Mathematics HL Core Worked Solutions · International Baccalaureate Mathematics HL oreMathematics HL ore(C ) Also suitable for HL & SL combined classes ... TRIGONOMETRY WITH RIGHTANGLED

Haese Harris Publications&

Roger Dixon

Valerie Frost

Robert Haese

Michael Haese

Sandra Haese

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RK

ED

SO

LU

TIO

NS

WO

RK

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LU

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HAESE HARRIS PUBLICATIONS&

for the international studentfor the international studentMathematicsMathematics

Paul Urban

John Owen

Robert Haese

Sandra Haese

Mark Bruce

Paul Urban

John Owen

Robert Haese

Sandra Haese

Mark Bruce

Diploma ProgrammeDiploma ProgrammeInternational BaccalaureateInternational Baccalaureate

Mathematics HL ore(C )Mathematics HL ore(C )Also suitable for HL & SL combined classesAlso suitable for HL & SL combined classes

IBHL_WS

MATHEMATICS FOR THE INTERNATIONAL STUDENTMathematics HL (Core) – WORKED SOLUTIONS

This book is copyright

Copying for educational purposes

Acknowledgements

Disclaimer

International Baccalaureate Diploma Programme

Roger DixonValerie Frost B.Sc., Dip.Ed.Robert HaeseMichael HaeseSandra Haese

Haese & Harris Publications3 Frank Collopy Court, Adelaide Airport, SA 5950, AUSTRALIATelephone: +61 8 8355 9444, Fax: + 61 8 8355 9471Email:

National Library of Australia Card Number & ISBN 1 876543 45 0

© Haese & Harris Publications 2005

Published by Raksar Nominees Pty Ltd3 Frank Collopy Court, Adelaide Airport, SA 5950, AUSTRALIA

First Edition 2005

Cover design by Piotr PoturajComputer software by David Purton

Typeset in Australia by Susan Haese (Raksar Nominees). Typeset in Times Roman 9/10

The textbook, its accompanying CD and this book of fully worked solutions have beendeveloped independently of the International Baccalaureate Organization (IBO). Thesepublications are in no way connected with, or endorsed by, the IBO.

. Except as permitted by the CopyrightAct (any fair dealing for the purposesof private study, research, criticism or review), no part of this publication may be reproduced, storedin a retrieval system, or transmitted in any form or by any means, electronic, mechanical,photocopying, recording or otherwise, without the prior permission of the publisher. Enquiries to bemade to Haese & Harris Publications.

: Where copies of part or the whole of the book are made underPart VB of the Copyright Act, the law requires that the educational institution or the body thatadministers it has given a remuneration notice to Copyright Agency Limited (CAL). Forinformation, contact the CopyrightAgency Limited.

: While every attempt has been made to trace and acknowledge copyright, theauthors and publishers apologise for any accidental infringement where copyright has proveduntraceable. They would be pleased to come to a suitable agreement with the rightful owner.

: All the internet addresses (URL’s) given in this book were valid at the time of printing.While the authors and publisher regret any inconvenience that changes of address may causereaders, no responsibility for any such changes can be accepted by either the authors or thepublisher.

B.Ed.

B.Sc.B.Sc.Hons., Ph.D.B.Sc.

\Qw_

[email protected]: www.haeseandharris.com.au

IBSL_WS

FOREWORD

This book gives you fully worked solutions for every question in each chapter of the Haese & Harris

Publications textbook which is one of three textbooks in our series

‘Mathematics for the International Student’. The other two textbooks are and

, and books of fully worked solutions are available for those textbooks

also.

Correct answers can sometimes be obtained by different methods. In this book, where applicable,

each worked solution is modeled on the worked example in the textbook.

Be aware of the limitations of calculators and computer modelling packages. Understand that when

your calculator gives an answer that is different from the answer you find in the book, you have not

necessarily made a mistake, but the book may not be wrong either.

We have a list of errata for on our website. Please contact us if you have

any additions to this list.

e-mail:

web:

Mathematics HL (Core)

Mathematics SL

Mathematical Studies SL

Mathematics HL (Core)

RLD VF

RCH PMH SHH

[email protected]

www.haeseandharris.com.au

IBSL_WS

TABLE OF CONTENTS

BACKGROUND KNOWLEDGE 5

Chapter 1 FUNCTIONS 33

Chapter 2 SEQUENCES & SERIES 48

Chapter 3 EXPONENTS 73

Chapter 4 LOGARITHMS 89

Chapter 5 NATURAL LOGARITHMS 105

Chapter 6 GRAPHING AND TRANSFORMING FUNCTIONS 114

Chapter 7 QUADRATIC EQUATIONS AND FUNCTIONS 128

Chapter 8 COMPLEX NUMBERS AND POLYNOMIALS 171

Chapter 9 COUNTING AND BINOMIAL THEOREM 218

Chapter 10 MATHEMATICAL INDUCTION 232

BACKGROUND KNOWLEDGE - TRIGONOMETRY WITH RIGHT ANGLED TRIANGLES 249

Chapter 11 THE UNIT CIRCLE AND RADIAN MEASURE 264

Chapter 12 NON RIGHT ANGLED TRIANGLE TRIGONOMETRY 278

Chapter 13 PERIODIC PHENOMENA 286

Chapter 14 MATRICES 323

Chapter 15 VECTORS IN 2 AND 3 DIMENSIONS 377

Chapter 16 COMPLEX NUMBERS 420

Chapter 17 LINES AND PLANES IN SPACE 450

Chapter 18 DESCRIPTIVE STATISTICS 496

Chapter 19 PROBABILITY 517

Chapter 20 INTRODUCTION TO CALCULUS 539

Chapter 21 DIFFERENTIAL CALCULUS 546

Chapter 22 APPLICATIONS OF DIFFERENTIAL CALCULUS 583

Chapter 23 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS 624

Chapter 24 DERIVATIVES OF CIRCULAR FUNCTIONS AND RELATED RATES 646

Chapter 25 INTEGRATION 672

Chapter 26 INTEGRATION (AREAS AND OTHER APPLICATIONS) 693

Chapter 27 CIRCULAR FUNCTION INTEGRATION 719

Chapter 28 VOLUMES OF REVOLUTION 731

Chapter 29 FURTHER INTEGRATION AND DIFFERENTIAL EQUATIONS 739

Chapter 30 STATISTICAL DISTRIBUTIONS 761

IBSL_WS

Background knowledge

EXERCISE A

1 ap3£p5

=p3£ 5

=p15

b (p3)2

=p3£p3

= 3

c 2p2£p2

= 2(p2£p2)

= 2£ 2= 4

d 3p2£ 2p2

= (3£ 2)(p2£p2)= 6£ 2= 12

e 3p7£ 2p7

= (3£ 2)(p7£p7)= 6£ 7= 42

f

p12p2

=

r12

2

=p6

g

p12p6

=

r12

6

=p2

h

p18p3

=

r18

3

=p6

2 a 2p2 + 3

p2

= (2 + 3)p2

= 5p2

b 2p2¡ 3p2

= (2¡ 3)p2= ¡p2

c 5p5¡ 3p5

= (5¡ 3)p5= 2

p5

d 5p5 + 3

p5

= (5 + 3)p5

= 8p5

e 3p5¡ 5p5

= (3¡ 5)p5= ¡2p5

f 7p3 + 2

p3

= (7 + 2)p3

= 9p3

g 9p6¡ 12p6

= (9¡ 12)p6= ¡3p6

hp2 +

p2 +

p2

= 3£p2= 3

p2

3 ap8

=p4£ 2

=p4£p2

= 2p2

bp12

=p4£ 3

=p4£p3

= 2p3

cp20

=p4£ 5

=p4£p5

= 2p5

dp32

=p16£ 2

=p16£p2

= 4p2

ep27

=p9£ 3

=p9£p3

= 3p3

fp45

=p9£ 5

=p9£p5

= 3p5

gp48

=p16£ 3

=p16£p3

= 4p3

hp54

=p9£ 6

=p9£p6

= 3p6

ip50

=p25£ 2

=p25£p2

= 5p2

jp80

=p16£ 5

=p16£p5

= 4p5

kp96

=p16£ 6

=p16£p6

= 4p6

lp108

=p36£ 3

=p36£p3

= 6p3

4 a 4p3¡p12

= 4p3¡p4£ 3

= 4p3¡ 2£p3

= 4p3¡ 2p3

= 2p3

b 3p2 +

p50

= 3p2 +

p25£ 2

= 3p2 + 5£p2

= 3p2 + 5

p2

= 8p2

c 3p6 +

p24

= 3p6 +

p4£ 6

= 3p6 + 2£p6

= 3p6 + 2

p6

= 5p6

d 2p27 + 2

p12

= 2p9£ 3 + 2p4£ 3

= 6p3 + 4

p3

= 10p3

ep75¡p12

=p25£ 3¡p4£ 3

= 5p3¡ 2p3

= 3p3

fp2 +

p8¡p32

=p2 +

p4£ 2¡p16£ 2

=p2 + 2

p2¡ 4p2

= ¡p2

IBHL_WS

6 Mathematics HL – BACKGROUND KNOWLEDGE

5 a 1p2

= 1p2£

p2p2

=p22

b 6p3

= 6p3£

p3p3

= 6p3

3

= 2p3

c 7p2

= 7p2£

p2p2

= 7p2

2

d 10p5

= 10p5£

p5p5

= 10p5

5

= 2p5

e 10p2

= 10p2£

p2p2

= 10p2

2

= 5p2

f 18p6

= 18p6£

p6p6

= 18p6

6

= 3p6

g 12p3

= 12p3£

p3p3

= 12p3

3

= 4p3

h 5p7

= 5p7£

p7p7

= 5p7

7

i 14p7

= 14p7£

p7p7

= 14p7

7

= 2p7

j 2p3p2

= 2p3p2£

p2p2

= 2p6

2

=p6

EXERCISE B

1 a 259

= 2:59£ 102= 2:59£ 102

b 259 000

= 2:590 00£ 105= 2:59£ 105

c 2:59

= 2:59£ 1= 2:59£ 100

d 0:259

= 02:59¥ 10= 2:59£ 10¡1

e 0:000 259

= 00002:59¥ 104= 2:59£ 10¡4

f 40:7

= 4:07£ 10= 4:07£ 101

g 4070

= 4:070£ 103= 4:07£ 103

h 0:0407

= 004:07¥ 102= 4:07£ 10¡2

i 407 000

= 4:070 00£ 105= 4:07£ 105

j 407 000 000

= 4:070 000 00£ 108= 4:07£ 108

2 a 149 500 000 000 m

= 1:49 500 000 000£ 1011= 1:495£ 1011 m

b 0:0003 mm

= 0003:£ 10¡4= 3£ 10¡4 mm

c 0:001 mm

= 001:£ 10¡3= 1£ 10¡3 mm

d 15 million degrees

= 15 000 000 oC

= 1:500 000 0£ 107 oC= 1:5£ 107 oC

e 300 000 times

= 3£ 100 000= 3£ 105 times

3 a 4£ 103= 4£ 1000= 4000

b 5£ 102= 5£ 100= 500

c 2:1£ 103

= 2:100£ 103= 2100

d 7:8£ 104

= 7:8000£ 104= 78 000

e 3:8£ 105

= 3:800 00£ 105= 380 000

f 8:6£ 101

= 8:6£ 10= 86

g 4:33£ 107

= 4:330 000 0£ 107= 43 300 000

h 6£ 107= 6£ 10 000 000= 60 000 000

k 0:000 040 7

= 000004:07¥ 105= 4:07£ 10¡5

IBHL_WS

Mathematics HL – BACKGROUND KNOWLEDGE 7

4 a 4£ 10¡3

= 004:¥ 103= 0:004

b 5£ 10¡2

= 05:¥ 102= 0:05

c 2:1£ 10¡3

= 002:1¥ 103= 0:0021

d 7:8£ 10¡4

= 0007:8¥ 104= 0:000 78

e 3:8£ 10¡5

= 00003:8¥ 105= 0:000 038

f 8:6£ 10¡1

= 8:6¥ 101= 0:86

g 4:33£ 10¡7

= 0000004:33¥ 107= 0:000 000 433

h 6£ 10¡7

= 0000006:¥ 107= 0:000 000 6

6 a (3:42£ 105)£ (4:8£ 104)= (3:42£ 4:8)£ (105 £ 104)= 16:416£ 109= 1:6416£ 1010= 1:64£ 1010 (2 d.p.)

b (6:42£ 10¡2)2= (6:42)2 £ (10¡2)2= 41:2164£ 10¡4= 4:121 64£ 10¡3= 4:12£ 10¡3 (2 d.p.)

c3:16£ 10¡106£ 107

=3:16

6£ 10

¡10

107

= 0:52¹6£ 10¡17= 5:2¹6£ 10¡18= 5:27£ 10¡18 (2 d.p.)

d (9:8£ 10¡4)¥ (7:2£ 10¡6)

=9:8£ 10¡47:2£ 10¡6

=9:8

7:2£ 10

¡4

10¡6

= 1:36¹1£ 102= 1:36£ 102 (2 d.p.)

e1

3:8£ 105

=1

3:8£ 10¡5

= 2:63£ 10¡6 (2 d.p.)

f (1:2£ 103)3= (1:2)3 £ (103)3= 1:728£ 109= 1:73£ 109 (2 d.p.)

