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The Actuarial Education Company © IFE: 2013 Examinations
FAC – P C – 13
Combined Materials Pack
ActEd Study Materials: 2013 Examinations
Foundation ActEd Course (FAC)
Contents
Study Guide for the 2013 exams Course Notes
Question and Answer Bank Summary Test
If you think that any pages are missing from this pack, please contact ActEd’s admin team by email at [email protected] or by phone on 01235 550005.
How to use the Combined Materials Pack
Guidance on how and when to use the Combined Materials Pack is set out in the Study Guide for the 2013 exams.
Important: Copyright Agreement
This study material is copyright and is sold for the exclusive use of the purchaser. You may not hire out, lend, give out, sell, store or transmit electronically or photocopy any
part of it. You must take care of your material to ensure that it is not used or copied by anybody else. By opening this pack you agree to these conditions.
© IFE: 2013 Examinations The Actuarial Education Company
All study material produced by ActEd is copyright and is sold for the exclusive use of the purchaser. The copyright is owned
by Institute and Faculty Education Limited, a subsidiary of the Institute and Faculty of Actuaries.
You may not hire out, lend, give out, sell, store or transmit electronically or photocopy any part of the study material.
You must take care of your study material to ensure that it is not used or copied by anybody else.
Legal action will be taken if these terms are infringed. In addition, we may seek to take disciplinary action through the
profession or through your employer.
These conditions remain in force after you have finished using the course.
FAC: Study Guide Page 1
The Actuarial Education Company © IFE: 2013 Examinations
Foundation ActEd Course (FAC)
Study Guide
Objectives of the Study Guide The purpose of this Study Guide is to give you the information that you should have before studying FAC.
0 Introduction
This document has the following sections: Section 1 Syllabus Page 2 Section 2 The FAC course Page 10 Section 3 Study skills Page 12 Section 4 Contacts Page 13 Section 5 Course index Page 15
Page 2 FAC: Study Guide
© IFE: 2013 Examinations The Actuarial Education Company
1 Syllabus
1.1 Syllabus
This syllabus has been written by ActEd to help in designing introductory tuition material for new students. The topics are all required within the Core Technical subjects. From past experience we know that some students can be a little rusty on mathematical techniques so we have designed a course to help students brush up on their knowledge. Unlike other actuarial subjects, there is no official syllabus or Core Reading written by the profession. (a) Mathematical Notation, Terminology and Methods (a)(i) Be familiar with standard mathematical notation and terminology, so as to be
able to understand statements such as the following:
1. , , , , 3 : n n na b c n n a b c
2. 2( ,0], {0}x x
3. { : 1,2,3, }x x
4. x x 163 5. “Zero is a non-negative integer; is a positive real number.”
6. “ f x( ) tends to 0 as x tends to , is not defined when x 0 , but
takes positive values for sufficiently large x .” (a)(ii) Know the representations and names of the letters of the Greek alphabet that are
commonly used in mathematical, statistical and actuarial work, including in particular, the following letters:
– lower-case: – upper-case:
(a)(iii) Understand the meaning of the following commonly used conventions:
– round brackets used to denote negative currency amounts, – K and m used as abbreviations for “thousand” and “million” – used to denote the change in a quantity – “iff” used as an abbreviation for “if and only if”.
FAC: Study Guide Page 3
The Actuarial Education Company © IFE: 2013 Examinations
(a)(iv) Understand the concept of a mathematical proof and the meaning of “necessary”, “sufficient” and “necessary and sufficient” as they are used in mathematical derivations.
(a)(v) Prove a result using the method of mathematical induction. (b) Numerical methods (b)(i) Evaluate numerical expressions using an electronic calculator with the
following features: arithmetic functions ( ), powers ( y x ) and roots
( yx ), exponential ( ex ) and natural log ( ln x ) functions. (The following
features are also useful but not essential: factorial function ( n!), combinations
( n rC ), hyperbolic tangent function and its inverse ( tanh x and tanh1 x ),
fraction mode, at least one memory and an “undo” facility. Statistical and financial functions are not required.) Students should be able to make efficient use of memories and/or brackets.
(b)(ii) Estimate the numerical value of expressions without using a calculator and
apply reasonableness tests to check the result of a calculation. (b)(iii) Quote answers to a specified or appropriate number of decimal places or
significant figures (using the British convention for representing numbers), and be able to assess the likely accuracy of the result of a calculation that is based on rounded or approximated data values.
(b)(iv) Be able to carry out consistent calculations using a convenient multiple of a
standard unit (eg working in terms of £000s). (b)(v) Express answers, where appropriate, in the form of a percentage (%) or as an
amount per mil (‰). (b)(vi) Calculate the absolute change, the proportionate change or the percentage
change in a quantity (using the correct denominator and sign, where appropriate) and understand why changes in quantities that are naturally expressed as percentages, such as interest rates, are often specified in terms of “basis points”.
(b)(vii) Calculate the absolute error, the proportionate error or the percentage error in
comparisons involving “actual” versus “expected” values or approximate versus accurate values (using the correct denominator and sign, where appropriate).
Page 4 FAC: Study Guide
© IFE: 2013 Examinations The Actuarial Education Company
(b)(viii)Determine the units of measurement (dimensions) of a quantity and understand the advantages of using dimensionless quantities in certain situations.
(b)(ix) Use linear interpolation to find an approximate value for a function or the
argument of a function when the value of the function is known at two neighbouring points.
(b)(x) Apply simple iterative methods, such as the bisection method or the Newton-
Raphson method, to solve non-linear equations. (b)(xi) Carry out simple calculations involving vectors, including the use of row/column
vectors and unit vectors, addition and subtraction of vectors, multiplication of a vector by a scalar, scalar multiplication (“dot product”) of two vectors, determining the magnitude and direction of a vector, finding the angle between two vectors and understanding the concept of orthogonality.
(b)(xii) Carry out calculations involving matrices, including transposition of a matrix,
addition and subtraction of matrices, multiplication of a matrix by a scalar, multiplication of two appropriately sized matrices, calculating the determinant of a matrix, calculating and understanding the geometrical interpretation of eigenvectors and eigenvalues, finding the inverse of a 2 2 matrix and using matrices to solve systems of simultaneous linear equations.
(c) Mathematical Constants and Standard Functions (c)(i) Be familiar with the mathematical constants and e . (c)(ii) Understand and apply the definitions and basic properties of the functions
x n (where n may be negative or fractional), c x (where c is a positive constant),
exp( )x [ ex ], and ln x [ loge x or log x ].
(c)(iii) Sketch graphs of simple functions involving the basic functions in (c)(ii) by
identifying key points, identifying and classifying turning points, considering the sign and gradient, and analysing the behaviour near 0, 1, or other critical
values. (c)(iv) Simplify and evaluate expressions involving the functions x (absolute value),
[ ]x (integer part), max( ) and min( ) , and understand the concept of a
bounded function. [The notation ( )x 100 will also be used as an
abbreviation for max( , )x 100 0 .]
FAC: Study Guide Page 5
The Actuarial Education Company © IFE: 2013 Examinations
(c)(v) Simplify and evaluate expressions involving the factorial function n! for non-negative integer values of the argument and the gamma function ( )x for
positive integer and half-integer values of the argument. (c)(vi) Understand the concept of a complex number and be able to simplify
expressions involving i 1 , including calculating the complex conjugate. (c)(vii) Calculate the modulus and argument of a complex number, represent a complex
number on an Argand diagram or in polar form ( z rei ).
(c)(viii) Apply Euler’s formula e ii cos sin and use the basic properties of the sine and cosine functions to simplify expressions involving complex numbers, including determining the real and imaginary parts of an expression.
(c)(ix) Understand the correspondence between the factors of a polynomial expression
and the roots of a polynomial equation and appreciate that a polynomial equation of degree n with real coefficients will, in general, have n roots consisting of conjugate pairs and/or real values.
(d) Algebra (d)(i) Manipulate algebraic expressions involving powers, logs, polynomials and
fractions. (d)(ii) Solve simple equations, including simultaneous equations (not necessarily
linear) by rearrangement, substitution, cancellation, expansion and factorisation.
(d)(iii) Solve an equation that can be expressed as a quadratic equation (with real
roots) by factorisation, by “completing the square” or by applying the quadratic formula, and identify which of the roots is appropriate in a particular context.
(d)(iv) Solve inequalities (“inequations”) in simple cases and understand the concept of
a “strict” or “weak” inequality. (d)(v) State and apply the arithmetic-geometric mean inequality, and know the
conditions under which equality holds. (d)(vi) Understand and apply the and notation for sums and products, including
sums over sets (eg i0
) and repeated sums.
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© IFE: 2013 Examinations The Actuarial Education Company
(d)(vii) Calculate the sum of a series involving finite arithmetic or geometric progressions or infinite geometric progressions using the formulae:
AP a n d a ln n 2 22 1( ( ) ) ( ) or ,
GPa r
r
n
( )1
1 and
GPa
r1,
and be able to determine when an infinite geometric series converges.
(d)(viii) Apply the formulae: k n nk
n
1
12
1( ) and k n n nk
n2
1
16
1 2 1 ( )( ) .
(d)(ix) Solve simple first or second order difference equations (recurrence relations),
including applying boundary conditions, by inspection or by means of an auxiliary equation.
(d)(x) Recognise and apply the binomial expansion of expressions of the form ( )a b n
where n is a positive integer, and ( )1 x p for any real value of p and, in the
latter case, determine when the series converges. (e) Calculus (e)(i) Understand the concept of a limit (including limits taken from one side) and
evaluate limits in simple cases using standard mathematical notation, including the use of “order” notation O x( ) and o x( ) , and the sup/ lub and inf / glb
functions (considered as generalisations of max and min ). (e)(ii) Understand the meaning of a derivative as the rate of change of a function when
its argument is varied (in particular, for functions dependent on t , the time measured from a specified reference point), including the interpretation of a derivative as the gradient of a graph.
(e)(iii) Differentiate the standard functions x n , c x , ex and ln x . (e)(iv) Evaluate derivatives of sums, products (using the product rule), quotients (using
the quotient rule) and “functions of a function” (using the chain rule). (e)(v) Understand the concept of a higher-order (repeated) derivative and be familiar
with the mathematical notation used to denote such quantities.
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(e)(vi) Use differentiation to find the maximum or minimum value of a function over a specified range (including the application of a monotone function, such as the natural log function, to simplify the calculation) and determine the nature of stationary points.
(e)(vii) Understand the meaning of a partial derivative and how to express a partial
derivative in standard mathematical notation, and be able to evaluate partial derivatives in simple cases. Find extrema of functions of two variables.
(e)(viii)Use the method of Lagrangian multipliers. (e)(ix) Understand the meaning of an indefinite integral as the anti-derivative of a
function and the meaning of a definite integral as the limit of a sum of infinitesimal elements, including the interpretation of a definite integral as the area under a graph.
(e)(x) Integrate the standard functions x n , c x and ex . (e)(xi) Evaluate indefinite and definite integrals by inspection, by identifying and
applying an appropriate substitution, by integration by parts, by using simple partial fractions or by a combination of these methods.
(e)(xii) Determine when a definite integral converges. (e)(xiii)Understand the meaning of a multiple integral and how to express a multiple
integral in standard mathematical notation, and be able to evaluate a double integral as a repeated integral in simple cases, including determining the correct limits of integration. Swap the order of integration.
(e)(xiv) Apply the trapezium rule to find the approximate value of an integral. (e)(xv) State and apply Taylor series and Maclaurin series in their simplest form,
including using these to determine the approximate change in a function when the argument is varied by a small amount. (Knowledge of the error terms is not required.)
(e)(xvi) Recognise and apply the Taylor series expansions for ex and ln( )1 x and, in
the latter case, determine when the series converges. (e)(xvii)Solve simple ordinary first-order differential equations, including applying
boundary conditions, by direct integration (which may involve a function of the dependent variable), by separation of variables or by applying an integrating factor.
Page 8 FAC: Study Guide
© IFE: 2013 Examinations The Actuarial Education Company
(e)(xvii)Differentiate expressions involving definite integrals with respect to a parameter, including cases where the limits of integration are functions of the parameter.
(g) General (g)(i) Be familiar with the currency systems of the United Kingdom (pounds and pence
sterling), the United States (dollars and cents), the European monetary system (Euro and cent) and other major economies, and be able to interpret and write down currency amounts using these systems.
(g)(ii) Be familiar with the Gregorian calendar, including determining when a
specified year is a leap year, the concepts of calendar years, quarters and tax years, and the abbreviations commonly used to represent dates in the United Kingdom, Europe and the United States.
(g)(iii) Understand the distinction between “expression”/“equation”/“formula” and
“term”/“factor”. (g)(iv) Understand the meaning of the words “gross” and “net”. (g)(v) Be able to spell the following words correctly: actuarial, appropriately,
basically, benefit, benefiting, bias(s)ed, calendar, cancelled, commission, consensus, correlation, cyclically, deferred, definitely, formatted, fulfil, gauge, hierarchy, immediately, independence, instalment (British spelling), lose, loose, millennium, necessary, occasion, occurred/occurring, offered, orthogonal, paid, particularly, pensioner, precede, proceed, receive, referred, relief, seize, separate, similarly, specifically, supersede, targeted, theorem, until, yield.
(g)(vi) Be able to determine the correct member of word pairs according to context: eg
affect/effect, principal/principle, dependant/dependent. (g)(vii) Be able to distinguish between the singular and plural forms of words of Latin
or Greek origin, including the following: criterion/criteria, formula/formulae, analysis/analyses. [The word “data” may be treated as singular or plural, according to the preferences of individual authors/speakers.]
(g)(viii)Be familiar with commonly used Latin expressions and abbreviations such as
“per annum”, “vice versa”, “status quo”, “pro rata”, “ie”, “eg”, “cf”, “sic” and “stet”.
FAC: Study Guide Page 9
The Actuarial Education Company © IFE: 2013 Examinations
1.2 The Profession’s Copyright
All of the course material is copyright. The copyright belongs to Institute and Faculty Education Ltd, a subsidiary of the Institute and Faculty of Actuaries. The material is sold to you for your own exclusive use. You may not hire out, lend, give, sell, transmit electronically, store electronically or photocopy any part of it. You must take care of your material to ensure it is not used or copied by anyone at any time. Legal action will be taken if these terms are infringed. In addition, we may seek to take disciplinary action through the profession or through your employer. These conditions remain in force after you have finished using the course.
Page 10 FAC: Study Guide
© IFE: 2013 Examinations The Actuarial Education Company
2 The FAC course
2.1 Course Notes
The Foundation ActEd Course consists of a set of eight chapters of notes covering the following ideas: Chapter 1 Notation Chapter 2 Numerical Methods I Chapter 3 Mathematical constants and standard functions Chapter 4 Algebra Chapter 5 Numerical Methods II Chapter 6 Differentiation Chapter 7 Integration Chapter 8 Vectors and matrices We recommend that you work through the sections that you are unsure of, completing the questions that are given. If you need further practice, there is a Question and Answer Bank and a Summary Test. These both cover material from all the chapters. Section 5 of this study guide gives you an index of the topics that are covered in order to identify quickly which chapters you need to look at. When you are working through the Core Technical subjects you can continue to use this course as a reference document if you come across areas of mathematics that you are still unhappy about.
FAC: Study Guide Page 11
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For further guidance, this grid shows which chapters from FAC are needed for Subjects CT1, CT3, CT4, CT5, CT6 and CT8: Chapter Section Course Notes
1 1-7 All This is just general information and notation. 2 1-4 All 3 1-3 All 4 1-9 All 5 1-6 All 5 7 CT6 Syllabus items (c)(vi), (c)(vii), (c)(ix) 5 8 CT6 Syllabus item (d)(ix) 6 1-8 All 6 9 CT8 Syllabus item e(viii) 7 1, 2 All 7 3.1-3.4 All 7 3.5 CT4, CT8 Syllabus item (e)(xviii) 7 4-7 All 7 8.1-8.3 CT4, CT8 Syllabus item (e)(xvii) 8 1, 2.1-2.4 CT4, CT6, CT8 Syllabus items (b)(xi), (b)(xii) 8 2.5 CT4 Syllabus item (b)(xii) (part)
2.2 Online Classroom
The Online Classroom is available to provide tuition on the material covered in FAC. It is a comprehensive, easily-searched collection of recorded tutorial units covering the same topics as the FAC Course Notes. These tutorial units are a mix of:
teaching, covering the relevant theory to help you get to grips with the course material, and
worked questions or examples, illustrating the various techniques you should be familiar with.
To find out more about the Online Classroom, and to watch example tutorial units, please visit the ActEd website at www.ActEd.co.uk.
Page 12 FAC: Study Guide
© IFE: 2013 Examinations The Actuarial Education Company
3 Study skills
All the mathematical techniques covered in this course will be used in the context of one of the Core Technical subjects and will therefore not be directly tested on their own. For example, in the Core Technical exams you will not be set a question saying “Integrate this function by parts” but you will be asked to work out expectations in Subject CT3 which may involve integration by parts. It is therefore essential that you feel really comfortable with each method. You should study this course actively. In particular we recommend the following: 1. Annotate your Notes with your own ideas and questions. This will make your
study more active. 2. Attempt the questions in the Notes as you work through the course. Write down
your answer before you check against the solution. 3. Attempt the Question and Answer Bank and the Summary Test on a similar
basis, ie write down or work out your answer before looking at the solution provided.
FAC: Study Guide Page 13
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4 Contacts on the course material
Queries From time to time, you may come across something in the study material that is not clear to you. If you cannot resolve the query through discussion with friends and colleagues, then you can use one of ActEd’s discussion forums: If you have access to the FAC Online Classroom, then you can post your query in
the forums within the Online Classroom itself.
Alternatively, you can post your query on our forum at www.ActEd.co.uk/forums (or use the link from our homepage at www.ActEd.co.uk). This forum includes a section for each actuarial exam – there’s one for “FAC and StatsPack”.
Our forums are dedicated to actuarial students so that you can get help from fellow students on any aspect of your studies from technical issues to general study advice. ActEd tutors monitor the forums to answer queries and ensure that you are not being led astray. If you are still stuck, then you can send queries by email to [email protected] (but we recommend you try a forum first). We will endeavour to contact you as soon as possible after receiving your query, but you should be aware that it may take some time to reply to queries, particularly when tutors are away from the office running tutorials. At the busiest teaching times of year, it may take us more than a week to get back to you. Corrections and feedback We are always happy to receive feedback from students, particularly concerning any errors, contradictions or unclear statements in the course. If you find an error, or have any comments on this course, please email them to [email protected].
Page 14 FAC: Study Guide
© IFE: 2013 Examinations The Actuarial Education Company
Further reading If you feel that you would find it useful to obtain a different viewpoint on a particular topic, or to have access to more information and further examples, then the best place to look would be a mathematics textbook. The level of mathematics covered in FAC is broadly similar to that covered by those examinations taken immediately prior to going to university (A-Level or Higher exams in the UK). You may still have your old textbooks, or know which ones you used and be able to track them down. If not, one set of textbooks published to help students prepare for A-Level exams is: Edexcel AS and A Level Modular Mathematics – Core Mathematics 1, 2, 3 & 4 These are available from internet retailers, including www.amazon.co.uk. Each textbook covers different topics, so you can choose which would be most suitable for you.
FAC: Study Guide Page 15
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5 Course index
Topic Chapter Page Absolute change 5 3 Arithmetic-geometric mean 4 20 Arithmetic progressions 4 24 Binomial expansion 4 32 Calculator, use of 2 4 Complex numbers 5 15 Convergence 7 16 Curve sketching 6 23 Determinants 8 12 Difference equations 5 22 Differential equations 7 31 Differentiating an integral 7 14 Differentiation, products and quotients 6 12 Differentiation, standard functions 6 11 Dimensions 5 6 Double integrals 7 18 Eigenvectors and eigenvalues 8 21 Errors 5 5 Estimation 2 9 Extrema 6 27 Factorial notation 3 12 Fractions, algebraic 4 5 Functions and graphs 3 2 Gamma function 3 13 Geometric progressions 4 24 Greek symbols 1 5 Indices 4 2 Induction 1 10 Inequalities 4 16 Infimum 6 7 Integer part 3 9 Integration, by parts 7 12 Integration, partial fractions 7 9 Integration, standard functions 7 5
Page 16 FAC: Study Guide
© IFE: 2013 Examinations The Actuarial Education Company
Integration, substitution 7 11 Interpolation 5 10 Iteration 5 13 Lagrangian multipliers 6 28 Leibniz’s formula 7 14 Limits 6 2 Logarithms 4 2 Maclaurin series 7 30 Mathematical notation 1 2 Matrices 8 8 Max and min notation 3 10 Modulus 3 8 Newton Raphson iteration 5 14 Order notation 6 3 Partial differentiation 6 25 Percentages 5 2 notation 4 21 Proof 1 7 Proportionate change 5 3 Quadratic equations 4 7 Rounding 2 2 Scalar product 8 5 Series 4 28 notation 4 21 Simultaneous equations 4 11 Stationary points 6 18 Supremum 6 7 Taylor series 7 27 Trapezium rule 7 23 Vectors 8 2
FAC-01: Notation Page 1
The Actuarial Education Company © IFE: 2013 Examinations
Chapter 1
Notation
You need to study this chapter to cover: ● standard mathematical notation and terminology
● the letters of the Greek alphabet
● conventions commonly used in financial and actuarial mathematics
● mathematical proof
● mathematical induction
● currencies, dates and ages.
0 Introduction
This chapter deals with the notation and terminology that you must be familiar with in order to study the actuarial exams. Much of this may be familiar to you already, in which case read the chapter quickly or use it as a reference guide.
Page 2 FAC-01: Notation
© IFE: 2013 Examinations The Actuarial Education Company
1 Mathematical notation
Symbol Meaning Further explanation/examples Types of numbers: Integers (whole numbers) {...,–3, –2, –1, 0, 1, 2, 3,...} Natural numbers {1, 2, 3, ...} (counting numbers)
Rational numbers 32 , 6
51.2 = , 14990.141414... = etc (all can
(fractions) be written as pq )
Real numbers rational numbers plus irrational numbers
(such as 2 , p and e) ie no imaginary component
Complex numbers can be written in the form a ib+ , where
1i = - Logic:
" For all (values) 2 x x " Œ $ Œ
: Such that (see next example) $ There exists : 1 5x x$ Œ + =
/$ There doesn’t exist 2 : 4x x/$ Œ = -
fi Implies 22 4x x= - fi =
¤ Implies and is implied by 32 8x x= ¤ = iff If and only if equivalent to ¤ Set Theory:
{ }1,2,3,... A set
etc ( ,0]-• A set containing numbers from -• to 0, not
including -• (since the bracket is round next to -• ), but including 0 (since the bracket is square next to 0)
{ } or ∆ Empty set the set of odd numbers divisible by 2
Œ Is a member of 2½ Œ A B» Union of two sets A or B or both, ie the things that are in one
or other or both
FAC-01: Notation Page 3
The Actuarial Education Company © IFE: 2013 Examinations
A B« Intersection of two sets A and B, ie only the things that are in both
A Complement of a set not A, ie the things that are not in A Miscellaneous: p Pi 3.14159...(the ratio of the circumference of
a circle to its diameter) e Euler’s constant used in 2.7182818...
the exponential function • Infinity Æ Tends to approaches eg x Æ• means that x is
becoming very large
Note: ● A superscript “+” or “ - ” on symbols such as refers to the positive or
negative numbers within the group ie + means 1, 2, 3,… (excluding zero), ie the same as .
● When a superscript “+” or “- ” is used in situations such as 1x +Æ , it means that x is approaching 1 “from above”, in other words x is taking values slightly bigger than 1.
● is the set of complex numbers. These can be written in the form a ib+ ,
where ,a b Œ and 2 1i = - , a being called the real part and b being called the
imaginary part. You may have seen j used for 1- rather than i.
Page 4 FAC-01: Notation
© IFE: 2013 Examinations The Actuarial Education Company
Example (i) Interpret the statement:
{ }: 4x x+Œ <
(ii) For the function 1( ) xf x = , describe what happens as x tends towards 0 and ±• .
Solution (i) This is the set of whole numbers which are less than 4 but are positive ie the set
{ }1,2,3=
(ii) ( )f x is not defined for 0x = , ( ) as 0f x x +Æ • Æ , ( ) as 0f x x -Æ -• Æ ,
( ) 0 as f x xÆ Æ ±• .
Question 1.1
(i) If { } { } and P prime numbers Q even numbers= = what is M P Q= « ?
(ii) What values are in the set 2{ : 10}x xŒ < ?
FAC-01: Notation Page 5
The Actuarial Education Company © IFE: 2013 Examinations
2 Greek symbols
You need to know the following Greek letters:
Letter Lower case
Used for Upper case Used for
alpha a parameter
beta b parameter B beta function
gamma g parameter G gamma function
delta d small change D difference
epsilon e small quantity
theta q parameter Q number of deaths
kappa k parameter
lambda l parameter
mu m mean and force of mortality
nu n force of mortality
when sick
pi p 3.14159= P product
rho r correlation coefficient and force of recovery
sigma s standard deviation and
force of sickness S sum
tau t parameter (pronounced
as in “torn”)
phi f probability density
function of standard normal distribution
F cumulative distribution
function of standard normal distribution
chi c
2c distribution
(pronounced as first syllable of “Cairo”)
psi y probability of ruin
omega w limiting age in a life
table
Page 6 FAC-01: Notation
© IFE: 2013 Examinations The Actuarial Education Company
3 Conventions
There are many conventions and short-hand notations that are used in mathematical and financial work. The following are used in the course notes: ● Abbreviations are used for thousands and millions to save writing many zeros.
For example, £9K means £9,000 (the “K” comes from the “kilo” prefix seen in words such as kilometre, meaning 1,000 metres) and $6.2m or $6.2M means $6,200,000. This is used in preference to using “standard form”, where
$6,200,000 would be written as 6$6.2 10¥ .
● D is used to denote a change in a quantity, for example D profit = £534K means that the profit has risen by £534,000.
● If interest rates were 6% in January, 8% in February and 347 % in March, this
would often be described as an increase of 2 percentage points, followed by a reduction of 25 basis points (one basis point being one-hundredth of a percentage point). Basis points are sometimes abbreviated to bps.
● In accounting, negative amounts of money are represented by placing them in brackets eg 5 (5)- = .
Example Here is an example of a very simple income statement for a company (sometimes called a profit and loss account), showing the negative cashflows in brackets. Don’t worry if you don’t understand what the individual items represent. £ Pre-tax profit 9.6m Tax (2.4m)
Net profit 7.2m Dividends (1.7m)
Retained profit 5.5m
Similarly you will see things like “calculate the profit (loss) made last year”, which means calculating the income minus the outgo and writing it as a positive number for a profit, or in brackets if it is negative, ie a loss.
FAC-01: Notation Page 7
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4 Proof
To prove that something is true in mathematics, it is not sufficient to show that something works for a particular case if you are trying to prove it in general. However you can prove that something is not true in general by showing that it doesn’t work for a particular case (this is called a counterexample). You need to be familiar with the terms sufficient, necessary and necessary and sufficient. If A is necessary for B, then B Afi (ie B implies A or B is true only if A is true). If A is sufficient for B, then A Bfi (ie A implies B or B is true if A is true). If A is necessary and sufficient for B, then A B¤ (ie A implies and is implied by B or A and B are equivalent statements or A is true if and only if B is true).
Example In a group of 50 people, there are 25 men, 11 people with beards (who are all male!) and 25 people who like football who are all men too. If we use the notation M for “is a man”, B for “has a beard” and F for “likes football”, then we have B Mfi and M F¤ . The first implication is true since if we know someone has a beard, they must be male. The second implication is true since if we know someone is male, they automatically like football and vice versa.
Example (i) If A is the statement “the integer x ends in a 5”, and B is the statement “the
number x is divisible by 5”, then A is sufficient for B, but not necessary (since a number ending in a 0 is also divisible by 5). So we can write A Bfi but not B Afi .
(ii) If x is a solution of the equation 2 0ax bx c+ + = , then if P is the statement
“ 2 4 0b ac- ≥ ” and Q is the statement “x is a real number”, then P is a necessary and sufficient condition for Q so that P Q¤ . (More about quadratic equations
later!)
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5 Expressions, equations and formulae
While you’re studying for the Core Technical exams you’ll use a lot of expressions, equations and formulas.
5.1 Expressions
A mathematical expression is any combination of mathematical symbols that can be evaluated to give an “answer”. Usually the expression involves more than one symbol, often it includes some letters ( , ,x y z etc), and usually the answer is numerical (but not
necessarily). For example, the following are all mathematical expressions:
2 2+ , 251.09 , 1
nt
t
v=Â ,
5(1 ) 1i
d+ -
, (½)G
In Subjects CT1 and CT5 you’ll meet some of the special symbols used in actuarial calculations, and your answers to assignment questions and exam questions will involve
expressions containing actuarial symbols such as |20a and 1|[30]:25A .
If a question says “Find an expression for …”, you should simplify your final expression as much as possible.
5.2 Equations
An equation is just a statement that two expressions are equal. For example, the following are all equations:
2 2 4+ = , 251.09 8.6231= , |1
nt
nt
v a=
=Â , 5
|5(1 ) 1i
sd
+ - = , (½) pG =
The issue is slightly confused by the fact that a lot of word processing packages use the word “equation” to refer to anything that contains mathematical symbols. A word processing “equation” may really be an expression, a formula, an inequality(!) or just
nonsense (eg * p ∆"≈ ).
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5.3 Formulas (or formulae)
A formula is just an expression that can be used specifically for calculating the answer to a particular type of problem. A formula may be stated in the form of an expression or as an equation, eg:
The quadratic formula is 2 4
2
b b ac
a
- ± -
The formula for calculating the present value of a unit annuity-certain payable annually
in arrears is |1 n
nv
ai
-= . You will meet this in Subject CT1.
A formula usually involves standard letters for the variables (eg you know that the
, ,a b c in the quadratic formula are the coefficients of 2x , x and the constant term).
Also the letters often stand for the quantities involved, as in F ma= (force = mass ¥ acceleration).
5.4 Terms and factors
A term is an element in an expression that is added or subtracted. A factor is an element in an expression that is multiplied or divided.
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6 Induction
One method of proving a general mathematical result is the method of (mathematical) induction. To prove that a result is true for all positive integers, we prove that if the result is true for any particular integer k then it must also be true for the next integer
1k + . If we can also show that it is true when 1k = , then it must be true for all positive integers 1,2,k = .
Example
Prove by induction that 121 2 3 ( 1)n n n+ + + + = + .
Solution Assume the result is true for n k= , ie:
121 2 3 ( 1)k k k+ + +◊◊◊ + = +
Adding the next term on to both sides it follows that:
[ ]
12
12
12
1 2 3 ( 1) ( 1) ( 1)
( 1)( 2)
( 1) ( 1) 1
k k k k k
k k
k k
+ + + + + + = + + +
= + +
= + + +
Since this is what the original equation “predicts” when 1n k= + , we have shown that if the result is true for n k= then it is also true for 1n k= + . Consider whether the result is true when 1n = :
LHS = 1 RHS = 1 So the result is true for 1n = and by the above result it must also be true for
2,3,4,n = ie for all positive integer values of n.
Question 1.2
Prove by induction that 2 2 2 2 161 2 3 ( 1)(2 1)n n n n+ + + + = + + .
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7 Some general knowledge
Currencies
Country Main currency unit Sub-division
United Kingdom Pound (£) Pence (£1 = 100p)
United States Dollar ($) Cent ($1 = 100¢) European Union Euro (€) Cent (€1 = 100¢)
Japan Yen (¥) –
Question 1.3
Today’s exchange rates are shown as: €/£ = 0.61 and £/$ = 1.44 How much would 1,000 Euro be worth in US dollars?
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Dates Gregorian calendar The calendar system used in Western Europe and America is officially called the Gregorian calendar (named after Pope Gregory XIII who introduced it). This system is recognised and understood worldwide, although a number of countries in other parts of the world have alternative calendar systems that they use as well. Leap years Calendar years usually have 365 days but, in order to prevent the seasons gradually drifting, an extra “leap” day is added at the end of February in some years. These leap years have 366 days, the extra day being 29 February. The general rule for determining whether a particular calendar year is a leap year is as follows:
LEAP YEAR OR NOT? A calendar year IS NOT a leap year …
… unless it divides exactly by 4, in which case it IS a leap year … … unless it also divides exactly by 100, in which case it IS NOT a leap year … … unless it also divides exactly by 400, in which case it IS a leap year!
Question 1.4
In actuarial calculations involving weekly payments it is often assumed that there are 52.18 weeks in an “average” year. Where does this figure comes from?
In a lot of actuarial applications the exact number of days in each month makes very little difference to the numerical answers. In these cases you can assume that the
months are of equal length ie each month is exactly 112 of a year long. This simplifies
the calculations considerably.
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Calendar years, quarters and tax years Many organisations divide each calendar year into four quarters for budgeting and accounting purposes. For example, the calendar year 2012 would be broken into the four quarters: 2012 Q1: 1 January 2012 – 31 March 2012 2012 Q2: 1 April 2012 – 30 June 2012 2012 Q3: 1 July 2012 – 30 September 2012 2012 Q4: 1 October 2012 – 31 December 2012 In actuarial calculations where payments are made quarterly it is normally sufficiently
accurate to assume that each quarter is exactly 14 of a year long.
In the UK the amount of tax payable by individuals is calculated based on the transactions during each tax year (sometimes also referred to as a “fiscal” year), which run from 6 April to 5 April. So, for example, the 2011/12 tax year is the period from 6 April 2011 to 5 April 2012 (both days inclusive). The actual dates will differ between countries, for example the New Zealand tax year runs from 1 April to 31 March. Fencepost errors
Question 1.5
If you need to erect a fence 10 metres long in an open field using 1 metre-long strips of wood, how many posts will you need to support it?
If you got the answer wrong, you’ll see that it’s very easy to make these fencepost errors. It’s particularly easy to make a mistake in calculations involving dates. Almost everyone gets one wrong at some point.
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Question 1.6
(i) Five payments are made at 9-month intervals with the first payment on 1 January 2012. On what date will the last payment be made?
(ii) A man was born on 9 September 1960. In New Zealand, how many complete
tax years are there between 1 May 1998 and his 60th birthday? (iii) How long is the period from 1 March 2005 to 28 February 2015?
Conventions for writing dates To save time, dates are often written in numbers, rather than in words. So make sure you know the numbers of the months (eg October = 10, November = 11). Also, just to make life difficult, there are two different conventions in use. In the UK and Europe we use the DD/MM/YY order, whereas Americans use MM/DD/YY. This can cause a lot of confusion since 01/11/12 would mean 1 November 2012 in the UK, but 11 January 2012 in the US. (The reason for this discrepancy is that in the UK we tend to say “the first of November”, whereas in the US they tend to say “November one”.) To decide which convention is being used, look which position contains numbers greater than 12. This must be the days bit. In actuarial symbols a fixed period of time is represented by using a right-angle symbol,
so that “5 years”, for example, is usually represented by 5 . Some of your actuarial colleagues may use this as a shorthand notation. For example, they might write: “The
pension incorporates a |5 guarantee” or they might even use 312 as an abbreviation for
“3 months”.
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Ages In life insurance work and pensions work, you’ll often have to work out people’s ages. This might sound easy, but there are actually three different ways commonly used to express ages: Age last birthday: This is the age one of your friends would tell you if you asked them how old they were. It’s just the number of candles they had on their last birthday cake. Age nearest birthday: This is the person’s age at their nearest birthday (which could be either the previous one or the following one). Pension fund calculations usually use this definition because, for a large group of people, age nearest birthday usually averages out at the true age. You’ll get some people who are slightly older and some who are slightly younger, and these will normally balance out. However, age last birthday will always understate the true age. Age next birthday: This is the person’s age at their next birthday. This definition is the one usually used by insurance companies. This will always overstate the age. These age definitions are often abbreviated to “age last”, “age nearest” and “age next”. You’ll use these ways of defining ages in Subject CT4. You may hear people in life offices referring to their policyholders as “a female aged 50 next” (say). Some people are born on one of the “leap days” eg 29 February 2012. For calculation purposes they are normally treated as if they were born on the following day ie 1 March 2012 in this example.
Question 1.7
A man was born on 6 May 1954. What will his age be on 1 January 2015 using each of these three age definitions?
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Chapter 1 Solutions Solution 1.1
(i) {2}M = . (Don’t forget that a prime number is one that has no factors other than
1 and itself!)
(ii) { }10 10x- < <
Solution 1.2
Assume the result is true for n k= , ie:
2 2 2 2 161 2 3 ( 1)(2 1)k k k k+ + + + = + +
Adding the next term on to both sides:
( )
2 2 2 2 2 216
16
216
216
16
16
1 2 3 ( 1) ( 1)(2 1) ( 1)
( 1)[ (2 1) 6( 1)]
( 1)(2 6 6)
( 1)(2 7 6)
( 1)( 2) 2 3
( 1)[( 1) 1][2( 1) 1]
k k k k k k
k k k k
k k k k
k k k
k k k
k k k
+ + + + + + = + + + +
= + + + +
= + + + +
= + + +
= + + +
= + + + + +
So we have shown that if the result is true for n k= then it is also true for 1n k= + . Consider when 1n = : LHS = 1 RHS = 1 So the result is true for 1n = and by mathematical induction it is also true for n = 2, 3, 4, … ie for all positive integer values of n.
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Solution 1.3
1,000 Euro must be equivalent to 1,000 0.61 £610¥ = .
£610 must be equivalent to 610 1.44 $878.40¥ = . The precise rules that banks have to use for converting currencies when Euro are involved are actually quite complicated eg you have to work to 6 decimal places. Solution 1.4
It’s 14365 7∏ (rounded to 2 DP).
Solution 1.5
11 The “obvious” answer was to divide 10 by 1 and say 10. But, because you need a post at each end of the fence, you actually need an extra one. If you said 10, you’ve made a “fencepost error”. To avoid these, you need to pay careful attention to which, if either, of the endpoints is included. This problem comes up when you are trying to work out the number of payments in a stream of payments. Solution 1.6
(i) 1 January 2015 (There are 4 gaps of 9 months between these 5 payments. This makes a total period of 36 months, which equals 3 years.)
(ii) 21 (The period from 1 April 1999 to 31 March 2020 consists of 21 complete tax
years ie 1999/2000, 2000/01, … , 2019/20.) (iii) 10 years (When you’re dealing with a period of time, it’s a straight subtraction.) Solution 1.7
Age last = 60 (because his last birthday would be 6 May 2014 and 2014 1954 60- = ). Age next = 61 (add 1 to his age last). Age nearest = 61 (because his nearest birthday would be 6 May 2015).
© IFE: 2013 Examinations The Actuarial Education Company
All study material produced by ActEd is copyright and issold for the exclusive use of the purchaser. The copyright
is owned by Institute and Faculty Education Limited, asubsidiary of the Institute and Faculty of Actuaries.
You may not hire out, lend, give out, sell, store or transmitelectronically or photocopy any part of the study material.
You must take care of your study material to ensure that itis not used or copied by anybody else.
Legal action will be taken if these terms are infringed. Inaddition, we may seek to take disciplinary action through
the profession or through your employer.
These conditions remain in force after you have finishedusing the course.
FAC-02: Numerical methods I Page 1
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Chapter 2
Numerical methods I
You need to study this chapter to cover: ● accurately rounding numbers
● the use of an electronic calculator
● estimation
● abbreviations.
0 Introduction
This chapter reminds you of the conventions about rounding and accuracy. Since the Core Technical subjects rely heavily on numerical accuracy, rounding is more important in your actuarial studies than perhaps it has been during your university studies.
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1 Rounding
Numerical answers should be rounded to a “suitable” number of decimal places (DP) or significant figures (SF). Marks can be lost in examinations if a final answer is not rounded to the accuracy that the question requires. Use common sense when deciding what “suitable” means. For example, when giving an answer in monetary terms, round to two decimal places. If a question is asking for an interest rate it is unlikely that you will need to work to more than three decimal places eg 6.125%. Do be aware though that we are talking about final answers here: if you are going on to use the interest rate later you will need an accurate figure to work with. More of that later!
Example Round the following numbers to two decimal places: 3.784, 15.239, 6.028, 6, 2002,
0.399- . Solution The answers are 3.78, 15.24, 6.03, 6.00, 2002.00, –0.40. Notice that all answers have two figures (digits) after the decimal point.
Example Round these numbers to two significant (ie non-zero) figures: 3.784, 15.239, 6.028, 6, 2002, 0.399- . Solution The answers are 3.8, 15, 6.0, 6.0, 2000, –0.40. Notice that the first “significant figure” is the first non-zero figure.
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Question 2.1
Round these numbers to three decimal places:
0.0678, 15.3489, 9.9999
Question 2.2
Round these numbers to three significant figures:
14.3678, 5.9879, 0.08006
Notes: ● When a number is written as 6.00, it implies that the author believes it is correct
to 2DP ie the exact value lies somewhere in the range [5.995,6.005) , using the
convention that a digit greater than or equal to 5 causes a number to be rounded up.
● In some countries (continental Europe in particular) commas and full stops in numbers are used with the opposite meaning from in the UK and the US ie the decimal point is written as a comma and full stops (or spaces) are used to separate a large number. For example 3,142p = and the population of the UK
is approximately 55.000.000. You should use the UK notation in the exam.
● Do not use accuracy that is not valid, for example quoting the price of something as £2.78643 is not appropriate.
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2 Use of a calculator
For a description of the functions you will need to have on your calculator, refer to the syllabus objectives for this course. In the exams you are not allowed graphical calculators. You will receive further guidance from the Institute and Faculty as to which types are permissible. You must be familiar with the calculator that you are going to use in the exam. Since there are so many calculators on the market, this section is not going to be about how to use your particular calculator but instead it will enable you to practise getting the correct answer with your calculator. For technical help see your manual (if you can find it!). It is really important for you to try each example below for yourself to get used to working with your calculator. Most numerical answers have been rounded to three decimal places. Always ensure that you copy numbers correctly when you are half way through a calculation or reading numbers from actuarial tables.
Example
Calculate 1.4 3.66.23 5.8+ . Solution
Using the power key ( or y xx y ), we get 573.1474.
Example
Calculate ( )5.141 1.723+ .
Solution
Using the root key ( x or 1
yx ) and the power key, we get 49.091.
If you don’t have this key, you can treat the expression as 14 5.1(1 1.723 )+ . Since
1xx y y= you can use your or y xx y key.
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Example
Calculate 2
21
3
.
Solution
The answer is 7
29
or 2.778. If your calculator has a fractions facility, be aware of the
order in which your calculator performs fractions!
Example
Calculate ( )247 ln 5.2+ .
Solution The answer is 4.929. Your calculator will have loge or ln for the natural logarithm
key. (We will deal with this function later in the course.) Most calculators have a
fraction key ( bca ) which can be helpful. Notice also that most calculators have a
squared key ( 2x ) to avoid you having to use the power key.
Example
Calculate 3.1243 e .
Solution
The answer is 1.961. On many calculators, xe and ln are on the same key. On recent
calculators, to get 3.1e , you would need to type 3.1xe (followed by = or Ans), whereas on older models you would need to type these in reverse. This is the same as for the square root key.
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Example Calculate 4! This is read as “four factorial”. Solution There should be a factorial key ( !n ) on your calculator. The answer is 24. Notice that !n means ( 1) ( 2) 1n n n¥ - ¥ - ¥◊◊◊¥ . So here 4! 4 3 2 1= ¥ ¥ ¥ .
Example
Calculate tanh 2 and 1tanh 0.4- . Solution These are hyperbolic tangents and inverse hyperbolic tangents respectively, and are needed for the Fisher transformation in Subject CT3. If your calculator is a recent model then you will need to press hyp then tan before the number; otherwise it will be
the number then hyp then tan. The answers are 1tanh 2 0.964, tanh 0.4 0.424-= = .
Example
Calculate 215.2 3.74 4 1.68 2.49
2.7 19.86 3
- - ¥±+
.
Solution The answers are 2.493 and –1.477. Notice here that you can use brackets (or stack) for the first part of the expression and the memory function for the second, or alternatively the memory for both parts.
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Question 2.3
Calculate 3.5 121 (1.04)
0.04 1.04
--.
Question 2.4
Calculate 1 1 0.1
ln2 1 0.1
.
Question 2.5
Calculate 2 4.5
3
2 (3.789 2.5)
5.5 2.1
+ +-
.
Question 2.6
Calculate 23 ( 3) 4 2 4
2 2
± - - ¥ ¥ -¥
.
Question 2.7
Calculate 10
10
117 1
100 1.035ln1.035(1.035)
.
Question 2.8
Calculate 10!
7!2!.
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When substituting a value for a variable that occurs several times within an expression, it can be helpful to use a shortcut called “nested multiplication”. The following example shows what this shortcut involves.
Example
Calculate 2 33 7v v v+ + , for 1
0.92591.08
v
.
Solution We can do this as a nested multiplication:
2 3 23 7 (3 7 )
[1 (3 7 )]
v v v v v v
v v v
+ + = + +
= + +
Start by multiplying v by 7, then add 3, then multiply by v, then add 1, and finally multiply by v. Using this version we get an answer of 9.055.
Question 2.9
Calculate 2 3 42 5 6v v v v+ + + , when 1
0.90911.1
v
.
Question 2.10
For Question 2.9, work out the number of keystrokes you need to make on your calculator using nested multiplication and also for an alternative method that you can think of.
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3 Estimation and accuracy
3.1 Estimation
We have looked at how to use calculators efficiently, but it is always possible to make a mistake. So it is important to have a rough idea of what the numerical answer to a calculation is likely to be. To do this, you can round the numbers involved to a convenient figure and then carry out the calculation without using a calculator (or you might make the same mistake again!).
Example
Estimate the value of 2 42.7 3.1
5.2 7.8
.
Solution
2 42.7 3.1
5.2 7.8
is roughly 2 43 3 9 81
305 8 3
.
The actual answer is –38.3.
Question 2.11
Find estimates for the answers to Question 2.5 and Question 2.6. Compare your estimates with the actual answers.
Note: You need to be very accurate with intermediate values in estimates and rounding when you are doing any of the following: ● subtracting two numbers of a similar size
● dividing by a small number
● raising a number to a high power.
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3.2 Accuracy
When numbers have been rounded, and are then subsequently used in a calculation, the final answer is affected by the rounding. Therefore, when answering questions, it is essential to realise how accurately you need to quote your final answer. For example, if all figures in the question are given to three significant figures, do not give your final answer to five significant figures.
Example
In the formula 101 (1 )i
pi
, the accurate value of i is 0.034724. Compare the
values of p obtained using: (i) the accurate value of i (ii) i rounded to three significant figures (iii) i rounded to one significant figure. Solution (i) 8.32819 (ii) 8.32920 (iii) 8.53020 Rounding to three significant figures gives a value of p that is still accurate to two significant figures. However rounding to one significant figure results in losing all accuracy.
