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Maths Extension 1

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Maths Extension 1

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succeeding in the exams, 2017

important notes

Our policy at TSFX is to provide students with the most detailed and comprehensive set of notes that will maximise student performance and reduce study time. These materials, therefore, include a wide range of questions and applications, all of which cannot be addressed within the available lecture time i.e. Due to time constraints; it is possible that some of the materials included in this booklet will not be addressed during the course of these lectures. Where applicable, fully worked solutions to the questions in this booklet will be handed to students on the last day of each subject lecture. Although great care is taken to ensure that these materials are mistake free, an error may appear from time to time. If you believe that there is an error in these notes or solutions, please let us know asap ([email protected]). Errors, as well as clarifications and important updates, will be posted at www.tsfx.com.au/hsc-updates. The views and opinions expressed in this booklet and corresponding lecture are those of the authors/lecturers and do not necessarily reflect the official policy or position of TSFX.

copyright notice

author(s)

These materials represent the collective effort of many teachers across the state. The principal authors of this booklet are:

Mr Geoffrey McLauchlan, B.Sc. (Hons)., B. Comm. Dip. Ed. (Maths Extension 2 Teacher – Queenwood School)

lecturer(s)

To ensure that students are afforded every possible advantage in their examinations, our lectures are prepared and delivered by qualified, currently practising HSC teachers and official HSC exam markers who possess the knowledge and experience to demonstrate the means by which students can achieve the higher ATAR scores. Further details regarding our teachers (including qualifications and experience) may be obtained at http://www.tsfx.com.au/what-is-tsfx/ourteachers/.

TSFX - voted number one for excellence and quality in HSC programs.

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STUDYING MATHEMATICS FOR THE HSC Welcome to TSFX. Our aim is to provide high quality additional instruction for your HSC exam as a supplement to your own school program. Our lectures are firmly based on the HSC syllabus and are only given by practicing HSC teachers and markers. We incorporate past HSC questions and stress examiner comments and suggestions.

SUCCESS IN MATHEMATICS Prelude:

We perform at our peak in sport, business, relationships and in educational matters if we are confident. Approach your studies in a confident fashion and prepare yourself so you will have good reason to be confident.

MATHEMATICAL PRACTICE I asked at three schools that I associate with, and at a university, what they considered to be the most important ingredients of successful study of Mathematics. The funny thing was that everyone said the same things: • Draw a diagram! This was the top of everyone’s list and if I may quote Professor

Michael Herschorn of UNSW who said: “It has been my experience that the better the Mathematician, the more often they draw a diagram” Professor Fred Chong of Sydney University is famous for his quote: “The only thing better than a big diagram is a bigger diagram”

• Don’t miss your Mathematics lesson – there is something to learn in every lesson. Maths should be done every day.

• Stay up to date – if you ever miss a lesson for a very good reason, and you have fallen behind, then do the work of the current lesson and catch up at home. Don’t run the risk of being behind for a week.

• Always put pen to paper. You cannot read Mathematics.

• Do plenty of exercises but be aware that some questions should be left out if they are

too easy while sometimes questions at the end of an exercise may be too hard for you to waste your precious time.

• Never be afraid to ask questions in class.

• Be prepared to help a friend – the best way of learning is to actually teach it.

• Study at a clean desk with a fresh sheet of paper to write on.

• Set your work out neatly and if you can’t see an error, start the question over.

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WORKING MATHEMATICALLY

1. If there is a formula, write it down.

2. Don’t skip lines.

3. Be careful of the minus sign – is there a way to get rid of that sign?

4. Try and avoid fractions wherever possible.

5. Look to factorise.

6. Always try and simplify.

7. Don’t use the calculator until you need to – can you tidy up the question a little bit before using that calculator?

STUDYING MATHEMATICS 1. One thing at a time – master each step as you go.

2. A summary should indeed be just that – don’t try and rewrite the textbook.

3. A summary should be done at the topic test time or at the end of each topic. It should

not be done the night before a major exam.

4. Allocate your study time – if the test is on 8 exercises; spend about 1/8 of the time on each exercise.

5. Use your textbook wisely – read the example, cover the solution and only look at it

when you are stuck.

6. Work on past exam questions regularly. Too many students leave this to the end because of the “I’m not quite ready yet” excuse.

7. Same for past papers – Do the entire paper then look at the answers. Don’t fall into the

lazy habit of turning quickly to the solution when stuck. Make your mind work a touch harder.

8. Though some rote learning is useful, it is so much better to understand where

something comes from.

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THE MAJOR EXAM

1. Have an early night before the exam.

2. Have a decent breakfast and don’t dress down – wear the type of clothes you usually do your work in at school.

3. Always arrive early – don’t time your travel to get there 5 minutes before the exam.

4. Only bring your topic summaries with you – you won’t have time to read through a

massive folder.

5. Don’t forget your ruler and compasses.

6. Use the reading time wisely. Make sure you see where the last question is. Check that last hard question – so often there are 1 or 2 easy marks there. If you spot a harder question, read it carefully – it is amazing how often your subconscious mind will work on that question.

7. Allocate your time. The multiple choice questions should be complete in 17 or 18

minutes. I have suggested to my students that they do this first but watch your time and once 20 minutes is reached, come back to them later.

8. Part marks – a one mark question requires one piece of information, a two mark

requires two pieces and so on. Don’t write too much or too little. 9. Follow on parts – be aware that if an HSC questions has three parts, they are likely to

follow on and the answer from an earlier part will most likely be used again.

10. Mark off each question or part as you do it – this will stop you leaving something out.

11. If stuck, read the question but go on. If you are lucky the method will come to you later; but be careful; if it jumps into your mind, it may just jump out again, so write something down.

12. Never panic if an exam is hard – it will be hard for everyone else.

13. The last question! If you have never got the last question out, don’t spend too much

time on it – better you go back and get one of the easier solutions.

14. Checking – wherever possible do the question another way or perhaps do it backwards. Don’t change things in the last few seconds – you will panic and may cross out the correct answer. Don’t cross out an answer until it is replaced (although markers are a bit generous here).