7 a 1 day = 24 hours

i.e., missile travels 5400£ 24= 129 600

= 1:296£ 105+ 1:30£ 105 km

b 1 week = 7 days

= 7£ 24 hours= 168 hours

i.e., missile travels 5400£ 168= 907 200

= 9:072£ 105+ 9:07£ 105 kmc 2 years = 2£ 365:25 days

= 730:5 days

= 730:5£ 24 hours= 17 532 hours

i.e., missile travels 5400£ 17 532= 94 672 800

= 9:467 28£ 107+ 9:47£ 107 km

5 a 9£ 10¡7 m= 0000009:¥ 107= 0:000 000 9 m

b 6:130£ 109 people= 6:130 000 000£ 109= 6130 000 000 people

c 1£ 105 light years= 1£ 100 000= 100 000 light years

d 1£ 10¡5 mm= 00001:¥ 105= 0:000 01 mm

IBHL_WS


8 a distance = speed £ timetime = 1 minute = 60 seconds

so, light travels (3£ 108)£ 60= 180£ 108= 1:80£ 1010 m

b distance = speed £ timetime = 1 day = 24 hours

= 24£ 60£ 60 seconds= 86 400 seconds

= 8:64£ 104 secondsi.e., light travels (3£ 108)£ (8:64£ 104)

= 3£ 8:64£ 1012= 25:92£ 1012+ 2:59£ 1013 m

c distance = speed £ timetime = 1 year = 365:25 days

= 365:25£ 8:64£ 104 sec ffrom bg= 3155:76£ 104+ 3:16£ 107 sec

i.e., light travels (3£ 108)£ (3:156£ 107)= 3£ 3:156£ 1015= 9:468£ 1015+ 9:47£ 1015 m

EXERCISE C

1 a fx : x > 5g reads ‘the set of all x such that x is greater than 5’b fx : x 6 3g reads ‘the set of all x such that x is less than or equal to 3’c fy : 0 < y < 6g reads ‘the set of all y such that y lies between 0 and 6’d fx : 2 6 x 6 4g reads ‘the set of all x such that x is greater than or equal to 2, but less

than or equal to 4’

e ft : 1 < t < 5g reads ‘the set of all t such that t lies between 1 and 5’f fn : n < 2 or n > 6g reads ‘the set of all n such that n is less than 2 or greater than or

equal to 6’

2 a fx : x > 2g b fx : 1 < x 6 5g c fx : x 6 ¡2 or x > 3gd fx : x 2 Z, ¡1 6 x 6 3g e fx : x 2 Z, 0 6 x 6 5g f fx : x < 0g

3 a b

c d

e

EXERCISE D

1 a 3x+ 7x¡ 10= 10x¡ 10

b 3x+ 7x¡ x= 9x

c 2x+ 3x+ 5y

= 5x+ 5y

d 8¡ 6x¡ 2x= 8¡ 8x

e 7ab+ 5ba

= 7ab+ 5ab

= 12ab

f 3x2 + 7x3

= 3x2 + 7x3

i.e., cannot be simplified

2 a 3(2x+ 5) + 4(5 + 4x)

= 6x+ 15 + 20 + 16x

= 22x+ 35

b 6¡ 2(3x¡ 5)= 6¡ 6x+ 10= 16¡ 6x

2 3 4 5 6 7 8 9 10 ��

��

�

��

IBHL_WS


c 5(2a¡ 3b)¡ 6(a¡ 2b)= 10a¡ 15b¡ 6a+ 12b= 4a¡ 3b

d 3x(x2 ¡ 7x+ 3)¡ (1¡ 2x¡ 5x2)= 3x3 ¡ 21x2 + 9x¡ 1 + 2x+ 5x2= 3x3 ¡ 16x2 + 11x¡ 1

3 a 2x(3x)2

= 2x£ 9x2= 18x3

b3a2b3

9ab4

=a

3b

cp16x4

=p16£

px4

= 4£p(x2)2

= 4x2

d (2a2)3 £ 3a4= 23 £ (a2)3 £ 3a4= 8£ a6 £ 3a4= 24a10

EXERCISE E

1 a 2x+ 5 = 25

) 2x = 20

) x = 10

b 3x¡ 7 > 11) 3x > 18

) x > 6

c 5x+ 16 = 20

) 5x = 4

) x = 45

dx

3¡ 7 = 10

)x

3= 17

) x = 51

e 6x+ 11 < 4x¡ 9) 2x < ¡20) x < ¡10

f3x¡ 25

= 8

) 3x¡ 2 = 40) 3x = 42

) x = 14

g 1¡ 2x > 19) ¡2x > 18) 2x 6 ¡18) x 6 ¡9

h 12x+ 1 = 2

3x¡ 2

) 36x¡ 46x = ¡3) ¡ 1

6x = ¡3

) x = 18

i2

3¡ 3x4= 1

2(2x¡ 1) Multiplying each term by

the LCD of 12 gives) 8¡ 9x = 6(2x¡ 1)) 8¡ 9x = 12x¡ 6

) 14 = 21x i.e., x = 23

2 a x+ 2y = 9 ..... (1)

x¡ y = 3 ..... (2)Multiplying (2) by 2 gives

x+ 2y = 9

2x¡ 2y = 6) 3x = 15 faddingg

) x = 5

Substituting x = 5 into (2) gives

5¡ y = 3) y = 2

) x = 5 and y = 2

b 2x+ 5y = 28 ..... (1)

x¡ 2y = 2 ..... (2)Multiplying (2) by ¡2 gives2x+ 5y = 28

¡2x+ 4y = ¡4) 9y = 24 faddingg) y = 24

9= 8

3

Substituting y = 83

into (2) gives

x¡ 2( 83 ) = 2 ) x¡ 163 = 2 and so x = 223) x = 22

3and y = 8

3

c 7x+ 2y = ¡4 ..... (1)3x+ 4y = 14 ..... (2)

Multiplying (1) by ¡2 gives¡14x¡ 4y = 83x+ 4y = 14

) ¡11x = 22 faddingg) x = ¡2

Substituting x = ¡2 into (2) gives3(¡2) + 4y = 14) ¡6 + 4y = 14

) 4y = 20 and ) y = 5

) x = ¡2 and y = 5

d 5x¡ 4y = 27 ..... (1)3x+ 2y = 9 ..... (2)

Multiplying (2) by 2 gives

5x¡ 4y = 276x+ 4y = 18

) 11x = 45 faddingg) x = 45

11

Substituting x = 4511

into (1) gives

5( 4511)¡ 4y = 27 ) 225

11¡ 27 = 4y

) 4y = ¡ 7211 and ) y = ¡ 1811

) x = 4511

and y = ¡ 1811

=3£ a£ a£ b£ b£ b

3£ 3£ a£ b£ b£ b£ b

IBHL_WS


e x+ 2y = 5 ..... (1)

2x+ 4y = 1 ..... (2)

Multiplying (1) by ¡2 gives¡2x¡ 4y = ¡102x+ 4y = 1

) 0 = ¡9 faddinggwhich is absurd

) there are no solutions

fx

2+y

3= 5 ..... (1)

x

3+y

4= 1 ..... (2)

Multiplying (1) by 18 and (2) by ¡24 gives9x+ 6y = 90 ..... (3)

¡8x¡ 6y = ¡24) x = 66 faddingg

Substituting x = 66 into (3) gives

9£ 66 + 6y = 90) 6y = 90¡ 594 = ¡504) y = ¡84

) x = 66 and y = ¡84

EXERCISE F

1 a 5¡ (¡11)= 5 + 11

= 16

b j5j ¡ j¡11j= 5¡ 11= ¡6

c j5¡ (¡11)j= j5 + 11j= j16j= 16

d¯̄(¡2)2 + 11(¡2)

¯̄= j4¡ 22j= j¡18j= 18

e j¡6j ¡ j¡8j= 6¡ 8= ¡2

f j¡6¡ (¡8)j= j¡6 + 8j= j2j= 2

2 a jaj = j¡2j= 2

b jbj = j3j= 3

c jaj jbj = j¡2j j3j= 2£ 3= 6

d jabj = j¡2£ 3j= j¡6j= 6

e ja¡ bj = j¡2¡ 3j= j¡5j= 5

f jaj ¡ jbj = j¡2j ¡ j3j= 2¡ 3= ¡1

g ja+ bj = j¡2 + 3j= j1j= 1

h jaj+ jbj = j¡2j+ j3j= 2 + 3

= 5

i jaj2 = j¡2j2

= 22

= 4

j a2 = (¡2)2 = 4 k¯̄̄c

a

¯̄̄=

¯̄̄¡4¡2¯̄̄= j2j = 2 l jcjjaj =

j¡4jj¡2j =

4

2= 2

3 a jxj = 3) x = §3

b jxj = ¡5but jxj > 0 for all x

(property of modulus)

) no solution

c jxj = 0) x = 0

d jx¡ 1j = 3) x¡ 1 = §3

) x = 1§ 3) x = ¡2 or 4

e j3¡ xj = 4) 3¡ x = §4) ¡x = ¡3§ 4) x = 3¨ 4) x = ¡1 or 7

f jx+ 5j = ¡1but jx+ 5j > 0 for all x

(property of modulus)

) no solution

IBHL_WS


g j3x¡ 2j = 1) 3x¡ 2 = §1

) 3x = 2§ 1) 3x = 3 or 1

) x = 13 or 1

h j3¡ 2xj = 3) 3¡ 2x = §3

) 2x = 3¨ 3) 2x = 0 or 6

) x = 0 or 3

i j2¡ 5xj = 12) 2¡ 5x = §12

) 5x = 2¨ 12) 5x = ¡10 or 14) x = ¡2 or 14

5

EXERCISE G

1 a (2x+ 3)(x+ 1)

= 2x2 + 2x+ 3x+ 3

= 2x2 + 5x+ 3

b (3x+ 4)(x+ 2)

= 3x2 + 6x+ 4x+ 8

= 3x2 + 10x+ 8

c (5x¡ 2)(2x+ 1)= 10x2 + 5x¡ 4x¡ 2= 10x2 + x¡ 2

d (x+ 2)(3x¡ 5)= 3x2 ¡ 5x+ 6x¡ 10= 3x2 + x¡ 10

e (7¡ 2x)(2 + 3x)= 14 + 21x¡ 4x¡ 6x2= ¡6x2 + 17x+ 14

f (1¡ 3x)(5 + 2x)= 5 + 2x¡ 15x¡ 6x2= ¡6x2 ¡ 13x+ 5

g (3x+ 4)(5x¡ 3)= 15x2 ¡ 9x+ 20x¡ 12= 15x2 + 11x¡ 12

h (1¡ 3x)(2¡ 5x)= 2¡ 5x¡ 6x+ 15x2= 15x2 ¡ 11x+ 2

i (7¡ x)(3¡ 2x)= 21¡ 14x¡ 3x+ 2x2= 2x2 ¡ 17x+ 21

j (5¡ 2x)(3¡ 2x)= 15¡ 10x¡ 6x+ 4x2= 4x2 ¡ 16x+ 15

k ¡(x+ 1)(x+ 2)= ¡(x2 + 2x+ x+ 2)= ¡(x2 + 3x+ 2)= ¡x2 ¡ 3x¡ 2

l ¡2(x¡ 1)(2x+ 3)= ¡2(2x2 + 3x¡ 2x¡ 3)= ¡2(2x2 + x¡ 3)= ¡4x2 ¡ 2x+ 6

2 a (x+ 6)(x¡ 6)= x2 ¡ 62= x2 ¡ 36

b (x+ 8)(x¡ 8)= x2 ¡ 82= x2 ¡ 64

c (2x¡ 1)(2x+ 1)= (2x)2 ¡ 12= 4x2 ¡ 1

d (3x¡ 2)(3x+ 2)= (3x)2 ¡ 22= 9x2 ¡ 4

e (4x+ 5)(4x¡ 5)= (4x)2 ¡ 52= 16x2 ¡ 25

f (5x¡ 3)(5x+ 3)= (5x)2 ¡ 32= 25x2 ¡ 9

g (3¡ x)(3 + x)= 32 ¡ x2= 9¡ x2

h (7¡ x)(7 + x)= 72 ¡ x2= 49¡ x2

i (7 + 2x)(7¡ 2x)= 72 ¡ (2x)2= 49¡ 4x2

j (x+p2)(x¡p2)

= x2 ¡ (p2)2= x2 ¡ 2

k (x+p5)(x¡p5)

= x2 ¡ (p5)2= x2 ¡ 5

l (2x¡p3)(2x+p3)= (2x)2 ¡ (p3)2= 4x2 ¡ 3

3 a (x+ 5)2

= x2 + 2(x)(5) + 52

= x2 + 10x+ 25

b (x+ 7)2

= x2 + 2(x)(7) + 72

= x2 + 14x+ 49

c (x¡ 2)2= x2 ¡ 2(x)(2) + 22= x2 ¡ 4x+ 4

d (x¡ 6)2= x2 ¡ 2(x)(6) + 62= x2 ¡ 12x+ 36

e (3 + x)2

= 32 + 2(3)(x) + x2

= x2 + 6x+ 9

f (5 + x)2

= 52 + 2(5)(x) + x2

= x2 + 10x+ 25

g (11¡ x)2= 112 ¡ 2(11)(x) + x2= x2 ¡ 22x+ 121

h (10¡ x)2= 102 ¡ 2(10)(x) + x2= x2 ¡ 20x+ 100

i (2x+ 7)2

= (2x)2 + 2(2x)(7) + 72

= 4x2 + 28x+ 49

j (3x+ 2)2

= (3x)2 + 2(3x)(2) + 22

= 9x2 + 12x+ 4

k (5¡ 2x)2= 52 ¡ 2(5)(2x) + (2x)2= 4x2 ¡ 20x+ 25

l (7¡ 3x)2= 72 ¡ 2(7)(3x) + (3x)2= 9x2 ¡ 42x+ 49

IBHL_WS


4 a y = 2(x+ 2)(x+ 3)

= 2(x2 + 3x+ 2x+ 6)

= 2(x2 + 5x+ 6)

= 2x2 + 10x+ 12

c y = ¡(x+ 1)(x¡ 7)= ¡(x2 ¡ 7x+ x¡ 7)= ¡(x2 ¡ 6x¡ 7)= ¡x2 + 6x+ 7

e y = 4(x¡ 1)(x¡ 5)= 4(x2 ¡ 5x¡ x+ 5)= 4(x2 ¡ 6x+ 5)= 4x2 ¡ 24x+ 20

g y = ¡5(x¡ 1)(x¡ 6)= ¡5(x2 ¡ 6x¡ x+ 6)= ¡5(x2 ¡ 7x+ 6)= ¡5x2 + 35x¡ 30

i y = ¡ 52(x¡ 4)2

= ¡ 52 (x2 ¡ 2(x)(4) + 42)= ¡ 52 (x2 ¡ 8x+ 16)= ¡ 5

2x2 + 20x¡ 40

b y = 3(x¡ 1)2 + 4= 3(x2 ¡ 2(x)(1) + 12) + 4= 3(x2 ¡ 2x+ 1) + 4= 3x2 ¡ 6x+ 3 + 4= 3x2 ¡ 6x+ 7

d y = ¡(x+ 2)2 ¡ 11= ¡(x2 + 2(x)(2) + 22)¡ 11= ¡(x2 + 4x+ 4)¡ 11= ¡x2 ¡ 4x¡ 4¡ 11= ¡x2 ¡ 4x¡ 15

f y = ¡ 12(x+ 4)2 ¡ 6

= ¡ 12(x2 + 2(x)(4) + 42)¡ 6

= ¡ 12(x2 + 8x+ 16)¡ 6

= ¡ 12x2 ¡ 4x¡ 8¡ 6

= ¡ 12x

2 ¡ 4x¡ 14h y = 12 (x+ 2)