Question 2.12
Using the same formula as in the example above (which is one that you will meet in Subject CT1) but with the accurate value of i being 0.04562378, how many significant figures can you round i to, in order to maintain three significant figures of accuracy in your final answer?
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4 Convenient abbreviations
Rather than writing several zeros when dealing with large numbers, it is often more convenient to work in, say thousands or millions. It is common to write £000s to mean thousands of pounds, and £m to mean millions of pounds.
Example
If 8
9 (1 )R vT Pv
i
, where
1£14,000, £700, , 0.05
1P R v i
i
, what is T?
(Work in £000s). Solution Working in £000s, 14, 0.7P R . Then 13.549T ie £13,549.
Example
If 4,000A , calculate 2 2,000 1,000,000A A
(i) directly (ii) working in units of 1,000 Solution
(i) 25,000,000 5,000
(ii) In units of 1,000:
1,000,000 becomes 21 1 , 2,000 becomes 2 and A becomes 4 The calculation is then done as follows:
24 2 4 1 25 5 ie the “real” answer is 5,000.
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This page has been left blank so that you can keep the chapter summaries together for revision purposes.
FAC-02: Numerical methods I Page 13
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Chapter 2 Summary You need to round off your final answers to a sensible degree of accuracy, but intermediate figures should not be rounded if that will mean an inaccurate final answer. You should be able to use your calculator efficiently and correctly. Remember to make a rough estimate of what the answer should be, in order to pick up careless numerical errors.
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This page has been left blank so that you can keep the chapter summaries together for revision purposes.
FAC-02: Numerical methods I Page 15
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Chapter 2 Solutions Solution 2.1
The rounded numbers are:
0.068, 15.349, 10.000 Solution 2.2
The rounded numbers are:
14.4, 5.99, 0.0801 Solution 2.3
The numerical answer is 0.295729. Solution 2.4
The numerical answer is 0.100. Solution 2.5
The numerical answer is –153.614. Solution 2.6
The numerical answer is 2.351 or –0.851. Solution 2.7
The numerical answer is 214.734. Solution 2.8
The numerical answer is 360.
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Solution 2.9
Using the nested multiplication method, we get:
2 3 4 2 3
2
2 5 6 2 (5 6 )
[2 (5 6 )]
(1 [2 (5 6 )])
v v v v v v v v
v v v v
v v v v
Now using 1
1.1v , we get the answer to be 10.42.
Solution 2.10
On the calculator I’m using, nested multiplication took 30 key presses. Using another method (just working from left to right) I took 43 key presses, or using the memory I took 25 key presses. Nested multiplication can be more efficient. Solution 2.11
For Question 2.5 one estimate is 2 4 2
3
2 (4 3) 2 19 2 400201
2 26 2
.
For Question 2.6 one estimate is 3 9 32 3 6
2.25 or 0.754 4
.
In each case you can see that the estimate is fairly close to the accurate answer.
FAC-02: Numerical methods I Page 17
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Solution 2.12
Using the accurate value of i we get p to be 7.888505805. Rounding i to 6SF we get p to be 7.888505. Rounding i to 5SF we get p to be 7.888497. Rounding i to 4SF we get p to be 7.888652. Rounding i to 3SF we get p to be 7.889427. Rounding i to 2SF we get p to be 7.873956. So i can be rounded off to 3SF to keep three significant figures of accuracy. Notice that your first guess might have been 4SF, in which case the preceding work would not be necessary.
© IFE: 2013 Examinations The Actuarial Education Company
All study material produced by ActEd is copyright and issold for the exclusive use of the purchaser. The copyright
is owned by Institute and Faculty Education Limited, asubsidiary of the Institute and Faculty of Actuaries.
You may not hire out, lend, give out, sell, store or transmitelectronically or photocopy any part of the study material.
You must take care of your study material to ensure that itis not used or copied by anybody else.
Legal action will be taken if these terms are infringed. Inaddition, we may seek to take disciplinary action through
the profession or through your employer.
These conditions remain in force after you have finishedusing the course.
FAC-03: Mathematical constants and standard functions Page 1
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Chapter 3
Mathematical constants and standard functions
You need to study this chapter to cover:
● the definitions and basic properties of the functions nx , xc , exp( )x , and ln x
● the functions x , max( ) and min( )
● the factorial function !n and the gamma function.
0 Introduction
This chapter covers some standard functions and notation that will be needed in the following chapters.
Page 2 FAC-03: Mathematical constants and standard functions
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1 Standard functions and graphs
1.1 Exponential function
The exponential function, xe , can be defined by:
lim 1n
x
n
xe
n
or the series expansion 2 3
12! 3!
x x xe x .
It is the inverse of the natural logarithmic function (which we will look at in Chapter 4),
so that ln xe x . Do remember that e is just a number which you can find from your calculators:
1 2.718...e
If the power is a long expression then a convenient alternative notation is exp( )xe x ,
for example 21
22
12
expx x
e
. This makes things clearer to read.
However don’t mix up these two notations by writing something like expx , as this is
meaningless.
The graph of xy e looks like this:
y=exp(x)
0
2
4
6
8
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
x
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Notice that xe can never be negative. This is because you can never get a negative answer if you raise a positive number to any power. Notice also that:
● 0e is 1
● as x gets large, so does xe
● and as x gets large but negative, 0xe .
Since xe is just “a number to the power of x”, then this graph is also the basic shape of
the graph of xy c , where c is a positive constant greater than 1. When c is a positive
constant less than 1, its graph is a reflection in the y-axis of the graph of xy c , with c
replaced by 1c . The diagram below shows the graphs of 0.5xy and 2xy :
0.5xy 2xy
Again notice that both graphs have an intercept of 1, ie they cross the y-axis at 1, but you must look at the value of c carefully to judge when the graph tends towards infinity or zero.
0
1
2
3
4
5
-3 -2 -1 0 1 2 3
x
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1.2 Log function
A logarithm is the inverse of a power. The graph of the natural logarithm function lny x is:
-6
-5
-4
-3
-2
-1
0
1
2
3
0 1 2 3 4 5 6 7
We will meet the idea of a base in the next chapter, but it is worth noting that this is the shape of the logarithmic graph whatever base is used. We will look at logarithms and bases in a later chapter. Notice that log1 0 , and that you can’t take the log of a negative number (or zero).
The limits are log as x x , and log as 0x x . These will be important
in statistical work in later subjects. You will see the notation log x and ln x used for
the natural logarithm function.
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1.3 Powers of x
The graph of ny x , where n is an even positive integer has the same general shape as 4y x which is shown below.
-202468
1012141618
-2 -1 0 1 2
x
y
Notice that the values of y for these functions can never be less than zero.
The graph of ny x , where n is an odd positive integer has the same general shape as 3y x which is shown below:
-10-8-6-4-202468
10
-3 -2 -1 0 1 2 3
x
y
You may notice that out of the last two graphs, one was symmetrical about the y-axis and the other one wasn’t. In general a function ( )f x which has the property
( ) ( )f x f x is called an even function. If ( ) ( )f x f x then ( )f x is called an odd
function.
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Remembering that 1nn
xx
, the graph of ny x will have a discontinuity (where the
function is not defined) at 0x . For example, look at the graph of 3y x :
-300
-200
-100
0
100
200
300
-2 -1 0 1 2
x
y
The graph of y x is (remembering that x represents the positive square root):
0
0.5
1
1.5
2
2.5
0 1 2 3 4 5
x
y
This is the inverse function of 2y x , so the graph of y x is the reflection of
2y x in the line y x . The same relationship applies to the graphs of log x and xe
which we have already looked at. Notice that this graph does not exist for negative values of x since a real value of the square root of a negative number cannot be found.
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When you see an expression involving a power, such as zt , you need to think carefully whether:
● it is like nx , ie t is a variable and z is a fixed number, or
● it is like xc , ie t is a fixed number and z is a variable.
1.4 Transformations
These standard graphs can be transformed. For example the graph of ( ) 1y f x is
found by translating (sliding!) the graph of ( )y f x up by one unit and the graph of
( 1)y f x is found by translating the graph of ( )y f x to the left by one unit.
There is also great similarity between the graphs of ( )y f x and ( )y af x or
( )y f x and ( )y f ax .
The following question illustrates these points.
Question 3.1
Sketch the graphs of the following functions:
(i) 1y x
(ii) 53y x
(iii) 4 1xy e
(iv) ln(1 )y x
(v) 4( 3)y x
Question 3.2
Describe what the graphs of 2 ( )f x and (2 )f x look like in relation to the graph of
( )f x .
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2 Other functions
2.1 Modulus function
x is the absolute value of x or the modulus of x. In practice, if x is a number, it means
ignore any negative sign, for example, 4 | 4 | 4 . Note that | |x c is the same as
saying c x c because x must either be a positive number in the range [0, )c or a
negative number in the range ( ,0)c .
Example
Write the expression 2 1 3x without the use of the modulus sign.
Solution
2 1 3 3 2 1 3x x
For those of you that can remember dealing with inequalities, we will further simplify this expression. For those that can’t, we will pick up on this again in the next chapter. Simplifying the inequality:
3 2 1 3 2 2 4
1 2
x x
x
Question 3.3
Write the expression 3 2x without the modulus sign.
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2.2 Integer part
x means the integer part of x, for example 2.89 is 2. This could be used, amongst
other contexts, to give the complete number of years that someone has lived. It is also used a lot in computer calculations.
When you divide m by n, where both are positive integers, you get m
n
as the quotient
with a remainder of m
m nn
.
When dealing with negative numbers you need to check whether x really means the
integer part, so that [ ] 3 , or whether it means the greatest integer not exceeding x,
so that [ ] 4 . The second definition is more common.
Example A baby boy was born on 26 November 1979. If x is defined to be his exact age, what is
x on 11 February 2012?
Solution
He is 32 on 26 November 2011, so x is 32.
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2.3 Max and min
The notation max( ) or min( ) is used to denote the largest or smallest of a set of
values. If the quantities involved are variables, the value of these functions may depend on the ranges of values that you are considering.
Example What is max( 2,10)x for the region 0 20x ?
Solution We could write max( 2,10)x as:
10 if 0 8
2 if 8 20
x
x x
So max( 2,10) 10 for 0 8x x , and max( 2,10) 2 for 8 20x x x .
The abbreviation ( 100)x is often used for max( 100,0)x .
Question 3.4
What is 2min( ,15)x for 0 6x ?
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Question 3.5
In the UK the calculation of an employee’s National Insurance contributions is based on their Upper Band Earnings (UBE), which is calculated from the formula: min( , ) min( , )UBE S UEL S LEL
In this formula, S is annual salary, and UEL and LEL are two published figures called the Upper and Lower Earnings Limits (which might for example be £30K and £4K). Describe in words what UBE represents and give an alternative mathematical formula in the form:
if
if
if
UBE
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3 Factorial and gamma functions
3.1 Factorial notation
We have already looked at how to evaluate !n on a calculator. The definition of !n is as follows: ! ( 1) ( 2) 1n n n n
where n is a non-negative integer. The factorial function satisfies the relationship ! ( 1)!n n n . If we put 1n , this
tells us that 1! 1 0! 0! 1 .
Question 3.6
Evaluate 5!
Example
Simplify the expression !(2 1)!
(2 )!( 2)!
n n
n n
.
Solution This expression can be written as:
! (2 1)!
( 2)! (2 )!
n n
n n
The first factor equals ( 1)n n , because !n contains an extra n and 1n in its
expansion, which are not contained in the expansion of ( 2)!n , and the second factor is
1
2n.
So we get 12
!(2 1)! ( 1)( 1)
(2 )!( 2)! 2
n n n nn
n n n
.
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Question 3.7
Simplify the expression (3 )!(2 1)!
(3 2)!(2 1)!
n n
n n
.
3.2 Gamma function
( )x , where 0x , is defined by the integral 10
x tt e dt . This is the gamma
function. This function is used in statistics in connection with the gamma distribution, and there are several properties of the function that you are going to need to know. In this chapter we will quote the results without proof. We will look at their proofs later in the course after we have covered the necessary integration techniques. Result 1: ( ) ( 1) ( 1)x x x where 1x
Result 2: ( ) ( 1)!n n where 1, 2, 3,...n
Result 3: 12
( )
These results can be found in the Tables page 5. The graph of the gamma function looks like this:
1 2 3 4
1
2
3
4
5
6
7
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Example Evaluate (6.5) .
Solution Using Result 1:
(6.5) 5.5 (5.5)
By repeating this process:
(6.5) 5.5 4.5 3.5 2.5 1.5 0.5 (0.5)
162.42
287.89
Question 3.8
Evaluate (5) .
Question 3.9
Evaluate (4.5) .
Question 3.10
Simplify ( 1)(2 1)!
(2 )( 1)!
n n
n n
where n is an integer.
Question 3.11
Show that, if n is a nonnegative whole number, 2
(2 )!( ½)
2 !n
nn
n .
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Stirling’s approximation Working out a factorial by multiplying all the numbers together would be very tedious for large values of n . The following approximations (called Stirling’s approximation) are sometimes useful in derivations involving large values of n :
½! 2n nn n e
and ½( ) 2n nn n e
(Don’t worry that these appear to be inconsistent with the relationship ( ) ( 1)!n n
Remember that for very large values of n , the ratio 1n
n
is very close to 1.)
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This page has been left blank so that you can keep the chapter summaries together for revision purposes.
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Chapter 3 Summary You should know the properties and be able to sketch the graphs of
, , , and logn x xx c e x .
The modulus of x, written as x , is the absolute value of x ie negative signs are ignored.
The integer part of x is written as x .
max( ) or min( ) is used to denote the largest or smallest of a set of values.
The definition of n factorial is ! ( 1) ( 2) 1n n n n .
0! 1
The gamma function is defined by 10
( ) x tx t e dt , when 0x .
It has the following properties: Result 1: ( ) ( 1) ( 1)x x x where 1x
Result 2: ( ) ( 1)!n n where 1, 2, 3,...n
Result 3: 12
( )
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This page has been left blank so that you can keep the chapter summaries together for revision purposes.
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Chapter 3 Solutions Solution 3.1
(i) The graph of 1y x is as follows:
-10
-5
0
5
10
-1.5 -1 -0.5 0 0.5 1 1.5
x
y
(ii) The graph of 53y x is as follows:
-100
-50
0
50
100
-2 -1 0 1 2
x
y
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(iii) The graph of 4xe is like that of xe , only steeper. In fact the whole graph is squeezed horizontally to a quarter of the original width. Adding 1 then shifts everything up by 1 unit.
0123456789
-1.5 -1 -0.5 0 0.5
x
y
(iv) This is like the graph of ln x shifted 1 unit to the left
-2.5
-2
-1.5
-1
-0.50
0.5
1
1.5
2
-1 -0.5 0 0.5 1 1.5
x
y
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(v) This is like the graph of 4x , but shifted 3 units to the right.
024
68
10121416
0 1 2 3 4 5
x
y
Solution 3.2
The graph of 2 ( )f x will have all vertical distances doubled.
The graph of (2 )f x will have all horizontal distances halved.
Solution 3.3
3 2x or 3 2x
If you wish to simplify these expressions further, proceed as follows:
22 3
3x x
or 2
2 33
x x
Here there are two separate ranges of values of x where the inequality holds.
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Solution 3.4
2 2min( ,15)x x for 0 3.87x
2min( ,15) 15x for 3.87 6x
Solution 3.5
UBE is the portion of the employee’s salary that falls in the “band” ( , )LEL UEL , which
would be (£4,000,£30,000) using the illustrative figures given.
The graphs in the diagram represent the two min functions. UBE is the difference between them, which corresponds to the height of the gap between the upper and lower lines.
LEL UEL
LEL
UEL
The alternative mathematical formula is:
0 if
if
if
S LEL
UBE S LEL LEL S UEL
UEL LEL S UEL
Solution 3.6
5! 5 4 3 2 1 120
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Solution 3.7
(3 )!(2 1)! (3 )! (2 1)! 1 2 (2 1)2 (2 1)
(3 2)!(2 1)! (3 2)! (2 1)! (3 1)(3 2) (3 1)(3 2)
n n n n n nn n
n n n n n n n n
Solution 3.8
(5) 4! 24
Solution 3.9
(4.5) 3.5 (3.5) 3.5 2.5 1.5 0.5 (0.5) 6.5625 11.63
Solution 3.10
( 1)(2 1)! ( 2)!(2 1)! 1
(2 )( 1)! (2 1)!( 1)! ( 1)( 1)
n n n n
n n n n n n n
or
3
1
n n
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Solution 3.11
We can prove this using mathematical induction.
When 0n , the equation says 0
0!(½)
2 0! , which is correct.
If we assume it’s true for a typical value of n , say n k , then we know that:
2
(2 )!( ½)
2 !k
kk
k
This would imply that:
2
(2 )!( 1½) ( ½) ( ½) ( ½)
2 !k
kk k k k
k
The last expression can be written as:
2 2 2 1
(2 )! 2 1 (2 )! (2 1)!( ½)
22 ! 2 ! 2 !k k k
k k k kk
k k k
This doesn’t quite match the formula we were hoping for. But, if we include an extra
factor of 2 2
2( 1)
k
k
(which won’t affect the answer), we get:
2 1 2 2
2 2 (2 1)! (2 2)!
2( 1) 2 ! 2 ( 1)!k k
k k k
k k k
This now matches the formula given when 1n k . So, if it is true for n k , it’s also true for 1n k , and by the principle of mathematical induction it must be true for all values of n .
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Chapter 4
Algebra
You need to study this chapter to cover: ● manipulating algebraic expressions involving powers, logs, polynomials and
fractions
● solving quadratic equations
● solving simultaneous equations
● solving inequalities
● the arithmetic-geometric mean inequality
● using the and notation for sums and products
● arithmetic and geometric progressions and other series
● the binomial expansion of expressions of the form ( )a b n where n is a positive
integer, and ( )1 x p for any real value of p .
0 Introduction
This chapter reminds you of the algebraic results that you need to be able to handle very easily in order to cope with the Core Technical subjects.
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1 Algebraic expressions
1.1 Indices
There are three laws governing indices (or powers):
( )
a b a b
a b a b
a b ab
x x x
x x x
x x
+
-
¥ =
=
=
Question 4.1
Simplify the following:
(i) 5 23 5x x
(ii) 16 62 7y y
(iii) ( )5 3 2b
1.2 Logarithms
A logarithm is the inverse of a power:
10 100 100 2210 log (read as log base 10 of 100)
So if we want to find a log, we need to answer the question: “To what power must we raise the base in order to get the number whose log we are trying to find?” They have bases (the 10 here), the most commonly used one being e. Logarithms to these bases can be calculated directly from a scientific calculator, eg log .e 15 2 708 .
An alternative abbreviation for log to base e is ln. On your calculator, the ln button gives you loge and the log button gives you log10 .
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There are three laws of logarithms that can be derived from the laws of indices:
log log log
log log log
log log
a a a
a a a
na a
x y xy
xx y
y
x n x
+ =
- =
=
In words, “the log of a product is the sum of the logs”, and “you can bring down the power”.
Question 4.2
Starting from the laws of indices, prove the second of these three laws.
Notice that the base of the logarithm has been missed out in the following examples. Here we are looking at the properties of logs in general. So the logarithm used can be to any base, but if you need to use a calculator then you will have to use base 10 or base e.
Example (i) Simplify log log log2 4 5x x x .
(ii) Write 101 2log x in the form log ( )f x .
Solution
(i) This simplifies to log log2 4
5
8
5
x x
x
x .
(ii) 2 210 10 10log log 10 log 10 x x .
Question 4.3
Simplify ln e.
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Logarithms are also used to solve equations where the unknown forms part of a power.
Example
Solve the equation 11 22 1. n . Solution Take logs:
2 1log1.1 log 2
(2 1) log1.1 log 2
log 2(2 1)
log1.1
1 log 21 3.136
2 log1.1
n
n
n
n
Question 4.4
Solve the equation 2 105 110252 2 1 . .t t .
Note:
The value of log
logb
b
x
y is independent of the base b used.
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1.3 Fractions
It is possible to manipulate algebraic fractions using the same rules that are used for numerical fractions. If you have forgotten how to factorise a quadratic you might want to look at the next section now!
Example Simplify the following:
(i) 2 3
1
3 2
6 1
x
x
x
x
(ii) x x
x x
2
2
3 4
2 7 4
(iii) x x
x x
x x
x x
2
2
2
2
2
2 5 12
3 2
2 5 3
Solution (i) Putting these over a common denominator:
2 2
2
2 3 3 2 (2 3)(6 1) (3 2)( 1)
1 6 1 ( 1)(6 1)
(12 16 3) (3 5 2)
( 1)(6 1)
9 11 5
( 1)(6 1)
x x x x x x
x x x x
x x x x
x x
x x
x x
+ + + - - + +- =+ - + -
+ - - + +=+ -
+ -=+ -
(ii) Factorising the numerator and denominator:
2
2
3 4 ( 1)( 4)
(2 1)( 4)2 7 4
1
2 1
x x x x
x xx x
x
x
- - + -=+ -- -
+=+
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(iii) Remembering that when you divide by a fraction you “invert and multiply”:
2 2
2 2
2 3 2 ( 2)( 1) ( 2)( 1)
( 4)(2 3) (2 3)( 1)2 5 12 2 5 3
( 2)( 1) (2 3)( 1)
( 4)(2 3) ( 2)( 1)
( 2)( 1)
( 4)( 2)
x x x x x x x x
x x x xx x x x
x x x x
x x x x
x x
x x
- - + + - + + +∏ = ∏- + + +- - + +
- + + += ¥- + + +
- +=- +
Example
Simplify
1 1
a ba
b
b
a
.
Solution Multiplying the numerator and denominator of a fraction by the same quantity leaves the value of the fraction unaffected. We can pick a suitable quantity to simplify the fraction:
1 1
2 2a ba
b
b
a
ab
ab
b a
a b
Question 4.5
Simplify the expression x x
x x
x x
x x
2
2
2
2
3 4
3 10 8
2 5 7
4 17 4
.
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2 Quadratic equations
2.1 Solution by factorisation
This is where you try to write the quadratic ax bx c2 0 as ( )( )dx e fx g 0
(the process being called factorisation) and then the solutions would be
x ed
gf or . Notice that by comparing the coefficients of x x2 , and the constant
term, a df , c eg , and b ef dg .
Example
Solve the equation 2 3 02x x by factorisation. Solution
22 3 0 (2 3)( 1) 0
1.5 or 1
x x x x
x
+ - = fi + - =
fi = -
Question 4.6
Solve the following equations by factorisation:
(i) 6 2 02x x
(ii) 16 1 02x
Some useful results:
a b a b a b2 2 ( )( )
a b a b a ab b3 3 2 2 ( )( )
Notice that in the last result to get the correct signs you need to read the top sign on and (or alternatively the bottom sign).
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2.2 Solution by completing the square
All quadratic equations can be written as ( )px q r 2 0 , where , ,p q r Œ . If the
coefficient of x2 is negative, you would have to multiply through by –1 first. If r 0 ,
then this method enables the quadratic to be solved (in the example here, xq r
p
).
Example Solve the following equations by completing the square:
(i) 4 8 1 02x x
(ii) 3 13 1 02x x Solution (i) There are several methods available to complete the square, for example you can
multiply out the expression ( )px q r 2 , giving p x pqx q r2 2 22 , and
compare coefficients to solve the equations:
( ) ( )
2 2
2
1 12 2
4 8 1 0 (2 2) 3 0
(2 2) 3
2 2 3
2 3 or 2 3 1.866 or 0.134
x x x
x
x
x
- + = fi - - =
fi - =
fi - = ±
fi = + - =
(ii) You need to multiply through by –1 first:
( )( )
2 2 13 13 3
213 169 16 36 3
213 1576 36
13 1576 36
3 13 1 0 3 0
3 ( ) 0
( )
4.255 or 0.078
x x x x
x
x
x
- + = fi - + =
fi - - + =
fi - =
fi = ± =
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Question 4.7
Solve the following equations by completing the square:
(i) x x2 4 8 0
(ii) 16 17 2 02x x
Question 4.8
If 2 2 2
1 2
1 2
x x x
, for all values of x, express and , in
terms of , , , , ,1 2 1 2 and .
2.3 Solution by formula
For the general quadratic equation, ax bx c2 0 the solutions are given by
xb b ac
a 2 4
2. This can be proved by applying the method of completing the
square to the general quadratic equation.
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Example
Solve the equation 3 8 2 02x x , using the quadratic formula. Solution If the equation is compared with the general quadratic, then: a b c 3 8 2, , so:
( ) ( )1 16 6
8 64 24
6
8 40 or 8 40
0.279 or 2.387
x- ± -=
= - + - -
= - -
Question 4.9
Solve the equation 1 4 7 02 x x .
Question 4.10
Solve the equation 3 4 3 4 02 1x x .
Whenever solving equations, especially quadratics, you should check that the answer that you give is reasonable. For example, you might be working out the number of people or another value that has to be positive, so state that a negative answer is impossible. There may also be situations when the value you require lies between particular limits, for example a probability must lie between 0 and 1. Under these circumstances reject any values which lie outside the limits, stating why you have rejected them.
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3 Simultaneous equations
When solving simultaneous equations it is important to use the most efficient way of solving them, in order to make best use of your time. Generally with two linear equations, rearrange them to a suitable form and then add or subtract to eliminate the variables. For non-linear equations, consider using substitution first, otherwise use cancelling or factorising.
Example (Linear equations solved by subtraction) Solve the simultaneous equations:
2 5 4 0
7 2 4
x y
y x
+ - =
= -
Solution Looking at the two equations in turn:
2 5 4 0 2 5 4x y x y (1)
7 2 4 4 7 2y x x y (2)
( ) ( )2 2 1 (or equation (2) take away 2 times equation (1)):
3 6 2y y
x 3
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Example (Non-linear equations solved by substitution) Solve the simultaneous equations:
x y2 2 20 2 3 14x y
Solution The second equation can be rearranged to give:
x y 12 14 3( )
Substituting this into the first equation gives:
2 214
2 2
2
(14 3 ) 20
196 84 9 4 80
13 84 116 0
y y
y y y
y y
+ + =
fi + + + =
fi + + =
Using the quadratic formula or otherwise:
5813
413
2 or
4 or
y
x
fi = - -
fi =
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Example (Non-linear equations solved by division) Solve the simultaneous equations:
2
3 4
8
30
ar
ar ar
=
+ =
Solution Dividing the second equation by the first equation:
r r 2 3 75. Solving this as a quadratic: r 15 2 5. . or a 32
93225 or
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Example (Non-linear equations solved by factorisation) Solve the simultaneous equations:
2 2
2 2
2 3 50
3 4 35
p pq q
p pq q
- - = -
+ + =
Solution Factorising both equations:
( )(2 3 ) 50
( )(3 ) 35
p q p q
p q p q
+ - = -
+ + =
Dividing:
3 7
2 3 10
10(3 ) 7(2 3 )
4
p q
p q
p q p q
p q
+ -=-
fi + = - -
fi =
Substituting:
2 2 23 16 16 35
1 or 1
4 or 4
p p p
p
q
+ + =
fi = -
fi = -
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Question 4.11
Solve the simultaneous equations:
(i) 2 4
4 6
5
20
ab ab
ab ab
+ =
+ =
(ii) 2 2
2 2
2 5 3 5 0
2 5 0
x xy y
x xy y
+ - + =
+ - - =
Example
If 2
1173
. , 2
9 3 . and k 41. , find , and k .
Solution Dividing the second equation by the first, we get:
9 3
117 21590
.
..
Substituting into the second equation gives us:
9 3 1590 23502. . . Finally substituting both of these into the final equation gives us: k 10 68.
Question 4.12
If e 12
2
3 , and e e2 2 2
1 112 ( ) . , what are and ?
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4 Inequalities
Inequalities are similar to equations, but rather than finding an exact value for the variable, we find a range in which the variable can lie. In some cases solving inequalities is like solving equations whereas other examples can be more complicated.
Example Solve the inequality 2 7 7 4x x . Solution Starting with 2 7 7 4x x , if we subtract 4 from both sides, and subtract 2x from both sides, we get: 11 5x Then dividing by 5 we get:
115 x
Note: ● You can add a constant to both sides (or subtract a constant from both sides).
● You can divide or multiply both sides by a positive constant.
● You can divide or multiply both sides by a negative constant, providing you reverse the inequality sign.
● You can take logs of both sides (provided both sides are positive).
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Example For what values of x is ( )( )2 11 7 57x x ?
Solution Manipulate the inequality to get zero on one side:
2
2
(2 11)( 7) 57
2 3 77 57
2 3 20 0
(2 5)( 4) 0
x x
x x
x x
x x
+ - < -
- - < -
- - <
+ - <
The inequality can be solved by drawing a graph of the function ( ) 22 3 20f x x x= - -
and working out when the graph lies below the x-axis. However this method relies upon the fact that the graph is easy to plot (or that you have a graphical calculator!). An alternative is as follows: The expression ( )( )2 5 4x x will change sign when x 2 5 4. or , so we can
consider whether the expression is positive or negative between these points. (See table below.) So the solution is 2 5 4. x , since we want the expression to be less than 0 ie negative.
x 2 5. 2 5 4. x x 4
2 5x – + +
x 4 – – + ( )( )2 5 4x x + – +
Example Solve the inequality | |x 1 3.
Solution This is equivalent to 3 1 3x , which simplifies to 2 4x .
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Example Solve the inequality:
8 5 7
3 5 22
2
2
x x
x x
Solution Getting zero on one side and using a common denominator, we get:
2 2
2
2
2
8 5 7 2(3 5 2)0
3 5 2
2 5 3 (2 1)( 3)0 0
(3 1)( 2)3 5 2
x x x x
x x
x x x x
x xx x
+ - - + - £+ -
- - + -£ fi £- ++ -
This will change sign at x 2 312
13, , , .
By considering the sign between each of these values (see table below), the solution
includes 2 12x or 1
3 3 x . However, we need also to see what happens at
x 2 312
13, , , . Substituting these values in, we can see that our solution is
2 312
13x x or .
x 2 2 12x 1
213x 1
3 3 x x 3
x 2 – + + + + 2 1x – – + + + 3 1x – – – + + x 3 – – – – +
( )( )
( )( )
2 1 3
3 1 2
x x
x x
+ – + – +
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Question 4.13
Solve the inequality 2 5 17 203 2x x x
Note: ● For a double inequality, for example 10 2 1 20 n , you can treat it as two
separate inequalities and then combine the answers at the end.
● Be careful that your answer matches the question eg or .
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5 Arithmetic-geometric mean inequality
The arithmetic mean of n numbers, a an1, , , is defined to be a a
nn1
. The
geometric mean of n positive numbers, a an1, , , is defined to be a ann
1 .
The Arithmetic-geometric mean inequality states that “the geometric mean of a set of positive numbers is less than or equal to the arithmetic mean of the same set of numbers” ie:
a aa a
nnn n
11
Example Show that the arithmetic-geometric mean inequality holds for the numbers 7 and 9. Solution
Arithmetic mean
7 9
28
Geometric mean 7 9 63 7 94. So it does hold for these two numbers.
Question 4.14
When does the arithmetic mean of two numbers equal the geometric mean?
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6 Sums and products
Earlier in the course, when we considered proof by induction, sums were written out in full eg 1 2 3 n . This section shows how to write sums and products of numbers using the (sigma) or (pi) notation as an abbreviation.
Example Write the following using the notation: (i) 3 3 3 3 3 n
(ii) 1 2 32 2 2 2 n
(iii) 1 2 3 13 3 3 3 ( )n
(iv) 1 3 5 71
(v) 1
2
1
4
1
82 2 2
Solution
(i) 3 31i
n
n (ii) i
i
n2
1 (iii) i
i
n3
1
1
(iv) 36
1
(2 1)i
i=
-Â (v) 1
221
ii
There is always more than one way to write a series in this notation. For example, (iv)
could be written as ( )2 10
35
nj
. However, it will often be that there is a ‘clearest’
approach which you should use in order to make life easier for your reader.
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Question 4.15
Write the following using the notation: (i) 2 5 10 226
(ii) 2 4 8 22 n
(iii) 13
15
17
12 1 n
(iv) 4 9 14 99
Example Write the following using the notation: (i) 3 5 7 2 1 ( )n
(ii) e e e e n 2 3 2 3
(iii) 1
4
1
9
1
16
1
2562 3 15
Solution
(i) ( )2 11
ii
n
(ii) ei
i
n
1
2 3
(iii) 1
1 21
15
( )i ii
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Question 4.16
Write this expression using the notation:
1 11 1
1 2 3
1 1 1 1n
t t t t
Example Simplify the expression:
2
1
(1 4 )n
i
i i=
+ +Â
Solution
2 2 2
1 1 1 1 1 1
(1 4 ) 1 4 4n n n n n n
i i i i i i
i i i i n i i= = = = = =
+ + = + + = + +Â Â Â Â Â Â
This can be further simplified, but we will deal with this later.
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7 Arithmetic and geometric progressions
7.1 Formulae
An arithmetic progression (AP) is a set of numbers, such as 3 7 1115 39, , , , , , in which the difference between each two successive numbers is a constant. A geometric progression (GP) is a set of numbers, such as 3 6 12 192, , , , , in which the ratio of each term to the preceding term is a constant. An infinite geometric progression is a GP
which does not end such as 12
14
18, , , , and we will consider the convergence, or
otherwise, of such series shortly. There are formulae to work out general terms and sums of all of these series. Remember that if the terms change by the same amount each time then the series is an AP. If they change by the same ratio then it is a GP. If a first term, n number of terms, l last term, d common difference, and r common ratio, then:
Type of series General term (or n th
term) Sum to n terms ( Sn )
Arithmetic progression a n d ( )1 ( )2 2 ( 1)n
nS a n d= + -
or ( )2n
nS a l= +
Geometric progression ar n1 ( )1
1
n
n
a rS
r
-=
-
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7.2 Convergence
In the case of infinite geometric progressions, the sums can either converge to a finite limit or diverge. For example consider the following two series:
1 2 4 1 12
14
18
The first one has r 2 , and clearly does not converge to a finite limit. The second has
r 12 , and does converge to a finite limit. In other words, however many terms you
take, the sum gets closer and closer to 2 without ever quite getting there. So we say that the sum of the infinite GP is 2. In general the sum of an infinite geometric progression converges to a finite limit if
1 1r , and the finite limit is given by a
r1. This is also written as S
a
r 1
.
Example A boy gets his father (who is not a mathematician) to agree to a new system for pocket money. In the first week the father puts 1p in the first square on a chessboard. In the second week the father puts 2p in the second square. In the third week the father puts 4p in the third square, and so it continues doubling each time. How much money will the father have to put on the last square? Solution
1, 2, 4, 8, …is a geometric progression with a r 1 2, and general term of ar n1 .
A chessboard has 64 squares, so the last square will need 1 263 pence, or (to 1 SF) £90,000,000,000,000,000!
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Example A woman made a New Year’s resolution to save money. She decided to put £1 in a savings account in the first week of the year, £3 in the second, £5 in the third and so on. Assuming that there are 52 weeks in a year, what is her last payment, and how much will she have put into the account by the end of the year? Solution 1, 3, 5,… is an arithmetic progression with a d 1 2, . Last payment a n d( ) £1031 1 51 2 .
Total 522 22 ( 1) 2 51 2 £2,704n a n d .
Example Write 1.231231231… as a fraction. Solution
1.231231231... = 1+ 2311,000
2311,000,000
2311,000,000,000
which is 1 plus an infinite geometric progression, which converges since the common
ratio is 11000 .
1.231231231... = 1+1
2311000
11000
1 1 1231999
231999
77333
Question 4.17
A mathematics student borrows some money from his parents (interest free) and agrees to pay it back as follows: the first monthly payment is £298, payments decrease by £28 per month, and the last payment is £18. How many payments were made? How much money did the student borrow?
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Question 4.18
If a x x 1 2 4 2 , where 1 2 1 x , b y y 1 3 9 2 , where 1 3 1 y ,
and 3 2 1y x , prove that ab a b .
Note the techniques used in the following example.
Question 4.19
The sum of the first n terms of a sequence of numbers (which is not an AP or a GP) is given by the formula:
S n n nn 16
4 5( )( ) ( n 1 2 3, , ,)
Find: (i) the sum of the first 10 terms (ii) the sum of the 11th to the 20th terms (inclusive) (iii) the 20th term (iv) a formula for un , the n th term
(v) the smallest term in the sequence.
Question 4.20
If n n 14 1
2 ( n 1 2 3, , ,) and 0 0 , find an explicit formula for n in terms
of n and 2 only.
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8 Further series work
8.1 Standard summations
There are two formulae for series that we proved as part of the work on proof by induction.
k n nk
n
1
21
1( ) and k n n nk
n2 1
61
1 2 1 ( )( )
These can be used to simplify more complicated expressions.
Example
Simplify the expression ( )1 4 2
1
i ii
n
.
Solution
( )
( )( )
2 2
1 1 1 1
2
1 1
4 12 6
16
216
216
(1 4 ) 1 4
4
( 1) ( 1)(2 1)
6 12( 1) ( 1)(2 1)
6 12 12 2 3 1
2 15 19
n n n n
i i i i
n n
i i
i i i i
n i i
n n n n n n
n n n n
n n n n
n n n
= = = =
= =
+ + = + +
= + +
= + + + + +
= + + + + +
= + + + + +
= + +
   Â
 Â
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Example
Sum the series 1 2 3 2 12 2 2 2 ( )n .
Solution
( )( )
2 12 2 2 2 2
1
16
16
13
1 2 3 (2 1)
(2 1) (2 1) 1 2(2 1) 1
(2 1)(2 2)(4 3)
(2 1)( 1)(4 3)
n
k
n k
n n n
n n n
n n n
+
=+ + + ◊◊◊ + + =
= + + + + +
= + + +
= + + +
Â
8.2 Swapping the order of summation
When we have summation over two variables, we are able to swap the order of the summation. This may make the actual summation easier.
Example
(i) Calculate 5
0 0
2y
y x
y x= =Â Â , where 0 5£ £ £x y .
(ii) Reverse the order of the summation and re-calculate the value.
You are given that 3 2 214
1
( 1)n
k
k n n=
= +Â .
Solution (i) Using the formula detailed in the last section:
5 5 5
3 2
0 0 0 0
12 2 ( 1) ( )
2
y
y x y y
y x y y y y y
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This can be simplified using the results for the sum of squares and the sum of cubes:
5
3 2 2 2
0
1 1( ) (5) (6) (5)(6)(11) 280
4 6y
y y=
+ = + =Â
(ii) To change the order of the summation, we consider the inequality 0 5£ £ £x y .
Previously we summed from 0x = to x y= , then from 0y = to 5y = .
Reversing this, we sum from y x= to 5y = then from 0x = to 5x = . To get
the limits for y , see what it is bounded by in the inequality 0 5£ £ £x y . So:
5 5 5
0 0 0
2 2y
y x x y x
y x x y= = = =
=Â Â Â Â
Using the formula detailed in the last section:
5 5 5 5 1 5
0 0 0 0 0
1 12 2 2 (5)(6) ( 1)( )
2 2
x
x y x x y y x
x y x y y x x x
Simplifying this:
5 5 5 5 5 5
3 2
0 0 0 0 0
2 (30 ( 1)) 30x y x x x x x
x y x x x x x x= = = = = =
= - - = - +Â Â Â Â Â Â
Again, using the standard summations:
5 5
2 2
0
1 1 12 30 (5)(6) (5) (6) (5)(6)(11) 280
2 4 6x y x
x y
Note that it is easier in this case to use the first version of the summation.
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Question 4.21
(i) If 0 10£ £ £t s , calculate 10 10
0
1.04 1t
t s t
-
= =Â Â .
(ii) Swap the order of summation to re-calculate the answer.
You may use the result 1
0
1 1 1.04
0.041.04 1 1.04 1.04
bb
t bt
t b
.
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9 Binomial expansions
The binomial theorem states that: If n is a positive integer, then, for all values of a and b:
1 2 2 1( )0 1 2 1
n n n n n r r n nn n n n n na b a a b a b a b ab b
r n n
Note that !
!( )!
n n
r r n r
, which appears as nrC on most calculators. The
n
r
expression comes in because there are many different ways of getting the term in n r ra b .
This can be used to expand expressions of the form ( )na b .
9.1 Positive powers
An alternative version of this states that ( )na b+ can be expanded as:
1 2 2 1( 1) ( 1)( 2) ( 1)
2! !n n n n r r n nn n n n n n r
a na b a b a b nab br
- - - -- - - ◊◊◊ - ++ + + ◊◊◊ + + ◊◊◊ + +
Notice that the sum of the powers of a and b is the power on the bracket being expanded, and that the coefficients of the terms can be obtained from Pascal’s triangle, which is:
1 1 1 1 2 1 1 3 3 1
1 4 6 4 1 Where the numbers are found by adding the two numbers in the row above it. For
example, the row 1, 4, 6, 4, 1 gives the coefficients for the expansion of 4( )a b+ , ie:
1, 4, 4 3
2!
¥,
4 3 2
3!
¥ ¥,
4 3 2 1
4!
¥ ¥ ¥.
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Example
Expand ( )2 3 7 x as far as the term in x3 .
Solution
7 7 6 1 5 2 4 3
2 3
7 6 7 6 5(2 3 ) 2 7 2 ( 3 ) 2 ( 3 ) 2 ( 3 )
2! 3!
128 1,344 6,048 15,120
x x x x
x x x
¥ ¥ ¥- = + ¥ ¥ - + ¥ ¥ - + ¥ ¥ - + ◊◊ ◊
= - + - +
Alternatively you could get the coefficients by looking at the row of Pascal’s triangle which begins 1, 7, … However you will still have to multiply in the appropriate powers of 2 and –3 to each term.
Example
Find the coefficient of the term in a4 in the expansion of ( )2 5 6a b .
Solution
The general term is 6 62 5C a brr r( ) ( ) , and we want a4 , so here r 2 .
The coefficient is 62
4 2 2 22 5 15 16 25 6000C b b b ( ) .
Question 4.22
Expand ( )1 2 8 x as far as the term in x4 .
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9.2 Fractional or negative powers
Very often in practice, the power on the bracket will not be a positive integer, so alternatively we can use:
( )( )
!
( )( )
!1 1
1
2
1 2
32 3
x nx
n nx
n n nxn
where n is negative or fractional, and 1 1x . Note that the first term in this type of series has to be 1.
Example
Expand ( )1 2 x as far as the term in x3 . For what values of x is this expansion
valid? Solution
12
31 1 1 12 32 2 2 2 21
2
2 31 12 2
( ) ( )( )(1 2 ) (1 2 ) 1 ( 2 ) ( 2 ) ( 2 )
2! 3!
1
x x x x x
x x x
- - -- = - = + - + - + - + ◊◊◊
= - - - + ◊◊◊
This expansion is valid for 1 2 1 x , or 1 12 2
x .
Question 4.23
Expand ( )112 x and hence approximate the following (without using a calculator!):
10412. , 110
14. , 0 99
12. ,
109
107
12
12
.
., 10200 ,
a b
a b
12
12
where a 111. and b 110. .
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Example
Expand 1
2 3
x
x
++
as far as the term in x3 . For what values of x is this expansion valid?
Solution We start by manipulating the expression, since we want a binomial expansion which starts with a 1:
1 1 132
2 31 3 3 32 2 2
1 2 33 9 272 4 8
2 3 2 33 9 27 3 912 2 4 8 2 4
2 33 91 12 4 8 16
1(1 )(2 3 ) 2 (1 )(1 )
2 3
( 1)( 2) ( 1)( 2)( 3)2 (1 ) 1
2! 3!
2 (1 ) 1
1
xx x x x
x
x x x x
x x x x
x x x x x x
x x x
This is valid for 321 1 x- < £ , ie 2 2
3 3x- < £ .
Question 4.24
Expand 1
1 4( ) x as far as the term in x3 . For what values of x is this expansion
valid? Use your expansion to estimate 0 9812. to 6DP.
Question 4.25
Simplify 10
1 k x
x
k xS p q
x
, and 2
0
nx n x
x
nS x p q
x
, given that p q 1 and
k is a positive integer.
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This page has been left blank so that you can keep the chapter summaries together for revision purposes.
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Chapter 4 Summary There are three laws of indices:
( )
a b a b
a b a b
a b ab
x x x
x x x
x x
+
-
¥ =
=
=
and three laws of logarithms:
log log log
log log log
log log
a a a
a a a
na a
x y xy
x y x y
x n x
+ =
- =
=
which can be used to simplify expressions. Logarithms are used to solve equations where the unknown is a power. There are three methods used to solve quadratics. It is important to look at the solutions and check that they are correct in the context of the question. The quadratic formula is:
xb b ac
a 2 4
2
When solving simultaneous equations, think about which method might be most efficient before proceeding. Inequalities can be solved by looking at graphs or by drawing tables. The and notations are used to write products and sums in shorthand.
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Number sequences which have a commom difference or a common ratio are called arithmetic and geometric progressions. The general terms and sums of these progressions are obtained by the use of formulae:
AP general term: a n d ( )1 AP sum: S a n dnn 2 2 1( )
GP general term: ar n1 GP sum: Sa r
rn
n
( )1
1 GP S
a
r 1
The binomial theorem can be used to expand bracketed expressions with integral or rational powers.
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Chapter 4 Solutions Solution 4.1
(i) 810x
(ii) 5
8
3y
(ii) 625b Solution 4.2
Let log mam x a x= ¤ = and let log n
an y a y= ¤ = .
Now consider m na - , using the second law of indices, this can be simplified as follows:
m n m na a a x y- = ∏ = ∏
Using the definition of logs:
log log logm na a a
x xa m n x y
y y- = ¤ = - = -
and hence we have proved the result. Solution 4.3
ln 1e =
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Solution 4.4
2 2 12 1.05 1.1025t t- -¥ = Taking logs:
2 2 1log 2 1.05 log1.1025
log 2 ( 2) log1.05 (2 1) log1.1025
t t
t t
- -¥ =
+ - = -
Gathering up the terms:
log 2 2log1.05 log1.1025 (2log1.1025 log1.05)
log 2 2log1.05 log1.1025
2log1.1025 log1.05
4.736
t
t
- + = -
- +=-
=
Solution 4.5
2 2
2 2
3 4 2 5 7 ( 4)( 1) ( 4)(4 1)
(3 2)( 4) ( 1)(2 7)3 10 8 4 17 4
( 4)(4 1)
(3 2)(2 7)
x x x x x x x x
x x x xx x x x
x x
x x
+ - + - + - - -∏ = ¥+ - - +- - - +
+ -=+ +
Solution 4.6
(i) (2 1)(3 2) 0x x+ - = so 1 22 3or x = -
(ii) (4 1)(4 1) 0x x+ - = so 14x = ±
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Solution 4.7
2(i) ( 2) 12 0
2 12
2 12
1.464 or 5.464
x
x
x
+ - =
+ = ±
= - ±
= -
( )
217 2898 64
17 1618 64
17 16114 8 64
(ii) (4 ) 2 0
(4 )
0.135 or 0.928
x
x
x
+ - + =
+ = ±
= - ±
= - -
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Solution 4.8
2 2 21 2
1 2
x x x
Comparing the coefficients of 2x and x:
2 2 21 2
1 1 as s s
+ = and 1 22 2 21 2
2 2 2m m mas s s
- - = -
Dividing the second equation by –2, and then dividing the equations:
1 22 21 2
2 21 2
1 1
m ms s m
s s
+=
+
So 2 2
1 2 2 12 22 1
m s m sms s
+=+
.