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TIPS, TRICKS, COMMON ERRORS AND KEEPING THE MARKERS HAPPY: TOPIC BY TOPIC

ARITHMETIC

(a) 33 2× is not equal to 36 .

(b) Be careful with the calculator display – do not write the entire display but do keep

enough digits to ensure sufficient accuracy of an answer. An answer correct to 2 decimal places generally requires you to use 3 places on the calculator.

(c) If asked for an exact answer, you don’t use the calculator.

ALGEBRA

(a) If 2 2x x= , you cannot cancel the x ’s.

(b) Simplify both sides of an equation first:

If 3 7 2 4 11 2 3x x x x− + = + + − , the next line should be 5 7 6 8x x− = + . (c) When solving equations get rid of fractions early:

If we have 2 23

4 3x x−+ = + then multiply EACH TERM by 12.

(d) Remember what perfect squares look like: Both 2 6 9x x− + and 2 26 9x xy y− + are

perfect squares.

(e) There are very few times when you should not factorise: How does 2 6 9x y xy y− + simplify?

(f) Inequalities rarely go the same way. If 2 4x < , do not write 2x < ± .

(g) *When solving 2 5

3x<

−, do not expand the brackets.

GEOMETRY (a) If using similar triangles, write the vertices in corresponding order. Then you can easily

see the sides which are in the same ratio.

(b) Give reasons but make sure they are abbreviated.

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30°

38m

30°

x m

CURVE SKETCHING (a) Make sure it is neat. Use a ruler and if a circle, use compasses.

(b) If a factor is squared, there is a turning point on the X axis, if cubed there is a horizontal

point of inflexion, and make sure you show it correctly. 3y x= − has a horizontal point of inflexion.

TRIGONOMETRY

(a) Set up a question before you use a calculator.

If 3sin 75x

= , you can go a little further before picking up the calculator.

(b) Angle of depression is often OUTSIDE the figure.

38tan 30

38tan 30

65.8

o

o

x

x

=

∴ =

(c) To use the Sine rule, you must know an angle and the opposite side.

(d) The cosine rule has a COS in it: 2 2 2 2a b c bcCosA= + − .

CO-ORDINATE GEOMETRY

(a) Draw the diagram so you can see which lines look equal, sign of the gradient etc.

(b) If a vertical or horizontal distance is required, don’t use a formula – similar for the equation for a vertical or horizontal line.

e.g. The distance between (3,4) and (3,-2) is simply 4 - - 2 = 6

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(c) Area of triangle questions can often be seen by looking from the side. If (0,2) is added to the diagram, what is the area of the triangle?

DIFFERENTIAL CALCULUS (a) Only differentiate by 1st principles if asked.

(b) Watch where an algebraic fraction can be split into two parts: 2

3 x xx+ =

(c) Remember the product rule if there are two functions. (d) Don’t write a quotient rule question as a product rule – that is, do not change

( )22 3

3

2 1 to (2 1)

xx x

x−+

+

(e) If asked to find a tangent, remember to substitute for the gradient. The following is often

seen:

Find the equation of the tangent to 2y x= at (2,4) Answer: 4 2 ( 2)y x x− = −

PROBABILITY

(a) Draw a tree diagram then multiply the line of branches and only add the results after the end branches.

(b) ‘At least’ questions are usually easiest when one (or two) result(s) are subtracted from 1.

EXAMPLE

A die has one red face and five black faces. The die is rolled twice, find the probability of obtaining:

(a) 2 red faces (b) 1 red and one black face (c) At least one red face

R

R

R

B

B

B

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SEQUENCES AND SERIES (a) Note the alternate formula for sum of an AP if the first and last term is known:

{ }2nnS a l= + .

(b) You are allowed to use common sense in these questions – it may be easier to write

out all the terms of something like:

Find the 6th term of 192,96,48...

(c) Be careful of the ‘-‘ sign in something like: 411

3 − −

QUADRATICS (a) Remember that the sum and product of roots can be used in many questions. If one

x value is known, then you can use the sum of the roots to find the other x value e.g. Solve 24 3 27 0x x− − = , given 3x = is one solution.

(b) If asked to find , ,A B C in 2 23 2 4 ( 2) ( 2)x x A x B x C+ + = − + − + , don’t expand the

brackets. (c) If the sum of the coefficients is 0, then ( 1)x − is a factor, e.g. factorise 24 45 49x x+ − .

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SAMPLE MATHEMATICS EXTENSION 1 NOTES

PERMUTATIONS AND COMBINATIONS

PERMUTATIONS A permutation is the total number of an ordered selection. This would include questions such as, find the number of ways: (a) 6 people to sit in 6 chairs in a row (b) 4 people to sit in 6 chairs (c) 3 people to sit in 6 chairs Where there is no restriction in the order of selection the symbol n

rP can be used where n is

total number and r is the selected number. ( )

!!

nr

nPn r

=−

Use the nrP button on your calculator to find:

(a) The number of ways 5 people can sit in 15 seats. (b) The number of ways of choosing 1st, 2nd and 3rd in a 10 horse race. Solution 360360, 720 Usually there is a restriction on the Permutation and n

rP will be of little use. Set out by listing a row of boxes and consider the number of ways to fit that condition. EXAMPLE 1 Find the number of ways 5 people stand in a row if one person A wants to stand at an end. (a) Draw the table:

(b) Fill in one of the two end boxes.

2

(c) Now fill in the other boxes with the remaining people.

2 4 3 2 1

Solution

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EXAMPLE 2 Find the number of ways 5 people stand in a row if two people A and B want to be next to each other. (a) Draw a row of 5 boxes.

(b) If two are next to each other, mark off 2 spaces.

(c) There are now 4 spaces so put in the numbers 4 down to 1.

(d) And there are 2 ways of filling the double space.

EXAMPLE 3 This time 3 of those same 5 people want to be together.

(a) Put a line through three of the spaces.

(b) Consider that there are 3 spaces left and then consider the number of ways of filling the

triple space.

