 푏 such that 푓(2) = 0 and 푓(−3) = 25 푓(2) = (2) + 푎(2) + 푏 = 8 + 2푎 + 푏 ⟹ 8 + 2푎 + 푏 = 0 ⟹ 2푎 + 푏 = −8 … … … (1) 푓(−3) = (−3) + 푎(−3) + 푏 = −27 − 3푎 + 푏 ⟹ −27 − 3푎 + 푏 = 25 ⟹ −3푎 + 푏 = 25 + 27 ⟹ −3푎 + 푏 = 52 … … … (2) Subtract (2) from (1) ⟹ 2푎 + 3푎 + 푏 − 푏 = −8 − 52 ⟹ 5푎 = −60

∴, 푎 = −12 Put 푎 = −12 into (1) ⟹ 2(−12) + 푏 = −8 ⟹ −24 + 푏 = −8 ∴,푏 = −8 + 24 = 16 Hence, 푓(푥) = 푥 − 12푥 + 16 b. Let 푥 − 3 = 0 ⟹ 푥 = 3 푓(3) = (3) − 12(3) + 16 = 27 − 36 + 16 = 7 ∴,the remainder when the polynomial is divided by (푥 − 3) is 7 c. let us first factorise the polynomial. Let (푥 − 푘 ) and (푥 − 푘 ) be the other factors of the polynomial. So we can write: 푥 − 12푥 + 16 ≡ (푥 − 2)(푥 − 푘 )(푥 − 푘 ) ≡ (푥 − 2)(푥 − 푘 푥 − 푘 푥 + 푘 푘 ) ≡ 푥 − (푘 + 푘 )푥 + 푘 푘 푥 − 2푥 + 2(푘 + 푘 )푥 − 2푘 푘 ≡ 푥 − (푘 + 푘 + 2)푥 + (푘 푘 + 2푘 + 2푘 )푥 − 2푘 푘 By comparing the LHS with the RHS, we get 푘 + 푘 + 2 = 0 … … … (1) 푘 푘 + 2푘 + 2푘 = −12 … … … (2) −2푘 푘 = 16 … … … (3) From (1) 푘 = −2 − 푘 Put 푘 = −2 − 푘 into (3) ⟹ −2푘 (−2 − 푘 ) = 16(Divide through by −2) ⟹ 푘 (−2 − 푘 ) = −8 ⟹ −2푘 − 푘 = −8 ⟹ 푘 + 2푘 − 8 = 0 ⟹ 푘 − 2푘 + 4푘 − 8 = 0 ⟹ (푘 − 2)(푘 + 4) = 0 ⟹ 푘 − 2 = 0 or 푘 + 4 = 0 ⟹ 푘 = 2 or 푘 = −4 Put the values of 푘 into (1) When 푘 = 2, 푘 = −2 − 2 = −4 When 푘 = −4, 푘 = −2 + 4 = 2 Hence, 푓(푥) = 푥 − 12푥 + 16 = (푥 − 2)(푥 − 2)(푥 + 4) 푓(푥) = 0 ⟹ (푥 − 2)(푥 − 2)(푥 + 4) = 0

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∴, 푥 = 2 (a double root) or 푥 = −4 {푥: 푥 = −4or푥 = 2}

Example 8.14 Given that 푔(푥) = 푥 + 푘푥 − 2푥 + 1 is divided by 푥 − 푘 the remainder is 푘,find the possible values of 푘.

Solution We have, 푓(푥) = 푥 + 푘푥 − 2푥 + 1 let: 푥 − 푘 = 0 ⟹ 푥 = 푘, We want to find 푘 such that 푓(푘) = 푘 푓(푘) = (푘) + 푘(푘) − 2(푘) + 1 = 푘 + 푘 − 2푘 + 1 ⟹ 2푘 − 2푘 + 1 = 푘 ⟹ 2푘 − 3푘 + 1 = 0 Let 푘 − 1 = 0 ⟹ 푘 = 1 Put 푘 = 1 into the polynomial 2푘 − 3푘 + 1 ⟹ 2(1) − 3(1) + 1 = 2 − 3 + 1 = 0 Therefore, (푘 − 1) is a factor of the polynomial 2푘 − 3푘 + 1 Let (푘 − 푎) and (푘 − 푏) be the other factors of the polynomial. So we can write: 2푘 − 3푘 + 1 ≡ 2(푘 − 1)(푘 − 푎)(푘 − 푏) ≡ 2(푘 − 1)(푘 − 푎푘 − 푏푘 + 푎푏) ≡ 2[푘 − (푎 + 푏)푘 + 푎푏푘 − 푘 + (푎 + 푏)푘 − 푎푏] ≡ 2푘 − 2(푎 + 푏 + 1)푘 + 2(푎푏 + 푎 + 푏)푘 − 2푎푏 By comparing the LHS with the RHS, we get 2(푎 + 푏 + 1) = 0 (Divide both sides by 2) ⟹ 푎 + 푏 + 1 = 0 … … … (1) 2(푎푏 + 푎 + 푏) = −3 … … … (2) −2푎푏 = 1 … … … (3) From (1) 푏 = −1− 푎 Put 푏 = −1 − 푎 into (3) ⟹ −2푎(−1 − 푎) = 1 ⟹ 2푎 + 2푎 = 1 ⟹ 2푎 + 2푎 − 1 = 0

⟹ 2푎 + 2푎 − 1 = 0 ⟹ 푎 = ± ( )( )( )

= ±√ = √

⟹ 푎 = √ = − − √ or 푎 = √ = − + √ Put the values of 푎 into (1)

⟹ 푏 = −1 − − − √ = − + √ or 푏 = −1 − − + √ = − − √

Hence, 2푘 − 3푘 + 1 ≡ 2(푘 − 1) 푘 − − − √ 푘 − − + √

2푘 − 3푘 + 1 = 0 ⟹ 2(푘 − 1) 푘 − − − √ 푘 − − + √ = 0

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(Divide both sides by 2) ⟹ (푘 − 1) 푘 − − − √ 푘 − − + √ = 0

Hence, 푘 − 1 = 0 or 푘 − − − √ = 0 or 푘 − − + √ = 0

∴, 푘 = 1 or 푘 = − − √ or 푘 = − + √

Example 8.15 푥 − 4 is a factor of the polynomial 푓(푥) = 푎푥 + 푏푥 + 푐푥 + 2. when the polynomial is divided by (푥 + 1) the remainder is 6 i. Find the values of 푎,푏 and 푐. ii. With these values factorise the polynomial completely. iii. Hence, solve the equation 푎푥 + 푏푥 + 푐푥 + 2 = 0

Solution Check We have, 푓(푥) = 푎푥 + 푏푥 + 푐푥 + 2 let: 푥 − 4 = 0 ⟹ 푥 = 4 ⟹ 푥 = ±2, 푥 + 1 = 0 ⟹ 푥 = −1 i. We want to find 푎,푏 and 푐 such that 푓(−2) = 0, 푓(2) = 0 and 푓(−1) = 6 푓(−2) = 푎(−2) + 푏(−2) + 푐(−2) + 2 = −8푎 + 4푏 − 2푐 + 2 ⟹ −8푎 + 4푏 − 2푐 + 2 = 0 ⟹ −8푎 + 4푏 − 2푐 = −2 (Divide through by 2) ⟹ −4푎 + 2푏 − 푐 = −1 … … … (1) 푓(2) = 푎(2) + 푏(2) + 푐(2) + 2 = 8푎 + 4푏 + 2푐 + 2 ⟹ 8푎 + 4푏 + 2푐 + 2 = 0 ⟹ 8푎 + 4푏 + 2푐 = −2 (Divide through by 2) ⟹ 4푎 + 2푏 + 푐 = −1 … … … (2) 푓(−1) = 푎(−1) + 푏(−1) + 푐(−1) + 2 = −푎 + 푏 − 푐 + 2 ⟹ −푎 + 푏 − 푐 + 2 = 6 ⟹ −푎 + 푏 − 푐 = 4 … … … (3)

Now, solve the system: −4푎 + 2푏 − 푐 = −1 … … … (1)4푎 + 2푏 + 푐 = −1 … … … (2)−푎 + 푏 − 푐 = 4 … … … (3)

Add (1) to (2) ⟹ 4푎 − 4푎 + 2푏 + 2푏 + 푐 − 푐 = −1 − 1 ⟹ 4푏 = −2 ∴, 푏 = − Add (2) to (3) ⟹ 4푎 − 푎 + 2푏 + 푏 + 푐 − 푐 = −1 + 4 ⟹ 3푎 + 3푏 = 3 (Divide both sides by 3) ⟹ 푎 + 푏 = 1 ∴,푎 = 1 − 푏 = 1 − − = Since 푏 = −

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Put 푎 = and 푏 = − into (3) ⟹ − + + 푐 = 4 ⟹ −1 + 푐 = 4 ∴, 푐 = 4 + 1 = 5 Hence, 푓(푥) = 푥 − 푥 + 5푥 + 2 ii. Let (푥 + 푘) be the third factor of 푓(푥). So we can write 푥 − 푥 + 5푥 + 2 ≡ (푥 − 4)(푥 + 푘)

≡ (푥 + 푘푥 − 4푥 − 4푘)

≡ 푥 + 푘푥 − 6푥 − 6푘

By comparing the LHS with the RHS we get 2 = −6푘 ∴,푘 = −

∴, 푓(푥) = 푥 − 푥 + 5푥 + 2 = (푥 − 4) 푥 −

iii. 푓(푥) = 0 ⟹ (푥 − 4) 푥 − = 0 Dividethroughby

⟹ (푥 − 4) 푥 − ∴, 푥 = ±2 or 푥 = −

Example 8.16

Given that the polynomial 푓(푥) = 6 − 푥 − 푥 is a factor of the polynomial 푔(푥) = 푝푥 + 5푥 + 푞푥 − 18, where 푝 and 푞 are constants, find i. the values of 푝 and 푞 ii. the remainder of 푔(푥) when it is divided by (푥 + 2) iii. the values of 푥 for which 푔(푥) = 푓(푥)

Solution We have, 푔(푥) = 푝푥 + 5푥 + 푞푥 − 18 let: 6 − 푥 − 푥 = 0 ⟹ 6− 3푥 + 2푥 − 푥 = 0 ⟹ 3(2 − 푥) + 푥(2− 푥) = 0 ⟹ (3 + 푥)(2 − 푥) = 0 ⟹ 푥 = −3 or 푥 = 2 i. We want to find 푝 and 푞 such that 푔(−3) = 0 and 푔(2) = 0 푔(−3) = 푝(−3) + 5(−3) + 푞(−3) − 18 = −27푝 + 45 − 3푞 − 18 ⟹ −27푝+ 45 − 3푞 − 18 = 0 ⟹ −27푝 − 3푞 + 27 = 0 (Divide both sides by −3 ) ⟹ 9푝 + 푞 = 9 … … … (1) 푔(2) = 푝(2) + 5(2) + 푞(2) − 18 = 8푝 + 20 + 2푞 − 18 ⟹ 8푝 + 20 + 2푞 − 18 = 0 ⟹ 8푝 + 2푞 + 2 = 0 (Divide both sides by 2) ⟹ 4푝+ 푞 = −1 … … … (2)

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Subtract (2) from (1) ⟹ 9푝 − 4푝 + 푞 − 푞 = 9 + 1 ⟹ 5푝 = 10 ∴, 푝 = 2 Put 푝 = 2 into (1) ⟹ 9(2) + 푞 = 9 ⟹ 18 + 푞 = 9 ∴,푞 = 9 − 18 = −9 Hence, 푔(푥) = 2푥 + 5푥 − 9푥 − 18 ii. Let 푥 + 2 = 0 ⟹ 푥 = −2 푔(−2) = 2(−2) + 5(−2) − 9(−2)− 18 = −16 + 20 + 18 − 18 = 4 ∴,the remainder when the polynomial is divided by (푥 + 2) is 4. iii. 푔(푥) = 푓(푥) ⟹ 2푥 + 5푥 − 9푥 − 18 = 6 − 푥 − 푥 ⟹ 2푥 + 5푥 + 푥 − 9푥 + 푥 − 18 − 6 = 0 ⟹ 2푥 + 6푥 − 8푥 − 24 = 0 (Divide through by 2) ⟹ 푥 + 3푥 − 4푥 − 12 = 0 let 푝(푥) = 푥 + 3푥 − 4푥 − 12 Let푥 + 2 = 0 ⟹ 푥 = −2 푝(−2) = (−2) + 3(−2) − 4(−2) − 12 = −8 + 12 + 8 − 12 = 0 Hence, (푥 + 2) is a factor of the polynomial 푝(푥) = 푥 + 3푥 − 4푥 − 12 Let (푥 − 푘 ) and (푥 − 푘 ) be the other factors of the polynomial. So, we can write 푥 + 3푥 − 4푥 − 12 ≡ (푥 + 2)(푥 − 푘 )(푥 − 푘 ) ≡ (푥 + 2)(푥 − 푘 푥 − 푘 푥 + 푘 푘 ) ≡ 푥 − (푘 + 푘 )푥 + 푘 푘 푥 + 2푥 − 2(푘 + 푘 )푥 + 2푘 푘 ≡ 푥 − (푘 + 푘 − 2)푥 + (푘 푘 − 2푘 − 2푘 )푥 + 2푘 푘 By comparing the LHS with the RHS, we get 푘 + 푘 − 2 = 3 … … … (1) 푘 푘 − 2푘 − 2푘 = −4 … … … (2) 2푘 푘 = −12 … … … (3) From (1) 푘 = 3 + 2 − 푘 = 5 − 푘 Put 푘 = 5 − 푘 into (3) ⟹ 2푘 (5 − 푘 ) = −12 (Divide through by 2) ⟹ 푘 (5− 푘 ) = −6 ⟹ 5푘 − 푘 = −6 ⟹ 푘 − 5푘 − 6 = 0 ⟹ 푘 − 6푘 + 푘 − 6 = 0 ⟹ (푘 − 6)(푘 + 1) = 0 ⟹ 푘 − 6 = 0 or 푘 + 1 = 0 ⟹ 푘 = 6 or 푘 = −1 Put the values of 푘 into (1) When 푘 = 6, 푘 = 5 − 6 = −1 When 푘 = −1, 푘 = 5 + 1 = 6 Hence, 푝(푥) = 푥 + 3푥 − 4푥 − 12 = (푥 + 2)(푥 − 6)(푥 + 1) 푝(푥) = 0 ⟹ (푥 + 2)(푥 − 6)(푥 + 1) = 0 ∴, 푥 = −2,−1 or 6 Therefore the values of 푥 for which 푔(푥) = 푓(푥) are −2,−1 and 6.

Example 8.17 If 푓(푥) = 6푥 +13푥 + 2푥 − 5,

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a. show that 푓(−1) = 0 b. find the factors of 푓(푥) Solution

We have, 푓(푥) = 6푥 + 13푥 + 2푥 − 5 i. 푓(−1) = 6(−1) + 13(−1) + 2(−1) − 5 = −6 + 13− 2 − 5 = 0 Since 푓(−1) = 0, (푥 + 1) is a factor of the polynomial Let (푥 − 푘 ) and (푥 − 푘 ) be the other factors of the polynomial. So, we can write 6푥 + 13푥 + 2푥 − 5 ≡ 6(푥 + 1)(푥 − 푘 )(푥 − 푘 ) ≡ 6(푥 + 1)(푥 − 푘 푥 − 푘 푥 + 푘 푘 ) ≡ 6[푥 − (푘 + 푘 )푥 + 푘 푘 푥 + 푥 − (푘 + 푘 )푥 + 푘 푘 ] ≡ 6푥 − 6(푘 + 푘 − 1)푥 + 6(푘 푘 − 푘 − 푘 )푥 + 6푘 푘 By comparing the LHS with the RHS, we get −6(푘 + 푘 − 1) = 13 … … … (1) 6(푘 푘 − 푘 − 푘 ) = 2 … … … (2) 6푘 푘 = −5 … … … (3) From (1) −6푘 − 6푘 + 6 = 13 ⟹ −6푘 = 13 − 6 + 6푘 ⟹ 푘 = −

Put 푘 = − into (3)

⟹ −6푘 = 5 ⟹ −푘 (6푘 + 7) = −5 ⟹ −6푘 − 7푘 = −5

⟹ 6푘 + 7푘 − 5 = 0 ⟹ 푘 = ± ( ) ( )( )( )

= √ = ±√ = ±

⟹ 푘 = = − = − or 푘 = = Put the values of 푘 into (1)

When 푘 =, 푘 = − = When 푘 = , 푘 = − = −

So, 6푥 + 13푥 + 2푥 − 5 = 6(푥 + 1) 푥 − 푥 + Hence, the factors of the 푓(푥) are (푥 + 1), (2푥 − 1) and (3푥 + 5) Try: 1. When 푓(푥) = 푃푥 − 3푥 + 8푥 + 36 is divided by 푥 + 2 it leaves no remainder i. Find the value 푃. ii. Hence express 푓(푥) in the form; 푓(푥) = (푥 + 2)(푎푥 + 푏푥 + 푐), where 푎, 푏 and 푐 are constants and find the zeros. 2. The polynomial 푓(푥) = 푎푥 + 6푥 + 푐, where 푎and 푏 are constants, has 푥 − 2as a

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factor. When it is divided by 푥 − 1, the remainder is 4. Find i. the values of the constants 푎 and 푏. ii. the zeros of the function 푓(푥) (SSSCE) 3. 훼. A factor of the polynomial ℎ(푥) = 푏푥 − 4푥 + 푐, where 푏 and 푐 are constants, is (1 − 2푥). When the polynomial it is divided by 1 + 푥 the remainder is 3. Find i. 푏 and 푐 ii. the zeros of ℎ(푥). 훽. If 푓(푥 − 2) = 푥 + 3푥 − 3푥 + 7, find 푓(1). (SSSCE) 4. When the polynomial 푓(푥) = 푥 + 6푥 + 푘 is divided by 푥 − 푎, the remainder is the same as when it is divided by 푥 − 2푎. Find; i. the values of 푎 ii. the value of 푘 if the remainder is 2. (SSSCE) 5. When the polynomial 푓(푥) = 푥 + 5푥 + 푝푥 − 1is divided by 푥 − 2, the remainder is 7. Find the value of 푝. (SSSCE) 6. If 6 − 푥 − 푥 is the factor of the polynomial 푔(푥) = 푚푥 + 5푥 + 푝푥 − 1, find the values of 푚 and 푝. (SSSCE) 7. When the polynomial 푓(푥) = 2푥 − 푥 −푚푥 − 2 is divided by (푥 − 1), the remainder is −6. i. Determine the value of the constant 푚, and hence factorise 푓(푥) completely. ii. Find the truth set of the equation 푓(푥) = 0 (SSSCE) 8. The polynomial 푓(푥) = 푥 + 푎푥 − 5푥 + 푏 has (푥 + 3) a factor and when it is divided by(푥 + 2) the remainder is 4. Find the values of 푎 and 푏. (SSSCE) 9. If (푥 + 2) and (푥 − 3) are factors of 2푥 + 푝푥 + 푞푥 + 6, find the values of the constants 푝 and 푞. (WASSCE) 10. Given that (푥 − 1) is a factor of the polynomial 푓(푥) = 2푥 + 9푥 + 푚푥 − 푛, where 푚 and 푛 are constants, find the values of 푚 and 푛. (WASSCE) 11. If 푥 + 푥 − 2 is a factor of the polynomial 푓(푥) = 2푥 + 9푥 + 푚푥 − 푛, where 푚 and 푛 are constants, find the values of 푚 and 푛. 12. If (푥 − 2) is a factor of the polynomial 푓(푥) = 4푥 − 19푥 + 푎푥 − 14, find i. the value of the constant, 푎; ii. the zeros of 푓(푥) (WASSCE)

Final Exercises 1. When 푓(푥) = 푃푥 − 3푥 − 2푥 + 18 is divided by 푥 − 2 it leaves no remainder i. Find the value 푃. ii. Hence express 푓(푥) in the form; 푓(푥) = (푥 − 2)(푎푥 + 푏푥 + 푐), where 푎, 푏 and 푐 are constants and find the zeros.

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2. The polynomial 푓(푥) = 푎푥 + 6푥 + 푐, where 푎and 푏 are constants, has 푥 + 2as a factor. When it is divided by 푥 + 1, the remainder is 9. Find i. the values of the constants 푎 and 푏. ii. the zeros of the function 푓(푥) 3. 훼. A factor of the polynomial ℎ(푥) = 푝푥 − 2푥 + 푞, where 푏 and 푐 are constants, is (1 + 2푥). When the polynomial it is divided by 1 − 푥 the remainder is 3. Find i. 푝 and 푞 ii. the zeros of ℎ(푥). 훽. If 푓(푥 − 2) = 푥 + 3푥 − 3푥 + 7, find 푓(1). 4. When the polynomial 푓(푥) = −3푥 + 푥 + 푘 is divided by 푥 − 푏, the remainder is the same as when it is divided by 푥 − 3푎. Find; i. the values of 푎 ii. the value of 푘 if the remainder is 2. 5. When the polynomial 푓(푥) = 2푥 + 푝푥 + 푥 − 1is divided by 푥 + 2, the remainder is 7. Find the value of 푝. 6. If 14 − 9푥 − 푥 is the factor of the polynomial 푔(푥) = 푚푥 + 5푥 + 푝푥 − 1, find the values of 푚 and 푝. 7. When the polynomial 푓(푥) = 3푥 − 푥 − 푛푥 − 2 is divided by (푥 + 1), the remainder is −6. i. Determine the value of the constant 푛, and hence factorise 푓(푥) completely. ii. Find the truth set of the equation 푓(푥) = 0 8. The polynomial 푓(푥) = 3푥 + 푎푥 − 5푥 + 푏 has (푥 − 3) a factor and when it is divided by(푥 + 2) the remainder is 4. Find the values of 푎 and 푏. 9. If (푥 + 2) and (푥 − 3) are factors of 2푥 + 푝푥 + 6푥 + 푞, find the values of the constants 푝 and 푞. 10. Given that (푥 − 4) is a factor of the polynomial 푓(푥) = 푥 + 9푥 + 푚푥 − 푛, where 푚 and 푛 are constants, find the values of 푚 and 푛. 11. If 푥 − 푥 − 6 is a factor of the polynomial 푓(푥) = 2푥 + 푚푥 + 푥 − 푛, where 푚 and 푛 are constants, find the values of 푚 and 푛. 12. If (푥 + 2) is a factor of the polynomial 푓(푥) = 푥 − 19푥 + 푏푥 − 14, find i. the value of the constant, 푏; ii. the zeros of 푓(푥) 13. Evaluate (푥 + 푥 + 4푥 − 5)− (푥 − 푥 +7푥 + 13) (CCE) 14. (푥 − 2) and (푥 − 1) are factors of the polynomial 푎푥 + 푏푥 − 4푥 + 푐. When the polynomial is divided by (푥 + 2) is −12. i. Find the values of 푎,푏and 푐. ii. With these values, find the remaining factor.

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iii. Hence, solve the equation 푎푥 + 푏푥 − 4푥 + 푐 = 0 15. If (푥 − 1) is a factor of the polynomial 푓(푥) = 2푥 + 2푥 − 푝. Find the value of 푝. Hence, factorise푓(푥) completely. 16. When the polynomial 푓(푥) = 2푥 + 푎푥 + 3푥 + 푐 is divided by (푥 + 1) it leaves a remainder of −6. If (2푥 − 3) is a factor of the polynomial, find the values of the

constants 푎and푏. Calculate 푓 − . 17. The polynomial 2푥 −푚푥 − 푛푥 − 1 has a double root at 푥 = −2. Find the values of 푚 and 푛. Find the third factor of the polynomial. 18. (2푥 − 1)divides the polynomial 푓(푥) = 푎푥 + 푏푥 + 푐. The polynomial leaves a remainder of 12 and −15 when divided by (푥 + 1) and (푥 − 5)respectively. Find the values of the constants 푎, 푏 and 푐. 19. Find the values of 푄 when the polynomial 푔(푥) = 3푥 − 2푄푥 + 3 is divided by (2푥 + 4) 20.When the polynomial 푝(푥) = 2 − 3푥 − 푥 − 푎푥 is divided by (푥 + 3), the remainder it leaves is twice that when the polynomial divided by (푥 − 6). Find the value of 푎. 21. A polynomial is defined as 푞(푥) = 푥 − 푏푥 + 푐푥 − 푑. 푥 − 4is a factor of the polynomial. If 푞(푥) is divided by (푥 + 2) it leavea a remainder of −5. Find the values of the constants 푏, 푐 and 푑. 22. Factorise the polynomial푓(푥) = 4푥 + 푥 − 2푥 + 3. 23. (푥 + 2) and (푥 − 3) are factors of the polynomial 푓(푥) = 푎푥 − 푏푥 + 푐푥, where 푎, 푏 and 푐 are constants. The remainder when 푓(푥) is divided by(푥 + 1) is 12. Find i. The values of the constant 푎 and 푏. ii. Factorize 푓(푥) completely. iii. The zeroes of푓(푥) 24. The polynomial 푓(푥) = 푥 + 푎푥 + 푏푥 + 푐 is exactly divisible by (푥 + 2)and(푥 − 5) and leaves a remainder of −16 when divided by (푥 − 2). Find the constants 푎, 푏 and 푐. 25. Given the polynomial 푝(푥) = 3 − 푐푥 + 5푥 − 푎푥 . Find the values of 푎 and 푐 if 푥 − 2 is a factor of the polynomial. 26. When the polynomial,푓(푥) is divided by(푥 − 1)the remainder is −3, and when it divided by(푥 − 2) the remainder is −8. Given that푓(푥) may be written in the form 푓(푥) = (푥 − 1)(푥 − 2)푄(푥) + 푎푥 + 푏. Find the values of the constants 푎 and 푐.

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27. Find the remainder when the polynomial 푓(푥) = −푥 +2푥 − 15푥 − 6 is divided by(푥 − 3). 28. If (푥 + 1) and (푥 − 5) are factors of the polynomial 푓(푥) = 푎푥 + 푥 + 푏, find i. the values of the constants 푎 and 푏 ii. the remainder when the polynomial is divided by (푥 − 1) 39. (푥 − 3)is a factor of the polynomial 푝(푥) = 푥 + 푎푥 + 푏푥 + 6. When the

polynomial is divided by (푥 + 1)the remainder is 8. i. Find the values of 푎 and 푏. ii. Find the remaining factors of the polynomial. iii. Hence, solve the equation 푥 + 푎푥 + 푏푥 + 6 = 0. 30. (3푥 − 2) is a factor of the polynomial 푟(푥) = 3푥 + 푎푥 + 푏푥 − 4. When the

polynomial is divided by (푥 − 1) the remainder is 6. i. Find the values of the constants 푎 and 푏. ii. Sketch 푟(푥). iii. Solve the equation 푟(푥) = 0.

QUADRATICS A 푞푢푎푑푟푎푡푖푐 expression is any expression of the form 푎푥 + 푏푥 + 푐, 푎 ≠ 0, with degree of 푥 being 2. Introduction of the equal sign transforms the expression to either an equation or function.

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Therefore, the function 푦 = 푎푥 + 푏푥 + 푐 is a quadratic function. This function becomes a quadratic equation if 푦 is a constant, a linear function or a quadratic function with different co-efficient of the highest power of 푥 at both side of the equation. The degree of the quadratic equation determines the roots of the equation. For instance, the 푔푒푛푒푟푎푙 quadratic equation 푎푥 + 푏푥 + 푐 = 0, where 푎,푏 and 푐 are constants, has two roots.

Methods of solving quadratic equation 1. Factorisation method This method transforms the general quadratic equation into four (4) terms. The procedure is describe below

Multiply the first term by the last term [푎푥 × 푐 = 푎푐푥 ] Find all the possible factors of 푎푐푥 Choose two possible factors that when added will give 푏푥. Replace 푏푥by the addition of the two factors Factorise the first two and the last two terms.

Example 9.1

Factorise the following: i. 2푥 + 3푥 + 1 = 0 ii. 푥 − 3푥 − 2 = 0 iii. 푥 − 9푥 + 14 = 0 iv. 3푥 − 17푥 + 10 = 0 v. 푥 + 5푥 = 0 vi. 푥 − 9 = 0

Solution i.2푥 + 3푥 + 1 = 0 ⟹ 2푥 + 푥 + 2푥 + 1 = 0 ⟹ (2푥 + 푥) + (2푥 + 1) = 0 ⟹ 푥(2푥 + 1) + 1(2푥 + 1) = 0 ⟹ (2푥 + 1)(푥 + 1) = 0 ⟹ 2푥 + 1 = 0 or 푥 + 1 = 0 ⟹ 2푥 = −1 ⟹ 푥 = − or 푥 = −1 ∴ 푥 = − or 푥 = −1 Hint:

2푥 × 1 = 2푥 factorsof2푥 :± 푥, +2푥Now, 푥 + 2푥 = 3푥

ii. 푥 − 3푥 − 2 = 0 ⟹ 푥 − 2푥 − 푥 − 2 = 0 ⟹ (푥 − 2푥)− (푥 − 2) = 0 ⟹ 푥(푥 − 2)− 1(푥 − 2) = 0 ⟹ (푥 − 2)(푥 − 1) = 0 ⟹ 푥 − 2 = 0 or 푥 − 1 = 0 ⟹ 푥 = 2 or 푥 = 1 ∴ 푥 = 2 or 푥 = 1

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Hint:푥 × −2 = −2푥 factorsof − 2푥 :± 푥, +2푥Now,−푥 − 2푥 = −3푥

iii. 푥 − 9푥 + 14 = 0 ⟹ 푥 − 7푥 − 2푥 + 14 = 0 ⟹ (푥 − 7푥)− (2푥 − 14) = 0 ⟹ 푥(푥 − 7)− 2(푥 − 7) = 0 ⟹ (푥 − 7)(푥 − 2) = 0 ⟹ 푥 − 7 = 0 or 푥 − 2 = 0 ⟹ 푥 = 7 or 푥 = 2 ∴ 푥 = 7 or 푥 = 2 Hint:푥 × 14 = 14푥 factorsof14푥 :± 푥, ±2푥Now,−푥 − 2푥 = −3푥

iv. 3푥 − 17푥 + 10 = 0 ⟹ 3푥 − 15푥 − 2푥 + 10 = 0 ⟹ (3푥 − 15푥) − (2푥 − 10) = 0 ⟹ 3푥(푥 − 5)− 2(푥 − 5) = 0 ⟹ (푥 − 5)(3푥 − 2) = 0 ⟹ 푥 − 5 = 0 or 3푥 − 2 = 0 ⟹ 푥 = 5 or 푥 = ∴ 푥 = 5 or 푥 =

Hint:

3푥 × 10 = 30푥 factorsof30푥 :± 푥, ±2푥, ±3푥, ±5푥, ±6푥, ±10푥, ±15, ±30푥Now,−15푥 − 2푥 = −17푥

v. 푥 + 5푥 = 0 ⟹ 푥 + 0푥 + 5푥 + 0 = 0 ⟹ (푥 + 0푥) + (5푥 + 0) = 0 ⟹ 푥(푥 + 0) + 5(푥 + 0) = 0 ⟹ (푥 + 0)(푥 + 5) = 0 ⟹ 푥 + 0 = 0 or 푥 + 5 = 0 ⟹ 푥 = 0 or 푥 = −5 ∴ 푥 = 0 or 푥 = −5 Hint:푥 × 0 = 0푥 factorsof0푥 :0푥andanytimes푥.Now, 0푥 + 5푥 = 5푥

Or

푥 + 5푥 = 0 ⟹ 푥(푥 + 5) = 0 푥 = 0 or 푥 + 5 = 0 ⟹ 푥 = −5 vi. 푥 − 9 = 0 ⟹ 푥 + 0푥 − 9 = 0 ⟹ 푥 − 3푥 + 3푥 − 9 = 0 ⟹ (푥 − 3푥) + (3푥 − 9) = 0 ⟹ 푥(푥 − 3) + 3(푥 − 3) = 0 ⟹ (푥 − 3)(푥 + 3) = 0 ⟹ 푥 − 3 = 0 or 푥 + 3 = 0 ⟹ 푥 = 3 or 푥 = −3 ∴ 푥 = 3 or 푥 = −3 Hint:푥 × −9 = −9푥 factorsof − 9푥 :± 푥, ±3푥Now,−3푥 + 3푥 = −3푥

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2. Method of completing the squares This method is used to solve all forms of quadratic equations. The procedure is describe below

Make sure the co-efficient of 푥 in the equation given to you is 1. If not, then divide of the equation by the value of the co-efficient.

Send to the right-hand side of the equation if any.

Find the ℎ푎푙푓 of , square it and add it to both sides of the equation.

Write the write left-hand side of the equation in the form (푥 + 푘) ,where 푘 =

Take the square root of both sides of the equation and find the values of 푥.

Example 9.2

Solve the equations by completing the squares: i. 푥 + 3푥 + 1 = 0 ii. 2푥 + 2푥 − 1 = 0 iii. 5푧 + 3푧 − 2 = 0

Solution

i. 푥 + 3푥 + 1 = 0 ⟹ 푥 + 3푥 = −1 ⟹ 푥 + 3푥 + = −1 +

⟹ 푥 + = −1 + ⟹ 푥 + = ⟹ 푥 + = ±

⟹ 푥 + = ± √ ⟹ 푥 = − ± √

∴ 푥 = − + √ or 푥 = − − √

ii. 2푥 + 2푥 − 1 = 0 ⟹ 푥 + 푥 − = ⟹ 푥 + 푥 =

⟹ 푥 + 푥 + = + ⟹ 푥 + = +

⟹ 푥 + = ⟹ 푥 + = ±

⟹ 푥 + = ± √ ⟹ 푥 = − ± √

∴ 푥 = − + √ or 푥 = − − √

iii. 5푧 + 3푧 − 2 = 0 ⟹ 푧 + 푧 − = 0 ⟹ 푧 + 푧 =

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⟹ 푧 + 푧 + = + ⟹ 푧 + = +

⟹ 푧 + = ⟹ 푧 + = ±

⟹ 푧 + = ± ⟹ 푧 = − ±

⟹ 푧 = − + = = = or 푧 = − − = = − = −1

∴ 푧 = or 푧 = −1 Try: Solve the following equations by completing the squares: i. 2푥 − 5푥 + 1 = 0 ii. 3푥 − 8푥 − 7 = 0 iii. (2푥 + 1) = 0 iv. 3푥 + 15푥 + 4 = 0

The formula When the method of completing the squares is applied to the general quadratic equation, the values of 푥 can be obtained by the formula:

푥 =−푏 ± √푏 − 4푎푐

2푎

Proof Let 푎푥 + 푏푥 + 푐 = 0 ⟹ 푥 + 푥 + =

⟹ 푥 + 푥 = − ⟹ 푥 + 푥 + = −

⟹ 푥 + = − ⟹ 푥 + =

⟹ 푥 + = ⟹ 푥 + = ±

∴ 푥 = ±√ ∎

Example 9.3

Use the formula to solve the following equations: i. 푥 + 4푥 − 77 = 0 ii. 2푦 − 7푦 − 15 = 0 iii. 푥 − 13푥 + 30 = 0

Solution i. We have, 푥 + 4푥 − 77 = 0 … … … (1). By comparing (1) with

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푎푥 + 푏푥 + 푐 = 0, we get, 푎 = 1, 푏 = 4and 푐 = −77

The values of 푥 is given by the formula: 푥 = ±√

⟹ 푥 = ±√ = ± ( ) ( )( )( )

= ±√ = ±√ = ±

∴ 푥 = = = 7 or 푥 = = = −11 ii. We have, 2푦 − 7푦 − 15 = 0 … … … (1). By comparing (1) with 푎푦 + 푏푦 + 푐 = 0, we get, 푎 = 2, 푏 = −7and 푐 = −15

The values of 푥 is given by the formula: 푥 = ±√

⟹ 푥 = ±√ = ( )± ( ) ( )( )( )

= ±√ = ±√ = ±

∴ 푥 = = = 5 or 푥 = = = − iii. We have, 푥 − 13푥 + 30 = 0 … … … (1). By comparing (1) with 푎푥 + 푏푥 + 푐 = 0, we get, 푎 = 1, 푏 = −13and 푐 = 30

The values of 푥 is given by the formula: 푥 = ±√

⟹ 푥 = ±√ = ( )± ( ) ( )( )( )

= ±√ = ±√

= ±

∴ 푥 = = = 10 or 푥 = = = 3 Try: Use the formula to solve for the following equations: a. 2푥 − 2푥 + 1 = 0 b. 3 − 푥−푥 = 0 c. 푥 + 3푥 = 0

The discriminant From the above worked examples, it can be seen that the roots of a quadratic equation is determined by the expression √푏 − 4푎푐. The following rules are valid 1. If 푏 − 4푎푐 = 0, then the quadratic equation has real equal roots. 2. If 푏 − 4푎푐 > 0, then the quadratic equation has different real roots. 3. If 푏 − 4푎푐 < 0,then the quadratic equation has no real (complex) roots.

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The diagram below shows the nature of the graphs of equations that have been listed above: 푦 푦 푦 푥 푥 푥 푏 − 4푎푐 = 0 푏 − 4푎푐 > 0 푏 − 4푎푐 < 0

Example 9.4 Find the value of 푏 for which the equation 푥 + 푏푥 + 1 = 0 has equal roots.

Solution We have, 푥 + 푏푥 + 1 = 0 … … … (1). By comparing (1) with 푎푥 + 푏푥 + 푐 = 0, we get, 푎 = 1, 푏 = 푏and 푐 = 1 For equal roots: 푏 − 4푎푐 = 0 ⟹ 푏 − 4(1)(1) = 0 ⟹ 푏 = 4 ⟹ √푏 = ±√4 = ±2 ∴ 푏 = 2 or 푏 = −2.

Example 9.5

Find the values of 푘 for which the equation 푥 + (푘 + 1)푥 + 4 = 0 has real roots Solution

We have, 푥 + (푘 + 1)푥 + 4 = 0 … … … (1). By comparing (1) with 푎푥 + 푏푥 + 푐 = 0, we get, 푎 = 1, 푏 = 푘 + 1and 푐 = 4 For real roots: 푏 − 4푎푐 > 0 ⟹ (푘 + 1) − 4(1)(4) > 0 ⟹ 푘 + 2푘 + 1 − 16 > 0 ⟹ 푘 + 2푘 − 15 > 0 ⟹ 푘 − 3푘 + 5푘 − 15 > 0 ⟹ (푘 − 3푘) + (5푘 − 15) > 0 ⟹ 푘(푘 − 3) + 5(푘 − 3) > 0 ⟹ (푘 − 3)(푘 + 5) > 0 ⟹ 푘 − 3 > 0 or 푘 + 5 > 0 ∴ 푘 > 3 or 푘 > −5

Example 9.6 Find the values of 푘 for which the expression 푘푥 + 3푘푥 + 6 is a positive definite.

Solution We have, 푘푥 + 3푘푥 + 6 … … … (1). By comparing (1) with

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푎푥 + 푏푥 + 푐, we get, 푎 = 푘, 푏 = 3푘 + 1and 푐 = 6 For positive definite: 푏 − 4푎푐 > 0 ⟹ (3푘) − 4(푘)(6) > 0 ⟹ 9푘 − 24푘 > 0 ⟹ 3푘(3푘 − 8) > 0 ⟹ 3푘 > 0 or 3푘 − 8 > 0 ∴,푘 > 0 or 푘 >

Example 9.7 Find the values of 푚 for which the quadratic equation 푥 − 2푚푥 + 9 = 0 has real and different roots.

Solution We have, 푥 − 2푚푥 + 9 = 0 … … … (1). By comparing (1) with 푎푥 + 푏푥 + 푐 = 0, we get, 푎 = 1, 푏 = −2푚and 푐 = 9 For real roots: 푏 − 4푎푐 > 0 ⟹ (−2푚) − 4(1)(9) > 0 ⟹ 4푚 − 36 > 0 ⟹ (2푚 − 6)(2푚 + 6) > 0 ⟹ 2푚− 6 > 0 or 2푚 + 6 > 0 ∴ 푘 > 3 or 푘 > −5 ⟹ 2푚 > 6 or 2푚 > −6 ∴ 푚 > 3 or 푚 > −3 Try: 1. If 푥 − 6푥 + 7 = 푘(2푥 − 3) has real roots, find the possible values of 푘. (SSSCE) 2. The roots of the quadratic equation 푥 − 2(3푘 + 1)푥 + 7(2푘 + 3) = 0,where 푘 is a constant, are equal. Find the values of 푘. 3. If 푚푥 + 4푥 + 2 = 0 has real roots, find the largest value of 푚. (푊퐴푆푆퐶퐸) 4. Given that 푘(푥 − 10푥 − 2) + (2푥 + 1) = 0, find the values of 푘 for which the quadratic equation has equal roots. (WASSCE)

Relationship between The Roots and Co-efficient Of a Quadratic Equation A quadratic equation has two roots. Therefore, if the general quadratic equation 푎푥 + 푏푥 + 푐 = 0 has roots 훼 and 훽 then it may be written in the form

푎(푥 − 훼)(푥 − 훽) = 0. It follows that: 푎푥 + 푏푥 + 푐 ≡ 푎(푥 − 훼)(푥 − 훽) ≡ 푎(푥 − 훼푥 − 훽푥 + 훼훽) ≡ 푎푥 − 푎(훼 + 훽)푥 + 푎훼훽 ⟹ 푥 + 푥 + ≡ 푥 − (훼 + 훽)푥 + 훼훽 [dividebothsidesbya] By comparing both sides of the equation; −(훼 + 훽) = ⟹ (훼 + 훽) = − and 훼훽 =

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훼 + 훽 = 푠푢푚표푓푟표표푡푠 훼훽 = 푝푟표푑푢푐푡표푓푟표표푡푠 A quadratic equation whose roots are known can be formed by:

푥 − (sumofroots)푥 + productofroots = 0 Example 9.8

Find a quadratic equation whose roots are 2 + √5 and 2 − √5 Solution

We have the roots: 2 + √5 and 2 − √5 Sum of roots = 2 + √5 + 2 − √5 = 2 + 2 + √5 − √5 = 4 Product of roots = 2 + √5 2 − √5 = 2 − √5 = 4 − 5 = −1 The equation whose roots are 2 + √5 and 2 − √5 is given as: 푥 − (푠푢푚표푓푟표표푡푠)푥 + 푝푟표푑푢푐푡표푓푟표표푡푠 = 0 ⟹ 푥 − 4푥 − 1 = 0 ∴, the equation: 푥 − 4푥 − 1 = 0

Example 9.9 Find the value of 푘 if one root of the equation 푥 + 9푥 + 푘 = 0 is double the other.

Solution We have, 푥 + 9푥 + 푘 = 0 … … … (1).Let one of the roots of the equation be 훼. So, the other root is 2훼. Sum of roots= 훼 + 2훼 = 3훼 Product of roots = 훼(2훼) = 2훼 The equation whose roots are 훼 and 2훼 is given as: 푥 − (푠푢푚표푓푟표표푡푠)푥 + 푝푟표푑푢푐푡표푓푟표표푡푠 = 0 ⟹ 푥 − 3훼푥 + 2훼 = 0 By comparing 푥 + 9푥 + 푘 = 0with 푥 − 3훼푥 + 2훼 = 0, we get −3훼 = 9 ⟹ 훼 = = −3 and 2훼 = 푘 ⟹ 푘 = 2(−3) = 18

∴ 푘 = 18. Example 9.10

Show that the roots of the equation 푚푥 + 푘푥 + 푚 = 0 are reciprocal of one another. Solution

We have, 푚푥 + 푘푥 + 푚 = 0 ⟹ 푥 + + = ⟹푥 + + 1 = 0 … … … (1). Let the roots of the equation be 훼 and 훽 Sum of roots= 훼 + 훽 Product of roots = 훼훽 The equation whose roots are 훼 and훽 is given as: 푥 − (푠푢푚표푓푟표표푡푠)푥 + 푝푟표푑푢푐푡표푓푟표표푡푠 = 0 ⟹ 푥 − (훼 + 훽)푥 + 훼훽 = 0 By comparing 푥 + + 1 = 0 with 푥 − (훼 + 훽)푥 + 훼훽 = 0, we get

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훼훽 = ⟹ 훼 = ∎

Example 9.11 Find the value of 푛 in the equation 2푥 − 3(푛 + 1)푥 + 4 = 0 if the roots are consecutive.

Solution We have, 2푥 − 3(푛 + 1)푥 + 4 = 0 ⟹ 푥 − ( ) + =

⟹ 푥 − ( ) + 2 = 0 … … … (1). Let the consecutive roots of the equation be 훼 and 푎 + 1 Sum of roots= 훼 + 훼 + 1 = 2훼 + 1 Product of roots = 훼(훼 + 1) The equation whose roots are 훼 and 훽 is given as: 푥 − (푠푢푚표푓푟표표푡푠)푥 + 푝푟표푑푢푐푡표푓푟표표푡푠 = 0 ⟹ 푥 − (2훼 + 1)푥 + 훼(훼 + 1) = 0 By comparing 푥 − ( ) + 2 = 0 with 푥 − (2훼 + 1)푥 + 훼(훼 + 1) = 0, we get

(2훼 + 1) = ( ) ⟹ 2훼 = ( ) − 1 ⟹ 2훼 = =

⟹ 훼 = ÷ 2 ⟹ 훼 = × = and

훼(훼 + 1) = 2 ⟹ + 1 = 2 ⟹ = 2

⟹ = 2 ⟹ 16 × = 2 × 16 ⟹ 9푛 + 18푛 + 5 = 32 ⟹ 9푛 + 18푛 + 5 − 32 = 0 ⟹ 9푛 + 18푛 − 27 = 0 (Divide both sides by 3) ⟹ 3푛 + 6푛 − 9 = 0 ⟹ 3푛 − 3푛 + 9푛 − 9 = 0 ⟹ 3푛(푛 − 1) + 9(푛 − 1) = 0 ⟹ (푛 − 1)(푛+ 9) = 0 ⟹ 푛 − 1 = 0 or 푛 + 9 = 0 ⟹ 푛 = 1 or 푛 = −9. But 푛 = −9 is not a true solution. ∴ 푛 = 1.

Hint:If푛 = −9, 훼 = ( ) = − .But − − + 1 ≠ 2

Try: 1. Find the equation of the curve with 푘 and −2 as zeros. 2. Find the set of values of 푥 that satisfy the inequality 푥 + 3푥 + 2 ≤ 0. (SSSCE) Note The Following Symmetric Functions 1. 훼 + 훽 = (훼 + 훽) − 2훼훽[Thisistruebecause(훼 + 훽) = 훼 + 훽 + 2훼훽] 2. (훼 − 훽) = (훼 + 훽) − 4훼훽

[Thisistruebecause(훼 − 훽) = 훼 + 훽 − 2훼훽푎푛푑훼 + 훽 = (훼 + 훽) − 2훼훽]

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3. 훼 − 훽 = (훼 + 훽) − 4훼훽 4. 훼 + 훽 = (훼 + 훽) − 3훼훽(훼 + 훽)[notethat(훼 + 훽) = 훼 + 훽 + 3훼 훽 + 3훼훽 ] 5. 훼 − 훽 = (훼 + 훽) + 3훼훽(훼 − 훽)

6. + = 7. + = = ( )( )

8. 훼 + 훽 = (훼 + 훽 ) − 2훼 훽 = [(훼 + 훽) − 2훼훽] − 2(훼훽)

Example 9.12 If 훼 and 훽 are the roots of the equation 푥 − 2푥 − 5 = 0, find the value of the following i. 훼 + 훽 ii. + iii. 훼 + 훽 iv. + v. (훼 − 훽) vi. 훼 − 훽

Solution We have, 푥 − 2푥 − 5 = 0 … … … (1). By comparing (1) with 푎푥 + 푏푥 + 푐 = 0, we get, 푎 = 1, 푏 = −2and 푐 = −5 Let the roots of the equation be 훼 and 훽 Sum of roots= 훼 + 훽 = − = 2 Product of roots = 훼훽 = = −5 i. 훼 + 훽 = (훼 + 훽) − 2훼훽 = (2) − 2(−5) = 4 + 10 = 14

∴ 훼 + 훽 = 14

ii. + = =( )

=( )

= ∴ + =

iii. 훼 + 훽 = (훼 + 훽) − 3훼훽(훼 + 훽) = (2) − 3(−5)(2) = 8 + 30 = 38 ∴ 훼 + 훽 = 38

iv. + = = − ∴ + = −

v. (훼 − 훽) = (훼 + 훽) − 4훼훽 = (2) − 4(−5) = 4 + 20 = 24 ∴ (훼 − 훽) = 24 vi. 훼 − 훽 = (훼 − 훽) = √24 = 3√2 ∴ 훼 − 훽 = 3√2

Example 9.13 The roots of the equation 3푥 + 8푥 + 16 = 0 are denoted by 훼 and 훽. Find the quadratic equation whose roots are i. 훼 + and 훽 + ii. and iii. 2훼 + 훽 and 훼 + 2훽

Solution We have, 3푥 + 8푥 + 16 = 0 … … … (1). By comparing (1) with

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푎푥 + 푏푥 + 푐 = 0, we get, 푎 = 3, 푏 = 8and 푐 = 16 Let the roots of the equation be 훼 and 훽 Sum of roots= 훼 + 훽 = − Product of roots = 훼훽 =

i. We want to find an equation whose roots are 훼 + and 훽 +

Sum of roots = 훼 + + 훽 + = 훼 + 훽 + + = (훼 + 훽) + = − +

⟹ Sum of roots = − − × = − − = = −

Product of roots= 훼 + 훽 + = 훼훽 + + + = 훼훽 + 2 +

⟹ Sum of roots = + 2 + = + = + = =

The equation whose roots are 훼 + and 훽 + is given as:

푥 − (푠푢푚표푓푟표표푡푠)푥 + 푝푟표푑푢푐푡표푓푟표표푡푠 = 0 ⟹ 푥 − − 푥 + = 0

⟹ 48푥 + 48 푥 + 48 = 0 ⟹ 48푥 + 152푥 + 361 = 0 ∴, the equation: 48푥 + 152푥 + 361 = 0 ii. We want to find an equation whose roots are and

Sum of roots = + = = = − × = −

Product of roots = × = = =

The equation whose roots are and is given as:

푥 − (푠푢푚표푓푟표표푡푠)푥 + 푝푟표푑푢푐푡표푓푟표표푡푠 = 0 ⟹ 푥 − − 푥 + = 0

⟹ 16푥 + 16 푥 + 16 = 0 ⟹ 16푥 + 8푥 + 3 = 0

∴, the equation: 16푥 + 8푥 + 3 = 0 iii. We want to find an equation whose roots are 2훼 + 훽 and 훼 + 2훽 Sum of roots = 2훼 + 훽 + 훼 + 2훽 = 3훼 + 3훽 = 3(훼 + 훽) = 3 × − = −8 Product of roots = (2훼 + 훽)(훼 + 2훽) = 2훼 + 4훼훽 + 훼훽 + 2훽 = 2(훼 + 훽 ) + 5훼훽 = 2[(훼 + 훽) − 2훼훽] + 5훼훽

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= 2(훼 + 훽) − 4훼훽 + 5훼훽 = 2((훼 + 훽) + 훼훽

⟹ Product of roots= 2 − + = 2 + = + = =

The equation whose roots are 2훼 + 훽 and 훼 + 2훽 is given as: 푥 − (푠푢푚표푓푟표표푡푠)푥 + 푝푟표푑푢푐푡표푓푟표표푡푠 = 0 ⟹ 푥 − (−8)푥 + = 0

⟹ 9푥 + 9(8)푥 + 9 = 0 ⟹ 9푥 + 72푥 + 160 = 0

∴, the equation: 9푥 + 72푥 + 160 = 0 Try: 1. If 훼 and 훽 are the roots of the equation 3푥 − 6푥 − 9 = 0, find the quadratic equation whose roots are and . (SSSCE)

2. Let 훼 and 훽 are the roots of the quadratic equation 푥 + 2푥 − 5 = 0. without solving the equation, i. evaluate 훼 + 훽 ii. evaluate (훼 − 훽) iii. obtain a quadratic equation whose roots are 훼 + 1,훽 +1 (SSSCE) 3. If 훼 and 훽 are the roots of the quadratic equation 2푥 − 10푥 + 7 = 0, evaluate .

(SSSCE) 4. If 훼 and 훽 are the roots of the equation 4푥 + 7푥 + 3 = 0, find the value of + .

(SSSCE) 5. The roots of the equation 푥 + 6푥 + 푝 = 0 are 훼 and 훼 − 1. Find the value of 푝. (SSSCE) 6. If 훼 and 훽 are the roots of the equation 3푥 − 푥 − 1 = 0, find the equation whose roots are and . (SSSCE)

7. The roots of the quadratic equation 2푥 + 3푥 − 4 = 0 are 훼 and 훽. Find the equation whose roots are 훼 and 훽 . (SSSCE) 8. The roots of the quadratic equation 3푥 + 5푥 − 2 = 0 are 훼 and 훽. Find the equation whose roots are 훼 and 훽 . (WASSCE) 9. If 훼 and 훽 are the roots of the quadratic equation 2푥 − 10푥 + 7 = 0, find a quadratic equation whose roots are and . (SSSCE)

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Quadratic Identities The quadratic function 푦 = 푎푥 + 푏푥 + 푐, where 푎, 푏 and 푐 are constants and 푎 ≠ 0 has a minimum value if 푎 > 0 and a maximum value if 푎 < 0. The minimum and maximum

value of the function is given by 푦 = − and occur at 푥 = − called 푎푥푖푠표푓푠푦푚푚푒푡푟푦.

Proof We have 푦 = 푎푥 + 푏푥 + 푐 = 푎 푥 + 푥 + (By factoring out 푎)

= 푎 푥 + 푥 + − + Add andsubtract

= 푎 푥 + − + (푥 + ) = 푥 + 푥 + ( )

= 푎 푥 + −

= 푎 푥 + − 푎

= 푎 푥 + −

Line/axis of symmetry minimum/ maximum value

Put 푎(푥 + ) = 0 ⟹ 푥 = − 푦 = −

Minimum/ maximum point − ,− ∎

Example 9.14

Find the minimum value of the function 푦 = 3푥 − 푥 − 5. Solution

We have, 푦 = 3푥 − 푥 − 15 … … … (1). By comparing (1) with 푎푥 + 푏푥 + 푐 = 0, we get, 푎 = 3, 푏 = −1and 푐 = −15

The minimum value of the function is defined as: 푦 = −

⟹ 푦 = − ( ) ( )( )( )

= − = −

∴,the minimum value of the function is 푦 = − Example 9.15

159

Determine the maximum value of the function 푦 = 5 − 푥 − 2푥 . Solution

We have, 푦 = 5 − 푥 − 2푥 … … … (1). By comparing (1) with 푎푥 + 푏푥 + 푐 = 0, we get, 푎 = −2, 푏 = −1and 푐 = 5

The minimum value of the function is defined as: 푦 = −( )

⟹ 푦 = − ( ) ( )( )( )

= − =

∴,the minimum value of the function is 푦 = Example 9.16

Find the line of symmetry and the minimum point of the curve 푦 = 2푥 − 푥 − 6. Solution

We have, 푦 = 2푥 − 푥 − 6 … … … (1). By comparing (1) with 푎푥 + 푏푥 + 푐 = 0, we get, 푎 = 2, 푏 = −1and 푐 = −6 The axis of symmetry of the function is defined as: 푥 = −

⟹ 푥 = −( )

=

The minimum value of the function is defined as: 푦 = −

⟹ 푦 = − ( ) ( )( )( )

= − = −

∴,the axis of symmetry is and the minimum value of the function is − Try 1. Find the maximum point of the following functions: a. 푓(푥) = 3 − 2푥 − 푥 b. 푓(푥) = 2 − 5푥 − 3푥 c. 푔(푥) = 7 + 푥 − 푥 2. Find the minimum value of the functions: a. 푓(푥) = 푥 − 2푥 + 5 b. 푔(푥) = 2푥 − 6푥 + 3

Example 9.17

Given that 4(푥 − 2) = 푘(푥 − 1), show that of roots and their product are equal. Find the values of the constant 푘 for which the difference between the roots is 3 .

Solution We have 4(푥 − 2) = 푘(푥 − 1) ⟹ 4(푥 − 4푥 + 4) = 푘푥 − 푘

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⟹ 4푥 − 16푥 + 16 = 푘푥 − 푘 ⟹ 4푥 − 16푥 − 푘푥 + 16 + 푘 = 0 ⟹ 4푥 − (16 + 푘)푥 + 16 + 푘 = 0 (1) By comparing 4푥 − (16 + 푘)푥 + 16 + 푘 = 0with 푎푥 + 푏푥 + 푐 = 0 It follows that 푎 = 4, 푏 = −(16 + 푘) and 푐 = 16 + 푘 Let 훼 and 훽 be the roots of (1)

훼 + 훽 = − = 훼훽 = = ∎ Now, 훼 − 훽 = 3 ⟹ (훼 + 훽) − 4훼훽 = ⟹ (훼 + 훽) − 4훼훽 =

⟹ − 4 = ⟹ − 4 =

⟹ = ⟹ = ⟹ =

⟹ 16푘 + 푘 = 225 ⟹ 푘 + 16푘 − 225 = 0 ⟹ 푘 − 9푘 + 25푘 − 225 = 0 ⟹ (푘 − 9)(푘 + 25) = 0 ⟹ 푘 − 9 = 0 or 푘 + 25 = 0 ∴ 푘 = 9 or 푘 = −25

Example 9.18

Find the values of 푚, 푝 and 푞 for which 3푥 − 푥 − 5 = 푚(푥 + 1) + 푝(푥 − 3) + 푞 − 4. Solution

We have the identity: 3푥 − 푥 − 5 ≡ 푚(푥 + 1) + 푝(푥 − 3) + 푞 − 4 ≡ 푚(푥 + 2푥 + 1) + 푝푥 − 3푝 + 푞 − 4 ≡ 푚푥 + 2푚푥 + 푚 + 푝푥 − 3푝+ 푞 − 4 ≡ 푚푥 + 2푚푥 + 푝푥 + 푚 − 3푝+ 푞 − 4 ≡ 푚푥 + (2푚 + 푝)푥 + 푚− 3푝+ 푞 − 4 By comparing the LHS with the LHS it follows that, 푚 = 3, 2푚 + 푝 = −1 … … … (1) ⟹ 2(3) + 푝 = −1 ⟹ 6 + 푝 = −1 ⟹ 푝 = −1 − 6 = −7 and 푚− 3푝+ 푞 − 4 = −5 … … … (2) ⟹ 3 − 3(−7) + 푞 − 4 = −5 ⟹ 3 + 21 + 푞 − 4 = −5 ⟹ 푞 = −5− 21 + 4 − 3 = −25 ∴,푚 = 3, 푝 = −7and푞 = −5.

Example 9.19 Express the following functions in the form 푝(푥 + 푞) + 푟 i. 푓(푥) = 2푥 − 푥 − 1 ii. 푦 = 3푥 − 푥 − 6

Solution

161

i. Method 1

Let 푓(푥) = 2푥 − 푥 − 1

= 2 푥 − 푥 − Hint:factorise2outind 표푓 − = −

= 2 푥 − − − − check:푥 − 푥 + − = 푥 −

= 2 푥 − − −

= 2 푥 − −

= 2 푥 − − 2 × [Multiplythroughby2]

= 2 푥 − −

Now, comparing 2 푥 − − with 푝(푥 + 푞) + 푟, it follows that

푝 = 2,푞 = − and푟 = −

Method 2 We have 푓(푥) = 2푥 − 푥 − 1 Let 2푥 − 푥 − 1 ≡ 푝(푥 + 푞) + 푟 ≡ 푝(푥 + 2푞푥 + 푞 ) + 푟 ≡ 푝푥 + 2푝푞푥 + 푝푞 + 푟 By comparing the LHS with the LHS it follows that, 푝 = 2, 푝푞 = −1 … … … (1) ⟹ 2(2)푞 = −1 ⟹ 푞 = −

and 푝푞 + 푟 = −1 … … … (2) ⟹ 2 − + 푟 = −1 ⟹ 2 + 푟 = −1

⟹ 푟 = −1 − = − ∴ ,2푥 − 푥 − 1 ≡ 2 푥 − − ii.

Method 1 Let 푓(푥) = 3푥 − 푥 − 6

162

= 3 푥 − 푥 − Hint:factorise2outind 표푓 − = −

= 3 푥 − − − − 2 check:푥 − 푥 + − = 푥 −

= 3 푥 − − − 2

= 3 푥 − −

= 3 푥 − − 3 × [Multiplythroughby3]

= 3 푥 − −

Now, comparing 3 푥 − − with 푝(푥 + 푞) + 푟, it follows that

푝 = 3,푞 = − and푟 = − Method 2

We have 푓(푥) = 3푥 − 푥 − 6 Let 3푥 − 푥 − 6 ≡ 푝(푥 + 푞) + 푟 ≡ 푝(푥 + 2푞푥 + 푞 ) + 푟 ≡ 푝푥 + 2푝푞푥 + 푝푞 + 푟 By comparing the LHS with the LHS it follows that, 푝 = 3, 푝푞 = −1 … … … (1) ⟹ 2(3)푞 = −1 ⟹ 푞 = −

and 푝푞 + 푟 = −6 … … … (2) ⟹ 2 − + 푟 = −6 ⟹ 3 + 푟 = −6

⟹ 푟 = −6 − = − ∴, 3푥 − 푥 − 6 ≡ 3 푥 − −

Try: 1. Write the function푦 = 2 − 4푥 − 푥 in the form 푦 = 푒(푥 + 푓) + 푔. 2. Express 푦 = 5− 2푥 − 4푥 in the form 푦 = 퐴 − 퐵(푥 + 퐶) , where A, B and C are constants. Hence state the maximum value of the function 푓:푥 → 5 − 2푥 − 4푥 . (SSSCE) 3. Express 3푥 − 6푥 + 10 in the form 푎(푥 − 푏) + 푐, where 푎,푏 and 푐 are integers. Hence state the minimum value of 3푥 − 6푥 + 10 and the value of 푥 for which it occurs.

163

(SSSCE) 4. Express 7 − 5푥 − 2푥 in the form 푎 − 푏(푥 + 푐) . State the greatest value of the expression. (SSSCE) 5. If 2푥 − 푥 − 3 = 퐴(푥 + 퐵) + 퐶, find the values of 퐴,퐵 and C. Hence, find the maximum value of 2푥 − 푥 − 3. 6. By writing 12푥 − 6푥 + 1 in the form 푝(푥 + 푚) + 푛,find the values of constants 푝,푚and 푛. Hence find the minimum value of 푦 = 12푥 − 6푥 + 1. (WASSCE)

Application Of Quadratic Equation The following problems involve the use of quadratic equation.

Example 9.20 Find the truth set of the following pairs of equation i. 푥 + 푦 = 5 and 푦 − 푥 = 1 ii. 3푥푦 − 푥 = 0 and 푥 + 3푦 = 2

Solution We have, 푥 + 푦 = 5 … … … (1) and −푥 + 푦 = 1 … … … (2) From (2) 푦 = 푥 + 1 Put 푦 = 푥 + 1into (1). ⟹ 푥 + (푥 + 1) = 5 ⟹ 푥 + 푥 + 2푥 + 1 − 5 = 0 ⟹ 2푥 + 2푥 − 4 = 0 (Divide both sides by 2) ⟹ 푥 + 푥 − 2 = 0 ⟹ 푥 − 푥 + 2푥 − 2 = 0 ⟹ (푥 − 1)(푥 + 2) = 0 ⟹ 푥 − 1 = 0 or 푥 + 2 = 0 ⟹ 푥 = 1 or 푥 = −2 Put the values of 푥 into (2) When 푥 = 1, 푦 = 1 + 1 = 2. When 푥 = −2, 푦 = −2 + 1 = −1 Truth set: {푥, 푦: 푥 = 1,푦 = 2or푥 = −2, 푦 = −1} ii. We have, 3푥푦 − 푥 = 0 … … … (1) and 푥 + 3푦 = 2 … … … (2) From (2) 푥 = 2 − 3푦 Put 푥 = 2 − 3푦into (1). ⟹ 3(2 − 3푦)푦 − (2 − 3푦) = 0 ⟹ 6푦 − 9푦 − 2 + 3푦 = 0 ⟹ −9푦 + 9푦 − 2 = 0 ⟹ 9푦 − 9푦 + 2 = 0 (Divide both sides by 2) ⟹ 9푦 − 6푦 − 3푦 + 2 = 0 ⟹ (9푦 − 6푦) − (3푦 − 2) = 0 ⟹ 3푦(3푦 − 2) − (3푦 − 2) = 0 ⟹ (3푦 − 2)(3푦 − 1) = 0 ⟹ 3푦 − 2 = 0 or 3푦 − 1 = 0 ⟹ 푦 = or푦 = Put the values of 푦 into (2)

When 푦 = , 푥 = 2 − 3 = 2 − 2 = 0.

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When 푦 = , 푥 = 2 − 3 = 2 − 1 = 1

Truth set: 푥,푦: 푥 = 0,푦 = or푥 = 1, 푦 = Try: Solve for the values of 푥 and 푦 in the pair of equation: 푥푦 = 2 and 푥 + 푦 = 3

Example 9.21 Find the value of 푥 in the equation (log 푥) + log 푥 + 6 = 0

Solution We have, (log 푥) + log 푥 + 6 = 0 ⟹ (log 푥) + 5 log 푥 + 6 = 0 Let log 푥 = 푦 ⟹ 푦 + 5푦 + 6 = 0 ⟹ 푦 + 2푦+ 3푦+ 6 = 0 ⟹ (푦 + 2)(푦+ 3) = 0 ⟹ 푦 + 2 = 0 or 푦 + 3 = 0 ⟹ 푦 = −2, 푦 = −3 But log 푥 = 푦 ⟹ log 푥 = −2 ⟹ 푥 = 4 = (By definition)

Or log 푥 = −3 ⟹ 푥 = 4 =

∴ 푥 = or 푥 = Example 9.22

Solve for 푥 if 3 − 10 ∙ 3 + 9 = 0 Solution

We have, 3 − 10 ∙ 3 + 9 = 0 ⟹ (3 ) − 10 ∙ 3 + 9 = 0 Let 3 = 푦 ⟹ 푦 − 10푦 + 9 = 0 ⟹ 푦 − 9푦 − 푦 + 6 = 0 ⟹ (푦 − 9)(푦 − 1) = 0 ⟹ 푦 − 9 = 0 or 푦 − 1 = 0

∴ 푦 = 9, 푦 = 1 But 3 = 푦 ⟹ 3 = 9 ⟹ 3 = 3 ⟹ 푥 = 2 Or 3 = 푦 ⟹ 3 = 1 ⟹ 3 = 3 ⟹ 푥 = 0 ∴ 푥 = 2 or 푥 = 0

Example 9.23 Solve for 푥 if 2푥 − + 1 = 0

Solution We have, 2푥 − + 1 = 0 ⟹ 푥(2푥)− 푥 + 푥 = 0

⟹ 2푥 − 3 + 푥 = 0 ⟹ 2푥 + 푥 − 3 = 0 ⟹ 2푥 − 2푥 + 3푥 − 3 = 0 ⟹ 2푥(푥 − 1) + 3(푥 − 1) = 0 ⟹ (푥 − 1)(2푥 + 3) = 0

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⟹ 푥 − 1 = 0 or 2푥 + 3 = 0 ∴ 푥 = 1 or 푥 = − Example 9.24

The perimeter of a rectangle is 34cm and the area is 52cm . Find the length and width of the rectangle.

Solution The perimeter of a rectangle is defined as: 푃 = 2(푙 + 푤) ⟹ 34 = 2(푙 + 푤) ⟹ 17 = 푙 + 푤… … … (1) The area of a rectangle is defined as: 퐴 = 푙 × 푤 ⟹ 52 = 푙푤… … … (2). From (1) 푙 = 17 −푤. Put 푙 = 17 −푤 into (2) ⟹ 52 = (17 − 푤)푤 ⟹ 52 = 17푤 −푤 ⟹ 푤 − 17푤 + 52 = 0 ⟹ 푤 − 17푤 + 52 = 0 ⟹ (푤 − 13)(푤− 4) = 0 ⟹ 푤 − 13 = 0 or 푤 − 4 = 0 ∴ 푤 = 13 or 푤 = 4 Put the values of 푤 into (1) When푤 = 13, 푙 = 17 − 13 = 4 When 푤 = 4, 푙 = 17 − 4 = 13 ∴ 푙 = 13and 푤 = 4.

Example 9.25 Find the values of 푘 for which the equation푥 + 푘푥 + 푘 + 3 = 0 has equal real roots.

Solution We have, 푥 + 푘푥 + 푘 + 3 = 0. By comparing (1) with 푎푥 + 푏푥 + 푐 = 0, we get, 푎 = 1, 푏 = 푘and 푐 = 푘 + 3 For equal roots: 푏 − 4푎푐 = 0 ⟹ (푘) − 4(1)(푘 + 3) = 0 ⟹ 푘 − 4푘 − 12 = 4 ⟹ 푘 − 6푘 + 2푘 − 12 = 0 ⟹ (푘 − 6)(푘 + 2) = 0 ⟹ 푘 − 6 = 0 or 푘 + 2 = 0 ∴ 푘 = 6 or 푘 = −2 Some Selected Examples

Example 9.26 Solve for the equation using the method of completing 푥 − 6푥 + 9 = 0

Solution i. 푥 − 6푥 + 9 = 0 ⟹ 푥 − 6푥 = −9 ⟹ 푥 − 6푥 + (−3) = −1 + (−3) ⟹ (푥 − 3) = −1 + 9 ⟹ (푥 − 3) = 8 ⟹ (푥 − 3) = ±√8 ⟹ 푥 − 3 = ±√8 ⟹ 푥 = 3 ± 2√2 ∴ 푥 = 3 + 2√2 or 푥 = 3 − 2√2

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Example 9.27 Find the minimum value of the function 푦 = 3푥 − 3푥 − 2 and the value of 푥 for which it occurs.

Method 1 Let 푦 = 3푥 − 3푥 − 2

= 3 푥 − 푥 − Hint:factorise3outind 표푓 − 1 = −

= 3 푥 − − − − check:푥 − 푥 + − = 푥 −

= 3 푥 − − −

= 3 푥 − −

= 3 푥 − − 3 × [Multiplythroughby3]

= 3 푥 − −

The minimum value of the function is − . This occurs at 푥 − = 0

∴ 푥 = Example 9.28

If 푥 − (푘 + 1)푥 + 5푘 − 19 = 0 has equal roots, find the values of 푘. Solution

We have, 푥 − (푘 + 1)푥 + 5푘 − 19 = 0. By comparing (1) with 푎푥 + 푏푥 + 푐 = 0, we get, 푎 = 1, 푏 = −(푘 + 1)and 푐 = 5푘 − 19 For equal roots: 푏 − 4푎푐 = 0 ⟹ [−(푘 + 1)] − 4(1)(5푘 − 19) = 0

⟹ 푘 + 2푘 + 1 − 20푘 + 76 = 0 ⟹ 푘 − 18푘 + 77 = 0 ⟹ 푘 − 11푘 − 7푘 + 77 = 0 ⟹ (푘 − 11)(푘 − 7) = 0 ⟹ 푘 − 11 = 0 or 푘 − 7 = 0 ∴ 푘 = 11 or 푘 = 7

Example 9.29 The roots of the equation 2푥 − 5푥 + 1 are 훼 and 훽. Find i. 훼 + 훽 ii. + iii. + 1 + ( + 1) iv. 훼 − 훽

Solution

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We have, 2푥 − 5푥 + 1 = 0 … … … (1). By comparing (1) with 푎푥 + 푏푥 + 푐 = 0, we get, 푎 = 2, 푏 = −5and 푐 = 1 Let the roots of the equation be 훼 and 훽 Sum of roots= 훼 + 훽 = − = Product of roots = 훼훽 = =

i. 훼 + 훽 = (훼 + 훽) − 2훼훽 = − 2 = − 1 = ∴ 훼 + 훽 =

ii. + = = = × 2 = ∴, + =

iii. + 1 + + 1 = + + 1 + 1 = + 2 = + 2 = × 2 + 2 =

∴ + 1 + + 1 =

iv. 훼 − 훽 = (훼 + 훽)(훼 − 훽) = (훼 + 훽) (훼 + 훽) − 4훼훽

⟹ 훼 − 훽 = × − 4 = × − 2 = × = √17

∴ 훼 − 훽 = √17

Example 9.30 If 훼 and 훽 are the roots of the quadratic equation 푥 − 21푥 + 1 = 0 and 훼 and 훽 are positive, find i. 훼 + 훽 ii. 훼훽 iii. the equation with roots and .

Solution We have, 푥 − 21푥 + 1 = 0 … … … (1). The roots of the equation are 훼 and 훽 Sum of roots= 훼 + 훽 = (훼 + 훽) − 2훼훽 Product of roots = 훼 훽 = (훼훽) The equation whose roots are 훼 and 훽 is given as: 푥 − (푠푢푚표푓푟표표푡푠)푥 + 푝푟표푑푢푐푡표푓푟표표푡푠 = 0 ⟹ 푥 − (훼 + 훽 )푥 + (훼훽) = 0 … … … (2) By comparing 푥 − 21푥 + 1 = 0 with 푥 − (훼 + 훽 )푥 + (훼훽) = 0, we get 훼 + 훽 = 21 ⟹ (훼 + 훽) − 2훼훽 = 21 … … … (푎) and (훼훽) = 1 … … … (푏) ⟹ (훼훽) = √1 ∴ 훼훽 = 1. Since both 훼 and훽are positive. Put 훼훽 = 1into (a)

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⟹ (훼 + 훽) − 2훼훽 = 21 ⟹ (훼 + 훽) − 2(1) = 21 ⟹ (훼 + 훽) = 21 + 2 ⟹ (훼 + 훽) = 23 ⟹ (훼 + 훽) = √23 ∴ 훼 + 훽 = √23 i. 훼 + 훽 = √23 ii. 훼훽 = 1 iii. We want to find an equation whose roots are and

Sum of roots = + = =( )

= √ = √23

Product of roots = × = =( )

= = 1

The equation whose roots are and is given as:

푥 − (푠푢푚표푓푟표표푡푠)푥 + 푝푟표푑푢푐푡표푓푟표표푡푠 = 0 ⟹ 푥 − √23푥 + 1 = 0 ∴, the equation: 푥 − √23푥 + 1 = 0

Example 9.31 If 훼 + 훽 = 5 and 훼훽 = 2, calculate + and hence determine the values of 푚 and 푛

such that the equation 푥 + 푚푥 + 푛 = 0 has roots and .

Solution We have, 훼 + 훽 = 5,훼훽 = 2 and 푥 + 푚푥 + 푛 = 0 … … … (1) Now, + = =

We want to find an equation whose roots are and

Sum of roots = + = = Product of roots = × = =

The equation whose roots are and is given as:

푥 − (푠푢푚표푓푟표표푡푠)푥 + 푝푟표푑푢푐푡표푓푟표표푡푠 = 0 ⟹ 푥 − 푥 + = 0 … … … (2)

By comparing 푥 + 푚푥 + 푛 = 0 with 푥 − 푥 + = 0

푚 = − and 푛 = Example 9.32

If 푝 and 푞 are the roots of the equation 2푥 + 3푥 + 푏 = 0, find the value of 푏 when 푝 − 푞 = 4.

169

Solution We have,2푥 + 3푥 + 푏 = 0 ⟹ 푥 + 푥 + = 0 … … … (1) and 푝 − 푞 = 4 … … … (2) We want to find an equation whose roots are 푝 and 푞 Sum of roots = 푝 + 푞 Product of roots = 푝푞 = The equation whose roots are 푝 and 푞 is given as: 푥 − (푠푢푚표푓푟표표푡푠)푥 + 푝푟표푑푢푐푡표푓푟표표푡푠 = 0

⟹ 푥 − (푝 + 푞)푥 + 푝푞 = 0 … … … (2) By comparing 푥 + 푥 + = 0 with 푥 − (푝+ 푞)푥 + 푝푞 = 0

푝 + 푞 = − … … … (3) and 푝푞 = … … … (4) Solve (2) and (3) Add (2) to (3) But푝 + 푝 + 푞 − 푞 = 4 − ⟹ 2푝 = ⟹ 푝 = × =

Put 푝 = into (3). ⟹ + 푞 = − ⟹ 푞 = − − = −4

Put 푝 = , 푞 = −4into (4). ⟹ (−4) = ⟹ −5 = ⟹ 푏 = −5 × 2 = −10 ∴ 푏 = −10.

Example 9.33 Given that the roots of the quadratic equation 3푥 − 2푥 + 1 = 0 are 훼 and 훽. Evaluate i. 훼 + 훽 ii 훼 − 훽 iii. Find the equation with roots 훼 and 훽 .

Solution We have, 3푥 − 2푥 + 1 = 0 ⟹ 푥 − 푥 + … … … (1). The roots of the equation are 훼 and 훽 Sum of roots= 훼 + 훽 Product of roots = 훼훽 The equation whose roots are 훼 and 훽 is given as: 푥 − (푠푢푚표푓푟표표푡푠)푥 + 푝푟표푑푢푐푡표푓푟표표푡푠 = 0 ⟹ 푥 − (훼 + 훽)푥 + 훼훽 = 0 … … … (2) By comparing 푥 − 푥 + = 0 with 푥 − (훼 + 훽)푥 + 훼훽 = 0,

we get 훼 + 훽 = − and 훼훽 =

i. 훼 + 훽 = (훼 + 훽) − 2훼훽 = − − 2 − = + = ii. We want to find an equation whose roots are 훼 and 훽 Sum of roots = 훼 + 훽 = (훼 + 훽) − 3훼훽(훼 + 훽)

170

⟹ Sum of roots= − − 3 − = − + =

Product of roots = 훼 × 훽 = (훼훽) = =

The equation whose roots are 훼 and 훽 is given as: 푥 − (푠푢푚표푓푟표표푡푠)푥 + 푝푟표푑푢푐푡표푓푟표표푡푠 = 0 ⟹ 푥 − 푥 + = 0

27푥 − 27 × 푥 + 27 × = 0 ⟹ 27푥 − 10푥 + 1 = 0 ∴, the equation: 27푥 − 10푥 + 1 = 0

Final Exercises

1. Factorise the following equations: i. 푥 − 3 = 0 ii. 2푥 − 푥 − 3 = 0 iii. 푥 + 푥 = 0 iv. 3푥 − 7푥 − 6 = 0 v. 10푥 + 11푥 + 1 = 0 vi. 3푥 − 13푥 + 10 = 0 vii. 1 − 3푥 − 2푥 viii. 16 − 15푥 − 푥 = 0 ix. 푥 + 3푥 + 1 = 0 2. Complete the squares of the following equations: i. 푥 + 2푥 − 3 = 0 ii. 2푥 + 푥 − 3 = 0 iii. 2푥 + 푥 = 0 iv. 2푥 + 7푥 − 6 = 0 v. 3푥 + 2푥 − 1 = 0 vi. 3푥 + 3푥 − 1 = 0 vii. 1 + 3푥 − 2푥 = 0 viii. 2 − 15푥 − 2푥 = 0 3. Use the quadratic formula to solve the following equations: i. 푥 − 4푥 − 3 = 0 ii. 5푦 + 푦 − 1 = 0 iii. 2푧 + 푧 − 10 = 0 iv. 3푥 + 4푥 − 6 = 0 v. 3푥 − 3푥 − 1 = 0 vi. 4푥 + 3푥 − 1 = 0 vii. 2 + 3푥 − 6푥 = 0 viii. 2 − 3푥 − 7푥 = 0 4. Solve the following equations: i. 푥 − 16 = 0 ii. 푥 + 3푥 − 16 = 0 iii. 푥 − 20푥 + 9 = 0 iv. 푥 + 푥 − 10 = 0 v. 2푥 − 3푥 + 1 v. 푥(푥 − 4) = 푥(2푥 + 5) + 10 5. Find the maximum or minimum value of the following functions and value of 푥 for which it occurs. i. 푓(푥) = −푥 − 3푥 − 3 ii. 푓(푥) = 6푥 + 12푥 − 15 iii. 푓(푥) = 푥 − 6푥 + 8 iv. 푓(푥) = 푥 + 4 v. 푓(푥) = 2푥 + 8푥 + 1 vi. 푓(푥) = 푥 − 3 vii. 푓(푥) = 3 − 푥 − 2푥 6. Find the maximum or minimum value of the following functions and value of 푥 for which it occurs.

171

i. 푓(푥) = 푥 + 2푥 + 3 ii. 푓(푥) = 6 − 푥 − 3푥 iii. 푓(푥) = 5푥 − 6푥 + 1 iv. 푔(푥) = 4푥 − 푥 v. 푔(푥) = 푥 − 푥 − vi. 푔(푥) = 6 − 2푥 − 푥

vii. 푓(푥) = √2푥 − 2√3푥 − 2 7. Write the following expression in the form 푎(푥 + 푏) + 푐, where 푎, 푏 and 푐 are constants. i. 푥 − 3푥 − 3 ii. 푥 − 6푥 + 8 iii. 푥 + 4푥 iv. 2푥 + 8푥 + 1 v. 푥 − 3 vi. 3 − 4푥 − 3푥 8. Write the following expression in the form 푝(푥 + 푞) + 푟, where 푝, 푞 and 푟 are constants. i. 푥 − 2푥 + 1 ii. 2푥 − 푥 iii. 5푥 − 6푥 + 8 iv. 2푥 + 5푥 v. 2푥 − 3푥 + 4 vi. (2 − 푥)(푥 + 3) vii. 푥 − √2 푥 + √3 9. Use the discriminant to determine whether he following equations will have different real roots, equal roots or complex roots. i. 4푥 + 2푥 − 1 = 0 ii. 3푥 − 2푥 − 21 = 0 iii. 푥 + √2푥 − 12 = 0 iv. 2푥 − 4푥 − 5 = 0 v. 6푥 + 2푥 − 2 = 0 vi. 푥 + 3푥 − 1 = 0 vii. 1 + 3푥 − 2푥 = 0 viii. 푥 − − + 푥 = 0 ix. (2푥 − 1) = 18

10. Solve the equation: i. + − = 0 ii. (log 푥) − 2 log 푥 = 3 iii. − = 9 11. A quadratic relation is defined as 푦 = 푎푥 − 푏푥 + 2. If the points (2, 3) and (−2, 1) lie on the curve of the relation, find the values of 푎 and 푏. Hence, determine the zeroes of the relation. 12. Find the quadratic equation whose roots are: i. 3 and 5 ii. 6 and −3 iii. 4 and −7 iv. √3 + 2 and √3 − 2 13 Find the constants 푎, 푏 and 푐 such that 2푥 − 6푥 + 5 ≡ 푎(푥 + 푏) + 푐. 14. Find the constants 푝,푞 and 푟 such that 3푥 − 2푥 − 2 ≡ 푝(푥 + 푞) + 푟 15. The roots of the equation 2푥 − 3푥 + 1 are 훼 and 훽. Find i. 훼 + 훽 ii. + iii. + 1 + ( + 1) iv. 훼 − 훽 v. 훼 − 훽

16. If 훼 and 훽 are the roots of the quadratic equation 푥 − 11푥 + 1 = 0 and 훼 and 훽 are positive, find i. 훼 + 훽 ii. 훼훽 iii. the equation with roots and .

17. Write the function 푦 = 3 − 4푥 − 2푥 in the form 푦 = 푒(푥 + 푓) + 푔.

172

18. Express 푦 = 2 − 3푥 − 5푥 in the form 푦 = 퐴 − 퐵(푥 + 퐶) , where A, B and C are constants. Hence state the maximum value of the function 푓:푥 → 2 − 3푥 − 5푥 . 19. Express 2푥 − 6푥 + 9 in the form 푎(푥 − 푏) + 푐, where 푎,푏 and 푐 are integers. Hence state the minimum value of 2푥 − 6푥 + 9 and the value of 푥 for which it occurs. 20. Express 13 − 8푥 − 24푥 in the form 푎 − 푏(푥 + 푐) . State the greatest value of the expression. 21. If 푥 − 푥 − 5 ≡ 퐴(푥 + 퐵) + 퐶, find the values of 퐴,퐵 and C. Hence, find the maximum value of 푥 − 푥 − 5. 22. By writing 10푥 − 푥 + 1 in the form 푝(푥 + 푚) + 푛,find the values of constants 푝,푚and 푛. Hence find the minimum value of 푦 = 10푥 − 푥 + 1. 23. Given that the roots of the quadratic equation 3푥 − 2푥 − 1 = 0 are 훼 and 훽. a. Evaluate i. 훼 + 훽 ii 훼 − 훽 iii. 훼 − 훽 b. Find the equation with roots 훼 and 훽 . 24. Find the truth set of the following pairs of equation i. 푥 + 푦 = 8 and 푥 − 푦 = −2 ii. 3푥푦 = 2 and 3푥 + 푦 = −1 25. If 훼 and 훽 are the roots of the equation 푥 − 3푥 − 2 = 0, find the quadratic equation whose roots are and .

26. Let 훼 and 훽 are the roots of the quadratic equation 푥 + 2푥 − 6 = 0. without solving the equation, i. evaluate 훼 + 훽 ii. evaluate (훼 − 훽) ii. 훼 + 훽 iii. obtain a quadratic equation whose roots are 훼 + 2,훽 +2 27. If 훼 and 훽 are the roots of the quadratic equation 2푥 − 6푥 + 2 = 0, evaluate + 1.

30. If 훼 and 훽 are the roots of the equation 4푥 + 7푥 + 3 = 0, find the value of i. + ii. 훼 + 훽 iii. 훼 − 훽.

31. The roots of the equation 푥 + 5푥 + 푝 = 0 are 훼 and 훼 + 1. Find the value of 푝. 32. If 훼 and 훽 are the roots of the equation 2푥 − 푥 − 8 = 0, find the equation whose roots are and .

33. The roots of the quadratic equation 2푥 − 푥 − 4 = 0 are 훼 and 훽. Find the equation whose roots are 훼 and 훽 .

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34. The roots of the quadratic equation 푥 + 5푥 − 2 = 0 are 훼 and 훽. Find the equation whose roots are 훼 and 훽 . 35. If 훼 and 훽 are the roots of the quadratic equation 2푥 − 13푥 + 7 = 0, find a quadratic equation whose roots are and .

36. Find The values of 푘 for which the quadratic equation 푥 + (푘 + 3)푥 − 15 = 0 has equal roots. 37. Find the values of 푝 for which the quadratic equation 푝푥 − 3(푝 + 1)푥 − 12 = 0 has distinct real roots. 38. Find the solution of the following quadratic equations i. 푚 + 3푚 − 18 = 0 ii. 2푥 + 5푥 − 12 = 0 (CCE) 39. Complete the square for the following expressions i. 푥 + 3푥 ii. 2푥 − 푥 (CCE) 40. Find the range of the function 푓(푥) = 푥 + 6푥 − 1 for the domain of 푥 ∈ ℝ. (CCE) 41. What is the maximum or minimum value of 3푥 − 2푥 + 1 and for what value of 푥 does it occur? (CCE) 42. Factorise completely: 푥 + 11푥 − 12 (CCE)

RATIONAL FUNCTIONS

A function of the form 푝(푥) = ( )( )

is a rational function in terms of 푥.

Types Of Rational Functions 1. Improper rational function

174

The rational function 푝(푥) = ( )( )

is improper if the degree of 푓(푥) is greater than that

of 푔(푥). For instance, 푝(푥) = is improper because the degree of the numerator is 2 while the degree of the denominator is 1. 2. Proper rational function The rational function 푝(푥) = ( )

( ) is proper if the degree of 푓(푥) is lesser than that of

푔(푥).For instance, 푝(푥) = Multiplication of rational functions

Note that for rational numbers × = . We will apply this rule to rational functions. Example 10.1

Multiply the following functions: i. 푓(푥) = , 푥 ≠ 2 and 푔(푥) = ,푥 ≠ 0 ii. 푝(푥) = and ℎ(푥) =

Solution i. We have, 푓(푥) = and 푔(푥) =

푓(푥) ∙ 푔(푥) = ∙ = ∴ 푓(푥) ∙ 푔(푥) =

ii. We have, 푝(푥) = and ℎ(푥) =

푝(푥) ∙ ℎ(푥) = ∙ = =

∴ 푝(푥) ∙ ℎ(푥) = Try:

Given that 푓(푥) = , 푥 ≠ 3 and 푔(푥) = , 푥 ≠ −1 , find 푓(푥) ∙ 푔(푥). Division of rational functions

Note that for rational numbers ÷ = . We will apply the same rule here.

Example 10.2 Simplify the following

i. ÷ ii. ÷ iii. ÷ Solution

i. ÷ = ∙ = ∙ =( )( )

∙ =( )

175

ii. ÷ = ∙ =( )

∙ =( )

iii. ÷ = ∙ = ( )( )( )

∙ ( )( )( )

= ( )( )( )( )

Try:

If 푓(푥) = ,푥 ≠ 49 and 푔(푥) = ,푥 ≠ 7 , find ( )

( ).

Addition and subtraction of rational functions Note that for rational functions + = and − = . We will apply these rules here.

Example 10.3 Simplify the following: i. + ii.

( )− iii.

( )− iv. −

Solution i. + =

( )( )+ =

( )( )=

( )( )=

( )

ii. ( )

− = ( )( )

=( )

iii. ( )

− = ( ) ( )( )( )( )

=( )( )

=( )( )

iv. − =( )

− = ( )( )

=( )

=( )

Try: Simplify the following: i.

( )+

( ) ii. + iii. −

Zeroes of a rational function

Let 푝(푥) be a rational function. The zeros of 푝(푥) is known if 푝(푥) is equated to 0. Example 10.4

Solve for the zeros of the following rational functions:

i. 푓(푥) = ( )( )( )

ii. 푔(푥) = iii. 푝(푥) = iv. ℎ(푥) =

Solution

i. We have, 푓(푥) =( )( )

. Set 푓(푥)to zero.

⟹ ( )( )

= 0 ⟹ 2푥(3 − 푥 ) = 0 ⟹ 2푥 = 0or (3 − 푥 ) = 0

176

⟹ 푥 = 0or 푥 = 3 ⟹ 푥 = ±√3 ∴ 푥 = √3, 0or−√3

ii. We have, 푔(푥) = . Set 푔(푥)to zero.

⟹ = 0 ⟹ 푥 + 푥 − 6 = 0 ⟹ (푥 − 2)(푥 + 3) = 0 ⟹ 푥 − 2 = 0or 푥 + 3 = 0 ∴ 푥 = −2or 푥 = −3

iii. We have, 푝(푥) = . Set 푝(푥)to zero.

⟹ = 0 ⟹ 6푥 − 11푥 + 4 = 0 ⟹ 6푥 − 3푥 − 8푥 + 4 = 0 ⟹ 3푥(2푥 − 1) − 4(2푥 − 1) = 0 ⟹ (2푥 − 1)(3푥 − 4) = 0 ⟹ 2푥 − 1 = 0or 3푥 − 4 = 0 ∴ 푥 = or 푥 =

iv. We have, ℎ(푥) = . Set ℎ(푥)to zero.

⟹ = 0 ⟹ 푥 + 푥 − 6푥 = 0 ⟹ 푥(푥 + 푥 − 6) = 0 ⟹ 푥(푥 − 2)(푥 + 3) = 0 ⟹ 푥 = 0or 푥 − 2 = 0or 푥 + 3 = 0 ∴ 푥 = 0or푥 = 2 or 푥 = −3

Identity functions Let 푝(푥) and 푓(푥) be two functions. If 푝(푥) is identical to 푓(푥), then all values of 푥 that satisfies 푝(푥) also satisfies 푓(푥). Symbolically we write

푝(푥) ≡ 푓(푥) to mean ′푝(푥) is identical to푓(푥). ′ For instance, 푥 − 9 ≡ (푥 − 3)(푥 + 3) also (푥 + 2) = 푥 + 4푥 + 4 But 푥 + 푎 ≢ 푥(푥 + 푎), where 푎 is constant. Methods of resolving identical functions 1. Covering up rule Under this method, we determine the value of a parameter by substituting a 푟푒푎푙 number that will make other parameter(s)0. For instance, if 푥 − 푑 ≡ 퐴(푥 − 푎) + 퐵(푥 − 푏), then to determine the value 퐴, we put in 푥 = 푏. 2. Expanding and Comparing Co-efficient Here, we expand both sides of the identity and compare co-efficient.

Example 10.5 Find the values of 푃, 푄 and 푅 for which 9푥 − 3 ≡ 푃(푥 + 2) + 푄(푥 − 1) + 푅(푥 + 4) using covering up rule.

177

Solution We have the identity, 9푥 − 3 ≡ 푃(푥 + 2) + 푄(푥 − 1) + 푅(푥 + 4) Put 푥 = 1 into the identity ⟹ 9(1)− 3 ≡ 푃(1 + 2) + 푄(1 − 1) + 푅(1 + 4) ⟹ 6 = 3푃 + 5푅… … … (1) Put 푥 = −1 into the identity ⟹ 9(−1) − 3 ≡ 푃(−1 + 2) + 푄((−1) − 1) + 푅(−1 + 4) ⟹−12 = 푃 + 3푅… … … (2) Solve equations (1) and (2) From (1) 푃 = −12 − 3푅. Put 푃 = −12 − 3푅 into (1) ⟹ 6 = 3(−12 − 3푅) + 5푅 ⟹ 6 = −36 − 9푅 + 5푅 ∴ 푅 = − = −

Put 푅 = 42 into (1) ∴ 푃 = −12 − 3 − = =

Put 푥 = −4 into the identity ⟹ 9(−4) − 3 ≡ 푃(−4 + 2) + 푄((−4) − 1) + 푅(−4 + 4) ⟹ −39 = −2푃 + 15푄… … … (3)

Put 푃 = into (3). ⟹ −39 = −2 + 15푄 ∴ 푄 = = 0

Example 10.6 Given that 2푥 + 푥 ≡ 푀(푥 + 1) + 푁(2푥 − 1) + 푅(3 − 푥), find the values of 푀,푁 and 푅 by comparing co-efficient. We have the identity, 2푥 + 푥 ≡ 푀(푥 + 1) + 푁(2푥 − 1) + 푅(3− 푥) ≡ 푀푥 + 푀 + 2푁푥 −푁 + 3푅 − 푅푥 ≡ 푀푥 + 2푁푥 − 푅푥 + 푀 − 푁 + 3푅 ≡ 푀푥 + (2푁 − 푅)푥 + 푀 −푁 + 3푅 By comparing coefficient of the identity we get 푀 = 2 2푁 − 푅 = 1 … … … (1) 푀−푁 − 3푅 = 0 … … … (2) Put 푀 = 2into (2) ⟹ 2 −푁 − 3푅 = 0 푁 + 3푅 = 2 … … … (2 ∗) From (2 ∗)푁 = 2 − 3푅 Put 푁 = 2 − 3푅 into (1) ⟹ 2(2 − 3푅) − 푅 = 1 ⟹ 4 − 6푅 − 푅 = 1 ∴ 푅 =

Put 푅 = into (2 ∗) ∴ 푁 = 2 − 3 = = Try: Solve for the values of 퐴,퐵 and 퐶 if i. 푥 + 1 ≡ (퐴푥 + 퐵)(푥 − 1) + 퐶(푥 − 1)

178

ii. 5푥 + 4 ≡ 퐴(푥 + 2) + 퐵(푥 − 1)(푥 + 2) + 퐶(푥 − 1)

Partial Fractions Let 푝(푥) and 푓(푥) be two rational functions. The sum of 푝(푥) and 푓(푥) is another rational function 푝(푥) + 푓(푥).the process of separating the function 푝(푥) + 푓(푥) into the main functions is referred to as 푟푒푠표푙푣푖푛푔푖푛푡표푝푎푟푡푖푎푙푓푟푎푐푡푖표푛푠. Note: To resolve a given rational function into partial fractions a. check whether or not the denominator can be factorised. b. check whether or not the function is improper or proper. Method of Resolving a Rational Function into Partial Fractions Let 푓(푥) = ( )

( ), where 푔(푥) and ℎ(푥) are polynomials in 푥. Then we can resolve 푓(푥)

into partial fraction through one of the following cases: Case 1 (denominator with linear factors) If degree of ℎ(푥) is greater than that of 푔(푥) and ℎ(푥) has non-repeated linear factors, for example, (푥 − 푎), (푥 − 푏) etc, then we can write 푓(푥) as 푓(푥) = ( )

( )( )≡

( )+

( ), for two linear factors and 퐴 and 퐵 are parameters.

Example 10.7 Express the following into partial fractions: i.

( )( ) ii.

( )( ) iii. iv. v.

Solution i. We have the function,

( )( )

Let ( )( )

≡( )

+( )

≡ ( ) ( )( )( )

⟹ 3푥 − 4 ≡ 퐴(3푥 + 1) + 퐵(푥 − 2) Put 푥 = 2 into the above identity. ⟹ 3(2)− 4 ≡ 퐴(3 ∙ 2 + 1) + 퐵(2 − 2) ⟹ 2 = 7퐴 ∴ 퐴 =

Put 푥 = − into the above identity.

⟹ 3 − − 4 ≡ 퐴 3 ∙ − + 1 + 퐵 − − 2 ⟹ −1 = − 퐵 ∴ 퐵 =

∴ ( )( )

≡(

+( )

≡( )

+( )

179

ii. We have the function, ( )( )

Let ( )( )

≡( )

+( )

≡ ( ) ( )( )( )

⟹ 6푥 + 1 ≡ 퐴(푥 − 1) + 퐵(2푥 − 1) Put 푥 = into the above identity.

⟹ 6 + 1 ≡ 퐴 − 1 + 퐵 2 ∙ − 1 ⟹ 4 = − 퐴 ∴ 퐴 = −8

Put 푥 = 1 into the above identity. ⟹ 6(1) + 1 ≡ 퐴(1 − 1) + 퐵(2 ∙ 1 − 1) ⟹ 7 = 퐵 ∴ 퐵 = 7 ∴

( )( )≡

( )+

( )≡

( )−

( )

iii. We have the function,

Let ≡( )( )

≡( )

+( )

≡ ( ) ( )( )( )

⟹ 푥 + 2 ≡ 퐴(푥 + 4) + 퐵(푥 − 4) Put 푥 = 4 into the above identity. ⟹ 4 + 2 ≡ 퐴(4 + 4) + 퐵(4 − 4) ⟹ 6 = 8퐴 ∴ 퐴 = = Put 푥 = −4 into the above identity. ⟹ −4 + 2 ≡ 퐴(−4 + 4) + 퐵(−4 − 4) ⟹ −2 = −8퐵 ∴ 퐵 = =

∴ ≡( )( )

≡( )

+( )

≡( )

+( )

iv. We have the function,

Let ≡( )

≡ +( )

≡ ( )( )

⟹ 7 ≡ 퐴(3푥 + 2) + 퐵푥 Put 푥 = 0 into the above identity. ⟹ 7 ≡ 퐴(3(0) + 2) + 퐵(0) ⟹ 7 = 2퐴 ∴ 퐴 =

Put 푥 = − into the above identity.

⟹ 7 ≡ 퐴 3 ∙ − + 2 + 퐵 − ⟹ 7 = − 퐵 ∴ 퐵 = −

∴ ≡( )

≡ +( )

≡ −( )

180

v. We have the function,

Let ≡ ≡( )( )

≡( )

+( )

≡ ( ) ( )( )( )

⟹ 푥 − 1 ≡ 퐴(2푥 − 5) + 퐵(푥 − 2) Put 푥 = 2 into the above identity. ⟹ 2 − 1 ≡ 퐴(2(2)− 5) + 퐵(2 − 2) ⟹ 1 = −퐴 ∴ 퐴 = −1 Put 푥 = into the above identity.

⟹ − 1 ≡ 퐴 2 − 5 + 퐵 − 2 ⟹ = 퐵 ∴ 퐵 = 3

∴ ≡( )( )

≡( )

+( )

≡( )

−( )

Try: a. Express the following in partial fractions:

i. ( )( )

ii. ( )( )

iii. ( )( )

iv. vi. ( )( )

v. ( )( )

b. Express ( )( )

in partial fractions. (SSSCE)

c. Find the value of 푦 for which the expression 푦 =( )( )

is undefined. Express 푦 in

partial functions. (SSSCE) d. Given that

( )( )≡

( )+

( ), find the value of 퐴.

e. Express in partial fractions.(SSSCE 2000 Question 3)

f. Express in partial fractions. (SSSCE)

g. Express in partial fractions. (WASSCE)

h. Express in partial fractions (WASSCE) Case 2 (denominator with repeated factors) If degree of ℎ(푥) is greater than that of 푔(푥) and ℎ(푥) has repeated linear factors, for example (푥 − 푎) , (푥 − 푏) etc, then we can write 푓(푥) as 푓(푥) = ( )

( ) ( )≡

( )+

( )+

( )+

( ), where 퐴,퐵,퐶and 퐷 are parameters.

