Cambridge International AS & A Level Further Mathematics Worked Solutions Manual sample

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Cambridge International AS & A Level

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Further Mathematics

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Worked Solutions Manual

Elevate Original material Š Cambridge University Press 2020. This material is not final and is subject to further changes prior to publication.

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We are working with Cambridge Assessment International Education towards endorsement of this title.

Original material Š Cambridge University Press 2020. This material is not final and is subject to further changes prior to publication.

We are working with Cambridge Assessment International Education towards endorsement of this title.

Lee Mckelvey, Martin Crozier and Muriel James

Cambridge International AS & A Level

Further Mathematics

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Worked Solutions Manual

Original material Š Cambridge University Press 2020. This material is not final and is subject to further changes prior to publication.

We are working with Cambridge Assessment International Education towards endorsement of this title.

University Printing House, Cambridge CB2 8BS, United Kingdom One Liberty Plaza, 20th Floor, New York, NY 10006, USA 477 Williamstown Road, Port Melbourne, VIC 3207, Australia 314–321, 3rd Floor, Plot 3, Splendor Forum, Jasola District Centre, New Delhi – 110025, India 79 Anson Road, #06–04/06, Singapore 079906 Cambridge University Press is part of the University of Cambridge.

It furthers the University’s mission by disseminating knowledge in the pursuit of education, learning and research at the highest international levels of excellence.

© Cambridge University Press 2020

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www.cambridge.org Information on this title: www.cambridge.org/9781108770187 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

First published 2020

20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 Printed in The United Kingdom by CPI

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ISBN 978-1-108-77018-7 Paperback with Digital Access

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Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate. Information regarding prices, travel timetables, and other factual information given in this work is correct at the time of first printing but Cambridge University Press does not guarantee the accuracy of such information thereafter.

Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers which are contained in this publication.

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All worked solutions within this resource have been written by the author.

In examinations, the way marks are awarded may be different. References to assessment and/or assessment preparation are the publisher’s interpretation of the syllabus requirements and may not fully reflect the approach of Cambridge Assessment International Education. Cambridge International recommends that teachers consider using a range of teaching and learning resources in preparing learners for assessment, based on their own professional judgement of their students’ needs.

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Original material © Cambridge University Press 2020. This material is not final and is subject to further changes prior to publication.

Contents We are working with Cambridge Assessment International Education towards endorsement of this title.

Contents

The items in orange are available on Cambridge GO.

How to use this book

vii

Further Pure Mathematics 1 1 Roots of polynomial equations

1 1

Exercise 1B

2

Exercise 1C

4

Exercise 1D

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Exercise 1A

6

End-of-chapter review exercise 1

2 Rational functions Exercise 2A

8 8

Exercise 2B

10

Exercise 2C

13

Exercise 2D

A

14

End-of-chapter review exercise 2

3 Summation of series Exercise 3A

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Exercise 3B

17 17 19

End-of-chapter review exercise 3

21

Exercise 4A

21

Exercise 4B

23

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4 Matrices 1

Exercise 4C

26

Exercise 4D

27

End-of-chapter review exercise 4

5 Polar coordinates

31

Exercise 5A

31

Exercise 5B

33

End-of-chapter review exercise 5

6 Vectors

35

Exercise 6A

35

Exercise 6B

36

Exercise 6C

38

End-of-chapter review exercise 6

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Original material Š Cambridge University Press 2020. This material is not final and is subject to further changes prior to publication.

Cambridge International AS & A Level Further Mathematics: Worked Solutions Manual We are working with Cambridge Assessment International Education towards endorsement of this title.

7 Proof by induction

41

Exercise 7A

41

Exercise 7B

43

End-of-chapter review exercise 7

Cross-topic review exercise 1

Further Probability & Statistics 8 Continuous random variables

46 46

Exercise 8B

49

Exercise 8C

55

Exercise 8D

58

End-of-chapter review exercise 8

9 Inferential statistics Exercise 9A

63 63

Exercise 9B

65

Exercise 9C

67 69

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Exercise 9D Exercise 9E

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Exercise 8A

71

End-of-chapter review exercise 9

10 Chi-squared tests Exercise 10A

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Exercise 10B

74 74 78

Exercise 10C

82

Exercise 10D

89

End-of-chapter review exercise 10

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11 Non-parametric tests

98

Exercise 11A

98

Exercise 11B

102

Exercise 11C

104

Exercise 11D

106

Exercise 11E

108

End-of-chapter review exercise 11

12 Probability generating functions

112

Exercise 12A

112

Exercise 12B

113

Exercise 12C

116

Exercise 12D

119

End-of-chapter review exercise 12

Cross-topic review exercise 2 iv

Original material Š Cambridge University Press 2020. This material is not final and is subject to further changes prior to publication.

Contents We are working with Cambridge Assessment International Education towards endorsement of this title.

Further Mechanics

13 Projectiles 121 Exercise 13A

121

Exercise 13B

122

End-of-chapter review exercise 13

14 Equilibrium of a rigid body

125 125

Exercise 14B

126

Exercise 14C

129

Exercise 14D

131

End-of-chapter review exercise 14

15 Circular motion Exercise 15A

136 138

Exercise 15C End-of-chapter review exercise 15

Exercise 16C

143

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16 Hooke’s Law Exercise 16B

135

135

Exercise 15B

Exercise 16A

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Exercise 14A

143 145 148

End-of-chapter review exercise 16

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17 Linear motion under a variable force

152

Exercise 17A

152

Exercise 17B

154

End-of-chapter review exercise 17

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18 Momentum 157 Exercise 18A

157

Exercise 18B

159

End-of-chapter review exercise 18

Cross-topic review exercise 3

Further Pure Mathematics 2 19 Hyperbolic functions

164

Exercise 19A

164

Exercise 19B

168

Exercise 19C

171

End-of-chapter review exercise 19

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Original material Š Cambridge University Press 2020. This material is not final and is subject to further changes prior to publication.

Cambridge International AS & A Level Further Mathematics: Worked Solutions Manual We are working with Cambridge Assessment International Education towards endorsement of this title.

20 Matrices 2

175

Exercise 20A

175

Exercise 20B

179

Exercise 20C

184

Exercise 20D

190

End-of-chapter review exercise 20

21 Differentiation

193 193

Exercise 21B

196

Exercise 21C

199

Exercise 21D

202

End-of-chapter review exercise 21

22 Integration

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Exercise 21A

210

Exercise 22A

210

Exercise 22B

213

Exercise 22C

217

Exercise 22D

222

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End-of-chapter review exercise 22

23 Complex numbers Exercise 23A Exercise 23B Exercise 23C

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Exercise 23D

225 225 227 231 235

End-of-chapter review exercise 23

24 Differential equations

239 239

Exercise 24B

242

Exercise 24C

244

Exercise 24D

249

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Exercise 24A

End-of-chapter review exercise 24

Cross-topic review exercise 4

Further Pure Mathematics 1 practice exam-style paper Further Probability & Statistics practice exam-style paper Further Mechanics practice exam-style paper Further Pure Mathematics 2 practice exam-style paper

vi

Original material Š Cambridge University Press 2020. This material is not final and is subject to further changes prior to publication.

How to use this book We are working with Cambridge Assessment International Education towards endorsement of this title.

How to use this book

This book contains worked solutions to the questions in the Cambridge International AS & A Level Further Mathematics Coursebook. The solutions to the chapter exercises are available in this book and also digitally. The solutions to the end-of-chapter review exercises, cross-topic review exercises and practice exam-style papers are only available digitally.