From the equation above for the coefficients of 2x , we get:
2 21 2
2 2 21 2
s s as s s+ =
Rearranging this equation gives us:
2 2 2 22 1 2 1 2
2 2 2 21 2 1 2
as s as ss ss s s s
= fi =+ +
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Solution 4.9
Rearranging the equation so that the 2x term is positive, and then using the formula:
4 16 280.76 or 0.19
14x
± += = -
Solution 4.10
The equation can be written as 23(3 ) 4(3 ) 4 0x x+ - = , which is a quadratic in 3x . If
3xy = , then the equation can be written as 23 4 4 0y y+ - = , so that:
23
23
(3 2)( 2) 0
or 2
3 or 2x
y y
y
- + =
= -
fi = -
so:
23ln
0.369ln 3
x = = -
Note that it is impossible to raise a positive number to any power and get a negative
number. So the alternative equation 3 2x = - has no solutions, and 0.369x = - is the only solution to the equation.
Page 44 FAC-04: Algebra
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Solution 4.11
(i) Dividing the second equation by the first 4 2
22 2
(1 )4
(1 )
ab b bab b
+ = =+
ie 2b = ± .
Substituting this back into the first equation:
14
4 16 5a a
a
+ =
=
So the two solutions are 14 and 2a b= = or 1
4 and 2a b= = - .
(ii) Factorising and dividing the second equation by the first:
(2 )( 3 ) 5
(2 )( ) 5
x y x y
x y x y
- + = -
- + =
13
3 2
x y
x y
x y x y x y
+ = -+
+ = - - fi = -
Substituting this into the first factorised equation:
( 5 )( ) 5
1
2
y y
y
x
- = -
fi = ±
fi =
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Solution 4.12
Since 21
2em s+
squared is 22e m s+ , we can substitute the first equation into the second:
2
9( 1) 11.2es - =
Rearranging this, we get:
22.244es =
Which can be rearranged to give:
0.899s = ± Substituting this into the first equation gives 0.694m = .
Solution 4.13
3 22 5 17 20 0
(2 5)( 4)( 1) 0
x x x
x x x
+ - - <
fi - + + <
4x < - 4 1x- < < - 1 2.5x- < < 2.5x > 4x + – + + + 1x + – – + +
2 5x - – – – + (2 5)( 4)( 1)x x x- + + – + – +
ie the solution is 4x < - and 1 2.5x- < < . Solution 4.14
The arithmetic and geometric means are equal when the two numbers are equal. This can be proved as follows:
22( )
( ) 42 4
a b a bab ab a b ab
+ += fi = fi + =
This simplifies to 2( ) 0a b- = , ie a b= .
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Solution 4.15
(i) 15
2
1
( 1)k
k=
+Â
(ii) 2
1
2n
k
k=Â
(iii) 1
1
2 1
n
k k= +Â
(iv) 20
1
( 1) (5 1)k
k
k=
- -Â
Solution 4.16
1
1
1n
kk
t
Solution 4.17
If the number of payments is n, then, using the formula for the general term of an arithmetic progression:
18 298 ( 1) 28
11
n
n
= + - ¥ -
=
So, the total amount of the payments can be found by using the sum of an arithmetic progression, ie:
( )112 2 298 (11 1) 28 £1,738¥ + - ¥ - =
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Solution 4.18
Since a and b are both infinite geometric progressions, 1
1 2a
x=
- and
1
1 3b
y=
-.
1
(1 2 )(1 3 )
1 1 1 3 1 2 2 (3 2 ) 1
1 2 1 3 (1 2 )(1 3 ) (1 2 )(1 3 ) (1 2 )(1 3 )
abx y
y x y xa b
x y x y x y x y
=- -
- + - - ++ = + = = =- - - - - - - -
Solution 4.19
(i) The sum of the first 10 terms is found by putting 10n = into the formula for nS :
1
10 6 10 6 5 50S = ¥ ¥ ¥ =
(ii) We have just calculated the sum of the first 10 terms to be 50. We can calculate
the sum of the first 20 terms to be:
120 6 20 16 15 800S = ¥ ¥ ¥ =
The sum of the 11th to the 20th terms is just the difference between these two, ie:
20 10 800 50 750S S- = - =
(iii) The 20th term can be calculated as the difference between 19S and 20S , ie:
20 19 800 665 135S S- = - =
(iv) Similarly, the n th term is:
1 11 6 6
16
1 16 2
( 4)( 5) ( 1)( 5)( 6)
( 5)[ ( 4) ( 1)( 6)]
( 5)[3 6] ( 5)( 2)
n n nu S S n n n n n n
n n n n n
n n n n
-= - = - - - - - -
= - - - - -
= - - = - -
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(v) To find the smallest term, we need to look at how two consecutive terms compare. The n th term will be at least as big as the one before it iff:
1 0n nu u -- ≥
ie 1 12 2( 5)( 2) ( 6)( 3) 0n n n n- - - - - ≥
This simplifies to: 4 0n - ≥ ie 4n ≥ So a given term will be at least as big as the previous term whenever 4n ≥ (with
equality only when 4n = ). This means that: 5 4u u≥ and 4 3u u= but 3 2u u≥/
Combining these, we get: 2 3 4 5u u u u> = <
So 3u and 4u are (jointly) the smallest terms. They each take the value 1- .
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Solution 4.20
Here is one way to tackle this problem. For the first few values of n , we know that: 0 0q =
211 04q q s= +
212 14q q s= +
213 24q q s= + etc
If we multiply these equations by successive powers of 4, like this: 0 0q =
21 04 4q q s= +
2 2 22 14 4 4q q s= +
3 2 3 23 24 4 4q q s= +
… 1 2
14 4 4n n nn nq q s-
-= +
we see that the terms on each side match up. For example, there is a 224 q term on the
LHS and the RHS. If we then add up all these equations, most of the terms will disappear, leaving just:
2 2 2 24 0 4 4 4n nn q s s s= + + + +
This trick is often referred to as a “telescoping sum” (because it’s like a sailor’s telescope, which folds up into something very small when you close it because the segments all fit together nicely). We now have a GP, which allows us to write down the required formula:
2434 (4 1)n n
nq s= - 243 (1 4 )n
nq s-fi = -
We can check 1n = . This formula predicts that 2 2341 3 4q s s= ¥ = , which matches that
given in the question.
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Solution 4.21
(i) We have:
10
1 11s t
t=
= -Â
So the summation is:
10 10 10
0 0 0
1.04 (11 ) 11 1.04 1.04t t t
t t t
t t- - -
= = =- = - ¥Â  Â
The first series is a geometric progression and the second series can be summed
using the formula given in the question:
11 1010
1 1 100
1 1.04 1 1 1.04 101.04 (11 ) 11 58.23
0.041 1.04 1 1.04 1.04t
t
t
(ii) Swapping the order of the summation:
10
0 0
1.04s
t
s t
-
= =ÂÂ
The sum over the variable t can be carried out using the formula for the sum of
a geometric progression:
10 10 10( 1)
10 0 0 0
1 1.04 1.04 1.041.04
0.041 1.04
s s st
s t s s
- + --
-= = = =
- -= =-
ÂÂ Â Â
Simplifying this and using the formula for the sum of a geometric progression:
10 10 11
10 0
1.04 1 11 1.04 1 1 1.041 1.04 58.23
0.04 0.04 0.04 0.04 1 1.04s
s s
--
-= =
¥ -- = - =-
 Â
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Solution 4.22
8 8 7 6 2 5 3 4 4
2 3 4
8 7 8 7 6 8 7 6 5(1 2 ) (1) 8(1) (2 ) (1) (2 ) (1) (2 ) (1) (2 )
2! 3! 4!
1 16 112 448 1,120
x x x x x
x x x x
¥ ¥ ¥ ¥ ¥ ¥+ = + + + + +
= + + + + + Solution 4.23
The binomial expansion tells us that:
½ 12(1 ) 1x x+ ª +
In other words, if you want to approximate the square root of a number close to 1, you just halve the extra bit. So we get:
½ 121.04 1 0.04 1.02ª + ¥ =
¼1.10 is the 4th root of 1.10 , which is the same as square rooting twice. So we get:
¼ ½ ½ ½1.10 (1.10 ) 1.05 1.025= ª ª
The same method works with a number a bit less than one and/or a negative power:
½ ½ 120.99 (1 0.01) 1 ( 0.01) 1.005- -= - ª - ¥ - =
The next one gives:
½
½
1.09 1.045
1.0351.07ª
To approximate this ratio, we can subtract 0.035 from the top and bottom (which doesn’t affect the answer very much):
½
½
1.09 1.045 0.035 1.011.01
1.035 0.035 11.07
-ª = =-
Note that the 0.01 bit is just half the difference between the 1.09 and the 1.07.
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We must have numbers close to 1 for our approximation to work. So, for the next one we need to take out a factor first:
10200 10000 1.02 100 1.02 100 1.01 101= ¥ = ¥ ª ¥ = And finally:
½ ½ 1.055 1.05 0.0050.5
1.11 1.10 0.01
a b
a b
- -ª = =- -
Calculations involving rates of return earned on investments often involve expressions similar to these. It is useful to be able to approximate them to check that the calculations look sensible. Solution 4.24
12
3 3 51 12 32 2 2 2 2
2 3
(1 4 ) 1 2 ( 4 ) ( 4 )2! 3!
1 2 6 20
x x x x
x x x
- - ¥ - - ¥ - ¥ -- = + + - + - +
= + + + +
This is valid for 1 1
4 4x- £ < .
If we now substitute in 0.005x = , we get:
120.98 1 0.01 0.00015 0.0000025
1.0101525
- = + + +
=
So to 6DP 120.98 1.010153
- = .
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Solution 4.25
If we write out the first sum without using sigma notation (and dividing through by kp ), it is:
2 31
2 3
1 1 2
0 1 2 3
( 1) ( 2)( 1)1
2 6
k
k k k kSq q q
p
k k k k kkq q q
It’s not obvious how to deal with this because the k factors are going up, rather than down, as in an ordinary binomial expansion. However, we can introduce some negatives to enable us to simplify the expression:
2 31 ( )( 1) ( )( 1)( 2)1 ( )( ) ( ) ( )
2 6
[1 ( )]
- -
- - - - - - - -= + - - + - + - +
= + - =
k
k k
S k k k k kk q q q
p
q p
So: 1 1S =
This is actually a calculation of the sum of the probabilities for a negative binomial distribution. If we write out the second sum without using sigma notation, we have:
1 2 22
0
0 21 2
nx n x n n n
x
n n n nS x p q pq p q n p
x n
If we take a typical coefficient in this series, we get for example:
1( 1)( 2) ( 1)( 2)
3 33 23 2 1 2 1
n nn n n n nn n
So the series is:
1 2 22
1 1 1
0 1 1n n nn n n
S n pq n p q n pn
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Dividing through by the n and a p factor:
1 2 12 1 1 1
0 1 1n n nn n nS
q pq pnnp
The RHS is just the binomial expansion of 1( )np q , which equals 1 (since 1p q ).
So:
2 1S
np 2S np
This is actually the calculation of the mean value of a binomial distribution.
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Chapter 5
Numerical methods II
You need to study this chapter to cover: ● expressing quantities as percentages or per mil
● absolute change, proportionate change and percentage change
● absolute error, proportionate error and percentage error
● dimensions
● linear interpolation
● iteration
● complex numbers
● difference equations.
0 Introduction
This chapter completes the numerical methods required for the Core Technical subjects.
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1 Percentages
A percentage means “out of a hundred”. Percentages are used a lot in actuarial work so you must be familiar with how to use them efficiently and accurately.
Example If prices have risen by 1.2% in a shop, what is the new price of an item that was £15.75 before the rise? Solution 1.012 15.75 £15.94¥ = Notice that the final answer has to be rounded off to two decimal places because we are working in pounds and pence.
Example At the start of year 1 a group consists of 51,890 people. By the start of year 2, there are 51,548 people. If the only reason for leaving the group is death, what percentage of the group died in year 1? Solution In the group, 51,890 51,548 342- = people have died which is
342100% 0.66%
51,890¥ = .
Sometimes you may be asked to express something “per mil”, or out of a thousand, which is written as ‰. This notation is often used in connection with insurance premiums which are often small percentages so writing them per thousand rather than per hundred makes them easier to interpret.
Question 5.1
Express the number of people who have died in the previous example as a rate per mil.
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2 Changes
The previous example is really asking about a percentage change in the population, and in this section we are going to look at different ways of expressing change.
2.1 Absolute change
An absolute change is asking for the difference between two figures, where the difference can be positive or negative. Absolute change should not be confused with absolute value which was covered in Chapter 4.
Example What is the absolute change in the population described in the last example? Solution The absolute change is –342, since the population decreased by 342 people.
2.2 Proportionate change
The disadvantage of looking at absolute changes is that it gives no indication of the size of the change compared to the original figure. For example, there is an absolute change of 1 when a population goes from 3 to 4 people, or from 340,025 to 340,026 people! Proportionate change looks at the change relative to the original number in the population, by dividing the absolute change by the original amount.
Example What is the proportionate change in the price of a car that goes from £9,950 to £9,550? Solution
The absolute change is –£400, so the proportionate change is 400
0.049,950
- = - . Notice
that the answer has no units of measurement, since we have divided pounds by pounds. We will return to this idea in a later section.
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2.3 Percentage change
A percentage change expresses the proportionate change as a percentage.
Example What is the percentage change in the previous example? Solution The proportionate change is –0.04 so the percentage change is –4%.
Question 5.2
A man’s wage has risen from £14,567pa to £15,034pa. What are the absolute, proportionate and percentage changes in his wage?
When you are calculating percentage changes you often need to divide two numbers that are close together. If only an approximate answer is needed, a small number can be added to the numerator and denominator to make the calculation easier, eg:
110 110 6 1041.04
106 106 6 100
-ª = =-
You can also apply this technique in reverse:
31.01 1.01 1.01 1.01
101 102 103 1031.03
100 101 102 100
= ¥ ¥
ª ¥ ¥ = =
This approximation becomes less accurate if the adjustment is not small.
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3 Errors
Errors are similar to changes in that they look at differences, but errors refer specifically to differences between “actual” and “expected” (or “approximate” and “accurate”) figures.
Example
When calculating 1
1v
i=
+, the true value of i is 0.0372534. Calculate the absolute,
proportionate, and percentage error that will be introduced in the value of v by rounding i to two decimal places. Solution
The true value of v is 1
1.0372534. When rounded to 2 DP, i becomes 0.04, which gives
v to be 1
1.04.
Absolute error 1 1
0.00251.04 1.0372534
= - = - (4 DP).
Proportionate error (absolute error divided by true value of v) 0.002641= - (6 DP). Percentage error (proportionate error multiplied by 100) 0.26 %= - .
Question 5.3
It was discovered that, in an actuarial calculation, a mistake was made copying down the last figure in the value of i . If i was written down as 0.274, what is the largest
percentage error that could have been made in the calculated value of 1
1v
i=
+?
You always calculate these in the order of “actual minus expected” or “approximate minus accurate”.
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4 Dimensions
In the example in Section 2.2, we mentioned that the answer had no units, since we divided a number of pounds by a number of pounds. We are going to look further at that situation now by considering dimensions. The units in which we measure a value, (eg pounds, metres, kilograms), affect the numerical value given, eg lengths of 2.7m and 270cm represent the same quantity. This creates problems when we wish to compare values, since we need to ensure that they are measured in the same units, and this conversion can be complicated (converting a speed in metres per second into one in miles per hour for example). Dimensions are used to show what a value represents, in other words we would say that 2.7m or 270cm both have the dimension of length. In actuarial work the dimensions usually involve currency, time and people. For example, if it was stated that the average salary in the UK was £20,000, the units of this would be pounds per year per person. Numbers (including constants such as p ) have a dimension of zero and they are referred to as dimensionless. If two values that have the same dimension are divided, then the resulting value is dimensionless. This is true of percentage or proportionate errors. Other values that you will meet in Subject CT3, which are dimensionless, are the correlation coefficient and the coefficient of skewness. Dimensions can give us a convenient way of telling if a formula is correct.
Example What is the dimension of the mean length of life of a light bulb? Solution The mean is defined to be the sum of a set of lifetimes, divided by the number of
lifetimes considered, ie 1
1 n
ii
x xn =
= Â . “Length of life” is measured in time, and n is
dimensionless, so the dimension of the mean is T, where T stands for time.
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Question 5.4
If the variance is defined to be ( )21
1
1
n
ii
x xn =
-- Â , what is the dimension of variance for
the previous example?
Question 5.5
Given the definition of the variance in Question 5.4, by considering dimensions, decide which of the following could be other possible formulae for variance.
(i) 2 2
1
1
1
n
ii
x xn =
-- Â
(ii) 2
1
( 1)n
ii
x n x=
- -Â
(iii) 2
1 1
1 1
1 1
n n
i ii i
x x xn n= =
-- -Â Â
The argument of functions such as xe and log x must be dimensionless.
Question 5.6
Why must they be dimensionless?
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Example
Use dimensional analysis to work out whether 0
tte dtll•
-Ú is equal to 21,1, or l l
l.
Solution The question says “Use dimensional analysis…”. This means that evaluating the integral is not a valid solution!
Using the fact that the argument of xe is dimensionless, we know that tl must be
dimensionless, in other words t has the same dimension as 1
l.
Since tl is dimensionless, the whole expression to be integrated is also dimensionless. When you integrate with respect to t, you increase the dimension of the expression you are integrating by the dimension of t. Therefore here when we have integrated we will have an expression of dimension t,
which is equivalent to a dimension of 1
l, giving
1
l as the correct expression.
Question 5.7
The integral 0
( )xx e dx
aa ll
a
•-
GÚ can be expressed in the form p qa l . Using dimensional
analysis, find the value of q.
Probabilities are dimensionless. Therefore probability density functions (PDFs), ( )f x ,
must have “inverse” units because ( )f x dx represents a probability. If you have not
met PDFs yet don’t worry. They appear in Subject CT3. The dimensions also tell you how the result of a formula will be affected if the values of the components are rescaled eg multiplied by 10.
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Example A discussion is being held about the wages for the workers in a small firm. The director has said “If I double all wages then the mean and variance of the wages will also double”. Is she correct? Solution The dimensions of the mean of wages will be in the currency unit eg pounds. However the dimensions of the variance will be in the currency unit squared, eg pounds squared. This means that the variance will be multiplied by 4 and not 2, so the director is wrong.
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5 Interpolation and extrapolation
Sometimes it is not possible to find an exact solution to an equation such as ( ) 0f x =
by algebraic methods. Under such circumstances an approximate value can be found if you know two values of the function, evaluated at values of x close to the true value, say 1 2 and x x . If the true value is in between these, it is called interpolation; otherwise
it is extrapolation. The method we are going to look at is called linear interpolation. As the name suggests, this method relies on the function being approximately linear between 1 2and x x .
The disadvantages of this method are: ● It can only give an approximate value (unless the function is truly linear)
● The answer obtained will only be close to the true value of x if 1 2 and x x are
relatively close to the true value
● The answer obtained will only be close to the true value of x if ( )f x is
approximately linear between 1 2and x x .
In practice, as long as you are aware that your answer may be slightly inaccurate, this is often an appropriate method to use and it is the method usually expected in exam solutions. To illustrate what we mean consider the following diagram:
0 x1 x
y
x2
error
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This shows the graph of ( )y f x= , with the close values 1 2and x x , and the error that
this interpolation has introduced. Do be aware however that we have drawn a large-scale diagram and not chosen two values close to the true values just for illustration! To show how interpolation works in practice, consider the following diagram:
x1 x x2
f(x1) f(x2)f(x)
By considering the ratio of ‘lengths’ above and below the line, we can write:
1 2 1
1 2 1( ) ( ) ( ) ( )
x x x x
f x f x f x f x
- -=- -
or 1 1
2 1 2 1
( ) ( )
( ) ( )
x x f x f x
x x f x f x
- -=- -
Consider the first version. This can be rearranged to give:
2 1 11
2 1
( )( ( ) ( ))
( ) ( )
x x f x f xx x
f x f x
- -= +-
This gives us a formula for x. It is better for you to understand how this formula is derived rather than trying to learn it!
Question 5.8
If x does not lie between 1 2and x x , ie we are dealing with extrapolation, derive a
formula for finding x.
To see how this works in practice, look at the next example.
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Example If (1.34) 0.523f = , and (1.65) 0.42f = - , what is an approximate value for x if
( ) 0f x = ?
Solution We know that x lies somewhere between 1.34 and 1.65, so:
1.65 1.65 1.34
0 ( 0.42) 0.42 0.523
x - -=- - - -
which gives 1.51x = .
Question 5.9
By finding two consecutive integer values which x lies between, and using linear
interpolation, find a value of x that satisfies the equation 3 25 3 7 0x x x- + + = .
Interpolation can be used to find the value of “x” or the value of the function.
Question 5.10
This is an extract from the tables of a function ( )F x :
x ( )F x
0.54 0.70540 0.55 0.70884 0.56 0.71226 Using linear interpolation, work out: (i) the value of ( )F x when 0.548x =
(ii) the value of x when ( ) 0.71F x = .
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6 Iteration
There are some equations that cannot be solved using ‘standard’ methods to find an exact solution, but an approximate solution can usually be found using a numerical method based on iteration.
6.1 Bisection
The bisection method is based on finding two values between which the solution lies and then using the midpoint of the values for the next approximation.
Example
Find an approximate negative solution to the equation 3 12 7 0x x- - = . Give your answer to 1 DP. Solution
Let 3( ) 12 7f x x x= - - , so we are trying to solve ( ) 0f x = .
(0) 7 and ( 1) 4 the root lies between 0 and 1
( 0.5) 1.125 the root lies between 0.5 and 1
( 0.75)=1.578125 the root lies between 0.5 and 0.75
( 0.625) 0.255859 the root lies between 0
f f
f
f
f
= - - = fi -
- = - fi - -
- fi - -
- = fi - .5 and 0.625
( 0.5625) 0.42798 the root lies between 0.5625 and 0.625f
-
- = - fi - -
Therefore the answer is –0.6.
Question 5.11
Using the bisection method, find i to 3DP if i satisfies the equation:
( )2
320 1 (1 )
25(1 ) 61.5i
ii
--
- ++ + =
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6.2 Newton-Raphson iteration
The Newton-Raphson iterative formula states that if nx is an approximate solution to the
equation ( ) 0f x = then a better approximation is 1( )
( )n
n nn
f xx x
f x+ = -¢
.
Example Using the Newton-Raphson formula, find an approximate positive solution to the
equation 5 24 7x x+ = , giving your answer to 2 DP. Solution
Let 5 2( ) 4 7f x x x= + - , so that:
4( ) 5 8
(1) 2
(2) 41
f x x x
f
f
Therefore there is a solution between 1 and 2. Try 1 1x = .
2
3
4
1.15385
1.13336
1.13290
x
x
x
=
=
=
Therefore the solution is 1.13 (to 2DP).
Question 5.12
Using the Newton-Raphson formula, find an approximate solution to the equation:
5
4 10(1 (1 ) )5(1 ) 41
ii
i
-- - ++ + =
Give your answer to 3 DP.
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7 Complex numbers
7.1 Basic algebra
A complex number is represented in the form a ib+ , where 1i = - , and a and b are real numbers. a is called the “real part” and b is called the “imaginary part”. You may also see complex numbers written with j’s, rather than i’s. Complex numbers can be added, subtracted and multiplied as follows: (i) ( ) ( ) ( ) ( )a ib c id a c i b d+ + + = + + +
(ii) ( ) ( ) ( ) ( )a ib c id a c i b d+ - + = - + -
(iii) ( )( ) ( ) ( )a ib c id ac bd i ad bc+ + = - + +
The rules for addition and subtraction just require the real and imaginary parts to be added separately. The logic of the rule for multiplication can be seen by multiplying
out the brackets and using the fact that 2 1i = - .
Question 5.13
Find: (i) (2 3 ) (3 4 )i i+ + - (ii) (4 6 ) (3 7)i i- - -
(iii) (3 4 )(6 2 )i i- + (iv) (4 2 )(4 2 )i i+ -
The answer to last part of this question, where the product of two complex numbers is a real number, leads to an important property of complex numbers. a ib+ and a ib- are called “complex conjugates”. Their product is the real number
2 2a b+ . The complex conjugate of the number a ib+ is a ib- , and vice versa. This enables us to divide complex numbers.
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Example
Simplify 2 3
1 2
i
i
-+
.
Solution If we multiply the numerator and denominator by the complex conjugate of the denominator, we get: 2 3 1 2 2 3 4 6 4 7 1
( 4 7 )1 2 1 2 1 4 5 5
i i i i ii
i i
- - - - - - -¥ = = = - -+ - +
This is in the form a ib+ , where 0.8a = - and 1.4b = - .
Question 5.14
Simplify 3 4
1 4
i
i
--
.
7.2 Argand diagrams
Complex numbers can be represented on an “Argand diagram”, where you plot the imaginary part against the real part. For example, the complex number 1 3z i= - can be shown as follows:
0
z
Re
Im
1
3
2
1 2 3-1-2-3
-1
-2
-3
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Notice that it is the point that represents the complex number, not the line. There is a reason for including the line though, which we will describe shortly. By representing a complex number on an Argand diagram, it is easy to picture two further properties of complex numbers.
The modulus, r, of a complex number a ib+ is given by 2 2r a b= + , ie the square
root of the sum of the real part squared and the imaginary part squared. It is equivalent to the length of the line shown on the Argand diagram. The argument, q , of a complex number a ib+ is defined to be the angle between the line and the positive x-axis on the Argand diagram. In radians, p q p- < £ . This diagram illustrates this for two complex numbers 1z and 2z , with moduli 1r and
2r , and arguments 1q and 2q respectively.
Note that in the diagram, 2q is negative. By convention, angles measured
anticlockwise from the positive real axis are positive, and angles measured clockwise are negative.
Re
Im
z2
z1
r1
1
r2
2
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Example Find the modulus and argument of the complex number 3 i- - . Solution
The modulus is 2 21 3 10 3.16+ = = .
The argument is 1 1tan 2.82
3
.
Question 5.15
Find the modulus and argument of the complex number 1 2
4 3
i
i
--
.
7.3 Euler’s formula
As we have seen, complex numbers can be represented as points on an Argand diagram.
The number ie q is represented by the point on the unit circle (the circle centred at the origin with radius 1) that forms an angle q with the real axis.
Re
Im
1
y
x
ie
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We can work out the real and imaginary parts of this number (ie the horizontal and vertical co-ordinates) by drawing the triangle shown in the diagram. From basic
trigonometry, we know that adjacent
cosinehypotenuse
= , so that cos1
xq = ie cosx q= .
Similarly siny q= . So ie q has real part cosq and imaginary part sinq . In other
words:
cos sinie iq q q= + This is known as “Euler’s formula” and it holds for any value of q (including complex ones). Note, however, that q must be measured in radians not degrees.
Question 5.16
What are the co-ordinates of the numbers exp4
ip and 2exp
6
ip on the Argand diagram?
Note that any complex number can be expressed in the form ire q , which is known as polar form. q is the argument and r is the modulus.
It is useful to look at the complex conjugate of ie q . The complex conjugate can be
found by replacing i by i- in the number, so here it is ie q- . Since cos sinie iq q q= + ,
we can replace i by i- to give cos sinie iq q q- = - .
Re
Im
ie
e -i
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Question 5.17
How can we deduce from this that cos( ) cosq q- = and sin( ) sinq q- = - ?
Question 5.18
Show that the product of ie q- and ie q does give a real number.
In many applied maths applications of complex numbers you will find that you need to
add ie q- and ie q . This gives:
cos sin cos sin 2cosi ie e i iq q q q q q q- + = - + + = Note that this also gives a purely real number.
Question 5.19
Show that 51 12 2 4(1 )(1 ) cosi ie ew w w-+ + = + .
Question 5.20
Use Euler’s formula to derive the angle sum formulae:
sin( ) sin cos cos sin
cos( ) cos cos sin sin
A B A B A B
A B A B A B
+ = +
+ = -
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7.4 Solution of polynomial equations
If a polynomial has real coefficients, then any complex roots must occur in complex conjugate pairs. Allowing complex roots means that any quadratic equation will have possible solutions.
Example
Find the roots of the equation 2 3 4 0z z- + = . Solution Using the quadratic formula, we get:
23 9 16 3 7
2 2
iz
± - ±= =
Simplifying this we get:
7 73 32 2 2 2 or z i i= + -
Notice that these roots are a complex conjugate pair.
Question 5.21
You are given that 1 2z i= + is a root of the cubic equation 3 24 9 10 0z z z- + - = . Find the other two roots.
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8 Difference equations
A difference equation is typically of the form 1 1 2 2 0t t ty y ya a- -+ + = , where 1a and
2a are constants. The equation has a solution of the form ( )ty f t= and it has an
associated auxiliary equation 21 2 0v va a+ + = .
The general solution of the difference equation depends on the roots, a and b, of the auxiliary equation. A particular solution can be found if some boundary conditions are given.
21 24a a- Solution of difference equation Note
0> t tty Aa Bb= +
The auxiliary equation has two real roots
0= ( ) tty At B a= +
There is only one (repeated) root
0< cos sint tty Ar t Br tq q= +
The complex roots of the auxiliary equation are
written in polar form ie ire q±
Example Solve the difference equations: (i) 1 25 6 0t t ty y y- -- + = given that 0 0y = and 1 1y =
(ii) 2 16 13 0t t ty y y+ +- + = .
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Solution
(i) The auxiliary equation is 2 5 6 0v v- + = which has roots 2 and 3.
The general solution is of the form 2 3t tty A B= ¥ + ¥ .
However we know that 0 0y = and 1 1y = , so:
0
1 2 3
A B
A B
= +
= +
These can be solved to give 1B = and 1A = - , so our particular solution is:
3 2t tty = -
(ii) The auxiliary equation is 2 6 13 0v v- + = , which has roots
6 36 523 2
2i
± - = ± .
The modulus of these complex numbers is 2 23 2 13+ = , and the argument is 1 2
3tan 0.588-± = ± , so the roots can be written in polar form as 0.58813 ie± .
The general solution of the difference equation is then
13 cos 0.588 13 sin 0.588t tty A t B t= + .
Question 5.22
Solve the difference equations: (i) 1 28 16 0t t ty y y- -- + = given that 0 1y = - and 1 3y =
(ii) 1 24 5 0t t ty y y- -- + = .
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This page has been left blank so that you can keep the chapter summaries together for revision purposes.
FAC-05: Numerical methods II Page 25
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Chapter 5 Summary The absolute change between two numbers is the new number minus the old number. The proportionate change is the absolute change divided by the old number. The percentage change is the proportionate change multiplied by 100. The absolute error is the approximate value minus the accurate value. The proportionate error is the absolute error divided by the accurate figure. The percentage error is the proportionate error multiplied by 100. Dimensions can be used to tell if formulae are correct. When an equation cannot be solved directly to get an exact solution, a numerical method such as interpolation or iteration needs to be used.
The formula for Newton-Raphson iteration is 1( )
( )n
n nn
f xx x
f x+ = -¢
.
Any complex number z can be written in the form z a ib= + , where 1i = - . The complex conjugate of z is a ib- . An Argand diagram can be used to represent complex numbers with the y-axis as the imaginary part and the x-axis as the real part.
The modulus of z is given by 2 2a b+ . The argument is the angle made by the line
representing z and the positive x-axis on an Argand diagram.
The complex number z can be written in polar form iz re q= , where r is the modulus and q is the argument.
Euler’s formula states that cos sinie iq q q= + . The cosine function is even ie cos( ) cosq q- = .
The sine function is odd ie sin( ) sinq q- = - .
Complex roots of polynomial equations with real coefficients must occur in complex conjugate pairs. Difference equations such as 1 1 2 2 0t t ty y ya a- -+ + = can be solved, their solution
depending on the roots of the auxiliary equation.
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Chapter 5 Solutions Solution 5.1
0.66% is equivalent to saying 6.6‰. Solution 5.2
Absolute change is £467. Proportionate change is 0.0321. Percentage change is 3.21%. Solution 5.3
The largest error will have been made if the true value of i was 0.279, in which case, the true value of v is 0.7818608. The value of v calculated when i is 0.274 is 0.7849294, which gives a percentage error of 0.39%. Solution 5.4
Since the variance is calculated by squaring the x values, the dimension of the variance
here is 2T , where T is the dimension of time. Solution 5.5
The other possible formulae are (i) and (iii). The second one can’t be correct because
the dimension of the first term is 2T , while that of the second term is T, and terms of different dimensions cannot be subtracted.
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Solution 5.6
Consider the series for xe which is 2
12!
xx + + + . If x had a dimension, then each
term of this series would have a different dimension and so this series would be meaningless. Therefore x must be dimensionless. A similar reasoning applies to log x .
Solution 5.7
Since the argument of xe is dimensionless, we know that x has the same dimension as 1
l.
( )aG is dimensionless.
When we integrate we get an expression of dimension equivalent to 1 1
,
which simplifies to 1
l.
Therefore 1q = - .
Solution 5.8
You use exactly the same formula for extrapolation as you do for interpolation!
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Solution 5.9
By trial and error, the consecutive integers can be found. They are 0 and –1, or 2 and 3, or 3 and 4.
Between 0 and –1, interpolating gives ( 1) 0 ( 2)
0.7780 ( 1) 7 ( 2)
xx
- - - -= fi = -- - - -
.
Between 2 and 3, interpolating gives 2 0 1
2.3333 2 2 1
xx
- -= fi =- - -
.
Between 3 and 4, interpolating gives 3 0 ( 2)
3.44 3 3 ( 2)
xx
- - -= fi =- - -
.
Solution 5.10
(i) Using the formula for interpolation:
(0.70884 0.70540)(0.548 0.54)(0.548) 0.70540
0.55 0.54
0.70815
F- -= +
-
=
(ii) Using the formula:
(0.56 0.55)(0.71 0.70884)0.55
0.71226 0.70884
0.553
x- -= +
-
=
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Solution 5.11
When 0.03i = the LHS gives 61.15, and when 0.02i = the LHS gives 62.39. Using the bisection method, we get:
Value of i Value of equation
0.025 61.76 0.0275 61.45 0.02625 61.61 0.026875 61.53
So the value of i is 0.027 to 3DP. Solution 5.12
Let 5
4 10(1 (1 ) )( ) 5(1 ) 41
if i i
i
-- - += + + - , so that:
6 5
52
6 55
2
10 5(1 ) 10(1 (1 ) )( ) 20(1 )
50 (1 ) 10 10(1 )20(1 )
i i if i i
i
i i ii
i
- --
- --
¥ ¥ + - - += - + +¢
+ - + += - + +
The Newton-Raphson formula gives the formula for the next approximation to be:
54
6 55
2
10(1 (1 ) )5(1 ) 41
50 (1 ) 10 10(1 )20(1 )
ii
iii i i
ii
--
- --
- ++ + --
+ - + +- + +
(0.1) 0.323f = and (0.11) 0.747f = - , so the root lies between 0.1 and 0.11.
Try 1 0.1i = .
The formula gives 2 0.1029557i = , 3 0.1029739i = , ie the root is 0.103 to 3 DP.
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Solution 5.13
(i) (2 3 ) (3 4 ) 5i i i+ + - = -
(ii) (4 6 ) (3 7) 11 9i i i- - - = -
(iii) 2(3 4 )(6 2 ) 18 24 6 8 26 18i i i i i i- + = - + - = -
(iv) 2(4 2 )(4 2 ) 16 8 8 4 20i i i i i+ - = - + - =
Solution 5.14
Multiplying the numerator and denominator by the complex conjugate of the denominator:
2
2
3 4 1 4 3 4 16 12 16 13
1 4 1 4 171 16
i i i i i i
i i i
- + - - + - -¥ = =- + -
Solution 5.15
We first need to simplify the complex number:
21 2 4 3 4 8 3 6 10 5 2
4 3 4 3 16 9 25 5
i i i i i i i
i i
- + - + - - -¥ = = =- + +
The modulus is then 2 2
2 10.4472
5 5
.
The argument is 1 1tan 0.46
2-- = - .
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Solution 5.16
Using Euler’s formula 4 4
1 1exp cos sin
4 2 2
ii ip pp = + = + so the co-ordinates are:
1 1,
2 2
Similarly 6 6
3 12exp 2(cos sin ) 2
6 2 2
ii i
, so the co-ordinates are ( 3,1) .
Solution 5.17
We could work out ie q- by thinking of it as ( )ie q- and applying the original form of Euler’s formula:
( ) cos( ) sin( )ie iq q q- = - + -
Comparing this with the other version for ie q- , we see that cos( )q- must equal cosq
and sin( )q- must equal sinq- .
These properties of cosines and sines can be described by saying that cosine is an “even” function and sine is an “odd” function. Solution 5.18
Consider their product:
2 2 2
2 2
(cos sin )(cos sin )
cos sin cos sin cos sin
cos sin 1
i ie e i i
i i i
q q q q q q
q q q q q q
q q
- = - +
= - + -
= + =
which is a real number. You could, of course, have used the basic properties of powers
to say that 0 1i i i ie e e eq q q q- - += = = .
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Solution 5.19
Multiplying out gives:
1 1 1 1 1 12 2 2 2 2 2
1 12 4
5 14 2
54
(1 )(1 ) 1
1 ( )
2cos
cos
i i i i i i
i i
e e e e e e
e e
w w w w w w
w w
w
w
- - -
-
+ + = + + + ¥
= + + +
= + ¥
= +
Solution 5.20
Consider ( )i A Be + . We can work this out in two different ways. Firstly, we can apply Euler’s formula directly to get:
( ) cos( ) sin( )i A Be A B i A B+ = + + +
But ( )i A Be + is also equal to iA iBe e¥ , which is:
(cos sin )(cos sin )
(cos cos sin sin ) (sin cos cos sin )
iA iBe e A i A B i B
A B A B i A B A B
¥ = + +
= - + +
If we compare the real and imaginary parts of these two calculations we get the required trigonometrical identities.
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Solution 5.21
We are told that 1 2i+ is a root, and since the polynomial has real coefficients, we know that 1 2i- must also be a root. The factors are therefore:
( (1 2 ))( (1 2 ))z i z i- + - -
and some other factor which is, as yet, unknown. Multiplying these factors out we get:
2 2 2( 1 2 )( 1 2 ) ( 1) 4 2 5z i z i z i z z- - - + = - - = - +
The cubic can then be written as:
3 2 24 9 10 ( 2 5)( 2)z z z z z z- + - = - + -
Therefore the other two roots are 2 and 1 2i- .
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Solution 5.22
(i) The auxiliary equation is 2 8 16 0v v- + = , which has a repeated root of 4.
The general solution is given by ( ) 4tty At B= + ¥ .
However we know that 0 1y = - and 1 3y = , so we have the simultaneous
equations:
1
4( ) 3
B
A B
= -
+ =
which are solved to give 1.75A = and 1B = - .
The particular solution is then (1.75 1) 4tty t= - ¥ .
We can check, for example, that 2 8 3 16( 1) 40y = ¥ - - = , which agrees with the
formula just found.
(ii) The auxiliary equation is 2 4 5 0v v- + = , which has roots:
4 16 202
2i
± - = ± .
These complex numbers have modulus 2 22 1 5+ = , and argument 0.464± ,
so they can be written in polar form as 0.4645 ie± .
The general solution of the difference equation is then:
5 cos 0.464 5 sin 0.464t tty A t B t= +
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FAC-06: Differentiation Page 1
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Chapter 6
Differentiation
You need to study this chapter to cover: ● limits and the order notation
● derivatives as rates of change of functions
● derivatives of the standard functions
● derivatives of sums, products, quotients and “functions of a function”
● higher-order derivatives
● maxima, minima and stationary points
● partial derivatives
● extrema of functions of two variables
● Lagrangian multipliers.
0 Introduction
This chapter covers all the methods of differentiation that you need in order to study the Core Technical subjects.
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1 Limits
1.1 Limits
The value of 1
( )1
f xx
=-
cannot be directly calculated at the point 1x = . However,
we can say that when x is slightly bigger than 1, say 1x e= + , where e is a small positive number, then ( )f x is going to be very large and positive. This is obviously a
long-winded and cumbersome way of saying what happens, so we use the “limit notation”:
1
1lim
1x x+Æ= •
-
Remember that 1x +Æ means that x is approaching 1 from above.
Question 6.1
What is 1
1lim
1x x-Æ -?
Question 6.2
What is 2
2
3 2lim
3 4x
x x
x xƕ
- +-
?
In order to find limits, you may want to consider the value of the function close to the limit. For example, in Question 6.2 you might wish to calculate the value of the function for larger and larger positive values of x. In Question 6.1 you could substitute in, say, 0.9, 0.99, 0.999 and so on. This will often give you some idea of whether the limit is finite or not (and if so, what its value is).
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1.2 The order notation
If you studied series at school you will have seen series such as:
2 31 12 61xe x x x = + + + +
This series goes on for ever, and in some contexts it is important to know how the “remainder” terms, the bits contained in the “” behave. One way to be a little more accurate would be to write this series as:
2 3121 terms in xe x x x= + + + etc
What we are thinking here is that, if x is a fairly small number (say 0.1) then the term
in 3x will be much smaller than the earlier terms, and the terms in even higher powers will be smaller still. In algebraic calculations where it is important to be clear about the behaviour of the remainder term, we can use a special notation where we would write:
2 3121 ( )xe x x O x= + + + as 0x Æ
or:
2 2121 ( )xe x x o x= + + + as 0x Æ
Note that the first equation uses a capital “O” while the second uses a small “o”. O() Notation
The 3( )O x in the first equation is read as “Big-Oh of x cubed”. You can think of this as
representing some function that is no bigger than a fixed multiple of 3x .
Mathematically speaking, a function ( )f x is 3( )O x if you can find a constant M such
that 3( )f x M x£ for all sufficiently small values of x .
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o() Notation
The 3( )o x in the second equation is read as “Small-Oh of x cubed”. You can think of
this as representing some function that is smaller than any fixed multiple of 3x .
Mathematically speaking, a function ( )f x is 3( )o x if 30
( )lim 0x
f x
xÆ= .
If you’re consistently using big O’s or small o’s in a derivation you can just say “order of x cubed” etc.
Example
Which of the following functions are 3( )O x and/or 3( )o x for small values of x ?
(i) 25x
(ii) 30.00001x
(iii) 4999,999,999x
(iv) x xe e
x
--
Solution
(i) 25x is neither 3( )O x nor 3( )o x . The inequality in the definition of 3( )O x
would be 2 35x M x£ . If x was a small positive number this would require
there to be a finite number M for which it was always true that 5
Mx£ .
(ii) 30.00001x is 3( )O x but not 3( )o x . We could use 0.00001M = here (or any
bigger number!) to establish 3( )O x , but the limit for the definition of 3( )o x
equals 0.00001. Notice that the 0.00001 factor doesn’t affect the orders here.
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(iii) 4999,999,999x is both 3( )O x and 3( )o x . The inequality in the definition of
3( )O x would be 4 3999,999,999x M x£ . If x was a small positive number
this would require there to be a finite number M for which 999,999,999
Mx £
was true for sufficiently small values of x . This will work for any positive number M .
Again the 999,999,999 factor doesn’t make any difference.
Note that being 3( )o x is a stronger condition than being 3( )O x . Any function
that is 3( )o x is automatically 3( )O x .
(iv) For x xe e
x
--, we have to do a “calculation” involving the order symbols. We
know that:
2 3121 ( )xe x x O x= + + +
and:
2 3121 ( )xe x x O x- = - + +
Notice that we’ve written 3( )O x+ rather than 3( )O x- in the second equation.
We can do this because these two symbols are equivalent. If we now evaluate the numerator, we get:
2 3 2 31 12 21 ( ) 1 ( )x xe e x x O x x x O x- È ˘ È ˘- = + + + - - + +Î ˚ Î ˚
Simplifying:
32 ( )x xe e x O x-- = +
Again notice that adding two 3( )O x terms just gives 3( )O x .
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We now need to divide by x :
3( )
2x xe e O x
x x
-- = +
This last term is a function that is no bigger than a fixed multiple of 3x divided
by x . This will give a function that is no bigger than a fixed multiple of 2x ie 2( )O x .
So:
22 ( )x xe e
O xx
-- = +
Since 2( )O x is “bigger” than 3( )O x , we can conclude that x xe e
x
-- is neither
3( )O x nor 3( )o x .
You may also see the order notation used to describe the behaviour of a function for large values of x .
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Example
Explain what the statement 211 ( )
1
xO x
x x-= + +
- as x Æ +• means.
Solution
This means that the behaviour of the function 1
x
x - is very similar to the behaviour of
the function 1
1x
+ for large values of x. The remainder (discrepancy) is a term that
approaches zero more quickly than any fixed multiple of 2
1
x.
You can see this if you expand 1
x
x - by writing it as:
( ) 111 2 3
1 1 1 11 1
1 1 xx
x
x x x x
-= = - = + + + +
- -
1.3 Supremums and infimums
For finite sets of numbers, you can always identify which element(s) is/are the largest/smallest. For infinite sets, this is not always possible, and we have to generalise the idea of maximum/minimum. The supremum (sup), or least upper bound of a set A is defined as the number a such that and a a A a a aa a≥ " Œ " < $ >¢ . In other words sup( )A is the number that is “never quite” exceeded by any member of
the set. It is literally the “least upper bound”.
For example, consider the sequence 31 22 3 4, , , . This sequence has no maximum,
but the supremum of the sequence is 1, since any number less than 1 will eventually be exceeded and any other upper bound would be greater than 1. Similarly the idea of the minimum element is generalised to give the infimum or greatest lower bound.
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The infimum (inf), or greatest lower bound of a set A is defined as the number a such that and a a A a a aa a£ " Œ " > $ <¢ . Note that any other lower bound a ¢ is such that a a£¢ . Hence a is the greatest lower bound.
For example, consider the sequence 1 1 12 4 8, , , . This sequence has no minimum, but
the infimum of the sequence is 0. Note that for finite sets A, sup( ) max( )
a AA A
Œ= and inf( ) min( )
a AA A
Œ= .