Example 10.7

181

Express the following in partial fractions:

i. ( )

ii. ( )( )

iii. ( )( )

iv. ( )

Solution i. We have the function,

( )

Let ( )

≡ + +( )

≡ ( ) ( )( )

⟹ 11푥 − 1 ≡ 퐴푥(푥 + 1) + 퐵(푥 + 1) + 퐶푥 Put 푥 = −1 into the above identity. ⟹ 11(−1)− 1 ≡ 퐴(−1)(−1 + 1) + 퐵(−1 + 1) + 퐶(−1) ∴ −12 = 퐶 Put 푥 = 0 into the above identity. ⟹ 11(0)− 1 ≡ 퐴(0)(0 + 1) + 퐵(0 + 1) + 퐶(0) ∴ −1 = 퐵 Put 푥 = 1 into the above identity. ⟹ 11(1)− 1 ≡ 퐴(1)(1 + 1) + 퐵(1 + 1) + 퐶(1) ⟹ 10 = 2퐴+ 2퐵 + 퐶 ⟹ 10 = 2퐴+ 2(−1)− 12 ⟹ 2퐴 = 10 + 2 + 12 ∴ 퐴 = = 12 ∴

( )≡ + +

( )≡ − −

( )

ii. We have the function, ( )( )

Let ( )( )

≡( )

+( )

+( )

≡ ( ) ( )( ) ( )( )( )

⟹ 푥 + 3푥 ≡ 퐴(푥 + 1) + 퐵(푥 + 2)(푥 + 1) + 퐶(푥 + 2) Put 푥 = −2 into the above identity. ⟹ (−2) + 3(−2) ≡ 퐴(−2 + 1) + 퐵(−2 + 2)(−2 + 1) + 퐶(−2 + 2) ∴ −2 = 퐴 Put 푥 = −1 into the above identity. ⟹ (−1) + 3(−1) ≡ 퐴(−1 + 1) + 퐵(−1 + 2)(−1 + 1) + 퐶(−1 + 2) ∴ −2 = 퐶 Put 푥 = 0 into the above identity. ⟹ (0) + 3(0) ≡ 퐴(0 + 1) + 퐵(0 + 2)(0 + 1) + 퐶(0 + 2) ⟹ 0 = 퐴 + 2퐵 + 2퐶 ⟹ 0 = −2 + 2퐵 + 2(−2) ⟹ 2퐵 = 2 + 4 ∴ 퐵 = = 2

182

∴ ( )( )

≡( )

+( )

+( )

≡( )

−( )

−( )

iii. We have the function,

( )( )

Let ( )( )

≡( )

+( )

+( )

≡ ( ) ( )( ) ( )( )( )

⟹ 11푥 + 2 ≡ 퐴(푥 − 3) + 퐵(푥 + 2)(푥 − 3) + 퐶(푥 + 2) Put 푥 = −2 into the above identity. ⟹ 11(−2) + 2 ≡ 퐴(−2 − 3) + 퐵(−2 + 2)(−2 − 3) + 퐶(−2 + 2) ⟹−22 + 2 = 25퐴 ∴ 퐴 = − = − Put 푥 = 3 into the above identity. ⟹ 11(3) + 2 ≡ 퐴(3 − 3) + 퐵(3 + 2)(3 − 3) + 퐶(3 + 2) ⟹ 33 + 2 = 5퐶 ∴ 퐶 = = 7 Put 푥 = 0 into the above identity. ⟹ 11(0) + 2 ≡ 퐴(0 − 3) + 퐵(0 + 2)(0 − 3) + 퐶(0 + 2)

⟹ 2 = 9퐴 − 6퐵 + 2퐶 ⟹ 2 = 9 − + 2퐵 + 2(7)

⟹ 2퐵 = 2 + − 14 ∴ 퐵 = =

∴ ( )( )

≡( )

+( )

+( )

≡( )

−( )

+( )

iv. We have the function, ( )

Let ( )

≡( )

+( )

+( )

≡ ( ) ( )( )

⟹ 3푥 + 16푥 + 15 ≡ 퐴(푥 + 3) + 퐵(푥 + 3) + 퐶 Put 푥 = −3 into the above identity. ⟹ 3(−3) + 16(−3) + 15 ≡ 퐴(−3 + 3) + 퐵(−3 + 3) + 퐶 ∴ 퐶 = 27 − 48 + 15 = −6 Put 푥 = 0 into the above identity. ⟹ 3(0) + 16(0) + 15 ≡ 퐴(0 + 3) + 퐵(0 + 3) + 퐶 ⟹ 15 = 9퐴 + 3퐵 + 퐶 ⟹ 15 = 9퐴 + 3퐵 − 6 ⟹ 9퐴 + 3퐵 = 21 … … … (1)

183

Put 푥 = −1 into the above identity. ⟹ 3(−1) + 16(−1) + 15 ≡ 퐴(−1 + 3) + 퐵(−1 + 3) + 퐶 ∴ 3 − 16 + 15 = 4퐴+ 2퐵 + 퐶 ⟹ 4 = 4퐴 + 2퐵 − 6 ⟹ 4퐴 + 2퐵 = 10 … … … (2) Solve equations (1) and (2) From (2) 퐵 = Put 퐵 = into (1)

⟹ 9퐴 + 3 = 21 ⟹ 18퐴 + 30 − 12퐴 = 21 ⟹ 6퐴 = 21− 30

∴ 퐴 = − = − Put 퐴 = − into (2) ∴ 퐵 = = =

∴ ( )

≡( )

+( )

+( )

≡( )

−( )

−( )

Example 10.8

If ( )( )

≡( )

+( )

+( )

, find (퐴 + 퐵 − 퐶)

Solution

We have, ( )( )

≡( )

+( )

+( )

Let ( )( )

≡( )

+( )

+( )

≡ ( ) ( )( ) ( )( )( )

⟹ 3푥 − 푥 − 1 ≡ 퐴(푥 + 1) + 퐵(푥 − 2)(푥 + 1) + 퐶(푥 − 2) Put 푥 = 2 into the above identity. ⟹ 3(2) − 2 − 1 ≡ 퐴(2 + 1) + 퐵(2 − 2)(2 + 1) + 퐶(2 − 2) ⟹ 12 − 3 = 9퐴 ∴ 퐴 = = 1 Put 푥 = −1 into the above identity. ⟹ 3(−1) − (−1) − 1 ≡ 퐴(−1 + 1) + 퐵(−1 − 2)(−1 + 1) + 퐶(−1 − 2)

⟹ 3 + 1 − 1 = −3퐶 ∴ 퐶 = − = −1 Put 푥 = 0 into the above identity. ⟹ 3(0) − 0 − 1 ≡ 퐴(0 + 1) + 퐵(0 − 2)(0 + 1) + 퐶(0 − 2) ⟹ −1 = 퐴 − 2퐵 − 퐶 ⟹ −1 = 1 − 2퐵 − 2(−1) ⟹ 2퐵 = 1 + 2 + 1 ∴ 퐵 = = 2 ∴ 퐴 + 퐵 − 퐶 = 1 + 2 − (−1) = 4

184

Try: a. Express the following into partial fractions:

i. ( )( )

ii. ( ) ( )

iii. ( )( )

iv. ( ) ( )

b. Express ( )

in partial fraction. (SSSCE)

Case 3 (denominator with quadratic factors) If the degree of ℎ(푥) is greater than that of 푔(푥) and ℎ(푥) has quadratic factors which cannot be factorised, then we can write 푓(푥) as 푓(푥) = ( )

( )( )≡

( )+

( ) , for two factors and

퐴,퐵,퐶and 퐷 are constants. Example 10.9 Express the following in partial fractions:

i.( )

ii. ( )( )

iii. ( )( )

iv. ( )( )

Solution i. We have the function,

( )

Let ( )

≡ +( )

≡ ( )( )

⟹ 2푥 + 1 ≡ 퐴(푥 + 1) + (퐵푥 + 퐶)푥 ≡ 퐴푥 + 퐴 + 퐵푥 + 퐶푥 ≡ (퐴 + 퐵)푥 + 퐶푥 + 퐴 By comparing the LHS with the RHS we get 퐴 + 퐵 = 0 … … … (1) 퐶 = 2 퐴 = 1 Put 퐴 = 1 into (1) ⟹ 1 + 퐵 = 0 ∴ 퐵 = −1

OR 2푥 + 1

푥(푥 + 1) ≡퐴푥 +

퐵푥 + 퐶(푥 + 1) ≡

퐴(푥 + 1) + (퐵푥 + 퐶)푥푥(푥 + 1)

⟹ 2푥 + 1 ≡ 퐴(푥 + 1) + (퐵푥 + 퐶)푥 Put 푥 = 0 into the identity above ⟹ 2(0) + 1 ≡ 퐴(0 + 1) + (퐵(0) + 퐶)(0) ∴ 1 = 퐴 Put 푥 = 1 into the identity above

185

⟹ 2(1) + 1 ≡ 퐴(1 + 1) + (퐵(1) + 퐶)(1) ⟹ 2 + 1 ≡ 2퐴 + 퐵 + 퐶 3 = 2(1) + 퐵 + 퐶 Since 퐴 = 1 ⟹ 퐵 + 퐶 = 3 − 2 ⟹ 퐵 + 퐶 = 1 … … … (1) Put 푥 = −1 into the identity above ⟹ 2(−1) + 1 ≡ 퐴((−1) + 1) + (퐵(−1) + 퐶)(−1) ⟹ −2 + 1 ≡ 2퐴 + 퐵 − 퐶 −1 = 2(1) + 퐵 − 퐶 Since 퐴 = 1 ⟹ 퐵 − 퐶 = −1 − 2 ⟹ 퐵 − 퐶 = −3 … … … (2) Now, solve equations (1) and (2) Add (2) to (1) ⟹ 퐵 + 퐵 + 퐶 − 퐶 = 1 − 3 ⟹ 2퐵 = −2 ∴ 퐵 = −1 Put 퐵 = −1into (1) ⟹ −1 + 퐶 = 1 ∴ 퐶 = 1 + 1 = 2 ∴

( )≡ +

( )≡ +

( )

iii. We have the function,

( )( )

Let ( )( )

≡( )

+( )

≡ ( )( )( )( )

⟹ 6푥 − 7 ≡ (퐴푥 + 퐵)(푥 − 1) + 퐶(푥 + 1) ≡ 퐴푥 − 퐴푥 + 퐵푥 − 퐵 + 퐶푥 + 퐶 ≡ (퐴 + 퐶)푥 + (퐵 − 퐴)푥 + 퐶 − 퐵 By comparing the LHS with the RHS we get 퐴 + 퐶 = 0 … … … (1) 퐵 − 퐴 = 6 … … … (2) 퐶 − 퐵 = −7 … … … (3) From (2) 퐵 = 6 + 퐴 Put 퐵 = 6 + 퐴 into (3) ⟹ 퐶 − (6 + 퐴) = −7 ⟹ 퐶 − 6 − 퐴 = −7 ⟹ −6 + 7 = 퐴 − 퐶 퐴 − 퐶 = 1 … … … (4) Now, solve equation (1) and (4) Add (4) to (1) ⟹ 퐴 + 퐴 + 퐶 − 퐶 = 1 ⟹ 2퐴 = 1 ∴ 퐴 =

Put 퐴 = into (1) ⟹ + 퐶 = 0 ∴ 퐶 = −

Put 퐴 = into (2) ⟹ 퐵 = 6 + = ∴ 퐵 =

OR 6푥 − 7

(푥 + 1)(푥 − 1) ≡퐴푥 + 퐵

(푥 + 1)+퐶

(푥 − 1) ≡(퐴푥 + 퐵)(푥 − 1) + 퐶(푥 + 1)

(푥 + 1)(푥 − 1)

186

⟹ 6푥 − 7 ≡ (퐴푥 + 퐵)(푥 − 1) + 퐶(푥 + 1) Put 푥 = 1into the identity above ⟹ 6(1)− 7 ≡ (퐴(1) + 퐵)(1 − 1) + 퐶(1 + 1) ⟹ −1 = 2퐶 ∴ 퐶 = − Put 푥 = 0 into the identity above ⟹ 6(0)− 7 ≡ (퐴(0) + 퐵)(0 − 1) + 퐶(0 + 1) ⟹ −7 ≡ −퐵 + 퐶 −7 = −퐵 − Since 퐶 = −

⟹ 퐵 = − + 7 ⟹ 퐵 = = ∴ 퐵 = Put 푥 = −1 into the identity above ⟹ 6(−1)− 7 ≡ (퐴(−1) + 퐵)(−1 − 1) + 퐶((−1) + 1) ⟹ −6 − 7 ≡ 2퐴 − 2퐵 + 2퐶 −13 = 2퐴 − 2 + 2 − Since 퐶 = − and퐵 = So, −13 = 2퐴 − 13 − 1 ⟹ 2퐴 = −13 + 13 + 1

∴ 퐴 =

∴( )( )

≡( )

+( )

=( )

+( )

=( )

−( )

iii. We have the function,

( )( )

Let ( )( )

≡( )

+( )

≡ ( )( )( )( )

⟹ 12푥 − 6 ≡ 퐴(푥 + 푥 + 1) + (퐵푥 + 퐶)(푥 − 4) ≡ 퐴푥 + 퐴푥 + 퐴 + 퐵푥 − 4퐵푥 + 퐶푥 − 4퐶 ≡ (퐴 + 퐵)푥 + (퐴 − 4퐵 + 퐶)푥 + 퐴 − 4퐶 By comparing the LHS with the RHS we get 퐴 + 퐵 = 0 … … … (1) 퐴 − 4퐵 + 퐶 = 12 … … … (2) 퐴 − 4퐶 = −6 … … … (3) From (1) 퐵 = −퐴 Put 퐵 = −퐴 into (2) ⟹ 퐴 − 4(−퐴) + 퐶 = 12 ⟹ 5퐴 + 퐶 = 12 … … … (4) Now, solve equation (3) and (4) From (4) 퐶 = 12− 5퐴 Put 퐶 = 12 − 5퐴into (3) ⟹ 퐴 − 4(12 − 5퐴) = −7 ⟹ 퐴 − 48 + 20퐴 = −7 ⟹ 21퐴 = −6 + 48

187

∴ 퐴 = = 2 Put 퐴 = 2into (4) ⟹ 퐶 = 12 − 5(2) = 12 − 10 ∴ 퐶 = 2 Put 퐴 = 2into (1) ⟹ 퐵 = −2 ∴ 퐵 = −2

OR 12푥 − 6

(푥 − 4)(푥 + 푥 + 1) ≡퐴

(푥 − 4)+퐵푥 + 퐶

(푥 + 푥 + 1) ≡퐴(푥 + 푥 + 1) + (퐵푥 + 퐶)(푥 − 4)

(푥 − 4)(푥 + 푥 + 1)

⟹ 12푥 − 6 ≡ 퐴(푥 + 푥 + 1) + (퐵푥 + 퐶)(푥 − 4) Put 푥 = 4 into the identity above ⟹ 12(4)− 6 ≡ 퐴(4 + 4 + 1) + (퐵(4) + 퐶)(4 − 4) ⟹ 48− 6 = 퐴(16 + 4 + 1) ⟹ 42 = 21퐴 ∴ 퐴 = = 2 Put 푥 = 0 into the identity above ⟹ 12(0)− 6 ≡ 퐴(0 + 0 + 1) + (퐵(0) + 퐶)(0 − 4) ⟹ −6 ≡ 퐴 − 4퐶 −6 = 3 − 4퐶 Since 퐶 = −

⟹ 4퐶 = 2 + 6 ⟹ 퐶 = = 2 ∴ 퐶 = 2 Put 푥 = 1 into the identity above ⟹ 12(1)− 6 ≡ 퐴(1 + 1 + 1) + (퐵(1) + 퐶)(1 − 4) ⟹ 12 − 6 ≡ 3퐴− 3퐵 − 3퐶 ⟹ 6 = 3(2)− 3퐵 − 3(2) Since 퐴 = 2 and퐶 = 2 So, 6 = 6 − 3퐵 − 6 ⟹ 6 = −3퐵 ∴ 퐵 = − = −2 ∴

( )( )≡

( )+

( )≡

( )+

( )

iv. We have the function,

( ) ( )

Let ( ) ( )

≡( )

+( )

+( )

≡ ( ) ( )( )( ) ( )

⟹ 3푥 − 5 ≡ 퐴(푥 − 1)(푥 + 1) + 퐵(푥 + 1) + (퐶푥 + 퐷)(푥 − 1) ≡ 퐴(푥 + 푥 − 푥 − 1) + 퐵푥 + 퐵 + (퐶푥 + 퐷)(푥 − 2푥 + 1) ≡ 퐴푥 + 퐴푥 − 퐴푥 − 퐴 + 퐵푥 + 퐵 + 퐶푥 − 2퐶푥 + 퐶푥 + 퐷푥 − 2퐷푥 + 퐷 ≡ (퐴 + 퐶)푥 + (−퐴 + 퐵 − 2퐶 + 퐷)푥 + (퐴 + 퐶 − 2퐷)푥 + (−퐴 + 퐵 + 퐷) By comparing the LHS with the RHS we get 퐴 + 퐶 = 0 … … … (1) −퐴 + 퐵 − 2퐶 + 퐷 = 0 … … … (2)

188

퐴 + 퐶 − 2퐷 = 3 … … … (3) −퐴 + 퐵 + 퐷 = −5 … … … (4) From (1) 퐴 + 퐶 = 0 ⟹ 퐴 = −퐶 Put 퐴 + 퐶 = 0 into (3) ⟹ 0 − 2퐷 = 3 ⟹−2퐷 = 3 ∴ 퐷 = −

Put 퐴 = −퐶and 퐷 = − into (2) and (4)

⟹ −(−퐶) + 퐵 − 2퐶 − = 0 ⟹ 퐵 − 퐶 = … … … (5)

and −(−퐶) + 퐵 − = −5 ⟹ 퐵 + 퐶 = −5 + ⟹ 퐵 + 퐶 = − … … … (6) Add (6) to (5) ⟹ 퐵 + 퐵 − 퐶 + 퐶 = − ⟹ 2퐵 = −2 ∴ 퐵 = −1

Put 퐵 = −1into (6) ⟹ −1 + 퐶 = − ⟹ 퐶 = − + 1 = − ∴ 퐶 = −

Put 퐶 = − into (1) ∴ 퐴 = OR

( ) ( )≡

( )+

( )+

( )≡ ( ) ( )( )

( ) ( )

⟹ 3푥 − 5 ≡ 퐴(푥 − 1)(푥 + 1) + 퐵(푥 + 1) + (퐶푥 + 퐷)(푥 − 1) Put 푥 = 1 into the identity above ⟹ 3(1)− 5 ≡ 퐴(1 − 1)(1 + 1) + 퐵(1 + 1) + (퐶(1) + 퐷)(1 − 1) ⟹ 3 − 5 = 퐵(1 + 1) ⟹ −2 = 2퐵 ∴ 퐵 = − = −1 Put 푥 = 0 into the identity above ⟹ 3(0)− 5 ≡ 퐴(0 − 1)(0 + 1) + 퐵(0 + 1) + (퐶(0) + 퐷)(0 − 1) ⟹ −5 ≡ −퐴 + 퐵 + 퐷 ⟹ −5 = −퐴 − 1 + 퐷 Since 퐵 = −1 So, 퐴 − 퐷 = −1 + 5 ⟹ 퐴 − 퐷 = 4 … … … (1) Put 푥 = −1 into the identity above 3 ∙ −1 − 5 ≡ 퐴(−1 − 1)((−1) + 1) + 퐵((−1) + 1) + (−퐶 + 퐷)(−1 − 1) ⟹ −3 − 5 ≡ −4퐴+ 2퐵 − 4퐶 + 4퐷 ⟹ −8 = −4퐴+ 2(−1) − 4퐶 + 4퐷 Since, 퐵 = −1.So, 4퐴 + 4퐶 − 4퐷 = −2 + 8 (Divide both sides by 2) ⟹ 2퐴 + 2퐶 − 2퐷 = 3 … … … (2) Put 푥 = 2 into the identity above ⟹ 3 ∙ 2 − 5 ≡ 퐴(2 − 1)(2 + 1) + 퐵(2 + 1) + (퐶 ∙ 2 + 퐷)(2 − 1) ⟹ 6 − 5 ≡ 5퐴 + 5퐵 + 2퐶 + 퐷 ⟹ 1 = 5퐴 + 5(−1) + 2퐶 + 퐷 Since, 퐵 = −1.So, 5퐴 + 2퐶 + 퐷 = 1 + 5 ⟹ 5퐴+ 2퐶 + 퐷 = 6 … … … (3)

189

Subtract (2) from (3) 5퐴 − 2퐴 + 2퐶 − 2퐶 + 퐷 − (−2퐷) = 6 −3 ⟹ 3퐴 + 3퐷 = 3 (Divide both sides by 3) ⟹ 퐴 + 퐷 = 1 … … … (4) Now, solve equation (1) and (4) Add (1) to (4) 퐴 + 퐴 + 퐷 − 퐷 = 1 + 4 2퐴 = 5 ∴ 퐴 =

Put 퐴 = into (4) ⟹ + 퐷 = 1 ∴ 퐷 = 1 − = −

Put 퐴 = and 퐷 = − into (2)

⟹ 2퐴 + 2퐶 − 2퐷 = 3 ⟹ 2 + 2퐶 − 2 − = 3

⟹ 5 + 2퐶 + 3 = 3 ⟹ 2퐶 = 3 − 3 − 5 ∴ 퐶 = −

∴( ) ( )

≡( )

+( )

+( )

≡( )

−( )

−( )

Example 10.10

If ( )( )

≡( )

+( )

, find 푃 + 푄 + 푅.

Solution

We have, ( )( )

≡( )

+( )

Let ( )( )

≡( )

+( )

≡ ( )( )( )( )

⟹ 3푥 + 4푥 ≡ 푃(푥 + 1) + (푄푥 + 푅)(푥 − 2) Put 푥 = 2 into the above identity. ⟹ 3(2) + 4(2) ≡ 푃(2 + 1) + (푄(2) + 푅)(2 − 2) ⟹ 12 + 8 = 5푃 ∴ 푃 = = 4 Put 푥 = 0 into the above identity. ⟹ 3(0) + 4(0) ≡ 푃(0 + 1) + (푄(0) + 푅)(0 − 2) ⟹ 0 = 푃 − 2푅 ⟹ 2푅 = 푃 ∴ 푅 = = = 2 Put 푥 = 1 into the above identity. ⟹ 3(1) + 4(1) ≡ 푃(1 + 1) + (푄(1) + 푅)(1 − 2) ⟹ 7 = 2푃 − 푄 − 푅 ⟹ 푄 = 2푃 − 푅 − 7 ∴ 푄 = 2(4)− 2 − 7 = −1 ∴ 푃 + 푄 + 푅 = 4 − 1 + 2 = 5

190

Case 4 (Improper Fractions) If the degree of ℎ(푥) is lesser than (sometimes equal to) that of 푔(푥), then we do the following:

Divide the function to get a quotient and a divisor remainder

푓(푥) =푔(푥)ℎ(푥) = 푝(푥) +

푟(푥)ℎ(푥)

Resolve ( )( )

into partial fractions using any of the above cases.

푓(푥) can be written as 푓(푥) = 푝(푥) + 푅 ( ( )( )

), where 푅 ( ( )( )

) is the resolved partial

fractions. Example 10.11

Resolve the following in partial fractions.

i. ( )( )

ii. iii.

Solution

i. We have, ( )( )

= = 푥 +( )( )

(see the division below)

2 3 2

3 2

0 ( 1) 0 0 3

0

3

xx x x x

x x x

x

Let

( )( )≡

( )+

( )≡ ( ) ( )

( )( )

⟹ 푥 + 3 ≡ 퐴(푥 + 1) + 퐵(푥 − 1) Put 푥 = 1 into the above identity. ⟹ 1 + 3 ≡ 퐴(1 + 1) + 퐵(1 − 1) ⟹ 4 = 2퐴 ∴ 퐴 = = 2 Put 푥 = −1 into the above identity. ⟹ −1 + 3 ≡ 퐴(−1 + 1) + 퐵(−1 − 1) ⟹ 2 = −2퐵 ∴ 퐵 = − = −1

∴ ( )( )

= 푥 +( )( )

≡ 푥 +( )

+( )

≡ 푥 +( )

−( )

ii. We have, = ( )( )

= 1 +( )( )

(see the division below)

191

2 2

2

1 ( 6) 9 8

6

8 14

x x x x

x x

x

Let

( )( )≡

( )+

( )≡ ( ) ( )

( )( )

⟹ 8푥 + 14 ≡ 퐴(푥 + 3) + 퐵(푥 − 2) Put 푥 = 2 into the above identity. ⟹ 8(2) + 14 ≡ 퐴(2 + 3) + 퐵(2 − 2) ⟹ 30 = 5퐴 ∴ 퐴 = = 6 Put 푥 = −3 into the above identity. ⟹ 8(−3) + 14 ≡ 퐴(−3 + 3) + 퐵(−3 − 2) ⟹ −10 = −5퐵 ∴ 퐵 = = 2

∴ == 1 +( )( )

≡ 1 +( )

+( )

≡ 1 +( )

+( )

iii. We have, = ( )( )

= 푥 − 2 +( )( )

(see the division below)

2 3 2

3 2

2

2

2 ( 2 8) 0 10 3

( 2 8 ) 2 18 3

( 2 4 16) 22 13

xx x x x x

x x xx x

x xx

Let ( )( )

≡( )

+( )

≡ ( ) ( )( )( )

⟹ 22푥 − 13 ≡ 퐴(푥 + 4) + 퐵(푥 − 2) Put 푥 = 2 into the above identity. ⟹ 22(2)− 13 ≡ 퐴(2 + 4) + 퐵(2 − 2) ⟹ 31 = 6퐴 ∴ 퐴 = Put 푥 = −4 into the above identity. ⟹ 22(−4)− 13 ≡ 퐴(−4 + 4) + 퐵(4 − 2) ⟹ −101 = 2퐵 ∴ 퐵 = −

∴ = 푥 − 2 +( )( )

≡ (푥 − 2) +( )

+( )

≡ (푥 − 2) +( )

−( )

Try:

a. Express ( )

in partial fraction.

192

b. Determine the values of the constants 푎,푏 and 푐 such that

( )( )= + (WASSCE)

Some Selected Examples

Example 10.12

A function is defined by 푓(푥) =( )( )( )

i. State the domain of the function. ii. Find the zeros of the function. iii. Find the values of the constants 퐴,퐵 and 퐶 such that 푓(푥) ≡

( )+

( )+

( )

Solution

i. We have the function, ( ( )( )

The function 푓 is defined except when (푥 + 1)(푥 − 2)(푥 − 3) = 0 ⟹ 푥 + 1 = 0or푥 − 2 = 0 or 푥 − 3 = 0 ⟹ 푥 = −1or푥 = 2 or 푥 = 3 퐷표푚푎푖푛,퐷 = {푥: 푥 ∈ ℝ, 푒푥푐푒푝푡푥 = −1표푟푥 = 2표푟푥 = 3} ii. For zeroes of 푓(푥) let 푓(푥) = 0 ⟹

( ( )( )= 0

⟹ 3푥 − 21푥 + 24 = 0 (Divide both sides by 3) ⟹ 푥 − 7푥 + 8 = 0 ⟹ (푥 − 8)(푥 + 1) = 0 ∴ 푥 = 8or 푥 = −1 iii.

Let ( ( )( )

≡( )

+( )

+( )

≡ ( )( ) ( )( ) ( )( )( ( )( )

⟹ 3푥 − 21푥 + 24 ≡ 퐴(푥 − 2)(푥 − 3) + 퐵(푥 + 1)(푥 − 3) + 퐶(푥 + 1)(푥 − 2) Put 푥 = −1 into the above identity. ⟹ 3(−1) − 21(−1) + 24 ≡ 퐴(−3)(−4) + 퐵(0)(−2) + 퐶(0)(−2) ⟹ 3 + 21 + 24 = −12퐴 ∴ 퐴 = − = −4 Put 푥 = 2 into the above identity. ⟹ 3(2) − 21(2) + 24 ≡ 퐴(0)(−1) + 퐵(3)(−1) + 퐶(3)(0) ⟹ 12 − 42 + 24 = −3퐵 ∴ 퐵 = = 2 Put 푥 = 3 into the above identity. ⟹ 3(3) − 21(3) + 24 ≡ 퐴(1)(0) + 퐵(4)(0) + 퐶(4)(1) ∴ 27 − 63 + 24 = 4퐶 ⟹ 4퐶 = −12 ∴ 퐶 = − = −3

193

∴ ( ( )( )

≡( )

+( )

+( )

≡( )

−( )

−( )

Example 10.13

Find the values of 퐴,퐵,퐶 and 퐷 such that 푓(푥) =( )( )

≡ 퐴 +( )

+( )

Solution

We have the function, 푓(푥) =( )( )

≡ 퐴 +( )

+( )

Let ( )( )

≡ 퐴 +( )

+( )

≡ ( ) ( )( )( )( )

2푥 − 푥 − 1 ≡ 퐴(푥 − 3)(푥 + 1) + 퐵(푥 + 1) + (퐶푥 + 퐷)(푥 − 3) ≡ 퐴(푥 + 푥 − 3푥 − 3) + 퐵푥 + 퐵 + 퐶푥 − 3퐶푥 + 퐷푥 − 3퐷 ≡ 퐴푥 − 3퐴푥 + 퐴푥 − 3퐴 + 퐵푥 + 2퐵 + 퐶푥 − 3퐶푥 + 퐷푥 − 3퐷 ≡ 퐴푥 + (−3퐴 + 퐵 + 퐶)푥 + (퐴 − 3퐶 + 퐷)푥 + (−3퐴 + 퐵 − 3퐷) By comparing the LHS to the RHS we get 퐴 = 2, −3퐴 + 퐵 + 퐶 = 0 … … … (1) 퐴 − 3퐶 + 퐷 = −1 … … … (2) −3퐴 + 퐵 − 3퐷 = −1 … … … (3) Put 푥 = 3 into the above identity. ⟹ 2(3) − 3 − 1 ≡ 퐴(3 − 3)(3 + 1) + 퐵(3 + 1) + (3퐶 + 퐷)(3 − 3) ⟹ 54 − 4 = 10퐵 ∴ 퐵 = = 5 Put 퐴 = 2 and 퐵 = 5into (1). ⟹ −3(2) + 5 + 퐶 ≡ 0 ⟹ −6 + 5 + 퐶 = 0 ∴ 퐶 = 1 Put 퐴 = 2 and 퐶 = 1into (2). ⟹ 2 − 3(1) + 퐷 ≡ −1 ⟹ −1 + 퐷 = −1 ∴ 퐷 = −1 + 1 = 0 ∴ 퐴 = 2, 퐵 = 5, 퐶 = 1and퐷 = 0

OR

Let 푓(푥) = ( )( )

= = 2 +( )( )

(see the division below)

194

3 2 3 3

3 2

2

2 ( 3 3) 2 0 1

2 6 2 6

6 3 5

x x x x x x

x x x

x x

Let ( )( )

≡( )

+( )

≡ ( )( )( )( )

⟹ 6푥 − 3푥 + 5 ≡ 퐵(푥 + 1) + (퐶푥 + 퐷)(푥 − 3) ≡ 퐵푥 + 퐵 + 퐶푥 − 3퐶푥 + 퐷푥 − 3퐷 ≡ (퐵 + 퐶)푥 + (−3퐶 + 퐷)푥 + (퐵 − 3퐷) By comparing the LHS and RHS we get 퐵 + 퐶 = 6 … … … (1) −3퐶 + 퐷 = −3 … … … (2) 퐵 − 3퐷 = 5 … … … (3) Put 푥 = 3 into the above identity. ⟹ 6(3) − 3(3) + 5 ≡ 퐵(3 + 1) + (3퐶 + 퐷)(3 − 3) ⟹ 54 − 9 + 5 = 10퐵 ⟹ 10퐵 = 50 ∴ 퐵 = = 5 Put 퐵 = 5into (1) and (3) . ⟹ 5 + 퐶 = 6 ⟹ 퐶 = 6 − 5 ∴ 퐶 = 1 and 5 − 3퐷 = 5 ⟹ −3퐷 = 5− 5 ∴ 퐷 = − = 0 Now,

푓(푥) = ( )( )

= 2 +( )( )

≡ 2 +( )

+( )

… … … (푎)

By comparing (푎) with

푓(푥) =( )( )

≡ 퐴 +( )

+( )

퐴 = 2,퐵 = 5,퐶 = 1and 퐷 = 0 Example 10.14 Convert the following in partial fractions

i. ( )( )( )

ii. ( ) ( )

iii. ( )( )

Solution i.

195

Let ( )( )( )

≡( )

+( )

+( )

≡ ( )( ) ( )( ) ( )( )( )( )( )

⟹ 6푥 − 9푥 − 25 ≡ 퐴(푥 + 2)(푥 + 3) + 퐵(푥 − 1)(푥 + 3) + 퐶(푥 − 1)(푥 + 2) Put 푥 = 1 into the above identity. ⟹ 6(1) − 9(1)− 25 ≡ 퐴(3)(4) + 퐵(0)(4) + 퐶(0)(3) ⟹ 6 − 9 − 25 = 12퐴 ⟹ −28 = 12퐴 ∴ 퐴 = − = − Put 푥 = −2 into the above identity. ⟹ 6(−2) − 9(−2) − 25 ≡ 퐴(0)(1) + 퐵(−3)(1) + 퐶(−3)(0) ⟹ 24 + 18 − 25 = −3퐵 ⟹ 17 = −3퐶 ∴ 퐵 = Put 푥 = −3 into the above identity. ⟹ 6(−3) − 9(−3) − 25 ≡ 퐴(−1)(0) + 퐵(−4)(0) + 퐶(−4)(−1) ⟹ 54 + 27 − 25 = 4퐶 ⟹ 4퐶 = 56 ∴ 퐶 = = 14

∴ ( )( )( )

≡( )

+( )

+( )

≡( )

−( )

+( )

ii. We have the function, ( ) ( )

Let ( ) ( )

≡( )

+( )

+( )

≡ ( ) ( )( )( )( )

⟹ 푥 + 4푥 + 20푥 − 7 ≡ 퐴(푥 − 1)(푥 + 8) + 퐵(푥 + 8) + (퐶푥 + 퐷)(푥 − 1) ≡ 퐴(푥 + 8푥 − 푥 − 8) + 퐵푥 + 8퐵 + (퐶푥 + 퐷)(푥 − 2푥 + 1) ≡ 퐴푥 + 8퐴푥 − 퐴푥 − 8퐴 + 퐵푥 + 8퐵 + 퐶푥 − 2퐶푥 + 퐶푥 + 퐷푥 − 2퐷푥 + 퐷 ≡ (퐴 + 퐶)푥 + (−퐴 + 퐵 − 2퐶 + 퐷)푥 + (8퐴 + 퐶 − 2퐷)푥 + (−퐴 + 8퐵 + 퐷) By comparing the LHS with the RHS we get 퐴 + 퐶 = 1 … … … (1) −퐴 + 퐵 − 2퐶 + 퐷 = 4 … … … (2) 8퐴 + 퐶 − 2퐷 = 20 … … … (3) −퐴 + 8퐵 + 퐷 = −7 … … … (4) Put푥 = 1 into the identity above ⟹ 1 + 4(1) + 20(1)− 7 = 퐴(0)(9) + 퐵(9) + (퐶 + 퐷)(0) ⟹ 1 + 4 + 20 − 7 = 9퐵 ⟹ 9퐵 = 18 ∴ 퐵 = = 2 From (1) 퐴 + 퐶 = 1 ⟹ 퐶 = 1 − 퐴 Put 퐶 = 1 − 퐴 into (3) ⟹ 8퐴 + 1 − 퐴 − 2퐷 = 20 ⟹ 7퐴 − 2퐷 = 20 − 1 ⟹ 7퐴 − 2퐷 = 19 … … … (5)

196

Put 퐵 = 2 into (4) ⟹ −퐴 + 8(2) + 퐷 = −7 ⟹−퐴 + 퐷 = −7 − 16 ⟹ −퐴 + 퐷 = −23 … … … (6) Now, solve equation (5) and (6) From (6) 퐷 = 퐴 − 23 Put 퐷 = 퐴 − 23 into (5) ⟹ 7퐴 − 2(퐴 − 23) = 19 ⟹ 7퐴 − 2퐴 + 46 = 19 ⟹ 5퐴 = 19 − 46 ∴ 퐴 = − Put 퐴 = − into (6)

⟹ 퐷 = − − 23 = = − ∴ 퐷 = −

Put 퐴 = − into (1). ⟹− + 퐶 = 1 ∴ 퐶 = 1 + = into(6)

∴ ( ) ( )

≡( )

+( )

+( )

≡( )

−( )

+( )

iii. We have the function,

( )( )

Let ( )( )

≡( )

+( )

+( )

≡ ( )( ) ( )( )( )

⟹ 푥 − 3 ≡ (퐴푥 + 퐵)(푥 − 1) + 퐶(푥 − 1)(푥 + 1) + 퐷(푥 + 1) ≡ (퐴푥 + 퐵)(푥 − 2푥 + 1) + 퐶(푥 + 푥 − 푥 − 1) + 퐷푥 + 퐷 ≡ 퐴푥 − 2퐴푥 + 퐴푥 + 퐵푥 − 2퐵푥 + 퐵 + 퐶푥 − 퐶푥 + 퐶푥 − 퐶 + 퐷푥 + 퐷 ≡ (퐴 + 퐶)푥 + (−2퐴 + 퐵 − 퐶 + 퐷)푥 + (퐴 − 2퐵 + 퐶)푥 + (퐵 − 퐶 + 퐷) By comparing the LHS with the RHS we get 퐴 + 퐶 = 0 … … … (1) −2퐴 + 퐵 − 퐶 + 퐷 = 0 … … … (2) 퐴 − 2퐵 + 퐶 = 1 … … … (3) 퐵 − 퐶 + 퐷 = −3 … … … (4) Put 푥 = 1 into the identity above ⟹ 1 − 3 = (퐴 + 퐵)(0) + 퐶(0)(2) + 퐷(2) ⟹ −2 = 2퐷 ∴ 퐷 = − = −1 From (1) 퐶 = −퐴 Put 퐶 = −퐴 into (3) ⟹ 퐴 − 2퐵 − 퐴 = 1 ⟹ −2퐵 = 1 ∴ 퐵 = −

Put 퐵 = − and 퐷 = −1 into (4)

⟹ − − 퐶 − 1 = −3 ⟹−퐶 = −3 + + 1 ∴ 퐶 =

197

Put 퐶 = into (1) ∴ 퐴 = − into(6)

∴ ( )( )

≡( )

+( )

+( )

≡( )

−( )

−( )

Final exercises

1. Resolve the following functions in partial fractions:

i. ( )( )

ii. iii. iv. ( )( )

v. vi.

vii. ( )( )

viii. ix. ( )( )( )

x.

2. A function 푓 is defined by 푓(푥) =( )

.

a. State the zeros of 푓(푥) b. Resolve 푓(푥) in partial fractions. 3. Resolve the following functions in partial fractions:

i. ( )( )

ii. ( )( )

iii. )( )

vi. ( )( )

v. ( )( )( )

4. If ( )( )

≡ + , find the values of 퐴,퐵 and 퐶.

5. Determine the zeros of the following functions. Hence, resolve each of the function in partial functions.

a. ( ) ( )

b. ( ) ( )

6. If ≡ 퐴 + ( ) + , find the values of 퐴,퐵and 퐶 . 7. Determine the values of the constants 퐴,퐵and 퐶 such that

≡ 퐴 + + 8. Resolve the following into partial fractions. i. ii. iii. iv. v.

9. If ( )( )

≡ + , find the value of 퐴 + 퐵 + 퐶.

10. Decompose into partial fractions.

11. Decompose 푓(푥) = into partial fractions. 12. Resolve the following into partial fractions.

198

i. ( )( )

ii. iii. ( )( )

vi. v. ( )( )( )

vi. ( )( )

vii. ( )( )( )

iii. ( )( )

vi. ( )( )

13. Find the values of A and B in terms of 훼푎푛푑훽 if ( )( )

=( )

+( )

14. Resolve 푓(푥) =( )

into partial function

15. Simplify the following. i. + ii. + iii. + iv. + 16. Simplify the following. i. − ii.

√+ iii. − iv. −

17. Simplify the following

i. ÷ ii. ÷ iii. ÷

199

SEQUENCES AND SERIES Definition: A sequence refers to a set of numbers written in some pattern or order. Each number in a sequence is called a 푡푒푟푚. For instance, the sequence: 1, 3, 5, … is a sequence of odd numbers which has a first term 1, second term 3, third term 5 and so on. The order in which the numbers are written can be obtained by a simple rule. Consider the numbers: 2, 4, 6, 8, … This represent a sequence of positive even numbers. We can say that the terms of the sequence has a rule 2푛 such that 2(1) = 2, 2(2) = 4, 2(3) = 6, ….

Forms of sequences 1. Finite sequence A sequence with a known last term is called a 푓푖푛푖푡푒푠푒푞푢푒푛푐푒. For instance, the sequence: 1, 4, 9, 16, 25 is a sequence of square numbers which has a last term 25. 2. Infinite sequence A sequence with an unknown number of last term is called an 푖푛푓푖푛푖푡푒푠푒푞푢푒푛푐푒. For instance, the sequence 1, 4, 16, 25, …. is a sequence with an unknown last term. The symbol ‘… ′ mean the sequence continue infinitely.

Notation In this book, we will use ′푎 ′and′푢 ′to denote the 푛푡ℎ term of a sequence interchangeably. We will sometimes also use ′ɭ′ to denote the last term of finite sequences. So the first term of a sequence will be denoted by 푢 , thesecond term 푢 ,the third term 푢 and so on. Note: Sequences are functions such that 푢 = 푢(푛).

Example 11.1 If 푢 = 3푛 + 1 denote the 푛푡ℎ term of a sequence, find 푢 ,푢 , 푢 and 푢 .

Solution We have, 푢 = 3푛 + 1 푢 = 3(1) + 1 ∴,푢 = 4 푢 = 3(2) + 1 ⟹ 푢 = 6 + 1 ∴,푢 = 7 푢 = 3(3) + 1 ⟹ 푢 = 9 + 1 ∴,푢 = 10 푢 = 3(4) + 1 ⟹ 푢 = 12 + 1 ∴,푢 = 13

Example 11.2 A sequence is given as 푢 = 4푛 − . Write down the first 5 term of the sequence.

200

Solution We have, 푢 = 4푛 − 푢 = 4(1)− ∴,푢 =

푢 = 4(2)− ⟹ 푢 = 8− ∴,푢 =

푢 = 4(3)− ⟹ 푢 = 12 − ∴,푢 =

푢 = 4(4)− ⟹ 푢 = 16 − ∴,푢 =

푢 = 4(5)− ⟹ 푢 = 20 − ∴,푢 = Example 11.3 Write down the first six terms of the sequence given as 푎 = 2 + 3

Solution We have, 푎 = 2 + 3 푎 = 2 + 3 ⟹푎 = 2 + 1 = 3 ∴,푎 = 3 푎 = 2 + 3 ⟹ 푎 = 4 + 3 ∴,푎 = 7 푎 = 2 + 3 ⟹ 푎 = 8 + 9 ∴,푎 = 17 푎 = 2 + 3 ⟹ 푎 = 16 + 27 ∴,푎 = 43 푎 = 2 + 3 ⟹ 푎 = 32 + 81 ∴,푎 = 113 푎 = 2 + 3 ⟹ 푎 = 64 + 243 ∴,푎 = 307 Try: A sequence is given by 푢 = , write down the first 3 terms of the sequence. Definition: A series refers to the sum of the terms of a sequence. For example, suppose we have the sequence: 푢 , 푢 , 푢 , … ,푢 , then we can obtain the series 푢 + 푢 + 푢 ⋯+ 푢 . We will denote the sum of the 푛푡ℎ term of a sequence by 푆 .

⟹푆 = 푢 + 푢 + 푢 + ⋯+ 푢 ⟹ 푆 = 푆 + 푢 ⟹ 푢 = 푆 − 푆 Example 11.4

A sequence is given by 1, 4, 9, … . Find 푠 , 푠 and 푠 Solution

We have the sequence, 1, 4, 9, … 푠 = 1 푠 = 1 + 4 ∴ 푠 = 5

201

푠 = 1 + 4 + 9 ∴ 푠 = 14 Example 11.5

Find the sum of the first four terms of the sequence with an 푛푡ℎ term 푢 = 4푛 + 1. Solution

We have, 푢 = 4푛 + 1 푢 = 4(1) + 1 ⟹ 푢 = 4 + 1 ∴ 푢 = 5 푢 = 4(2) + 1 ⟹ 푢 = 8 + 1 ∴ 푢 = 9 푢 = 4(3) + 1 ⟹ 푢 = 12 + 1 ∴ 푢 = 13 푢 = 4(4) + 1 ⟹ 푢 = 16 + 1 ∴ 푢 = 17 푠 = 5 + 9 + 13 + 17 ∴ 푠 = 44

Types of sequences 1. Arithmetic/ linear sequence (Arithmetic progression (AP)) An arithmetic sequence is a sequence where each successive term after the first term is obtained by adding a constant number to the preceding term. The constant number is called 푐표푚푚표푛푑푖푓푓푒푟푒푛푐푒. Now, let 푎 stands for the first term of linear sequence and let 푑 stands for the common difference between successive term of the sequence. For example the sequence

2, 4, 6, … can be written as 2, 2 + 2, 2 + 2 + 2, …. It follows that our sequence is of the form 푎, 푎 + 푑, 푎 + 2푑, … . where 푎 = 2 and 푑 = 2 From this sequence, we can conclude that: If there are푛 terms in an AP, then there are (푛 − 1) common difference between the successive terms. Again, consider the sequence 1, 3, 5. This sequence can be written as

1, 1 + 2, 1 + 2 + 2. So we see that as there are three terms in the sequence, there are two common differences (2 two’s) between the successive terms of the sequence. So the 푛푡ℎterm of an arithmetic progression is given by

푢 = ɭ = 푎 + (푛 − 1)푑… … …(1) Now, from the sequence 푎, 푎 + 푑, 푎 + 2푑 , we have 푑 = (푎 + 푑) − 푎 = (푎 + 2푑)− (푎 + 푑) = 푢 −푢 = 푢 −푢 In general, for all AP

202

푑 = 푢 − 푢 … … … (2) How many common differences are there between the successive terms of the sequence −2,−4,−6,−8,−10 ? A. 5 B. 6 C. 4 D. 1 Example 11.6 Write down the first four terms of the AP with first term 2 and common difference 5.

Solution We have, 푎 = 2 and 푑 = 5. The first four terms of an AP is given as: 푎, 푎 + 푑,푎 + 2푑,푎 + 3푑 ⟹ 2, 2 + 5, 2 + 2(5), 2 + 3(5) 2, 7, 2 + 10, 2 + 15 ⟹ 2, 7, 12, 17

Example 11.7 Find the common difference between the linear sequence 7, 4, 1, …

Solution We have the sequence: 7, 4, 1, … The common difference is defined as 푑 = 푢 − 푢 = 4 − 7 = −3. ∴the common difference for the sequence is −3.

Example 11.8 Find the 10th term of an AP with the first term 5 and common difference −4.

Solution We have, first term, 푎 = 5 and common difference, 푑 = −4. The 푛th term of the sequence is given by: 푢 = 푎 + (푛 − 1)푑 ⟹ 푢 = 5 + (10− 1)(−4) ⟹ 푢 = 5 + (9)(−4) ⟹ 푢 = 5 − 36 ⟹ 푢 = −31 ∴,the 10th term of the sequence is −31

Example 11.9 An AP is given by 푥, 2푥 + 1, 3푥 + 2, … i. Determine the fourth term ii. Find the nth term . iii. If the 15th term of the sequence is equal to 44, find the value of 푥.

Solution We have the sequence: 푥, 2푥 + 1, 3푥 + 2, … The first term of the sequence is 푎 = 푥

203

The common difference is defined as 푑 = (2푥 + 1)− 푥 = 2푥 − 푥 + 1 = 푥 + 1. i. The fourth term of the sequence is given by: 푢 = 푎 + 3푑 = 푥 + 3(푥 + 1) = 푥 + 3푥 + 3 = 4푥 + 3 ∴ 푢 = 4푥 + 3

ii. The 푛th term of the sequence is given by: 푢 = 푎 + (푛 − 1)푑 ⟹ 푢 = 푥 + (푛 − 1)(푥 + 1) ⟹ 푢 = 푥 + 푛푥 + 푛 − 푥 − 1 ⟹ 푢 = 푥 − 푥 + 푛푥 + 푛 − 1 ⟹ 푢 = 푛푥 + 푛 − 1 ∴the푛th term of the sequence is 푢 = 푛푥 + 푛 = 푛(푥 + 1)− 1 iii. We are given the 15th term, 푢 = 44. 푢 = 푎 + 14푑 ⟹ 43 = 푥 + 14(푥 + 1) ⟹ 43 = 푥 + 14푥 + 14 ⟹ 44− 14 = 15푥 ⟹ 15푥 = 30 ⟹ 푥 = = 2 ∴ 푥 = 2

Or Using the 푛th term, 푢 = 푛푥 + 푛 − 1 ⟹ 푢 = 15푥 + 15− 1 But 푢 = 44 ⟹ 44 = 15푥 + 14 ⟹ 44 − 14 = 15푥 푥 = = 2 ∴ 푥 = 2

Example 11.10 Find the values of 푥 and 푦 in the following arithmetic sequences. i. : 푥, 푦, 2,푥 − 1, … ii. 2,−푥, 2푦 − 푥, 6, …

Solution i. We have the sequence: 푥, 푦, 2, 푥 − 1, … The sequence is an AP with common difference: 푑 = 푢 − 푢 = 푢 − 푢 = 푢 − 푢 ⟹ 푑 = 푦 − 푥 = 2 − 푦 = 푥 − 1 − 2 ⟹ 푦 − 푥 = 2 − 푦 ⟹ 푥 − 푦 − 푦 = −2 ⟹ 푥 − 2푦 = −2 … … … (1) Also, 푦 − 푥 = 푥 − 1 − 2 ⟹ 푥 + 푥 − 푦 = 1 + 2 ⟹ 2푥 − 푦 = 3 … … … (2) Now, solve equation (1) and (2) From (1), 푥 = −2 + 2푦 Put 푥 = −2 + 2푦 into (2). ⟹ 2(−2 + 2푦) − 푦 = 3 ⟹ −4 + 4푦 − 푦 = 3 ⟹ 3푦 = 3 + 4 ∴ 푦 = Put 푦 = into (1)

⟹ 푥 = −2 + 2 = −2 + = = ∴ 푥 =

204

ii. We have the sequence: 2,−푥, 2푦 − 푥, 6, … The sequence is an AP with common difference: 푑 = 푢 − 푢 = 푢 − 푢 = 푢 − 푢 ⟹ 푑 = −푥 − 2 = 2푦 − 푥 − (−푥) = 6 − (2푦 − 푥) ⟹ −푥 − 2 = 2푦 − 푥 − (−푥) ⟹ −푥 − 2 = 2푦 − 푥 + 푥 ⟹ 푥 + 2푦 = −2 … … … (1) Also, −푥 − 2 = 6 − (2푦 − 푥) ⟹ −푥 − 2 = 6 − 2푦 + 푥 ⟹ −2 − 6 = 푥 + 푥 − 2푦 ⟹ −8 = 2푥 − 2푦 (Divide both sides by 2) ⟹ 푥 − 푦 = −4 … … … (2) Now, solve equation (1) and (2) From (1), 푥 = −2 + 2푦 Put 푥 = −2 + 2푦 into (2). ⟹ (−2 + 2푦) − 푦 = −4 ⟹ 2푦 − 푦 = −4 + 2 ∴ 푦 = −2 Put 푦 = −2into (1) ⟹ 푥 = −2 + 2(−2) = −2 − 4 = −6 ∴ 푥 = −6

Example 11.11 The eighth term of an AP is 15 and the second term is . Find the first term and the common difference.

Solution The 푛th term of the sequence is given by: 푢 = 푎 + (푛 − 1)푑 푢 = 푎 + (8 − 1)푑 ⟹ 푢 = 푎 + 7푑 ⟹ 푎 + 7푑 = 15 … … … (1) 푢 = 푎 + (2 − 1)푑 ⟹ 푢 = 푎 + 푑 ⟹ 푎 + 푑 = … … … (2) Now, solve equation (1) and (2) Subtract (2) from (1) ⟹ 푎 − 푎 + 7푑 − 푑 = 15− ⟹ 6푑 = ⟹ 푑 = × =

Put 푑 = into (2) ⟹ 푎 + = ⟹ 푎 = − = = −

∴,first term, 푎 = − and common difference, 푑 =

The sum of an arithmetic series Let 푎,푑 and 푢 be the first term, common difference and last term of an AP respectively. We can proceed to form the sequence:

푎, (푎 + 푑), (푎 + 2푑), … , (푢 − 2푑), (푢 − 푑),푢 Summing the sequence gives the series

205

푆 = 푎 + (푎 + 푑) + (푎 + 2푑) + ⋯+ (푢 − 2푑) + (푢 − 푑) + 푢 … … … (3) Let’s rewrite the series in the reverse order:

푆 = 푢 + (푢 − 푑) + (푢 − 2푑) + ⋯+ (푎 + 2푑) + (푎 + 푑) + 푎… … … (4) Adding (3) to (4) gives

푆 + 푆 = (푎 + 푢 ) + (푎 + 푢 ) + (푎 + 푢 ) + ⋯ (푎 + 푢 ) + (푎 + 푢 ) + (푎 + 푢 ) So we have n 푎′s and n 푢 ′s. This gives a sum

2푆 = 푛푎 + 푛푢 = 푛(푎 + 푢 ) ⟹ 푆 =푛2

(푎 + 푢 )

But 푢 = ɭ = 푎 + (푛 − 1)푑. ⟹ 푆 = (푎 + 푎 + (푛 − 1)푑) = 2푎 + (푛 − 1)

Example 11.12 The sixth term of an arithmetic sequence is two-third the eighth term. Find i. the first term. ii. the common difference. iii the sum of the first 20 terms of the sequence .

Solution The 푛th term of the sequence is given by: 푢 = 푎 + (푛 − 1)푑 We have, 푢 = 푢 ⟹ 푎 + 5푑 = (푎 + 7푑) ⟹ 3(푎 + 5푑) = 2(푎 + 7푑) ⟹ 3푎 + 15푑 = 2푎 + 14푑 ⟹ 3푎 − 2푎 + 15푑 − 14푑 = 0 ⟹ 푎 + 푑 = 0 … … … (1) And 푢 = 12 ⟹ 푎 + 3푑 = 12 … … … (2) Now, solve equation (1) and (2) Subtract equation (1) from (2) 푎 − 푎 + 3푑 − 푑 = 12 − 0 ⟹ 2푑 = 12 ⟹ 푑 = = 6 Put 푎 = −6into (1). ⟹ 푎 + 6 = 0 ⟹ 푎 = −6 i.The first term of the AP is −6 and ii. The common difference is 6. iii. The sum of the 푛th term of an AP is defined as: 푆 = [2푎 + (푛 − 1)푑]

⟹ 푆 = [2(−6) + (20 − 1)6] = 10(−12 + 114) = 10(102) = 1020 ∴the sum of the first twenty terms of the sequence is 1020.

푆 = (푎 + 푢 ) 푆 = 2푎 + (푛 − 1)

206

Or The 푛th term of the sequence is given by: 푢 = 푎 + (푛 − 1)푑 We can calculate the 20th term of the sequence as: 푢 = −6 + (20 − 1)(6) = −6 + 114 = 108 The sum of the 푛th term of an AP is defined as: 푆 = (푎 + 푢 )

⟹ 푆 = [−6 + 108] = 10(102) = 1020 ∴the sum of the first twenty terms of the sequence is 1020.

Example 11.13 Find the sum of the series 3 + 7 + ⋯+ 63.

Solution We are given the series: 3 + 7 + ⋯+ 63. The first term of the series, 푎 = 3, the common difference, 푑 = 7 − 3 = 4 and the 푙푎푠푡 term of the series, 푙 = 63 The last term of an arithmetic series is defined as: 푙 = 푎 + (푛 − 1)푑 ⟹ 63 = 3 + (푛 − 1)(4) ⟹ 63 − 3 = 4(푛 − 1) ⟹ 60 = 4(푛 − 1) ⟹ 푛 − 1 = ⟹ 푛 = 15 + 1 = 16 Therefore, the nth term of the series is 16. The sum of the 푛th term of an AP is defined as: 푆 = (푎 + 푢 )

⟹ 푆 = [3 + 63] = 8(66) = 528. ∴, the sum of the series is 528.

Example 11.14 The first term of an AP is 3. The sum of the first six terms is twice the sum of the first two terms. Find the common difference.

Solution The sum of the 푛th term of an AP is defined as: 푆 = [2푎 + (푛 − 1)푑]

We have, 푆 = 2푆 ⟹ [2푎 + (6 − 1)푑] = 2 × [2푎 + (2 − 1)푑] 3[2푎 + 5푑] = 2[2푎 + 푑] ⟹ 6푎 + 15푑 = 4푎 + 2푑 15푑 − 2푑 = 4푎 − 6푎 ⟹ 13푑 = −2푎 But the first term, 푎 = 3 So, 13푑 = −2(3) ⟹ 푑 = −

207

∴, the common difference of the AP is −

Example 11.15 The third term and the eleventh term of an AP are 19 and 63 respectively. Find i. the common difference ii. the sum of the first 28 terms of the sequence.

Solution The 푛th term of the sequence is given by: 푢 = 푎 + (푛 − 1)푑 We have, 푢 = 푎 + 2푑 ⟹ 19 = 푎 + 2푑… … … (1) And 푢 = 푎 + 10푑 ⟹ 63 = 푎 + 10푑… … … (2) Now, solve equation (1) and (2) Subtract equation (1) from (2) 63 − 19 = 푎 − 푎 + 10푑 − 2푑 ⟹ 44 = 8푑 ⟹ 푑 = =

Put 푑 = into (1). ⟹ 19 = 푎 + 2 ⟹ 19 = 푎 + 11

푎 = 19 − 11 = 8 i. The common difference is 푑 = = 5.5

ii. The sum of the 푛th term of an AP is defined as: 푆 = [2푎 + (푛 − 1)푑]

⟹ 푆 = [2(8) + (28 − 1)5.5] = 14(16 + 148.5) = 14(164.5) = 2303 ∴,the sum of the first 28 terms of the sequence is 2303. Example 11.16 The sum of the three consecutive term of an AP is 12 and their product is 52. Find the first term and the common difference.

Solution

Method 1 Let the three consecutive terms of the AP be: 푎, 푎 + 푑,푎 + 2푑 So, 푎 + 푎 + 푑 + 푎 + 2푑 = 12 ⟹ 3푎 + 3푑 = 12 (Divide both sides by 3) ⟹ 푎 + 푑 = 4 … … … (1) And 푎(푎 + 푑)(푎+ 2푑) = 52 … … … (2) From (1) 푑 = 4 − 푎 Put 푑 = 4 − 푎 into (2) ⟹ 푎(푎 + 4 − 푎) 푎 + 2(4 − 푎) = 52 ⟹ 푎(4)(푎+ 8− 2푎) = 52 ⟹ 4푎(8− 푎) = 52 (Divide both sides by 4) ⟹ 푎(8 − 푎) = 13 8푎 − 푎 = 13 푎 − 8푎 + 13 = 0

208

푎 = ( )± ( ) ( )( ) = ±√( )

= ±√ = ± √ = ± √

∴ 푎 = 4 + √3 or 푎 = 4 − √3 Put the values of 푎, into (1) When 푎 = 4 + √3, 푑 = 4 − 4 + √3 = 4 − 4 − √3 = −√3 When 푎 = 4 − √3, 푑 = 4 − 4 − √3 = 4 − 4 + √3 = √3 Therefore, the sequence is: 4 + √3, 4 + √3 − √3, 4 + √3 − 2√3 ⟹ 4 + √3, 4, 4 − √3 or 4 − √3, 4 − √3 + √3, 4 − √3 + 2√3 ⟹ 4 − √3, 4, 4 + √3

Method 2 By symmetry, we can let the three consecutive term of the AP be:

푎 − 푑,푎,푎 + 푑 So, 푎 − 푑 + 푎 + 푎 + 푑 = 12 ⟹ 3푎 = 12 (Divide both sides by 3) ⟹ 푎 = 4 And (푎 − 푑)(푎)(푎 − 푑) = 52 But 푎 = 4 So, 4(4− 푑)(4 + 푑) = 52 (Divide both sides by 4) (4 − 푑)(4 + 푑) = 13 16 − 푑 = 13 16 − 13 = 푑 푑 = 3 푑 = ±√3 Therefore, the sequence is: 4 − √3, 4, 4 + √3 For 푑 = √3 or 4 + √3, 4, 4 − √3 For 푑 = −√3

Example 11.16 The sum of an AP is 45. If the first term and the common difference are −5 and 5 respectively, find the number of terms of the sequence.

Solution The sum of the 푛th term of the sequence is given by: 푆 = [2푎 + (푛 − 1)푑] We have, 푎 = −5, 푑 = 5and 푆 = 45 ⟹ 45 = [2(−5) + (푛 − 1)(5)] ⟹ 45 = [−10 + (푛 − 1)5]

⟹ 45 = + (푛 − 1) ⟹ 2(45) = 2 + 2 (푛 − 1) ⟹ 90 = −10푛 + 5푛(푛 − 1) ⟹ 90 = −10푛 + 5푛 − 5푛 ⟹ 5푛 − 10푛 − 5푛 − 90 = 0 ⟹ (Divide both sides by 5) ⟹ 푛 − 3푛 − 18 = 0 ⟹ 푛 − 6푛 + 3푛 − 18 = 0

209

⟹ (푛 − 6)(푛 + 3) = 0 ⟹ 푛 − 6 = 0 or 푛 + 3 = 0 ⟹ 푛 = 6or푛 = −3 But 푛 > 0

Example 11.17 The first term of an AP is 5 and the last term is 35. If the sum of the series is 420, find the common difference.

Solution We have the first term, 푎 = 5, the last term 푢 = 35and the sum of the sequence 푆 = 420 The sum of the 푛th term of an AP is defined as: 푆 = (푎 + 푢 )

⟹ 420 = (5 + 35) ⟹ 420(2) = 2 × (40) ⟹ 840 = 40푛

⟹ 푛 = = 21 The 푛th term of the sequence is given by: 푢 = 푎 + (푛 − 1)푑 ⟹ 35 = 5 + (21 − 1)(푑) ⟹ 35 − 5 = 20푑 ⟹ 푑 = =

∴the common difference of the AP is Try: 1. The nth term, 푈 , of a sequence is given by 푈 = 3 + 5 + 7 + 9 + ⋯+ (2푛 + 1) a. Express 푈 in the form 푈 = 푝푛(푞푛 + 푟),where 푝,푞 and 푟 are constants. Hence, find 푈 ,푈 ,푈 ,푈 , and푈 . b. Find 푎 , 푎 , 푎 , 푎 given that 푎 = 푈 − 푈 . Deduce an expression for 푎 in terms of 푛 and hence find 푈 . (SSSCE) 2. Find the number of terms of the arithmetical progression, 4 + 6 + 9 + 11 + ⋯ needed to make a total of 126. (SSSCE) 3. a. Given that 푎 = 25 and 푎 = 푎 + 4 for 푛 > 1, find : i. 푎 ,푎 , and푎 ii. a formula for 푎 in terms of 푛. b. Use your formula for 푎 to find a formula for 푆 , the sum of the first 푛 terms of the sequence 푎 ,푎 , … , 푎 . Hence find 푆 (SSSCE) 4. A linear sequence has a common difference 3 and fifteenth term is 50, find the first term. (SSSCE) 5. The thirteenth term of an arithmetic progression is 27 and the seventh term is three times the second term. Find;

210

i. The first term, ii. The common difference, iii. The sum of the first 10 terms. (SSSCE) 6. The sum of the first twelve terms of an arithmetic sequence is 168. If the third term is 7. Find the value of i. the common difference; ii. the first term. (SSSCE) 7. The second, fourth and eighth terms of an arithmetic sequence (A. P) forms three consecutive terms of a geometric sequence. The sum of the third and fifth terms of the A. P is 20. Find i. the first four terms; ii. sum of the first ten terms; of the arithmetic sequence. (SSSCE) 8. The sum of the first 푛 terms of an arithmetic sequence is −45 . If the first term is 38 and the common difference is −7, find the value of 푛. (SSSCE) 9. The sum of the first n terms of the series 4 + 7 + 10 + ⋯ is 209. Find 푛. (WASSCE) 10. The sum of the first eight terms of an arithmetic progression is 164. If the sum of the next eight terms is 333. Find the i. first term; ii. common difference. (WASSCE) 11. The sum of the first 푛 terms of a linear sequence (A. P) is −27 . If its first term and

common difference are 1 and − respectively. Find the possible value of 푛. (WASSCE) 12. The sum of the first 푛 terms of a series is 8푛 − 4푛. i. Find the first three terms of the series. ii. Show that the series is a linear sequence. iii. Find the nth term of the series. (WASSCE)

Application of AP Example 11.18

Every new employee in an oil company receives an annual salary of GH₵20,000. An employee is given a salary increase of GH₵2,000 any time he stays in the company up to

211

a maximum of GH₵108,000perannum.Howmuchdoesanewemployeeearnintotalup to the year of maximum salary.

Solution We can form the arithmetic series: 20,000 + 20,000 + 2,000 + ⋯+ 108,000 20,000 + 22,000 + ⋯+ 108,000 First tem of the series, 푎 = 20,000common difference, 푑 = 2,000 and the 푙푎푠푡 term of the series, 푙 = 108,000 The last term of an arithmetic series is defined as: 푙 = 푎 + (푛 − 1)푑 ⟹ 108,000 = 20,000 + (푛 − 1)(2,000) ⟹ 108,000 − 20,000 = 2,000(푛 − 1) ⟹ 88,000 = 2,000(푛− 1) ⟹ 푛 − 1 = ,

, ⟹ 푛 − 1 = 44 푛 = 44 + 1 = 45

Therefore, for an employee to earn a maximum salary of GHC 108,000, he has to stay in the business for 45 years. The sum of the 푛th term of an AP is defined as: 푆 = (푎 + 푢 )

⟹ 푆 = [20,000 + 108,000] = 22.5(128,000) = 2,880,000. ∴, an employee will earn in total GHC 2,880,000 up to the year of maximum salary. Try: A man started a project with $1, 160.00. If he increases his value with $ 40.00 every month until he last invested $ 1,760.00, calculate the total amount he invested in the project at the time he invested $ 1,760.00. (WASSCE)

2. Geometric sequence/Geometric progression (GP) A geometric progression is a sequence where each new term after the first term is obtained by multiplying the preceding term by a constant number. The constant number is called 푐표푚푚표푛푟푎푡푖표. Let 푎 and 푟 stand for the first term and common ratio of a GP respectively. Then we can form the sequence: 푎, 푎 ∙ 푟,푎 ∙ 푟 ∙ 푟,푎 ∙ 푟 ∙ 푟, …

⟹ 푎, 푎푟,푎푟 , 푎푟 , … This follows that the nth term of a GP is given by

푢 = 푎푟

212

This is true because, for a GP with 푛 terms, there are 푛 − 1 common ratios between the successive terms.

Example 11.19 If the sequence 2 − 푥,−1 + 푥, 2푥 + 3, … is a geometric sequence, find the values of 푥.

Solution We have the sequence: 2 − 푥,−1 + 푥, 2푥 + 3, … Since the sequence is a GP, the common ratio: 푟 = = ⟹ 푟 = = ⟹ =

⟹ (−1 + 푥)(−1 + 푥) = (2푥 + 3)(2 − 푥) ⟹ (−1 + 푥) = 4푥 − 2푥 + 6 − 3푥 ⟹ 1 − 2푥 + 푥 = 6 + 푥 − 2푥 ⟹ 푥 + 2푥 − 2푥 − 푥 + 1 − 6 = 0 ⟹ 3푥 − 3푥 − 5 = 0. By comparing this equation with the general equation

푎푥 + 푏푥 + 푐 = 0, 푎 = 3, 푏 = −3 and 푐 = −5

푥 = ±√ = ( )± ( ) ( )( )( )

= ±√ = ±√ = ± √ = ± √2

∴ 푥 = + √2 or 푥 = − √2

Example 11.20 How many terms are there in the sequence 2, 4, 8, … , 256

Solution We have the sequence: 2, 4, 8, … ,256 The first term of the sequence, 푎 = 2 The sequence is a GP with common ratio: 푟 = = = 2

The last term of the GP, 푙 = 256 The 푛th term of a GP is defined as: 푢 = 푎푟 ⟹ 256 = (2)(2) ⟹ 256 = 2 ⟹ 2 = 2 ∴ 푛 = 8 There are 8 terms in the sequence.

Example 11.21 Find the 13th of the following GPs: i. 0.3, 0.6, 1.2, … ii. 6, 18, 24, …

Solution

213

i. We have the sequence: 0.3, 0.6, 1.2, … The first term of the sequence, 푎 = 0.3 The sequence is a GP with common ratio: 푟 = = .

.= 2

The 푛th term of a GP is defined as: 푢 = 푎푟 ⟹ 푢 = 0.3(2) = (0.3)(2) = 1228.8 ∴the 13th term of the sequence is 1228.8 ii. We have the sequence: 6, 18, 24, … The first term of the sequence, 푎 = 6 The sequence is a GP with common ratio: 푟 = = = 3

The 푛th term of a GP is defined as: 푢 = 푎푟 ⟹ 푢 = 6(3) = (6)(3) = 3188646. ∴the 13th term of the sequence is 3,188,646.

Sum of a geometric series Given the GP: 푎,푎푟, 푎푟 ,푎푟 , … ,푎푟 We can perform the sum of the series as:

푆 = 푎 + 푎푟 + 푎푟 + 푎푟 + ⋯+ 푎푟 ………….(1) Let’s multiply 푆 by 푟

⟹ 푟푆 = 푎푟 + 푎푟 + 푎푟 + ⋯+ 푎푟 ………….(2) Subtracting (1) from (2) gives

푟푆 − 푆 = 푎푟 − 푎 ⟹ 푆 (푟 − 1) = 푎(푟 − 1) ⟹ 푆 = 푎(푟푛−1)푟−1

Subtracting (2) from (1) gives

푆 − 푟푆 = 푎 − 푎푟 ⟹ 푆 (1 − 푟) = 푎(1 − 푟 ) ⟹ 푆 = 푎 1−푟푛

1−푟

Example 11.22 Find the sum of the following GPs to the number of terms stated.

푆 = ( ) , 푟 > 1 푆 = ( ) , 푟 < 1

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i. 4 + 12 + ⋯ 8 terms. ii. + + ⋯ 7 terms. iii. −1.2 − 1.4 − 4.2−⋯ 10 terms. Solution

i. We have the series: 4 + 12 + ⋯ The first term of the series, 푎 = 4 The common ratio of the series: 푟 = = = 3 3 > 1

The sum of the 푛th term of a GP is defined as: 푆 = ( )

⟹ 푆 = ( ) = ( ) = 13,120 ∴the sum of the first 8 terms of the GP is 13,120 ii. We have the series: + + ⋯

The first term of the series, 푎 =

The common ratio of the series: 푟 = = = × 4 = < 1

The sum of the 푛th term of a GP is defined as: 푆 = ( )

⟹ 푆 = = = , = × ×,

=

∴the sum of the first 7 terms of the GP is iii. We have the series: −1.2− 1.4 −⋯ The first term of the series, 푎 = −1.2 The common ratio of the series: 푟 = = .

.≈ 1.167 1.167 > 1

The sum of the 푛th term of a GP is defined as: 푆 = ( )

⟹ 푆 ≈ . ( . ).

≈ . ( . ).

≈ 26.4 ∴the sum of the first 10 terms of the GP is approximately 26.4

Example 11. 23 How many terms in the GP 5, 1, , … are needed so that the sum exceed 45.

Solution

215

We have the sequence: 5, 1, … The first term of the sequence, 푎 = 5 The common ratio of the series: 푟 = = = 0.2 0.2 < 1

The sum of the 푛th term of a GP is defined as: 푆 = ( ) We want to find 푛 such that 푆 > 45

⟹ > 45 ⟹ × > 45 ⟹ 5 − 5 > (45)

⟹ 5− 5 > 36 ⟹ −5 > 36 − 5 ⟹ 5 < −31 (Take log of both sides) log 5 < −log 31 (1 − 푛) log 5 < − log 31 (1 − 푛) < −푛 < −2.1336− 1 ∴ 푛 > 3.1336

∴, since 푛 > 3.1336, the number of terms needed for the sum of the sequence to be greater than 45 is 4.

Example 11.24

The first term of a GP is 3 and the common ratio is . How many terms of this sequence must be added to obtain 339?

Solution The first term of the sequence, 푎 = 3 The common ratio: 푟 =

The sum of the 푛th term of a GP is defined as: 푆 = ( ) We want to find 푛 such that 푆 = 339

⟹ 399 = ⟹ 339 = × ⟹ 339 × = 3 − 3

⟹ 3 = 3 − 226 (Take log of both sides) ⟹ log 3 = −223 ⟹ (1 − 푛) log 3 = − log 223 ⟹ (1 − 푛) = ⟹ −푛 ≈ −4.92 − 1 ∴ 푛 ≈ 5.92

∴, approximately 6 terms of the sequence must be added to obtain a 399

Example 11.25 The sum of the first two terms of a GP is 4√2. If the first term is √2. Find the common ratio.

216

Solution The first term of the sequence, 푎 = √2

The sum of the 푛th term of a GP is defined as: 푆 = ( )

We have, 푆 = 4√2 ⟹ = 4√2 ⟹ 푎(1 + 푟) = 4√2

⟹ √2(1 + 푟) = 4√2 ⟹ (1 + 푟) = √√

⟹ 1 + 푟 = 4

⟹ 푟 = 4 − 1 ∴ 푟 = 3

Example 11.26 The first, third and sixth term of an AP are the three consecutive terms of a GP. The sum of the second and tenth term of the AP is 36. Find i. the common ratio. ii. the sum of the first 10 terms of the GP.

Solution The 푛th term of an AP is defined as: 푢 = 푎 + (푛 − 1)푑 The 1st , 3nd and 6th terms of an AP are given respectively as: 푎, 푎 + 2푑,푎 + 5푑 If the terms are the three consecutive terms of a GP, then 푟 = = 푟 = = =

⟹ (푎 + 2푑)(푎+ 2푑) = 푎(푎 + 5푑) ⟹ (푎 + 2푑) = 푎 + 5푎푑 ⟹ 푎 + 4푎푑 + 4푑 = 푎 + 5푎푑 ⟹ 푎 − 푎 + 5푎푑 − 4푎푑 − 4푑 = 0 ⟹ 푎푑 − 4푑 = 0 ⟹ 푑(푎 − 4푑) = 0 ⟹ 푑 = 0 or 푎 − 4푑 =0 ⟹ 푎 = 4푑… … … (1). But 푑 ≠ 0 Also, the 2nd term and the 10th term of the AP are given respectively as: 푎 + 푑,푎 + 9푑 ⟹ 푎 + 푑 + 푎 + 10푑 = 36 ⟹ 2푎 + 10푑 = 36 … … … (2) ⟹ 2(4푑) + 10푑 = 32 ⟹ 8푑 + 10푑 = 36 ⟹ 18푑 = 36 ∴ 푑 = = 2 Put 푑 = 2 into (1) ⟹ 푎 = 4(2) = 8 i. The common ratio, 푟 = = ( ) = = ∴ 푟 =

ii. The sum of the 푛th term of a GP is defined as: 푆 = ( )

⟹ 푆 = ≈ ( . ) ≈ 2 × 8(56.6650) ≈ 906.64

217

∴, the sum of the first 10 terms of the GP is 906.64 (corrected to 2 places of decimal). Try: The 1st , 5th and 10th term of a linear sequence are in geometric progression. If the sum of the 2nd and 8th terms of the linear sequence is 30, find the i. 1st term ii. non-zero common difference of the linear sequence. (WASSCE)

Application of GP Example 11.27

The shrubs in a garden grows at 1.85 cm in the first week. After that week, the shrubs grows 4% more successively each week. By how much does it grow in 8 weeks.

Solution The shrub grows: 푟 = = 0.04 more successfully each week. 푛 = 8. We can form the sequence: ⟹ 1.85, 1.85 + 0.04(1.85), … ⟹ 1.85, 1.924, 1.924 + 0.04(1.924) … ⟹ 1.85, 1.924, 2.00096, … This can be written as a series:

1.85 + 1.85(1 + 0.04) + 1.85(1 + 0.04) + 1.85(1 + 0.04) + ⋯+ 푎(1 + 푟 ) The first term of the sequence, 푎 = 1.85 The sequence is a GP with common ratio, 푟 = (1 + 푟 ) = = .

.= 1.04

The sum of the 푛th term of a GP is defined as: 푆 = ( )

⟹ 푆 = . ( . ).

= . ( . ).

≈ . ( . ).

≈ 17.0478 ∴, the shrub will grow approximately 17.0478more successfully in 8 weeks. Note: An 푟%increase in the value of 푥 is given by 푥 + 푥푟% = 푥(1 + 푟%)

Example 11.28

Mr. Appiah invested GH₵120,000intoabeebreedingproject.Theinvestmentcapitalgrows at 5% per annum. By how many years will his capital exceed GH₵1,000,000.

Solution The investment capital grows: 푟 = = 0.05 per annum

218

The initial capital invested= GH₵120,000 We can form the sequence ⟹ 120,000, 120,000 + 0.05(120,000), … ⟹ 120,000, 126,000, 126,000 + 0.05(126,000) … ⟹ 120,000, 126,000, 132,300, … This can be written as a series:

120,000 + 120,000(1 + 0.05) + 120,000(1 + 0.05) + 120,000(1 + 0.05) + ⋯+ 푎(1 + 푟 )

The first term of the sequence, 푎 = 120,000 The sequence is a GP with common ratio, 푟 = (1 + 푟 ) = = ,

,= 1.05

We want to find 푛, such that 푎(1 + 푟 ) > 1,000,000 ⟹ 120,000(1 + 0.05) > 1,000,000 (1 + 0.05) > , ,

,

⟹ (1.05) > 8.3333 (Take log of both sides) ⟹ log(1.05) > log 8.3333 푛 log 1.05 > log 8.3333 ⟹ 푛 > .

. 푛 > 43.46. Since, 푛 > 43.46 ∴ 푛 = 44.

∴, the investment capital will exceed GHC1,000,000 in 44 years.

Recursive sequences A sequence written such that the nth term depends on some of the terms preceding it is called 푟푒푐푢푟푠푖푣푒푠푒푞푢푒푛푐푒. The recursive sequence of an AP and a GP are respectively written as 푢 = 푢 + 푑 and 푢 = 푟푢

Example 11.29 The nth term of a sequence is defined by as 푢 = 푢 − , 푢 = 1. Find 푢 ,푢 and 푢

Solution We have, 푢 = 푢 − , 푢 = 1

⟹ 푢 = 푢 − = 푢 − = (1)− = = ∴,푢 =

⟹ 푢 = 푢 − = 푢 − = − = − = = − ∴,푢 = −

⟹ 푢 = 푢 − = 푢 − = − − = − − = = − ∴,푢 = −

219

Example 11.30 A sequence is defined as 푢 = 푢 + 푢 ,푢 = 3, 푢 = −2. Write down the first 6 terms of the sequence.

Solution We have, 푢 = 푢 + 푢 , 푢 = 3 푢 = −2 ⟹ 푢 = 푢 + 푢 = 푢 + 푢 = −2 + 3 = 1 ⟹ 푢 = 푢 + 푢 = 푢 + 푢 = 1 − 2 = −1 ⟹ 푢 = 푢 + 푢 = 푢 + 푢 = −1 + 1 = 0 ⟹ 푢 = 푢 + 푢 = 푢 + 푢 = 0 − 1 = −1 ∴,the first six terms of the sequence are 3,−2, 1,−1, 0,−1 respectively.

Sigma notation Assuming we are to list the series of the first 1000 natural numbers. This may seem to be easy but it involves a lot of time. A Mathematician came up with a formula which involve the Greek upper case letter ∑(sigma) to help us in such calculations. The sigma notation has three components: 1. the nth term of the sequence. 2. Upper limit. 3. Lower limit. The lower limit is the argument that output the terms of our series of interest while the upper limit is the argument that output the last term of our series of interest. So our problem can be simplified as

10003211000

1

nn

In general,

b

annu , is read ‘The summation of 푢 from 푛 = 푎 to 푛 = 푏. ′푛 is called the

index of summation. Note the following properties

1.

b

ann

b

ann

b

annn uaua

2.

b

ann

b

ann

b

annn uaua

3.

b

nncc

1, where c is constant. For instance, 20545

4

1

n

220

4.

b

ann

b

ann uccu

Example 11.31 Simplify the following:

i.

5

1)1(

nn ii.

7

1)1(2

nnn iii.

14

0 31

n

n iv.

1)12(

nn v. 1

1(3 )n

n

Solution

i.

5

1)1(

nn = (1 + 1) + (2 + 1) + (3 + 1) + (4 + 1) + (5 + 1)

= 2 + 3 + 4 + 5 + 6

= 20 ∴

5

1)1(

nn = 20

ii.

7

1)1(2

nnn = 2 ∙ 1(1 + 1) + 2 ∙ 2(2 + 1) + 2 ∙ 3(3 + 1) + 2 ∙ 4(4 + 1)

+2 ∙ 5(5 + 1) + 2 ∙ 6(6 + 1) + 2 ∙ 7(7 + 1) = 2(2) + 4(3) + 6(4) + 8(5) + 10(6) + 12(7) + 14(8) = 4 + 12 + 24 + 40 + 60 + 84 + 112 = 336

∴,

7

1)1(2

nnn = 336

iii.

4

0

1

31

n

n

= + + + +

= + + + +

= + + + +

= ∴

4

0

1

31

n

n

=

221

iv.

1)12(

nn = (2 ∙ 1 − 1) + (2 ∙ 2 − 1) + ⋯ = 2(∞)− 1 = ∞− 1 = ∞

∴

1)12(

nn = ∞

v. 1

1(3 )n

n

= (3 ) + (3 ) + 3 … = 1 + 3 + 9 + ⋯ = ∞

∴ 1

1(3 )n

n

= ∞

Example 11. 32

Find the sum of the first 30 terms of the terms of the sequence 1, 4, 9 … Solution

We have the sequence: 1, 4, 9, …

The sum of the sequence: 1 + 4 + 9 + ⋯ =

30

1

2

nn

Note that the sequence is a sequence of square numbers.

Graphing sequences Sequences are graphed using the number of a term and the value of the term as an ordered. In fact, sequences are function with domain ℕ, the set of natural numbers. Let 푎 be the first term of a sequence. Then we have the ordered pair (1,푎). For instance, the sequence 1, 3, 5, 7 forms ordered pairs (1, 1), (2, 3), (3, 5), (4, 7). We use these points to draw the graph below:

222

푢 8 ⦁ 6 ⦁ 4 ⦁ 2 ⦁ 0 푛 1 2 3 4 The slope/gradient of the graph is estimated as 푚 = ∆

∆= = 2 = 푑

Try: Graph the following sequences: i. 2, 4, 6, 8, 10 ii. 3, 6, 9, 12, 15 iii. 5, 15, 45, 135 iv. 10, 100, 1000, 10000

Convergence of a of a GP

(Sum to Infinity) Consider the sum of the nth term of the geometric progression:

푆 = ( ) = − As the value of 푛 becomes large and 푟 < 1, the value of 푟 gets closer to zero. Symbolically, as 푛 → ∞, 푟 → 0if – 1 < 푟 < 1. Therefore, the sum to infinity of a geometric series is given by:

푆 = 푎1−푟

Example 11.33 Find the sum to infinity of the GP with first term 4 and common ratio .

Solution We have first term, 푎 = 4 and common ratio 푟 =

푆 =푎

1 − 푟 , 푟 ≠ 1

223

The sum to infinity of a GP is defined as: 푆 =

⟹ 푆 = = = 4 × 2 = 8. ∴, the sum to infinity of the GP is 8.

Example 11.34 The sum to infinity of a GP is 3 times the sum of the first two terms. Find the common ratio.

Solution The sum to infinity of a GP is defined as: 푆 =

The sum of the푛th term of a GP is defined as: 푆 = , where 푟 is the common ratio.

We have, 푆 = 3푆 ⟹ = 3 ∙ ⟹ 푎 = 3푎(1 − 푟 ) ⟹ 1 = 3(1 − 푟 ) ⟹ 1 = 3 − 3푟 ⟹ 3푟 = 3− 1

⟹ 3푟 = 2 ∴ 푟 = ±

Example 11.35

Given the relation 푢 = 9 + 3 , for 푢 , 푢 …. Show that 푢 = 3(푢 − 6). Proof

푢 = 9 + 3 ⟹ 푢 = 9 + 3( ) = 9 + 3 = 3(3 + 3 ) = 3(9 + 3 − 6) = 3(푢 − 6) ∎ Try: A sequence 푈 is defined by 푈 = 2,푈 = 2푈 − 1,푛 = 1, 2, 3, … i. Find first five terms of the sequence. ii. Show that 푈 = 3푈 − 2푈 (SSSCE) Final exercises 1. The ninth term of an arithmetic progression is 52 and the sum of the first twelve terms is 414. Find the first term and the common difference. (AEB) 2. In an arithmetic progression the eight term is twice the third term and the twentieth term is 100.

224

a. Find the common difference. b. Determine the sum of the first 100 terms. 3. The first term of a geometric progression is 8 and the sum to infinity is 400. a. Find the common ratio. b. Determine the least number of terms with a sum greater than 399. (AEB) 4. The sum of the first and second term of a geometric progression is 108 and the sum of the third and fourth term is 12. Find the possible values of the common ration and the corresponding number of first term. 5. The first term of a geometric sequence is 7 and the third term is 63. Find the second term. 6. Consider the series 5 + 9 + 13 + 17 + ⋯ a. How many terms of this series are less than 1000? b. What is the value of 푛 for which 푆 > 1000? 7. Evaluate the following

i.

5

12

n ii.

20

0)32(

nn iii.

1

1)5.0(n

n iv.

0 323

n

n

8. An AP has a common difference of 2. If the 19th term is 39, find the common difference. Hence, write down the first four terms of the sequence. 9. The 15th term of an AP is 103 and the common difference is 4. Find a formula for the nth term of the sequence. 10. An AP has first term 2 and a common difference of 1.5. if the last term of the sequence is 64, find the number of terms of the sequence. 11. For what value of 푛 does the sequence 20 + 60 + 100 + ⋯+ 20(3) exceed 1500? 12. New employees joining a firm in logistics receive an annual salary of ₦85,000. Every year they stay with the firm they have a salary increase of ₦8,000 up to a maximum of ₦133, 000 per annum. How much does a mew employee earn in total up to and including the year of maximum salary. 13. Find the first five terms of the following sequences.

i. 푢 = 3 ∙ (2) ii. 푢 = −5(5) iii. 푢 = 3 −

iv. 푎 = 푎 , 푎 = 2 14. Find the 12th term of the a GP with the following first term and common ratio:

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i. 푎 = 2,푟 = ii. 푎 = 9,푟 = iii. 푎 = 1,푟 = 15. Find the 10th term of the following GP : i. 2, 10, 50, … ii. 16, 4, 1, …. iii. 5, 15, 45, … iy. 3,−6, 12, … v. 2, 1, , … 16. Find i. the first term. ii. common difference. iii. the number of terms of a sequence with last three terms 82, 85, 88 17. Find the sum of the series i. −1 + 2 + 5 + ⋯+ to 14 terms. ii. 22 + 17 + 12 + ⋯+ to 10 terms iii. 7.2 + 7.8 + 8.4 + ⋯+to 17 terms. iv. 8 + 16 + 32 + ⋯+ to 9 terms. 18. How many terms are there in the series if i. 16 + 32 + 64 + ⋯+= 824? ii. 0.25 + 0.5 + 0.75 + ⋯+= 25 iii. 1 + 2 + ⋯+= 100 19. The sum of the first three of a GP is . the sum of the first six terms is . Find the first term and the common ratio. (Math Centre) 20. Find the first five terms of the following: i. 푢 = −13 + 푛 ii. 푢 = 푢 − 3, 푢 = −3 iii. 푎 = , 푎 = 푎 + . 21. Find the nth term of the following arithmetic sequences: i. −2, 5, … ii. 2,−5, … iii. 0.2, 0.4, … 22. Find the sum of the first seventeen terms of the linear sequence 2, 1, 0, … 23. The first term of an AP is 1.5 and the common difference is 6. Find i. the last term of the sequence. ii. the sum of the first 16 terms of the sequence. 24. The first and second terms of an AP are 6 and 9 respectively. Find i. the common difference. ii. the number of terms of the sequence if the 13th term is 39. 25. Write down the first eight terms of the Fibonacci sequence obtained by 푢 = 푢 + 푢 ,푢 = 1 and 푢 = 1. 26. An AP is given by 푘, , , i. Find the sixth term. ii. Find the nth term. iii. If the 20th of the is equal to 15, find 푘. (Math Centre) 27. Write down the first 7 term of the following sequences with the nth terms: i. 푢 = (−푎) ii. 푢 = 5 + 2 iii. 푢 = +

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iv. 푢 = 2푛 v. 푎 = ( ) vi. 푎 = ( )( ) vii. 푎 = (−3) + 1 viii. 푎 = 6(3 − 2) 28. Find the sum of the first 20 terms of the sequence 1, 4, 9, … 29. Write down the first five term of an AP with first term 2 and common difference 3. 30. Write down the 10th and the 15th of the following linear sequences i. 6, 2,−2, … ii. 8, 5, 2, … iii. 1, 7, … iv. 0.2, 0.4, … 31. An arithmetic series has first term 4 and common difference . Find i. the sum of the first 20 terms ii. the sum of the first 100 terms. (Math Centre) 32. The sum of the first 20 terms of an arithmetic series is identical to the sum of the first 22 terms. If the common difference is −2, find the first term. (Math Centre)

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TRIGONOMETRY The word ‘trigonometry’ comes from two Greek words namely, 푡푟푖푔표푛표푛 meaning 푡푟푖푎푛푔푙푒 and 푚푒푡푟표푛 meaning 푚푒푎푠푢푟푒.Trigonometry started with the study of triangles. A triangle is a plane figure formed by three lines segment joined together. Definition: An angle is formed when two rays meet at a point called the 푣푒푟푡푒푥 . 훼

Angles are measured in either 푑푒푔푟푒푒 or 푟푎푑푖푎푛. An angle equal to 90° is called a 푟푖푔ℎ푡푎푛푔푙푒 . An angle equal to 180° is called a 푠푡푟푎푖푔ℎ푡푎푛푔푙푒. Two angles add up to 90° are said to be 푐표푚푝푙푒푚푒푛푡푎푟푦푎푛푔푙푒푠. Two angles add up to 180° are said to be 푠푢푝푝푙푒푚푒푛푡푎푟푦푎푛푔푙푒푠. An angle of revolution is an angle equal to 360°. An angle between 0° and 90° is called an 푎푐푢푡푒푎푛푔푙푒. An angle between 90° and 180° is called an 표푏푡푢푠푒푎푛푔푙푒. An angle whose value is between 180° and 360° is called a 푟푒푓푙푒푥푎푛푔푙푒. The sum of angles in a triangle is equal to 180°. A triangle whose one angle is 90° is called a푟푖푔ℎ푡푡푟푖푎푛푔푙푒.

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The longest side of a triangle is called ℎ푦푝표푡푒푛푢푠푒. If the angle we are interested in is angle휃, then the side of the triangle far away from 휃is called the 표푝푝표푠푖푡푒. The side near to 휃 is called 푎푑푗푎푐푒푛푡.

ℎ푦푝표푡푒푛푢푠푒 Opposite 휃 Adjacent Let푎,푏and푐 the lengths of the hypotenuse, opposite and adjacent respectively. Then

푎 = 푏 + 푐 (This result is called the 푃푦푡ℎ푎푔표푟푒푎푛푇ℎ푒표푟푒푚; named after the Greek philosopher and mathematician Pythagoras) The area of a triangle is calculated by the formula

퐴 =12 × 푏푎푠푒 × ℎ푒푖푔푡

The height is also known as 푎푙푡푖푡푢푑푒.

Pythagorean triple A Pythagorean triple is a combination of three numbers (푎, 푏,푐) that obeys the Pythagorean Theorem. That is, a Pythagorean triple form the sides of a right triangle. Given any two numbers 푚 and 푛,푚 > 푛, the combination (푚 + 푛 , 2푚푛,푚 − 푛 ) forms a Pythagorean triple. For example, given the numbers 1 and 2, the combination (2 + 1 , 2 ∙ 2 ∙ 1, 2 − 1 ) = (5, 4, 3) forms a Pythagorean triple. Also, given the numbers 3 and 2, the combination (3 + 2 , 2 × 3 × 2, 3 − 2 ) = (13, 12,5) form a Pythagorean triple. Similarly, given the numbers and , the combination

+, 2 × × , − = , , forms a Pythagorean triple.

Example 12.1 The numbers 4 and 5 form a Pythagorean triple (41, 푥, 9) find the value of 푥.

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Solution The two numbers form a Pythagorean principle from the combination: (푚 + 푛 , 2푚푛,푚 − 푛 ) We have, 푚 = 5 and 푛 = 4 So, (5 + 4 , 2(5)(4), 5 − 4 ) = (25 + 16, 40, 25 − 16) = (41, 40, 9) But, (41, 40, 9) = (41,푥, 9) ∴ 푥 = 40.

Congruent and Similar triangle Two triangles are said to be 푐표푛푔푟푢푒푛푡 if they have shape and size. Two triangles are said to be similar if they have the same shape but different size.

Congruent triangles 푎 푏 퐴 퐵 Similar triangles 푐 퐶 The corresponding sides of similar triangles have the same proportion. from the similar triangles above

퐴푎 =

퐵푏 =

퐶푐

Trigonometric Ratios The three basic trigonometric ratios are 푠푖푛푒 (sin), 푐표푠푖푛푒(cos) and 푡푎푛푔푒푛푡(tan). Consider the triangle below: A sin 휃 = =

cos휃 = =

tan 휃 = =

휃

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B C To help remember these ratios you can use the acronym (SOHCAHTOA).

Reciprocal ratios The reciprocal ratios for sine, cosine and tangent are 푐표푠푒푐푎푛푡(cosec/csc), 푠푒푐푎푛푡(sec) and푐표푡푎푛푔푒푛푡(cot)respectively. That is,

cosec 휃 = sec 휃 = cot 휃 = Inverse ratios5

In general, if sin 푥 = 푘,then 푥 = sin 푘. The inverse ratios for sine, cosine and tangent are:

sin-1=arc sin cos-1=arccos and tan-1=arctan respectively. Example 12.2

Given that sin 푥 = , 0° < 푥 < 90°, find cos푥, tan 푥and sec 푥. Solution

We have, sin 푥 = , 0° < 푥 < 90° A |퐴퐶| = |퐴퐵| + |퐵퐶| ⟹ 17 = 15 + 푑 ⟹ 289 = 225 + 푑 ⟹ 289 − 225 = 푑 15 17 ⟹ 푑 = 64 ⟹ 푑 = √64 =8 푥 ∴ cos 푥 = ∴ tan 푥 = ∴ sec 푥 = = B 푑 C

Example 12.3 Given that tan퐴 = , 0° < 퐴 < 90°, find sin퐴,cos퐴 and cosec퐴.

Solution We have, tan퐴 = , 0° < 퐴 < 90° A |퐴퐶| = |퐴퐵| + |퐵퐶| ⟹ 푑 = 1 + 2 ⟹ 푑 = 1 + 4 ⟹ 푑 = 5 1 푑 ⟹ 푑 = √5 퐴 B 2 C

5 The inverse ratios and the reciprocal ratios are not equal. For instance,cos-1≠ .

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∴ sin퐴 =√

= √ ∴ cos퐴 =√

= √ cosec 퐴 = = √5

In trigonometry, angles are measured in an anticlockwise direction. The diagram below shows the direction of basic trigonometric ratios.

90°

Positive measurement 180°0°/360°

Negative measurement

270° Let an angle of interest be 휃. Then from the diagram: If 0° < 휃 < 90°, all the basic ratios give a positive value. If 90° < 휃 < 180°, only sin gives a positive value. If 180° < 휃 < 270°, only tan gives a positive value. If 270° < 휃 < 360°, only cos is positive.

Example 12.4

Given that 푐표푠훼 = − , 180° < 훼 < 270°, find sin훼 and cos훼. Solution

We have, cos훼 = − , 180° < 훼 < 270° A |퐴퐶| = |퐴퐵| + |퐵퐶| ⟹ 13 = 푑 + 12 ⟹ 169 = 푑 + 144 ⟹ 푑 = 169 − 144 푑 13 ⟹ 푑 = 25 ⟹ 푑 = √25 = 5 ∴ sin 훼 = − ∴ tan 훼 = 훼 B 12 C

Example 12.5 If sin휑 = , 90° < 휑 < 180°, find cos휑and tan휑.

sin 푎푙푙 tan cos

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Solution We have, sin휑 = , 90° < 훼 < 180° A |퐴퐶| = |퐴퐵| + |퐵퐶| ⟹ 3 = 2 + 푑 ⟹ 9 = 4 + 푑 ⟹ 푑 = 9 − 4 2 3 ⟹ 푑 = 5 ⟹ 푑 = √5 휑

∴ cos휑 = − √ ∴ tan휑 = −√

= − √ B 푑 C

Example 12.6

Given that tan훽 = − , 90° < 훽 < 180°, find sin 훽and cos훽. Solution

We have, tan훽 = − , 90° < 훼 < 180° A |퐴퐶| = |퐴퐵| + |퐵퐶| ⟹ 푑 = 4 + 5 ⟹ 푑 = 16 + 25 ⟹ 푑 = 41 4 푑 ⟹ 푑 = √41 훽 B 5 C

∴ sin 훽 =√

= √ ∴ cos훽 = −√

= − √

Finding an acute angle

Let 휃 be a given angle and let 훼 be an acute angle. The relationship 휃 and 훼 and the positive 푥 −axis in the various quadrants is shown below: 푦 푦

훼 훼 푥 푥 훼 = 180°− 휃 훼 = 휃

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푦 푦 훼 푥 푥 훼 훼 = 휃 − 180° 훼 = 360°− 휃

Example 12.7

Find the acute angle for the following trigonometric ratios: i. sin 120°, cos 120° and tan 135° ii. Sin 210°, cos 230° and tan 220° iii. sin 315°, cos 330° and tan 330°

Solution i. sin 120° = sin(180°− 120°) = sin 60°, where 90° < 120° < 180° cos 120° = −cos(180° − 120°) = −cos 60°, where 90° < 120° < 180° tan 135° = − tan(180°− 135°) = −tan 45°, where 90° < 135° < 180° ii. sin 210° = −sin(210° − 180°) = −sin 30°, where 180° < 210° < 270° cos 230° = −cos(230° − 180°) = −cos 50°, where 180° < 230° < 270° tan 220° = tan(220°− 180°) = tan 40°, where 180° < 220° < 270° iii. sin 315° = −sin(360°− 315°) = −sin 45°, where 270° < 210° < 360° cos 330° = cos(360° − 330°) = cos 30°, where 270° < 330° < 360° tan 330° = −tan(360°− 330°) = −tan 30°, where 270° < 330° < 360°

Converting Angles from degree to radian and vice versa

Definition: A central angle is an angle whose vertex is the centre of a circle.

푎푟푐푙푒푛푔푡ℎ

푟 휃 푟

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Definition: A radian is a central angle whose arc length is equal to the radius of the circle. From the definition; 180° = 휋 rads. 1° = rads. 1 rad= °

Example 12.8

Change the following from degrees to radians: i. 60° ii. 135° iii. 75° iv. 90° vi. 180° v. 45° vi. 300°

Solution i. If 1° =

°rad, then 60° =

°× 60° = rads ∴ 60° = rads

ii. 135° =°

× 135° = rads ∴, 135° = rads

iii. 75° =°

× 75° = rads ∴, 75° = rads

iv. 90° =°

× 90° = rads ∴, 90° = rads

v. 45° =°

× 45° = rads ∴, 45° = rads

vi. 300° =°

× 300° = rads ∴, 300° = rads Try: Convert the following from degrees to radians. i. 270° ii. 315° iii. 22.5° iv. 250°

Example 12.9

Convert the following from radians to degrees: i. 휋rads ii. 휋rads iii. 휋rads iv. 휋rads v. rads vi. 휋rads

Solution i. If 1rad = °,then 휋rads = ° × 휋 = 180° × = 120° ∴, 휋 = 120° ii. 휋rads = ° × 휋rads = 180° × = 240° ∴, 휋 = 240°

iii. 휋rads = ° × 휋rads = 180° × = 67.5° ∴, 휋 = 67.5°

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O 푟 휃 푟

iv. 휋rads = ° × 휋rads = 180° × = 135° ∴, 휋 = 135°

v. rads = ° × rads = 180° × = 30° ∴, = 30°

vi. 휋rads = ° × 휋 = 180° × = 270° ∴, 휋 = 270° Try: Convert the following from radians to degrees i. 휋rads ii. 휋rads iii. 휋rads

Circular measures Consider a sector of a circle which forms an angle of 휃 (measured in degrees) at the centre of a circle with a minor arc 푙.

Angle 휃 =°

× 휃푟푎푑푠 1. The length of 푚푖푛표푟푎푟푐 generated by the Sector is given by: 푙 =

°× 2휋푟 =

°× 휋푟 =

°× 휃푟

∴ 푙 = 푟휃 (휃 is now converted to radians) A B 2. The length of the major sector left is given by: 퐿 = 훼푟, where훼 = 2휋 − 휃 Now consider triangle AOB sin 휃 = = ⟹ 푟 sin휃 = ℎ 푂

퐴푟푒푎표푓∆퐴푂퐵 = 푏푎푠푒 × ℎ푒푖푔ℎ푡 휃 6 ∗

= 푟 × ℎ푟푟

= 푟 × 푟 sin휃 A B

= 푟 sin휃 3. Area of the sector is given by: 퐴 =

°× 휋푟 =

( °)× 휃푟 ∴ 퐴 = 푟 휃

6 Try to prove that the length of chord AB is given by |퐴퐵| = 2푟 sin 휃

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4. The area of the major sector left is given by: 훢 = 푟 훼 , 훼 = 2휋 − 휃 (훼푎푛푑휃 are measured in radians)

5. The area of the 푚푖푛표푟 segment by the chord AB and the minor arc is given by: 퐴 = 푎푟푒푎표푓푠푒푐푡표푟 − 푎푟푒푎표푓∆퐴푂퐵 = 푟 휃 − 푟 sin휃 = 푟 (휃 − sin 휃) 6. The area of the major segment left is given by:

퐴 = 푎푟푒푎표푓푚푎푗표푟푠푒푐푡표푟 + 푎푟푒푎표푓∆퐴푂퐵 = 푟 훼 + 푟 sin 휃 = 푟 (훼 + sin 휃) where, 훼 = 2휋 − 휃

Example 12.10 A circle of radius 3.5푐푚subtends an angle of 푟푎푑푠 at the centre. Find, leaving your answers in 휋, i. the length of arc generated by the sector. ii. the area of the sector iii. the area of the minor segment; of the circle.

Solution We have, radius, 푟 = 3.5cm and 휃 =

i. The length of arc is defined as: 푙 = 휃푟 ∴ 푙 = 3.5 × = 휋cm

ii. The area of the minor sector is defined as: 퐴 = 푟 휃

∴ 퐴 = (3.5) × = . 휋 = 휋cm

iii. The area of the minor segment is defined as: 퐴 = 푟 − sin

∴ 퐴 = (3.5) − sin = . − = × − 1 = − 1 cm

Example 12.11 The diameter of a circle is 12cm. If the circle makes an angle of 휃 at the centre, find i. the value of 휃 if the minor length arc generated the sector is 16.23cm. ii. the area of the minor and major sector of the circle, leaving your answer in 휋. iii. the area of the minor segment, leaving your answer in 휋.

Solution We have, diameter, 푑 = 12cm. But 푟 = = = 6cm i. The length of arc is defined as: 푙 = 휃푟 ⟹ 2.23휋 = 6휃

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O 푟 휃 푟 푁

∴ 휃 = . ≈ 0.372휋

ii. The area of the minor sector is defined as: 퐴 = 푟 휃

∴ 퐴 ≈ (6) × 0.372휋 ≈ 18 × 0.372휋 ≈ 6.696휋cm

The area of the major sector is defined as: 퐴 = 푟 훼, 훼 ≈ 2휋 − 0.372휋 ≈ 1.628휋 ∴ 퐴 ≈ (6) × 1.628휋 ≈ 18 × 1.628휋 ≈ 29.304휋cm

iii. The area of the minor segment is defined as: 퐴 = 푟 (휃 − sin휃)

∴ 퐴 = (6) (0.372휋 − sin 0.372휋) = 18(0.372휋 − 0.920)cm Example 12.12

A chord AB generated by a sector of a circle of radius 푟 is 2√3푐푚.if the perpendicular from the centre 푂 to 퐴퐵 is 4cm, find i. the radius of the circle correct to 2 decimal places. ii. the angle 휃 at the centre of circle subtended by the sector, leaving your answer in 휋. iii. the area of the sector.

Solution We have, |퐴퐵| = 2√3cm |푂푀| = 4cm (see the figure below)

i. |푂퐴| = |퐴푁| + |푂푁| ⟹ 푟 = √3 + 4 푟 = 3 + 16 ⟹ 푟 = √19 = 4.36cm ii. From ∆푂퐴푁 tan 휃 = | |

| |

⟹ tan 휃 = √ 휃 = tan √

⟹ 휃 ≈ 23.4132° 휃 = 2 × 23.4132° ≈ 46.8264° A B iii. The area of the minor sector is defined as: 퐴 = 푟 휃

∴ 퐴 = (4.36) × 0.2601휋 = 9.5048 × 0.2601휋 ≈ 2.4722휋cm

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Try: 1. A sector of a circle of radius 5푐푚 subtends an angle of 75° at the centre. Find, leaving your answer in 휋, i. the area of the sector. ii. the perimeter of the sector. iii. The length of the minor segment generated by the sector. 2. A sector of a circle of radius 7푐푚 subtends an angle of 휃 at the centre. If the area of the sector is 푐푚 Find, leaving your answer in 휋, i. the value of 휃 ii. the length of the chord generated by the sector. iii. the area of the major segment of the circle.

Trigonometric ratios of negative angles Consider the diagram below: 푦 B In this figure 퐵퐴퐶 = 퐶퐴퐸 but are measured in the opposite directions 푥 cos(−휃) = cos휃 = AC sin(−휃) = 퐶퐸 = −퐵퐶 = − sin휃 D ∴ sin(휃) = − sin휃 note that 퐵퐸 = 퐵퐶 + 퐶퐸

Example 12.13 Simplify the following negative trigonometric ratios: i. sin(−30°) ii. cos(−120°) iii. tan(−300°) iv. cos(−225°)

Solution i. sin(−30°) = − sin 30 ° ii. cos(−210°) = cos 210° = cos(210°− 180°) = − cos 30° iii. tan(−300°) = −tan 300 ° = −[−tan (360° − 300°)] = tan 30° iv. cos(−225°) = cos 225 ° = −cos (225°− 180°) = −cos 45° Try:

1 휃 A -휃 C 1

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Simplify the following negative trigonometric ratios: i. cos(−60°) ii. tan(−105°) iii. cos(−300°) iv. sin(−230°)

Trigonometric ratios of some special angles (a) By constructing a square of side 1 unit: A 1 B Take ∆퐵퐷퐶. |퐵퐷| = |퐵퐶| + |퐷퐶| 1 1 푥 = 1 + 1 = 2 푥 = √2 D 1 C

sin 45° = =√

= √ cos 45° = =√

= √ tan 45° = = 1

(b) By constructing an equivalent triangle of side 1 unit. A Take ∆퐴퐵푀. 30° 30° |퐴퐵| = |퐵푀| + |퐴푀| 1 푥 1 2|퐵푀| = |퐵퐶| ∠퐵퐴푀 = ∠퐶퐴푀 60° 60° B M C

|퐴퐵| = |퐵푀| + |퐴푀| ⟹ 1 = + 푥 ⟹ 푥 = 1 − = ⟹ 푥 = = √

sin 30° = = cos 30° = = √ tan 30° = = ÷ √ = ×√

=√

= √

sin 60° = = √ cos 60° = = tan 60° = = √ ÷ = √ × 2 = √3 (c) By constructing a unit circle: A unit circle is a circle whose centre is the origin and a radius of 1 unit. The equation of the unit circle is given by 푥 + 푦 = 1.