Each solution shows you step-by-step how to solve the question. You will be aware that often questions can be solved by multiple different methods. In this book, we provide a single method for each solution. Do not be disheartened if the working in a solution does not match your own working; you may not be wrong but simply using a different method. It is good practice to challenge yourself to think about the methods you are using and whether there may be alternative methods.

Some questions in the coursebook go beyond the syllabus. We have indicated these solutions with a red line to the left of the text.

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Additional guidance is included in Commentary boxes throughout the book. These boxes often clarify common misconceptions or areas of difficulty.

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All worked solutions within this resource have been written by the author. In examinations, the way marks are awarded may be different.

vii

Original material Š Cambridge University Press 2020. This material is not final and is subject to further changes prior to publication.

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R

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We are working with Cambridge Assessment International Education towards endorsement of this title.

Original material Š Cambridge University Press 2020. This material is not final and is subject to further changes prior to publication.

We are working with Cambridge Assessment International Education towards endorsement of this title.

Chapter 1 Roots of polynomial equations EXERCISE 1A 1 a a + b = −5, ab = 9

b a + b = −b, ab = c

b a + b = 4, ab = 8

Since a = b − 2 , a + b = ( b − 2) + b = −b 2−b which means b = 2 Similarly, a b = ( b − 2)b = c or b 2 − 2b = c

c Rewrite the equation as x + 3 x − 7 = 0 2 2 2

Hence a + b = − 3 , ab = − 7 2 2

( )

2−b 2 − (2 − b) = c 2 So (2 − b) 2 − 4(2 − b) = 4c Hence b 2 − 2b =

Don’t forget to write your expression in the form b c x 2 + a x + a = 0.

FT

⇒ b 2 − 4b + 4 − 8 + 4b = 4c ⇒ b 2 = 4c + 4 = 4(c + 1)

2 a Let x 2 + 4 x + 4 = 0 3 Hence a + b = − 4 , ab = 4 3 2 2 b Using a + b = (a + b ) 2 − 22ab

( )

we have a 2 + b 2 = − 4 3

− 2 × 4 = − 56 9

Similarly ab = − 16 p = −8, so p = 2 Substituting for p gives q = 1

A

3 Let roots be a , a + 1

2

q 6 Rewrite the equation as x 2 + p x − 16 p =0 q Then a + b = − p = − 1 , so p = 2q 2

So the sum of the roots is a + a + 1 = 2 + p

7 a + b = −2, ab = −6

The product of the roots is a (a + 1) = 7 + p

Since a 2 + b 2 = (a + b ) 2 − 2ab ,

Equating expressions for p: 2a − 1 = a (a + 1) − 7

substituting values gives:

2a − 1 = a + a − 7

a 2 + b 2 = (−2) 2 − 2(−6) = 16

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2

Therefore a 2 − a − 6 = 0

Then (a − b ) 2 = a 2 + b 2 − 2ab

⇒ (a − 3)(a + 2) = 0

which gives (a − b ) 2 = 16 − 2(−6) = 28

⇒ a = −2 or a = 3

a2 + b2 Next we note that 12 + 12 = a b a 2b 2 = 16 2 = 16 = 4 36 9 (−6)

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Using 2a − 1 = p, and given that p > 0, leads to a = 3 and p = 5

4 Using a 2 + b 2 = (a + b) 2 − 2ab

Hence we have the solutions (a − b ) 2 = 28 and 1 + 1 =4 a2 b2 9

7 = 9 − 2ab, ∴ ab = 1

The quadratic equation with roots a, b is x 2 − (a + b)x + ab = 0

a +b 1 8 Since a1 + 1 = = 2 b ab then 2(a + b ) = ab .......................................................[1]

Substituting values: x 2 + 3x + 1 = 0 5 a a + b = −b, ab = c

Using a 2 + b 2 = (a + b ) 2 − 2ab

Since a = 3b , a + b = 4b = −b which means b = − b 4 Then a × b = 3b × b = 3b 2 = c

( )

So c = 3b 2 = 3 − b 4

2

gives (a + b ) 2 − 2ab = 12..............................................[2] Now let u = a + b and v = ab Substituting these new variables in [2] gives u 2 − 2v = 12 and substituting in [1] gives 2u = v

2 = 3b 16

Hence b 2 = 16 c 3

1

Original material © Cambridge University Press 2020. This material is not final and is subject to further changes prior to publication.

Cambridge International AS & A Level Further Mathematics: Worked Solutions Manual We are working with Cambridge Assessment International Education towards endorsement of this title.

So u − 4u − 12 = 0 10 a Let x 2 − 1 x + 3 = 0 2 4 ⇒ (u − 6)(u + 2) = 0 1 So a + b = , ab = 3 ⇒ either u = −2 and v = −4, or u = 6 and v = 12 2 4 2 2 Using a + b = (a + b ) 2 − 2ab In terms of a and b : 2 In the first case we have a + b = −2 and ab = −4 1 − 2 × 3 = − 47 2 2 a + b = − leads to 2 4 16 which yields x 2 + 2x − 4 = 0 2

( )

The other case, a + b = 6 and ab = 12, yields x 2 − 6x + 12 = 0

b Since a 2 + b 2 < 0 this means the roots must be complex.

2 x− 4 =0 3 3 2 So a + b = − , ab = − 4 . These can then be used to 3 3 find S1, S2 and S−1

9 Let x 2 +

2 3

28 S2 = a 2 + b 2 = (a + b ) 2 − 2ab = 4 + 8 = 9 3 9

EXERCISE 1B 1 a a + b + g = −3, abg = 5

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2 a +b −3 1 1 1 S−1 = a + = = = b ab −4 2 3

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S1 = a + b = −

The only way the sum of the squares of two numbers can be negative is if at least one of them is complex. Using that a + b = 1 , we then have 4 that both numbers must be complex.

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b We divide the equation by 2 to have a leading coefficient equal to 1: x3 + 5 x2 − 3 = 0 2

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Then a + b + g = − 5 , abg = 3 2 c a + b + g = 0, abg = 9

Σ ab Then, since S−1 = a1 + 1 + g1 = , we have b Σ abg 4 S−1 = 3 = 4 1 3

c a + b + g = −3, ab + ag + bg = 5, abg = 7 Using S2 = ( Σ a ) 2 − 2 Σ ab gives S2 = (−3) 2 − 2 × 5 = −1 Σ ab Then since S−1 = a1 + 1 + g1 = , we have b Σ abg S−1 = 5 7

2 a a + b + g = 3, ab + ag + bg = 0 b Using Σ a 2 = (Σ a ) 2 − 2ab gives Σ a 2 = (3) 2 − 2 × 0 = 9

Hence a 2 + b 2 + g 2 = 9

4 Σ a = a + b + g = 0 and ab + ag + bg = −1 Then Σ a 2 = ( Σ a ) 2 − 2Σ ab = (0) 2 − 2 × (−1) = 2

3 a a + b + g = 2, ab + ag + bg = 0, abg = −5

5 First rearrange as x 3 + 5 x 2 + 1 = 0 2 2

Using S2 = ( Σ a ) 2 − 2 Σ ab gives S2 = (2) 2 − 2 × 0 = 4 Σ ab Since S−1 = a1 + 1 + g1 = we have S−1 = 0 = 0 −5 b Σ abg

Then a + b + g = − 5 , ab + ag + bg = 0, abg = − 1 2 2

b First let x 3 + 4 x − 1 = 0 3 3

Using S2 = a 2 + b 2 + g 2 = ( Σ a ) 2 − 2 Σ ab we have

( )

S2 = − 5 2

Then a + b + g = 0, ab + ag + bg = 4 , abg = 1 3 3 4 2 2 Using S2 = ( Σ a ) − 2 Σ ab gives S2 = (0) − 2 × = − 8 3 3 2

2

−2×0=

25 4

Original material © Cambridge University Press 2020. This material is not final and is subject to further changes prior to publication.