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EXAMPLE 4 Five people want to sit in an old fashioned railway carriage. The carriage has 8 seats, 4 facing each way. Find the number of seating permutations if: (a) No restriction where people sit. (b) Two people want to face the engine. (c) One person wants a window seat. (d) Of the two people that want to face the engine, one also

wants the window seat. Solution

EXERCISE 1 QUESTION 1 How many ways can 7 seats be occupied by: (a) 7 people (b) 6 people (c) 5 people

Solution

QUESTION 2 If there are 12 horses in a race, in how many ways can the first 3 places be filled? Solution

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QUESTION 3 A child has 5 different coloured blocks and wants to build a tower of 2 or 3 or 4 or 5 blocks. (a) How many different 3 block towers can be made? (b) How many different (2, 3, 4 or 5) block towers can be made? Solution QUESTION 4 How many ways can 7 people stand in a line if: (a) 2 people stand next to each other (b) 3 people stand next to each other Solution

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QUESTION 5 How many different 5 digit numbers can be made from 1, 2, 3, 4, 5 if the numbers are not repeated and: (a) No other restriction (b) The number is odd Solution QUESTION 6 How many different numbers can be made from 1, 2, 3, 4, 5 if the numbers are not repeated and is less than 3000? Solution

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QUESTION 7 William, Kate and 6 other people need to go through a door. In how many ways can this be done if: (a) Kate is immediately followed by William. (b) Kate goes through the door before William. Solution QUESTION 8 How many ways can Amy, Emma, Katherine, Lauren, Isabel and Rebecca occupy 6 seats if: (a) They sit in alphabetical order (b) No restrictions (c) A & L sit at the ends (d) E & K sit next to each other (e) L, I, R sit next to each other (f) K and R don’t sit next to each other (g) There are at least 2 seats between E and K (h) L and I sit next to each other and A has an end seat Solution

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PERMUTATIONS WITH CONDITIONS

REPEATED NUMBERS OR LETTERS There are 5! or 120 ways of placing ABCDE in order. If all of the B’s are replaced by an A, then ABCDE and BACDE will become AACDE, thus halving the permutation. If the C’s are also replaced by an A, then the permutation is reduced by a factor of 3!

EXAMPLE 5 Find the number of permutations of: (a) AAABCD (b) AABBCD (c) AABBBB (d) WOOLLOOMOOLOO Solution

SEATED AT A TABLE Here we just consider the order of seating. If Alicia, Emma and Katherine sit at a table, then Alicia has either Emma on her left or her right with one place left for Katherine. Hence there are just two permutations. Now Rebecca turns up and they have to move to a larger table. We let Alicia take a seat, Emma has the choice of 3 seats, Katherine 2 seats, and Rebecca sits at the last place – 3 x 2 x 1 = 6 combinations.

Rule: If there are n people seated around a table, there are ( )1 !n − seating combinations.

EXAMPLE 6 Four men and four women sit alternatively around a table. How many permutations are possible if: (a) No other restriction. (b) One man and one woman are not to be placed together.

Solution

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EXERCISE 2 QUESTION 9 How many letter combinations are there of: (a) ALGEBRA (b) EERIE (c) AABBBB (d) SYDNEY (e) WOOLLONGONG (f) USEELANANTHAN Solution QUESTION 10 How many ways can 4 white, 3 red and 1 blue counters be arranged in a line? Solution

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QUESTION 11 How many different 7 digit numbers can be made from: (a) 2,2,2,2,3,3,3 (b) 2,2,3,3,4,4,4 (c) 2,2,3,3,4,4,5 Solution QUESTION 12 There are 5 garage doors in a row that need painting. Two are to be painted red, one white, one green and the other blue. How many arrangements are possible if: (a) No other restriction? (b) The red doors are next to each other? Solution

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QUESTION 13 In a rowing 8, how many rowing combinations are possible if all 8 can row on the right side but: (a) Only 4 can row on the left side (b) 5 can row on the left side Solution QUESTION 14 How many ways can 3 men and 3 ladies be placed around a table if: (a) No restrictions. (b) Seated alternatively male and female. (c) Seated alternatively and one man and one woman don’t sit together. (d) Seated alternatively and one man and one woman do sit together. Solution

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QUESTION 15 How many ways can 6 people sit around a table if three people sit together? Solution QUESTION 16 How many ways can a white, red, yellow and blue bead be arranged in a circle? Solution

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QUESTION 17 Given that a bracelet can be turned over, in how many ways can a white, red, yellow and blue bead be arranged in a bracelet? Solution QUESTION 18 If 6 120rP = , find the value of r . Solution QUESTION 19 How many arrangements of the letter of the word ISOSCELES can be made if: (a) The S’s do not occupy the first or last place. (b) The S’s are together. Solution

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QUESTION 20 Five travellers arrive in a town with five hotels. (a) How many different accommodation arrangements can be made if there are no

restrictions on where the travellers stay? (b) How many different accommodation arrangements can be made if each traveller stays

in a different hotel? (c) Suppose one pair is a husband and wife and they go to the same hotel while the others

can go to any of the remaining hotels. How many arrangements now? Solution QUESTION 21 Three identical blue and four identical yellow marbles are arranged in a row. (a) How many different arrangements? (b) How many different arrangements of just 5 of these marbles? Solution

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COMBINATIONS A combination is the number of ways of choosing an unordered selection. EXAMPLE 7 Choose two people out of 5. There are 20 (5 x 4) ways of choosing two people in order (say ABCDE). Here AB is the same as BA. So there are 10 possible choices. The calculator key n

rC where 5n = and 2r = also works this out – though the C stands for combination, we often say “5 choose 2”. EXAMPLE 8 How many ways can 4 people be chosen from a group of 8: Solution 8

4 70C =

Note that !