푥

45°

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90° (0, 1) 푃(푥,푦) = cos 휃, sin휃) 180° (-1, 0) 0°(1, 0) 270° (0, -1)

From ∆푃푂푄 sin휃 = = 푦 cos휃 = = 푥 tan 휃 = =

The point that corresponds to 0°/360° is (1, 0).

∴, sin 0° = 0 cos 0° = 1 tan 0° = = 0 The point that corresponds to 90° is (0, 1).

∴, sin 90° = 1 cos 90° = 0 tan 90° = = notde ined The point that corresponds to 180° is (−1, 0).

∴, sin 180° = 0 cos 180° = −1 tan 180 = = 0 The point that corresponds to 270° is (0,−1) ∴, sin 270° = −1 cos 270° = 0 tan 270° = = notde ined

휃 0° = 0 30° =

휋6 45° =

휋4 60° =

휋3

90° =휋2

180° = 휋 270° =3휋2

360° = 2휋

sin휃 0 12 √2

2 √32

1 0 −1 0

cos휃 1 √32

√22

12 0 −1 0 1

1 휃 푦 O(0, 0) 푥 Q

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tan휃 0 √33

1 √3 − 0 − 0

Example 12.14 Simplify the following leaving your answer as a surd. i. sin 150° ii. Cos 150° iii. tan 120° iv. cos 210° v. tan(−315°)

Solution i. sin 150° = sin(180°− 150°) = sin 30° = ∴ sin 150° =

ii. cos 150° = −cos(180°− 150°) = −cos 30° = √ ∴ cos 150° = √

iii. tan 120° = − tan(180°− 120°) = −tan 60° = √3 ∴ tan 120° = √3 iv. cos 210° = −cos(210°− 180°) = −cos 30° = − ∴ cos 210° = − v. tan(−315)° = −tan 315 ° = −[−tan (360°− 315°)] = tan 45° ∴ tan 45° = 1 Try: Simplify the following leaving your answer as a surd i. sin 120° ii. sin(−210°) iii. cos(−315°)

Simple trigonometric identities

Consider the right triangle below: A sin 휃 = cos휃 = tan 휃 =

= ÷ = × = = tan 휃

푥 푟 휃 B 푦 C

cot 휃 = = From the Pythagorean Theorem:

|퐴퐵| +|퐵퐶| = |퐴퐶| ⟹ 푥 + 푦 = 푟 … … …(1)

By dividing both sides by 푟 we get: + = ⟹ + = 1

tan 휃 =sin 휃cos휃

242

⟹ + = 1 But sin휃 = cos휃 = ⟹ sin θ + cos θ = 1 … … …(2)7

∴ sin θ = 1 − cos θ cos θ = 1 − sin θ

Divide (2) by cos θ : + = ⟹ tan θ+ 1 = sec θ ∴ sec θ − tan θ = 1

Divide (2) by sin θ : + = ⟹ 1 + cot θ = cosec θ ∴ cosec θ − cot θ=1

Example 12.14

Given that sin휃 = − , find cos휃, tan휃and sec 휃. Solution

We have, sin휃 = − . But sin 휃 + cos 휃 = 1 ⟹ cos 휃 = 1 − sin 휃

⟹ cos 휃 = 1 − − ⟹ cos 휃 = 1 − ⟹ cos 휃 =

⟹ cos휃 = ± = ± √ . But 180° < 휃 < 270°

∴, cos휃 = − √ , tan휃 = = √ = ×√

=√

= √ and

cosec 휃 = = = −4 [Hint: If 푓: 푥 → 푥 ,푓(푎) = 푎 ]

Try: The functions 푓 and 푔 are defined on the set of real numbers by 푓: 푥 → 1 − 푥 and 푔: 푥 → 1 + 푥 Find i. 푓(cos푥) ii. 푔(tan 푥) iii. 푓 ∘ 푔(푥) (SSSCE)

Trigonometric ratios of Complementary angles

7 In general, for trigonometric ratios, for example, (sin 푥) = sin 푥

sin θ + cos θ = 1

tan θ + 1 = sec θ

1 + cot θ = cosec θ

243

If 훼 + 훽 = 90° then 훼 and 훽 are said to be complementary angles. Consider the right triangle below: A 훽 sin 훼 = sin훽 =

cos 훼 = cos훽 =

푥 푟 tan훼 = tan훽 =

훼 + 훽 + 90° = 180° 훼 훽 = 90°− 훼 B 푦 C sin 훼 = cos훽 ⟹ sin훼 = cos(90°− 훼) cos훼 = sin 훽 ⟹ cos훼 = sin(90°− 훼) tan훼 = = cot훽 ⟹ tan 훼 = cot(90°− 훼)

Example 12.15

Find the complementary ratios for the following ratios. i. sin 70° ii. cos 30° iii. cos 45° iv. tan 15°

Solution i. sin 70° = cos(90° − 70°) = cos 20° ∴ sin 70° = cos 20° ii. cos 30° = sin(90°− 30°) = sin 60° ∴ cos 30° = sin 60° iii. cos 45° = sin(90°− 45°) = sin 45° ∴ cos 45° = sin 45° iv. tan 75° = tan(90°− 75°) = tan 15° ∴ tan 75° = tan 15°

Example 12.16 Given that sin훼 = , where 훼 is obtuse, find the value of cos훼 and tan 훼.

Solution We have, sin훼 = , where 90° < 훼 < 180°. But sin 훼 + cos 훼 = 1 ⟹ cos 훼 = 1 − sin 훼

sin훼 = cos(90°− 훼) cos훼 = sin(90°− 훼) tan훼 = cot(90°− 훼)

244

cos 훼 = 1 − ⟹ cos 휃 = 1 − ⟹ cos 훼 =

⟹ cos훼 = ± = ± . ∴ cos훼 = −

tan훼 = = = ×− = −

Example 12.17 Solve the following equations leaving your answer to the nearest degree. i. sin 푥 = ii. 3 tan휃 = 1 iii. 3cos θ − 2cos휃 = 1 iv. 5cosec α − 3cotα = 2

Solution i. sin 푥 = ⟹ 푥 = sin ⟹ 푥 ≈ 36.87° . But sin 푥 is also positive in the 2nd quadrant, where 훼 = 180°− 푥 ⟹ 푥 = 180°− 훼 ⟹ 푥 ≈ 180°− 36.87 ≈ 143.13° ∴ 푥 = 36.87°or푥 = 143.13° (Corrected to 2 places of decimal) ii. 3 tan 휃 = 1 ⟹ 휃 = tan ⟹ 휃 ≈ 18.43° But tan 휃 is also positive in the 3rd quadrant, where 훼 = 휃 − 180° ⟹ 휃 = 180° + 훼 ⟹ 휃 ≈ 180° + 18.43° ≈ 198.43° ∴ 휃 = 18.43° or 휃 = 198.43° (Corrected to 2 places of decimal) iii. 3 cos 휃 − 2 cos휃 = 1 ⟹ 3 cos 휃 − 2 cos휃 − 1 = 0. Let cos휃 = 푥 ⟹ 3푥 − 2푥 − 1 = 0 ⟹ 3푥 − 3푥 + 푥 − 1 = 0 ⟹ 3푥(푥 − 1) + (푥 − 1) = 0 ⟹ (푥 − 1)(3푥 + 1) = 0 ⟹ 푥 − 1 = 0 or 3푥 + 1 = 0 ⟹ 푥 = 1 or 푥 = − But cos 휃 = 푥 ⟹ cos휃 = 1 ⟹ 휃 = cos 1 = 0° orcos휃 = − ⟹ 휃 = cos − ⟹ 휃 ≈ 109.47°

But cos 휃 is also positive in the 4th quadrant, where , 훼 = 360°− 휃 ⟹ 휃 = 360°− 훼 ⟹ 휃 = 360°− 0° = 360° ∴ 휃 = 0°, 109.47°or360° iv. 5 cosec 휃 − 2 cot휃 = 8 ⟹ 5 (1 + cot 휃) −2 cot = 8. 5 +5cot 휃 −2 cot = 8 ⟹ 5푐표푡 휃−2 푐표푡+5 − 8 = 0 ⟹ 5cot 휃−3 cot +5 − 8 = 0 Let cot휃 = 푥 ⟹ 5푥 − 2푥 − 3 = 0 ⟹ 5푥 − 3푥 + 5푥 − 3 = 0

245

⟹ 푥(5푥 − 3) + (5푥 − 3) = 0 ⟹ (5푥 − 3)(푥 + 1) = 0 ⟹ 5푥 − 3 = 0or푥 + 1 = 0 ⟹ 푥 = or 푥 = −1

But cot 휃 = 푥 ⟹ = 푥 ⟹ tan 휃 = ⟹ tan휃 = =

⟹ 휃 = tan ⟹ 휃 ≈ 59.04

Or tan휃 = = −1 ⟹ 휃 = tan (−1) ⟹ 휃 = −45° But tan 휃 is also positive in the 3rd quadrant, where, 훼 = 휃 − 180° ⟹ 휃 = 훼 + 180° ⟹ 휃 ≈ 59.04° + 180° ≈ 239.04° Or 휃 = −45 + 180° = 135°

∴, 휃 = 59.04°, 135°or 239.04° Try: Solve the following equations leaving your answer in two significant figures. i. cos 훾 = 0.345 ii. 2sec α − 3tan훼 = 11 iii. 3 cot 휃 − tan휃 = 0 iv. Find the largest possible domain of the function 푓(푥) =

v. If sin휃 = , 0 < 휃 < 휋,find the value of .

Trigonometric ratios of compound angles If훼 and 훽are two angles, then the 푠푢푚,훼 + 훽, and difference, 훼 − 훽, of 훼 and 훽 are called compound angles. Consider the figure below: D 훼 E C 1 훼 훽 훼 A F B In the figure, ∠퐵퐴퐶 = 훼, ∠퐶퐴퐷 = 훽 , |퐴퐷| = 1 and |AB| = |AF| + |FB|. cos(훼 + 훽) = |퐴퐹| = |퐴퐵| − |퐹퐵| cos훼 = | |

| | ⟹ |퐴퐵| = |퐴퐶| cos훼

cos훽 = |퐴퐶| ⟹ |퐴퐵| = cos 훼 cos훽 Also ∠CDE = ∠ACE = 훼

246

sin 훽 = |퐷퐶| From ∆퐷퐶퐸 sin 훼 = | || |

⟹ |퐸퐶| = |퐷퐶| sin훼 = sin 훼 sin 훽

But |퐹퐵| = |퐸퐶| ⟹퐹퐵 = sin 훼 sin훽 ∴ cos(훼 + 훽) = cosαcosβ − sin 훼 sin훽 Now, sin(훼 + 훽) = |퐹퐷| = |퐹퐸| + |퐸퐷| = |퐵퐶| + |퐸퐷| sin 훼 = | |

| |

⟹ |퐵퐶| = |퐴퐶| sin 훼 But cos 훽 = |퐴퐶| ⟹ |퐵퐶| = sin 훼 cos 훽 Also cos훼 = | |

| | ⟹ |퐸퐷| = |퐷퐶| cos훼

But sin 훽 = |퐷퐶| ⟹ |ED| = cosα sin β ∴ sin(훼 + 훽) = sin 훼 cos훽 + cosα sin β cos(훼 − 훽) = cos 훼 + (−훽) = cos훼 cos(−훽) − sin훼 sin(−훽) = cos훼 cos훽 + sin 훼 sin훽 sin(훼 − 훽) = sin 훼 + (−훽) = sin훼 cos(−훽) + cos훼 sin(−훽) = sin 훼 cos훽 − cos훼 sin 훽 Note that cos(−휃) = cos(휃) and sin(−휃) = − sin 휃

tan(훼 + 훽) = ( )( )

= =

= =

tan(훼 − 훽) = ( )( )

= = =

=

(Divide by cos훼 cos훽)

sin(훼 + 훽) = sin 훼 cos훽 + cosα sinβ sin(훼 − 훽) = sin훼 cos 훽 − cos훼 sin 훽

cos(훼 + 훽) = cosαcosβ − sin 훼 sin훽 cos(α − β) = cos훼 cos훽 + sin 훼 sin 훽

tan(훼 − 훽) =tan 훼 − tan훽

1 + tan 훼 tan훽

tan(훼 + 훽) =

247

Example 12.18 Given that sin훼 = and cos훽 = − , find sin(훼 + 훽) and cos(훼 + 훽), where 훼 and 훽 are both obtuse.

Solution We have, sin훼 = , cos훽 = − , 90° < 훼 < 180° and 90° < 훽 < 180°.

sin 훼 + cos 훼 = 1 ⟹ cos 훼 = 1 − sin 훼 ⟹ cos 훼 = 1 −

⟹ cos 훼 = 1 − ⟹ cos 훼 = ⟹ cos훼 = ± = ±

Since 훼 is obtuse cos훼 = −

sin 훽 + cos 훽 = 1 ⟹ sin 훽 = 1 − cos 훽 ⟹ sin 훽 = 1 − −

⟹ sin 훽 = 1 − ⟹ sin 훽 = ⟹ sin훽 = ± = ±

Since 훽 is obtuse sin훽 = sin(훼 + 훽) = sin훼 cos훽 + cos훼 sin 훽 cos(훼 + 훽) = cos훼 cos훽 − sin훼 sin훽 = ×− − × = − × − − ×

= − − = −

= − = −

Example 12.19 Given that cos(α− 휙) = cos훼 cos휙 + sin훼 sin휙, find the value of cos 15°, leaving your answer as a surd.

Solution We have, cos(훼 − 휙) = cos훼 cos휙 + sin 훼 sin휙 ⟹ cos15° = cos(45°− 30°) = cos 45° cos 30° + sin 45° sin 30°

= √ × √ + √ ×

= √ ×√ + √ = √ √

248

∴ cos15° = √ √

Example 12.20 Given that sin(푥 − 푦) = sin 푥cos푦 − cos푥sin푦, find the value of sin 135°, leaving your answer as a surd.

Solution We have, sin(푥 − 푦) = sin 푥 cos 푦 − cos 푥 sin 푦 ⟹ sin135° = sin(180°− 45°) = sin 180° cos 45°− cos 180° sin 45°

= 0 × √ + 1 × √

= √ ∴, sin135° = √

Example 12.21 Without using a table or calculator ° °

° °

Solution The value of ° °

° °= tan(75° − 15°) = tan 60 = √3

Try: Evaluate tan 105°, leaving the value as a surd.

Example 12.22 Express

° as a surd.

Solution

The value of °

= √ = √ =√

=√∙ √

√= √

√= √

∴,°

= √ = + √ = + √ = √

Example 12.23 By writing 훼 + 훽 + 훾 = (훼 + 훽) + 훾,prove that

sin(훼 + 훽 + 훾) = sin 훼 (cos훽 cosγ− sin훽 sin 훾) + cos훼(sin훽 cos훾 + cos훽 sin 훾) Proof

sin(훼 + 훽 + 훾) = sin (훼 + 훽) + 훾

249

= sin(훼 + 훽) cos훾 + cos(훼 + 훽) sin 훾 =(sin훼 cos훽 + cos훼 sin훽) cos 훾 + (cos훼 cos훽 − sin훼 sin훽) sin훾 = sin 훼 cos훽 cos훾 + cos훼 sin훽 cos 훾 + cos 훼 cos 훽 sin 훾 − sin훼 sin 훽 sin 훾 = sin 훼 cos훽 cosγ−sin 훼 sin훽 sin훾 + cos훼 sin훽 cos 훾 + cos훼 cos 훽 sin 훾 = sin 훼 (cos훽 cosγ− sin 훽 sin 훾) + cos훼(sin훽 cos 훾 + cos훽 sin 훾)

∎ Try: 1. Given that sin퐴 = and cos퐵 = , find sin(퐴 + 퐵), where 퐴 is obtuse and 퐵 is acute. 2. Find the value of cos 15°, leaving your answer as a surd. 3. Given that sin(퐴 + 퐵) = sin퐴 cos퐵 + cos퐴 sin퐵, find the value of sin 75°, leaving your answer as a surd. 4. Given that sin(퐴 + 퐵) = sin퐴 cos퐵 + cos퐴 sin퐵, find the value of sin 105°, leaving your answer as a surd. 5. using a table or calculator ° °

° °

6. Given that tan(퐴 + 퐵) = , evaluate tan 75°, leaving your answer as a surd.

7. Given that sin 훾 = and sin휙 = , find the value of cos(훾 + 휙) if 훾 is acute and 휙 is obtuse. 8.writing 훼 + 훽 + 훾 = (훼 + 훽) + 훾,simplify cos(훼 + 훽 + 훾).

9. Given that tan 30° = √ , evaluate tan 15°, leaving your answer in surd form. (SSSCE)

10. Given that; sin(푥 + 푦) = sin 푥 cos 푦 + cos푥 sin푦 sin(푥 − 푦) = sin 푥 cos푦 − cos 푥 cos푦

Show that: sin(60° + 휃) + sin(60°− 휃) = √3 cos휃 (SSSCE)

11. Given that sin퐴 = and cos퐴 = , find the value of cos(퐴 + 퐵), leaving your answer in surd form.

(SSSCE) 12. Given that sin퐴 =

√ and sin퐵 =

√ , where 퐴and퐵 are acute angles, evaluate

sin(퐴 + 퐵)your answer in surd form. (SSSCE)

250

13. Given that sin퐴 = and cos퐵 = , where 퐴 and 퐵 are acute angles, evaluate, without using tables, sin(퐴 − 퐵).

(WASSCE) 14. Given that tan퐴 = − , 90° < 퐴 < 180°and sin퐵 = − ,270 ° < 퐵 < 360°, evaluate cos(퐴 + 퐵).

(WASSCE)

Double angle identities If 퐴 is a number then 2퐴 is the double of 퐴. Now, we know that

sin(훼 + 훽) = sin 훼 cos훽 + cos α sin β, Then, sin(훼 + 훼) = sin훼 cos훼 + cos훼 sin훽. ⟹ sin(2훼) = sin(훼 + 훼) = sin 훼 cos훼 + cos훼 sin훽 = 2 sin 훼 cos훼 ⟹ sin훼 cos훼 = Similarly,cos(2훼) = cos(훼 + 훼) = cos훼 cos훼 − sin훼 sin 훼 = cos 훼 − sin 훼 But sin 훼 = 1 − cos 훼 and cos 훼 = 1 − sin 훼 ⟹ cos(2훼) = cos 훼 − (1 − cos 훼) = 2cos 훼 − 1 and cos(2훼) = 1 − sin 훼 − sin 훼 = 1 − 2sin 훼 ∴ cos 훼 = ( ) andsin 훼 = ( )

tan(2훼) = tan(훼 + 훼) =tan훼 + tan 훼

1 − tan훼 tan 훼 =2 tan훼

1 − tan 훼

⟹ 1 − tan 훼 =2 tan훼tan 2훼 ⟹ tan 훼 =

tan 2훼 −2 tan훼tan 2훼

Example 12.24 Given that sin훼 = , find the value of sin 2훼 and cos 2훼.

Solution We have, sin훼 =

sin 훼 + cos 훼 = 1 ⟹ cos 훼 = 1 − sin 훼 ⟹ cos 훼 = 1 −

sin 2훼 = 2 sin훼 cos훼 cos 2훼 = cos 훼 − sin 훼

tan 2훼 =2 tan훼

1 − tan 훼

cos 훼 =1 + cos(2훼)

2

sin 훼 =1 − cos(2훼)

2

= 2cos 훼 − 1 =1 − 2sin 훼

251

⟹ cos 훼 = 1 − ⟹ cos 훼 = ⟹ cos훼 = ± = ±

Since, 훼 is acute cos훼 =

sin 2훼 = 2 sin 훼 cos훼 = 2 × × = ∴ sin 2훼 =

cos 2훼 = cos 훼 − sin 훼 = − = − = ∴ cos 2훼 =

Example 12.25

Given that cos휑 = , 푥 ≠ −1, find in terms of 푥 the value of sin 2휑 , cos 2휑and tan 2휑. Solution

We have, sin훼 =

sin 훼 + cos 훼 = 1 ⟹ cos 훼 = 1 − sin 훼 ⟹ cos 훼 = 1 −

⟹ cos 훼 = 1 − ⟹ cos 훼 = ⟹ cos훼 = ± = ±

Since, 훼 is acute cos훼 =

sin 2훼 = 2 sin 훼 cos훼 = 2 × × = ∴ sin 2훼 =

cos 2훼 = cos 훼 − sin 훼 = − = − = ∴ cos 2훼 =

Example 12.26

Find the truth set of the equation 2 sin 2휔 − cos휔 = 1, 0° ≤ 휔 ≤ 360°. Solution

We have, 2 sin 2휔 − cos휔 = 1 ⟹ 2(2 cos 휔 − 1) − cos휔 = 1 4 cos 휔 − 2 − cos휔 = 1 ⟹ 4 cos 휔 − cos휔 − 2 − 1 = 0 Let cos휔 = 푥 ⟹ 4푥 − 푥 − 3 = 0 ⟹ 4푥 − 4푥 + 3푥 − 3 = 0 ⟹ 4푥(푥 − 1) + 3(푥 − 1) = 0 ⟹ (푥 − 1)(4푥 + 3) = 0 ⟹ 푥 − 1 = 0or4푥 + 3 = 0 ⟹ 푥 = 1 or 푥 = − But cos휔 = 푥 ⟹ cos휔 = 1 ⟹ 휔 = cos 1 = 0° Or cos휔 = − ⟹ 휔 = cos − ≈ 138.59° But cos휃 is also positive in the 4th quadrant, where 훼 = 360°− 휃

252

⟹ 휃 = 360°− 훼 ⟹ 휃 = 360°− 0° ∴ 휃 = 0°, 138.59°or 360°

Example 12.27

Given that sin퐴 = − and cos퐵 = √ ,180° ≤ 퐴 ≤ 360°, evaluate cos(2퐴 − 퐵). Solution

We have, sin퐴 = − , cos퐵 = √ and 180° < 퐴 < 270°.

sin 퐴 + cos 퐴 = 1 ⟹ cos 퐴 = 1 − sin 퐴 ⟹ cos 퐴 = 1 − −

⟹ cos 퐴 = 1 − ⟹ cos 퐴 = ⟹ cos퐴 = ± = ± √

Since 180° < 퐴 < 270°,cos퐴 = − √

sin 퐵 + cos 퐵 = 1 ⟹ sin 퐵 = 1 − cos 퐵 ⟹ sin 퐵 = 1 − √

⟹ sin 퐵 = 1 − ⟹ sin 퐵 = ⟹ sin퐵 = ± = ±

Since 퐵 is acute sin퐵 = cos 2퐴 = cos 퐴 − sin 퐴 sin 2퐴 = 2 sin퐴 cos퐴

= − √ − − = 2 × − ×− √

= − = √

=

cos(2퐴 − 퐵) = cos 2퐴 cos퐵 − sin2퐴 sin퐵 = × √ − √ × = √ − √ = 0 ∴, cos(2퐴 − 퐵) = 0

Try: 1. Given that cos 푥 = , find the value of .

2. Given that tan훽 = , find the value of sin 2훽, cos 2훽and tan 2훽.

3. Given that sin 푥 = − and cos 푥 = − √ , both 푥 and 푦 obtuse, find the value of i. cos(2푥 − 푦)ii. sin(푥 + 2푦)iii. tan 2푥 − tan푦. 4. Given that, sin(훼 − 훽) sin 훼 cos훽 − cos훼 sin 훽, show that; sin 2훼 = 2 sin 훼 cos훼

253

(SSSCE)

Half angle identities By using our knowledge of double angles we have

tan훼 = tan + = If we let tan = 푡 thentan훼 =

Also cos훼 = cos + = cos − sin let divide by cos + sin

⟹ cos훼 =cos2훼

2−sin2훼2

cos2훼2+sin2훼

2=

cos2훼2

cos2훼2−

sin2훼2

cos2훼2

cos2훼2

cos2훼2

+sin2훼

2cos2훼

2

= 1−tan2훼2

1+tan2훼2

= 1−푡2

1+푡2 and

hint: cos + sin = 1

sin 훼 = sin + = 2 sin cos = = = =

From cosine identity cos훼 = cos + = cos − sin

⟹ cos훼 = 2 cos − 1 ⟹ cos = ⟹ cos = ±

Similarly, cos훼 = 1 − 2 sin ⟹ sin = ⟹ sin = ±

tan = = =/

÷/

= ÷/

⟹ tan = ×/

=/

= ×/

= ( ) /

⟹ tan = ( ) / =

sin 훼 = 2푡1+푡2 cos훼 = 1−푡2

1+푡2 tan훼 =

Where tan = 푡

sin = ± cos = ± tan =

254

Try: Given that tan휃 = − , without using tables or calculator, find the values of tan .

(WASSCE)

Multiple angle identities A trigonometric equation which involves ratios of angles is called multiple angle.

Example 12.28 By writing 3훼 = 2훼 + 훼, show that sin 3훼 = 3 sin 훼 − 4 sin 훼,

cos 3훼 = 4 cos 훼 − 3 cos훼 and tan 3훼 = 3 tan훼−tan3 훼1−3 tan2훼

Proof

ℎ푖푛푡: cos 훼 = 1 − sin 훼cos 2훼 = cos 훼 − sin 훼

sin 2훼 = 2 sin 훼 cos훼

sin 3훼 = sin(2훼 + 훼) = sin 2훼 cos훼 + cos 2훼 sin훼 = 2 sin 훼 cos훼(cos훼) + (cos 훼 − sin 훼) sin훼 = sin훼[2 cosα cos훼 + cos 훼 − sin 훼] =sin훼[2 cos 훼 + cos 훼 − sin 훼] = sin훼[3 cos 훼 − sin 훼] = sin훼[3(1 − sin 훼) − sin 훼] = sin훼[3 − 4 sin 훼] = 3 sin훼 − 4 sin 훼

cos 3훼 = cos(2훼 + 훼) = cos 2훼 cos훼 − sin 2훼 sin 훼 =(2 cos 훼 − 1) cos훼 − (2 sin 훼 cos 훼) sin 훼 = cos훼[(2 cos 훼 − 1) − (2 sin훼) sin 훼] = cos훼[2 cos 훼 − 1 − 2(1 − cos 훼)] = cos훼[4 cos 훼 − 3] =4 cos 훼 − 3 cos훼

255

hint: sin 훼 = 1 − cos 훼cos 2훼 = 2 cos 훼 − 1sin 2훼 = 2 sin훼 cos훼

⎣⎢⎢⎢⎢⎢⎡ ℎ푖푛푡: =

= sec 훼1 + tan 훼 = sec 훼

= ∙= tan훼 sec 훼⎦

⎥⎥⎥⎥⎥⎤

∎

Try: 1. By writing 4휃 = 2휃 + 2휃 or 4휃 = 3휃 + 휃, find an expression for sin 4휃, cos 4휃푎푛푑 tan 4휃 in terms of only sin, cos and tan respectively.

2. Using the fact that tan(퐴 + 퐵) =

, prove that tan 3퐴 =

Proving/ simplifying other trigonometric identities The first step in proving trigonometric identities is to know what you want to prove and where to start from. You can start from the 푅퐻푆8 of an identity and ends at the 퐿퐻푆9and vice versa depending on how easy to get what you want.

8 Right hand side

tan 3α = sin 3훼cos 3훼 = 3 sin훼−4 sin3훼

4 cos3훼−3 cos훼

=

=

= ( )( )

=

=

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We use conjugate in getting expression like 1 ± cos 푥 , 1 ± sin푥푎푛푑 sin 푥 ± cos 푥 in the denominator out of fractional form. Recall that the conjugate of 푎 + 푏 is 푎 − 푏 such that (푎 + 푏)(푎 − 푏) = 푎 − 푏

Example 12.29

Simplify the following: i. ii. iii. + iv.

Solution

i. = = ( ) = ( ) = = −

= cosec 휃 − cot휃 ii. = = ( ) = ( ) = 2 − 2

= 2 − 2∙

=2 cosec 푥 − 2 cot푥 cos ec푥 = 2 cosec푥(1 − cot푥)

iii. + = + = ( ) +

= ( ) +

= +

= = = cosec 휃

iv. = = × = =

= ( )

= ( )

=

9 Left hand side

257

= + = sec 푥 + tan 푥

Example 12.30 Verify the following: i. sec 푥 − tan 푥 sin 푥 = cos푥 ii. + = 2 cosec 푥

iii. tan 휃 + cot 휃 = 2 cosec 2휃 iv. = 1 + cos 휃

v. = sec 2푥 + tan 2푥 vi. = Proof

i. To verify that sec 푥 − tan푥 sin 푥 = cos 푥 take the푅퐻푆 and simplify to get the 퐿퐻푆.

sec 푥 − tan 푥 sin 푥 = − ∙ sin 푥 = = = cos 푥 ∎

ii. To verify that + = 2 cosec 푥 take the 푅퐻푆 and simplify to get the 퐿퐻푆.

+ = + = + ( )

= + ( )

= +

= = = 2 cosec 푥 ∎

iii. To verify that tan 휃 + cot휃 = 2 cosec 2휃 take the 푅퐻푆 and simplify to get the 퐿퐻푆.

tan휃 + cot 휃 = + = =

= = = 2 cosec 휃

∎

iv. To verify that = 1 + cos 휃 take the 푅퐻푆 and simplify to get the 퐿퐻푆.

= + 2 ÷ 1 + = ×

= sin 휃 + 2 cos 휃 = 1 − cos 휃 + 2cos 휃=1 + cos 휃

∎

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v. To verify that = sec 2푥 + tan 2푥 take the 푅퐻푆 and simplify to get the 퐿퐻푆.

= = =

= + = sec 2푥 + tan 2푥

∎ vi. To verify that = take the 퐿퐻푆 and simplify to get the 푅퐻푆.

= = ( ) = ( ) = ∎

Example 12.31

Prove that tan (45− 훼) = . Proof

tan (45°− 훼) = (tan(45°− 훼)) = ( ° )( ° )

= √ √

√ √

⟹ tan (45°− 훼) = ° °° °

=√ ( )

√ ( )=

⟹ tan (45°− 훼)= = ∎

General solution of the basic trigonometric ratios In general, If 훼 = sin 푘, then 훼 = 180°푛 + (−1) 훼 or 휃 = 휋푛 + (−1) 훼 , where 훼 is the principal acute angle of the sine inverse function and 푛 is an integer. If 훼 = cos 푘, then 훼 = 360°푛 ± 훼 or 휃 = 2휋푛 ± 훼 , where 훼 is the principal acute angle of the cosine inverse function and 푛 is an integer. If 훼 = tan 푘, then 훼 = 180°푛 + 훼 or 휃 = 휋푛 + 훼 , where 훼 is the principal acute angle of the tangent inverse function and 푛 is an integer.

Example 12.32 Find the general solution of the following:

259

i. sin 푥 = √ ii. tan(휃 − 45°) =√

iii. tan 3푥 = √3 iv. 3 tan 휃 + sec 휃 + 1 = 0

Solution

i. We have, sin 푥 = √ ⟹ 푥 = sin √ = 45° ⟹ 푥 = 45° General solution for sine: 푥 = 180°푛+ (−1) 푥 ⟹ 푥 = 180°푛 + (−1) 45° When 푛 = 0, 푥 = 180°(0) + (−1) 45° = 45° When 푛 = 1 푥 = 180°(1) + (−1) 45° = 180°− 45° = 135° When 푛 = 2 푥 = 180°(2) + (−1) 45° = 360° + 45° = 405° 405°is out of our range of solution. ∴ 푥 = 45°or 푥 = 135° ii. We have, tan(휃 − 45°) =

√ Let (휃 − 45°) = 훼

⟹ 훼 = tan√

= 30° ⟹ 훼 = 30°

General solution for tangent: 훼 = 180°푛 + 훼 ⟹ (휃 − 45°) = 180°푛 + 30° ⟹ 휃 = 180°푛 + 30° + 45° ⟹ 휃 = 180°푛 + 75° When 푛 = 0, 휃 = 180°(0) + 75° = 75° When 푛 = 1 휃 = 180°(1) + 75° = 180° + 75° = 255° ∴ 푥 = 75°or푥 = 255° iii. We have, tan 3푥 =√3 Let 3푥 = 훼 ⟹ 훼 = tan √3 = 60° ⟹ 훼 = 60° General solution for tangent: 훼 = 180°푛 + 훼 ⟹ 3푥 = 180°푛 + 60° ⟹ 푥 = ° 푛 + ° ⟹ 휃 == 60°푛 + 20°

When 푛 = 0, 휃 = 60°(0) + 20° = 20° When 푛 = 1, 휃 = 60°(1) + 20° = 80° When 푛 = 2, 휃 = 60°(2) + 20° = 120° + 20° = 140° When 푛 = 3, 휃 = 60°(3) + 20° = 180° + 20 = 200° When 푛 = 4, 휃 = 60°(4) + 20° = 240° + 20° = 260° When 푛 = 5, 휃 = 60°(5) + 20° = 300° + 20° = 320° ∴,휃 = 20°, 80°, 140°, 200°, 260°, 320°

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iv. We have, 3 tan 휃 + sec 휃 + 1 = 0 ⟹ 3 (sec 휃 − 1)+ sec 휃 + 1 = 0. ⟹ 3sec 휃 − 3 + sec 휃+1 = 0 ⟹ 3sec 휃+ sec 휃 −2 = 0 Let sec 휃 = 푥 ⟹ 3푥 + 푥 − 2 = 0 ⟹ 3푥 − 2푥 + 3푥 − 2 = 0 ⟹ 푥(3푥 − 2) + (3푥 − 2) = 0 ⟹ (푥 + 1)(3푥 − 2) = 0 ⟹ 푥 + 1 = 0or3푥 − 2 = 0 ⟹ 푥 = −1 or 푥 =

But sec 휃 = 푥 ⟹ = 푥 ⟹ cos휃 =

⟹ cos휃 = = −1 ⟹ 휃 = cos (−1) = 180° ⟹ 휃 = 180°

orcos휃 = = ⟹ 휃 = cos = not defined.

General solution for cosine: 휃 = 360°푛 ± 휃 ⟹ 푥 = 360°푛 ± 180° When 푛 = 0, 푥 = 360°(0) ± 180° = ±180° When 푛 = 1 , 푥 = 360°(1) ± 180° = 360° ± 180° ⟹ 푥 = 180°or푥 = 540° −180°and 540° are out of our range of solution.

∴, 푥 = 180° Try: Find the general solution of the following equations:

1. i. sin 3휃 = √ ii. cos 2훼 = iii. tan 푥 = − vi. sec 푥 − tan 푥 = 4 2. solve the equation tan 푥 + 3 cot푥 − 5 sec 푥 = 0, 0° ≤ 푥 ≤ 180°.

(푆푆푆퐶퐸) 3. Solve for 푥 in the equation 2 sin 푥 − 5 sin푥 − 3 = 0, 0° ≤ 푥 ≤ 270°

(SSSCE) 4. Find the truth set of the equation cos 푥 + cos 푥 = , 0° ≤ 푥 ≤ 270° .

(SSSCE)

5. Given that + = 1, 0° ≤ 휃 ≤ 360°,find the values of 휃. (WASSCE)

The expression of the form 푎 cos 푥 + 푏 sin 푥 The expression 푎 sin 푥 + 푏 cos 푥, where 푎and are constants, is identical to either 푅 cos(푥 ± 훼) or 푅 sin(푥 ± 훼). That is, 푎 cos 푥 + 푏 sin 푥 ≡ 푅 cos(푥 ± 훼) ≡ 푅 sin(푥 ± 훼)

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푅 > 0, −푅 ≤ 푅 cos(푥 ± 훼) ≤ 푅 and 훼 is acute. The maximum and minimum value of the expression 푎 cos 푥 + 푏 sin푥 푅 and –푅 respectively. The maximum and minimum value of the expression 푎 cos푥 + 푏 sin 푥 occur at the value of 푥 for which 푅 cos(푥 ± 훼) = 푅 and 푅 cos(푥 ± 훼) = −푅. The values of 푅 and 훼 are given by:

푅 = √푎 + 푏 and 훼 = tan = sin = cos Proof

Let 푎 cos 푥 + 푏 sin 푥 ≡ 푅 cos(푥 − 훼) ≡ 푅(cos푥 cos훼 + sin 푥 sin 훼) ≡ 푅 cos 푥 cos훼 + 푅 sin 푥 sin 훼 By comparing the 푅퐻푆 and LHS we have 푎 = 푅 cos훼 (1) and 푏 = 푅 sin 훼 (2)⟹ 훼 = cos = sin By squaring both sides of equations (1) and (2) we have 푎 = 푅 cos 훼 (3) and 푏 = 푅 sin 훼 (4) (3) + (4) gives 푎 + 푏 = 푅 cos 훼 + 푅 sin 훼 ⟹ 푎 + 푏 = 푅 (cos 훼 + sin 훼) ⟹ 푎 + 푏 = 푅 ⟹ 푅 = √푎 + 푏 (1) ÷ (2) gives

= ⟹ tan 훼 = ∴ 훼 = tan = sin = cos ∎

Example 12.33 Express 2√3 cos휙 + 2 sin휙 in the form 푅 cos(휙 − 훼), 푅 > 0 and 0° ≤ 훼 ≤ 90°.

Solution Let 2√3 cos휙 + 2 sin휙 ≡ 푅 cos(휙 − 훼) ≡ 푅 cos휙 cos훼 + 푅 sin휙 sin 훼 By comparing the RHS and LHS we have, 2√3 = 푅 cos훼… … … (1) 2 = 푅 sin훼… … … (2)

푎 = 2√3 and 푏 = 2 푅 = √푎 + 푏 = 2√3 + (2) = √12 + 4 = 4

훼 = tan = tan√

= 30° ∴ ,2√3 cos휙 + 2 sin휙 ≡ 4 cos(휙 − 30°)

Example 12.34

Express 4 sin 푥 − √6 cos 푥 in the form 푅 sin(푥 − 훼), 푅 > 0 and 훼 is acute. Find the maximum and minimum value of the expression and the value of 푥 for which they occur.

Solution

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Let 4 sin 푥 − √6 cos 푥 ≡ 푅 sin(푥 − 훼) ≡ 푅 sin 푥 cos훼 − 푅 cos푥 sin훼 By comparing the RHS and LHS we have, 4 = 푅 cos훼… … … (1) √6 = 푅 sin훼… … … (2)

푎 = 4 and 푏 = √6 푅 = √푎 + 푏 = (4) + √6 = √16 + 6 = √22

훼 = tan = tan √ = 31.48° (Corrected to 2 places of decimal)

∴ 4 sin 푥 − √6 cos 푥 ≡ √22 sin(푥 − 31.48°) The maximum value of expression= 푅 = √22 The minimum value of expression= −푅 = −√22 The maximum value of the expression occurs at the value of 푥 for which sin(푥 − 31.48°) = 1 ⟹ (푥 − 31.48°) = sin 1 ⟹ (푥 − 31.48°) = 90° ⟹ 푥 = 90° + 31.48° = 121.48° ∴,the maximum value of expression occurs at 푥 = 121.48° The minimum value of the expression occurs at the value of 푥 for which sin(푥 −31.48°) = −1 ⟹ (푥 − 31.48°) = sin −1 ⟹ (푥 − 31.48°) = −90° ⟹ 푥 = −90° + 31.48° = −58.52° ∴,the minimum value of expression occurs at 푥 = −58.52°

Example 12.35 Express 2√5 cos훽 − 4 sin 훽 in the form 푅 cos(훽 + 훼), 푅 > 0,and 훼 is acute. Find the minimum value of the expression and the value of 훽 for which it occurs.

Solution Let 2√5 cos훽 − 4 sin훽 ≡ 푅 cos(훽 + 훼) ≡ 푅 cos훽 cos훼 − 푅 sin 푥 sin훼 By comparing the RHS and LHS we have, 2√5 = 푅 cos훼… … … (1) 4 = 푅 sin훼… … … (2)

푎 = 2√5 and 푏 = 4 푅 = √푎 + 푏 = 2√5 + (4) = √20 + 16 = 6

훼 = tan√

= tan√

= 41.81° (Corrected to 2 places of decimal)

∴ 2√5 cos훽 − 4 sin 훽 ≡ 6 cos(훽 − 41.81°) The minimum value of expression= −푅 = −6 The minimum value of the expression occurs at the value of 푥 for which cos(훽 − 41.81°) = −1 ⟹ (훽 − 41.81) = cos −1 ⟹ (훽 − 31.48°) = 180°

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⟹ 훽 = 180° + 41.81° = 221.81° ∴,the minimum value of expression occurs at 푥 = 221.81° Try: 1. Express 푦 = 4 cos푥 + 3 sin 푥 in the form푅 cos(푥 − 훼),푅 > 0 and 0° ≤ 푥 ≤ 360°. 2. Express 7 sin 푥 + 24 cos푥 in the form 푅 sin(푥 + 훼) ,푅 > 0 and 0° ≤ 푥 ≤ 360°. Hence, find the minimum value of the expression and the value of 푥 it occurs.

Equation of the form 푎 cos푥 + 푏 sin 푥 = 푐 The equation of the form 푎 cos 푥 + 푏 sin 푥 = 푐, where 푎,푏푎푛푑푐 are constants, is solved by

Taking one term of the expression at the 푅퐻푆 to the 퐿퐻푆 and squaring both sides afterwards. (It is always advisable to test for the values of 푥 that satisfies the equation.) or

Writing the 푅퐻푆 of the equation in the form 푅 cos(푥 ± 훼) or 푅 sin(푥 ± 훼) and equating the result to the value of 푐. or

Putting cos푥 = and sin 푥 = , where tan = 푡 Example 12.36

Solve the equation sin 푥 + cos푥 = 1, 0° ≤ 푥 ≤ 360°. Solution

Method 1 We have sin 푥 + cos 푥 = 1 ⟹ sin 푥 = 1 − cos 푥 (by squaring both sides) ⟹ (sin 푥) = (1 − cos 푥) ⟹sin 푥 = 1 − 2 cos푥 + cos 푥 But sin 푥 = 1− cos 푥 ⟹ 1 − cos 푥 = 1 − 2 cos 푥 + cos 푥 Grouping like terms gives 2 cos 푥 − 2 cos푥 = 0 ⟹ 2 cos 푥(cos푥 − 1) = 0 ⟹ 2 cos 푥 = 0 or cos 푥 − 1 = 0 ⟹ cos푥 = 0 or cos푥 = 1 ⟹ 푥 = cos (0) = 90° or푥 = cos (1) = 0° General solution for cosine: 푥 = 360°푛 ± 푥 , for some integers 푛. ⟹ 푥 = 360°푛 ± 90° or 푥 = 360°푛 ± 0° If 푛 = 0, then 푥 = 90° or푥 = 0° If 푛 = 1,then 푥 = 270° or 푥 = 360° (choose (−)) When we check, we see that sin 270° + cos 270° ≠ 1

∴ 푥 = 0°, 90°and360° Method 2

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We have sin 푥 + cos 푥 = 1 let sin 푥 + cos 푥 = 푅 sin(푥 + 훼) orcos 푥 + sin 푥 = 푅 cos(푥 −훼) Take sin 푥 + cos푥 = 푅 sin(푥 + 훼) = 푅 sin 푥 cos 훼 + 푅 cos 푥 sin 훼 By comparing co-efficient gives 1 = 푅 cos훼 and 1 = 푅 sin훼 ⟹ 푅 = √1 + 1 = √2훼 = tan (1) = 45° ⟹ sin 푥 + cos 푥 = √2 sin(푥 + 45°) ⟹ √2 sin(푥 + 45°) = 1

⟹ sin(푥 + 45°) =√

= √ ⟹ 푥 + 45° = sin √ = 45°

⟹ 푥 = 45°− 45° = 0° General solution of sine: 푥 = 180°푛 + (−1) 푥 , for some integers 푛. ⟹ 푥 = 180°푛 + (−1) 0° If 푛 = 0, then 푥 = 0°. if 푛 = 1 then 푥 = 180°. If 푛 = 2, then 푥 = 360° Take cos푥 + sin 푥 = 푅 cos(푥 −훼) = 푅 cos 푥 cos훼 + 푅 sin 푥 sin 훼 By comparing co-efficient gives 1 = 푅 cos훼 and 1 = 푅 sin훼 ⟹ 푅 = √1 + 1 = √2훼 = tan (1) = 45° ⟹ sin 푥 + cos 푥 = √2 cos(푥 − 45°) ⟹ √2 cos(푥 − 45°) = 1

⟹ cos(푥 − 45°) =√

= √ ⟹ 푥 − 45° = cos √ = 45°

⟹ 푥 = 45° + 45° = 90° General solution of cosine: 푥 = 360°푛 ± 푥 , for some integers 푛. ⟹ 푥 = 360°푛 ± 90° If 푛 = 0, then 푥 = 90°. if 푛 = 1 then 푥 = 270°. (Choose (−)) When we check, we will see that sin 270° + cos 270° ≠ 1 or sin 180° + cos 180° ≠ 1 Hence, 푥 = 180°, 270° gives a false solution.

∴ 푥 = 0°, 90°and360° Method 3

We have sin 푥 + cos 푥 = 1 let sin 푥 = and cos 푥 =

⟹ + = 1 ⟹ = 1 ⟹ 2푡 + 1 − 푡 = 1 + 푡 Grouping like terms gives 2푡 − 2푡 = 0 ⟹ 2푡(푡 − 1) = 0 ⟹ 2푡 = 0 or 푡 − 1 = 0 ⟹ 푡 = 0 or 푡 = 1 But tan = 푡 ⟹ tan = 0 or tan = 1 let = 훼 ⟹ tan 훼 = 0 or tan훼 = 1 ⟹ α = tan (0) = 0°표푟훼= tan (1) = 45°

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General solution for tangent: 훼 = 180°푛 + 훼 , for some integers 푛. ⟹ 훼 = 180°푛 + 0° or 훼 = 180°푛+ 45° ⟹ = 180°푛 or = 180°푛 + 45° ⟹ 푥 = 360°푛 or 푥 = 360°푛 + 90° If 푛 = 0, then 푥 = 0° or 푥 = 90° If 푛 = 1, then 푥 = 360° or 푥 = 450°

∴ 푥 = 0°, 90°and360°

Example 12.37 Express 푓(푥) = √3 cos 휃 + sin 휃 in the form 푅 cos(휃 − 훼),푅 > 0 and 훼 is acute. Hence, solve the equation √3 cos휃 + sin휃 = 0.3, 0° ≤ 휃 ≤ 360°.

Solution We have, 푓(푥) = √3 cos휃 + sin 휃. Let √3 cos휃 + sin휃 ≡ 푅 cos(휃 − 훼) ≡ 푅 cos휙 cos훼 + 푅 sin휙 sin훼 By comparing the RHS and LHS we have, √3 = 푅 cos훼… … … (1) 1 = 푅 sin훼… … … (2)

푎 = √3 and 푏 = 1 푅 = √푎 + 푏 = √3 + (1) = √3 + 1 = 2

훼 = tan√

= tan√

= 30° ∴ √3 cos휃 + sin 휃 ≡ 2 cos(휃 − 30°)

The equation √3 cos휃 + sin 휃 = 0.3 ⟹ 2 cos(휃 − 30°) = 0.3 ⟹ 2 cos(휃 − 30°) = 0.3 ⟹ cos(휃 − 30°) = . ⟹ cos(휃 − 30°) = 0.15. Let (휃 − 30°) = 훼. ⟹ cos훼 = 0.15 ⟹ 훼 = cos 0.15 ≈ 81.37° ⟹ 훼 = 81.37° General solution of cosine: 훼 = 360°푛 ± 푎 ⟹ 훼 ≈ 360°푛 ± 81.37° When 푛 = 0,푥 ≈ 360°(0) ± 81.37° ≈ ±81.37° When 푛 = 1 푥 ≈ 360°(1) ± 81.37° ≈ 360° ± 81.37° ⟹ 푥 ≈ 278.63° or 푥 ≈ 441.37° −81.37°and 441.37° are out of our range of solution.

∴,푥 ≈ 81.37°or푥 ≈ 278.63° Try:

450° is not within our range of solution

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1. a. Express 3 cos 푥 + √2 sin 푥 in the form 푅 cos(푥 − 훼) ,푅 > 0and 훼is acute. Hence, find the maximum and minimum value of the expression 3 cos 푥 + √2 sin 푥. b. Solve the equation 3 cos 푥 + √2 sin 푥 = 1.5 2. a. Express 6 sin 푥 + 8 cos푥 in the form 푅 sin(푥 + 훼), 푅 > 0and 훼 is acute.Hence, find the maximum and minimum value of the expression 6 sin 푥 + 8 cos 푥and the values of 푥 for which they occur. b. Solve the equation 6 sin 푥 + 8 cos 푥 = 5.5, 0° ≤ 푥 ≤ 360°.

Basic trigonometric functions a. The sine function The sine function is an odd and periodic function. It has a period of 2휋 or 360°. The Sine function lies between −1 and 1 inclusive.That is, sin(−푥) = − sin(푥) and sin(푥 + 2휋) = sin(푥) − 1 ≤ sin 푥 ≤ 1. A sketch of 푓(푥) = sin 푥, for some positive values of 푥 is shown below: 푓(푥) 1 0.5 0 휋/2 휋 3휋/2 2휋 푥 -0.5 -1 b. The cosine function The cosine function is an even and periodic function. It has a period of 2휋 or 360°. The cosine function lies between −1 and 1 inclusive. That is,

267

cos(−푥) = cos(푥) cos(푥 + 2휋) = cos(푥) −1 ≤ cos 푥 ≤ 1. A sketch of 푓(푥) = cos 푥, for some positive value of 푥, is shown below: 푓(푥) 1 0.5 0 휋/2 휋 3휋/2 2휋 푥 0.5 1 c. The tangent function The tangent function is an odd and periodic function. It has a period of 휋 or 90°. The tangent function is ratio of sine function on cosine function. The tangent function is not defined when the output of the cosine function is zero. This fact makes the tangent curve asymptotic to some values of 푥. That is tan(−푥) = − tan(푥) and tan(푥 + 휋) = tan(푥) tan 푥 = , cos푥 ≠ 0. A sketch of 푓(푥) = tan 푥, for some positive values of 푥 is shown below:

268

푓(푥) 2 1 0 휋/2 휋 3휋/2 2휋 -2 -1 Example 12.38 a. Draw the graph of the function 푦 = sin 푥 + cos푥, for the values of 푥 from 0° to 120° at intervals of 30° b. Use your graph to find the value of 푥 for which i. sin 푥 + cos 푥 = 0 ii. 2 sin 푥 + 2cos푥 = 1.2.

Solution a. The table below gives the values of the relation 푦 = sin 푥 + cos 푥for the given 푥 interval. (The values of 푦 is obtained by substituting the given value of 푥 into the relation).

푥 0° 30° 60° 90° 120° 150° 180° 210° 푦 = sin 푥 + cos 푥 1.00 1.37 1.37 1.00 0.37 −0.37 −1.00 −1.37

The graph of the relation 푦 = sin 푥 + cos푥is shown below. b. i. We have the relation 푦 = sin 푥 + cos 푥⋯⋯ (1) and sin푥 + cos 푥 = 0⋯⋯ (2) Comparing (1) and (2) ⟹푦 = 0. We see from the graph that line 푦 = 0 is the 푥-axis. The value of 푥 for which sin 푥 + cos 푥 = 0 is the point of intersection between line 푦 = 0 and the relation y = sin 푥 + cos푥 ∴,푥 = 135° ii. We have the relation 푦 = sin 푥 + cos푥 … … … (1) and 2 sin 푥 + 2 cos 푥 = 1.2 … … … (2) dividing (2) by 2 gives sin 푥 + cos푥 = 0.6⋯⋯ (3) comparing (1) and (3) gives 푦 = 0.6. we the proceed to draw line 푦 = 0.6 to cut

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the relation 푦 = sin 푥 + cos푥. The value of 푥 for which 2 sin 푥 + 2 cos 푥 = 1.2 is the point of intersection between line 푦 = 0.6 and the relation 푦 = sin 푥 + cos푥 ∴,푥 = 108°

scale:푥-axis− 2 cm to 30°푦-axis− 2cm to 0.3 unit 푦 1.5 1.2 0.9 0.6 0.3 0 푥 -0.3 -0.6 -0.9 -1.2 -1.5 Try: 1. Draw the graph of 푦 = 4 + 2 cos 푥for the values of 푥 from 0to2휋, at the intervals of .

30° 60° 90° 120° 150° 180° 210° 240° 270°

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Use your graph to; a. find the values of 푥 for which 푦 = 5. b. solve cos 푥 + 1 = 0

(푆푆푆퐶퐸) 2. Draw the graph of 푦 = 2 sin 푥° + 3 cos 푥°, for 0° ≤ 푥 ≤ 180°, at the intervals of 30°. Using your graph, find the solution set of the equation 4 sin 푥° + 6 cos푥° = −3.

(SSSCE) 3. a. Using the same coordinate axes for the values of 푥 from 0°to90° at the intervals of 15°,draw the graph of i. 푦 = 2 cos 2푥 ii. 푦 = − sin 2푥 (Use 2 cm to 15° on 푥 −axis and 2 cm to 1 unit on 푦 −axis). b. Use your graph in (a) to solve i. 8 cos 2푥 + 4 sin 2푥 = 0 ii. 3 − 12 sin 2푥 = 0,correct to one decimal place.

(SSSCE) 4. i. Copy and complete the table of values for the relation 푦 = 2 cos 푥 + sin 푥, for the interval 0 ≤ 푥 ≤ 180°.

ii. Using a scale of 2푐푚to 30° on the 푥 −axis and 2푐푚 to 1 unit on the 푦 −axis, draw the graph of the relation 푦 = 2 cos 푥 + sin 푥, for the given interval. iii. Use your graph to solve

훼. 2 cos 푥 + sin 푥 = 1 훽. 2 cos 푥 + sin 푥 = 0. (WASCCE)

Triangular rules 1. Sine rule Consider the figures below B B 훾 푎 푏 푎 ℎ ℎ 훽 훼 훼 A 푐 C A 푐 C

푥 0° 30° 60° 90° 120° 150° 180° 푦 = 2 cos 푥 + sin 푥 2.000 1.866 −1.232

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Not drawn to scale The Figures 푎 and 푏 represent the same ⧍퐴퐵퐶.in Figure푎 a perpendicular ℎ is drawn from 퐵 to meet 퐴퐵 at 푁 while in 푏 the perpendicular ℎ is drawn from 퐴 to meet 퐵퐶 at 푃. Let |퐴퐵| = 푎, |퐵퐶| = 푏, |퐴퐶| = 푐,∠퐵퐴퐶 = 훽,∠퐴퐵퐶 = 훾and ∠퐴퐶퐵 = 훼 Let’s take Figure 푎

sin 훼 = ⟹ ℎ = 푏 sin 훼 sin 훽 = ℎ푎 ⟹ ℎ = 푎 sin 훽

If ℎ = 푏 sin 훼 and ℎ = 푎 sin훽 then, ⟹ 푏 sin 훼 = 푎 sin훽 ⟹ = (Divide both side by sin 훼 sin훽)

Let’s take Figure 푏 sin 훼 = ⟹ ℎ = 푐 sin훼 sin 훾 = ⟹ ℎ = 푎 sin 훾

⟹ 푐 sin훼 = 푎 sin 훾 ⟹ = (Divide both side by sin 훼 sin 훾)

Since ℎ = 푏 sin훼 = 푎 sin훽 and ℎ = 푐 sin훼 = 푎 sin 훾, we write = = (Sine rule)

This rule can be used to calculate i. the length of a triangle given the length of one side and two included angles opposite to the sides. ii. an angle between two adjacent sides of a triangle given the length of two sides and an included angle opposite to one of the side given.

Example 12.39 B 푎 훼 푏 63° 30° A 13 cm C In the ∆퐴퐵퐶, |퐴퐵| = 푎, |퐵퐶| = 푏, |퐴퐶| = 13cm,∠퐵퐴퐶 = 63°,∠퐴퐵퐶 = 훼and ∠퐴퐶퐵 = 30°. Find, correct to 3 significant figures, the values of 푎, 푏and훼.

Solution We have, |퐴퐵| = 푎, |퐵퐶| = 푏 |퐴퐶| = 13cm, 퐵퐴퐶 = 63°,퐴퐵퐶 = 훼 and 퐴퐶퐵 = 30° In ∆퐴퐵퐶: 퐵퐴퐶 + 퐴퐵퐶 + 퐴퐶퐵 = 180° ⟹ 63° + 훼 + 30° = 180° 훼 = 180°− (63° + 30°) ⟹ 훼 = 180°− 93° = 87°

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The sine rule states that: | |

°= | |

°= | | ⟹

°=

°=

°

⟹ °

=° ⟹ 푎 =

°× sin 30° ⟹ 푎 ≈

.× 0.5 ≈ 6.51

and

°=

° ⟹ 푏 =

°× sin 63° ⟹ 푏 =

.× 0.8910 ≈ 11.60

∴ 푎 ≈ 6.51 cm, 푏 ≈ 11.6 cm (Corrected to 3 significant figures) and 훼 = 87°

Example 12.40 In a triangle 푃푄푅, ∠푃푄푅 = 68°, ∠푃푅푄 = 42°and |푃푄| = 2√5units. Find |푃푅| and ∠푄푃푅.

Solution We have, |푃푄| = 2√5units,푃푄푅 = 68° and 푃푅푄 = 42° In ∆푃푄푅: ∠푃푄푅 + ∠푃푅푄 + ∠푄푃푅 = 180° 푄 ⟹ 68° + 42° + 푄푃푅 = 180° 푄푃푅 = 180°− (68° + 42°) 2√5 68° ⟹ 푄푃푅 = 180°− 110° = 70° 휃 42° 푃 푅 The sine rule states that: | |

°= | |

°= | |

° ⟹ √

°= | |

°= | |

°

⟹ √°

= | |° ⟹ √

°× sin 68° = |푃푅|

⟹ |푃푅| ≈ √.

× 0.9271 ≈ 6.20 ∴, |푃푅| ≈ 6.20units (Corrected to 2 places of decimal) and 푄푃푅 = 70° Try: 1. XYZ is a triangle with |푋푌| = 12, |푌푍| = 15.5 |푋푍| = 푚,∠푋푌푍 = 훼,∠푋푍푌 = 휃and ∠푌푋푍 = 55°. Find the values of 푚,훼and휃. 2. Cosine rule Consider the figure below: B 훽 푐 ℎ 푎 훼 훾

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A C 푥 N 푏 − 푥 푏 Let |퐴퐵| = 푐, |퐵퐶| = 푎, |퐴퐶| = 푐 |퐵푁| = ℎ, ∠퐵퐴퐶 = 훼, ∠퐴퐶퐵 = 훽 and ∠퐴퐵퐶 = 훾 sin 훼 = ⟹ 푐 sin훼 = ℎ cos 훼 = ⟹ 푐 cos 훼 = 푥 Now, consider ∆퐵푁퐶 Pythagorean theorem:푎 = ℎ + (푏 − 푥) = (푐 sin 훼) + (푏 − 푐 cos훼) = 푐 sin 훼 + 푏 − 2푏푐 cos훼 + 푐 cos 훼

= 푏 + 푐 sin 훼 + 푐 cos 훼 − 2푏푐 cos훼 = 푏 + 푐 (sin α + cos α)− 2푏푐 cos훼 = 푏 + 푐 − 2푏푐 cos훼

⟹ 푎 = 푏 + 푐 − 2푏푐 cos훼… … …(1) If 훼 = 90°, then 푎 = 푏 + 푐 . This shows that the triangle is right triangle. Similarly,

푏 = 푎 + 푐 − 2푎푐 cos훽… … …(2) and 푐 = 푎 + 푏 − 2푎푏 cos훾… … …(3) Note: 1. The cosine rule can be used to find a length of a triangle given two sides and an included angle. 2. The cosine rule can be used to find the angle between two adjacent sides of a triangle..

Example 12.41 Q 120° 17.8 m 훽 R 훼 23.7 m Not drawn to scale P In the diagram above, |푃푅| = 23.7m, |푄푅| = 17.8 m,∠푃푄푅 = 120°∠푃푅푄 = 훽and ∠푄푃푅 = 훼. Find correct to 3 decimal places, the values of |푃푄|, 훼and훽.

Solution We have, |푃푅| = 23.7m, |푄푅| = 17.8°m, 푃푅푄 = 120°,푃푅푄 = 훽 and ∠푄푃푅 = 훼 The sine rule states that: | | = | | = | |

° ⟹ | | = . = .

°

274

⟹ . = .° ⟹ 17.8 sin 120° = 23.7 sin훼 ⟹ sin 훼 = . °

.

⟹ sin훼 = 0.6504 훼 = sin 0.6504 ≈ 40.5717° In ∆푃푄푅 훼 + 훽 + 120° = 180° ⟹ 40.5717 + 훽 + 120° ≈ 180° ⟹ 훽 = 180°− (40.5717° + 120°) ⟹ 훽 ≈ 180°− 160.5717° ≈ 19.4283° The cosine rule states that: |푃푄| = |푄푅| + |푃푅| − 2|푄푅||푃푅| cos 훽 ⟹ |푃푄| = (17.8) + (23.7) − 2(17.8)(23.7) cos(19.4283°) = 316.84 + 561.69 − 795.6773 = 82.8527 ⟹ |푃푄| = √82.8527 = 9.1023 ∴, |푃푄| = 9.10m,훼 ≈ 40.57° and 훽 = 19.43°

(Corrected to 2 places of decimal) Example 12.42

Find the values of 휃,휑and휔in the triangle below. A 휔 23 28 B 휃 휑 35 C

Solution The cosine rule states that: |퐴퐵| = |퐴퐶| + |퐵퐶| − 2|퐴퐶||퐵퐶| cos휑 ⟹ (23) = (28) + (35) − 2(28)(35) cos휑 ⟹ 529 = 784 + 1,225− 1,960 cos휑 ⟹ 1,960 cos휑 = 2,009− 529 ⟹ 1960 cos휑 = 1,480 ⟹ cos휑 = ,

,≈ 0.7551 ⟹ 휑 ≈ cos 0.7551

⟹ 휑 ≈ 40.9659° The cosine rule states that: |퐴퐶| = |퐴퐵| + |퐵퐶| − 2|퐴퐵||퐵퐶| cos휃 ⟹ (28) = (23) + (35) − 2(23)(35) cos휃 ⟹ 748 = 529 + 1,225− 1,610 cos휃 ⟹ 1,610 cos휃 = 1,754 − 748 ⟹ 1,610 cos휃 = 1,006 ⟹ cos휃 = ,

,≈ 0.6248 ⟹ 휃 ≈ cos 0.6248

⟹ 휃 ≈ 51.3292°

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In ∆퐴퐵퐶 휃 + 휑 + 휔 = 180° ⟹ 40.9659° + 51.3292° + 휔 ≈ 180° ⟹ 휔 = 180°− (40.9659° + 51.3292°°) ⟹ 휔 ≈ 180°− 92.2951° ≈ 87.7049°

∴, 휃 ≈ 51.33°,휑 ≈ 40.97°and 휔 ≈ 87.70°

Formulas for calculating the Area of triangles The area of a figure is the space occupied by its boundaries. The area of quadrilaterals such as rectangle, square and parallelogram is calculated from the product of their base and height/altitude. For example, the area of a rectangle is given by: Area= base×height=length× width. Geometrically, a triangle can be constructed by drawing a diagonal to divide certain quadrilaterals( for example square, rectangle) into two equal half. Then we will conclude that the area of each triangle is half the area of the quadrilateral. ℎ푦푝표푡푒푛푢푠푒 ℎ푒푖푔ℎ푡 푏푎푠푒 1. The area of a triangle given the 푏푎푠푒 and ℎ푒푖푔ℎ푡 is given by

퐴 =12 푏푎푠푒 × ℎ푒푖푔ℎ푡

Now, consider the triangle below: A From triangle 퐴퐵퐶 sin 휃 = ℎ = 푎 sin 훼 ℎ 푏

퐴푟푒푎 = 푏푎푠푒 × ℎ푒푖푔ℎ푡 B 휃

= 푏 × 푎 sin 휃 푎 C

= 푎푏 sin휃 2. The area of a triangle given two adjacent sides푎 and푏 and an angle 휃 between the sides is given by:

276

퐴 =12 푎푏 sin휃

3. The area of a triangle given all the sides of the triangle is given by the 퐻푒푟표푛 푠 Formula10:

퐴 = 푠(푠 − 푎)(푠 − 푏)(푠 − 푐), where 푠 = and 푎,푏and푐 are the sides of the triangle.

Example 12.43 Find the area of a triangle whose base is 8cm and height is 6cm.

Solution The area of the triangle is defined as 퐴 = 푏푎푠푒 × ℎ푒푖푔ℎ푡

⟹ 퐴 = × 8 × 6 = 4 × 6 = 24 ∴ 퐴푟푒푎 = 24cm (Corrected to 2 places of decimal) Example 12.44

The length of two adjacent sides of a triangle are 13m and 15m. If the angle between the two sides is 135°, find the area of the triangle.

Solution The area of the triangle is defined as 퐴 = 푎푏 sin휃 ⟹ 푎 = 13cm, 푏 = 15cmand휃 = 135° ⟹ 퐴 = × 13 × 15 sin 135° = sin 135° ≈ 68.9429 ∴ 퐴푟푒푎 = 68.94cm (Corrected to 2 places of decimal)

Example 12.45 Calculate, correct to the nearest meter, the area of a triangle whose sides are 19m, 12mand9mrespectively.

Solution The area of the triangle is defined as 퐴 = 푠(푠 − 푎)(푠 − 푏)(푠 − 푐) ⟹ 푎 = 19m, 푏 = 12mand 푐 = 9m. 푠 = = = 20m

⟹ 퐴 = 20(20 − 19)(20− 12)(20− 9) = 20(1)(8)(11) = √1,760 ∴ 퐴 = √1,760 = 41.95m (Corrected to 2 places of decimal)

10 The proof of this formula− see 퐶표푙푙푒푔푒푇푟푖푔표푛표푚푒푡푟푦,푉푒푟푠푖표푛⌊휋⌋퐶표푟푟푒푐푡푒푑푒푑푖푡푖표푛(2013)푏푦퐶. 푆푡푖푡푧&퐽.푍푒푎푔푒푟

277

Example 12.46 In a ∆푃푄푅, |푃푄| = 8푐mand∠푃푄푅 = 120°. If the area of the triangle is 12√5cm , calculate, correct to one decimal place, the sides |푃푅|and|푄푅. |

Solution

The area of the triangle is defined as 퐴 = |푃푄||푄푅| sin휃 푃 ⟹ |푃푄| = 8cm, ∠푃푄푅 = 휃 = 120° And 퐴 = 12√5cm 8푐m ⟹ 12√5 = × 8 × |푄푅| sin 120° 휃

⟹ 12√5 = 4|푄푅| sin 120 푄 푅

⟹ 12√5 = √ |푄푅| ⟹ |푄푅| = × √√

= √√

= √ = 2√15

∴ |푄푅| = 2√15cm The cosine rule states that: |푃푅| = |푄푅| + |푃푄| − 2|푄푅||푃푄| cos 휃

⟹ |푃푅| ≈ 2√15 + (8) − 2 2√15 (8) cos(120°) ≈ 60 + 64 + 61.9677 ≈ 185.9677 ⟹ |푃푅| = √185.9677 ≈ 13.3500 ∴, |푃푅| = 13.35cm (Corrected to 2 places of decimal) Try: Determine the area of the triangles below: A R Z 9 6 cm 10 m 11 30° 48° 50° 68° Y B 5 cm C P 푄 X 19

Angle of elevation and depression The angle of elevation of an object is the angle between the horizontal and the line connecting your position to the object given that the object is above you. The angle of depression of an object is the angle between the horizontal and the line connecting your position to the object given that the object is below.

278

Object You 휃 Angle of depression Angle of elevation 휃 You object Now consider the following situations: 1. Suppose a body X is standing at a distance 푑 away from a cliff C. If at the standing position the angle of elevations at the top of and bottom of a tower T on C are 훼 and 훽 respectively, then the height 퐻, of the tower can be calculated using the figure below: T 퐻 C ℎ 훽 훼 푑 2. Suppose a body X is moving towards a mountain M. If at the starting position the angle of elevation at the top of M is 훼 and after moving a distance 푑 the angle of elevation at the top of M is 훽, then the height of M can be calculated using the figure below: ℎ 훼 훽 푑

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Example 12.47 A boy stands at a distance of of 100 m away from a tree. The angle of elevation at the top of the tree is 25°. How tall is the tree?

Solution (See the figure below) 퐵 From ∆퐴퐵퐶 tan 25° = | |

| |=

⟹ 100tan 25° = ℎ ∴ ℎ ≈ 46.63m ℎ (Corrected to 2 places of decimal) 25° 퐴 100 m 퐶

Example 12.48 A boy is walking towards a mountain. The angle of elevation at the top of the mountain is 15°. After walking moving 50 m, the angle of elevation at the top of the pole is 23°. How tall is the mountain?

Solution (See the figure below) Let |퐵퐷| = 퐻 퐵 From ∆퐵퐶푁 tan 23° = | |

| |=

⟹ 푥 tan 23° = ℎ… … … (1) From ∆퐴퐵퐶 tan 15° = | |

| |= ℎ

15° 23° 퐴 50 m 퐶 푥 cm 푁 ⟹ (50 + 푥) tan 15° = ℎ… … … (2) Put ℎ = 푥 tan 23° into (2). ⟹ (50 + 푥) tan 15° = 푥 tan 23° (50 + 푥) ≈ 0.2679 0.4245푥 ⟹ 50 × 0.2679 + 0.2679푥 ≈ 0.4245푥 13.3950 ≈ 0.4245푥 − 0.2679푥 ⟹ 13.3950 ≈ 0.1566푥 푥 ≈ .

.≈ 85.5364

Put 푥 ≈ 85.5364 into (1) ⟹ ℎ ≈ 85.5364 tan 23° ≈ 36.3080 ∴, the height of the mountain is 36.31cm

(Corrected to 2 places of decimal)

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Try: In the figure, ∠퐵퐴푁 = 18°,∠퐵푁퐶 = 28°, B |퐴푁| = 35푚, |푁퐶| = 푥 , |퐵퐶| = ℎ. Find the values of 푥 and ℎ. 18° 28° A N C 35푚 푥

Factor formulae11 By the identities: sin(훼 + 훽) = sin 훼 cos훽 + cosα sin β sin(훼 − 훽) = sin 훼 cos훽 − cos훼 sin훽

cos(훼 + 훽) = cosαcosβ − sin훼 sin 훽 cos(α − β) = cos훼 cos훽 + sin훼 sin훽 Let 푥 = 훼 + 훽 … … …(1) and 푦 = 훼 − 훽… … … (2)

Add (1) to (2) ⟹ 푥 + 푦 = 훼 + 훼 ⟹ 훼 =

Subtract (2) from (1) ⟹ 푥 − 푦 = 훽 − (−훽) ⟹ 훽 = Now, sin(훼 + 훽) + sin(훼 − 훽) = sin 훼 cos훽 + cosα sin β+ sin훼 cos훽 − cos훼 sin 훽 = 2 sin 훼 cos훽 ⟹ sin 푥 + sin푦 = 2 sin cos

cos(훼 + 훽) + cos(α − β) = cosαcosβ − sin훼 sin 훽+cos 훼 cos훽 + sin 훼 sin훽 = 2cosα cos β ⟹ cos푥 + cos 푦 = 2 cos cos

sin(훼 + 훽) − sin(훼 − 훽) = sin훼 cos훽 + cosα sin β− sin훼 cos훽 + cos훼 sin 훽 = 2 cosα sin β ⟹ sin 푥 − sin푦 = 2 cos sin cos(훼 + 훽) − cos(α − β) = cosαcosβ − sin훼 sin 훽−cos훼 cos훽 − sin훼 sin 훽 = −2 sin 훼 sin훽 ⟹ cos 푥 − cos 푦 = −2 sin sin = 2 sin sin 11∗we will use these formulae under calculus

sin 푥 + sin 푦 = 2 sin cos cos푥 + cos 푦 = 2 cos cos

sin 푥 − sin 푦 = 2 cos sin cos푥 − cos 푦 = 2 sin sin

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Example 12.50 Simplify the following: i. sin 4푥 + sin 2푥 ii. cos3푥 + cos 푥 iii. sin 5푥 − sin 3푥

Solution i. sin 4푥 + sin 2푥 = 2 sin cos = 2 sin 3푥 cos푥

ii. cos3 푥 + cos푥 = 2 cos cos = 2 cos 2푥 cos 푥

iii. sin 5푥 − sin 3푥 = 2 cos sin = 2 cos 4푥 sin(−2푥) = −2 cos 4푥 sin 2푥

Final exercises

1. Solve the following equations:

i. tan 휃 = √3 , 0° ≤ 휃 ≤ 180° ii. Cos휃 = − √ , 0° ≤ 휃 ≤ 180°

2. Given that cos휃 = , 푦 > 1, find sin 휃, tan 휃 and cot휃.

3. Find an angle complementary to the following angles i. 45° ii. 30° iii. 63° iv. 72° v. 15° vi. 25° 4. Find the third size of a triangle with the following two sizes i. 45°, 63° ii. 32°, 45° iii. 10°, 22.5° iv. 20°, 90° 5. Find the Pythagorean triple of the following two numbers: i. 1, 3 ii, 5, 6 iii. , iv. √3, 2√2 6. Convert the following from radians to degrees i. 휋 ii. 휋 iii. 휋 iv. 휋 v. vi. 휋 vi. 휋 vii. 휋 7. Convert the following from degrees to radians i. 15° ii. 22.5° iii. 25° iv. 35° v. 85° vi. 108° vii. 225° vii. 325° 8. A sector of a circle of radius 3.5 cm subtends angle of 푟푎푑푠 at the centre. Calculate, leaving your answers in terms 휋, i. the length of the minor arc generated by the sector. ii. the area of the sector.

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iii. the area of the minor segment. iv. the perimeter of the minor segment. 9. The length of the minor arc generated by a sector of a circle of radius 12 m is 6√2휋 m. Find, leaving your answers in 휋, i. the value of the angle, 휃, formed at the centre of the circle. ii. the area of the major sector left. iii. the length of chord 퐴퐵 generated by the minor sector. iv. the perimeter of the major sector. 10. The length of chord AB which subtend an angle of 휋 at the centre 푂 of a circle of radius 푟 is 8cm. calculate, i. the value of 푟, correct to 1 decimal place. ii. the area of the major segment cut off by AB, leaving your answer in 휋. iii. the perimeter of the major segment, leaving your answer in 휋. 11. Simplify the following

i. + tan 푥 ii. − iii. + iv. (sin 푥 + cos 푥) + (sin푥 − cos 푥) v. sec 푥−sin 푥−cos 푥 vi. + tan 푥 vii. + 12. Solve for the values of 푥 in the following equations for the interval 0° ≤ 푥 ≤ 360° i. tan 푥 + cot푥 = 1.5 ii. cos 푥 − cos 푥 − 6 = 0 iii. sin 푥 − sin 푥 − 2 = 0 13. Express the following in the form 푅 cos(푥 − 훼),푅 > 0 and 훼 is acute. i. 9 cos 푥 + 12 sin 푥 ii. 4 cos 푥 + 3 sin 푥 iii. 4 cos푥 + 2√3 sin 푥 14. Express the following in the form 푅 cos(푥 + 훼),푅 > 0and 훼 is acute. i. 12 cos푥 − 5 sin 푥ii. 6 cos 푥 − 2 sin 푥iii. 7 cos 푥 − 6 sin 푥 15. Express the following in the form 푅 sin(푥 + 훼),푅 > 0 and 훼 is acute. i. 3 sin 푥 + 4 cos푥 ii. 3 sin 푥 + cos푥 iii. 33 sin 푥 + 3 cos 푥 16. Express the following in the form 푅 sin(푥 − 훼),푅 > 0 and 훼 is acute. i. 15 sin 푥 − 8 cos 푥ii. 2 sin 푥 − 3 cos 푥 iii. 7 sin 푥 − 12 cos푥 iv. 2 sin 푥 − 21 cos 푥 iii. 5 sin 푥 − √5 cos 푥 17. By writing the expression 7 sin 푥 + 3 cos 푥 in the form 푅 cos(푥 + 훼) ,푅 > 0 and 0° ≤ 훼 ≤ 90°, solve the equation 7 sin 푥 + 3 cos 푥 = 5 18. Given that 3 cos휃 + 4 sin 휃 = 푅 cos(휃 − 훼),where 푅 > 0,and 0 ≤ 훼 ≤ , state the values of 푅and tan훼.

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19. For each of the following equation, solve for 휃, on the interval 0 ≤ 휃 ≤ 2휋 and give your answers in radians, correct to 1 decimal place. i. 3 cos휃 + 4 sin휃 = 2 ii. 3 cos휃 + 4 sin휃 = 0.5 20. Starting from the identity cos(퐴 + 퐵) = cos퐴 cos퐵 − sin퐴 sin퐴, prove that i. cos 2휃 = 2 cos 휃 − 1ii. cos 2휃 = 1− 2 sin 휃 21. Find the general solution of the equation sin 휃 + tan휃 cos 2휃 = 0, giving your answer in radians. 22. By writing 3휃 = 2휃 + 휃 prove that i. sin 3휃 = 3 sin 휃 − 4 sin 휃 ii. cos 3휃 = 4 cos 휃 − 3 cos휃 23. Solve the following equations i. 2 sin 휃 = 1 + 2 sin 휃,0° ≤ 휃 ≤ 360° ii. 3 cos 2휃 + sin휃 = 1,0° ≤ 휃 ≤ 360° iii. sin 푥 = sin 2푥,0 ≤ 푥 ≤ 2휋 iv. 4 tan 휃 + sec 휃 + 1 = 0,0° ≤ 휃 ≤ 360° 24. Find all the solutions in the interval 0° ≤ 휃 ≤ 360° of the equation sin 휃 − cos휃 = 푘 when i. 푘 = 1 ii. 푘 = 0 25. Find the general solution of the equation sin 2푥 + cos 푥 = 0 , for 푥 radians. 26. Express 3 cos 푥 − 4 sin 푥in the form 퐴 cos(푥 + 훼),where 퐴 > 0and 훼 is acute, Stating the value of 훼 to the nearest 0.1°. 27. Solve the equation √3 tan휃 − sec 휃 = 1, giving all the solutions in the form

0° ≤ 휃 ≤ 360° 28. Express 5 cos휃 + 2 sin 휃 in the form 푅 sin(휃 + 훼), where 푅 > 0,and 0° ≤ 훼 ≤ 90° 29. A function 푓 is defined by 푓(푥) = 6 − 5 cos푥 − 2 sin 푥, for 0° ≤ 푥 ≤ 360°. Find the values of 푥 for which 푓(푥) = 0

30. By using the identities sin 푥 = , cos 푥 = and tan = 푡, solve the following equations, for 0° ≤ 푥 ≤ 360°. i. 4 cos 푥 + sin 푥 = 2 ii. 3 sin 푥 − 2 cos 푥 = 1 iii. sec 푥 = 2 31. Solve the following equations for 0° ≤ 푥 ≤ 270°. i. 7 cos 푥 − 6 sin 푥 = 4 ii. 6 sin 푥 + 8 cos 푥 = 7 iii. 4 cos푥 + 2 sin 푥 = √5 iv. sec 푥 + tan 푥 + 12 = 0. 32. Prove that

i. sin(휋 − 휃) = sin 휃 ii. + = 2 sec 푥 iii. + = 1

iv. (sin 푥 + cos 푥) = 1 + sin 2푥 v. sin 푥 − tan 푥 + sin 푥 tan 푥 = 0 vi. cot 휃 + tan 휃 = cosec 2휃

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33. Find the general solution of the following for 푥 radians i. sin 2푥 = cos푥 ii. sin 푥 = sin 푥 iii. 2 sin4 푥 = 1 iv. √3 tan 2푥 = 1 34. By referring to the factor formulae, simplify the following i. cos 7푥 − cos 5푥 ii. sin 7푥 − sin 2푥 iii. cos 2푥 + cos 8푥 iv. sin 3푥 + sin 5푥 v. cos 114° + sin 240° vi. sin 75° + sin 15°(leaving your answer as a surd) 35. If a 25 m pool cast a shadow of 18 m, what is the angle of elevation of the sun. 36. Find the area of the triangles below, correct, to the nearest 0.1 square units. a. b. 13 12 63° 56° 15 17 37. Solve the following equations for the value of 푥, 0° ≤ 푥 ≤ 360°. i. sin 푥 = cos 푥 ii. 2 sin푥 = cos푥 iii. sin 푥 − 1 = cos 푥 iv. cos푥 − 1 = sin 푥 iv. sin(5푥 − 80°) = 0.2 vi. 4 sin 2푥 = 3. 38. Use the Heron’s formula to find the area of the triangle below. a. b. c. 13 17 17 cm 12 8 cm 8 15 13 cm 6 39. Evaluate cos 20° cos 5° − sin 20° sin 5°, leaving your answer in surd form. 40. Find the values of ℎ and 푥 in the figure below. ℎ 18° 23° 50m 푥

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COORDINATE GEOMETRY (STRAIGHT LINES)

Consider the number line below: Figure 13.1

푥 푥 푥 푥 푥 푥

Every 푥on the number line is called a point or a coordinate. In the Cartesian coordinate system (named after mathematician René Descartes) a point can be specified by two coordinates using two perpendicular axes/planes (traditionally called number lines). The horizontal axis and the vertical axis are called 푥-axis and 푦-axis respectively. A point 푃(푥, 푦) in the plane is called an ordered pair; where 푥is called abscissa/푥-coordinate and 푦 is called ordinate/푦- coordinate. The point of contact between the two perpendicular axes is called the origin labeled 푂(0, 0). Note: in the ordered pair statement푃(푥,푦) ≠ 푃(푦, 푥).

The length/ distance/modulo between two points Consider the diagram below: 푦 푃(푥 ,푦 )

Figure 13.2 (푦 −푦 )

푄(푥 ,푦 ) 푥

( 푥 − 푥 )

From diagram |푃푄| = (푥 − 푥 ) + (푦 −푦 ) … … …(1)

(Pythagorean Theorem)

Example 13.1

Determine the distance between the following points:

i. A(8, 2)and퐵(3,−1) and ii.퐵(10,−3)andC(7, 0)

Solution

i. |퐴퐵| = (푥 − 푥 ) + (푦 −푦 ) ii. |퐴퐵| = (푥 − 푥 ) + (푦 −푦 )

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= (3 − 8) + (−1 − 2) = (7 − 10) + (0 − (−3))

= (−5) + (−3) = (−3) + (3)

= √25 + 9 = √9 + 9

= √34 = 3√2

Example 13.2

The distance between the points 퐴(−1, 1)and퐵(−푎, 2) is2√6.Find the value of 푎.

Solution

We have |퐴퐵| = 2√6 but

|퐴퐵| = (푥 − 푥 ) + (푦 −푦 ) = (−푎 − (−1)) + (2 − 1) = (−푎 + 1) + 1

So, 2√6 = (−푎 + 1) + 1 ⟹ 2√6 = (−푎 + 1) + 1

⟹ 4(6) = (−푎 + 1) + 1 ⟹ 24 = 푎 − 2푎 + 1 + 1

⟹ 푎 − 2푎 − 22 = 0

By using the quadratic formula 푎 = ( )± ( ) ( )( )( )

= ±√ = ± √

∴ 푎 = 1 − √23 or 푎 = 1 + √23

The midpoint between two points

Let 푃and푄be two points joining a straight line and let 푀be the midpoint of the two

points

Consider the diagram below:

푦 푄(푥 ,푦 )

(푦 − 푏) 푦

푀(푎,푏) 푏 Figure 13.3 (푏−푦 ) 푃(푥 ,푦 )푎 − 푥 푥 − 푎 푦 O 푥 푎 푥 푥 From the diagram: 푎 − 푥 = 푥 − 푎 ⟹ 2푎 = 푥 + 푥 ∴푎 = .

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Also 푦 − 푏 = 푏−푦 ∴푏 =

Hence,푀(푎, 푏) = 푥1+푥22 , 푦1+푦2

2 … … … (2) This gives the coordinates of the midpoint between two points.

Example 13.3 Find the midpoint of the following points. i.퐴(2, 1)and퐵(−1, 3) ii. 푃(10, 0)and푄(−2, 4)

Solution i. We have 퐴(2, 1)and퐵(−1, 3) midpoint M is given by

푀(푎, 푏) = 푥2+푥12 , 푦2+푦1

2 = 2+(−1)2 , 1+3

2 = , 2 ii. We have 푃(10, 0)and퐵(−2, 4) midpoint M is given by

푀(푎, 푏) = 푥2+푥12 , 푦2+푦1

2 = 10+(−2)2 , 0+4

2 = (4, 2)

Example 13.4 The midpoint of the line joining 푋(푖, 푗) and 푌 1, is 푅(3, 0). determine the values of 푖

and 푗. Solution

We have 푋(푖, 푗) and 푌 1, and midpoint 푅(3, 0) but 푀(푎,푏) = 푥2+푥12 , 푦2+푦1

2

So 푅(3, 0) = 푖+12 , 푗+

32

2 by comparing coordinates 3 = 푖+12 and 0 =

푗+32

2

⟹ 3(2) = 푖 + 2 ∴, 푖 = 6 − 2 = 4 and 2(0) = 푗 + ∴, 푗 = 0 − = − The gradient/slope of a line

Let 푃(푥 ,푦 ) and 푄(푥 ,푦 ) be any two points on a straight. Let 푚denote the gradient of the line. Consider the diagram below: 푦 푃(푥 ,푦 ) Figure 13.4 (푦 −푦 ) 푚 휃

푄(푥 ,푦 ) ( 푥 − 푥 ) 푥

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푚 = 푦1−푦2푥1−푥2

also tan휃 = 푦1−푦2푥1−푥2

So, tanθ = 푚

Example 13.5

Determine the gradient of the lines passing through the following points:

i. 퐴(−1, 1)and퐵(−1, 3) ii.푃(12, 1)and푄(−2, 3)

iii.푅(15,17) and 푆(−2,−2) iv.퐽(11, 1) and 퐾(3, 1)

Solution

i. Gradient of 퐴퐵 = 푦1−푦2푥1−푥2

=( )

= =not defined

ii. Gradient of 푃푄 = 푦1−푦2푥1−푥2

= = = −

iii. Gradient of 푅푆 = 푦1−푦2푥1−푥2

= = =

iv. Gradient of 퐴퐵 = 푦1−푦2푥1−푥2

= = = 0

It can be seen that the gradient can be negative, positive, zero or undefined (infinity).

Example 13.6

The gradient of the line joining the points (8, 2)and(3,푥) is . Find the value of 푥.

Solution

We have 퐴(8, 2) and 퐵(3,푥) and Gradient of 퐴퐵 =

But Gradient of 퐴퐵 = 푦1−푦2푥1−푥2

= so = ⟹ 5(−5) = 푥 − 2

∴ 푥 = −25 + 2 = −23

It can be seen that the gradient can be negative, positive, zero or undefined (infinity).

Collinear points Collinear points are points that lie on the same line. Therefore, if three or more points are collinear then the gradients between two pair of points are equal.

Example 13.6

Show that the following points are collinear. 푂(0, 0),푃(2, 2) and푄(4, 4).

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Solution We have 푂(0, 0),푃(2, 2) and푄(4, 4). If the points are collinear then Gradient of 푂푃 = Gradientof푂푄 = Gradientof푃푄 Gradient of 푂푃 = 푦1−푦2

푥1−푥2= = = 1 Gradient of 푂푄 = 푦1−푦2

푥1−푥2= = = 1

Gradient of 푂푃 = 푦1−푦2푥1−푥2

= = = 1 Therefore, points O, P and Q are collinear

Example 13.7 The points (−1,푏), (3 − 푏, 4) and (2푏,−푏) are collinear. Find the possible value of 푏.

Solution We have 푃(−1, 푏),푄(3 − 푏, 4) and 푅(2푏,−푏) If the points are collinear then Gradient of 푃푄 = Gradientof푃푅 = Gradientof푄푅 Gradient of 푃푄 = 푦1−푦2

푥1−푥2=

( )= = 1 Gradient of 푃푅 = 푦1−푦2

푥1−푥2=

( )=

Gradient of 푄푅 = 푦1−푦2푥1−푥2

=( )

=

So, 1 = = For the possible value of 푏 we set the equation

= 1 ⟹ −2푏 = 2푏 + 1 ⟹ −2푏 − 2푏 = 1 ∴ 푏 = − Try a. Which of the following set of points are collinear? i. (3,−1), (−3, 1) and (0, 0) ii. (2, 5), (−1,−1)and (−1,−7) iii. (1,−1), (−2, 4) and (0, 1). b.If the points (−1, 푡 − 1), (푡, 푡 − 3) and (푡 − 6, 3) are collinear, find the values of 푡.

Division of a line in a given ratio A line can be divided in a given ratio either internally or eternally. Let 푚 ∶ 푛 be the ratio that the point 푃(푝, 푞) divides a given line in both means. Then 푃(푝, 푞) = 푚푥2+푛푥1

푚+푛 ,푚푦2+푛푦1푚+푛 if 푃 divides the line joining 퐴 and 퐵internally and

푃(푝,푞)= , , 푚 > 푛if 푃 divides the line joining 퐴 and 퐵externally.

Proof Consider the following diagrams:

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Figure 13.5 Figure 13.6 푦 푦 푦 퐵(푥 ,푦 ) 푞 푃(푝, 푞) 푞 푃(푝, 푞) 푛 퐸 푦 퐵 푥 ,푦 푛 퐸 푚 푞 − 푦 푚 푞 − 푦 푦 퐴 푥 ,푦 푝 − 푥 푥 − 푝퐷 푦 퐴(푥 ,푦 ) 퐶 푝 − 푥 퐷 푥 푝 푥 푥 푥 푥 푝 푥

푝 − 푥 Internal division external division

In Figure 13.5 = (푏푦푠푖푚푖푙푎푟푖푡푖푒푠) ⟹ = ⟹ 푛(푝 − 푥1) = 푚(푥2 − 푝)

⟹ 푛푝 − 푛푥 = 푚푥 − 푚푝 ⟹ 푛푝 + 푚푝 = 푛푥 + 푚푥 ⟹ 푝(푛 + 푚) = 푚푥 + 푛푥 ∴,푝 = 푚푥2+푛푥1

푚+푛 Also = ⟹ = ⟹ 푛(푞 − 푦 ) = 푚(푦 − 푞) ⟹ 푛푞 − 푛푦 = 푚푦 −푚푞

⟹ 푛푞 + 푚푞 = 푛푦 + 푚푦 ⟹ 푞(푚 + 푛) = 푚푦 + 푛푦 ∴, 푞 = 푚푦2+푛푦1푚+푛

Hence, 푃(푝,푞) = 푚푥2+푛푥1푚+푛 ,푚푦2+푛푦1

푚+푛 Similarly, in Figure 13.6 = (푏푦푠푖푚푖푙푎푟푖푡푖푒푠) ⟹ 푝−푥1

푝−푥2= 푚

푛 ⟹ 푛(푝 − 푥 ) = 푚(푝 − 푥 ) ⟹ 푛푝 − 푛푥 = 푚푝 −푚푥 ⟹ 푛푞 − 푛푦 = 푚푞 −푚푦

⟹푚푝 − 푛푝 = 푚푥 − 푛푥 ⟹ 푝(푚− 푛) = 푚푥 − 푛푥 ∴,푝 = 푚푥2−푛푥1푚−푛

Also, = ⟹ = ⟹푛(푞 − 푦 ) = 푚(푞 − 푦 )

⟹ 푚푞 − 푛푞 = 푚푦 − 푛푦 ⟹ 푞(푚 − 푛) = 푚푦 − 푛푦 ∴, 푞 = 푚푦2−푛푦1푚−푛

Hence,푃(푝,푞) = 푚푥2−푛푥1푚−푛 ,푚푦2−푛푦1

푚−푛 ,푚 > 푛

Example 13.8 The point 푃(푥,푦) divides the line joining the following points internally and externally in the given ratios. Determine the coordinates of 푃 i. 퐴(3,−1), 퐵(7, 1) and 3: 1 ii.푋(−3, 2),퐵(13, 4) and 5: 2 iii. 푅(11, 8),푆 , 5 and 2: 3

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Solution i. We have,퐴(3,−1), 퐵(7, 1) and 푚 = 3and 푛 = 1

Internal division 푃(푥, 푦) = 푚푥2+푛푥1

푚+푛 ,푚푦2+푛푦1푚+푛 = 3(7)+1(3)

3+1 , 3(1)+1(−1)3+1 = 21+3

4 , 3−13+1 = 6,

External division

We have, 퐴(3,−1), 퐵(7, 1) and 푚 = 3and 푛 = 1, since 3 > 2 푃(푥, 푦) = 푚푥2−푛푥1

푚−푛 ,푚푦2−푛푦1푚−푛 = 3(7)−1(3)

3−1 , 3(1)−1(−1)3−1 = 21−3

2 , 3+12 = (9, 2)

ii. We have, 푋(−3, 2),퐵(13, 4) and 푚 = 5and 푛 = 2 Internal division 푃(푥, 푦) = 푚푥2+푛푥1

푚+푛 ,푚푦2+푛푦1푚+푛 = 5(13)+2(−3)

5+2 , 5(4)+2(2)5+2 = 65−6

7 , 20+47 = ,

External division We have, 푋(−3, 2),퐵(13, 4) and 푚 = 5and 푛 = 2, since 5 > 2 푃(푥, 푦) = 푚푥2−푛푥1

푚−푛 ,푚푦2−푛푦1푚−푛 = 5(13)−2(−3)

5−2 , 5(4)−2(2)5−2 = 65+6

3 , 20−43 = ,

iii. We have, 푅(11, 8),푆 , 5 and 푚 = 2and푛 = 3 Internal division

푃(푥, 푦) = 푚푥2+푛푥1푚+푛 ,푚푦2+푛푦1

푚+푛 = 2 12 +3(11)

2+3 , 2(5)+3(8)2+3 = 1+33

5 , 10+245 = ,

External division We have, 푅(11, 8),푆 , 5 and 푚 = 3and 푛 = 2, since 3 > 2

푃(푥, 푦) = 푚푥2−푛푥1푚−푛 ,푚푦2−푛푦1

푚−푛 = 3 12 −2(11)

3−2 , 3(5)−2(8)3−2 = − 22, 15− 16

= − ,−1

Example 13.9 The point 퐷(푥,푦) divides the line joining the points퐴(12,−1) and 퐵(−1,−7)in the ratio 2:−1. Find the values of 푚 and푛.

Solution We have, 퐴(12,−1) and 퐵(−1,−7) ratio 2:−1

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Since the question is silent about which division we assume that 퐷(푥,푦) divide line AB internally. So, 퐷(푥, 푦) = 푚푥2+푛푥1

푚+푛 ,푚푦2+푛푦1푚+푛 = 2(−1)−1(12)

2−1 , 2(−7)−1(−1)2−1 = (−2 − 12,−14 + 1)

=(−14,−13)

Example 13.9 The point 푃(4,−3) divides the line joining the points 퐾(−5, 6) and 퐿(푎, 푏) externally in the ratio 4: 1. Find the coordinates of 퐿.

Solution We have,푃(4,−3),퐾(−5, 6) and 퐿(푎,푏) and 푚 = 4and 푛 = 1,since 4 > 1 푃(푥,푦) = 푚푥2−푛푥1

푚−푛 ,푚푦2−푛푦1푚−푛 = 4(푎)−1(−5)

4−1 , 4(푏)−1(6)4−1 = 4푎+5

3 , 4푏−63

⟹푃(4,−3) = 4푎+53 , 4푏−6

3 By comparing coordinates

4 = 4푎+53 ⟹ 4(3) = 4푎 + 5 ⟹ 12 − 5 = 4푎 ∴,푎 = and

−3 = 4푏−63 ⟹ 3(−3) = 4푏 − 6 ⟹−9 + 6 = 4푏 ∴,푏 = −

The equation of a line

In general, an equation of a line can be written in the form 푎푥 + 푏푦+ 푐 = 0, where 푎, 푏and푐 are parameters. The coordinates of the 푥- and 푦-intercepts are , 0 and

0, respectively. The 푥-intercept refers to any point on the 푥-axis at which the 푦 coordinate is 0. The 푥-intercept can be calculated as follows:

푎푥 + 푏(0) + 푐 = 0 ⟹ 푎푥 + 푐 = 0 ∴, 푥 = The 푦-intercept can be defined and calculated in a similar way. The equation of a line can be formed if the following are given. i. A point on the line and the gradient of the line. ii. Two points on the line. iii. The 푥- and 푦-intercepts of the line.

a.The equation of a line given a point on the line and the gradient of the line Let 퐴(푥 ,푦 ) and 푃(푥,푦) be a given point and arbitrary point on a line respectively. Let 푚 be the gradient of the line. Then

= 푚 ⟹ 푦 − 푦 = 푚(푥 − 푥 ) … … …(a)

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This is the equation of a line given a point on the line and the gradient of the line. Example 13.10

For each of the following, find the equation of the line given the point on the line and the gradient of the line. i.푇(5,−6), gradient of ii. 푈(3.2, 5.3), gradient of 0.25

iii. 푉 , ,gradient of 8

Solution i. We have, 푇(5,−6) and 푚 = . The equation of the line passing through of 푇(5,−6) is given by 푦 − 푦 = 푚(푥 − 푥 ) ⟹ 푦— 6 = − (푥 − 5) ⟹ 4(푦 + 6) = −7(푥 − 5) and 4푦 + 24 = −7푥 + 35 ⟹ 7푥 + 4푦 + 24 − 35 = 0 ∴ 7푥 + 4푦 − 11 = 0 ii. 푈(3.2, 5.3)and 푚 = 0.25 The equation of the line passing through of 푈(3.2, 5.3)is given by

푦 − 푦 = 푚(푥 − 푥 ) so, 푦 − 5.3 = 0.25(푥 − 3.2) ⟹ 푦 − 5.3 = 0.25(푥 − 3.2) and 푦 − 5.3 = 0.25푥 − 0.8 ⟹ 0.25푥 − 푦 − 0.8 + 5.3 = 0 ∴ 0.25푥 − 푦 + 4.5 = 0 iii. 푉 , and 푚 = 8

The equation of the line passing through of 푉 , is given by 푦 − 푦 = 푚(푥 − 푥 )

so, 푦 − = 8 푥 − ⟹ 21 푦 − = 21 8푥 − and

21푦 − 2 = 168푥 − 21 × ⟹ 21푦 − 2 = 168푥 − 24 ⟹ 168푥 − 21푦 − 24 + 2 = 0 ∴, 168푥 − 21푦 − 22 = 0

b. The equation of a line given two points on the line Let 퐴(푥 ,푦 ) and 퐵(푥 ,푦 ) be two points on a line. And let 푃(푥, 푦) be an arbitrary point on the line. Since the points are collinear points then

= = ⟹ =

or = .

But

= 푚 ⟹ = 푚hence, 푦 − 푦 = 푚(푥 − 푥 )

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Or = 푚 ⟹ 푦 − 푦 = 푚(푥 − 푥 ) … … …(b)

Either (a) or (b) can be used to find the equation of a line given two points on the line.

Example 13.11 For each of the following, find the equation of the line given the two points on the line. i. (9, 11) and (2, 16) ii. , and ,−1 iii. (−8, 19) and(4,−17)

Solution i. Let 퐴(9, 11) and 퐵(2, 16) 푚 = 푦2−푦1

푥2−푥1= 16−11

2−9 = − The equation of the line passing through point 퐴(9, 11) is given by

푦 − 푦 = 푚(푥 − 푥 ) So, 푦 − 11 = − (푥 − 9) ⟹ 7(푦 − 11) = −5(푥 − 9) and 7푦 − 77 = −5푥 + 45 ⟹ 5푥 + 7푦 − 77− 45 = 0 ∴, 5푥 + 7푦 − 122 = 0 Similarly, The equation of the line passing through point 퐵(2, 16) is given by

푦 − 푦 = 푚(푥 − 푥 )

So, 푦 − 16 = − (푥 − 2) ⟹ 7(푦 − 16) = −5(푥 − 2) and 7푦 − 112 = −7푥 + 10 ⟹ 5푥 + 7푦 − 112 − 10 = 0 ∴ ,7푥 + 5푦 − 122 = 0

ii. Let 퐴 , and 퐵 ,−1 푚 = = = = ÷ = × = The equation of the line passing through point 퐴 , is given by 푦 − 푦 = 푚(푥 − 푥 ) So, 푦 − = 푥 − ⟹ 푦 − = 푥 − and

30 푦 − = 30 푥 − ⟹ 30푦 − 30 × = 30 × 푥 − 30 × 푥

⟹ 30푦 − 75 = 378푥 + 126 ∴ 378푥 − 30푦 + 201 = 0

iii. Let 퐴(−8, 19) and퐵(4,−17) 푚 = 푦2−푦1푥2−푥1

= −17−194−(−8) = − = −3

The equation of the line passing through point 퐴(−8, 19) is given by 푦 − 푦 = 푚(푥 − 푥 )

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So, 푦 − 19 = −3(푥 − (−8))⟹ 푦 − 19 = −3(푥 + 8) and 푦 − 19 = −3푥 − 24 ⟹ 3푥 − 푦 − 19 + 24 = 0 ∴ 3푥 + 푦 + 5 = 0

c. The equation of a line given the 푥- and 푦-intercepts Let 퐴(푎, 0) and 퐵(0,푏) the points of the 푥- and 푦-intercepts of a line respectively. Also let 푃(푥, 푦) be an arbitrary point on the line. Then the equation of a line passing through these points is given by:

+ = 1. Proof

Consider the diagram below: 푦

Since 퐴,퐵 and 푃 are collinear points 퐵(0,푏)Figure 2.7 ⟹ = = .

푃(푥,푦) Take = ⟹ 푎푦 = −푏푥 + 푎푏 ⟹ 푏푥 + 푎푦 = 푎푏 퐴(푎, 0) 푥 ⟹ + =

∴, + = 1

Example 13.12 Find the equation of the line which passes through the intercepts (3, 0) and (0, 4)

Solution We have 퐴(3, 0) and 퐵(0, 4)푥 −intercept푎 = 3 and 푦 − intercept푏 = 4 The equation of the line passing through the intercepts is given by + = 1 ⟹ + = 1 ⟹ 12 × + 12 × = 1 × 12 ⟹ 4푥 + 3푦 = 12 ∴ ,4푥 + 3푦 − 12 =0

Example 13.13

Find the equation of the line that cut the lines 푥 = −3 and 푦 = 2 Solution

Given 푥 = −3 implies 푦 = 0, so let 퐴(−3, 0). Also, given 푦 = 2, implies 푥 = 0 so, let 퐵(0, 2). Now, 푥 −intercept 푎 = −3 and 푦 −intercept 푏 = 2 The equation of the line passing through the intercepts is given by + = 1 ⟹ + = 1 ⟹ 6 × + 6 × = 1 × 6

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⟹ −2푥 + 3푦 = 6 ∴ ,2푥 − 3푦 + 6 =0

Determining the gradient and 풙- and 풚- intercepts of a line Let 퐴(0, 푐) and 푃(푥,푦) be point of the 푦-intercept and arbitrary point on a line respectively. And let 푚be the gradient of the line.

Then = 푚 ⟹ 푦 − 푐 = 푚푥 ∴푦 = 푚푥 + 푐… … …(3) Now, let 푎푥 + 푏푦 + 푘 = 0 be an equation of a line.

Then 푏푦 = −푎푥 − 푘 ⟹ 푦 = 푥 − … … …(4)

Comparing (3) with (4)푚 = and푐 = .

From (4) when푦 = 0, 푥 = − . Multiplying through by푏, we have 푎푥 = −푘

∴,푥 = − .

Hence, the coordinates of the 푥- and 푦- intercepts of (3) are ( , 0) and 0, , where 푎, 푏 and 푐 are parameters. (3) is the general equation used to compare with any given equation to determine the gradient and 푦- intercept of any given line. The 푥-intercept is determined by putting the 푦in the equation to 0.