Chapter 1: Roots of polynomial equations We are working with Cambridge Assessment International Education towards endorsement of this title. b Considering a 3 (a + b + g ) + b 3 (a + b + g ) Since a, b, g all satisfy x 3 + 5 x 2 + 1 = 0, 2 2 + g 3 (a + b + g ), which is equal to S3S1, we see that 5 5 1 1 a 3 ( b + g ) + b 3 (a + g ) + g 3 (a + b ) = S3S1 − S4 a 3 + a 2 + = 0, b 3 + b 2 + = 0 and 2 2 2 2 Since S1 = 0, g 3 + 5g 2 + 1 = 0 a 3 ( b + g ) + b 3 (a + g ) + g 3 (a + b ) = −S4 = −2 2 2 Possible alternative wording:

Adding these equations gives S3 + 5 S2 + 3 = 0 2 2 ⇒ S3 = − 5 × 25 − 3 = − 137 2 2 8 4 Recall that Sn = a + b + g n

n

We consider a 3 ( b + g ) + b 3 (a + g ) + g 3 (a + b ) + S4, since this is equal to a 3 (a + b + g ) + b 3 (a + b + g ) + g 3 (a + b + g ) = S3S1 Then a 3 ( b + g ) + b 3 (a + g ) + g 3 (a + b ) = S3S1 − S4 = S3 (0) − 2 = −2

n

6 a a + b + g = −a,

ab + ag + bg = b,

FT

8 a Since 2x 3 − x 2 + x − 5 = 0 has roots a, b, g , we see that 2x n+3 − x n+2 + x n+1 − 5x n = 0 must also have roots a, b, g .

abg = −a

So:

So Σ a = −a

2a n+3 − a n+2 + a n+1 − 5a n = 0 2b n+3 − b n+2 + b n+1 − 5b n = 0 2g n+3 − g n+2 + g n+1 − 5g n = 0

Σ ab b =− b Using Σ a1 = gives Σ a1 = −a a Σ abg

b From part a, Σ a = −a and Σ a1 = − ab . If these are equal, then –a = − ab so that b = a 2

Adding these equations gives 2Sn+3 − Sn+2 + Sn+1 − 5Sn = 0

b Let x 3 − 1 x 2 + 1 x − 5 = 0 2 2 2

A

Using Σ a 2 = ( Σ a ) 2 − 2 Σ ab we can calculate that

Then a + b + g = 1 , ab + ag + bg = 1 , abg = 5 2 2 2

Σ a 2 = (−a) 2 − 2 × b = a 2 − 2b Then Σ a 2 = a 2 − 2a 2 = −a 2

1 Σ ab 2 1 1 1 Since S−1 = a + + g = , we have S−1 = = 1 5 5 b Σ abg 2

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Therefore a 2 + b 2 + g 2 = −a 2, hence we have complex roots.

Then, using 2Sn+3 − Sn+2 + Sn+1 − 5Sn = 0 with n = −2 gives:

Here we have the sum of the squares of the roots being equal to a value less than 0, and this can only mean complex roots.

2S1 − 3 + S−1 − 5S−2 = 0

(

)

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So S−2 = 1 (2S1 − 3 + S−1) = 1 2 × 1 − 3 + 1 = − 9 2 5 5 5 25

7 a S1 = Σ a = a + b + g = 0

Σ ab = ab + ag + bg = −1

Note that S0 = 1, but since we have 3 roots this becomes 3.

S2 = ( Σ a ) − 2 Σ ab = (0) − 2 × (−1) = 2 2

2

Multiplying x 3 − x + 3 = 0 by x gives q 9 a Rewrite as x 3 + p x 2 + pr = 0, then q S1 = Σ a = a + b + g = − p

x 4 − x 2 + 3x = 0 for which a , b, g must be the roots.

So we get: a 4 − a 2 + 3a = 0 b 4 − b 2 + 3b = 0 g 4 − g 2 + 3g = 0

b First note that Σ ab = ab + ag + bg = 0 as there is no term in x.

Then using S2 = ( Σ a ) 2 − 2 Σ ab gives

Adding these equations gives S4 − S2 + 3S1 = 0

( )

q S2 = − p

So S4 = S2 − 3S1 = 2 − 3(0) = 2

3

2

−2×0=

q2 p2

Original material © Cambridge University Press 2020. This material is not final and is subject to further changes prior to publication.

Cambridge International AS & A Level Further Mathematics: Worked Solutions Manual

FT

We are working with Cambridge Assessment International Education towards endorsement of this title. q 10 S1 = a + b + g = −p c Since x 3 + p x 2 + pr = 0 has roots a, b, g we have: Since S1 = 0 we have p = 0 q 2 r 3 From the form of our cubic we have: a + pa + p = 0 Σ ab = ab + ag + bg = q q 2 r 3 b + pb + p = 0 So S2 = ( Σ a ) 2 − 2 Σ ab = (−p) 2 − 2 × q = −2q q Since S2 = −2, we know that q = 1 and g 3 + p g 2 + pr = 0 Σ ab q Using S−1 = a1 + 1 + g1 = , we have S−1 = −r q 3r b Σ abg Adding these equations gives S3 + p S2 + p = 0 q q q q 2 3r But S−1 = −r = 1 = − × So S3 = − p S2 − 3r − 5 2 p p p p Substituting q = 1 gives r = −5 3 q S3 = − 3 − 3r p Hence p = 0, q = 1, r = −5 p EXERCISE 1C

3 S1 = Σ a = a + b + g + d = 0

1 a a + b + g + d = 2, Σ ab = 5

Σ abg S−1 = a1 + 1 + g1 + 1 = =−1 2 d b Σ abgd

c Recall that Σ ab = ab + ag + ad + bg + bd + gd = a

Note that Σ ab = 0 because there is no term in x 2 Using S2 = ( Σ a ) 2 − 2 Σ ab

b Dividing the equation by 2: x + 5 x 3 − 3 x + 2 = 0 2 2

gives S2 = (0) 2 − 2 × 0 = 0

A

4

Then a + b + g + d = − 5 , Σ ab = 0 2 2 c Dividing the equation by 3: x 4 − x 2 + 3x − 11 = 0 3 3

Then S1 = Σ a = − 1 and Σ ab = 0 2

2 3

R

Then a + b + g + d = 0, Σ ab = −

4 Let x 4 + 1 x 3 − 1 x + 7 = 0 2 2 2

2 a Dividing the equation by 5: x 4 − 3 x 3 + 1 x − 13 = 0 5 5 5

D

3 Then Σ a = a + b + g + d = −b a =5 and Σ ab = 0

Since a, b, g, d are the roots of x 4 + 1 x 3 − 1 x + 7 = 0, 2 2 2 we have: a4 + 1a3 − 1a + 7 = 0 2 2 2 b4 + 1b3 − 1b + 7 = 0 2 2 2 g 4 + 1g 3 − 1g + 7 = 0 2 2 2

Using Σ a 2 = ( Σ a ) 2 − 2 Σ ab gives: 2 Σa2 = 3 − 2 × 0 = 9 5 25

()

d 4 + 1d 3 − 1d + 7 = 0 2 2 2

e 13 1 b Σ abg = −d a = − 5 and Σ abgd = a = − 5

Adding these equations gives S4 + 1 S3 − 1 S1 + 14 = 0 2 2 1 1 So S4 = − S3 + S1 − 14 2 2 We are given that S3 = 11 , so substituting for S3 and S1: 8 239 S4 = − 11 − 1 − 14 = − 16 4 16

−1 Σ abg 1 Then Σ a = = 5 = 1 Σ abgd − 13 13 5 Recall that Σ abg = abg + abd + agd + bgd

Since we have 4 roots, we add all 4 equations together; this gives the +14 in our recurrence relation.