( )! !n

rnC

n r r=

EXAMPLE 9 How many ways can 3 Year 11 and 4 Year 12 students be chosen from a group of 6 Year 11 and 10 Year 12? Solution

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EXERCISE 3 QUESTION 22 In how many ways can I choose 5 objects from a set of 8? Solution QUESTION 23 How many ways can 3 girls and 2 boys be chosen from 6 girls and 6 boys? Solution QUESTION 24 There are 14 people available to play in a soccer team of which two people only like to play goal keeper. How many ways can this team of 11 be chosen? Solution

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QUESTION 25 A premier has to choose a cabinet of 10 people from a list of 15. In how many ways can this be done if both the premier and deputy are automatic choices? Solution QUESTION 26 There are 7 batsmen, 2 wicket keepers and 6 bowlers from which a cricket team of 6 batsmen, 1 wicket keeper and 4 bowlers is to be selected. In how many ways can this be done?

Solution

QUESTION 27 In how many ways can a committee of 5 be chosen from 4 ALP and 6 Liberals if it is to have: (a) 2 ALP and 3 Liberals (b) Majority Liberal Solution

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QUESTION 28 In how many ways can a committee of 6 be chosen from 10 available members if two people refuse to serve if the other is chosen?

Solution

QUESTION 29 In a bowl there are 5 bananas, 3 apples and 2 pears. How many ways are there of choosing: (a) 2 bananas, 1 apple and 1 pear (b) 4 fruit, but at least one of each (c) 5 fruit, at least one of each Solution

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QUESTION 30 How many different poker hands are possible (5 cards out of 52)? Solution QUESTION 31 How many poker hands have: (a) 3 aces and 2 kings (b) 4 aces (c) 4 of a kind (d) 5 of the one suit Solution

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QUESTION 32 I have 6 maths books, 5 science books and 4 history books in a box and have space for 7 books in a bookshelf. How many ways can I select 7 books if: (a) No other restriction (b) There are 3 maths, 2 science and 2 history (c) There are majority maths books Solution QUESTION 33 In how many ways can a jury of 12 be chosen from 10 men and 7 women so there are at least 6 men and not more than 4 women? Solution

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QUESTION 34 In how many ways can n things be shared between two people? Solution QUESTION 35 In how many ways can a committee of 3 men and 4 boys be chosen from 7 men and 6 boys so as to not include the youngest boy if the eldest man is selected? Solution

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QUESTION 36 How many: (a) Selections (b) Arrangements can be made if selecting 3 consonants and 2 vowels from 8 consonants and 4 vowels? Solution QUESTION 37 Containers are coded by different arrangements of dots in a row. The colours used are red, white and blue. At most, three dots are red, two are white and one is blue. (a) If 6 dots are used, find the number of different codes. (b) If 5 dots are used, find the number of different codes. Solution

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QUESTION 38 If 8 6

n nC C= , find the value of n. Solution QUESTION 39 How many ways can 5 people be placed in 3 rooms so no room is left empty.

Solution

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QUESTION 40 John and Robin engage in an archery competition with the first to hit the target 5 times wins. One possible result would be recorded as R J R R R R meaning Robin hits the target, then John and then Robin 4 times in a row. Find: (a) The number of ways Robin can win 5 targets to 1. (b) The total number of different ways in which this competition can be recorded.

Solution

QUESTION 41 In how many ways can a set of 3 or more be selected from 8 people?

Solution

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*QUESTION 42 In how many ways can: (a) 4 boys be divided into 2 sets of 2? (b) 6 boys be divided into 3 sets of 2? (c) 8 boys be divided into 4 sets of 2? Solution

PERM AND COMB PROBABILITY Probability questions on Permutations and Combinations follow with using the standard

probability formula ( )( )( )

n EP En S

= .

EXAMPLE 10 If 6 students out of the 20 in a Maths class are chosen at random, what is the probability that: (a) Alan and Francis are picked. (b) Just one of the 7 chess players is chosen? Solution (a) ( )n E means A and F are included so 4 more are chosen from the remaining 18.

184( ) 3060n E C∴ = = 20

6( ) 38760n S C= = 3060 3( , )38760 38

P A F∴ = =

(b) Now 1 of the 7 chess players are picked along with 5 of the remaining 13.

7 131 5( ) . 7 x 1287 9009n E C C= = =

9009 3003(1/ 2)38760 12920

P E∴ = =

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EXAMPLE 11 If the letters AABCCC are picked at random, find the probability that: (a) The A’s are together. (b) The A’s are not together. (c) The C’s are together. Solution

(a) Join two spaces – this leaves 5 spaces. (b) (c)

( ' together)P C s∴ =

EXAMPLE 12 An urn contains 3 red, 3 blue and 3 white balls. If three balls are drawn at random, find the probability of drawing: (a) One of each colour (b) Three of the same colour (c) Two of one colour and one of another colour Solution (a) We need to choose 1 ball of each of the 3 colours (b) We can choose any of 3 Red, 3 Blue of 3 White

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(c) This choice completes all the possibilities. EXAMPLE 13 An urn contains 6 blue and 6 red balls. If 6 balls are drawn at random (without replacement) find, correct to three decimals places: (a) Probability of 3 blue and 3 red balls (b) Choosing more red balls than blue balls Solution (a) 6 6

3 3( )n E C C= × , 126( )n S C=

(b) There is an equal chance of more red and less red. (majorityRed)P∴ ≈

EXERCISE 4 QUESTION 43 If there are 6 runners of equal ability in a race, what is the probability of guessing: (a) The first two places in order. (b) The first two places but not in order. Solution

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QUESTION 44 Two girls and 4 boys sit on 6 seats in a row. What is the (theoretical) probability that the 4 boys will sit together?