Example 13.14 Determine the gradient and the coordinates of the 푥- and 푦-intercepts of the following equations. i. 푦 = √5푥 + 17 ii. 6푦 + 3푥 − 1 = 0 iii. 푦 + 푥 = −2.

Solution i. We have 푦 = √5푥 + 17. By comparing this relation with푦 = 푚푥 + 푐 we get 푚 = √5and 푐 = 17. For the 푥 −intercept set 푦 = 0 ⟹ 0 = √5푥 + 17 ⟹ √5푥 = −17 ∴ 푥 = −

√= − √

∴ 푚 = √5,푥 −intercept= − √ , 0 and 푦 −intercept= (0, 17) ii. We have, 6푦 + 3푥 − 1 = 0 ⟹ 6푦 = −3푥 + 2 ⟹ 푦 = − 푥 + = − 푥 + … … … (1). By comparing (1) with 푦 = 푚푥 + 푐

we get 푚 = − and 푐 = . For the 푥 −intercept set 푦 = 0 in (1)

⟹ 0 = − 푥 + ⟹ − 푥 = ∴ 푥 = × −2 = −

∴ 푚 = − , 푥 −intercept= − , 0 and 푦 −intercept= 0,

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iii. We have, 푦 + 푥 = −2 ⟹ 6 푦 + 6 푥 = −2(6) ⟹ 2푦 + 33푥 = −12 ⟹ 2푦 = −33 − 12 푦 = − 푥 − = − 푥 − 6 … … … (1). By comparing (1) with 푦 = 푚푥 + 푐

we get 푚 = − and 푐 = −6. For the 푥 −intercept set 푦 = 0 in (1)

⟹ 0 = − 푥 − 6 ⟹ − 푥 = 6 ∴,푥 = 6 × − = −

∴,푚 = − , 푥 −intercept= − , 0 and 푦 −intercept= (0,−6)

Intersection of two lines Let 푙 and 푙 be two independent and inconsistent lines. If 푙 and 푙 intersect each other at a point 푃 then the coordinates of 푃 is the simultaneous solution of the two lines.

Example 13.15 The lines 2푥 − 푦 − 8 = 0 and 3푥 + 푦 = 2 intersect at point푃, find the coordinates of푃.

Solution We have,2푥 − 푦 − 8 = 0 ⟹ 2푥 − 푦 = 8 … … … (1) and 3푥 + 푦 = 2 … … … (2) Now, solve equation (1) and (2) Add (1) to (2) ⟹ 3푥 + 2푥 + 푦 − 푦 = 2 + 8 ⟹ 5푥 = 10 ∴ 푥 = = 2 Put 푥 = 2 into (1) ⟹ 3(2) + 푦 = 2 ⟹ 6 + 푦 = 2 ∴ 푦 = 2 − 6 = −4 ∴ 푃(2, 4)

Example 13.16 Find the coordinates of the point of intersection of the lines 5푦 = 3푥 − 6and 4푦 + 2푥 = −1.

Solution We have, 5푦 = 3푥 − 6 ⟹ 3푥 − 5푦 = 6 … … … (1) and 2푥 + 4푦 = −1 … … … (2) Now, solve equation (1) and (2) From (2) 푥 = Put 푥 = into (1)

⟹ 3 − 5푦 = 6 ⟹ 2 × 3 − 2 × 5푦 = 6 × 2 3(−1− 4푦)− 10푦 = 12 ⟹ −3 − 12푦 − 10푦 = 12 ⟹ −22푦 = 12 + 3 ∴ 푦 = − Put 푦 = − into (2)

⟹ 푥 = ⟹ 푥 = ∴ 푥 = =

∴the coordinates of the point of the point of intersection is ,−

Example 13.17

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The point 푄(−8, 6) is the point of intersection of the lines 푦 = 푎푥 − 7 and푦 − 2푏푥 = 3. Find the values of 푎and푏.

Solution Method 1

We have, 푄(−8, 6), 푦 = 푎푥 − 7 … … … (1) and푦 − 2푏푥 = 3 … … … (2) Put 푄(−8, 6) into both equations. ⟹ 6 = 푎(−8)− 7 ⟹ 6 + 7 = −8푎 ⟹ −8푎 = 13 ∴,푎 = − and 6 − 2푏(−8) = 3 ⟹ 16푏 = 3 − 6 ∴,푏 =

Method 2 We have, 푄(−8, 6), 푦 = 푎푥 − 7 … … … (1) and푦 − 2푏푥 = 3 … … … (2) Solve equation (1) and (2) for 푥 and 푦. Put 푦 = 푎푥 − 7 into (2) ⟹ 푎푥 − 7− 2푏푥 = 3 ⟹ 푎푥 − 2푏푥 = 3 + 7 ⟹ (푎 − 2푏)푥 = 10 ⟹ 푥 = Put 푥 = into (1)

⟹ 푦 = 푎 − 7 ⟹ 푦 = ( ) = = Hence, the coordinates of the point of intersection for the two lines is ,

So, = −8 ⟹ 10 = −8푎 + 16푏… … … (3) and = 6 ⟹ 3푎 + 14푏 = 6푎 − 12푏 ⟹ 14푏 + 12푏 = 6푎 − 3푎 ⟹ 26푏 = 3푎 ⟹ 푎 = 푏… … … (4) Now, solve equation (3) and (4) for 푎 and 푏. Put 푎 = 푏 into (3) ⟹ −8 푏 + 16푏 = 10 ⟹ = 10

⟹ 푏 = 10 ∴,푏 = 10 × − = −

Put 푏 = − into (4) ∴,푎 = ×− = −

Condition for a point to satisfy the equation of a line Let 푎푥 + 푏푦 + 푐 = 0 an equation of a line. If a point 퐴 푥 ,푦 lies on this line then

푎푥 + 푏푦 + 푐 = 0

Example 13.18 Show whether or not the points푅(0, 7), 푆(−3,4) and 푇(7, 7) lie on the line

2푥 + 푦 − 7 = 0 Solution

We have the points 푅(0, 7), 푆(−3,4) and 푇(7, 7) and the line 2푥 + 푦 − 7 = 0 Put 푅(0, 7)into the line.

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2(0) + 7 − 7 = 0 + 7 − 7 = 0 ∴ 푅(0, 7) lies on the line. Put 푆(−3, 4)into the line. 2(−4) + 4− 7 = −12 + 4 − 7 = −15 ≠ 0 ∴,푆(−3, 4) does not lie on the line. Put 푇(7, 7)into the line. 2(7) + 7 − 7 = 14 + 7 − 7 = 14 ≠ 0 ∴,푇(7, 7) does not lie on the line.

Parallel and perpendicular lines Let 푙 and 푙 be two lines and let 푚 and 푚 be the gradients of 푙 and 푙 . If 푙 and 푙 are parallel to each other, then 푚 = 푚 . If 푙 and 푙 are perpendicular to each other, then 푚 푚 = −1.

Proof Consider the following diagrams:

Figure 13.7 Figure 13.8 푦 푦 푄(1, 2푚 ) 푃(0,푚 ) 푙 퐶(0,푚 −푚 ) 푙 푅(1, 2푚 ) 퐵(1,푚 ) 푆(0,푚 ) 푙 푂(0, 0) 푙 푥 푥

Parallel lines perpendicular lines In Figure 2.7,|푃푆| = |푄푅| (use 푑 = (푥 − 푥 ) + (푦 − 푦 ) ) ⟹ (0 − 0) + (푚 −푚 ) = (1 − 1) + (2푚 − 2푚 ) ⟹ (푚 −푚 ) = (2푚 − 2푚 ) ⟹ 푚 −푚 = 2푚 − 2푚 ⟹ 2푚 −푚 = 2푚 −푚 ∴,푚 = 푚 In Figure 2.8, |푂퐶| = |퐵퐶| + |퐵퐶| This follows that

300

(0) + (푚 −푚 − 0) = (1 − 0) + (푚 − 0) + (0− 1) + (푚 −푚 −푚 ) ⟹ (푚 −푚 ) = 1 + 푚 + 1 + 푚 ⟹ 푚 − 2푚 푚 + 푚 =1 + 푚 + 1 + 푚 ⟹ −2푚 푚 = 2 + 푚 −푚 + 푚 −푚 ⟹ −2푚 푚 = 2 (Divide both sides by −2) ∴, 푚 푚 = −1

Example 13.19 i. Find the equation of the line parallel to the line 7푥 − 2푦 = 16and passing through the point (1,−3).

Solution We have, 7푥 − 2푦 = 16 ⟹ 2푦 = 7푥 − 16 ⟹ 푦 = 푥 − = 푥 − 8 … … … (1). By comparing (1) with 푦 = 푚 푥 + 푐

we get 푚 = 푚 = (since the two lines are parallel they have the same gradient.) The equation of the line parallel to the line 7푥 − 2푦 = 16and passing through the point (1,−3) is given by: 푦 − 푦 = 푚 (푥 − 푥 ) ⟹ 푦 − (−3) = (푥 − 1) ⟹ 2(푦 + 3) = 7(푥 − 1) ⟹ 2푦 + 6 = 7푥 − 7 ⟹ 7푥 − 2푦 − 7 − 6 = 0 The line: 7푥 − 2푦 − 13 = 0 ii. Find the equation of the line perpendicular to the line 10푥 + 11푦 − 3 = 0 .and passing through the point(7, 8). Solution We have, 10푥 + 11푦 − 3 = 0 ⟹ 11푦 = −10푥 + 3 ⟹ 푦 = − 푥 + = 푥 − 8 … … … (1). By comparing (1) with 푦 = 푚 푥 + 푐

we get 푚 = − and 푚 = − = − =

(since the two lines are perpendicular to each other the gradient of one is the negative reciprocal of the other.) The equation of the line perpendicular to the line 10푥 + 11푦 − 3 = 0and passing through the point (7, 8) is given by: 푦 − 푦 = − (푥 − 푥 )

⟹ 푦 − (8) = (푥 − 7) ⟹ 10(푦 − 8) = 11(푥 − 7) ⟹ 10푦 − 80 = 11푥 − 77 ⟹ 11푥 − 10푦 − 77 + 80 = 0

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The line: 11푥 − 10푦 + 3 = 0 From the two solutions given above we can deduce that if 푎푥 + 푏푦 + 푐 = 0is a line, then 1. a line parallel to the this line and passing through the point (푥 ,푦 ) is given by:

푎푥 + 푏푦 + 푘 = 0 2. a line perpendicular to this line and passing through the point (푥 , 푦 ) is given by:

푏푥 − 푎푦 + 푘 = 0 In both cases k is parameter. Using the above idea, example 13.19 can be solved as follow: i. We have, 7푥 − 2푦 = 16 ⟹ 7푥 − 2푦 − 16 = 0 … … … (1).By comparing (1) with the general equation푎푥 + 푏푦 + 푐 = 0 we get 푎 = 7, 푏 = −2 The equation of the line parallel to the line 7푥 − 2푦 = 16 and passing through the point (1,−3) is given by: 푎푥 + 푏푦 + 푘 = 0 ⟹ 7푥 − 2푦 + 푘 = 0 … … … (2) Put (1,−3) into (2) ⟹ 7(1)− 2(−3) + 푘 = 0 ⟹ 7 + 6 + 푘 = 0 ∴ 푘 = −13 The line: 7푥 − 2푦 − 13 = 0 ii. We have, 10푥 + 11푦 − 3 = 0 … … … (1).By comparing (1) with the general equation 푎푥 + 푏푦 + 푐 = 0 we get 푎 = 10, 11 = −2 The equation of the line perpendicular to the line 10푥 + 11푦 − 3 = 0 and passing through the point (7,8) is given by: 푏푥 − 푎푦 + 푘 = 0 ⟹ 11푥 − 10푦+ 푘 = 0 … … … (2) Put (7, 8) into (2) ⟹ 11(7)− 10(8) + 푘 = 0 ⟹ 77 − 80 + 푘 = 0 ∴ 푘 = 3 The line: 11푥 − 10푦 + 3 = 0

Example 13.20 Find the equation of the line parallel to 13푦 = 2푥 − 17 and passing through the point 퐴(16,−1).

Solution We have, 13푦 = 2푥 − 17 ⟹ 2푥 − 13푦 − 17 = 0 … … … (1).By comparing (1) with the general equation 푎푥 + 푏푦 + 푐 = 0 we get 푎 = 2,푏 = −13 The equation of the line parallel to the line 13푦 = 2푥 − 17and passing through the point 퐴(16,−1) is given by: 푎푥 + 푏푦 + 푘 = 0 ⟹ 2푥 − 13푦 + 푘 = 0 … … … (2) Put 퐴(16,−1) into (2)

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⟹ 7(16)− 2(−1) + 푘 = 0 ⟹ 112 + 2 + 푘 = 0 ∴ 푘 = −114 The line: 2푥 − 13푦 − 114 = 0

Example 13.21 Determine the equation of the line that passes through the midpoint of the line joining

the points (9,−1) and (5, 5). Solution

Let 퐴(9,−1) and 퐵(5, 5). The midpoint of 퐴퐵 is given by 푀(푎, 푏) = , , = (7, 2)

The gradient of 퐴퐵 is given by 푚 = =( )

= − = − The equation of the line passing through midpoint 푀(7, 2) is given by 푦 − 푦 = 푚(푥 − 푥 ) So, 푦 − 2 = − (푥 − 7) ⟹ 3(푦 − 2) = −2(푥 − 7) and 3푦 − 6 = −2푥 − 14 ⟹ 2푥 + 3푦 − 6 + 14 = 0 The line: 2푥 + 3푦 + 8 = 0 Try: The lines 4푥 + 6푦 − 5 = 0 and 2푥 + 4푦 − 3 = 0 intersect at 푁.Find the equation of the line through 푁 perpendicular to the line 푥 + 2푦 = 0 (SSSCE)

The perpendicular distance from a point to a line Let 푑 be the perpendicular distance from a point 푃(푥 ,푦 ) to the line 푎푥 + 푏푦 + 푐 = 0, Then 푑 =

√

Proof Consider the following diagram: 푦 푎푥 + 푏푦+ 푐 = 0 푄(푥, 푦) Figure 13.10 푦 − 푦 푑 휃 푃(푥 ,푦 ) 푁 푥 푥 − 푥

In Figure 13.10. sin휃 = 푦−푦1푑 and cos휃 =

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⟹ 푑 sin휃 = 푦 − 푦 ⟹ 푦 = 푑 sin 휃 + 푦 and푑 cos휃 = 푥 − 푥 ⟹ 푥 = 푑 cos휃 + 푥 So, 푄(푥,푦) = (푑 cos휃 + 푥 , 푑 sin휃 + 푦 ) Substitute푥 = 푑 cos휃 + 푥 and푦 = 푑 sin 휃 + 푦 into 푎푥 + 푏푦 + 푐 = 0 ⟹ 푎(푑 cos휃 + 푥 ) + 푏(푑 sin 휃 + 푦 ) + 푐 = 0 ⟹ 푎푑 cos 휃 + 푎푥 + 푏d sin θ + by + 푐 = 0 (Expand the brackets) ⟹ 푎푥 + 푏푦 + 푐 = −푎푑 cos휃 − 푏푑 sin 휃 (Group like terms) ⟹−푑(푎 cos휃 + 푏 sin 휃) = 푎푥 + 푏푦 + 푐(Factorise–푑out) From trigonometry, we learnt that 푎 cos휃 + 푏 sin 휃 ≡ 푅(cos휃 ± 훼) We proved that 푅 = √푎 + 푏 and that the maximum value of 휃occurs at 푅(cos휃 ± 훼) = 푅

⟹ −푑(푅) = 푎푥 + 푏푦 + 푐 ⟹ −푑 √푎 + 푏 = 푎푥 + 푏푦 + 푐(Divide through by √푎 + 푏 )

⟹−푑 = 푎푥1+푏푦1+푐

푎2+푏2

⟹ |−푑| = 푎푥1+푏푦1+푐

푎2+푏2∴ 푑 = 푎푥1+푏푦1+푐

푎2+푏2(Take the absolute value of both sides)

Example 13.22

Find the perpendicular distance from the point (−1, 2)to the line 푥 − 2푦 − 7 = 0 Solution

Let 푃(−1, 2). We have the line 푥 − 2푦 − 7 = 0 … … … (1). By comparing (1) with 푎푥 + 푏푦 + 푐 = 0, we get 푎 = 1, 푏 = −2 and 푐 = −7 The perpendicular distance from 푃(−1, 2) to the line 푥 − 2푦 − 7 = 0 is given by 푑 =

√= ( ) ( )

( ) ( )=

√=

√= √

∴ 푑 = √

Example 13.23 A point 푃(−3, 푟) lies in the plane such that such that the perpendicular distance from 푃 to the line 푦 = 3푥 − 6 is √5,find the value of푟.

Solution We have, 푃(−3, 푟)and 푦 = 3푥 − 6 ⟹ 3푥 − 푦 − 6 = 0 … … … (1). By comparing (1) with 푎푥 + 푏푦 + 푐 = 0, we get 푎 = 3, 푏 = −1 and 푐 = −6 The perpendicular distance from 푃(−3, 푟) to the line 3푥 − 푦 − 6 = 0 is given by

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푑 = √

= ( ) ( )( )

=√

= −√

=√

√5 =√

√5 √10 = 푟 + 15 5√2 − 15 = 푟

∴ 푟 = 5√2 − 15

Reflection of a point in a line If a point 푃(푥, 푦) is reflected by a line, an image point is formed. Note the following about reflection of point in front of a line: 1. The point is perpendicular to the line. 2. The distance from the point to the line is equal to the distance from the image point to the line 3. Both the point and the image point formed are at the opposite sides of the line. 푦 푃(푥, 푦) 푙 푙 Figure 13.11 푀(Midpoint of 퐴 and 푃 or point of intersection of 푙 and푙 ) 퐴(푎, 푏) Image point 푥

Example 13.24 A point 푃(5,−1) is reflected in the line 푥 − 2푦 = 8. Find the image point.

Solution The point (5,−1) is reflected in the line 푥 − 2푦 = 8 … … … (1). Let the image point be 푄(푚,푛) First, find the equation of the line passing through푃.That line is perpendicular to the line 푥 − 2푦 = 8. By comparing (1) with the general equation 푎푥 + 푏푦 + 푐 = 0, we get 푎 = 1, 푏 = −2 The equation of the line perpendicular to the line 푥 − 2푦 = 8 and passing through the point (5,−1) is given by: 푏푥 − 푎푦 + 푘 = 0 ⟹ −2푥 − 푦 + 푘 = 0 … … … (2) Put (5,−1) into (2) ⟹ −2(5)− (−1) + 푘 ⟹ −10 + 1 = 푘 = 0 ∴ 푘 = 9 The line: −2푥 − 푦 + 9 = 0 Solve equation (1) and (2)to get the midpoint or the point of intersection of the two lines.

305

From (1) 푥 = 8 + 2푦 put 푥 = 8 + 2푦 into (2) −2(8 − 2푦) − 푦 + 9 = 0 −16 + 4푦 − 푦 + 9 = 0 3푦 = 16 − 9 푦 = Put 푦 = into (1)

푥 = 8 + 2 = ( ) = =

Hence, 푀 , is the point of intersection of the two lines.푀 , is also the point of the midpoint of 푃(5,−1) and the image point 푄(푚,푛)

The midpoint between the two points is defined as: 푀(푎,푏) = , = ,

푀 , = , By comparing the LHS with RHS we get, = and = ⟹ 2(70) = 3(푚 + 5) 2(7) = 3(푛 − 1) ⟹ 140 = 3푚 + 15 14 = 31 − 3 ⟹ 3푚 = 140 − 15 3푛 = 14 + 3 ∴ 푚 = 푛 =

∴ the image point is 푄 , Example 13.25

The point 퐵(2, 3) is the image point of the point 퐴 reflected in the line5푥 = 6푦 − 4. Find the coordinates of퐴.

Solution The point 퐵(2, 3)is the image formed when퐴 is reflected in the line 5푥 = 6푦 − 4 … … … (1). Let the point be 퐴(푚,푛) First, find the equation of the line passing through 퐵(2, 3). That line is perpendicular to the line 5푥 = 6푦 − 4 ⟹ 5푥 − 6푦 + 4 = 0 By comparing (1) with the general equation 푎푥 + 푏푦 + 푐 = 0, we get 푎 = 5, 푏 = −6 The equation of the line perpendicular to the line 5푥 = 6푦 − 4and passing through the point 퐵(2, 3) is given by: 푏푥 − 푎푦 + 푘 = 0 ⟹ −6푥 + 5푦 + 푘 = 0 … … … (2) Put (2, 3) into (2) ⟹ −6(2) + 5(3) + 푘 ⟹ −12 + 15 = 푘 = 0 ∴ 푘 = −3 The line: −6푥 + 5푦 − 3 = 0 Solve equation (1) and (2)to get the midpoint or the point of intersection of the two lines. From (1) 푥 = put 푥 = into (2)

306

−6 + 5푦 − 3 = 0 −6 × 5 + 5 × 5푦 − 3 × 5 = 0 −6(6푦 − 4) + 25푦 − 15 = 0 −36푦 + 24 + 25푦 − 15 = 0 −36푦 + 25푦 = 15 − 24 −11푦 = −9 푦 = Put 푦 = into (1)

푥 = = = = −

Hence, 푀 − , is the point of intersection of the two lines. 푀 , is also the point of the midpoint of 퐴(푚,푛) and the image point 푄(2, 3)

The midpoint between the two points is defined as: 푀(푎,푏) = , = ,

푀 − , = , By comparing the LHS with RHS we get, − = , and = ⟹ 2(−8) = 55(푚 + 2) 2(9) = 11(푛 + 3) ⟹ −16 = 55푚 + 110 18 = 11푛 + 33 ⟹ 55푚 = 110 + 16 11푛 = 33 − 18 ∴ 푚 = 푛 =

∴ 퐴 ,

The acute angle between two lines Let 푚 and 푚 be the gradients of the lies 푙 and 푙 respectively. The angle 휃 that lies between 푙 and 푙 is given by the formula tan휃 = 푚2−푚1

1+푚1푚2 , 푚 > 푚

Proof Consider the diagram below: 푦 푙 푙 푚 휃 푚 Figure 13.12 훼 180°− 훽 훽

푥

307

In the figure above,

훼 + 휃 = 훽 ⟹ 휃 = 훽 − 훼 tan 훼 = 푚 tan 훽 = 푚

⟹ tan휃 = tan(훽 − 훼) = tan훽−tan훼1+tan훼 tan훽 = 푚2−푚1

1+푚1푚2

∴ tan휃 = 푚2−푚11+푚1푚2

Since 푚 > 푚 and 휃 must be acute.

Example 13.26 Find the acute angle between the lines 3푥 − 4푦 = 11 and 푥 + 2푦 = 4

Solution We have the lines, 3푥 − 4푦 = 11 ⟹ 푦 = 푥 − … … … (1) By comparing (1)

with 푦 = 푚푥 + 푐, we get 푚 = and

푥 + 2푦 = 4 ⟹ 푦 = − 푥 + 2 … … … (2) By comparing (1) with 푦 = 푚푥 + 푐, we get

푚 = −

Now, choose 푚 = and 푚 = − , since > − . The acute angle between the two lines is defined as: tan휃 = 푚2−푚1

1+푚1푚2

⟹ tan휃 =34− −1

21+3

4×−12

=34+1

21−3

8=

5458

= × = 2

∴ 휃 = tan (2) = 63.43°(2 places of decimal)

Example 13.27 If the acute angle between the lines 2푦 = 6푥 + 3 and 3푦 = 5 − 푎푥 is 45°, find the possible value of 푎.

Solution We have the lines, 2푦 = 6푥 + 3 ⟹ 푦 = 2푥 + … … … (1) By comparing (1) with 푦 = 푚푥 + 푐, we get 푚 = 2 and 3푦 = 5− 푎푥 ⟹ 푦 = − 푥 + … … … (2) By comparing (1) with 푦 = 푚푥 + 푐, we get 푚 = − Now, choose 푚 = 2 and 푚 = − The acute angle between the two lines is defined as:

308

tan휃 =2− −푎3

1+2 −푎3

⟹ tan 45° =2+푎

31−2푎

3=

6+푎3

3−2푎3

=6+푎

33−2푎

3= 6+푎

3 × 33−2푎 =

⟹ 1 = ⟹ 3 − 2푎 = 6 + 푎 ⟹ 3 − 6 = 푎 + 2푎

⟹ 3푎 = −3 ∴ 푎 = − = −1 Try: Determine the acute angle between the lines 푥 − 푦 = and − 푥 − 1 = 0

The equation of the bisectors of angles between two lines Let 푎 푥 + 푏 푦 + 푐 = 0 and 푎 푥 + 푏 푦 + 푐 = 0 be two intersecting lines with gradients 푚 and 푚 respectively. Assume that휃 is one of the angles between the two lines, then the equation of the line that bisect 휃 can be derived from the equation:

푎1푥+푏1푦+푐1

푎12+푏1

2= 푎2푥+푏2푦+푐2

푎22+푏2

2 or

= ±

Proof

Consider the diagram below: 푦 푃(푥,푦) Figure 푑 푑 푎 푥 + 푏 푦 + 푐 = 0 푎 푥 + 푏 푦 + 푐 = 0

푥

309

In the diagram the perpendicular distances from 푃 to the lines 푎 푥 + 푏 푦 + 푐 = 0 and 푎 푥 + 푏 푦 + 푐 = 0are 푑 and 푑 respectively.

푑 = 푎1푥+푏1푦+푐1

푎12+푏1

2 and 푑 = 푎2푥+푏2푦+푐2

푎22+푏2

2 But 푑 = 푑

⟹ 푎1푥+푏1푦+푐1

푎12+푏1

2= 푎2푥+푏2푦+푐2

푎22+푏2

2

∴ 푎1푥+푏1푦+푐1

푎12+푏1

2= 푎2푥+푏2푦+푐2

푎22+푏2

2 (By using the definition of absolute value)

= ±

Example 13.28 Find the equations of the lines that bisect the angle between the lines 6푥 − 8푦 = 11 and

3푥 + 4푦 = 2 Solution

We have, 6푥 − 8푦 = 11 ⟹ 6푥 − 8푦 − 11 = 0 … … … (1). By comparing (1) with 푎 푥 + 푏 푦 + 푐 = 0 we get, 푎 = 6, 푏 = −8 and 푐 = −11. Also, 3푥 + 4푦 = 2 3푥 + 4푦 − 2 … … … (2). By comparing (2) with 푎 푥 + 푏 푦 + 푐 = 0 we get, 푎 = 3, 푏 = 4 and 푐 = −2. The equations of the lines that bisect the angle between the two lines is defined as:

= ±

⟹ ( ) ( )

= ±( ) ( )

⟹ √

= ±√

⟹ = ± ⟹ 5(6푥 − 8푦 − 11) = ±10(3푥 + 4푦 − 2) (Divide both sides by 5) ⟹ (6푥 − 8푦 − 11) = ±2(3푥 + 4푦 − 2) Now, either (6푥 − 8푦 − 11) = 2(3푥 + 4푦 − 2) or (6푥 − 8푦 − 11) = −2(3푥 + 4푦 − 2) 6푥 − 8푦 − 11 = 6푥 + 4푦 − 4 or 6푥 − 8푦 − 11 = −6푥 − 8푦 + 4 6푥 − 6푥 − 8푦 − 4푦 − 11 + 4 = 0 or 6푥 + 6푥 − 8푦 + 8푦 − 11 − 4 = 0 −12푦 − 7 = 0 or 12푥 − 15 = 0 (Divide both sides by 3)

310

12푦+ 7 = 0 or 4푥 − 5 = 0 ∴ the lines are: 12푦 + 7 = 0 and 4푥 − 5 = 0 Try: Find the equations of the lines which bisect the angles between the lines 푥 + 푦 + 2 = 0 and푥 − 7푦 + 26 = 0

Solved application questions Example 13.29

The vertices of triangle 퐴퐵퐶 are 퐴(−16, 0), 퐵(9, 0) and 퐶(0, 12). i. Find the distances퐴퐵, 퐵퐶 ,퐴퐶 and the area of the triangle. ii. Prove that the equation of the internal bisector of angle 퐴 of the triangle is

푥 − 3푦 + 16 = 0 iii. Find the equation of the internal bisector of angle 퐵 of the triangle.

Solution i. We have, 퐴(−16, 0), 퐵(9, 0) and 퐶(0, 12). |퐴퐵| = (푥 − 푥 ) + (푦 −푦 ) = (9 + 16) + (0 − 0) = (25) = 25units |퐵퐶| = (푥 − 푥 ) + (푦 −푦 ) = (0 − 9) + (12 − 0) = (−9) + (12) = √81 + 144 = √225 = 15units |퐴퐶| = (푥 − 푥 ) + (푦 −푦 ) = (0 + 16) + (12− 0) = (16) + (12) = √256 + 144 = √400 = 20 units The triangle 퐴퐵퐶is a scalene triangle. The are of the triangle is defined as: 퐴 = 푠(푠 − 푎)(푠 − 푏)(푠 − 푐) Where, 푠 = = = = 30 units

퐴 = 30(30 − 25)(30 − 15)(30− 20) 퐴 = 30(5)(15)(10) 퐴 = 22,500

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퐴 = 150units ii. Let us find the equation of line 퐴퐵 and 퐴퐶. Gradient of 퐴퐵 = = = 0. The equation of the line 퐴퐵 is given by:

푦 − 푦 = 푚(푥 − 푥 ) ⟹ 푦 − 0 = 0(푥 + 16) ⟹ 푦 = 0 … … … (1) Gradient of 퐴퐶 = = = . The equation of the line 퐴퐶 is given by:

푦 − 푦 = 푚(푥 − 푥 ) ⟹ 푦 − 0 = (푥 + 16) ⟹ 4푦 = 3푥 + 48 ⟹ 3푥 − 4푦 + 48 = 0 … … … (2) The equation of the angle bisector of angle 퐴 is defined as:

= ±

⟹ ( )√

= ±( )

⟹ 푦 = ±√

⟹ 푦 = ± ⟹ 5푦 = ±(3푥 − 4푦 + 48) Now, either 5푦 = 3푥 − 4푦 + 48 or 5푦 = −(3푥 − 4푦 + 48) 3푥 − 4푦 − 5푦 + 48 = 0 or 3푥 + 5푦 − 4푦 + 48 = 0 3푥 − 9푦 + 48 = 0 or 3푥 + 푦 + 48 = 0 푥 − 3푦+ 16 = 0 ∎ iii. Let us find the equation of line 퐴퐵 and 퐵퐶. Gradient of 퐴퐵 = = = 0. The equation of the line 퐴퐵 is given by:

푦 − 푦 = 푚(푥 − 푥 ) ⟹ 푦 − 0 = 0(푥 + 16) ⟹ 푦 = 0 … … … (1) Gradient of 퐵퐶 = = = − . The equation of the line 퐵퐶 is given by:

푦 − 푦 = 푚(푥 − 푥 ) ⟹ 푦 − 0 = − (푥 − 9) ⟹ 3푦 = −4푥 + 36 ⟹ 4푥 + 3푦 − 36 = 0 … … … (2) The equation of the angle bisector of angle 퐴 is defined as:

= ±

⟹ ( )√

= ± 4푥+3푦−36

√ ⟹ 푦 = ± 4푥+3푦−36

√

312

⟹ 푦 = ± 4푥+3푦−36 ⟹ 5푦 = ±(4푥 + 3푦 − 36) Now, either 5푦 = 4푥 + 3푦 − 36 or 5푦 = −(4푥 + 3푦 − 36) 4푥 + 3푦 − 5푦 − 36 = 0 or 4푥 + 5푦 + 3푦 − 36 = 0 4푥 − 2푦 − 36 = 0 or 4푥 + 8푦 − 36 = 0 2푥 − 푦 − 18 = 0 or 푥 + 2푦 − 9 = 0

Try: If 퐴(-3, -4) and 퐶(5, 4) are the two vertices of a rhombus퐴퐵퐶퐷, find the equation of 퐵퐷. Given that the gradient of the side 퐵퐶 is 2, calculate the coordinates of 퐵 and 퐷.

Final Exercises 1. Find the distance between each pair of points in the following: i.퐴(2,2)and퐵(3,4) ii. 퐴(3,3)and퐵(−2,2) iii. 퐴(11,4)and퐵(5,2) iv. 퐴(−10,3)and퐵(−5,6) v. 퐴(7,3)and퐵(−5,−6) vi. 퐴(0, 0) and 퐵(2,3) 2. The vertices of a ∆ABC are 퐴(4,5)퐵(2,−2) and 퐶(7,6) respectively. i. Find the sides of the triangle. ii. Find the perimeter of ∆퐴퐵퐶. 3.The vertices of a quadrilateral ABCD are (13,2), (3,−5), (5,3)and(3,−2) respectively. i. Find the length of each side of the quadrilateral. ii. Find the length of each diagonal of the quadrilateral. iii. What is the name of this quadrilateral ABCD? 4. Find the coordinates of which divide the line segment AB in the given ratio in each of the following: i. 퐴(2,9),퐵(−3, 2) 4:1 externally. ii. 푃(−3,−2),푄(−3,2), 3: 1internally. iii. X(2, -3), N (11, 2), 6:2 externally. 5. Find the equation of the line passing through the point i. P(4, 5) and parallel to the line 푦 = 2푥 + 3. ii P(3, 5) and parallel to the – 푥 − 푦 + 4 = 0. iii. P(1,-6) and parallel to the line 푥 + 2푦 − 6 = 0. iv. 푃( , 2) and parallel to 푥 + 2푦 = 0. 6. Find the equation of the line passing through the point

313

i. Q(-3,-1) and perpendicular to the lines. 2푥 − 푦 + 5 = 0 ii. Q(3, 2) and perpendicular to the line 푥 − 푦 = 3. iii. Q(3, √2) and perpendicular to the line 푥 − 푦 − 1 = 0. 7. Proof that the lines 2푥 − 푦 + 3 = 0, 2푥 − 푦 = 0,푥 + 2푦 = 0 and 푥 + 2푦 = 5 form a rectangle. 8. The points 퐴(4,−3)퐵(6,−3)and퐶(3,−17)are the vertices of a triangle. P is the midpoint of AB and Q is the midpoint of AC .Find the: i. Coordinates of P and Q. ii. Equation of 푃푄 iii. Equation of the perpendicular line from A to BC. 9. Find the equation of the straight line joining the 푃(1, 1) and the intersection of the lines 푦 = 3푥 + 4 and5푦 − 푥 = 10. 10. Find the perpendicular distance from the point (2,−9) to the line i. 푥 − 3푦 − 5 = 0 ii. 2푥 + 3푦 = −3iii. 푦 = 0 iv. 푥 = 0 v. 푥 − 푦 − 2 = 0 vi. 3푥 + 8푦 − 3 = 0 11. Calculate the area of triangle with vertices푋(2, 15)(6, 8)and(−3,−). What type of triangle is this? 12. Determine, by comparing gradients whether the following three points lie on the same straight line (i.e. are collinear) i.(0,−3)(1,−4), , ii. , 1 , 1, (4,−4) iii. (0,2), (3,5), (1,7)

13. Find the image point if the point i. 푃(−3, 4) is reflected in the line 푦 = 2푥 + 2 ii. 퐴(3, 4) is reflected in the line 푦 = −3푥 + 4 iii. 푄 , is reflected in the line 푦 = 12푥 − 5

14. Determine whether the given points lie on the line 푦 − 8푥 − 10 = 0 i. 퐴(2, 3) ii. 퐵(1, 10) iii. 퐶(1, 18) 15. Find the equation of the line that passes through the point 푃(−2, 1) and parallel to the line i. 2푥 + 푦 = 4 ii. 3푥 − 2푦 = 2 iii. 푦 = 2 iv. 푥 = 4 16. The point 푃(2,−1) lies on the line 푦 = 푎푥 + 4. Find the value of 푎. 17. ABCD is a rectangle whose vertices are A(2,1),B(7,0)andC(12,4)respectively. Find the coordinates of the point D. Hence, determine the perimeter and area of the triangle.

314

18.The point A(3,−1)B(6,0)C(7,3)andD(4,2)are the vertices of a quadrilateral. Show that the diagonals of a quadrilateral and bisect each other. 19. The equation 푎푥 + 푏푦 + 푐 = 0 passes through the point (−2, 3), (7, 3)and(1,−3) i. Find the values of a, b and c. ii. Determine the 푥and푦 − intercept respectively and equation of the line passing through the point (3, 3) and parallel to this line. 20. The equation 3푥 + 푏푦 + 푐 = 0 passes through the point(−2,−3)and(7,3). i. Find the value of 푏 and c. ii. Determine the 푥푎푛푑푦 intercept and the gradient of the line. 21. What is the equation of the perpendicular bisector of AB where A is (−3, 9) and B is (−7, 1)? 22. Find the equation of the line i. parallel to 3푥 + 푦 − 1 = 0 and passes through the point (−2, 6). ii. perpendicular to the line 푥 − 3푦 = 2 and has vertices (−2, 3). iii. that the line joining P (1,-3) and Q (-2, 4). 23. Find the perpendicular distance from the given fixed point to the given line in each of the following: i. A (5, 3);2푥 − 푦 + 4 = 0 ii. B (3, 4); 2푥 − 푦 − 6 = 0 iii. (−2,−1);4푥 − 3푦 − 1 = 0 iv. D(4,−1); 푦 = −푥 + 6 v. 푃(3, 2); 푥 − 2푦 − 3 = 0 vi. 푄(푎, 푏); 푎푥 + 푏푦 + 푐 = 0 24. Find the acute angle between each of the following pair of straight lines. i. 푥 = 푦 + 1,2푥 = 3푦 − 1 ii. 2푥 + 푦 = −1,푦 = 2푥 + 1 iii. 푦 = 푥 − 6,푦 = 2푥 + iv. √2푥 + √3푦 = 1, 2푦 = −푥 − 3 v. 4푦 + −1 = 0, −3푦 = 2푥 + 4 vi 푥 − 푦 = 0,2푦 = 3푥 − 4 vii. 푥 + 3푦 = 7,3푥 − 2푦 = 1 viii. 2푥 − 푦 + 5 = 0,3푥 = 5푦 25. Find the equation of the line that passes through the midpoint of 퐴(2, 4) and 퐵(−6,1) and make and angle of 30° with positive 푥 −axis. 26. Find the equation of the line that passes through the intersection of the lines 푥 + 4푦 − 1 = 0and 3푥 − 7푦 = 2 and parallel to the line 6푥 − 3푦 = 8 27. Given that 퐴(−4,−2), 퐵(4, 2) and 퐶(2, 6). i. Show that the triangle 퐴퐵퐶is right triangle. ii. Find the equations of the three sides of the triangle.

315

28. The lines 푥 − 2푦 = 3, 푥 + 3푦 = −1, and 푥 = −푦 + 2 form the three sides of triangle 푃푄푅. i. Find the coordinates of 푃,푄 and 푅. ii. Find the equations of the angle bisectors of point 푄. iii. Determine the area of the triangle. 29. The perpendicular distance from the point (3, 1) to the line 푥 + 2푦 − 푐 = 0 is √5. Determine the value of 푐. 30. The distance between the point 푃(2,푎) and 푄(3, 1) is √13. Find the value of 푎. 31. Find all the values of 푟 such that the gradient of the line through (푟, 4) and (1, 3 − 푟) is less than 5. 32. For what values of 푘 is the distance between (푘, 3) and (5, 2푘) greater than √26. 33. Find the equations of the lines through the point (2, 3) which makes an angle of 45° with the line 푥 − 2푦 = 1. 34. If the midpoint of the line 퐴퐵 segment has the coordinates (4, 8),find the coordinates of 퐵 if the A has the coordinates (2, 3) 35. Find the equation of the perpendicular bisector of 푃푄 whose end points are 푃(3, 2) and 푄(5, 7) 36. Prove that the points 퐴(16,−13),퐵(−2,−2),퐶(13, 10) and 퐷(21,−5) are the vertices of square. 37. If the vertices of a parallelogram are 퐴(−1,−3),퐵(푝, 푞),퐶(−7, 5) and 퐷(−12, 0), find the values of 푝 and 푞. Find the |퐴퐷|

316

r C(푎, 푏)

COORDINATE GEOMETRY (THE CIRCLE) A circle is a locus of a point moving so that its distance at any point is equidistant from a fixed point. The fixed point is called the centre and the distance from any movement to the centre is called the radius of the circle. This is illustrated below: 푦 푦 P(푥, 푦)

(푦 − 푏) 푏∙ Figure 1.1 figure 14.1. 푎 푥 (푥 − 푎)

From the above, the triangle gives:

(푥 − 푎) + (푦 − 푏) = 푟 (Pythagorean Theorem) This equation gives the equation of a circle given the centre (푎, 푏)and the radius푟, where 푎,푏, 푟 ∈ ℝ and푟 ≥ 0

Example 14.1 Find the equation of with i. Centre (1, 7) and radius 9 units. ii. Centre ( , )and radius units.

iii. Centre (11, ) and radius 3√2 units.

iv. Centre(√3,√3) and passing through the point (−1, 1) Solution

The equation of a circle with centre(푎,푏) and radius 푟 is defined as: (푥 − 푎) + (푦 − 푏) = 푟

i. Centre (1, 7)and radius9 units. ⟹ (푥 − 1) + (푦 − 7) = 9 ⟹ 푥 − 2푥 + 1 + 푦 − 14푦 + 49 = 81 ⟹ 푥 + 푦 − 2푥 − 14푦 + 49 + 1 − 81 = 0 ∴, the equation: 푥 + 푦 − 2푥 − 14푦 − 31 = 0

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ii. Centre − , and radius unit.

⟹ 푥 + + 푦 − = ⟹ 푥 + 푥 + + 푦 − 푦 + =

⟹ 푥 + 푦 + 푥 − 푦 + + − = 0 ⟹ 푥 + 푦 + 푥 − 푦 + = 0

⟹ 144푥 + 144푦 + 144 × 푥 − 144 × 푦 + 144 × = 0 ∴, the equation: 144푥 + 144푦 + 192푥 − 96푦+ 71 = 0 iii. Centre 11, and radius3√2 units.

⟹ (푥 − 11) + 푦 − = 3√2

⟹ 푥 − 22푥 + 121 + 푦 − 푦 + = 18

⟹ 푥 + 푦 − 22푥 − 푦 + 121 + − 18 = 0

⟹ 푥 + 푦 − 22푥 − 푦 + = 0

∴, the equation: 푥 + 푦 − 22푥 − 푦 + = 0 iv. Centre √3,√3 and the point (−1, 1) The distance between the centre and any point on the circle is the radius.

⟹ 푟 = (푥 − 푥 ) + (푦 − 푦 ) = −1 − √3 + 1 − √3

= 1 + 2√3 + 3 + 1 − 2√3 + 3 = √8 = 2√2

⟹ 푥 − √3 + 푦 − √3 = 2√2 ⟹ 푥 − 2√3푥 + 3 + 푦 − 2√3푦 + 3 = 8 ⟹ 푥 + 푦 − 2√3푥 − 2√3푦+ 3 + 3 − 8 = 0 ∴, the equation: 푥 + 푦 − 2√3푥 − 2√3푦 − 2 = 0

Properties of equation of a circle 1. The highest degree or power of 푥 and 푦is 2.

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2. The co-efficient of 푥 and 푦 are the same. 3. There is no 푥푦term. 4.There are 푥and 푦terms but either one of them or both can be omitted. Example, 푥 + 푦 = 푟 is an equation of a circle with centre (0, 0).

In general, the equation of a circle can be represented by equation: 푥 + 푦 + 2푔푥 + 2푓푦 + 푐 = 0 Where 푔,푓and 푐 are real parameters. From the general equation, it can be shown that the centre (푎, 푏) = (−푔,−푓)and 푟 = 푔 + 푓 − 푐

Proof

We know that (푥 − 푎) + (푦 − 푏) = 푟 ⟹ 푥 − 2푎푥 + 푎 + 푦 − 2푏푦 + 푏 = 푟 ⟹ 푥 + 푦 − 2푎푥 − 2푏푦+ 푎 + 푏 − 푟 = 0 … … …(1) By comparing (1) with 푥 + 푦 + 2푔푥 + 2푓푦 + 푐 = 0 It follows that −2푔 = 2푎 ⟹ 푎 = −푔 and −2푓 = 2푏 ⟹ 푏 = −푓 Also, 푐 = 푎 + 푏 − 푟 ⟹ 푟 = 푎 + 푏 – 푐 ⟹ 푟 = √푎 + 푏 − 푐 = 푔 + 푓 − 푐 Since 푎 = −푔and 푏 = −푓 The general equation is used to find the centre and radius of a given circle.

Example 14.2 Determine the centre and radius of the following circles: i. 푥 + 푦 + 2푥 − 6푦 − 1 = 0 ii. 3푥 + 3푦 − 2푥 − 5푦 − 3 = 0 iii. 11푥 + 11푦 − 10푥 − 9 = 0

Solution i. We have, 푥 + 푦 + 2푥 − 6푦 − 1 = 0 … … … (1) By comparing (1) with 푥 + 푦 + 2푔푥 + 2푓푦 + 푐 = 0, we get 2푔 = 2 ⟹ 푔 = = 1,

2푓 = −6 ⟹ 푓 = − = −3 and푐 = −1.

The radius of a circle is defined as: 푟 = 푔 + 푓 − 푐 ⟹ 푟 = (1) + (−3) − (−1) = √1 + 9 + 1 = √11 ∴,thecentre and radius of the circle are (−푔,−푓) = (−1, 3) and √11 respectively.

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ii. We have, 3푥 + 3푦 − 2푥 − 5푦 − 3 = 0 (Divide both sides by 3) 푥 + 푦 − 푥 − 푦 − 1 = 0 … … … (1) By comparing (1) with

푥 + 푦 + 2푔푥 + 2푓푦 + 푐 = 0, we get 2푔 = − ⟹ 푔 = − × = − ,

2푓 = − ⟹ 푓 = − × = − and푐 = 1.

The radius of a circle is defined as: 푟 = 푔 + 푓 − 푐

⟹ 푟 = − + − − (−1) = + + 1 = = √

∴,the centre and radius of the circle are (−푔,−푓) = , and √

respectively. iii. We have, 11푥 + 11푦 − 10푥 − 9 = 0 (Divide both sides by 11) 푥 + 푦 − 푥 + 0푦 − = 0 … … … (1) By comparing (1) with

푥 + 푦 + 2푔푥 + 2푓푦 + 푐 = 0, we get 2푔 = − ⟹ 푔 = − × = − ,

2푓 = 0 ⟹ 푓 = 0 and푐 = .

The radius of a circle is defined as: 푟 = 푔 + 푓 − 푐

⟹ 푟 = − + (0) − − = + = = √

∴,the centre and radius of the circle are (−푔,−푓) = − , 0 and √ respectively.

Example 14.3 The radius of a circle 푥 + 푦 + 2푥 + 4푦+ 푐 = 0, where 푐 is a constant, is 3√2 units. Find the value of 푏.

Solution We have, 푥 + 푦 + 2푥 + 4푦 + 푏 = 0 … … … (1) By comparing (1) with 푥 + 푦 + 2푔푥 + 2푓푦 + 푐 = 0, we get 2푔 = 2 ⟹ 푔 = = 1,

2푓 = 4 ⟹ 푓 = = 2 and 푐 = 푏

The radius of a circle is defined as: 푟 = 푔 + 푓 − 푐 ⟹ 3√2 = (1) + (2) − 푏 ⟹ 3√2 = √5 − 푏

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⟹ 3√2 = √5 − 푏 ⟹ 18 = 5 − 푏 ⟹ 푏 = 5 − 18 = −13 ∴,푏 = −13.

Example 14.4

The equation of a circle is defined as 푥 + 푦 + 2푎푥 − 4푦 + 푐 = 0. The circle passes through the points퐴(−1. 3) and 퐵(3, 2). Find the values of the constants 푎 and 푐.

Solution We have, 푥 + 푦 + 2푎푥 − 4푦 + 푐 = 0 If 퐴(−1, 3) lies on the circle, then (−1) + (3) + 2푎(−1) − 4(3) + 푐 = 0 ⟹ 1 + 9 − 2푎 − 12 + 푐 = 0 ⟹ 2푎 − 푐 = 1 + 9 − 12 ⟹ 2푎 − 푐 = −2 … … … (1) If 퐵(3, 2) lies on the circle, then (3) + (2) + 2푎(3)− 4(2) + 푐 = 0 ⟹ 9 + 4 + 6푎 − 8 + 푐 = 0 ⟹ 6푎 + 푐 = 8 − 9 − 4 ⟹ 6푎+ 푐 = −5 … … … (1) Solve (1) and (2) Add (1) to (2) ⟹ 6푎 + 2푎 + 푐 − 푐 = −5− 2 ⟹ 8푎 = −7 ⟹ 푎 = −

Put 푎 = − into (1). ⟹ 2 − − 푐 = −2 ⟹ 푐 = − + 2 ⟹ 푐 =

∴, 푎 = − and 푐 = . Try: 1. The equation of the circle of the form 푥 + 푦 + 2푥 − 6푦+ 푐 = 0, where 푐is a constant if the circle has radius 2,find the value of 푐. (SSSCE) 2. The radius of the circle 푥 + 푦 + 푝푥 + 6푦 − 3 = 0 is 4. Find the value of 푝.

(SSSCE) 3. A circle is defined by 푥 + 푦 + 2푔푥 + 2푓푦 − 13 = 0. If the circle passes through the points 푃(2, 1) and푄(3,−2), find the values of 푓and푔.

The equation of a circle passing through three points (Concyclic equation)

If a circle passes through three points say A (푥 ,푦 ), B (푥 ,푦 ) and C (푥 ,푦 ) then, all the points satisfies the equation of the circle. To get the circle we proceed as follow: 1. Put each point into the general equation of a circle.

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2. Simplify the various equations. 3. Solve the equations simultaneously.

Example 14.5 Find the equation of the circle passing the following points i. 퐴(3, 3), 퐵(1,−1) and 퐶(4, 5) .ii. 퐴(−2, 2), 퐵(2, 3) and 퐶(−2,−4) iii. (−1,−1), (6.5, 5.5)and (0, 2)

Solution Let 푥 + 푦 + 2푔푥 + 2푓푦 + 푐 = 0 be the general equation of the circle. Let 퐴(3, 3), 퐵(1,−1) and 퐶(4, 5) If 퐴(3, 3) lies on the circle, then` (3) + (3) + 2푔(3) + 2푓(3) + 푐 = 0 ⟹ 9 + 9 + 6푔 + 6푓 + 푐 = 0 ⟹ 6푔 + 6푓 + 푐 = −18 … … … (1) If 퐵(1,−1) lies on the circle, then (1) + (−1) + 2푔(1) + 2푓(−1) + 푐 = 0 ⟹ 1 + 1 + 2푔 − 2푓 + 푐 = 0 ⟹ 2푔 − 2푓 + 푐 = −2 … … … (2) If 퐵(4,−5) lies on the circle, then (4) + (−5) + 2푔(4) + 2푓(−5) + 푐 = 0 ⟹ 16 + 25 + 8푔 − 10푓 + 푐 = 0 ⟹ 8푔 − 10푓 + 푐 = −41 … … … (3)

Now, solve the system: 6푔 + 6푓 + 푐 = −18 … … … (1)2푔 − 2푓 + 푐 = −2 … … … (2)

8푔 − 10푓 + 푐 = −41 … … … (3)

From (1) 푐 = −18− 6푔 − 6푓. Put 푐 = −18 − 6푔 − 6푓into (2) and (3) ⟹ 2푔 − 2푓 − 18− 6푔 − 6푓 = −2 ⟹ −18 + 2 = 6푔 − 2푔 + 6푓 + 2푓 ⟹ 4푔 + 8푓 = −16 (Divide through by 4) ⟹ 푔 + 2푓 = −4 … … … (4) and ⟹ 8푔 − 10푓 + −18− 6푔 − 6푓 = −41 ⟹ 2푔 − 16푓 = −41 + 18 ⟹ 2푔 − 16푓 = −23 … … … (5) Solve (4) and (5) (Make 푔 the subject of (4) ) ⟹ 푔 = −4 − 2푓 Put 푔 = −4 − 2푓 into (5) ⟹ 2(−4 − 2푓)− 16푓 = −23 ⟹ −8 − 4푓 − 16푓 = −23 ⟹ −20푓 = −23 + 8 ⟹ −20푓 = −15 ⟹ 푓 = =

Put 푓 = into (4) 푔 = −4 − 2 = −4 − = −

Put 푔 = − and 푓 = into (1) 푐 = −18 − 6 − − 6

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= −18 + 33 −

= Put the values of 푔,푓and 푐 into the general equation. ⟹ 푥 + 푦 + 2 푥 + 2 푦 + = 0 ⟹ 푥 + 푦 + 11푥 + 푦 + = 0

⟹ 2푥 + 2푦 + 2 × 11푥 + 2 × 푦 + 2 × = 2 × 0 ⟹ 2푥 + 2푦 + 22푥 + 3푦 + 21 = 0 The equation of the circle passing through the points 퐴(3, 3), 퐵(1,−1) and 퐶(4, 5) is: 2푥 + 2푦 + 22푥 + 3푦 + 21 = 0 ii.

Solution Let 푥 + 푦 + 2푔푥 + 2푓푦 + 푐 = 0 be the general equation of the circle. Let 퐴(−2, 2), 퐵(2, 3) and 퐶(−2,−4) If 퐴(−2, 2) lies on the circle, then (−2) + (2) + 2푔(−2) + 2푓(2) + 푐 = 0 ⟹ 4 + 4 − 4푔 + 4푓 + 푐 = 0 ⟹ −4푔 + 4푓 + 푐 = −8 … … … (1) If 퐵(2, 3) lies on the circle, then (2) + (3) + 2푔(2) + 2푓(3) + 푐 = 0 ⟹ 4 + 9 + 4푔 + 6푓 + 푐 = 0 ⟹ 4푔 + 6푓 + 푐 = −13 … … … (2) If 퐶(−2,−4) lies on the circle, then

(−2) + (−4) + 2푔(−2) + 2푓(−4) + 푐 = 0 ⟹ 4 + 16 − 4푔 − 8푓 + 푐 = 0 ⟹ −4푔 − 8푓 + 푐 = −20 … … … (3)

Now, solve the system: −4푔 + 4푓 + 푐 = −8 … … … (1)4푔 + 6푓 + 푐 = −13 … … … (2)−4푔 − 8푓 + 푐 = −20 … … … (3)

Add (1) to (2) and Add (2) to (3) ⟹ 4푔 − 4푔 + 6푓 + 4푓 + 푐 + 푐 = −13− 8 ⟹ 10푓 + 2푐 = −21 … … … (4) and−4푔 + 4푔 − 8푓 + 6푓 + 푐 + 푐 = −20 − 13 ⟹ −2푓 + 2푐 = −33 … … … (5) Subtract (5) from (4) ⟹ 10푓 − (−2푓) + 2푐 − 2푐 = −21− (−33) ⟹ 12푓 = 12 ⟹ 푓 = = 1 Put 푓 = 1into (5) ⟹ −2(1) + 2푐 = −33 ⟹ 2푐 = −33 + 2 ⟹ 2푐 = −31

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⟹ 푐 = − Put 푐 = − and 푓 = 1into (1).

⟹ −4푔 + 4(1) − = −8 ⟹ −4푔 = −8 − 4 + −4푔 = −

⟹ 푔 = − × − = Put the values of 푔,푓and 푐 into the general equation. ⟹ 푥 + 푦 + 2 푥 + 2(1)푦 − = 0 ⟹ 푥 + 푦 + 푥 + 2푦 − = 0

⟹ 4푥 + 4푦 + 4 × 푥 + 4 × 2푦 + 4 × = 4 × 0 ⟹ 4푥 + 4푦 + 11푥 + 4푦 − 62 = 0 The equation of the circle passing through the points 퐴(−2, 2), 퐵(2, 3) and 퐶(−2,−4)is: 4푥 + 4푦 + 11푥 + 4푦 − 62 = 0 iii.

Solution Let 푥 + 푦 + 2푔푥 + 2푓푦 + 푐 = 0 be the general equation of the circle. Let 퐴(−1,−1), 퐵(6.5, 5.5) and 퐶(0, 2) If 퐴(−1,−1) lies on the circle, then (−1) + (−1) + 2푔(−1) + 2푓(−1) + 푐 = 0 ⟹ 1 + 1 − 2푔 − 2푓 + 푐 = 0 ⟹ −2푔 − 2푓 + 푐 = −2 … … … (1) If 퐵(6.5, 5.5) lies on the circle, then (6.5) + (5.5) + 2푔(6.5) + 2푓(5.5) + 푐 = 0 ⟹ 42.25 + 30.25 + 13푔 + 11푓 + 푐 = 0 ⟹13푔 + 11푓 + 푐 = −72.5 … … … (2) If 퐶(0, 2) lies on the circle, then (0) + (2) + 2푔(0) + 2푓(2) + 푐 = 0 ⟹ 0 + 4 + 0 + 4푓 + 푐 = 0 ⟹ 4푓 + 푐 = −4 … … … (3)

Now, solve the system: −2푔 − 2푓 + 푐 = −2 … … … … (1)

13푔 + 11푓 + 푐 = −72.5 … … … (2)4푓 + 푐 = −4 … … … (3)

From (3) 푐 = −4 − 4푓. Put 푐 = −4 − 4푓into (1) and (2) ⟹ −2푔 − 2푓 − 4 − 4푓 = −2 ⟹ −2푔 − 6푓 = −2 + 4 ⟹ −2푔 − 6푓 = 2 (Divide through by −2) ⟹ 푔 + 3푓 = −1 … … … (4) and ⟹ 13푔 + 11푓 + −4 − 4푓 = −72.5 ⟹ 13푔 + 7푓 = −72.5 + 4

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⟹ 13푔 + 7푓 = −68.5 … … … (5) Solve (4) and (5) (Make 푔 the subject of (4) ) ⟹ 푔 = −1 − 3푓 Put 푔 = −1 − 3푓 into (5) ⟹ 13(−1− 3푓) + 7푓 = −68.5 ⟹ −13 − 39푓 + 7푓 = −68.5 ⟹ −32푓 = −68.5 + 13 ⟹ −32푓 = −55.5 ⟹ 푓 = . =

Put 푓 = into (4) ⟹ 푔 = −1 − 3 = −1− = −

Put 푓 = into (3) ⟹ 푐 = −4 − 4 = −4 − = − Put the values of 푔,푓and 푐 into the general equation. ⟹ 푥 + 푦 + 2 − 푥 + 2 푦 − = 0

⟹ 푥 + 푦 − 푥 + 푦 − = 0 (Multiply both sides by 32) ⟹ 32푥 + 32푦 − 397푥 + 111푦 − 350 = 0 The equation of the circle passing through the points 퐴(−1,−1), 퐵(6.5, 5.5) and 퐶(0, 2)is: 32푥 + 32푦 − 397푥 + 111푦 − 350 = 0

Example 14.6 Solution

Consider the figure below: Find the midpoint of 퐴퐵: 퐵(2, 3) 푀 = , = , = (1, 2) Not drawn to scale

Find the midpoint of 퐴퐶: 푀 푃 푁 = , = , = (−1, 1)

Find the midpoint of 퐵퐶: 퐴(0, 0) 푁 퐶(−2, 2)

푃 = , = , = (0, 3) Let 푥 + 푦 + 2푔푥 + 2푓푦 + 푐 = 0 be the general equation of the inscribed circle. The inscribed circle passes through the midpoints 푀(1, 2), 푁(−1, 1) and 푃(0, 3) If 푀(1, 2) lies on the circle, then (1) + (2) + 2푔(1) + 2푓(2) + 푐 = 0 ⟹ 1 + 4 + 2푔 + 4푓 + 푐 = 0 ⟹ 2푔 + 4푓 + 푐 = −5 … … … (1)

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If 푁(−1, 1) lies on the circle, then (−1) + (1) + 2푔(−1) + 2푓(1) + 푐 = 0 ⟹ 1 + 1 − 2푔 + 2푓 + 푐 = 0 ⟹ −2푔 + 2푓 + 푐 = −2 … … … (2) If 푃(0, 3) lies on the circle, then (0) + (3) + 2푔(0) + 2푓(3) + 푐 = 0 ⟹ 0 + 9 + 0 + 6푓 + 푐 = 0 ⟹ 6푓 + 푐 = −9 … … … (3)

Now, solve the system: 2푔 + 4푓 + 푐 = −5 … … … (1)−2푔 + 2푓 + 푐 = −2 … … … (2)

6푓 + 푐 = −9 … … … (3)

From (3) 푐 = −9 − 6푓. Put 푐 = −9 − 6푓into (1) and (2) ⟹ 2푔 + 4푓 − 9 − 6푓 = −5 ⟹ 2푔 − 2푓 = −5 + 9 ⟹ 2푔 − 2푓 = 4 (Divide through by 2) ⟹ 푔 − 푓 = 2 … … … (4) and ⟹ −2푔 + 2푓 − 9 − 6푓 = −2 ⟹ −2푔 − 4푓 = −2 + 9 ⟹ −2푔 − 4푓 = 7 … … … (5) Solve (4) and (5) (Make 푔 the subject of (4) ) ⟹ 푔 = 2 + 푓 Put 푔 = 2 + 푓 into (5) ⟹ −2(2 + 푓) − 4푓 = 7 ⟹ −4 + 2푓 − 4푓 = 7 ⟹ −2푓 = 7 + 4 ⟹ −2푓 = −11 ⟹ 푓 =

Put 푓 = into (4) ⟹ 푔 = 2 + =

Put 푓 = into (3) ⟹ 푐 = −9 − 6 = −9 − 33 = −42 Put the values of 푔,푓and 푐 into the general equation. ⟹ 푥 + 푦 + 2 푥 + 2 푦 − 42 = 0 ⟹ 푥 + 푦 + 15푥 + 11푦 − 42 = 0 The equation of the inscribed circle of ∆퐴퐵퐶 is: 푥 + 푦 + 15푥 + 11푦 − 42 = 0

Equation of a circle given the ends of a diameter Given the points of the ends of a diameter of a circle, we can find the equation of the circle by using one of the following methods:

Method 1 a. Find the midpoint of the two points. This gives the centre of the circle. b. Find the distance between the midpoint and any of the points. This gives the radius of the circle.

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c. With the centre and the radius, find the equation of the circle. Method 2

a. Indicate an arbitrary point say 푃(푥, 푦) on the circle. b. Draw straight lines to join P and the two points. From elementary geometry the angle at P is90 . c. Find the gradient between each point and P. Multiply the two gradients function and equate the result to−1. This gives the equation of the circle. Method two is illustrated below: 푦 P(푥,푦)

figure 14.2.

A (푎, 푏) B (푗, 푘)

푥 From the diagram:

Gradient of AP = and gradient of BP= since AP⊥BP ⟹ ∙ = −1

Example 14.7 Find the equation of the circle with the following points at the ends of the diameter. i. (휋, ) and ( , ) where π= ii. (9, 0) and (−3,−5) iii. (−2,−1) and (3, 2)

Solution

Method 1 i. Let 퐴 휋, = , and 퐵 , = , . Let 퐶 be the midpoint of the diameter퐴퐵.

퐶 = , = , = , =×

, = ,

The point 퐶 , is the centre of the circle.

The distance between 퐶 and point 퐴 or 퐵 is the radius of the circle.

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푟 = |퐴퐶| = (푥 − 푥 ) + (푦 − 푦 ) = − + −

= + −

= +

= √ units The equation of a circle with centre(푎, 푏) and radius 푟 is defined as: (푥 − 푎) + (푦 − 푏) = 푟

⟹ 푥 − + 푦 − = √

⟹ 푥 − 푥 + + 푦 − 푦 + =

⟹ 푥 + 푦 − 푥 − 푦 + + − = 0

⟹ 푥 + 푦 − 푥 − 푦 + = 0 ∴, the equation of the circle passing through the ends of the diameter퐴 휋, and퐵 , is: 푥 + 푦 − 푥 − 푦 + = 0

Method 2 The equation of the circle passing through the ends of the diameter 퐴 휋, = , and퐵 , = , is defined by:

∙ = −1 ⟹ ∙ = −1

⟹ 푦 − 푦 − = − 푥 − 푥 −

⟹ 푦 − 푦 − 푦 + = − 푥 − 푥 − 푥 +

⟹ 푦 − 푦 + = −푥 + 푥 −

⟹ 푥 + 푦 − 푥 − 푦 + + = 0

⟹ 푥 + 푦 − 푥 − 푦 + = 0 ∴, the equation of the circle passing through the ends of the diameter퐴 휋, and퐵 , is: 푥 + 푦 − 푥 − 푦 + = 0

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ii. Method 1

Let 퐴(9, 0) and 퐵(−3,−6) Let 퐶 be the midpoint of the diameter퐴퐵. 퐶 = , = , = , = (3,−3) The point 퐶(3,−3) is the centre of the circle. The distance between 퐶 and point 퐴 or 퐵 is the radius of the circle. 푟 = |퐴퐶| = (푥 − 푥 ) + (푦 − 푦 ) = (3 − 9) + (−3− 0) = (−6) + (−3) = √36 + 9 = √45 = 3√5units The equation of a circle with centre(푎, 푏) and radius 푟 is defined as: (푥 − 푎) + (푦 − 푏) = 푟

⟹ (푥 − 3) + (푦 + 3) = 3√5 ⟹ 푥 − 6푥 + 9 + 푦 + 6푦 + 9 = 45 ⟹ 푥 + 푦 − 6푥 + 6푦 + 9 + 9 − 45 = 0 ⟹ 푥 + 푦 − 6푥 + 6푦 − 27 = 0 ∴, the equation circle passing through the end of the diameter퐴(9, 0) and 퐵(−3,−6) is: 푥 + 푦 − 6푥 + 6푦 − 27 = 0

Method 2

The equation of the circle passing through the ends of the diameter 퐴(9, 0)and퐵(−3,−6) is defined by: ∙ = −1 ⟹ ∙ = −1

⟹ (푦)(푦 + 6) = −(푥 − 9)(푥 + 3) ⟹ 푦 + 6푦 = −(푥 + 3푥 − 9푥 − 27) ⟹ 푦 + 6푦 = −푥 + 6푥 + 27 ⟹ 푥 + 푦 − 6푥 + 6푦 − 27 = 0 ∴, the equation of the circle passing through the ends of the diameter퐴(9, 0) and 퐵(−3,−6)is: 푥 + 푦 − 6푥 + 6푦 − 27 = 0 Try: Find the equation of the circle with the following points at the ends of the diameter. (−2,−1) and (3, 2)

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Points inside or outside a circle Let the points (푗, 푘)and (푠, 푡) lie inside and outside a circle respectively. And let 푑 , 푑 and 푟 be the distance between (푗,푘) and the centre, the distance between (푠, 푡) and the centre and the radius of the circle respectively. Then the following conditions are true: 1. 푑 ˂푟so far as (푗,푘) lies inside the circle and 2. 푑 ˃푟푠o far as (푠, 푡) lies outside the circle.

Example 14.8

Show whether the points A (16,1), B (10,3), C (−7, 1)and 퐷(−8, 2) lie inside or outside the circle 푥 + 푦 + 13푥 − 2푦 + 1 = 0.

Solution We have, 푥 + 푦 + 14푥 − 2푦 + 1 = 0 … … … (1) By comparing (1) with 푥 + 푦 + 2푔푥 + 2푓푦 + 푐 = 0, we get 2푔 = 14 ⟹ 푔 = = 7,

2푓 = −2 ⟹ 푓 = − = −1 and 푐 = 1.

The radius of a circle is defined as: 푟 = 푔 + 푓 − 푐 ⟹ 푟 = (7) + (−1) − 1 = √49 + 1 − 1 = √49 = 7units The centre and radius of the circle are (−푔,−푓) = (−7, 1) and 7 respectively. The distance between two points is defined as: 푑 = (푥 − 푥 ) + (푦 − 푦 ) Find the distance between point 퐴(16, 1) and the centre퐶(−7, 1) of the circle. |퐴퐶| = (−7 − 16) + (1 − 1) = (−23) = √529 = 23units Point 퐴(16, 1) lies outside the circle since |퐴퐶| > 푟. Find the distance between point 퐵(10, 3) and the centre 퐶(−7, 1) of the circle. |퐵퐶| = (−7 − 10) + (1 − 3) = √289 + 4 = √293units Point 퐵(10, 3) lies outside the circle since |퐵퐶| > 푟. Find the distance between point 퐷(−8, 2) and the centre 퐶(−7, 1) of the circle. |퐷퐶| = (−7 + 8) + (1 − 2) = √1 + 1 = √2units Point 퐷(−8, 2) lies inside the circle since |퐷퐶| < 푟.

Example 14.9 Obtain a condition for the origin to lie inside the circle 푥 + 푦 + 2푔푥 + 2푓푦 + 푐 = 0.

Solution We have, 푥 + 푦 + 2푔푥 + 2푓푦 + 푐 = 0

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The radius of a circle is defined as: 푟 = 푔 + 푓 − 푐 The centre of the circle is(−푔,−푓) = (−7, 1) The distance between two points is defined as: 푑 = (푥 − 푥 ) + (푦 − 푦 ) Find the distance between the origin푂(0, 0) and the centre퐶(−푔,−푓) of the circle.

|푂퐶| = (−푔 − 0) + (−푓 − 0) = (−푔) + (−푓) = 푔 + 푓 The origin will always lie inside the circle if |푂퐶| < 푟 ⟹ 푔 + 푓 < 푔 + 푓 − 푐 (Square both sides) ⟹ 푔 + 푓 < 푔 + 푓 − 푐 ⟹ 푔 + 푓 − 푔 − 푓 + 푐 < 0 ⟹ 푐 < 0 ∴,for the origin to lie inside the circle, 푐 < 0.

Touching circles Two circles can touch each other either internally or externally. Consider the following diagrams.

푦 figure 14.3. 푦 figure 14.4. S O 푥 O 푥 External touch internal touch

Let P and Q denote the centre of the external and R and S denote the centre of the internal touching circles respectively. The following conditions are true:

1. |푃푄| = 푟 + 푟 2. |푅푆| = 푟 − 푟 , 푟 > 푟 . Example 14.10

Find how the circle 푥 + 푦 + 14푥 − 2푦 + 46 = 0 and the circle 푥 + 푦 + 4푥 − 2푦 = 4touch each other and determine the coordinates of their points of intersection.

Solution We have, 푥 + 푦 + 14푥 − 2푦 + 46 = 0 … … … (1) By comparing (1) with 푥 + 푦 + 2푔푥 + 2푓푦 + 푐 = 0, we get 2푔 = 14 ⟹ 푔 = = 7,

P 푟

푟 Q

푟 푟 R

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2푓 = −2 ⟹ 푓 = − = −1 and푐 = 46.

The radius of a circle is defined as: 푟 = 푔 + 푓 − 푐 ⟹ 푟 = (7) + (−1) − 46 = √49 + 1 − 46 = √4 = 2units The centre and radius of the circle are 퐶 (−푔,−푓) = (−7, 1) and 2units respectively. We also have ,푥 + 푦 + 4푥 − 2푦 = 4 ⟹ 푥 + 푦 + 4푥 − 2푦 − 4 = 0… … … (2) By comparing (2) with 푥 + 푦 + 2푔푥 + 2푓푦 + 푐 = 0, we get 2푔 = 4 ⟹ 푔 = = 2

2푓 = −2 ⟹ 푓 = − = −1 and푐 = −4.

The radius of a circle is defined as: 푟 = 푔 + 푓 − 푐 ⟹ 푟 = (2) + (−1) + 4 = √4 + 1 + 4 = √9 = 3units The centre and radius of the circle are 퐶 (−푔,−푓) = (−2, 1) and 3units respectively. Find the distance between the centre퐶 (−7, 1) and 퐶 (−2, 1) |퐶 퐶 | = (푥 − 푥 ) + (푦 − 푦 ) = (−2 + 7) + (1 − 1) = √25 = 5units The two cycles touches each other internally since |퐶 퐶 | = 푟 + 푟 = 5 units Point of intersection (contact) From (1) 푥 + 푦 = −14푥 + 2푦 − 46. Put 푥 + 푦 = −14푥 + 2푦 − 46 into(2) ⟹−14푥 + 2푦 − 46 + 4푥 − 2푦 − 4 = 0 ⟹ −10푥 − 50 = 0 ⟹ −10푥 = 50 ⟹ 푥 = − = −5 Put 푥 = −5into (2) . ⟹(−5) + 푦 + 4(−5)− 2푦 − 4 = 0 ⟹ 25 + 푦 − 20 − 2푦 − 4 = 0 ⟹ 푦 − 2푦 + 1 = 0 ⟹ (푦 − 1) = 0 ⟹ (푦 − 1) = √0 ⟹ 푦 − 1 = 0 ⟹ 푦 = 1 ∴,the point of contact is (−5, 1)

Example 14.11 Show that the circles 푥 + 푦 − 4푥 − 2푦 + 3 = 0 and 푥 + 푦 + 2푥 + 4푦 − 3 = 0 touch each other. Obtain coordinates of the point of contact and the equation of the common tangent at that point.

Solution We have, 푥 + 푦 − 4푥 − 2푦 + 3 = 0 … … … (1) By comparing (1) with

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푥 + 푦 + 2푔푥 + 2푓푦 + 푐 = 0, we get 2푔 = −4 ⟹ 푔 = − = −2,

2푓 = −2 ⟹ 푓 = − = −1 and푐 = 3.

The radius of a circle is defined as: 푟 = 푔 + 푓 − 푐 ⟹ 푟 = (−2) + (−1) − 3 = √4 + 1 − 3 = √2units The centre and radius of the circle are 퐶 (−푔,−푓) = (2, 1) and √2 units respectively. We also have ,푥 + 푦 − 2푥 − 4푦 − 3 = 0 … … … (2) By comparing (2) with 푥 + 푦 + 2푔푥 + 2푓푦 + 푐 = 0, we get 2푔 = −2 ⟹ 푔 = − = −1

2푓 = −4 ⟹ 푓 = − = −2 and푐 = −3.

The radius of a circle is defined as: 푟 = 푔 + 푓 − 푐 ⟹ 푟 = (−1) + (−2) + 3 = √1 + 4 + 3 = √8 = 2√2units The centre and radius of the circle are 퐶 (−푔,−푓) = (1, 2) and 2√2 units respectively. Find the distance between the centre퐶 (2, 1) and 퐶 (1, 2) |퐶 퐶 | = (푥 − 푥 ) + (푦 − 푦 ) = (1 − 2) + (2 − 1) = √2units The two cycles touches each other externally since |퐶 퐶 | = 푟 − 푟 = √2 units Point of intersection (contact) From (1) 푥 + 푦 = 4푥 + 2푦 − 3. Put 푥 + 푦 = 4푥 + 2푦 − 3. into (2) ⟹ 4푥 + 2푦 − 3 − 2푥 − 4푦 − 3 = 0 ⟹ 2푥 − 2푦 − 6 = 0 (Divide both sides by 2) 푥 − 푦 − 3 = 0. ⟹ 푥 = 푦 + 3 … … … (3) Put 푥 = 푦 + 3into (1). ⟹ (푦 + 3) + 푦 − 4(푦 + 3) − 2푦 + 3 = 0 ⟹ 푦 + 2푦 + 9 + 푦 − 4푦 − 12 − 2푦 + 3 = 0 ⟹ 2푦 − 4푦 = 0 ⟹ 2푦(푦 − 2) ⟹ 2푦 = 0 or 푦 − 2 = 0 ⟹ 푦 = 0 or 푦 = 2. Put the values of 푦 into (3). When 푦 = 0, 푥 = 0 + 3 = 3 When 푦 = 2, 푥 = 2 + 3 = 5 ∴,the point of contact is (3, 0) since, (2, 5) is a false solution.

[check:(5) + 2 − 4(5)− 2(2) + 3 ≠ 0] Try:

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푑 퐶(푎,푏)

1. i. Obtain the equation of the circle which has the points A(2, 5) and B(−1, 8) at ends of its diameter. ii. Show the point P(1, 2) lies outside the circle. 2. A circle passes through the points, A(3, 2), B(−1,−2)and C(5,−4). i. Calculate the coordinates of the centre and the radius of the circle. ii. Find the equation of the circle.

Tangent and normal to a circle

A tangent to a circle is a line drawn to touch a circle at only one point. At that point the tangent line is perpendicular to the radius of the circle. Therefore, for a line to be tangent to a circle, the radius of the circle must be equal to the distance between the centre and the point of tangency. If 푚 is the gradient of the line between the centre and point of tangency, then the equation of the tangent line is defined as:

푦 − 푦 = (푥 − 푥 ), where (푥 ,푦 )is any point on the tangent line. Let 푑,푘 and 푙 be the distance between the centre and the point of tangency, the distance between the centre and any point outside the circle but on the tangent line and the length of the tangent line respectively.

Then 푘 = 푑 + 푙 . This is illustrated below: 푦 푙 푃(푥 ,푥 ) 푘

O 푥 figure 14.5.

A normal to a circle is a drawn to pass through the centre of a circle at one point and perpendicular to the tangent line. The equation of the normal to a circle is equal to the equation of the line passing through the centre and any point on the circle. This is defined as:

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푦 − 푦 = 푚(푥 − 푥 ), where (푥 ,푦 ) is any point on the normal line.

Example 14.12 Find the equation of the tangent and normal to the circle 푥 + 푦 + 4푥 − 16 = 0and passing through the point 푃(2,−2)

Solution We have, 푥 + 푦 + 4푥 − 16 = 0 … … … (1). By comparing (1)with 푥 + 푦 + 2푔푥 +2푓푦+ 푐 = 0, we get 2푔 = 4 ⟹ 푔 = = 2 ⟹ 2푓 = 0 ⟹ 푓 = = 0 and 푐 = −16. The centre and radius of the circle is퐶(−푔,−푓) = (−2, 0) Find the gradient between 퐶(−2, 0) and 푃(2,−2)

푚 =푦 − 푦푥 − 푥 =

−2 − 02 − (−2) = −

24 = −

12

The equation of the tangent to the circle passing through point 푃(2,−2) is defined as: 푦 − 푦 = − (푥 − 푥 )

⟹ 푦 − (−2) = − (푥 − 2) ⟹ 푦 + 2 = 2(푥 − 2) ⟹ 푦 + 2 = 2푥 − 4

⟹ 2푥 − 푦 − 6 = 0 ∴,the equation of the tangent: 2푥 − 푦 − 6 = 0 The equation of the normal to the circle passing through point 푃(2,−2) is defined as: 푦 − 푦 = 푚(푥 − 푥 ) ⟹ 푦 − (−2) = − (푥 − 2) ⟹ 2(푦 + 2) = −(푥 − 2) ⟹ 2푦+ 4 = −푥 +2 ⟹ 푥 + 2푦 + 2 = 0 ∴,the equation of the normal: 푥 + 2푦 + 2 = 0

Example 14.13 The equation of the tangent to the circle at point, 푃(−1,−1)is 푥 + 푦 + 2 = 0. If the line 3푥 − 2푦 = 6 passes through the centre of the circle, find i. the coordinates of the centre, 퐶. ii. the equation of the circle.

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iii. the points of intersection of the circle and the 푥 −axis. iv. the distance between the points in iii above.

Solution We have the equation of the tangent line: 푥 + 푦 + 2 = 0 3푥 + 2푦 = 6 Let us find the equation of the line between 푦 = 푥 the line: 3푥 + 2푦 = 5 … … … (1) That line is perpendicular to the tangent line and passes through the point 푃(−1,−1) 푥 + 푦 + 2 = 0 푃(−1,−1) (Consider the sketch) So, first, let us find the gradient 푚 of the the tangent line. ⟹ 푥 + 푦 + 2 = 0 ⟹ 푦 = −푥 − 2 By comparing푦 = −푥 − 2 with 푦 = 푚푥 + 푐 we get, 푚 = −1. The equation of the line perpendicular to the tangent line and passing through 푃(−1,−1) is defined as: 푦 − 푦 = − (푥 − 푥 ) 푦 + 1 = (푥 + 1) ⟹ 푥 − 푦 + 1 − 1 = 0 푥 = 푦… … … (2) i. The centre is the point of intersection of the lines: 3푥 + 2푦 = 6 … … … (1)and푥 = 푦… … … (2) Put 푥 = 푦 into (1) ⟹ 3푦 − 2푦 = 6 ⟹ 푦 = 6 ∴ 푥 = 6. ∴ 퐶(6, 6) ii. The radius of the circle is the distance between the centre퐶(6, 6) and the point푃(−1,−1). 푟 = (푥 − 푥 ) + (푦 − 푦 ) = (−1 − 6) + (−1 − 6) = √98 = 7√2units.

The equation of a circle with centre퐶(푎, 푏) and radius 푟 units is defined as: (푥 − 푎) + (푦 − 푏) = 푟

⟹ (푥 − 6) + (푦 − 6) = 7√2 ⟹ 푥 − 12푥 + 36 + 푦 − 12푥 + 36 = 98

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⟹ 푥 + 푦 − 12푥 − 12푦 + 36 + 36 − 98 = 0 ⟹ 푥 + 푦 − 12푥 − 12푦 − 26 = 0 … … … (1) ∴,the equation of the circle: 푥 + 푦 − 12푥 − 12푦 − 26 = 0 iii. At any point on the 푥 −axis, 푦 = 0. Put 푦 = 0 into (1) . ⟹ 푥 + 0 − 12푥 − 12(0)− 26 = 0 ⟹ 푥 − 12푥 − 26 = 0 The values of 푥 can be determined by the formula:

푥 = ±√ = ± ( ) ( )( )( )

= ±√ = ±√ = ± √

푥 = √ ≈ 6 + √62 or 푥 = √ = 6 − √62

∴,the points of intersection are 푃 6 + √62, 0 and 푃 6 − √62, 0 iv. The distance between the points is defined as:

|푃 푃 | = (푥 − 푥 ) + (푦 − 푦 ) = 6 − √62 − 6 − √62 + (0 − 0)

= −2√62

= 2√62 Try: 1. The equation of the tangent to a circle at 푃(4, 2) is 3푦 = 푥 + 2. the line 푦 = 3푥 passes through the centre of the circle 퐶 of the circle. Find i. the coordinates of 퐶. ii. the equation of the circle. iii. the point 푇 and 푇 where the circle cut the 푥-axis. (SSSCE). iv. the equation of the normal to the circle. 2. A circle touches the line푥 = 0, 푥 = 4, 푦 = 0 and푦 = 4. Find the equation of i. the circle. ii. the line passing through (0, 4)and the circle. (WASSCE) 3 i. Show that the line 4푥 + 7푦 = 25is a tangent to the circle 푥 + 푦 + 2푥 + 푦 − 15 = 0 ii. Find the coordinates of the point of contact. (WASSCE) iii. Determine the length of the tangent.

Example 14.14 Find the equation of the two circles which satisfy the following conditions: i. The 푥-axis is a tangent to the circle.

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ii. The centre of the circle lies on the line 2푦 = 푥. iii.The point (14, 2)lies on the circle. prove that the line 3푦 = 4푥 is a common tangent to these circles.

Solution Consider the sketch. Let 퐶(푎, 푏) be the centre of the circle. 푦 Since the centre of the circle lies on the 푃(14, 2) line 2푦 = 푥 ⟹ 2푏 = 푎… … … (1) The radius of the circle is the distance between the centre 퐶(푎, 푏) and the point 푃(14, 2). 3푦 = 4푥 푄(푎, 0) 푥 ⟹ 푟 = |퐶푃| = (푥 − 푥 ) + (푦 − 푦 ) = (14 − 푎) + (2 − 푏)

= √19 − 28푎 + 푎 + 4 − 4푏 + 푏 = √푎 + 푏 − 28푎 − 4푏+ 200

The radius can also be seen as the distance between the point 푄(푎, 0) and the centre of the circle. ⟹ 푟 = |퐶푄| = (푥 − 푥 ) + (푦 − 푦 ) = (푎 − 푎) + (푏 − 0) = √푏 = 푏 So, |퐶푄| = |퐶푃| ⟹ 푏 = √푎 + 푏 − 28푎 − 4푏 + 200 (Square both sides) ⟹ 푏 = 푎 + 푏 − 28푎 − 4푏 + 200 ⟹ 푎 − 28푎 − 4푏 + 200 = 0 … … … (2) Put 푎 = 2푏 into (2) ⟹ (2푏) − 28(2푏)− 4푏 + 200 = 0 ⟹ 4푏 − 60푏 + 200 = 0 (Divide both sides by 4) ⟹ 푏 − 15푏 + 50 = 0 ⟹ (푏 − 10)(푏 − 5) = 0 ⟹ 푏 − 10 = 0 or 푏 − 5 = 0 ⟹ 푏 = 10or푏 = 5 Put the value of 푏 into (1) ⟹ 푎 = 2(10) = 20 or 푎 = 2(5) = 10 ∴,the centre 퐶(20, 10), 푟 = 10 and 퐶(10, 5),푟 = 5 The equation of a circle with centre퐶(푎,푏) and radius 푟 units is defined as: (푥 − 푎) + (푦 − 푏) = 푟 ⟹ (푥 − 20) + (푦 − 10) = (10) ⟹ 푥 − 40푥 + 400 + 푦 − 20푦 + 100 = 100

(푎,푏) 푟

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⟹ 푥 + 푦 − 40푥 − 20푦 + 400 + 100 − 100 = 0 ⟹ 푥 + 푦 − 40푥 − 20푦 + 400 = 0 … … … (3) ∴,the equation of the circle: 푥 + 푦 − 40푥 − 20푦 + 400 = 0

And ⟹ (푥 − 10) + (푦 − 5) = (5) ⟹ 푥 − 20푥 + 100 + 푦 − 10푦 + 25 = 25 ⟹ 푥 + 푦 − 20푥 − 10푦 + 100 + 25 − 25 = 0 ⟹ 푥 + 푦 − 20푥 − 10푦 + 100 = 0 … … … (4) ∴,the equation of the circle: 푥 + 푦 − 20푥 − 10푦 + 100 = 0 If the line 3푦 = 4푥… … … (5) is a tangent to the circles, then the line intersect the circle at a point. From (5),푦 = . Put 푦 = into (3).

⟹ 푥 + − 40푥 − 20 + 400 = 0

⟹ 푥 + − 40푥 − 20 + 400 = 0

⟹ 9푥 + 9 × − 9 × 40푥 − 20 × 9 + 9 × 400 = 9 × 0

⟹ 9푥 + 16푥 − 360푥 − 240푥 + 3600 = 0 ⟹ 25푥 − 600푥 + 3600 = 0 (Divide both sides by 25) ⟹ 푥 − 24푥 + 144 = 0 ⟹ (푥 − 12) = 0 ⟹ (푥 − 12) = √0 ⟹ 푥 − 12 = 0 ⟹ 푥 = 12. Put 푥 = 12 into (5) 푥 = ( ) = 16 The point of tangency of the line 3푦 = 4푥 and the circle 푥 + 푦 − 40푥 − 20푦 + 400 = 0 is (16, 12) Put 푦 = into (3).

⟹ 푥 + − 20푥 − 10 + 100 = 0

⟹ 푥 + − 20푥 − 10 + 100 = 0

⟹ 9푥 + 9 × − 9 × 20푥 − 10 × 9 + 9 × 100 = 9 × 0

⟹ 9푥 + 16푥 − 180푥 − 120푥 + 900 = 0

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⟹ 25푥 − 300푥 + 900 = 0 (Divide both sides by 25) ⟹ 푥 − 12푥 + 36 = 0 ⟹ (푥 − 6) = 0 ⟹ (푥 − 6) = √0 ⟹ 푥 − 6 = 0 ⟹ 푥 = 6. Put 푥 = 12 into (5) 푥 = ( ) = 8 The point of tangency of the line 3푦 = 4푥 and the circle 푥 + 푦 − 20푥 − 10푦 + 100 = 0is(8, 6)

Orthogonal circles Two circles are said to be orthogonal if and only if the square of the distance between their centre is equal to the sum of the squares of their radii. Let 퐶 , 퐶 and 푟 , 푟 be the centres and the radii of the two circles respectively. Then from the definition: |퐶 퐶 |2= 푟 + 푟 .

Example 14.15 Show that the circles 푥 + 푦 − 8푥 − 6푦 + 19 = 0 and 푥 + 푦 − 4푥 − 2푦 + 3 = 0 orthogonate.

Solution We have, 푥 + 푦 − 8푥 − 6푦 + 19 = 0 … … … (1) By comparing (1) with 푥 + 푦 + 2푔푥 + 2푓푦 + 푐 = 0, we get 2푔 = −8 ⟹ 푔 = − = −4,

2푓 = −6 ⟹ 푓 = − = −3 and푐 = 19. The radius of a circle is defined as: 푟 = 푔 + 푓 − 푐 ⟹ 푟 = (−4) + (−3) − 19 = 16 + 9 − 19 = 6units The centre and radius of the circle are 퐶 (−푔,−푓) = (4, 3) and √6 units respectively. We also have, 푥 + 푦 − 4푥 − 2푦 + 3 = 0… … … (2) By comparing (2) with 푥 + 푦 +2푔푥 + 2푓푦 + 푐 = 0, we get 2푔 = −4 ⟹ 푔 = − = −2

2푓 = −2 ⟹ 푓 = − = −1 and푐 = 3. The radius of a circle is defined as: 푟 = 푔 + 푓 − 푐 ⟹ 푟 = (−2) + (−1) − 3 = 4 + 1 − 3 = 2units The centre and radius of the circle are 퐶 (−푔,−푓) = (2, 1)and 3 units respectively. |퐶 퐶 | = (푥 − 푥 ) + (푦 − 푦 ) = (2 − 4) + (1 − 3) = 4 + 4 = 8units The two circles orthogonate since|퐶 퐶 | = 푟 + 푟 = 8 units

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C(푎, 푏)

푟

Try: The circles 푥 + 푦 − 푥 − 2푦 + 푐 = 0 and 5푥 + 5푦 − 4푥 − 2푦 − 1 = 0 orthogonate. Find the value of 푐.

Polar/ parametric equation of a circle Consider the diagram below: 푦 푦 푃(푥, 푦) 푦 − 푏 푏 휃 푁 figure 14.6.

O 푎 푥 푥 푥 − 푎 From the diagram, sin 휃 = and cos휃 = ⟹푦 = 푏 + 푟 sin 휃… … …(1) and 푥 = 푎 + 푟 cos휃… … …(2)

These two equations are known as the parametric equations of a circle given by the arbitrary point 푃(푥, 푦).

tan 휃 = = ∙ = , whereθistheacuteangletheradiusmakeswiththelineCNparalleltothepositive푥 −axis.

Example 14.16 Find the parametric equations of a circle with centre (2,−3) and radius 5units.

Solution We have the centre퐶(2,−3) and radius,푟 = 5 units. The parametric equation of a circle with centre(푎,푏) and radius, 푟 is defined as: 푥 = 푎 + 푟 cos휃 and 푦 = 푏 + 푟 sin휃. ∴, 푥 = 2 + 5 cos휃 and 푦 = −3 + 5 sin휃

Example 14.17 Determine the centre and the radius of the circle 푥 = −1 + √3 cos훼 ,푦 − 5 = √3 sin 훼, where 훼the angle is the radius makes with the line parallel to the 푥-axis.

341

Solution We have, 푥 = −1 + √3 cos훼… … …(i) and 푦 − 5 = √3 sin 훼 ⟹ 푦 = 5 + √3 sin훼… … …(ii) The parametric equation of a circle with centre(푎,푏) and radius,푟 is defined as: 푥 = 푎 + 푟 cos휃… … …(1) and 푦 = 푏 + 푟 sin 휃… … … (2) By comparing (1) with (i), we get 푎 = −1, 푏 = 5 and 푟 = √3 units. ∴,centre퐶(−1, 5) and radius, 푟 = √3 units.

Example 14.18 A circle passes through the point (7,1). If the centre of the circle is (1, 1),find i. the radius of the circle. ii. The acute angle radius with the line parallel to the 푥-axis.

Solution i. Let 푃(푥, 푦) = (7, 2) and the centre 퐶(푎, 푏) = (1, 1). The radius of the circle is the distance between point 푃 and the centre. 푟 = |퐶푃| = (푥 − 푥 ) + (푦 − 푦 ) = (7 − 1) + (2 − 1) = √36 + 1 = √37units. ii. tan휃 = ⟹tan휃 = ⟹ tan 휃 = ⟹ 휃 = tan 휃 ≈ 9.46°

∴,the acute angle is 9.46° (corrected to 2 places of decimal)

Example 14.19 Find the equation of the normal and tangent to the circle 푥 = 2 + cos훼 ,푦 = 3 + sin 훼, where훼 = 60 .

Solution We have, 푥 = 2 + cos훼 ⟹ 푥 = 2 + cos 60° = 2 + = and 푦 = 3 + sin 훼

⟹ 푦 = 3 + sin 60° = 3 + √ = √ The centre of the circle is 퐶(2, 3) and the radius is 푟 = 1 unit.

The point 푃 , √ is the a point on the circle. The equation of the tangent to the circle and passing through point 푃(푥, 푦) is defined as: 푦 − 푦 = − (푥 − 푥 )

푚 = =√

=√

= √ × 2 = √3

342

⟹ 푦 − √ = −√

푥 − ⟹ √ = −√

푥 −

⟹ √3 2푦 − 6 − √3 = −2 푥 − ⟹ 2√3푦 − 6√3− 3 = −2푥 + 5

⟹ 2푥 + 2√3푦 − 6√3 − 3 − 5 = 0 ⟹ 2푥 + 2√3푦 − 6√3− 8 = 0 ∴,the equation of the tangent: 2푥 + 2√3푦 − 6√3− 8 = 0 The equation of the normal to the circle and passing through point 푃(푥,푦) is defined as: 푦 − 푦 = 푚(푥 − 푥 )

⟹ 푦 − √ = √3 푥 − ⟹ √ = √3 푥 −

⟹ 2푦 − 6 − √3 = 2√3 푥 − ⟹ 2푦 − 6 − √3 = 2√3푥 − 5√3

⟹ 2√3푥 − 2푦 − 5√3 + √3 + 6 = 0 ⟹ 2√3푥 − 2푦 − 4√3 + 6 = 0. ∴,the equation of the tangent: 2푥 + 2√3푦 − 6√3− 8 = 0

The locus of a point The locus of a point can be defined as the relationship between the coordinates of a point moving in the 푂푥푦 plane under certain conditions.

Example 14.20 Find and describe the locus of the set of points P in the 표푥푦plane that satisfy the following conditions. i. 푃 is equidistant from the point (3,−5) and the line 푥 = −1 ii. 푃 is twice as far as the point (3, 0) and the line 푦 = 5. iii. 푃is equidistant from (5, 2) and (−5,−2). iv. 푃is equidistant from the lines 3푥 + 3푦 = −9 and 12푥 − 4푦+ 12 = 0.

Solution i. Let 푃(푥, 푦)and 퐴(3,−5). We also have, 푥 = −1 ⟹ 푥 + 1 = 0 the locus is the set of points that satisfies the condition that the distance between P and A is equal to the perpendicular distance from the point P to the line 푦 + 5 = 0. The distance between two points is defined as: 푑 = (푥 − 푥 ) + (푦 − 푦 ) The distance between two points is defined as: |푃퐴| = (푥 − 푥 ) + (푦 − 푦 ) = (푥 − 3) + (푦 + 5) The perpendicular distance from a point to the line 푎푥 + 푏푦 + 푐 = 0 is

343

defined as: 푑 =√

By comparing 푎푥 + 푏푦 + 푐 = 0with 푥 + 1 = 0. 푎 = 1, 푏 = 0and푐 = 1. Therefore, the perpendicular distance from the point 푃(푥, 푦) to the line 푥 + 1 = 0 푑 =

√= ( )

√= 푥 + 1

⟹ (푥 − 3) + (푦 + 5) = 푥 + 1 (Square both sides) ⟹ (푥 − 3) + (푦 + 5) = (푥 + 1) ⟹ 푥 − 6푥 + 9 + 푦 + 10푦 + 25 = 푥 + 2푥 + 1 ⟹ 푥 − 푥 + 푦 − 6푥 − 2푥 + 10푦 + 25 − 1 = 0 ⟹ 푦 − 8푥 + 10푦 + 24 = 0 ∴, the locus is the parabola 푦 − 8푥 + 10푦 + 24 = 0 ii. Let 푃(푥, 푦)and 퐴(3, 0). We also have, 푦 = 5 ⟹ 푦 − 5 = 0 the locus is the set of points that satisfies the condition that the distance between P and A is equal to the perpendicular distance from the point P to 푦 − 5 = 0 The distance between two points is defined as: 푑 = (푥 − 푥 ) + (푦 − 푦 ) The distance between two points is defined as: |푃퐴| = (푥 − 3) + (푦 − 0) = (푥 − 3) + 푦 The perpendicular distance from a point to the line 푎푥 + 푏푦 + 푐 = 0 is defined as: 푑 =

√

By comparing 푎푥 + 푏푦 + 푐 = 0with 푦 − 5 = 0. 푎 = 0, 푏 = 1and푐 = −5. Therefore, the perpendicular distance from the point 푃(푥, 푦) to the line 푦 − 5 = 0 푑 =

√= ( )

√= 푦 − 5

⟹ (푥 − 3) + 푦 = 푦 − 5 (Square both sides) ⟹ (푥 − 3) + 푦 = (푦 − 5) ⟹ 푥 − 6푥 + 9 + 푦 = 푦 − 10푦 + 25 ⟹ 푥 + 푦 − 푦 − 6푥 + 10푦 + 9 − 25 = 0 ⟹ 푥 − 6푥 + 10푦 − 16 = 0 ∴, the locus is the parabola 푥 − 6푥 + 10푦 − 16 = 0 iii. Let 푃(푥, 푦) and 퐴(5, 2) and 퐵(−5,−2).

344

The distance P and A is equal to the distance between P and B. The distance between two points is defined as: 푑 = (푥 − 푥 ) + (푦 − 푦 ) ⟹ |푃퐴| = |푃퐵| ⟹ (푥 − 5) + (푦 − 2) = (푥 + 5) + (푦 + 2) ⟹ 푥 − 10푥 + 25 + 푦 − 4푦 + 4 = 푥 + 10푥 + 25 + 푦 + 4푦 + 4 ⟹ 푥 + 푦 − 10푥 − 4푦+ 29 = 푥 + 푦 + 10푥 + 4푦 + 29 (Square both sides) ⟹ 푥 + 푦 − 10푥 − 4푦 + 29 = 푥 + 푦 + 10푥 + 4푦 + 29 ⟹ 푥 − 푥 + 푦 − 푦 − 10푥 − 10푥 − 4푦 − 4푦 + 29 − 29 = 0 ⟹ −20푥 − 8푦 = 0 ⟹ 20푥 + 8푦 = 0 ⟹ 5푥 + 2푦 = 0 ∴,the locus is a straight line whose 푔푟푎푑푖푒푛푡 is − and 푦 −intercept (0, 0) iv. We have the line 3푥 + 4푦 = −9 and 12푥 − 5푦 + 12 = 0 ⟹ 3푥 + 4푦 + 9 = 0 The locus of points equidistance between two intersecting straight lines is the line bisecting the angle between the two lines. The angle bisector between two straight lines is defined as:

= ±

The equation of the line bisecting the straight line 3푥 + 4푦 = −9 and 12푥 − 5푦 + 12 = 0 ⟹

√= ±

( ) ⟹

√= ±

√

⟹ = ± ⟹ 13(3푥 + 4푦 + 9) = ±5(12푥 − 5푦 + 12) ⟹ 39푥 + 52푦+ 117 = 5(12푥 − 5푦 + 12) ⟹ 39푥 + 52푦+ 117 = 60푥 − 25푦+ 60 ⟹ 60푥 − 39푥 − 25푦 − 52푦 + 60 − 117 = 0 ⟹ 21푥 − 77푦 − 57 = 0 or ⟹ 39푥 + 52푦+ 117 = −5(12푥 − 5푦 + 12) ⟹ 39푥 + 52푦+ 117 = −60푥 + 25푦 − 60 ⟹ 39푥 + 60푥 + 52푦 − 25푦 + 117 + 60 = 0

345

⟹ 99푥 + 27푦+ 177 = 0 (Divide both sides by 3) ⟹ 33푥 + 9푦 + 59 = 0 ∴,the locus is the line 21푥 − 77푦 − 57 = 0 with 푔푟푎푑푖푒푛푡 and

푦 −intercept 0,− or the line 33푥 + 9푦 + 59 = 0 with 푔푟푎푑푖푒푛푡 −

푦 −intercept 0,−

Final exercises

1. Find the equation of the circle given the centre and the radius r.

i. (3, 5);푟 = 4 푖푖. (−3, 2); 푟 = √2 iii. (−1, 6); 푟 = √ iv. (4,−1); 푟 =

v. √2, 3 ; 푟 = √ 2. Find the centre and radius of the following circle: i. 푥 + 푦 − 2푥 + 4푥 = 5 ii. 푥 + 푦 − 4푥 − 6푥 − 1 = 0 iii. 2푥 + 2푦 + 6푥 + 2 = 0 iv. √3푥 + √3푦 − 6푦 − √61 = 0 v. 푥 + 푦 − 4푥 + 푦 − 4 = 0 vi. 10푥 + 10푦 − 100푥 − 20푦 − 5 = 0. 3. In each case, find the equation of a circle passing through the following three points; i. 퐴(1, 3),퐵(3,−2)and퐶(4, 5) ii. 푃(−1, 4)푄 , 4 and푅(−1,−1)

iii. 푋(2,−3),푌(−1,−4)and푍(−7,−8) iv. 푈(6, 1),푉( ,− )and푊, (7,−2)

v. 퐼(−2,−3), 퐽(6, 3) and 퐾 √ , 4 4. The point 푃(2, 7) lies on the circle 푥 + 푦 − 2푥 + 4푏푦 + 1 = 0. find the value of b. Determine the radius of the circle. 5. The points 퐴(−3,2)and퐵 , lie on the circle 푥 + 푦 + 푎푥 + 2푦 + 푐 = 0. Find the values of 푎and 푐. Hence, find the radius of the circle. 6. In each of the case, find the equation of the circle with the points at the ends of diameter. a. 퐴(3,−2)and퐵(4, 5) 푏.푃(−1, 4)and푄(2,−1) 푐.푋(3,−2)푎푛푑푌(−7, 5) d. 푈(6, 1)and푉 ,− 푒. 퐼(6,−2)and퐽(4, 2)

7. Show whether or not the following points lies on, inside or outside the circle 푥 + 푦 − 5푥 + 8푦 − 5 = 0

346

i.퐴(2, 5) ii.퐵(3,−3) iii.퐶(7,−2) iv.퐷 , v.퐸 4,− vi.퐹(6, 1) vii.퐺(−8,−3) viii. 퐻(−6,−2) 8. The line 푥 + 푦 + 3 − 0intersect the circle 푥 + 푦 − 4푥 + 6푦 − 3 = 0 at the points P and Q. Find the points. Hence, determine the line that passes the points. Determine|푃푄|. 9. Find the points of intersection between the circles 푥 + 푦 + 2푥 + 4푦 − 3 = 0 and 푥 + 푦 − 푥 − 4푦 − 1 = 0. Hence, determine the perpendicular distance from the point 푃(1,−3) to the line that passes through the points of the intersection. 10. The radius of a circle푥 + 푦 + 2푥 + 6푦 + 푐 = 0. Find the value of c. 11. Find the equation of the tangent and normal to the circle. i.푥 + 푦 − 2푥 + 4푦 = 0atthepoint(1, 1) ii.푥 + 푦 − 2√3푥 + √3푦 − 3 = 0atthepoint √3,√3 iii.푥 + 푦 + 2푥 − 6푦 − 7 = 0atthepoint(3, 2) 12. Show whether or not the following circles are diagonal. i. 푥 + 푦 − 2푥 + 6푦 + 1 = 0 and 푥 + 푦 + 4푥 + 8푦 + 16 = 0 ii. 푥 + 푦 = 16 and 푥 + 푦 − 8푥 + 6푦 + 4 = 0 iii. 2푥 +2푦 + 10푥 − 14푦 − 1 = 0 and 푥 + 푦 − 2푥 + 4푦 = 17. 13. Find the value of the constant c such that the circles 푥 + 푦 + 2푥 − 4푦 + 16 = 0 and 푥 + 푦 − 10푥 − 5푦 + 푐 = 0 are orthogonal. 14. Find the equation of the circumscribed circle and the inscribed circle of the triangle with vertices 퐴(−5, 3),퐵(−1,−1)and퐶(−3,−5). 15. A circle is parameterize as푥 = −3 + 3cos휃푎푛푑푦 = −5 + sin 휃 . Find the center and radius of the circle 16. Find the equation of the inscribed of triangle ABC where A is (2, 6), B is (−2, 4) and C is (−12, 6) 17. Find the centre and radius of the following circles: i. 푥 + 푦 = 4 ii. 푥 +푦 − 4푥 + 6푦 = 3 iii. 푥 +푦 − 6푥 − 8푦 − 6 = 0 iv. 푥 +푦 − 푥 + 3푦 − 2 = 3 v. 푥 +푦 − 2푦 − 3 = 0 18. Find the equation of the circle that passes through the points i. 퐴(0, 3),퐵(1, 1) and 퐶(4, 1) ii. 퐴(1, 2),퐵(3, 0) and 퐶(3,−1) 19. Find the equation of the circle with centre 퐶(3, 2) and tangent to the line 푥 + 푦 = 6. 20. Find the point of intersection of the circles 푥 +푦 = 4 and 푥 +푦 − 2푥 − 2푦 = 0

347

21. Find the equation of the circle inscribed in the a triangle with vertices 퐴(2, 4),퐵(−4,−6) and 퐶(0, 8) 22. Find the equation of the circles that satisfy the following conditions: i. the 푦 −axis is a tangent to the circles. ii. the 푃(3,6) lies of the circles. iii. the line 푦 − 3푥 + 2 = 0 passes through the centres of the circles. 23. Find the equation of the tangent and normal to the circle 푥 +푦 − 2푥 + 8푦 − 8 = 0 that passes through the point 푃(1, 1) 24. Find the equation of the two circles of radius 5 units which passes through the origin and whose centres lie on the line 푥 + 푦 − 1 = 0. 25. Find the equation of the circle with centre (−2, 5) and tangent to the 푥 −axis. 26. Find the equation of the circle with centre (4, 6) and tangent to the 푦 −axis. 27. The length of the tangent from 퐴(6, 2) to the circle with centre (2, 3) is 2√6 units. Find the radius of the circle.