2 ¡ 6= 12 (x

2 + 2(x)(2) + 22)¡ 6= 1

2(x2 + 4x+ 4)¡ 6

= 12x2 + 2x+ 2¡ 6

= 12x2 + 2x¡ 4

5 a 1 + 2(x+ 3)2

= 1 + 2(x2 + 2(x)(3) + 32)

= 1 + 2(x2 + 6x+ 9)

= 1 + 2x2 + 12x+ 18

= 2x2 + 12x+ 19

b 2 + 3(x¡ 2)(x+ 3)= 2 + 3(x2 + 3x¡ 2x¡ 6)= 2 + 3(x2 + x¡ 6)= 2 + 3x2 + 3x¡ 18= 3x2 + 3x¡ 16

c 3¡ (3¡ x)2= 3¡ (9¡ 2(3)(x) + x2)= 3¡ (x2 ¡ 6x+ 9)= 3¡ x2 + 6x¡ 9= ¡x2 + 6x¡ 6

d 5¡ (x+ 5)(x¡ 4)= 5¡ (x2 ¡ 4x+ 5x¡ 20)= 5¡ (x2 + x¡ 20)= 5¡ x2 ¡ x+ 20= ¡x2 ¡ x+ 25

e 1 + 2(4¡ x)2= 1 + 2(42 ¡ 2(4)(x) + x2)= 1 + 2(x2 ¡ 8x+ 16)= 1 + 2x2 ¡ 16x+ 32= 2x2 ¡ 16x+ 33

f x2 ¡ 3x¡ (x+ 2)(x¡ 2)= x2 ¡ 3x¡ (x2 ¡ 22)= x2 ¡ 3x¡ x2 + 4= ¡3x+ 4

g (x+ 2)2 ¡ (x+ 1)(x¡ 4)= x2 + 2(x)(2) + 22 ¡ (x2 ¡ 4x+ x¡ 4)= x2 + 4x+ 4¡ (x2 ¡ 3x¡ 4)= x2 + 4x+ 4¡ x2 + 3x+ 4= 7x+ 8

h (2x+ 3)2 + 3(x+ 1)2

= (2x)2 + 2(2x)(3) + 32 + 3(x2 + 2(x)(1) + 12)

= 4x2 + 12x+ 9 + 3(x2 + 2x+ 1)

= 4x2 + 12x+ 9 + 3x2 + 6x+ 3

= 7x2 + 18x+ 12

IBHL_WS


i x2 + 3x¡ 2(x¡ 4)2= x2 + 3x¡ 2(x2 ¡ 2(x)(4) + 42)= x2 + 3x¡ 2(x2 ¡ 8x+ 16)= x2 + 3x¡ 2x2 + 16x¡ 32= ¡x2 + 19x¡ 32

j (3x¡ 2)2 ¡ 2(x+ 1)2= (3x)2 ¡ 2(3x)(2) + 22 ¡ 2(x2 + 2(x)(1) + 12)= 9x2 ¡ 12x+ 4¡ 2(x2 + 2x+ 1)= 9x2 ¡ 12x+ 4¡ 2x2 ¡ 4x¡ 2= 7x2 ¡ 16x+ 2

EXERCISE H

1 a 3x2 + 9x

= 3x(x+ 3)

b 2x2 + 7x

= x(2x+ 7)

c 4x2 ¡ 10x= 2x(2x¡ 5)

d 6x2 ¡ 15x= 3x(2x¡ 5)

e 9x2 ¡ 25= (3x)2 ¡ 52= (3x+ 5)(3x¡ 5)

f 16x2 ¡ 1= (4x)2 ¡ 12= (4x+ 1)(4x¡ 1)

g 2x2 ¡ 8= 2(x2 ¡ 4)= 2(x2 ¡ 22)= 2(x+ 2)(x¡ 2)

h 3x2 ¡ 9= 3(x2 ¡ 3)= 3(x2 ¡ (p3)2)= 3(x+

p3)(x¡p3)

i 4x2 ¡ 20= 4(x2 ¡ 5)= 4(x2 ¡ (p5)2)= 4(x+

p5)(x¡p5)

j x2 ¡ 8x+ 16= x2 ¡ 2(x)(4) + 42= (x¡ 4)2

k x2 ¡ 10x+ 25= x2 ¡ 2(x)(5) + 52= (x¡ 5)2

l 2x2 ¡ 8x+ 8= 2(x2 ¡ 4x+ 4)= 2(x2 ¡ 2(x)(2) + 22)= 2(x¡ 2)2

m 16x2 + 40x+ 25

= (4x)2 + 2(4x)(5) + 52

= (4x+ 5)2

n 9x2 + 12x+ 4

= (3x)2 + 2(3x)(2) + 22

= (3x+ 2)2

o x2 ¡ 22x+ 121= x2 ¡ 2(x)(11) + 112= (x¡ 11)2

2 a x2 + 9x+ 8

= (x+ 1)(x+ 8)

fas sum = 9, product = 8g

b x2 + 7x+ 12

= (x+ 3)(x+ 4)

fas sum = 7, product = 12g

c x2 ¡ 7x¡ 18= (x¡ 9)(x+ 2)fas sum = ¡7, product = ¡18g

d x2 + 4x¡ 21= (x+ 7)(x¡ 3)fas sum = 4, product = ¡21g

e x2 ¡ 9x+ 18= (x¡ 6)(x¡ 3)fas sum = ¡9, product = 18g

f x2 + x¡ 6= (x+ 3)(x¡ 2)fas sum = 1, product = ¡6g

g ¡x2 + x+ 2= ¡(x2 ¡ x¡ 2)= ¡(x¡ 2)(x+ 1)fas sum = ¡1, product = ¡2g

h 3x2 ¡ 42x+ 99= 3(x2 ¡ 14x+ 33)= 3(x¡ 3)(x¡ 11)fas sum = ¡14, product = 33g

i ¡2x2 ¡ 4x¡ 2= ¡2(x2 + 2x+ 1)= ¡2(x2 + 2(x)(1) + 12)= ¡2(x+ 1)2

j 2x2 + 6x¡ 20= 2(x2 + 3x¡ 10)= 2(x+ 5)(x¡ 2)fas sum = 3, product = ¡10g

IBHL_WS


k 2x2 ¡ 10x¡ 48= 2(x2 ¡ 5x¡ 24)= 2(x¡ 8)(x+ 3)fas sum = ¡5, product = ¡24g

l ¡2x2 + 14x¡ 12= ¡2(x2 ¡ 7x+ 6)= ¡2(x¡ 6)(x¡ 1)fas sum = ¡7, product = 6g

m ¡3x2 + 6x¡ 3= ¡3(x2 ¡ 2x+ 1)= ¡3(x2 ¡ 2(x)(1) + 12)= ¡3(x¡ 1)2

n ¡x2 ¡ 2x¡ 1= ¡(x2 + 2x+ 1)= ¡(x2 + 2(x)(1) + 12)= ¡(x+ 1)2

o ¡5x2 + 10x+ 40= ¡5(x2 ¡ 2x¡ 8)= ¡5(x¡ 4)(x+ 2)fas sum = ¡2, prod. = ¡8g

3 a 2x2 + 5x¡ 12= 2x2 + 8x¡ 3x¡ 12= 2x(x+ 4)¡ 3(x+ 4)= (2x¡ 3)(x+ 4)

has ac = 2£¡12 = ¡24Factors of ¡24 which add to 5 are 8 and ¡3.

b 3x2 ¡ 5x¡ 2= 3x2 ¡ 6x+ x¡ 2= 3x(x¡ 2) + (x¡ 2)= (3x+ 1)(x¡ 2)

has ac = 3£¡2 = ¡6Factors of ¡6 which add to ¡5 are ¡6 and 1.

c 7x2 ¡ 9x+ 2= 7x2 ¡ 7x¡ 2x+ 2= 7x(x¡ 1)¡ 2(x¡ 1)= (7x¡ 2)(x¡ 1)

has ac = 7£ 2 = 14Factors of 14 which add to ¡9 are ¡7 and ¡2.

d 6x2 ¡ x¡ 2= 6x2 + 3x¡ 4x¡ 2= 3x(2x+ 1)¡ 2(2x+ 1)= (3x¡ 2)(2x+ 1)

has ac = 6£¡2 = ¡12Factors of ¡12 which add to ¡1 are 3 and ¡4.

e 4x2 ¡ 4x¡ 3= 4x2 + 2x¡ 6x¡ 3= 2x(2x+ 1)¡ 3(2x+ 1)= (2x¡ 3)(2x+ 1)


f 10x2 ¡ x¡ 3= 10x2 + 5x¡ 6x¡ 3= 5x(2x+ 1)¡ 3(2x+ 1)= (5x¡ 3)(2x+ 1)


g 2x2 ¡ 11x¡ 6= 2x2 ¡ 12x+ x¡ 6= 2x(x¡ 6) + (x¡ 6)= (2x+ 1)(x¡ 6)


h 3x2 ¡ 5x¡ 28= 3x2 ¡ 12x+ 7x¡ 28= 3x(x¡ 4) + 7(x¡ 4)= (3x+ 7)(x¡ 4)


IBHL_WS


i 8x2 + 2x¡ 3= 8x2 ¡ 4x+ 6x¡ 3= 4x(2x¡ 1) + 3(2x¡ 1)= (4x+ 3)(2x¡ 1)


j 10x2 ¡ 9x¡ 9= 10x2 ¡ 15x+ 6x¡ 9= 5x(2x¡ 3) + 3(2x¡ 3)= (5x+ 3)(2x¡ 3)


k 3x2 + 23x¡ 8= 3x2 ¡ x+ 24x¡ 8= x(3x¡ 1) + 8(3x¡ 1)= (x+ 8)(3x¡ 1)


l 6x2 + 7x+ 2

= 6x2 + 3x+ 4x+ 2

= 3x(2x+ 1) + 2(2x+ 1)

= (3x+ 2)(2x+ 1)

has ac = 6£ 2 = 12Factors of 12 which add to 7 are 4 and 3.

m ¡4x2 ¡ 2x+ 6= ¡2(2x2 + x¡ 3)= ¡2(2x2 ¡ 2x+ 3x¡ 3)= ¡2[2x(x¡ 1) + 3(x¡ 1)]= ¡2(2x+ 3)(x¡ 1)


n 12x2 ¡ 16x¡ 3= 12x2 ¡ 18x+ 2x¡ 3= 6x(2x¡ 3) + (2x¡ 3)= (6x+ 1)(2x¡ 3)


o ¡6x2 ¡ 9x+ 42= ¡3(2x2 + 3x¡ 14)= ¡3(2x2 ¡ 4x+ 7x¡ 14)= ¡3[2x(x¡ 2) + 7(x¡ 2)]= ¡3(2x+ 7)(x¡ 2)


p 21x¡ 10¡ 9x2= ¡(9x2 ¡ 21x+ 10)= ¡(9x2 ¡ 6x¡ 15x+ 10)= ¡[3x(3x¡ 2)¡ 5(3x¡ 2)]= ¡(3x¡ 5)(3x¡ 2)


q 8x2 ¡ 6x¡ 27= 8x2 + 12x¡ 18x¡ 27= 4x(2x+ 3)¡ 9(2x+ 3)= (4x¡ 9)(2x+ 3)


r 12x2 + 13x+ 3

= 12x2 + 4x+ 9x+ 3

= 4x(3x+ 1) + 3(3x+ 1)

= (4x+ 3)(3x+ 1)


IBHL_WS


s 12x2 + 20x+ 3

= 12x2 + 2x+ 18x+ 3

= 2x(6x+ 1) + 3(6x+ 1)

= (2x+ 3)(6x+ 1)


t 15x2 ¡ 22x+ 8= 15x2 ¡ 10x¡ 12x+ 8= 5x(3x¡ 2)¡ 4(3x¡ 2)= (5x¡ 4)(3x¡ 2)


u 14x2 ¡ 11x¡ 15= 14x2 ¡ 21x+ 10x¡ 15= 7x(2x¡ 3) + 5(2x¡ 3)= (7x+ 5)(2x¡ 3)


4 a 3(x+ 4) + 2(x+ 4)(x¡ 1)= (x+ 4)[3 + 2(x¡ 1)]= (x+ 4)(3 + 2x¡ 2)= (x+ 4)(2x+ 1)

b 8(2¡ x)¡ 3(x+ 1)(2¡ x)= (2¡ x)[8¡ 3(x+ 1)]= (2¡ x)(8¡ 3x¡ 3)= (2¡ x)(5¡ 3x)

c 6(x+ 2)2 + 9(x+ 2)

= (x+ 2)[6(x+ 2) + 9]

= (x+ 2)(6x+ 12 + 9)

= (x+ 2)(6x+ 21)

= (x+ 2)£ 3(2x+ 7)= 3(x+ 2)(2x+ 7)

d 4(x+ 5) + 8(x+ 5)2

= (x+ 5)[4 + 8(x+ 5)]

= (x+ 5)(4 + 8x+ 40)

= (x+ 5)(8x+ 44)

= (x+ 5)£ 4(2x+ 11)= 4(x+ 5)(2x+ 11)

e (x+ 2)(x+ 3)¡ (x+ 3)(2¡ x)= (x+ 3)[(x+ 2)¡ (2¡ x)]= (x+ 3)(x+ 2¡ 2 + x)= (x+ 3)(2x)

= 2x(x+ 3)

f (x+ 3)2 + 2(x+ 3)¡ x(x+ 3)= (x+ 3)[(x+ 3) + 2¡ x]= (x+ 3)(x+ 3 + 2¡ x)= (x+ 3)(5)

= 5(x+ 3)

g 5(x¡ 2)¡ 3(2¡ x)(x+ 7)= 5(x¡ 2) + 3(x¡ 2)(x+ 7)= (x¡ 2)[5 + 3(x+ 7)]= (x¡ 2)(5 + 3x+ 21)= (x¡ 2)(3x+ 26)

h 3(1¡ x) + 2(x+ 1)(x¡ 1)= ¡3(x¡ 1) + 2(x+ 1)(x¡ 1)= (x¡ 1)[¡3 + 2(x+ 1)]= (x¡ 1)(¡3 + 2x+ 2)= (x¡ 1)(2x¡ 1)

5 a (x+ 3)2 ¡ 16= (x+ 3)2 ¡ 42= (x+ 3 + 4)(x+ 3¡ 4)= (x+ 7)(x¡ 1)

c (x+ 4)2 ¡ (x¡ 2)2= [(x+ 4) + (x¡ 2)][(x+ 4)¡ (x¡ 2)]= (x+ 4 + x¡ 2)(x+ 4¡ x+ 2)= (2x+ 2)(6)