Solution

QUESTION 45 An urn contains 3 red, 4 blue and 5 yellow balls and three balls are chosen at random. Find the probability of picking: (a) Three yellow balls (b) Three balls of the same colour (c) Three balls of different colours (d) Two balls of one colour and 1 ball of another colour Solution

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QUESTION 46 A four person team is to be selected at random from 9 girls and 7 boys. What is the probability that the team will consist only of girls? Solution QUESTION 47 A committee of 5 is to be chosen (at random) from 5 boys and 6 girls. Find the probability that the committee contains: (a) 3 boys and 2 girls (b) Majority boys Solution

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QUESTION 48 Find the probability that a poker hand contains: (a) 4 of a kind (b) A flush (5 cards of the one suit) Solution QUESTION 49 A bridge hand has 13 cards dealt from a pack of 52. Find the probability that a bridge hand is dealt and contains no 10s, jacks, queens, kings or aces. Solution

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QUESTION 50 Amy, Brian, Colin, Don and Emma sit in 5 adjoining seats randomly. Find the probability that: (a) Amy is sitting next to Don. (b) Amy is not sitting next to Don. Solution QUESTION 51 If 6 people are seated at a table, what is the probability that two particular people are seated next to each other? Solution

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QUESTION 52 A total of 5 players are selected from four sporting teams where each of the teams consists of 6 players numbered 1 to 6. (a) What is the probability of three being number 4 and two numbered 6? (b) What is the probability that four players are from the one team? Solution QUESTION 53 4 white, 3 red and 1 blue counter are arranged in a line. Find the probability that: (a) The 4 white counters are together. (b) The 4 white counters are in alternate positions. (c) A red counter is at each end. (d) The order from left to right is 4 white, 3 red and then 1 blue. Solution

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QUESTION 54 The letters of the word FORTY are selected at random. Find the probability that: (a) The F is chosen before the T. (b) The letters are chosen in alphabetical order. Solution QUESTION 55 The letters of the word FIFTY are selected at random. Find the probability that the T is chosen somewhere in between the two F’s. Solution

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QUESTION 56 Eight people are divided into two groups. Find the probability that there will be four people in each group. Solution QUESTION 57 The letters of the word INDEPENDENCE are arranged in a row. Find the probability of the E’s being together. Solution QUESTION 58 An urn contains 5 red and 4 white cubes. If 3 cubes are drawn out, find the probability of two of the cubes being red and one cube being white. Solution

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QUESTION 59 A box contains 10 tennis balls, 4 new and 6 used. For the first match, two balls are selected at random from the box and returned, then another two balls are selected for the second match. (a) Find the probability that exactly one of the balls for the first match is new. (b) Find the probability that both balls for the first match are new while both balls for the

second match are not new. Solution *QUESTION 60 Urn A contains 6 white and 4 black balls while Urn B contains 2 white and 2 black ones. Two balls are selected at random from Urn A and placed into Urn B. Two further balls are now selected from Urn B. What is the probability that both of the finally selected balls will be white? Solution

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*QUESTION 61 An urn contains n red, n blue and n white balls. If three balls are drawn at random, find: (a) The probability of drawing:

i. One of each colour ii. Three of the same colour iii. Two of one colour and one of another colour

(b) The limiting probability of each case as n → ∞ . Solution

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SAMPLE 2 UNIT MATHS NOTES

QUADRATICS

SKETCHING AND SOLVING QUADRATICS Easiest to sketch a factorised quadratic. EXAMPLE 1 Sketch: (a) ( 1)( 3)y x x= − +

(b) 2( 2)y x= −

(c) (4 )y x x= − Solution

FACTORISING QUADRATICS EXAMPLE 2 Factorise: (a) 2 5 14x x− − Product: Sum: Factors: (b) 23 5 12x x− − Product: Sum: Factors:

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SOLVING QUADRATICS Write in the form 2ax bx c+ + =0 and attempt to factorise before using quadratic formula. EXAMPLE 3 Solve: (a) 2 8 20x x= +

(b) 2 and 2 3y x y x= = + Solution

PAST HSC QUESTIONS QUESTION 1 Solve 22 5 12 0x x+ − = . Solution QUESTION 2 Solve 23 2x x= − . Solution

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QUESTION 3 Find the points of intersection of 2 4 and 4y x x y x= − = − . Solution QUESTION 4 Find the points of intersection of 26y x x= − and 2y x= . Solution

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THE DISCRIMINANT (DENOTED Δ )

If 2 0ax bx c+ + = then 2 4

2b b acx

a− ± −= .

The part under the square root is called the discriminant and it determines the nature of the roots. If Δ > 0 there are 2 roots. If Δ = 0 there is 1 root (sometimes called 2 equal roots). If Δ < 0 there are no roots. Also, if 2NΔ = the roots are rational (and the quadratic can be factorised). EXAMPLE 4 (a) Find the value of k if 24 10 0x x k− + = has equal roots. (b) Find the value of k if 24 3 0x kx+ + = has no roots.

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PAST HSC QUESTIONS QUESTION 5 For what values of k does: (a) 2 4 0x kx− + =

(b) 23 2 0x x k+ + = have no real roots. Solution QUESTION 6 Find the values of k if ( ) ( )2 4 7 0x k x k− + + + = has equal roots. Solution QUESTION 7 (a) Find the value of the discriminant of ( )22 2 8 0x k x+ − + = .

(b) Hence find the values of k for which 22 9y x kx= + + does not intersect 2 1y x= + . Solution

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QUESTION 8 Graphs 2 4y x= + and y x k= + have just one point in common. Find that point.

Solution

QUESTION 9 Show that 23y mx m= − touches 2 12x y= for all values of m . Solution

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SUM AND PRODUCT OF ROOTS

If 2 0ax bx c+ + = has the roots ,α β then , b ca a

α β αβ+ = − = .

EXAMPLE 5 If 22 8 1 0x x− − = has roots α and β , find the value of: (a) α β+ (b) αβ (c) 1 1

α β+

(d) ( )( )1 1α β− −

(e) 2 2α β+

Solution

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PAST HSC QUESTIONS QUESTION 10 If 22 4 1 0x x− − = has roots and α β , find the value of: (a) α β+ (b) αβ

(c) 1 1α β

+

Solution QUESTION 11

(a) Show that 2 2 2( ) 2a b ab a b+ − = + . (b) If 2 4 6 0x x− − = has roots and α β , find the value of: (i) α β+ (ii) αβ (iii) 2 2α β+ Solution

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QUESTION 12

If 2 3 1 0x x− + = has roots and α β , find the value of: (a) αβ

(b) 1αα

+

Solution

QUESTION 13

(a) Find the equation of the axis of symmetry on 2 4 1y x x= − − and show this is equal to the average of the 2 roots.