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2 Differentiation
Consider the function ( )y f x= , with two points lying on the graph of this function,
( )1 , ( )P x f x= and ( )2 , ( )P x h f x h= + + . Then the average rate of change of ( )y f x=
from 1P to 2P is given by ( ) ( )f x h f x
h
+ -.
We then define 0
( ) ( )limh
f x h f x
hÆ
+ - to be the rate of change of y with respect to x at the
point 1P , which is written as dy
dx or ( )
df x
dx or ( )f x¢ , and is known as the derivative
of y with respect to x.
Since ( ) ( )f x h f x
h
+ - is also the gradient of the straight line joining 1P to 2P ,
(technically known as a “chord”) then 0
( ) ( )limh
f x h f x
hÆ
+ - is the gradient of the curve
( )y f x= at the point 1P , since as 0h Æ , the chord gets closer and closer to the tangent
at 1P , and the gradient of the tangent is the same as the gradient of the curve at 1P .
P1 (x, f(x))
P2 (x+h, f(x+h))
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In summary then:
0
( ) ( )( ) lim
h
f x h f xf x
hÆ
+ -=¢
which is the gradient of the graph of ( )y f x= at the point ( , )x y .
Example
By considering the gradient of the chord joining two points on the curve 2y x= , find
dy
dx.
Solution
Consider the points 21( , )P x x and 2
2(( ), ( ) )P x h x h+ + . The gradient of the chord
joining them is given by 2 2( )x h x
h
+ -.
Simplifying this gives:
2 2 2 2 2 2( ) 2 22
x h x x xh h x xh hx h
h h h
+ - + + - += = = +
Since 2 2
0
( )lim 2h
x h xx
hÆ
+ - = , we have:
2dy
xdx
=
Note that the method used in the above example is called differentiation from first principles. For most work we will develop rules for finding derivatives that avoid the need for differentiation from first principles. However, it is important for you to understand the concept of differentiation first.
Question 6.3
Differentiate 22 3 4y x x= + + from first principles.
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3 Differentiation of standard functions
Some standard functions and their derivatives are shown in the table below:
Function Derivative nx 1nnx - 0n π xc lnxc c
kxe kxke
ln x 1
x
ln ( )f x ( )
( )
f x
f x
¢
Example Differentiate the following with respect to x:
(i) 7y x= (ii) 9xy =
(iii) ln(3 2)y x= +
Solution
(i) 121
2
77
2
dyx
dx x
-= ¥ = (Remember that 12x x= )
(ii) 9 ln 9xdy
dx=
(iii) 3
3 2
dy
dx x=
+
Question 6.4
Differentiate the following functions with respect to x:
(i) 2
3
x (ii) 3 42 x (iii) 22 ln(3 4)x - (iv) 4 23 xe -
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4 Products and quotients
Very often the functions that you need to differentiate are more complicated than those dealt with in Section 3.
4.1 Products
If u and v are functions of x, and ( )f x uv= , then ( )dv du
f x uv vu u vdx dx
= + = +¢ ¢ ¢ . This
is called the product rule.
Example If ( ) lnf x x x= , what is ( )f x¢ ?
Solution
1( ) 1 ln 1 lnf x x x x
x= ¥ + ¥ = +¢
To differentiate a product you differentiate each factor in turn (leaving the other factor the same) and add the results together. This rule also works when there are more than two factors involved, provided you do it properly. For example if y uvw= ,
( )dy d du
u vw vwdx dx dx
= ¥ + ¥ .
Question 6.5
A sum of money is paid continuously at a rate 2( ) 100 tt ter = , for 0 2t< < . At what
rate is the payment increasing at time 1.5t = ?
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4.2 Quotients
If u and v are functions of x, and ( )u
f xv
= , then 2 2
( )du dvdx dxv uvu uv
f xv v
--¢ ¢= =¢ . This is
called the quotient rule.
Example
If ( )3 1
xf x
x=
+, what is ( )f x¢ ?
Solution
2 2
(3 1) 1 3 1( )
(3 1) (3 1)
x xf x
x x
+ ¥ - ¥= =¢+ +
4.3 Chain rule
This is used to differentiate expressions involving nested functions. These might
include functions such as 2 5(3 2 1)x x+ - , where multiplying out the bracket is
impractical, or functions such as 2log(3 2 1)x x+ - where there is no method of
expanding the argument of the log. The chain rule states that:
dy dy dt
dx dt dx= ¥
So if 2 5 2(3 2 1) and 3 2 1y x x t x x= + - = + - ie 5y t= , then:
45 (6 2)dy dy dt
t xdx dt dx
= ¥ = ¥ +
or replacing t:
2 45(6 2)(3 2 1)dy
x x xdx
= + + -
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In practice these stages are not written out in full as the answer can be written out immediately. When differentiating a bracketed expression raised to a power, think of differentiating the bracket as “normal” ie bring the power down and decrease the power by one, then multiply by the derivative of the bracket.
Example
If 2 5(3 4)p r= + , what is dp
dr?
Solution
2 4 2 45(3 4) 6 30 (3 4)dp
r r r rdr
= + ¥ = +
Similarly for expressions involving the exponential function, if 24 2and 4xy e t x= = ie
ty e= , then 8tdye x
dx= ¥ , or replacing t,
248 xdyxe
dx= . Again think of this as
differentiating the power and multiplying by the original function. This is really a method of “differentiation by substitution”. We have to make a substitution so that the expression becomes one that we can differentiate.
Question 6.6
Differentiate the following expressions with respect to x:
(i) 23xy e-=
(ii) { }22exp ( 1)xy el= - .
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Example
Differentiate 2
log(1 2 )xeÈ ˘+Î ˚ .
Solution Using the rules we have used so far we have:
2log(1 2 ) 2 log(1 2 ) log(1 2 )x x xd d
e e edx dx
È ˘ È ˘+ = + ¥ +Î ˚ Î ˚
Consider log(1 2 )xde
dx+ . This can be found as follows:
2log(1 2 ) (1 2 ) (1 2 )
(1 2 )
xx x x
x
d ede e e
dxdx e+ = + + =
+
Returning to our original expression, the required answer is:
2 2 4log(1 2 ) 2 log(1 2 ) log(1 2 )
(1 2 ) (1 2 )
x xx x x
x x
d e ee e e
dx e eÈ ˘ È ˘+ = + ¥ = +Î ˚ Î ˚ + +
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Combinations of these rules can be used in the same question:
Example
If 4
2 1
(3 4)
xy
x
+=-
, find dy
dx.
Solution Using the quotient rule and chain rule:
{ } { }
( )
4 3
8
3
8
5
(3 4) 2 (2 1) 4(3 4) 3
(3 4)
2(3 4) (3 4) 6(2 1)
(3 4)
2(9 10)
(3 4)
x x xdy
dx x
x x x
x
x
x
- ¥ - + ¥ - ¥=
-
- - - +=
-
+= --
Question 6.7
Differentiate the following with respect to x:
(i) 2( 1) xy x e= +
(ii) 2 2
ln
( 1)x
xy
e=
+
(iii) 4 2(2 7) ln(4 3)y x x= + +
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5 Higher order derivatives
Expressions can be differentiated repeatedly. The notation for higher order derivatives is:
2 3 4
2 3 4, ,
d y d y d y
dx dx dx etc
where 2
2
d y
dx is pronounced “dee-2-why-by-dee-x-squared” etc.
or, in function notation, ( ), ( )f x f x¢¢ ¢¢¢ etc.
The second derivative corresponds to the rate of change of the rate of change etc. The idea of a second or third derivative may not seem to have many practical applications. However, we shall see in the next section that the second derivative at least is often a very useful function.
Example
If ln(2 3)x t= + , what is 2
2
d x
dt?
Solution
21
2 2
2 42(2 3)
2 3 (2 3)
dx d xt
dt t dt t- -= = + =
+ +
Question 6.8
What is ( )f x¢¢ in each of the following cases?
(i) 2 2( ) (3 2 3)f x x x= + +
(ii) ( )1
xf x
x=
+
(iii) 1( ) ln xf x x +=
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6 Stationary points
6.1 Maxima, minima and points of inflexion
A stationary point is one where the tangent to a graph is horizontal. Stationary points can be maxima, minima or points of inflexion, as illustrated in the diagram below. They can also be called turning points.
0.2 0.4 0.6 0.8 1
– 0.01
0.01
0.02
0.03
0.04
This is the graph of the function 3 2(1 )y x x= - , and it has 3 stationary points, a
minimum at 1x = , a maximum at 0.6x = , and a point of inflexion (neither a maximum or a minimum) at 0x = . A function ( )f x has a local maximum at the point x b= , if the values of ( )f x close to
x b= are less than ( )f b , or more formally, ( ) ( )f b f be± < , where e is a small
positive number. Similarly, ( )f x has a local minimum at the point x b= , if
( ) ( )f b f be± > .
Generally at these points, ( ) 0f x =¢ , and this is the method used to find maxima and
minima. However, the following graph illustrates that this is not always the case:
– 2 – 1 1 2
1
2
x
This is the graph of | |y x= , and it has a minimum at 0x = .
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Once the points have been found, it is necessary to check whether they are maxima, minima or points of inflexion by finding the second derivative, and substituting in the value of x. If the second derivative is negative, the point is a maximum, if it is positive, the point is a minimum. If the second derivative is zero, further investigation is required.
Example
Find the maxima and minima of the function 3 2( ) 8 12f x x x x= - - + .
Solution
3 2
2
( ) 8 12
( ) 3 2 8 (3 4)( 2)
f x x x x
f x x x x x
= - - +
= - - = + -¢
The stationary points are at ( ) 0f x =¢ which occurs when 43 or 2x = - .
( ) 6 2f x x= -¢¢
If 43 , ( ) 10 0x f x= - = - <¢¢ , so there is a maximum point at ( )4 14
3 27,18- .
If 2, ( ) 10 0x f x= = >¢¢ , so there is a minimum point at (2,0) .
In order to understand why the second derivative is positive for a minimum, it is
necessary to consider the sign of the gradient of the graph (which we know is just dy
dx
or ( )f x¢ ). For a minimum point, the gradient of the graph changes from negative, to
zero, to positive, in other words the gradient is increasing. Since 2
2
d y
dx is the rate of
change of dy
dx,
2
2
d y
dxmust be positive for a minimum. By similar reasoning, it must be
negative for a maximum. If 2
20
d y
dx= , you need to check the sign of
dy
dx to see how the
graph is behaving. For example, 4y x= gives both dy
dx and
2
2
d y
dx to be zero at 0x = .
However dy
dx is negative just below zero, and positive just above zero, so 0x = gives a
minimum.
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To find a point of inflexion, it is necessary to set ( ) 0f x =¢¢ .
You can have points of inflexion that are not horizontal. For example, the graph of the standard normal distribution has a maximum at 0x = and points of inflexion (at angles of about 13.6∞ ) at 1x = ± .
Example
Find the stationary points of 3 2( ) 2 3 72 3f x x x x= + - + , and determine their nature.
Solution To find stationary points, set ( )f x¢ equal to zero.
2
2
( ) 6 6 72
6( 12)
6( 3)( 4)
f x x x
x x
x x
= + -¢
= + -
= - +
The stationary points are at ( ) 0f x =¢ which is when 3 or 4x = - .
To find the points of inflexion, set ( )f x¢¢ equal to zero.
( ) 12 6f x x= +¢¢
The points of inflexion are at ( ) 0f x =¢¢ which is when 0.5x = - .
If 3, ( ) 42 0x f x= = >¢¢ , so there is a minimum point at (3, 132)- .
If 4, ( ) 42 0x f x= - = - <¢¢ , so there is a maximum point at ( 4, 211)- .
There is a point of inflexion at ( )0.5, 39.5- .
Question 6.9
Find any turning points on the curve 3 24 3 90 6y x x x= - - + , and determine their
nature.
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If you are trying to find the maxima and minima of a positive function ( )f x , then you
can find the maxima and minima of ln ( )f x . Since ln x is a steadily increasing
function, the x values at which the maxima/minima occur will be the same. This technique will be used frequently in the statistical subjects. It is useful when the functions to be differentiated are algebraically complex. Taking logs makes the algebra easier.
Example Find the maximum value of:
30 2 15 2 5( ) (1 ) ( ) ( )f e e e el l l ll - - - -= - -
( )f l is a positive-valued function. So we can take logs:
{ }{ }{ }
30 2 15 2 5
30 15 15 10
45 25
ln ( ) ln (1 ) ( ) ( )
ln (1 ) (1 )
ln (1 )
25 45ln(1 )
f e e e e
e e e e
e e
e
l l l l
l l l l
l l
l
l
l
- - - -
- - - -
- -
-
= - -
= - -
= -
= - + -
Now differentiating with respect to l , we get:
45ln ( ) 25
1
d ef
d e
l
lll
-
-= - +-
and this must be zero at a maximum, so:
2570
4525
1
25 25 45
25 70
ln 1.0296
e
e
e e
e
l
l
l l
l
l
-
-
- -
-
=-
- =
=
= - =
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Checking that this is a maximum (and writing ln ( )d
fd
ll
as 45
251el
- +-
to simplify
calculations):
2
2
2
45ln ( ) 25
1
45
1
d df
dd e
e
e
Since this is always negative, the point we have found must be a maximum, and the
value here gives 20ln ( ) 45.6230 ( ) 1.535 10f fl l -= - fi = ¥ .
Question 6.10
Find the maximum value of:
( ) p nf e ll l -=
by first finding ln ( )f l .
You don’t always have to look at the second derivative when the nature of the turning points is obvious from the shape of the graph. For example, in the previous question
( )f l is always positive, (0) 0f = and ( ) 0f • = . So any stationary point we find must
be a “hump in the middle” ie a maximum, because the function is continuous in the range of values that we are interested in.
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6.2 Curve sketching
It is often useful to be able to sketch the graphs of functions, in order to see what is happening. If you have a graphical calculator, this can be straightforward, although you are not allowed to take graphical calculators into the exams. If not you may be able to deduce the shape by transforming the graph of a standard function, as already seen. However, there are some useful techniques available if sketching by hand! Standard functions In Chapter 3 we looked at the standard functions and their graphs, including variations
of those functions, for example, 21 xy e= + .
Other curve sketching techniques The ideas in this chapter are also helpful when trying to decide how a particular function behaves. For example, in order to sketch the graph of a curve, the following techniques may be useful: 1. Find where the function crosses the x and y axes 2. Find any stationary points and their nature 3. Consider the sign and gradient of the function and the ranges of values for which
the function is positive or negative 4. Consider the behaviour of the function at extreme values ie when x or y tend to
zero or infinity, or at “impossible” values, such as where the denominator of a fraction would become zero.
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Example
Sketch the graph of 21
2x
y e-= .
Solution
212
xdyxe
dx-= -
So 0dy
dx= when 0x = for max/min.
2 2 21 1 1
2 2 2
22 2
2( 1)
x x xd yx e e e x
dx
- - -= - = -
When 0x = , 2
20
d y
dx< , so there is a maximum at (0, 1) .
As , 0x yÆ ± • Æ .
Graph:
0
0.2
0.4
0.6
0.8
1
1.2
-2 -1 0 1 2
x
y
Question 6.11
Sketch the graph of 2
2
4
4 3
x xy
x x
-=- +
.
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7 Partial differentiation
If ( , )f x y is a function of two independent variables x and y, then the partial derivatives
of ( , )f x y with respect to x and y are defined to be:
0
( , ) ( , )limh
f f x h y f x y
x h
∂∂ Æ
+ -= 0
( , ) ( , )limh
f f x y h f x y
y h
∂∂ Æ
+ -=
respectively. Effectively they can be calculated by differentiating f with respect to x
treating y as a constant for f
x
∂∂
, and differentiating f with respect to y treating x as a
constant for f
y
∂∂
.
f
x
∂∂
, the partial derivative of f with respect to x, tells you the rate of change of the
function f when x is varied but all other variables are kept constant.
Example
Find f
x
∂∂
and f
y
∂∂
for the function 2 3( , ) 2 ( 2 )f x y x y x y= + + .
Solution
24 3( 2 )f
xy x yx
∂∂
= + + 2 22 6( 2 )f
x x yy
∂∂
= + +
Higher derivatives can be found in a similar way. The notation used here is:
2 3
2 3,
f f
x x
∂ ∂∂ ∂
etc 2 3
2 3,
f f
y y
∂ ∂∂ ∂
etc 2 f
x y
∂∂ ∂
etc
2 f
x y
∂∂ ∂
means partially differentiate f
y
∂∂
with respect to x.
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Example
If ( , ) ( ) xyf x y x y e= + , show that 2
2xyf f
y exx
.
Solution
( ) xy xyfx y ye e
x
∂∂
= + + 2
22
( ) xy xy xyfx y y e ye ye
x
∂∂
= + + +
2( )xy xy xy xyfy e x y y e ye ye
x
so the relationship is true.
Question 6.12
Find the 2 2 2
2 2, , , , and
f f f f f
x y x yx y
∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂∂ ∂
for the function:
4 2 2 3( , ) (3 ) 2 (4 7 )f x y x y x y x= + - + -
Question 6.13
Show that 2 2f f
x y y x
∂ ∂∂ ∂ ∂ ∂
= for 1
( , )xye
f x yx
--= .
When x and y are both functions of some other variable t, the partial derivatives can be used to find the change in the function f when t is varied by calculating the total derivative using the relationship:
df f dx f dy
dt x dt y dt
∂ ∂∂ ∂
= +
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8 Extrema of functions of two variables
Extrema in three-dimensional space are equivalent to turning points in two-dimensional space. Functions of two variables can have three types of extrema; maxima, minima and saddle points. Maxima and minima have their usual interpretations whereas a saddle point can be thought of as a horse’s saddle ie such a point is a minimum in one direction and a maximum in the other. To find these turning points we need to set the two partial derivatives:
,f f
x y
∂ ∂∂ ∂
equal to zero to find the turning point 0 0,x x y y= = .
To discover the nature of the turning point, the roots of the following equation for l need to be found:
0 0 00 0 0
2
2 2 2
2 20
x x x x x xy y y y y y
f f f
y xy x
The following conditions then apply. If the roots are: both positive there is a local minimum both negative there is a local maximum of differing signs there is a saddle point
Question 6.14
Find the turning points and their nature for the function 2 2( , ) 2f x y x y= + .
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9 Lagrange multipliers
We are now going to extend the idea of maximisation to more than two variables. To
find the local extrema of a function ( )1, , nf x x we solve the equations:
1 20, 0, , 0
n
f f f
x x x∂ ∂ ∂
∂ ∂ ∂= = =
However the variables may be constrained in some way. We can then use the method of Lagrangian multipliers. To explain this method we will refer to this simple problem.
Example A farmer wishes to enclose an area within a rectangular field. He wants the area enclosed to be as large as possible given the fact that he has 100m of fencing available. A straight river forms one of the boundaries of the field. What dimensions should he make the field in order to maximise the area?
In this example, we can set up the problem by letting the length and width of the rectangle be x and y respectively. Here x is the length of side perpendicular to the river and y is the length of the side parallel to the river. The problem is to maximise xy
subject to the condition that 2 100x y+ = .
In this very simple case it is fairly easy to see the solution: From the constraint 100 2y x= - , we are effectively maximising (100 2 )x x- .
Differentiating this we get 100 4x- , which gives 25x = when we set it equal to zero. By differentiating again we can see that 25x = does give a maximum, so the solution is to have a field which is 25 metres by 50 metres. However we can approach this in a different way, using Lagrangian functions. We will look at the specific case here and then generalise what you would need to do to tackle any problem. The Lagrangian function is defined to be (2 100)L xy x yl= - + - . (Note that the
constraint equation can be written as 2 100 0x y+ - = .)
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We then solve the problem by solving 0, 0, 0L L L
x y
∂ ∂ ∂∂ ∂ ∂ l
= = = .
Question 6.15
Solve the problem using the Lagrangian function.
For more than two variables, any constraint equation can be written in the form
( )1 2, , , 0ng x x x = , in other words, we want to solve:
Find the (local) extrema of ( )1, , nf x x subject to ( )1 2, , , 0ng x x x =
Lagrangian theory tells us to construct the Lagrangian function:
( ) ( )1 1, , , ,n nL f x x g x x l= -
and then solve the equations:
1 20, 0, , 0, 0
n
L L L L
x x x∂ ∂ ∂ ∂
∂ ∂ ∂ ∂ l= = = = .
However, there may be more than one constraint. Generally, if we want to solve the problem:
Find the (local) extrema of ( )1, , nf x x subject to the m constraints:
( ) ( )1 1 2 1 2, , , 0, , , , , 0n m ng x x x g x x x = =
Lagrangian theory tells us to set up the Lagrangian function:
( ) ( ) ( )1 1 1 1 1, , , , , ,n n m m nL f x x g x x g x x l l= - - -
and then solve the equations:
1 2 10, 0, , 0, 0, , 0
n m
L L L L L
x x x ∂ ∂ ∂ ∂ ∂
∂ ∂ ∂ ∂ l ∂ l= = = = =
We shall provide no justification for this method.
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Question 6.16
Find the extrema of ( ) 2 2 2, ,f x y z x y z= + + subject to 2 3x y z- + = .
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Chapter 6 Summary The limit notation is a short hand way of describing the behaviour of functions. One very important limit involves the definition of differentiation, namely:
0
( ) ( )( ) lim
h
f x h f xf x
hÆ
+ -=¢
Differentiating a function gives the gradient of the function at a general point. There are several rules governing differentiation: The product rule:
If u and v are functions of x, and ( )f x uv= , then ( )f x uv vu= +¢ ¢ ¢
The quotient rule:
If u and v are functions of x, and ( )u
f xv
= , then 2
( )vu uv
f xv
-¢ ¢=¢
The chain rule:
dy dy dt
dx dt dx= ¥
To find the maxima and minima of a function, differentiate the function and set the derivative equal to zero. To find points of inflexion, set the second derivative equal to zero. In some situations to find the maximum or minimum of a function it is easier to find the maximum or minimum of the log of the function. To sketch a curve, use your knowledge of standard functions. You can also find where the function crosses the x and y axes, find any stationary points and their nature, consider the sign and gradient of the function, and consider the behaviour of the function at extreme values or at impossible values.
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Partial derivatives for ( , )f x y can be calculated by differentiating ( , )f x y with respect
to x treating y as a constant for f
x
∂∂
, and differentiating ( , )f x y with respect to y
treating x as a constant for f
y
∂∂
.
To find the extrema of functions of two variables, set ,f f
x y
∂ ∂∂ ∂
equal to zero.
To discover the nature of the turning point, find the roots of the following equation:
0 0 00 0 0
2
2 2 2
2 20
x x x x x xy y y y y y
f f f
y xy x
If the roots are: both positive there is a local minimum both negative there is a local maximum of differing signs there is a saddle point
To find the (local) extrema of ( )1, , nf x x subject to:
( ) ( )1 1 2 1 2, , , 0, , , , , 0n m ng x x x g x x x = =
Lagrangian theory tells us to set up the Lagrangian function:
( ) ( ) ( )1 1 1 1 1, , , , , ,n n m m nL f x x g x x g x x l l= - - -
and then solve the equations:
1 2 10, 0, , 0, 0, , 0
n m
L L L L L
x x x ∂ ∂ ∂ ∂ ∂
∂ ∂ ∂ ∂ l ∂ l= = = = =
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Chapter 6 Solutions Solution 6.1
As x gets closer to 1 (from below), 1x - is going to get very small and negative.
Thus 1
1lim
1x x-Æ= -•
-.
Solution 6.2
We can find this limit if we divide the top and bottom through by 2x :
2 2
2
3 21
3 2 1lim lim
4 33 4 3x x
x x x xx x
xƕ ƕ
- +- + = =- -
This is because all the terms with x ’s in their denominator tend to zero. Solution 6.3
Consider the points 1 2( , ( )) and ( , ( ))P x f x P x h f x h+ + , where 2( ) 2 3 4f x x x= + + .
The gradient of the chord joining them is ( ) ( )f x h f x
h
+ -, which when expanded and
simplified gives:
2 2 2 2 2
2
2( ) 3( ) 4 (2 3 4) 2 4 2 3 3 4 (2 3 4)
4 2 34 3 2
x h x h x x x xh h x h x x
h h
xh h hx h
h
+ + + + - + + + + + + + - + +=
+ += = + +
Now taking the limit as 0h Æ , we get that 4 3dy
xdx
= + .
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Solution 6.4
(i) ( )2 32 3
3 63 6
d dx x
dx dxx x- -Ê ˆ = = - = -Á ˜Ë ¯
(ii) 4 13 3
33 4 8 8
2 23 3
d d xx x x
dx dx
(iii) ( )22 2
2 6 122ln(3 4)
3 4 3 4
d x xx
dx x x
¥- = =
- -
(iv) ( )4 2 4 23 12x xde e
dx- -=
Solution 6.5
To find the rate of increase, we want the derivative ( )tr ¢ . Using the product rule:
2 2
2
( ) 100 100 2
100 (1 2 )
t t
t
t e t e
e t
r = ¥ + ¥¢
= +
So at 1.5t = , 2 1.5(1.5) 100 (1 2 1.5) 8,034er ¥= + ¥ =¢ .
The payment is increasing at a rate of 8,034 at this point. We cannot specify the units since the units of the payment were not given in the question.
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Solution 6.6
(i) Using the rule that we differentiate the power and then multiply by the original function, we get:
236 xdy
xedx
-= -
(ii) We need to differentiate the power:
2 2 22 2 2( 1) 4 4x x xde xe xe
dxl l l- = ¥ =
But we then need to multiply by the original function, so our answer is:
{ }2 22 24 exp ( 1)x xdyxe e
dxl l= -
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Solution 6.7
(i) Using the product rule we get:
{ }
2 2( 1) ( 1) 2( 1)
( 1) 1 2
( 1)( 3)
x x x
x
x
dx e x e e x
dx
x e x
x x e
È ˘+ = + + +Î ˚
= + + +
= + +
(ii) Using the quotient rule we get
2 2 2 2
2 2 2 4
2 2
2 3
2
2 3
1( 1) 2( 1)2 lnln
( 1) ( 1)
1( 1) 4 ln
( 1)
1 4 ln 1
( 1)
x x x
x x
x x
x
x
x
e e e xd x xdx e e
e e xx
e
e x x
x e
(iii) Using the product rule we get
( )
{ }
4 2 4 3 2
32
3
8(2 7) ln(4 3) (2 7) 4(2 7) 2 ln(4 3)
4 3
8(2 7)(2 7) (4 3) ln(4 3)
4 3
2 78(2 7) 2 ln(4 3)
4 3
dx x x x x
dx x
xx x x
x
xx x
x
+ + = + ¥ + + ¥ ¥ ++
+= + + + ++
+Ï ¸= + + +Ì ˝+Ó ˛
We have used the fact that 2ln(4 3) 2 ln(4 3)x x+ = + to make life easier.
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Solution 6.8
(i) Differentiating gives:
2( ) 2(3 2 3)(6 2)f x x x x= + + +¢
Differentiating again, this time using the product rule, we get:
2
2 2
2
2
( ) 2(3 2 3) 6 (6 2) 2(6 2)
36 24 36 72 48 8
108 72 44
4(27 18 11)
f x x x x x
x x x x
x x
x x
= + + ¥ + + ¥ +¢¢
= + + + + +
= + +
= + +
(ii) Differentiating using the quotient rule we get:
{ }
12
12
32
12
12
( 1) ( 1)( )
1
( 1) 2( 1)
1
2
2( 1)
x x xf x
x
x x x
x
x
x
-
-
+ - ¥ +=¢
+
+ + -=
+
+=+
Differentiating again using the quotient rule gives:
{ }
3 12 2
12
52
3
3
2( 1) ( 2) 3( 1)( )
4( 1)
( 1) 2( 1) 3( 2)
4( 1)
4
4( 1)
x x xf x
x
x x x
x
x
x
+ - + ¥ +=¢¢+
+ + - +=
+
+= -+
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(iii) The expression can be rewritten as ( ) ( 1) lnf x x x= + .
Differentiating using the product rule, we get:
1
( ) ( 1) ln
11 ln
f x x xx
xx
= + ¥ +¢
= + +
Differentiating again we get 2
1 1( )f x
xx= - +¢¢ .
Solution 6.9
To find the turning points we have to differentiate:
212 6 90dy
x xdx
= - -
Setting this equal to zero we get:
2 212 6 90 0 2 15 0 (2 5)( 3) 0
3 or 2.5
x x x x x x
x x
- - = fi - - = fi + - =
fi = = -
We need to differentiate again to find out what sort of turning points we have:
2
224 6
d yx
dx= -
Setting this equal to zero for points of inflexion, we get 0.25x = .
If 2
23, 0
d yx
dx= > , so there is a minimum point at (3, 183)- .
If 2
22.5, 0
d yx
dx= - < , so there is a maximum point at ( 2.5,149.75)- .
There is a point of inflexion at (0.25, 16.625)- .
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Solution 6.10
( )ln ( ) ln ln ln
ln
p n p nf e e
p n
l ll l l
l l
- -= = +
= -
Differentiating with respect to l , we get:
ln ( )d p
f nd
ll l
= -
So for maxima or minima, setting this equal to zero, we get p
nl = .
Differentiating again:
2
2 2ln ( ) 0
d pf
dl
l l= - < (assuming that 0p > )
So p
n gives a maximum value of
ppp
en
or p
p
ne
.
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Solution 6.11
2
2
4 ( 4)
( 3)( 1)4 3
x x x xy
x xx x
- -= =- -- +
, so when 0y = , 0 or 4x x= = . There is no need to
consider when 0x = , since that possibility has already been covered.
2
41
4 31
xy
x x
-=
- +, so as x Æ• , 1y -Æ .
The graph has a discontinuities when the denominator is zero, ie at 3 and 1x x= = . To find maxima and minima, we need to differentiate. Using the quotient rule we get:
2 2
2 2
( 4 3)(2 4) ( 4 )(2 4)
( 4 3)
dy x x x x x x
dx x x
- + - - - -=- +
Multiplying out the numerator:
2 2
6 12
( 4 3)
dy x
dx x x
-=- +
Setting this equal to zero gives 2x = . Differentiating again:
2 2 2 2
2 2 4
( 4 3) 6 (6 12) 2( 4 3)(2 4)
( 4 3)
d y x x x x x x
dx x x
- + ¥ - - ¥ - + -=- +
so when 2x = , 2
26 0
d y
dx= > , ie there is a minimum at (2, 4) .
There are no solutions to the equation 2
20
d y
dx= , so there are no points of inflexion.
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The graph is: You could also have deduced the general shape by writing the function in the form
31
( 1)( 3)y
x x= -
- -.
Solution 6.12
3 2 2
3 2 2
4(3 ) 3 4 3(4 7 ) 7
12(3 ) 4 21(4 7 )
fx y xy x
x
x y xy x
∂∂
= + ¥ - + - ¥ -
= + - - -
2
2 22
2 2
36(3 ) 3 4 42(4 7 ) 7
108(3 ) 4 294(4 7 )
fx y y x
x
x y y x
∂∂
= + ¥ - - - ¥ -
= + - + -
3 24(3 ) 4f
x y x yy
∂∂
= + - , therefore 2
2 22
12(3 ) 4f
x y xy
∂∂
= + -
x-4 -2 2 4 6
y
-4
-2
2
4
6
8
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2 f
x y
∂∂ ∂
means partially differentiate f
y
∂∂
with respect to x, so:
2
236(3 ) 8f
x y xyx y
∂∂ ∂
= + -
Solution 6.13
To do this we need to find the first partial derivatives and then partially differentiate them again.
xyxyf xe
ey x
∂∂
--= = and
2 2
( ) (1 ) (1 ) 1xy xy xyf x ye e xy e
x x x
∂∂
- - -- - + -= =
Differentiating again, doing f
x y
for simplicity we get:
2
xyfye
x y
∂∂ ∂
-= -
Similarly:
2 2
2 2
(1 ) xy xy xy xy xyxyf xy xe e x xe xe x ye
yey x x x
∂∂ ∂
- - - - --+ ¥ - + ¥ - -= = = -
Therefore we can see that 2 2f f
x y y x
∂ ∂∂ ∂ ∂ ∂
= .
This result is generally true.
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Solution 6.14
Partially differentiating:
4 2f f
x yx y
∂ ∂∂ ∂
= =
Setting these equal to zero, we get 0x = and 0y = , ie (0,0) is a turning point.
Finding the second derivatives:
2 2 2
2 24 2 0
f f f
y xx y
∂ ∂ ∂∂ ∂∂ ∂
= = =
Substituting these into the required equation:
(4 )(2 ) 0 2,4l l l- - = fi =
Since both these values are positive, the turning point is a local minimum. Solution 6.15
2 0
0
2 100 0
Ly
x
Lx
y
Lx y
∂ l∂
∂ l∂
∂∂ l
= - =
= - =
= + - =
Substituting the first two equations into the third, we get:
2 2 100 0 25l l l+ - = ¤ =
This gives 25x = and 50y = as required.
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Solution 6.16
We need to find the partial derivative equations:
0, 0, 0, 0L L L L
x y z
∂ ∂ ∂ ∂∂ ∂ ∂ ∂ l
= = = =
This gives us:
2 2 0, 2 0, 2 0, 2 3x y z x y zl l l- = + = - = - + =
Substituting the first three equations into the fourth:
1 12 3 1
2 2l l l l+ + = fi =
Hence:
1 11, ,
2 2x y z= = - =
1l = doesn’t mean a lot in this situation. Often this method is applied in mechanics to
real physical systems. In such problems the Lagrangian multiplier, l , does have physical meaning.
FAC-07: Integration Page 1
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Chapter 7
Integration
You need to study this chapter to cover: ● indefinite and definite integrals, including the area under a graph
● integrating standard functions
● integrating by inspection, by substitution, and by parts
● convergence of definite integrals
● multiple integrals
● numerical methods for integration
● Taylor series and Maclaurin series
● simple ordinary differential equations.
0 Introduction
This chapter covers all the methods of integration that you need in order to study the Core Technical subjects.
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1 Integrals
Suppose we have a function ( )f x , and that ( ) ( )=¢F x f x . Then ( )F x is an anti-
derivative of ( )f x . The general form of the anti-derivative is the indefinite integral of
( )f x which is written as ( ) Ú f x dx . Integration and differentiation are connected in
that they are the “reverse” of each other.
Since ( )( ) ( ) ( )+ = =¢d
F x c F x f xdx
, then, in general, ( ) ( )= +Ú f x dx F x c , where c is
known as the constant of integration. Integrals with limits are called definite integrals. They are evaluated between two
values. For example [ ]( ) ( ) ( ) ( )= = -Úb
ba
a
f x dx F x F b F a . This is read as “the integral
from a to b of ( )f x ”. The numbers a and b are called the lower and upper limits of
integration respectively. For such integrals there is no “+ c ”, since the constants cancel when the integral is evaluated. The practical application of integration is that it also gives the sum of infinitesimally small elements and, as illustrated below, this can be used to find areas under curves. Consider the area enclosed by the curve ( )=y f x , between =x a , and =x b , as shown
in the diagram below. We are going to look at the small section of area under the curve contained between x and +x xd . 0
y
xb
y = f(x)
a x x + x
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If ( )A x is the area under the curve from =x a to x, then the area bounded by x and
+x xd is given by ( ) ( )+ -A x x A xd . Looking at the diagram and considering the
areas of the smaller and bigger rectangles in the diagram:
( ) ( ) ( ) ( )+ £ + - £x f x x A x x A x x f xd d d d
Dividing by xd , we get:
( ) ( )( ) ( )
+ -+ £ £A x x A xf x x f x
x
ddd
As 0Æxd , ( ) ( )+ -A x x A x
x
dd
lies between two values which get closer and closer
together, so:
0
( ) ( )lim ( )Æ
+ - =x
A x x A xf x
xd
dd
The LHS is just the definition of dA
dx, so:
( )=dAf x
dx or ( ) ( ) = ÚA x f x dx
This is a general expression and we wanted the area enclosed between =x a and =x b ,
so the area is actually given by ( ) ( ) ( )= -Úb
a
f x dx A b A a .
In the diagram given, the curve was sloping down between x and +x xd . You get the same result if the function is sloping up. It is also important to note that if the area is below the x-axis then the integral, when evaluated, will give rise to a negative number. If the graph of ( )=y f x crosses the
x-axis between =x a and =x b , then ( ) Úb
a
f x dx will not give the total area because
some of the area will be below the x-axis and some above. Under these circumstances it is necessary to split up the integral. For example to find the area enclosed between the
x-axis, 2= -x , 2=x and the curve 3( ) =f x x , you would evaluate 2 0
3 3
0 2
-
-Ú Úx dx x dx ,
using “–” since the second integral will be negative.
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You also need to be careful if the function you are integrating tends to ±• in the range you are looking at. In this case the integral may be infinite or it may have a finite value.
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2 Integration of standard functions
Some standard functions and their integrals are shown in the table below:
Function Integral k +kx c
nx 1
1
++
+
nxc
n, 1π -n
1
x ln x c+
( )
( )
¢f x
f x ln ( )f x c+
xa ln
+xa
ca
kxe +kxe
ck
Example Integrate the following functions:
(i) x
(ii) 27 x
(iii) 43 xe
(iv) 2
1+x
Solution
(i) Writing x as 12x and using the formula above:
32
3233
2
+ = +xc x c
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(ii) This is the same as 2(7 )x . So using the formula with 27=a :
27
2ln 7+
x
c
(iii) 434 +xe c
(iv) 2ln 1x c+ +
If you need to make it clear which quantity in an expression is the variable that is changing, you can say “integrating with respect to x” etc.
Question 7.1
Integrate the following functions with respect to x:
(i) 3 22
24+ x
x
(ii) 212
xe
(iii) 126
x
(iv) 2
2 5
5 7
++ +x
x x
When evaluating a definite integral, the integrated function is written in square brackets with the limits on the right. This notation means evaluate the function shown at the top and bottom limits and subtract. American textbooks use a line after the function with limits at the top and bottom, rather than the square brackets notation.
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Example
Evaluate ln 2 5
02Ú xe dx .
Solution
ln 2ln 2 5 5 5ln 2 02 2 25 5 50 0
2 È ˘= = -Î ˚Ú x xe dx e e e
Since 55ln 2 ln 2 52 32= = =e e , our expression simplifies to be:
64 6225 5 5- =
Constant factors can be pulled out of the square brackets if it helps to simplify the working.
Question 7.2
Evaluate 3 2
13Ú x dx . What does this integral represent?
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3 Further integration
3.1 Integration by inspection
Unlike differentiation, where there are foolproof procedures for differentiating any function, integration requires you to be aware of the techniques available and think through each one until you find the correct method. As yet we have not covered the
methods needed to integrate functions such as 2xxe or 2 52 ( 6)+x x . These can be done
by inspection. This includes spotting an exact derivative as part of your integral. If you are not convinced by any of the answers given below don’t forget you can check your answers by differentiating.
Example Find the following:
(i) 2
3Ú xxe dx (ii) 4( 2)+Ú x dx (iii) 2 5( 6)+Ú x x dx
Solution
(i) If the integral was 2
2 xxe , we could integrate it directly since the expression on the front would be the derivative of the power. So we pull out the 3, insert a 2 in the integral and divide by 2 outside:
2 2 23 3
3 22 2
= = +Ú Úx x xxe dx xe dx e c
(ii) The derivative of the bracket is 1 so we don’t have to multiply or divide by
factors in this part:
4 51( 2) ( 2)
5+ = + +Ú x dx x c
(iii) The derivative of the bracket is 2x so we insert a 2 into the integral and divide
by 2 outside:
2 6 2 6
2 5 2 51 1 ( 6) ( 6)( 6) 2 ( 6)
2 2 6 12
+ ++ = + = + = +Ú Úx x
x x dx x x dx c c
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This technique fails if you try to “compensate” with anything other than numbers! You cannot pull outside the integral factors containing x or any function of x.
Question 7.3
Find the following:
(i) 2 5
4 6
( 3 4)
++ +Ú
xdx
x x
(ii) 23 47 -Ú xxe dx
(iii) 22Ú xx e dx
3.2 Integration using partial fractions
Partial fractions is the name of the technique used to split up a single fraction into
separate parts. For example 7
( 1)( 2)
++ -
x
x x is equivalent to
3 2
2 1-
- +x x. Using this
technique, expressions that cannot be integrated directly can be written as expressions that can.
Example
Find 9 2
(3 2)(3 2)
-+ -Ú
xdx
x x.
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Solution We need to split up the single fraction into two fractions that are added together. Using A and B as the numerators we get:
9 2 (3 2) (3 2)
(3 2)(3 2) 3 2 3 2 (3 2)(3 2)
- - + += + =+ - + - + -
x A B A x B x
x x x x x x
Comparing the numerators from the first and last terms we get 9 2 (3 3 ) 2 2- = + - +x A B x A B , and comparing coefficients we get 2, 1= =A B .
Substituting this into our integral we get:
1 2
3 2 3 2+
- +Ú dxx x
We can now evaluate the integral:
2
1 2 1 2ln 3 2 ln 3 2
3 2 3 2 3 3
1ln (3 2)(3 2)
3
dx x x cx x
k x x
+ = - + + +- +
= - +
Ú
The two constants in the last two lines are not the same, which is why we have changed the letter.
Question 7.4
What is the relationship between c and k in the last example?
The denominator in the last example could have been given in the question as 29 4-x , so you will sometimes have to be careful to spot that partial fractions are required.
Question 7.5
Integrate 2
7
1
--
x
x.
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3.3 Integration by substitution (change of variable)
This method involves the replacement of one variable by another, enabling expressions to be simplified and hence integrated. The steps you need to follow are: ● Decide what substitution to use, making u your new variable. Generally you
will make u the “complicated” part of the expression.
● Find dx
du, and hence change dx into du .
● Express the function in terms of u.
● If you have limits, change them so that they correspond to u.
● Evaluate the simpler integral.
Example
Evaluate 1
4
0
(2 3)+Ú x x dx .
Solution
Let 2 3= +u x , then 2=du
dx which means that
1
2=dx
du, and
3
2
-= ux . We need to
change the limits, when 0=x , 3=u , and when 1=x , 5=u . The integral then becomes:
1 54 4
0 3
5 5 4
3
56 5
3
6 5 6 5
3(2 3)
2
3 1
2 2
1 3
4 6 5
1 5 3 5 1 3 3 3 11188
4 6 5 4 6 5 30
u dxx x dx u du
du
u udu
u u
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This example could also have been carried out using integration by parts. For details of this method, see the next section.
Question 7.6
Find 2
( 1)-Ú
xdx
x.
Question 7.7
Evaluate 1
30
4
2 1+Ú xdx
e, by using the substitution 32 1= +xu e .
3.4 Integration by parts
This method is used to integrate products. The product is broken down into one part that can be integrated and one part that can be differentiated. The formula for integration by parts is:
= -Ú Údv du
u dx uv v dxdx dx
This formula is in the Tables page 3. You may see this written as = -Ú Úu dv uv v du .
Note that the formula only replaces one integral with another integral. So you should aim to replace an integral that you cannot do by one that you can do, ie you should try to get a simpler integral than the one you started with.
Example
Find Ú xxe dx .
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Solution Both parts of this integral can be differentiated and integrated, so we have to pick u and v so that the resulting integral is possible (and simpler than the original one!)
Let =u x , and = xdve
dx, so that 1=du
dx and = xv e . So:
( 1)
= -
= - +
= - +
Ú Úx x x
x x
x
xe dx xe e dx
xe e c
x e c
Sometimes the choice of u and dv
dx can be less obvious.
Example
Find 23Ú xx e dx
Solution
The 2xe term cannot be directly integrated, and putting 3=dv
xdx
will mean that the
second integral will be more complicated than the first, so we need to split up the 3x
term. The expression 2xxe can be integrated, so we can proceed as follows. Let
2=u x , and 2
= xdvxe
dx, so that 2=du
xdx
and 2
12= xv e . Then:
2 2
2 2
2 2
2
3 2
212
21 12 2
21( 1)
2
= ¥
= -
= - +
= - +
Ú Ú
Ú
x x
x x
x x
x
x e dx x xe dx
x e xe dx
x e e c
x e c
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Notice that this example used integration by inspection. Often a combination of several techniques will be required to evaluate an integral.
Question 7.8
Find:
(i) 6( 1)+Ú x x dx
(ii) ln Ú x x dx
(iii) 2
1
lnÚ x dx (Hint: think of it as 1 ln¥ x )
There is not always a unique way of integrating. Part (i) of this question could be done by substitution. Question 7.6 could have been carried out using integration by parts.
3.5 Differentiating an integral (Leibniz’s formula)
We have already mentioned the concept of integration being the reverse of differentiation. We will now look at differentiating an integral.
This section uses the result that ( , ) ( , ) =Ú Úb b
a a
fdf x t dt x t dt
dx x
∂∂
, where a and b are
constants. This can be thought of as taking the d
dx inside the integral. This can be
generalised to:
( ) ( )
( ) ( )
( , ) ( ) ( , ( )) ( ) ( , ( )) ( , ) = - +¢ ¢Ú Úb x b x
a x a x
fdf x t dt b x f x b x a x f x a x x t dt
dx x
∂∂
This formula can be found on page 3 of the Tables. The proof of this result is beyond the scope of this course, but it is really just an application of the function of a function rule.
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The formula is very useful in cases where the integral cannot easily be evaluated
directly, for example 2
0
•-Ú xtd
e dtdx
.
Example
Evaluate 2
0
+Úx
dx t dt
dx.
Solution
Here ( ) 0=a x , ( ) =b x x , 2( , ) = +f x t x t , so:
2 2
0 0
2 2
2
+ 1( ) 0 2
2
3
= + - +
= + +
= +
Ú Úx x
dx t dt x x x dt
dx
x x x
x x
In this case we can show that this is the same as integrating directly:
2 2 2 3 2
00
1 1
2 2
xx
x t dt x t t x x
So 2 3 2 2
0
13
2
xd d
x t dt x x x xdx dx
.
Question 7.9
Evaluate 2 3
2
0
( 1) + +È ˘+Î ˚Ú
xd
x tx dtdx
.
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4 Convergence
So far we have only dealt with integrals that have finite limits. We define:
( ) lim ( ) •
Æ•=Ú Ú
k
a akf x dx f x dx and ( ) lim ( )
Æ-•-•
=Ú Úb b
kk
f x dx f x dx
If these limits exist, the integrals are said to converge, otherwise the integrals do not converge. Convergence is also an issue when the function being integrated tends to ±• at a
certain point eg 1
x when 0=x .
Example
Does 21
1
•Ú dx
x converge? If so, what is its value?
Solution
2 21 11
1 1 1 1 lim lim lim 1
kk
k k kdx dx
x kx x
Since this limit exists, the integral converges, and its value is 1.