348

CALCULUS (DIFFERENTIATION)

Differentiation is the process of finding a general expression for the gradient of a curve at any point on the curve. The process is also used to find the gradient of straight lines.

The idea of limits Let 푓(푥) = 푎 푥 + 푎 푥 + 푎 푥 + ⋯+ 푎 be an 푛푡ℎ degree polynomial function. The limit of 푓(푥) as 푥 approaches 푐 which is written as lim → 푓(푥) refers to the value of 푓(푥) as 푥 gets near to 푐. Limits of polynomial functions In general, if ℎ(푥) = 푓(푥) ± 푔(푥), then lim → ℎ(푥) = lim → 푓(푥) ± lim → 푔(푥) Also lim → 푎 = 푎 where 푎 is constant.

Example 15.1 Evaluate the following: i. 2

1(2 3 1)lim

xx x

ii. 3 2

2(4 2 5 7)lim

xx x x

iii. 4 2

3(5 3 6)lim

xx x

Solution

i. 5)1)1(3)1(2(1)1(3)1(21lim3lim2lim)132(lim 2211

21

21

xxxx

xxxx

ii. 87108327)2(5)2(2)2(4)7524(lim 23232

xxxx

iii. 102627135)6)3(3)3(5)635(lim 23243

xxx

Try: i.

32

2(2 5 1)limx

x x

ii. 2

1( 43 20)lim

xx x

iii. 3 2

2( 7)lim

xx x

iv. 4 2

3(3 3 1)lim

xx x x

Limits of rational functions

In general, if ℎ(푥) = 푓(푥)푔(푥), then

( )( ( ))

limlim lim ( )

x c

x cx c

f xh x

g x

Note: If ( ) 0limx c

g x

then there will be the need to simplify the entire function.

Example 15.2

349

Evaluate the following:

i. 2

0

3 61lim

x

xx

ii. 2

21

3 62 1

limx

x xx x

iii. 3

6

2 53lim

x

x xx

iv. 2

2

42lim

x

xx

v. 3

0limx

x xx

vi. 4

24lim

x

xx

vii. 25

255lim

x

xx

Solution

i. 610

6)0(3)1(lim

)63(lim

163lim

2

0

20

2

0

x

x

xx

x

xx

ii. 2112631

1)1()1(26)1(3)1(

1263

2

2

2

2

1lim

xxxx

x

iii. 3

22336

51221636

5)6(2)6(3

52 33

6lim

xxx

x

iv. First, simplify the function. So

422)2(2

)2)(2(2

42limlimlim

22

2

2

x

xxx

xx

xxx∴ 4

242 2

2lim

xx

x

v. First simplify the function. So

11)0(1)1( 22

0

2

0

3

0limlimlim

x

xxx

xxx

xxx ∴ 1

3

0lim

xxx

x

vi. First, simplify the function. So

41

241

21

)2)(2(2

42

limlimlim444

xxxx

xx

xxx

∴41

42

lim4

xx

x (Hint: Apply difference of two squares)

vii. First, simplify the function. So

350

1025555

)5)(5(525

limlimlim252525

x

xxx

xx

xxx

∴, 10525

lim25

xx

x (Hint: Apply difference of two squares)

Alternatively

xxx

xxx

xxxx

xx

xxxx 25)5)(25(

)()5()5)(25(

)5)(5()5)(25(

525

limlimlimlim25

22252525

= xx

5lim25

= 10 (Hint: Rationalise the denominator)

Try: 3

21

11

limn

nn

Limits of trigonometric functions

In general, if 휃 is a small angle measured in radians, then

0 0 0sin cos 1 and tanlim lim lim

Example 15.3 Evaluate the following:

0 0 0 0

0 0

sin tan 3 seci. ii. iii. iv. cos sin 2 sec3

1 cos2 sin 4v. vi. sin 3

lim lim lim lim

lim limx

x xx

Solution

i. 1sin

sin

lim

limlim

0

0

0

∴, 1sin

lim0

351

ii. 1tan

tan

lim

limlim

0

0

0

∴, 1sin

lim0

iii. 301

31

3sincos

3limlim

00

xx∴

13

sincos3

lim0x

iv. 22

21

)1(211

3cos12

cos1

3sec2sec

limlim00

∴

22

cos2sec

lim0

v.x

xx

xxx

xxx

xx

xx

xxx

sinsin2sinsin2sin2)sin21(12cos1 22

00limlim

0)1)(1)(0(2

∴ 02cos1lim

0

xx

x

vi. 34

34

33sin3

44sin4

3sin4sin

limlim11

x

x

xxx

xxx

xx

xx∴, 3

43sin4sin

lim1

x

xx

Try: i. 20 0

sin 4 1 cos 2 ii. tan 7lim lim

x x

x xx x

Limits to infinity

Let 푥 be a positive integer. The larger the value of 푥, the smaller the value of . That is, as

the value of 푥 keeps on increasing, the value of tends to 0. This is expressed by

saying ‘as 푥 tends to infinity(∞), tends to 0’. This is summarised as: 1limx x

In general 0lim

ax xb , where 푎 > 0and 푏 are parameters.

Limits to infinity of polynomial functions Let 푓(푥) = 푎 푥 + 푎 푥 + 푎 푥 + ⋯+ 푎 , 푛 ≥ 1.

352

if 푛 is even, then lim ( )x

f x

is +∞ or −∞ depending on the nature of the function.

If 푛 is odd, lim ( )x

f x

lim ( )x

f x

depends on the nature of the function.

Example 15.4 Evaluate the following:

i. 3lim 2 2 1x

x x

ii. 3lim 2x

x

solution i.

123 2lim xx

x ii.

23lim x

x

Limits to infinity of rational functions

In general, if 푓(푥) = 푎푥푛+푏푥푛−1

푐푥푛−2+푑푥푛−3, then 1

1

2 3 2 3lim ( ) lim lim

n n

n n n n

n n n nx x xn n

x xa bax bx x xf xcx dx x xc d

x x

, 푛 ≥ 1

Example 15.5 Evaluate the following:

4 3 3

4 33 1 3 2 2 7 3i. lim ii. lim iii. lim2 1 5 2 1 3x x y

x x x y yx x x y

Solution

i. 23

)0(2)0(3

2

31213

1

1

12

13

limlimlim

x

x

xxxx

xxx

xx xx

∴23

1213

lim

xx

x

ii. 53

)0()0(5)0()0(3

5

3

12523

42

43

44

2

4

4

44

3

4

4

12

21

125

23

24

34

limlimlim

xx

xx

xxx

xxx

xxx

xx

xx xxxx

53

12523

24

34

lim

xxxx

x

353

iii. 20102

3372

33

3

333

3

3

372

3

3

limlim

yy

y

yyy

yy

xx yyy ∴ 2

3372

3

3

lim

yyy

x

Try: 2

23lim

3x

n nn n

Differentiating from the first principles

Consider the diagram below: 푦

푄(푥 + ∆푥,푦 + ∆푦) 푦 + ∆푦 ∆푦 푦 = 푓(푥) figure 15.1. 푃(푥,푦) 푦

푥 ∆푥 푥 + ∆푥 푥

Let ∆푥 and ∆푦 be small changes in 푥 and 푦 respectively. In the diagram, as point 푄 moves towards 푃the gradient of curve 푃푄changes at any point on the curve. As 푄 move very close to 푃, ∆푥 → 0

Gradient of 푃푄 = 푦+∆푦−푦푥+∆푥−푥 = ∆푦

∆푥 = 푑푦푑푥

Since point 푄 lies on the curve 푦 = 푓(푥) ⟹ 푦 + ∆푦 = 푓(푥 + ∆푥) ⟹ ∆푦 = 푓(푥 + ∆푥)− 푦

∆푦 = 푓(푥 + ∆푥)− 푓(푥)

⟹∆푦∆푥 =

푓(푥+ ∆푥)− 푓(푥)∆푥

⟹x

xfhxfxy

dxdy

x

)()(lim

0 , where denote the gradient function of the

curve. This is often read as ‘푑푦on 푑푥’ or simply ‘푑푦, 푑푥’.

For functions of푓, = 푓 (푥). If we let∆푥 = ℎ, then h

xfhxfdxdy

h

)()(lim

0

354

The process for finding the gradient of a curve in this way is called differentiating from the first principles.

Example 15.6 Differentiate the following functions from the first principles. i) 푦 = 푥 + 3푥 − 4 ii) 푦 = 푥 iii) 푦 = 푥 + iv) 푓(푥) = 2푥

v) 푦 = √푥 vi) 푦 =√

Solution i. We have 푦 = 푥 + 3푥 − 4. Let ∆푥and∆푦be small changes in 푥 and 푦 respectively. ⟹ 푦 + ∆푦 = (푥 + ∆푥) + 3(푥 + ∆푥)− 4 = 푥 + 2푥∆푥 + (∆푥) + 3푥 + 3∆푥 − 4 ⟹ ∆푦 = 푥 + 2푥∆푥 + (∆푥) + 3푥 + 3∆푥 − 4 − 푦 = 푥 + 2푥∆푥 + (∆푥) + 3푥 + 3∆푥 − 4 − (푥 + 3푥 − 4) = 푥 + 2푥∆푥 + (∆푥) + 3푥 + 3∆푥 − 4 − 푥 − 3푥 + 4 = 2푥∆푥 + (∆푥) + 3∆푥

⟹ ∆∆

= ∆ (∆ ) ∆∆

= ∆∆

+ (∆ )∆

+ ∆∆

= 2푥 + ∆푥 + 3

323)0(2lim0

xxxy

dxdy

x

ii. We have 푦 = 푥 Let ∆푥and∆푦be small changes in 푥 and 푦 respectively. ⟹ 푦 + ∆푦 = (푥 + ∆푥) = 푥 + 3푥 ∆푥 + 3푥(∆푥) + (∆푥) ⟹ ∆푦 = 푥 + 3푥 ∆푥 + 3푥(∆푥) + (∆푥) − 푦 = 푥 + 3푥 ∆푥 + 3푥(∆푥) + (∆푥) − (푥 ) = 3푥 ∆푥 + 3푥(∆푥) + (∆푥)

⟹ ∆∆

= 3푥2∆푥+3푥(∆푥)2+(∆푥)3

∆푥=

3푥2∆푥

∆푥+

3푥(∆푥)2

∆푥+

(∆푥)3

∆푥= 3푥2 + 3푥∆푥 + (∆푥)2

22

00)0(33lim

xxxy

dxdy

x

iii. We have 푦 = 푥 + . Let ∆푥and∆푦be small changes in 푥 and 푦 respectively.

⟹ 푦 + ∆푦 = (푥 + ∆푥) +∆

355

= 푥 + 2푥∆푥 + (∆푥) +∆

⟹ ∆푦 = 푥 + 2푥∆푥 + (∆푥) +∆− 푦

= 푥 + 2푥∆푥 + (∆푥) +∆− 푥 +

= 푥 + 2푥∆푥 + (∆푥) +∆− 푥 −

= 2푥∆푥 + (∆푥) +∆−

⟹ ∆푦 = 2푥3∆푥+2푥2(∆푥)2+푥2(∆푥)2+푥(∆푥)3+2푥−2푥−2∆푥푥(푥+∆푥)

∆∆

= 2푥3∆푥+3푥2(∆푥)2+푥(∆푥)3−2∆푥푥(푥+∆푥)(∆푥) = 2푥3

푥(푥+∆푥) + 3푥2(∆푥)푥(푥+∆푥) + 푥(∆푥)2

푥(푥+∆푥)−2

푥(푥+∆푥)

∴ 222

3223

0

2222)0(

2)0(

)0()0()0(3

)0(2

limx

xxx

xxxxx

xxxx

xxx

xy

dxdy

x

iv. We have 푓(푥) = 2푥 = Let ℎ be the small change in 푥.

⟹ 푓(푥 + ℎ) =( )

⟹ 푓(푥 + ℎ) − 푓(푥)=( )

− 푓(푥) =( )

− = ( )( )

⟹ 푓(푥 + ℎ) − 푓(푥) =2푥2−2 푥2+2푥ℎ+ℎ2

푥2(푥+ℎ)2 =( )

=( )

⟹ 푓(푥+ℎ)−푓(푥)ℎ = −4푥ℎ−2ℎ2

푥2ℎ(푥+ℎ)2 = − 4푥ℎ푥2ℎ(푥+ℎ)2 −

2ℎ2

푥2(푥+ℎ)2

∴ 3322220

44)0(

)0(2)0(

4)()(lim

xxx

xxxxx

hxfhxf

dxdy

h

v. We have 푦 = √푥 Let ∆푥and∆푦be small changes in 푥 and 푦 respectively. ⟹ 푦 + ∆푦 = √푥 + ∆푥 ⟹ ∆푦 = √푥 + ∆푥 − 푦 = √푥 + ∆푥 − √푥 (By rationalising the numerator of∆푦,we have)

∆푦 = √푥 + ∆푥 − √푥 √ ∆ √√ ∆ √

=√ ∆ √ ∆ √ √ √ ∆ √√ ∆ √

= ∆√ ∆ √

356

⟹ ∆∆

= ∆푥

∆푥 √푥+∆푥+√푥=

1

√푥+∆푥+√푥

∴xx

xxxxxxy

dxdy

x 2211

01

lim0

vi. We have, 푦 = √

Let ∆푥and∆푦be small changes in 푥 and 푦 respectively.

⟹ 푦 + ∆푦 = 1푥+∆푥

⟹ ∆푦 = 1푥+∆푥

− 푦 = 1푥+∆푥

− 1√푥

∆푦 = √ √ ∆√ ∆ √

(By rationalising the numerator of ∆푦,we have)

∆푦= √ √ ∆√ ∆ √

√ √ ∆

√ √ ∆

= ( ∆ )√ ∆ √ √ √ ∆

⟹ ∆∆

= 푥−푥−∆푥

∆푥 √푥+∆푥 √푥 √푥+√푥+∆푥=

−1

√푥+∆푥 √푥 √푥+√푥+∆푥

∴

30 2

12

12

100

1lim

xxxxxxxxxxxy

dxdy

x

Try: 1. Differentiate from the first principles i. 푦 = (푥 − 3) ii. 푦 = 2. Find from first principles, the derivative of 푓: 푥 = (푥 + 2) (SSSCE) 3. Find from first principles, the derivative of 푦 = 푥 + 1 (SSSCE) 4. Find the derivative of 3푥 + 1, from first principles. (SSSCE) 5. Find, from first principles the derivative of the function 푓(푥) = 2푥 (SSSCE) 6. Find from first principles the derivative of푥 + 2푥 (SSSSCE) 7. Find, from first principles, the derivative of푥 + 푥 (SSSCE) 8. Find, from first principles, the derivative of푓(푥) = (2푥 + 3) (SSSCE) 9. Obtain from first principles the derivative 푓(푥) = 2푥 + 푥 (SSSCE) 9. Find, from first principles, the derivative of 푓:푥 = 3푥 + 4푥 (WASSCE) 10. Find from first principles the derivative of 푦 = (3푥 + 푥 )푥 , where 푥 ≠ 0.

357

(WASSCE)

The derivative of a function In general, if 푦 = 푎푥 , then = 푛 ∙ 푎푥 or 푓 (푥) = 푛 ∙ 푎푥 for functions of 푓. This is known as the first derivative. Proof We are proving from first principles. Given 푦 = 푎푥 let 푄(푥 + ∆푥,푦 + ∆푦) be a point on the curve.

⟹ 푦 + ∆푦 = 푎(푥 + ∆푥) ⟹ ∆푦 = 푎(푥 + ∆푥) − 푦 ⟹ ∆푦 = 푎(푥 + ∆푥) − 푎푥

⟹ ∆푦 = 푎 푥 + 푛푥 ∆푥 + ( ) (∆ )!

+ ⋯ − 푎푥 (By binomial theorem, expand(푥 + ∆푥) )

⟹ ∆푦 = 푎푥 + 푎푛푥 ∆푥 + 푎푛(푛 − 1)푥 (∆푥)

2! + ⋯− 푎푥

⟹ ∆푦 = 푎푛푥 ∆푥 + 푎푛(푛 − 1)푥 (∆푥)

2! + ⋯

⟹ = 푎푛푥 + 푎 ( ) ∆!

+ ⋯ (Factorise ∆푥 out)

⟹Δ푦Δ푥 = 푎푛푥 + 푎

푛(푛 − 1)푥 ∆푥2! + ⋯

푑푦푑푥 = lim

∆ →

Δ푦Δ푥 = 푎푛푥 + 푎

푛(푛 − 1)푥 (0)2! + ⋯

∴푑푦푑푥 = 푎푛푥 = 푛 ∙ 푎푥

The second and the third derivatives are as follow: 푑 푦푑푥 = 푓 (푥) = (푛 − 1)(푛)푎푥

푑 푥푑푥 = 푓 (푥) = (푛 − 2)(푛 − 1)(푛)푎푥

Example 15.7

Find the derivative of the following functions: i) 푦 = 푥 + 푥 ii) 푦 = 6푥 − 3푥 iii) 푦 = 푥 − iv) 푦 = 푥 + iv) 푦 = √푥

i. If 푦 = 푥 + 푥 then = 5푥 + 1 ∙ 푥 = 5푥 + 1

ii. If 푦 = 6푥 − 3푥 then = 2 ∙ 6푥 − 3 ∙ 1푥 = 12푥 − 3

358

iii. If 푦 = 푥 − = 푥 − 3푥 then = 2푥 − 3(−2)푥 = 2푥 + 6푥 = 2푥 +

iv. If 푦 = 푥 + = 푥 + 2푥 then = 3푥 + 2(−1)푥 = 3푥 − 2푥 = 3푥 −

v. If 푦 = √푥 = 23

x then = 푥 = 푥 = √푥

Example 15.8 Show from first principles that the derivative of a constant is zero.

Proof Let푦 = 푐, where 푐 is constant. Let 푄(푥 + ∆푥,푦 + ∆푦) be a point on the line. ⟹ 푦 = 푐푥 ⟹ 푦 + ∆푦 = 푐(푥 + ∆푥) ⟹ ∆푦 = 푐(푥 + ∆푥) − 푦 ∆푦 = 푐(푥 + ∆푥) − 푐 ⟹ =

∆ ⟹ = 0 ∴ = 0 Example 15.9

Find, with respect to 푥 the second and third derivatives of the following functions: i. 푦 = 2푥 + 5푥 − 푥 ii. 푓(푥) = 3푥 iii.푔(푥) = 4푥 +

Solution

i. If 푦 = 2푥 + 5푥 − 푥, then = 10푥 + 15푥 − 1, = 40푥 + 30푥 = 10푥(4푥 + 3)

and = 120푥 + 30 = 30(4푥 + 1) ii. If 푓(푥) = 3푥 , then 푓 ′(푥) = 126푥 ,and 푓 (푥) = 756푥 iii. If 푔(푥) = 4푥 + , then 푔 (푥) = 16푥 − , 푔 (푥) = 48푥 + and

푔 (푥) = 96푥 −

Try: Find if i) 푦 = √푥 + 3 ii) 푦 = iii) 푦 =

Differentiation of product of two functions The product rule

Let푦 = 푢푣, where 푢and 푣 are functions of 푥. Then = 푣 + 푢 Proof

We are proving from first principles. Let∆푥,∆푦, ∆푢 and ∆푣 be small changes in 푥, 푦,푢 and 푣 respectively. Given 푦 = 푢푣 ⟹ 푦 + ∆푦 = (푢 + ∆푢)(푣 + ∆푣) ⟹ 푦 + ∆푦 = 푢푣 + 푢∆푢 + 푣∆푣 + ∆푢∆푣 ⟹ ∆푦 = 푢푣 + 푢∆푢 + 푣∆푣 + ∆푢∆푣 − 푦 ⟹ ∆푦 = 푢푣 + 푢∆푢 + 푣∆푣 + ∆푢∆푣 − 푢푣

359

⟹ ∆푦 = 푢∆푢 + 푣∆푣 + ∆푢∆푣 ⟹ = ∆∆

+ ∆∆

+ ∆ ∆∆

⟹ = lim∆ ∆ → = 푣 + 푢 ∴, = 푣 + 푢 Example 15.10

Differentiate the following functions with respect to 푥. i. 푦 = (푥 − 1)(2푥 + 3) ii. 푦 = 푥(푥 − 1)(푥 + 2) iii. 푓(푥) = 3 + √2푥 (1 − 푥 ) iv. 푦 = 푥 / (3푥 − 1)

Solution i. We have 푦 = (푥 − 1)(2푥 + 3) Let 푢 = 푥 − 1 ⟹ = 2푥 and 푣 = 2푥 + 3 ⟹ = 6푥

= 푣 + 푢 ∴, = (2푥 + 3)(2푥) + (푥 − 1)(6푥 ) = 4푥 + 6푥 + 6푥 − 6푥 = 10푥 − 6푥 + 6푥 ii. We have 푦 = 푥(푥 − 1)(푥 + 2) = (푥 − 푥)(푥 + 2) Let 푢 = 푥 − 푥 ⟹ = 2푥 − 1 and 푣 = 푥 + 2 ⟹ = 1

= 푣 + 푢 ⟹ = (푥 + 2)(1) + (푥 − 푥)(2푥 − 1) = 푥 + 2 + 2푥 − 푥 − 2푥 + 푥 = 2푥 − 3푥 + 2푥 + 2 iii. We have 푓(푥) = 3 + √2푥 (1− 푥 )

Let 푢 = 3 + √2푥 ⟹ = √√

and 푣 = 1 − 푥 ⟹ = −2푥

푓 (푥) = 푣 + 푢 ⟹ = (1 − 푥 ) √√

+ 3 + √2푥 (−2푥) =

= √√− √

√− 6푥 − 2푥√2푥

= √ √ √ √ √√

= √ ( √ / / / )√

= √ √ √

√

iv. We have 푦 = 푥 / (3푥 − 1) Let 푢 = 푥 / ⟹ = 푥 / = 푥 / and 푣 = 3푥 − 1 ⟹ = 3

= 푣 + 푢 ⟹ = (3푥 − 1) 푥 / + 푥 / (3)

360

= 푥 / [(3푥 − 1)푥 + 3] = 푥 / (3푥 − 푥 + 3)

Differentiation of rational functions The quotient rule

Let푦 = , where 푢 and 푣are functions of 푥. Then = Proof

We are proving from first principles. Let∆푥,∆푦, ∆푢 and ∆푣 be small changes in 푥, 푦,푢 and 푣 respectively. Given 푦 =

⟹ 푦 + ∆푦 = ∆∆

⟹ ∆y= ∆∆− 푦 ⟹ ∆푦 = 푢+∆푢

푣+∆푣−푢푣

⟹ ∆푦 = (푢+∆푢)푣−(푣+∆푣)푢(푣+∆푣)푣 ⟹ ∆푦 = 푢푣+푣∆푢−푢푣−푢∆푣

(푣+∆푣)푣 = ∆ ∆( ∆ )

⟹ = 푣∆푢−푢∆푣(∆푥)(푣+∆푣)푣 = 푣∆푢∆푥−푢

Δ푣Δ푥

(푣+∆푣)푣 = lim0v

=( )

∴ =푣푑푢푑푥−푢

푑푣d푥

푣2 Example 15.11

Differentiate the following functions with respect to 푥.

i) 푦 = ii. 푓(푥) = iii. 푦 = Solution

i. We have 푦 =

Let 푢 = 푥 + 3 ⟹ = 2푥 and 푣 = 3푥 − 1 ⟹ = 3

∴, = ⟹ = (3푥−1)( ) (푥2+3)( )( ) = ( ) = ( )

ii. We have 푦 = 푓(푥) =

Let 푢 = 푥 − 3 ⟹ = 1 and 푣 = 푥 + 4 ⟹ = 2푥

푓 (푥) =푣푑푢푑푥−푢

푑푣d푥

푣2 ⟹ 푓 (푥) = (1)−( )(2푥)

푥2+4 2 = 푥2+4−2푥2+6푥푥2+4 2 = 4+6푥−푥2

푥2+4 2

iii. We have 푦 = 푓(푥) =

361

Let 푢 = 푥 − 1 ⟹ = 2푥 and 푣 = 푥 + 푥 ⟹ = 2푥 + 1

= ⟹ = ( )( ) ( )( )( ) = ( )

= ( )

= ( )( )

=

Differentiation of a function of a function Chain rule

Let 푦 = (푢) , where 푢 = 푓(푥) and 푛 ∈ ℝ. Then = ∙ Example 20.11

Find the derivative of the following functions. i. 푦 = (6푥 − 3) ii. 푦 = (푥 + 2푥) iii. 푔(푥) = (4푥 − 3) iv.푦 = (1 − 3푥 )

Solution i.푦 = (6푥 − 3) Let 푢 = 6푥 − 3 ⟹ 푦 = 푢 = 7푢 and = 6

= ∙ = 7푢 (6) = 42푢 (but 푢 = 6푥 − 3) ∴, = 42(6푥 − 3) ii. 푦 = (푥 + 2푥) Let 푢 = 푥 + 2푥 ⟹ 푦 = 푢 = 5푢 and = 2푥 + 2

= ∙ = 5푢4(2푥 + 2) = 10(푥 + 1)푢4 (but 푢 = 푥 + 2푥)

∴, = 10(푥 + 1)(푥 + 2푥)

iii. 푔(푥) = (4푥 − 3) Let 푢 = 4푥 − 3 ⟹ 푦 = 푢 = 푢 / and = 4

= ∙ =3

2푢1/2(4) = 6푢1/2 (but 푢 = 4푥 − 3) ∴ = 6(4푥 − 3) /

iv. 푦 = (1 − 3푥) Let 푢 = 1 − 3푥 ⟹ 푦 = 푢 = − 푢 / and = −3

= ∙ = −1

3푢−4/3(−3) = / (but 푢 = 4푥 − 3)

∴ = 1(1−3푥)4/3 = 1

(1−3푥)43

362

Example 15.2 Find the derivative of the following functions. i.푦 = (2푥 − 3)(푥 − 5) ii.푦 =

( )

Solution i. We have 푦 = (2푥 − 3)(푥 − 5) Let 푢 = 2푥 − 3 ⟹ = 2 and 푣 = (푥 − 5)

(By using the chain rule we have) = 4(1)(푥 − 5) = 4(푥 − 5)

= 푣 + 푢 ∴, = (푥 − 5) (2) + (2푥 − 3)4(푥 − 5) = (푥 − 5) [2(푥 − 5) + 4(2푥 − 3)] = (푥 − 5) [2푥 − 10 + 8푥 − 12] = (푥 − 5) (10푥 − 22) = 2(푥 − 5) (5푥 − 11) ii. We have 푦 =

( ) Let 푢 = 7푥 + 4 ⟹ = 7 and 푣 = (5푥 + 2)

(By using the chain rule we have) = 15푥 (3)(5푥 + 2) = 451푥 (5푥 + 2)

= ∴, = (5푥 + 2) (7)− (7푥 + 4)45푥 (5푥 + 2)

= (5푥 + 2) [7(5푥 + 2) − 45푥 (7푥 + 4)] = (5푥 + 2) [35푥 + 14− 315푥 − 180푥 ] = (5푥 + 2) (14− 180푥 − 280푥 ) = 2(5푥 + 2) (7− 90푥 − 140푥 )

Derivative of trigonometric, exponential and logarithmic functions

푦 푑푦푑푥

sin 푥 cos푥 cos 푥 − sin 푥 tan 푥 sec 푥

cosec 푥 − cosec푥 cot푥 sec 푥 sec 푥 tan 푥 cot푥 cosec 푥

sin(푢) 푑푢.푑푥 ∙ cos(푢)

363

Where u is a function of 푥.

Example 15.13 Differentiate the following functions with respect to 푥. i. 푦 = 푒 ii. 푦 = 푒 iii. 푦 = cot(3푥 − 4) iv. 푦 = ln 푥 v. 푦 = ln(2푥 + 푥) vi. 푦 = 푒 ln 2푥 vii. 푦 = 푥 sin 3푥 viii. 푦 = sin 3푥

ix. 푦 = Solution

i. We have 푦 = 푒 Let 푢 = 3푥 ⟹ 푦 = 푒 = 푒 and = 3

= ∙ ∴, = 푒 (3푥) = 3푥푒

ii. We have 푦 = 푒 Let 푢 = sin 푥 ⟹ 푦 = 푒 = 푒 and = cos 푥

= ∙ ∴, = 푒 (cos 푥) = cos푥 푒

iii. We have 푦 = cot(3푥 − 4) Let 푢 = 3푥 − 4 ⟹ 푦 = cot 푢 = cosec 푢 and

= 3푥 = ∙ ∴, = cosec 푢 (3푥) = 3푥 cosec (3푥 − 4)

cos(푢) −푑푢푑푥 . sin(푢)

tan(푢) 푑푢푑푥 ∙ sec (푢)

cos(푢) −푑푢푑푥 ∙ cos푒푐(푢) cot(푢)

푠푒푐(푢) 푑푢푑푥

∙ 푠푒푐(푢)푡푎푛(푢)

푐표푡푢 −푑푢푑푥 ∙ 푐표푠푒푐 (푢)

푒 푒 푒( )

푑푢푑푥 ∙ 푒

( )

ln푔(푥) 푔 (푥)

푔(푥)

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iv. We have 푦 = ln푥 Let 푢 = 푥 ⟹ 푦 = ln푢 = and = 1

= ∙ ∴, = (1) = 1푥

v. We have 푦 = ln 2푥 + 푥Let 푔(푥) = 2푥 + 푥 ⟹ 푔 (푥) = 4푥 + 1

= ( )( )

∴, 푑푦푑푥 = 4푥+12푥2+푥

vi. We have 푦 = 푒 ln 2푥Let푢 = 푒 ⟹ = 2푥푒 and 푣 = ln 2푥

⟹ = = = 푣 + 푢

∴, 푑푦푑푥 = ln 2푥 2푥푒 + 푒 = 2푥푒 ln 2푥 + = 푒 2푥 ln 2푥 +

= (2푥 ln 2푥 + 1) vii. We have 푦 = 푥 sin 3푥Let 푢 = 푥 ⟹ = 3푥 and 푣 = sin 3푥

⟹ = 3 cos 3푥 = 푣 + 푢

∴, 푑푦푑푥 = sin 3푥 (3푥 ) +푥 (3 cos 3푥) = 3푥 sin 3푥 + 3푥 cos 3푥 = 3푥 (sin 3푥 + cos 3푥) viii. We have 푦 = sin 3푥 = (sin 3푥) Let 푢 = sin 3푥 ⟹ = 3 cos 3푥

Now,푦 = 푢 so = 3푢 = ∙

∴, 푑푦푑푥 = 3푢 (3 cos 3푥) = 9 sin 3푥 cos 3푥 = 9(1 − cos 3푥) cos 3푥 = 9 (cos 3푥 − cos 3푥)

ix. We have 푦 = 푦 = Let 푢 = 푒 ⟹ = 2푒 and

푣 = sin 푥 + cos푥 ⟹ = cos푥 − sin 푥 =

∴, = ( )( ) ( )( )

= ( )( )

= 푒2푥(3sin푥+cos푥)(sin푥+cos푥)2

Try: Differentiate the function ℎ(푥) = sin 푥 + cos 푥

Differentiation Implicit relations The relation 푦 = 푓(푥) is an explicit relation of 푦 in terms of 푥. A relation such as 푓(푥) + 푔(푦) = 푎 is referred to as an implicit relation between 푥 and 푦, where 푎 is

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constant. To find the derivative of 푦 with respect to 푥, we take 푦 as a function of 푥. Then we differentiate the function term by term.

That is (푦) = , (푦 ) = 2푦 , (푦 ) = 3푦 and so on. Example 15.14

i. Find if 푦 + 푥 − 3푥 = 3 ii. If 푦 + 푥 + 푥푦 = 2, find .

iii. Find if (푥 + 푦) = 13푥 Solution

i. If 푦 + 푥 − 3푥 = 3, then 푦 + 2푥 − 3 = 0 ⟹ 푦 = 3 − 2푥 ∴, =

ii. If푦 + 푥 + 푥푦 = 2, then + 3푥 + 푦(1) + 푥 = 0 ⟹ + 푥 = −푦 − 3푥

⟹ (1 + 푥) = −푦 − 3푥 ∴ = = − [Hint: Treat푥푦asproductoftwofunctionsothat푢 = 푥and푣 = 푦] iii. (푥 + 푦) = 13푥 then 2(푥 + 푦) 1 + = 13 ⟹ 2푥 + 2푥 + 2푦 + 2푦 = 13

⟹ 2푥 + 2푦 = 13 − 2푥 − 2푦 or (2푥 + 2푦) = 13 − 2푥 − 2푦

∴, =

[Hint: Treat(푥 + 푦) asa푓uctionoffunctionssothat푢 = 푥 + 푦푎푛푑푦 = 푢 ] Alternatively, If (푥 + 푦) = 13푥 or 푥 + 2푥푦+ 푦 = 13푥 then 2푥 + 푦(2) + 2푥 + 2푦 = 13

⟹ 2푥 + 2푦 + 2푥 + 2푦 = 13 or (2푥 + 2푦) = 13 − 2푥 − 2푦

∴, =

Hint: expand(푥 + 푦) andtreat2푥푦asproductoftwofunctionsothat푢 = 2푥and푣 = 푦

Try: Differentiate the function: (푥 + 푦) − 푥푦 + 4푥 = 0

Differentiation of parametric equations in time t

Certain functions can be parameterised in time t 푡. Let 푥 = 푓(푡) and 푦 = 푔(푡). Then = ∙

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Example 15.15 i. Given that 푥 = 푡 + 3 and 푦 = 푡 − 푡, find .

ii. If 푥 = , 푡 ≠ 1 and 푦 = , 푡 ≠ 2, find . Solution

i. 푥 = 푡 + 3 ⟹ = 2푡 and 푦 = 푡 − 푡 ⟹ = 2푡 − 1 But = =

= ∙ ∴, = (2푡 − 1) =

ii. 푥 = ⟹ = ( )( ) ( )( )( )

=( )

=( )

and

푦 = 푡 − 푡 ⟹ = ( )( ) ( )( )( )

=( )

= −( )

But = = ( )

= ∙ ∴, = −( )

∙ ( ) = Try: If 푥 = , 푡 ≠ and 푦 = , 푡 ≠ −1, find .

Application of differentiation a. The equation of the tangent and normal to a curve

A tangent to a curve is a line that touches a curve at only one point while a normal to a curve is a line that intersects the tangent at right-angle. Consider the diagram below: 푦 푦 = 푓(푥) Tangent 푃(푥 ,푦 ) figure 15.2. Normal 휃 푥 The gradient of a curve at a point 푥 = 푐 is the value of the derivative function at 푥 = 푐, where 푐 is constant. That is, if 푦 = 푓(푥) then (read ‘푑푦표푛 푑푥 at 푥 equal to푐’) is

the value of the gradient of the curve at a point 푥 = 푐.

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Let 푚 be the gradient of the tangent. Then tan휃 = 푚, where 휃 is the acute angle the tangent makes with the 푥-axis. If the tangent touches the curve 푦 = 푓(푥) at 푃(푥 ,푦 ),then the equation of the tangent to the curve is given by:

푦 − 푦 =푑푦푑푥 (푥 − 푥 )

The equation of the normal to the curve is given by: 푦 − 푦 = (푥 − 푥 ) ⟹ (푦 − 푦 ) = −(푥−푥 )

Note: = tan휃

Example 15.16 Find the equation of the equation of the tangent and normal to the curve 푦 = 푥 − 5 at point 푃(1,−2). Find the coordinates of the point 푄 where this tangent meets the 푦-axis.

Solution We have푃(1,−2) and 푦 = 푥 − 5 ⟹ = 2푥 = 2(1) = 2

The equation of the tangent is given by: 푦 − 푦 = (푥 − 푥 )

That is, 푦 − (−2) = 2(푥 − 1) or 푦 + 2 = 2푥 − 2 or 2푥 − 푦 − 4 = 0 The equation of the normal is given by: (푦 − 푦 ) = −(푥−푥 )

That is, 2 푦 − (−2) = −(푥 − 1) or 2푦 + 4 = −푥 + 1 or 푥 + 2푦 + 3 = 0 At the point where the tangent 2푥 − 푦 − 4 = 0 meet the 푦 −axis, 푥 = 0 ⟹ 2(0)− 푦 − 4 = 0 or 푦 = −4 ∴ 푄(0,−4)

Example 15.17 Find equation of the tangent and normal to the curve 푦 = 푥 − 6푥 − 8 at point 퐴(2, 3).

Solution We have 퐴(2, 3) and 푦 = 푥 − 6푥 − 8 ⟹ = 2푥 − 6 = 2(2)− 6 = −2

The equation of the tangent is given by: 푦 − 푦 = (푥 − 푥 )

That is, 푦 − 3 = −2(푥 − 2) or 푦 − 2 = −2푥 + 4 or 2푥 + 푦 − 4 = 0 The equation of the normal is given by: (푦 − 푦 ) = −(푥−푥 )

That is, −2(푦 − 3) = −(푥 − 2) or −2푦 + 6 = −푥 + 2 or 푥 − 2푦 + 4 = 0

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Example 15.18 Find the equations of the tangent to the curve 푦 = 3푥 − 5푥 + 1 at the points of intersection of the line 푦 = −푥 and the curve.

Solution We have 푦 = 3푥 − 5푥 + 1 … … … (1) and 푦 = −푥… … … (2) At the point of intersection 3푥 − 5푥 + 1 = −푥 ⟹ 3푥 − 4푥 + 1 = 0 3푥 − 3푥 − 푥 − 1 = 0 or 3푥(푥 − 1)− (푥 − 1) = 0 or (푥 − 1)(3푥 − 1) = 0 ⟹ 푥 − 1 = 0 or 3푥 − 1 = 0 ∴ 푥 = 1or푥 = Put the values of 푥 into (2)

When 푥 = 1, 푦 = −1 and when 푥 = ,푦 = − so the points of intersection

are 퐴(1,−1) and 퐵 ,−

Now, = 6푥 − 5 = 6(1)− 5 = 1 and = 6 − 5 = −3

The equations of the tangent are given by: 푦 − 푦 = (푥 − 푥 ) on 퐴(1,−1)

That is, 푦 − (−1) = (푥 − 1) or 푦 + 1 = 푥 − 1 or 푥 − 푦 − 2 = 0 and 푦 − 푦 = (푥 − 푥 ) on 퐵 ,−

That is, 푦 − − = −3 푥 − or 푦 + = −3푥 + 1 or 3푦 + 1 = −9푥 + 3or

9푥 + 3푦 − 2 = 0 Example 15.19

Find the angle of slope of the tangent to the curve 푦 = 1 − 푥 − 2푥 at the point 푃(1,−1) Solution

We have 퐴(1,−1) and 푦 = 1 − 푥 − 2푥 ⟹ = −1 − 4푥 푑푦푑푥 = −(−1) − 4(−1) = 5

The angle of slope is given by: tan 휃 = 푚, where 푚 =

That is, tan 휃 = 5 ⟹ 휃 = tan (5) or 휃 ≈ 78.69°

Example 15.20 Find the equations of the normal to the curve 푦 = 푥 − 푥 at the points (16,−3) and (−2, 2) and find their point of intersection.

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Solution Let 푃(16,−3) and 푄(−2, 2). Given 푦 = 푥 − 푥 = 2푥 −

= 2(16)− = = 2(−2) − = −

The equation of the normal at point 푃(16,−3) is given by: (푦 − 푦 ) = −(푥−푥 )

That is, 푦 − (−3) = −(푥 − 16) or 127푦 + 381 = −4(푥 − 16) or

127푦 + 381 = −4푥 + 64 or 4푥 + 127푦+ 317 = 0 or 푦 = − 푥 −

The equation of the normal at point 푄(−2, 2) is given by: (푦 − 푦 ) = −(푥−푥 )

That is, − (푦 − 2) = −(푥 − (−2)) or −17푦 + 34 = −4(푥 + 2) or

4푥 − 17푦 + 42 = 0 or 푦 = 푥 +

At the point of intersection − 푥 − = 푥 + ⟹ 푥 + 푥 = − −

푥 = − ⟹ 푥 = − ≈ −13.3048 ⟹ 푦 = − + = −0.6610 Therefore, the point of intersection is (−13.3048,−0.6610)

Example 15.21 Find the coordinates of the points on the curve 3푥 + 2푥푦+ 3푦 = 16 where the gradient is −1.

Solution If 3푥 + 2푥푦 + 3푦 = 16⋯⋯⋯ (1) then 6푥 + 2푥 + 2푦 + 6푦 = 0 or

2푥 + 6푦 = 6푥 + 2푦 ⟹ = = ( )( )

=

If the gradient is −1, then = −1 ⟹ 3푥 + 푦 = −푥 − 3푦 or 4푥 = −4푦 ⟹ 푦 = −푥… … … (2)

Put 푦 = −푥 into (1) 3푥 + 2푥(−푥) + 3(−푥) = 16 or 6푥 = 16

⟹ 푥 = = 푥 = ± = ± √√

= ± √

When 푥 = √ , 푦 = − √ and when 푥 = − √ ,푦 = √

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Therefore, the coordinates of the point on the curve are √ ,− √ and √ ,− √

Example 15.22 Find the value of 푟 for which the line 푦 = 3푥 + 푟 is tangent to the curve 푦 = 3 − 2푥 .

Solution We have 푦 = 3푥 + 푟… … … (1) and 푦 = 3 − 2푥 … … … (2) If (1) is tangent to (2) then at the point of tangency the gradient of (1) is equal to that of (2). For (1) = 3 and for (2) = −4푥. So 3 = −4푥 ⟹ 푥 = −

Put 푥 = − into (2). That is, 푦 = 3 − 2 − = 3 − =

Put the values of 푥 and 푦into (1). That is, = 3 − + 푟 ⟹ 푟 = + ∴ 푥 = Alternatively, We have 푦 = 3푥 + 푟⋯⋯⋯ (1) and 푦 = 3 − 2푥 ⋯⋯⋯ (2) If (1) is tangent to (2) then it intersect (2) at a point say 푃. At the point of intersection, 3푥 + 푟 = 3 − 2푥 ⟹ 2푥 + 3푥 + 푟 − 3 = 0 Since, the point of tangent is one and only one, the equation 2푥 + 3푥 + 푟 − 3 = 0 has equal roots. That is, 푏 − 4푎푐 = 0, where 푎 = 2, 푏 = 3and푐 = 푟 − 3 ⟹ (3) − 4(2)(푟 − 3) = 0 or 9− 8푟 + 24 = 0 or 8푟 = 33 ∴ 푟 =

b. maxima and minima Consider the diagram below: 푦 Stationary point ( = 0) 푦 = 푓(푥) Decreasing Increasing Decreasing < 0 > 0 < 0푥

Stationary/turning point = 0

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훼. Increasing and decreasing functions. Let 푦 = 푓(푥). If 푦 is decreasing at a given interval, then < 0. If 푦 is increasing at a

given internal, then > 0. Example 15.23

Find the values of 푥 for which the curve 푦 = 푥 − 4푥 + 1 is decreasing. Solution

We have 푦 = 푥 − 4푥 + 1 = 2푥 − 4. If 푦 is decreasing, then < 0

That is, 2푥 − 4 < 0 ⟹ 2푥 < 4 or 푥 < or 푥 < 2 Therefore, 푦 is decreasing for all values of 푥 < 2

Example 15.24 Find the range of values of 푥 for which the curves 푦 = 3푥 − 3푥 + 2 and 푦 = 1 − 3푥 − 푥 are increasing.

Solution We have 푦 = 3푥 − 3푥 + 2 = 6푥 − 3. If 푦 is increasing, then > 0

That is, 6푥 − 3 > 0 ⟹ 6푥 > 3 or 푥 > or 푥 >

Therefore, 푦 is increasing for all values of 푥 >

For 푦 = 1 − 3푥 − 푥 = −3 − 2푥.If 푦 is increasing, then > 0

That is, −3 − 2푥 > 0 ⟹ −2푥 > 3 or 푥 < −

Therefore, 푦 is increasing for all values of 푥 < − Example 15.25

If 푦 = 3 + 5푥 − 푥 − 푥 , find the interval at which 푦 is increasing and decreasing. Solution

We have 푦 = 3 + 5푥 − 푥 − 푥 = 5 − 2푥 − 3푥 . If 푦 is increasing, then > 0 That is, 5 − 2푥 − 3푥 > 0 ⟹ 5 − 5푥 + 3푥 − 3푥 > 0 or 5(1 − 푥) + 3푥(1 − 푥) > 0 or (1 − 푥)(5 + 3푥) > 0 ⟹ 1 − 푥 > 0 or 5 + 3푥 > 0 So −푥 > −1 ⟹ 푥 < 1 or 5 + 3푥 > 0 ⟹ 푥 > −

Therefore, 푦 is increasing for all values of 푥 < 1or푥 > −

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훽. Maximum and minimum points Let 푦 = 푓(푥). The maximum/minimum/stationary/turning point is the point at which

= 0. We may distinguish between a maximum and a minimum point.

If > 0, then we have a minimum point. If < 0, then we have a maximum point.

If = 0, then we have point of inflexion. (This analysis is sometimes called second derivative test) If 푃(푎,푏) is the stationary point of a curve, then line 푦 = 푏 and line푥 = 푎 are the stationary value and the axis of symmetry of the curve12 respectively.

Example 15.26 Find the turning point on the curve 푦 = 2푥 + 2푥 − 5 and determine its nature. axis of symmetry is often used for quadratic relation.

Solution We have 푦 = 2푥 + 2푥 − 5 = 4푥 + 2. At the turning point = 0.

That is, 4푥 + 2 = 0 or 푥 = − = − . Put the value of 푥 into 푦 = 2푥 + 2푥 − 5

⟹ 푦 = 2 − + 2 − − 5 = 2 − 1 − 5 = −

Therefore the turning point is − ,−

= 4 > 0 .Since > 0, the turning point − ,− is minimum.

Example 15.27 Find the turning points on the curve푦 = 1 + 3푥 − 4푥 − 푥 and determine their nature.

Solution We have 푦 = 1 + 3푥 − 4푥 − 푥 = 3 − 8푥 − 3푥 .

At the turning points = 0. That is, 3 − 8푥 − 3푥 = 0 or 3− 9푥 + 푥 − 3푥 = 0 or 3(1 − 3푥) + 푥(1 − 3푥) = 0 or (1 − 3푥)(3 + 푥) = 0 ⟹ (1 − 3푥) = 0 or (3 + 푥) = 0 ⟹ 푥 = or 푥 = −3 Put the values of 푥 into 푦 = 1 + 3푥 − 4푥 − 푥

12axis of symmetry is often used for quadratic relation.

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⟹ 푦 = 1 + 3 − 4 − = 1 + 1 − − = = or

푦 = 1 + 3(−3)− 4(−3) − (−3) = 1 − 9 − 36 + 27 = −17 Therefore, the turning point are , and (−3, 53)

= −8 − 6푥 = −8− 6 = −10 < 0

Since < 0, the turning point , is maximum.

= −8 − 6푥 = −8 − 6(−3) = 10 > 0

Since > 0, the turning point (−3, 53) is minimum. Example 15.28

Find the turning points on the curve 푦 = − 푥 − 8푥 + 1 and determine their nature. Solution

We have 푦 = − 푥 − 8푥 + 1 and = 푥 − 2푥 − 8.

At the turning points = 0. That is, 푥 − 2푥 − 8 = 0 or 푥 − 4푥 + 2푥 − 8 = 0 or 푥(푥 − 4) + 2(푥 − 4) = 0 or (푥 − 4)(푥 + 2) = 0 ⟹ (푥 − 4) = 0 or (푥 + 2) = 0 ⟹ 푥 = 4 or 푥 = −2 Put the values of 푥 into 푦 = 1 + 3푥 − 4푥 − 푥

⟹ 푦 = ( ) − (4) − 8(4) + 1 = − 16 − 32 + 1 = − 47 = − or

푦 = ( ) − (−2) − 8(−2) + 1 = − − (−4) + 16 + 1 = − + 21 =

Therefore, the turning point are 4,− and −2,

= 2푥 − 2 = 2(4)− 2 = 6 > 0

Since > 0, the turning point 4,− is minimum.

= 2푥 − 2 = 2(−2)− 2 = −6 < 0

Since < 0, the turning point −2, is maximum.

Example 15.29 The function 푦 = 푝푥 + 푞푥 + 푟 has a gradient 4푥 − 1 and a stationary value 3. Find the values of 푝, 푞 and 푟.

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Solution If 푦 = 푝푥 + 푞푥 + 푟 = 2푝푥 + 푞. But = 4푥 − 1 ⟹ 2푝푥 + 푞 = 4푥 − 1 By comparing coefficients: 2푝 = 4 ⟹ 푝 = 2 and 푞 = −1 At the stationary, = 0. That is, 4푥 − 1 = 0 or 푥 = Stationary value, 푦 = 3. Put the values of 푝,푞,푥and푦into 푦 = 푝푥 + 푞푥 + 푟

That is, 3 = (2) + (−1) + 푟 ⟹ 3 = − + 푟 ⟹ 푟 = 3 − + =

∴ 푝 = 2, 푞 = −1, 푟 =258 .

Example 15.30 The curve 푦 = 푥 + 푎푥 + 푏푥 + 7 has a stationary point at (2, 3). Find the values of 푎 and 푏.

Solution We have 푦 = 푥 + 푎푥 + 푏푥 + 7 = 3푥 + 2푎푥 + 푏

Stationary point (2, 3) ⟹ 푥 = 2, 푦 = 3 At the stationary point = 0 ⟹ 3푥 + 2푎푥 + 푏 = 0 or 3(2) + 2푎(2) + 푏 = 0 ⟹ 4푎 + 푏 = −12⋯⋯⋯ (1) Also, 3 = (2) + 푎(2) + 푏(2) + 7 or 3 = 8 + 4푎 + 2푏 + 7 or 3 − 8 − 7 = 4푎 + 2푏 ⟹ 4푎 + 2푏 = −12 (Dividing both sides by 2) We have 2푎 + 푏 = −6⋯⋯⋯ (2)From (1) 푏 = −12 − 4푎.Put 푏 = −12 − 4푎 into (2) That is, 2푎 + (−12 − 4푎) = −6 ⟹ 푎 = −3 and 푏 = −12 − 4(−3) = 0

∴ 푎 = −3and푏 = 0

Example 15.31 The curve 푦 = 푛푥 + 2푥 − 9푥 + 1 has an inflexion where 푥 = 1. Find the value of 푛.

Solution

We have 푦 = 푛푥 + 2푥 − 9푥 + 1 = 3푛푥 + 4푥 − 9 = 6푛푥 + 4

At the stationary point of inflexion = 0

⟹ 6푛푥 + 4 = 0 or 6푛(1) + 4 = 0or6푛 = −4 ∴ 푛 = − = − Example 15.32

The perimeter of a rectangle is 20 units. Find its dimensions so that the area the area is maximised.

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Solution Let 푙푎푛푑푤 be the length and width of a parallelogram respectively: Perimeter, 푃 = 2(푙 + 푤)⟹ 2푏 + 2푠 = 20 (dividing both sides by 2 gives) 푙 + 푤 = 10 or 푙 = 10− 푤 Area, 퐴 = 푙푤 ⟹ 퐴 = (10 −푤)푤or 퐴 = 10푤 − 푤 Now, = 10 − 2푤

At the maximum area = 0 ⟹ 10 − 2푤 = 0 or 2푤 = 10 ⟹ 푤 = = 5 When 푤 = 5, 푙 = 10 − 5 = 3 ∴ the dimensions are 5unitsby5units

Example 15.33 A rectangular sheet of metal is 8cm by 5cm. Equal squares of side 푥푐푚 are removed from each corner and the edges are the turned up to make an open box of volume 푉푐푚 . Show that 푉 = 40푥 − 26푥 + 4푥 and find the maximum possible value of 푥.

Solution (8 − 2푥)

푥 푥 푥 푥 5cm

(5 − 2푥) (5 − 2푥) cm 푥 푥

8 cm 푥 푥 (8 − 2푥) cm

푥cm (5 − 2푥)cm (8− 2푥)cm Volume of the box, 푉 = 푥(5 − 2푥)(8− 2푥) = 푥(40 − 10푥 − 16푥 + 4푥 ) = 40푥 − 26푥 + 4푥

∎ At the maximum volume = 0 ⟹ = 40 − 52푥 + 12푥 = 0 ⟹ 40 − 52푥 + 12푥 = 0 or 3푥 − 13푥 + 10 = 0 or 3푥 − 10푥 − 3푥 + 10 = 0

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푟 푟

or 푥(3푥 − 10)− (3푥 − 10) = 0 or (푥 − 1)(3푥 − 10) = 0 ⟹ 푥 = 1 or 푥 =

= −52 + 24푥 = −52 + 24(1) = −28 = −52 + 24 = 28

Since < 0, the maximum value of 푥 is 1.

Example 15.34 A right circular cone of vertical angle 2휃 is inscribed in a sphere of radius 푎 units. Show that the area of the curved surface of the cone is 휋푎 (sin3휃 + sin휃) and find the value of 휃 at which the area of the curved surface is maximised.

Solution Curved surface area of cone 푃푄푅, 퐴 = 휋푟푙 Q From ∆푃푂푁 sin 2휃 = ⟹ 푎 sin 2휃 = 푟 휃휃 ⟹ 푟 = 푎 sin 2휃 2휃 From ∆푃푄푁 sin 휃 = ⟹ 푙 = 푙 푂 2휃2휃 P N R Now,

퐴 = 휋푟 ∙ = 휋 = 휋 = 휋푎 ( ) = 휋푎 ( ) = 휋푎 4 sin2θcos2휃

sin휃 ⟹ 퐴 = 휋푎 [4sin휃 cos 휃] = 휋푎 [4sin휃(1 − sin 휃)] = 휋푎 [4sin휃 − sin 휃] ⟹ 퐴 = 휋푎 [4sin휃 − (3 sin 휃 − sin 3휃)] = 휋푎 [4sin휃 − 3 sin 휃 + sin 3휃] = 휋푎 (sin3휃 + sin휃) ∎ [Recallthat sin 3휃 = 3 sin휃 − 4 sin 휃or4 sin 휃 = 3 sin 휃 − sin 3휃]

= 휋푎 (3 cos 3휃 + cos휃) At the maximum curved surface area = 0 That is, 휋푎 (3 cos 3휃 + cos휃) = 0 (Dividing both sides by 휋푎 we have) 3 cos 3휃 + cos휃 = 0 ⟹ 3(4 cos 휃 − 3 cos휃) + cos휃 = 0 or 12 cos 휃 − 8 cos휃 = 0 or 4 cos휃(3cos 휃 − 2) = 0 ⟹ 4 cos휃 = 0 ⟹ 휃 = cos ( 0) = 90°or3 cos 휃 − 2 = 0

⟹ cos휃 = ± ⟹ 휃 = cos = 35.2644° or

휃 = cos − = 144.7356° = 휋푎 (−9 sin 3휃 − sin휃)

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푑 퐴푑휃

°= 휋푎 (−9 sin 3(90°)− sin 90°) = 8휋푎

푑 퐴푑휃

. °≈ 휋푎 (−9 sin 3(35.2644°)− sin 35.2644°) ≈ −9.2376휋푎

푑 퐴푑휃

. °≈ 휋푎 (−9 sin 3(144.7356°)− sin 144.7356°) ≈ −9.2376휋푎

Since

. °< 0 and

. °< 0, the value of 휃 is either 35.2644° or

144.7356° Try: 1. A ball is thrown vertically upwards from the ground level and its height after 푡 seconds is given by ℎ = 2푡 − 푡 , where ℎ is measured in meters. Find the greatest height the ball reaches and the time it takes to reach there. 2. A farmer has 60푚 length of wire for fencing a rectangular plot against an existing

wall which provides the fourth side. Find the maximum area he can enclose the sides of the rectangular plot.

c. curve sketching Let 푦 = 푓(푥). To sketch y, we need:

Coordinates of the 푥- and 푦 intercepts. The gradient function and coordinates of the turning point(s). The nature/behaviour of the curve using the second derivative test.

Example 15.35 Sketch the curve 푦 = 2푥 − 3푥 − 5 indicating the turning point. Hence, state the range value of 푥 for which the curve is increasing.

Solution We have: 푦 = 2푥 − 3푥 − 5 = 4푥 − 3 Intercepts: For 푥 −intercepts, let 푦 = 0

378

⟹ 2푥 − 3푥 − 5 = 0 or 2푥 − 5푥 + 2푥 − 5 = 0 푦 or (2푥 − 5)(푥 + 1) = 0 ⟹ 푥 = or 푥 = −1 10

푃 (−1, 0)푃 , 0 For 푦 −intercepts, let 푥 = 0 5 ⟹ 푦 = 2(0) − 3(0) − 5 = −5 푃 (0,−5) 푦 = 2푥 − 3푥 − 5 Turning point: At the turning point = 0 −4 −2 2 4 푥

⟹4푥 − 3 = 0 푥 = When 푥 = , −5 푇푃 ,−

푦 = 2 − 3 − 5 = − − 5 = = − −10

푇푃 ,−

= 4 > 0. The turning point is minimum. A푠푥 → ±∞,푦 → ∞ Example 15.36

Sketch the curve 푓(푥) = 3 + 2푥 − 푥 and state the range of values of 푥 for which the curve is decreasing.

Solution We have: 푓(푥) = 3 + 2푥 − 푥 푓 (푥) = = 2 − 2푥 Intercepts: For 푥 −intercepts, let 푓(푥) = 0 ⟹ 3 + 2푥 − 푥 = 0 or 3 − 푥 + 3푥 − 푥 = 0 푓(푥) or (3 − 푥)(1 + 푥) = 0 ⟹ 푥 = 3 or 푥 = −1 4 푃 (−1, 0)푃 (3, 0) 푇푃(1, 4) For 푓(푥) −intercepts, let 푥 = 0 2 ⟹ 푓(푥) = 3 + 2(0)− (0) = 3 푃 (0, 3) 푓(푥) = 3 + 2푥 − 푥 Turning point: At the turning point = 0 −4 −2 2 4 푥 ⟹ 2 − 2푥 = 0 푥 = 1 When 푥 = 1, −2 푓(푥) = 3 + 2(1)− (1) = 3 + 2 − 1 = 4 푇푃(1, 4) −4

= −2 < 0. The turning point is a maximum. A푠푥 → ±∞,푦 → −∞

Example 15.37 Sketch the curves 푓(푥) = 푥 − 푥 and 푔(푥) = 푥 − 푥 in the same axes and find their points of intersection.

379

Solution Take: 푓(푥) = 푥 − 푥 푓 (푥) = = 1 − 2푥 푦 Intercepts: For 푥 −intercepts, let 푓(푥) = 0 ⟹ 푥 − 푥 = 0 or 푥(1 − 푥) = 0 2 ⟹ 푥 = 0 or 푥 = 1 푃 (0, 0) 푃 (1, 0) 푔(푥) 1 푇푃 , For 푓(푥) −intercepts, let 푥 = 0 ⟹ 푓(푥) = 0 푃 (0, 0) -2 -1 1 2 Turning point: At the turning point 푓 (푥) = 0 -1 푇푃 ,− 푓(푥) ⟹ 1 − 2푥 = 0 or 푥 = When 푥 = , -2

푓(푥) = − = − = 푇푃 , 푓 (푥) = −2 < 0. The turning point is a maximum. A푠푥 → ±∞,푦 → −∞ Now, take: 푔(푥) = 푥 − 푥 푔 (푥) = = 2푥 − 1Intercepts: For 푥 −intercepts, let 푔(푥) = 0⟹ 푥 − 푥 = 0 or 푥(푥 − 1) = 0⟹ 푥 = 0 or 푥 = 1 푃 (0, 0) 푃 (1, 0) For 푔(푥) −intercepts, let 푥 = 0 ⟹ 푔(푥) = 0 푃 (0, 0) Turning point: At the turning point 푔 (푥) = 0⟹ 2푥 − 1 = 0 or 푥 =

When 푥 = ,푔(푥) = − = − = − 푇푃 ,− 푔 (푥) = 2 < 0. The turning point is a minimum. A푠푥 → ±∞,푦 → ∞

Example 15.38 Sketch the curve 푦 = 2푥 + 푥 − 4푥 + 1.

Solution We have 푦 = 2푥 + 푥 − 4푥 + 1 = 6푥 + 2푥 − 4 Intercepts: For 푥 −intercept, 푦 = 0 ⟹ 2푥 + 푥 − 4푥 + 1 = 0 or (푥 − 1)(2푥 + 3푥 − 1) = 0 or (푥 − 1)(푥 − 0.2808)(푥 + 1.7808) = 0 ⟹ 푥 = 1 or 푥 = 0.2808or 푥 = −1.7808

푃 (−1.7808, 0) 푃 (0.2808, 0) 푃 (1, 0) For 푦 −intercept, let 푥 = 0 ⟹ 푦 = 2(0) + (0) − 4(0) + 1 = 1 푃 (0, 1)

380

Turning point: At the turning point = 0 6푥 + 2푥 − 4 = 0 (Dividing both sides by we have) 3푥 + 푥 − 2 = 0 or 3푥 − 2푥 + 3푥 − 2 = 0 or (3푥 − 2)(푥 + 1) = 0 ⟹ 푥 = or 푥 = −1

When 푥 = , 푦 = 2 + − 4 + 1 = + − + 1 = = − When 푥 = −1 푦 = 2(−1) + (−1) − 4(−1) + 1 = −2 + 1 + 4 + 1 = 4

푇푃 (−1, 4)푇푃23 ,−

1727

= 12푥 + 1 = −12 + 1 = −11 < 0 = 12 + 1 = 9 > 0

The turning point 푇푃 (−1, 4) is a maximum. The turning point 푇푃 ,− is a minimum.

푦 푇푃 (−1, 4) 4 2 푦 = 2푥 + 푥 − 4푥 + 1. -2 -1 1 2 푥 -2 푇푃 ,− -4

Example 15.39 Sketch the curve 푦 = −푥 − 5푥 − 6푥 indicating the turning points and their nature.

Solution We have 푦 = −푥 − 5푥 − 6푥 = −1 − 10푥 − 18푥 Intercepts: For 푥 −intercept, 푦 = 0 ⟹ −푥 − 5푥 − 6푥 = 0 or −푥(1 + 5푥 + 6푥 ) = 0 or −푥(1 + 2푥 + 3푥 + 6푥 ) = 0 or −푥(3푥 + 1)(2푥 + 1) = 0 ⟹ 푥 = 0 or 푥 = − or 푥 = − 푃 − , 0 푃 (0, 0) 푃 − , 0 For 푦 −intercept, let 푥 = 0 ⟹ 푦 = 0 푃 (0, 0)

381

Turning point: At the turning point = 0 ⟹ −1 − 10푥 − 18푥 = 0 or 18푥 + 10푥 + 1 = 0 The values of 푥 can be determined by the formula:

푥 =−푏± 푏2−4푎푐

2푎 ⟹ 푥 =−10± (10)2−4(18)(1)

2(18)= −10±√28

36 = −10±2√736

⟹ 푥 = √ ≈ −0.1308 or 푥 = √ ≈ −0.4248 When 푥 ≈ −0.1308, 푦 ≈ −(−0.1308) − 5(−0.1308) − 6(−0.1308) ≈ 0.0587 When 푥 ≈ −0.4248, 푦 ≈ −(−0.4248) − 5(−0.4248) − 6(−0.4248) ≈ −0.0175

푇푃 (−0.4248,−0.0175)푇푃 (−0.1308, 0.0587)

= −10 − 36푥 .

= −10 − 36(−0.4248) ≈ 5.2928 > 0

푑 푦푑푥

.= −10− 36(−0.1308) ≈ −5.2912 < 0

The turning point 푇푃 (−0.4248,−0.0175) is a minimum. The turning point 푇푃 (−0.1308, 0.0587) is a minimum. 푦 0.6 0.3 푇푃 (−0.13, 0.06)

-0.6 -0.3 0.3 0.6 푥 푇푃 (−0.42,−0.02) -0.3 푦 = −푥 − 5푥 − 6푥 -0.6

Example 15.40

vi) Sketch the curve 푓(푥) = 푥 . Solution

We have 푓(푥) = 푥 푓 (푥) = 3푥

382

Intercepts: For 푥 −intercept, 푓(푥) = 0 푓(푥) ⟹ 푥 = 0 or 푥 = 0 푃 (0, 0) For 푓(푥)−intercept, let 푥 = 0 ⟹ 푓(푥) = 0 푃 (0, 0) = 푃 푓(푥) = 푥 Turning point: At the turning point 푓 (푥) = 0 ⟹ 2푥 = 0 ⟹ 푥 = 0 When 푥 = 0, 푦 = (0) 푇푃(0, 0) = 푃 = 푃 푓 (푥) = 6푥 푓 (푥)| = 0 The turning point 푇푃(0, 0) is an inflexion point

As푥 → ∞,푦 → ∞and푎푠푥 → −∞,푦 → −∞

d. Small changes Let 푦 = 푓(푥) and ∆푥 be a small change in 푥 such that the point (푥 + ∆푥, 푦 + ∆푦) lies on the curve 푦 = 푓(푥). Then the small change in 푦, ∆푦, as a result of ∆푥 is given by: 푦 + ∆푦 = 푓(푥 + ∆푥) or ∆푦 = 푓(푥 + ∆푥)− 푦 ⟹ ∆푦 = 푓(푥 + ∆푥)− 푓(푥). This can also be calculated from the fact that as

∆푥 → 0 ≈ ⟹ ∆푦 ≈ ∙ ∆푥

Example 15.41 Given that 푦 = 4푥 − 푥,푥 = 3 and ∆푥 = 0.2. Find ∆푦, where ∆푥 and ∆푦 are the small changes in 푥 and 푦 respectively.

Solution We have: 푦 = 4푥 − 푥, 푥 = 3 and ∆푥 = 0.2 ⟹ 푦 + ∆푦 = 4(푥 + ∆푥) − (푥 + ∆푥) or ∆푦 = 4(푥 + ∆푥) − (푥 + ∆푥)− 푦 ⟹ ∆푦 = 4푥 + 8푥∆푥 + (∆푥) − 푥 − ∆푥 − 4푥 + 푥 or ∆푦 = 8푥∆푥 + (∆푥) − ∆푥 ∴ ∆푦 = 8(3)(0.2) + (0.2) − 0.2 = 4.64 Alternatively, We have: 푦 = 4푥 − 푥, 푥 = 3 and ∆푥 = 0.2 = 8푥 − 1

As ∆푥 → 0 ≈ ⟹ ∆푦 ≈ ∙ ∆푥 ∴ ∆푦 ≈ (8푥 − 1)(∆푥) ≈ (8(3) − 1)(0.2) ≈ 4.6

Example 15.42 If 푝 = 푥 − 3 and 푥 increases from 3 to 3.05, find the approximate change in the value of 푝.

383

Solution We have: 푝 = 푥 − 4,푥 = 3and∆푥 = 3.05 − 3 = 0.05

⟹ 푝 + ∆푝 = (푥 + ∆푥) − 4 or ∆푝 = (푥 + ∆푥) − 4 − 푝

⟹ ∆푝 = 푥 + 푥∆푥 + (∆푥) − 4 − 푥 + 4 or ∆푝 = 푥∆푥 + (∆푥)

∴ ∆푝 = (3)(0.05) + (0.05) = 0.3525 Alternatively, We have: 푝 = 푥 − 4,푥 = 3and∆푥 = 3.05 − 3 = 0.05 = 푥

As ∆푥 → 0 ≈ ⟹ ∆푦 ≈ ∙ ∆푥

∴ ∆푦 ≈ 푥(∆푥) ≈ (3)(0.05) ≈ 0.35 Example 15.43

That radius of a cylinder increased by 0.2 cm. If the radius of the cylinder is 3.5 cm and the height is 10 cm, find the in terms of 휋 the increased in the volume of the cylinder.

Solution The volume of a cylinder is given by: 푉 = 휋푟 ℎ But ℎ = 10 ⟹ 푉 = 10휋푟 = 20휋푟

As ∆푟 → 0 ≈ ⟹ ∆푉 ≈ ∙ ∆푟,where ∆푟 = 0.2 ∴,∆푉 ≈ 10휋푟 ∆푟 ≈ 10휋(3.5) (0.2) ≈ 24.5휋

Example 15.44

Find the approximate error in using sin 휃 instead of sin(휃 + 훼), where 훼 is a small change.

Solution Let 푦 = sin휃 푦 + ∆푦 = sin(휃 + 훼) and ∆휃 = 훼 = cos휃

As ∆휃 → 0 ≈ ⟹∆푦 ≈ ∙ ∆푟 ∴ ∆푦 ≈ 훼 cos휃 Example 15.45

Given the sides 푎, 푏 of a triangle and an included angle C, its area is calculated. Prove that the area of the triangle is 푎푏 sin C and find the approximate change in this area if (C + ∆C) is the true angle atC.

384

Solution The area of a triangle is given by: 퐴 = 푏푎푠푒 × ℎ푒푖푔ℎ푡 Now, consider the triangle below: A From triangle 퐴퐵퐶 sin 휃 = ℎ = 푎 sin훼 ℎ 푏

퐴 = 푏푎푠푒 × ℎ푒푖푔ℎ푡 B

= 푏 × 푎 sin 휃 푎 C

= 푎푏 sin C ∎

Now, 퐴 = 푎푏 sin C = 푎푏 cos C

As ∆C → 0 ≈ ⟹∆퐴 ≈ ∙ ∆C ∴ ∆퐴 ≈ 푎푏 ∆Ccos C Try: Find the approximate error in using cos휃 instead of cos(휃 + 훼), where 훼 is a small change.

e. rate of change The rate of change in any quantity refers to the rate at which that quantity changes with time. For example, the rate of change in radius 푟 is , the rate of change in area 퐴 is

and the rate of change of change in volume 푉 is . To find the rate of change in a quantity with respect to time 푡, it may be helpful to apply the chain rule. For instance the rate of change in volume 푉 can be: = ∙ ∙

Example 15.46

The radius of a circle increases at a rate of 0.7 cms . Find the rate at which the area is increasing if the radius of the circle is 0.3cm.

Solution We have = 0.7cms 푟 = 0.3.

The area of a circle is given by: 퐴 = 휋푟 ⟹ = 2휋푟 or = 2휋(0.3) = 0.6휋 The rate of increase in the area can be calculated as:

385

= ∙ = 0.6휋(0.7) = 0.42cm s

Example 15.47 The circumference of a circle is 25mm. If the radius of the circle is increasing at a rate of 5mms , find the radius of the circle and the rate at which the area is increasing. [푇푎푘푒휋 = 3.142]

Solution The circumference of a circle is given by: 퐶 = 2휋푟 ⟹ 25 = 2(3.142)푟 ⟹ 푟 =

( . )≈ 3.9784

The area of a circle is given by: 퐴 = 휋푟 ⟹ = 2휋푟 = 퐶 or = 25 The rate of increase in the area can be calculated as:

= ∙ = 25(5) = 125mm s Example 15.48

The radius of a cone increases at rate of 0.6푐ms . Half the height is the radius and the volume is 18휋cm . Find the rate at which the volume increases.

Solution We have = 0.7ms 푟 = ℎ 푉표푙푢푚푒,푉 = 18휋cm

The volume of a cone is given by: 푉 = 휋푟 ℎ

⟹ 푉 = 휋 ℎ ℎ or 푉 = 휋ℎ ⟹ = 휋ℎ . Also =

Now, 18휋 = 휋ℎ ⟹휋ℎ = 18휋 × 12 (Dividing both sides by 휋wehave)

ℎ = 216 ∴ ℎ = √216 = 6. Hence, 푟 = (6) = 3 The rate of increase in the volume can be calculated as: = ∙ ∙ = 휋ℎ (2)(0.6) = 휋(6) (2)(0.6) = 10.8휋cm s

Example 15.49 If water is poured into an inverted hollow cone whose semi-vertical angle is60° so that its depth increases at the rate of 25ms . Find correct to the nearest meter the rate at which the volume of water is increasing if the depth is 2m.

386

Solution We have = 25ms tan 60° =

⟹ ℎ tan 60° = 푟 ∴ 푟 = 2 tan 60° = 2√3

Volume,푉 = 푟 ℎ = 2√3 ℎ = 4ℎ ⟹ = 8ℎ 퐻

= ∙ = 8ℎ(25) = 8(2)(25) = 400m s 푟 60° ℎ

f. percentage rate of change The percentage rate of change compares the small change in a quantity with the size of that quantity and the result is express as a percentage. If 푦 = 푓(푥), then the percentage rate of change in 푦 is given by:

푃 = ∆ × 100%

Note: as ∆푥 → 0 ≈ ⟹ ∆푦 ≈ ∙ ∆푥 Example 15.50

Find the percentage error in volume if an error of 0.5% is made in increasing the radius of a sphere.

Solution We have, ∆ × 100% = 0.5% ⟹ ∆ = . = 0.005 ⟹ ∆푟 = 0.005푟

The volume of a sphere is defined as: 푉 = 휋푟 ⟹ = 4휋푟

as ∆푟 → 0 ≈ ⟹ ∆푉 ≈ ∙ ∆푟 ⟹ ∆푉 = 4휋푟 (0.005푟) = 0.02휋푟

Now, 푃 = ∆ × 100% = .

× 100% = 0.02 × × 100% = 1.5%

Example 15.51 The radius of a sphere decreases by 6%. Find the percentage decrease in the surface area of the sphere.

Solution We have, ∆ × 100% = −6% ⟹ ∆ = − = −0.06 ⟹ ∆푟 = −0.06푟

The surface area of a sphere is defined as: 푉 = 4휋푟 ⟹ = 8휋푟

387

as ∆푟 → 0 ≈ ⟹ ∆퐴 ≈ ∙ ∆푟 ⟹ ∆퐴 = 8휋푟(−0.06푟) = −0.48휋푟

Now, 푃 = ∆ × 100% = .

× 100% = −012.× 100% = −12% Try: i. The radius of a circle increases by 0.25%. Find the percentage increase in the area of the sphere. ii. The radius of a circle diminishes by 3.2%. Find the percentage decrease in the volume of the sphere.

g. Differential kinematics kinematics deals with the study of a body’s displacement(푠), velocity(푣) and acceleration(푎).

Example 15.52 The displacement of a body relative to 푂 is given by 푠 = 푡 − 푡 − 푡 − 1 a. i. Find the velocity of the body. ii. Find the value of 푡 at which the velocity is maximised. b. Find the acceleration of the body.

Solution We have, 푠 = 푡 − 푡 − 푡 − 1 a. i. The velocity of the body: 푣 = = 3푡 − 2푡 − 1.

ii. At the maximum velocity = 0 = 6푡 − 2. ⟹ 6푡 − 2 = 0

⟹ 푡 = = . ∴, the maximum velocity occurs at time 푡 = second.

b. i. The acceleration of the body: 푎 = = 6푡 − 2 = 2(3푡 − 1) Final Exercises

1. Evaluate the following:

i. 422

0lim

xxx

ii. 13

1lim

xx

iii. 2

5221lim xx

x

iv. 34 2

0lim

xx

v. 62

6lim

xx

vi. 1123

4lim

yyy

2. Evaluate the following:

388

i.

xx

x

cos1lim

0 ii.

xx

x 5tan3sin

lim0

iii.

xx

x 3sin2cos

lim0

iv.

xx

x 2sin12sin1

lim0

v.

xxx

x cos2sin1

lim0

vi.

3cos3

lim0 ec

3. . Evaluate the following:

i.

1532

1lim x

xx

x ii.

1

62

2

3lim

xx

x iii.

xxxx

x 2

432

2

5lim

iv.

112

1lim x

x

x

v.

xx

x 93

lim9

vi.

2652

2lim x

xx

x

4. Evaluate the following:

i. 14lim

xx

ii. xxx

22 2lim

iii. xxx

53lim

iv. 25lim xxx

v. 42lim xxx

vi. 1123lim

yyy

5. . Evaluate the following:

i.

xx

xx

x 2

2 124lim ii.

4

23

1

432lim

x

xxx

x

iii.

1327

lim xx

x iv.

2

2

1

1lim

x

x

x

v.

xx

xx

x 2

23 2lim

vi.

13

92

2lim

xx

xx

x

6. Differentiate the following functions from the first principles. i.푦 = 푥 + 2푥 ii.푓(푥) = 푥 + 푥 iii.푓(푥) = 1 − 푥 − 푥 iv. 푔(푥) = (푥 + 1) v. 푔(푥) = 1 − 2푥 vi. 푦 =

vii.푓(푥) = vii. 푦 = 푥√푥 viii. 푦 = 7. Find the derivative of the following functions

389

i.푦 = 푥 + 2푥 ii.푦 = 4푥 − 푥 − 푥 iii.푔(푥) = 1 − 3푥 iv. 푓(푥) = (푥 − 3) v. 푔(푥) = + 푥 vi. 푦 = + 6푥

vii.푓(푥) =√

+√푥 vii.푓(푥) = √푥

8.Differentiatethe following with respect to 푥. i.푦 = (푥 + 2)(푥 + 3) ii.푦 = (푥 + 1)(푥 − 1)

iii.푦 = (3푥 + 1)(2푥 − 6) iv.푦 =√

(푥 +1)

v. 푦 = (3푥 + 1)(2푥 + 6) vi. 푦 = (푥 + 3푥 + 1)

9.Differentiate the following with respect to 푥.

i.푦 = ii.푦 = iii.푓(푥) = vi. 푔(푥) =√

v.푦 = vi.푓(푥) vii. ℎ(푥) = 10. Differentiate the following functions with respect to 푥.

i. 푦 = (푥)(푥 + 3) ii. 푦 =( )

iii.푔(푥) =( )

vi.푦 = (3푥 − 4) vii. 푓(푥) = 2푥 (2푥 + 7) viii.푦 = (1 − 6푥) ix.푦 = (푥 + 2푥 )(푥 + 3) vii.푦 =

( )

11. Differentiate the following with respect to 푥. i.ln(3푥 + 1) ii.푦 = 푒( ) iii.푦 = sin(푥 + 3푥) iv. 푓(푥) = tan

v.푓(푥) = sec(3푥) vi.푓(푥) = cos 2푥 −√

vii.푔(푥) = 푒( )

12. Find the equation of tangent to the following curves at point P (-2, 3). i.푦 = 4푥 ii.푦 = 푥 − 3푥 iii.푦 = 2푥 + 6푥 + 7 iv.푦 = 3푥 v.푦 = 4푥 + 2푥 − 푥 vi.푦 = 푥 − 푥 vii. 푦 = 1 − 2푥 − 푥 − 푥 13. Find if i.푥 + 푦 + 푥푦 = 0 ii.(푥 + 2푦) +푦 = 푥 iii.2푥 + 3푥 + 3푥푦=푥 iv.(2푥 + 푥푦) + 푥 = 푦 v.푥푦 + 푦 = 2푥푦 vi.푥 + + 3 = 푥푦

14. Find the equation of the tangent to the following curves at the point 푃(−2, 3)

i.푦 = 4푥 ii.푦 = 푥 − 3푥 iii.푦 = 2푥 + 6푥 + 7 iv.푦 = 3푥 v.푦 = 4푥 + 2푥 − 푥 vi.푦 = 1 − 2푥 − 푥 − 푥

390

15. Find the equation of normal to the following curve at point Q(−3, 2) i.푦 = 푥 − 푥 ii.푦 = 3푥 iii.푦 = 1 − 푥 iv.푦 = 1 − 2푥 − 3푥

v.푦 = vi.푦 = 2푥 + 2푥 + 3푥

16. If푦 = 1 − 푡 and푥 = 푡 + 2, ind

17. If푦 = 푡 − 푡 and푥 = 푡 − 2, ind

18. If푦 = 2푡 and 푥 = 푡 − 4, ind 19. Find the equations of tangent and normal to the curve 푦 = 3푥 + 2푥푎푡퐴(−1,−2) 20. Find the equation of tangent and normal to the curve 푦 = 푥(푥 − 3) atpointwherethecurvecutthe푥 −axis 21. Find the equation of tangent to the curve 푦 = 푥 − 4푥 + 5 atthepointofintersectionbetweenthecurveandline 푦 = 3 − 푥 22. Determine the point of intersection between the curve 푦 = 1 − 2푥 − 3푥 and the y axis. Hence, find the equation of the normal to the curve at this point. 23. Find the range of values of y for which i. The function of 푦 = 1 − 2푥 − 푥 isdecreasing ii. The function f(푥) = 2푥 − 푥 + 3 isincreasing iii. The function g(x) =1 − 4푥 = 2푥 − 푥 is decreasing iv. The function =푓(푥) = 1 − 3푥 + 2푥 − 4푥 isincreasing v.푦 = 푥(푥 − 1)(푥 − 2)(푥 − 3) isincreasing 22. Find the stationary value of the curve푦 = 2푥 − 3푥 + 6 23. Find the stationary point of the curve 푦 = 3푥 + 5푥 − 6 24. Find the turning point of the following curves (i)푦 = 6푥 + 푥 − 6 (ii)푦 = 1 − 3푥 − 2푥 (iii)푔(푥) = 7푥 − 5푥 − 3 (iv)푓(푥) = 3 − 푥 − 2푥 25. Determine the turning point of the following curves (i)푦 = 1 − 3푥 − 2푥 − 푥 (ii)푦 = 2푥 + 3푥 − 5푥 − 6 (iii)푓(푥) = 3− 푥 − 2푥 − 푥 (iv)푔(푥) = 3푥 − 푥 − 6 26. Determine the maximum value of the curves (i)푦 = 1 − 3푥 − 4푥 (ii)푦 = 10 − 푥 − 푥

391

27. The relationship between the height and width of a parallelogram is ℎ = − 푏. Find the value of b and h which will maximize the area. 28. The relationship the cost of a manufacturing item c and the quantity of item produced is given as 퐶 = 1 + 2푥 − 4푥 . (i)Find; the value of 푥 that gives the minimum cost. (ii)The minimum cost. 29. The perimeter of a rectangle is 32m find the possible values of the width and length that would give a maximum area. 30. The profit function of a firm producing an item 푥 is defined as 푝(푥) = −1 = 2푥 + 푥 − 4푥 . Find (i)The values of 푥 at which the profit is maximum. (ii)The value of 푥that will maximize profit. 31. Sketch the following curves. (i)푦 = 푥 − 11푥 + 10 (ii) 푦 = 2푥 − 푥 (iii)푦 = 1 − 푥 − 6푥 (iv)푦 = 1 − 2푥 − 3푥 (v) 푓(푥) = 6푥 − 푥 − 푥 32. If 푦 = 푥 − 4푥, 푥 = 1and ∆푥 = 0.2 ind∆푦. 33. Find the small charge in y if 푥 increase from 2.5 to 2.75 and 푦 = 푥 − 2푥 + 1.

34. If 푞 = 푥 − and 푥 increase s by 0.03. Find the approximate change in the value of 푞. 35. Find the approximate error in using sin 휃 instead of sin(휃 + 푟),푤ℎ푒푟푒푟푖푠푎푠푚푎푙푙푐ℎ푎푛푔푒. 36. The radius of a sphere increases at a rate of 0.3m푠 ,Find the rate at which the volume is increasing if the radius of the circle is 3.5m. 37. The radius of a cone increase at a rate of1.5cms .Find the rate at which the volume is increasing if the height is 10cm. Take휋 =

38. Find if

(i)푦 = 푥 sin 2푥 (ii) 푦 = (iii) 푦 = (푥 + sin 푥)

(iv)푔(푥) = (1 − cos 2푥) (v)푓(푥) = 39. Find the percentage error in the volume if an error of 0.75% is made in increasing the radius of a sphere.

392

40. The radius of a circle increases by 7%.Find the percentage increase in the area of the semi-circle. 41. The radius of a sphere diminishes by 2.2%.Find the percentage decrease in the volume of a sphere. 42. The displacement of a body relative to 푂 is given by 푠 = 2푡 − 7푡 − 3푡 − 1. a. i.Find the velocity of the body. ii. Find the value of t at which the velocity is maximised. b. Find the acceleration of the body. 43. The displacements of a body at time t is given by 푠 = 1 − 20푡 − 푡 . Find the velocity of the body and the value of t which the body velocity is maximised. 44. The velocity of a body at time t is given by 푣 = 1 − 2푡 −3푡 Find the acceleration of the body at time t and the maximum velocity. 45. If 푦 = 푥 − 4푥, 푥 = 1 and ∆푥 = 0.2. Find ∆푦. 46. Find the small change in 푦 if 푥 increases from 2.5 to 2.75 and 푦=푥 − 2푥 + 1

47. If 푞 = 푥 − and 푥 increases by 0.03. Find the approximate change in the value of 푞. 48. Find the approximate error in using cos휃 instead of cos(휃 + 훾) where 훾 is a small change. 49. The radius of a sphere increases at a rate of 0.3ms , Find the rate at which the volume is increasing if the radius of the circle is 3.5m 50. The radius of a cone increases at a rate of 1.5cms . Find the rate at which the volume is increasing if the height is 10cm.

[푇푎푘푒:휋 = ]

51. Find the if Take :

(i) 푦 = 푥 푠푖푛2푥 (ii) 푦 = (iii) 푦 = (푥 + sin 푥)

(iv) 푓(푥) = (v) 푔(푥) = (1 − cos 2푥) 52. Find the percentage error in volume if an error of 0.75% is made in increasing the radius of a sphere. 53. The radius of a circle increases by 5%. Find the percentage increase in the

393

area of the semi-circle. 54. The radius of a sphere diminishes by 2.2%. Find the percentage decrease in the volume of the sphere. 55. The displacement of a body relative to 푂 is given by 푠 = 푡 − 푡 − 푡 − 1 a. i. Find the velocity of the body. ii. Find the value of 푡 at which the velocity is maximised. b. i. Find the acceleration of the body. ii. Find the value of 푡 at which the acceleration is maximised. 56. The displacement of a body at time 푡 is given by 푠 = 1 − 2푡 − 푡 Find the velocity of the body and the value of 푡 at which the velocity is maximised. 57. The velocity of a body at time 푡 is given by 푣 = 1 − 2푡 − 3푡 . Find the acceleration of the body and the maximum velocity.

394

INTEGRATION

Integration refers to the reverse of the process of differentiation. Let 푦 = 푓(푥). Then = 푓 (푥) ⟹ 푑푦 = 푓 (푥)푑푥 ⟹ ∫ 푑푦 = ∫푓 (푥)푑푥 Here, our goal is to get our function back. The symbol ,∫,istheintegralsignandthefunction, 푓 (푥), to be integrated is called the integrand. 푑푥 means that we are integrating with respect to the variable 푥. ∫ 푓 (푥)푑푥 is read ‘ The integral of 푓 (푥) d x’.

Indefinite integrals

In general ∫ 푎푥 푑푥 = + 푐, where 푐 is arbitrary constant called the constant of integration and 푛 ≠ 1. Also∫푎푑푥 = 푎푥 + 푐 and∫푎푓(푥)푑푥 = 푎 ∫ 푓(푥)푑푥 + 푐, where 푎 is constant.

Example 16.1 Integrate the following functions with respect to 푥.

i. = 4푥 ii. 푓 (푥) = 3푥 + 1 iii. 푔 (푥) = 푥 − 푥 iv. =

v. ℎ (푥) = √푥 −√

Solution i. We have, = 4푥 ⟹ 푑푦 = 4푥 푑푥 ⟹ ∫푑푦 = ∫4푥 푑푥

⟹ 푦 = 4 ∙ + 푐 = 4 ∙ + 푐 = 푥 + 푐 ∴ ∫ 4푥 = 푥

ii. We have, 푓 (푥) = 3푥 + 1 ⟹ 푓(푥) = ∫(3푥 + 1) = 푑푥 = ∫ 3푥 푑푥 + ∫ 푑푥

⟹ 푓(푥) = 3 ∙ + + 푐 = 3 ∙ + 푥 + 푐 = 푥 + 푥 + 푐 ∴ ∫3푥 + 1 = 푥 + 푥 + 푐

iii. We have, 푔 (푥) = 푥 − 푥 ⟹ 푔(푥) = ∫ 푥 − 푥 푑푥 = ∫푥푑푥 − ∫ 푥 푑푥

⟹ 푔(푥) = − ∙ + 푐 = − ∙ + 푐 = − 푥 + 푐

∴ ∫푥 − 푥 = − 푥 + 푐

iv. We have, = = + − = 5푥 + 3푥 − 1

⟹ 푑푦 = (5푥 + 3푥 − 1)푑푥 ⟹ ∫푑푦 = ∫ 5푥 푑푥 + 3푥푑푥 − 푑푥

395

⟹ 푦 = 5 ∙ + 3 ∙ − + 푐 = 5 ∙ + 3 ∙ − 푥 + 푐 = 푥 + 푥 − 푥 + 푐

∴ ∫ = 푥 + 푥 − 푥 + 푐

v. We have, ℎ (푥) = √푥 −√

=2132 2x x ⟹ ℎ(푥) =

2132( 2 )x x dx

⟹ ℎ(푥) = 2132 11

1 22 3

21 1

x x c

= 3 1

32

3 132

2x x c = 3 12 23 6

2x x c

∴ ∫√푥 −√

=2132 2x x =

3 12 23 6

2x x c

Try:

Integrate the function with respect to 푥. =

Determination of constant of integration

If 푦 = ∫ 푎푥 = + 푐,then we can determine the value of 푐 if we know the value of 푥 and 푦.

Example 16.2 The gradient of a curve is given as 3푥 − 1. If the curve passes through the point (−2, 3), find the equation of the curve.

Solution We have, = 3푥 − 1 ⟹ 푑푦 = (3푥 − 1)푑푥 ⟹ 푦 = 푥 − 푥 + 푐 Let 푃(푥, 푦) = (−2, 3). Since the curve passes through the point 푃, then 3 = (−2) − (−2) + 푐 ⟹ 3 = −8 + 2 + 푐 ⟹ 푐 = 3 + 8 − 2 = 9 Therefore, the equation of the curve is 푦 = 푥 − 푥 + 9

Example 16.3

The gradient of a curve is given as 1 − 2푥 − 3푥 . If 1 is a root of the equation, find the equation of the curve.

Solution We have, = 1 − 2푥 − 3푥 ⟹ 푑푦 = (1 − 2푥 − 3푥 )푑푥 ⟹ 푦 = 푥 − 푥 − 푥 + 푐. At root 푥 = 1, 푦 = 0. ⟹ 0 = 1 − 1 − 1 + 푐 ⟹ 0 = 1 − 1 − 1 + 푐 ⟹ 푐 = 1

396

Therefore, the equation of the curve is 푦 = 푥 − 푥 − 푥 + 1

Example 16.4 if 푓 (푥) = 푥 − 푥 is the gradient of 푓(푥), find 푓(푥) if 푥 = −2 is a root of 푓(푥)

solution We have, 푓 (푥) = 푥 − 푥 ⟹ 푓(푥) = ∫ 푥 − 푥 푑푥 ⟹ 푓(푥) = 푥 − 푥 + 푐

At root 푥 = −2, 푓(푥) = 0 ⟹ 0 = (−2) − (−2) + 푐 ⟹ 0 = 4 + + 푐

⟹ 푐 = −4− = −

Therefore, the equation of the curve is 푓(푥) = 푥 − 푥 − Try: The gradient of a curve at any point 푃(푥, 푦) on it is given as 6푥 − 4. If (1, 3) lies on the curve , find the equation of the curve. Integration by substitution (Change of variables) Let = 푔(푥). Then 푑푦 = 푔(푥)푑푥 ⟹ 푦 = ∫푔(푥)푑푥 + 푐. If we let 푥(the argument of the function 푔) be a function 푢, we can work our way back to get our function in terms of 푥. By using the chain rule: = ∙ ⟹ = 푔(푥) ∙

⟹ 푑푦 = 푔(푥)푑푥 ⟹ 푦 = ∫ 푔(푥)푑푥 Now, consider the following: 1. ∫ 푓(푎푥 + 푏)푑푥 . If we let 푢 = 푎푥 + 푏, where 푎and 푏 are constant, then = 푎

⟹ 푑푢 = 푎푑푥 and 푑푥 = ⟹ ∫ 푓(푎푥 + 푏)푑푥 = ∫ 푓(푢) = ∫푓(푢)푑푢 + 푐

2. ∫ 푎[푓(푥)] 푓 (푥)푑푥. If we let 푢 = 푓(푥), then = 푓 (푥) ⟹ 푑푢 = 푓 (푥)푑푥

and 푑푥 =( )

⟹ ∫푎[푓(푥)] 푓 (푥)푑푥 = 푎 ∫푢 푓 (푥)( )

= 푎 ∫푢 푑푢 = + 푐

3. ∫(푎푥 + 푏) 푑푥 . If we let 푢 = 푎푥 + 푏, then = 푎 ⟹ 푑푢 = 푎푑푥 and 푑푥 =

⟹ ∫(푎푥 + 푏) 푑푥 =∫ 푢 = ∫ 푢 푑푢 =( )

+ 푐 *13

13∗For our level these three forms of functions are usually integrated by substitution.

397

Example 16.5

The following are the gradient functions of some functions; integrate each of them with respect to 푥.

i. (2푥 − 7) ii. (2 − 3푥) iii. 4푥 (푥 + 3) iv. 푥√1 + 푥 Solution

i. We are to find ∫(2푥 − 7) 푑푥. Let 푢 = 2푥 − 7 ⟹ = 2 ⟹ 푑푢 = 2푑푥

⟹ 푑푥 = . Now, ∫(2푥 − 7) 푑푥 = ∫ 푢 = ∫푢 푑푢 = ∙ + 푐 = + 푐

But 푢 = 2푥 − 7 ∴ , ∫(2푥 − 7) 푑푥 = 푢 + 푐 = (2푥 − 7) + 푐

ii. We are to find ∫(2 − 3푥) 푑푥. Let 푢 = 2 − 3푥 ⟹ = −3 ⟹ 푑푢 = −3푑푥

⟹ 푑푥 = − . Now, ∫(2 − 3푥) 푑푥 = ∫푢 = − ∫푢 푑푢 = − ∙ + 푐 = − 푢 + 푐

But 푢 = 2 − 3푥 ∴ ,∫(2 − 3푥) 푑푥 = − 푢 + 푐 = − (2− 3푥) + 푐

iii. We are to find ∫4푥 (푥 + 3) 푑푥. Let 푢 = 푥 + 3 ⟹ = 4푥

⟹ 푑푢 = 4푥 푑푥 ⟹ 푑푥 = . Now, ∫4푥 (푥 + 3) 푑푥 = ∫ 4푥 푢 ∙

⟹ ∫ 4푥 (푥 + 3) 푑푥 = ∫푢 푑푢 = + 푐

But 푢 = 푥 + 3 ∴, ∫4푥 (푥 + 3) 푑푥 = + 푐 = (푥 + 3) + 푐

iv. We are to find ∫푥√1 + 푥 푑푥 = ∫ 푥(1 + 푥 ) . Let 푢 = 1 + 푥 ⟹ = 2푥

⟹ 푑푢 = 2푥푑푥 ⟹ 푑푥 = . Now, ∫ 푥(1 + 푥 ) 푑푥 = ∫푥(푢) ∙

⟹ ∫ 푥(1 + 푥 ) 푑푥 = ∫(푢) 푑푢 = × (푢) + 푐 = (푢) + 푐

But 푢 = 1 + 푥 ∴, ∫푥√1 + 푥 푑푥 = (푢) + 푐 = (1 + 푥 ) + 푐 Try:

Find, i.∫√

ii. ∫(푥 + 1)(2푥 + 푥 )푑푥 iii. √

1 + √푥

Integration of some trigonometric, exponential and logarithmic functions

398

Some proofs 1. To find ∫ sin(푎푥 + 푏) 푑푥,let 푢 = 푎푥 + 푏 ⟹ = 푎 and 푑푥 =

⟹∫ sin(푎푥 + 푏) 푑푥 = ∫ sin푢 = ∫ sin 푢 푑푢 = (− cos푢) + 푐 = − cos(푎푥 + 푏) + 푐

∴ sin(푎푥 + 푏) 푑푥 = −1푎 cos(푎푥 + 푏) + 푐

2. To find ∫ cos(푎푥 + 푏)푑푥,let 푢 = 푎푥 + 푏 ⟹ = 푎 and 푑푥 =

⟹∫ cos(푎푥 + 푏)푑푥 = ∫푐표푠푢 = ∫ cos푢 푑푢 = sin푢 + 푐 = sin(푎푥 + 푏) + 푐

∴ ∫ cos(푎푥 + 푏)푑푥 = sin(푎푥 + 푏) + 푐

3. To find ∫ 푠푒푐 (푎푥 + 푏)푑푥, let 푢 = 푎푥 + 푏 ⟹ = 푎 and 푑푥 =

⟹ ∫ 푠푒푐 (푎푥 + 푏)푑푥 = ∫ 푠푒푐 푢 = ∫ 푠푒푐 푢푑푢 = 푡푎푛푢 + 푐 = tan(푎푥 + 푏) + 푐

∴ ∫ 푠푒푐 (푎푥 + 푏)푑푥 = tan(푎푥 + 푏) + 푐

푦푑푥

푠푖푛푥 −푐표푠푥 + 푐 푐표푠푥 푠푖푛푥 + 푐 푠푒푐 푥 푡푎푛푥 + 푐

sec 푥 tan 푥 sec 푥 + 푐 푐표푠푒푐 푥 − cot 푥 + 푐

cosec 푥 cot푥 −푐표푠푒푐푥 + 푐 sin(푎푥 + 푏) −

1푎 푐표푠

(푎푥 + 푏) + 푐

cos(푎푥 + 푏) 1푎 푠푖푛

(푎푥 + 푏) + 푐

푠푒푐 (푎푥 + 푏) 1푎 tan(푎푥 + 푏) + 푐

푒 푒 + 푐 1푥 ln푥 + 푐

푒( ) 1푎 푒

( ) + 푐

399

Example 16.6

Evaluate the following i. ∫ sin 4푥 푑푥 ii. ∫푒( ) 푑푥 iii. ∫ cosec 3푥 cot 3푥 푑푥 iv. ∫(4푥 − 2푥 + cos 푥)푑푥

Solution i. To find ∫ sin 4푥 푑푥, let 푢 = 4푥 ⟹ = 4 and 푑푥 =

⟹ ∫ sin 4푥 푑푥 = ∫ sin푢 = ∫ sin 푢 푑푢 = − cos푢 + 푐 = − cos 4푥 + 푐

∴ ∫ sin 4푥 푑푥 = − cos 4푥 + 푐

ii. To find ∫푒( ) 푑푥 , let 푢 = 3푥 + 5 ⟹ = 3 and 푑푥 =

⟹ ∫ 푒( ) 푑푥 = ∫푒 ∙ = ∫푒 = 푒 + 푐 But 푢 = 3푥 + 5

∴ 푒( ) 푑푥 =13 푒 + 푐 =

13 푒

( ) + 푐

iii. To find ∫ cosec 3푥 cot 3푥 푑푥, let 푢 = 3푥 ⟹ = 3 and 푑푥 =

⟹ ∫ cosec 3푥 cot 3푥 푑푥 = ∫ cosec 푢 cot 푢 = ∫ cosec 푢 cot 푢 = − cosec푢 + 푐 But 푢 = 3푥

∴ cosec 3푥 cot 3푥 푑푥 = −13 cosec 푢 + 푐 = −

13 cosec 3푥 + 푐

iv. ∫(4푥 − 2푥 + cos푥)푑푥 = ∫4푥 푑푥 − ∫ 2푥푑푥 + ∫ cos푥 푑푥 = 푥 − 푥 + sin 푥 + 푐 ⟹

∴ (4푥 − 2푥 + cos푥)푑푥푥 − 푥 + sin 푥 + 푐

Try:

1. ∫ sin 푥 푑푥 2. ∫(sec 3푥 + 3 sin 푥)푑푥 3) ∫(sin 3푥 + 푒 )푑푥 Integration by parts

Let 푦 = 푢푣, where 푢 and 푣 are functions of 푥. Then, ∫ 푣푑푢 = 푢푣 − ∫ 푢푑푣 + 푐 Proof

Given 푦 = 푢푣, = 푣 + 푢 ⟹ 푑푦 = 푣 ∙ 푑푥 + 푢 ∙ 푑푥 ⟹ 푑푦 = 푣푑푢 + 푢푑푣 ⟹ ∫ 푑푦 = ∫푣푑푢 + ∫푢푑푣 ⟹ 푦 = ∫ 푣푑푢 + ∫푢푑푣 ⟹ 푢푣 = ∫푣푑푢 + ∫ 푢푑푣 ⟹ 푢푣 − ∫ 푢푑푣 = ∫ 푣푑푢

400

∴ 푣푑푢 = 푢푣 − 푢푑푣 + 푐

Example 16.7 Integrate the following with respect to 푥. i. 푥 sin푥 ii. 푥푒 iii. 푥 sin 2푥 iv. 푒 푠푖푛푥

Solution i. We are to find ∫ 푥 sin 푥. Let 푣 = 푥 and 푑푢 = sin 푥푑푥 ⟹ = 1 ⟹ 푑푣 = 푑푥 and ∫푑푢 = ∫ sin 푥푑푥 ⟹ 푢 = − cos 푥

Now, ∫ 푣푑푢 = 푢푣 − ∫ 푢푑푣 + 푐 = −푥 cos 푥 −∫(− cos푥)푑푥 + 푐 = −푥 cos푥 + sin푥 + 푐 ∴ ∫푥 sin 푥 = −푥 cos 푥 + sin 푥 + 푐

ii. We are to find ∫ 푥푒 . Let 푣 = 푥 and 푑푢 = e 푑푥 ⟹ = 1 ⟹ 푑푣 = 푑푥 and ∫ 푑푢 = ∫푒 푑푥 ⟹ 푢 = 푒

Now, ∫ 푣푑푢 = 푢푣 − ∫ 푢푑푣 + 푐 = 푥푒 − ∫ 푒 푑푥 + 푐 = 푥푒 − 푒 + 푐 ∴ ∫푥푒 = 푥푒 − 푒 + 푐 = 푒 (푥 − 1) + 푐

iii. We are to find ∫푥 sin2 푥. Let 푣 = 푥 and 푑푢 = sin 2푥푑푥 ⟹ = 2푥 ⟹ 푑푣 = 2푥푑푥 and ∫ 푑푢 = ∫ sin 2푥푑푥

(Use integration by substitution to find ∫ sin 2푥) ⟹ 푢 = − cos2푥

Now, ∫푣푑푢 = 푢푣 − ∫푢푑푣 + 푐 = −푥 ∙ cos2 푥 −∫(− ∙ 2푥 cos2 푥)푑푥 + 푐

= − cos2푥 + ∫ 푥 cos2 푥푑푥 + 푐

⟹ ∫푥 sin2 푥 = − cos2 푥 + ∫푥 cos2 푥푑푥 + 푐 Now, let us find ∫ 푥 cos2 푥푑푥 Let 푣 = 푥 and 푑푢 = cos 2푥푑푥 ⟹ = 1 ⟹ 푑푣 = 푑푥 and ∫ 푑푢 = ∫ cos 2푥푑푥 ⟹ 푢 = sin2 푥

∫ 푣푑푢 = 푢푣 − ∫푢푑푣 + 푐 = sin 2푥 − ∫ sin 2푥푑푥 + 푐 = sin 2푥 + cos 2푥 + 푐 So, ∫ 푥 sin2 푥 = − cos2 푥 + ∫푥 cos2 푥푑푥 + 푐 = cos2 푥 + sin 2푥 + cos 2푥 + 푐

∴ ∫ 푥 sin2 푥 = cos2푥 + sin 2푥 + cos 2푥 + 푐 = cos 2푥 + sin 2푥 + 푐

iv. We are to find ∫푒 sin 푥. Let 푣 = 푒 and 푑푢 = sin 푥푑푥

401

⟹ = 푒 ⟹ 푑푣 = 푒 푑푥 and ∫ 푑푢 = ∫ sin 푥푑푥 ⟹ 푢 = − cos 푥

Now, ∫푣푑푢 = 푢푣 − ∫푢푑푣 + 푐 = −푒 cos 푥 −∫(−푒 cos푥)푑푥 + 푐 = −푒 cos 푥 + ∫(푒 cos푥)푑푥 + 푐 ⟹ ∫푒 sin 푥 = −푒 cos 푥 + ∫(푒 cos 푥)푑푥 + 푐 Now, let us find ∫(푒 cos 푥)푑푥 푣 = 푒 and 푑푢 = cos푥푑푥 ⟹ = 푒 ⟹ 푑푣 = 푒 푑푥 and ∫푑푢 = ∫ cos 푥푑푥 ⟹ 푢 = sin 푥 ∫ 푣푑푢 = 푢푣 − ∫푢푑푣 + 푐 = 푒 sin 푥 − ∫ 푒 sin 푥푑푥 + 푐 So, ∫푒 sin푥 = −푒 cos푥 + ∫(푒 cos 푥)푑푥 + 푐 = −푒 cos푥 + 푒 sin 푥 − ∫푒 sin 푥푑푥 + 푐 ⟹ ∫ 푒 sin 푥 = −푒 cos 푥 + 푒 sin푥 − ∫ 푒 sin 푥푑푥 + 푐 ⟹ ∫ 푒 sin 푥 + ∫푒 sin 푥 = −푒 cos푥 + 푒 sin 푥

⟹ 2∫ 푒 sin 푥 = −푒 cos푥 + 푒 sin 푥 ⟹ ∫ 푒 sin 푥 = + 푐

∴ ∫ 푒 sin 푥 = + 푐 = (cos푥 − sin푥) Try: Evaluate i. ∫푒 cos푥 푑푥 ii. ∫ (2푥 + 1)푒 푑푥

Integration by partial fractions14* Let 푓(푥) = ( )

( ), where 푔(푥) and ℎ(푥) are polynomials in 푥. Then we can integrate

푓(푥)through one of the following cases: Case 1 If degree of ℎ(푥) is greater than that of 푔(푥) and ℎ(푥) has non-repeated linear factors, for example, (푥 − 푎), (푥 − 푏) etc, then we can write 푓(푥) as 푓(푥) = ( )

( )( )≡

( )+

( ), for two linear factors and 퐴 and 퐵 are parameters. This

follows that ∫ 푓(푥)푑푥 = ∫( )

푑푥 + ∫( )

푑푥 + 푐

Case 2

14∗for our level we consider only two cases.