(b) 4 1y x= + is a chord AB on the parabola 2y x= as shown in the diagram. Find the

midpoint of that chord.

Solution

x-2 2 4 6

y

5

10

15

20

A

B

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SOLUTIONS

SAMPLE MATHS EXTENSION 1 NOTES

QUESTION 1

(a) 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040

(b) 7 x 6 x 5 x 4 x 3 x 2 = 5040

(c) 7 x 6 x 5 x 4 x 3 = 2520

QUESTION 2 123 1320P =

QUESTION 3

(a) 53 60P =

(b) 5 5 5 55 4 3 2 320P P P P+ + + =

QUESTION 4

(a)

2

6

1

5 4 3 2 1

2 x 6! = 1440 (b)

3

5

2

1

4 3 2 1

3! x 6! = 720

QUESTION 5

(a) 5! = 120 (b) There are 3 ways of placing the last digit: 3 x 4! = 72

4 3 2 1 3

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QUESTION 6 5 5 5

1 2 3 2 x 4 x 3 x 2 x 1 133P P P+ + + = , noting the 5 digit number must start with 1 or 2.

QUESTION 7

(a) Regard Kate and William as one and the answer is 7! = 5040. (b) There are 8! ways that 8 people can pass through the door and in half of those cases,

Kate will be first; hence 8!/2 = 20160.

QUESTION 8

(a) 2 (b) 720 (c) 48 (d) 240 (e) 144 (f) 480 (g) 288 (h) 96

QUESTION 9

(a) 7! 25202!

=

(b) 5! 203!

=

(c) 6! 15

2!4!=

(d) 6! 3602!

=

(e) 11!

2!2!4!= 207900

(f) 13!

2!3!3!= 86486400

QUESTION 10 8!3!4!

= 280

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QUESTION 11

(a) 7! 35

3!4!=

(b) 7!

2!2!3!= 210

(c) 7!

2!2!2!= 735

QUESTION 12

(a) 5!2!

= 60

(b) 2 4!

2!× = 24

QUESTION 13

(a) 4 44 4P P× = 576

(b) 5 44 4P P× = 2880

QUESTION 14

(a) 5!= 120

(b) 1 2! 3!× × = 12

(c) Seat 1st man, then next 2 men, his ex wife then other 2 women: 1 2! 1 2!× × × = 4

(d) 1st man, next 2 men, chosen lady, next 2 ladies: 1 2! 2 2!× × × = 8

QUESTION 15

Place the 3 that sit together, then the other 3, so consider 4 places only: 1 3! 3!× × = 36.

QUESTION 16 3!= 6 QUESTION 17 This halves the possibilities of Q16 3!/2 = 3 QUESTION 18 3

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QUESTION 19

(a)

6 7 6 5 4 3 2 1 5 There are 6 then 5 letters than can occupy the 1st and last place then 7! For the other places, and then divide by the repeated letters. 5 6 7! 12600

3!2!× × =

3

7

2

1

6 5 4 3 2 1

(b) A line through 3 boxes leaves 7 spaces, 1st consider the S’s as different so multiply by

3!

At the end divide by the repeated letters: 3! 7!3!2!× = 2520.

QUESTION 20

(a) 55 = 3125

(b) 5!= 120

(c) 35 4× = 320

QUESTION 21

(a) 7! 35

3!4!=

(b) 5! 5! 5!

3!2! 2!3! 1!4!+ + = 25

QUESTION 22 85C = 56

QUESTION 23 6 6

3 2C C× = 300 QUESTION 24 2 12

1 10C C× = 132 QUESTION 25 13

8C = 1287 QUESTION 26 7 2 6

6 1 4C C C× × = 210

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QUESTION 27

(a) 6 43 2C C× = 120

(b) 6 4 6 4 6

3 2 4 1 5C C C C C× + × + = 186

QUESTION 28 10 8

6 4C C− = 140. All of the possibilities less the ones that are excluded.

QUESTION 29

(a) 5 3 22 1 1C C C× × = 60

(b) 5 3 2 5 3 2 5 3 22 1 1 1 2 1 1 1 2C C C C C C C C C× × + × × + × × = 105

(c) Now we can have 3:1:1, 2:2:1, 2:1:2, 1:3:1, 1:2:2 which eventually adds to give 175.

QUESTION 30 52

5C = 2598960 In HSC exams, this answer would be left as 525C . Note though, that

questions involving packs of playing cards are not asked in the HSC.

QUESTION 31

(a) 4 43 2C C× = 24

(b) 4 484 1C C× = 48

(c) 4 484 113 C C× × = 624

(d) 13

54 C× = 5148

QUESTION 32

(a) 97C = 6435

(b) 7 5 4

3 2 2C C C× × = 1200 (c) 6 9 6 9 6 9

6 1 5 2 4 3C C C C C C× + × + × = 1485

QUESTION 33 10 7 10 7 10 78 4 9 3 10 2C C C C C C× + × + × = 1946

QUESTION 34 2 2n − QUESTION 35 7 6 6 5

3 4 2 3C C C C× − × = 375

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QUESTION 36

(a) 8 43 2C C× = 336

(b) 8 4

3 2 5!C C× × = 40320

QUESTION 37

(a) 6!3!2!

= 60

(b) Removing the last dot makes no difference if just 6 dots, so answer remains 60.

QUESTION 38

Since , 14n nr n rC C r n r−= + − =

QUESTION 39

Can be 3:1:1, 2:2:1 and 3 ways of doing each one = 5! 5!33! 2!2!

× + =

150

QUESTION 40

(a) Can’t include RRRRRJ, so 6! 15!

− = 5

(b) Consider Robin wins and John keeps firing arrows until 9 shots are made. Hence total

shots are 9! 126

5!4!= . A further 126 shots could be made for John to win, so answer is

252.