= 2(x+ 1)(6)

= 12(x+ 1)

b 4¡ (1¡ x)2= 22 ¡ (1¡ x)2= [2 + (1¡ x)][2¡ (1¡ x)]= (2 + 1¡ x)(2¡ 1 + x)= (3¡ x)(x+ 1)

d 16¡ 4(x+ 2)2= 4[4¡ (x+ 2)2]= 4[2 + (x+ 2)][2¡ (x+ 2)]= 4(x+ 4)(¡x)= ¡4x(x+ 4)

IBHL_WS


e (2x+ 3)2 ¡ (x¡ 1)2= [(2x+ 3) + (x¡ 1)][(2x+ 3)¡ (x¡ 1)]= (2x+ 3 + x¡ 1)(2x+ 3¡ x+ 1)= (3x+ 2)(x+ 4)

g 3x2 ¡ 3(x+ 2)2= 3[x2 ¡ (x+ 2)2]= 3[x+ (x+ 2)][x¡ (x+ 2)]= 3(x+ x+ 2)(x¡ x¡ 2)= 3(2x+ 2)(¡2)= ¡6(2x+ 2)= ¡12(x+ 1)

i 12x2 ¡ 27(3 + x)2= 3[4x2 ¡ 9(3 + x)2]= 3[(2x)2 ¡ 32(3 + x)2]= 3[(2x)2 ¡ (3(3 + x))2]= 3[2x+ 3(3 + x)][2x¡ 3(3 + x)]= 3[(2x+ 9 + 3x)(2x¡ 9¡ 3x)]= 3(5x+ 9)(¡x¡ 9)= ¡3(5x+ 9)(x+ 9)

f (x+ h)2 ¡ x2= [(x+ h) + x][(x+ h)¡ x]= (x+ h+ x)(x+ h¡ x)= (2x+ h)(h)

= h(2x+ h)

h 5x2 ¡ 20(2¡ x)2= 5[x2 ¡ 4(2¡ x)2]= 5[x2 ¡ 22(2¡ x)2]= 5[x2 ¡ (2(2¡ x))2]= 5[x+ 2(2¡ x)][x¡ 2(2¡ x)]= 5(x+ 4¡ 2x)(x¡ 4 + 2x)= 5(¡x+ 4)(3x¡ 4)= ¡5(x¡ 4)(3x¡ 4)

EXERCISE I

1 a a+ x = b

) x = b¡ ab ax = b

) x =b

a

c 2x+ a = d

) 2x = d¡ a) x =

d¡ a2

d c+ x = t

) x = t¡ ce 5x+ 2y = 20

) 5x = 20¡ 2y) x =

20¡ 2y5

f 2x+ 3y = 12

) 2x = 12¡ 3y) x =

12¡ 3y2

g 7x+ 3y = d

) 7x = d¡ 3y) x =

d¡ 3y7

h ax+ by = c

) ax = c¡ by) x =

c¡ bya

i y = mx+ c

) mx = y ¡ c) x =

y ¡ cm

2 a az =b

c

) z =b

ac

ba

z= d

)z

a=1

d

) z =a

d

c3

d=2

z

) 3z = 2d

) z =2d

3

3 a F = ma

) a =F

m

b C = 2¼r

) r =C

2¼

c V = ldh

) d =V

lh

d A =b

K

) KA = b

) K =b

A

IBHL_WS


4 a A = ¼r2

)A

¼= r2

) r =

rA

¼

b N =x5

a

) aN = x5

) x = 5paN

c V = 43¼r3

) 34V = ¼r3

)3V

4¼= r3

) r = 3r3V

4¼

d D =n

x3

) Dx3 = n

) x3 =n

D

) x = 3qn

D

5 a d =

pa

n

) dn =pa

) a = (dn)2

) a = d2n2

b T = 15

pl

) 5T =pl

) l = (5T )2

) l = 25T 2

c c =pa2 ¡ b2

) c2 = a2 ¡ b2) a2 = c2 + b2

) a = §pb2 + c2

d T = 2¼

rl

g

)T

2¼=

rl

g

)

³T

2¼

´2=l

g

)T 2

4¼2=l

g

) l =gT 2

4¼2

e P = 2(a+ b)

)P

2= a+ b

) a =P

2¡ b

f A = ¼r2 + 2¼rh

) A¡ ¼r2 = 2¼rh

) h =A¡ ¼r22¼r

g I =E

R+ r

) E = I(R+ r)

)E

I= R+ r

) r =E

I¡R

h A =B

p¡ q) A(p¡ q) = B

) p¡ q = BA

) q = p¡ BA

6 a k =d2

2ab

) d2 = k(2ab)

)d2

k= 2ab

) a =d2

2kb

b When k = 112, d = 24, b = 2,

a =242

2£ 112£ 2

=576

448

+ 1:29

7 a V = 43¼r3

) 34V = ¼r3

) r3 =3V

4¼

) r = 3r3V

4¼

b When V = 40,

r = 3r3£ 404¼

+ 2:122 cm

8 a S = 12at2

) t2 =2S

a

) t =

r2S

a

b 10 m ´ 10£ 100 cm = 1000 cmi.e., we need to find t when a = 8,

S = 1000

t =

r2£ 1000

8

) t + 15:81 seconds

IBHL_WS


9 a1

f=1

u+1

v)

1

f¡ 1u=1

v

)u¡ fuf

=1

vand so ) v =

uf

u¡ fb i When u = 50, f = 8

so v =50£ 850¡ 8

) v + 9:52 cm

ii When u = 30, f = 8

so v =30£ 830¡ 8

) v + 10:9 cm

10 a m =m0r

1¡³v

c

´2) m

r1¡

³v

c

´2= m0

)

r1¡

³v

c

´2=m0

m

) 1¡³v

c

´2=³m0

m

´2)

³v

c

´2= 1¡ m

20

m2

)v

c=

rm2 ¡m 20m2

) v =c

m

pm2 ¡m 20

b When m = 3m0,

v =c

3m0

p(3m0)2 ¡m 20

=c

3m0

p9m 20 ¡m 20

=c

3m0

p8m 20

=c

3m0

p8m0

=p83c

c When m = 30m0, c = 3£ 108,

v =3£ 10830m0

p(30m0)2 ¡m 20

=30£ 10730m0

p900m 20 ¡m 20

=107

m0

p899m 20

=107

m0£p899£m0

= 107 £p899+ 2:998£ 108 m/s

EXERCISE J

1 a 3 +x

5

= 3£ 55+x

5

=15

5+x

5

=x+ 15

5

b 1 +3

x

= 1£ xx+3

x

=x

x+3

x

=x+ 3

x

c 3 +x¡ 22

= 3£ 22+x¡ 22

=6

2+x¡ 22

=6 + x¡ 2

2

=x+ 4

2

IBHL_WS


d 3¡ x¡ 24

= 3£ 44¡ x¡ 2

4

=12

4¡ x¡ 2

4

=12¡ x+ 2

4

=14¡ x4

e2 + x

3+x¡ 45

=2 + x

3£ 55+x¡ 45

£ 33

=5(2 + x)

15+3(x¡ 4)15

=10 + 5x+ 3x¡ 12

15

=8x¡ 215

f2x+ 5

4¡ x¡ 1

6

=2x+ 5

4£ 33¡ x¡ 1

6£ 22

=3(2x+ 5)

12¡ 2(x¡ 1)

12

=6x+ 15¡ 2x+ 2

12

=4x+ 17

12

2 a 1 +3

x+ 2

= 1£ x+ 2x+ 2

+3

x+ 2

=x+ 2 + 3

x+ 2

=x+ 5

x+ 2

b ¡2 + 3x¡ 4

= ¡2£ x¡ 4x¡ 4 +

3

x¡ 4

=¡2(x¡ 4) + 3

x¡ 4

=¡2x+ 8 + 3

x¡ 4

=11¡ 2xx¡ 4

c ¡3¡ 2x¡ 1

= ¡3£ x¡ 1x¡ 1 ¡

2

x¡ 1

=¡3(x¡ 1)¡ 2

x¡ 1

=¡3x+ 3¡ 2

x¡ 1

=1¡ 3xx¡ 1

d2x¡ 1x+ 1

+ 3

=2x¡ 1x+ 1

+ 3£ x+ 1x+ 1

=2x¡ 1 + 3(x+ 1)

x+ 1

=2x¡ 1 + 3x+ 3

x+ 1

=5x+ 2

x+ 1

e 3¡ xx+ 1

= 3£ x+ 1x+ 1

¡ xx+ 1

=3(x+ 1)¡ x

x+ 1

=3x+ 3¡ xx+ 1

=2x+ 3

x+ 1

f ¡1 + 41¡ x

= ¡1£ 1¡ x1¡ x +

4

1¡ x

=¡(1¡ x) + 4

1¡ x

=x¡ 1 + 41¡ x

=x+ 3

1¡ x

3 a3x

2x¡ 5 +2x+ 5

x¡ 2

=3x

2x¡ 5 £x¡ 2x¡ 2 +

2x+ 5

x¡ 2 £2x¡ 52x¡ 5

=3x(x¡ 2) + (2x+ 5)(2x¡ 5)

(2x¡ 5)(x¡ 2)

=3x2 ¡ 6x+ (4x2 ¡ 52)(2x¡ 5)(x¡ 2)

=3x2 ¡ 6x+ 4x2 ¡ 25(2x¡ 5)(x¡ 2)

=7x2 ¡ 6x¡ 25(2x¡ 5)(x¡ 2)

b1

x¡ 2 ¡1

x¡ 3

=1

x¡ 2 £x¡ 3x¡ 3 ¡

1

x¡ 3 £x¡ 2x¡ 2

=(x¡ 3)¡ (x¡ 2)(x¡ 2)(x¡ 3)

=x¡ 3¡ x+ 2(x¡ 2)(x¡ 3)

= ¡ 1(x¡ 2)(x¡ 3)

IBHL_WS


c5x

x¡ 4 +3x¡ 2x+ 4

=5x

x¡ 4 £x+ 4

x+ 4+3x¡ 2x+ 4

£ x¡ 4x¡ 4

=5x(x+ 4) + (3x¡ 2)(x¡ 4)

(x+ 4)(x¡ 4)

=5x2 + 20x+ 3x2 ¡ 12x¡ 2x+ 8

(x+ 4)(x¡ 4)

=8x2 + 6x+ 8

(x+ 4)(x¡ 4)

d2x+ 1

x¡ 3 ¡x+ 4

2x+ 1

=2x+ 1

x¡ 3 £2x+ 1

2x+ 1¡ x+ 42x+ 1

£ x¡ 3x¡ 3

=(2x+ 1)2 ¡ (x+ 4)(x¡ 3)

(x¡ 3)(2x+ 1)

=(2x)2 + 2(2x)(1) + 12 ¡ (x2 ¡ 3x+ 4x¡ 12)

(x¡ 3)(2x+ 1)

=4x2 + 4x+ 1¡ (x2 + x¡ 12)

(x¡ 3)(2x+ 1)

=4x2 + 4x+ 1¡ x2 ¡ x+ 12

(x¡ 3)(2x+ 1)

=3x2 + 3x+ 13

(x¡ 3)(2x+ 1)EXERCISE K.1

1 ]ECA = ]DCB fopposite anglesgAlso, EC = CD and AC = BC fgiveng) ¢s AEC and BDC are congruent (SAS)

) AE = BD

2 MP = NP fgivengAP is common to both ¢AMP and ¢ANP

) ¢s AMP and ANP are congruent (RHS)

) ]MAP = ]NAP fcorresponding anglesgi.e., AP bisects ]BAC

) P lies on the bisector of ]BAC

3 OA = OB fboth radii of circleg]OPA = ]OPB = 90o

fsince AB is a tangent to inner circlegOP is common to both ¢AOP and ¢BOP

) ¢s AOP and BOP are congruent (RHS)

) AP = BP fcorresponding sidesg) P is the midpoint of AB

EXERCISE K.2

1 a ]ABC = ]ADE fcorresponding anglesg]ACB = ]AED fcorresponding anglesgi.e., ¢s ABC and ADE are equiangular (AAA)

and hence similar.

)x

2=

6

2 + 3=6

5

) x = 125= 2:4

A B

C

E D

A

B

C

P

M

N

AP

BO

2 cm

3 cm

6 cm

x cm

A

B C

D E

IBHL_WS


b ]PRQ = ]TRS fopposite anglesg) ¢s PQR and TSR are equiangular (AAA)

and hence similar.

)5

7=2

x

) 5x = 14

) x = 145 = 2:8

c ]VUZ = ]YXZ fcorresponding anglesg) ¢s UVZ and XYZ are equiangular (AAA)

and hence similar.

)x

6=

6

5 + 6

)x

6=

6

11

) x = 3611 = 3311

d ]ABE = ]DBC fopposite anglesg) ¢s ABE and DBC are equiangular (AAA)

and hence similar.

)x

5=4

3

) x = 203= 6 2

3

e ]XYZ = ]XUV fcorresponding anglesg]XZY = ]XVU fcorresponding anglesg) ¢s XYZ and XUV are equiangular (AAA)

and hence similar.

)x

2 + 5=5

5

)x

7= 1 and so x = 7

f ]SQR = ]PQT fopposite anglesg) ¢s SQR and PQT are equiangular (AAA)

and hence similar.

)x

9=

8

10

) x = 7210= 7:2

2 ]BAC = ]STC fcorresponding anglesg) ¢s ABC and TSC are equiangular (AAA)

and hence similar.