402

If degree of ℎ(푥) is greater than that of 푔(푥) and ℎ(푥) has repeated factors, for example (푥 − 푎) , (푥 − 푏) etc, then we can write 푓(푥) as 푓(푥) = ( )

( ) ( )≡

( )+

( )+

( )+

( ), where 퐴,퐵,퐶and 퐷 are parameters.

This follows that ∫ 푓(푥)푑푥 = ∫( )

푑푥 + ∫( )

푑푥 + ∫( )

푑푥 + ∫( )

푑푥 + 푐

Example 16.8

Evaluate the following by partial fractions:

i. ∫ ( )( )푑푥 ii. ∫ ( )( )

푑푥 iii. ∫ ( )( )푑푥

Solution i. To find ∫ ( )( )

푑푥, first resolve the function ( )( )

into partial fractions.

Let ( )( )

≡( )

+( )

≡ ( ) ( )( )( )

⟹ 6푥 ≡ 퐴(푥 − 4) + 퐵(푥 − 3) Put 푥 = 3 into the above identity. ⟹ 3(3) ≡ 퐴(3 − 4) + 퐵(3 − 3) ⟹ 9 = −퐴 ∴ 퐴 = −9 Put 푥 = 4 into the above identity. ⟹ 3(4) ≡ 퐴(4 − 4) + 퐵(4 − 3) ⟹ 12 = 퐵 ∴ 퐵 = 12 ∴

( )( )≡

( )+

( )≡

( )−

( )

So, ∫ ( )( )푑푥 = ∫ ( )

−( )

푑푥 = ∫ ( )푑푥 − ∫

( )푑푥

Now, ∫ ( )푑푥 = 12∫ ( )

푑푥 Let 푢 = 푥 − 3 ⟹ = 1 ⟹ 푑푢 = 푑푥

⟹ ∫ ( )푑푥 = 12∫ ( )

푑푥 = 12∫ 푑푢 = 12 ln푢 = 12 ln(푥 − 3)

Similarly, ∫ ( )푑푥 = 9∫ ( )

푑푥 Let 푢 = 푥 − 4 ⟹ = 1 ⟹ 푑푢 = 푑푥

⟹ ∫ ( )푑푥 = 9∫ ( )

푑푥 = 9∫ 푑푢 = 9 ln푢 = 9 ln(푥 − 4)

⟹ ∫ ( )( )푑푥 = ∫ ( )

푑푥 − ∫( )

푑푥 = 12 ln(푥 − 3)− 9 ln(푥 − 4)

∴ ∫ ( )( )푑푥 = 12 ln(푥 − 3)− 9 ln(푥 − 4) + 푐

ii. To find ∫ ( )( )

푑푥, first resolve the function ∫ ( )( )푑푥 into partial

fractions.

403

Let ( )( )

≡( )

+( )

≡ ( ) ( )( )( )

⟹ 푥 − 7 ≡ 퐴(1 − 3푥) + 퐵(1 − 2푥) Put 푥 = into the above identity.

⟹ − 7 ≡ 퐴 1 − 3 ∙ + 퐵 1 − 2 ∙ ⟹ − = 퐴 − ∴ 퐴 = 13

Put 푥 = into the above identity.

⟹ − 7 ≡ 퐴 1 − 3 ∙ + 퐵 1 − 2 ∙ ⟹ − = 퐵 ∴ 퐵 = −20

⟹ ( )( )

≡( )

−( )

So, ∫ ( )( )푑푥 = ∫ ( )

−( )

푑푥 = ∫ ( )푑푥 − ∫ ( )

푑푥

Now, ∫ ( )푑푥 = 13∫ ( )

푑푥 Let 푢 = 1− 2푥 ⟹ = −2 ⟹ 푑푢 = −2푑푥

⟹ ∫ ( )푑푥 = 13∫ ( )

푑푥 = 13∫ ∙ −2푑푢 = −26 ln푢 = −26 ln(1 − 2푥)

Similarly, ∫ ( )푑푥 = 10∫ ( )

푑푥 Let 푢 = 1 − 3푥 ⟹ = −3 ⟹ 푑푢 = −3푑푥

⟹ ∫ ( )푑푥 = 20∫ ( )

푑푥 = 20∫ ∙ −3푑푢 = −60 ln푢 = −60 ln(1− 3푥)

⟹ ∫ ( )( )푑푥 = ∫

( )푑푥 − ∫

( )푑푥 = −26 ln(1 − 2푥) + 60 ln(1− 3푥)

∴ ∫ ( )( )푑푥 = 60 ln(1 − 3푥)− 26 ln(1 − 2푥) + 푐

iii. To find ∫ ( )( )푑푥, first resolve the function

( )( ) into partial fractions.

Let ( )( )

≡( )

+( )

+( )

≡ ( ) ( )( ) ( )( )( )

⟹ 푥 + 1 ≡ 퐴(푥 − 1) + 퐵(푥 − 2)(푥 − 1) + 퐶(푥 − 2) Put 푥 = 2 into the above identity. ⟹ (2) + 1 ≡ 퐴(2 − 1) + 퐵(2− 2)(2 − 1) + 퐶(2 − 2) ∴ 퐴 = 5 Put 푥 = 1 into the above identity. ⟹ (1) + 1 ≡ 퐴(1 − 1) + 퐵(1− 2)(1 − 1) + 퐶(1 − 2) ∴ 퐶 = −1 Put 푥 = 0 into the above identity. ⟹ (0) + 1 ≡ 퐴(0 − 1) + 퐵(0 − 2)(0 − 1) + 퐶(0 − 2) ⟹ 1 = 퐴 + 2퐵 − 2퐶 ⟹ 1 = 5 + 2퐵 + 2(−1) ⟹ 2퐵 = 1 − 5 + 2 ∴ 퐵 = − = −1

404

⟹ ( )( )

≡( )

−( )

−( )

So, ∫ ( )( )푑푥 = ∫ ( )

−( )

−( )

푑푥 = ∫ ( )푑푥 − ∫

( )푑푥 − ∫ ( )

푑푥

Now, ∫ ( )푑푥 = 5∫ ( )

푑푥 Let 푢 = 푥 − 2 ⟹ = 1 ⟹ 푑푢 = 푑푥

⟹ ∫ ( )푑푥 = 5∫ ( )

푑푥 = 5∫ 푑푢 = 5 ln푢 = 5 ln(푥 − 2)

Similarly, ∫( )

푑푥 = ln(푥 − 1)

Also, ∫ ( )푑푥 = ∫(푥 − 1) 푑푥 Let 푢 = 푥 − 1 = 1 ⟹ 푑푢 = 푑푥

⟹ ∫ ( )푑푥 = ∫(푥 − 1) 푑푥 = ∫ 푢 푑푢 = −푢 = − = −

( )

∴ ∫ ( )( )푑푥 = 5 ln(푥 − 2) − ln(푥 − 1) +

( )+ 푐

Try:

Find, i.∫ ( )( )푑푥 ii. ∫ 푑푥 iii. ∫ ( )( )

푑푥 iv. iv. ∫( ) ( )

푑푥

Definite integrals

Theorem: If 퐹 (푥) = 푓(푥) and 푓(푥) is continuous over the range 푎 ≤ 푥 ≤ 푏 then ∫ 푓(푥)푑푥 = 퐹(푏) − 퐹(푎), where 푎 and 푏 are the lower and upper limits bounded by

푓(푥). For example ∫ 3푥 푑푥 = 3 b

ax = 푏 − 푎 . This is called the definite integral of

2푥 within the intervals ([푎, 푏] for 푥, where 푏 > 푎.

Note the following properties of definite integrals: 1. ∫ 푓(푥)푑푥 = ∫ 푓(푢)푑푢 , where 푢 is a function of 푥. (Integrating by substitution)

2. ∫ 푓(푥)푑푥 = −∫ 푓(푥)푑푥 Example 16.9

Evaluate the following: i. ∫ (2푥 + 1)푑푥 ii. ∫ 푑푥 iii. ∫ 푥(푥 − 3)푑푥 iv. ∫ 푥(2푥 + 3) 푑푥

405

v. ∫ 푥 − 푑푥 vi. ∫ 3푥 sin 3푥 푑푥

Solution

i. ∫ (2푥 + 1)푑푥 = 32

1

22x x

=

321

x x

= [3 + 3]− [(−1) − 1] = 12

∴ ∫ (2푥 + 1)푑푥 = 12

ii. ∫ 푑푥 = ∫ 3푥 ∙ 푑푥. Let 푢 = 1 + 푥 ⟹ = 2푥 ⟹ 푑푥 =

⟹ ∫ 푑푥 = ∫ 3푥 ∙ ∙ = ∫ 푑푢 = ∫ 푑푢 = 421

ln(1 )x

⟹ ∫ 푑푥 = [ln(1 + 4 ) − ln(1 + 1 )] = [ln 17 − ln 2] = ln

∴, ∫ 푑푥 = ln

iii. ∫ 푥(푥 − 3)푑푥 = ∫ (푥 − 3푥)푑푥 =1

3 2

1

1 33 2

x x

⟹ ∫ 푥(푥 − 3)푑푥 = (1) + (1) − (−1) + (−1) = + + − =

∴, ∫ 푥(푥 − 3)푑푥 = iv. ∫ 푥(2푥 + 3) 푑푥. Let 푢 = 2푥 + 3 ⟹ = 4푥 ⟹ 푑푥 =

⟹ ∫ 푥(2푥 + 3) 푑푥 = ∫ 푥푢 ∙ = ∫ 푢 푑푢 = ∫ 푢 푑푢 = 22 40

1 (2 3)16

x

⟹ ∫ 푥(2푥 + 3) 푑푥 = [(2(2) + 3) − 0] = (14641) =

∴, ∫ 푥(2푥 + 3) 푑푥 =

v. ∫ 푥 − 푑푥 =9

2

1

1 ln2

x x = (9) + ln 9 − (1) + ln 1

⟹ ∫ 푥 − 푑푥 = + ln 9 − − 0 = − + ln 3 = 40 + 3 ln 3 ∴, ∫ 푥 − 푑푥 = 40 + 3 ln 3

vi. To find ∫ 3푥 sin 3푥 푑푥, let 푣 = 3푥 and 푑푢 = sin 3푥푑푥

⟹ = 3 ⟹ 푑푣 = 3푑푥 and ∫ 푑푢 = ∫ sin 3푥푑푥

406

(Use integration by substitution to find ∫ sin 3푥) ⟹ 푢 = − cos3푥

Now, ∫ 푣푑푢 = [푢푣] − ∫ 푢푑푣

⟹ ∫ 3푥 sin 3푥 푑푥 = −3푥 ∙ cos 3푥 − ∫ −3 ∙ cos3푥푑푥

= [−푥 cos 3푥] + ∫ cos3푥푑푥

⟹ ∫ 3푥 sin 3푥 푑푥 = [−푥 cos 3푥] + [sin 3푥] = [(−휋 cos 3휋)− 0] + [sin 3휋 − 0]

= − (−1) + (0)

∴ ∫ 3푥 sin 3푥 푑푥 = Example 16.10

Find the value of 푘 if ∫ (푘푥 − 2푥 ) = 0. Solution

We have, ∫ (푘푥 − 2푥 ) = 0 ⟹ − = 0

⟹ − − − = 0 ⟹ − − + = 0

⟹ 8푘 − 2푘 − + = 0 ⟹ 6푘 − = 0 ⟹ 6푘 =

∴ 푘 = × = Try: i) Evaluate ∫ 푥 + 2푥 + 푑푥 (SSSCE)

ii) Evaluate, ∫ ( )( 푑푥 (SSSCE)

iii) Evaluate ∫ √ 푑푥 (SSSCE)

iv) Evaluate ∫ (2푥 − 5푥)푑푥 (SSSCE)

v) Evaluate ∫ ( ) 푑푥 (SSSCE)

vi) Evaluate ∫ 푑푥 (SSSCE)

vii) Evaluate ∫ 푑푥, 푥 ≠ 0. (SSSCE)

viii) If ∫ 2(푥 + 2)푑푥 = 15, where 푎 > 0, find the value of 푎. (SSSCE)

ix) Using the substitution 푢 = 푥 + 1, evaluate ∫( )

푑푥 (WASSCE)

407

x) By putting 푢 = 1 + 2푥 , evaluate ∫√

푑푥 (WASSCE)

Application of integration

a. Area under and between curves a. Area enclosed between a curve and the 푥 −axis.

I. If 푦 = 푎 푥 + 푏 푥 + 푐 , where 푎 < 0 such as the diagram below 푦 푦 = 푎 푥 + 푏 푥 + 푐 , 푎 < 0 푨 푎 푏 푥 Then the area enclosed by this curve and the 푥-axis is given by:

퐴 = ∫ 푦푑푥, 푏 > 푎 II. If 푦 = 푎 푥 + 푏 푥 + 푐 , where 푎 > 0 such as the diagram below 푦 푦 = 푎 푥 + 푏 푥 + 푐 , 푎 > 0 푎 푏 푥

푨

Then the area enclosed between this curve and the 푥-axis is given by:

퐴 = 푦푑푥 ,푏 > 푎

408

III. Consider the graph of the cubic function 푦 = 푎 푥 + 푏 푥 + 푐 푥 + 푑 below: 푦 푦 = 푎 푥 + 푏 푥 + 푐 푥 + 푑 푨ퟏ 푎 푏 푐 푥 푨ퟐ The area enclosed by this curve and the 푥-axis is given by:

퐴 = 퐴푟푒푎표푓푡ℎ푎푡푝푎푟푡표푓푡ℎ푒푐푢푟푣푒푎푏표푣푒푡ℎ푒푥 − 푎푥푖푠 + 퐴푟푒푎표푓푡ℎ푎푡푝푎푟푡표푓푡ℎ푒푐푢푟푣푒푏푒푙표푤푡ℎ푒푥 − 푎푥푖푠 = ∫ 푦푑푥 − ∫ 푦푑푥 ,푐 > 푏 > 푎, where 푎,푏 and 푐 are intercepts on the 푥 − 푎푥푖푠.

Example 16.11 Sketch and find the area enclosed between the curve 푦 = 푥 − 푥 and the 푥-axis.

Solution We have: 푦 = 푥 − 푥 = 2푥 − 1 Intercepts: For 푥 −intercepts, let 푦 = 0 ⟹ 푥 − 푥 = 0 ⟹ 푥(푥 − 1) = 0 ⟹ 푥 = 0 or 푥 = 1 푦 푃 (0, 0) 푃 (1, 0) 2 For 푦 −intercepts, let 푥 = 0 1 ⟹ 푦 = (0) − 0 = 0 푃 (0, 0) 푦 = 푥 − 푥 Turning point: At the turning point = 0 -2 -1 퐴 1 2 푥

⟹ 2푥 − 1 = 0 푥 = When 푥 = , -1 푇푃 ,−

푦 = − = − = − 푇푃 ,− -2

= 2 > 0. The turning point is minimum. A푠푥 → ±∞,푦 → ∞

The area under the curve and the 푥 −axis is defined as: 퐴 = −∫ 푦푑푥

∴ 퐴 = −∫ (푥 − 푥)푑푥 == − = − − − 0 = − − = unit

409

Example 16.12 Sketch and find area enclosed between the curve 푦 = 1 − 6푥 − 7푥 and the 푥-axis.

Solution We have: 푦 = 1 − 6푥 − 7푥 = −6 − 14푥 Intercepts: For 푥 −intercepts, let 푓(푥) = 0 ⟹ 1 − 6푥 − 7푥 = 0 ⟹ 1 + 푥 − 7푥 − 7푥 = 0 ⟹ (1 + 푥)− 7푥(1 + 푥) = 0 ⟹ (1 + 푥)(1 − 7푥) = 0 ⟹ 1 + 푥 = 0 or 1 − 7푥 = 0 ⟹ 푥 = −1 or 푥 = 푦

푃 (−1, 0) and 푃 , 0 2 For 푦 −intercept, let 푥 = 0 1 푇푃 − , ⟹ 푦 = 1 − 6(0)− 7(0) = 1 푃 (0, 1) 푦 = 1 − 6푥 − 7푥 퐴 Turning point: At the turning point = 0 -2 -1 1 2

⟹ −6 − 14푥 = 0 ⟹ 푥 = − = − When 푥 = − , -1

푦 = 1 − 6 − − 7 = 1 + − =

푇푃 − ,

= −14 < 0. The turning point is a maximum. A푠푥 → ±∞,푦 → −∞

The area under the curve and the 푥 −axis is defined as: 퐴 = ∫ 푦푑푥

∴ 퐴 = ∫ (1 − 6푥 − 7푥 )푑푥 = 푥 − 3푥 −

= − 3 − − 1 − 3(1) − (1)

= − − − 1 + 3 +

= units

Example 16.13 Sketch the curve 푦 = 2푥 + 푥 − 4푥 + 1. Hence, find the area enclosed between the curve and the 푥 −axis.

Solution We have 푦 = 2푥 + 푥 − 4푥 + 1 = 6푥 + 2푥 − 4

410

Intercepts: For 푥 −intercept, 푦 = 0 ⟹ 2푥 + 푥 − 4푥 + 1 = 0 or (푥 − 1)(2푥 + 3푥 − 1) = 0 or (푥 − 1)(푥 − 0.2808)(푥 + 1.7808) = 0 ⟹ 푥 = 1 or 푥 = 0.2808or 푥 = −1.7808 푃 (−1.7808, 0)푃 (0.2808, 0)푃 (1, 0) For 푦 −intercept, let 푥 = 0 ⟹ 푦 = 2(0) + (0) − 4(0) + 1 = 1 푃 (0, 1) Turning point: At the turning point = 0 6푥 + 2푥 − 4 = 0 (Dividing both sides by we have) 3푥 + 푥 − 2 = 0 or 3푥 − 2푥 + 3푥 − 2 = 0 or (3푥 − 2)(푥 + 1) = 0 ⟹ 푥 = or 푥 = −1

When 푥 = , 푦 = 2 + − 4 + 1 = + − + 1 = = −

When 푥 = −1 푦 = 2(−1) + (−1) − 4(−1) + 1 = −2 + 1 + 4 + 1 = 4

푇푃 (−1, 4)푇푃23 ,−

1727

= 12푥 + 1 = −12 + 1 = −11 < 0 = 12 + 1 = 9 > 0

The turning point 푇푃 (−1, 4) is a maximum. The turning point 푇푃 ,− is a minimum. 푦 푇푃 (−1, 4) 4 푨ퟏ 2 푦 = 2푥 + 푥 − 4푥 + 1. -2 -1 푨ퟐ 1 2 푥 -2 푇푃 ,− -4

The area enclosed by this curve and the 푥-axis is given by: 퐴 = 푨ퟏ + 푨ퟐ = ∫ 푦푑푥 − ∫ 푦푑푥, 푐 > 푏 > 푎

퐴 = ∫ (2푥 + 푥 − 4푥 + 1)푑푥.. − ∫ (2푥 + 푥 − 4푥 + 1)푑푥.

411

= 푥 + 푥 − 2푥 + 푥.

.− 푥 + 푥 − 2푥 + 푥

.

Now, take 푥 + 푥 − 2푥 + 푥.

.

푥 + 푥 − 2푥 + 푥.

.

= 푥 + 푥 − 2푥 + 푥 − 푥 + 푥 − 2푥 + 푥

= (0.28) + (0.28) − 2(0.28) + 0.28 − (−1.78) + (−1.78) − 2(−1.78) −

1.78

= [(0.00307 + 0.00732− 0.1568 + 0.28) − (5.0194 − 1.8799− 6.3368− 1.78)] = [0.1336 + 4.97773] = 5.1109

Take, 푥 + 푥 − 2푥 + 푥.

푥 + 푥 − 2푥 + 푥.

= 푥 + 푥 − 2푥 + 푥 − 푥 + 푥 − 2푥 + 푥

= (1) + (1) − 2(1) + 1 − (0.28) + (0.28) − 2(0.28) + 0.28

= [(0.5 + 0.3333 − 2 + 1) − (0.00307 + 0.00732− 0.1568 + 0.28)] = [−0.1667 − 0.1336] = −0.3003

∴, 퐴 = 푥 + 푥 − 2푥 + 푥.

.− 푥 + 푥 − 2푥 + 푥

.

= 5.1109 − (−0.3003) = 5.4112 units

Example 16.14 Sketch and find the area enclosed by the curve 푓(푥) = 1 + 9푥 − 푥 − 푥 and the 푥-axis.

Solution iii. We have: 푓(푥) = 1 + 9푥 − 푥 − 푥 푓′(푥) = = 9 − 3푥 − 2푥

412

Intercepts: For 푥 −intercept, 푓(푥) = 0 ⟹ 1 + 9푥 − 푥 − 푥 = 0 ⟹ (푥 + 0.11)(푥 + 0.49)(푥 − 2.78) ≈ 0 ⟹ 푥 ≈ −0.11 or 푥 = −0.49 or 푥 = 2.78

푃 (−049. , 0) 푃 (0.11, 0) 푃 (2.78, 0) For 푓(푥)−intercept, let 푥 = 0 ⟹ 푓(푥) = 1 + 9(0)− (0) − (0) = 1

푃 (0, 1) Turning point: At the turning point = 0 ⟹ 9 − 3푥 − 2푥 = 0 ⟹ 2푥 + 3푥 − 9 = 0 ⟹ 2푥 − 3푥 + 6푥 − 9 = 0 ⟹ 푥(2푥 − 3) + 3(2푥 − 3) = 0 ⟹ (푥 + 3)(2푥 − 3) = 0 ⟹ 푥 = −3 or 푥 =

When 푥 = −3, 푦 = 1 + 9(−3)− (−3) − (−3) = 1 − 27− + = −

When 푥 = , 푦 = 1 + 9 − − = 1 + − − =

푇푃 3,− 푇푃 , = −3 − 4푥

= −3 − 4(−3) = 9 > 0 = −3 − 4 = −9 < 0

The turning point 푇푃 −3, is minimum.

The turning point 푇푃 , is a maximum.

[Sketch this curve and find the area. 퐴 = −∫ 푦푑푥 + ∫ 푦푑푥 , 푐 > 푏 > 푎] Example 16.15

Sketch and find the area enclosed by the curve 푦 = 푥 − 7푥 + 10 and the 푥-axis. Solution

We have: 푦 = 푥 − 7푥 + 10 = 3푥 − 14푥 Intercepts: For 푥 −intercept, 푦 = 0 ⟹ 푥 − 7푥 + 10 = 0 ⟹ (푥 − 6.78)(푥 + 1.11)(푥 − 1.33) ≈ 0 ⟹ 푥 ≈ −6.78 or 푥 = −1.11 or 푥 = 1.33

푃 (−1.11. , 0) 푃 (1.33, 0) 푃 (6.78, 0) For 푦 −intercept, let 푥 = 0 ⟹ 푦 = 0 − 7(0) + 10 = 10

푃 (0, 10) Turning point: At the turning point = 0 ⟹ 3푥 − 14푥 = 0

⟹ 푥(3푥 − 14) = 0 ⟹ 푥 = 0 or 푥 = When 푥 = 0, 푦 = 0

413

When 푥 = , 푦 = − 7 + 10 = − + 10 ≈ −40.8 ≈ 1

푇푃 (0, 0) 푇푃 (4.67,−40.81) = 6푥 − 14

= 6(0)− 14 = −14 < 0 = 6 − 14 = 14 > 0

The turning point 푇푃 (0, 0) is maximum. The turning point 푇푃 , is a maximum.

[Sketch this curve and find the area. 퐴 = ∫ 푦푑푥 − ∫ 푦푑푥 , 푐 > 푏 > 푎]

Example 16.16 Sketch and find the area enclosed between the curve 푦 = 3 + 푥 , the lines 푥 = 1, 푥 = −1 and the 푥-axis.

Solution We have: 푦 = 3 + 푥 = 2푥 Intercepts: For 푥 −intercepts, let 푦 = 0 푦 ⟹ 3 + 푥 = 0 ⟹ 푥 = −3 ⟹ 푥 = ±√−3 But 푥 ∉ ℝ 푦 = 3 + 푥 6 For 푦 −intercepts, let 푥 = 0 3 푇푃(0, 3) ⟹ 푦 = 3 + 0 = 3 푃 (0, 3) 푥 = −1 푨 푥 = 1 Turning point: At the turning point = 0 -2 -1 1 2 푥 ⟹ 2푥 = 0 ⟹ 푥 = 0 When 푥 = 0,푦 = 3 -3 푇푃 ,− 푦 푇푃(0, 3) -6

= 2 > 0. The turning point is minimum. A푠푥 → ±∞,푦 → ∞

The area between the curve and the lines 푥 = 1,푥 = −1 is defined as: 퐴 = −∫ 푦푑푥

∴ 퐴 = ∫ (3 + 푥 )푑푥 = 3푥 + = 3(1) + − 3(−1) + ( )

= 3 − + 3 + = 6 units b. Area enclosed between a curve and the 푦-axis Consider the graph of 푥 = 푓(푦) below.

414

푦 The area enclosed between this curve and the 푦-axis is given by: 푏 퐴 = ∫ 푥푑푦 푥 = 푓(푦) where 푎 and 푏 are the 푥 intercepts on the 푦-axis. 푎

Example 16.17

Sketch and find the area enclosed between the curve 푥 = 푦 − 9 and the 푦-axis. Solution

We have: 푥 = 푦 − 9 = 2푦

Intercepts: For 푦 −intercepts, let 푥 = 0 푦 ⟹ 푦 − 9 = 0 ⟹ 푦 = 9 ⟹ 푥 = ±√9 = ±3 푥 = 푦 − 9 푃 (0,−3) 푃 (0, 3) 6 For 푥 −intercepts, let 푦 = 0 3 ⟹ 푥 = 0 − 9 = −9 푃 (0,−9) 푇푃(0,−9) 푨 Turning point: At the turning point = 0 -12 -6 6 12 푥

⟹ 2푦 = 0 ⟹ 푦 = 0 When 푦 = 0,푥 = −9 -3 푇푃(0,−9) -6

= 2 > 0 The turning point 푇푃(0,−9)is minimum.

A푠푦 → ±∞,푥 → ∞ The area between the curve and the 푦 −axis is defined as: 퐴 = −∫ 푥푑푦

∴ 퐴 = −∫ (푦 − 9)푑푥 = − − 9푦 = − − 9(3) − ( ) − 9(−3)

= −[9 − 27 + 9 − 27] = 36 units

Example 16.18

Sketch and find the area enclosed between the curve 푦 = 3푦 − 푦 and the 푦-axis.

415

Solution We have: 푥 = 3푦 − 푦 = 3 − 2푦

Intercepts: For 푦 −intercepts, let 푥 = 0 푦 ⟹ 3푦 − 푦 = 0 ⟹ 푦(3 − 푦) = 0 ⟹ 푦 = 0 or 푦 = 3 3 푥 = 3푦 − 푦 푃 (0, 0) 푃 (0, 3) For 푥 −intercepts, let 푦 = 0 1.5 A 푇푃 ,

⟹ 푥 = 3(0)− 0 = 0 푃 (0, 0) 푇푃(0,−9) 퐴 Turning point: At the turning point = 0 -4 -2 2 4 푥

⟹ 3 − 2푦 = 0 ⟹ 2푦 = 3 ⟹ 푦 = -1.5

When 푦 = ,푥 = 3 − = − = -3

푇푃 ,

= −2 < 0 The turning point 푇푃 , is maximum.

A푠푦 → ±∞,푥 → −∞ The area between the curve and the 푦 −axis is defined as: 퐴 = ∫ 푥푑푦

∴ 퐴 = ∫ (3푦 − 푦 )푑푥 = − = ( ) − ( ) − (0) = − = units

Try: Sketch and find the area enclosed between the curve 푦 = 5− 푦 − 6푦 and the 푦-axis. c. Area enclosed between a curve and a line consider the diagrams below: 푦 푦 푦 푦 퐴 푦 푦 퐴 푎 푏 푥 푎 푏 푥

416

The area between the curve and The area enclosed between the curve and the line in Figure 4.5푎 is given by: the line in Figure 4.5푏 is given by: 퐴 = ∫ (푦 − 푦 )푑푥 퐴 = ∫ (푦 − 푦 )푑푥 In both cases, 푎 and 푏 are the points of intersection between the curve and the line. The area enclosed between the curve and the line in Figure 4.5푐 is given by:

퐴 = (푦 − 푦 )푑푥 + (푦 − 푦 )푑푥

Where 푎, 푏 and 푐 are the point intersection between the curve and the line. 푦 푦 . 푦 푎 푏 푐 푥

Example 16.19

Sketch and find the area enclosed between the curve 푦 = 2− 푥 − 푥 and the line 푦 = 2푥.

Solution We have: 푦 = 2 − 푥 − 푥 = −1 − 2푥 Intercepts: For 푥 −intercepts, let 푦 = 0 ⟹ 2 − 푥 − 푥 = 0 ⟹ 2 + 푥 − 2푥 − 푥 = 0 ⟹ (2 + 푥)− 푥(2 + 푥) = 0 ⟹ (2 + 푥)(1 − 푥) = 0 푦 = 2 − 푥 − 푥

417

⟹ 2 + 푥 = 0 or 1 − 푥 = 0 ⟹ 푥 = −2 or 푥 = 1 푦

푃 (−2, 0) and 푃 (1, 0) 4 For 푦 −intercept, let 푥 = 0 2 푇푃 − , ⟹ 푦 = 2 − (0)− 2(0) = 2 푃 (0, 2) 푨 푦 = 푥 Turning point: At the turning point = 0 -4 -2 2 4 푥

⟹ −1 − 2푥 = 0 ⟹ 푥 = − When 푥 = − , -2

푦 = 2 − − − − = 2 + − = -4

푇푃 − , 푦 = 2 − 푥 − 푥

= −2 < 0. The turning point is maximum. A푠푥 → ±∞,푦 → −∞ We also have, 푦 = 푥. For 푥 −intercept, 푦 = 0 푃 (0, 0) For 푦 −intercept, 푥 = 0. Same as 푥-intercept. = 22 The line has a positive slope. Point of intersection/contact: At the point of contact 2 − 푥 − 푥 = 2푥

⟹ 푥 + 3푥 − 2 = 0 (By using the formula 푥 = ±√ ) 푥 ≈ 0.56 or 푥 ≈ −3.56

The area under the curve and the line is defined as: 퐴 = ∫ (푦 − 푦 )푑푥

퐴 ≈ ∫ [2 − 푥 − 푥 − (2푥)].. ≈ ∫ [2 − 3푥 − 푥 ].

. ≈ 2푥 − 푥 − 푥.

.

∴, 퐴 ≈ 2(0.56)− (0.56) − (0.56) − 2(−3.56)− (−3.56) − (−3.56)

≈ [1.12− 0.4704 − 0.0585 + 7.12 + 19.0104 − 15.0393] ≈ 11.68 units Try: i. Sketch and find the area enclosed between the curve 푓(푥) = 3푥 − 12 and the line 푦 = −푥. ii. Sketch and find the area enclosed between the curve 푦 = 푥(2푥 − 1)(푥 − 5) and the l ine 푦 = 푥 + 2. d. Area enclosed between two curves

418

Consider the graph of 푦 = 푎 푥 + 푏 푥 + 푐 , 푎 < 0 and 푦 = 푎 푥 + 푏 푥 + 푐 ,푎 > 0 on the same axes.

푦 The area enclosed between the two curves is given by: 퐴 = ∫ 푦 푑푥 − ∫ 푦 푑푥

= ∫ (푦 − 푦 )푑푥, where 푎 and 푏 are the points of

푨 intersection of the curves. 푎 푏 푥

Example 16.20 Sketch the curves 푓(푥) = 푥 − 푥 and 푔(푥) = 푥 − 푥 in the same axes and find their points of intersection. Hence, find the area between the two curves.

Solution

Take: 푓(푥) = 푥 − 푥 푓 (푥) = = 1 − 2푥 푦 Intercepts: For 푥 −intercepts, let 푓(푥) = 0 ⟹ 푥 − 푥 = 0 or 푥(1 − 푥) = 0 2 ⟹ 푥 = 0 or 푥 = 1 푃 (0, 0) 푃 (1, 0) 푔(푥) 1 푇푃 , For 푓(푥) −intercepts, let 푥 = 0 ⟹ 푓(푥) = 0 푃 (0, 0) -2 -1 푨 1 2 Turning point: At the turning point 푓 (푥) = 0 -1 푇푃 ,− 푓(푥) ⟹ 1 − 2푥 = 0 or 푥 = When 푥 = , -2

푓(푥) = − = − = 푇푃 , 푓 (푥) = −2 < 0. The turning point is a maximum. A푠푥 → ±∞,푦 → −∞

419

Now, take: 푔(푥) = 푥 − 푥 푔 (푥) = = 2푥 − 1Intercepts: For 푥 −intercepts, let 푔(푥) = 0⟹ 푥 − 푥 = 0 or 푥(푥 − 1) = 0⟹ 푥 = 0 or 푥 = 1 푃 (0, 0) 푃 (1, 0) For 푔(푥) −intercepts, let 푥 = 0 ⟹ 푔(푥) = 0 푃 (0, 0) Turning point: At the turning point 푔 (푥) = 0⟹ 2푥 − 1 = 0 or 푥 =

When 푥 = ,푔(푥) = − = − = − 푇푃 ,− 푔 (푥) = 2 < 0. The turning point is a minimum. A푠푥 → ±∞,푦 → ∞ The area between the two curves is defined as: 퐴 = ∫ 푦 푑푥 − ∫ 푦 푑푥

⟹ 퐴 = ∫ 푓(푥)푑푥 − ∫ 푔(푥)푑푥 = ∫ (푥 − 푥 )푑푥 − (푥 − 푥)∫ 푑푥

⟹ 퐴 = ∫ (푥 − 푥 − 푥 + 푥)푑푥 = ∫ (2푥 − 2푥 )푑푥 = 푥 − 푥

∴, 퐴 = (1) − (1) − (0) = 1 − = unit

Try: 1. Sketch and find the area enclosed between the curves 푓(푥) = 2푥 − 3푥 − 5 and 푔(푥) = 2 − 푥 − 푥 2.i Sketch the curves 푦 = 푥 − 5푥 and 푦 = 2푥 − 6푥 on the same axes. ii. With the aid of the curves sketched in (i), find the finite area bounded by the two curves. (SSSCE) 3. A function ,푓, defined by 푓(푥) = 2푥 + 3푥 . a. Find the turning points and distinguish between them. b. Sketch the curve, 푓(푥) = 2푥 + 3푥 . c. Calculate the finite area enclosed by the curve, the 푥-axis and the line 푥 + 2 = 0.

(SSSCE) 3. Find the area of the finite region enclosed by the curve 푦 = 2푥 − 푥 − 1 and the 푥- axis. (SSSCE) 4. Find the area of the enclosed by the curve 푦 = 푥 − 3푥 − 4 and the line 푦 = 0.

(SSSCE) 5. The finite area enclosed by the curve 푦 = 푘푥 − 푥 and the 푥-axis is 288 square units. Find the value of the constant 푘. (WASSCE) 6. Sketch the graphs of 푓(푥) = 푥 − 3푥 and 푔(푥) = 푥 on the same axes. Find the area of the bounded region bounded by 푓(푥) and 푔(푥). (WASSCE)

420

b. Area under curve− an approximate method.

(Trapezium rule) 푦

Ordinate/ coordinate on 푦 −axis

푦 푦 푦 푦 푦 ⋯ 푦 푦

푎 푏 푥

Consider the area lying under the curve 푦 = 푓(푥) from 푥 = 푎 to 푥 = 푏. A decomposition of the base of this curve into little shapes yields a decomposition of the region into vertical strips. Each shape (strip) represent a trapezium of equal height ℎ. The summation of the area of all the trapeziums gives an approximate value of the area under this curve. For 푛 strips, 푛ℎ = 푏 − 푎 ⟹ ℎ = (that is, the sum of the heights of the trapeziums=푏 − 푎) Total area of the trapeziums: 퐴 = ℎ(푦 + 푦 ) + ℎ(푦 + 푦 ) + ℎ(푦 + 푦 ) + ℎ(푦 + 푦 ) + ⋯+ ℎ(푦 + 푦 )

= ℎ[(푦 + 푦 ) + (푦 + 푦 ) + (푦 + 푦 ) + (푦 + 푦 ) + ⋯+ (푦 + 푦 ]

= ℎ[(푦 + 푦 ) + 2(푦 + 푦 + 푦 + 푦 + 푦 + ⋯+ 푦 )] But ℎ =

∴,퐴 =푏 − 푎

2푛[(푦 + 푦 ) + 2(푦 + 푦 + 푦 + 푦 + 푦 + ⋯+ 푦 )]

This gives an approximate value of the area, ∫ 푦푑푥, under the curve. Note: i. The area of a trapezium equal half the product of the parallel sides and the height. ii. number of strips,푛 = number of ordinates−1

Example 16.21

421

Calculate, using the trapezium rule with the ordinates 1, 2, 3, 4, 5, 6, the approximate

value of ∫ 푑푥 Solution

The approximate area under the curve is define as:

퐴 = [(푦 + 푦 ) + 2(푦 + 푦 + 푦 + 푦 )]

⟹ 퐴 ≈( )

[(2 + 6.167) + 2(2.5 + 3.33 + 4.25 + 5.2)] ≈ [(8.167 + 30.56] ≈ 19.36

∴, the approximate area under the curve 푦 = is 19.36 units .

Example 16.22 Using the trapezium rule with the interval of 0.2, find the approximate value of

∫ 푑푥.

Solution

The approximate area under the curve is define as:

퐴 = [(푦 + 푦 ) + 2(푦 + 푦 + 푦 )]

⟹ 퐴 ≈( )

[(0.775 + 1) + 2(0.837 + 0.894 + 0.947)] ≈ [(1.775 + 5.356] ≈ 0.89

∴, the approximate area under the curve 푦 = is 0.89 unit . Example 16.23

Using the trapezium rule with 5 ordinates, calculate correct to two decimal places ∫ (푥 + 2)푑푥

푥 1 2 3 4 5 6

푦 =푥 + 1푥

2 2.5 3.33 4.25 5.2 6.167

푥 0.2 0.4 0.6 0.8 1.0

푦 = 0.775 0.837 0.894 0.947 1

422

Solution

The approximate area under the curve is define as:

퐴 = [(푦 + 푦 ) + 2(푦 + 푦 + 푦 )]

⟹ 퐴 ≈( )

[(6 + 102) + 2(18 + 38 + 66)] ≈ 1[(108 + 244] ≈ 352

∴, the approximate area under the curve 푦 = (푥 + 2) is 352 units .

Example 16.24 Using the trapezium rule with five strips, find an approximate value of ∫ 푥(푥 + 1)푑푥.

Solution

The approximate area under the curve is define as:

퐴 = [(푦 + 푦 ) + 2(푦 + 푦 + 푦 + 푦 )]

⟹ 퐴 ≈ .( )

[(0.9086 + 2.2894) + 2(1.2599 + 1.5536 + 1.8171 + 2.0606)]

≈ 0.25[(3.1980 + 13.3824] ≈ 4.15 ∴, the approximate area under the curve 푦 = 푥(푥 + 1) is 4.15 units . Try: i. Copy and complete the table below for the value at the ordinates 0, 0.5, 1.0, 1.5, 2.0 and 2.5

푥 0 0.5 1.0 1.5 2.0 2.5 푥

푥 + 5 0.80

0.67

0.57

0.53

푥 2 4 6 8 10 푦 = (푥 + 2) 6 18 38 66 102

푥 0.5 1 1.5 2 2.5 3 푦 = 푥(푥 + 1) 0.9086 1.2599 1.5536 1.8171 2.0606 2.2894

423

Use the trapezium rule to find the approximate value of ∫ 푑푥.

(SSSCE) 2. A river is 30푚 wide. The depth at a distance from one bank is given in the table below. 푿(풎) 0 3 6 9 12 15 1.8 21 24 27 30

푫(풎) 0 1.5 2.2 3.0 4.2 4.8 4.1 3.2 3.0 2.0 1.1

Using the trapezium rule, find approximately the area of the cross section of the river, correct to one decimal place.

(SSSCE) 3. using the trapezium rule with intervals of 0.5, calculate, the approximate value of

∫ √푥 + 1푑푥

(SSSCE) 2. Volumes of solids of revolution15 If a region in a plane is revolved about an axis in the plane, the resulting solid formed is called 푎푠표푙푖푑표푓푟푒푣표푙푢푡푖표푛. a. Rotation of a curve above the 푥-axis Let 푦 = 푓(푥),where 푎 ≤ 푥 ≤ 푏. Then the volume 푉, of the solid generated by revolving the region bounded by the graph of 푓, 푥 = 푎,푥 = 푏 and the 푥-axis is given by:

푉 = 휋 ∫ 푦 푑푥, 푎 and 푏 are the 푥-intercepts b. Rotation of a curve above the 푦-axis Let 푥 = 푓(푦), where 푎 ≤ 푦 ≤ 푏. then the volume 푉 of the solid generated by revolving the region bounded by the graph of 푥, 푦 = 푎, 푦 = 푏 and the 푦-axis is given by:

푉 = 휋∫ 푥 푑푦 ,푎 and 푏 are the 푦-intercepts of the curve. c. Rotation of two curves

15Herewemeanacompleterevolutionoffourrightangles(360°). fortworightangleswe

dividethevolumeby2.

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Let 푦 = 푓(푥) and 푦 = 푔(푥) be two curves bounded by the interval 푎 ≤ 푥 ≤ 푏. If 푦 ≥ 푦 ,then the volume 푉, of the region enclosed between the graph of 푥 = 푎,푥 = 푏 and the two curves is given by:

푉 = 휋 ∫ 푦 푑푥 − 휋 ∫ 푦 = 휋 ∫ (푦 − 푦 )푑푥 푎 and 푏 are the points of intersection of the two curves

Example 16.25 Determine the volume of the solid generated if the curve 푦 = 3푥 is rotated above the 푥-axis between the limits 푥 = 1 and 푥 = 4

Solution The volume of the solid generated by the curve and the 푥 −axis is defined as

푉 = 휋 ∫ 푦 푑푥 , 푎 = 1 and 푏 = 4

푉 = 휋 ∫ (3푥 ) 푑푥 = 휋 ∫ 9푥 푑푥 = 휋 푥 = 휋 (4) − (1) = 휋 (1024)−

∴, 푉 = 1841.4휋 units Example 16.26

The region bounded by the curve 푓(푥) = 1 − 3푥 is rotate above the 푥-axis. Find the volume of the solid generated.

Solution Let 푓(푥) = 푦 ⟹ 푦 = 1 − 3푥 . For 푥 −intercept, put 푦 = 0 ⟹ 1 − 3푥 = 0 ⟹ 3푥 = 1 ⟹ 푥 = 푥 = ±

√

The volume of the solid generated by the curve and the 푥 −axis is defined as 푉 = 휋∫ 푦 푑푥 , 푎 = −

√ and 푏 =

√

푉 = 휋 (1 − 3푥 ) 푑푥√

√

= 휋 (1 − 6푥 + 9푥 )푑푥 = 휋 푥 − 2푥 +59 푥

√

√√

√

⟹ 푉 = 휋√− 2

√+

√− −

√− 2 −

√+ −

√

= 휋√−

√+

√+

√−

√+

√

= 휋√

= 휋√

∴, 푉 = √ 휋 unit

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Example 16.27 Calculate the volume of the region enclosed between the curves 푓(푥) = 4 − 푥 and 푔(푥) = 푥 − 4

Solution Point of intersection: At the point of intersection 푓(푥) = 푔(푥) ⟹ 4 − 푥 = 푥 − 4 ⟹ 2푥 − 8 = 0 (Divide both sides by 2) ⟹ 푥 − 4 = 0 ⟹ 푥 = 4 ⟹ 푥 = ±2 Let 푓(푥) = 푦 and 푔(푥) = 푦 The volume of the finite region generated by the two curves is defined as 푉 = 휋∫ (푦 − 푦 )푑푥, 푎 = −2 and 푏 = 2

푉 = 휋 [(4 − 푥 ) − (푥 − 4) ]푑푥 = 휋 [(16 − 8푥 + 푥 ) − (푥 − 8푥 + 16)]푑푥

= 휋 ∫ [0]푑푥 ∴, 푉 = 0

Example 16.28 Find the volume of the solid generated if the region bounded by the curve 푦 = 푥 , the lines 푥 = 1,푥 = 3 and 푥-axis is rotated through a right angle.

Solution The volume of the solid generated by the curve and the 푥 −axis is defined as

푉 = ∫ 푦 푑푥 , 푎 = 1 and 푏 = 3

푉 = ∫ (푥 ) 푑푥 = ∫ 푥 푑푥 = 푥 = (3) − (1) = (243)−

∴, 푉 = 12.1휋 units Example 16.29

The area of the finite region enclosed by the curve 푥 = 푦 − 1 from 푦 = 1 to 푦 = 2 is rotated above the 푦-axis. Find the volume of the solid generated.

Solution The volume of the finite region enclosed by the curve and the 푦 −axis is defined as:

푉 = 휋 ∫ 푥 푑푦 , 푎 = 1 and 푏 = 2

푉 = 휋 ∫ (푦 − 1)푑푦 = 휋 푦 − 푦 = 휋 (2) − 2 − (1) − 1 = − 2 − + 1

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∴, 푉 = 휋 units Try: 1. Find the volume of the solid generated when the area of enclosed by the curve 푓(푥) = 푥 − 2 and line 푥 = 0 and 푥 = 1 is rotated through 360° about the 푥-axis. (SSSCE ) 3. Integral kinematics As already mentioned kinematics deals with the study of a body’s displacement(푠), velocity(푣) and acceleration(푎). Let 푎푚푠 be the acceleration of a body measured along time 푡 in seconds. We know that the rate of change in velocity of a body with respect to time is acceleration. That is, = 푎 ⟹ 푑푣 = 푎푑푡 ⟹ ∫ 푑 푣 = ∫푎푑푡 ∴,푣 = ∫푎푑푡 + 푐 Let 푣푚푠 be the velocity of a body measured along time 푡 in seconds. We know that the rate of change in displacement with respect to time is velocity. That is, = 푣 ⟹ 푑푠 = 푣푑푡 ⟹ ∫푑푠 = ∫푣푑푡 ∴, 푠 = ∫ 푣푑푡 + 푐

Example 16.30 A particle 푃 moves in a straight line so that at time 푡 seconds its acceleration is (2푡 + 푡 )ms . If 푃 passes through the origin at 푡 = 0 with a velocity of 2ms , find the velocity of 푃 and the distance from 푂 when 푡 = 3 seconds.

Solution We have, 푎 = (2푡 + 푡 )ms 푣 = ∫푎푑푡 + 푐 ⟹ 푣 = ∫(2푡 + 푡 )푑푡 + 푐 ⟹ 푣 = 푡 + 푡 + 푐… … … (1). At 푡 = 0, 푣 = 2

Put the values of 푡 and 푣 into (1). ⟹ 2 = 0 + (0) + 푐 ⟹ 푐 = 2

∴, 푣 = 푡 + 푡 + 2.

푠 = ∫푣푑푡 + 푐 ⟹ 푠 = ∫ 푡 + 푡 + 2 푑푡 + 푐 ⟹ 푠 = 푡 + 푡 + 2푡 + 푐… … … (2) At O 푠 = 0 and 푡 = 0. Put the values of 푡 and 푠 into (2). ⟹ 0 = (0) + (0) + 2(0) + 푐

⟹ 푐 = 0 ∴, 푠 = 푡 + 푡 + 2푡

At 푡 = 3 푣 = 3 + (3) + 2 = 20ms and 푠 = (3) + (3) + 2(3) =

427

Example 16.31

The acceleration of a particle at time 푡 seconds is (4푡 − 2)ms . When 푡 = 2, the velocity is 13.8푚푠 and its displacement is 6m. Find the velocity of the particle at time 푡 = 5 and its displacement at time 푡 = 4.

Solution We have, 푎 = (4푡 − 2)ms 푣 = ∫푎푑푡 + 푐 ⟹ 푣 = ∫(4푡 − 2)푑푡 + 푐 ⟹ 푣 = 2푡 − 2푡 + 푐… … … (1). At 푡 = 2, 푣 = 13.8 Put the values of 푡 and 푣 into (1). ⟹ 13.8 = 2(2) − 2(2) + 푐 ⟹ 푐 = 13.8 − 8 + 4 = 9.8 ∴, 푣 = 2푡 − 2푡 + 9.8 푠 = ∫푣푑푡 + 푐 ⟹ 푠 = ∫(2푡 − 2푡 + 9.8)푑푡 + 푐 ⟹ 푠 = 푡 − 푡 + 9.8푡 + 푐… … … (2) At 푡 = 2, 푠 = 6. Put the values of 푡 and 푠 into (2). ⟹ 6 = (2) − (2) + 9.8(2) + 푐 ⟹ 푐 = 6 − + 4 − 19.6 ≈ −14.93

⟹ 푐 = 0 ∴, 푠 = 푡 − 푡 + 9.8푡 − 14.93 At 푡 = 5 푣 = 2(5) − 2(5) + 9.8 = 49.8ms And 푡 = 4 푠 = (4) − (4) + 9.8(4)− 14.93 ≈ 50.94 Try: 1. The acceleration of a particle at time 푡 seconds was 6(2푡 − 푡 )푚푠 . When 푡 was 1s the velocity was 2ms and its displacement was 3 푚. Calculate i. The velocity at time 푡 = 2s. ii) The displacement at time 푡 = 3s.

(SSSCE) 2. A particle starts from rest and its acceleration, 푎 at time 푡 seconds, is given by 푎 = 푡 + 1. Find the; a. velocity of the particle after1 second. b. position of the particle at any time 푡.

(SSSCE) 3. The velocity of a body moving in a straight line from a point 푂 in time 푡 seconds is given by 푣 = (3푡 + 2푡 − 1)푚푠 . find the distance from 푂 at time 푡 = 3.3 seconds.

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VECTORS A vector quantity is any quantity that has a magnitude and a direction. Examples of such quantities include velocity, acceleration, moment, momentum, force, displacement etc. Definition: A nonzero vector is a directed line segment drawn from a point 푃(called its 푖푛푖푡푖푎푙푝표푖푛푡) to a point 푄 (called its 푡푒푟푚푖푛푎푙푝표푖푛푡), with 푃 and 푄 being distinct points. The vector is denoted by 푃푄⃗.

Ways of denoting a vector 1. A free vector is denoted by a pair of upper case letters and an arrow on them. For instance 푃푄⃗ 2. A vector can also be represented by a lower case letter with an underscore. For instance 푎 3. A similar form is by using bold lower case letters. For instance 풂,풃 etc. 4. Another way of denoting a vector is the use of lower case letter with a v-cap on it. For example 풂. This form is normally used to denote unit vectors.

Representation of vectors A vector can be represented analytically by one of the following forms: i) the ordered pair form. For example 푃푄⃗(푥, 푦)

ii) The column form . for example 푃푄⃗ , where 푥 and 푦 are called the component of 푃푄⃗ iii) The Cartesian coordinate form. For example 푎 = 푥퐢 + 푦퐣, where are unit vectors in the direction of 푥 and 푦. We define 퐢 as and 퐣as iv) The magnitude-direction form. For example, we can represent acceleration as (푎푚푠 ,m°), where 푎 is constant and 0° ≤ 푚 ≤ 360° Definition: A zero vector is a point whose magnitude is 0. We denote a zero vector as 풐 . Note: To indicate the direction of a vector we draw an arrow from its initial point to its terminal point. Consider the diagram below: 푦 푄 푡푒푟푚푖푛푎푙푝표푖푛푡 푃푄⃗ figure 17.1. 푃 푖푛푖푡푖푎푙푝표푖푛푡 푥

429

A vector in two dimensional space

Definition: A position vector is a vector whose initial point is the origin. If a point 푃 lies in the plane then the position of 푃 relative to the origin is defined by the vector 푂푃⃗ = 푃.

Expressing two given points as a vector Let 푃(푥 ,푦 ) and 푄(푥 ,푦 ) be two point in a plane. Then 푃푄⃗ = 푂푄⃗ − 푂푃⃗ = (푥 ,푦 ) − (푥 ,푦 ) = (푥 − 푥 ,푦 − 푦 ). In component form, 푃푄⃗ =

Example 17.1 a. Express the following points as vectors: i. 퐴(3, 5) and 퐵(6, 6) ii. 퐶(−1, 2) and 퐷(12, 3) b. The points 푃(−2, 1) and 푄(푎, 푏) are points in the 푂푥푦 plane such that 푃푄⃗ = . Find the coordinates of 푄.

Solution a. i. We have, 퐴(3, 5) and 퐵(6, 6). 퐴퐵⃗ = 푂퐵⃗ − 푂퐴⃗ = (푥 − 푥 , 푦 − 푦 ) ⟹ 퐴퐵⃗ = (6 − 3, 6 − 5) = (3, 1) ∴,퐴퐵⃗ = (3, 1) ii. We have, 퐴(3, 5) and 퐵(6, 6). 퐶퐷⃗ = 푂퐷⃗ − 푂퐶⃗ = (푥 − 푥 ,푦 − 푦 ) ⟹퐶퐷⃗ = (12 + 1, 3 − 2) = (13, 1) ∴,퐶퐷⃗ = (13, 1) b. We have, 푃(−2, 1) and 푄(푎, 푏). 푃푄⃗ = 푂푄⃗ − 푂푃⃗ =

⟹ 푃푄⃗ = But ,푃푄⃗ = ⟹ = ⟹ 푎 + 2 = 3 ⟹ 푎 = 3 − 2 = −1 and 푏 − 1 = −1 ⟹ 푏 = 1 − 1 = 0 ∴,푄(−1, 0) Try: 퐴(−1, 5), 퐵(−1, 2) and 퐶(3, 0) are points in the 푥 − 푦 plane. Find i) 퐵퐴⃗ ii) 퐵퐶⃗ in the

form . (SSSCE)

Magnitude/ length of a vector The magnitude of a vector is the length of the directed line segment drawn from its initial point to its terminal point. We will denote the magnitude of a vector 푃푄⃗ as 푃푄⃗ .

430

Recall that if 푃(푥 ,푦 ) and 푄(푥 ,푦 ) are two points in a plane, then the distance 푑 between 푃 and 푄 is given by 푑 = (푥 − 푥 ) + (푦 − 푦 ) . By this formula, for a vector 푃푄⃗ with the initial point 푃(푥 ,푦 ) and the terminal point 푄(푥 ,푦 ) the magnitude is given by 푃푄⃗ = (푥 − 푥 ) + (푦 − 푦 ) . If we let 푥 − 푥 = 푎 and 푦 − 푦 = 푏, then 푃푄⃗ = √푎 + 푏 . In general, for a vector 풗(푎,푏) in two dimensional space the magnitude of 푣 is given by:

|풗| = 푎 + 푏

Example 17.2 a. i. Given that 푃(12,−5) and 푄(2, 0). Find 푃푄⃗ . ii. Given that 푎 = (−3,−4), find |풂|. b. Given that 퐴퐵⃗ = and 퐴퐵⃗ = √13, find the value of 푎.

Solution a. i. We have, 푃(12,−5) and 푄(2, 0). 푃푄⃗ = 푂푄⃗ − 푂푃⃗ = (푥 − 푥 ,푦 − 푦 ) ⟹ 푃푄⃗ = (2 − 12, 0 + 5) = (−10, 5) 푃푄⃗ = √푎 + 푏 ⟹ 푃푄⃗ = (−10) + (5) = √100 + 25 = √125 = 5√5 units a. We have, 퐴퐵⃗ = . 퐴퐵⃗ = √푎 + 푏 = √푎 + 3 = √푎 + 9

But 퐴퐵⃗ = √13 ⟹ √푎 + 9 = √13. (Square both sides) ⟹ 푎 + 9 = 13 ⟹ 푎 = 13 − 9 = 4 푎 = ±√4 = ±2 ∴,푎 = ±2. Try: Find the magnitude of each of the following: i. 풗(0, 2) ii. 풘(7, 1) iii. 풖 = iv. 푃 = (2, 2)and푄 = (3,5) Definition 1: A scalar quantity is any quantity that has only magnitude. Examples of such quantities are time, electric current, mass, amount of substance, speed etc. Definition 2: A scalar is a quantity that can be represented by a single number. We will choose our scalar from the set of real numbers.

Multiplication of a vector by a scalar

431

Definition: For any real number 푘 and vector 풗, their product is the vector 푘풗 whose magnitude is |푘||푣|. The direction of vector 푘풗 is the same as 푣 if 푘 > 0 and opposite to 풗 if 푘 < 0. vector 푘풗 is the zero vector if 푘 = 0. Let 풗(푎, 푏),then 푘풗 = (푘푎,푘푏)

Example 17.3 i. If 풂 = (−2, 1), find 3풂 and |3풂| ii. Find −2푄푅⃗ and −2푄푅⃗ if 푄푅⃗ = iii. Given that 풗(푎,푏) and 풘(3푎, 3푏), show that |풗| = 3|풘|.

Solution i. if 푎 = (−2, 1), then 3풂 = 3(−2, 1) = (−6, 3). |3풂| = (−6) + (3) = √36 + 9 = √45 = 3√5 units

ii. If 푄푅⃗ = , then −2푄푅⃗ = −2 = 6 × −2−2 × −2 = −12

4

푄푅⃗ = (−12) + (4) = √144 + 16 = √160 = 4√10 units iii. We have, 풗(푎, 푏) and 풘(3푎, 3푏). |풗| = √푎 + 푏 |풘| = (3푎) + (3푏) = √9푎 + 9푏 = 9(푎 + 푏 ) = 3√푎 + 푏 = 3|풗| ∎

Inverse of a vector The inverse of any vector is defined as the multiplication of the vector by the scalar −1. That is, the inverse of 푃푄⃗ is −푃푄⃗ = 푄푃⃗ . A vector and its inverse have the same magnitude but opposite in direction.

Example 17.4 Given that 푃(2,−1) and 푄(5, 4), find i. 푃푄⃗ ii. 푄푃⃗ iii.−푃푄⃗ iv. 푃푄⃗ v. −푃푄⃗ . What conclusion can you draw about i. and iii?

Solution i. We have, 푃(2,−1) and 푄(5, 4). 푃푄⃗ = 푂푄⃗ − 푂푃⃗ = (푥 − 푥 ,푦 − 푦 ) ⟹ 푃푄⃗ = (5 − 2, 4 + 1) = (3, 5) ii. 푄푃⃗ = (2 − 5,−1 − 4) = −(3,−5) iii. −푃푄⃗ = −(5 − 2, 4 + 1) = (−3,−5) iv. 푃푄⃗ = √푎 + 푏 = (3) + (5) = √9 + 25 = √34 units

v. −푃푄⃗ = √푎 + 푏 = (−3) + (−5) = √9 + 25 = √34 units i. and iii have the same magnitudes but opposite in direction.

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Addition/summation of vectors

Definition: The sum of two vectors 푎 and 푏 is the vector 풂+ 풃 formed by placing the initial point of 풃 to the terminal point 푎 and joining the initial point of 풂 and terminal point of 풃. Consider the line segments below: 풂 풂 풃 풃

풂 + 풃 (Adding two vectors)

Let 풂(푎 , 푎 ) and 풃(푏 , 푏 ). Then 푎 + 푏 = + =

Example 17.5

Given that 풗(1, 1) and 풘(2, 9), find i. 풗 + 풘 ii. 2풗+ 3풘 iii. |2풗+ 3풘| Solution

i. We have, 풗(1, 1) and 풘(2, 9). 풗 + 풘 = (푎 + 푏 ,푎 + 푏 ) = (1 + 2, 2 + 9) = (3, 11)

Triangular law of vectors (Resultant of two vectors)

The triangular law of vectors states that: Given the vertices of triangle 퐴,퐵 and 퐶 in a plane 퐴퐵⃗ + 퐵퐶⃗ = 퐴퐶⃗

Parallelogram law vectors

This is an equivalent definition to addition of two vectors which states that: The resultant vector 풂 + 풃 can be obtained by constructing the diagonal of the parallelogram having the vectors 풂 and 풃 as adjacent sides. Consider the diagram below: Recall that the opposite Sides of a parallelogram are equal.

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푦 풃 풂 + 풃

figures 17.3. 풂

푥

Subtraction of vectors Let 푎 and 푏 be two vectors. Since −1 ∙ 풃 = −풃 is well-defined, we will define subtraction as: 풂 + (−풃) = 풂 − 풃.

풂 − 풃 −풃

풂

Example 17.6

a. Given that 풂 = and 풃 = , find i. 풂 + 풃 ii. 풂 − 풃 iii. |풂 − 풃|

b. In a triangle 퐴퐵퐶,퐴퐵⃗(0, 4) and 퐵퐶⃗(5,−6).퐹ind 퐴퐶⃗ and 퐴퐵⃗ c. In a triangle 푃푄푅,푅푄⃗ = and 푃푄⃗ = .Find 푃푅⃗ and 푃푅⃗

d. In a triangle 푋푌푍, 푋푌⃗(7, 3), 푋푍⃗(11, 1) and 푍(2, 2). Find the coordinates of 푌 if 푌푍⃗(4,−2). e. Evaluate 푃푄⃗ + 5푋푌⃗ + 7푄푃⃗ + 7푌푋⃗

Solution a. i. We have, 풂 = and 풃 = . 풂+ 풃 = ∴, 풂 + 풃 = =

ii. 풂 − 풃 = ∴, 풂 − 풃 = =

iii. |풂 − 풃| = (−10) + (−1) = √100 + 1 = √101 units. b. We have, 퐴퐵⃗(0, 4) and 퐵퐶⃗(5,−6). From the triangular law of vectors 퐴퐶⃗ = 퐴퐵⃗ + 퐵퐶⃗ ∴,퐴퐶⃗ = (0, 4) + (5,−6) = (0 + 5, 4 − 6) = (5,−2)

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퐴퐵⃗ = (0) + (4) = √4 = 4 units. c. We have, 푅푄⃗ = and 푃푄⃗ = . From the triangular law of vectors

푃푅⃗ = 푃푄⃗ + 푄푅⃗. 푄푅⃗ = −푅푄⃗ = ∴,푃푅⃗ = + = 144

d. Let 푌(푥,푦). From the triangular law of vectors 푋푍⃗ = 푋푌⃗ + 푌푍⃗ We have 푋푌⃗(7, 3), 푋푍⃗(11, 1) and 푌푍⃗(4,−2).

⟹ 푋푍⃗ = 푋푌⃗ + 푂푍⃗ − 푂푌⃗ ⟹ = 73 + 2 − 푥

2 − 푦

⟹ = 9− 푥5 − 푦 ⟹ 11 = 9− 푥 ⟹ 푥 = 9 − 11 = −2 and

1 = 5 − 푦 ⟹ 푦 = 5 − 1 = 4 ∴,푌(−2, 4) Alternatively,

푌푍⃗ = 푂푍⃗ − 푂푌⃗ ⟹ 4−2 = 2 − 푥

2 − 푦 ⟹ 4 = 2 − 푥 ⟹ 푥 = 2 − 4 = −2

and −2 = 2 − 푦 ⟹ 푦 = 2 + 2 = 4 ∴,푌(−2, 4) e. 푃푄⃗ + 5푋푌⃗ + 7푄푃⃗ + 7푌푋⃗ = 푃푄⃗ − 7푃푄⃗ + 5푋푌⃗ − 7푋푌⃗ = −6푃푄⃗ − 2푋푌⃗ = −2(3푃푄⃗ + 푋푌⃗) ∴, 푃푄⃗ + 5푋푌⃗ + 7푄푃⃗ + 7푌푋⃗ = −2(3푃푄⃗ + 푋푌⃗) Try: 1. What is the sum of the following five vectors; 3푂퐴⃗, 6퐵푍⃗, 2퐴푂⃗,퐴퐵⃗and 5푂퐵⃗?

(SSSCE) 2. Two vectors are given by 풂 = ퟐ

ퟏ and풃 = ퟑퟒ . Find|풂 − ퟐ풃|

(SSSCE) 3. if 푃푄⃗ = and 푅푄⃗ = , express 푃푅⃗ as a column vector.

(WASSCE) Equality of vectors

Definition: Two vectors are equal if they have the same magnitude and direction. Let 풂(푎 , 푎 ) and 풃(푏 , 푏 ). Then 풂 = 풃 if and only if 푎 = 푎 and 푏 = 푏 . Example 17.7 i. Given that 풂(2, 4) and 풃(푥 − 1, 푦), find the values of 푥 and 푦 for which 풂 = 풃.

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ii. Given that 풑 and 풒 , for what values of 푥 and 푦 is 풑 = 풒.

iii. If 풖 and 풗 , find the possible values of 푥 and 푦 for which 풖 = 풗. Solution

i. We have, 풂(2, 4) and 풃(푥 − 1,푦). If 풂 = 풃, then 2 = 푥 − 1 ⟹ 푥 = 2 + 1 = 3 and 4 = 푦 ∴,푥 = 3,푦 = 4 ii. We have, 풑 and 풒 . If 풑 = 풒, then 푥 = 푦 + 3 ⟹ 푥 − 푦 = 3 … … … (1)

and 푦 = 3푥 − 2 ⟹ 3푥 − 푦 = 2 … … … (2) Subtract (1) from (2) ⟹ 3푥 − 푥 − 푦 + 푦 = 2 − 3 ⟹ 2푥 = −1 ⟹ 푥 = − . Put 푥 = − into (1) ⟹ − − 푦 = 3 ⟹ 푦 = − − 3 = −

∴,푥 = − . ,푦 = −

iii. We have, 풖 and 풗 . If 풖 = 풗 then −1 = 푥 − 8푥 + 11

⟹ 푥 − 8푥 + 12 = 0 ⟹ 푥 − 6푥 − 2푥 + 12 = 0 ⟹ (푥 − 6)(푥 − 2) = 0 ∴,푥 = 6 or 푥 = 2. And 3 = 푦 + 2푦 ⟹ 푦 + 2푦 − 3 = 0 ⟹ 푦 + 3푦 − 푦 − 3 = 0 ⟹ (푦+ 3)(푦 − 1) = 0 ∴,푦 = −3 or 푦 = −1. Try: if 퐴퐵⃗ , 푃푄⃗ and 퐴퐵⃗ = 푃푄,⃗ find 푥 and 푦.

Unit vector A unit vector is a vector whose magnitude is 1. For any vector 풂, the unit vector which points to the same direction of (parallel to) 풂 is given by: 푎 = 풂

| |. The unit vectors, also

called basis vectors, in the direction of 푥 –and푦-axes in the coordinate system are: 퐢(1, 0) and 퐣(0, 1). These unit vectors are perpendicular to each other. Every vector 풂(푥 ,푦 )can be written as a scalar combination of the basis vectors as: 풂(푥 ,푦 ) = 푥 퐢+ 푦 퐣 For two vectors 풂 = 푥 퐢+ 푦 퐣 and 풃 = 푥 퐢 + 푦 퐣, addition and subtraction of 풂 and 풃 is defined as 풂 + 풃 = 푥 퐢 + 푦 퐣 + 푥 퐢 + 푦 퐣 = (푥 + 푥 )퐢 + (푦 + 푦 )퐣풂 − 풃 = (푥 − 푥 )퐢+ (푦 − 푦 )퐣. The magnitude of 풂is still |풂| = 푥 + 푥 ⟹ |풂| = 푥 + 푥 푦 푦

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풂 = 푥 퐢 + 푦 퐣 1 푦 퐣 퐣 |풂|

퐢 1 푥 푥 퐢 푥

Basis vectors 퐢and 퐣

Example 17.8 Write the following vectors as scalar combination of 퐢 and 퐣: i. 풂(2,−1) ii. 풃(0, 4) iii. 풄(−7, 0)

Solution i. . 풂(2,−1) ⟹ 풂 = 2퐢 − 퐣 ii. 풃(0, 4) ⟹ 풃 = ퟎi + ퟒ퐣 = 4퐣 iii. 풄(−7, 0) ⟹ 풄 = −7퐢+ 0퐣 = −7퐢

Example 17.9 Given that 풖 = 3퐢+ 4퐣and 풗 = 4퐢 − 6퐣. Find i. 풖 + 풗 ii. the unit vector in the direction of 풖.

Solution i. We have. 풖 = 3퐢 + 4퐣and 풗 = 4퐢 − 6퐣 풖 + 풗 = 3퐢+ 4퐣 + 4퐢 − 6퐣 = (3 + 4)퐢+ (4 − 6)퐣 = 7퐢 − 2퐣 ∴,풖 + 풗 = 7퐢 − 2퐣 ii. |풖| = (3) + (4) = √9 + 16 = 5 units. The unit vector in the direction 풖 is defined as: 풖 = 풖

|풖|= 퐢 퐣 = 퐢 + ퟒ

ퟓ퐣

∴, 풖 = 퐢 퐣 = 퐢 + ퟒퟓ퐣

Example 17.10 Given that 풓 = 12퐢 − 퐣 and 풔 = 퐢 − 퐣, find i. 풓 − ퟏ

ퟐ풔 ii. 풓 − 풔 iii. the unit vector parallel to 풔 − 풓.

Solution We have, that 풓 = 12퐢 − 퐣 and 풔 = 퐢 − 퐣 풔 = ퟏ

ퟐ퐢 − 퐣

i. 풓 − 풔 = 12퐢 − 퐣 − ퟏퟐ퐢 − 퐣 = 12퐢 − 퐢 − 퐣 + 퐣 = 퐢 − 퐢 ∴,풓 − 풔 = 퐢 − 퐢

ii. 풓 − 풔 = + − = + = √ units.

풓 − 풔 = 12퐢 − 퐣 − (퐢 − 퐣) = 11퐢 Let 풓 − 풔 = 풂 iii. The unit vector in the direction of 풓 − 풔 is defined as: 풂 = 풂

|풂|

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|풂| = √11 + 0 = 11 units. ∴, 풂 = 퐢 = 퐢 Example 17.11

Given that 풂 = 5퐢 − 2퐣, 풃 = 8퐢 − 퐣 and 풄 = 14퐢 − 3퐣. Find the values of 푥 and 푦 such that 풄 = 풂푥 + 풃푦

Solution We have, 풂 = 5퐢 − 2퐣, 풃 = 8퐢 − 퐣 and 풄 = 14퐢 − 3퐣 풄 = 풂푥 + 풃푦 ⟹ 14퐢 − 3퐣 = (5퐢 − 2퐣)푥 + (8퐢 − 퐣)푦 ⟹ 14퐢 − 3퐣 = (5푥풊 − 2푥풋) + (8푥풊 − 푦퐣) = (5푥 + 8푦)퐢 + (−2푥 − 푦)퐣 ⟹ 14 = 5푥 + 8푦… … … (1) and −3 = −2푥 − 푦 3 = 2푥 + 푦… … … (2) From (2) 푦 = 3 − 2푥 Put 푦 = 3− 2푥 into (1) ⟹ 14 = 5푥 + 8(3 − 2푥) ⟹ 14 = 5푥 + 24− 16푥 ⟹ 14 − 24 = −11푥 ⟹ 푥 = Put 푥 = into (2) 푦 = 3 − 2 = =

∴,푥 = , y = Try: 1. If 풂 = 6퐢 − 5퐣,풃 = 2퐢 − 7퐣 and풄 = 8퐢 + 2퐣, find the values of 푝 and 푞 such that 풄 = 푝풂+ 푞풃 (SSSCE) 2. The position vectors of the points 푃,푄,푅and 푆 relative to fixed point 푂 are 풑 = 2퐢 + 3퐣,풒 = 6퐢+ 23퐣, 풓 = 2퐢 − 7퐣, 풔 = 7퐢+ 4퐣. Find the scalars 푚 and 푛 such that 풔 = 푚풑 + 푛풒+ 풓 The ratio theorem Let 퐴 and 퐵 be two points in a plane such that 푂퐴⃗ = 풂(position vector of 퐴)and 푂퐵⃗ =풃(position vector of 퐵). If 푂푃⃗ = 풑(position vector of 푃)divides 퐴퐵⃗ internally in the ratio 푚:푛, then 풑 = 풂 풃

Proof Consider the Figure below:

푦 퐴 푚 풂 푃 푛 퐵 푂 풃 푥

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Since 퐴푃⃗:푃퐵⃗ = 푚:푛 and 퐴푃⃗ and 푃퐵⃗ are in the same direction

⟹⃗⃗ = ⟹ 퐴푃⃗ = 푃퐵⃗ We know that 퐴푃⃗ = 푂푃⃗ − 푂퐴⃗ = 풑 − 풂

and 푃퐵⃗ = 푂퐵⃗ − 푂푃⃗ = 풃 − 풑 ⟹ 풑− 풂 = 풎풏

(풃 − 풑) ⟹ 푛(풑 − 풂) = 푚(풃− 풑) ⟹ 푛풑− 푛풂 = 푚풃 −푚풑 ⟹ 푛풑+ 푚풑 = 푛풂 + 푚풃 ⟹ 풑(푚 + 푛) = 푛풂+ 푚풃

∴ 풑 = 풂 풃 Note:

If 푃 divides 퐴퐵 externally in the ratio given above then 풑 = 풃 풂, 푚 > 푛.

If 푃 is the midpoint of 퐴퐵, 푚 = 푛 = 1 and 풑 = 풂 풃. Example 17.12

Given that 풂 = 2퐢+ 3퐣 and 풃 = 2퐢 − 퐣, find the position vector풄which divides풂and풃in the ratio 5 : 3

Solution We have, 풂 = 2퐢 + 3퐣 and 풃 = 2퐢 − 퐣 푚:푛 = 5: 3 풄 = 풂 풃 = ( 퐢 퐣) ퟓ( 퐢 퐣) = 퐢 퐣 퐢 퐣 = 퐢 퐣 = 2퐢+ 퐣 ∴, 풄 = 2퐢+ 퐣

Example 17.13 If 푂퐴⃗ = 3퐢 − 6퐣and 푂퐵⃗ = 5퐢+ 퐣,find the position vector 푂푃⃗ which divides 푂퐴⃗ and 푂퐵⃗ ii. internally in the ratio 2:1 ii. externally in the ratio 4:3

Solution We have, 푂퐴⃗ = 3퐢 − 6퐣and 푂퐵⃗ = 5퐢+ 퐣

i. 푂푃⃗ =⃗ ⃗ = ( 퐢 퐣) ( 퐢 퐣) = 퐢 퐣 퐢 퐣 = 퐢 퐣 ∴,푂푃⃗ = 퐢 − 퐣

ii. 푂푃⃗ =⃗ ⃗ = ( 퐢 퐣) ( 퐢 퐣) = 20퐢+ 4퐣 − 9퐢 + 18퐣 = 11퐢 + 22퐣 ∴,푂푃⃗ = 11퐢+ 22퐣

Example 17.14 If 퐹is the midpoint of 퐴퐵⃗, where 퐴(2, 6) and 퐵(5, 4), find in terms of퐢and퐣, the position vector푂퐹⃗

Solution Let 푂퐴⃗ = 2퐢 + ퟔ퐣 and 푂퐵⃗ = 5퐢+ 4퐣

푂푃⃗ =⃗ ⃗ = 퐢 ퟔ퐣 퐢 퐣 = 퐢 퐣 ∴,푂퐹⃗ = 퐢+ 5퐣

Try: 1. Find the midpoint of 풖 = 13퐢 − 퐣 and 풗 = 5퐢

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2. A point 퐴 has position vector 풂and 퐵 has position vector 풃. Prove that 푀 the midpoint of 퐴퐵⃗ has the position vector 풎 = (풂+ 풃)

(SSSCE) Dot/scalar/inner product of two vectors

Let풂(푎 ,푎 ) and 풃(푏 ,푏 ) be two vectors. The dot product also called inner or scalar product of 풂 and 풃, denoted by 풂 • 풃, is given by:

풂 • 풃 = 푎 푏 + 푎 푏 Notice that the dot product of two vectors is a scalar. Using the definition given here we have the following results:

풂 • 풂 = 풂 = 푎 ∙ 푎 + 푎 ∙ 푎 = 푎 + 푎 = 푎 + 푎ퟐ

= |풂|

퐢 • 퐢 = 1 ∙ 1 + 0 = 1 퐣 • 퐣 = 0 + 1 ∙ 1 = 1 퐢 • 퐣 = 퐣 • 퐢 = 1 ∙ 0 + 0 ∙ 1 = 0 If 풂 = 푎 퐢+ 푎 퐣 and 풃 = 푏 퐢 + 푏 퐣, then 풂 • 풃 = (푎 퐢 + 푎 퐣) • (푏 퐢 + 푏 퐣) = 푎 퐢 • (푏 퐢 + 푏 퐣) + 푎 퐣 • (푏 퐢+ 푏 퐣) = 푎 푏 퐢 • 퐢+ 푎 푏 퐢 • 퐣 + 푎 푏 퐣 • 퐢+ 푎 푏 퐣 • 퐣 = 푎 ∙ 푏 + 푎 ∙ 푏 , Since, 퐢 • 퐢 = 퐣 • 퐣 = ퟏand퐢 • 퐣 = 퐣 • 퐢 = 0 The following properties are valid: 1.풂 • 풃 = 풃 • 풂 (Commutative property) 2. 풂 • (풃 + 풄) = 풂 • 풃+ 풂 • 풄 (Distributive property) Note: The dot product of three vectors is not defined.

Parallel and perpendicular vectors Two vectors 풂 and 풃 are parallel if the absolute value of their dot product equal the product of their magnitudes; i.e.|풂 • 풃| = |풂||풃| Also if vector 풂and 풃 are parallel, then there exist a scalar 푘 such that

풂 = 푘풃 Two vector 풂 and 풃 are perpendicular (orthogonal) if their dot product is 0; i.e. 풂 • 풃 = 0 Parallel vectors moves in the same direction.

Example 17.14

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i. Given that 풂 = and풃 = ퟓퟐ . Find 풂 • 풃.

ii. Given that 풖 = 2퐢 − 5퐣 풗 = 퐢 + 3퐣. Find 풖 • 풗 iii. If the vectors 푎퐢 + 푏퐣 and 2퐢 − 3퐣 are perpendicular, find 푎 if 푏 = −1. iv. Show that the vectors 퐢 − 3퐣 and −2퐢+ 6퐣 are parallel.

Solution i. We have, 풂 = and 풃 = ퟓ

ퟐ 풂 • 풃 = 푎 푏 + 푎 푏 ⟹ 풂 • 풃 = 3 ∙ 5 + (−2) ∙ 2 = 15 − 4 = 11 ii. . We have, 풖 = 2퐢 − 5퐣 풗 = 퐢 + 3퐣 풖 • 풗 = 푎 푏 + 푎 푏 ⟹ 풖 • 풗 = 2 ∙ 1 + (−5) ∙ 3 = 2 − 15 = 13 iii. We have, 풖 = 푎퐢+ 푏퐣 풗 = 2퐢 − 3퐣 if 풖 and 풗are perpendicular then 풖 • 풗 = 푎 푏 + 푎 푏 = 0 ⟹ 푎 ∙ 2 + 푏 ∙ (−3) = 0 ⟹ 2(1) − 3푏 = 0 ⟹ 3푏 = 2 ∴,푏 = iv. Method 1 We have, 풖 = 퐢 − 3퐣 풗 = −2퐢 + 6퐣 풖 • 풗 = 푎 푏 + 푎 푏 ⟹ 풖 • 풗 = 1 ∙ (−2) + (−3) ∙ 6 = −2 − 12 = −20 |풖| = (1) + (−3) = √1 + 9 = √10 |풗| = (−2) + (6) = √4 + 36 = 2√10 If 풖 and 풗 are parallel to each other, then |풖 • 풗| = |풖||풗| Now, |−20| = 20 |풖||풗| = √10 × 2√10 = 20 ∎

Method 2 If 풖 and 풗 are parallel to each other, then there exist a scalar 푘 such that 풖 = 푘풗 ⟹ 퐢 − 3퐣 = 푘(−2퐢+ 6퐣) ⟹ 퐢 − 3퐣 = −2푘퐢+ 6푘퐣 ⟹ 1 = −2푘 ⟹ 푘 = − and −3 = 6푘 ⟹ 푘 = − = − ∎

Example 17.15 Given the vectors 푥 퐢 + 푥퐣 and 2퐢 + 퐣are perpendicular, find the possible values of 푥.

Solution Let 풖 = 푥 퐢 + 푥퐣 풗 = 2퐢+ 퐣 if 풖 and 풗are perpendicular then 풖 • 풗 = 푎 푏 + 푎 푏 = 0 ⟹ 푥 ∙ 2 + 푥 ∙ 1 = 0 ⟹ 2푥 + 푥 = 0 ⟹ 푥(2푥 + 1) = 0 ⟹ 푥 = 0 or 2푥 + 1 = 0 ∴,푥 = 0 or 푥 = −

Example 17.16 Given that 풖 = 푎퐢 + 푏퐣,퐯 = 퐢, 퐰 = 4퐢 − 3퐣, where 푎 and 푏 are scalars. Find풖 such that i. 풖 has a magnitude twice that of 풗 and is perpendicular to 풘. ii. 풖 has a magnitude twice that of 풗 and is parallel to 풘.

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Solution We have 풖 = 푎퐢 + 푏퐣, 퐯 = 퐢, 퐰 = 4퐢 − 3퐣 i. We want to find a vector 풖 such that |풖| = 2|풗| and 풖 • 풘 = 0

|풖| = √푎 + 푏 |풗| = + 0 = ⟹ √푎 + 푏 = 2 = 1

(Square both side) 푎 + 푏 = 1 … … … (1) And 풖 • 풘 = 0 ⟹ 푎 ∙ 4 + 푏 ∙ (−3) = 0 4푎 − 3푏 = 0 … … … (2)

From (2) 푎 = 푏. Put 푎 = 푏 into (1). ⟹ 푏 + 푏 = 1

⟹ 푏 + 푏 = 1 (Multiply both sides by 16) ⟹ 9푏 + 16푏 = 16

⟹ 25푏 = 16 ⟹ 푏 = 푏 = ± = ± Put 푏 = ± into (2)

⟹ 푎 = ∙ ± = ± ∴, 푢 = 퐢+ 퐣 or 푢 = − 퐢 − 퐣 ii. We want to find a vector 풖 such that |풖| = 2|풗| and |풖 • 풘| = |풖||풘|

|풖| = √푎 + 푏 |풗| = + 0 = ⟹ √푎 + 푏 = 2 = 1

(Square both side) ⟹ 푎 + 푏 = 1 … … … (1) |풘| = 4 + (−5) = 5 And |풖 • 풘| = |풖||풘| ⟹ |푎 ∙ 4 + 푏 ∙ (−3)| = 5√푎 + 푏 ⟹ |4푎 − 3푏| = 5√푎 + 푏 ⟹ |4푎 − 3푏| = 5 (Square both sides) ⟹ (4푎 − 3푏) = 25 ⟹ (4푎 − 3푏) = 25 = 5 ⟹ 4푎 − 3푏 = 5 … … … (2)

From (2) 푎 = 푏. Put 푎 = into (1). ⟹ + 푏 = 1

⟹ + 푏 = 1 (Multiply both sides by 16) ⟹ 25 + 30푏 + 9푏 + 16푏 = 16 ⟹ 25푏 + 30푏 + 25 − 16 = 0 ⟹ 25푏 + 30푏 + 9 = 0 ⟹ 25푏 + 15푏 + 15푏 + 9 = 0 ⟹ 5푏(5푏+ 3) + 3(5푏 + 3) = 0 ⟹ (5푏 + 3)(5푏 + 3) = 0 ⟹ 푏 = −

Put 푏 = − into (1) 푎 = = = =

∴, 푢 = − 퐢+ 퐣 Try: 1. If the vectors 푥퐢 + 푦퐣 and 3퐢 − 퐣 are perpendicular, find 푦 when 푥 = 2.

(SSSCE)

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2. Given that 풑 = , 풒 = and풓 = , find a vector s such that i. |풔| = 35 is and풔 is in the direction of 풒 − ퟓ풓 ii. 풔is parallel to풑 − 풒 and is half the size of 풑 − 풒.

(SSSCE) 3. A vector 푎퐢+ 푏퐣, where 푎 and 푏 are scalars, has its magnitude twice that of the vector 퐢 + 3퐣 and is parallel to the 3퐢 − 4퐣. Find the vector.

(SSSCE) Cosine rule

Observe the figure below carefully: 퐷 C 풂 − 풃 풃 풂 + 풃

훼 휃 훼 퐴 풂 B

The Figure is a parallelogram 퐴퐵퐶퐷. If we take ⧍퐴퐵퐷 and draw the perpendicular ℎ from 퐷 to 퐴퐵 we have: From this Figure 퐴퐵⃗ + 퐵퐷⃗ = 퐴퐷⃗ 퐷 ⟹ 퐵퐷⃗ = 퐴퐷⃗ − 퐴퐵⃗ = 풃 − 풂 풃 풂 − 풃 but 퐷퐵⃗ = −퐵퐷⃗ = 풂 − 풃 ℎ 휃 풙 풂 − 풙 퐴 푁 퐵 풂 Let 퐴퐵⃗ = |풂|, 퐴퐷⃗ = |풃|, 퐷퐵⃗ = |풂 − 풃| = |풄| 퐴푁⃗ = |풙| and ∠퐵퐴퐷 = 휃, where 휃 is the angle between the initial points of 풂and 풃and 풄 is a vector. sin 휃 =

|풃| ⟹ |풃| sin휃 = ℎ cos휃 = |풙|

|풃| ⟹ |풃| cos휃 = |풙|

From Pythagoras theorem: |풄| = ℎ + (|풂|− |풙|) = (|풃| sin 휃) + (|풂| − |풃| cos휃) = |풃| sin 휃 + |풂| − 2|풂||풃|cos휃 + |풃| cos 휃

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= |풂| + |풃| sin 휃 + |풃| cos 휃 − 2|풂||풃|cos휃 = |풂| + |풃| (sin 휃 + cos 휃) − 2|풂||풃|cos휃 = |풂| + |풃| − 2|풂||풃|cos휃

⟹ |풂 − 풃| = |풂| + |풃| − 2|풂||풃|푐표푠휃… … …(1) Recall that |풂| = 풂 , |풃| = 풃 , and |풄| = 풄 ⟹ |풂| = 풂 and |풃| = 풃 and |풄| = 풄

∴, (풂 − 풃) = 풄 = 풂 + 풃 − ퟐ풂풃풄풐풔휽 Now, we take ⧍퐴퐵퐶 and draw the perpendicular 퐻 from 퐶 to the line 퐴퐵.

퐶 From this Figure 퐴퐵⃗ + 퐵퐶⃗ = 퐴퐶⃗ ⟹ 풂+ 풃 = 퐴퐶⃗

풂 + 풃 풃 퐻

풙 풂 − 풙 휶 퐴 푃 퐵 풂

Let 퐴퐵⃗ = |풂|, 퐵퐶⃗ = |풃|, 퐴퐶⃗ = |풂+ 풃| = |풅|, |퐴푃| = |풙|and∠퐴퐵퐶 = 훼, where훼is the angle between terminal point of 풂and the initial point of 풃 and 풅 is a vector. sin 훼 =

|풃| ⟹ |풃|sin훼 = 퐻cos훼 = |풂| |풙|

|풃| ⟹ |풃|cos훼 = |풂| − |풙| and |풙| = |풂| − |풃|cos훼

From Pythagoras theorem: |풅| = |풙| + 퐻 = (|풂| − |풃|푐표푠훼) + (|풃|푠푖푛훼) = |풂| − 2|풂||풃|푐표푠훼 + |풃| 푐표푠 훼 + |풃| 푠푖푛 훼 = |풂| + |풃| (푐표푠 훼 + 푠푖푛 훼)− 2|풂||풃|푐표푠훼 On the straight line 퐴퐵훼 + 휃 = 180° ⟹ 훼 = 180°− 휃

⟹ |풅| = |풂| + |풃| (cos 훼 + sin 훼)− 2|풂||풃|cos(180°− 훼) ⟹ |풂+ 풃| = |풂| + |풃| (푐표푠 훼 + 푠푖푛 훼) − 2|풂||풃| cos(180°− 훼) … … … (2)

But |풂| = 풂 , |풃| = 풃 , and |풅| = 풅 ⟹ |풂| = 풂 and |풃| = 풃 and |풄| = 풄 ∴, (풂+ 풃) = 풅 = 풂 + 풃 − ퟐ풂풃cos(180°− 훼)

If 휃 = 훼 = 90°, then in both cases we have 풄 = 풂 + 풃 and 풅 = 풂 + 풃 , since 푐표푠90° = 0

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These mean that 풂and 풃are perpendicular to each other. Finding the angle between two vectors

The angle between two nonzero vector with the same initial point is the smallest angle between them. Let 풂, 풃be two vectors and let 휃 be the smallest angle between them. Then

푐표푠휃 =풂 • 풃|풂||풃|

= 풂 • 풃

Proof Let 풂(푎 , 푎 ), 풃(푏 , 푏 )and let 풂 − 풃 = (푎 − 푏 , 푎 − 푏 )

|풂 − 풃| = (푎 − 푏 ) + (푎 − 푏 ) By squaring both sides, we have |풂 − 풃| = (푎 − 푏 ) + (푎 − 푏 ) = 푎 − 2푎 푏 + 푏 + 푎 − 2푎 푏 + 푏 = 푎 + 푎 + 푏 + 푏 − 2푎 푏 − 2푎 푏 = 푎 + 푎 + 푏 + 푏 − 2(푎 푏 + 푎 푏 ) Recall that 풂 • 풃 = 푎 푏 + 푎 푏 , |풂| = 푎 + 푎 and |풃| = 푏 + 푏

⟹ |풂 − 풃| = |풂| + |풃| − 2풂 • 풃… … …(3) From (1) above we have

|풂 − 풃| = |풂| + |풃| − 2|풂||풃|cos휃 By comparing (1) and (3) we have

−2풂 • 풃 = −2|풂||풃|푐표푠휃 ⟹ 풂 • 풃 = |풂||풃|cos휃… … …(4)

cos휃 =풂 • 풃|풂||풃| = 풂 • 풃

Geometrically, the dot product of two nonzero vectors 풂and 풃is defined by

풂 • 풃 = |풂||풃|푐표푠휃 If 휃 = 90°, we have 풂 • 풃 = 0. Definition: Two vectors are perpendicular if the angle between them is 90°.

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Note: i. (1) and (2) can be used to find a length of a triangle given two side and an included angle. ii. (1), (2) and (3) can be used to find the angle between two vectors.

Example 17.17 Vectors 풂 and 풃 defined such that |풂| = 5units and |풃| = 4units and |풂 + 풃| = 3√3 units. Find i. the angle between 풂and 풃ii) the scalar product of 풂and 풃.

Solution |푎 + 푏| = |푎| + |푏| − 2|푎||푏| cos(180°− 훼)

⟹ 3√3 = 5 + 4 − 2(4)(5) cos(180°− 훼) ⟹ 27 = 25 + 16 − 40 cos(180°−훼) ⟹ 40 cos(180°− 훼) = 25 + 16− 27 ⟹ cos(180°− 훼) =

⟹ (180°− 훼) = cos 180°− 훼 = 69.5127°

∴, 훼 = 180° − 69.5127° = 110.4877° 풂⦁풃 = |풂||풃| cos훼 ⟹ 풂⦁풃 = 5 ∙ 4 cos 110.4877° = −17.2404 ∴,풂⦁풃 = −17.2 (Corrected to 1 place of decimal)

Example 17.18 Vectors 풑 and 풒 are defined such that |풑| = 9units and |풒| = 7units and |풑 − 풒| = √23units. Find i. the angle between 풑and 풒ii) the dot product of 풑and 풒.

solution |푝 − 푞| = |푝| + |푞| − 2|푝||푞| cos휃

⟹ √23 = 9 + 7 − 2(9)(7) cos 휃 ⟹ 23 = 81 + 49 − 126 cos휃 ⟹ 126 cos휃 = 81 + 49− 223 ⟹ cos휃 = = cos휃 ≈ 0.8497 ⟹ 휃 = cos (0.8497) ∴,휃 = 31. 82° 풑⦁풒 = |풑||풒| cos휃 ⟹ 풑⦁풒 = 9 ∙ 7 cos31.8209 ° = 53.53 ∴,풑⦁풒 = 53.53

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Example 17.19 Given that 풖 =3i+4j and 풗 =i−2j. Find the angle between 풖 and 풗.

Solution We have, 푢 = 3퐢+ 4퐣 and 푣 = 퐢 − 2퐣 풖⦁풗 = 푎 푏 + 푎 푏 ⟹ 풖⦁풗 = 3 ∙ 1 + 4 ∙ (−2) = 3 − 8 = −5 |풖| = (3) + (4) = √9 + 16 = 5 |풗| = (1) + (−2) = √1 + 4 = √5 풖⦁풗 = |풖||풗| cos 휃

⟹ −5 = 5 × √5 cos휃 ⟹ cos휃 = −√

= −√

⟹ 휃 = cos − √ = 63.4349° ∴,the angle between 푢 and 푣 is 63.43° (corrected to 2 places of decimal).

Example 17.20 The points 푃(7,−2),푄(−3, 2) and 푅(0, 5)are the points of the vertices of a rectangle in the 푂푥푦 plane. 푀(푟, 푠) is the foot of the perpendicular from 푅 to 푃푄. Using dot product, find i. ∠푃푄푅 ii. an expression in terms of 푠 and 푟.

Solution We have,푃(7,−2) 푄(−3, 2) 푅(0, 5) 푀(푟, 푠)

푃푄⃗ = 푂푄⃗ − 푂푃⃗ = −32 − 7

−2 = −104 푀푅⃗ = 푂푅⃗ − 푂푀⃗ = 0

5 − 푟푠 = −푟

5 − 푠

Since, 푂푀 is perpendicular to 푃푄⃗, it follows that 푀푅⃗⦁푃푄⃗ = 풐

⟹ −104 ⦁ −푟

5 − 푠 = 00 ⟹ 10푟 + 4(5 − 푠) = 0 ⟹ 10푟 + 20 − 4푠 = 0

(Divide both sides by 2) ⟹ 5푟 − 2푠 + 10 = 0

푄푅⃗ = 푂푅⃗ − 푂푄⃗ = 05 − −3

2 = 33

Now, 푃푄⃗ = (−10) + (4) = √100 + 16 = √116 = 4√29 units

푄푅⃗ = 3 + (3) = √9 + 9 = √18 = 3√2 units Try: 1. 퐴퐵퐶 is a triangle in which 퐵퐶⃗ = 풂,퐶퐴⃗ = 풃,퐵퐴⃗ = 풄and∠퐴퐵퐶 = 휃 i. If one of the vectors can be expressed in terms of the other two, write down the expression ii. Using dot product show that 풄 = 풂 + 풃 − 2풂풃푐표푠휃, where 풂 = |풂|,풃 = |풃|풄 = |풄| iii. For what value of 휃 will the result in (ii) represent the Pythagoras theorem.

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[Hint: see(3)] (SSSCE) 2. Using the scalar (dot) product, find, correct to the nearest degree, the angle between the vectors 풂 = 3i+j and 풃 = 8i+9j (SSSCE) 3. Vectors 풖and 풗are such that |풖| = 3푐푚, |풗| = 10푐푚and |풖+ 풗| = √139. Find i. the angle between 풖and 풗 ii. the scalar (dot) product of 풖and 풗. (SSSCE) 4. The points 퐴(4, 1),퐵(−4, 4) and 퐶(4, 6). A point 푃(푎, 푏)is the foot of the perpendicular from 퐶 to 퐴퐵. By using scalar product, find i. ∠퐴퐵퐶 ii. An expression in terms of 푎 and 푏. (SSSCE) 5. Given that 풑and 풒are position vectors of 푂푃⃗ and 푂푄⃗ respectively, show that |풒 − 풑| = |풑| + |풒| − 2|풑||풒|푐표푠휃, where 휃 is the angle between 푂푃⃗and 푂푄⃗. (SSSCE) 6. Find correct to the nearest degree, the angle between the vectors 풂 = ퟑ and 풃 = ퟖ . (SSSCE) 7. Given that 풂 = 3i−4j and 풃 =6i+4j, find, the angle between the two vectors, correct to the nearest degree. (SSSCE) 8. The vectors 풑and 풒are such that |풑| = 5푐푚, |풒| = 12푐푚 and |풑+ 풒| = 15.7푐푚. Find i. the angle between 풑and 풒; ii. the scalar (dot) product of 풑and 풒. (SSSCE) 9. a. If 푃푄⃗ = and 푅푄⃗ = , express 푃푅⃗ as a column vector. b. The position vectors of points 퐴,퐵,퐶 and 퐷 are , and respectively.

i. Show that 퐴퐵⃗ is perpendicular to 퐶퐷⃗. ii. Calculate, correct to the nearest degree, angle 퐴퐷퐵. (WASSCE) 10. Given that the angle between 풑 = 2퐢+ 3퐣 and 풒 = −2퐢 + 푚퐣 is cos − , find the

constant 푚.

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Sine rule Observe the Figures below carefully:

B 퐵

훾 풃 풃 풂 ℎ 풂ℎ 훽 훼 훼 퐴 풄 퐶 퐴 풄 퐶

Not drawn to scale The Figures 푎 and 푏 represent the same ⧍퐴퐵퐶.in Figure푎 a perpendicular ℎ is drawn from 퐵 to meet 퐴퐵 at 푁 while in 푏 the perpendicular ℎ is drawn from 퐴 to meet 퐵퐶 at 푃. Let 퐴퐵⃗ = |풂|, 퐵퐶⃗ = |풃|, 퐴퐶⃗ = |풄|,∠퐵퐴퐶 = 훽,∠퐴퐵퐶 = 훾and ∠퐴퐶퐵 = 훼 Let’s take Figure 푎

sin 훼 =|풃|

⟹ ℎ = |풃| sin 훼 sin 훽 =|풂|

⟹ ℎ = |풂| sin훽

If ℎ = |풃| sin 훼 and ℎ = |풂| sin 훽 then, ⟹ |풃| sin 훼 = |풂| sin 훽 ⟹ |풂| = |풃| (Divide both side by sin훼 sin 훽)

Let’s take Figure 푏 sin 훼 =

|풄| ⟹ ℎ = |풄| sin훼 sin 훾 =

|풂| ⟹ ℎ = |풂| sin 훾

⟹ |풄| sin훼 = |풂|푠푖푛훾 ⟹ |풂| = |풄| (Divide both side by sin 훼 sin 훾)

Now, since ℎ = |풃| sin훼 = |풂| sin훽 and ℎ = |풄| sin 훼 = |풂| sin 훾, we write |풂| = |풃| = |풄|

… … …(5)

Recall that |풂| = 풂 , |풃| = 풃 , and |풄| = 풄 ⟹ |풂| = 풂 and |풃| = 풃 and |풄| = 풄 ∴ 풂 = 풃 = 풄

This rule can be used to calculate i. a length of a triangle given the length of one side and two included angles opposite to the sides. ii. an angle between two adjacent sides of a triangle given the length of two sides and an included angle opposite to one of the given sides.

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Example 17.21 If 푋(2,−1),푌(6,−7) and 푍(−3, 6) are the vertices of a triangle in the 푂푥푦 plane. Find i. 푋푌⃗ , 푋푍⃗ and 푌푍⃗ ii. the angle ∠푋푌푍 using the inner product. iii. ∠푌푋푍

Solution We have, 푋(2,−1), 푌(6,−7)and 푍(−3, 6) i. 푋푌⃗ = 푂푌⃗ − 푂푋⃗ = (6 − 2,−7 + 1) = (4,−6) 푌(6,−7) 푋푌⃗ = (4) + (−6) = √16 + 36 = √52 = 2√13units. 훼 푋푍⃗ = 푂푍⃗ − 푂푋⃗ = (−3 − 2, 6 + 1) = (−5, 7) 휃 푍(−3, 6) 푋(2,−1) 푋푍⃗ = (−5) + (7) = √25 + 49 = √74units. 푌푍⃗ = 푂푍⃗ − 푂푌⃗ = (−3 − 6, 6 + 7) = (−9, 13) 푌푍⃗ = (−9) + (13) = √81 + 169 = √250 = 5√10units. ii.푋푌⃗⦁푌푍⃗ = 푋푌⃗ 푌푍⃗ cos훼 ⟹ (4,−6)⦁(−9, 13) = 2√13 × 5√10 cos훼 4(−9) + (−6)(13) = 2√13 × 5√10 cos훼 ⟹ −36− 78 = 2√13 × 5√10 cos 훼 cos훼 = −

√ × √≈ −0.9998 ⟹ 훼 ≈ cos (−0.9998) ≈ 178.49°

∴, ∠푋푌푍 = 178.49° iii.푋푌⃗⦁푋푍⃗ = 푋푌⃗ 푋푍⃗ cos휃 ⟹ (4,−6)⦁(−5, 7) = 2√13 × √74 cos휃 4(−5) + (−6)(7) = 2√13 × √74 cos 휃 ⟹ −20 − 42 = 2√13 × √74 cos휃 cos휃 = −

√ ×√≈ −0.9995 ⟹ 휃 ≈ cos (−0.9995) ≈ 178.20°

∴,∠푌푋푍 = 178.20° Try: 1. In a triangle 푃푄푅, ∠푃푄푅 = 78°, ∠푃푅푄 = 32°and 푃푄⃗ = 5√5units. Find 푃푅⃗ . 2. b. A triangle 퐴퐵퐶 lies in the 푂푥푦 plane such that 퐴퐵⃗ = and 퐵퐶⃗ = . i) Use the scalar product to find ∠퐴퐵퐶 ii) Find ∠퐴퐶퐵.