4

Original material © Cambridge University Press 2020. This material is not final and is subject to further changes prior to publication.

Chapter 1: Roots of polynomial equations We are working with Cambridge Assessment International Education towards endorsement of this title.

2 5 Σ a = a + b + g + d = 1, Σ ab = 0, Σ abg = −1 and 7 a Let x 4 + x 3 + 7 x 2 + 4 = 0 3 3 3 Σ abgd = 2 2 7 Σ a = − , Σ ab = , Σ abg = 0, Σ abgd = 4 So Σ a 2 = ( Σ a ) 2 − 2 Σ ab = (1) 2 − 2 × 0 = 1 3 3 3 ⇒ Σa2 = 1 2 Then S1 = − 3 Σ abg Then Σ a1 = =−1 2 2 2 Σ abgd S2 = ( Σ a ) 2 − 2 Σ ab = − − 2 × 7 = − 38 3 3 9 Dividing the original equation by x, we have 2 2 b Dividing x 4 + x 3 + 7 x 2 + 4 = 0 by x gives x3 − x2 + 1 + x = 0 3 3 3 2 which has roots a, b, g, d . x 3 + x 2 + 7 x + 4 = 0 which has roots a, b, g, d .

3 3 3x 2 3 2 So a − a + 1 + a = 0 2 Then a 3 + a 2 + 7 a + 4 = 0 3 3 3a 2 3 2 2 b −b +1+ =0 b3 + b2 + 7b + 4 = 0 b 3 3 3b 2 2 7 3 2 3 2 g −g +1+ g =0 g + g + g + 4 =0 3 3 3g 2 2 d 3 + d 2 + 7d + 4 = 0 d3−d2+1+ =0 3 3 3d d

FT

( )

Adding these equations gives 2 S3 + S2 + 7 S1 + 4 S−1 = 0 3 3 3

Adding these equations gives S3 − S2 + 4 + 2S−1 = 0

A

From above, S2 = Σ a 2 = 1 and S−1 = Σ a1 = − 1 2

⇒ S3 = − 2 S2 − 7 S1 − 4 S−1 3 3 3

Then Σ a 3 = S3 = S2 − 4 − 2S−1

( )

=1−4−2 −1 2 = −2

Since S−1 = S3 = −

R

6 S1 = Σ a = 0, Σ ab = a, Σ abg = −b, Σ abgd = 1 Using S2 = ( Σ a ) 2 − 2 Σ ab , we have

Σ abg = 0, Σ abgd

( ) ( )

2 38 2 − −7 − = 118 3 9 3 3 27

2 Using x 4 + x 3 + 7 x 2 + 4 = 0 with roots a, b, g, d , 3 3 3 we can deduce that

S2 = (0) 2 − 2 × a = −2a

Σ abg S−1 = Σ a1 = = − b = −b 1 Σ abgd

S4 +

D

Since S2 = S−1 we have b = 2a

2 S + 7 S + 16 = 0 3 3 3 2 3

( ) ( )

⇒ S4 = − 2 118 − 7 − 38 − 16 = 130 3 27 3 9 3 81

Dividing x 4 + ax 2 + bx + 1 = 0 by x gives x 3 + ax + b + 1x = 0, which also has roots a, b, g, d .

c Yes there are complex roots since S2 < 0

So a 3 + aa + b + a1 = 0

8 S1 = Σ a = a + b + g + d = −a, hence a = −2 Σ abgd = abgd = c, hence c = 1

b 3 + ab + b + 1 = 0 b 1 g 3 + ag + b + g = 0

Σ a 2 = ( Σ a ) 2 − 2 Σ ab = 0, where Σ ab = b Then (−2) 2 − 2b = 0 ⇒ b = 2 Solution: a = −2, b = 2, c = 1

d 3 + ad + b + 1 = 0 d

9 Σ a = S1 = 2, Σ ab = 1, Σ abg = 0, Σ abgd = −4

Adding these equations gives S3 + aS1 + 4b + S−1 = 0

So S2 = ( Σ a ) 2 − 2 Σ ab = (2) 2 − 2 × 1 = 2

Substituting the values found above:

Dividing x 4 − 2x 3 + x 2 − 4 = 0 by x gives

S3 = −aS1 − 4b − S−1 = −a(0) − 4b + b = −6a

x 3 − 2x 2 + x − 4x = 0 which has roots a, b, g, d .

5

Original material © Cambridge University Press 2020. This material is not final and is subject to further changes prior to publication.

Cambridge International AS & A Level Further Mathematics: Worked Solutions Manual We are working with Cambridge Assessment International Education towards endorsement of this title.

We already know that S1 = 2 and S2 = 2.

So a 3 − 2a 2 + a − a4 = 0 Σ abg Similarly, S−1 = = 0 =0 −4 b 3 − 2b 2 + b − 4 = 0 Σ abgd b So S3 = 2(2) − 2 + 4 × 0 = 2 g 3 − 2g 2 + g − g4 = 0 Using x 4 − 2x 3 + x 2 − 4 = 0 we can obtain S4 − 2S3 + S2 − 16 = 0

d 3 − 2d 2 + d − 4 = 0 d

So S4 = 2(2) − 2 + 16 = 18

Adding these equations gives S3 − 2S2 + S1 − 4S−1 = 0

Therefore S4 = 9S3

⇒ S3 = 2S2 − S1 + 4S−1 EXERCISE 1D y 3

Substituting for x gives So

⇒ 3 − 2(y − 1) + 9(y − 1) 2 = 0

() () y 3

2

+5

y +3=0 3

y + 5 y + 3 = 0, or y 2 + 15y + 27 = 0 9 3 2

FT

1 Let y = 3x, so that x =

So 3 − 2y + 2 + 9y 2 − 18y + 9 = 0

Therefore the equation is 9y 2 − 20y + 14 = 0

1 4 Let y = 1 x ⇒x= y

2 a Let y = x 2

Substituting for x in the equation:

() ()

Rearranging 2x − 4x + 7 = 0 gives 2x + 7 = 4x 2

2

⇒ 4x 4 + 28x 2 + 49 = 16x 2 ⇒ 4x 4 + 12x 2 + 49 = 0 Substituting y = x 2:

Therefore the equation is 9y 2 − 4y + 1 = 0

5 Let y = x 2

Rearranging first lowers the number of cross terms we need to calculate. It also means that all terms are an even power of x.

y+3 2

D b Let y = 2x − 3, so that x =

2

−4

2x 3 − 5x = −1 ⇒ (2x 3 − 5x) 2 = (−1) 2 ⇒ 4x 6 − 20x 4 + 25x 2 = 1.................................................[1] Substituting for x 2 in the equation: 4y 3 − 20y 2 + 25y − 1 = 0...................................................[2] Comparing terms, S4 for [1] is equal to S2 for [2].