You can usually tell whether an integral converges by trying to work it out and seeing whether you end up dividing by zero in the calculation. Note that the following integrals do not exist:
20
1
•Ú dx
x
0
1 Ú
Ndx
x
1
1•Ú dx
x
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Question 7.10
Explain why these integrals do not exist.
You will find in practice that any integral that has a real-life meaning will have a finite value, and that the integrated function will work out to be zero at the infinity end.
Question 7.11
Evaluate 2
0
• -Ú xxe dx .
Question 7.12
Use integration by parts to show that ( ) ( 1) ( 1)G = - G -x x x . Remember that the
definition of the gamma function is 1
0
( )•
- -G = Ú x tx t e dt .
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5 Double integrals
If you have a function of two variables, ( , )f x y , the graph of the function gives you a
surface, which might look like this:
-1
0
1x
-2
-1
0
1
2
y
0
1
z
-1
0
1x
The two horizontal axes are the x- and y-axes, and the z axis goes vertically upwards. We can extend the concept of an integral to calculate the volume of the solid region under a part of the surface. In order to calculate this volume we need to evaluate a double integral of the function
over the region R. This is written as ( , ) ÚÚR
f x y dx dy .
The mathematical definition of a double integral is just an extension of the definition for an ordinary integral. It involves taking the limit of the sum of the volumes of a large number of rectangular blocks. In practice you work out a double integral by first integrating with respect to one of the variables (x or y) then integrating with respect to the other variable. The limits of integration must be chosen so that they describe the boundaries of the region R. This is the footprint of the volume on the x-y plane.
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To evaluate these integrals, we treat x as a constant when integrating with respect to y ,
and y as a constant when integrating with respect to x .
The case where the region R is a rectangle parallel to the axes is straightforward.
Question 7.13
Evaluate the double integral 15 5
10 0
2
3000= =
+Ú Ú
x y
x ydy dx .
However, in other cases the limits will be a function of x when first integrating with respect to y , and of y when first integrating with respect to x . If you need to work
out the limits, you will need to consider the required region, as illustrated in the following example.
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Example
Integrate 2xy over the region R shown below:
(i) by first integrating with respect to y and then integrating with respect to x (ii) by first integrating with respect to x and then integrating with respect to y. Solution (i) In this part x is the variable in the “outside” integral and the range of x values is
0 2£ £x . For any given value of x the range of values of y is 1 3 1£ £ +y x .
This gives us our limits of integration.
2 3 12
0 1
3 12 3
0 1
2 3
0
24 3 2
0
25 43
0
3
(3 1)
3 3
9 9 3
9 9
5 4
101.6
x
x
I xy dy dx
xydx
x x xdx
x x x dx
x xx
0
y
x
y = 3x + 1
2
1
7
R
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(ii) In this part y is the variable in the “outside” integral and the range of y values is
1 7£ £y . For any given value of y the range of values of x is 1
23
- £ £yx .
This gives us our limits of integration.
7 22
113
27 2 2
113
7 2 22
1
73 5 4 3
1
2
( 1)2
18
2 1 2
3 18 5 4 3
102.26 0.66 101.6
y
y
I xy dx dy
x ydy
y yy dy
y y y y
The following results can be used. The proofs are beyond the scope of this course.
( , ) ( , ) = ( , ) + ( , ) +ÚÚ ÚÚ ÚÚR R R
af x y bg x y dx dy a f x y dx dy b g x y dx dy
1 2
( , ) = ( , ) + ( , ) ÚÚ ÚÚ ÚÚR R R
f x y dx dy f x y dx dy f x y dx dy
Here R is the combined region consisting of the separate regions 1R and 2R , ie
1 2= »R R R and 1 2« = ∆R R .
Question 7.14
Evaluate the double integral 3 2
2 3
0
+Ú Ú x y
y
xe dxdy .
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Sometimes double integrals with explicit limits are written with the dx and dy at the
front next to the integral sign each belongs to. In this case the last question would be
written as 3 2
2 3
0
+Ú Ú x y
y
dy dx xe .
The results for double integrals can be extended to triple integrals.
Question 7.15
Evaluate 3 3 2
1 0 1
++Ú Ú Ú
z zx ye dx dy dz .
If the integrand (the expression being integrated) factorises into functions of the three separate variables, and the limits are constants, you can multiply together the individual integrals. This is not obvious but it is true. This would mean:
2 2 2 2
1 1 1 1
( ) ( ) ( ) ( )=Ú Ú Ú Úx y x y
x y x y
dx dy f x g y f x dx g y dy
Question 7.16
Evaluate 1 2 3
0 0 0Ú Ú Údx dy dz xyz .
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5.1 Swapping the order of integration
In a similar way to when we changed the order of summation in Chapter 4, we can change the order of integration.
Example
Given that 0 5t s< < < , calculate the value of 5
2
0 0
s
s t
s t dt ds= =Ú Ú , first by using this order
and then by swapping the order of the two variables. Solution Firstly, we will use the order given:
55 5 52 4 52 2
0 0 0 00 0
312.52 2 10
ss
s t s s
t s ss t dt ds s ds ds
Previously we integrated from 0t = to t s= , then from 0s = to 5s = . Reversing this, we integrate from s t= to 5s = then from 0t = to 5t = . To get the limits for s , see what it is bounded by in the inequality 0 5t s< < < . So:
5 5 5
2 2
0 0 0
s
s t t s t
s t dt ds t s ds dt= = = =
=Ú Ú Ú Ú
Carrying out this integration:
5 55 5 5 53 3 2 52
0 0 0 0
125 1 125312.5
3 3 3 2 5t s t t tt
s t t tt s ds dt t dt t dt
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Question 7.17
(i) Carry out the integration 30
0 0
yy x x
y x
e e e dx dym n dm- - -
= =Ú Ú , where 0 30x y< < < .
(ii) Confirm that you get the same answer if you reverse the order of integration and
use the values 0.001, 0.06, 0.005n d m= = = .
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6 Numerical methods for integration
Sometimes it is not possible to find the area under the curve ( )=y f x by using
integration techniques because you cannot integrate the function ( )f x . In such cases
we can find an approximation to the integral by using a numerical method. The method that we will look at here is called the trapezium rule. The technique involves approximating the area under the curve by finding the sum of the areas of several trapezia.
6.1 The trapezium rule
In the following diagram, we have split the area under the curve into trapeziums.
The area of a trapezium is given by the sum of the lengths of the two parallel sides, multiplied by the distance between them, divided by 2, ie it is the base multiplied by the average height. Therefore the total area of the trapezia shown in the diagram (ie the estimate of the area under the curve between 1x and 5x ) is:
1 2 2 3 3 4 4 51 1 1 1
( ) ( ) ( ) ( )2 2 2 2
+ + + + + + +y y d y y d y y d y y d
This can be rewritten as:
1 2 3 4 51
( 2 2 2 )2
+ + + +y y y y y d
y1 y2 y3 y4 y5
x
y
d d d d
x1 x5
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This can also be expressed in words: “The sum of the first and last ordinate (or y value), plus twice the sum of the
other ordinates, multiplied by 12 d .”
We can choose the number of ordinates to suit the area we are trying to estimate, and obviously the smaller the value of d , the more accurate the approximation will be.
Example
Use the trapezium rule to estimate the area under the curve 2=y x between 1=x and
3=x . Use 9 ordinates. Also find the true value of the area and comment on your answers. Solution Using 9 ordinates means that we need to split the area into 8 sections, each of width 0.25. The trapezium rule will then give the approximate area to be:
( )2 2 2 2 2 2 2 2 211 3 2(1.25 1.5 1.75 2 2.25 2.5 2.75 ) 0.25
2
8.6875
+ + + + + + + + ¥
=
The true area is:
3 3
2 3 3 3
11
1 1(3 1 ) 8.67
3 3È ˘= = - =Í ˙Î ˚Ú x dx x
You can see that here the trapezium rule has slightly over estimated the true area under
the curve. If you plotted an accurate graph of the function 2=y x and drew on the
trapeziums you should be able to see that the over-estimation has occurred due to the shape of the curve.
Question 7.18
Estimate the area under the curve 2=yx
between 2=x and 3=x , using 5 ordinates.
Will this be an overestimate or an underestimate of the true value.
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7 Taylor and Maclaurin series
You may need to expand a function as a power series. Taylor and Maclaurin series can be used to do this expansion.
7.1 Taylor series
If you want to approximate the value of ( )f x when ªx a , then Taylor’s series states
that:
2 3
( ) ( ) ( ) ( ) ( )2! 3!
+ = + + + + ◊◊ ◊¢ ¢¢ ¢¢¢h h
f a h f a hf a f a f a
Here we are thinking that = +x a h , where h is a small quantity.
Example
Expand ln x for ªx e in ascending powers of -x e , up to the term in 3( )-x e .
Solution Applying the formula for Taylor’s series with =a e and = -h x e so that
( ) ln ( ( ))= = + -f x x f e x e , we get:
( ) ln=f x x ( ) 1=f e
1( ) =¢f x
x
1( ) =¢f e
e
2
1( )
-=¢¢f xx
2
1( )
-=¢¢f ee
3
2( ) =¢¢¢f x
x
3
2( ) =¢¢¢f e
e and so on
So:
2 32 3
1 1 1ln 1 ( ) ( ) ( )
2 3= + - - - + - +x x e x e x e
e e e
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Question 7.19
Expand 2 5(1 ) (1 )- -+ + +i i as a series about 0.1=i .
What can you say about the value of this function if i increases from 0.1 to a slightly higher value (for example 0.11)?
We can also have a Taylor’s series expansion when there are two variables. If you want to approximate the value of ( , )f x y about ( , )a b then Taylor’s series states
that:
( , ) ( , )
2 2 22 2
2 2( , ) ( , ) ( , )
1( , ) ( , ) ( ) ( )
1!
1( ) 2 ( )( ) ( )
2!
È ˘Í ˙= + - + -Í ˙Î ˚
È ˘Í ˙+ - + - - + -Í ˙Î ˚
+
a b a b
a b a b a b
f ff x y f a b x a y b
x y
f f fx a x a y b y b
x yx y
∂ ∂∂ ∂
∂ ∂ ∂∂ ∂∂ ∂
As for the single variable case, a Taylor series is often used to look at the effect of a small change in the value of one of the arguments. So, for example, if = +x a h and = +y b k (where h and k are small) then we would have:
( , ) ( , )
2 2 22 2
2 2( , ) ( , ) ( , )
( , ) ( , )
12
2
+ + = + +
È ˘Í ˙+ + +Í ˙Î ˚
a b a b
a b a b a b
f ff a h b k f a b h k
x y
f f fh hk k
x yx y
∂ ∂∂ ∂
∂ ∂ ∂∂ ∂∂ ∂
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Example
Expand 3xye about (2,1) as far as the terms in 2( 2)-x , 2( 1)-y etc.
Solution The required partial derivatives in their general form and then evaluated at 2=x and
1=y are:
3 6
3 6
22 3 6
2
22 3 6
2
23 3 6
3 3
3 6
9 9
9 36
3 3 3 21
= =
= =
= =
= =
= ¥ + =
xy
xy
xy
xy
xy xy
fye e
x
fxe e
y
fy e e
x
fx e e
y
fy xe e e
y x
∂∂
∂∂
∂∂
∂∂
∂∂ ∂
So the expansion is:
6 6 6 6 2 6 6 23 ( 2) 6 ( 1) 4.5 ( 2) 21 ( 2)( 1) 18 ( 1)+ - + - + - + - - + -e e x e y e x e x y e y
Question 7.20
Expand 1
-x y about (3, 2) as far as the terms in 2( 3)-x , 2( 2)-y etc.
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7.2 Maclaurin series
A Maclaurin series is just a Taylor series about 0=x ie change a to 0 and h to x. If ( )f x can be expanded as an infinite, convergent series of powers of x, then:
2 3(0) (0)( ) (0) (0)
2! 3!
¢¢ ¢¢¢= + + + + ◊◊ ◊¢ f ff x f f x x x
Maclaurin’s series are most useful for finding the series expansions for basic functions
such as xe , ln(1 )+ x etc.
Example
Obtain the expansion of xe as far as the term in 3x . Solution
Let ( ) = xf x e , then ( ) ( ) =n xf x e , where ( ) ( )nf x means the n th derivative, and ( ) (0) 1=nf , so:
2 3
( ) 12! 3!
= + + + +x xf x x
Question 7.21
Obtain the expansion of ln(1 )+ x as far as the term in 3x .
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8 Differential equations
Any equation which contains n
n
d y
dx in some form, for example, 2 3= +dy
x ydx
, is called
a differential equation. In this section we are going to consider some of the many techniques used to solve these equations. Solving differential equations means finding
( )=y f x .
8.1 Solution by direct integration
If we integrate dy
dx, then we get y and if we integrate
2
2
d y
dx, then we get
dy
dx, etc. This
enables us to solve differential equations of the form ( )=n
n
d yf x
dx.
Example
Solve the differential equation 2
22
2=d rt
dt.
Solution Integrating once with respect to t:
323= +dr
t cdt
Integrating again:
416= + +r t ct d
As seen in this example, we can end up with several constants in the solution. This is called a “general solution”. You can find a “particular solution” if there are “boundary conditions” given in the question, giving you particular values of the variables involved.
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Example
Solve the differential equation = xdye
dx, given that 1=y when ln 2=x .
Solution
The general solution of the differential equation is = +xy e c , but we know that when
ln 2, 1= =x y .
So:
1 2 1= + fi = -c c and the particular solution is:
1= -xy e
Question 7.22
Find the general solution of the differential equation 2
22=d x
atdt
. Given that when
0, 4= =t x , when 97121, = =t x , and when 1
121, = - =t x , find a and the particular
solution.
8.2 Solution by separation of variables
If ( , )=dyf x y
dx, and ( , )f x y can be expressed as ( ) ( )p x q y , then the differential
equation can be solved by separating the variables, ie by writing it in the form:
1
( ) ( )
=Ú Údy p x dxq y
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Example
Solve the differential equation 2 1
=+
dr r
dt t, where 0r > and 0t > .
Solution
We can rearrange this in the form 2 1
=+
dr dt
r t, so:
( )12
1 1
2 1
ln ln(2 1) ln ln 2 1
dr dtr t
r t c r t c
=+
= + + fi = + +
Ú Ú
Note that we do not need to include modulus signs in the integrated expressions as both r and 2 1t + are positive. Note also that we need to include a constant on one side of the equation, not both. Taking exponentials of both sides:
( )ln 2 1ln 2 1 2 1t cr ce e r e t r k t+ +
= fi = + fi = +
where ck e= .
Question 7.23
Find the particular solution of the differential equation 2( 9) 2+ = +dyx xy x
dx, where
0y > , given that when 4, 0.5= =x y .
Question 7.24
Find the general solution to the differential equation ( )= -dNa bN N
dt, where
0a
Nb
< < . This is known as the logistic equation or the Verhulst equation after the
mathematician who used it in his study of populations.
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8.3 Solution by integrating factor
A differential equation of the form ( ) ( )+ =dyf x y g x
dx can be solved by use of an
integrating factor. The integrating factor here is exp ( )È ˘Î ˚Ú f x dx , and the solution of the
differential equation is then the solution of:
exp ( ) ( ) exp ( )È ˘ È ˘=Î ˚ Î ˚Ú Ú Úy f x dx g x f x dx dx
Example
Find the general solution of 34 2+ = xdyy e
dx.
Solution
The integrating factor is 4exp 4È ˘ =Î ˚Ú xdx e . So we multiply through by 4xe which
gives 4 4 74 2x x xdye ye e
dx+ = .
We see that the LHS is the derivative of 4xye , derived from the product rule. So
4 7( ) 2=x xdye e
dx.
We then need to solve 4 72x xye e dx= Ú , which gives 4 72
7= +x xye e c . The solution of
our differential equation is 3 42
7-= +x xy e ce , where c is an arbitrary constant.
Question 7.25
Find the particular solution of the differential equation 23 2= +dyxy y xy
dx, where 0x > ,
given that 1=y when 27=x .
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Chapter 7 Summary Integration and differentiation are the reverse of each other. Integration can be used to find the area under a curve. A definite integral has limits, in which case your answer does not require a constant of integration. There are several methods of integration, including integration by inspection, by substitution, by partial fractions, and by parts. The formula for integration by parts is:
= -Ú Údv du
u dx uv v dxdx dx
To evaluate integrals with infinite limits, you need to consider if they converge. You need to be able to evaluate double integrals by treating them as 2 nested integrals. Taylor and Maclaurin series are used to expand functions as infinite series. The formula for a Taylor series (in one variable) is:
2 3
( ) ( ) ( ) ( ) ( )2! 3!
+ = + + + + ◊◊ ◊¢ ¢¢ ¢¢¢h hf a h f a hf a f a f a
The formula for a Taylor series (in two variables) is:
( , ) ( , )
2 2 22 2
2 2( , ) ( , ) ( , )
1( , ) ( , ) ( ) ( )
1!
1( ) 2 ( )( ) ( )
2!
a b a b
a b a b a b
f ff x y f a b x a y b
x y
f f fx a x a y b y b
x yx y
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An alternative form for a Taylor series is:
( , ) ( , )
2 2 22 2
2 2( , ) ( , ) ( , )
( , ) ( , )
12
2
a b a b
a b a b a b
f ff a h b k f a b h k
x y
f f fh hk k
x yx y
The formula for a Maclaurin series is:
2 3(0) (0)( ) (0) (0)
2! 3!
¢¢ ¢¢¢= + + + + ◊◊ ◊¢ f ff x f f x x x
Differential equations can be solved by a variety of techniques, including solution by direct integration, by separation of variables and by using an integrating factor. The general solution of a differential equation contains unknown constants. If you are given appropriate boundary conditions, you can find a particular solution.
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Chapter 7 Solutions Solution 7.1
(i) The function can be written as 2322 4- +x x , which can then be integrated to give:
5 53 31 12 2 12
25 5
-- + + = - + +x x c x cx
(ii) 214 +xe c
(iii) This is equivalent to 12(6 )x , so using the formula given in the notes, we get:
126
2ln 6
¥ +x
c
(iv) We have the derivative of the denominator on the numerator, so this integrates to
give:
2ln 5 7x x c+ + +
Check this by differentiating if you are not sure.
Solution 7.2
332 3
11
3 26È ˘= =Î ˚Ú x dx x
This represents the area between the curve 23=y x , and the x-axis from 1=x to 3=x .
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Solution 7.3
2 52 5
2 4
2 4
4 6(i) 2(2 3)( 3 4)
( 3 4)
2( 3 4)
4
1
2( 3 4)
-
-
+ = + + ++ +
+ += +-
= - ++ +
Ú Úx
dx x x x dxx x
x xc
cx x
(ii) 2 2 23 4 3 4 3 47 7
6 67 6 - - -= = +Ú Úx x xxe dx xe dx e c
(iii) 2 2 2 22 1 1
2 2 2 = = = +Ú Ú Úx x x xx e dx xe dx xe dx e c
Solution 7.4
Expanding the last line of the solution, we get:
21 1 1ln 3 2 ln(3 2) ln
3 3 3x x k- + + +
So it can be seen that 1
ln3
c k= .
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Solution 7.5
2
7
1
--
x
x can be written as
7
( 1)( 1)
-+ -x
x x.
Using partial fractions, we can split it as follows:
7 ( 1) ( 1)
( 1)( 1) 1 1 ( 1)( 1)
- - + += + =+ - + - + -x A B A x B x
x x x x x x
Comparing coefficients, we get 4, 3= = -A B .
The integral then becomes:
4
3
4 3 ( 1)4ln 1 3ln 1 ln
1 1 ( 1)
xdx x x c c
x x x
+- = + - - + = ++ - -Ú
Solution 7.6
One possibility is to let 1= -u x , then the integral becomes:
31 1 12 2 2 2
12
43
4 43 3
2( 1) (2 2 ) 4
( 3) ( 2) 1
-+ = + = + +
= + + = + - +
Ú Úu
du u u du u u cu
u u c x x c
It is also possible to solve this by making the substitution 1= -u x .
Page 40 FAC-07: Integration
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Solution 7.7
If 32 1= +xu e , then 36 3( 1)= = -xdue u
dx. When 0=x , 3=u and when 1=x ,
32 1= +u e . Substituting gives:
3 32 1 2 1
3 3
4 1 4 1
3( 1) 3 ( 1)
+ +
¥ =- -Ú Ú
e e
du duu u u u
Splitting up the integral using partial fractions gives:
( ) ( )
3 3
3
2 1 2 1
3 3
2 1
3
3 3
3
3
4 1 4 1 1
3 ( 1) 3 ( 1)
4ln 1 ln
3
4 4ln 2 ln(2 1) ln 2 ln3
3 3
4 3ln
3 2 1
e e
e
du duu u u u
u u
e e
e
e
+ +
+
= -- -
È ˘= - -Î ˚
= - + - -
=+
Ú Ú
FAC-07: Integration Page 41
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Solution 7.8
(i) Let =u x , and 6( 1)= +dvx
dx, ie 1=du
dx, and 71
7 ( 1)= +v x , then we get:
6 7 71 1
7 7
7 81 17 56
7156
( 1) ( 1) ( 1)
( 1) ( 1)
( 1) (7 1)
+ = + - +
= + - + +
= + - +
Ú Úx x dx x x x dx
x x x c
x x c
It is also possible to use the substitution 1= +u x to do this integral.
(ii) Let ln=u x , and =dvx
dx, ie
1=du
dx x, and 21
2=v x . So:
2 21 1 1
2 2
21 12 2
2 21 12 4
ln ln
ln
ln
= - ¥
= -
= - +
Ú Ú
Ú
xx x dx x x x dx
x x x dx
x x x c
This example is unusual since it is more common to set u equal to a power of x when integrating by parts.
(iii) Let ln=u x , and 1=dv
dx, ie
1=du
dx x, and =v x .
[ ] [ ]2 2
2 21 1
1 1
ln ln 1 ln= - = -Ú Úx dx x x dx x x x
This can be evaluated to give:
(2 ln 2 2) (1ln1 1) 2 ln 2 1 ln 4 1- - - = - = -
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Solution 7.9
We need to find 2 3
2
0
( 1) + +
+Úx
dx tx dt
dx, so comparing with the general equation:
2( ) 0, ( ) 2 3, and ( , ) ( 1)= = + = + +a x b x x f x t x tx
giving:
( ) 0, ( ) 2, and 2( 1)= = = + +¢ ¢f
a x b x x tx
∂∂
.
So:
( )
2 3 2 32 2
0 0
2 32 212 0
2 212
2 12
( 1) + 2 ( 1) (2 3) 0 2( 1)
2(3 5 1) 2( 1)
2(3 5 1) 2( 1)(2 3) (2 3)
12 26 12
+ +
+
È ˘+ = + + + - + + +Î ˚
È ˘= + + + + +Î ˚
= + + + + + + +
= + +
Ú Úx x
x
dx tx dt x x x x t dt
dx
x x x t t
x x x x x
x x
Solution 7.10
The first one does not exist because 0
1limƕ
È ˘-Í ˙Î ˚
k
k x is not defined for the lower limit.
The second one does not exist because 0
lim lnN
kkx
ÆÈ ˘Î ˚ is not defined for the lower limit.
The last one does not exist because 1
lim lnk
kx
Æ•È ˘Î ˚ does not converge.
One consequence of these integrals not existing is that the area under the graphs of these functions is infinite.
FAC-07: Integration Page 43
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Solution 7.11
We will do this using integration by parts.
Let =u x , and 2-= xdve
dx, ie 1=du
dx, and 21
2-= - xv e . Then:
2 2 21 12 20
0 0
2 21 12 40 0
2 21 1 12 4 4
14
lim
lim
lim
•- - -
Æ•
- -Æ•
- -Æ•
È ˘È ˘Í ˙= - - -Î ˚Í ˙Î ˚
È ˘È ˘ È ˘= - -Í ˙Î ˚ Î ˚Î ˚
È ˘= - - +Î ˚
=
Ú Úk
kx x x
k
k kx x
k
k k
k
xe dx xe e dx
xe e
ke e
We have used the fact that exponential functions dominate polynomials for large x, ie
2lim 0-Æ•
=k
kke .
In practice when evaluating such integrals, you consider the limit, but don’t write it, as seen in the next solution. Solution 7.12
From the definition of the gamma function 1
0
( )•
- -G = Ú x tx t e dt ( 0)>x .
So let 1-= xu t , ie 2( 1) -= - xdux t
dt, and -= tdv
edt
, ie -= - tv e . This gives:
1 1 2
00 0
2
0
( ) ( 1)
( 1) ( 1) ( 1)
• ••- - - - - -
•- -
È ˘G = = - - - -Î ˚
= - = - G -
Ú Ú
Ú
x t x t x t
x t
x t e dt t e x t e dt
x t e dt x x
where 1>x .
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Solution 7.13
15 5 15521
2 010 0 10
15
10
152
10
2 1 2
3000 3000
110 12.5
3000
15 12.5
3000
11
48
= =
+ È ˘= +Î ˚
= +
È ˘= +Î ˚
=
Ú Ú Ú
Ú
x y
x ydy dx xy y dx
x dx
x x
Solution 7.14
We need to integrate by parts.
Use =u x , and 2 3+= x ydve
dx so that 1=du
dx and 2 31
2+= x yv e , to give:
23 2 3 2
2 3 2 3 2 3
0 0
233 4 5 2 3
0
33 4 3 4 5
0
33 4 5
0
1 1
2 2
1 1
2 4
1 1 1
4 4 2
3 1 1
4 4 2
x y x y x y
yy y
y y x y
y
y y y
y y
xe dxdy xe e dx dy
e ye e dy
e e y e dy
e y e dy
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Again using integration by parts, this time with 1 1
4 2= -u y , we get:
33 35
3 4 5
0 00
33 4 5 5
0
13 15 4
1 1 1 1
4 4 2 5 10
1 1 1 1 1
4 5 4 2 50
1 23 1 7
4 100 4 100
641,284
yy y
y y y
ee y e dy
e y e e
e e e
Solution 7.15
3 3 2 3 32
11 0 1 1 0
3 32 1
1 0
332 1
01
32 4 1 3 2
1
32 4 1 3 21 14 3 1
14 10 6 5 4 31 1 1 14 3 4 3
2
293,103
+ ++ +
+ + +
+ + +
+ + +
+ + +
È ˘= Î ˚
= -
È ˘= -Î ˚
= - - +
È ˘= - - +Î ˚
= - - - + + +
=
Ú Ú Ú Ú Ú
Ú Ú
Ú
Ú
z z zzx y x y
zz y y
zz y y
z z z
z z z
e dx dy dz e dy dz
e e dy dz
e e dz
e e e e dz
e e e ez
e e e e e e e
Solution 7.16
1 2 3 1 2 3
0 0 0 0 0 0
1 9 12 4
2 2 2= ¥ ¥ = ¥ ¥ =Ú Ú Ú Ú Ú Údx dy dz xyz x dx y dy z dz
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Solution 7.17
(i) Integrating first with respect to x :
30
0 0
30( )
00
30( )
00
30( )
0
1
1
1(1 )
yy x x
y x
yy x
y
yy x
y
y y
y
e e e dx dy
e e dy
e e dy
e e dy
Then integrating with respect to y , we get:
30( )
0
30( )
0
30 30( )
1 1
1 1 1 1
y y
y
y y
e e dy
e e
e e
(ii) Evaluating (i) numerically, we get 1.213. Changing the order of integration, consider 0 30< < <x y . So:
30 30 30
0 0 0
yy x x x x y
y x x y x
e e e dx dy e e e dy dxm n d n d mm m- - - - - -
= = = =
=Ú Ú Ú Ú
FAC-07: Integration Page 47
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Then integrating with respect to y , we get:
30 30
0
3030
0
30( ) ( ) 30
0
30( ) ( ) 30
0
30( ) 30( ) 30
1 1
1 1(1 ) ( )
x x y
xx
x x x
x
x x
x
x x
e e e dx
e e e e dx
e e e dx
e e e
e e e
Substituting in the given values, we also get 1.213, confirming that the two
methods give the same numerical value. Solution 7.18
Using 5 ordinates means that we need to split the area into 4 sections, each of width 0.25. The trapezium rule will then give the approximate area to be:
1 2 4 4 4 2
0.25 0.8122 2 2.25 2.5 2.75 3
This will give an over estimate of the true value since the gradient of the curve is increasing over the range of values. We can check the exact value by integrating:
3
3
22
22ln 2ln3 2ln 2 0.811dx x
xÈ ˘= = - =Î ˚Ú
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Solution 7.19
2 5( ) (1 ) (1 )- -= + + +f i i i , so (0.1) 1.447=f , and:
3 6
4 7
5 8
( ) 2(1 ) 5(1 )
( ) 6(1 ) 30(1 )
( ) 24(1 ) 210(1 )
- -
- -
- -
= - + - +¢
= + + +¢¢
= - + - +¢¢¢
f i i i
f i i i
f i i i
giving:
(0.1) 4.325
(0.1) 19.493
(0.1) 112.869
= -¢
=¢¢
= -¢¢¢
f
f
f
Using the Taylor series, this gives:
2 32 5 ( 0.1) ( 0.1)
(1 ) (1 ) 1.447 4.325( 0.1) 19.493 112.8692! 3!
- - - -+ + + = - - + -i ii i i
(as far as the term in 3( 0.1)-i ).
Because the term 4.325( 0.1)- -i has a negative coefficient, increasing i would reduce
the value of the function. In fact, changing i to 0.11 (ie 1 0.01- =i ) would increase the value of the function by approximately 4.325 0.01 0.04325- ¥ = - .
FAC-07: Integration Page 49
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Solution 7.20
The required partial derivatives in their general form and then evaluated at 3=x and 2=y are:
2
2
23
2
23
2
23
( ) 1
( ) 1
2( ) 2
2( ) 2
2( ) 2
-
-
-
-
-
= - - = -
= - =
= - =
= - =
= - - = -
fx y
x
fx y
y
fx y
x
fx y
y
fx y
y x
∂∂
∂∂
∂∂
∂∂
∂∂ ∂
So the expansion is:
2 2
2 2
11 ( 3) ( 2) 2( 3) 2 2( 3)( 2) 2( 2)
2
2 ( 3) 2( 3)( 2) ( 2)
È ˘- - + - + - + ¥ - - - + -Î ˚
= - + + - - - - + -
x y x x y y
x y x x y y
Solution 7.21
( ) ln(1 )= +f x x , so (0) 0=f , and:
2 3
1 1 2( ) , ( ) , ( )
1 (1 ) (1 )= = - =¢ ¢¢ ¢¢¢
+ + +f x f x f x
x x x
giving:
(0) 1 (0) 1 (0) 2= = - =¢ ¢¢ ¢¢¢f f f
Using Maclaurin’s series, this gives:
2 31 12 3ln(1 )+ = - +x x x x (as far as the term in 3x )
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Solution 7.22
Integrating twice gives:
3
3= +dx at
cdt
4
12= + +at
x ct d
But using the boundary conditions:
0 4= fi =t d
971 4
12 12= fi = + +a
t c
1
1 412 12
= - fi = - +at c
Solving these simultaneously, we get 1, 4= =a c , so the particular solution is:
4
4 412
= + +tx t
FAC-07: Integration Page 51
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Solution 7.23
Separating the variables gives us:
2
1
2 9=
+ +Ú Úx
dy dxy x
Integrating both sides gives us:
212ln( 2) ln( 9)y c x+ = + +
Note that no modulus signs are required here as 2 9x + is always positive, and 2y + is
always positive since we are given that 0y > .
So:
2ln( 2) ln 9 2 22 9 9 2y c x ce e y e x y k x+ + += fi + = + fi = + -
where ck e= . But if 0.5=y when 4=x , then 0.5k = , so the particular solution is:
20.5 9 2y x= + -
Solution 7.24
The differential equation can be solved by separating the variables:
( )
1
1 1ln ln( )
dNdt
a bN N
a b adN dt
N a bN
N a bN c ta a
=-
+ =-
- - + =
Ú Ú
Ú Ú
Note that 0N > and 0a bN- > from the condition in the question, so no modulus signs are required here.
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Setting 1
lnc ka
= , for some positive constant k , this can be simplified to give the
general solution:
1ln = fi = fi =
- - +
atat
at
kN kN aet e N
a a bN a bN k be
Solution 7.25
The differential equation can be rewritten in the form 1 2
3 3- =dy
ydx x
, providing that
0πy . 0=y is a trivial solution to the differential equation.
Using this version, we can see that the integrating factor is
13
1 1exp exp ln
3 3
-È ˘ È ˘- = - =Í ˙ Í ˙Î ˚ Î ˚Ú dx x xx
.
Note that since 0x > , we don’t need to include modulus signs here. Multiplying through by the integrating factor and integrating both sides with respect to x gives the general solution to the differential equation:
1 1 23 3 3
2
3
- -= = +Úyx x dx x c
But we know that 1=y when 27=x , so:
23
19 8
3= + fi = -c c
So the particular solution is 1 23 3 2
38- = -yx x or
132
38= -y x x .
FAC-08: Vectors and matrices Page 1
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Chapter 8
Vectors and matrices
You need to study this chapter to cover: ● calculations involving vectors
● multiplication by a scalar
● scalar product of two vectors
● magnitude
● finding the angle between vectors
● orthogonality
● transposition of a matrix
● addition and subtraction of matrices
● multiplication of matrices
● determinants and inverses.
0 Introduction
You will meet the work covered in this chapter in Subjects CT4, CT6 and CT8. If you have not studied vectors and matrices before then you will need to look at this chapter very carefully.
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1 Vectors
1.1 Notation and arithmetic
Vectors are defined as quantities that have a magnitude (or length) and a direction. They can be thought of as representing the position of a point in two dimensions, three dimensions etc. We will deal with two and three-dimensional vectors in this chapter but the results can easily be extended to the n-dimensional case. All vectors in three-dimensional space can be written in terms of the base vectors i, j and k, which are the unit vectors (ie the vectors of length 1) in the x, y and z directions of cartesian space. There are different ways of writing vectors:
2
2 3 2 3 2 3 2
2
Ta
a i j k
(we’ll see what the T means shortly). In this chapter we will use the convention that vectors are in bold. In handwritten work a calculation involving the scalar l , the vector v and the matrices M and I (matrices will be introduced in Section 2) might look like this:
(M I )v 0l- =
where underlining is used to represent a vector or a matrix.
Note that the column vector
2
3
2
shows the coefficients of i, j and k, so that
2
3 2 3 2
2
i j k where
1
0
0
i ,
0
1
0
j , and
0
0
1
k . Column vectors are usually
used in preference to row vectors because matrix equations then have a similar format to the corresponding set of simultaneous equations.
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Vectors can be added, subtracted or multiplied by a scalar:
a d a d
b e b e
c f c f
a d a d
b e b e
c f c f
a ka
k b kb
c kc
ie you add, subtract or multiply the individual elements. Showing addition and multiplication pictorially:
Example
If
2
4
4
a and
3
5
1
b , what are:
(i) 2a (ii) a b (iii) 3 4a b ? Solution
(i)
4
2 8
8
a (ii)
5
1
3
a b (iii)
6
3 4 32
16
a b
ba
a + b
b
a2
2
Page 4 FAC-08: Vectors and matrices
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Question 8.1
(i) If
3
2
8
a and
1
6
2
b what is 3 2b a ?
(ii) What are the values of p and q such that
5
50
46
p q
a b ?
1.2 Magnitude
The magnitude of a vector is just its length, which can be calculated using an extended version of Pythagoras’ theorem. The magnitude of vector a is written as a or | |a .
In general, if
p
q
r
a , then 2 2 2a p q r= + + .
Example
Find the magnitude of the vector
1
2
4
a .
Solution
2 2 21 ( 2) 4 21a = + - + =
This gives us a way of finding a unit vector ie a vector with length 1. For example if a
is as defined in the last example, then the unit vector in the direction of a is
11
221
4
.
FAC-08: Vectors and matrices Page 5
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Question 8.2
Find the magnitude of the vector 2 5 3i j k- + .
Question 8.3
Find the unit vector in the direction of 3 4 2i j k+ - .
1.3 Scalar product
There are different types of product you can work out for vectors: the scalar (or dot) product and the vector (or cross) product. We will only look at the scalar product here, as the vector product is not needed for the actuarial exams. The scalar product of two vectors can be defined to be cosaba . b q= where q is the angle between them in the plane containing and a b , as shown below. It can also be calculated by multiplying the corresponding coefficients of the unit vectors, i, j and k, and then summing.
b
a
Page 6 FAC-08: Vectors and matrices
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Example
Find a . b , where
1
4
5
a and
2
3
6
b .
Solution
(1)(2) ( 4)(3) (5)( 6) 2 12 30 40a . b = + - + - = - - = -
Note the answer here is a scalar not a vector, hence the name “scalar product”.
The scalar product can be used to find the angle between two vectors:
Example
Find the angle between the vectors
3
6
1
a and
3
3
4
b .
Solution
9 18 4 5a . b = - + - = , 9 36 1 46a = + + = , and 9 9 16 34b = + + = So:
cos 5 46 34 cosaba . b q q= fi =
5cos
46 34
83
q
q
\ =
=
Question 8.4
Find the angle between the vectors 2 3 5a i j k= + - and 3 2 8b i j k= - + + .
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Since the scalar product is related to the angle between vectors, an important result
occurs when 0a . b = , since this means that the angle between the vectors is 90 , ie they are perpendicular. When this occurs, the vectors are called “orthogonal”. Note that in the following question, the vectors are in two-dimensional space.
Question 8.5
Find p such that a and b are perpendicular, where 1
7
a , 3
4
p
b .
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2 Matrices
Matrices are arrays of numbers whose size is referred to as the number of rows by the number of columns. Notice that a vector is a special case of a matrix where the number of columns is one. A square matrix is one where the number of columns equals the number of rows.
Example
Describe the size of this matrix 1 1 3 5
3 4 6 4
.
Solution It is a 2 4¥ matrix.
Matrices have many uses, such as describing transformations, solving simultaneous equations, carrying out calculations involving multivariate statistical distributions, and showing state transition probabilities. We will look here at the techniques that you will need for the actuarial exams.
2.1 Basic arithmetic
The transpose of a matrix is found by swapping the rows and the columns. The
transpose of a matrix A is written as TA or A¢ .
Example
If
1 3
2 5
0 9
A , what is TA ?
Solution
1 2 0
3 5 9T
A
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Transposing converts a column vector to a row vector and vice versa, and an m n¥ matrix to an n m¥ matrix. Column vectors are often written as the transpose of row
vectors in textbooks so that they can be written on one line eg (1 2 3)T .
Addition and subtraction are performed by adding or subtracting corresponding elements in the matrix.
So for 2 2¥ matrices a b e f a e b f
c d g h c g d h
.
Example
If
2 4 5
4 2 6
7 1 3
A and
4 3 2
8 5 1
3 6 0
B , what is A B ?
Solution
2 4 5 4 3 2 6 7 3
4 2 6 8 5 1 12 3 5
7 1 3 3 6 0 10 7 3
A B
The matrix with zero as all of its elements is called the zero matrix. For example the
2 2¥ zero matrix is 0 0
0 0
0 . A matrix such as
4 0 0
0 3 0
0 0 1
, where all the elements
not on the main diagonal are zero, is called a diagonal matrix.
Question 8.6
If a matrix is equal to its transpose, what can you say about the matrix?
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2.2 Multiplication
Matrices can be multiplied by a scalar or by another matrix. When multiplying by a scalar, multiply each element in the matrix by that number.
Example
If 3 3
5 6
A , what is 2A ?
Solution
6 62
10 12
A
Multiplying two matrices together gives another matrix. The elements in the rows of the first matrix are multiplied individually by the elements in the columns of the second and then summed.
So for 2 2¥ matrices a b e f ae bg af bh
c d g h ce dg cf dh
.
Example
If 3 1
5 4
A and 4 2
3 1
B what is AB ?
Solution
3 1 4 2 (3 4) (1 3) (3 2) (1 1) 15 7
5 4 3 1 ( 5 4) ( 4 3) ( 5 2) ( 4 1) 32 14
AB
Note: Not all matrices can be multiplied together – the number of columns in the first matrix must be the same as the number of rows in the second. An m n¥ matrix multiplied by an n q¥ matrix gives an m q¥ matrix. You will soon realise if it is
impossible to multiply your matrices as you will run out of numbers.
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Question 8.7
If
1 2 4
3 2 3
0 1 1
A and
4 1 1
5 0 2
1 5 3
B , what are AB and BA ?
Here AB and BA are not equal. Unlike in ordinary arithmetic, matrix multiplication is not commutative ie you get different answers if you multiply matrix B in front by matrix A (called pre-multiplying by A ) or if you multiply behind by matrix A (called post-multiplying by A ).
Question 8.8
What is
0 2 3 1
4 2 1 2
0 5 1 1
?
These could not have been multiplied the other way round.
Question 8.9
What is 1
2 1 3 3
2
?
Page 12 FAC-08: Vectors and matrices
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Example
What do you get if you multiply the matrix a b
c d
M by 1 0
0 1
I ?
Solution
a b
c d
MI and a b
c d
IM
Note that multiplying by 1 leaves the original matrix unchanged. Because of this property the matrix with 1’s along the leading diagonal and 0’s elsewhere is called the identity matrix and is written as I .
2.3 Determinants and inverses
Determinants A determinant is a scalar quantity associated with a square matrix. The determinant of a 2 2¥ matrix is equal to the product of the numbers on the leading diagonal (top left corner to bottom right corner) minus the product of the numbers on the other diagonal. It is written as det , | | , or A A D when it is clear which matrix is
involved.
Example
What is det A if 2 6
4 3
A ?
Solution det (2 3) (4 6) 30A = ¥ - ¥ - =
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Question 8.10
What is the determinant of 3 5
2 7
P ?
This is a specific definition, and we need to generalise. For any matrix M , if ijc is the
element in the matrix corresponding to row i and column j, and ijM is the determinant
of the matrix formed when we strike out row i and column j, then the determinant is
defined as 11 1
1
( 1)n
jj j
j
c M+
=-Â .
This looks quite daunting, but in practice it is quite simple to use.
Example
Calculate the determinant of
1 1 2
3 0 4
2 3 1
.
Solution The determinant is given by:
0 4 3 4 3 01 ( 1) ( 1) 2
3 1 2 1 2 3
1(0 12) 1(3 8) 2( 9 0)
12 11 18
19
- -D = ¥ + - ¥ - ¥ + ¥
- -
= - + + + - -
= - + -
= -
We have taken the elements in the top row and multiplied every other one by –1. These are then multiplied by the determinant of the 2 2¥ matrix that is left when the row and column containing the element are removed. Finally the results are summed.
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Question 8.11
Find the determinant of
2 1 0
0 1 1
4 3 6
.
Inverses
The inverse of a matrix A , written as 1A- , is such that 1 1AA A A I- -= = . It is the matrix equivalent to a reciprocal. To find the inverse of a 2 2¥ matrix, swap the elements on the leading diagonal, change the sign of the elements on the other diagonal and divide by the determinant, ie
11a b d b
c d c aad bc
.
If the determinant is 0 then the matrix is said to be singular and the inverse does not exist. This is equivalent to trying to divide by zero with ordinary numbers.
Example
If 3 2
5 4
A , what is 1A ?
Solution
1 4 21
5 32
A
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Example
Using the matrix in the last example, show that 1 1AA A A I- -= = . Solution Note that the factor can be brought to the front.
1 3 2 4 2 2 0 1 01 1
5 4 5 3 0 2 0 12 2
AA
1 4 2 3 2 2 0 1 01 1
5 3 5 4 0 2 0 12 2
A A
Question 8.12
Find the inverse of 1 9
2 10
B , and check it by finding 1BB- .
There is no simple rule for finding the inverse of a 3 3¥ matrix. You need to use a more complicated method, such as solving a set of simultaneous equations.
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2.4 Simultaneous equations
Matrices can be used to solve simultaneous linear equations, by first writing them in matrix form and then pre-multiplying by the inverse.
Example (Method 1) Solve the simultaneous equations (using matrices):
2 3 14
3 4 4
x y
x y
- =
+ =
Solution Note firstly that these simultaneous equations can be written as the following matrix equation. If you need convincing, multiply out the matrices.
2 3 14
3 4 4
x
y
Pre-multiplying by the inverse matrix on both sides, we get:
4 3 2 3 4 3 141 1
3 2 3 4 3 2 417 17
x
y
This gives the identity matrix on the left hand side, so we simplify to get:
1 0 681
0 1 3417
4
2
x
y
x
y
So the solution is 4x = and 2y = - .
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Question 8.13
Solve the simultaneous equations (using matrices):
2 4 14
2 3 13
p q
q p
- = -
- =
Matrices can also be used in a different way to solve simultaneous equations. This method more closely resembles the method used to solve them algebraically that we covered earlier in the course. We will look at the last example again, this time looking at an elimination method.
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Example (Method 2) Solve the simultaneous equations (using matrices):
2 3 14
3 4 4
x y
x y
- =
+ =
Solution We can write this in the abbreviated form:
2 3 14
3 4 4
x
y
This can in turn be written as:
2 3 14
3 4 4
We would like to get zero in the bottom left hand corner. We are allowed to replace any row with a linear combination of the rows. We will do this in two stages. Replacing the first row by second row minus first row ( (2) (1) ):
1 7 10
3 4 4
Replacing the second row by (2) 3 (1) :
1 7 10 1 7 10
0 17 34 0 1 2
This tells us that 2y = - , and so 10 7 4x y= - - = .
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Example Solve the simultaneous equations (using the elimination method):
2 2 11
3 2 4 8
4 3 15
x y z
x y z
x y z
+ - =
- + = -
+ - =
Solution These can be written as:
2 1 2 11
3 2 4 8
1 4 3 15
This time we would like to get zeros in the bottom left hand corner. Replacing the first row by (1) (3) and the second row by (2) 3 (3) we get:
1 3 1 4
0 14 13 53
1 4 3 15
Replacing the third row by 2 ((3) (1)) (2) , we get:
1 3 1 4 1 3 1 4
0 14 13 53 0 14 13 53
0 0 5 15 0 0 1 3
Replacing the second row by (2) 13 (3) , we get:
1 3 1 4
0 14 0 14
0 0 1 3
so 3, 1, 2z y x= - = = .
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When solving simultaneous equations, you do not always have as many different equations as it first appears. Any equation that can be written as a linear combination of the other equations (ie you can obtain this equation from the others by adding, subtracting or multiplying) is said to be linearly dependent on the others. When solving simultaneous equations you should ensure that the equations you are working with form a linearly independent set. You only need to solve linearly independent equations. For example, suppose you were given the simultaneous equations:
3 4 5
2 3 8
7 13
x y
x y
x y
- = -
+ =
- = -
Here you would only need to consider, say, the first two equations since the third is a linear combination of them (equation 3 is equation 1 minus equation 2). You could equally ignore the first or second equation. Careful choice of which equations to use can make things considerably easier for you, so do watch out!