Vector projections Consider the diagram below.

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퐴 풂 푂 휽 풃 푁 퐵 Let the position vectors 푂퐴⃗ = 풂 and 푂퐵⃗ = 풃. Let 푂퐴⃗ = |풂| and 푂퐵⃗ = |풃|

From ⧍푂퐴푁 cos휃 = | || |

⟹ |푂푁| = |푂퐴| cos 휃 = |풂| cos휃

But we know that cos휃 = 풂•풃|풂||풃|

⟹ |푂푁| = |풂| ∙ 풂•풃|풂||풃|

= 풂•풃|풃|

⟹ |푂푁| = 풂 • 풃

This refers to the projection of 풂in the direction of 풃. This is a scalar and is denoted by

푝푟표푗풃풂 =풂•풃

|풃| = 풂 • 풃

The vector component of 풂onto 풃(along 풃)is given by

푐표푚푝풃 = 푝푟표푗풃풂 ∙ 풃

|풃|= 풂•풃

|풃|∙

|풃|= 풂 • 풃 풃

This is a vector. Example 17.22

Given that 풂 = 3퐢+ 5퐣 and 풃 = 퐢 − 3퐣,find i. the vector component of 풂in the direction of 풃. ii. the projection of 풂along 풃 iii. the angle between 풂and 풃.

Solution We have, 풂 = 3퐢+ 5퐣 and 풃 = 퐢 − 3퐣

|풃| = (1) + (−3) = √1 + 9 = √10 풃 = 풂|풃|

= 풊 퐣√

= √ 풊 − √ 퐣

풂⦁풃 = (3퐢+ 5퐣)⦁ √ 풊 − √ 퐣 = √ − √ = − √

i. comp풃풂 = 풂⦁풃(풃) ⟹ comp풃풂 = −6√10

5

√10

10퐢 −

3√10

10퐣 = −

60

50퐢 +

180

50퐣 = −

6

5퐢 +

18

5퐣

∴, comp풃풂 = −6

5퐢 +

18

5퐣

ii. proj풃풂 = 풂⦁풃 ∴, proj풃풂 = − √

iii. 풂⦁풃 = (3퐢 + 5퐣)⦁(풊 − 3퐣) = 3 − 15 = −12 |풂| = (3) + (5) = √9 + 25 = √34

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풂⦁풃 = |풂||풃| cos휃 ⟹ −12 = √10 × √34 cos휃 ⟹ cos휃 = −√ ×√

⟹ cos휃 ≈ −0. 6508 ⟹ 휃 ≈ cos (−0.6508) ≈ 130.60° ∴,the angle between 풂 and 풃 is 130.60° Try: 1. Given that 풂 = ퟔ풊+ ퟒ풋 and 풃 = 풊 − ퟐ풋,find i the projection of 풂in the direction of 풃ii) the vector component of 풃 along 풂. 2. b. if 풗 = 퐢 + ퟔ퐣and풘 = ퟓ퐢 − 퐣,find i. the projection of 풘in the direction of 풗. Ii. the vector component of 풘along 풗. Definition: A bearing is an angle measured in a clockwise direction. As we have already noted the vector, acceleration, can be written in the form (푎ms ,휃), where 푎ms denotes the magnitude and 훼 denotes the angle direction of the vector respectively. A particular vector can move in one of the following directions: North(N), North-East (NE), East(E), South-East(SE), South(S), South-West(SW), West (W) and North-West(NW). 푁 360°/0° 푁푊 푁퐸 Clockwise 푊 퐸 270°90° 푆푊 푆퐸 푆 180°

Exercise

452

Draw the following bearings: i. 30° ii. 145° iii. 210° iv. 330° Try: Draw the following bearings: i. 20° ii. 300°

Changing a component vector to its magnitude-direction/bearing form

Let 풂(푎 , 푎ퟐ) be a vector in the 푂푥푦plane. Consider the following figures: 푦 푦 풂 푎 휃 푎 휃 훼 −푥 푥 −푥 푎 푥 푎 풂 −푦 −푦

In this Figure 푎 > 0 and 푎 > 0 In this Figure 푎 > 0 but 푎 < 0 |풂| = 푎 + 푎 |풂| = 푎 + 푎

The angle direction of 풂 is 휃. This can be The angle direction of 풂is 휃. This can

calculated as follow: calculated as follow: 퐬퐢퐧휽 = 풂ퟏ

|풂| ⟹ 휽 = 퐬퐢퐧 ퟏ 풂ퟏ|풂| 퐬퐢퐧(휽 − ퟗퟎ°) = − 풂ퟐ

|풂| ⟹ 휽 = 퐬퐢퐧 ퟏ − 풂ퟐ|풂| +90°

퐜퐨퐬 휽 = 풂ퟐ|풂| ⟹ 휽 = 퐜퐨퐬 ퟏ 풂ퟐ

|풂| 퐜퐨퐬(휽 − ퟗퟎ°) = 풂ퟏ|풂| ⟹ 휽 = 퐜퐨퐬 ퟏ 풂ퟏ

|풂| + ퟗퟎ°

퐭퐚퐧휽 = 풂ퟏ풂ퟐ

⟹ 휽 = 퐭퐚퐧 ퟏ 풂ퟏ풂ퟐ

퐭퐚퐧(휽− ퟗퟎ°) = − 풂ퟐ풂ퟏ

⟹ 휽 = 퐭퐚퐧 ퟏ − 풂ퟐ풂ퟏ

+ ퟗퟎ°

푦 푦

풂

푎 −푥 휃 푥 −푥 푎 휃 푥 푎

풂 푎

−푦 −푦 In this Figure 푎 < 0 and 푎 < 0 In this Figure 푑 where 푎 < 0 and 푎 > 0

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|풂| = 푎 + 푎 |풂| = 푎 + 푎 The angle direction of 풂is 휃. This can be The angle direction of 풂is 휃. This can be

calculated as follow: calculated as follow: 퐬퐢퐧(휽 −ퟏퟖퟎ°) = − 풂ퟏ

|풂| ⟹ 휽 = 퐬퐢퐧 ퟏ − 풂ퟏ|풂| +180° 퐬퐢퐧( 휽 −ퟐퟕퟎ°) = 풂ퟐ

|풂| ⟹ 휽 = 퐬퐢퐧 ퟏ 풂ퟐ|풂| +270°

퐜퐨퐬(휽 − ퟏퟖퟎ°) = − 풂ퟐ|풂| ⟹ 휽 = 퐜퐨퐬 ퟏ − 풂ퟐ

|풂| + ퟏퟖퟎ° 퐜퐨퐬(휽 −ퟐퟕퟎ°) = − 풂ퟏ|풂| ⟹ 휽 = 퐜퐨퐬 ퟏ − 풂ퟏ

|풂| + ퟐퟕퟎ°

퐭퐚퐧(휽 − ퟏퟖퟎ°) = 풂ퟏ풂ퟐ

⟹ 휽 = 퐭퐚퐧 ퟏ 풂ퟏ풂ퟐ

+ ퟏퟖퟎ° 퐭퐚퐧(휽− ퟐퟕퟎ°) = − 풂ퟐ풂ퟏ

⟹ 휽 = 퐭퐚퐧 ퟏ − 풂ퟐ풂ퟏ

+ ퟐퟕퟎ°

Note16. Example 17.23

Find the magnitude and direction of the following vectors: i. 풂 = 6퐢 + 8퐣 ii. 퐴퐵⃗ = iii. 풑 = −2퐢 − 퐣 iv. 푃푄⃗ = v. 풃 = 7퐢 iv. 풄 = −3퐣

Solution i. We have, 풂 = 6퐢+ 8퐣 Let 푎 = 6 and 푎 = 8 |풂| = (6) + (8) = √36 + 64 = √100 = 10 units

Since, 풂 lies in the positive 푥 and 푦 direction, 휃 = tan ⟹ 휃 = tan

⟹ 휃 ≈ 36.87° ∴,풂 = (10units, 36.87°)

ii. We have, 퐴퐵⃗ = 1−3 Let 푎 = 1 and 푎 = −3

퐴퐵⃗ = (1) + (−3) = √1 + 9 = √10units

Since, 퐴퐵⃗ lies in the positive 푥 and negative 푦 direction, 휃 = tan − + 90°

⟹ 휃 = tan − + 90° ≈ 71.56° + 90° ≈ 161.56°

∴, 퐴퐵⃗ = √10units, 108.43° iii. We have, 풑 = −2퐢 − 퐣 Let 푎 = −2 and 푎 = −1 |풑| = (−2) + (−1) = √4 + 1 = √5units

Since, 풑 lies in thenegative푥 and 푦 direction, 휃 = tan + 180°

16In this book we will mostly use tangent to calculate the angle direction

454

⟹ 휃 = tan + 180° ≈ 63.43° + 180° ≈ 243.43°

⟹ 휃 ≈ 243.43° ∴,풑 = √5units, 243.43°

iv. We have, 푃푄⃗ = −17 Let 푎 = −1 and 푎 = 7

푃푄⃗ = (−1) + (7) = √1 + 49 = √50 = 5√2units

Since, 푃푄⃗ lies in the negative 푥 and positive푦 direction, 휃 = tan − + 270°

⟹ 휃 = tan − + 270° ≈ 81.87° + 270° ≈ 351.87°

∴, 푃푄⃗ = 5√2units, 351.87° v. We have, 풃 = 7퐢 Let 푎 = 7 and 푎 = 0 |풃| = (7) + (0) = √49 + 0 = 7units

Since, 풃 lies in the positive푥 and 푦 direction, 휃 = sin|풃|

⟹ 휃 = sin = 90° ⟹ 휃 = 90° ∴,풃 = (7units, 90°)

vi. We have, 풄 = −3퐣 Let 푎 = 0 and 푎 = −3 |풄| = (0) + (−3) = √0 + 9 = 3units

Since, 풄 lies in the positive 푥 and negative푦 direction, 휃 = sin −|풄|

+ 90°

⟹ 휃 = sin − + 90° = 90° + 90° = 180° ⟹ 휃 = 180° ∴, 풄 = (3units, 180°)

(Refer to the figures above) Try: 1. Two vectors are defined as; 풑 =6i+8j 풒 = −8i+ퟕ

ퟑj. Find the

i. magnitudes and direction of 풑and 풒; ii. angle between 풑and 풒. (SSSCE)

2. Find, correct to three significant figures, the magnitude and direction of the vectors 풂 = 퐢 − 퐣 and 풃 = − 퐢 − 퐣.

455

Changing a vector from its magnitude-direction form to its component form Let 풂 = (munits,휃) be a vector in the 푂푥푦plane, where 푚 is a scalarand 휃 is the angle direction of the vector. We can resolve the vector into its component form by one of the following methods: Method 1 i. Draw the direction of the vector either by a rough sketch or by an appropriate scale. ii. Choose the acute angle that the length of the vector mixes with the 푥-axis. iii. Write the 푥 component using the 푐표푠푖푛푒of the acute angle and 푦 component using the 푠푖푛푒 of the acute angle. iv. Apply appropriate signs corresponding to the 푥-and-y axes in the direction of the vector.

Method 2 Given the vector 풂write: 풂 = as a component vector.

Example 17.24 Find the components of the following vectors using the methods described in this book. i. 푃푄⃗ = (12km, 060°) ii. 퐹퐺⃗(6N, 150°) iii. 퐴퐶⃗(5ms , 210°) iv) 풂(15units, E)

Solution

i. We have, 푃푄⃗ = (12km, 060°) 푃푄⃗ = 12 sin 60°12 cos 60 ° =

12 × √

12 ×= 6√3

6

∴,푃푄⃗ = 6√36

ii. We have, 퐹퐺⃗ = (6N, 150°) 퐹퐺⃗ = 6 sin 150°6 cos 150° =

6 ×

6 × − √= 6√3

6

∴,퐹퐺⃗ = 3−3√3

iii. We have, 퐴퐵⃗ = (5ms , 210°) 퐴퐵⃗ = 5 sin 210°5 cos 210° =

5 × −

5 × − √=

−

− √

456

∴, 퐴퐵⃗ =−

− √

iv. We have, 풂 = (15units, 060°) = (15units, 090°) 풂 = 15 sin 90°12 cos 90 ° = 15

0

∴,풂 = 150

Try: 1. Find the components of the following vectors. i. 퐴퐵⃗ = (16km, 030°) ii. 퐹퐺⃗(13N, 160°) iii. 푉퐶⃗(5푚푠 , W) iv. 풂(10units,푁)

Resultant of vectors Definition: A resultant vector is a vector that result from the addition of two or more vectors. Let 풓 = (푎unit,훼), 풔(푏unit,훼) and 풕(푐unit,훼) be three vectors. The resultant of the vectors, denote d by 푅⃗ is given by 푅⃗ = 풓 + 풔 + 풕 = (푎 + 푏 + 푐unit,훼), where 푎,푏 and 푐 are scalars and 훼 is the angle direction of the three vectors.

Example 17.25 Given that 풂(2m, 060°),풃(5m, 060°) and 풄(7m, 060°). Show that resultant of the vectors is (14m, 060°)

Solution We have, 풂(2m, 060°), 풃(5m, 060°) and 풄(7m, 060°)

풂 = 2 sin 60°2 cos 60 ° = √3

1 풃 = 5 sin 60°

5 cos 60 ° =√

풄 = 7 sin 60°7 cos 60 ° =

√

푅⃗ = 풂 + 풃 + 풄 ⟹ 푅⃗ = √31

+√

+√

= 7√37

Now, 푅⃗ = 7√3 + (7) = √147 + 49 = √196 = 14m

Let us find the angle of the resultant vector.

457

Let 푎 = 7√3 and 푎 = 7. Since the resultant vector is in the direction of the

positive 푥 and 푦 direction, 휃 = tan ⟹ 휃 = tan √ = 60°

푅⃗ = (14m, 060°) Example 17.26

Find the resultant of the vectors 풓(13km, 150°)and 풔(7km, 220°). Solution

We have, 풓(13km, 150°), 풔(7km, 220°)

풓 = 13 sin 150°13 cos 150 ° ≈ 6.5

−11.2583 풔 = 7 sin 220°7 cos 220 ° ≈ −4.4995

−5.3623

푅⃗ = 풓 + 풔 ⟹ 푅⃗ ≈ 6.5−11.2583 + −4.4995

−5.3623 ≈ 2.0005−16.6206

푅⃗ ≈ (2.0005) + (−16.6206) ≈ √4.0020 + 276.2443 ≈ √280.2463 ≈ 16.74units Let us find the angle of the resultant vector. Let 푎 = 2.0005 and 푎 = −16.6206. Since the resultant vector is in the direction of the positive푥 and negative 푦 direction, 휃 = tan − + 90°

⟹ 휃 ≈ tan − ..

+ 90° ≈ 6.8633° + 90° ≈ 96.86

푅⃗ = (16.74units, 96.86°) Example 17.27

Three forces 3i+2j, 7i−j and 4i−8j act on a body. Find the magnitude and direction of the resultant force.

Solution Let 푭 = 3퐢 + 2퐣, 푭 = 7퐢 − 퐣 and 푭 = 4퐢 − 8퐣 푅⃗ = 푭 + 푭 + 푭 ⟹ 푅⃗ = 3퐢 + 2퐣 + 7퐢 − 퐣 + 4퐢 − 8퐣 = 14퐢 − 7퐣 푅⃗ = (14) + (−7) = √196 + 49 = √245 = 7√5N

휃 = tan − + 90° ⟹ 휃 = tan − + 90° ≈ 26.57° + 90° ≈ 116.57°

∴, 푅⃗ = 7√5N, 116.57° Try: 1. Three forces 푭 = 2퐢 − 퐣,푭 = −5퐢+ 3퐣and푭ퟑ = 6퐢 − 퐣act on a particle of mass 0.001푘푔. Find ; i. the acceleration of the particle

458

ii. the acute angle between the resultant force on the particle and the vector i+j , correct to one decimal place. [HintF = ma]

(SSSCE) 2. A body is acted upon by three forces 푭 = (10푁, 090°),푭 = (15푁, 180°)and푭 (5푁, 270°). Find, correct to three significant figures, the magnitude of the resultant force.

(WASSCE) Graphical representation of vectors

Given the magnitudes and direction of vectors we can represent them graphically using the following procedures: i. Choose a scale for the magnitudes of the vector. ii. Use a protractor to measure the direction of the vectors from 0° north one after the other. iii. Draw the magnitude for each vector using a ruler. iv. Join the terminal point terminal point of each vector and measure the magnitude of the resultant vector with the ruler. v. Measure the direction of the resultant vector from 0° north with the protractor.

Example 17.28 A boy travel 50km East and 30km South. Using a scale of 2cm to 10km, find the magnitude and direction of the resultant displacement of the boy.

Example 17.29

A car travel at a velocity of 120kms at 35° and then 220kms at 125°. Using a scale of 2cm to 30kms . Find the magnitude and direction of the resultant displacement. Try: 1. Points 푃,푄and 푅 lies in the plane such that 푃푄⃗(16푘푚, 030°) and 푄푅⃗(15푘푚, 150°). Find the magnitude and direction of the resultant vector 퐴퐶⃗. 2. A wind is blowing from the south at 75knots, and an airplane flies heading east at 100knots. Find the resultant velocity of the airplane. (SSSCE)

General Problems. Example 17.30

459

Two ships 퐴 and 퐵 leave a port 푂 at the same time. Ship 퐴 moves at 16푘푚ℎ on a bearing of 75° and ship 퐵 move on a bearing of 130°. After 2 hours, the bearing of 퐵 from 퐴 is 160°. Calculate, correct to one place of decimal, i. the initial speed of ship 퐵; ii. the distance between the position of the ships at that instance.

Example 17.31 An airplane is travelling at 530mph on a bearing of 125°. The wind is blowing on a bearing of70° at 23mph17*. What is the ground speed and the actual bearing of the airplane.

Example 17.32 An airplane flew from town 푃 at 600mphon a bearing of 150° to town 푄. It then flew 430mph on a bearing of 65° to town 푅. Calculate, correct to one place of decimal. i. the distance of 푃 from 푅 ii. the bearing of 푃 from R. Try: Two ships 푆 and 푇 leave a port 푂 at the same time. Ship 푆 moves at 22푘푚ℎ on a bearing of 125° and 푇 on a bearing of 200°. After 1 ℎ푟푠, the bearing of ship 푇 from 푆 is 240°. Calculate correct to one decimal place, i. the speed of the ship 푇; ii. the distance between the positions of the ships at that instance.

Application of vectors on plane figures 1. Triangle

A triangle is a plane figure with three sides. Let 퐴,퐵 and 퐶 be the vertices of a triangle in the plane. Let ∠퐵퐴퐶 = 휃 and draw a perpendicular ℎ from 퐵 to 퐴퐶. 퐵 ℎ 휃 퐶 퐴 17∗ mile per hour

460

In the Figure above sin 휃 = ⃗ ⟹ 퐴퐵⃗ 푠푖푛휃 = ℎ

Area of a triangle is given by: 퐴 = 푏푎푠푒 × ℎ푒푖푔ℎ푡

∴, Area of the ⧍퐴퐵퐶 = 퐴퐶⃗ ∙ 퐴퐵⃗ 푠푖푛휃 Perimeter of a plane figure is the distance around its edges. ∴ Perimeter of ⧍퐴퐵퐶= 퐴퐵⃗ + 퐵퐶⃗ + 퐴퐶⃗ We can prove the following about the triangle: i. Given that the position vector of 퐴 is 풂, and position vector of 퐵 is 풃, then the midpoint 푀 of 퐴퐵⃗ with position vector 풎 is 풂 풃 ii. If the midpoint of 퐴퐵is 퐷 and midpoint of 퐵퐶 is 퐸 with position vectors then, 퐷퐸⃗ = 풄−풂

2 where, 풃and 풄are the position vectors of 퐴 and 퐵 respectively.

Proof i. We have 푂퐴⃗ = 풂, 푂퐵⃗ = 풃 and 푂푀⃗ = 풎 If 푀 is the midpoint of 퐴퐵, then it divides 퐴퐵 in the ratio 1:1 Using the ratio formula, we have

푂푀⃗ = ∙풂 ∙풃 = 풂 풃 ∴,풎 = 풂 풃 ii. Let position vectors of 퐴,퐵 and 퐶 be 풂,풃and 풄 respectively.

The position vector of 퐷 = 풂+풃2 , since 퐷 is the midpoint of 퐴퐵

The position vector of 퐸 = 풃+풄2 , since 퐸 is the midpoint of 퐵퐶

퐷퐸⃗ = 푂퐸⃗ − 푂퐷⃗ = 풃 풄− 풂 풃=풄 풂 ∎ 2. Quadrilaterals A quadrilateral is a plane figure with four sides. Some quadrilaterals are discussed below. a. Parallelogram A parallelogram is a quadrilateral with equal opposite sides and angle. Example of a parallelogram includes a square, rectangle and rhombus. Consider the Figures below:

461

퐵 퐶 In this Figure, 퐴퐷⃗ = 퐵퐶⃗ ⟹ 퐴퐷⃗ = 퐵퐶⃗ 퐴퐵⃗ = 퐷퐶⃗ ⟹ 퐴퐵⃗ = 퐷퐶⃗ 휃 퐴 퐷 푝푎푟푎푙푙푒푙표푔푟푎푚 Area of parallelogram 퐴퐵퐶퐷 = 푏푎푠푒 × ℎ푒푖푔ℎ푡 = 2(Area of ⧍ABD) = 2 퐴퐵⃗ ∙ 퐴퐷⃗ 푠푖푛휃

= 퐴퐵⃗ ∙ 퐴퐷⃗ 푠푖푛휃 Perimeter of parallelogram 퐴퐵퐶퐷 = 2( 퐴퐵⃗ + 퐴퐷⃗ )

푄 푅 퐵 퐶 퐴 퐷 푟푒푐푡푎푛푔푙푒 푃 푠푞푢푎푟푒 푆

In this Figure, In this Figure, 푃푆⃗ = 푄푅⃗ ⟹ 푃푆⃗ = 푄푅⃗ 퐴퐷⃗ = 퐵퐶⃗ = 퐴퐵⃗ = 퐷퐶⃗

푃푄⃗ = 푆푅⃗ ⟹ 푃푄⃗ = 푆푅⃗ ⟹ 퐴퐷⃗ = 퐵퐶⃗ = 퐴퐵⃗ = 퐷퐶⃗ 푃푆⃗ • 푃푄⃗ = 0 since, ∠푄푃푆=90° 퐴퐵⃗ • 퐴퐷⃗ = 0 since,∠퐵퐴퐷 = 90°

Area of rectangle 푃푄푅푆 = 푙푒푛푔ℎ푡 × 푤푖푑푡ℎ Area of square 퐴퐵퐶퐷= 푙푒푛푔푡ℎ

= 푃푆⃗ ∙ 푃푄⃗ = 푃푆⃗

462

Perimeter of rectangle 푃푄푅푆 = 2( 푃푆⃗ + 푃푄⃗ ) Perimeter of square 퐴퐵퐶퐷=4 퐴퐵⃗ 퐵 퐶 퐴 퐷 푟ℎ표푚푏푢푠 In this Figure,

퐴퐷⃗ = 퐵퐶⃗ = 퐴퐵⃗ = 퐷퐶⃗ ⟹ 퐴퐷⃗ = 퐵퐶⃗ = 퐴퐵⃗ = 퐷퐶⃗ Area of rhombus 퐴퐵퐶퐷 = 퐴퐶⃗ ∙ 퐵퐷⃗ Perimeter of rhombus 퐴퐵퐶퐷 = 4 퐴퐷⃗ We can prove that the diagonals 퐴퐶⃗ and 퐵퐷⃗ of the rhombus bisect at 90°.

Proof Let 퐴퐵⃗ = 퐷퐶⃗ = 풂퐴퐷⃗ = 퐵퐶⃗ = 풃 퐴퐶⃗ = 퐴퐵⃗ + 퐵퐶⃗=풂 + 풃퐵퐷⃗ = 퐵퐴⃗ + 퐴퐷⃗ = −풂+ 풃 = 풃 − 풂 퐴퐶⃗ • 퐵퐷⃗ = (풂 + 풃) • (풃− 풂) = 풂 • 풃 − 풂 • 풂 + 풃 • 풃 − 풂 • 풃 = 풃 − 풂 = |풃| − |풂| By definition 퐴퐷⃗ = 퐵퐶⃗ ⟹ 풂 = 풃 ∴ 퐴퐶⃗ • 퐵퐷⃗ = |풂| − |풂| = 0 ∎

Example 17.33 The vertices 퐴,퐵 and 퐷 of a rectangle 퐴퐵퐶퐷 are (푥, 2), (−1, 3) and (1,−2푥) respectively. i. Find the possible value of 푥 ii. Calculate the area and perimeter of rectangle.

Solution We have, 퐴(푥, 2), 퐵(−1, 3) and 퐷(1,−2푥). 퐴퐵⃗ + 퐵퐷⃗ = 퐴퐷⃗ 퐷(1,−2푥) 퐶 퐴퐵⃗ = 푂퐵⃗ − 푂퐴⃗ = (−1, 3) − (푥, 2) = (−1 − 푥 , 1) 퐵퐷⃗ = 푂퐷⃗ − 푂퐵⃗ = (1,−2푥)− (−1, 3) = (2,−2푥 − 3) 퐴퐷⃗ = 푂퐷⃗ − 푂퐴⃗ = (1,−2푥)− (푥, 2) = (1 − 푥,−2푥 − 2) 퐴(푥, 2) 퐵(−1, 3) 퐴퐵⃗⦁퐴퐷⃗ = ퟎ ⟹ (−1 − 푥, 1)⦁(1 − 푥,−2푥 − 2) = 0 ⟹ (−1 − 푥)(1− 푥) + 1(−2푥 − 2) = 0 ⟹ −1 + 푥 − 푥 + 푥 − 2푥 − 2 = 0 ⟹ 푥 − 2푥 − 3 = 0 ⟹ 푥 + 2푥 + 3 = 0 ⟹ (푥 − 3)(푥 + 1) = 0

463

∴,푥 = 3 or 푥 = −1 ii. When 푥 = 3 퐴퐵⃗ = (−4, 1) 퐵퐷⃗ = (2,−9) 퐴퐷⃗ = (−2,−8) 퐴퐵⃗ = (−4) + (1) = √16 + 1 = √17units

퐴퐷⃗ = (−2) + (−8) = √4 + 64 = √68 = 2√17units Area= 퐴퐵⃗ × 퐴퐷⃗ = √17 × 2√17 = 34units Perimeter = 2 퐴퐵⃗ + 퐴퐷⃗ = 2 √17 + 2√17 = 2 3√17 = 6√17units iii. When 푥 = −1 퐴퐵⃗ = (0, 1) 퐵퐷⃗ = (2,−1) 퐴퐷⃗ = (2, 0) 퐴퐵⃗ = (0) + (1) = √0 + 1 =1 unit

퐴퐷⃗ = (2) + (0) = √4 + 0 = √4 = 2 units Area= 퐴퐵⃗ × 퐴퐷⃗ = 1 × 2 = 2units Perimeter = 2 퐴퐵⃗ + 퐴퐷⃗ = 2(1 + 2) = 2(3) = 6units

Example 17.34 The vertices of a triangle 푃푄푅 are (5,−2), (2, 7) and (4, 6) respectively. Find i. the perimeter ii. ∠푄푃푅 iii. the area of the rectangle

Solution We have, 푃(5,−2), 푄(2, 7) and 푅(4, 6) 푃푄⃗ = 푂푄⃗ − 푂푃⃗ = (2, 7) − (5,−2) = (−3, 9) 푄(2, 7) 푃푅⃗ = 푂푅⃗ − 푂푃⃗ = (4, 6) − (5,−2) = (−1, 8) 휃 푄푅⃗ = 푂푅⃗ − 푂푄⃗ = (4, 6) − (2, 7) = (2,−1) 푃푄⃗ = (−3) + (9) = √9 + 81 = √90 = 3√10units 푅(4, 6)

푃푅⃗ = (−1) + (8) = √1 + 64 = √65units 푃(5,−2) 푄푅⃗ = (2) + (−1) = √4 + 1 = √5units

i. Perimeter= 푃푄⃗ + 푃푅⃗ + 푄푅⃗ = 3√10 + √65 + √5 = √5 3√2 + √13 + 1 units ii. 푃푄⃗⦁푄푅⃗ = 푃푄⃗ 푄푅⃗ cos휃 ⟹ (−3, 9)⦁(2,−1) = 3√10 × √5 cos휃

⟹ −6 − 9 = 15√2 cos휃 ⟹ cos휃 = −√

⟹ 휃 = cos −√

⟹ 휃 = 135°

∴,∠푃푄푅 = 135° iii. Area= 푃푄⃗ 푄푅⃗ sin휃 = 3√10 × √5 sin 135° = 15√2

√= 7.5units

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Example 17. 35

If 푊(−1, 1),푋(9,−2),푌(푟, 푠)and푍(3, 3) are the vertices of a parallelogram, find the coordinates of 푌 and the area of the parallelogram.

Solution We have, 푊(−1, 1), 푋(9,−3) 푌(푟, 푠) and 푍(3, 3). 푍(3, 3) 푌(푟, 푠) 푊푋⃗ = 푂푋⃗ − 푂푊⃗ = (9,−3)− (−1, 1) = (10,−4) 푍푌⃗ = 푂푌⃗ − 푂푍⃗ = (푟, 푠) − (3, 3) = (푟 − 3, 푠 − 3) 푊푍⃗ = 푂푍⃗ − 푂푊⃗ = (3, 3) − (−1, 1) = (4, 2) 푋푌⃗ = 푂푌⃗ − 푂푋⃗ = (푟, 푠) − (9,−3) = (푟 − 9, 푠 + 3) 휃 푊(−1, 1) 푋(9,−3) 푊푋⃗ = 푍푌⃗ ⟹ (10,−4) = (푟 − 3, 푠 − 3) ⟹ 10 = 푟 − 3 ⟹ 푟 = 10 + 3 = 13 and −4 = 푠 − 3 ⟹ 푠 = −4 + 3 = −1 ∴, 푌(13,−1) 푊푍⃗ = (4) + (2) = √16 + 4 = √20 = 2√5 units

푊푋⃗ = (10) + (−4) = √100 + 16 = √116 = 2√29units 푊푋⃗⦁푊푍⃗ = 푊푋⃗ 푊푍⃗ cos휃 ⟹ (10,−4)⦁(4, 2) = 2√5 × 2√29 cos휃

⟹ 40 − 8 = 4√145 cos휃 ⟹ cos휃 =√

⟹ 휃 = cos√

⟹ 휃 ≈ 48.3664°

Area= 푊푋⃗ 푊푍⃗ sin휃 ≈ 2√5 × 2√29 sin 48.3664 = 36.0units Try: 1. a. 푆 is the mid-point of the side of 푄푅 of a triangle 푃푄푅. If 푃푅⃗ = 풖and 푃푄⃗ = 풗, express 푃푆⃗in terms of 풖and 풗 b. 푆(1, 2), 퐵(3, 5), 퐶(3,−6) and 퐷(푥, 푦) are the vertices of the parallelogram 퐴퐵퐶퐷. Find the values of 푥 and 푦. Hence, find, correct to the nearest degree, the angle between 퐴퐷⃗ and 퐷퐶⃗

(SSSCE) 2. a. A point 퐴 has position vector 풂and 퐵 has position vector 풃. Prove that 푀, the midpoint of 퐴퐵⃗has the position vector 풎 = ퟏ

ퟐ(풂+ 풃).

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b. A triangle 퐴퐵퐶 is such that 푀 is the mid-point of 퐴퐵, 푁 is the mid-point of 퐴퐶 and 푃 is the 퐵퐶. i. Prove that 푀푁푃퐵 is a parallelogram. ii. Show that area of ⧍퐴퐵퐶 = 4times area of ⧍푀푁푃

(SSSCE) 3. If 퐴(2, 6),퐵(2, 2),퐶(7, 3) and 퐷(푥, 푦) are the vertices of a parallelogram 퐴퐵퐶퐷, calculate the coordinates of 퐷.

Final Exercises 1. Express the given points as vectors. i. 퐴(3, 5) and 퐵(4,−4) ii. 퐴(−2,−6)and퐵(6,−4) iii. 푃(1, 7)and푄(−8, 4) iv. 푃 , 4 and푄 − , 6 v. 푀 √3, 2√3 and 푁 −√3, √3

vi. 푀 , and푁 ,

2. Express the following points as vectors in the form 푥푦 .

i. 퐴(−1, 0) and 퐵(−2,−3) ii. 푃(−6, 4)and푄(13, 2) iii. 푀(10, 11)and푁(16, 3) 3. The points 퐴(−1, 3), 퐵(3, 4) and 퐶(8, 12) are points in the 푂푥푦 plane. a. Find i.퐴퐵⃗ ii. 퐴퐶⃗ iii. 퐵퐶⃗ iv. 퐵퐴⃗ b. Find 2.퐴퐵⃗ − 3.퐵퐶⃗ 4. Given that i. 퐴(푥, 푦), 퐵(7, 2) and 퐴퐵⃗(3, 2).Find the values of 푥 and 푦. ii. 푃(푎, 푏),푄(−1,−2)and 푃푄⃗(−5, 3). Find the values of 푎 and 푏. iii. 푀(8, 11),푁(푟, 푠) and 푀푁⃗(1, 10). Find 푁. 5. Given that i. 퐴(−1,−4) and 퐵(3, 9). Find 퐴퐵⃗ ii. 푃 3, and푄(−2, 3). Find 푃푄⃗ .

iii. 퐴(0,−7)and퐵(1,−4). Find 퐴퐵⃗ . iv. 풘(1, 12). Find |풘|

6. if퐴퐵⃗ = 푚−2 and 퐴퐵⃗ = 6√2 units. Find the possible values of 푚.

7. If 푃푄⃗ = 3푎 and 퐴퐵⃗ = √14. Find the values of 푎.

8. Given that 풂 = (−3, 2) and 풃 = (4,−1). Find |2풂|, |풃 − 풂| and |2풃 − 풂| 9. 퐴(18, 23), 퐵(14, 2)and퐶(7, 9) are the vertices of ∆퐴퐵퐶. Find the perimeter of the triangle.

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10. If 푝 = (9, 16),푞 = (11, 3) and 푟 = (1,−6), find i. 풑 + 풒 + 풓 ii. |풑+ 풒 + 풓| iii. 2풒 − 풑 − 2풓 iv. |2풒 − 풑 − 2풓|

11. Given that 풂 = 91 and 풃 = 2 + 푦

−3 + 푥 . Find the values of 푥 and 푦 if 풂 = 풃.

12. P, Q and R are vertices of a triangle. Given that 푃푄⃗ = 48 and 푅푄⃗ = −2

3 . Find 푃푅⃗.

13. Given that 풎 = (−8, 6) and 풏 = (11,−2). Find the values of 푥 and 푦 if

풎 + 풏 =푥 − 푦

2푥 + 푦 .

14. Write down the following vectors as a scalar combination of 퐢 and 퐣. i. 풂 = (4, 6) 풃 = (13, 2) and 풄 = (−1, 3) ii. 풑 = √3, 2 풒 = , and 풓 = ,√2 .

15. Given that 푎 = 2퐢+ 6퐣. Find the unit vector in the direction of vector 푎.

16. If 풑 = 125 . Find 풑.

17. Given that 푝 = 10퐢+ 퐣 and 푞 = 3퐢 − 2퐣. Find i. 풑 + 풒 ii. |풑 + 풒| iii. the unit vector in the direction of 풑 + 풒 18. Given that 풂 = −퐢 + 4퐣,풃 = 3퐢 − 2퐣. Find the values of 푥 and 푦 such that 2풄 = 풂푥 + 풃푦 19. Find the position vector 푐 that divides the vectors i. 풂 = 17퐢 + 9퐣 and 풃 = 37퐢 + 12퐣 in the ratio 3: 1 ii. 풂 = 35퐢+ 11퐣 and 풃 = 13퐢 + 20퐣 in the ratio 4: 2 iii. 풂 = 4퐢 + 10퐣 and 풃 = 15퐢 + 3퐣 in the ratio 8: 4 iv. 풂 = 7퐢+ 12퐣 and 풃 = 8퐢 − 퐣 in the ratio 2: 3 v. 풂 = 2퐢 and 풃 = 퐣 in the ratio 1: 4. 20. Find the midpoint of the following vectors: i. 풂 = 3퐢+ 4퐣 and 풃 = 7퐢 + 2퐣 ii. 풑 = 20퐢 − 2퐣 and 풒 = 10퐢 + 5퐣 iii. 풖 = 퐢+ 퐣 and 풗 = 7퐢 + 9퐣 iv. 풎 = 2퐢 and 풏 = −4퐣 21. Find the vector that is of magnitude 5 units and parallel to the vector 푏 = 퐢 + 퐣.CCE

22. If 풂 = 35 풃 = 7

3 and 풄 = −3−2 , calculate 푟 if 풓 = 풂 − 2풃+ 풄. CCE

23. The coordinates of the vertices of a parallelogram QRST are 푄(1, 6),푅(2, 2),푆(5, 4) and 푇(푥, 푦). a. i. Find vectors 푄푅⃗ and 푇푆⃗ and hence, determine the values of 푥 and 푦. ii. Calculate the magnitude of 푅푆⃗

467

iii. Express 푅푆⃗ in the form (푘, 휃°), where 푘 is the magnitude and 휃 the bearing. b. Find the values of 푥 and 푦 in the equation

푥 + 32 −

푦푥 + 푦 = 2

1 CCE

24.a. A ship is initially at position A and travels 6 km on a bearing of 055°followed by 8 km on a bearing of 150°to reach a final position B. Find the distance and the bearing of B from A.

b. If 퐴퐵⃗ = 푎and 퐶퐷⃗ = 3푎. Which of the following statement are true? Justify your answer. i. 퐴퐵⃗ is parallel to 퐶퐷⃗. ii. 퐴퐵⃗ is equal to 퐶퐷⃗. iii. 퐴퐵⃗ is three times as long as 퐶퐷⃗. CCE

25.Given that 푎 = 4−3 and 푏 = 2

4 and 푐 = 22−11 . Find

i. a unit vector perpendicular to 풂. ii. a vector that is of magnitude 6 units and parallel to the vector. iii. the values of the constants 푚 and 푛 for which 푚풂 + 푛풃 = 풄 CCE

Hint:Avectorperpendiculartoavector푥푦 canbeeither −푦

푥 or 푦−푥

26. The captain of a ship measures the angle between two lighthouses as seen from his ship to be 48.5°. His radar tells him that Lighthouse A is 23 km away from him and Lighthouse B is 31 km away from him. He also found out that the bearing of lighthouse B from the ship is 090°. i. How far away are the two lighthouses from each other. ii. What is the bearing of lighthouse A from the ship. CCE 27. A triangle PQR has vertices 푃(−4,−2), 푄(1, 10)and 푅(8, 3). Show that the triangle is isosceles and find the size of the base angles. 27. A triangle has vertices 퐴(3, 2),퐵(−1, 5)and 퐶(2, 1). a. Find the lengths of the triangle. b. Use scalar product to find the angles of the triangle. 28. The vertices of a parallelogram ABCD are 퐴(−4, 4),퐵(−2, 1), 퐶(4, 5)and 퐷(푝,푞). i. Write down 퐴퐵⃗ and 퐷퐶⃗ as column vectors and deduce the values of 푝 and 푞. ii. Use the dot product to show that ABCD is a parallelogram. 29. Find the angle between the pair of vectors

468

i. 5퐢 + 12퐣 and 4퐢+ 3퐣 ii. 2퐢 + 3퐣 and 6퐢 + 4퐣 iii. 퐢 + 3퐣 and 2퐢+ 6퐣 iv. 5퐢+ 2퐣 and 2퐢 + 5퐣 30. The three vertices of a parallelogram PQRS are 푃(−1, 3),푄(6,−3)and 푅(−2,−4). Find the coordinates of 푆(푥, 푦). 31. Express the following vectors in magnitude bearing forms.

풂 = 12 풃 = 5

3 풄 = −512 풅 = −8

15 풇 = −43

32. Express the following as column vectors. a. (12N, 045°) b. (13km, 330°) c. (8m, 075°) d. (15cm, 030°) f. (9Nm, 270°) 33. Express each of the following vectors in terms of the unit vectors 푖and 푗. a. (8, 060°) b. (12, 300°) c. (10, 120°) d. (16, 180°) f. (13, 210°) 34. Find the dot product of the following vectors

i. 풂 = −31 and 풃 = 5

−4 ii. 풗 = 616 and 풘 = 17

12

iii. 풖 = −32 and 풗 = 5

−6 iv. 풎 = √2√3

and 풏 = 2√2−√3

35. The bearing angle of town 퐴 from 퐵 is 055°. Town 퐶 is on a bearing of 180° from town 퐴.if town 퐴 is 20 km from town 퐵 and town 퐶 is 40 km from town 퐴, calculate the distance of town 퐵 from 퐶 . CCE 36. The position vectors of the points 퐴 and 퐵 with respect to the origin 푂 are 2퐢+ 3퐣 and −퐢+ 5퐣 respectively. Find the position vector of the vector of the point 퐶 such that 퐴퐶⃗ = 2퐴퐵⃗. CCE

37. Given that 푎 = 45 and 푏 = −1

3 . Find

i. 풂⦁풃 ii. The angle between 풂 and 풃.

38. The coordinates of the vertices of triangle ABC is given by 퐴(4, 7),퐵퐴⃗ = 53 and

퐴퐶⃗. Find the coordinates of 퐵.

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MATRICES

An 푚 × 푛 (read ‘m by n’)matrix is rectangular array of elements arranged in 푚 rows and 푛 columns. 푚 and 푛are called the dimension of the matrix. We denote a matrix with bold upper case letters. An 푚 × 푛 matrix can be written in the form

푨( × ) =푎 푎 … 푎푎 ...

푎 ...… 푎 ...

푎 푎 … 푎

For instance, a 2× 2 and 3 × 3 will be written respectively as

푨( × ) =푎 푎푎 푎 and 푨( × ) =

푎 푎 푎푎 푎 푎푎 푎 푎

The elements 푎 , 푎 and 푎 are called the principal or main diagonal elements. We can find the element of a matrix by connecting the dimensions of the matrix with equations.

Example 18.1 i. A 2 × 2 matrix is defined by 푨( × ) = 푚 + 푛. Find the elements of the matrix. ii. A 3 × 3 matrix is defined by 푩( × ) = 2푛 − 푚. Find the element of the matrix.

Solution i. We have 푨( × ) = 푚+ 푛. The elements of the matrix are: 푎 = 1 + 1 = 2 푎 = 1 + 2 = 3 푎 = 2 + 1 = 3 푎 = 2 + 2 = 4

Hence, 푨( × ) = 2 33 4

ii. We have 푩( × ) = 2푛 −푚. The elements of the matrix are: 푏 = 2(1) − 1 = 1 푏 = 2(2)− 1 = 3 푏 = 2(3)− 1 = 5 푏 = 2(1) − 2 = 0 푏 = 2(2)− 2 = 2 푏 = 2(3)− 2 = 4 푏 = 2(1) − 3 = −1 푏 = 2(2)− 3 = 1 푏 = 2(3)− 3 = 3

Hence, 푩( × ) =1 3 50 2 4−1 1 3

Try: Write the 3 × 3 matrix whose elements are defined by 푪( × ) = 푚 − 푛

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Types of matrices 1. A square matrix This is a matrix with 푚 rows and 푚 columns (same number of rows and columns). Examples of such matrix are 2 × 2 and 3 × 3 matrices. 2. Row matrix This is a matrix with only one row. Examples of a row matrix are 1 × 2matrix defined with elements (푎 푎 ) and 1 × 3 matrix defined with elements(푎 푎 푎 ). 3. Column matrix This is a matrix with only one column. Examples of a column matrix are 2 × 1 matrix

defined with elements 푎푎 and 3 × 1 matrix defined with elements

푎푎푎

.

4. Zero (null) matrix This is a matrix that has 0 for all its elements. Examples are the matrices

0 00 0 and

0 0 00 0 00 0 0

5. Diagonal matrix This is a matrix that has 0 for all its elements off the main diagonal. Examples are the matrices

2 00 3 and

−1 0 00 2 00 0 3

6. Unit matrix (I) This is a diagonal matrix that has 1 as element of the leading diagonal. The unit matrix is denoted by 푰.

The 2 × 2 and 3 × 3 unit matrix is defined by 푰 = 1 00 1 and 푰

Equality of matrices Two matrices 푨and 푩 are equal if they have the same dimension and their corresponding entries are the same.

Let 푨 =푎 푎푎 푎 and 푩 = 푏 푏

푏 푏 . Then 푨 = 푩 if and only if

푎 = 푏 ; 푎 = 푏 ; 푎 = 푏 ; 푎 = 푏 By the definition given above (1, 2) ≠ (2, 1)

Example 18.2

471

i. Given that 푨 = 1 푥푦 2 and 푩 = 1 −3

5 2 , find the values of 푥 and 푦 if 푨 = 푩

ii. Given that 푪 = 0 12푎 − 푏 푎 + 푏 and 푫 = 0 1

2 − 푎 4 , find the values of 푎 and 푏 if

푪 = 푫.

iii. Given that 푷 =1 푥 + 2 30 2 84 2 푑

and 푸 =1 3푥 30 2 84 2 푑 + 6

, find the possible values

of 푑 and 푥 if 푷 = 푸

Solution

i. We have 푨 = 1 푥푦 2 and 푩 = 1 −3

5 2 .

If 푨and 푩are equal then 푥 = −3 and 푦 = 5. ∴ 푥 = −3and푦 = 5.

ii. We have 푪 = 0 12푎 − 푏 푎 + 푏 and 푫 = 0 1

2− 푎 4

If 푪 = 푫, then 2푎 − 푏 = 2 − 푎 ⟹ 3푎 − 푏 = 2 … … … (1) and 푎 + 푏 = 4 … … … (2) From (1) 푏 = 3푎 − 2. Put 푏 = 3푎 − 2into (2) ⟹ 푎 + 3푎 − 2 = 4 ⟹ 4푎 = 6 ⟹ 푎 = = . Put 푎 into (2)

⟹ 푏 = 3 − 2 = − 2 = ∴ 푎 = and 푏 =

iii. We have푷 =1 푥 + 2 30 2 84 2 푑

and 푸 =1 3푥 30 2 84 2 푑 + 6

If 푷 = 푸, then 푑 = 푑 + 6 ⟹ 푑 − 푑 − 6 = 0 ⟹ 푑 − 3푑 + 2푑 − 6 = 0 ⟹ (푑 − 3)(푑 + 2) = 0 ∴ 푑 = 3 or 푑 = −2 Also, 푥 + 2 = 2푥 ⟹ 푥 − 3푥 + 2 = 0 ⟹ (푥 − 2)(푥 − 1) = 0 ∴ 푥 = 2표푟푥 = 1 Try:

Given that 푴 = 2 3 4푥 − 푦 1 2푥 and 푵 = 2 3 4

푦 − 2 1 푦 + 2 , find the values of 푥 and 푦

472

Multiplication of a matrix by a scalar

Let 푨( × ) =푎 푎 … 푎푎 ...

푎 ...… 푎 ...

푎 푎 … 푎be an 푚 × 푛 matrix. The multiplication of the matrix

by a scalar 푘 is by

푘푨( × ) =푘푎 푘푎 … 푘푎푘푎 ...

푘푎 ...… 푘푎 ...

푘푎 푘푎 … 푘푎

i.e. multiplying the scalar by the elements of the matrix. Example 18.3

i. Given that 푨 = 5 61 −3 , find 2푨.

ii. Given that 푩 = 1 푥푦 7 and 푪 = 2 3

−5 14 , find the values of 푥 such that 푩 = 푪.

Solution

i. We have, 푨 = 5 61 −3 2푨 = 2 × 5 2 × 6

2 × −1 2 × −3 = 10 12−2 −6

∴ ,2푨 = 10 12−2 −6

ii. 푩 = 1 푥푦 7 and 푪 = 2 3

−5 14 ⟹ 푪 =−

=1

− 7

If 푩 = 푪, then푥 = , 푦 = − . ∴= , 푦 = − Try:

Given that 푫 =2

0 4 11 3

, find 푫

Addition of matrices

Let 푨and 푩be two 푚 × 푛 matrices. The sum of 푨and 푩is the 푚 × 푛 matrix 푨 + 푩 formed by adding the corresponding elements of 푨and 푩.

If 푨 =푎 푎푎 푎 and푩 = 푏 푏

푏 푏 ,then푨 + 푩 = 푎 + 푏 푎 + 푏푎 +푏 푎 + 푏

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Addition of 3 × 3 matrices is done in a similar way. Note that we add matrices of the same dimension.

Additive inverse of a matrix Let 푨 be an 푚 × 푛 matrix. The additive inverse of 푨, denoted as –푨, is the matrix derived by each element of 푨 by scalar −1. Thus, the additive inverse of

푨 =푎 푎푎 푎 is –푨 =

−푎 −푎−푎 −푎

Subtraction of matrices Let 푨and 푩 be 푚 × 푛 matrices. The 푚 × 푛matrix 푨 −푩is the matrix formed by adding the elements of 푨and additive inverse of 푩 in the corresponding order.

If 푨 =푎 푎푎 푎 and 푩 = 푏 푏

푏 푏 ,then 푨 − 푩 = 푎 − 푏 푎 − 푏푎 −푏 푎 − 푏

Subtraction of 3 × 3 matrices is done in a similar way. Note: 푨 − 푩 ≠ 푩− 푨

Example 18.4

a. Given that 푨 = 8 05 2 and 푩 = 11 4

3 9 , find

i. 푨 + 푩 ii. additive inverse of A iii. 푨 − 푩.

b. Given that 푷 =−5 12−3 76 3

and 푩 =2 9

18 62 1

, find

i. 푨 + 푩 ii. additive inverse of A iii. 푨 − 푩. Solution

a. We have, 푨 = 8 05 2 and 푩 = 11 4

3 9

i. 푨 + 푩 = 8 05 2 + 11 4

3 9 = 8 + 11 0 + 45 + 3 2 + 9 = 19 4

8 11

∴ 푨 + 푩 = 19 48 11

ii. −푨 = −8 0−5 −2 ∴ −푨 = −8 0

−5 −2

iii. 푨 −푩 = 8 05 2 − 11 4

3 9 = 8 − 11 0 − 45 − 3 2 − 9 = −3 −4

2 −7

∴ 푨 − 푩 = −3 −42 −7

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b. We have, 푷 =−5 12−3 76 3

and 푸 =2 9

18 62 1

i. 푷+ 푸 =−5 12−3 76 3

+2 9

18 62 1

=−5 + 2 12 + 9−3 + 18 7 + 6

6 + 2 3 + 1=

−3 2115 138 4

∴ 푷 + 푸 =−3 2115 138 4

ii. −푨 =5 −123 −7−6 −3

∴ −푨 =5 −123 −7−6 −3

iii. 푷 −푸 =−5 12−3 76 3

−2 9

18 62 1

=−5 − 2 12 − 9−3 − 18 7 − 6

6 − 2 3 − 1=

−7 −3−21 1

4 2

∴ 푷 − 푸 =−7 −3−21 1

4 2

Try:

If 푴 =−1 1 34 6 0−1 3 −2

and 푩 =−2 7 −83 3 1

10 11 0, find

i. 푨 + 푩 ii. additive inverse of A iii. 푨 − 푩. Note the following properties of matrix addition:

1. 푨 + 푩 = 푩 + 푨(commutativeproperty) 2. 푨 + 푶 = 푨(additiveidentity) 3. 푨 + (푩 + 푪) = (푨 + 푩) + 푪(associatveproperty) 4. 푨 + −푨 = 푶(additiveinverse), where 푶is the zero matrix.

Multiplication of two matrices Let 푨be an 푚 × 푛matrix and 푩 be an 푛 × 푝 matrix. The product of 푨and 푩is an 푚 ×푝푨푩formed multiplying every element in the rows of matrix 푨 by every element in the

columns of matrix 푩in a well defined order. That is , if 푨 =푎 푎푎 푎 and푩 =

푏 푏푏 푏 ,then푨푩 = 푎 푏 + 푎 푏 푎 푏 + 푎 푏

푎 푏 + 푎 푏 푎 푏 + 푎 푏

Multiplication of 3 × 3 matrices is done in a similar way. We can multiply two matrices of different dimensions if and only if the number of columns of the first equals number of rows of the second. In general 푨푩 ≠ 푩푨

475

Example 18.5

a. Given that 푨 = −3 54 1 and 푩 = −2 2

−2 7 .

Find i. 푨푩 ii. 푩푨. What conclusion can you draw about 푨푩and푩푨?

b. Find the matrix 푷푸if 푷 =7 0 53 1 −12 4 −2

and 푸 =5 6 −2−3 5 110 2 1

.

c. Given that 푨 =4 1−2 −21 4

, 푩 = 1 2 03 −1 5 and푪 = 2 7

1 −4 . Find if possible

i. 푨푩 ii. 푩푨 iii. 푨푪 iv. 푪푨

d. Find 2 × 2 matrices 푨and 푩such that 푨 + 푩 = 2 01 1 and 2푨 − 푩 = 1 4

−6 2 .

Solution

a. We have, 푨 = −3 54 1 and 푩 = −2 2

−2 7

푨푩 = −3 54 1

−2 2−2 7 = −3 ∙ −2 + 5 ∙ −2 −3 ∙ 2 + 5 ∙ 7

4 ∙ −2 + 1 ∙ −2 4 ∙ 2 + 1 ∙ 7 = 6 − 10 −6 + 35−8 − 2 8 + 7

= −4 29−10 15

푩푨 = −2 2−2 7

−3 54 1 = −2 ∙ −3 + 2 ∙ 4 −2 ∙ 5 + 2 ∙ 1

−2 ∙ −2 + 7 ∙ 4 −2 ∙ 5 + 7 ∙ 1 = −6 + 8 −10 + 2−4 + 28 −10 + 7

= 2 −824 −3

We can conclude 푨푩 ≠ 푩푨.

b. We have, 푷 =7 0 53 1 −12 4 −2

and 푸 =5 6 −2−3 5 110 2 1

푷푸 =7 0 53 1 −12 4 −2

5 6 −2−3 5 110 2 1

=7 ∙ 5 + 0 ∙ −3 + 5 ∙ 10 7 ∙ 6 + 0 ∙ 5 + 5 ∙ 2 7 ∙ −2 + 0 ∙ 1 + 5 ∙ 1

3 ∙ 5 + 1 ∙ −3 + −1 ∙ 10 3 ∙ 6 + 1 ∙ 5 + −1 ∙ 2 3 ∙ −2 + 1 ∙ 1 + −1 ∙ 12 ∙ 5 + 4 ∙ −3 + −2 ∙ 10 2 ∙ 6 + 4 ∙ 5 + −2 ∙ 2 2 ∙ −2 + 4 ∙ 1 + −2 ∙ 1

=35 + 0 + 50 42 + 0 + 10 −14 + 0 + 515 − 3 − 10 18 + 5− 2 −6 + 1 − 1

10 − 12 − 20 12 + 20 − 4 −4 + 4 − 2

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=85 52 −92 5 −6

−22 28 −2

∴,푷푸 =85 52 −92 5 −6

−22 28 −2

c. We have, =4 1−2 −21 4

,푩 = 1 2 03 −1 5 and 푪 = 2 7

1 −4

i.

푨푩 =4 1−2 −21 4

1 2 03 −1 5 =

4 ∙ 1 + 1 ∙ 3 4 ∙ 2 + 1 ∙ −1 4 ∙ 0 + 1 ∙ 5−2 ∙ 1 + −2 ∙ 3 −2 ∙ 2 + −2 ∙ −1 −2 ∙ 0 + −2 ∙ 5

1 ∙ 1 + 4 ∙ 3 1 ∙ 2 + 4 ∙ −1 1 ∙ 0 + 4 ∙ 5

=4 + 3 8 − 1 0 + 5−2 − 6 −4 + 2 0 − 101 + 12 2 − 4 0 + 20

=7 7 5−8 −2 −1013 −2 20

∴,푨푩 =7 7 5−8 −2 −1013 −2 20

ii. 푩푨 = 1 2 03 −1 5

4 1−2 −21 4

= 1 ∙ 4 + 2 ∙ −2 + 0 ∙ 1 1 ∙ 1 + 2 ∙ −2 + 0 ∙ 43 ∙ 4 + −1 ∙ −2 + 5 ∙ 1 3 ∙ 1 + −1 ∙ −2 + 5 ∙ 4

= 0 −119 25

∴ 푩푨 = 0 −319 25

iii. 푨푪 =4 1−2 −21 4

2 71 −4 =

8 + 1 28 − 4−4 − 2 −14 + 82 + 4 7 − 16

=9 24−6 −66 −9

∴ 푨푪 =9 24−6 −66 −9

iv. We cannot calculate 푨푪 since the number of column in 푪 is not equal to the number of row in 푨.

d. Let 푨 =푎 푎푎 푎 and 푩 = 푏 푏

푏 푏

477

We have,푨 + 푩 = 2 01 1 ⟹

푎 푎푎 푎 + 푏 푏

푏 푏 = 2 01 1

⟹ 푎 + 푏 푎 + 푏푎 + 푏 푎 + 푏 = 2 0

1 1

⟹ 푎 + 푏 = 2 … … … (1) 푎 + 푏 = 0 … … … (2) 푎 + 푏 = 1 … … … (3) 푎 + 푏 = 1 … … … (4)

Also, 2푨 −푩 = 1 4−6 2 ⟹ 2

푎 푎푎 푎 − 푏 푏

푏 푏 = 1 4−6 2

⟹ 2푎 − 푏 2푎 − 푏2푎 − 푏 2푎 − 푏 = 1 4

−6 2

⟹ 2푎 − 푏 = 1 … … … (i) 2푎 − 푏 = 4 … … … (ii) 2푎 − 푏 = −6 … … … (iii) 2푎 − 푏 = 2 … … … (iv) Solve equations (1) and (i). Add (i) to (1) ⟹ 푎 +2푎 + 푏 − 푏 = 2 + 1 ⟹ 3푎 = 3 ∴ 푎 = 1 Put 푎 = 1 into (1) ⟹ 1 + 푏 = 2 ∴ 푏 = 2 − 1 = 1 Now, solve equations (2) and (ii). Add (ii) to (2) ⟹ 푎 +2푎 + 푏 − 푏 = 0 + 4 ⟹ 3푎 = 4 ∴ 푎 =

Put 푎 = into (2) ⟹ +푏 = 0 ∴ 푏 = 0 − = − Now, solve equations (3) and (iii). Add (iii) to (3) ⟹ 푎 +2푎 + 푏 − 푏 = 1 − 6 ⟹ 3푎 = −5 ∴ 푎 = −

Put 푎 = − into (2) ⟹ − +푏 = 1 ∴ 푏 = 1 + = Now, solve equations (4) and (iv). Add (iv) to (4) ⟹ 푎 +2푎 + 푏 − 푏 = 1 + 2 ⟹ 3푎 = 3 ∴ 푎 = 1 Put 푎 = 1 into (2) ⟹ 1+푏 = 1 ∴ 푏 = 1 − 1 = 0

478

∴ 푨 =1

− 1 and 푩 =

1 −

0

Try:

1. If 푪 = −2 54 3 and 푫 = 3 1

−2 −3 , find the values of 푎 and 푏 such that;

푫푪 = 2 −3 72 푎 + 푏 4

−12 −3

(SSSCE) 2. Find the two by two matrices 푨and 푩such that

푨 + 2푩 = ퟏ −ퟏퟎ ퟏ and 2푨 + 3푩 = 2 1

−1 0 .

(SSSCE) Note the following properties of matrix multiplication: 1. 푨푰 = 푨, where 푰 is the unit(identity) matrix. 2. 푨(푩푪) = (푨푩)푪 (associativeproperty) 3. 푨(푩 + 푪) = 푨푩 + 푨푪 (distributiveproperty)

Multiplication of a matrix by a vector Let 푨be an 푚 × 푛 matrix and 풗be a column vector such that the columns of 푨equals the

rows of 풗. If 푨 =푎 푎푎 푎 and 풗 = , then the multiplication of 푨 and 풗 is defined

by:

푨풗 = 푨푎 푎푎 푎

푥푦 =

푎 푥 + 푎 푦푎 푥 + 푎 푦

In general, if we multiply a matrix by a vector as defined above we get a vector. Example 18.6

i. Given that 푨 = 3 16 1 and 풗 = ퟐ

ퟑ . Find 푨풗

ii. Given that 푷 = 3 −1푥 1 and 풗 = ퟓ

ퟔ . Find 푥 if 푷풗 =

Solution

i. We have, 푨 = 3 16 1 and 풗 =

479

푨풗 = 3 16 1 = ∙ ∙

∙ ∙ = = ∴, 푨풗 =

ii. We have, 푷 = 3 −1푥 1 and 풗 =

푷풗 = 3 −1푥 1 = ∙ ∙

∙ ∙ = = But 푷풗 =

So, = ⟹ 5푥 + 6 = −1 ⟹ 5푥 = −1 − 6 = −7 ∴, 푥 = − Try

Find 푨풗 if 푨 =0 5 4−2 3 19 2 2

and풗 =3−22

.

Transpose of a matrix

Let푨 be an 푚 × 푛 matrix. The transpose of 푨, denotedby 푨 , is the matrix formed by

interchanging the rows of 퐴 for its columns. Thus, if 푨 =푎 푎푎 푎 , then the transpose

matrix 푨 =푎 푎푎 푎

Example 18.7 Transpose the following matrices:

푨 = 5 78 3 푩 = 1 4 3

2 −1 2 푪 =2 135 −91 11

and 푫 =2 4 74 6 50 4 4

Solution

i. 푨 = 5 78 3 푨 = 5 8

7 3 ii. 푩 = 1 4 32 −1 2 푩 =

1 24 −13 2

iii. 푪 =2 135 −91 11

푪 = 2 5 13 −9 11 iv. 푫 =

2 4 74 6 50 4 4

푫 =2 4 04 6 47 5 4

Try:

Transpose the matrix 푨 =8 4 −55 −4 −57 1 −11

480

Signs attached to the elements of a given matrix Let 푨 be an 푚 × 푛 matrix and let 푎 be an element of 푨. The sign attached to the element 푎 is defined by the expression

(−1)

If 푨 =푎 푎푎 푎 , then the signs attached to the elements are calculated as follow:

For 푎 we have (−1) = (−1) = 1; For 푎 we have (−1) = (−1) = −1 푎 gives (−1) = (−1) = −1; 푎 gives (−1) = (−1) = 1.

In general a 2 × 2 matrix has the signs + −− + attached to the respective elements.

Similarly, by following the calculation above, a 3 × 3 matrix has the signs + − +− + −+ − +

attached to the respective elements. Adjoint of a ퟐ × ퟐ matrix

Let 푨 =푎 푎푎 푎 defines the 2 × 2 matrix. The adjoint of 푨, denoted by 푨 , is the

matrix formed by interchanging the positions of the elements and attaching the sign factor to the respective elements. Thus the adjoint of 푨 is defined as

푨 =푎 −푎−푎 푎

For instance, the adjoint of the matrix 4 35 2 is 2 −3

5 4

Determinant of ퟐ × ퟐ matrix

Let 푨 =푎 푎푎 푎 defines the 2 × 2 matrix. The determinant of 푨, denoted by det 푨

or simply |푨|, is given by

det 퐴 = |퐴| =푎 푎푎 푎 = 푎 푎 − 푎 푎

Thus, the determinant is seen as the product of the leading diagonal minus the product of the other diagonal.

For instance the determinant of the matrix 4 35 2 defined as

4 35 2 = 4(2)− 3(5) = 8 − 15 = −7

481

Try: Write down the adjoint and find the determinant of the following 2 × 2 matrices:

푨 = 1 6−2 −2 푩 = 3 −2

7 −4 푪 = 4 −58 4

Minors and cofactors of a ퟑ × ퟑ matrix

Let 푨( × ) =푎 푎 푎푎 푎 푎푎 푎 푎

defines the 3 × 3 matrix and let 푎 be an element in

row 푚 and column 푛 of 푨. The 푚푖푛표푟 of the element 푎 , denoted by 푴 , is the determinant of the 2 × 2 matrix formed be deleting the row 푚 and column 푛 of 푨. The 푐표푓푎푐푡표푟of the element 푎 , denoted by 퐶 , is the product of the sign factor (−1) and the minor 푴 . That is, 푪 = (−1) 푴 If 푚 + 푛 is even, then 푪 = 푴 and if 푚 + 푛 is odd then 푪 = −푴 A 3 × 3matrix has nine minors and hence, nine cofactors.

Example 18.8

Identify the minor and cofactor of each of the element of the matrix 푨 =1 −3 17 2 58 −2 6

Solution

We have, 푨 =1 −3 17 2 58 −2 6

Minors Cofactors

푴 = 2 5−2 6 = 2(6)− 5(−2) = 12 + 10 = 22 푪 = 2 5

−2 6 = 22

푴 = 7 58 6 = 7(6)− 5(8) = 42 − 40 = 2 푪 = − 7 5

8 6 = −2

푴 = 7 28 −2 = 7(−2)− 8(2) = −14 − 16 = 30 푪 = 7 2

8 −2 = 30

푴 = −3 1−2 6 = −3(6)− 1(−2) = −18 + 2 = −16 푪 = − −3 1

−2 6 = 16

푴 = 1 18 6 = 1(6)− 1(8) = 6 − 8 = −2 푪 = 1 1

8 6 = −2

푴 = 1 −38 −2 = 1(−2)− −3(8) = −2 + 8 = −6 푪 = − 1 −3

8 −2 = 6

482

푴 = −3 12 5 = −3(5)− 1(2) = −15 − 2 = −17 푪 = −3 1

2 5 = −17

푴 = 1 17 5 = 1(5)− 1(7) = 5 − 7 = −2 푪 = − 1 1

7 5 = 2

푴 = 1 −37 2 = 1(2)−−3(7) = 2 + 21 = 22 푪 = 1 −3

7 2 = 22

Try:

Identify the minor and cofactor of each of the element of the matrix 푷 =6 10 38 −7 42 −1 4

Determinant of a ퟑ × ퟑ matrix

Let 푨( × ) =푎 푎 푎푎 푎 푎푎 푎 푎

defines the 3 × 3 matrix. The determinant of 푨is the sum

of the of any three elements in a row or column and their cofactor. By taking the elements in the first row of 푨, the determinant of 푨 is defined by

det 푨 = |푨| = 푎 푪 + 푎 푪 + 푎 푪

Example 18.9

i. Find the determinant of the matrix 푨 =1 −3 17 2 58 −2 6

ii. Given that −1 1 34 −2 36 푥 2

= 23, find the value of 푥.

Solution

i. We have, 푨 =1 −3 17 2 58 −2 6

det푨 = 푎 푪 + 푎 푪 + 푎 푪 From the above solution

푪 = 2 5−2 6 = 22 푪 = 7 5

8 6 = −2 푪 = 7 28 −2 = 30

∴ det푨 = 푎 푪 + 푎 푪 + 푎 푪 = 6(22) + 10(−2) + 3(30) = 132 − 20 + 90

483

= 202

ii. −1 1 34 −2 36 푥 2

= 23 ⟹ −1 −2 3푥 2 − 1 4 3

6 2 + 3 4 −26 푥 = 23

⟹ −1(−4 − 3푥)− 1(8 − 18) + 3(4푥 + 12) = 23 ⟹ 4 + 3푥 + 10 + 12푥 + 36 = 23 ⟹ 15푥 = 23 − 36 − 4 − 10 = 27 ∴,푥 = = Try:

1. Find the determinant of the matrix 푷 =6 10 38 −7 42 −1 4

2. Given that 2 −5 −13 1 −21 푘 3

= 58, find the value of the constant 푘.

(WASSCE) Adjoint of a ퟑ × ퟑ matrix

The adjoint of a 3 × 3 is formed by: 1. Writing down the cofactors of each of the element as a representative matrix of the original one and 2. Transposing the representative matrix.

Example 18.10

Find the adjoint of the matrix 푨 =1 −3 17 2 58 −2 6

.

Solution

We have, 푨 =1 −3 17 2 58 −2 6

Minors Cofactors

푴 = 2 5−2 6 = 2(6)− 5(−2) = 12 + 10 = 22 푪 = 2 5

−2 6 = 22

푴 = 7 58 6 = 7(6)− 5(8) = 42 − 40 = 2 푪 = 7 5

8 6 = −2

푴 = 7 28 −2 = 7(−2)− 8(2) = −14 − 16 = 30 푪 = 7 2

8 −2 = 30

푴 = −3 1−2 6 = −3(6)− 1(−2) = −18 + 2 = −16 푪 = −3 1

−2 6 = 16

484

푴 = 1 18 6 = 1(6)− 1(8) = 6 − 8 = −2 푪 = 1 1

8 6 = −2

푴 = 1 −38 −2 = 1(−2)− −3(8) = −2 + 8 = −6 푪 = 1 −3

8 −2 = 6

푴 = −3 12 5 = −3(5)− 1(2) = −15 − 2 = −17 푪 = −3 1

2 5 = −17

푴 = 1 17 5 = 1(5)− 1(7) = 5 − 7 = −2 푪 = 1 1

7 5 = 2

푴 = 1 −37 2 = 1(2)−−3(7) = 2 + 21 = 22 푪 = 1 −3

7 2 = 22

Write the cofactors as a representative of matrix 푨

So, 푨 =22 −2 3016 −2 6−17 2 22

∴, 푨 =22 16 −17−2 −2 230 6 22

Try:

Find the adjoint of the following matrices: i. 푷 =6 10 38 −7 42 −1 4

ii. 푸 =3 5 34 −2 01 4 −1

Multiplicative inverse of a square matrix

let 푨be an 푛 × 푛 matrix. Provided |푨| ≠ 0, then the inverse of 푨, denoted by 푨 is defined by

푨 =1

|푨| ∙ 푨

If |푨| = 0, the matrix is said to be singular and the inverse does not exist. If |푨| ≠ 0, the matrix is said to be non-singular and the inverse is defined above. Note the following properties: 1. 푨푨 ퟏ = 푰 2. (푨 ) = 푨 3. If 푨푩 = 푰, then 푩 = 푨 ퟏ 4. (푨 + 푩) = 푨 + 푩 푰 is the identity matrix.

Example 18.11 Find the multiplicative inverses of the following matrices:

485

i.푨 = 4 35 2 ii. 푩 = 2 3

4 −5 iii.푷1 −3 17 2 58 −2 6

Solution

i. We have, 푨 = 4 35 2 |푨| = 4 3

5 2 = 4(2)− 3(5) = 20 − 15 = 5

푨 = 2 −3−5 4 ∴,푨 =

|푨|∙ 푨 = 2 −3

−5 4 =−

−1

ii. We have, 푩 = 2 34 −5 |푩| = 2 3

4 −5 = 2 ∙ −5 − 3 ∙ 4 = −10 − 12 = −22

푩 = 2 34 −5 ∴ 푩 =

|푩|∙ 푩 = − 2 3

4 −5 =− −

−

iii. We have, 푨 =1 −3 17 2 58 −2 6

Minors Cofactors

푴 = 2 5−2 6 = 2(6)− 5(−2) = 12 + 10 = 22 푪 = 2 5

−2 6 = 22

푴 = 7 58 6 = 7(6)− 5(8) = 42 − 40 = 2 푪 = − 7 5

8 6 = −2

푴 = 7 28 −2 = 7(−2)− 8(2) = −14 − 16 = 30 푪 = 7 2

8 −2 = 30

푴 = −3 1−2 6 = −3(6)− 1(−2) = −18 + 2 = −16 푪 = − −3 1

−2 6 = 16

푴 = 1 18 6 = 1(6)− 1(8) = 6 − 8 = −2 푪 = 1 1

8 6 = −2

푴 = 1 −38 −2 = 1(−2)− −3(8) = −2 + 8 = −6 푪 = − 1 −3

8 −2 = 6

푴 = −3 12 5 = −3(5)− 1(2) = −15 − 2 = −17 푪 = −3 1

2 5 = −17

푴 = 1 17 5 = 1(5)− 1(7) = 5 − 7 = −2 푪 = − 1 1

7 5 = 2

푴 = 1 −37 2 = 1(2)−−3(7) = 2 + 21 = 22 푪 = 1 −3

7 2 = 22

Write the cofactors as a representative of matrix 푨

486

So, 푨 =22 −2 3016 −2 6−17 2 22

∴, 푨 =22 16 −17−2 −2 230 6 22

det푨 = 푎 푪 + 푎 푪 + 푎 푪

푪 = 2 5−2 6 = 22 푪 = 7 5

8 6 = −2 푪 = 7 28 −2 = 30

∴, det푨 = 푎 푪 + 푎 푪 + 푎 푪 = 6(22) + 10(−2) + 3(30) = 202

∴ 푨 =|푨|∙ 푨 =

22 16 −17−2 −2 230 6 22

=

⎝

⎜⎛

−

− −

⎠

⎟⎞

Try: Find the multiplicative inverses of the following matrices.

1.푃 = 1 −1−2 2 2.푸 =

16

3 3.푺 =

3 2 −26 −3 0

11 2 4

4. 푼 =1 4 0−1 −1 4−3 2 1

Application of matrices 1. 퐹푖푛푑푖푛푔푠표푙푢푡푖표푛푡표푠푦푠푡푒푚표푓푙푖푛푒푎푟푒푞푢푎푡푖표푛푠푖푛푡푤표푎푛푑푡ℎ푟푒푒푣푎푟푖푎푏푙푒푠 훼. Using the Cramer’s rule The Cramer’s rule (named after the Swiss mathematician Gabriel Cramer) involves the use of determinants to solve system of linear equations. Let: 푎 푥 + 푏 푦 = 푐 푎 푥 + 푏 푦 + 푐 푧 = 푑 푎 푥 + 푏 푦 = 푐 and 푎 푥 + 푏 푦+푐 푧 = 푑 푎 푥 + 푏 푦+푐 푧 = 푑 be system of linear equation in 푡푤표 and 푡ℎ푟푒푒 unknown variables respectively. For the two unknowns,

푎 푥 + 푏 푦 = 푐푎 푥 + 푏 푦 = 푐 ⟺ 푎 푏

푎 푏푥푦 =

푐푐

487

The matrix 푎 푏푎 푏 is called coefficient matrix. The matrix

푥푦 is called variable

matrix and the matrix 푐푐 is called constant matrix.

Similar meaning is given to those of the three unknowns. We will consider the first part of the Cramer’s rule. The first part of the Cramer’s rule states that:

Provided 푎 푏푎 푏 ≠ 0 in the system 푎 푥 + 푏 푦 = 푐

푎 푥 + 푏 푦 = 푐

then the solution is given by

푥 =푐1 푏1푐2 푏2푎1 푏1푎2 푏2

and 푦 =푎1 푐1푎2 푐2푎1 푏1푎2 푏2

푐 푏푐 푏 is formed by replacing

푎푎 of 푎 푏

푎 푏 with 푐푐 .

Similarly 푎 푐푎 푐 is formed by replacing 푏푏 of 푎 푏

푎 푏 with 푐푐 .

Let 푎 푏푎 푏 = 퐷, 푐 푏

푐 푏 = 퐷 and 푎 푐푎 푐 = 퐷 . Then 푥 = and 푦 =

Solution of three unknowns uses a similar procedure. (An example will help explain the rules in Three unknowns)

Example 18.12 a. Find the solution of the following systems of equations using determinants: i. 푥 + 푦 = 4 and ii. 푥 + 2푦 = 17 and iii. 2푥 + 3푦 − 푧 − 8 = 0 푥 − 푦 = 6 2푥 − 푦 = 9 −푥 + 푦 + 2푧 + 4 = 0 and 3푥 + 푦 − 2푧 − 2 = 0

Solution

i. We have the system, 푥 + 푦 = 4푥 − 푦 = 6 ⟹ 1 1

1 −1 =

Let 푨 = 1 11 −1 , 퐱 = and 퐛 =

So, we are solving the system, 푨퐱 = 퐛.

푥 = = ( )( )

= = 5and푦 = = ( ) ( )( )

= = −1

∴ 푥 = 5, 푦 = −1

488

ii. We have the system, 푥 + 2푦 = 172푥 − 푦 = 9 ⟹ 1 2

2 −1 =

Let 푨 = 1 22 −1 , 퐱 = and 퐛 =

So, we are solving the system, 푨퐱 = 퐛.

푥 = = ( )( )

= = 7and푦 = = ( ) ( )( )

= = 5

∴ 푥 = 7, 푦 = 5

iii. We have the system, 2푥 + 3푦 − 푧 − 8 = 0−푥 + 푦 + 2푧 + 4 = 03푥 + 푦 − 2푧 − 2 = 0

⟹ 2푥 + 3푦 − 푧 = 8−푥 + 푦 + 2푧 = −4

3푥 + 푦 − 2푧 = 2

⟹ 2 3 −1−1 1 23 1 −2

= Let 푨 =2 3 −1−1 1 23 1 −2

퐱 = and 퐛 =

푥 = = = ( ) ( ) ( )( ) ( ) ( )

=

∴, 푥 = − = −

푦 = = = ( ) ( ) ( )( ) ( ) ( )

` ∴, 푦 = = =

푧 = = = ( ) ( ) ( )( ) ( ) ( )

` ∴, 푧 = = − = − Example 18.13

The cost of a pencil and a pen is GH₵1.5 and the difference between the cost of 4 pencils and 2 pens is GH₵4.5. Find the number of pencils and pens using determinants.

Solution

489

Let 푝 = price of pencil and 푞 = price of pen. We have, 푝 + 푞 = 1.5 … … … (1) and 4푝 − 2푞 = 4.5 … … … (2)

We have the system, 푝 + 푞 = 1.54푝 − 2푞 = 4.5 ⟹ 1 1

4 −2 = ..

Let 푨 = 1 14 −2 , 퐱 = and 퐛 = .

.

So, we are solving the system, 푨퐱 = 퐛.

푝 =.. = . = . = 0.25and푦 =

.

. = . = . ≈ 0.42

∴, 푝 = 0.25, 푞 ≈ 0.42 Try: Find the solution of the following system of equations using determinants: i. 푥 − 3푦 − 1 = 0 and ii. 2푥 − 3푦 = 8 and iii. 푥 + 푦 = 3 −2푦 + 3푥 = 7 3푥 + 2푦 = 42 푥 − 3푦 + 푧 = 10 and 푦 − 3푥 − 2푧 = 17 훽. The inverse approach Given any system of linear equations, we write the equations in the form 푨퐱 = 푩, where 푨 is the coefficient matrix, 퐱 the variable matrix and 푩 the constant matrix. We are interested in finding 퐱 therefore we proceed as follow

푨퐱 = 푩 ⟹ 푨 푨퐱 = 푨 푩⟹ 푰퐱 = 푨 푩⟹ 퐱 =1

|푨| ∙ 푨풂풅풋푩

Since 푨 푨 = 푰, 푰퐱 = 퐱 and 푨 ퟏ = ퟏ|푨|∙ 푨풂풅풋

Find the solution of the following systems of equations using the inverse approach: i. 푥 + 푦 = 4 and ii. 푥 + 2푦 = 17 and iii. 2푥 + 3푦 − 푧 − 8 = 0 푥 − 푦 = 6 2푥 − 푦 = 9 −푥 + 푦 + 2푧 + 4 = 0 and 3푥 + 푦 − 2푧 − 2 = 0

Solution

i. We have the system, 푥 + 푦 = 4푥 − 푦 = 6 ⟹ 1 1

1 −1 =

Let 푨 = 1 11 −1 , 퐱 = and 퐛 = |푨| = 1 1

1 −1 = −1 − 1 = −2

So, we are solving the system, 푨퐱 = 퐛. Note that푨 푨퐱 = 푨 퐛 ⟹ 푰퐱 = 푨 퐛 ∴ 퐱 = 푨 퐛

490

푨 = −1 −1−1 1 푨 =

|푨|∙ 푨 = −1 −1

−1 1 =−

⟹ 퐱 = 푨 퐛 =−

=× ×

× ×= 2 + 3

2 − 3 = 5−1

⟹ 퐱 = = ∴ 푥 = 5, 푦 = −1

ii. We have the system, 푥 + 2푦 = 172푥 − 푦 = 9 ⟹ 1 2

2 −1 =

Let 푨 = 1 22 −1 , 퐱 = and 퐛 = |푨| = 1 2

2 −1 = −1 − 4 = −5

So, we are solving the system, 푨퐱 = 퐛.

푨 = −1 −2−2 1 푨 =

|푨|∙ 푨 = −1 −2

−2 1 =−

⟹ 퐱 = 푨 퐛 =−

=× ×

× ×=

+

−= 7

5

⟹ 퐱 = = ∴ 푥 = 7, 푦 = 5

iii. We have the system, 2푥 + 3푦 − 푧 − 8 = 0−푥 + 푦 + 2푧 + 4 = 03푥 + 푦 − 2푧 − 2 = 0

⟹ 2푥 + 3푦 − 푧 = 8−푥 + 푦 + 2푧 = −4

3푥 + 푦 − 2푧 = 2

⟹ 2 3 −1−1 1 23 1 −2

= Let 푨 =2 3 −1−1 1 23 1 −2

퐱 = and 퐛 =

So, we are solving the system, 푨퐱 = 퐛.

|푨| =2 3 −1−1 1 23 1 −2

= 2 1 21 −2 − 3 −1 2

3 −2 − 1 −1 13 1 = 8

Minors Cofactors

푴 = 1 21 −2 = −2 − 2 = −4 푪 = 1 2

1 −2 = −4

푴 = −1 23 −2 = 2 − 6 = −4 푪 = − −1 2

3 −2 = 4

491

푴 = −1 13 1 = −1 − 3 = −4 푪 = −1 1

3 1 = −4

푴 = 3 −11 −2 = −6 + 1 = −5 푪 = − 3 −1

1 −2 = 5

푴 = 2 −13 −2 = −4 + 3 = −1 푪 = 2 −1

3 −2 = −1

푴 = 2 33 1 = 2 − 9 = −7 푪 = − 2 3

3 1 = 7

푴 = 3 −11 2 = 6 + 1 = 7 푪 = 3 −1

1 2 = 7

푴 = 2 −1−1 2 = 4 + 1 = 5 푪 = − 2 −1

−1 2 = −5

푴 = 2 3−1 1 = 2 + 3 = 5 푪 = 2 3

−1 1 = 5

Write the cofactors as a representative of matrix 푨

So, 푨 =−4 4 −45 −1 77 −5 5

∴ 푨 =−4 5 74 −1 −5−4 7 5

∴ 푨 =|푨|∙ 푨 =

−4 5 74 −1 −5−4 7 5

=

⎝

⎜⎛−

− −

− ⎠

⎟⎞

⟹ 퐱 = 푨 퐛 =

⎝

⎜⎛−

− −

− ⎠

⎟⎞ 8

−42

=

⎝

⎜⎛− ∙ 8 + ∙ −4 + ∙ 2

∙ 8 − ∙ −4 − ∙ 2

− ∙ 8 + ∙ −4 + ∙ 2⎠

⎟⎞

=

⎝

⎜⎛−4 − +

4 + −

−4 − + ⎠

⎟⎞

⟹ 퐱 =푥푦푧

=

⎝

⎜⎛−

− ⎠

⎟⎞

∴ 푥 = − , 푦 = and 푧 = −

Try: Find the solution of the following system of equations using the inverse approach: i. 푥 − 3푦 − 1 = 0 and ii. 2푥 − 3푦 = 8 and iii. 푥 + 푦 = 3 −2푦 + 3푥 = 7 3푥 + 2푦 = 4 2푥 − 3푦 + 푧 = 10 and

492

푦 − 3푥 − 2푧 = 17 2. 퐹푖푛푑푖푛푔푡ℎ푒퐴푟푒푎표푓푎푡푟푖푎푛푔푙푒푎푛푑푝푎푟푎푙푙푒푙표푔푟푎푚푠 Let 퐴(푥 ,푦 ),퐵(푥 ,푦 ) and 퐶(푥 ,푦 ) be the vertices of a triangle 퐴퐵퐶. The Area of ⧍퐴퐵퐶 is defined by:

퐴푟푒푎,퐴 =푥 푥 푥푦 푦 푦1 1 1

푢푛푖푡푠 = |푥 (푦 − 푦 ) − 푥 (푦 − 푦 ) + 푥 (푦 − 푦 )|푢푛푖푡푠

Let 푃(푥 ,푦 ),푄(푥 ,푦 ),푅(푥 ,푦 ) and 푆(푥 ,푦 ) be the vertices of a parallelogram such as a square, a rectangle or a rhombus. The area of a parallelogram is given by:

퐴푟푒푎,퐴 =푥 푥 푥푦 푦 푦1 1 1

푢푛푖푡푠 = |푥 (푦 − 푦 ) − 푥 (푦 − 푦 ) + 푥 (푦 − 푦 )|푢푛푖푡푠

Example 18.14

Calculate the area of a triangle 푃푄푅 with vertices 푃(2, 2),푄(3, 4) and 푅(5, 6). Solution

We have, 푃(2, 2), 푄(3, 4)and 푅(5, 6). The area of triangle 푃푄푅 is given by 퐴 = |푥 (푦 − 푦 ) − 푥 (푦 − 푦 ) + 푥 (푦 − 푦 )|units

= |2(4− 6) − 3(2 − 6) + 5(2 − 4)|

= |−4 + 12 − 10| = 1unit

Example 18.15 Calculate the area of a parallelogram 퐴퐵퐶퐷 with vertices 퐴(−1, 3)퐵(−7, 5),퐶(3, 6) and 퐷(4, 4)

Solution We have, 퐴(−1, 3), 퐵(−7, 5),퐶(3, 6) and 퐷(4, 4) The area of parallelogram퐴퐵퐶퐷 is given by 퐴 = |푥 (푦 − 푦 ) − 푥 (푦 − 푦 ) + 푥 (푦 − 푦 )|units = |−1(5 − 6)− −7(3 − 6) + 3(3− 5)| = |1 − 21 − 10| = 26unit

493

Try: a. Calculate the area of a triangle 푋푌푍 with vertices 푋(8, 0),푌(16, 4) and 푍(15,−6). b. Calculate the area of a parallelogram 퐴퐵퐶퐷 with vertices 퐴(−1,−4)퐵(−7, 7),퐶(−11, 6) and 퐷(3,−1)

(see linear transformation in vol.2)

Final Exercises 1. Find the two by two matrix defined as: i. 푨( × ) = 푚 − 푛 ii. i. 푩( × ) = 3푛 −푚 iii. i. 푪( × ) = 푚 + 2푛 푚 iv. i. 푫( × ) = (푚 − 2푛) 2. Find the 3 × 3 matrix defined as: i. 푨( × ) = 2푚− 4푛 + 1 ii. i. 푩( × ) = 1 −푚 푛 iv. i. 푪( × ) = 푚 iv. i. 푨( × ) = 푛 3. Solve for 푥 and 푦 if i.푨 = 푩.

i. 푨 = 푥 4푦 −1 and 푩 = −2 4

−3 −1 ii. 푨 = −1 푥푦 2 and 푩 = −1 4

2 2

ii. 푨 = −1 푥 − 4푦 − 4 3 and 푩 = −1 0

5 3

iv. 푨 =1 푥 − 3 34 0 2푦5 5 4

and 푩 =1 −2푥 34 0 푦 + 15 5 4

v. 푨 = 2 1 6푥 − 푦 1 2푥 and 푩 = 2 1 6

푦 1 푦+ 2

vi. 푨 = 2푥 − 1 00 푥 + 푦 and 푩 = 푥푦 0

표 0

4. Given that 푨 = 4 6−8 −12 and 푩 = 1 3

4 5 . Find

i. 푷 ii. –푸 iii. . 푷 –푸 vi. 2푸 −ퟐ푷 v. 푷ퟐ = 푷 ∙ 푷

5. Given that 푷 =−1 −1 21 3 4−4 2 6

and 푸 =2 3 45 6 87 4 −1

. Find

i. 푷 + 푸 ii. 푷–푸 iii. . 푷푸 vi. 푸푷 − 2푷 v. 푸ퟐ = 푸 ∙ 푸

6. Given that 푨 =4 3−2 and 푩 = 5 4

2 6 . Find the values of 푎 and 푏 such that

494

푨푩 = ퟏퟐ

8 푎−2 −10 −

4 2푏 −

7. Find the 2 × 2 matrix 푨 and B such that

–푨 + 2푩 = 1 −10 1 and −2푨 + 푩 = 6 1

2 3

8. Find the 3 × 3 matrix 푨,푩 and C such that

푨 − 푩 + 푪 =1 0 −14 2 31 0 1

, ퟐ푨 + 푩− 푪 =2 3 41 2 4−1 −2 −1

and

푨 + 푩 + 푪 =2 1 1−1 0 53 2 2

9. Find the product of the matrix and the vectors;

i. 푨 = 2 −34 −2 and 풗 = ii. 푨 = 2 3

4 5 and 풗 =

iii. 푨 = 11 −27 −6 and 풗 = iv. 푷 =

−1 −1 21 3 4−4 2 6

and 풗 =

v. 푷 =1 −4 −62 4 −56 8 2

and 풗 = vi. . 푴 =2 3 −7

10 12 −156 −3 4

and 풗 =

10. Find the transpose of the following matrices:

i. 푨 = 푎 −푎−푏 푏 ii. 푩 = 1 2

−3 −4 iii. 푪 = 2 4 78 2 5

iv. 푫 =2 6 74 0 23 −2 4

v. 푩 = [3 1 3] vi. 푸 =2 133 −9−1 −3

11. Write down the sign matrix of the element 푎 attached to a 4 × 4 matrix. 12. Find the adjoint of the following matrices.

i. 푨 = 2 −3−4 5 ii. 푩 = −1 2

4 −3 iii. 푷 =1 0 −4−1 0 −32 −3 −4

vi. 푸 =1 2 −4−3 −4 34 6 −4

13. Find the determinants of the matrices in (12) above. 14. Find the determinant of the following matrices:

495

i. 푨 = ii. 푩 = 1 24 2 iii. 푷 =

1 2 57 2 −38 2 4

iv. 푸 =2 −2 4−4 −8 2−4 −4 5

v. 푹 =4

vi. 푺 = −1 1 −1

15. Find the multiplicative inverses of the matrices in (14) above. 16. Identify the minors and cofactors of the following matrices:

i. 푴 =3 5 24 −5 24 −5 −3

ii. 푵 =1 3 7−2 −3 51 4 5

iii. 푷 =3 5 24 −5 24 −5 −3

iv. 푸 =1 2 0−1 −1 −3−3 2 1

17. Given that 푨 =3 2 25 −2 0

11 2 −4 and 푩 =

5 3 3−2 2 0−1 4 1

. Show that

i. 푨푨 ퟏ = 푰 ii. (푨+ 푩) = 푨 + 푩 18. Use determinants to solve the following system of linear equations.

i. 푥 + 푦 = 12푥 − 푦 = 6 ii. 2푥 − 푦 = 6

푦 + 3푥 = 4 iii. 2푝+ 3푞 = 133푝 − 2푞 = 17 iv. 3푚 + 푛 = 11

5푚 + 2푛 = 73

19. Use determinants to solve the following system of linear equations.

i. 푥 + 푦 − 푧 = 19

2푥 − 푦 + 푧 = 20푥 − 푦 + 3푧 = 23

ii. 푥 − 푦 = 13

2푥 + 푦 + 푧 = 172푥 − 푦 + 3푧 = 19

iii. 2푝 + 푞 − 푟 = 14푝+ 2푞 − 3푟 = 162푝 − 3푞 + 푟 = 18

iv.

푞 − 푟 + 4푠 = 17푞 + 푟 + 6 = 푠

푞 + 푟 − 푠 = 0

20. Given that 푘 −1 40 2 44 −2 3

= 13. Find the possible value of 푘.

21. Given that 푘 − 1 −2 4

0 푘 52 4 3

= −2. Find the possible values of 푘.

22. Use the inverse approach to solve the system of linear equations in (18)and (19) above.

496

23. Find the possible value 푥 can take given that

푨 = 푥 31 3푥

and 푩 = 3 62 푥 and 푨푩 = 푩푨

푨푩 = 3푥 + 6 6푥 + 3푥3 + 6푥 6 + 3푥

and푩푨 = 3푥 + 6 9 + 18푥2푥 + 푥 6 + 3푥

CCE

497

FUNDAMENTAL PRINCIPLES OF COUNTING, PERMUTATIONS AND COMBINATIONS PRINCIPLE 1 Suppose one operation can be done in 푚 ways and another operation can be done in 푛 ways, then the number of possible outcomes when performing the 푎푛푑 the second operation is given by 푚 × 푛. Here, 푎푛푑 is understood to mean 푚푢푙푡푖푝푙푦.Also, we assume that the two outcomes are independent of each other. This principle can be extended to three or more operations.

Example 19.1 A fair die is thrown once. How many outcomes is possible.

Solution A die has six faces dotted 1, 2, 3, 4, 5, 6. Therefore the number of possible outcomes is 6.

Example 19.2 A die is cast and a coin is tossed. How many different outcomes are possible.

Solution A die has six faces dotted 1, 2, 3, 4, 5, 6.So there are 6 possible outcomes for casting a die. A coin has two faces namely: Head and Tail. So there are 2 possible outcomes for tossing a coin. Therefore the number of different outcomes for casting a die and throwing a coin is 6 × 2 = 12. PRINCIPLE 2 Suppose one operation has 푚 possible outcomes and that a second operation has 푛 outcomes, then the number of possible outcomes of the first operation 표푟 the second operation is given by 푚 + 푛. Here, 표푟 is understood to mean 푎푑푑.We also assume that it is not possible for both operation to occur. This principle can also be extended to three or more operations.

Example 19.3 A box contain 10 balls numbered from 0to 10. A ball is drawn from the bag. If the number is even, then a coin is tossed. If the number is odd, then a die is cast. How many outcomes are possible.

Solution The balls are numbered as: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Even numbers:2, 4, 6, 8, 10 So, there are 5 possible outcomes. Odd numbers: 1, 3, 5, 7, 9 So, there are 5 possible outcomes.

498

Tossing a coin reveals 2 possible outcome; a head and a tail. Casting a die reveals 6 possible outcomes; 1, 2, 3, 4, 5, 6 The number of possible outcomes for drawing any of the even numbers and tossing a coin is 5 × 2 = 10 The number of possible outcomes for drawing any of the odd numbers and casting a die is 5 × 6 = 30. Therefore there are 10 + 30 = 40 possible outcomes.

Permutations/Arrangements

A permutation is an arrangement of a number of objects/items in a definite order. In a permutation or arrangements the order of the objects chosen is very important.

Example 19.4 In how many ways can three item labelled A, B and C be arranged.

Solution We have three positions to arrange the items. An item can occupy a first position, second or third position. For instance, if we take item A, it can be place at the 1st, 2nd or 3rd position as shown below

We can see that we have 3 possible ways of arranging the first item (A) chosen. If we take a second item, we are left with two positions to place that item. For instance, if we take item B, we can place it as follow:

The last item left, in our case item C would be placed in the positions left. So there is only one way to arrange the last item left as shown below:

A A A

A B B A B A

A B A B B A

A B C B C A B A C

A C B C A B C B A

499

Therefore there are six possible arrangements. Using the fundamental principle of counting: The first item chosen can be arranged in 3 different ways and the second item chosen can be arranged in 2 different ways and the last item can be arranged in only one way. Thus, the items can be arranged in 3 × 2 × 1 = 6ways. If we have n different object to arrange, then: the total number of arrangement is푛!. 푛! = 푛(푛 − 1)(푛 − 2)(푛 − 3) … × 3 × 2 × 1 For example,5! = 5 × 4 × 3 × 2 × 1 = 120

Example 19.5 In how many ways can five different objects be arranged taken three objects at a time.

Solution Let us label the objects as A, B, C, D, and E. We are making arrangement of three of the items at a time so we have three positions but five different items. Any of the five items can be chosen to take a position in the 3positions. So there are 5 possible outcomes. Then any of the 4 items left can be arranged in 4 different ways. Then any of the 3 objects left can be arranged in 3 different ways. So we have5 × 4 × 3 = 60푤푎푦푠. By using factorials, we have 5 × 4 × 3 = × × × ×

×= !

!= !

( )! .

Number of arrangements= ( )!( )!

The number of arrangement of n different objects taking r at a time is given by

nrP = !

( )!= 푛(푛 − 1)(푛 − 2) × … × (푛 − 푟 + 1)

Example 19.6 How many different arrangements can be made from the letters A, B, C, D, E, taking at a time, if B must be second and E cannot be last.

Solution B must occupy the second position so there is only one way of arranging the letter B. The letter E can occupy either the 1st, 3rd or 4thpositions so there are 3 ways of arranging the letter E. Any of the letters chosen can also be arranged in three different ways. The next letter can be arranged in two different ways and the last letter can be arranged in only one way so we have

500

1 × 3 × 3 × 2 × 1 = 18표푢푡푐표푚푒푠 Example 19.7

In how many ways can the letters P, Q and R be arranged if there are no restrictions? In how many ways can the letters be arranged if P and Q must always be together? In how many ways can the letters be arranged if P and Q must not be together?

Solution The letters P, Q and R can be arranged in 3! = 3 × 2 × 1 = 6푤푎푦푠 if there are no restrictions as shown below.

In the above arrangements there are 4 different arrangements in which P and Q are together. We can think about this in this way: The letters P and Q can be arranged together in 2! = 2 × 1 = 2 ways; either PQ or QP. If we assume that P and Q are seen as one word, then we have two letters⟦P, Q⟧, R (two letters) to arrange and this can be done in 2! = 2푤푎푦푠. And we can calculate the number of way in which P and Q are together as

푁푢푚푏푒푟표푓푤푎푦푠푖푛푤ℎ푖푐ℎ푃푎푛푑푄푎푟푒푡표푔푒푡ℎ푒푟

=푃표푠푠푖푏푙푒푤푎푦푠

푖푛푤ℎ푖푐ℎ푃푎푛푑푄푐푎푛푏푒푡표푔푒푡ℎ푒푟푡푎푘푒푛푎푙표푛푒

×푃표푠푠푖푏푙푒푤푎푦푠푖푛푃푎푛푑푄푡푎푘푒푛푎푠표푛푒푤표푟푑푏푒

푎푟푟푎푛푔푒푑푤푖푡ℎ푡ℎ푒표푡ℎ푒푟푙푒푡푡푒푟푠

Therefore, the number of ways in which P and Q are together= 2! × 2! = 4푤푎푦푠 There are 2 = 6 − 4 in which P and Q are not together. We can deduce the following facts from the above example: 1. =

+

= 푐! × 푘! Where, 푐 = the numbers of objects that must be together and 푘 =the number of other objects +1

P Q R Q R P Q P R

P R Q R P Q R Q P

501

2.

=

푃표푠푠푖푏푙푒푤푎푦푠푖푛푤ℎ푖푐ℎ푡ℎ푒표푢푡푐표푚푒푠푐푎푛푏푒푡표푔푒푡ℎ푒푟푡푎푘푒푛푎푙표푛푒

×푃표푠푠푖푏푙푒푤푎푦푠푡ℎ푒표푢푡푐표푚푒푠푡푎푘푒푛푎푠표푛푒표푢푡푐표푚푒푐푎푛푏푒푎푟푟푎푛푔푒푑푤푖푡ℎ푡ℎ푒표푡ℎ푒푟푙푒푡푡푒푟푠

Example 19.8

In how many ways can the letters A, B, C, D, E be arranged if i. there are no restrictions ii. A and B must always be together iii. A and B must not be together. iv. A and E must be at the ends of each arrangement.

Solution i. Arranging the letters without restrictions is the same as arranging the letters taking all at a time. ∴the 5 letters can be arranged in 5! = 5 × 4 × 3 × 2 × 1 = 120푤푎푦푠 ii. The letters A and B can be arranged togetherin 2! = 2 × 1푤푎푦푠. That is either AB or BA. Now, taking A and B as on letter, the rest of the letters can be arranged with A, B in 4! = 4 × 3 × 2 × 1 = 24푤푎푦푠. ∴the numbers ways of arranging the 5 letters if A and B must always be together = 2! × 4! = 48푤푎푦푠 iii.

= −

= 120 − 48 = 72푤푎푦푠 iv. The letters C, D, E can be arranged together in 3! = 3 × 2 × 1 = 6푤푎푦푠. Now, taking C, D, E as on letter, the rest of the letters can be arranged with C, D, E in 3! = 3 × 2 × 1 = 6푤푎푦푠. ∴the numbers ways of arranging the 5 letters if A and B must always be together = 3! × 3! = 36푤푎푦푠 v. Now, A and E must be at the ends. In the arrangements, this can be done in 2! = 2푤푎푦푠. That is, either A appearing first and E appearing last or E appearing first and A appearing last in each arrangement. The rest of the three letters can be arranged in 3! = 3 × 2 × 1 = 6푤푎푦푠 ∴the number of arrangements with A and E at the ends= 2 × 3! = 12푤푎푦푠

Permutations with repetitions

502

In permutations with repetitions each element is repeated. Given a set of 푛 objects, the number of permutations form by 푟 elements of subsets with repetition is defined by

rn r rP n

If 푟 = 푛, then n

n r nP n Example 19.9

How many 3 letter words can be formed from the letters A, B, C if repetition of letters is allowed.

Solution There are three letter so 푛 = 3 and we are forming three letter word each allowing repetitions of letters. The numbers of possible words is defined as: n r n

nP n 3 3

3 3 27P

Permutations with repetition of indistinguishable object Indistinguishable objects are objects that are repeated in the original set. For example, in the word PERMUTATION, the two T’s are indistinguishable from each other. The number of permutations of 푛 where there are 푟 indistinguishable objects of objects of kind 1, 푟 indistinguishable objects of kind 2 and so on and 푟 indistinguishable objects of kind 푘 is given by

푛!푟 ! × 푟 ! × ⋯× 푟 !

Example 19.20 How many different 12 letter words can be formed from the word PERMUTATIONS.

Solution Number of letters 푛 = 12. There are two T’s so 푟 = 2. ∴, there are !

!= 239500800 words.

Example 19.21 How many different 5-digits numerals can be formed from the digits 1, 1, 2, 2, 3, 3

Solution Number of digits 푛=5. There are two 1’s and two 2’s. So 푟 = 2 and 푟 = 2. ∴, there are !

!× != 30numeral.

503

Circular and Ring Permutations When objects are arranged in a circle there is ordering such as first objects, second object and so on. Any of the objects can be first. The permutation of 푛 different objects in a circle is given by:

(푛 − 1)! The permutation of 푛 different objects in a ring is given by:

(푛 − 1)!2

Example 19.22 In how many different ways can eight players group in a circle.

Solution Number of players 푛 = 8. ∴, there are (8 − 1)! = 7! = 5040 ways.

Example 19.23 In how many ways can 8 beads be arrange on a ring.

Solution Number of beads 푛 = 8. ∴, there are ( )! = 2520 ways.

Combinations/Selections

A combination is a selection of a number of objects in any order. Here, the order in which the objects are selected is not important. That is, in selection of objects, AB and BA represent the same selection. The number of ways of choosing 푟 objects from 푛 different objects is given by

rnC = !

( )! !

Other notations for rnC are 푛푟 and rn C . r

nC is pronounced ‘n- choose- r’ By definition 0! = 1. For example the number of ways of selecting 3 shirts out of 5 different shirts is given by

35C = !

( )! != × × × ×

× × × ×= 10푤푎푦푠

Some Basic Properties

1. 10 Cn Since 0Cn = !( )! !

= !!

= 1. That is, there is only one way of

choosing no object out of 푛 objects.

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2. 1nnC Since n

nC = !( )! !

= !! !

= 1. That is, there is only one way

of choosing 푛 objects out of 푛 objects.

3. rnn

rn CC Since rn

nC = !( ) !( )!

= !( )!( )!

= !( )! !

= rnC

This identity often used for 푟 >

For instance, 1012C = 2

12C

4. rnC =

! rPr

n Since

! rPr

n=

!( )!

!= !

( )! != r

nC

Example 19.23 In how many ways can a committee of 3 people be selected from a group of 10 people?

Solution We need to choose 3 people from a group of 10 people to form a committee. So, 푛 = 10and 푟 = 3.

∴the number of ways= 310C = !

( )! != × × × !

!× × ×= × ×

× ×= 120푤푎푦푠

Example 19.24

i. In how many ways can a committee of 3 persons be chosen from 4 persons? ii. If a particular person must always be on the committee, how many ways can the committee be chosen? iii. If a particular person must not be on the committee, how many ways can the committee be chosen?