25 y− 1 =0 4 4 25 Σ a = 5, Σ ab = 4

y 3 − 5y 2 +

( ) ( ) y+3 2

Rearrange the equation:

Rewrite the equation as:

Substituting for x in the equation: 2

− 4 1y + 9 = 0

So 1 − 4y + 9y 2 = 0

R

4y 2 + 12y + 49 = 0

2

A

Squaring both sides: (2x 2 + 7) 2 = (4x) 2

1 y

y+3 +7=0 2

⇒ 2 (y 2 + 6y + 9) − 4 (y + 3) + 7 = 0 2 4

For [2], S2 = ( Σ a ) 2 − 2 Σ ab = (5) 2 − 2 ×

⇒ y 2 + 6y + 9 − 4y − 12 + 14 = 0

=

Therefore the equation is y + 2y + 11 = 0 2

1 3 Let y = x + x . Then xy = x + 1 ⇒ xy − x = 1 ⇒ x(y − 1) = 1 ⇒ x =

Hence for [1], S4 =

2

−2

25 2

6 Let y =

( ) ( )

( ) ( ) 1 y−1

25 2

x+2 2 x . Then xy = x + 2 ⇒ x = y − 1 Substituting for x in the equation:

1 y−1

Substituting for x in the equation: 3

25 4

2 y−1

1 +9=0 y−1 6

3

+3

2 y−1

2

−1=0

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Chapter 1: Roots of polynomial equations We are working with Cambridge Assessment International Education towards endorsement of this title.

So 8 + 12(y − 1) − (y − 1) 3 = 0 ⇒ S2 = ( Σ a ) 2 − 2 Σ ab = 1 + 9 = 289 64 2 64 3 2 This gives 8 + 12y − 12 − y + 3y − 3y + 1 = 0 Similarly, S−3 for [1] is equal to S−1 for [2] which then simplifies to y 3 − 3y 2 − 9y + 3 = 0 So S−3 = − 1 (a + 2)( b + 2)(g + 2) 2 a The expression is the product abg of the roots of our cubic equation 8 Let y = x 2 So

Rearranging the equation gives:

(a + 2)( b + 2)(g + 2) = −3 abg

x 3 − x = −4 Squaring both sides:

b g a + + is the a +2 b +2 g +2 reciprocal of the sum of the roots, so for our cubic Σ ab equation, this is Σ a1 = = −9 = 3 −3 Σ abg

b The expression

Substituting for x 2 in the equation:

y 3 − 2y 2 + y − 16 = 0.......................................................[2]

S6, S8, S10 for [1] are equal to S3, S4, S5 for [2].

Rearranging the equation gives: 2x = x + 6 4

⇒ x 6 − 2x 4 + x 2 = 16......................................................[1]

FT

7 Let y = x 3

(x 3 − x) 2 = (−4) 2

3

Cubing both sides: (2x 4) 3 = (x 3 + 6) 3 Expanding the brackets: 8x 12 = x 9 + 18x 6 + 108x 3 + 216

S1 = Σ a = 2, Σ ab = 1, S2 = ( Σ a ) 2 − 2 Σ ab = (2) 2 − 2 × 1 = 2

For y 3 − 2y 2 + y − 16 = 0, the roots are such that S3 − 2S2 + S1 − 48 = 0 ⇒ S3 = 2S2 − S1 + 48 = 50 Multiplying by y gives

y 4 − 2y 3 + y 2 − 16y = 0, which leads to

Substituting for x 3 in the equation:

S4 − 2S3 + S2 − 16S1 = 0

8y 4 − y 3 − 18y 2 − 108y − 216 = 0.....................................[2]

⇒ S4 = 2S3 − S2 + 16S1

So S6 for [1] is equal to S2 for [2]

So S4 = 100 − 2 + 32 = 130

Dividing [2] by 8:

Multiplying by y again leads to

A

⇒ 8x 12 − x 9 − 18x 6 − 108x 3 − 216 = 0.............................[1]

S5 − 2S4 + S3 − 16S2 = 0

Then Σ a = 1 , Σ ab = − 9 and 8 4 27 Σ abg 2 Σ a1 = = =−1 2 −27 Σ abgd

= 260 − 50 + 32 = 242

R

27 y4 − 1 y3 − 9 y2 − y − 27 = 0 8 2 4

⇒ S5 = 2S4 − S3 + 16S2

D

Hence S6 = 50, S8 = 130, S10 = 242

7

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Chapter 2 Rational functions EXERCISE 2A 1 a Factorising the denominator 2x + 3 2x + 3 = x 2 + 3x + 2 (x + 1)(x + 2)

Differentiating using the quotient rule, dy (x 2 − 2x − 3) .1 − x(2x − 2) = dx (x 2 − 2x − 3) 2

So the vertical asymptotes are x = −2 , x = −1 .

So

2x . x2 Hence the horizontal asymptote is y = 0 As |x | → ∞, y →

dy ≠ 0. dx Therefore there are no turning points.

Since −x 2 − 3 ≠ 0,

FT

b Vertical asymptote is x = 2 As |x | → ∞, y → 4x x .

Hence the horizontal asymptote is y = 4

b Differentiating using the quotient rule, dy (x 2 + x + 4)(2x + 2) − (x 2 + 2x)(2x + 1) = dx (x 2 + x + 4) 2

c Factorising the denominator x 2 − 2x − 3 x 2 − 2x − 3 = , and so the vertical 2 x − 2x − 8 (x + 2)(x − 4) asymptotes are x = −2, x = 4

dy 2x 3 + 4x 2 + 10x + 8 − (2x 3 + 5x 2 + 2x) = dx (x 2 + x + 4) 2 2 = −x2 + 8x + 82 (x + x + 4) Then

dy = 0, then x 2 − 8x − 8 = 0 dx The discriminant is b 2 − 4ac = 64 − 4(−8) = 96 If

A

2 As |x | → ∞, y → x 2 . x Hence the horizontal asymptote is y = 1

2 dy = 2 −x − 3 2 dx (x − 2x − 3)

2 Factorising the denominator, 2x 2 2x 2 = 2 (x − 6)(x + 1) x − 5x − 6

Hence there are two turning points.

4 First note that the vertical asymptote is x = 1

R

Writing the expression in partial fraction form, 2x 2 = A + B + C , and note that as x−6 x+1 (x − 6)(x + 1) 2 2x |x | → ∞ , y → 2 . So A = 2 x

D

Simplifying to find the coefficients,

2x Then |x | → ∞ means that y → x The horizontal asymptote is y = 2

5 First find the asymptotes.

The vertical asymptote is x = −2, from the denominator.

As |x | → ∞, y → 3x x . So the horizontal asymptote is y = 3.

2x 2 = 2(x + 1)(x − 6) + B(x + 1) + C(x − 6) 2 72 x = −1 ⇒ C = − and x = 6 ⇒ B = 7 7 72 2 Hence y = 2 + − 7(x − 6) 7(x + 1)

Points of intersection: when x = 0, y = − 1 and y = 0 2 occurs when x = 1 3

Horizontal asymptote is y = 2, vertical asymptotes are x = −1, x = 6.

y

Since the curve is tending to infinity at positive and negative infinity, it is quite a popular technique to obtain the constant A by looking at large values of x.

3

–2

3 a Multiplying out the brackets in the denominator, x x y= = (x + 1)(x − 3) x 2 − 2x − 3

–1 2

8

O

1 3

x

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Chapter 2: Rational functions We are working with Cambridge Assessment International Education towards endorsement of this title. y 6 a Let x − 1 = A + B x−2 x−2 So x − 1 = A(x − 2) + B Setting x = 2 yields B = 1 and similarly when x = 3 we see that A = 1 So y = 1 + 1 x−2 b Horizontal asymptote: y = 1

–1 O

1

x

3

Vertical asymptote: x = 2 c For both points of intersection it is simpler to use the original equation. When x = 0, we have y = 1 and 2 y = 0 when x = 1

FT

y

–3

8 a We first expand the denominator in order to be able to use the quotient rule to differentiate the function:

1

1

2

x

2x 2x = (x − 1)(x + 3) x 2 + 2x − 3

So

dy (x 2 + 2x − 3) .2 − 2x (2x + 2) = dx (x 2 + 2x − 3) 2

2x 2 + 4x − 6 − 4x 2 − 4x −2x 2 − 6 = 2 2 2 (x + 2x − 3) (x + 2x − 3) 2 dy We then consider = 0 to find any turning points. dx dy Since −(2x 2 + 6) ≠ 0, we have that ≠ 0 for every dx value of x. So the function has no turning points. =

A

O

1 2

y=

R

3−x 7 a Factorising the denominator, 32 − x = x − 1 (x − 1)(x + 1) The vertical asymptotes are x = −1, x = 1 When |x | → ∞, y → − x2 . The horizontal asymptote x is y = 0

b We find the points of intersection and the asymptotes.