Question 8.14
Solve the simultaneous equations:
3 2 13
7 6 9
x y
x y
+ =
- =
using the method of the last example.
Question 8.15
Solve the simultaneous equations:
2 4
2 3 2 1
3 3 17
x y z
x y z
x y z
+ - = -
- + =
+ - = -
using the method of the last example.
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2.5 Eigenvectors and eigenvalues
There is an important aspect of matrix theory that arises from the fact that matrices can be used to represent transformations. If a vector v is transformed by a matrix A, then the resulting vector is Av. If the direction of the vector is unchanged once it has been transformed we can write: Av vl= where l is a constant. The vector v is called an “eigenvector” of the matrix, and the corresponding value of l is called an “eigenvalue”. To find the eigenvectors and eigenvalues, you have to work with the equation Av vl= : ( )Av v A I v 0l l= fi - =
where 0 is the zero matrix. Notice that we had to insert the identity matrix I into the equation since we cannot subtract a scalar from a matrix. Thinking of our ways of solving “ordinary” equations, we see that this equation would be true if either ( )A I 0l- = , or v 0= , but these are not going to be very helpful since
this is just the trivial solution and is not the one we are interested in. To solve a general matrix equation Bx C= , we would normally find the inverse of B
and calculate 1x B C-= . Using this technique to try to solve the equation
( )A I v 0l- = , then if ( )A Il- has an inverse we get 1( )v A I 0l -= - , ie v 0= . Since
we are looking for a non-trivial solution, we must prevent this method from working. The only thing that would stop us getting v 0= would be if A Il- does not have an inverse ie it is singular. Remembering that singular matrices have a determinant of zero this gives us a way to find the eigenvalues and eigenvectors. In summary, to find the eigenvalues of a matrix A we must solve the equation det( ) 0A Il- = . The equation obtained from det( ) 0A Il- = is called the
“characteristic equation of the matrix”. Once you have the eigenvalues, you can return to the equation ( )A I v 0l- = , to find
the eigenvectors. This method will work for a general n n¥ matrix, but due to the complication of calculating the determinant we will only look at 2 2 and 3 3¥ ¥ matrices here.
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Example
Find the eigenvectors and corresponding eigenvalues of the matrix 2 1
2 5
B .
Solution First we use the equation det( ) 0 B I to find the eigenvalues:
2
2 1det 0
2 5
(2 )(5 ) 2 0
7 12 0
( 3)( 4) 0
so 3 or 4 .
We now use the equation ( ) 0 B I v to find the eigenvectors, letting x
y
v . There
is a separate eigenvector corresponding to each eigenvalue.
If 3 , then 2 1 0
2 5 0
x
y
, ie 1 1 0
2 2 0
x
y
, ie
0
2 2 0
x y
x y
.
Any value of x and y such that 0x y ie y x will make both rows work out.
So the eigenvector corresponding to 3 is 1
1k
, where k is a constant.
If 4 , then 2 1 0
2 1 0
x
y
, so the eigenvector corresponding to 4 is
1
2k
.
Notice that we include k in the eigenvector, since any value of the constant k will give an eigenvector.
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Question 8.16
Find the eigenvectors and corresponding eigenvalues of the matrix 1 2
3 4
B .
Question 8.17
Show that 1l = is an eigenvalue of the matrix
2 2 3
1 1 1
1 3 1
, and hence find all the
eigenvalues and eigenvectors.
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This page has been left blank so that you can keep the chapter summaries together for revision purposes.
FAC-08: Vectors and matrices Page 25
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Chapter 8 Summary Vectors have both a magnitude and a direction. They can be added, subtracted and multiplied by a scalar.
The magnitude of a column vector
p
q
r
a , is defined by 2 2 2a p q r= + + .
The scalar product of two vectors is given by cosaba . b q= , and can also be calculated by multiplying the corresponding coefficients of the unit vectors, i, j and k, and then summing. A matrix is a rectangular array of numbers. Provided that the sizes are compatible, matrices can be added, subtracted and multiplied. The transpose of a matrix can be found by swapping the rows and the columns. To find the inverse of a 2 2¥ matrix, swap the elements on the leading diagonal, change the sign of the elements on the other diagonal and divide by the determinant.
The determinant of a matrix is given by 11 1
1
( 1)n
jj j
j
c M+
=-Â .
A singular matrix has no inverse, ie the determinant is zero. Matrices can be used to solve simultaneous equations. The eigenvalues of a matrix A can be found by solving the equation det( ) 0A Il- = .
The corresponding eigenvectors v can then be found by solving the equation ( )A I v 0l- = .
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This page has been left blank so that you can keep the chapter summaries together for revision purposes.
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Chapter 8 Solutions Solution 8.1
(i)
3
3 2 22
22
b a
(ii) If
5
50
46
p q
a b then 3 5 and 2 6 50p q p q+ = - + = - . Solving these
equations simultaneously, we get 4 and 7p q= = - .
You should also check that the third equation, 8 2 46p q- = , also works out with
these values of p and q. Otherwise there are would be no solutions. Solution 8.2
The magnitude is 2 2 22 ( 5) 3 38+ - + = .
Solution 8.3
The magnitude of the vector is 2 2 23 4 ( 2) 29+ + - = , so the unit vector is
( )13 4 2
29i j k+ - .
Solution 8.4
To find the angle, we take the dot product.
cosaba . b q= ie 2 2 2 2 2 240 2 3 ( 5) ( 3) 2 8 cosq- = + + - - + +
Hence 40
cos or 13838 77
q q-= = .
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Solution 8.5
If a and b are perpendicular, then their dot product is zero.
3 28 0pa . b = - = , hence 283p = .
Solution 8.6
The matrix must be a square matrix, otherwise the matrix and its transpose would be different sizes and could therefore not be equal. Also elements reflected about the leading diagonal must be equal. Such a matrix is called symmetric. Solution 8.7
(1 4) ( 2 5) (4 1) (1 1) ( 2 0) (4 5) (1 1) ( 2 2) (4 3)
(3 4) (2 5) ( 3 1) (3 1) (2 0) ( 3 5) (3 1) (2 2) ( 3 3)
(0 4) (1 5) ( 1 1) (0 1) (1 0) ( 1 5) (0 1) (1 2) ( 1 3)
10 21 9
25 12 2
6 5
AB
1
Similarly:
1 11 20
5 12 22
16 11 14
BA
Solution 8.8
0 2 3 1 7
4 2 1 2 7
0 5 1 1 11
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Solution 8.9
1
2 1 3 3 1 1
2
A 1 1¥ matrix is just a scalar, so note that multiplying a row vector by a column vector in that order will give you a scalar. Solution 8.10
det 21 ( 10) 11P = - - - = - .
Solution 8.11
The determinant is 2(6 3) 1(0 ( 4)) 0 2- - - - + = .
Solution 8.12
1 10 91
2 18
B
To check:
1 1 9 10 9 8 0 1 01 1
2 10 2 1 0 8 0 18 8
BB
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Solution 8.13
Noticing that the letters are in different orders, these equations can be written in the form:
2 4 14
3 2 13
p
q
Pre-multiplying both sides by the inverse of 2 4
3 2
, we get:
2 4 141
3 2 138
24 31
16 28
p
q
So 3p = - and 2q = .
Solution 8.14
These equations can be written as:
3 2 13
7 6 9
Replacing (1) by (2) 2 (1) :
1 10 17
7 6 9
Replacing (2) by (2) 7 (1) :
1 10 17
0 64 128
So 3x = and 2y = .
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Solution 8.15
These equations can be written as:
1 2 1 4
2 3 2 1
3 1 3 17
Replacing (2) by (2) 2 (1) , and (3) by (3) (2) (1) , we get:
1 2 1 4
0 7 4 9
0 2 4 14
Replacing (2) by (2) (3) , we get:
1 2 1 4 1 2 1 4
0 5 0 5 0 1 0 1
0 2 4 14 0 2 4 14
Replacing (3) by (3) 2 (2) , we get:
1 2 1 4 1 2 1 4
0 1 0 1 0 1 0 1
0 0 4 16 0 0 1 4
This gives 4, 1, 2z y x= = = - .
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Solution 8.16
For eigenvalues, 1 2
det 03 4
, so we have the characteristic equation:
2(1 )( 4 ) 6 0 3 10 0
This gives 2, 5 .
If 2 , then 1 2 0 1 2 0
3 4 0 3 6 0
x x
y y
.
So the eigenvector corresponding to 2 is 2
1k
.
If 5 , then 1 2 0 6 2 0
3 4 0 3 1 0
x x
y y
.
So the eigenvector corresponding to 5 is 1
3k
.
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Solution 8.17
For the eigenvalues we must solve the equation:
2 2 3
det 1 1 1 0
1 3 1
So we have the equation:
{ } { } { }(2 ) (1 )( 1 ) 3 2 ( 1 ) 1 3 3 (1 ) 0l l l l l- - - - - + - - - + - - =
and this equation simplifies to 3 22 5 6 0l l l- - + = . Factorising the given cubic equation completely gives:
( 1)( 3)( 2) 0l l l- - + =
The eigenvalues are 1, 3, and –2.
When 1 ,
2 2 3 0 1 2 3 0
1 1 1 0 1 0 1 0
1 3 1 0 1 3 2 0
x x
y y
z z
, which gives
the eigenvector of
1
1
1
k
.
When 3 ,
2 2 3 0 1 2 3 0
1 1 1 0 1 2 1 0
1 3 1 0 1 3 4 0
x x
y y
z z
, which
gives the eigenvector of
1
1
1
k
.
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When 2 ,
2 2 3 0 4 2 3 0
1 1 1 0 1 3 1 0
1 3 1 0 1 3 1 0
x x
y y
z z
, which gives
the eigenvector of
11
1
14
k
.
FAC: Glossary Page 1
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Glossary
0 Introduction
This glossary contains useful information for your studies. As well as being mathematically correct your answers should demonstrate accurate use of language and grammar. This is especially important in subjects where you will have to produce essay style answers, such as Subject CT7 and the later subjects. The topics have been picked out from looking at assignment scripts and working out what reminders are necessary! The contents are: 1. Commonly misspelt words 2. Confusing word pairs 3. Latin and Greek 4. Words you will meet in actuarial work
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1 Commonly mis-spelt words
These are the correct spellings of words that have been spelt wrongly on assignments: actuarial ageing appropriately basically benefit benefiting biased calendar cancelled commission consensus correlation cyclically deferred definitely formatted fulfil gauge hierarchy immediately independence instalment interest millennium necessary occasion occurred occurring offer offered offering orthogonal paid particularly pensioner precede proceed receive referred referring relief seize separate similarly supersede targeted theorem until Notes: ● Single or double letters. With –ED and –ING words the rule is that you double
up the letter if the stress in the original word was on the last syllable (as in oc-CUR and re-FER) but not if the stress comes earlier in the word (as in OFF-er, TAR-get, BEN-e-fit). Words like “FOR-MAT”, which have equally stressed syllables, are considered to be stressed on the last syllable. With BIAS you can spell it either way – BIASED is more common, but you can use BIASSED if you prefer.
● British and American English treat L’s differently. In British spelling an L is always doubled-up before –ED and –ING, whereas the Americans don’t. So the British spelling is CANCELLED and the American spelling is CANCELED. Also in American English INSTALLMENT is spelt with two L’s. (In the exams it is best to use the British spelling conventions.)
● “I before E except after C” usually works provided the combination in question sounds like “EE”. However, the rule doesn’t work in plurals (eg POLICIES) or names (eg NEIL) or in the word SEIZE. (If the sound isn’t “EE”, you just have to remember the correct spelling eg HIERARCHY.)
● Adverbs derived from –IC words (eg CYCLICALLY) always have a silent AL in the ending, even if the corresponding –ICAL word doesn’t exist (eg BASICALLY, SPECIFICALLY). However, the word PUBLICLY is an exception for some reason.
● For some reason people (especially non-actuaries!) often miss the middle A out of ACTUARIAL.
● Make sure you put the right number of O’s in the words LO(O)SE. LOSE rhymes with “booze” and is the opposite of “find”. LOOSE rhymes with “moose” and is the opposite of “tight”.
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2 Confusing word pairs
The following pairs of words sound similar but are spelt differently according to the context: Advice/advise (and practice/practise) ADVICE and PRACTICE are nouns. Example: He took the advice he was given about passing the exams. ADVISE and PRACTISE are verbs. Example: I advised him how to pass the exams. Affect/effect AFFECT is a verb meaning “to influence”. Example: Studying will affect your exam performance. EFFECT can be a noun meaning a “noticeable change”. Example: Taking this drug causes a strange effect. EFFECT can also be a verb meaning “to put into effect”. (This is rather a specialist meaning, but it comes up often in actuarial work.) Example: She effected her car insurance policy on 1 July. Dependant/dependent DEPENDANT is a noun meaning “someone who is supported by someone else”. Example: He has two dependants: his son and his sick mother. DEPENDENT is an adjective meaning “being supported by someone else” or “reliant on something else”. Example: This formula is dependent on the assumption of equal variances. In fact, there is some flexibility with this pair, but these are the conventions usually adopted.
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Precede/proceed PRECEDE is a verb meaning “to come before”. Example: Will payment precede or follow delivery? PROCEED is a verb meaning “to continue”. Example: Do you wish to proceed with reformatting your hard disk? Principal/principle PRINCIPAL is an adjective meaning “main” or “primary”. Example: The principal reason was financial. This spelling is also used for the head of a school or college. (This is really an abbreviation for “principal teacher” or “principal director”.) PRINCIPLE is a noun meaning a “rule” or “moral guidance”. Example: He was a man of principle. This spelling is also used in the phrases “on principle” and “in principle”, which are derived from this meaning.
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3 Latin and Greek
Plurals A lot of words that come directly from Latin and Greek keep their original Latin and Greek plural endings when they’re used in English. The patterns you’ll see are:
Singular Plural Examples
LA
TIN
–US –I stimulus/stimuli
–UM –A maximum/maxima
medium/media
–A –AE formula/formulae
–IX or –EX –ICES index/indices
appendix/appendices
GR
EE
K –ON –A
criterion/criteria phenomenon/phenomena
–IS –ES basis/bases
analysis/analyses crisis/crises
Notes: ● A lot of Latin and Greek words use the regular English –(E)S ending. So you
should say: SURPLUSES, CENSUSES, SYLLABUSES, STATUSES, PREMIUMS, LEMMAS.
● With a lot of words you have the choice of a classical or an English plural. For example, with formula you can use either FORMULAE or FORMULAS. (Americans tend to use formulas.) You can also say TRAPEZIA or TRAPEZIUMS.
● The plural of INDEX is INDICES when you’re talking about an economic index, but INDEXES if you mean the pages at the back of a book. The plural of APPENDIX is APPENDICES when you’re talking about the section in a book, but APPENDIXES if you mean the thing that gives people appendicitis.
● The plural of SERIES is the same as the singular.
● The word DATA can be treated as plural (eg “The data are complete”). However, many people think of DATA as singular (eg “The data has arrived”). In the course notes you will probably come across both uses.
● The word DICE also causes problems. Strictly speaking, the thing with six sides you play games with is called a DIE and DICE is the plural form.
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Words and abbreviations PER ANNUM is a common phrase meaning a year eg “Some actuaries are paid more than £100,000 per annum.” The abbreviation pa is universally recognised. We also use pm (for a month), pcm (for a calendar month) and pq (for a quarter), but these are not so well known. PRO RATA means in proportion. For example, “Your pension will be calculated as 1/60th of your salary for each year of service, with months counting pro rata” means
that, if you worked for 5 years and 5 months (say), this would be counted as 5 512
in the
calculation. It is also used as a verb eg “We can pro rata the payments to allow for holidays”. SIC. This is the Latin word for thus. People use it to mean “it really said this” when they’re quoting a passage from a document that has a mistake in. For example, “In his letter the client said that the company would be increasing its contribution rate from 6½% to 6¼% [sic].” I’ve put “sic” to show that I realised that the client’s secretary had probably typed these numbers the wrong way round, but this was what the letter actually said. STATUS QUO means the current position. So “trying to maintain the status quo” means trying to keep things as they are. STET. This means “Let it stand” in Latin. If you cross something out or change something, but then realise it was right after all and you wish you hadn’t changed it, you can write stet next to it. This tells the person reading it that you crossed it out by accident. This one can be useful in exams too. VICE VERSA means the other way round. eg. This means for example and is used where you want to give an example that could have been one of several. ie. This means that is and is used when you want to pin down exactly what you mean. cf. This just means compare. For example, you might write, “Using the approximation, I got £73.98 (cf £74.02 when calculated accurately).”
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4 Words you will meet in actuarial work
The BASIS for an actuarial calculation is used to mean the set of assumptions (eg mortality rates, interest rates) used in the calculation. A STRONG basis is one with very pessimistic assumptions. A WEAK basis is one with optimistic assumptions. A LIFE just means a person. A FIRST-CLASS LIFE is a person in perfect health. Otherwise, they are IMPAIRED. IMMEDIATE is the opposite of “deferred”. It doesn’t necessarily mean “straight away”. A “deferred pension” would normally start making payments a number of years in the future. An “immediate pension” would make the first payment at some time during the coming year, but not necessarily at the start of that year. LEVEL means constant eg “level payments” are for the same amount each time. A NET payment is one where something has been deducted. Net monthly pay generally refers to the amount of your “take home” pay after your employer has deducted any amounts due in tax, pension contributions etc. Your GROSS monthly pay ignores these deductions. In actuarial contexts the words “net” and “gross” are used a lot and need not refer to tax. So you should always ask yourself “net of what?” ie what is it that has been deducted? An OFFICE (short for LIFE OFFICE) just means an insurance company. OUTGO is (very logically) the opposite of INCOME. The word PAYABLE means “must be paid” rather than “may be paid”. For example: “£1000 tax is payable on 31 January” doesn’t mean you have an option. The word SECULAR is used to mean, “in relation to time measured by reference to the calendar”. It clarifies the meaning when time could be measured relative to some other reference point eg the time since you were born or since you took out your life insurance policy (which is called the DURATION). So, for example, if I said “Mortality can be expected to improve over your lifetime because of secular effects”, I would mean that in the future they will find a cure for cancer etc, so you are likely to live longer than your parents’ generation. STOCHASTIC means allowing for random variation over time. It is the opposite of DETERMINISTIC.
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There are some words that put the S in a strange place when you make them plural eg SUMS ASSURED, ANNUITIES-CERTAIN, CLAIMS EXPERIENCE and NO
CLAIMS DISCOUNT.
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Question & Answer Bank – Questions
Chapter 1 Question A1
Interpret the statement { }: 1 5x xŒ - £ < .
A { }1,0,1,2,3,4,5xŒ -
B { }1,0,1,2,3,4xŒ -
C { }0,1,2,3,4xŒ
D { }1,2,3,4xŒ [1]
FAC 1 1 Question A2
If {2,4,6,8}A = and {1,2,3,4}B = and {1,3,6,10}C = which are subsets of
{1,2,3,4,5,6,7,8,9,10} , which one of the following is ( )A B C« » ?
A {1,3,6}
B {6}
C {1,3,6,8,10}
D {1,3,6,10} [1]
FAC 1 1 Question A3
If {1,3,5,7,9}A = and {1,3,6,10}B = which are subsets of , which one of the
following would be in A B« ? A 3 B 5 C 6 D 8 [1] FAC 1 1
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Question A4
What are the Greek letters used to represent a sum and a product? A ,S P
B ,s p
C ,rD
D ,pS [1]
FAC 1 2 Question A5
In June, interest rates were 2%. They rose by 40 basis points in July. What was the interest rate after this change? A 42% B 6% C 2.4% D 2.04% [1] FAC 1 3 Question A6
A is the statement “the integer x is odd”, B is the statement “the integer x is divisible by 3”. Which of the following statements is TRUE? A A is necessary for B B A is sufficient for B C A is necessary and sufficient for B D none of the above [2] FAC 1 4 Question A7
Which of the following was a leap year? A 1900 B 1910 C 1920 D 1950 [1] FAC 1 7
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Question A8
Using the exchange rates below, how much would £100 be worth in US dollars?
£/€ = 1.227 and €/$ = 1.314 A $62.02 B $93.38 C $107.09 D $161.23 [1] FAC 1 7 Question A9
Ten payments of $100 are made at half-yearly intervals. The first payment is on 1 September 2011. On what date was the last payment made? A 1 March 2016 B 1 September 2015 C 31 August 2015 D 1 September 2016 [1] FAC 1 7 Question A10
For each of the following statements, either state that it is true or state that it is false and write down a correct version of the statement. (i) Ending in a 5 is a necessary condition for an integer to be a multiple of 5. [1]
(ii) 2{ : , 4, 2}Œ = π = ∆x x x x [1]
(iii) 2 5 7 0,- + > " Œx x x [2]
(iv) The natural logarithm function is defined on the set of values (0, ]• . [1]
[Total 5]
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Question A11
A life office operates an investment fund that allows UK investors to buy investment units whose value is denominated in US dollars. Investors who wish to purchase units pay in sterling and an initial charge of 2½% is deducted before the payment is converted to US dollars and used to purchase the investment units. Last December, when the exchange rate between Sterling and the US dollar was £1 $1.67= , a woman invested £64,000 in this fund. Since then the quoted unit price has gone up by 5% and the exchange rate is now 1.60. What is the current sterling value of her investment? [3] FAC 1 7
Chapter 2 Question A12
What is 2.89951 rounded to 3 decimal places? A 2.899 B 2.900 C 2.90 D 2.810 [1] FAC 2 1 Question A13
What is 4,716 rounded to 3 significant figures? A 472 B 716 C 4,700 D 4,720 [1] FAC 2 1
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Question A14
What is 0.020581 rounded to 2 significant figures? A 0.02 B 0.021 C 0.0206 D 0.02058 [1] FAC 2 1 Question A15
Calculate 1tanh 0.6- . A 0.0105 B 0.537 C 0.693 D 30.963 [1] FAC 2 2 Question A16
Calculate 1 (1 )
1
ni
i i
-- ++
where 0.062i = and 10n = .
A 14.130- B 12.528- C 6.865 D 7.743 [2] FAC 2 2
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Question A17
Calculate 2 3 46 8 10 12v v v v where 1
1.08v = . Give your answer to 3 significant
figures. A 29.173 B 29.2 C 33.3 D 33.333 [2] FAC 2 2 Question A18
Calculate the value of 2
1 1 ln5 1.02exp
2 0.80.8 2
to 3 decimal places.
A 0.112 B 0.210 C 0.345 D 0.380 [2] FAC 2 2 Question A19
Calculate 5 12 46!4 20
3+ + .
A 5.897 B 243.9 C 259.7 D 327.4 [1] FAC 2 2
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Question A20
Evaluate
1- -n
nvnv
di
, given that 0.10=i , 10=n , 1(1 )-= +v i and =d iv . [2]
FAC 2 2 Question A21
The interest rate i charged for a financial arrangement satisfies the following equation:
10 543,600,000(1 ) 12 10 366,000(1 ) 60,192,000+ = ¥ ¥ + +i i
By working in units of 1 million, or otherwise, calculate the value of i , expressing your answer as a percentage to three significant figures. [4] FAC 2 2
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Question A22
An independent financial advisor (IFA) uses the form shown in the diagram to calculate the contribution rates for clients with personal pension policies. A colleague has carried out the calculation shown, which shows that, if Mr Smith wishes to retire at age 60 with an index-linked pension of 50% of his salary at retirement, he will have to make payments of 12.7% of his salary into his pension plan. This figure has been communicated to the client.
(i) Study the form and check that the calculations are correct. Then answer the
following additional questions that Mr Smith has raised:
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(a) What contribution rate would you recommend if I made a one-off extra contribution now of £10,000 to my existing fund?
(b) If I decide to pay 15% of my salary instead, what percentage pension can
I expect to receive at retirement? (c) If I decide to retire at age 55 instead, what contribution rate would you
recommend if I still want a 50% pension? [6] (ii) Ms Jones, another client, has telephoned. She is aged 42 years and 6 months, has
a current salary of £35,000 and an existing fund of £175,000, and wishes to
retire at age 60 with a 23 rds pension. Calculate the recommended contribution
rate for her. [3] [Total 9] Question A23
The rules of a pension scheme state that employees who leave the company before retirement age are entitled to receive an annual pension payable monthly from retirement age. The amount of each payment is calculated using the formula:
Pension payment = Pension entitlement at date of leaving 1.05¥ t where t is the number of complete calendar years between the date of leaving and the date of payment. Pension payments are made on the first day of each month, starting in the month following the member’s 60th birthday. One member’s details are as follows: Sex = Male Date of birth = 14.04.1956 Date of leaving = 23.07.2003 Pension entitlement at date of leaving = £3,620 per annum (i) Calculate the amount of the first pension payment this member will receive and
write down the date on which this will be paid. (Assume that all months are of equal length.) [4]
(ii) Calculate the total amount of pension this member will receive during the first
ten years of retirement assuming that he is alive throughout that period. Apply a reasonableness check to your answer. [5]
[Total 9]
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Chapter 3 Question A24
Which of these is the graph of xy e-= ?
A
x
y
B
x
y
C
x
y
D
x
y
[1] FAC 3 1
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Question A25
Which of these is the graph of 1n
yx
= where n is an odd number?
A
x
y
B
x
y
C
x
y
D
x
y
[1] FAC 3 1
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Question A26
If the dotted line represents the graph of ( )y f x= , which of these is the graph of
(2 1)y f x= - ?
A
x
y
B
x
y
C
x
y
D
x
y
[2] FAC 3 1
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Question A27
If a function ( )f x has the property ( ) ( )f x f x- = - , the function is known as:
A an odd function B a symmetrical function C an even function D an inverse function [1] FAC 3 1 Question A28
A person was born on 15 June 1955. If x is defined to be his exact age, what is [x] on 21 March 2011? A 55 B 55.75 C 55.83 D 56 [1] FAC 3 2 Question A29
Write the expression 4 1 5x + < without the use of the modulus sign.
A 1x < B 1.5x < - C 1.5 1x- < < D 1 1x- < < [1]
FAC 3 2 Question A30
Calculate V, where 0.8x = and min( , 4)xV e p-= .
A 0.443 B 0.449 C 0.886 D 2.226 [1] FAC 3 2
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Question A31
Calculate the value of 100!
97!3! to 2 decimal places.
A 161,700 B 323,400 C 15,684,900 D 31,369,800 [1] FAC 3 3 Question A32
Evaluate (5.5)G . You may use the fact that (½) pG = .
A 52.34 B 104.69 C 230.31 D 287.89 [2] FAC 3 3 Question A33
Simplify the expression( 1)(3 1)!
(3 )( 1)!
n n
n n
G + +G -
, where n is an integer.
A ( 1)(3 1)n n n+ +
B ( 1)(3 1)
3 ( 1)
n n
n n
+ +-
C (3 1)
(3 1)( 1)
n n
n n
+- -
D 23 (3 1)n n + [2]
FAC 3 3
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Question A34
Evaluate each of the following, quoting your answers to 2 significant figures:
(i) e e [1] (ii) log 0.00001e [1]
(iii) 1tanh 0.9- [1] (iv) (12)G [1]
[Total 4]
Chapter 4 Question A35
Which of the following is the correct simplification of ( )2 32 2¥ +bb bx x x x ?
A 62 bx
B 2 32 2+ ++b b bx x
C 34 bx
D 3 32 2 ++b bx x [1] FAC 4 1 Question A36
Simplify 33 2x xe e .
A ½e
B 33 8x xe -
C 31
xe
D 33 8x xe [1]
FAC 4 1
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Question A37
If loga x A= then:
A xa A=
B ax A=
C Aa x=
D aA x= [1] FAC 4 1 Question A38
Simplify 2log log .a ax
yy
A 2loga x
B 2
logax
y
C loga yx
D 2
logax
y
[1]
FAC 4 1
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Question A39
Express 3 5 2
3 1 1
x x
x x
as a single fraction:
A 25 15 8
(3 1)( 1)
x x
x x
B 27 11 2
(3 1)( 1)
x x
x x
C 27 11 8
(3 1)( 1)
x x
x x
D 25 15 2
(3 1)( 1)
x x
x x
[2]
FAC 4 1 Question A40
Simplify 2
2 2
3 6
2 3 2 2 3 1
x x x
x x x x
+ + -∏- - + +
.
A 3 2
4 3 2
2 9 10 3
2 17 16 12
x x x
x x x x
B 2
1
( 2)
x
x
C 2
2
4 3
4 4
x x
x x
+ +- +
D 2( 3)
(2 1)( 1)
x
x x
[5]
FAC 4 1
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Question A41
Solve the quadratic equation 22 5 3 0x x . A 0.5 and 3x = - B 0.5 and 3x = - C 0.5 and 3x = D The equation has no real solutions [2] FAC 4 2 Question A42
Find the solution to 2 12 12 0x x in surd form.
A 6 2 6
B 6 4 3
C 12 6
D 12 2 3 [2] FAC 4 2 Question A43
Solve the simultaneous equations:
95, 432 64,717 57,963
92, 404 64,522 80,068
+ =
+ =
x y
x y [2]
FAC 4 3 Question A44
Solve the simultaneous equations:
20.75 and 0.0225( ) ( 1)
a aba b a b a b
= =+ + + +
[2]
FAC 4 3
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Question A45
Solve the simultaneous equations:
91=
-l
a
2
2162
( 1) ( 2)=
- -a l
a a [2]
FAC 4 3 Question A46
If 2 3 7x - < then:
A 5 2x- < < B 2 5x- < < C 5 5x- < < D 5x < [1] FAC 4 4 Question A47
If 2 12 0x x- - < then: A 4x < B 3 and 4x x< - >
C 3x < - D 3 4x- < < [1] FAC 4 4 Question A48
Find the range(s) of values of x for which 2
( 2)(9 8)1
2 4
- - >+ -
x x
x x. [5]
FAC 4 4
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Question A49
12 2
0 1
n n
k k
k k
A 2
0
2n
k
k
B 1
2
0
n
k
k
C 2 1 2
1
2n
n
k
k k
D 2 2
0
1 2n
k
n k
[1]
FAC 4 6 Question A50
A child receives pocket money of £5 in the first week, then £6 in the second week, increasing by £1 each week. How much does he (she) receive in the first year? A £1,456 B £1,586 C £1,612 D £3,172 [1] FAC 4 7 Question A51
Evaluate 1
210
4kk
•
=Â
A 70 B 210 C 280 D 840 [2] FAC 4 7
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Question A52
Simplify 2
1
(1 2 )n
i
i i=
+ +Â . You may use the results 12
1
( 1)n
k
k n n=
= +Â and
2 16
1
( 1)(2 1)n
k
k n n n=
= + +Â .
A 3 22 4 3n n n+ +
B 3 22 9 7 6
6
n n n+ + +
C 28 15 1
6
n n+ +
D 3 22 9 13
6
n n n+ + [2]
FAC 4 8 Question A53
20 202
1y x y
y x= =
=Â Â
A 20 20
2
1 1x y
x y= =Â Â
B 20 20
2
1x y x
x y= =Â Â
C 20 20
2
1x y y
x y= =Â Â
D 20
2
1 1
x
x y
x y= =Â Â [1]
FAC 4 8
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Question A54
Calculate 6 6
1
2 ( 1)x y x
x y= =
-Â Â . You may use the results 12
1
( 1)n
k
k n n=
= +Â ,
2 16
1
( 1)(2 1)n
k
k n n n=
= + +Â and 3 2 214
1
( 1)n
k
k n n=
= +Â .
A 350 B 490 C 540 D 630 [5] FAC 4 8 Question A55
What is the coefficient of 5 3x y in the expansion of the expression 8(2 5 )x y+ ?
A 56 B 224,000
C 448,000
D 5,600,000 [2]
FAC 4 9 Question A56
Use the result ( 1) ( 1)( 2)2 32! 3!(1 ) 1 p p p p ppx px x x to simplify 2
(2 5 )
(3 2)
x
x
by
expanding as far as the term in 3x .
A 2 31 11 57 243
2 4 8 16x x x
B 2 357 2432 11
2 4x x x
C 2 323 28 2442
3 3 27x x x
D 2 32 23 28 244
9 27 27 243x x x [5]
FAC 4 9
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Question A57
The interest rate i for a financial arrangement satisfies the following equation:
½ ¼20,000(1 ) 5,000(1 ) 2,500(1 ) 30,000+ + + + + =i i i
(i) Find an approximate value for i by using the approximate relationship
(1 ) 1+ ª +ti ti , quoting your answer to the nearest 0.01%. [3]
(ii) By refining the approximate relationship to include an additional term, calculate
a more accurate value of i . [4] [Total 7] Question A58
You are given the following results involving series:
0 !
•
== Â
kx
k
xe
k
1
1
( 1)log (1 )
• +
=
-+ = Âk k
ek
xx
k
1
( 1)...( 1)(1 ) 1
!
•
=
- - ++ = +Âp k
k
p p p kx x
k
Use these to simplify the following series expressions:
(i) 2 3 41 1 12 6 24x x x x- + - +
[2]
(ii) 2 41 12 241 x x+ + +
[2]
(iii) 2 3 41 1 12 3 4x x x x+ + + +
[2]
(iv) 2 34 5 6 7
0 1 2 3x x x
Ê ˆ Ê ˆ Ê ˆ Ê ˆ+ + + +Á ˜ Á ˜ Á ˜ Á ˜Ë ¯ Ë ¯ Ë ¯ Ë ¯
[2]
[Total 8]
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Question A59
Let i be a positive quantity and let n be an integer greater than 1. (i)(a) By considering the arithmetic-geometric mean inequality as it would apply to the
quantities 1 and 1+ i , show that ½1 (1 )2
+ > +ii .
(i)(b) Show that 1/1 (1 )+ > + nii
n. [5]
(ii) Show that 1 (1 )+ < + nin i . [2]
[Total 7] Question A60
Let 2 3( ) 2 3= + + + + nnS x x x x nx .
(i) By considering the expression ( )
( )-nn
S xS x
x, or otherwise, show that:
1
2
(1 )( )
(1 )(1 )
n n
nx x nx
S xxx
+-= ---
[3]
(ii) Write down an expression for lim ( )
Æ•n
nS x , stating the range of values of x for
which convergence occurs. [2] [Total 5]
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Chapter 5 Question A61
The population of a country at the start of the year is 62,400,000. The population at the end of the year is 63,760,000. Express the change in population as a rate per mil. A 2.13‰ B 2.18‰ C 21.8‰ D 21.3‰ [1]
FAC 5 1 Question A62
An item has increased in cost by 30% to $58.50. What was the original cost? A $38.50 B $40.95 C $45 D $76.05 [1] FAC 5 1 Question A63
What is the proportionate change in the price of a house that goes from £180,000 to £210,000?
A 14.29% B 16.67% C 83.33% D 116.7% [1] FAC 5 2
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Question A64
When calculating 3y x= , a student uses 50x = to 2 significant figures. What is the
maximum absolute error possible here? A 0.0122 B 0.0123 C 0.119 D 0.127 [1] FAC 5 3 Question A65
You know that a equals 5,000 correct to one significant figure and b equals 0.20 correct to two decimal places. You calculate /a b to be 25,000. What is the greatest percentage by which you might be overestimating the true value? A 8.78% B 11.4% C 12.8% D 13.9% [2] FAC 5 3 Question A66
The following values from a standard normal distribution table were obtained: ( 1.20) 0.88493 ( 1.21) 0.88686P Z P Z< = < =
Use linear interpolation to calculate the value of ( 1.207)P Z < .
A 0.88507 B 0.88551 C 0.88590 D 0.88628 [1] FAC 5 5
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Question A67
The Newton-Raphson method ( )( )1
nn n
n
f xx x
f x+ = -¢
is being used to find the solution
between 1 and 2 to the equation ( ) 2 3 5 0f x x x= + - = . What is 4x if 1 1x = ?
A 1.2 B 1.1925 C 1.1926 D 1.2080 [2] FAC 5 6 Question A68
Expand (2 4)(3 )i i- - and express it in its simplest form, where 1i = - .
A 22 10 12i i- + - B 10 14i - C 10 10i - D 2 14i - [1] FAC 5 7 Question A69
What is the result if ( )2 3i- is multiplied by its complex conjugate?
A 5- B 5 12i- - C 13 D 13 12i- [1] FAC 5 7
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Question A70
Simplify 2
2
i
i
-+
, where 1i = - .
A 3 45 5
i-
B 431 i-
C 45
1 i-
D 5 43 3
i- [2]
FAC 5 7 Question A71
What is the modulus, r , and argument, q , in radians, of the complex number ( )5 3i- ?
A 4, 0.540r q= =
B 5.831, 0.540r q= = -
C 4, 30.96r q= =
D 5.831, 30.96r q= = - [2]
FAC 5 7 Question A72
Find the roots of 2 3 2.5 0x x+ + = . A 0.5 1.5 or 0.5 1.5x i i= - + - - B 0.5 1.5 or 0.5 1.5x i i= + - C 1.5 0.5 or 1.5 0.5x i i= + - D 1.5 0.5 or 1.5 0.5x i i= - + - - [2]
FAC 5 7
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Question A73
In the following equations, k is measured in £ and a is a dimensionless quantity:
1 = + kakl
, 2 2= akl
, 3 3
2= akl
(i) Determine the units of measurement of 1 2 3, ,k k k . [3]
(ii) Find the value of the constant c that would make the quantity 3
2c
kk
dimensionless. [1] [Total 4] Question A74
Let 1 2 4= +z i and 2 1 2= -z i .
(i) Write each of the following in the form +a bi where , Œa b R :
1 2+z z , 1 2-z z , 1 2z z , 1
2
z
z and 2 2
1 2+z z [5]
(ii) Calculate 1 2z z and 1 2z z , and comment on your answers. [3]
(iii) Calculate 1 2z z and 1 2z z , and comment on your answers. [3]
Note that z denotes the complex conjugate of z .
(iv) Express 1 23 2+z z in the form ire q . [2]
[Total 13]
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Question A75
Consider the equation 3 22 9 14 5 0- + - =z z z . (i) Show that 2 + i is a root of this equation. [3] (ii) Write down the other complex root. [1] (iii) Hence find all the roots of the equation. [2]
(iv) Which of these roots would lie outside the circle 1=z when represented on an
Argand diagram? [1] [Total 7] Question A76
Find the solution of the difference equation: 1 26 8 0- -+ + =t t ty y y
given that 0y and 1y are both equal to 5. [4]
FAC 5 8
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Chapter 6 Question A77
Which of the options best describes the function 23 5x x ? A ( ) and ( )O x o x
B ( ) but not ( )O x o x
C ( ) but not ( )o x O x
D none of the above [1] FAC 6 1 Question A78
What is the supremum of the sequence 1 2 3
, , ,...4 7 10
?
A 1 3
B C 0.333 D 0 [1] FAC 6 1 Question A79
Which of the following best describes what dy dx represents?
A y xd d for small values of xd and yd
B the change in the value of the graph between 0 and x C the gradient of ( )y f x=
D the rate of change of a chord between x a= and x a h= + [1] FAC 6.2
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Question A80
If ( ) 5f x x= , which of the following gives 0
( ) ( )lim
h
dy f x h f x
dx h+Æ
+ -= ?
A 0 B 1
C 5 D x [1] FAC 6.2 Question A81
Differentiate ( ) 5(4 )tf t with respect to t .
A 120t
B 5(4 ) ln 4t
C 15 4tt D 20ln t [1] FAC 6 3 Question A82
Find the gradient of 3 22 3y x x= - + when 2x = .
A 15 B 20 C 23 D 44 [1] FAC 6 3
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Question A83
Find the derivative of 3 5
4( )f x
x .
A 8 5
12
5x
B 2 5
12
5x
C 2 53
5
4
x
D 8 53
5
4
x [1]
FAC 6 3 Question A84
Find ( )f u¢ where 2 9( ) (5 7)f u u .
A 2 109(5 7)u -
B 2 89(5 7)u -
C 2 845 (5 7)u u -
D 2 890 (5 7)u u - [1]
FAC 6 4 Question A85
Find (0)M ¢ if 2 2½( ) t tM t e .
A
B em
C e
D 1 [2] FAC 6 4
Page 34 FAC: Question & Answer Bank – Questions
© IFE: 2013 Examinations The Actuarial Education Company
Question A86
Find (0)M ¢ if ( ) ln 1t
M t
.
A a-
B a l
C al
D a l- [2]
FAC 6 4 Question A87
Find ( )f x¢ where 2
2
4( ) ln 2 x
xf x e
e
.
A 2
2
14
2 xx
ee
B
2
2
2
2
42
42
xx
xx
xee
ee
C
2
2
2
2
84
42
xx
xx
xee
ee
D
2
2
2
2
42
2
xx
xx
xee
ee
[2]
FAC 6 4
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Question A88
Find ( )f x¢ where 26(5 3)( ) ln 4 xf x e .
A 60(5 3)x
B 26(5 3)60(5 3)ln 4 xx e
C 26(5 3)
60(5 3)
4 x
x
e
D
2
2
6(5 3)
6(5 3)
60(5 3)
4
x
x
x e
e
[5]
FAC 6 4 Question A89
Differentiate 2 5(4 3 )( 2 1)y x x x with respect to x .
A 2 4 2( 2 1) (3( 2 1) 5(4 3)(2 2))x x x x x x
B 2 4 2( 2 1) ( 33 76 43)x x x x
C 2 5 43( 2 1) 5(4 3 )(2 2)x x x x
D 2 415(2 2)( 2 1)x x x [5]
FAC 6 4
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Question A90
Find ( )f x¢¢ where 2 4( ) 32 xf x x e .
A 2 464(1 8 ) xx e
B 2 464(1 8 8 ) xx x e
C 41,024 xe
D 2 4 464(1 8 8 ) xx x e [5]
FAC 6 5 Question A91
Differentiate 23
x
xy
e twice with respect to x .
A 26 12 3
x
x x
e
- +
B 26 12 3
x
x x
e
- -
C 26 12 3
x
x x
e
+ +
D 26 12 3
x
x x
e
+ - [5]
FAC 6 5 Question A92
Calculate the maximum value of the function 2( ) 5 2 3f x x x .
A 6 B 4
C 163
D 13- [2]
FAC 6 6
FAC: Question & Answer Bank – Questions Page 37
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Question A93
Which of the following is a graph of ( 3)( 5)y x x x= + - ?
A
x-5 -4 -3 -2 -1 1 2 3 4 5
y
B
x-5 -4 -3 -2 -1 1 2 3 4 5
y
C
x-5 -4 -3 -2 -1 1 2 3 4 5
y
D
x-5 -4 -3 -2 -1 1 2 3 4 5
y
[2] FAC 6 6
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Question A94
Find and distinguish between the turning points of the function: 3 2( ) 7f x x x x .
A maximum 1,6 and minimum 513 27
( , 7 )
B maximum 1, 6 and minimum 1613 27
( ,6 )
C maximum 513 27
( , 7 ) and minimum 1,6
D maximum 1613 27
( ,6 ) and minimum 1, 6 [5]
FAC 6 6 Question A95
Using log differentiation or otherwise, find the value of x for which 4 3902( ) xf x x e-=
is a maximum.
A 43-
B 34-
C 34
D 43
[2]
FAC 6 6 Question A96
What is ( ) ( )( )2 3axy b xy c xy
x
∂ + +∂
where a , b and c are constants?
A ( ) ( )22 3a b yx c xy+ +
B 2 2 3 3 4
2 3 4
ax y by x cy x+ +
C 22 3ay byx cyx+ +
D 2 3 22 3ay by x cy x+ + [1]
FAC 6 7
FAC: Question & Answer Bank – Questions Page 39
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Question A97
For the function ( ) ( )2, ,f x y z xyz= , what is 3 f f
x y z x
∂ ∂¥∂ ∂ ∂ ∂
?
A ( )28 2xyz x yz-
B 8yz
C ( )328x yz
D ( )3216x yz [2]
FAC 6 7 Question A98
Find the turning points and their nature for the function 3 2 2( , ) 3 2f x y x x xy y
by considering the roots of the equation:
0 0 00 0 0
2
2 2 2
2 20
x x x x x xy y y y y y
f f f
x yy x
A Local maximum at 0, 0x y
B Saddle point at 4 43 3,x y
C Local maximum at 0, 0x y and saddle point at 4 43 3,x y
D Local minimum at 0, 0x y and maximum at 4 43 3,x y [5]
FAC 6 8
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Question A99
Determine the extrema of the function ( , ) 5 3f x y x y subject to the constraint 2 2 136x y .
You may make use of the Lagrangian function: 1 1( , , ) ( , , )n nL f x x g x x .
A maximum at (10, 6)f -
B minimum at ( 5,3)f -
C maximum at (10, 6)f - and minimum at ( 10,6)f -
D maximum at (5, 3)f - [5]
FAC 6 9 Question A100
Evaluate:
(i) 3 3
0lim
-
-Æ
--
x x
x xx
e e
e e [2]
(ii) 1
lim 12
n
n n
[2]
[Total 4] Question A101
Differentiate the following function with respect to i :
10
10
100 1 (1 )5
(1 )
-- +++
i
ii
and evaluate the derivative at 0.05=i . [3] FAC 6 4
FAC: Question & Answer Bank – Questions Page 41
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Question A102
Find a formula for the n th derivative of the function 1
2 +x
x. [4]
FAC 6 5 Question A103
By considering log y , or otherwise, find an expression in terms of l for the maximum
value attained by the function 1 1( )- - -= +y x xa g g aagl l when 1=a and 4=g . [5]
FAC 6 6 Question A104
A function is defined by:
( ) (1 )= +f x x xa b , [0,1]Œx
where , ŒRa b are parameters.
By considering the shape of the graphs, or otherwise, find the range of values attained by the function ( )f x for values of x in the range [0,1] in each of the following cases:
(i) 1, 3= = -a b [3]
(ii) 2= =a b [3]
(iii) ½, 1½= - =a b . [3]
[Total 9] Question A105
Find the positions of the extrema of 3 2 2( , ) 2 2= - +f x y x x y and determine their
nature. [5] FAC 6 8
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Question A106
Find the positions of the extrema of ( , , ) = + +f x y z x y z , subject to the constraints
2 0+ =x y and 2 2 2 1+ + =x y z . [5]
FAC 6 9 Question A107
According to a mortality table, the instantaneous rate of mortality at age x , which is denoted by ( )xm , is calculated from the formula:
0 1 0 170 70
( ) exp50 50
x xx a a b b
, 17≥x
and 0 0.00338415= -a , 1 0.00386512= -a , 0 3.352236= -b and 1 4.656042=b .
(i) Show that (20) 0.000814=m and find the value of (40)m . [3]
(ii) Find the age in the range 20 40£ £x at which the function m is stationary,
rounding your answer to the nearest month, and indicate the nature of this point. [4] (iii) Another mortality function ( )q x is related to ( )xm by the relationship:
1
0( ) 1 exp ( )q x x t dt
Find the value of (20)q . [5]
[Total 12]
FAC: Question & Answer Bank – Questions Page 43
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Chapter 7 Question A108
Which of the following is NOT true?