Solution i. We need to choose 3 people from a group of 4 people to form a committee. So, 푛 =4and 푟 = 3. ∴the number of ways is given by:

14

34 CC = !

( )! != × !

!×= = 4푤푎푦푠

ii. If one particular person must be on the committee, then we are left with choosing 2 persons (3 − 1 = 2) from 3 persons (4 − 1 = 3) to join that person on the committee. So, 푛 = 3and 푟 = 2. ∴the number of ways is given by:

505

23C = !

( )! != × ×

× ×= 3푤푎푦푠

iii. If one particular person must not be on the committee, then we are still choosing 3persons from3 persons (4 − 1 = 3) to form the committee. So, 푛 = 3and 푟 = 3. ∴the number of ways is given by:

33C =1푤푎푦

Example 19.25 i. In how many ways can a committee of 5 persons be chosen from 12 persons? ii. If a particular person must always be on the committee, how many ways can the committee be chosen? iii. If a particular person must not be on the committee, how many ways can the committee be chosen?

Solution i. We need to choose 5 people from a group of 12 people to form a committee. So, 푛 = 12and 푟 = 5. ∴,the number of ways is given by:

512C = !

( )! != × × × × × !

!× × × × ×= × × × ×

× × × ×= 792푤푎푦푠

ii. If one particular person must be on the committee, then we are left with choosing 4 persons (5− 1 = 4) from 11 persons (12− 1 = 11) to join that person on the committee. So, 푛 = 11and 푟 = 4. ∴,the number of ways is given by: 11

4C = !( )! !

= × × × !!× × × ×

= × ×× × ×

= 330푤푎푦푠

iii. If one particular person must not be on the committee, then we are still choosing 5 persons from11 persons (12 − 1 = 11) to form the committee. So, 푛 = 11and 푟 = 5. ∴,the number of ways is given by: 11

5C = !( )! !

= × × × × !!× × × × ×

= × × ×× × × ×

= 462푤푎푦푠

Final Exercises

1. Evaluate the following:

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i. ii. iii. iv. v. vi. 2. In how many ways can a committee of four people be chosen from nine people? 3. There are 15 students in a class. i. How many teams of 8 can be selected from the class? ii. If one person in the class is made a captain and must always be included in each team, how teams can now be selected? iii. If two students in a class refuse to play, how many teams can now be selected, if the captain must still be on every team. 4. In how many ways can a committee of six people be chosen from a group of ten people if i. Any person may be selected. ii. The oldest person must be selected. iii. The youngest person must be selected? iv. The youngest and the oldest must both be selected. 5. A third year student has to choose four subjects from the following list. Accounting, Biology, Physics, French, Applied Math’s and Classical Studies. i. How many different choices are possible? ii. How many choices include French? iii. How many choices do not include French? iv. How many choices include Accounting and Biology? v. How many choices include Applied Math’s but not Chemistry. 6. Three delegates to form a committee are to be selected from ten members of a club. How many committees can be formed if: i. There are no restrictions? ii. A certain member must be on each committee? iii. Two particular members cannot both be on the committee? 7. From a set of 6 different coins, in how many ways can four or more coins be selected? 8. In how many ways can twelve different objects be divided into groups 6,4 and 2? 9. Five Female students, four Male students and six teacher Germans are available for selection to form a disciplinary club of four. If each group has to be represented on the committee, in how many ways can the committee be selected? 10. Find the number of different selection of five letters that can be made from the letters of the word ARISE.

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11. In how ways can a committee of 5 people be selected from four men and six women, if the committee must have at least four women on it? 12. A committee of eight is to be formed from eight students and six teachers. How many different committees can be formed if there are to be more teachers than students? 13. An examination consists of 12 questions 4 in section A and 8 in section B, a candidates must attempt six questions at least three of which must be from each section. In how many different ways may the candidate select the six questions? 14. In how many ways can a committee of six be selected from five men and four women; if each committee consists of i. An equal number of men and women? ii. At least three men? 15. A team of five players is to be chosen from six boys and five girls. If there must be more boys than girls, how many different teams can be formed? 16. A group consists of five men and seven women. A committee can be chosen in which there odd number of men? 17. How many different arrangement a can be made using all the letters of the word RAISED? i. How many arrangement begin with the letter D? ii. How many arrangement begin with E and end in R? iii. How many arrangement begin with a vowel? vi. How many arrangement begin and end with a vowel? v. How many arrangement ends with RAI? vi. How many arrangement begin with D and end in LIN? 18. Taking all the letters of the word ALGEBRA, in how many different arrangements are possible. 19 i. In how many ways can three girls and two boys be seated in a row of five seats? ii. In how many ways can this be done if the boys must be together. iii. In how many ways can this be done if the boys must not sit together? 20. How many arrangement of the letters of the word QUADRATICS are possible if all the three vowels must come together in the arrangement. 21 .Six children are to be seated in a row on a bench i. How many arrangements are possible?

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ii. How many arrangement are possible if the youngest child must sit at the left-hand end and the oldest child must sit at the right –hand end? If two of the children are twins, in how many ways can the children be arranged if ∝.There twins are together. 훽. The twins are not together. 22. In how many ways can four letters of the word COUNT be arranged in a row if i. no repetition of letters is allowed? ii. repetition of letters is allowed? 23. How many different three digits number can be formed using the digits 2, 4, 6, 8, 10, if 2 cannot be the first digit and i. no digit may be repeated ii. repetitions are allowed? 24. How many four digits can be formed from the digits 1, 3, 5, 7, 9 if: i. There are no restrictions and repetitions are allowed; ii. The number is divided by 3 and repetitions are not allowed. iii. The number is greater than five 5,000 no repetition are allowed? 25. How many numbers between 3,000and4,000 can be made with the digits 1, 2, 3 and 4 if no digit may be repeated. 26. How many number between 100and1000 use only odd digits, if no digits are repeated? 27. How many odd numbers between 2,000and3000 can be formed from the digits 2, 3, 4, 5 and 6 if i. Repetitions are allowed ii. Repetitions are not allowed 27. How many odd numbers between 4,000and6,000 can be formed from the digits 3,4, 5, 6,7 if no digit can be repeated? 28. How many different three- digits can be formed using the digit 0 to 9 inclusive, if no digit can be used more than once and zero cannot be the first digit? If the digits 2 and 5 cannot be used together, how many different numbers can be formed? 29. A bag contains five disc numbered from one to five inclusive. A disc is drawn from the bag. If the number is even, then a die is thrown. If the number is odd, then a coin is tossed, how many outcomes are possible. 30. Evaluate the following:

5 6 13 9 42 3 5 5 3i. ii. iii. iv. v. rP P P P P

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31. Solve the equations:

1 43 2

2

!i. ii. 220

nn

n nn CP

C

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THE BINOMIAL THEOREM A binomial expression is an expression of the form 푥 + 푦,푎 + 푏, 푢 + 푣푎푛푑푠표표푛. Under this topic in finding a general formula for the binomial expansion of

(푥 + 푛) , 푛 ∈ ℤ. Now let us consider the following expansion. (푥 + 푦) = 1. (푥 + 푦) = 푥 + 푦. (푥 + 푦) = 푥 + 2푥푦 + 푦 . (푥 + 푦) = 푥 + 3푥 푦 + 3푥푦 + 푦 . (푥 + 푦) = 푥 + 4푥 푦 + 6푥 푦 + 4푥푦 + 푦 . (푥 + 푦) = 푥 + 5푥 푦 + 10푥 푦 + 10푥 푦 + 5푥푦 + 푦 . When we write down the coefficients to the terms in the expansion we have: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 This called the PASCAL’S TRIANGLE. After the first ………………. The following characteristics can be made from the expansion above: 1. There are (푛 + 1) terms in the expansion of(푥 + 푦) . For example the expansion of(푥 + 푦) has 4 terms 2. The first and the last term of the expansion are 푥 and 푦 respectively. This is the symmetric property of the expansion of(푥 + 푦) . 3. The powers or exponents of푥 decreases exponents of푦 increases by 1 from term to term. 4. The sum of the exponents of 푥푎푛푑푦 in each term in the expansion is푛. Now consider the expansion: (푥 + 푦) = 푥 + 3푥 푦 + 3푥푦 + 푦 . The coefficients of each term for a combination 3C0 3 C1 3C2 3C3

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So we can write (푥 + 푦) =3C0푥 +3C1푥 푦 +3C2푥푦 +3C3푦 In general the expansion of(푥 + 푦) can be written as (푥 + 푦) =nC0푥 +nC1푥 푦 +nC2푥 푦 + ⋯+nCn푦 = 푥 +nC1푥 푦 +nC2푥 푦 + ⋯+ 푦 But nCr= !

( )! ! so nC0= !

( )! != 1 nC1= !

( )! != ( )!

( )!= 푛

nC2= !( )! !

= ( )( )!( )! !

= ( )!

⋯ nCn= !( )! !

= !!

= 1

So the expansion becomes (푥 + 푦) = 푥 + 푛푥 푦 + ( )

!푥 푦 + ⋯+ 푦

Example 20.1 Expand the following:

i. (푥 + 푦) ii. (2푥 − 푦) iii. (3푥 + 2푦) iv. √푎 + √푏 solution

i. We have: (푥 + 푦) . The expansion of (푥 + 푦) is given by : (푥 + 푦) = 푥 + 푛푥 푦 + ( )

!푥 푦 + ⋯+ 푦

∴, (푥 + 푦) = 푥 + 4푥 푦 + ( )

!푥 푦 + ( )( )

!푥 푦 + ( )( )( )

!푥 푦

= 푥 + 4푥 푦 + 6푥 푦 + 4푥 푦 + 푦 ii. We have: (2푥 − 푦) . The expansion of (푥 + 푦) is given by :

(푥 + 푦) = 푥 + 푛푥 푦 + ( )!

푥 푦 + ⋯+ 푦

∴, (2푥 − 푦) = (2푥) + 5(2푥) (−푦) + ( )!

(2푥) (−푦) + ⋯+ (−푦) = 32푥 − 80푥 푦 + 80푥 푦 + ⋯− 푦 iii. We have: (3푥 + 2푦) . The expansion of (푥 + 푦) is given by :

(푥 + 푦) = 푥 + 푛푥 푦 + ( )!

푥 푦 + ⋯+ 푦

∴, (3푥 + 2푦) = (3푥) + 4(3푥) (2푦) + ( )!

(3푥) (2푦) + ⋯+ (2푦) = 81푥 + 216푥 푦 + 216푥 푦 + ⋯+ 16푦

iv. We have: √푎 + √푏 . The expansion of (푥 + 푦) is given by :

(푥 + 푦) = 푥 + 푛푥 푦 + ( )!

푥 푦 + ⋯+ 푦

512

∴,

(√푎 + √푏) = √푎 + 3(√푎) √푏 + ( )! √푎 √푏 + ( )( )

! √푎 √푏

= 푎√푎 + 3푎√푏 + 3푏√푎 + 푏√푏 Following the expansion of (푥 + 푦) we can also expand the binomial (1 + 푥) as (1 + 푥) =nC0(1) +nC1(1) 푥 +nC2(1) 푥 + ⋯+nCn푥 = 1 +nC1푥 +nC2푥 + ⋯+ 푥 = 1 + 푛푥 + ( )

!푥 + ⋯+ 푥

Note: (푎 + 푥) = 푎 1 + Example 20.2

Expand the following

i. (1 + 푥) ii. (1 − 푎) iii. 1 − iv. (3푥 + 3) solution

i. We have: (1 + 푥) . The expansion of (1 + 푥) is given by : (1 + 푥) = 1 + 푛푥 + ( )

!푥 + ⋯+ 푥

∴, (1 + 푥) = 1 + 5푥 + ( )!

푥 + ( )( )!

푥 + ( )( )( )!

푥 + 푥

= 1 + 5푥 + 10푥 + 10푥 + 5푥 + 푥 ii. We have: (1 − 푎) . The expansion of (1 + 푥) is given by :

(1 + 푥) = 1 + 푛푥 + ( )!

푥 + ⋯+ 푥

∴, (1− 푎) = 1 + 6(−푎) + ( )!

(−푎) + ( )( )!

(−푎) + ⋯+ (−푎) = 1 − 6푎 + 15푎 − 20푎 + 15푎 − 6푥 + 푎

iii. We have: 1 − . The expansion of (1 + 푥) is given by :

(1 + 푥) = 1 + 푛푥 + ( )!

푥 + ⋯+ 푥

∴, (1− ) = 1 + 4 − + ( )!

− + ( )( )!

− + −

= 1 − 2푥 + 푥 − 푥 + 푥 iv. We have: (3푥 + 3) = 3 (1 + 푥) . The expansion of (1 + 푥) is given by :

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(1 + 푥) = 1 + 푛푥 + ( )!

푥 + ⋯+ 푥

3 (1 + 푥) = 3 1 + 4(푥) + ( )!

(푥) + ( )( )!

(푥) + (푥)

= 81[1 + 4푥 + 6푥 + 4푥 + 푥 ] = 81 + 324푥 + 486푥 + 324푥 + 81푥 ∴, (3푥 + 3) = 81 + 324푥 + 486푥 + 324푥 + 81푥

Linear Approximation of the Binomial (1 + 푥) Consider the expansion (1 + 푥) = 1 + 푛푥 + ( )

!푥 + ⋯+ 푥

As the value of 푥 increases and the values gets closer to zero, the expansion of (1 + 푥) becomes equivalent to1 + 푛푥. That is, (1 + 푥) ≈ 1 + 푛푥

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PROBABILITY Probability implies lack of certainty and lack of certainty is attributed to chance. The theory of probability measures how near our certainty is. Probability theory is used in the various field of Science, Arts, Economics and business related studies such as Commerce, marketing, supply management and so on.

Definition of some Important terms Random Experiment

A random experiment is the one which outcome comes from one of the possible outcomes of an experiment performed again and again under the same condition.

Trial A trial is the act of performing a random experiment.

Event An event refers to one or more outcome of a random experiment. An event is denoted by upper case letters such as A, B, C etc.

Types of events 1. Sure/ certain event An event is sure if it is certain that it will occur. The probability of occurrence of a sure event is unity (1). 2. Impossible/ Absurd events An event is impossible if it will not occur. The probability of occurrence of an impossible event is 0. 3. Uncertain event An event is said to be uncertain if we cannot tell whether it will occur or not. The probability of occurrence of an uncertain event lies between 0 and 1. Thus, if A is an uncertain event, then

0 < 푃(퐴) < 1 Where, P(A) denotes the probability of an event A.

An event is said to be simple or elementary if it cannot be decompose into more than one event. For example, the event of getting either a head or a tail when a coin is tossed is a simple event.

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An event is said to be composite or compound if it can be decomposed into more than one simple event. For example, the event of a multiple of 3 on each die when two dice are thrown is a compound event. Two or more events associated with a random experiment are said to equally likely if neither of them occurs in preference to the other. Two or more event associated with a random experiment are said to be mutually exclusive if they are such that when one occurs the other(s) cannot occur. Two events A and B are said to be mutually exclusive if they cannot occur simultaneously. By illustration, if we have 20 balls numbered 1 to 20, the probability or chance of getting a number divisible by 4 would be the addition of the separate probabilities of getting balls numbered 4, 8, 12, 16 or 20. Let A be the event of getting a number divisible by 4. Then P(A)= + + + + = = And the probability of getting a number divisible by 5 would be the addition of the separate probabilities of getting balls numbered 5, 10, 15 or 20. Let B be the event of getting a number divisible by 5. Then P(B)= + + + = = It is wrong to assume that that the probability of getting a number divisible by 4 or 5 is

+ = .

That is, P(A or B)≠ + , since 퐴 ∩ 퐵 ≠ ∅. Two or more events associated with a random experiment are said to be exhaustive if they include all the possible outcomes of the experiment. The outcomes of a random experiment which are favourable to a particular event are called favourable cases to the event. A random event is usually defined as the event which occurs in an undesigned or unforseen way.

Sample (outcome) space A sample space of an experiment is defined as the set S of all possible (simple events) outcomes. It should be noted that the null set, ∅,denotes the impossible events and the sample space S denotes the sure event. The number of all possible points in the sample

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space S is denoted by 푛(푆) and the number of sample point of an event A is denoted by 푛(퐴). A sample space can be finite or infinite. A sample space is said to be finite if all its simple events is known. For example, the experiment of throwing a die once has a finite sample space. In this experiment, the performer can get either a 1, 2, 3, 4, 5 or 6. That is, S= {1, 2, 3, 4, 5, 6}. A sample space is infinite if some of it simple elements is unknown. For example, the experiment of tossing a coin repeatedly till a head appears has an infinite sample space. If 푒 , 푒 , 푒 , … represent the outcomes of the experiment, then the infinite sample space S={푒 , 푒 , 푒 , … }.

Classical (Mathematical) Approach to Probability Definition: Probability is the proportion represented by the possible events containing a desired attribute out of the number of all possible equally likely events in the same set. If there are 푛 equally likely, mutually exclusive and exhaustive cases and 푚 of them are favourable to an event 퐴, then the probability (chance) of the occurrence (success) of the event 퐴 is defined as the ratio . The classical or theoretical approach to probability is based on the assumptions that all outcomes of a random experiment are equally likely. The total number of all possible outcome of a random experiment is finite. Symbolically, P(A)=

Where P(A) denotes the probability of occurrence of the event A. The probability of non-occurrence of event A is denoted by P(퐴̅) and it is given by

푃(퐴̅) =

This follows that P(A)+P(A) = + = 1

Example 21.1 A box contain 5 black, 4 white and 2 red balls. If a ball is selected at random from the box, what is the probability that a. it is white b. it is not white.

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Solution Black ball White balls Red balls Total balls 5 + 4 + 2 = 11 a. Let W=white balls n(W) = 4 ∴, P(W) =

b. n(W) = 5 + 2 = 7 ∴, P(W) =

Example 21.2 There are 20 marbles in a bag numbered from 1 to 20. If a marble is drawn at random, find the probability that the number on the marble is i. a multiple of 2 or 5 ii. a multiple of 3 and 7.

Solution the marbles are numbered as: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, … , 20 i. Event of multiples of 2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 Event of multiples of 5: 5, 10, 15, 20 P(a multiple of 2 or 5)= P(a multiple of 2) + P(a multiple of 5)= + = = ii. Event of multiples of 3: 3, 6, 9, 12, 15, 18 Event of multiples of 7: 7, 14 P(a multiple of 3 and 7)= P(a multiple of 3) × P(a multiple of 7)= + = =

Example 21.3 Three fair coins are tossed once. What is the probability of obtaining a. exactly two heads? b. at least one head? c. not more than one tail? d. no head?

Solution If a coin is tossed either a Head (H) or a Tail (T) will appear. Therefore the sample space for tossing a coin once is {H, T} The sample space for tossing the coin a second time=tossing two coins once is shown below:

S H T H HH HT T TH TT

S = {HH, HT, TH, TT}

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The sample space for tossing the coin a third time=tossing three coins once is shown below:

S HH HT TH TT H HHH HHT HTH HTT T THH THT TTH TTT

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} n(S) = 8 a. Event of exactly two heads: {HHT, HTH, THH} 푛(event) = 3 P(exactly two heads)= b. Event of at least one head: {HHH, HHT, HTH, HTT, THH, THT, TTH} 푛(event) = 7 P(at least one head)= c. Event of not more than one tail {HHH, HHT, HTH, THH} 푛(event) = 4 P(not more than one tail)= =

Example 21.4

A fair die is thrown twice. What is the chance of obtaining sum of points from the thrown a. 7 points? b. more than 8? c. at least 10? d. at most 5?

Solution

S 1 2 3 4 5 6 1 1, 1 1, 2 1, 3 1, 4 1, 5 1, 6 2 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6 3 3, 1 3, 2 3, 3 3, 4 3, 5 3, 6 4 4, 1 4, 2 4, 3 4, 4 4, 5 4, 6 5 5, 1 5, 2 5, 3 5, 4 5, 5 5, 6 6 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6

a. Event of sum of 7 points {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} P(sum of 7 points)= = b. Event of sum of more than 8 points {(3, 6), (4, 5), (5, 4), (6, 3), (6, 4), (4, 6), (5, 5), (5, 6), (6, 5)(6, 6)} P(sum of more than 8 points)= = c. Event of sum of at least 8 points {(6, 4), (4, 6), (5, 5), (5, 6), (6, 5)(6, 6)}

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P(sum of more than 8 points)= = d. Event of sum of at most 5 points {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1)(2, 2), (2, 3), (3, 1), (3, 2), (4, 1)} P(sum of at most 5 points)= =

Example 21.5 A fair die is thrown twice. What is the chance of obtaining a a. sum of 3 points? b. sum of 3 points or a multiple of 6 points? c. a multiple of 3 or a sum of 6 points? d. sum of 3 points and a multiple of 6 points? e. a multiple of 3 and a sum of 6 points

Solution

S 1 2 3 4 5 6 1 1, 1 1, 2 1, 3 1, 4 1, 5 1, 6 2 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6 3 3, 1 3, 2 3, 3 3, 4 3, 5 3, 6 4 4, 1 4, 2 4, 3 4, 4 4, 5 4, 6 5 5, 1 5, 2 5, 3 5, 4 5, 5 5, 6 6 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6

a. Event of sum of 3 points {(1, 2), (2, 1)} P(sum of 3 points)= = b. Event of sum of 3 points {(1, 2), (2, 1)} Event of multiple of 6 points {(6, 6)} Event of sum of 3 points or a multiple of 6 points {(1, 2), (2, 1), (6, 6)} P(sum of 3 points or a multiple of 6 points)= + = = Note that the two events are mutually exclusive. c. Event of multiple of 3 points {(3, 3), (3, 6), (6, 3), (6, 6)} Event of sum of 6 points {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}

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Event of multiple of 3 points or a sum of 6 points {(1, 5), (2, 4), (3, 3), (3, 6), (4, 2), (5, 1), (6, 3), (6, 6)}

P(multiple of 3 points or sum of 6 points)= = Note that the two events are not mutually exclusive. d. Event of sum of 3 points and a multiple of 6 points { } P(sum of 3 points and a multiple of 6 points)= = 0 e. Event of multiple of 3 points and a sum of 6 points {(3, 3)} P(multiple of 3 points and a sum of 6 points)=

Example 21.6

A box 푃 contains 4 red and 3 blue balls. Another box 푄 contains 4 green and 6 blue balls. A ball is picked at random from each box. Find the probability that, i. One is green and the other is red. ii. They are of the same colour.

Solution For box 푃. Red balls Blue balls Total balls 4 + 3 = 7 Let 푅 =Red balls in box P. For box 푄. Green balls Blue balls Total balls 4 + 6 = 10 a. Let W=white balls n(W) = 4 ∴, P(W) =

b. n(W) = 5 + 2 = 7 ∴, P(W) = Note: An unbiased/fair die is the one which is perfectly balanced so that the probabilities of each of the sides are equal. Try: 1. A coin is tossed twice and a die is rolled one. Find the probability of the following

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events. a. Two heads and a 5. b. Exactly one head and a 1 or 2. c. At least one head and not a two.

SSSCE 2. Two fair coins are tossed and a die is thrown. What is the probability of obtaining a head, a tail and a 4?

(SSSCE) 3. A purse contains 3 five-thousand, 4 two thousand and 6 one-thousand cedis notes, all of the same size. A currency note is selected from the purse at random. What is the probability that either a one thousand or a two thousand cedi note is selected? (SSSCE)

4. A blue die and a green die are thrown simultaneously. a. Write a sample space for this experiment, expressing each element of the sample space as an ordered pair. b. Determine the set representation of each of the set listed below; 퐴 = the sum of the score is divisible by 3; 퐵= the sum of the two scores is odd; 퐶 = the score differ by at least 3. 퐷 = an odd score is obtained on the green die. c. Calculate i. 푛(퐴 ∩ 퐵) ii. 푃(퐴 ∪ 퐵).

(SSSCE) 5. A box 퐴 contains 3 white and two blue balls. Another box 퐵 contains 4 green and 5 blue balls. A ball is picked at random from each box. Find the probability that, i. One is green and the other is white. ii. They are of the same colour.

(SSSCE)

Statistical (Frequency) Approach to Probability The statistical approach to probability is based on experiment performed in the past. If in the 푛 trials of an experiment, an event A occurs 푓 times, then the probability of A, P(A), is given by 푃(퐴) = , where 푛 = ∑푓

Example 21.5

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The table below gives the frequency distribution of marks of students recorded in a test. Marks 30 40 50 60 70 80 90 Frequency 1 4 6 2 7 3 5 If a student is selected, find the probability that he scored a. 60 marks b. more than 60 marks c. between 50 to 80 marks inclusive.

Solution

푓 = 1 + 4 + 6 + 2 + 7 + 3 + 5 = 28

a. The frequency of scoring 60 marks is 2 ∴, P(Scoring 60 marks)= = b. The frequency of scoring more than 60 marks= 7 + 3 + 5 = 15 ∴,P(Scoring more than 60 marks)= c. The frequency of scoring between 50 and 80 marks inclusive= 6 + 2 + 7 + 3 = 18 ∴,P(Scoring between 50 and 80 marks inclusive)= =

Subjective Approach to probability Subjective approach to probability refers to to the act of assigning probabilities to events based one’s belief or subjective assessment that the events will occur.

Example 20.6 A sum is given to five students A, B, C, D and E. Their respective chances of solving it are

, , , and . What is the chance that at least one of them solves the sum? Solution

P(A)= P(퐴) = 1 − = P(B)= P(퐵)= 1 − = 푃(퐶) = 푃 퐶 = 1 − =

P(D)= 푃 퐷 = 1 − = P(E)= 푃 퐸 = 1 − =

P(none of them solve the sum)=P(퐴 ∩ 퐵 ∩ 퐶 ∩ 퐷 ∩ 퐸) = × × × × = P(at least one of them solves the sum)=1−P(none of them solve the sum)

= 1 −3

20

=

523

Set theory (Modern) Approach to probability

This approach to probability is based on axioms∗ of set theory and metric theory of functions. It sometimes include the classical and statistical definitions of probability but it is free from its drawbacks. Note the following properties of sets: 퐴 = (퐴 ∩ 퐵 ) + (퐴 ∩ 퐵) ⟹ P(퐴) = 푃(퐴 ∩ 퐵 ) + 푃(퐴 ∩ 퐵) 퐵 = (퐴′ ∩ 퐵) + (퐴 ∩ 퐵) ⟹ P(퐴) = 푃(퐴′ ∩ 퐵) + 푃(퐴 ∩ 퐵) (퐴′ ∪ 퐵′) = (퐴 ∩ 퐵)′ ⟹ P(퐴′ ∪ 퐵′) = 푃(퐴 ∩ 퐵)′=1−P(A∩ 퐵) Definition: To every event A, there corresponds a real valued function 푃(퐴) called the probability of the happening of the event A which satisfies the following three (3) axioms i. 0 < 푃(퐴) < 1 ii. 푃(푆) = 1 iii. If 퐴 ,퐴 , … ,퐴 are mutually exclusive (disjoint) events, then

P(A orA or… orA ) = P(A ) + P(A ) + … + P(A ) P(A)=

= ( )

( )

Example 21.7

In a class of 100 students, 60 offer Mathematics and 50 offer English. What is the probability that a. a student picked at random offer both Mathematics and English b. a student picked at random offer only Mathematics. c. a English student picked at random offer English.

Solution Number of students in class, 푛(푈) = 100 Number of Mathematics students, 푛(푀) = 60 Number of English Students, 푛(퐸) = 50 a. 푛(푈) = 푛(푀) + 푛(퐸) − 푛(푀 ∩ 퐸) ⟹ 푛(푀 ∩ 퐸) = 푛(푀) + 푛(퐸) − 푛(푈) ⟹ 푛(푀 ∩ 퐸) = 60 + 50 − 100 = 10 ∴, 푃(푀 ∩ 퐸) = ( ∩ )

( )= =

b. 푛(푀) = 푛(푀 ∩ 퐸 ) + 푛(푀∩ 퐸) ⟹ 푛(푀 ∩ 퐸′) = 푛(푀) − 푛(푀 ∩ 퐸)

⟹ 푛(푀 ∩ 퐸′) = 60− 10 = 50 ∴, 푃(푀 ∩ 퐸′) = ( ∩ )( )

= =

c. 푃(퐸) = ( )( )

= = ∴,푃(퐸) =

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Example 21.8

If P(A)= , P(B)= , P(A∪ 퐵) = , find P(A∩ 퐵′) and P(A’∪ 퐵′),where A and B are not mutually exclusive events.

Solution If P(A)= , P(B)= , P(A∪ 퐵) =

P(A ∩ B) = P(A) + P(B) − P(A∪ B) ⟹ P(A ∩ B) = + − = =

P(A ∩ B ) = P(A)− P(A ∩ B) ∴, P(A ∩ B ) = − = =

P(A′ ∪ B′) =1−P(A∩ B) ∴, P(퐴′ ∪ 퐵′) =1− =

Mutually Exclusive Events If A and B are disjoint then 퐴 ∩ 퐵 = ∅⟹ P(퐴 ∩ 퐵) = 푃(∅) = 0.

Independent and Dependent Events (Conditional Probability)

Two events A and B are said to be independent if the occurrence or non-occurrence of one does not affect the occurrence or non-occurrence of the other. If A and B are independent events ,then

P(퐴 ∩ 퐵) = 푃(퐴) ∙ 푃(퐵) Two events are said to be dependent if one can occur if and only if the other is known to have occurred. If A and B are dependent events, then A can occur if it is known that B have occurred and vice versa. The probability associated with such events is called conditional probability. The probability of event A given B is denoted by P(A\B) and is defined by

P(A\B)= ( ∩ )( )

Similarly, 푃(퐵\퐴) = ( ∩ )( )

denotes the probability of event B given A.

Again, if A and B are independent events, then 푃(퐴\퐵) = 푃(퐴) and P(B\A)=P(B)

Example 21.9

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A and B are two independent events such that P(A)= , P(A∩B)= . Find i. P(B) ii. P(A∪ 퐵)

solution i. If A and B are two independent events then P(퐴 ∩ 퐵) = 푃(퐴) ∙ 푃(퐵)

⟹ ( ∩ )( )

= P(B) ∴, P(B) = × =

ii. P(A∪ 퐵) = 푃(퐴) + 푃(퐵) − 푃(퐴 ∩ 퐵) ∴, P(A∪ 퐵) = + − = =

Example 21.10 For two events A and B, let P(A)=0.6, P(A∪ 퐵) = 0.8 and P(B)=0.35. Check whether A and B are independent.

Solution If A and B are independent, then P(A∩ 퐵) = 푃(퐴) ∙ 푃(퐵) But P(A∪ 퐵) = 푃(퐴) + 푃(퐵) − 푃(퐴 ∩ 퐵) ⟹ 0.8=0.6+0.35−푃(퐴 ∩ 퐵) ⟹ 푃(퐴 ∩ 퐵) = 0.6 + 0.35 − 0.8 = 0.15 P(A)∙ 푃(퐵) = 0.6 × 0.35 = 0.21 A and B are not independent since 푃(퐴 ∩ 퐵) ≠ 푃(퐴) ∙ 푃(퐵)

Example 21.11

If P(A)= , P(B)= and P(A∪ 퐵) = , find i. P(A∩ 퐵) ii. P(A\B) iii. Show whether or not A and B are independent.

Solution We have, P(A)= , P(B)= and P(A∪ 퐵) =

i. P(A∩ 퐵) = 푃(퐴) + 푃(퐵) − 푃(퐴 ∪ 퐵) ∴, P(A∩ 퐵) = + − = =

ii. P(A\B)= ( ∩ )( )

∴, P(A\B)= = × =

iii. A and B are not independent events since P(A\B)≠ P(A

Addition law of probability 훼If A and B are mutually exclusive and exhaustive events, then the probability that either A or B occurs is equal to the sum of the separate probabilities of the two events. That is, P(A or B)=푃(퐴 ∪ 퐵) = 푃(퐴) + 푃(퐵).

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For three mutually exclusive and exhaustive events A, B and C, 푃(퐴or퐵or퐶) = 푃(퐴 ∪ 퐵 ∪ 퐶) = 푃(퐴) + 푃(퐵) + 푃(퐶)

훽.If A and B are two events which are not mutually exclusive, then the probability that either A or B will occur is given by

P(A or B)= 푃(퐴) + 푃(퐵) − 푃(퐴and퐵) OR P(A∪ 퐵) = 푃(퐴) + 푃(퐵) − 푃(퐴 ∩ 퐵)

For events A, B and C which are not mutually exclusive P(A∪ 퐵 ∪ 퐶) = 푃(퐴) + 푃(퐵) + 푃(퐶)− 푃(퐴 ∩ 퐵) − 푃(퐴 ∩ 퐶)− 푃(퐵 ∩ 퐶) + 푃(퐴 ∩ 퐵 ∩ 퐶)

Multiplication law of probability

훼.If A and B are mutually exclusive, independent and exhaustive events, then the probability that both A and B occurs is equal to the sum of the separate probabilities of the two events. That is, P(A 푎푛푑B)=푃(퐴 ∩ 퐵) = 푃(퐴) ∙ 푃(퐵). For three mutually exclusive and exhaustive events A, B and C,

푃(퐴and퐵and퐶) = 푃(퐴 ∩ 퐵 ∩ 퐶) = 푃(퐴) ∙ 푃(퐵) ∙ 푃(퐶) 훽.If A and B are two events which are not mutually exclusive, then the probability that both A and B will occur is given by

P(A and B)= 푃(퐴) + 푃(퐵) − 푃(퐴or퐵) OR P(A∪ 퐵) = 푃(퐴) + 푃(퐵) − 푃(퐴 ∪ 퐵)

For events A, B and C which are not mutually exclusive P(A∩ 퐵 ∩ 퐶) = 푃(퐴) + 푃(퐵) + 푃(퐶) − 푃(퐴 ∩ 퐵) − 푃(퐴 ∩ 퐶) − 푃(퐵 ∩ 퐶) + 푃(퐴 ∪ 퐵 ∪

퐶) 훾. If A and B are dependent events, then

P(A∩ 퐵) = 푃(퐴)푃(퐴\퐵) = 푃(퐵)푃(퐵\퐴)

Example 21.12 If A, B and C are three mutually exclusive and exhaustive events, find P(B) if P(C)= 푃(퐴) = 푃(퐵).

Solution Since the events are mutually exclusive we have P(A∪B∪C)=P(A)+P(B)+P(C) But P(C)= P(A), P(C)=P(B) and P(A)=P(B)

527

Since the events are exhaustive we have P(A∪B∪C)=1 ⟹ P(A)+P(B)+P(C)=1 ⟹ P(A)+ P(A)+ P(A)=1 ⟹ 2P(A)=1 ⟹ P(A)=

∴ P(B)= × =

Example 20. Two boys and one girl can be selected by taking one child from each group in the following mutually exclusive ways i. 1 girl from first group, 1 boy from second group and 1 boy from third group. ii. 1 boy from first group, 1 girl from second group and 1 boy from third group. iii. 1 boy from first group, 1 boy from second group and 1 girl from third group. Let 퐺 ,퐺 and 퐺 represent the events of selecting a girl from 1st, 2nd and 3rd group respectively.

Solution Let 퐵 ,퐵 and 퐵 represents the events of selecting a boy from 1st, 2nd and 3rd group respectively. P(퐺 ) = P(퐺 ) = = P(퐺 ) = P(퐵 ) = P(퐵 )= = P(퐵 ) = P(1 girl and 2 boys)=P(퐺 ∩ 퐵 ∩ 퐵 ) + 푃(퐵 ∩ 퐺 ∩ 퐵 ) +P(퐵 ∩ 퐵 ∩ 퐺 ) = ∙ ∙ + ∙ ∙ + ∙ ∙

= + +

=

Example 21.13 If A, B and C are mutually exclusive and exhaustive events and P(A)= 푃(퐵),

P(B)= P(C), find P(B) and P(C). Solution

Since the events are mutually exclusive we have P(A∪B∪C)=P(A)+P(B)+P(C) But P(A)= P(B) P(B)= P(C)

⟹P(B)=2P(A) and P(C)=3P(A) Hint:푃(퐴) = ∙ 푃(퐶) Since the events are exhaustive we have P(A∪B∪C)=1 ⟹ P(A)+P(B)+P(C)=1 ⟹ P(A)+2P(A)+3P(A)=1 ⟹ 6P(A)=1 ⟹ P(A)=

528

∴ P(B)=2 × = and P(C)=3 × =

Odd Against and Odd in Favour of an Event

The odd in favour of an event E is given by the ratio of the number of ways E can occur to the number of ways E cannot occur. The odd against an event E is given by the ratio of the number of ways E cannot occur to the number of ways E can occur. If the odd in favour of an event E is in the ratio 푎: 푏 then P(E)=

Example 21.14

The odd against student X solving a Business Statistics problem are 8:6 and the odd in favour of student Y solving the same problem is 14:16. a. What is the probability that the problem will be solved if they both try independent of each other? b. What the probability that neither of them solve the problem? c. What is the probability that one of them solved the problem?

Solution P(X)= = P(푋) = 1 − = P(Y)= = P(푌) = 1 − = a. P(at least one of them solved the problem) =P(X∩ 푌) + 푃 푋 ∩ 푌 + 푃(푋 ∩ 푌) = 1 − 푃(푋 ∩ 푌) = × + × + × = 1 − × =

b. P(none of them solved the problem)=푃 푋 ∩ 푌 = × =

c. P(one of them solved the problem)=P(X∩ 푌) + 푃 푋 ∩ 푌 = × + × =

Selection of Items Without Replacement Suppose there are 푛 items in a bag of which 푟 items are favourable to an event E. Then the chance of selecting a first item from the bag that favours E is . If the first item selected is not replaced, then there are 푟 − 1 items left that are favourabe to events E and the number of items in the bag is reduced to 푛 − 1. The chance of selecting a second item which favours E is . So far as the item selected is not replaced, the number of

529

items favourable to event E is now 푟 − 2 and the number of items in the bag is 푛 − 2. this can be done till all the 푟 items are selected.

Example 21.15 . A bag contains 4 green and 6 red balls. A ball is drawn at random and without replacing it a second ball drawn. What is the probability of drawing a. two green balls? b. no green ball ?

Solution Green balls Red balls Total balls 4 + 6 = 10 a. P(two green balls)=P(1st green ball and 2nd green ball)= ∙ = or

P(two green balls)=

=4C2

10C2= 2

15

b. P(no green ball)=P(two red balls)= ∙ = or

P(no green ball)=

=6C2

10C2

Example 21.16

A bag contains 8 red and 5 white balls. Two successive draws of three balls are made without replacement. Find the probability that a. the first drawing gives 3 white balls and the second drawing 3 red balls. b. the first and the second drawing gives balls of the same colour.

Solution Red balls white balls Total balls 8 + 5 = 11 a. P(3 white and 3 red balls)= × × × × × = . Or P(3 white and 3 red balls) =

and

=5C3

13C3×

8C3

10C3=

b. P(same colour) =P(1st drawing are white and second red) or P(1st drawing are red and second white) = × × × × × + × × × × ×

530

= Or

P(same colour)=5C3

13C3×

8C3

10C3+

8C3

13C3×

5C3

10C3=

Example 21.17

A man draws at random 3 balls from a bag containing 6 red and 5 white balls. What is the chance that a. all the balls are red b. two are red and 1 is white.

Solution Red ball White balls Total balls 6 + 5 = 11 a. P(all red)=

=

6C3

11C3= Or

P(all red)=P(1st red and 2nd red and 3rd red)= × × = (withoutreplacement)

b. P(two red and one black)

=

=6C2×5C1

11C3= Or

P(two red and one black)=P(1st red and 2nd red and 3rd white)+P(1st white and 2nd red and 3rd red)+P(1st red and 2nd white and 3rd red)

= × × + × × + × ×

= (withoutreplacement)

Example 21.18 What is the probability of getting 3 white balls in a draw of 3 balls from a box containing 5 white and 7 black balls?

Solution White balls Black balls Total balls 5 + 7 = 12

P(all white)=

=5C3

12C3= or

531

P(all white)=P(1st white and 2nd white and 3rd white)= × × = (withoutreplacement)

Final Exercises

1. Two fair dice are thrown once. What is the probability of getting a sum of 7 points or 11 points on the dice? 2. A bag contains 4 red and 3 blue balls. Two drawings of 2 balls each are made. Find the probability that the first drawing gives 2 red balls and the second drawings 2 blue balls if a. the balls are returned to the bag after the first draw. B. the balls are not returned. 3. If A, B and C are mutually exclusive and exhaustive events, and P(A)= 푃(퐵) and

P(B)= P(C). Find P(A), P(B) and P(C). 4. An urn contains 8 black and 4 red balls. Three balls are drawn at random from the box one after the other without replacement. Find the probability that 5. The first two are red and the second is black. b. the first and the third are black and the second is red. 6. The first and the third are of the same colour and the second is of the opposite colour. 7. A bag contains 7 red balls, 4 white balls and 9 black balls. Three balls are drawn at random. Find the probability that a. all the three balls are red b. two are red and one is black. c. one red, one white and one black. 8. A problem in Economics is given to three students A, B and C whose probability of solving it are , and respectively. What is the probability that at least one of them solved the problem if they try independently? 9. Three groups of children contain respectively 3 girls and 1 boy, 2 girls and 2 girls, 1 girl and 3 boys. One child is selected at random from each group. Find the probability that the three selected consist of 1 girl and 2 boys 10. For two events A and B, let P(A)=0.5 P(A∪ 퐵) = 0.58 and P(B)=0.3. Check whether or not A and B are independent.

532

THE BINOMIAL PROBABILITY DISTRIBUTION A experiment that consist of repeated trials with two possible outcomes at each trial is classified as a binomial experiment. Each trial is labelled as a success or failure. A binomial experiment has the following characteristics The experiment consists of 푛 identical trials. Each trial results in one of two outcomes. One outcome is labelled as 푠푢푐푐푒푠푠 and

the other a 푓푎푖푙푢푟푒. The probability of success on a single trial is labelled 푝 and remains the same

from trial to trial. The probability of failure is labelled (1− 푝) = 푞. The trials are independent. We are interested in 푟 sucesses observed during the 푛 trials, for 0 ≤ 푟 ≤ 푛,wher

푟 is an integer. The binomial distribution is given by:

푃(푋 = 푟) = nCr푝푟푞푛−푟

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Example 22.1 A fair coin is tossed 8 times. Find the probability of getting i. 5 heads ii. more than 2 heads. iii. At least 3 head.

Solution

The binomial probability distribution is defined as: 푃(푋 = 푟) = rnrr

n qpC If we toss a coin, the possible outcome can be a head or a tail. Therefore, the number of possible outcome is 2. i. We are interested in the number of heads. So, 푃(푎ℎ푒푎푑) = 푝 = 푞 = 1 − 푝 = 1 − = 푛 = 8 푟 = 5

푃(푋 = 5) =585

58

21

21

C =

355

821

21

C =

∴ 푃(5heads) = (Use your calculator) ii. We are interested in the number of heads. So, 푃(푎ℎ푒푎푑) = 푝 = 푞 = 1 − 푝 = 1 − = 푛 = 8 푟 > 2 푃(푋 > 2) = 푃(푋 = 3) + 푃(푋 = 4) + 푃(푋 = 5) + 푃(푋 = 6) + 푃(푋 = 7) + 푃(푋 = 8)

푃(푋 > 2) =53

38

21

21

C +

444

821

21

C +

355

821

21

C +

26

68

21

21

C +

177

821

21

C +

088

821

21

C

= + + + + +

=

∴ 푃(morethan2heads) =

Or

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Note: P(morethan2heads) + P(notmorethantwoheads) = 1 ⟹ P(morethan2heads) = 1 − P(notmorethantwoheads) ⟹ P(X > 2) = 1− P(X ≤ 2) ⟹ 푃(푋 > 2) = 1 − [푃(푋 = 0) + 푃(푋 = 1) + 푃(푋 = 2)]

= 1 −80

08

21

21

C +

711

821

21

C +

622

821

21

C

= 1 − + + (Use your calculator) = 1 −

= iii. We are interested in the number of heads. So, 푃(푎ℎ푒푎푑) = 푝 = 푞 = 1 − 푝 = 1 − = 푛 = 8 푟 ≤ 6 푃(푋 ≤ 6) = 푃(푋 = 0) + 푃(푋 = 1) + 푃(푋 = 2) + 푃(푋 = 3) + 푃(푋 = 4) + 푃(푋 = 5) + 푃(푋 = 6)

푃(푋 ≤ 6) =80

08

21

21

C +

711

821

21

C +

622

821

21

C +

53

38

21

21

C +

444

821

21

C +

355

821

21

C +

26

68

21

21

C

= + + + + + +

=

∴ 푃(atmost6heads) =

Or Note: 푃(atmost6heads) + 푃(morethamsixheads) = 1 ⟹ 푃(atmost6heads) = 1 − 푃(morethamsixheads) ⟹ 푃(푋 ≤ 6) = 1 − 푃(푋 > 6)

535

⟹ 푃(푋 > 2) = 1 − [푃(푋 = 7) + 푃(푋 = 8)]

= 1 −17

78

21

21

C +

088

821

21

C

= 1 − + (Use your calculator) = 1 −

= Example 22.2

In a Mathematics test, 80% of the student passed. If 10 students are selected at random, calculate the probability that i. exactly 5 passed. ii. at most 2 failed.

Solution

The binomial probability distribution is defined as: 푃(푋 = 푟) = rnrr

n qpC i. We are interested in the number of students who passed. So, 푃(studentswhopassed) = 푝 = = 0.8 푞 = 1 − 푝 = 1 − 0.2 = 0.8 푛 = 10 푟 = 5

푃(푋 = 5) = 51055

10 2.08.0 C = 555

10 2.08.0C ≈ 0.0264 ∴ 푃(푒푥푎푐푡푙푦5푠푡푢푑푒푛푡푠푝푎푠푠푒푑) = 0.0264 (Use your calculator) ii. We are interested in the number of students who failed. So, 푃(푠푡푢푑푒푛푡푠푤ℎ표푓푎푖푙푒푑) = 푝 = = 0.2 푞 = 1 − 푝 = 1 − 0.2 = 0.8 푛 = 10 푟 ≤ 2 푃(푋 ≤ 2) = 푃(푋 = 0) + 푃(푋 = 1) + 푃(푋 = 2)

푃(푋 ≤ 2) = 1000

10 8.02.0C + 911

10 8.02.0C + 822

10 8.02.0C = 0.1074 + 0.2684 + 0.3020 = 0.6778 ∴ 푃(atmost2studentsfailed) = 0.6778

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Try: 1. In an examination, 90% of the candidates passed. If 8 of the candidates are selected at random, calculate the probability that i. exactly 5 passed. ii. at most 3 failed.

(WASSCE) 2. For every ten persons in a city one is left handed. If six persons are selected at random from the city, find the probability that i. exactly 3. ii. more than half. iii. at least two are left-handed

(WASSCE) 3. A survey carried out among a group of secondary school students showed that 2 out of every 10 students buy textbooks. If 15 students are randomly selected from the group, find, correct to three decimal places the probability that i. exactly 3. ii. less than 4. iii. at least 2; students buy textbooks.

(WASSCE) 4. The probability that a pupil will be canned daily in school is . If a school week consists of 5 days, find, correct to three decimal places, the probability that the pupil will i. be caned in 2 days only; ii. not be caned in a week; iii. by all means be caned during the week.

(WASSCE) 5. In an examination 4% of the candidates passed with distinction. If 7 of the candidates are selected at random, what is the probability that i. 4 of them obtained distinction; ii. at most three of them obtained distinction?

(WASSCE) 6. A fair die is thrown 5 times. Calculate, correct to three decimal places, the probability of obtaining i. at least 3 sixes ii. exactly 2 sixes.

(WASSCE) 7. Six fair coin of the same size are tossed once. Find, correct to three decimal places, the probability that i. a head is obtained ii. at least four heads are obtained.

(WASSCE)

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Final Exercises 1. A fair coin is tossed 6 times. Find the probability of getting i. 4 heads ii. At least 3 tails. 2. A fair die is thrown 5 times. Find the probability of getting i. exactly 2 fives. ii. at least 4 sixes 3. In an examination 20% of the students failed. If 6 students are selected at random, calculate the probability that i. exactly 3 students passed. ii. exactly 4 students passed. iii. at most 2 students failed. 4. A report shows that 3 out of every 10 students in a school are very punctual. If 15 students are selected from a group, find, correct to two places of decimal, the probability that i. Exactly 5 students are punctual. ii. less tan 6 students are punctual. iii. at most 13 students are not punctual.

MULTIPLE CHOICE QUESTIONS 1. State the domain of the function ℎ(푥) =

A. {푥:푥휖ℝ,푥 ≠ 4} B. {푥: 푥휖ℝ, 푥 ≠ −4} C. 푥: 푥휖ℝ,푥 ≠ 2

3 D. {푥: 푥휖ℝ} 2. A binary operation ∗ is defined on the set real numbers by 푎 ∗ 푏 = 푎 . Find the value of 3 ∗ −2.

A. −9 B. − C. D. −6

3. An exponential sequence is given by , , 27,⋯. Find the common ratio.

A. 퐵. 퐶. 퐷. 6

4. Find the truth set of log − 푥 = 0

A. {푥: 푥 = 4} B. 푥: 푥 = C. 푥: 푥 = D. 푥: 푥 = − 5. Find the coefficient of 푥 in the binomial expansion of (3푥 − 2)

A. -15 B. 15 C. 8640 D. 2160 6. Find the inverse ℎ of the function ℎ: 푥 =

A. ℎ : 푥 = 1 B. ℎ :푥 = 푥 − 1 C. ℎ : 푥 = ,푥 ≠ 1 D. ℎ : 푥 = ,푥 ≠ 1 7. If 푥푦 = 9 and 푥 = 4, find 푥(푦 + 1) A. 4 B. 13 C. 20 D. 31

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8. Solve for 푛 if 2 = 32 A. 4 B. 4 C. 퐷.

9. If sin 푥 = , find tan 푥 A. B. 푞 − 1 C. D.

푈 = {푥: 0 ≤ 푥 ≤ 6}, 퐴 = {0, 2, 4, 6} 퐵 = {1, 2, 3, 4} and 퐶{1, 3},

퐴,퐵 and 퐶 are subsets of 푈. Use the information above to answer questions 10 and 11. 10. Find 퐴 ∪ (퐵 ∪ 퐶) A. {6} B .{0, 6} C. {1, 3} D.{} 11. Find 퐴 ∩ (퐵 ∪ 퐶)′ A. {6} B. {} C. {2, 4} D. {0, 6} 12. A function 푓 is said to be odd if for all values of 푥 it satisfies the domain A. 푓(푥) = 푓(−푥) B. 푓(−푥) = 푓 (푥) C. 푓(−푥) = −푓(푥) D. 푓(−푥) = 푓 (−푥)

13. If 푑 = , express 푃 in terms of 푑,퐾and 푄.

A. B. C. D. 푑 푄 + 퐾

14. Simplify, + A. B. C. D.

15. Simplify, A. B. C. D.

16. Simplify, √√

A. 1 + √6 B. 1 + √6 C. 6 + √6 D. 6 + 2√6

17. Given that 2 log 푥 = 3, find the relationship between 푥 and 푦. A. 푦 = 푥 B. 푦 = 푥 C. 푦 = 푥 D. 푦 = 푥 18. The equation of a curve is given by 3푥 +6푥 + 3푦 − 9푦 − 3 = 0. Find the coordinates of the centre.

A. −2, B. −1,− C. 1,− D. (2,−3)

19. If 훼 and 훽 are the roots of the equation 4푥 − 9푥 − 16 = 0,which of the following is/are true? I. 훼 + 훽 = II. 훼훽 = −4 III. 훼 + 훽 = −

A. I only B. II only C. I and II only D. II and III only 20. Given that 푓(푥) = 푥 − 3푥 − 7푥 + 11, find 푓(−1) A. 2 B. 8 C. 14 D. 16

21. If √√

is expressed in the form 푎 + 푏√3, where 푎 and 푏 are real numbers, find the

value of 푏. A. 10 B. 7 C. -7 D. -10

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22. The second and fourth term of an arithmetic sequence are 9 and 17 respectively. Find the common difference.

A. 2 B. 4 C. 5 D. 8 23. Given that 퐴 = {2, 4, 6, 8}. Find the number of subsets of 퐴. A. 15. B. 61 C.16 D. 14 24. Which of the following binary operations is not commutative.

A. 푎⨂푏 = 푎 + 푏 − 푎푏 B. 푎⨂푏 = 2푎 + 2푏 − 푎푏 C. 푎⨂푏 = + D. 푎⨂푏 = 푎 − 푏 + 푎푏

25. Evaluate lim → A. 0 B. 2 C. 3 D. 4

26. If ≡ + , find the value of 퐴 + 퐵. A. −3 B. 2 C. 3 D. 4 27. Given that 푓(푥) = 1 + 2푥 and 푔(푥) = 푥 + 1, find 푓표푔. A. 3− 2푥 B. 3 + 2푥 C. 3 + 4푥 D. 3푥 + 2 28. If 푓(푥) = −2푥 + 푞푥 + 2, find 푞 if 푓(−1) = 2. A. -1 B. -2 C. 2 D. 6

29. Simplify, 2 + 3√5 − 2 − 3√5 A. 24√5 B. 90 C. 0 D. 4 + 18√5 30. The curve 푓(푥) = 푎푥 + 푥 + 푥 + 1 has a point of inflexion at (−1, 8). Find 푎.

A. -2 B. -1 C. 0 D. 4 A function is defined by 푦 = 3푥 + 5푥 + 6. Use this information to answer questions 31 and 32 31. Find the minimum value of 푦 A. B. − C. D. −

32. Find the axis of symmetry. A. B. C. − D. − 33. Which of the following is identical to 푥 + 2푥 + 1

A. (1 + 2푥) B. (1 − 푥) C. (푥 − 1) D.(푥 + 1) 34. Find the equivalent acute angle of 푐표푠170°

A. 푠푖푛10° B. cos(−10°) C. – 푠푖푛10° D.푐표푠10°

35. If 푡푎푛푥 + 푠푖푛푦 = 1, find A. B. − C. − D.

36. If a quadratic equation has two real and different roots then A. 푏 − 4푎푐 = 0 B. 푏 − 4푎푐 > 0 C. 푏 − 4푎푐 < 0 D. 푏 − 4푎푐 ≥ 0 37. Find the equation of a circle with ends of diameter 퐴(3, 4) and 퐵(4, 5). A. 푥 +푦 + 7푥 − 9푦 + 32 = 0 B. 푥 +푦 − 7푥 + 9푦 + 32 = 0 C. 푥 +푦 − 7푥 − 9푦 + 32 = 0 D. 푥 +푦 − 7푥 − 9푦 − 32 = 0 38. Convert 80° to radians. A. 휋rad B. 휋rad C. 휋rad D. 휋rad

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39. If sin휃 = − , 180° ≤ 휃 ≤ 270°, find the value of cos휃. A. B. − C. D. − 40. The sum of the three consecutive terms of an 퐴푃 is 24. Find the first term.

A. 12 B. 21 C. 4 D. 3 41. If 푥 + 푦 = and −푦 = , find the value of 2푦 + 푥. A. -2 B. -1 C. D. 1 42. Simplify (푃 ∩ 푄 ) ∪ (푃 ∩ 푄),where 푃 and 푄 are subsets of the universal set 푈.

A. ∅ B. 푄 C. 푃 D. 푈 The polynomial 푝(푥) = 푥 + 푎 푥 + 푎 푥 + 푎 has zeros 푟 , 푟 and푟 .

Use this information to answer questions 43 and 44 43. Find the relationship between the sum of the zeros and the coefficients of 푝(푥).

A. 푟 + 푟 + 푟 = 푎 B. 푟 + 푟 + 푟 = −푎 C. 푟 + 푟 + 푟 = −푎 D. 푟 + 푟 + 푟 = 푎 44. Which of the following relationship is correct? A. 푟 + 푟 푟 + 푟 푟 = 푎 B. 푟 + 푟 푟 + 푟 푟 = 푎 C. 푟 푟 + 푟 푟 + 푟 푟 = 푎 D. 푟 푟 푟 = 푎

45. Given that 푃(푥) = (푥 − 푎) , the following are all true except that A. 푥 = 푎 has multiplicity 1 B. (푥 − 푎) is a factor of 푃(푥) C. 푃(푥) is of degree 3 D. (푥 − 푎) is a factor of 푃(푥) 46. Find two positive numbers whose product is 100 and whose is minimum.

A. 20, 5 B. 50, 2 C. 25, 4 D. 10, 10 47. A spherical balloon is being inflated. Find the rate of increase of the surface area (푆 = 4휋푥 )With respect to the radius 푟 when 푟 is 1 foot. A. 8휋 B. 16휋 C. 32휋 D. 64휋 48. Find an equation of the tangent line to the curve 푦 = 2푥푒 at the point (0, 0)

49. Find the coefficient of the term in 푥 in 푥 − . A. −220 B. 495 C. −1760 D. 1980

50. What is the total number of terms in the expansion of [(푎 + 푏) ] A. 52 B. 52 C. 50 D. 20

51. Find 푥 if cos(푥 − 30°) = 0.5. A. 60° B. 240° C. 300° D. 330°

52. Find the determinant of 푨 =8 4 −45 −5 −57 1 −11

. A. 240 B. 400 C. 480 D. 560

53. Find the exact value of (1.02) . A. 0.02 B. 0.014 C. 1.1041 D. 0.0014. 54. If 푴and 푵 are non-singular matrices, then for 푴푵

A. |푴푵| ≠ 0 B. its inverse does not exist C. its transpose does not exist. D. |푴푵| = 0 55. Evaluate the integral ∫ 푠푒푐푥푡푎푛푥 푑푥.

541

A. tan푥sec푥 + 푐 B. cosec푥 + 푐 C. se푐푥 + 푐 D. – sec푥 + 푐 56. Find if 푦 = sin(푒 ). A. cos(푒 ) B. – cos(푒 ) C. −푒 cos(푒 ) D. 푒 cos(푒 )

A binomial expression is given as (푥 + 2) . Use it to answer questions 57 and 58

57. Find the 5 term of the expansion of the expression in descending powers of 푥. A. 5050푥 B. 360푥 C. 3360푥 D. 13440푥

58. Find the 5 term of the expansion of the expression in ascending powers of 푥. A. 3360푥 B. 13440푥 C. 360푥 D. 130푥

59. Find the truth set of log (2푥 + 5푥 + 97) = 2. A. − ,−3 B. , 3 C. ,−1 D. ,−3

60. For what value of 푘 does the equation 4푥 − 8푥 + 푘 = 0 has equal roots? A. 푘 = 0 B. 푘 = 2 C. 푘 = 4 D. 푘 = 8

61. Solve log √2 = 푥. A. 푥 = − B. 푥 = − C. 푥 = 1 D. 푥 = 2

62. If 휃 is obtuse, and 휃 = , find 푐표푠휃. A. − B. − C. D. 63. Find the values of 푘 if the distance between 퐴(푘,−2) and 퐵(2, 1) is 5. A. -2, 6 B. 2,−6 C. −2,−6 D. −2, 6 64. What value of ∝ makes the slope of the line through 푃(−4,∝) and 푄(−1, 18)

A. -6 B. 0 C. 6 D. 10 65. Find an equation for the perpendicular bisector of the line segment from 퐴(1, 0) and 퐵(3, 4).

A. −2푦 + 푥 − 6 = 0 B. 2푦 + 푥 − 6 = 0 C. 2푦 − 푥 + 6 = 0 D. 2푦 + 푥 + 6 = 0 66. Find the real number 푘 such that the point 푃(−1, 2) is on the line 푘푥 + 2푦 − 7 = 0.

A. 3 B. −3 C. 4 D. −4 67. For what value of 푟 will the line 5푥 + 푟푦 − 3 = 0 have a 푦-intercept of −5?

A. − B. 0 C. − D. 4 68. Find the centre of the circle with equation 2푥 +2푦 − 푥 + 푦 − 3 = 0.

A. (1,−1) B. (−1, 1) C. ( ,− ) D. (− , )

69. If the distance of 푃(3,∝) from the line 12푥 + 5푦 − 11 = 0 is , find the value of ∝. A. 1 B. 2 C. 3 D. 4

70. Find an equation of the tangent to the circle with centre (1,−2) at (3, 2). A. 푥 − 2푦 − 7 = 0 B. – 푥 + 2푦 − 7 = 0 C. 푥 + 2푦 − 7 = 0 D. 푥 + 2푦 + 7 = 0

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71. The length of the tangent from 퐴(7, 2) to the circle with centre (3, 4) is 4. Find the radius of the circle.

A. 1 B. 2 C. 3 D. 4 72. Solve 푥 + 2푥 − 35 ≤ 0. A. −7 ≤ 푥 ≤ 5 B. −5 ≤ 푥 ≤ 7 C. 푥 ≤ −5 or 푥 ≥ 7 D. 푥 ≤ −7or 푥 ≥ 5 73. If (푥 − 1) is a factor of the polynomial 푓(푥) = 푥 − 3푥 + 푝푥 − 6푥 + 1, find the value of 푝.

A. 5 B. 7 C. 9 D. 10 74. The 푦-intercept of a line is . If the line is parallel to the line 3푦+ 4푥 − 9 = 0, find its equation.

A. 20푥 + 15푦 + 12 = 0 B. 20푥 + 15푦 − 12 = 0 C. 20푥 + 15푦 + 6 = 0 D. 20푥 + 15푦 + 12 = 0 75. The equation of a circle is 푥 +푦 + 4푥 − 6푦 + 4 = 0. Find the coordinate of the centre.

A. (2, 3) B. (2,−3) C. (−2, 3) D. (−2,−3) 76. Given that sin휃 = 1

2 , where 90° < 휃 < 180°, find 푐표푠휃.

A. − √ B. − 12 C. 1 2 D. √

77. Find the coefficient of 푥 in the binomial expansion of (1 + 2푥) . A. 60 B. 100 C. 160 D. 180

78. A binary operation ∗ is defined on the set 푍 of integers by 푥 ∗ 푦 = 2푥 − 푥푦 + 푦 . Evaluate 5∗ (−2)

A. 2 B. 4 C. 11 D. 24 79. Simplify

√. A. 4(√3− 1) B. 4(√3 + 1) C. 8(√3− 1) D. 8 √3 + 1

80. If 훼 and 훽 are the roots of the equation 2푥 + 5푥 + 8 = 0. Find the value of (훼 + 훽 ). A. − B. − C. D.

81. Given that 2 = 0.5, find the value of 푥. A. −2 B. −1 C. 1 D. 2 82. The gradient at the point 푃 on the curve 푦 = 3푥 − 2푥 + 4 is 4. Find the coordinates of 푃.

A. (−5,−1) B. (−1, 5) C. (1, 5) D. (5, 1) 83. An arc subtends an angle of 0.74 radians at the centre of a circle. If the diameter of the circle is 24푐푚, calculate, correct to the nearest whole number, the area of the sector.

543

84. The 1 , 3 and 6 terms of a linear sequence are the first 3 terms of an exponential sequence (G.P). Find the common ratio of the (G. P).

A. B. C. 2 D. 3 85. Let 푝 and 푞 be the statements; 푝:퐻푒푖푠푟푖푐ℎ. 푞:퐻푒푖푠푔푒푛푒푟표푢푠 Which of the following symbols represents ‘He is poor but generous’ ?

A. 푝⋀~푞 B. 푝⋀푞 C. ~푝⋀~푞 D. ~푝⋀푞 86. Find the values of 푘 for which the equation 푘푥 + (푘 − 2)푥 + 푘 = 0 has equal roots.

A. −2 and B. −1 and C. 1 and − D. 2 and −

87. If = 3푥 − , find A. 6푥 + 4 B. 6푥 − 4 C. 6푥 − D. 6푥 +

88. Simplify 8 A. 2 B. 3 C. 4 D. 6 89. Given that log (2푥 + 3)− log (2푥 − 3) = 1, find an expression for 푦 in terms of 푥

A. B. 4푥 − 9 C. 4푥 − 3 D. 4푥 − 9 90. Evaluate 8C6−5C3. A. 4 B. 6 C. 18 D. 33. 91. 퐴 (푥 + 2)cm 푦cm 45° 퐵 퐶 In the diagram 퐴퐵퐶 is a right-angled triangle in which |퐴퐵| = 푦 cm, |퐴퐶| = (푥 + 2) cm and angle 퐴퐶퐵 = 45°. Find 푦 in terms of 푥.

A. 푦 = 2(푥 + 2) B. 푦 = √ (푥 + 2) C. √ ( )

D. √( )

92. If the distance between the points 푃(−2, 4) and 푄(2,푘) is √58, find the possible value of the constant 푘.

A. −3 or 11 B. −3 or −11 C. 3 or 11 D. 3 or −11 93. Evaluate ∫ 1 − 푑푥. A. 0 B. 1 2 C. 1 D. 2

94. Evaluate 3 4−1 2

−11 . A. (1, 3) B. (1, 1) C. D.

95. Simplify log 27 + log 64 − log 49 . A. 7 B. 4 C. 3 D. 2

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96. Which of the following graph is used to determine the mode of a grouped data? A. Histogram B. Bar chart C. Pie chart D. Ogive Use the information to answer Questions 97 and 98. An object is thrown into the air. Its height ℎ meters, after 푡 seconds is given by ℎ = 36푡 − 4푡 97. Find the velocity after at time 푡 = 0. A. 8푚푠 B. 12푚푠 C. 18푚푠 D. 36푚푠 98. Find the time taken to reach the maximum height.

A. 3.5푠 B. 4.0푠 C. 4.5푠 D. 6.0푠 99. There are 6 nurses and 5 doctors in a clinic. 4 nurses and 2 doctors are to be selected at random. How many different selections are possible if a particular doctor must not be included?

A. 60 B. 90 C. 120 D. 150 100. In triangle 퐴퐵퐶,퐴퐵⃗ = and 퐴퐶⃗ = . Find 퐵퐶⃗

A. 4√2 B. 6√2 C. 2√10 D. 4√10 101. Two unbiased dice with faces numbered 1 to 6 are thrown once. What is the probability of obtaining a total score of at most 8? A. B. C. D. 102. Find the angle between (5퐢 + ퟑ퐣)and (3퐢 − 5퐣). A. 0° B. 45° C. 60° D. 90°

103. Given that 푴 = 2 −7−5 9 and 푷 = −4 3

7 1 , find the matrix 푵such that

푴 + 2푵 = 푷.

A. −1 −21 4 B. −1 −2

6 4 C. −3 56 −4 D. −3 5

−6 −4

104. The deviations from the mean of a set of numbers are (푘 + 3) , (푘 + 7),−2, (푘 + 2) , where 푘 is a constant. Find the value of 푘.

A. −3 B. −2 C. 2 D. 3 105. A man has one each of the following currency notes in his bag: ₦5,₦10,₦20,₦50,₦100, ₦200,₦500 and ₦1,000. How many different sums

of money can be taken from the bag, if only 4 currency are taken for each sum? A. 28 B. 35 C. 70 D. 210.

106. A box contains 25 identical balls numbered 1, 2, 3,…, 25. If a ball is selected from the box what is the probability it has prime numbers written on them?

A. 0.32 B. 0.36 C. 0.48 D. 0.52

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107. A force of magnitude 24 N acts on a body of mass 6kg which is at rest. How far will the body move in 4 seconds?

A. 16m B. 24m C. 32m D. 40m 108. A force of magnitude10N acts in the direction of the 3퐢 + 4퐣. Write the force in the form푎퐢 + 푏퐣, where 푎 and 푏 are real numbers.

A. 6퐢 + 8퐣 B. 3푖 + 4퐣 C. 퐢 + 4퐣 D. 퐢 + 퐣 109. In a class of 50 students, 30 like English and 35 like Mathematics. What is the probability of selecting who likes both English and Mathematics?

A. B. C. D.

110. What is the variance of 1, 2, 0, −3, 5, −2, 4. A. B. C. D. 111. A body is acted upon by three forces (5N, 030°), (2N, 090°), (3N, 180°). Calculate, correct to 2 significant figures, the magnitude of the resultant force.

A. 4.40N B. 4.69N C. 4.79N D. 5.79N 112. Convert 225° to radians. A. 휋 B. 휋 C. 휋 D. 휋 113. In a class of 25 students, 6 study French (퐹), 14 study Physics (푃) and 3 study both French and Physics. Find n(퐹 ∩ 푃 ).

A. 2 B. 3 C. 5 D. 8 114. If log 푥 = − , find the value of 푥. A. B. C. 3 D. 9

115. If (푥) = 푥 − , 푥 ≠ 0. Find its derivative with respect to 푥.

A. 2푥 − B. 2푥 − C. 2푥 + D. 2푥 + 116. Find the remainder when 푥 − 5푥 + 5푥 − 6 is divided by (푥 + 1).

A. 0 B. 1 C. 3 D. 4 117. Find the minimum value of 2푥 − 8푥 + 3. A. −7 B. −5 C. −2 D. 3 118. Simplify: 3√5− 7 2√5 + 3 . A. −9 − 5√5 B. −4√5 C. 4√5 D. 9 − 5√5 119. Which of the following lines passes through the points (4, 1) and (−2, 3)?

A. 푥 − 3푦 − 7 = 0 B. 푥 − 3푦 + 7 = 0 C. 푥 − 2푦 + 7 = 0 D. 푥 + 3푦 − 7 = 0 120. Find the equation of the normal to the curve 푥 +푦 = 13 at point (3, 2).

A. 푦 = − 푥 B. 푦 = 푥 C. 푦 = − 푥 + 4 D. 푦 = 푥 + 4

121. Evaluate: ∫ 푥 (4푥 + 3)푑푥. A. 106 B. 110 C. 2182 D. 2192

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122. If ℎ(g(푥)) = 푥 + 2푥 + 1 and ℎ(푥) = 푥 , find g(푥). A. 1 − 푥 B. 1 + 푥 C. 2푥 + 1 D. 푥

123. Given that sin = , 0° ≤ 푥 ≤ 90°, find, correct to three decimal places,

. A. 0.089 B. 0.097 C. 0.019 D. 0.020

124. Given that ( ) = 27( ), express 푦 in terms of 푥.

A. 푦 = B. 푦 = C. D.

125. Find the co-efficient of 푥 in the expansion 1 + . A. B. C. D.

126. Find the domain of g: 푥 → .

A. {푥:푥 ∈ 푅,푥 ≠ −1} B. 푥: 푥 ∈ 푅, 푥 ≠ − C. 푥: 푥 ∈ 푅, 푥 ≠ D. {푥: 푥 ∈ 푅, 푥 ≠ 1}

127. Given that 퐏 = 3 27 5 and 퐐 = 7 2

9 3 , find 2퐐퐏.

A. 70 2590 30 B. 108 88

126 72 C. 105 72144 99 D. 135 24

148 33

128. Solve: 3 − 2푥 − 푥 < 0. A. −3 < 푥 < 1 B. 푥 ≤ 1,푥 > 3 C. −1 < 푥 < 3 D. 푥 < −3,푥 > 1

129. If cos2푃 +sin푄 + 1 = 0 and 푃 = 45°, find the maximum value of 0° ≤ 푄 ≤ 360°. A. 210° B. 270° C. 300° D. 330°

130. The sum of the first five terms of a linear sequence (퐴.푃) is 35. If the first term is 3, find the Common difference.

A. 1 B. 2 C. 3 D. 4 131. Find the radius of circle 3푥 + 3푦 − 12푥 + 18푦 − 27 = 0.

A. 2 B. 4 C. √22 D. 22 132. The probabilities of 푃,푄 and 푅 being selected as school prefects are 0.2, 0.3,and 0.4 respectively. What is the probability that two of them will be selected?

A. 0.19 B. 0.26 C. 0.50 D. 0.70 132. Given that 퐏 = (5퐢 + 12퐣), find the unit vector in the direction of 퐏.

A. (5퐢 − 12퐣) B. (5퐢 − 12퐣) C. (5퐢 + 12퐣) D. (5퐢 + 12퐣) 133. If nC2= 15,find the value of 푛. A. 5 B. 6 C. 13 D. 30 134. A man weighing 60 kg stands on a lift which moves upward with an acceleration of 8ms-2. Find the reaction of the floor. [Takeg = 10ms ].

A. 120N B. 600 N C. 1000N D. 1080 N

547

135. How many three digit numbers can be formed using the digits 1, 2, 3, 4, 5, if repetition is not allowed?

A. 60 B. 80 C. 115 D. 125 136. If the resultant, 퐑, of vectors (퐢+ 2푦퐣), (1 − 푥)퐢+ (3푦+ 푥)퐣, (2푥 − 푦)퐢 + (푥 + 1)퐣 is zero Find the values of 푥 and 푦. A. 푥 = − ,푦 = − B. 푥 = − ,푦 = C. 푥 = −1,푦 = − D. 푥 = 1,푦 = −

137. Express 퐅 = (20N, 120°) in the form . A. √ B. √ C. √ D. √

138. Solve: (4푥 − 7)(푥 + 3) ≤ 0. A. −3 ≤ 푥 ≤ B. − ≤ 푥 ≤ 3 C. ≤ 푥 ≤ 3 D. −3 < 푥 ≤

In a class of 40 students, 27 passed in Literature test, 26 passed in Music test and each student passed at least one of the test. Use this information to answer Questions 139 and 140. 139. Find the probability that a student selected at random passed both tests.

A. B. C. D. 140. What is the probability that a student selected at random passed the Music test given that he/she passed Literature test?

A. B. C. D. 141. Forces 퐅 (3퐢+ 7퐣)N,퐅 (5퐢+ 3퐣)Nand퐅 (푛퐢+ 푚퐣)N keep an object in equilibrium. Find 푛 and 푚

A. 푛 = −10,푚 = 12 B. 푛 = −2,푚 = 10 C. 푛 = 2,푚 = −10 D. 푛 = 102,푚 = 90 142. If log 3 + log 27 = 2, find the value of 푥. A. 2 B. 8 C. 9 D. 10 143. Given that 퐏퐐⃗ = and 퐏퐑⃗ = , find 퐐퐑.⃗ A. B. C. D. The table below shows the scores of students in an examination.

scores 1− 5 6 − 10 11 − 15 16− 20 21− 25 26 − 30 31− 35 Frequency 5 13 8 10 5 3 1

Use this table to answer Questions 144 to 146. 144. What is the median class interval?

A. 6 − 10 B. 11 − 15 C. 16 − 20 D. 21 − 25 145. What is the lower class boundary of the modal class?

548

A. 5.5 B. 6.5 C. 16.5 D 30.5 146. Calculate, correct to the nearest whole number, the mean of the distribution.

A. 10.0 B. 14.0 C. 24.0 D. 34.0 147. If 퐫 = (20N, 030°) and 퐧 = (5N, 210°), find (퐫+ 퐧)

A. (15N, 030°) B. (15N, 240°) C. (25N, 030°) D. (25N, 240°) 148. An object of mass 12 kg was thrown with a velocity of (4퐢+ 3퐣)ms-1. If the final velocity of the object was (−5퐢+ 2퐣)ms-1, find correct to two decimal places, the magnitude of the change in momentum.

A. 108.66 kgms-1 B. 109.66 kgms-1 C. 110.66 kgms-1 D. 111.66 kgms-1 149. A binary operation ∗ is defined by 푥 ∗ 푦 = 2푥 + 푥푦. Find 2 ∗ 3

A. −10 B. −2 C. 2 D. 10.

150. If the determinant of matrix 4 푥 − 12푥 3푥 + 4 = 0, calculate the value of 푥,푥 > 0.

A. 1 B. 2 C. 4 D. 8 151. Find the ratio of the coefficient of 푥 to the coefficient of 푥 in the binomial expansion of (2푥 + 3) .

A. 2 : 3 B. 2 ∶ 1 C. 1 ∶ 2 D. 1 ∶ 4 152. Solve the simultaneous equations 2 = 8 and 2 = 128. A. 푥 = −2,푦 = −1 B. 푥 = −2,푦 = 1 C. 푥 = 2,푦 = −1 D. 푥 = 1, 푦 = −2 153. Find the value of 푥 in the equations log 푦 = 3 and 푥푦 = 16.

A. 1 B. 2 C. 3 D. 4 154. The functions 푓 and g are defined by 푓: 푥 → 푥 − 1 and g→ 3푥 + 2,where 푥 is a real number Find g표푓.

A. 3푥 + 2 B. 3푥 − 2 C. 3푥 − 1 D.3푥 + 1 155. Find the zeros of the polynomial 푓(푥) = 푥 −2푥 − 푥 + 2.

A. 1,−1and 2 B. 1, −2 and 3 C. 1, −3 and 4 D. 1,−4 and 5 156. If 훼 and 훽 are the roots of the equation 2푥 − 10푥 + 5 = 0, find the value of

+ .

A. B. C. 2 D. 4

157. The 푛th term of a sequence is given by 푈 = 3 . Find .

A. 27 B. 9 C. 3 D. 158. Solve the equation 2cos θ − cosθ = 0. A. 20° B. 30° C. 60° D. 75°

549

159. Find the gradient of the curve 푦 = 푥 + 2푥 + 2 at 푥 = −1. A. -1 B. 0 C. 1 D. 2

160. The gradient of a curve is given by = 2푥 + 5. If (−1, 2) lies on the curve, find its equation.

A. 푥 + 5푥 + 6 B. 푥 + 5푥 + 6 C. 푥 + 5푥 − 6 D. 푥 − 5푥 + 6 161. The ages, in years, of four boys are 10, 12, 14 and 18. What is the average age of the boys?

A. 12 years B. 12 years C. 13 years D. 13 years

162. Simplify: √12(√3 − √48) A. −18 B. −16 C. −14 D. −12 163. If X = {0, 2, 4, 6}, Y = {1, 2, 3, 4}andZ = {1, 3} are the subsets of 푈 = {푥: 0 ≤ 푥 ≤ 6} find X ∩ (Y ∪ Z). A. {0, 2, 6} B. {1, 3} C. {0, 6} D. { } 164. Find the truth set of the equation 푥 − 6푥 − 18 = 0

A. {푥: 푥 = 3,푥 = 9} B. {푥: 푥 = −3,푥 = −9} C. {푥: 푥 = 3,푥 = −9} D. {푥: 푥 = −3,푥 = 9} 165. Find the gradient of the line joining the points (2,−3) and (2, 6).

A. −9 B. 9 C. − D. 푛표푡푑푒푓푖푛푒푑

166. Simplify ××

A. 16 B. 8 C. 4 D. 1

167. Find the truth set of the quadratic equation 3푥 + 5푥 − 2 = 0 A. , 2 B. − , 2 C. {−2, 3} D. {2, 3}

168. Find the gradient of the line 푦 + 2푥 − 2 = 0 A. −2 B. 0 C. D. 4

169. A function 푓 is defined by 푓(푥) = . For what value of 푥 is 푓 not defined.

A. 1 B. C. 0 D. − 170. Solve the simultaneous equation 푥 − 푦 = −1 and 3푥 + 2푦 = 7.

A. 푥 = 2,푦 = 1 B. 푥 = 1,푦 = 2 C. 푥 = 2, 푦 = 3 D. 푥 = 3,푦 = 2 171. If log 3 = 푎 and log 5 = 푏, find in terms of 푎 and 푏, the value of log 75.

A. 푎 + 2푏 B. 2푎 + 푏 C. 푎 + 푏 D. 2푎푏 172. Find the image of −2 under the mapping 푥 → 2푥 − 푥

A. −8 B. −2 C. −6 D. 8 173. Find the value of (푥 + 푦) from the equations 4푥 − 푦 + 1 = 0 and 2푦 − 5푥 = −1.

550

A. −3 B. −4 C. −1 D. 4 174. If log (3푥 − 1)− log 2 = 0, find the value of 푥. A. −1 B. 1 C. D.

−

175. If 27 = , find the value of 푥. A. −1 B. 1 C. D. − 176. Which of the following could be used in frequency distribution when the class size are different?

A. Class mid-point B. Frequency density C. Cumulative frequency D. Class boundary 177. In how many ways can three books be selected from nine books?

A. 3 B. 12 C. 27 D. 84

178. Given that nC3

nP2= , find the value of 푛?. A. 5 B. 6 C.7 D.8

179. Given that 푃(퐴) = 0.3 and 푃(퐴\퐵) = 0.15, find 푃(퐴 ∩ 퐵) A. 0.045 B. 0.15 C. 0.45 D. 0.50

180. Write down the quadratic equation whose roots are 1and .

A. 푥 − (푥 − 1)푥 + = 0 B. 훼푥 − (푥 + 1)− 1 = 0 C. 훼푥 − (훼 + 1)푥 + 1 = 0 D. 푥 − (훼 + 1)푥 + 1 = 0

181. Find the equation of the straight line passing through the point (−3,−1) and parallel to the line 3푥 − 4푦 − 1 = 0

A. 3푥 − 4푦 + 5 = 0 B. 3푥 + 4푦 − 13 = 0 C. 3푥 − 4푦 + 13 = 0 D. 3푥 − 4푦 − 5 = 0

182. Evaluate lim → A. B. C. − D. −

183. If = 243√3, find the value of 푥. A. B. − C. − D. − √3

The table below shows the distribution of marks student in a test.

푈푠푒푖푡푡표푎푛푠푤푒푟푄푢푒푠푡푖표푛184푡표186 184. What is the upper class boundary of the model class?

A. 29.5 B. 59.5 C. 44.5 D. 30.5 185. Find the mean of the distribution. A. 47.1 B. 47.7 C. 50.0 D. 30.5

Marks 15− 29 30 − 44 45− 59 60− 74 75 − 89 Number of students 5 6 10 5 4

551

186. What is the probability that a student selected at random scored at least 45 marks. A. 0.3000 B. 0.3333 C. 0.6333 D. 0.6667

187. A function 푓 is defined by 푓(푥) = ,푥 ∈ ℝ. State the largest possible domain of 푓 , the inverse of 푓.

A. {푥: 푥 ∈ ℝ,푥 ≠ 3} B. {푥: 푥 ∈ ℝ,푥 ≠ −3} C. {푥: 푥 ∈ ℝ,푥 ≠ 1} D. {푥: 푥 ∈ ℝ, 푥 ≠ −1} 188. Evaluate lim → A. 1 B. −3 C. 3 D. −1 189. When the polynomial 푓(푥) is divided by (푥 + 2), the quotient is 푥 + 2푥 + 1 and the remainder is 5. Find 푓(푥).

A. 푥 + 4푥 + 5푥 + 7 B. 푥 + 4푥 + 푥 + 5 C. 푥 + 푥 + 푥 + 7 D. 14푥 + 56푥 + 170푥 190. The mean of the numbers 푥 ,푥 ,푥 and 푥 is 64. Find the mean of the numbers 푥 + 10,푥 + 10,푥 + 10and푥 + 10.

A. 16.4 B. 12.8 C. 8.2 D. 6.4 191. If (1 − 푥) ≈ 1 + 푎푥 + 푏푥 , find the value of (푎 − 푏).

A. 9 B. 3 D. 1 D. −3 192. 푓(푥) = 3푥 + 8푥 + 6푥 + 푘, If 푓(2) = 1,find 푘.

A. 67 B. −61 C. −67 D. 61 193. Find the derivative of 푥 + with respect to 푥.

A. 3푥 − B. 3푥 + C. 푥 − D. 푥 + 194. Which of the following points lies on the straight line 푦 = 2푥 − 3?

A. (0,−3) B. (0, 3) C. (3, 0) D. (−3, 0) 195. If 푓: 푥 → 2푥 + 1, find 푓 , the inverse of 푓.

A. 2(푥 + 1) B. C. D. 2(푥 − 1)

196. If 푓(푥) = 2푥 + 5 and 푔(푥) = 푥, find 푓표푔(2). A. 6 B. 7 C. 2 D. 3 197. The arithmetic mean of ten numbers is 36. If one of the numbers is 18, what is the mean of the other nine.

A. 18 B. 27 C. 38 D. 54 198. The mean of the numbers 12, 24,and 푦 is the same as the mean of the numbers 9, 12, 18 and 21. Find the value of 푦.

A. 9 B. 15 C. 18 D. 24 199.Divide by A. B. C. D.

552

200. Which of the following set is equivalent to {푝푟푖푚푒푛푢푚푏푒푟푠} ∩ {푒푣푒푛푛푢푚푏푒푟푠}? A. ∅ B. {2} C. {3} D. {4} 201. In a class of 25 students, 15 take history, 17 take history and three take neither subject. How many class members take both subject?

A. 14 B. 5 C. 10 D. 8 202. How many subsets can be formed from a set with 8 members?

A. 256 B. 64 C. 128 D. 16 203. What is the value of 푟 which makes 푥 + 10푥 + 푟 a perfect square?

A. 1 B. 5 C. 10 D. 25 204. If (푎 ) = 푎 , find 푛. A. −6 B. −2 C. 2 D. 6 205. Simplify: 1 × 1 − 5 ÷ 2 A. B. C. D. − 206. One of the roots of the equation 푥 − 5푥 + 6 = 0 is 3, find the other root.

A. 1 B. 2 C. 3 D. 4 207. If 3 log(2 + 푥) = log 27 find 푥. A. −1 B. 1 C. 5 D. 6 208. Solve the equation + = 1 A. 6 B. 4 C. −4 D. −6 209. From a point on the ground, the angle of elevation of the top of a flagpole is 35°. If the flagpole is 200 m away from the point, find, correct to two decimal places, the height of the pole.

A. 11.47 B. 15.40 C. 24.42 D. 28.56 210. If √74 + √27 + 푘√3 = 0, find the value of 푘.

A. −16 B. −8 C. 8 D. 16 211. Which of the following quadratic equations has − and as its roots.

A. 6푥 − 푥 − 1 = 0 B. 2푥 − 푥 − 3 = 0 C. 3푥 − 푥 − 2 = 0 D. 푥 − 푥 − 6 = 0

212. The complement of a given set is the set of all elements in the: A. set but not in the universal set. B. set and also in the universal set. C. universal set but not in the given set. D. empty set but not in the given set. 213. If 2 tan푥 = 1.5,find cos푥 + 2 sin 푥 . A. 2 B. 2 C. 1 D.

214. Find the value of 푥 for which is not defined?

A. 2,−1 B. 2,−2 C. 0,−1 D. 0,−2 215. The universal set 푈 = {1, 2, 3, … , 13}. If 푇 = {evennumbers} and 푅 = {multiplesof3} of 푈, find (푅 ∪ 푇)′.

553

A. {1, 5, 7, 11, 13} B. {5, 7, 11, 13} C. {1, 7, 11, 13} D. {1, 2, 5, 11, 13} 216. The bearing of E from F is 20° and that of G from E is 140°. IF G is directly east of F, find ∠퐹퐺퐸.

A. 80° B. 70° C. 60° D. 50° 217. A fair die is tossed twice. What is the probability that the sum of the digits is a perfect square? A. B. C. D. The sets 푃 = {1, 2, 4, 5, 7} and 푄 = {3, 4, 7, 8} are subsets of the universal set

푈 = {1, 2, 3, … , 9, 10} 푈푠푒푡ℎ푖푠푖푛푓표푟푚푎푡푖표푛푡표푎푛푠푤푒푟푞푢푒푠푡푖표푛푠218and 219

218. Find 푃′ A. {3, 6, 8} B. {3, 6, 7, 8} C. {3, 6, 7, 9, 10} D. {3, 6, 7, 8, 9, 10} 219. Find 푃 ∩ 푄 .

A. {4, 7} B. {1, 2, 3} C. {6, 9, 10} D. {1, 2, 3, 4, 5, 7, 8}

220. Simplify √45 −√

A. √ B. C. √5 D. √5

221. If log푦 = 3 log 2 + log 3 − log 6, find the value of 푦. A. 2 B. 3 C. 4 D. 8

222. If 3 = 푦, find 3 A. 3푦 B. C. − D. 푦

223. What is the equation of the line whose gradient is zero and intersects the 푦-axis at −2?

A. 푦 = −2 B. 푦 = 2 C. 푦 = −2푥 D. 푦 = 2푥 224. The following statements are true about students in form three students in a certain school: 푝:All 3A students are intelligent. 푞: No intelligent student is lazy. What conclusion can you make about Kofi, a form three student in the school who is not intelligent? A. Kofi is lazy B. Kofi is not in 3A C. Kofi is not lazy. D. Kofi is in 3A A survey was made of 100 families each with 5 children to find the number of boys in the family. The result is as shown in the table below.

No. of boys 0 1 2 3 4 5

554

푈푠푒푡ℎ푒푡푎푏푙푒푡표푎푛푠푤푒푟푄푢푒푠푡푖표푛푠225and 226 225. What is the model number of boys for a family of 5 children?

A. 3 B. 4 C. 5 D. 30 226. Estimate the probability that a family with 5 children will have at least 2 boys.

A. 0.20 B. 0.28 C. 0.52 D. 0.80 227. Find 푠 and 푡 from the simultaneous equations: 2푠 + 푡 = 53푠 − 2푡 = 4

A. 푠 = −1, 푡 = 2 B. 푠 = 1, 푡 = 2 C. 푠 = 2, 푡 = −1 D. 푠 = 2, 푡 = 1 228. For what values of 푥 is

( )not defined?

A. − and 6 B. −1 and − C. 0 and −1 D. 6 and −1 229. The mean mark of 16 girls in a test is 20 and that of the 14 boy in the class is 25. Find the mean mark of the whole class.

A. 21.67 B. 22. 33 C. 33.50 D. 45.00 230. There are 8 boys and 4 girls in a lift. What is the probability that the first person who steps out of the lift will be a boy?

A. B. C. D. 231. Find the truth set of 4푥 − 16푥 + 15 = 0.

A. 1 12 ,−2 1

2 B. 1 12 , 2 1

2 C. 2 12 ,−1 1

2 D. {3,−5}

232. Express 3√2 − 2√3 in the form 푎 + 푏√6, where 푎 and 푏 are integers. A. 6 − 12√6 B. 30 − 5√6 C. 30 − 6√6 D. 30− 12√6

233. Given that 퐩 = ,퐪 = and 퐫 = , find 2퐩+ 3퐪 − 4퐫. A. B. C. D.

234. Which of the following vectors is parallel to ? A. B. C. D.

Frequency 3 17 28 30 20 2

555

235. 푦 N 8 O 5 M 푥 What is the gradient of the line MN diagram above?

A. − B. − C. D.

236. Simplify 2 + 1 ÷ 2 A. 1 B. 3 C. 4 D. 4 237. If log 16 = log 36 , find 푥. A. 3 B. 4 C. 6 D. 8 The sets 퐻 = {푎, 푏, 푐} and 푀 = {푐,푑, 푒, 푓} are subsets of 푈 = {푎, 푏, 푐,푑, 푒,푓}.

푈푠푒푡ℎ푒푖푛푓표푟푚푎푡푖표푛푡표푎푛푠푤푒푟푄푢푒푠푡푖표푛ퟐퟑퟖ푎푛푑ퟐퟑퟗ. 238. Find 퐻′ ∩푀. A. {푐} B. {푎,푏} C. {푑, 푒,푓} D. {푐, 푑, 푒,푓} 239. Find (퐻 ∩ 푀)′ A. {푎,푏,푑} B. {푎,푏, 푐, 푑} C.{푎,푏, 푑, 푒, 푓} D. {푎,푏, 푐,푑, 푒, 푓} 240. Find the 푛푡ℎ term of the sequence 1, 2 , 4, 5 , 7, …

A. B. C. D. 4푛 − 1

241. Simplify √ A. 1 B. √ C. D. √

242. Which of the following sets of numbers is the set of the sides of a right angled triangle?

A. (2, 4, 6) B. (3, 4, 5) C. (3, 5, 7) D. (4, 6, 7) 243. From the top T of a vertical tower ST, the angle of depression of point R is 67°. If R is on the same horizontal ground as S and |푅푆| = 14.3 m, calculate |푅푇|.

A. 36.60 B. 28.60 C. 15.53 D. 13.16 244. If tan 푥 = and 푥 is acute, find the value of sin 푥.

A. B. C. D.

556

245. Q 040° P R The diagram above shows the positions of three point 푃,푄,푅. The bearing of 푄 from 푃 is 040°.푅 is directly under 푄 and |푃푄| = |푃푅|. Find the bearing of 푅 from P.

A. 080° B. 100° C. 130° D. 140° 246. The mean of 40, 50 and 90 is 푝. If the mean of 40, 50, 90 and 푝 is 푞. What is the value of 푞?

A. 45 B. 50 C. 60 D. 65 Three sets 푃,푄 and 푅 contain integers as shown below.

푈푠푒푡ℎ푒푖푛푓표푟푚푎푡푖표푛푡표푎푛푠푤푒푟푞푢푒푠푡푖표푛푠ퟐퟒퟕ푎푛푑ퟐퟒퟖ 247. If a number is selected at random, what is the probability that it is in 푃 only?

A. B. C. D. 248. If a number is chosen at random, what is the probability that it is in 푄 or 푅 but not in 푃?

A. B. C. D. The table below gives the score of a group of students in a test.

557

푈푠푒푡ℎ푒푖푛푓표푟푚푎푡푖표푛푡표푎푛푠푤푒푟푞푢푒푠푡푖표푛푠ퟐퟒퟗ푎푛푑ퟐퟓퟎ 249. How many students scored above the mean score?

A. 7 B. 9 C. 11 D. 15 250. Find the median score.

A. 2.5 B. 3.0 C. 3.5 D. 4.0

251. Simplify √ √√ ( √ )

A. √3 B. √5 C. 6√3 D. 6√5

252. A particle moving uniformly covers a distance 푆 at time 푡 seconds, given by 푆 = 24푡 − 6푡 . Find the value of 푡 when the particle covers a distance of 24 meters.

A. 4 seconds B. 3 seconds C. 2 seconds D. 1 second 253. Solve the equation 1 푎 − 4 = 1 + 푎.

A. 8 B. 2 C. −2 D. −1

254. Simplify / × /

/ A. B. √3 C. 3 D. 9 255. 푦 3 2 1 0 1 2 3 4 푥 푦 + 푥 = 3 Which of the following inequalities represent the shaded portion? A. 푦 + 푥 ≤ 3, 푥 ≥ 0,푦 ≥ 0 B. 푦 + 푥 ≥ 3,푥 ≥ 0, 푦 ≥ 0 C. 푦 + 푥 ≤ 3, 푥 ≥ 0,푦 ≤ 0 D. 푦 + 푥 ≥ 3,푥 ≤ 0,푦 ≥ 0 256. P 30° 10 cm

Score 0 1 2 3 4 5 Frequency 2 3 4 2 7 2

558

Q 푥 cm R In the diagram above, 푃푄푅 is a right-angled triangle. |푃푄| = 푥 cm, |푅푄| = 10 cm and ∠푃푄푅 = 30°. Find the value of |푃푄|

A. 6 B.5 C. 4 D. 3 257. Find the values of 푥 in the simultaneous equations 2푥 + 푦 = 4;푥 − 2푦 = 2.

A. 6 B. 4 C. 3 D. 2 258. Which of the following statement about matrix operation is not true? A. Matrix addition is commutative. B. Matrix multiplication is commutative.

C. Matrix multiplication is associative. D. Matrix addition is associative. 259. let 푎 = (−1) 푛 , find 푎 A. 16 B. 10 C. 11 D.13 260. Find the truth set of 푥 if log(3푥 + 8푥 − 1) = 1.

A. 푥: 푥 = −3,− B. 푥: 푥 = 3, C. 푥: 푥 = −3, D. 푥: 푥 = 3,−

261. Evaluate the determinant of the matrix 퐴 =3 −2 02 5 85 3 5

A. −97 B. 63 C. 3 D. −57 262. Evaluate tan( − 휃). A. cot휃 B. – tan 휃 C. tan D. cot 263. Find the turning point of the parabola 푥 = 푥 − 5.

A. , B. − , C. − , D. ,−

264. Decompose ( )( )

into partial fractions.

A. + B. − C. − D. − −

265. Find the common difference of the arithmetic sequence 푥 − 5푏,푥 − 3푏,푥 − 푏, … A. −2푏 B. 8푏 C. 2푏 D. −8푏

266. Evaluate the determinant of the matrix 퐵 =−

A. − B. − C. − D. − 267. Find the coefficient of 푥 in the expansion of the binomial expression (1 + 3푥) .

A. −648 B. 108 C. 648 D. −108 268. Find the value of 푥 if 푥 = log 0.001.

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A. 푥 = −10 B. 푥 = −3 C. 푥 = −0.001 D. 푥 = −2

269. Solve for 푋 in the equation 2푋 + 퐴 = 퐵. Where 퐴 = 1 −20 3 and 퐵 = −3 4

2 1

A. −2 31 −1 B. −4 6

2 −2 C. −8 124 −4 D. −2 2

2 4

270. Find the common difference for the arithmetic sequence 푥 + √3, 푥 + √12,푥 + √27, …

A. √9 B. √15 C. √3 D. √12 271. What is the value of 푥 in 2 + 2 = 2?A. 0 B. 1 C. 2 D. 3

272. Find the value of 푦 if 2 − 푦 푦 − 45 − 푦 푦 − 1 = 15

A. 푦 = B.푦 = 2 C. 푦 = D. 푦 = 4 273. How many terms would the expansion of (30푥 + 20푏) have?

A. 20 B. 30 C. 20 D. 21 274. Given that tan휃 = − and cos휃 > 0,find sec 휃.

A. √ B. −√

C. √ D. − √

275. Find the sum of the first three terms of the sequence with the 푛 term equal to 2. A. 0 B. 4 C. 6 D. 2

푈푠푒푡ℎ푒푖푛푓표푟푚푎푡푖표푛푡표푎푛푠푤푒푟푞푢푒푠푡푖표푛푠276푎푛푑277.

Given that 퐴 = 1 21 6 and 퐵 = −5 3

0 2 ,

276. Determine the adjoint of (퐴 + 퐵).

A. 8 5−3 −4 B. 4 −5

−3 8 C. −4 53 8 D. 8 5

3 −4

277. Evaluate the inverse of matrix 퐴.

A. 0 B. 0 00 0 C. −4 5

3 8 D. the inverse does not exist.

278. Find the domain of the function, 푔(푥) = √4 − 푥 . A. −2 < 푥 < 2 B. 0< 푥 < 2 C. 0 ≤ 푥 ≤ 2 D. −2 ≤ 푥 ≤ 2

푈푠푒푡ℎ푒푖푛푓표푟푚푎푡푖표푛푡표푎푛푠푤푒푟푞푢푒푠푡푖표푛푠279푎푛푑280

Given that the first term of an A. P. is 3 and the 20 term is 117, 279. Find the common difference of the sequence.

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A. 3 B. 4 C. 5 D. 6 280. Determine the sum of the first 19 terms.

A. 111 B. 112 C. 113 D. 114 281. Which of the following is a rational number? √0.16, 휋,√2,and √50.

A. √0.16 B. 휋 C. √2 D. √50 282. Find the coefficient of 푎 푏 in the expansion of (푎 − 3푏) .

A. −3240 B. −120 C. 120 D. 3240 283. Find the truth set of cos휃 = 0for 0 ≤ 휃 ≤ 2휋.

A. 휃:휃 = 0, B. 휃:휃 = 0, C. 휃: 휃 = , D. 휃:휃 = ,

284. Find the solution set of 3 − 푥 푥 + 15 − 푥 푥 + 2 = −6

A. 푥 = B. 푥 = 1 C. 푥 = D. 푥 =

285. If log 2 = 0.3010 and log 3 = 0.4770, find log.

. A. −1.875 B. −0.875 C. 0.875 D. 1.875

286. Find 푥 such that log 푥 = log 8 + log 9 − log 6 A. 5 B. 3 C. 2 D. 1

287. Solve, 푥 − 5 = √푥 − 3 A. {−4,−7} B. {4,−7} C. {4, 7} D. {−4, 7} 288. Solve for 푥 if log 푥 = A. 16 B. 32 C. 64 D. 128 289. If 푓(푡) = 푡 + 1, find 푓(푡 − 1)

A. 푡 B. 푡 − 1 C. 푡 − 2푡 + 2 D. 푡 + 푡 − 2 290. Determine the value of 푘 for which 2푘 푥 + 2푘푥 − 2 is divisible by (푥 − 1)

A. 2, B. −2, C. , 1 D. {−2, 2} 291. Find 휃 if cos(휃 + 60) = 0, 0° ≤ 휃 ≤ 180°

A. 120° B. 90° C. 160° D. 45° 292. What is the minimum value of 푦 = 푥 − 6푥

A. −11 B. −9 C. 3 D. 9 293. 푥 − 푎푥 − 3푥 − 5 has a remainder of 5 when divided by 푥 − 2. What is the value of 푎?

A. −6 B. −3 C. −2 D. 1 294. If 푛 − 2,푛and 푛 + 2 are consecutive terms of a geometric progression, find 푛.

A. −6 B. −4 C. 4 D. 5

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295. What is the equation of a circle with centre (2,−3) and radius √5 ? A. 푥 + 푦 − 4푥 + 6푦 + 8 = 0 B. 푥 + 푦 − 6푥 + 4푦 + 8 = 0 C. 푥 + 푦 + 6푥 − 4푦 + 8 = 0 D. 푥 + 푦 − 4푥 − 6푦 − 8 = 0 296. What is the equation of the perpendicular bisector of the line joining the point 푋(−4,−3) and 푌(1, 3)? A. 10푥 + 12푦 − 15 = 0 B. 10푥 + 12푦 + 15 = 0 C. 10푥 − 12푦 + 15 = 0 D. 10푥 + 12푦 − 5 = 0 297. Find the gradient of the curve 푦 =

√ at (4, 1)

A. − B. − C. D.

298. Evaluate ∫ (푣 + 3)푑푣 A. B. 14 C. D. 36 299. The binary operation ∗ is defined over the set of real numbers by 푎 ∗ 푏 = 푎 + 푏 − 1. What is the inverse of the element? A. 1 − 푎 B. 푎 − 2 C. – 푎 D. 2 − 푎 300. The probability that 퐴,퐵and 퐶 solved a problem are respectively , and . What is the probability that none of them solved the problem? A. B. C. D. 301. If five fair coins are tossed simultaneously, find the probability that all the five come down tail.

A. B. C. D. 302. An investigation was made on the number of defective component in a distribution shop. The table below shows a sample of defective components.

Numberofdefectives 0 1 2 3 4 5 Numberofsamples 1 2 3 5 푥 1

푈푠푒푡ℎ푒푖푛푓표푟푚푎푡푖표푛푡표푎푛푠푤푒푟푞푢푒푠푡푖표푛푠303푎푛푑304

303. If the mean of the distribution is ,find the unknown frequency 푥. A. 1 B. 2 C. 3 D. 4

304. What is the mode of the distribution? A. 0 B. 2 C. 3 D. 4

305. If 퐴 = 1 0−1 2 and 퐵 = 2 −1

1 0 find 2퐴 + 2퐵

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A. 2 3−2 5 B. 2 3

5 10 C. 3 2−5 10 D. 3 2

10 −5

306. A linear transformation 푃 → (푥 + 3푦,−2푥 + 푦). What is the image of the point (2,−3) under the transformation.

A. (−7, 7) B. (1,−7) C. (−7, 1) D. (−7,−7) 307. Given that 퐫 = and 퐬 = , evaluate 퐫⦁2퐬

A. −4 B. 0 C. 4 D. 8 308. A particle of mass 2푘푔 has an initial velocity of 푚푠 . If the change in momentum is 푁푆,find the final velocity of the particle?

A. B. C. D. 309. Simplify, sin(휃 − 180°) A. cos휃 B. – sin휃 C. – cos휃 D. sin 휃 A racing car starts from rest and its acceleration after 푡 seconds is 푘 − 푡 푚푠 until it

reaches a velocity of 60푚푠 at the end of 1 minute. 푈푠푒푡ℎ푒푖푛푓표푟푚푎푡푖표푛푡표푎푛푠푤푒푟푞푢푒푠푡푖표푛푠310푎푛푑311

310. Find the value of 푘. A. −6 B . 5 C. 14 D. 12 311. What is the distance travelled in the first minute?

A. 240푚 B. 2400푚 C. 4800푚 D. 5600푚 312. What is the degree of the polynomial 푥 +2푥 − 5푥 − 3푥 + 7?

A. 5 B. 4 C. 3 D. 2 313. The point (−3,−7) is in the quadrant …………

A. 1st quadrant B. 2nd quadrant C. 3rd quadrant D. 4th quadrant 314. If 8 cos푥 = 3, calculate 푥 to the nearest degree.

A. 32.4 cm B. 18.2 cm C. 16.2 cm D.8.1 cm

315. For what value(s) of 푥 is the function 푓(푥) = equal to zero? A. 1 B. 0, −1 C. 0 D. −1

316. What is the value of 1° in radians? A. B. C. D.

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i God Knows all See answers to all Final Exercises and Multiple Choice Questions in Vol. 2

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