First, when x = 0, we see that y = 0

D

We have that x = −3 and x = 1 are the vertical asymptotes.

2x Finally as |x | → ∞, y → 2 , so the horizontal x asymptote is y = 0

b To find the turning points, we differentiate y = 32 − x x −1 using the quotient rule.

So

dy (x 2 − 1)(−1) − (3 − x)(2x) −x 2 + 1 − 6x + 2x 2 = = dx (x 2 − 1) 2 (x 2 − 1) 2 2 = x −2 6x +21 (x − 1)

We have So x =

y

dy = 0 exactly when x 2 − 6x + 1 = 0 dx

6 ± √36 − 4 = 3 ± 2√2 2

–3

O

1

x

c Before sketching, we need to also know the points of intersection.

Setting x = 0 implies that y = −3.

Similarly, y = 0 when x = 3 9

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Cambridge International AS & A Level Further Mathematics: Worked Solutions Manual We are working with Cambridge Assessment International Education towards endorsement of this title.

2 c All that remains is to identify the points of 9 a As |x | → ∞, y → x 2 , so the horizontal asymptote x intersection. When x = 0, y = 5 and when 3 is y = 0 y = 0, x = ±√5 The two vertical asymptotes are x = −3 and x = 1. y

b Multiplying out the brackets in the denominator: 2 x2 − 5 y= = x − 5 . This is a quotient, (x − 1)(x + 3) x 2 + 2x − 3 so we can differentiate using the quotient rule.

5 3

dy (x 2 + 2x − 3)(2x) − (x 2 − 5)(2x + 2) = dx (x 2 + 2x − 3) 2 3 2 3 2x + 4x − 6x − (2x + 2x 2 − 10x − 10) = (x 2 + 2x − 3) 2 dy 2x 2 + 4x + 10 and so = dx (x 2 + 2x − 3) 2 Then

1 –√5

1

√5

x

FT

–3

O

Since 2x 2 + 4x + 10 = 2(x + 1) 2 + 8 > 8, there are no turning points. Alternatively we could calculate the discriminant to see the derivative is always non-zero.

EXERCISE 2B

2 a U sing the quotient rule, dy (x + 3) .2x − (x 2 − 5) .1 2x 2 + 6x − x 2 + 5 = = dx (x + 3) 2 (x + 3) 2 2 = x + 6x +2 5 (x + 3)

A

1 a W e use the partial fraction form 3x 2 + x + 3 = Ax + B + C x+1 x+1

2 As |x | → ∞, y → 3x x . Hence A = 3

Then 3x 2 + x + 3 = 3x(x + 1) + B(x + 1) + C

D

R

Setting x = −1 yields C = 5 and similarly when x = 0, we see that 3 = B + C which implies that B = −2 So y = 3x − 2 + 5 x+1 Hence the oblique asymptote is y = 3x − 2 and the vertical asymptote is x = −1 Again, note that the first coefficient for top-heavy fractions can easily be determined by considering large values of x. This method saves time by avoiding unnecessary calculations.

dy = 0 exactly when x 2 + 6x + 5 = 0, dx which factorises as (x + 1)(x + 5) = 0 We have

Therefore there are two turning points. c Using the quotient rule, dy (2x − 1)(2x + 5) − (x 2 + 5x − 4) .2 = dx (2x − 1) 2 =

4x 2 + 8x − 5 − (2x 2 + 10x − 8) 2x 2 − 2x + 3 = (2x − 1) 2 (2x − 1) 2

We have

2 b Let x + 3x − 31 = Ax + B + C x−4 x−4

dy = 0 exactly when 2x 2 − 2x + 3 = 0 dx

Using the discriminant, b 2 − 4ac = 4 − 24 = −20, and so there are no turning points.

2 As |x | → ∞, y → xx . Hence A = 1

2 3 Let 6x + x − 6 = Ax + B + C 2x − 1 2x − 1

Then x 2 + 3x − 31 = x(x − 4) + B(x − 4) + C Setting x = 4 yields C = −3 and setting x = 0 means −31 = −4B − 3. Hence B = 7

2 Then as |x | → ∞, y → 6x . So A = 3 2x

6x 2 + x − 6 = 3x(2x − 1) + B(2x − 1) + C

3 x−4 Hence the oblique asymptote is y = x + 7 and the vertical asymptote is x = 4 So y = x + 7 −

When x = 1 , C = −4 and when x = 0, −6 = −B + C 2 which means that B = 2 10

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Chapter 2: Rational functions We are working with Cambridge Assessment International Education towards endorsement of this title. b The vertical asymptote is x = −1 So y = 3x + 2 − 4 2x − 1 Setting x = −1: 1 − λ = C ⇒ C = −2 Hence the oblique asymptote is y = 3x + 2, and the 2 So y = x + 2 − vertical asymptote is x = 1 x + 1 2 Points of intersection: when x = 0, y = 0 and y = 0 2 x + 3 C when x = −3 4 Let = Ax + B + x−2 x−2 y

x

+

2

2 As |x | → ∞, y → xx . So A = 1

y=

Then x 2 + 3 = x(x − 2) + B(x − 2) + C When x = 2, C = 7 and when x = 0, 3 = −2B + 7 ⇒ B = 2 7 x−2 Hence the oblique asymptote is y = x + 2 and the vertical asymptote is x = 2 Points of intersection: when x = 0, y = − 3 but in order 2 for y = 0, we would need x 2 + 3 = 0. Hence there are no points of intersection with the x-axis.

So y = x + 2 +

FT –3

–2 –1

O

x

y=

A

x

+

2

y

2

x 2 − 2x + 1 = Ax + B + C x−4 x−4 2 x As |x | → ∞, y → x . So A = 1

6 a Let

Then x 2 − 2x + 1 = x(x − 4) + B(x − 4) + C

R

2

2

When x = 4, C = 9 and when x = 0 we have

x

1 = −4B + 9 ⇒ B = 2 9 x−4 Hence the oblique asymptote is y = x + 2, and the vertical asymptote is x = 4 So y = x + 2 +

D

–2 O –3 2

9 , dy = 1 − 9 x − 4 dx (x − 4) 2 dy At the turning point = 0, so (x − 4) 2 = 9. dx x − 4 = ±3 ⇒ x = 1 or x = 7.

b Since y = x + 2 +

5 a W e rewrite our unknown equation as a partial fraction: x 2 + λx = Ax + B + C x+1 x+1 Using that the oblique asymptote is x + 2, we can state that A = 1 and B = 2

Substituting these x values into the function yields the turning points (1, 0) and (7, 12). So the other turning point is at (7, 12).

Then x 2 + λx = x(x + 1) + 2(x + 1) + C = x 2 + 3x + 2 + C, giving λx = 3x So λ = 3

You don’t need to label the turning points unless asked to, but if you know where they are it can help you determine the shape of the curve.

The value of C can also be determined at this stage. By equating the constant coefficients, we see that 0 = 2 + C meaning C = −2.