QUESTION 41 8 8 8 8 8 83 4 5 6 7 8C C C C C C+ + + + + = 291 or ( )8 8 8 8

0 1 22 C C C− + + = 219 QUESTION 42 (a) AB,AC,AD covers all possibilities because once chosen the other pairs are CD,BD,BC

or using the notation: 4 2

2 2

2!C C× = 3

(b) 6 4 2

2 2 2

3!C C C× × = 15

(c) 8 6 4 2

2 2 2 2

4!C C C C× × × = 105

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QUESTION 43

(a) 62

1( ) 1, ( ) 30, ( )30

n E n S P P E= = = ∴ =

(b) 62

2 1( ) 2, ( ) 30, ( )30 15

n E n S P P E= = = ∴ = =

QUESTION 44 66

144 1( ) 4! 3! 144, ( ) 720, ( )720 5

n E n S P P E= × = = = ∴ = =

QUESTION 45

(a) 5 123 3

1( ) 10, ( ) 220, ( )22

n E C n S C P E= = = = ∴ =

(b) 5 4 3 123 3 3 3

3( ) 15, ( ) 220, ( )44

n E C C C n S C P E= + + = = = ∴ = 122

(c) 5 4 3 121 1 1 3

3( ) 60, ( ) 220, ( )11

n E C C C n S C P E= × × = = = ∴ =

(d) Parts (b), (c), (d) are all the possibilities, 3 3 29 ( ) 144 11 44

P E∴ = − − =

QUESTION 46

9 164 4

126 9( ) 126, ( ) 1820, ( )1820 130

n E C n S C P E= = = = ∴ = =

QUESTION 47

(a) 5 6 113 2 5

150 25( ) 150, ( ) 462, ( )462 77

n E C C n S C P E= × = = = ∴ = =

(b) 5 6 5 6 5 6 113 2 4 1 5 6 5

181( ) 181, ( ) 462, ( )462

n E C C C C C C n S C P E= × + × + × = = = ∴ =

QUESTION 48

(a) 4 48 524 1 5( ) 13 624, ( ) 2598960, ( )n E C C n S C P E= × × = = = ∴ ≈ 0.000923

(b) 13 52

5 5( ) 4 5148, ( ) 2598960, ( )n E C n S C P E= × = = = ∴ ≈ 0.00198

QUESTION 49

3232 52 13

13 13 5213

1( ) , ( ) , ( ) 0.0005471828

Cn E C n S C P EC

= = ∴ = = ≈

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QUESTION 50

(a) 2( ) 4! 2! 48, ( ) 5! 120, ( )5

n E n S P E= × = = = ∴ =

(b) 2 3( ) 1 (together) 15 5

P E P= − = − =

QUESTION 51

2( ) 1 2! 4! 48, ( ) 5! 120, ( )5

n E n S P E= × × = = = ∴ =

(a) 4 4 243 2 5

1( ) 24, ( ) 42504, ( ) 0.0005651771

n E C C n S C P E= × = = = ∴ = ≈

(b) 6 18 244 1 5

45( ) 4 1080, ( ) 42504, ( ) 0.02541771

n E C C n S C P E= × × = = = ∴ = ≈

QUESTION 53

(a)

4

3

5

2

1

4

3

2

1

5! 4! 8! 1( ) 20, ( ) 280, ( )4! 3! 4! 3! 14

n E n S P E×= = = = ∴ =× ×

QUESTION 53 (b) The whites can be at 1,3,5,7 or 2,4,6,8 so

2 4! 4! 8! 1( ) 8, ( ) 280, ( )

4! 3! 4! 3! 35n E n S P E× ×= = = = ∴ =

× ×

(c) Fill in the two ends with red balls first and then 6! of filling the rest, divided by 4!, 3!

3

6

5

4

3 2 1 2

3 2 6! 8! 3( ) 30, ( ) 280, ( )

4! 3! 4! 3! 28n E n S P E× ×= = = = ∴ =

× ×

(d) 8! 1( ) 1, ( ) 280, ( )

4! 3! 280n E n S P E= = = ∴ =

×

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QUESTION 54

(a) There is an equal chance of F being before T as being after T, 1( )2

P E∴ =

(b) 1( ) 1, ( ) 5! 120, ( )120

n E n S P E= = = ∴ =

QUESTION 55

5!( ) 602!

n S = = . The F’s can be together, a single space between, 2 spaces between or 3

spaces between. There are 4! = 24 ways of 0 space, 3 x 3! = 18 ways of 1 space, 2 x 3! =12 ways of 2 spaces and 1 x 3! = 6 ways of 3 spaces.

Hence 1 2 1 20 1( ) 18 12 6 20, ( )3 3 3 60 3

n E P E= × + × + × = ∴ = = .

QUESTION 56

8 8 8 8 84 4 3 2 1

35( ) 70, ( ) 192, ( )81

n E C n S C C C C P E= = = + + + = ∴ =

QUESTION 57

9!4! 12! 1( ) 30240, ( ) 21621600, ( )4!3!2! 4!3!2! 55

n E n S P E= = = = ∴ =

QUESTION 58

5 4 92 1 3

10( ) 40, ( ) 84, ( )21

n E C C n S C P E= × = = = ∴ =

QUESTION 59

(a) 6 4 101 1 2

8( ) 24, ( ) 45, ( )15

n E C C n S C P E= × = = = ∴ =

(b) Many students lost a mark in this HSC question when they didn’t realise that once a new tennis ball is used then it is a used ball. We need to find the 1st probability then multiply by the 2nd probability.

4 101 2 2 1

2( ) 6, ( ) 45, ( )15

n E C n S C P E= = = = ∴ =

8 102 2 2 2

28( ) 28, ( ) 45, ( )45

n E C n S C P E= = = = ∴ = 1 256( ) ( )675

P E P E∴ × =

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QUESTION 60

Need to consider the 3 cases of 2 white, 1 white and 0 white balls being selected and in each case multiply by a different second probability.