)x

2:4=1:8

3:2

) x =2:4£ 1:83:2

= 1:35

i.e., the son is 1:35 m tall

5 cm

2 cm7 cm

x cm

P

QR

S

T

6 cm

5 cm 6 cm

x cm

U

V Y

X

Z

��

5 cm3 cm

4 cm x cm

AD

B

EC

2 cm

5 cm

x cm

X

Y Z

UV

x cm

8 cm

9 cm

10 cm

50°

50°

S

R

Q

TP

A

B S

T

C3.2 m

2.4 m

x m

1.8 m

IBHL_WS


EXERCISE L

1 a AB =p(xB ¡ xA)2 + (yB ¡ yA)2

=p(4¡ 1)2 + (5¡ 3)2

=p9 + 4

=p13 units

b OC =p(xC ¡ xO)2 + (yC ¡ yO)2

=p(3¡ 0)2 + (¡5¡ 0)2

=p9 + 25

=p34 units

c PQ =p(xQ ¡ xP)2 + (yQ ¡ yP)2

=p(1¡ 5)2 + (4¡ 2)2

=p16 + 4

=p20 units

d ST =p(xT ¡ xS)2 + (yT ¡ yS)2

=p(¡1¡ 0)2 + (0¡¡3)2

=p1 + 9

=p10 units

2 a xM =xA + xB2

=3 + 1

2= 2 and yM =

yA + yB2

=6 + 0

2= 3

) the midpoint is at (2, 3)

b xM =xA + xB2

=5¡ 12

= 2 and yM =yA + yB2

=2¡ 42

= ¡1) the midpoint is at (2, ¡1)

c xM =xA + xB2

=7 + 0

2= 72 and yM =

yA + yB2

=0 + 3

2= 32

) the midpoint is at ( 72 ,32 )

d xM =xA + xB2

=5¡ 12

= 2 and yM =yA + yB2

=¡2¡ 32

= ¡ 52

) the midpoint is at (2, ¡ 52

)

3 a slope =y-step

x-step

= 23

bslope =

y-step

x-step

= 2¡5

= ¡ 25

c slope =y-step

x-step

= 33

= 1

dslope =

y-step

x-step

= 07

= 0

eslope =

y-step

x-step

= 4¡1

= ¡4

fslope =

y-step

x-step

= 30

which is undefined

4 a gradient =y2 ¡ y1x2 ¡ x1

=7¡ 34¡ 2

= 2

b gradient =y2 ¡ y1x2 ¡ x1

=8¡ 25¡ 3

= 3

c gradient =y2 ¡ y1x2 ¡ x1

=5¡ 2

¡1¡ (¡1)= 3

0

which is undefined

x-step

y-step 2 y-step 2

y-step 0

x-step

y-step 3

x-step �

y-step 3

x-step

y-step 4

x-step ��

IBHL_WS


d gradient =y2 ¡ y1x2 ¡ x1

=¡3¡ (¡3)¡1¡ 4

= 0¡5

= 0

e gradient =y2 ¡ y1x2 ¡ x1

=4¡ 0¡1¡ 0

= ¡4

f gradient =y2 ¡ y1x2 ¡ x1

=¡2¡ (¡1)¡1¡ 3

= ¡1¡4

= 14

5 a The lines have the same slope ( 25

), so are parallel.

b The lines have the same slope (¡ 17

), so are parallel.

c The two lines have slopes 12

and ¡21

, so the product of the

slopes is ¡1.Hence the lines are perpendicular.

d The two lines have slopes 13

and ¡41


slopes is ¡ 43

.

Hence the lines are neither parallel nor perpendicular.

e The two lines have slopes 27

and 15


slopes is 235

.

Hence they are neither parallel nor perpendicular.

f The two lines have slopes ¡12 and21 , so the product of the

slopes is ¡1.Hence the lines are perpendicular.

6 a gradient = ¡ 1( 34)= ¡ 43 b gradient = ¡

1

( 113)= ¡ 311 c gradient = ¡ 14

d gradient = ¡ 1(¡ 1

3)= 3 e gradient = ¡ 1

(¡5) =15

f gradient = ¡ 10

which is undefined

7 a The equation of the line is

y ¡ 1x¡ 4 = 2

) y ¡ 1 = 2(x¡ 4)) y ¡ 1 = 2x¡ 8

) y = 2x¡ 7

b The equation of the line is

y ¡ 2x¡ 1 = ¡2

) y ¡ 2 = ¡2(x¡ 1)) y ¡ 2 = ¡2x+ 2

) y = ¡2x+ 4

52

52

��77

��77

��

1

�

1

��

1

3

1

2

1

5

��

2

2

�

IBHL_WS


c The equation of the line is

y ¡ 0x¡ 5 = 3) y = 3(x¡ 5)) y = 3x¡ 15

d The equation of the line is

y ¡ 7x¡ (¡1) = ¡3

) y ¡ 7 = ¡3(x+ 1)) y ¡ 7 = ¡3x¡ 3

) y = ¡3x+ 4e The equation of the line is

y ¡ 5x¡ 1 = ¡4

) y ¡ 5 = ¡4(x¡ 1)) y ¡ 5 = ¡4x+ 4

) y = ¡4x+ 9

f The equation of the line is

y ¡ 7x¡ 2 = 1

) y ¡ 7 = x¡ 2) y = x+ 5

8 a The equation of the line is

y ¡ 1x¡ 2 =

3

2

) 2(y ¡ 1) = 3(x¡ 2)) 2y ¡ 2 = 3x¡ 6) 3x¡ 2y = 4

b The equation of the line is

y ¡ 4x¡ 1 = ¡

3

2

) ¡2(y ¡ 4) = 3(x¡ 1)) ¡2y + 8 = 3x¡ 3) 3x+ 2y = 11

c The equation of the line is

y ¡ 0x¡ 4 =

1

3

) 3y = x¡ 4) x¡ 3y = 4

d The equation of the line is

y ¡ 6x¡ 0 = ¡4

) y ¡ 6 = ¡4x) 4x+ y = 6

e The equation of the line is

y ¡ (¡3)x¡ (¡1) = 3

) y + 3 = 3(x+ 1)

) y + 3 = 3x+ 3) 3x¡ y = 0

f The equation of the line is

y ¡ (¡2)x¡ 4 = ¡

4

9

) ¡9(y + 2) = 4(x¡ 4)) ¡9y ¡ 18 = 4x¡ 16) 4x+ 9y = ¡2

9 a The slope of the line is2¡ 13¡ 0 =

1

3) its equation is

y ¡ 1x¡ 0 =

1

3) 3(y ¡ 1) = x) 3y ¡ 3 = x) x¡ 3y = ¡3

b The slope of the line is¡1¡ 40¡ 1 = 5 ) its equation is

y ¡ (¡1)x¡ 0 = 5

) y + 1 = 5x) 5x¡ y = 1

c The slope of the line is¡4¡ (¡1)¡1¡ 2 =

¡3¡3 = 1 ) its equation is

y ¡ (¡1)x¡ 2 = 1

) y + 1 = x¡ 2) x¡ y = 3

d The slope of the line is2¡ (¡2)5¡ 0 =

4

5) its equation is

y ¡ (¡2)x¡ 0 =

4

5

) 5(y + 2) = 4x

) 5y + 10 = 4x

) 4x¡ 5y = 10

IBHL_WS


e The slope of the line is2¡ 0

3¡ (¡1) =2

4=1

2) its equation is

y ¡ 0x¡ (¡1) =

1

2

) 2y = x+ 1

) x¡ 2y = ¡1

f The slope of the line is¡3¡ (¡1)2¡ (¡1) =

¡23

) its equation isy ¡ (¡1)x¡ (¡1) = ¡

2

3

) ¡3(y + 1) = 2(x+ 1)) ¡3y ¡ 3 = 2x+ 2) 2x+ 3y = ¡5

10 a Since both points on the line have y-coordinate ¡2, it must be horizontal.) its equation is y = ¡2.

b Since both points on the line have x-coordinate 6, it must be vertical.

) its equation is x = 6.

c Since both points on the line have x-coordinate ¡3, it must be vertical.) its equation is x = ¡3.

11 a 2x¡ 3y = 6) 3y = 2x¡ 6) y = 2

3x¡ 2) slope = 2

3, y-intercept = ¡2

When y = 0, 2x = 6

) x-intercept = 3

b 4x+ 5y = 20

) 5y = ¡4x+ 20) y = ¡ 45x+ 4

) slope = ¡ 45 , y-intercept = 4When y = 0, 4x = 20

) x-intercept = 5

c y = ¡2x+ 5) slope = ¡2, y-intercept = 5When y = 0, ¡2x+ 5 = 0 ) x = 5

2

i.e., the x-intercept = 52

d x = 8

) the line is vertical

) slope is undefined,

no y-intercept, x-intercept = 8

e y = 5

) the line is horizontal

) slope = 0,

y-intercept = 5, no x-intercept

f x+ y = 11

) y = 11¡ x) slope = ¡1, y-intercept = 11When y = 0, x = 11

) x-intercept = 11

g 4x+ y = 8

) y = ¡4x+ 8) slope = ¡4, y-intercept = 8When y = 0, 4x = 8

) x-intercept = 2

h x¡ 3y = 12) 3y = x¡ 12) y = 1

3x¡ 4

) slope = 13

, y-intercept = ¡4When y = 0, x = 12

) x-intercept = 12Summary of results:

Equation Slope x-int. y-int.

a 2x¡ 3y = 6 23

3 ¡2b 4x+ 5y = 20 ¡ 4

55 4

c y = ¡2x+ 5 ¡2 52

5

d x = 8 undefined 8 none

Equation Slope x-int. y-int

e y = 5 0 none 5

f x+ y = 11 ¡1 11 11g 4x+ y = 8 ¡4 2 8h x¡ 3y = 12 1

312 ¡4

IBHL_WS


12 a Substituting (3, 4) into 3x¡ 2y = 1 gives 3(3)¡ 2(4) = 1i.e., 1 = 1 which is true

) (3, 4) lies on the line

b Substituting (¡2, 5) into 5x+ 3y = ¡5 gives 5(¡2) + 3(5) = ¡5i.e., 5 = ¡5 which is false

) (¡2, 5) does not lie on the line

c Substituting (6, ¡ 12

) into 3x¡ 8y = 22 gives 3(6)¡ 8(¡ 12) = 22

i.e., 22 = 22 which is true) (6, ¡ 1

2) lies on the line

13 a For x+ 2y = 8,

when x = 0, y = 4when y = 0, x = 8

For y = 2x¡ 6,when x = 0, y = ¡6when y = 0, x = 3

x 0 8

y 4 0

x 0 3

y ¡6 0

The lines meet at (4, 2).

b For y = ¡3x¡ 3,when x = 0, y = ¡3when y = 0, x = ¡1

For 3x¡ 2y = ¡12,when x = 0, y = 6when y = 0, x = ¡4

x 0 ¡1y ¡3 0

x 0 ¡4y 6 0

The lines meet at (¡2, 3).

c For 3x+ y = ¡3,when x = 0, y = ¡3when y = 0, x = ¡1

For 2x¡ 3y = ¡24,when x = 0, y = 8when y = 0, x = ¡12

x 0 ¡1y ¡3 0

x 0 ¡12y 8 0

The lines meet at (¡3, 6).

d For 2x¡ 3y = 8,when x = 0, y = ¡ 8

3

when y = 0, x = 4

For 3x+ 2y = 12,

when x = 0, y = 6when y = 0, x = 4

x 0 4

y ¡ 83

0

x 0 4

y 6 0

The lines meet at (4, 0).

�

�

��y

x�

��

82 �� yx

62 �� xy

��

��

�

y

x

33 �� xy

1223 �� yx

y

x

��

��

��

33 �� yx

2432 �� yx

y

x�

�

-\Ie_-\Ie_

1223 �� yx 832 �� yx

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e For x+ 3y = 10,

when x = 0, y = 103

when y = 0, x = 10

For 2x+ 6y = 11,

when x = 0, y = 116

when y = 0, x = 112

x 0 10

y 103 0

x 0 112

y 116

0

The lines are parallel, so never meet.

f For 5x+ 3y = 10,

when x = 0, y = 103

when y = 0, x = 2

For 10x+ 6y = 20,

when x = 0, y = 206= 10

3

when y = 0, x = 2

x 0 2

y 103

0

x 0 2

y 103

0

The lines are coincident.

14 a Since the line is horizontal, its equation is y = ¡4.b Since the line is vertical, its equation is x = 5.

c Since the line is vertical, its equation is x = ¡1.d Since the line is horizontal, its equation is y = 2.

e The x-axis corresponds to y = 0.

f The y-axis corresponds to x = 0.

15 a The line has equationy ¡ 4

x¡ (¡1) =3

4) 4(y ¡ 4) = 3(x+ 1)) 4y ¡ 16 = 3x+ 3) 3x¡ 4y = ¡19

b The line has slope =0¡ (¡5)7¡ 2 =

5

5= 1 ) its equation is

y ¡ 0x¡ 7 = 1) y = x¡ 7

) x¡ y = 7c y = 3x¡ 2 has slope 3, so this line has slope 3 also.

It passes through (0, 0), so its equation isy ¡ 0x¡ 0 = 3 i.e., y = 3x

d Now a line parallel to 2x+ 3y = 8 has equation 2x+ 3y = k, where k is a constant.

Since (¡1, 7) lies on the line, 2(¡1) + 3(7) = k) k = 19

) the line is 2x+ 3y = 19

e y = ¡2x+ 5 has slope ¡2 ) lines perpendicular to it have slope 12

.

But this line must pass through (3, ¡1). ) y ¡ (¡1)x¡ 3 =

1

2

) 2(y + 1) = x¡ 3) 2y + 2 = x¡ 3) x¡ 2y = 5

y

x

Qd_P_Qd_P_Qh_Q_Qh_Q_

Qs_Q_Qs_Q_1010

1162 �� yx

103 �� yx

y

x

Qd_P_Qd_P_

22

20610 �� yx

1035 �� yx

IBHL_WS


f If 3x¡ y = 11, then y = 3x¡ 11) this line has slope 3 ) lines perpendicular to it have slope ¡ 13 .

But this line must pass through (¡2, 5). ) y ¡ 5x¡ (¡2) = ¡

1

3

) ¡3(y ¡ 5) = x+ 2) ¡3y + 15 = x+ 2) x+ 3y = 13

16 a Keach Avenue passes through (5, 11) and (13, 12)

) its line has slope12¡ 1113¡ 5 =

1

8

) its equation isy ¡ 12x¡ 13 =

1

8) 8(y ¡ 12) = x¡ 13) 8y ¡ 96 = x¡ 13) x¡ 8y = ¡83

b Peacock Street is perpendicular to Keach Avenue, so its line has slope ¡8.But this line also passes through (3, 17). )

y ¡ 17x¡ 3 = ¡8

) y ¡ 17 = ¡8(x¡ 3)) y ¡ 17 = ¡8x+ 24) 8x+ y = 41

c Diagonal Road runs from (5, 11) to (7, 20), but ends at these points.

Hence there is a restricted domain 5 6 x 6 7.