A ( ) ( )b a
a b
f x dx f x dx= -Ú Ú
B ( ) ( ) ( )0 0
= -Ú Ú Úb b a
a
f x dx f x dx f x dx
C ( ) ( ) ( )= -Úb
a
df x dx f b f a
dx
D ( )b
a
df x dx
dxÚ is used to find the area under the ( )f x curve [1]
FAC 7 1 Question A109
What is the integral of 26 6 6x x+ + ? A 12 6x c+ +
B 3 22 3 6x x x c+ + +
C 3 26 6 6x x x c+ + +
D 3 23 6 6x x x c+ + + [1] FAC 7 2
Question A110
What is the integral of 2
0.75
1
x dxÚ ?
A 1.351 B 0.636 C 0.477 D 0.119- [1]
FAC 7 2
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Question A111
What is the integral of 2 2x x-+ ?
A 2 2
2
x x-+c+
B 3 3
3
x x-+c+
C 3 13
3
x x--c+
D 3 3
3
x x--c+ [1]
FAC 7 1 Question A112
What is 4
2
5 x dxÚ ?
A 1,260
ln6
B 600
ln5
C 13583
D 625 25
ln 4 ln 2- [1]
FAC 7 2
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Question A113
What is xBc dxÚ ?
A ln
xBcconst
x+
B xBc const+
C 1
1
xBcconst
x
++
+
D ln
xBcconst
c+ [1]
FAC 7 2 Question A114
Evaluate 3
0
-Ú xe dxll .
A 3( 1)e ll - -
B 31 e l--
C 3(1 )e ll --
D 4.51 e l-- [1] FAC 7 2
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Question A115
What is the integral of 5
1x +?
A ( )5ln 1x + c+
B ( )2
10
1x
-+
c+
C ( )2
5
1x
-+
c+
D ( )5
ln 1x +c+ [1]
FAC 7 2 Question A116
Evaluate 43 5
0
• -Ú xx e dx .
A 0 B 0.05 C 1 D the integral does not converge [2]
FAC 7 2 Question A117
Evaluate 51
6 30 (10 )+Úx
dxx
.
A 0.00000132 B 0.00000792 C 0.000145 D 0.000868 [2] FAC 7 2
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Question A118
What is the integral of2
3
3 3
2 6 1
x
x x
++ +
, for 0x > ?
A 3
4 2
3
0.5 3
x xc
x x x
+ ++ +
B ( )2ln 3 3x c+ +
C ( )30.5ln 2 6 1x x c+ + +
D ( ) ( )3 33 3 ln 2 6 1x x x x c+ + + + [2]
FAC 7 2 & 3 Question A119
Use partial fractions to find the integral of ( )( )2 5
2 4 1
x
x x
+- +
.
A ( )( )
31
2
2 4ln
1
xk
x
Ï ¸-Ô ÔÌ ˝+Ô ÔÓ ˛
B 3 ½ln ( 1) (2 4)k x x+ -
C 123ln( 1) ln(2 4)x x c+ - - +
D 123ln(2 4) ln( 1)x x c- + + + [5]
FAC 7 3 Question A120
What is 1
2
0
xxe dx ?
A 10.778- B 0.5 C 1 D 2.097 [2] FAC 7 3
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Question A121
Leibniz’s formula states that:
( ) ( )
( ) ( )
( , ) ( ) ( , ( )) ( ) ( , ( )) ( , ) b x b x
a x a x
d ff x t dt b x f x b x a x f x a x x t dt
dx x
∂∂
= - +¢ ¢Ú Ú .
What is 2
0
3 2x
dx t dt
dx-Ú ?
A 23x
B 26x
C 26 8x x-
D 29 2x x- [2] FAC 7 3
Question A122
Find the value of 15 10
5 5
(5 )x y
x y dy dx= =
+Ú Ú .
A 2,875 B 3,250 C 2,812.5 D 1,406.25 [2]
FAC 7 5
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Question A123
Which of the following does NOT converge?
A 2
1
xe dx•
-Ú
B 1
0
ln x dxÚ
C 2
1
x dx•
-Ú
D 31
6dx
x
•
Ú [1]
FAC 7 4 Question A124
Using the trapezium rule and 7 ordinates, what is the approximate area under the curve 2(7 )= -y x between 1=x and 4=x ?
A 40.1875 B 63.125 C 80.375 D 138.25 [2] FAC 7 6 Question A125
If xe is expanded to its fourth term, using the Maclaurin expansion
2(0)( ) (0) (0)
2!
ff x f f x x
¢¢= + + + ◊◊◊¢ , what is the value of 2e based on this?
A 5 B 6.333 C 7.389 D 7.667 [2]
FAC 7 7
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Question A126
Which of the following is the correct first five terms of the Taylor expansion of
[ ]ln (3 )(3 )x y+ + about (0,0) ? You are given that the Taylor’s expansion of ( , )f x y
about ( , )a b is:
( , ) ( , )
2 2 22 2
2 2( , ) ( , ) ( , )
1( , ) ( , ) ( ) ( )
1!
1( ) 2 ( )( ) ( )
2!
a b a b
a b a b a b
f ff x y f a b x a y b
x y
f f fx a x a y b y b
x yx y
∂ ∂∂ ∂
∂ ∂ ∂∂ ∂∂ ∂
È ˘Í ˙= + - + -Í ˙Î ˚
È ˘Í ˙+ - + - - + -Í ˙Î ˚
+
A 2 21 1 1 1ln9
3 3 18 18x y x y+ + - -
B 2 21 1 1 1ln9
3 3 9 9x y x y+ + - -
C 2 21 1 1 1ln9
3 3 9 9x y x y+ + + +
D 2 21 1 1 1ln9
3 3 18 18x y x y+ + + + [5]
FAC 7 7 Question A127
Given that 3 1
dy y
dx x=
+, 2y e= when 0x = , what is the value of y when 2x = ?
A 51.723 B 7 C 14.135 D 6 [2] FAC 7 8
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Question A128
The gamma function is defined by 1
0( )
• - -G = Ú x tx t e dt .
(i) Show that ( ) ( 1) ( 1)G = - G -x x x for values of x for which both gamma functions
are defined. [3] (ii) Show that (1) 1G = . [2]
(iii) You are given that (½)G = p . By considering (½)G and applying the
substitution 212=t z , or otherwise, show that
212
11
2
• --•
=Ú ze dz
p. [4]
[Total 9] Question A129
Find the area of the region enclosed by the x axis and the curve 26 2= + -y x x . [4]
FAC 7 3 Question A130
Evaluate ( , )ÚÚA
f x y dxdy where ( , ) 2 6= +f x y x y and the region of integration is the
shaded area shown in the diagram. [6] FAC 7 5
1x
1y
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Question A131
Show that 2
210.035103
(5 )= =
+Úx
I dxx
using at least three different methods.
[3 marks for each correct method] Question A132
(i) Calculate the value of the following integral, where 0.5 1x y< < < :
2
12
0.5 0.5
(4 1)y
y x
y x
y e xe dx dy= =
+Ú Ú [4]
(ii) Explain why it may be considered easier to integrate with respect to x first
rather than integrating with respect to y first. [5]
[Total 9] Question A133
(i) Write out the first four terms in the Taylor series for xe and 1(1 )-+ y , stating
the range of values of x and y for which these series are valid. [2]
(ii) Hence determine the coefficients ,a b in the series expansion:
2 31 ( )1= + + +
-x
xax bx O x
e [5]
[Total 7]
FAC: Question & Answer Bank – Questions Page 53
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Question A134
(i) Find constants A and B such that 1
(2 5 ) 2 5∫ +
- -A B
P P P P. [2]
(ii) Hence, or otherwise, solve the differential equation:
1
2 5= -dPP
P dt
for the function ( )P t , where 250 ( )P t< < , subject to the boundary condition that
215(0) =P . [4]
[Total 6] Question A135
Find the particular solution of the differential equation:
( 1) ( 1)+ - + =dyx x x y
dx
where 0x ≥ , given that 0=y when 0=x . [5]
FAC 7 8
Chapter 8 Question A136
If
2
1
4
a
and
2
5
3
b
then the values x and y such that
10
7
18
x ya b
are:
A 1x and 6y
B 4x and 1y
C 3x and 2y
D 2x and 1y [2]
FAC 8 1
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Question A137
The unit vector in the direction of 3 2 4i j k is given by:
A 29(3 2 4 )i j k
B
31
221
4
C
31
229
4
D 3 2 421 21 21
i j k [1]
FAC 8 1 Question A138
The angle between
4
1
3
and
1
3
2
is:
A 65.4º B 74.8º C 105.2º D 114.6º [2] FAC 8 1 Question A139
Which one of the vectors is orthogonal to 3 2i j k ?
A 4 4 4 i j k
B 4 4 4 i j k
C 4 4 4 i j k
D 4 4 4 i j k [2]
FAC 8 1
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Question A140
If 1 2
3 4A
and 2 4
3 5B
then AB is given by:
A 2 8
9 20
B 8 14
6 32
C 8 14
18 32
D 14 20
18 26
[2]
FAC 8 2 Question A141
If
3 2 4
1 0 3
3 1 2
A
then det A is given by:
A 33 B 5
C 1- D 19- [2] FAC 8 2
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Question A142
If 2 1
4 3A
then 1A is given by:
A 3 11
4 22
B 0.3 0.1
0.4 0.2
C 1 31
2 42
D 2 1
1.5 0.5
[2]
FAC 8 2 Question A143
The eigenvalues, , of matrix A solve: A det( ) 0 A I
B det( ) 0 A I
C det( ) 0 A I
D det( ) 0A [1]
FAC 8 2
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Question A144
Which of the following is NOT an eigenvector of 3 2
6 1A
:
A 2
6
B 3
1
C 1
3
D 2
2
[2]
FAC 8 2 Question A145
(i) By considering expressions involving scalar products, find a unit vector of the form + +a b ci j k that is perpendicular to the displacement vectors 2 3- +i j and
10 +i k . [4] (ii) Write down the other unit vector perpendicular to 2 3- +i j and 10 +i k . [1]
[Total 5] Question A146
By solving a set of simultaneous linear equations, or otherwise, find the inverse of the
matrix
0.2 0.2 0.2
0.2 0 0.2
0 0.4 0.2
P . [5]
FAC 8 2
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Question A147
The probability density function of the multivariate normal distribution is given by:
11 1( ) exp ( ) ( )
2(2 ) det( )
T
nf
x x x
Evaluate ( )f x when 2n , 1
1
, 2 1
1 4
and 2
0
x . [5]
FAC 8 2
Glossary Question A148
The abbreviation eg means: A for example B compared with C and so on D that is to say [1] FAC Gloss Question A149
What is the plural of matrix? A matrices B matrixes C matrises D matrix [1] FAC Gloss
FAC: Question & Answer Bank – Questions Page 59
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Question A150
A weak basis for an actuarial calculation: A makes many assumptions B makes optimistic assumptions C makes few assumptions D makes pessimistic assumptions [1] FAC Gloss Question A151
A deterministic model has: A optimistic assumptions B assumptions which allow for random variation C many assumptions D assumptions which remain fixed over time [1] FAC Gloss Question A152
Which of the following word pairs should be used to complete the blanks in this sentence? The ___________ from ActEd is to ___________ lots of past papers before the exam. A advise, practise B advise, practice C advice, practise D advice, practice [1] FAC Gloss
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Question A153
A friend has a full-year salary equivalent to £18,000 per annum. The company she works for has a 6 day working week. She is paid pro rata as she only works 2½ days per week. How much does she earn each year? A £18,000 B £9,000 C £7,500 D £6,000 [1] FAC Gloss
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Question & Answer Bank – Solutions Solution A1
D The natural numbers are 1, 2, 3, 4, 5, ... Solution A2
C
( )A B« is everything in A and not in B which is {6, 8}. So ( )A B C« » is these two
values and all the values in C which is {1, 3, 6, 8 ,10}.
Solution A3
D
A B« means not in A and not in B.
Solution A4
A
Solution A5
C 40 basis points is 0.40%. Solution A6
D Both odd and even numbers are divisible by 3.
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Solution A7
C Solution A8
D £100 100 1.227 €122.7 122.7 1.314 $161.23= ¥ = = ¥ = Solution A9
A The first payment is 1 Sept 2011, the second payment is 1 Mar 2012, the third payment is 1 Sept 2012, the fourth payment is 1 Mar 2013, the fifth payment is 1 Sep 2013, the sixth payment is 1 March 2014, the seventh payment is 1 Sept 2014, the eighth payment is 1 Mar 2015, the ninth payment is 1 Sept 2015 and the tenth payment is 1 Mar 2016. Solution A10
(i) FALSE. Should say: “Ending in a 5 is a sufficient condition for an integer to be a multiple of 5.” [1]
(ii) FALSE. 2= -x is a member of this set. Various corrections are possible eg
changing to + or , adding the extra condition 2π -x or changing = ∆ to π ∆ (!). [1]
(iii) TRUE. If we “complete the square”, we can write 2 25 32 45 7 ( )- + = - +x x x ,
which always takes positive values. [2] (iv) FALSE. The natural logarithm function doesn’t have a finite value when the
argument is either 0 or • . So the set of values should be (0, )• . [1]
FAC: Question & Answer Bank – Solutions Page 3
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Solution A11
The investor invests £64,000. After the initial charge is deducted this would leave £64,000 0.975¥ .
When converted to dollars (at the exchange rate effective at that time) this would give $64,000 0.975 1.67¥ ¥ . [1]
If the unit price then was $P per unit, this would buy 1
64,000 0.975 1.67¥ ¥ ¥P
units.
The unit price has now increased to $1.05P . So the dollar value of the units would be
164,000 0.975 1.67 1.05 64,000 0.975 1.67 1.05¥ ¥ ¥ ¥ = ¥ ¥ ¥P
P. [1]
Applying the current exchange rate to find the current sterling value gives
1£64,000 0.975 1.67 1.05 £68,386.50
1.60¥ ¥ ¥ ¥ = . [1]
Solution A12
B Solution A13
D Solution A14
B Significant figures are counted from the first non-zero digit. Solution A15
C
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Solution A16
D
101 1.0627.743
0.062 1.062
-- =
Solution A17
B
2 3 4
6 8 10 1229.2
1.08 1.08 1.08 1.08 to 3 SF
Solution A18
D
2
21 1 ln5 1.02 1 1exp exp 0.7367974
2 0.8 20.8 2 0.8 2
1exp( 0.2714352) 0.380
0.8 2
Solution A19
B
5 12 5 124 46! 7204 20 4 20 240 1.782 2.115 243.9
3 3+ + = + + = + + =
FAC: Question & Answer Bank – Solutions Page 5
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Solution A20
Using the formulas given:
1 1(1 ) 1.1 ( 0.90909)- -= + = =v i [1]
and: 0.1/1.1 ( 0.09091)= = =d iv [1]
So:
10101 1 1.1
10 1.10.1/1.1 29.04
0.10
--- -- - ¥
= =
nnv
nvd
i [1]
Comment We’ve kept the exact figures throughout the calculation here to ensure that the final answer is accurate. However, you could equally use the calculated values of v and d , provided you retain enough decimals. Usually 5 significant figures is about right for intermediate calculations. You’ll meet these symbols when you study compound interest in Subject CT1.
Solution A21
Expressing the coefficients in units of 1 million and bringing all the terms onto the LHS gives:
10 543.6(1 ) 43.92(1 ) 60.192 0+ - + - =i i [1]
This is a quadratic equation in 5(1 )+ i . Using the quadratic formula, we get:
2
5 43.92 (43.92) 4(43.6)( 60.192) 43.92 111.474(1 )
2 43.6 87.2
± - - ±+ = =¥
i [1]
So:
5(1 ) 1.782 or 0.775+ = -i [1]
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Disregarding the second possibility (which corresponds to a negative interest rate), we find that:
1/5(1.782) 1 0.12248= - =i ie 12.2% [1]
Solution A22
While checking the calculation you should have noticed that factors needed for intermediate ages have been calculated using linear interpolation. For example, the
term of 1421 years is a quarter of the way from 20 to 25. So F1 is calculated as:
1416.5 (19.7 16.5) 17.3+ - =
(i)(a) Contribution rate This would bring the existing fund to £82,500 and the calculation would be: 4.125 17.3 50% 0.657 17.7+ ¥ = ¥ ¥k 9.8%fi =k [2] (i)(b) Percentage pension We now have: 3.625 15% 17.3 0.657 17.7+ ¥ = ¥ ¥t 53.5%fi =t [2] (i)(c) Contribution rate
We now have 31216=n and the calculation is:
3.625 13.875 50% 0.7255 19.9+ ¥ = ¥ ¥k 25.9%fi =k [2]
Comment In any calculation you should ask yourself whether you believe the answer you’ve come up with. You can do this here by checking that the alterations have the effect you would expect. For example, the contribution rate in part (a) falls slightly, which makes sense. In other questions, you can apply a reasonableness check by simplifying the calculation using approximations and “ball park” figures.
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(ii) Recommended contribution rate The calculation for Ms Jones (with 17½=n ) is:
235 14.75 0.708 19.9+ ¥ = ¥ ¥k 29.8%fi =k [3]
Note that you need to use the female retirement factor here. The m and f in the columns for F3 stand for male and female.
Comment Although we chose factors that were fairly realistic, there is no guarantee that this calculation form would be appropriate for a real life calculation. There are also limits as to how much you’re allowed to contribute to a personal pension plan.
Solution A23
(i) Pension first payment and date Pension payments will start on 1 May 2016. The complete calendar years between leaving and retirement are the twelve years 2004, 2005, … , 2015. So 12=t when the payments start, but it will increase by 1 on 1 January each year. [2]
So the member’s first payment will be 1213,620 1.05 £541.75
12¥ ¥ = . [2]
(ii) Total pension during first 10 years The payments the member will receive during the first ten years of retirement will consist of: 8 payments of £541.75 during 2016 12 payments of £541.75 1.05¥ during 2017
12 payments of 2£541.75 1.05¥ during 2018 …
12 payments of 9£541.75 1.05¥ during 2025
4 payments of 10£541.75 1.05¥ during 2026 [2]
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The total amount is:
9 108 541.75 12 541.75 (1.05 1.05 ) 4 541.75 1.05¥ + ¥ ¥ + + + ¥ ¥ [1]
The middle terms can be summed as a GP, which gives:
10
101.05 1.058 541.75 12 541.75 4 541.75 1.05 £83,132
1 1.05
-¥ + ¥ ¥ + ¥ ¥ =-
[1] (The exact amount will depend on how rounding is applied by the scheme’s administrators.) As a reasonableness check, we can say that the typical payment will be
5541.75 1.05 £691.43¥ = and there are 120 payments, which would make a total of 120 691.43 £82,971¥ = . This is close to our calculated answer. [1]
Solution A24
B
A is the graph of xy e=
C is the graph of lny x=
D is the graph of 2y x=
Solution A25
B
A is the graph of 1
yx
= -
C is the graph of 2
1y
x=
D is the graph of 2y x=
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Solution A26
A
(2 1)y f x= - squashes the graph horizontally by a factor of 2 and shifts it 1 unit to the
right. B is (0.5 1)y f x= +
C is (2 1)y f x= +
D is (0.5 1)y f x= -
Solution A27
A Solution A28
A The person is 55 years and approximately 9 months. Solution A29
C
4 1 5
5 4 1 5
6 4 4
1.5 1
x
x
x
x
+ <
- < + <
- < <
- < <
Solution A30
A
0.8 0.449e- = and 4 0.443p = so min( , 4) 0.443xe p- =
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Solution A31
A 100! 100 99 98
161,70097!3! 3!
¥ ¥= =
Solution A32
A
(5.5) 4.5 (4.5) 4.5(3.5) (3.5) 4.5(3.5)(2.5) (2.5)
4.5(3.5)(2.5)(1.5) (1.5) 4.5(3.5)(2.5)(1.5)(0.5) (0.5)
4.5(3.5)(2.5)(1.5)(0.5) 52.34p
G = G = G = G
= G = G
= =
Solution A33
D
( 1)(3 1)! !(3 1)!(3 1)(3 )
(3 )( 1)! (3 1)!( 1)!
n n n nn n n
n n n n
G + + += = +G - - -
Solution A34
To 2 significant figures, these work out as follows:
(i) 1.5 4.5= =e e e [1] (ii) log 0.00001 12= -e [1]
(iii) 1tanh 0.9 1.5- = [1] (iv) (12) 11! 40,000,000G = = [1]
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Solution A35
D
( )2 3 2 3 3 32 2 2 2 2 2bb b b b b b bx x x x x x x x x x +¥ + = + = +
Solution A36
C
33 2 3 6 33
1x x x x xx
e e e e ee
Solution A37
C See the definition in Chapter 4 Section 1.2 of the notes. Solution A38
D
2 2
2log log log log loga a a a ax x x
y yy y y
Solution A39
C
2 2
2
3 5 2 ( 3)( 1) (5 2 )(3 1)
3 1 1 (3 1)( 1) ( 1)(3 1)
2 3 6 13 5
(3 1)( 1) ( 1)(3 1)
7 11 8
(3 1)( 1)
x x x x x x
x x x x x x
x x x x
x x x x
x x
x x
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Solution A40
B
2
2 2
3 6 3 ( 3)( 2)
(2 1)( 2) (2 1)( 1)2 3 2 2 3 1
3
x x x x x x
x x x xx x x x
x
+ + - + + -∏ = ∏+ - + +- - + +
+=(2 1)x +
(2 1)
( 2)
x
x
+¥
-( 1)
( 3)
x
x
++
2
( 2)
( 1)
( 2)
x
x
x
-
+=-
Solution A41
A
Factorising gives (2 1)( 3) 0x x+ - = . Hence 0.5,3x = - . Solution A42
A
212 ( 12) 4(1)(12) 12 96 12 4 66 2 6
2(1) 2 2x
± - - ± ±= = = = ±
Solution A43
The coefficients in these equations are so horrible that it is better to represent them by letters, find a general solution, and then substitute the numbers at the end. If we write the equations as: ax by e cx dy f
we can then find x by multiplying them by d and b respectively and subtracting, to get: ( )ad bc x de bf
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64522 57963 64717 80068
8.13095432 64522 64717 92404
de bfx
ad bc
[1]
Similarly (ie if we just swap /a b and /c d in this formula), we get:
92404 57963 95432 80068
12.88464717 92404 95432 64522
ce afy
bc ad
[1]
These calculations are much easier if you have a calculator that can store variables to memory A, B, C etc. (An alternative method would be to find the inverse of the 2 2 matrix.)
Comment We put this question in to illustrate the point that you’re allowed to introduce your own symbols and abbreviations if it makes the calculations more tractable.
Solution A44
From the first equation:
3
4=
+a
a b 4 3( )fi = +a a b 3fi =a b [½]
Changing all the a ’s to b ’s in the second equation then gives:
2
2
30.0225
(4 ) (4 1)=
+b
b b
30.0225
16(4 1)fi =
+b [½]
Rearranging:
1 3
1 1.8334 16 0.0225
[½]
3 5.5fi = =a b [½]
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Solution A45
The easiest way to solve these equations is to square the first equation and substitute it into the second equation:
2
2162 9 1621 2 2
Ê ˆ = fi =Á ˜Ë ¯- - -l a a
a a a [½]
This gives:
22=
-a
a [½]
From which: 2( 2) 4= - fi =a a a [½]
This gives 27=l . [½] Solution A46
B
2 3 7
7 2 3 7
4 2 10
2 5
x
x
x
x
- <- < - <- < <- < <
Solution A47
D Factorising gives ( 4)( 3) 0x x- + < . So either 4 0x - < and 3 0x + > (ie 3 4x- < < )
or 4 0x - > and 3 0x + < (but it is not possible for both of these to be true at the same time).
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Solution A48
Rearranging the inequality so that we have zero on the RHS:
2
( 2)(9 8)1 0
2 4
- - - >+ -
x x
x x
Expressing the LHS as a single fraction:
2
2
( 2)(9 8) ( 2 4)0
2 4
- - - + - >+ -
x x x x
x x [½]
Simplifying:
2 2
2
(9 26 16) ( 2 4)0
2 4
- + - + - >+ -
x x x x
x x
2
2
8 28 200
2 4
- + >+ -
x x
x x [1]
We can factorise the numerator:
2
4(2 5)( 1)0
2 4
- - >+ -
x x
x x [½]
The denominator equals zero when 1 5= - ±x (ie 3.236- and 1.236). [1] If we call these ,a b (say), the graphs of the numerator and the denominator look like
this:
a b1 2.5
[1] So the ratio will be positive when both graphs are positive or both are negative, ie when
1 5 ( )< - - =x a or 1 1 5 ( )< < - + =x b or 2.5>x . [1]
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Solution A49
D
12 2 2 2 2 2 2
0 1 0 0 0
( 1) 2 ( 1)n n n n n
k k k k k
k k k k n k n
Solution A50
B This is an arithmetic series with 5a = , 1d = and 52n = . Hence:
152 2 (52)(2 5 51 1) £1,586S = ¥ + ¥ =
Solution A51
A
2 31
210 210 210 210
44 4 4kk
•
== + + +Â
This is an infinite geometric series with 210
4a = and
1
4r = . Hence:
210 4
701 1 4
S• = =-
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Solution A52
D
( )
2 2
1 1 1 1
1 12 6
2 3 216
3 2
(1 2 ) 1 2
2 ( 1) ( 1)(2 1)
(2 3 )
2 9 13
6
n n n n
i i i i
i i i i
n n n n n n
n n n n n n
n n n
= = = =+ + = + +
= + + + + +
= + + + + +
+ +=
   Â
Solution A53
D Considering the order of x and y we have 1 20y x£ £ £ . So if x sums over the
numbers 1 to 20, then looking at the order y must sum from 1 to x .
Solution A54
C Swapping the order of summation gives:
6 6 6
1 1 1
612
1
63 2
1
6 63 2
1 1
2 21 14 6
2 ( 1) 2( 1)
2( 1) ( 1)
6 (6 1) 6(6 1)(2(6) 1)
350
y
x y x y x
y
y
y y
x y y x
y y y
y y
y y
= = = =
=
=
= =
- = -
= - +
= -
= -
= + - + +
=
   Â
Â
Â
 Â
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Solution A55
B
The term will be 8 5 3 5 3 5 3 5 35(2 ) (5 ) 56 2 5 224,000C x y x y x y= ¥ ¥ =
Solution A56
A
2 2 232
2314 2
( 2)( 3) ( 2)( 3)( 4)2 33 3 314 2 2! 2 3! 2
2 327 2714 4 2
2 3 2 327 13514 2 4
571 112 4 8
(2 5 ) (2 5 )
(3 2) 2 ( 1)
(2 5 )(1 )
(2 5 ) 1 ( 2)( ) ( ) ( )
(2 5 )(1 3 )
(2 6 27 ) ( 5 15 )
x x
x x
x x
x x x x
x x x x
x x x x x x
x x
2 324316
x
Solution A57
(i) Approximate value for i Using the approximate relationship, the equation given becomes:
1 12 420,000(1 ) 5,000(1 ) 2,500(1 ) 30,000+ + + + + =i i i [1]
ie 27,500 23,125 30,000+ =i 10.81%fi =i [2]
We could have divided this through by 100 to make the figures easier. (ii) Improved approximate value for i If we included the quadratic term in the approximate relationship, it would become:
212(1 ) 1 ( 1)+ ª + + -ti ti t t i [1]
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The equation given would then become:
( )( )( )( )
21 1 1 12 2 2 2
231 1 14 2 4 4
20,000(1 ) 5,000 1
2,500 1
30,000
È ˘+ + + + -Î ˚
È ˘+ + + -Î ˚
=
i i i
i i
[1]
ie:
2
2
27,500 23,125 859.375 30,000
2,500 23,125 859.375 0
+ - =
fi - + - =
i i
i i [1]
Solving this quadratic equation, we get:
223,125 (23,125) 4( 859.375)( 2,500) 23,125 22,938.4
2( 859.375) 1,718.75
- ± - - - - ±= =- -
i
To get the root consistent with part (i), we need to take the + sign, which gives
10.85%ªi . [1] Solution A58
(i) 2 3 41 1 12 6 24 1 xx x x x e-- + - + = - [2]
(ii) 2 41 1 12 24 21 ( )x xx x e e-+ + + = + [2]
Comment When you add up the two series the odd powers cancel. You might recognise this as the hyperbolic cosine function cosh x .
(iii) 2 3 41 1 12 3 4 log (1 )ex x x x x+ + + + = - - [2]
(iv) 2 3 5 24 5 6 7 ( 5)( 6)(1 ) 1 5 ( )
0 1 2 3 2!x x x x x x
[2]
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Comment
This last one isn’t at all obvious, but if you write out the terms in the series 5(1 )-- x
and cancel all the minus signs, you’ll see that they do match up. You’ll need this type of result when you study the properties of the negative binomial distribution in Subjects CT3 and CT6.
Solution A59
(i)(a) Arithmetic-geometric mean inequality 1 and 1+ i are unequal positive quantities. So the AGM inequality applies, and tells us that:
1 (1 )
1(1 )2
+ + > +ii ie ½1 (1 )
2+ > +i
i [2]
(i)(b) Show that If we apply the AGM inequality to the n quantities consisting of 1-n 1’s and a 1+ i , we get:
( 1) 1 (1 )
1 1 1 (1 )- ¥ + + > ¥ ¥ ¥ ¥ +nn i
in
ie 1/1 (1 )+ > + nii
n [3]
(ii) Show that If we expand the RHS of the expression in the question using a binomial expansion, we get:
2(1 ) 11 2
n nn ni i i i
[1]
Since 0>i , all the terms in this series are positive. So, if we just keep the first two, we know that:
(1 ) 1+ > +ni ni ie 1 (1 )+ < + nin i [1]
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Solution A60
(i) Show that We have:
2 1
2 1
( )1 2 3
( ) 2 ( 1)
-
-
= + + +
= + + + - +
nn
n nn
S xx x nx
x
S x x x n x nx
Subtracting gives:
2 1( )( ) 1 -- = + + + + -n nn
nS x
S x x x x nxx
[1]
The LHS can be factorised and the terms except the last on the RHS can be summed as a GP:
1 1 1 1
1 ( ) ( )1 1
n nn n
n nx x x
S x nx S x nxx x x x
[1]
Multiplying both sides by 1
x
x
, then gives the required result:
1
2
(1 )( )
1(1 )
n n
nx x nx
S xxx
+-= ---
[1]
Comment This multiply-and-subtract “trick” can be a useful for summing certain types of series. This method is used for deriving formulae for annuity functions in Subject CT1.
(ii) Limit and convergence
Provided 1 1- < <x , nx will vanish as Æ•n and we will have:
2lim ( )(1 )
nn
xS x
xƕ=
- [2]
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Solution A61
C
The change is 1,360,000
0.0218 21.8‰62,400,000
= =
Solution A62
C $58.50 1.3 $45∏ = . Solution A63
B
The change is 30,000
16.67%180,000
= .
Solution A64
B 50 to 2 SF gives a range of 49.5 to 50.5.
3 350.5 50 3.6963 3.6840 0.0122- = - =
3 350 49.5 3.6840 3.6717 0.0123- = - = So the maximum absolute error is 0.0123.
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Solution A65
D The true value of a could be anywhere in the range [4500,5500) .
The true value of b could be anywhere in the range [0.195,0.205) .
The least possible true value is therefore 4500
21,9510.205
= , which is less than our
estimate by 3,049 ie 3,049
13.9%21,951
= .
Solution A66
D
[ ]( 1.207) ( 1.20) 0.7 ( 1.21) ( 1.20)
0.88493 0.7(0.88686 0.88493)
0.88628
P Z P Z P Z P Z< = < + < - <= + -=
Solution A67
C
( )2
21
3 53 5 0 ( ) 2 3
2 3n n
n nn
x xf x x x f x x x x
x++ -= + - = fi = + fi = -¢
+
21 1
2 11
22 2
3 22
23 3
4 33
3 5 11 1.2
2 3 5
3 5 0.041.2 1.1926
2 3 5.4
3 5 0.000054861.1926 1.1926
2 3 5.3852
x xx x
x
x xx x
x
x xx x
x
+ - -= - = - =+
+ -= - = - =+
+ -= - = - =+
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Solution A68
C
2(2 4)(3 ) 2 10 12 2 10 12 10 10i i i i i i- - = - + - = + - = -
Solution A69
C
( )( ) 22 3 2 3 4 9 4 9 13i i i- + = - = + = Solution A70
A
23 4
2 5 5
2 2 2 4 4 4 4 1 3 4
2 2 2 4 1 54
i i i i i i ii
i i i i
- - - - + - - -= ¥ = = = = -+ + - +-
Solution A71
B The Argand diagram is as follows:
r
5
3
2 25 ( 3) 34 5.831r = + - = =
13 35 5tan tan 0.540q q -- -= fi = = - radians
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Solution A72
D
23 3 4(1)(2.5) 3 1 31.5 0.5
2(1) 2 2
ix i
- ± - - ± - - ±= = = = - ±
Solution A73
(i) In the calculation of 1k , k is added to al
. So these must have the same
dimensions. Since k is measured in £ and a is a dimensionless quantity, this
means that l must have dimensions of 1£- , and 1k has dimensions of £. [1]
The dimensions of 2k and 3k are then seen to be 22
1£
£-= and 3
3
1£
£-= ,
respectively. [2]
(ii) 3
2c
kk
has dimensions 3
3 22
££
£-= c
c. This will be dimensionless if 3 2 0- =c ie
32=c . [1]
Comment
1 2 3, ,k k k and 33 2
2
kk
are actually the mean, variance, skewness and coefficient of
skewness of the translated gamma distribution.
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Solution A74
(i) Write in the form a + bi
1 2 (2 4 ) (1 2 ) 3 2+ = + + - = +z z i i i [1]
1 2 (2 4 ) (1 2 ) 1 6- = + - - = +z z i i i [1]
2
1 2 (2 4 )(1 2 ) 2 8 2 8 10= + - = - = + =z z i i i [1]
1
2
2 4 2 4 1 2 2 8 8 6 81.2 1.6
1 2 1 2 1 2 1 4 5
+ + + + - - += = ¥ = = = - +- - + +
z i i i i ii
z i i i [1]
2 2 2 2
1 2 (2 4 ) (1 2 ) ( 12 16 ) ( 3 4 ) 15 12+ = + + - = - + + - - = - +z z i i i i i [1]
(ii) Calculate and comment
1 2 2 4 1 2 20 5 100 10= + - = ¥ = =z z i i 1 2 10 10= =z z [2]
These two quantities are always equal, whatever the values of 1z and 2z . [1]
(iii) Calculate and comment
1 22 4
1.2 1.61 2
-= = - -+
iz z i
i 1 2 1.2 1.6 1.2 1.6= - + = - -z z i i [2]
Again, these two quantities are always equal, whatever the values of 1z and 2z . [1]
(iv) Express in the form iθre
1 23 2 3(2 4 ) 2(1 2 ) 8 8 8(1 )+ = + + - = + = +z z i i i i [1]
This has magnitude 2 28 1 1 8 2+ = and an argument of 45∞ or 4
p radians.
So / 41 23 2 8 2+ = iz z e p . [1]
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Solution A75
Consider the equation 3 22 9 14 5 0- + - =z z z . (i) Show 2 + i is a root When 2= +z i :
2 2(2 ) 3 4= + = +z i i 3 2 (2 )(3 4 ) 2 11= = + + = +z zz i i i [2]
So the LHS of the equation is: 2(2 11 ) 9(3 4 ) 14(2 ) 5 0+ - + + + - =i i i [1]
(ii) The other complex root Since this is a polynomial equation with real coefficients, the complex roots come in conjugate pairs. So the other complex root is 2= -z i . [1] (iii) Find the third root The third root (a , say) of this cubic equation must be real. So the equation must take the form: ( )[ (2 )][ (2 )] 0- - + - - =C z z i z ia
Multiplying out the two complex factors in square brackets, we must have the identity:
2 3 2( )( 4 5) 2 9 14 5- - + = - + -C z z z z z za [1]
For the 3z terms to match up, we must have 2=C . For the constant term to match up, we must have 5 5- = -Ca ½fi =a . So the third factor is ½-z and the third root is ½=z . [1] (iv) Which roots lie outside the unit circle
Both the complex roots have magnitude 5 and so lie outside the unit circle. The real root has magnitude ½, and so lies inside. [1]
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Solution A76
The auxiliary equation is 2 6 8 0+ + =v v which has roots 4- and 2- . [1]
The general solution is of the form ( 2) ( 4)= ¥ - + ¥ -t tty A B . [1]
However we know that 0 5=y and 1 5=y , so:
5
5 2 4
= +
= - -
A B
A B [1] These can be solved to give 7.5= -B and 12.5=A , so our particular solution is:
12.5( 2) 7.5( 4)= - - -t tty [1]
Check: 2y should equal 1 06 8 70- - = -y y
The formula gives 2 212.5 ( 2) 7.5 ( 4) 70¥ - - ¥ - = - .
Solution A77
B
23 53 5
x xx
x
- = -
3 5 3x- < so it is ( )O x but
0lim (3 5 ) 0x
xÆ
- π so it is not ( )o x .
Solution A78
A
The general term is 3 1
n
n + which will be bounded above by
1
3 3
n
n= .
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Solution A79
C Solution A80
C
0 0 0 0
( ) ( ) 5( ) 5 5lim lim lim lim 5 5
h h h h
dy f x h f x x h x h
dx h h h+ + + +Æ Æ Æ Æ
+ - + -= = = = =
Solution A81
B Solution A82
B
26 2dy
x xdx
= - which is 20 when 2x = .
Solution A83
A
3 5 8 5 8 53 123 5 5 5 8 5
4 12( ) 4 ( ) 4( )
5f x x f x x x
x x
Solution A84
D
Using the chain rule 2 8 2 8( ) 9(5 7) 10 90 (5 7)f u u u u u .
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Solution A85
A
Using the chain rule 2 22 ½( ) ( ) t tM t t e . Hence 0(0)M em m= =¢ .
Solution A86
B
Using the chain rule 1
( )1 1
M tt t
. Hence (0)Mal
=¢ .
Solution A87
D
Now 2 2 22
4( ) ln 2 ln 2 4x x x
xf x e e e
e
. Using the chain rule gives:
2
2
2 2
22
22
42
1( ) (4 8 )
22 4
xxx x
x x xx
xeef x xe e
e e ee
--
-= ¥ - =¢
+ +
Solution A88
D Using the chain rule gives:
22
2 2
6(5 3)1 6(5 3)
6(5 3) 6(5 3)
1 60(5 3)( ) 6(2)(5 3) 5
4 4
xx
x x
x ef x x e
e e
- -- -
- - - -
-= ¥ - - ¥ = -¢+ +
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Solution A89
B Using the product and chain rules:
2 5 2 4
2 4 2
2 4 2 2
2 4 2
( 3)( 2 1) (4 3 )5( 2 1) (2 2)
( 2 1) 3( 2 1) 5(4 3 )(2 2)
( 2 1) ( 3 6 3) 5( 6 14 8)
( 2 1) ( 33 76 43)
dyx x x x x x
dx
x x x x x x
x x x x x x
x x x x
Solution A90
B Using the product rule:
4 2 4 4 2 4( ) 64 32 ( 4 ) 64 128x x x xf x xe x e xe x e
4 4 4 2 4
4 4 2 4
2 4
( ) 64 64 ( 4 ) 256 128 ( 4 )
64 512 512
64(1 8 8 )
x x x x
x x x
x
f x e x e xe x e
e xe x e
x x e
Solution A91
A
Now 2
233 x
x
xy x e
e . So using the product rule gives:
26 3x xdyxe x e
dx- -= -
2 2
2 22
6 12 36 6 6 3 (6 12 3 )x x x x x
x
d y x xe xe xe x e x x e
dx e- - - - - - += - - + = - + =
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Solution A92
C
Now ( ) 2 6f x x . Setting this equal to zero gives a turning point at 13x = - .
Since ( ) 6f x this turning point is a maximum.
So the maximum value of this function is 2 161 1 13 3 3 3( ) 5 2( ) 3( )f - = - - - - = .
Solution A93
D Since ( 3)( 5)y x x x= + - we can see that this crosses the x -axis when 0, 3x = - and 5.
This means it is either C or D. Since we have a positive 3x term then y Æ• as
x Æ• which means it is D. Alternatively, we could calculate and determine the turning points using differentiation. Solution A94
C Differentiating and setting the derivative equal to zero gives:
2 13
( ) 3 2 1 (3 1)( 1) 0 ,1f x x x x x x
The second derivative is: ( ) 6 2f x x
When 1x = we have (1) 0f >¢¢ so we have a minimum. When 13x = - we have
13( ) 0f - <¢¢ so we have a maximum.
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Solution A95
D
We have 902ln ( ) ln 4 ln 3f x x x= + - , so:
4
ln ( ) 3d
f xdx x
= -
Setting this equal to zero and solving gives 43x = .
Solution A96
D
( ) ( )( ) ( )2 3 2 2 3 3
2 2 32 3
axy b xy c xy axy bx y cx yx x
ay bxy cx y
∂ ∂+ + = + +∂ ∂
= + +
Solution A97
D
( ) ( )2 2 2 2, ,f x y z xyz x y z= =
2 22f
xy zx
∂fi =∂
224
fxyz
x y
∂fi =∂ ∂
3
8f
xyzx y z
∂fi =∂ ∂ ∂
Hence:
32 2 2 3 38 2 16
f fxyz xy z x y z
x y z x
∂ ∂¥ = ¥ =∂ ∂ ∂ ∂
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Solution A98
C Finding the first derivatives and setting them equal to zero and solving gives:
23 6 2 0f
x x yx
(1)
2 2 0f
x yy
(2)
Equation (2) tells us that y x= . Substituting this into equation (1) gives:
2 43
3 4 0 0,f
x x xx
So our two turning points are 0x = , 0y = and 43x = , 4
3y = .
Calculating the second derivatives gives:
2
2
2
2
2
6 6
2
2
fx
x
f
y
f
x y
So for 0x = , 0y = we have:
0 0 00 0 0
2
2 2 22
2 2
2
( 2 )( 6 ) 2
8 8 0
x x x x x xy y y y y y
f f f
x yy x
Solving this gives roots of 1.172x = - and 6.828x = - . Since both of these roots are negative this means that we have a maximum at 0x = , 0y = .
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For 43x = , 4
3y = we have:
0 0 00 0 0
2
2 2 22
2 2
2
( 2 )(2 ) 2
8 0
x x x x x xy y y y y y
f f f
x yy x
Solving this gives roots of 2.828x = ± . Since these roots have different signs this
means that we have a saddle point at 43x = , 4
3y = .
Solution A99
C
The Lagrangian function is 2 2(5 3 ) ( 136)L x y x yl= - - + - . Finding the derivatives
and setting them equal to zero gives:
5
5 2 02
Lx x
xl
l∂ = - = fi =∂
(1)
3
3 2 02
Ly y
yl
l∂ = - - = fi = -∂
(2)
2 2 136 0L
x yl∂ = - - + =∂
(3)
Substituting (1) and (2) into (3) gives:
2 2
2
5 3 34 1136 0 136
2 2 44l
l l l-Ê ˆ Ê ˆ- - + = fi = fi = ±Á ˜ Á ˜Ë ¯ Ë ¯
When 14l = we have 10x = and 6y = - .
When 14l = - we have 10x = - and 6y = .
We could then check to see what kind of turning points we have, but only one answer has these two values – so it must be the correct one.
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Solution A100
(i) Limit
We can’t evaluate 3 3
0lim
-
-Æ
--
x x
x xx
e e
e e by substituting 0=x because this gives
0
0.
If we use the exponential series, we find that the denominator is:
( ) ( )2 2 31 12 21 1 2 ( )-- = + + + - - + - = +x xe e x x x x x O x [½]
The numerator is the same, but with x replaced by 3x :
3 3 36 ( )-- = +x xe e x O x [½]
So the limit is:
3 3 3 2
3 20 0 0
6 ( ) 6 ( )lim lim lim 3
2 ( ) 2 ( )
-
-Æ Æ Æ
- + += = =- + +
x x
x xx x x
e e x O x O x
e e x O x O x [1]
(ii) Limit
This limit is the special case of the result lim 1n
x
n
xe
n
when ½x .
So:
½1lim 1 1.649
2
n
ne e
n
[2]
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Solution A101
Differentiating, we get:
11 10
112
10(1 ) (1 (1 ) ) 110 100(1 ) 5
- -- ¥ + - - + ¥- ¥ + + i i i
ii
[2]
Evaluating this when 0.05=i gives:
11 10
112
0.5(1.05) 1 (1.05)1,000(1.05) 5 772.17
0.05
- -- - +- + = - [1]
Solution A102
Let 1
2= +xyx
. [1]
If we work out the first few derivatives, we get:
22 log 2 -= -xdyx
dx
2
2 32
2 (log 2) 2 -= +xd yx
dx
3
3 43
2 (log 2) 6 -= -xd yx
dx [2]
The pattern is then apparent:
12 (log 2) ( 1) ! - -= + -n
x n n nn
d yn x
dx [1]
Note the trick of using ( 1)- n to obtain the alternating signs in the second term. We
could use mathematical induction to show that we have guessed the pattern correctly.