6 42 2

1 2 1 210 62 2

1 2 2(2 white) ( )3 5 15

C CP P E PPC C

= = ∴ = = → =

6 4 31 1 2

1 2 1 210 62 2

8 1 4(1 white) ( )15 10 75

C C CP P E PPC C×= = ∴ = = → =

4 22 2

1 2 1 210 62 2

2 1 2(0 white) ( )15 15 225

C CP P E PPC C

= = ∴ = = → =

Hence ( ) 1 4 2 442 white15 75 225 225

P∴ = + + =

QUESTION 61

(a) (i) ( )3

3 21

33

3! 2(3 different)3 (3 1)(3 2) 3! (3 1)(3 2)

n

n

C n nPC n n n n n

×= = =− − × − −

(ii) ( )( ) ( )( )3

33

3 1 2 3! 1 23(3 same)3 (3 1)(3 2) 3! (3 1)(3 2)

n

n

n n n n nCPC n n n n n

× − − × − −×= = =− − × − −

(iii) ( )( )2 1 22(2 and 1) 1

(3 1)(3 2) (3 1)(3 2)n nnP

n n n n− −

= − −− − − −

(no need to simplify)

(b) 2 1 6, ,9 9 9

. The 3rd solution is gained by subtracting the 1st two solutions from 1.

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SAMPLE 2 UNIT MATHS NOTES

EXERCISE 1 QUESTION 1

22 5 12 0x x+ − =

Product: 2 x -12 = -24 Sum: 5 Factors: 8, -3

2 22 5 12 2 8 3 12 2 ( 4) 3( 4) (2 3)( 4)

(2 3)( 4) 03 , 42

x x x x xx x xx x

x x

x

∴ + − = + − −= + − += − +

∴ − + =

= −

QUESTION 2

23 2x x= −

23 2 0x x∴ + − =

Product: 3 x -2 = -6 Sum: 1 Factors: 3, -2

23 3 2 2 03 ( 1) 2( 1) 0 (3 2)( 1) 0

2 , 13

x x xx x x

x x

x

∴ + − − =+ − + =

− + =

∴ = −

QUESTION 3

2 4 and 4y x x y x= − = −

Sub for 2: 4 4 y x x x∴ − = −

2 5 4 0( 1)( 4) 0

1,4 5,8. . points are (1,5), (5,8)

x xx xxy

i e

∴ − + =− − =

∴ ==

© The School For Excellence 2017 Succeeding in the HSC – Mathematics Extension 1 Page 63

QUESTION 4

26 and 2y x x y x= − =

Sub for 2: 2 6y x x x∴ = − 2 4 0

( 4) 00,4 and 0,8

points are (0,0), (4,8)

x xx x

x y

∴ − =− =

∴ = =

EXERCISE 2 QUESTION 5

(a) 2 4 0x kx− + = (b) 23 2 0x x k+ + =

2

016 0

( 4)( 4) 04 4

kk k

k

Δ <

∴ − <+ − <

∴− < <

04 12 012 4

13

kk

k

Δ <∴ − <− < −

∴ >

QUESTION 6

( ) ( )2 4 7 0x k x k− + + + =

( ) ( )

2

2

2

2

0, 4 0

4 4 7 0

8 16 4 28 04 12 0

( 6)( 2) 06, 2

b ac

k k

k k kk kk kk

Δ = → − =

∴ + − + =

+ + − − =+ − =+ − =

∴ = −

QUESTION 7

(a) ( )22 2 8 0x k x+ − + =

( )

2

2

2

2

0, 4 0

2 4 2 8

4 4 64 4 60

b ac

k

k kk k

Δ = → − =

∴Δ = − − × ×

= − + −= − −

© The School For Excellence 2017 Succeeding in the HSC – Mathematics Extension 1 Page 64

(b) 22 9y x kx= + + and 2 1y x= + sub for 2: 2 1 2 9y x x kx∴ + = + + ( )22 2 8 0x k x∴ + − + = If no intersection, 0Δ <

2 4 60 0( 6)( 10) 0

6 10

k kk k

k

∴ − − <+ − <

∴− < <

QUESTION 8

2 4y x= + and y x k= + sub for 2: 4y x k x∴ + = +

2 4 0x x k− + − =

( )20 4 0

1 4 4 01 16 4 0

4 15 015 4

b ack

kk

k

Δ = → − =∴ − − =

− + =∴ − =

=

2 1 Quadratic equation is: 04

x x∴ − + =

21 02

1 1, 42 4

x

x y

∴ − =

∴ = =

QUESTION 9

23y mx m= − and 2 12x y=

sub for ( )2 2: 12 3y x mx m∴ = − 2 2 12 36 0x mx m∴ − + =

2 2144 4 36 0m mΔ = − × = Since 0Δ = , there is one solution only so the line and curve touch once.

© The School For Excellence 2017 Succeeding in the HSC – Mathematics Extension 1 Page 65

QUESTION 10

22 4 1 0x x− − =

(i) 4 2,2

ba

α β+ = − = =

(ii) 1 , 2

ca

αβ = = −

(iii) 1 1 12 4

2β α

α β αβ++ = = ÷ − = −

QUESTION 11 (a) 2 2 2( ) 2a b a ab b+ = + + 2 2 2( ) 2a b ab a b∴ + − = + (b) 2 4 6 0x x− − =

(i) 4 41

ba

α β+ = − = =

(ii) 6 6

1ca

αβ −= = = −

(iii) ( )22 2 2 16 12 28α β α β αβ+ = + − = + =

QUESTION 12

2 3 1 0x x− + =

(i) 1 11

ca

αβ = = =

(ii) If 11, αβ βα

= = and 1 3 3

1ba

α α βα

+ = + = − = =

QUESTION 13

(a) Axis of symmetry is 4 2

2 2bxa

−= − = − =

Average of roots is 4 2,

2 2 2ba

α β+ −= − = − = which is the same as the axis of

symmetry (b) 4 1y x= + , 2y x= Sub for 2 2: 4 1 4 1 0y x x x x∴ = + → − − = Average of the roots is 2 from part (a) Midpoint of chord is ( ) ( )2, 2 4 1 2,9× + =