Its slope is20¡ 117¡ 5 =

9

2

) its equation isy ¡ 11x¡ 5 =

9

2) 2(y ¡ 11) = 9(x¡ 5)) 2y ¡ 22 = 9x¡ 45) 9x¡ 2y = 23, 5 6 x 6 7

d Plunkit Street has x = 8, so it meets Keach Avenue when 8¡ 8y = ¡83 fusing ag) 8y = 91

) y = 918

) they intersect at (8, 918

)

17 a The line through (¡1, 5) and (0, 2)has slope =

5¡ 2¡1¡ 0 = ¡3

) the slope of the tangent is 13

) its equation isy ¡ 5

x¡ (¡1) =1

3

) 3(y ¡ 5) = x+ 1) 3y ¡ 15 = x+ 1) x¡ 3y = ¡16

(-1' 5)

2

y

x

IBHL_WS


b The line through (¡1, 1) and (3, ¡1)has slope =

¡1¡ 13¡ (¡1) = ¡

1

2

) the slope of the tangent is 2

) its equation isy ¡ 1

x¡ (¡1) = 2

) y ¡ 1 = 2(x+ 1)) y ¡ 1 = 2x+ 2) 2x¡ y = ¡3

c The line through (2, ¡2) and (5, ¡2)

has slope =¡2¡ (¡2)5¡ 2 = 0

i.e., it is horizontal

) the tangent must be vertical

) its equation is x = 5

18 Suppose the outlet is at T(x, 8).

Then AT = BT, sop(x¡ 5)2 + (8¡ 5)2 =

p(x¡ 7)2 + (8¡ 10)2

) (x¡ 5)2 + (8¡ 5)2 = (x¡ 7)2 + (8¡ 10)2) x2 ¡ 10x+ 25 + 9 = x2 ¡ 14x+ 49 + 4

) ¡10x+ 34 = ¡14x+ 53) 4x = 19

) x = 194

) the outlet is at (4 34

, 8)

19 a Suppose C is at (x, 7)

Then AC= BC, sop(x¡ 2)2 + (7¡ 3)2 =

p(x¡ 5)2 + (7¡ 4)2

) (x¡ 2)2 + (7¡ 3)2 = (x¡ 5)2 + (7¡ 4)2) x2 ¡ 4x+ 4 + 16 = x2 ¡ 10x+ 25 + 9

) ¡4x+ 20 = ¡10x+ 34) 6x = 14

) x = 73 ) the pumping station is at (73

, 7)

b The length of each pipe =p(x¡ 2)2 + (7¡ 3)2

=p( 73¡ 2)2 + 16

=p16 1

9) the total length = 2

p16 1

9+ 8:03 km

c Now CD is horizontal, so BD is vertical ) the x-coordinate of D is 5

) D is at (5, 7)

Then AB + BD =p(5¡ 2)2 + (4¡ 3)2 +

p(5¡ 5)2 + (7¡ 4)2

=p9 + 1 +

p0 + 9

=p10 + 3

+ 6:162 km Hence yes, it would be much cheaper.

y

x

(-1' 1)

(3'-1)

y

x

(2'-2) (5'-2)

IBHL_WS

(-1' 5)

(3' 1)

C r

C(2'-3)

(4' 1)P


20 Suppose Jason’s girlfriend lives at G(x, 8)

Then GJ =p(x¡ 4)2 + (8¡ 1)2 = 11:73) (x¡ 4)2 + 49 = 11:732

) (x¡ 4)2 = 88:5929) x¡ 4 + §9:412

) x + ¡5:412 or 13:412i.e., Jason’s girlfriend lives at (¡5:412, 8) or (13:41, 8)

21 If the centre is C(a, b) then PC is always r units.

)p(x¡ a)2 + (y ¡ b)2 = r fdistance formulag

) (x¡ a)2 + (y ¡ b)2 = r2

22 a (x¡ 4)2 + (y ¡ 3)2 = 52i.e., (x¡ 4)2 + (y ¡ 3)2 = 25

b (x¡¡1)2 + (y ¡ 5)2 = 22i.e., (x+ 1)2 + (y ¡ 5)2 = 4

c (x¡ 0)2 + (y ¡ 0)2 = 102i.e., x2 + y2 = 100

d C is³¡1 + 3

2,5 + 1

2

´i.e., (1, 3)

and r =p(3¡ 1)2 + (1¡ 3)2

=p4 + 4

=p8

) equation is (x¡ 1)2 + (y ¡ 3)2 = (p8)2i.e., (x¡ 1)2 + (y ¡ 3)2 = 8

23 a The centre is (1, 3) and radiusp4 = 2 units.

b The centre is (0, ¡2) and radius p16 = 4 units.c The centre is (0, 0) and radius

p7 units.

24 a (x¡ 2)2 + (y + 3)2 = 20 has centre (2, ¡3) and radiusp20 units.

b If x = 4 and y = 1, LHS = (4¡ 2)2 + (1 + 3)2= 22 + 42

= 20= RHS X

c Slope of radius CP is1¡¡34¡ 2 =

42= 2

1

) slope of tangent is ¡ 12

) equation isy ¡ 1x¡ 4 = ¡

12

i.e., 2y ¡ 2 = ¡x+ 4i.e., x+ 2y = 6

IBHL_WS


25

M is

³5 + 7

2,7 + 1

2

´i.e., (6, 4)

N is

³7 +¡12

,1 + 5

2

´i.e., (3, 3)

Slope of PQ =1¡ 77¡ 5 = ¡

62= ¡3

) slope of line 1 = 13

fnegative reciprocalg

) equation of line 1 isy ¡ 4x¡ 6 =

13

i.e., y ¡ 4 = 13x¡ 2

i.e., y = 13x+ 2 ...... (1)

Slope of QR =5¡ 1¡1¡ 7 =

4¡8 = ¡ 12

) slope of line 2 is 2 fnegative reciprocalg

) equation of line 2 isy ¡ 3x¡ 3 = 2

i.e., y ¡ 3 = 2x¡ 6i.e., y = 2x¡ 3 ...... (2)

We now solve (1) and (2) simultaneously

Now 13x+ 2 = 2x¡ 3) x+ 6 = 6x¡ 9) 15 = 5x

) x = 3 and so y = 2(3)¡ 3 = 3) the centre is at (3, 3).

(5' 7)(5' 7)

(7' 1)(7' 1)(-1' 5)(-1' 5)

(6' 4)(6' 4)centre

N (3' 3)(3' 3)

M

R Q

P

line 1

line 2

IBHL_WS

EXERCISE 1A

1 a (1, 3), (2, 4), (3, 5), (4, 6) is a function since no two ordered pairs have the same x-coordinate.

b (1, 3), (3, 2), (1, 7), (¡1, 4) is not a function since two of the ordered pairs, (1, 3) and(1, 7), have the same x-coordinate of 1.

c (2, ¡1), (2, 0), (2, 3), (2, 11) is not a function since each ordered pair has the samex-coordinate of 2.

d (7, 6), (5, 6), (3, 6), (¡4, 6) is a function since no two ordered pairs have the samex-coordinate.

e (0, 0), (1, 0), (3, 0), (5, 0) is a function since no two ordered pairs have the same x-coordinate.

f (0, 0), (0, ¡2), (0, 2), (0, 4) is not a function since each ordered pair has the same x-coordinateof 0.

2 a i.e., each linecuts the graph

no more than

once ) it

is a function

b i.e., each linecuts the graph

no more than

once ) it

is a function

c i.e., each linecuts the graph

no more than

once ) it

is a function

d i.e., the linescut the graph

more than

once ) it is

not a function

e i.e., each linecuts the graph

no more than

once ) it

is a function

f i.e., the linescut the graph

more than

once ) it is

not a function

g i.e., each linecuts the graph

no more than

once ) it

is a function

h i.e., one linecuts the graph

more than once

) it is not a

function

3 The graph of a straight line is not a function if the

graph is a vertical line, i.e., x = a for all a.

The vertical line through x = a cuts the graph at

every point ) it is not a function.

4 x2 + y2 = 9 is the equation of a circle, centre (0, 0) and radius 3.

Now x2 + y2 = 9

) y2 = 9¡ x2) y = §p9¡ x2 Hence y has two real values for any value of x where ¡3 < x < 3.

Chapter 1FUNCTIONS

y

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x

x a�

IBHL_WS

34 Mathematics HL, Chapter 1 – FUNCTIONS

EXERCISE 1B

1 a Domain is fx: ¡1 < x 6 5gRange is fy: 1 < y 6 3g

b Domain is fx: x 6= 2gRange is fy: y 6= ¡1g

c Domain is fx: x 2 RgRange is fy: 0 < y 6 2g

d Domain is fx: x 2 RgRange is fy: y > ¡1g

e Domain is fx: x > ¡4gRange is fy: y > ¡3g

f Domain is fx: x 6= §2gRange is fy: y 6 ¡1 or y > 0g

2 a Domain is

fx: x > 0gRange is

fy: y > 0g

b Domain is

fx: x 6= 0gRange is

fy: y > 0g

c Domain is

fx: x 6 4gRange is

fy: y > 0g

d Domain is

fx: x 2 RgRange is

fy: y > ¡2 14g

e Domain is

fx: x 2 RgRange is

fy: y 6 2512g

f Domain is

fx: x 6= 0gRange is

fy: y 6 ¡2or y > 2g

g Domain is

fx: x 6= 2gRange is

fy: y 6= 1g

h Domain is

fx: x 2 RgRange is

fy: y 2 Rg

i Domain is

fx: x 6= ¡1,x 6= 2g

Range is

fy: y 6 13

or y > 3g

j Domain is

fx: x 6= 0gRange is

fy: y > 2g

k Domain is

fx: x 6= 0gRange is

fy: y 6 ¡2or y > 2g

l Domain is

fx: x 2 RgRange is

fy: y > ¡8g

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�

y

x

xy �

��

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y

x

xy �� 4

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y

x

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Mathematics HL, Chapter 1 – FUNCTIONS 35

EXERCISE 1C

1 a f(0) = 3(0) + 2

= 2

b f(2) = 3(2) + 2

= 8

c f(¡1) = 3(¡1) + 2= ¡1

d f(¡5) = 3(¡5) + 2= ¡13

e f(¡ 13) = 3(¡ 1

3) + 2

= 1

2 a f(0) = 3(0)¡ 02 + 2= 2

b f(3) = 3(3)¡ 32 + 2= 9¡ 9 + 2= 2

c f(¡3) = 3(¡3)¡ (¡3)2 + 2= ¡9¡ 9 + 2= ¡16

d f(¡7) = 3(¡7)¡ (¡7)2 + 2= ¡21¡ 49 + 2= ¡68

e f( 32) = 3( 3

2)¡ ( 3

2)2 + 2

= 92¡ 9

4+ 2

= 174

3 a f(a) = 7¡ 3a b f(¡a) = 7¡ 3(¡a)= 7 + 3a

c f(a+ 3) = 7¡ 3(a+ 3)= 7¡ 3a¡ 9= ¡3a¡ 2

d f(b¡ 1) = 7¡ 3(b¡ 1)= 7¡ 3b+ 3= 10¡ 3b

e f(x+ 2) = 7¡ 3(x+ 2)= 7¡ 3x¡ 6= 1¡ 3x

4 a F (x+ 4)

= 2(x+ 4)2 + 3(x+ 4)¡ 1= 2(x2 + 8x+ 16) + 3x+ 12¡ 1= 2x2 + 16x+ 32 + 3x+ 11

= 2x2 + 19x+ 43

b F (2¡ x)= 2(2¡ x)2 + 3(2¡ x)¡ 1= 2(4¡ 4x+ x2) + 6¡ 3x¡ 1= 8¡ 8x+ 2x2 + 5¡ 3x= 2x2 ¡ 11x+ 13

c F (¡x)= 2(¡x)2 + 3(¡x)¡ 1= 2x2 ¡ 3x¡ 1

e F (x2 ¡ 1)= 2(x2 ¡ 1)2 + 3(x2 ¡ 1)¡ 1= 2(x4 ¡ 2x2 + 1) + 3x2 ¡ 3¡ 1= 2x4 ¡ 4x2 + 2 + 3x2 ¡ 4= 2x4 ¡ x2 ¡ 2

d F (x2)

= 2(x2)2 + 3(x2)¡ 1= 2x4 + 3x2 ¡ 1

5 a i G(2) =2(2) + 3

2¡ 4= 7¡2

= ¡ 72

ii G(0) =2(0) + 3

0¡ 4= 3¡4

= ¡ 34

iii G(¡ 12) =

2(¡ 12) + 3

¡12 ¡ 4

=¡1 + 3¡ 92

=2

(¡ 92)

= ¡ 49

b G(x) =2x+ 3

x¡ 4 is undefined when x¡ 4 = 0i.e., when x = 4

So, when x = 4, G(x) does not exist.

c G(x+ 2) =2(x+ 2) + 3

(x+ 2)¡ 4 =2x+ 4 + 3

x+ 2¡ 4 =2x+ 7

x¡ 2

d G(x) = ¡3 i.e., 2x+ 3x¡ 4 = ¡3 ) 2x+ 3 = ¡3(x¡ 4)

) 2x+ 3 = ¡3x+ 12) 5x = 9 and so x = 9

5

IBHL_WS


6 f is the function which converts x into f(x) whereas

f(x) is the value of the function at any value of x.

7 a V (4) = 9650¡ 860(4)= 9650¡ 3440= 6210

i.e., the value of the photocopier 4 yearsafter purchase is 6210 Yen.

b If V (t) = 5780,

then 9650¡ 860t = 5780) 860t = 3870

) t = 4:5

i.e., the value of the photocopier

4 12

years after purchase is 5780 Yen.c Original purchase price is when t = 0,

i.e., V (0) = 9650¡ 860(0)= 9650 i.e., the original purchase price was 9650 Yen.

8 First sketch the linear function which passes through

the two points (2, 1) and (5, 3).

Then sketch two quadratic functions which also pass

through the two points.