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Solution A103
Substituting the values given for the parameters gives:
3 4 24 ( )-= +y x xl l [1]
Taking logs, as suggested:
4log log 4 log 3log 2log( )= + + - +y x xl l [1]
Since the log function is monotonic, this will have its maximum value for the same value of x as the original function. So, differentiating using the function-of-a-function rule, and equating to zero:
3
4
3 2(4 )log 0= - =
+d x
ydx x xl
[1]
Rearranging:
4 43( ) 8x x 43 5x
1
43
5x
[1]
Substituting this back into the original function to find the maximum value of y :
3 32 2
4 4 3/ 4 5/ 4 1/ 43 3 3 8 14 4 3 5
5 5 5 5 16y
[1]
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Solution A104
(i) α = 1,β = -3
Here:
3( ) (1 )-= +f x x x
so that:
(0) 0=f and 18(1) =f [1]
The derivative (found using the product rule) is:
3 4( ) (1 ) 3 (1 )- -= + - +¢f x x x x
This equals zero when:
3 4(1 ) 3 (1 ) 0- -+ - + =x x x
ie (1 ) 3 0+ - =x x 12fi =x and 4
27(½) =f [1]
So the graph has a maximum value at ½=x and looks like this:
½ 1
427
The range of values of the function here is:
4270 ( )£ £f x [1]
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(ii) α = β = 2
Here:
2 2( ) (1 )= +f x x x
So that: (0) 0=f and (1) 4=f [1]
The derivative is:
2 2( ) 2 (1 ) 2 (1 ) 0= + + + >¢f x x x x x when [0,1]Œx [1]
So the function increases steadily over the range and the graph looks like this:
1
4
The range of values of the function here is: 0 ( ) 4£ £f x [1]
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(iii) α = -½,β = 1½
Here:
½ 1½( ) (1 )-= +f x x x
So that:
0
lim ( )Æ +
= +•x
f x and (1) 2 2=f [1]
The derivative is:
1½ 1½ ½ ½312 2( ) (1 ) (1 )- -= - + + +¢f x x x x x
This equals zero when:
1½ 1½ ½ ½312 2(1 ) (1 ) 0- -- + + + =x x x x
ie (1 ) 3 0- + + =x x 12fi =x and
3 3(½)
2=f [1]
So the graph looks like this:
½ 1
4
The range of values of the function here is:
3 3
( )2
≥f x [1]
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Solution A105
If 3 2 2( , ) 2 2= - +f x y x x y , then:
23 4
4
= -
=
fx x
x
fy
y
∂∂
∂∂
[1]
Setting these equal to zero we get:
4
0 or 3
=x , 0=y
So the extrema occur at 0, 0= =x y and at 4
0,3
= =x y . [1]
To determine their nature, the second partial derivatives are needed:
2
2
2
2
2
6 4
4
0
= -
=
=
fx
x
f
y
f
y x
∂∂
∂∂
∂∂ ∂
[1]
Setting up the required equations: For 0, 0= =x y the equation is (4 )( 4 ) 0- - - =l l , ie 4= ±l . This means that
0, 0= =x y is a saddle point. [1]
For 4
, 03
x y= = the equation (4 )(4 ) 0l l- - = , ie 4l = , twice. This means that
4, 0
3x y= = is a local minimum. [1]
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Solution A106
The Lagrangian function is:
2 2 2( 2 ) ( 1)L x y z x y x y zl m= + + - + - + + - [1]
Finding the partial derivatives:
2 2 2
1 2 (1) 2 (4)
1 2 2 (2) 1 (5)
1 2 (3)
L Lx x y
x
L Ly x y z
y
Lz
z
∂ ∂l m∂ ∂l
∂ ∂l m∂ ∂m
∂ m∂
= - - = - -
= - - = - - - +
= - [1]
Setting these equal to zero, we get:
1 1 2(1) (2)
2 2
1(3)
2
1 2 4(4) 0 3 5 0
2 2
x y
z
l lm m
m
l l lm m
- -fi = fi =
fi =
- + - +fi + = fi - + = [1]
So 0.6l = . Finally substituting into (5) we get:
2 2 2
2
0.4 0.2 11
2 2 2
1.2 4
0.3
[1]
This gives the positions of the extrema as:
1 1 1
, ,5 0.3 10 0.3 2 0.3
x y z= - = = -
and: 1 1 1
, ,5 0.3 10 0.3 2 0.3
x y z= = - = [1]
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Solution A107
(i) Show µ(20) and obtain µ(40)
20x 70
150
x
0 1 0 1(20) exp 0.000814a a b b [1]
40x 70
0.650
x
0 1 0 1(40) 0.6 exp 0.6 0.001077a a b b [2]
(ii) Find and determine the turning point The derivative of this function is:
11 0 1
1 70( ) exp
50 50 50
b xx a b b
[1]
Equating this to zero to find the stationary point gives:
1 1 0 170
exp 050
xa b b b
Rearranging:
10
1 1
50log 70 29.82e
ax b
b b
To the nearest month, this is 29 years and 10 months. [2] To determine the nature of this point (ie whether it is a minimum, a maximum or point of inflexion), we can look at the second derivative:
2
10 1
70( ) exp
50 50
È ˘-Ê ˆ Ê ˆ= +¢¢ Á ˜Á ˜ Í ˙Ë ¯Ë ¯ Î ˚
b xx b bm
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This must be a positive quantity since the squared factor and the exponential function must both be positive. So this point is a minimum. [1]
Comment Mortality reaches a low point in the late 20s before starting to increase into “old age”.
(iii) Find q(20) The integral in the formula for (20)q is:
1 10 1 0 10 0
10 1 0 10
20 70 20 70(20 ) exp
50 50
1 exp 150 50
t tt dt a a b b dt
t ta a b b dt
[2]
Although this looks complicated, we can integrate it directly:
121
0 1 0 101 0
0 1 0 1 0 11 1
50(20 ) exp 1
100 50
1 50 1 501 exp 1 exp
100 50
0.000791
t tt dt a t a t b b
b
a a b b b bb b
[2]
We then find that:
0.000791(20) 1 0.000791-= - =q e [1]
Comment You will meet the mortality functions featuring in this question in Subject CT4 and CT5. They are usually written as xq and xm .
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Solution A108
D
( )b
a
f x dxÚ is used to find the area under the ( )f x curve.
Solution A109
B We simply use the rule “raise the power by 1 and divide by the new power”. Solution A110
A
22 1.75 1.750.75
1 1
2 11.351
1.75 1.75
xx dx
È ˘ -= = =Í ˙Í ˙Î ˚
Ú
Solution A111
C
3 12 2 3 11 3
3 3
x xx x dx x x c c
-- - -+ = - + = +Ú
Solution A112
B
44 4 2
2 2
5 5 5 6005
ln5 ln5 ln5
xx dx
È ˘ -= = =Í ˙Í ˙Î ˚
Ú
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Solution A113
D This is a standard result, but can be proved as follows:
ln
ln
ln ln
¥¥= = + = +Ú Ú
x c xx x c Be Bc
Bc dx Be dx const constc c
Solution A114
B 3
3 3 0 3
00
( ) 1x xe dx e e e el l l ll - - - -È ˘= - = - - - = -Î ˚Ú
Solution A115
A Solution A116
B
Using the substitution 45u x= , we see that 320dudx x= , so 3
20du x dx= . Hence we get:
43 5
0 0 0
0 ( 1)0.05
20 20 20
u ux e e
x e dx du
•• • - -- È ˘ - -= = - = =Í ˙
Í ˙Î ˚Ú Ú
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Solution A117
C
Using the substitution 610= +w x (so that 56=dwx
dx), we get:
115 21 11 36 30 10
10
2 2
1 1
6 6 2(10 )
1 1 10.000145
12 11 10
x wdx w dw
x
Solution A118
C
Using the substitution 32 6 1u x x= + + (so that 26 6dudx x= + ), we get:
2
33
3 3 10.5 0.5ln 0.5ln(2 6 1)
2 6 1
xdx du u c x x c
ux x
+ = = + = + + ++ +Ú Ú
Solution A119
A
Let ( )( )2 5
2 5 ( 1) (2 4)2 4 1 2 4 1
x A Bx A x B x
x x x x
+ ∫ + fi + = + + -- + - +
.
Substituting 1x = - gives 3 6 ½B B= - fi = - .
Substituting 2x = gives 9 3 3A A= fi = .
( )( )
3 12 2
312
2 5 3 ½ln(2 4) ln( 1)
2 4 1 2 4 1
(2 4)ln
( 1)
xdx dx x x c
x x x x
xk
x
+ = - = - - + +- + - +
Ï ¸-Ô Ô= Ì ˝+Ô ÔÓ ˛
Ú Ú
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Solution A120
D Using integration by parts gives:
1 112 2 21 12 20
0 0
12 21 12 4 0
2 21 1 12 4 4
2.097
x x x
x
xe dx xe e dx
e e
e e
Solution A121
D Using Leibniz’s formula gives:
( ) ( )
[ ]
2 2 2
0 0
2
0
20
2 2
2
3 2 1 3 2 0 3 2
(3 2 ) 6
(3 2 ) 6
(3 2 ) 6
9 2
x x
t x
x
x
dx t dt x t x t dt
dx x
x x x dt
x x xt
x x x
x x
=
∂- = - - + -∂
= - +
= - +
= - +
= -
Ú Ú
Ú
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Solution A122
A 15 10 15
102
55 5 5
15
5
152
5
(5 ) 5 0.5
25 37.5
12.5 37.5
2,875
yx y x
x
x
x y dy dx xy y dx
x dx
x x
== = =
=
=
È ˘+ = +Î ˚
= +
È ˘= +Î ˚
=
Ú Ú Ú
Ú
Solution A123
B Using integration by parts we get:
[ ]1
10
0
ln lnx dx x x x= -Ú
But ln x is undefined when 0x = . To be honest it is much easier to eliminate the alternatives:
2 21 1 12 2 21
1
0 ( )x xe dx e• •- -È ˘= - = - - =Î ˚Ú
2 1
11
0 ( 1) 1x dx x• •- -È ˘= - = - - =Î ˚Ú
3 2
11
6 30 ( 3) 3dx
x x
• •È ˘= - = - - =Í ˙Î ˚Ú
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Solution A124
B Working with 7 ordinates, the area under the curve is approximately:
( )2 2 2 2 2 2 216 2 5.5 5 4.5 4 3.5 3 0.5 63.125
2È ˘+ + + + + + ¥ =Î ˚
Solution A125
B Using the Maclaurin expansion, we have:
2 3
12! 3!
x x xe x= + + + +
So:
2 3
2 2 21 2 6.333
2! 3!e + + + =
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Solution A126
A First note that ln(3 )(3 ) ln(3 ) ln(3 )x y x y+ + = + + + . Hence:
0, 00, 0
0, 00, 0
2
2 20, 00, 0
2
2 20, 00, 0
2
0, 0
1 1
3 3
1 1
3 3
1 1
9(3 )
1 1
9(3 )
0
x yx y
x yx y
x yx y
x yx y
x y
f
x x
f
y y
f
x x
f
y y
f
y x
∂∂
∂∂
∂∂
∂∂
∂∂ ∂
= == =
= == =
= == =
= == =
= =
= =+
= =+
= - = -+
= - = -+
=
So the expansion is:
2 2 2 21 1 1 1 1 1 1 1 1ln9 ln9
3 3 2 9 9 3 3 18 18x y x y x y x y
È ˘+ + + - - = + + - -Í ˙Î ˚
Solution A127
C
13ln ln(3 1)
3 1 3 1
dy y dy dxy x c
dx x y x= fi = fi = + +
+ +Ú Ú
When 0x = , 2y e= which means:
2 13ln ln1 2 0 2e c c c= + fi = + fi =
Hence when 2x = , we have 13ln ln 7 2 14.135y y= + fi = .
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Solution A128
(i) Show Γ( ) = ( - 1)Γ( - 1)x x x
Starting from the definition given, and integrating by parts with 1-= xu t and -= tdv e dt (so that 2( 1) -= - xdu x t dx and -= - tv e ), we get:
1 1 2
0 00( ) ( ) ( 1)
•• •- - - - - -È ˘G = = - + -Î ˚Ú Úx t x t x tx t e dt t e x t e dt [2]
The term in square brackets is zero. So we get:
2
0( ) ( 1) ( 1) ( 1)
• - -G = - = - G -Ú x tx x t e dt x x [1]
(This relationship will be valid provided the integral for ( 1)G -x converges, which
requires 1 0- >x ie 1>x .) (ii) Show Γ(1) = 1
Using the integral definition with 1=x , we find that:
0
0 0 0(1) 0 ( 1) 1
•• •- - -È ˘G = = = - = - - =Î ˚Ú Út t tt e dt e dt e [2]
(iii) Obtain integral Using the integral definition with ½=x , we know that:
½
0(½)
• - -G = =Ú tt e dt p [1]
If we apply the substitution 212=t z (so that =dt
zdz
), we find that:
( ) 212
½2120
-• - =Ú zz e zdz p [1]
Simplifying, and taking the constants to the RHS:
21
2
0 2
• - =Ú ze dz
p [1]
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Since the function 21
2- z
e is an even function (ie it takes the same values when the sign of z is reversed), the area under the graph for negative values of z is the same as for the positive values. So:
2 21 1
2 2
02 2
• •- --•
= =Ú Úz ze dz e dz p
212
11
2
• --•
fi =Ú ze dz
p [1]
Comment This result is important in connection with the standard normal distribution.
Solution A129
The curve 26 2y x x crosses the x axis when:
2 32
6 2 0 (2 )(3 2 ) 0 or 2x x x x x [2]
So the required area is:
22 2 2 3 71 2
2 3 243 2 3 2
343(6 2 ) 6 14
24x x dx x x x
[2]
Solution A130
The shaded region can be specified as {( , ) : 0 1,0 }£ £ £ £x y x y x . So the integral is:
1
0 0( , ) (2 6 )= +ÚÚ Ú Ú
x
A
f x y dxdy dx dy x y [2]
The inner integral is:
2 2
0 0(2 6 ) 2 3 5È ˘+ = + =Î ˚Ú
xxx y dy xy y x [2]
11 2 3530 0
5( , ) 5
3A
f x y dxdy x dx xÈ ˘fi = = =Î ˚ÚÚ Ú [2]
Note that you could also evaluate the integral in the opposite order as
1 1
0(2 6 )+Ú Úy
dy dx x y .
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Solution A131
(1) Substitution Using the substitution 5= +t x , we get:
7
7 7
2 26 66
5 1 5 5 7 5log log 0.035103
6 42
tI dt dt t
t tt t
[3]
(2) Integration by parts
Integration by parts, with =u x and 2
1
(5 )=
+dv
dx x (so that =du dx and
1
5= -
+v
x),
gives:
2
2 2
111
1 1 2 1 5 7log 5 log
5 5 7 6 42 6I x dx x
x xÈ ˘ È ˘= ¥ - - - = - + + + = - +Î ˚Í ˙+ +Î ˚ Ú [3]
(3) Partial fractions If we add and subtract 5 from the numerator of the integrand, it then takes a form that we can integrate directly:
2
2 2
2 21 11
(5 ) 5 1 5 5log 5
5 5(5 ) (5 )
xI dx dx x
x xx x
È ˘+ - È ˘= = - = + + =Í ˙ Í ˙+ ++ + Î ˚Î ˚Ú Ú [3]
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Solution A132
(i) Calculate the integral Integrating with respect to x :
2
2
2
2
12
0.5 0.5
12
0.50.5
12 0.5
0.5
1 12 0.5
0.5 0.5
(4 1)
1(4 1)
4
1 1(4 1)
4 4
1 1(4 1) (4 1)
4 4
yy x
y x
yy x
y
y y
y
y y y
y y
y e xe dx dy
y e e dy
y e e e dy
y e dy e y e dy
= =
=
=
+
= =
+
È ˘= + Í ˙Î ˚
È ˘= + -Í ˙Î ˚
= + - +
Ú Ú
Ú
Ú
Ú Ú [2]
The first integral can be done by inspection, the second one needs to be done by parts:
{ }
{ }
211 12 0.5
0.50.5 0.5
3 1 0.5 1 0.5 10.5
3 1 0.5 1 0.5 1 0.5
1 1(4 1) 4
4 4
1 1( ) 5 3 4[ ]
4 4
1 1( ) 5 3 4( ) 2.54
4 4
y y y y
y
y
e e y e e dy
e e e e e e
e e e e e e e
+
=
Ï ¸Ô ÔÈ ˘ È ˘- + -Ì ˝Î ˚Í ˙Î ˚ Ô ÔÓ ˛
= - - - -
= - - - - - =
Ú
[2]
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(ii) Calculate the integral by reversing the order We can reverse the order of integration:
2 2
1 1 12 2
0.5 0.5 0.5
(4 1) (4 1)y
y x x y
y x x y x
y e xe dx dy xe y e dy dx= = = =
+ = +Ú Ú Ú Ú [1]
To integrate with respect to y , we integrate by parts. We get:
{ }
{ }
{ }
2
2
2
2
2 2 2
1 112
0.5
12 1 1
0.5
12 1 1
0.5
12 1
0.5
12 1 2 2 2
0.5
(4 1) 4
5 (4 1) [4 ]
5 (4 1) 4 4
4 3
( 4 3 )
x y y
y xx y x
x x yx
x
x x x
x
x x x
x
x x x x x
x
xe y e e dy dx
xe e x e e dx
xe e x e e e dx
xe e xe e dx
xe x e xe dx
== =
=
=
=
+ + +
=
Ï ¸Ô ÔÈ ˘+ -Ì ˝Î ˚Ô ÔÓ ˛
= - + -
= - + - +
= - +
= - +
Ú Ú
Ú
Ú
Ú
Ú [3]
This would be complicated to integrate, as we would need to integrate numerically. This confirms that integrating with respect to x first is the best option. [1]
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Solution A133
(i) Taylor’s series The Taylor series for these functions are:
2 31 12 61= + + + +xe x x x ( -• < < •x ) [1]
and:
1 2 3(1 ) 1-+ = - + - +y y y y ( 1 1- < <y ) [1]
(ii) Determine the coefficients
Using the series for xe , we find that the series for the denominator is:
2 3 41 12 61 ( )- = + + +xe x x x O x
So the original function is:
2 3 41 1
2 61 ( )=
- + + +x
x x
e x x x O x [1]
Cancelling an x on the top and bottom:
{ } 12 31 1
2 62 31 12 6
11 ( )
1 1 ( )
-È ˘= = + + +Î ˚- + + +x
xx x O x
e x x O x [1]
We can now use the series for 1(1 )-+ y with 2 31 12 6 ( )= + +y x x O x , which gives:
22 3 2 3 31 1 1 1
2 6 2 61 ( ) ( ) ( )1
È ˘ È ˘= - + + + + + +Î ˚ Î ˚-x
xx x O x x x O x O x
e [1]
Multiplying out the terms and ignoring any terms of power greater than 2, we get:
2 2 3 2 31 1 1 1 12 6 4 2 121 ( ) 1 ( )
1È ˘ È ˘= - + + + = - + +Î ˚ Î ˚-x
xx x x O x x x O x
e [1]
So 12= -a and 1
12=b . [1]
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Comment If you follow the calculations through keeping more terms you’ll find that the
coefficient of the 3x term is zero. So 21 12 121
1ª - +
-x
xx x
e is actually a very good
approximation when x is small.
Solution A134
(i) Partial fractions The coefficients A and B satisfy the identity:
1
(2 5 ) 2 5∫ +
- -A B
P P P P
Multiply through by (2 5 )-P P :
1 (2 5 )∫ - +A P BP
ie 1 2 ( 5 )∫ + -A P B A [1]
Equating the coefficients of the constant terms and the P terms:
12=A
and 5 0- =B A 525fi = =B A
So:
1 1/ 2 5 / 2
(2 5 ) 2 5∫ +
- -P P P P [1]
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(ii) Solve the differential equation Rearranging the differential equation and integrating:
1
(2 5 )=
-Ú ÚdP dtP P
Using the identity in part (i):
1/ 2 5 / 2
2 5dP dt
P P
ie 1 12 2ln ln(2 5 )P P t c- - = +
12 ln
2 5
Pt c
P= +
- [1]
Note that 0P > and 2 5 0P- > , so no modulus signs are required here.
We are told that, when 0=t , 215=P . So:
12
2 /15ln
2 5(2 /15)c=
- 1 1 1
2 10 2ln ln10cfi = = - [1]
So:
1 12 2ln ln10
2 5
Pt
P= -
- [1]
We can make P the subject of this as follows. Doubling and rearranging:
ln ln10 22 5
Pt
P+ =
-
10
ln 22 5
Pt
P=
-
210
2 5=
-tP
eP
210 (2 5 )= - tP P e
2 2(10 5 ) 2+ =t tP e e [1]
2
2( )
5 10 tP te-
fi =+
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Solution A135
The differential equation can be re-written in the form 1
1- =
+dy
y xdx x
. [1]
This can be solved using an integrating factor, where the integrating factor is:
[ ]1 1exp exp ln(1 )
1 1È ˘- = - + =Í ˙+ +Î ˚Ú dx x
x x [1]
Note that no modulus signs are required here since we know 0x ≥ , so 1 0x+ > . Multiplying through by the integrating factor and integrating both sides with respect to x , we obtain:
1 1
1 1= ¥
+ +Úy x dxx x
[1]
Since 1
11 1= -
+ +x
x x, we find that:
1 1
1 ln( 1)1 1
= - = - + ++ +Úy dx x x c
x x [1]
where c is a constant. Since 0=y when 0=x , we find that 0=c , so the particular solution is
[ ]( 1) ln( 1)= + - +y x x x . [1]
Solution A136
C We need to solve the simultaneous equations:
2 2 10
5 7
4 3 18
x y
x y
x y
- + =+ =- = -
Rearranging the first equation gives 5y x= + . Substituting this into the second
equation gives 6 18 3x x= - fi = - . Hence 2y = .
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Solution A137
C
The magnitude of 3 2 4i j k is 2 2 23 ( 2) 4 29+ - + = . So the unit vector in the
same direction is:
33 2 4 1
229 29
4
i j k
Solution A138
C The scalar product is: 4 1 ( 1) 3 3 ( 2) 5¥ + - ¥ + ¥ - = -
Hence:
2 2 2 2 2 25 4 ( 1) 3 1 3 ( 2) cos
26 14 cos
q
q
- = + - + + + -
=
So:
5
cos 105.2º26 14
q q= - fi =
FAC: Question & Answer Bank – Solutions Page 63
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Solution A139
B The easiest way to do this is to see which of these vectors has a scalar product of zero with 3 2i j k . (4 4 4 ) (3 2 ) 12 8 4 0i j k i j k ( 4 4 4 ) (3 2 ) 12 8 4 0i j k i j k ( 4 4 4 ) (3 2 ) 12 8 4 0i j k i j k (4 4 4 ) (3 2 ) 12 8 4 0i j k i j k Solution A140
C
1 2 2 4 2 6 4 10 8 14
3 4 3 5 6 12 12 20 18 32AB
Solution A141
C
0 3 1 3 1 0det 3 ( 2) 4
1 2 3 2 3 1
3 3 2 ( 7) 4 1
1
- -= - - +
- -= ¥ + ¥ - + ¥= -
A
Solution A142
B
If a b
c d
Ê ˆ= Á ˜Ë ¯
A then 1 3 11 1
4 26 ( 4)
d b
c aad bc- - -Ê ˆ Ê ˆ= =Á ˜ Á ˜-- - -Ë ¯ Ë ¯
A . So we have:
1 3 1 3 1 0.3 0.11 14 2 4 2 0.4 0.26 ( 4) 10
- - - -Ê ˆ Ê ˆ Ê ˆ= = =Á ˜ Á ˜ Á ˜- - Ë ¯ Ë ¯ Ë ¯
A
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Solution A143
B Solution A144
B
The easiest way to check this is to find out which vectors satisfy 3 2
6 1
x xk
y y
Ê ˆ Ê ˆ Ê ˆ=Á ˜ Á ˜ Á ˜-Ë ¯ Ë ¯ Ë ¯
.
Those that do are eigenvectors.
3 2 2 6 23
6 1 6 18 6
-Ê ˆ Ê ˆ Ê ˆ Ê ˆ= = -Á ˜ Á ˜ Á ˜ Á ˜- - -Ë ¯ Ë ¯ Ë ¯ Ë ¯
3 2 3 7 3
6 1 1 19 1k
Ê ˆ Ê ˆ Ê ˆ Ê ˆ= πÁ ˜ Á ˜ Á ˜ Á ˜- - -Ë ¯ Ë ¯ Ë ¯ Ë ¯
3 2 1 3 13
6 1 3 9 3
-Ê ˆ Ê ˆ Ê ˆ Ê ˆ= = -Á ˜ Á ˜ Á ˜ Á ˜- - -Ë ¯ Ë ¯ Ë ¯ Ë ¯
3 2 2 10 25
6 1 2 10 2
Ê ˆ Ê ˆ Ê ˆ Ê ˆ= =Á ˜ Á ˜ Á ˜ Á ˜-Ë ¯ Ë ¯ Ë ¯ Ë ¯
Solution A145
(i) Unit perpendicular vector The scalar (dot) product of two perpendicular vectors is zero. So: ( ).( 2 3 ) 0+ + - + =a b ci j k i j and ( ).(10 ) 0+ + + =a b ci j k i k [1]
ie 2 3 0- + =a b and 10 0+ =a c So:
23=b a and 10= -c a [1]
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For a unit vector we also need 2 2 2 1+ + =a b c , which gives:
( )22 223 ( 10 ) 1+ + - =a a a
3
913fi =a [1]
and the required unit vector is:
3 2 30
913 913 913+ -i j k [1]
Comment If the question hadn’t directed you which method to use (and your knowledge of vectors is good) you could also have done this by working out the cross product ( 2 3 ) (10 )- + Ÿ +i j i k , then rescaled to get a unit vector. You will not need cross
products in the actuarial exams.
(ii) Other unit perpendicular vector The other unit vector perpendicular to these two vectors is the unit vector pointing in the opposite direction, which is just the negative of the one we’ve found ie:
3 2 30
913 913 913- - +i j k [1]
Solution A146
If we think of the matrix P as the matrix that transforms the column vector ( )Tx y z
into the column vector Ta b c , then we have the relationship:
0.2 0.2 0.2
0.2 0 0.2
0 0.4 0.2
a x
b y
c z
(*)
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Premultiplying both sides by 1P , we see that:
1
a x
b y
c z
P
Thinking in terms of simultaneous equations, Equation (*) corresponds to:
0.2 0.2 0.2
0.2 0.2
0.4 0.2
+ - =+ =+ =
x y z a
x z b
y z c
[1]
To solve these, we need to turn them round and express , ,x y z in terms of , ,a b c . We
can simplify these equations by multiplying through by 5:
5
5
2 5
+ - =+ =+ =
x y z a
x z b
y z c
[1]
We can use the last two equations to express x and y in terms of z :
5= -x b z 2.5 0.5= -y c z
From the first equation, we then get: (5 ) (2.5 0.5 ) 5- + - - =b z c z z a 2 2fi = - + +z a b c
and so: 5 ( 2 2 ) 2 3= - - + + = + -x b a b c a b c
2.5 0.5( 2 2 ) 2= - - + + = - +y c a b c a b c [1]
Writing these in the form of simultaneous equations:
2 3
2
2 2
+ - =- + =
- + + =
a b c x
a b c y
a b c z
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In matrix notation this is:
2 3 1
1 1 2
2 2 1
a x
b y
c z
[1]
So:
1
2 3 1
1 1 2
2 2 1
P [1]
Solution A147
Working out the various parts of this expression first, we find that:
2 1
det( ) (2)(4) (1)(1) 71 4
[1]
1
1 2 1 4 11
1 4 1 27
[2]
and:
2 1 1
( )0 1 1
x and ( ) 1 1T x
Putting these together, we get:
1 4 1 11 1 1( ) ( ) 1 1
1 2 12 2 7
311 1
114
4
14
2
7
T
x x
[1]
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and
2/7
2
1( ) 0.0452
(2 ) 7f ex
p-= =
¥ [1]
Comment Note that a 1 1¥ matrix is really just a scalar. So when we calculate
11( ) ( )
2-- - -Tx x we can treat the answer as a simple number.
Solution A148
A Solution A149
A Solution A150
B Solution A151
D Solution A152
C Solution A153
C
½£18,000 £7,500
6
2 ¥ =
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Summary Test – Questions Question T1
What is 0.040783 rounded to 2 significant figures? A 0.04 B 0.041 C 0.0408 D 0.04078 [1] FAC 2 1 Question T2
What is na a
xlog x log
y
Ê ˆ- Á ˜Ë ¯
A ( 1) a an log x log y- +
B ( )a x ynx a --
C na
xlog x
y
Ê ˆ-Á ˜Ë ¯
D 1n
ax
logy
-Ê ˆÁ ˜Ë ¯
[1]
FAC 4 1 Question T3
Solve 11 3 2 15x- £ - < A 7 6x- £ < B 7x ≥ and 6x < - C 7 6x≥ > - D 7x £ - and 6x > [1] FAC 4 4
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Question T4
Which of these is the graph of ny x= where n is an even number:
A
x
y
B
x
y
C
x
y
D
x
y
[1] FAC 3 1
FAC: Summary Test – Questions Page 3
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Question T5
Interest rates at the start of the year are set at 2.5%. At the end of the year they are 2.75%. What is the absolute change in the interest rate over the year? A 10%
B 250 basis points
C 9.09%
D 25 basis points [1]
FAC 5 2 Question T6
What is the result if (2 3 )i+ is multiplied by its complex conjugate?
A 5 12i- + B 5- C 13
D 5 12i- - [1] FAC 5 7 Question T7
Differentiate 5y x= with respect to x .
A 25x--
B ½52 x
C 3 252 x
D 52 x
[1]
FAC 6 3
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Question T8
Find ( )f x¢ where 2
3( ) xf x
e= .
A 2
3
2 xe
B 2
3
2 xe-
C 2
6xe
-
D 2
6xe
[1]
FAC 6 4 Question T9
What is 1
4
0
3 xe dxÚ ?
A 19.17 B 40.20 C 160.79 D 643.18 [1] FAC 7 2
FAC: Summary Test – Questions Page 5
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Question T10
What is the integral of 3
1 1
x x- ?
A 4
41 c
x- +
B 4
1ln
4x c
x- +
C 2
21 c
x- +
D 2
1ln
2x c
x+ + [1]
FAC 7 1 Question T11
If
1
2
2
Ê ˆÁ ˜= -Á ˜Ë ¯
a and
2
3
5
-Ê ˆÁ ˜=Á ˜Ë ¯
b then 2-b a is:
A 4.24 B 5.83 C 8.12 D 12.37 [1] FAC 8 1 Question T12
If 3 4
2 5
- -Ê ˆ= Á ˜Ë ¯
B then | |B is given by:
A 23- B 7- C 7 D 23 [1] FAC 8 2
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Question T13
The abbreviation etc means: A for example B compared with C that is to say D and so on [1] FAC Gloss Question T14
Simplify the expression(3 1)! (2 2)
(3 1)(2 )!
n n
n n
- G +G +
, where n is an integer.
A (2 2)(2 1)
(3 1)(3 )
n n
n n
+ ++
B (2 2)(2 1)
3
n n
n
+ +
C (2 1)
(3 1)(3 )
n
n n
++
D (2 1)
3
n
n
+ [2]
FAC 3 3 Question T15
Calculate 21 (5 4.712)
2.306 0.106310 2.82596
Ï ¸-Ô Ô+ ¥Ì ˝Ô ÔÓ ˛
.
A 0.088 B 0.270 C 0.338 D 0.741 [2] FAC 2 2
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Question T16
Solve the quadratic equation 27 4 0x x- - = giving your solutions to 1 DP. A 5.3 and 1.3x = - B 0.2 and 0.8x = - C 10.6 and 2.6x = - D The equation has no real solutions [2] FAC 4 2 Question T17
Use the result ( 1) ( 1)( 2)2 32! 3!(1 ) 1 p p p p ppx px x x- - -+ = + + + + to expand the
expression ( ) 1.21 x
-+ as far as the term in 3x and hence evaluate it when 0.4x = - .
A 0.641088 B 1.358912 C 1.781312 D 1.845944 [2] FAC 4 9 Question T18
Solve for x and y : 2 22 33x y+ =
3x y+ =-
A 21x = - and 18y =
B 4x = - and 1y =
C 5x = and 2y = - or 1x = - and 4y =
D 5x = - and 2y = or 1x = and 4y = - [2]
FAC 4 3
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Question T19
Calculate the sum of the first 10 terms in the sequence 5, 6, 7.2, 8.64, ... A 103.995 B 129.793 C 155.752 D 160.752 [2] FAC 4 7 Question T20
A population increases by 5% every year. Calculate the minimum whole number of years until the population has doubled in size. A 9 B 12 C 15 D 18 [2] FAC 5 1 Question T21
Given 3x i= + is a complex root of the cubic 3 24 2 20 0x x x- - + = , find the other two roots. A 3 or 2x i= - -
B ( ) ( )3 or 2x i i= - - +
C 3 or 3x i= - D 3 or 2x i= - - - [2]
FAC 5 7
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Question T22
Find dy
diwhere
41 (1 )iy
i
-- += .
A 4 5
2
(1 ) 4 (1 ) 1i i i
i
- -+ - + -
B 4 3
2
(1 ) 4 (1 ) 1i i i
i
- -+ - + -
C 4 5
2
(1 ) 4 (1 ) 1i i i
i
- -+ + + -
D 4 3
2
(1 ) 4 (1 ) 1i i i
i
- -+ + + - [2]
FAC 6 4 Question T23
Find the second derivative of ( 1)( )teM t em -= evaluated at 0t = , where m is a constant.
A (0)M m=¢¢
B 2(0)M m=¢¢
C 2(0)M m m= +¢¢
D 2(0)M emm=¢¢ [2]
FAC 6 5
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Question T24
What is ( ) ( )( )22 3
axy b xy c xyx y
∂ + +∂ ∂
?
A ( ) ( )22 3a b yx c yx+ +
B ( ) ( )24 9a b yx c yx+ +
C 2 3
2 3
yx yxaxy b c
Ê ˆ Ê ˆ+ +Á ˜ Á ˜Ë ¯ Ë ¯
D 2 3 4
2 3 4
yx yx yxa b cÊ ˆ Ê ˆ Ê ˆ+ +Á ˜ Á ˜ Á ˜Ë ¯ Ë ¯ Ë ¯
[2]
FAC 6 7 Question T25
What is 2 48 (3 2)x x dx+Ú ?
A 2 58 (3 2)
5
x xc
+ +
B 2 58(3 2)
5
xc
+ +
C 2 54(3 2)
15
xc
+ +
D 3 58 ( 2)
5
x xc
+ + [2]
FAC 7 3
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Question T26
What is 5 2
1 1
(3 4 )x y
x y dy dx= =
+Ú Ú ?
A 24
B 60
C 66
D 84 [2]
FAC 7 5 Question T27
Using the trapezium rule and 6 ordinates, what is the value of 1
3
0
xe dxÚ ?
A 4.330 B 6.552 C 13.103 D 65.516 [2]
FAC 7 6 Question T28
If 1 2
3 4
Ê ˆ= Á ˜-Ë ¯
A then TA A is given by:
A 5 5
5 20
Ê ˆÁ ˜-Ë ¯
B 2 16
9 2
-Ê ˆÁ ˜- -Ë ¯
C 1 0
0 1
Ê ˆÁ ˜Ë ¯
D 10 10
10 20
-Ê ˆÁ ˜-Ë ¯
[2]
FAC 8 2
Page 12 FAC: Summary Test – Questions
© IFE: 2013 Examinations The Actuarial Education Company
Question T29
The eigenvalues of 1 2
2 2
Ê ˆÁ ˜-Ë ¯
are:
A 2 and 3- B 1 and 6- C 4.37 and 1.37- D do not exist [2] FAC 8 2 Question T30
You are given that ( ) ( ) ( ) ( )4 3 2
6.6 2 2 2
11 1 1PV
ii i i
-= + + +++ + +
. By trial and error and
interpolation, calculate the value of i to 2 significant figures such that 0PV = .
Evaluate the formula (1 ) 1n
n
is
i
+ -= , using the value of i obtained, where 4n = .
A 4.246
ns =
B 4.297n
s =
C 4.310n
s =
D 4.375n
s = [5]
FAC 5 5 Question T31
Find and distinguish between the turning points of the function 3 5( ) 15f x x x= - .
A minima at 1.225x = ± and point of inflexion at 0x = B minimum at 3x = - , point of inflexion at 0x = and maximum at 3x = C maximum at 3x = - , point of inflexion at 0x = and minimum at 3x = D maxima at 1.225x = ± , point of inflexion at 0x = [5] FAC 6 6
FAC: Summary Test – Questions Page 13
The Actuarial Education Company © IFE: 2013 Examinations
Question T32
What is 2
2 0.5
1
2 xx e dx-Ú ?
A 8.437 B 4.218 C 2.109 D 0.875- [5] FAC 7 3
© IFE: 2013 Examinations The Actuarial Education Company
All study material produced by ActEd is copyright and issold for the exclusive use of the purchaser. The copyright
is owned by Institute and Faculty Education Limited, asubsidiary of the Institute and Faculty of Actuaries.
You may not hire out, lend, give out, sell, store or transmitelectronically or photocopy any part of the study material.
You must take care of your study material to ensure that itis not used or copied by anybody else.
Legal action will be taken if these terms are infringed. Inaddition, we may seek to take disciplinary action through
the profession or through your employer.
These conditions remain in force after you have finishedusing the course.
FAC: Summary Test – Solutions Page 1
The Actuarial Education Company © IFE: 2013 Examinations
Summary Test – Solutions Solution T1
B We start counting significant figure from the first non-zero digit, hence 0.041 is 2 significant figures. Solution T2
A Using the log rules we get:
( )log log ( 1)log logna a a a a a a
xlog x log nlog x x y n x y
y
Ê ˆ- = - - = - +Á ˜Ë ¯
Solution T3
C We have:
11 3 2 15
14 2 12
7 6
x
x
x
- £ - <- £ - <
≥ > -
Solution T4
D Even powers of numbers always give positive answers. Solution T5
D The absolute change is 0.25%. This is equivalent to 25 basis points.
Page 2 FAC: Summary Test – Solutions
© IFE: 2013 Examinations The Actuarial Education Company
Solution T6
C We have:
2(2 3 )(2 3 ) 4 9 4 9( 1) 13i i i+ - = - = - - = since 1i = - .
Solution T7
D
0.5 0.50.5
2.5 55 5 2.5
2
dyy x x x
dx xx-= = fi = = =
Solution T8
C
2 2 22 2
3 6( ) 3 ( ) 3( 2) 6x x x
x xf x e f x e ee e
- - -= = fi = - = - = -¢
Solution T9
B
( )1 1
4 4 4 0
00
3 33 40.20
4 4x xe dx e e e
È ˘= = - =Í ˙Î ˚Ú
Solution T10
D
21 3
3 2
1 1 1ln ln
2 2
xdx x x dx x c x c
x x x
-- -- = - = - + = + +
-Ú Ú
FAC: Summary Test – Solutions Page 3
The Actuarial Education Company © IFE: 2013 Examinations
Solution T11
C
2 2 2
2 1 4
2 3 2 2 7 2 ( 4) 7 1 66 8.12
5 2 1
- -Ê ˆ Ê ˆ Ê ˆÁ ˜ Á ˜ Á ˜- = - - = fi - = - + + = =Á ˜ Á ˜ Á ˜Ë ¯ Ë ¯ Ë ¯
b a b a
Solution T12
B
3 4( 3)(5) ( 4)(2) 15 8 7
2 5
- -Ê ˆ= = - - - = - + = -Á ˜Ë ¯
B
Solution T13
D etc is short for the Latin phrase et cetera which means “and so on”. Solution T14
D (3 1)! (2 2) (3 1)!(2 1)! (2 1)
(3 1)(2 )! (3 )!(2 )! 3
n n n n n
n n n n n
- G + - + += =G +
Solution T15
B
21 (5 4.712)2.306 0.1063 2.306 0.12935 0.1063
10 2.82596
2.306 0.01375
0.270
Ï ¸-Ô Ô+ ¥ = ¥Ì ˝Ô ÔÓ ˛
==
Page 4 FAC: Summary Test – Solutions
© IFE: 2013 Examinations The Actuarial Education Company
Solution T16
A
For 27 4 0x x- - = we have 1a = - , 4b = - and 7c = . So the solutions are:
24 ( 4) 4( 1)(7) 4 44
5.317,1.3172( 1) 2
x± - - - ±= = = -
- -
Solution T17
C
( 1.2)( 2.2) ( 1.2)( 2.2)( 3.2)1.2 2 32! 3!(1 ) 1 ( 1.2)x x x x- - - - --+ + - + + +
When 0.4x = - we get:
( 1.2)( 2.2) ( 1.2)( 2.2)( 3.2)1.2 2 32! 3!(1 ( 0.4)) 1 ( 1.2)( 0.4) ( 0.4) ( 0.4)
1 0.48 0.2112 0.090112
1.781312
- - - - --+ - + - - + - + -
= + + +=
Solution T18
D Since 3x y+ = - we have 3x y= - - . Substituting this into the first equation gives:
2 2
2 2
2
2
( 3 ) 2 33
9 6 2 33
3 6 24 0
2 8 0
( 2)( 4) 0
4,2
y y
y y y
y y
y y
y y
y
- - + =
+ + + =
+ - =
+ - =- + =
= -
When 4y = - then 3 ( 4) 1x = - - - = and when 2y = then 3 2 5x = - - = - .
FAC: Summary Test – Solutions Page 5
The Actuarial Education Company © IFE: 2013 Examinations
Solution T19
B This is a geometric series with 5a = and 1.2r = so the sum of the first 10 terms is:
10
105(1 1.2 )
129.7931 1.2
S-= =-
Solution T20
C If the starting population is p then we require the n such that:
1.05 2
1.05 2
ln1.05 ln 2
14.207
n
n
p p
n
n
¥ =
===
So 15 years are needed. Solution T21
A If 3x i= + is a root then its complex conjugate 3x i= - is also a root. This narrows the answer down to either answer A or answer C.
2( (3 ))( (3 )) 6 10x i x i x x- + - - = - +
We can see that for answer A:
2 3 2( 2)( 6 10) 4 2 20x x x x x x+ - + = - - +
Page 6 FAC: Summary Test – Solutions
© IFE: 2013 Examinations The Actuarial Education Company
Solution T22
C
Using the quotient rule with 41 (1 )u i -= - + and v i= we get:
5 4 4 5
2 2
4(1 ) (1 (1 ) ) 1 (1 ) 4 (1 ) 1dy i i i i i i
di i i
- - - -+ ¥ - - + ¥ + + + -= =
Solution T23
C
0
( 1) ( 1)
0 0
0
2
( ) ( )
( ) (1 )
(0) (1 )
(1 )
t t t
t
e t e t e
t t e
e
M t e M t e e e
M t e e
M e e
e
m m m m
m m
m m
m m
m m
m m
m m
m m
- - + -
+ -
+ -
= fi = =¢
fi = +¢¢
fi = +¢¢
= +
= +
Solution T24
B
( )( )
2 2 3 3 2 2 3
22 2 3 3 2 2
2 3
4 9
axy bx y cx y ay bxy cx yx
axy bx y cx y a bxy cx yx y
∂ + + = + +∂
∂fi + + = + +∂ ∂
Solution T25
C
Using the substitution 23 2u x= + we get 66 " "du dudx x xdx= fi = . Hence our
integral becomes:
4 4 5 2 54 4 46 3 15 158 (3 2)duu u du u c x c= = + = + +Ú Ú
FAC: Summary Test – Solutions Page 7
The Actuarial Education Company © IFE: 2013 Examinations
Solution T26
B
5 2 5 5 522
11 1 1 1 1
52
1
(3 4 ) 3 2 (6 8) (3 2) 3 6
1.5 6 67.5 7.5 60
yx y x x x
x
x y dy dx xy y dx x x dx x dx
x x
== = = = =
=
È ˘+ = + = + - + = +Î ˚
È ˘= + = - =Î ˚
Ú Ú Ú Ú Ú
Solution T27
B Using the ordinates 0, 0.2, 0.4, 0.6, 0.8 and 1 we get:
1
3 0 0.6 1.2 1.8 2.4 312
0
0.2 2 2 2 2 6.552xe dx e e e e e eÈ ˘¥ ¥ + + + + + =Î ˚Ú
Solution T28
D
1 2 1 3
3 4 2 4T -Ê ˆ Ê ˆ
= fi =Á ˜ Á ˜-Ë ¯ Ë ¯A A
Hence:
1 3 1 2 1 9 2 12 10 10
2 4 3 4 2 12 4 16 10 20T - + - -Ê ˆ Ê ˆ Ê ˆ Ê ˆ
= = =Á ˜ Á ˜ Á ˜ Á ˜- - + -Ë ¯ Ë ¯ Ë ¯ Ë ¯A A
Page 8 FAC: Summary Test – Solutions
© IFE: 2013 Examinations The Actuarial Education Company
Solution T29
A We require the values of l such that:
2
1 2 1 0 1 2det 0 det 0
2 2 0 1 2 2
(1 )( 2 ) 4 0
6 0
( 2)( 3) 0
3,2
ll
ll l
l ll l
l
Ê ˆ -Ê ˆ Ê ˆ Ê ˆ- = fi =Á ˜ Á ˜ Á ˜Á ˜- - -Ë ¯ Ë ¯ Ë ¯Ë ¯
fi - - - - =
+ - =- + =
= -
Solution T30
B Solving using trial and error gives 0.048i = . Hence:
4
4
1.048 14.297
0.048s
-= =
Solution T31
B
3 5 2 4( ) 15 ( ) 45 5f x x x f x x x= - fi = -¢
So we have turning points when:
2 4 2 2( ) 45 5 0 5 (9 ) 0 0, 3,3f x x x x x x= - = fi - = fi = -¢
Looking at the second derivative:
3( ) 90 20f x x x= -¢¢
When 3x = - we have ( 3) 270 0 minf - = > fi¢¢ .
When 3x = we have (3) 270 0 maxf = - < fi¢¢ .
When 0x = we have (0) 0f =¢¢ so we can’t tell. Looking at the gradient either side of
0x = we see that ( 0.5) 0f - >¢ and (0.5) 0f >¢ so we have a point of inflexion.
FAC: Summary Test – Solutions Page 9
The Actuarial Education Company © IFE: 2013 Examinations
Solution T32
C
Using integration by parts with 22u x= and 0.5xdvdx e-= , we see that 4du
dx x= and
0.52 xv e-= - . Hence:
2 222 0.5 2 0.5 0.5
11 1
21 0.5 0.5
1
2 4 8
16 4 8
x x x
x
x e dx x e xe dx
e e xe dx
- - -
- - -
È ˘= - +Î ˚
= - + +
Ú Ú
Ú
Using integration by parts with 8u x= and 0.5xdvdx e-= , we see that 8du
dx = and
0.52 xv e-= - . Hence:
2 220.5 0.5 0.5
11 1
21 0.5 0.5
1
1 0.5 1 0.5
1 0.5
8 16 16
32 16 32
32 16 32 32
64 48
x x x
x
xe dx xe e dx
e e e
e e e e
e e
- - -
- - -
- - - -
- -
È ˘= - +Î ˚
È ˘= - + + -Î ˚
= - + - +
= - +
Ú Ú
So our final answer is:
1 0.5 1 0.5 1 0.516 4 64 48 80 52 2.109e e e e e e- - - - - -- + - + = - + =
© IFE: 2013 Examinations The Actuarial Education Company
All study material produced by ActEd is copyright and issold for the exclusive use of the purchaser. The copyright
is owned by Institute and Faculty Education Limited, asubsidiary of the Institute and Faculty of Actuaries.
You may not hire out, lend, give out, sell, store or transmitelectronically or photocopy any part of the study material.
You must take care of your study material to ensure that itis not used or copied by anybody else.
Legal action will be taken if these terms are infringed. Inaddition, we may seek to take disciplinary action through
the profession or through your employer.
These conditions remain in force after you have finishedusing the course.