11

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Cambridge International AS & A Level Further Mathematics: Worked Solutions Manual We are working with Cambridge Assessment International Education towards endorsement of this title. y c Points of intersection: when x = 0, y = − 1 and y = 0 4 when x = 1 5 1 x + –4 – y= 2

4

2

y

y=

x+

–√ 5 – 1

2

–

8 a Using long division:

1

4

1 4

x

−x − 2 2 ⟌ 3−x x −x+1 x 2 − 3x 2x + 1 2x − 6 7

2 7 a Let x + ax + b = Ax + B + C 2x − 1 2x − 1 Since Ax + B = 1 x + 5 , we have A = 1 , B = 5 2 2 4 4 Multiplying by (2x − 1) gives x 2 + ax + b = 1 x(2x − 1) + 5 (2x − 1) + C 2 4

Hence y = −x − 2 +

7 , where a = −1, b = −2, g = 7 3−x

A

Long division can sometimes be used to determine unknown coefficients.

Then ax = − 1 x + 5 x ⇒ a = 2 2 2 From the given point (0, 4), we have that b = −4

R

b From y = −x − 2 +

D

b To find the turning points, differentiate x 2 + 2x − 4 y= using the quotient rule 2x − 1 dy (2x − 1)(2x + 2) − (x 2 + 2x − 4) .2 So = dx (2x − 1) 2

7 , dy = −1 + 7 3 − x dx (3 − x) 2

dy = 0 implies that (x − 3) 2 = 7, so x = 3 ± √7 dx Hence there are two turning points.

Setting

c Points of intersect ion: y = 1 when x = 0 3 and y = 0 when x 2 − x + 1 = 0 For x 2 − x + 1 = 0, the discriminant is b 2 − 4ac = −3, so the curve does not cross the x-axis.

4x 2 + 2x − 2 − (2x 2 + 4x − 8) 2x 2 − 2x + 6 = (2x − 1) 2 (2x − 1) 2 dy Setting = 0 means that x 2 − x + 3 = 0 dx Find the discriminant: b 2 − 4ac = 1 − 12 = −11 =

y

1 3

Since this is < 0, there are no turning points. c Points of intersection: when x = 0, y = 4 and y = 0 when x 2 + 2x − 4 = 0. This means x=

x

√5 – 1

FT

–2 O

O 1 2

–2 O –2

x

3

−2 ± √4 + 16 = −1 ± √5 2 y=

The vertical asymptote is x = 1 2

–x – 2

12

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Chapter 2: Rational functions We are working with Cambridge Assessment International Education towards endorsement of this title.

EXERCISE 2C ⇒ 4x 2 + 11x + 7 > 0

2x − 3 > 3, consider (2x − 3)(x + 2) > 3(x + 2) 2 x+2 So 2x 2 + x − 6 > 3x 2 + 12x + 12

1 For ⇒

x 2 + 11x + 18 < 0

⇒

(x + 9)(x + 2) < 0

⇒ (4x + 7)(x + 1) > 0 Hence x < − 7 or x > −1 4 You should make a quick sketch of your quadratic to see which values you need.

Hence −9 < x < −2 You don’t have to sketch curves in these cases, but it will aid you greatly, especially the consideration of any vertical asymptotes.

Then x 2 − xy − 3y = 0

Then xy − 3y = x 2 + x − 3, or x 2 + (1 − y)x + (3y − 3) = 0

FT

2 First let y(x + 3) = x 2, so that x 2 = xy + 3y

5 a First let y(x − 3) = x 2 + x − 3

The discriminant, b 2 − 4ac, is negative exactly when (1 − y) 2 − (4)(1)(3y − 3) < 0...................[1]

This has a negative discriminant exactly when (−y) 2 − 4(1)(−3y) < 0

Simplifying, we see that y 2 − 14y + 13 < 0 This factorises as (y − 1)(y − 13) < 0 which means 1 < y < 13 are the values that the curve cannot take.

Simplifying yields y 2 + 12y < 0, which factorises as y(y + 12) < 0

So −12 < y < 0 represents the values that cannot occur.

Again, a simple sketch will help you to check your solution.

A

Thus, y < −12 and y > 0 are the possibly values for y.

You can also just consider b 2 − 4ac > 0 for the values you can have.

b By substituting y = 1 into [1], we obtain x 2 = 0, therefore (0, 1) is a turning point.

Also from [1]: y = 13 ⇒ x 2 − 12x + 36 = 0 ⇒ (x − 6) 2 = 0

R

3 First determine the number of turning points. x 2 − 2x − 4 To do this, we differentiate y = using the 3x − 2 quotient rule.

So there is another turning point at (6, 13) 2 3x − 2 − 4x 2 , then as |x | → ∞, y → −4x 2 x − 2 − 3x 2 −3x So the horizontal asymptote is y = 4 3 Hence k = 4 3 b From the denominator, 3x 2 − x + 2 = 0

6 a If y =

dy (3x − 2)(2x − 2) − (x 2 − 2x − 4) .3 = dx (3x − 2) 2 2 2 6x − 10x + 4 − (3x − 6x − 12) 3x 2 − 4x + 16 = = (3x − 2) 2 (3x − 2) 2

D

So

( )

2 2 44 + Since the numerator is 3x 2 − 4x + 16 = 3 x − , 3 3 dy > 0. we have dx This tells us that the numerator is always positive, hence it is never zero.

Then b 2 − 4ac = 1 − 24 = −23. Since the value of the discriminant is < 0, there are no solutions.

Hence there are no vertical asymptotes. c Let (x − 2 − 3x 2)y = 3x − 2 − 4x 2 Then (3y − 4)x 2 + (3 − y)x + (2y − 2) = 0

So there are no turning points.

This function has positive discriminant exactly when

Since there are also no horizontal asymptotes we conclude that all y values are possible. So y ∈R

(3 − y) 2 − 4(3y − 4)(2y − 2) > 0 ⇒ 9 − 6y + y 2 − 24y 2 + 56y − 32 > 0 ⇒ 23y 2 − 50y + 23 < 0

x−2 < 5 consider (x − 2)(x + 1) < 5(x + 1) 2 x+1 So x 2 − x − 2 < 5x 2 + 10x + 5

4 For

For the critical values, y= 13

50 ± √502 − 4 × 23 × 23 25 ± 4√6 = 2 × 23 23

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Cambridge International AS & A Level Further Mathematics: Worked Solutions Manual We are working with Cambridge Assessment International Education towards endorsement of this title.

Hence the range for the function is Hence the values of x such that y > 1 are exactly those where −2 < x < −√2 or 1 < x < √2 25 − 4√6 25 + 4√6 <y< 23 23 8 Let y = 2. Then 2x 2 + 6x − 20 = 1 − 3x ⇒ 2x 2 + 9x − 21 = 0 7 a F actorising the denominator, Using the quadratic formula, x y= 2 x = −9 ± √92 − 4(2)(−21) −9 ± √249 x + x − 2 (x + 2)(x − 1) x= = 2×2 4 For the horizontal asymptote we need to consider These are the values of x for which the curve meets the large values of x. line y = 2 As |x | → ∞, y → x2 . So the horizontal asymptote Factorising the denominator of the equation for C, x is y = 0 x 2 + 3x − 10 = (x + 5)(x − 2) Hence there are two vertical asymptotes, x = −5 and x=2

FT

From the denominator, the vertical asymptotes are x = −2 and x = 1

Next, differentiating using the quotient rule:

b To find the turning points, differentiate using the quotient rule. y= 2 x x +x−2 dy (x 2 + x − 2) .1 − x(2x + 1) x 2 + x − 2 − 2x 2 − x So = = dx (x 2 + x − 2) 2 (x 2 + x − 2) 2 2 −x − 2 = 2 (x + x − 2) 2

dy (x 2 + 3x − 10)(−3) − (1 − 3x)(2x + 3) = dx (x 2 + 3x − 10) 2

2 2 3x 2 − 2x + 27 = −3x − 9x2+ 30 + 6x +2 7x − 3 = 2 (x + 3x − 10) (x + 3x − 10) 2

( )

Since – (x 2 + 2) ≠ 0, there are no turning points.