9 f(x) = ax+ b where f(2) = 1 and f(¡3) = 11i.e., a(2) + b = 1

) 2a+ b = 1

) b = 1¡ 2a ..... (1)

and a(¡3) + b = 11) ¡3a+ b = 11

) b = 11 + 3a ..... (2)

Solving (1) and (2) simultaneously, 1¡ 2a = 11 + 3a) 5a = ¡10) a = ¡2

Substituting a = ¡2 into (1) gives b = 1¡ 2(¡2) = 5 i.e., a = ¡2, b = 5Hence f(x) = ¡2x+ 5

10 T (x) = ax2 + bx+ c where T (0) = ¡4, T (1) = ¡2 and T (2) = 6i.e., a(0)2 + b(0) + c = ¡4

) c = ¡4Also, a(1)2 + b(1) + c = ¡2

) a+ b+ c = ¡2and a(2)2 + b(2) + c = 6

and ) 4a+ 2b+ c = 6

Substituting c = ¡4 into both equations givesa+ b+ (¡4) = ¡2

) a+ b = 2

) a = 2¡ b ..... (1)

and 4a+ 2b+ (¡4) = 6) 4a+ 2b = 10 ..... (2)

Substituting (1) into (2) gives 4(2¡ b) + 2b = 10 ) 8¡ 4b+ 2b = 10) ¡2b = 2) b = ¡1

Now, substituting b = ¡1 into (1) gives a = 2¡ (¡1) = 3 i.e., a = 3, b = ¡1, c = ¡4

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EXERCISE 1D

1 a (f ± g)(x)= f(g(x))

= f(1¡ x)= 2(1¡ x) + 3= 2¡ 2x+ 3= 5¡ 2x

b (g ± f)(x)= g(f(x))

= g(2x+ 3)

= 1¡ (2x+ 3)= 1¡ 2x¡ 3= ¡2x¡ 2

c (f ± g)(¡3)= f(g(¡3))= f(1¡ (¡3))= f(4)

= 2(4) + 3

= 11

2 (f ± g)(x) = f(g(x))= f(2¡ x)= (2¡ x)2

Domain is fx: x 2 RgRange is fy: y > 0g

(g ± f)(x) = g(f(x))= g(x2)

= 2¡ x2Domain is fx: x 2 RgRange is fy: y > 2g

3 a (f ± g)(x)= f(g(x))

= f(3¡ x)= (3¡ x)2 + 1= 9¡ 6x+ x2 + 1= x2 ¡ 6x+ 10

b (g ± f)(x)= g(f(x))

= g(x2 + 1)

= 3¡ (x2 + 1)= 3¡ x2 ¡ 1= ¡x2 + 2

c (g ± f)(x) = f(x)i.e., ¡x2 + 2 = f(x) ffrom bgi.e., ¡x2 + 2 = x2 + 1

) 2x2 = 1

) x2 = 12

) x = § 1p2

4 a ax+ b = cx+ d is true for all x fgivengLet x = 0,

) a(0) + b = c(0) + d

) b = d ..... (1)

Let x = 1, ) a(1) + b = c(1) + d

) a+ b = c+ d

but b = d (from (1))

) a+ d = c+ d

) a = cb (f ± g)(x) = x for all x fgiveng

) f(g(x)) = x

) f(ax+ b) = x

) 2(ax+ b) + 3 = x

) 2ax+ 2b+ 3 = x for all x

When x = 0,

2a(0) + 2b+ 3 = 0

) 2b = ¡3) b = ¡ 3

2

When x = 1, 2a(1) + 2b+ 3 = 1

) 2a+ 2b = ¡2) 2a = ¡2¡ 2(¡ 3

2) = 1

) a = 12i.e., a = 1

2and b = ¡ 3

2as required.

c If (g ± f)(x) = xthen g(f(x)) = x

i.e., g(2x+ 3) = x

i.e., a(2x+ 3) + b = x

) 2ax+ 3a+ b = x

When x = 0,

2a(0) + 3a+ b = 0

) 3a = ¡b) a = ¡ 1

3b ..... (1)

When x = 1, 2a(1) + 3a+ b = 1

) 2a+ 3a+ b = 1

) 5a+ b = 1

) 5(¡ 13b) + b = 1 (from (1))

) ¡ 23b = 1

) b = ¡ 32

Substituting b = ¡ 32

into (1) gives

a = ¡ 13(¡ 3

2) = 1

2

i.e., a = 12 and b = ¡ 32 as required.) the result in b is also true if

(g ± f)(x) = x for all x.

IBHL_WS


EXERCISE 1E

1 f(x), g(x) and h(x) are all reciprocal functions

which are all asymptotic about the x- and y-axes.

The graphs all lie in the 1st and 3rd quadrants.

The smaller the numerator, the closer is the graph to

the axes. Thus the graph of f(x) =1

xis closer to the

axes than g(x) =2

xfor corresponding values of x,

and g(x) =2

xis closer to the axes than h(x) =

4

x.

2 f(x), g(x) and h(x) are all reciprocal functions

which are all asymptotic about the x- and y-axes.

The graphs all lie in the 2nd and 4th quadrants.

The smaller the numerator, the closer is the graph to

the axes. Thus the graph of f(x) = ¡ 1x

is closer

to the axes than g(x)=¡ 2x

for corresponding val-

ues of x, and g(x) = ¡ 2x

is closer to the axes

than h(x) = ¡ 4x

.

EXERCISE 1F

1 a i ii f(x) passes through (0, 1) and (¡ 13

, 0)

) f¡1(x) passes through (1, 0) and (0, ¡ 13 )

f¡1(x) has slope¡ 13 ¡ 00¡ 1 =

¡ 13

¡1 =1

3

So, its equation isy ¡ 0x¡ 1 =

1

3

i.e., y =x¡ 13

i.e., f¡1(x) =x¡ 13iii f is y = 3x+ 1

so f¡1 is x = 3y + 1) x¡ 1 = 3y

) y =x¡ 13

i.e., f¡1(x) =x¡ 13

b i ii f(x) passes through (0, 12

) and (¡2, 0)) f¡1(x) passes through ( 1

2, 0) and (0, ¡2)

f¡1(x) has slope¡2¡ 00¡ 1

2

=¡2¡ 12

= 4

So, its equation isy ¡ 0x¡ 1

2

= 4i.e., y = 4x¡ 2

i.e., f¡1(x) = 4x¡ 2iii f is y =

x+ 2

4

so f¡1 is x =y + 2

4

) 4x = y + 2

) y = 4x¡ 2i.e., f¡1(x) = 4x¡ 2

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2 a i f is y = 2x+ 5

so f¡1 is x = 2y + 5

) x¡ 5 = 2y) y =

x¡ 52

i.e., f¡1(x) =x¡ 52

ii f(x) passes through

(0, 5) and (¡ 52

, 0)

) f¡1(x) passesthrough (5, 0) and

(0, ¡ 52

)

iii (f¡1 ± f)(x)= f¡1(2x+ 5)

2x+ 5¡ 52

=2x

2

= x

and (f ± f¡1)(x)= f(f¡1(x))

= f³x¡ 52

´= 2

³x¡ 52

´+ 5

= x¡ 5 + 5= x

b i f is y =3¡ 2x4

so f¡1 is x =3¡ 2y4

) 4x = 3¡ 2y) 4x¡ 3 = ¡2y

) y = ¡2x+ 32

i.e., f¡1(x) = ¡2x+ 32


(0, 34 ) and (

32 , 0)

) f¡1(x) passes

through ( 34 , 0) and

(0, 32

)

iii (f¡1 ± f)(x)= f¡1(f(x))

= f¡1³3¡ 2x4

´= ¡2

³3¡ 2x4

´+ 3

2

=3¡ 2x¡2 +

32

= ¡32+ x+ 3

2

= x

and (f ± f¡1)(x)= f(f¡1(x))

= f¡¡2x+ 3

2

¢=3¡ 2(¡2x+ 32 )

4

=3 + 4x¡ 3

4

=4x

4= x

c i f is y = x+ 3

so f¡1 is x = y + 3) y = x¡ 3

i.e., f¡1(x) = x¡ 3


(0, 3) and (¡3, 0)) f¡1(x) passes

through (3, 0) and

(0, ¡3)

iii (f¡1 ± f)(x)= f¡1(f(x))

= f¡1(x+ 3)

= (x+ 3)¡ 3= x

and (f ± f¡1)(x)= f(f¡1(x))

= f(x¡ 3)= (x¡ 3) + 3= x

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3 a b c

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4 a b Using the ‘horizontal line test’, f does not havean inverse function as a horizontal line through

y = x2 ¡ 4 cuts it more than once.c For x > 0, any horizontal line cuts it only once,

i.e., f does have an inverse function for x > 0.

5

EXERCISE 1G

1 a f(x) =1

xhas graph No vertical line cuts the graph more

than once, so it is a function.

No horizontal line cuts the graph

more than once.

Hence, f(x) =1

x, x 6= 0 has

an inverse function.

b f(x) =1

xi.e., y =

1

xhas an inverse function x =

1

yand so y =

1

x

So, f¡1(x) =1

xIt is a self-inverse function.

2 a f(x) =3x¡ 8x¡ 3 has graph The vertical line test shows it to

be a function.

The horizontal line test shows it

has an inverse function.

Symmetry about y = x showsit is a self-inverse function.

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b f(x) =3x¡ 8x¡ 3 i.e., y =

3x¡ 8x¡ 3 has an inverse function x =

3y ¡ 8y ¡ 3

) x(y ¡ 3) = 3y ¡ 8) xy ¡ 3x = 3y ¡ 8) y(x¡ 3) = 3x¡ 8

) y =3x¡ 8x¡ 3

i.e., f(x) = f¡1(x) ) it is a self-inverse function.

3 a

b i This graph satisfies the ‘horizontal line test’ and therefore has an inverse function.

ii, iii

c ii Domain fx: x > 1g (or fx: x 6 1g) iii Domain fx: x > 1g (or fx: x 6 ¡2g)

4 a f is y = x2, x 6 0

so f¡1 is x = y2, y 6 0

) y = ¡pxi.e., f¡1(x) = ¡px

b

5 a

b

c f is y = x2 ¡ 4x+ 3, x > 2so f¡1 is x = y2 ¡ 4y + 3, y > 2

i.e., x = (y ¡ 2)2 ¡ 4 + 3, y > 2= (y ¡ 2)2 ¡ 1, y > 2

) x+ 1 = (y ¡ 2)2, y > 2) y ¡ 2 = px+ 1, y > 2

) y = 2 +p1 + x, y > 2

i.e., f¡1(x) = 2 +p1 + x

as required

e f ± f¡1 = f(f¡1)=¡2 +

p1 + x

¢2 ¡ 4 ¡2 +p1 + x¢+ 3= 4 + 4

p1 + x+ 1 + x¡ 8¡ 4p1 + x+ 3

= x

f¡1 ± f = f¡1(f)= 2 +

p1 + x2 ¡ 4x+ 3

= 2 +p(x¡ 2)2

= 2 + x¡ 2= x

d i domain of f is fx: x > 2g,range is fy: y > ¡1g

ii domain of f¡1 is fx: x > ¡1g,range is fy: y > 2g

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If has an inverse function, then the inverse function must also be a function. Thus, itmust satisfy the ‘vertical line test’, i.e., no vertical line can cut it more than once. This conditionfor the inverse function cannot be satisfied if the original function does not satisfy the ‘horizontalline test’. Thus, the ‘horizontal line test’ is a valid test for the existence of an inverse function.

y f x= ( )

These graphs both fail the ‘horizontal line test’ so neither of these have inverse functions.

f x x x satisfies the ‘vertical line test’ so is thereforea function. It does not however satisfy the horizontal line test as anyhorizontal line above the vertex cuts the graph twice. Therefore itdoes not have an inverse function.

: 4 + 3! ¡2

For , all horizontal lines cut the graph no more thanonce. Therefore has an inverse function for .

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6 a f is y = (x+ 1)2 + 3, x > ¡1so f¡1 is x = (y + 1)2 + 3, y > ¡1

i.e., x¡ 3 = (y + 1)2, y > ¡1) y + 1 =

px¡ 3, y > ¡1, x > 3

) y =px¡ 3¡ 1, y > ¡1, x > 3

b

c i Domain fx: x > ¡1g, Range fy: y > 3gii Domain fx: x > 3g, Range fy: y > ¡1g

7 a g is y =8¡ x2

so g¡1 is x =8¡ y2

) 2x = 8¡ y) y = 8¡ 2x

i.e., g¡1(x) = 8¡ 2xNow g¡1(¡1) = 8¡ 2(¡1) = 10

b (f ± g¡1)(x) = 9) f(g¡1(x)) = 9

) f(8¡ 2x) = 9) 2(8¡ 2x) + 5 = 9) 16¡ 4x+ 5 = 9

) ¡4x = ¡12) x = 3

8 a i f is y = 5x

so, f(2) = 52

= 25

ii g is y =px

so g¡1 is x =py

) y = x2

i.e., g¡1(x) = x2, x > 0

) g¡1(4) = 42

) g¡1(4) = 16

b (g¡1 ± f)(x) = 25) g¡1(f(x)) = 25

) g¡1(5x) = 25

) (5x)2 = 25 fas g¡1(x) = x2, x > 0g

and so 52x = 52

) 2x = 2

) x = 1

9 Show: (f¡1 ± g¡1)(x) = (g ± f)¡1(x)f is y = 2x

so f¡1 is x = 2y

) y =x

2

i.e., f¡1(x) =x

2

g is y = 4x¡ 3so g¡1 is x = 4y ¡ 3

) 4y = x+ 3

) y =x+ 3

4

i.e., g¡1(x) =x+ 3

4

(g ± f)(x) = g(f(x))= g(2x)

= 4(2x)¡ 3i.e., (g ± f)(x) = 8x¡ 3

i.e., g ± f is y = 8x¡ 3so (g ± f)¡1 is x = 8y ¡ 3

) y =x+ 3

8

i.e., (g ± f)¡1(x) = x+ 38

Now, (f¡1 ± g¡1)(x) = f¡1(g¡1(x))

= f¡1³x+ 3

4

´

=

³x+ 3

4

´2

) (f¡1 ± g¡1)(x) = x+ 38

= (g ± f)¡1(x) as required

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10 a f is y = 2x

so f¡1 is x = 2y

) y =x

2

i.e., f¡1(x) =x

26= 2x

So, f¡1(x) 6= f(x)

b f is y = x

so f¡1 is x = y) y = x

i.e., f¡1(x) = x

So, f¡1(x) = f(x)

c f is y = ¡xso f¡1 is x = ¡y

) y = ¡xi.e., f¡1(x) = ¡xSo, f¡1(x) = f(x)

d f is y =2

x

so f¡1 is x =2

y

) y =2

x

i.e., f¡1(x) =2

x

So, f¡1(x) = f(x)

e f is y = ¡ 6x

so f¡1 is x = ¡6y

) y = ¡ 6x

i.e., f¡1(x) = ¡ 6x

So, f¡1(x) = f(x)

i.e., f¡1(x) = f(x) is true for parts b, c, d and e.