Since 3 x − 1 3

A

Moreover the numerator of the derivative is always negative and the denominator is always positive, which means that the curve is always decreasing.

Hence y ∈R.

( )

From the numerator, 3x 2 − 2x + 27 = 3 x − 1 3 2

2

+ 80 3

+ 80 > 80 , we have that the numerator 3 3

is always positive.

Also, the denominator is always positive, and so the gradient is always positive.

R

If a rational function has no turning points, and it is valid for x ∈R, then it must be valid for all y-values.

x x2 + x − 2 Then x 2 + x − 2 = x simplifies to x 2 = 2

D

c Let y = 1 so that 1 =

By considering y > 1 we get x 2 > 2 and so x < −√2 or x > √2

Hence, for y < 2, we require x <

−5 < x <

−9 + √249 or x > 2 4

−9 − √249 , 4

For harder examples such as this one, a sketch will help to ensure that all of the required regions are found.

By factorising the denominator, we see that the curve has vertical asymptotes at x = −2 and x = 1

EXERCISE 2D 1 a B ecause the equation has an absolute value sign around every x term, the curve must be symmetrical about the y-axis. This means that we can work with exclusively positive x: for each solution x that we then find, there will be another solution –x. Since every solution has a negative pair, we will also find all solutions by restricting our search to x > 0.

The left hand side of the equation is 0 exactly when the numerator is 0. We therefore want to solve x 2 − 2x − 3 = 0, which factorises as (x − 3)(x + 1) = 0.

With our restriction that x > 0, we now have just one solution for our quadratic. This is x = 3.

14

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Chapter 2: Rational functions We are working with Cambridge Assessment International Education towards endorsement of this title.

So the two solutions are (3, 0) and (−3, 0) Hence the curve has one turning point, occurring when x = 1 2 Recall that all curves of the form y = f (|x |) have reflective symmetry in the line x = 0, i.e. they are symmetrical about the y-axis.

Points of intersection: when x = 0, y = − 1 and y ≠ 0 for 6 any value of x. y

b Since we are dealing with y = f (|x |) , we can restrict to |x |2 − 3|x | + 2 x 2 − 3x + 2 x > 0 and rewrite as x−1 |x | − 1 (x − 1)(x − 2) x 2 − 3x + 2 = 0 factorises as =0 x−1 x−1 Since the denominator contains |x | − 1, the equation is not defined for x = 1 or x = −1. This implies that x = 2 is the only positive solution to our equation.

Now

O –1 6

3

FT

–2

x

Since the curve is symmetrical about the y-axis, the two solutions are (2, 0) and (−2, 0) 2 a From the denominator, x 2 − 5x − 6 = (x + 1)(x − 6) Thus there are two values of x such that the denominator x 2 − 5x − 6 is 0

Since the intersection values for y = x 2 − x − 6 are (−2, 0), (3, 0) and (0, −6), we use these to aid us with the sketch of the reciprocal of this curve.

Hence there are two vertical asymptotes.

A

b From the denominator, x 2 + 2x + 3 = (x + 1) 2 + 2

Since (x + 1) 2 + 2 > 2, there are no values of x such that the denominator is 0.

5 Points of intersection: when x = 0, y = ±√7 and when y = 0, x = 7 3 Then consider the curve as y = √7 − 3x and y = −√7 − 3x

Hence there are no vertical asymptotes.

1 = 1 x 2 − 4 (x + 2)(x − 2) so there are two vertical asymptotes, x = −2 and x = 2

R

3 a Factorising the denominator,

y

As |x | → ∞, y → 12 . The horizontal asymptote is y = 0. x 1 1 b Since 2 = , the denominator x + 2x + 5 (x + 1) 2 + 4 is always positive.

D

√7

Therefore the function has no vertical asymptotes.

O

As |x | → ∞, y → 12 . So the horizontal asymptote x is y = 0

7 3

x

–√ 7

1 4 Factorising the denominator, 2 1 = x − x − 6 (x − 3)(x + 2) So there are two vertical asymptotes, x = −2 and x = 3 As |x | → ∞, y → 12 . So the horizontal asymptote is y = 0 x Differentiating using the quotient rule: dy (x 2 − x − 6) .0 − 1. (2x − 1) 1 − 2x = = 2 dx (x 2 − x − 6) 2 (x − x − 6) 2

x x . So the vertical = x 2 − x − 2 (x − 2)(x + 1) asymptotes are x = −1 and x = 2 As |x | → ∞, y → x2 . So the horizontal asymptote x is y = 0

6 a Let

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Cambridge International AS & A Level Further Mathematics: Worked Solutions Manual We are working with Cambridge Assessment International Education towards endorsement of this title.

Points of intersection: when x = 0, y = 0 x 2 + 2x − 5 b Let = Ax + B + C x + 3 x+3 Recall that y = | f(x)| makes all negative branches 2 As |x | → ∞, y → xx . So A = 1.

positive y

Then x 2 + 2x − 5 = x(x + 3) + B(x + 3) + C Setting x = −3 yields C = −2 and similarly setting x = 0 we see that −5 = 3B − 2 ⇒ B = −1 2 x+3 So the oblique asymptote is y = x − 1 and the vertical asymptote is x = −3

Hence y = x − 1 −

–1

O

x

2

c Recall that y = | f(x)| makes all negative branches positive.

FT

b Recall that y = | f(x)| is the graph y = f(x) for x > 0 and has the other x values defined by reflecting this graph in the y-axis.

Points of intersection: when x = 0, y = − 5 , but since 3 this is | f(x)| the intersection point is 0, 5 3

( )

From a we recall that we have an intersection point at (0, 0). We also know what the graph of y = f(x) looks like. Using this graph for x > 0 we can then reflect it in the y-axis.

When y = 0, x 2 + 2x − 5 = 0, so x = = −1 ± √6

For the asymptotes, we then have that y = 0 is the horizontal asymptote and x = −2, x = 2 are the vertical asymptotes.

A

y

y=

R 7 a U sing the quotient rule dy (x + 3)(2x + 2) − (x 2 + 2x − 5) = dx (x + 3) 2 =

1

5 3

x

D

2

x–

1

O

+

–2

y=

–x

y

−2 ± √4 + 20 2

– √6 – 1 – 3

√6 – 1

x

First sketch the graph in rough, noting any negative branches that are present, then reflect all negative sections, including any oblique asymptotes.

2x 2 + 8x + 6 − x 2 − 2x + 5 x 2 + 6x + 11 = (x + 3) 2 (x + 3) 2

Since the numerator is x 2 + 6x + 11 = (x + 3) 2 + 2, we see that dy (x + 3) 2 + 2 (x + 3) 2 2 = = + dx (x + 3) 2 (x + 3) 2 (x + 3) 2 2 =1+ >1 (x + 3) 2 Hence there are no turning points.

16

Original material © Cambridge University Press 2020. This material is not final and is subject to further changes prior to publication.

D

R

A

FT

We are working with Cambridge Assessment International Education towards endorsement of this title.

Original material Š Cambridge University Press 2020. This material is not final and is subject to further changes prior to publication.

D

R

A

FT

We are working with Cambridge Assessment International Education towards endorsement of this title.

Original material Š Cambridge University Press 2020. This material is not final and is subject to further changes prior to publication.