11 a f is y = 3x+ 1

so f¡1 is x = 3y + 1

) y =x¡ 13

i.e., f¡1(x) =x¡ 13

(f ± f¡1)(x) = f(f¡1(x))

= f³x¡ 13

´= 3

³x¡ 13

´+ 1

= x¡ 1 + 1i.e., (f ± f¡1)(x) = x

(f¡1 ± f)(x) = f¡1(f(x))= f¡1(3x+ 1)

=3x+ 1¡ 1

3

=3x

3

i.e., (f¡1 ± f)(x) = x = (f ± f¡1)(x) as required

b f is y =x+ 3

4so f¡1 is x =

y + 3

4

) 4x = y + 3

) y = 4x¡ 3i.e., f¡1(x) = 4x¡ 3

(f ± f¡1)(x) = f(f¡1(x))= f(4x¡ 3)

=4x¡ 3 + 3

4

=4x

4

i.e., (f ± f¡1)(x) = x

(f¡1 ± f)(x) = f¡1(f(x))

= f¡1³x+ 3

4

´= 4

³x+ 3

4

´¡ 3

= x+ 3¡ 3i.e., (f¡1 ± f)(x) = x = (f ± f¡1)(x) as required

IBHL_WS


c f is y =px for x > 0 so f¡1 is x =

py

) y = x2

i.e., f¡1(x) = x2 for x > 0

(f ± f¡1)(x) = f(f¡1(x))= f(x2)

=px2

i.e., (f ± f¡1)(x) = x

(f¡1 ± f)(x) = f¡1(f(x))= f¡1(

px)

= (px)2

i.e., (f¡1 ± f)(x) = x = (f ± f¡1)(x) as required

12 a f(x) passes through A(x, f(x)), so f¡1(x) passes through B(f(x), x)

b Substitute the coordinates of B(f(x), x) into y = f¡1(x) :

i.e., x = f¡1(f(x)) = (f¡1 ± f)(x)c B has coordinates (x, f¡1(x)) since it lies on y = f¡1(x),

so A has coordinates (f¡1(x), x) as f(x) is the inverse of f¡1(x).

Substitute the coordinates of A(f¡1(x), x) into y = f(x):

i.e., x = f(f¡1(x))

i.e., f(f¡1(x)) = x as required

REVIEW SET 1A

1 a f(x) = 2x¡ x2f(2) = 2(2)¡ 22

= 0

b f(¡3) = 2(¡3)¡ (¡3)2= ¡6¡ 9= ¡15

c f(¡ 12) = 2(¡ 1

2)¡ (¡ 1

2)2

= ¡1¡ 14

= ¡ 54

2 a i range is fy: y > ¡5g, domain is fx: x 2 Rgii x-intercepts are ¡1 and 5; y-intercept is ¡ 25

9

iii The graph passes the ‘vertical line test’ so is therefore a function.

iv No, as it fails the horizontal line test.

b i range is fy: y = 1 or ¡3g, domain is fx: x 2 Rgii there are no x-intercepts; y-intercept is 1iii The graph passes the ‘vertical line test’ so is therefore a function.iv No, as it fails the horizontal line test.

3 a domain is fx: x > ¡2g, range is fy: 1 6 y < 3gb domain is fx: x 2 Rg, range is fy: y = ¡1, 1 or 2g

4 a h(x) = 7¡ 3xh(2x¡ 1) = 7¡ 3(2x¡ 1)

= 7¡ 6x+ 3= 10¡ 6x

b h(2x¡ 1) = ¡2) 7¡ 3(2x¡ 1) = ¡2) 7¡ 6x+ 3 = ¡2

) ¡6x = ¡12) x = 2

5 f(x) = ax2 + bx+ c, where f(0) = 5, f(¡2) = 21 and f(3) = ¡4When f(0) = 5,

5 = a(0)2 + b(0) + c

) 5 = c

) c = 5 ..... (1)

When f(¡2) = 21,21 = a(¡2)2 + b(¡2) + c= 4a¡ 2b+ c= 4a¡ 2b+ 5 fusing (1)g

) 4a¡ 2b = 16) 2a¡ b = 8 and so b = 2a¡ 8 ..... (2)

IBHL_WS


When f(3) = ¡4, ¡4 = a(3)2 + b(3) + c) ¡4 = 9a+ 3b+ c) ¡4 = 9a+ 3b+ 5 fusing (1)g) ¡4 = 9a+ 3(2a¡ 8) + 5 fusing (2)g) ¡9 = 9a+ 6a¡ 24) 15 = 15a and so a = 1

Now, substituting a = 1 into (2) gives b = 2(1)¡ 8 = ¡6i.e., a = 1, b = ¡6, c = 5

6 a f(x) =1

x2is meaningless when x2 = 0

i.e., when x = 0

b c domain of f(x) is fx: x 6= 0grange of f(x) is fy: y > 0g

7 a f(g(x)) = f(x2 + 2)

= 2(x2 + 2)¡ 3= 2x2 + 4¡ 3= 2x2 + 1

b g(f(x)) = g(2x¡ 3)= (2x¡ 3)2 + 2= 4x2 ¡ 12x+ 9 + 2= 4x2 ¡ 12x+ 11

8 a i (f ± g)(x) = f(g(x))= f(

px)

= 1¡ 2px

ii (g ± f)(x) = g(f(x))= g(1¡ 2x)=p1¡ 2x

b The domain of f ± g is fx: x > 0g for px to be defined.The range of f ± g is fy: y 6 1g.The domain of g ± f is obtained by noticing that 1¡ 2x must be > 0.

) 2x 6 1

i.e., x 6 12

i.e., fx: x 6 12gi.e., x 2 ]¡1, 12 ]

The range of g ± f is fy: y > 0g i.e., y 2 [ 0, 1 [

9 a f(g(x)) =p1¡ x2

= f(1¡ x2)i.e., f(x) =

px ,

g(x) = 1¡ x2

b g(f(x)) =³x¡ 2x+ 1

´2= g

³x¡ 2x+ 1

´i.e., g(x) = x2,

f(x) =x¡ 2x+ 1

REVIEW SET 1B

1 g(x) = x2 ¡ 3xa g(x+ 1) = (x+ 1)2 ¡ 3(x+ 1)

= x2 + 2x+ 1¡ 3x¡ 3= x2 ¡ x¡ 2

b g(x2 ¡ 2) = (x2 ¡ 2)2 ¡ 3(x2 ¡ 2)= x4 ¡ 4x2 + 4¡ 3x2 + 6= x4 ¡ 7x2 + 10

y

x

2

1

xy �

IBHL_WS


2 a f(x) = 7¡ 4xi.e., y = 7¡ 4x

so f¡1(x) is x = 7¡ 4y) y =

7¡ x4

i.e., f¡1(x) =7¡ x4

b f(x) =3 + 2x

5

i.e., y =3 + 2x

5

so f¡1(x) is x =3 + 2y

5

) 5x = 3 + 2y

) y =5x¡ 32

i.e., f¡1(x) =5x¡ 323 a y = (x¡ 1)(x¡ 5)

i.e., x-intercepts are x = 1 and 5

) vertex is at x = 3, y = (3¡ 1)(3¡ 5) = 2£ (¡2) = ¡4i.e., vertex is at (3, ¡4)domain is fx: x 2 Rg, range is fy: y > ¡4g i.e., y 2 [¡4, 1[

b From the graph, domain is fx: x 6= 0, 2g, range is fy: y 6 ¡1 or y > 0g

4 a b

5 a f(x) = 4x+ 2

i.e., y = 4x+ 2

so f¡1(x) is x = 4y + 2

) y =x¡ 24

i.e., f¡1(x) =x¡ 24

b f(x) =3¡ 5x4

i.e., y =3¡ 5x4

so f¡1(x) is x =3¡ 5y4

) 4x = 3¡ 5y) y =

3¡ 4x5

i.e., f¡1(x) =3¡ 4x5

6 a b f(x) = 2x¡ 7i.e., y = 2x¡ 7

so f¡1(x) is x = 2y ¡ 7) y =

x+ 7

2

i.e., f¡1(x) =x+ 7

2

c f ± f¡1= f

¡f¡1(x)

¢= f

³x+ 7

2

´= 2

³x+ 7

2

´¡ 7

= x+ 7¡ 7= x

and f¡1 ± f= f¡1 (f(x))

= f¡1(2x¡ 7)=2x¡ 7 + 7

2

=2x

2

= x So f ± f¡1 = f¡1 ± f = e

y

x

2

5

2 5

y x�ƒ

�1

ƒ

y

x

2

2

y x�

ƒ��

ƒ

y

x3\Qw_

3\Qw_

�

�

ƒ��

ƒy x�

IBHL_WS


7 a, d b If x 6 ¡3, we have the graph to the left ofx = ¡3, and any horizontal line through thegraph cuts it no more than once.

Therefore it has an inverse function.

c g(x) = x2 + 6x+ 7, x 6 ¡3i.e., y = x2 + 6x+ 7, x 6 ¡3

so g¡1(x) is x = y2 + 6y + 7, y 6 ¡3= (y + 3)2 ¡ 9 + 7

) x+ 2 = (y + 3)2

) y + 3 = §px+ 2) y = ¡3§px+ 2

but y 6 ¡3, so y = ¡3¡px+ 2

e The range of g is fy: y > ¡2g, so the domain of g¡1 is fx: x > ¡2gand the range of g¡1 is fy: y 6 ¡3g

8 a h(x) = (x¡ 4)2 + 3, x > 4i.e., y = (x¡ 4)2 + 3, x > 4

so h¡1(x) is x = (y ¡ 4)2 + 3, y > 4) x¡ 3 = (y ¡ 4)2) y ¡ 4 = §px¡ 3

) y = 4§px¡ 3but y > 4, so y = 4 +

px¡ 3

i.e., h¡1(x) = 4 +px¡ 3, x > 3

b h ± h¡1= h

¡h¡1(x)

¢= h(4 +

px¡ 3)

= (4 +px¡ 3¡ 4)2 + 3

= (px¡ 3)2 + 3

= x¡ 3 + 3= x

h¡1 ± h= h¡1(h(x))

= h¡1¡(x¡ 4)2 + 3

¢= 4 +

p(x¡ 4)2 + 3¡ 3

= 4 +p(x¡ 4)2

= 4 + x¡ 4 as x > 4= x

9 f(x) = 3x+ 6

i.e., y = 3x+ 6

so f¡1(x) is x = 3y + 6

) y =x¡ 63

i.e., f¡1(x) =x¡ 63

h(x) =x

3

i.e., y =x

3

so h¡1(x) is x =y

3

) y = 3x

i.e., h¡1(x) = 3x

Now (f¡1 ± h¡1)(x) = f¡1(h¡1(x))= f¡1(3x)

=3x¡ 63

= x¡ 2

(h ± f)(x) = h(f(x))= h(3x+ 6)

=3x+ 6

3

i.e., y = x+ 2

i.e., (h ± f)¡1(x) is x = y + 2) y = x¡ 2

i.e., (h ± f)¡1(x) = x¡ 2i.e., (f¡1 ± h¡1)(x) = (h ± f)¡1(x) as required

g��

g

y x�x��

�� A

x

y

IBHL_WS

EXERCISE 2A

1 a 4, 13, 22, 31, .... b 45, 39, 33, 27, .... c 2, 6, 18, 54, .... d 96, 48, 24, 12, ....

2 a The sequence starts at 8 and each term is 8 more than the previous term. The next two termsare 40 and 48.

b The sequence starts at 2 and each term is 3 more than the previous term. The next two termsare 14 and 17.

c The sequence starts at 36 and each term is 5 less than the previous term. The next two termsare 16 and 11.

d The sequence starts at 96 and each term is 7 less than the previous term. The next two termsare 68 and 61.

e The sequence starts at 1 and each term is 4 times the previous term. The next two terms are256 and 1024.

f The sequence starts at 2 and each term is 3 times the previous term. The next two terms are162 and 486.

g The sequence starts at 480 and each term is half the previous term. The next two terms are 30and 15.

h The sequence starts at 243 and each term is one third of the previous term. The next two termsare 3 and 1.

i The sequence starts at 50 000 and each term is one fifth of the previous term. The next twoterms are 80 and 16.

3 a Each term is the square of the number of the term. The next three terms are 25, 36 and 49.

b Each term is the cube of the number of the term. The next three terms are 125, 216 and 343.

c

EXERCISE 2B

1 a f2ng generates the sequence 2, 4, 6, 8, 10, ...... (letting n = 1, 2, 3, 4, 5, ......)b f2n+ 2g generates the sequence 4, 6, 8, 10, 12, ...... (letting n = 1, 2, 3, 4, 5, ......)c f2n¡ 1) generates the sequence 1, 3, 5, 7, 9, ...... (letting n = 1, 2, 3, 4, 5, ......)d f2n¡ 3g generates the sequence ¡1, 1, 3, 5, 7, ...... (letting n = 1, 2, 3, 4, 5, ......)e f2n+ 3g generates the sequence 5, 7, 9, 11, 13, ...... (letting n = 1, 2, 3, 4, 5, ......)f f2n+ 11g generates the sequence 13, 15, 17, 19, 21, ...... (letting n = 1, 2, 3, 4, 5, ......)g f3n+ 1g generates the sequence 4, 7, 10, 13, 16, ...... (letting n = 1, 2, 3, 4, 5, ......)h f4n¡ 3g generates the sequence 1, 5, 9, 13, 17, ...... (letting n = 1, 2, 3, 4, 5, ......)

2 a f2ng generates the sequence 2, 4, 8, 16, 32, ...... (letting n = 1, 2, 3, 4, 5, ......)b f3£ 2ng generates the sequence 6, 12, 24, 48, 96, ...... (letting n = 1, 2, 3, 4, 5, ......)c f6£ ( 1

2)ng generates the sequence 3, 3

2, 34

, 38

, 316

, ...... (letting n = 1, 2, 3, 4, 5, ......)

d f(¡2)ng generates the sequence ¡2, 4, ¡8, 16, ¡32, ...... (letting n = 1, 2, 3, 4, 5, ......)

3 f15¡ (¡2)ng generates the sequence with first five terms:t1 = 15¡ (¡2)1 = 17, t2 = 15¡ (¡2)2 = 11, t3 = 15¡ (¡2)3 = 23,t4 = 15¡ (¡2)4 = ¡1, t5 = 15¡ (¡2)5 = 47

Chapter 2SEQUENCES AND SERIES

Each term is where is the number of the term. The next three terms are , and .n n n£ ( + 1) 30 42 56

IBHL_WS

Mathematics HL, Chapter 2 – SEQUENCES AND SERIES 49

EXERCISE 2C

1 a 17¡ 6 = 1128¡ 17 = 1139¡ 28 = 1150¡ 39 = 11

So, assuming that the pattern continues, consecutive terms differ by 11.

) the sequence is arithmetic with u1 = 6, d = 11.

b un = u1 + (n¡ 1)d= 6 + (n¡ 1)11

i.e., un = 11n¡ 5

c u50 = 11(50)¡ 5= 545

d Let un = 325 = 11n¡ 5) 330 = 11n

) n = 30

So, 325 is a member, i.e., u30.e Let un = 761 = 11n¡ 5

) 766 = 11n

) n = 69 711

, but n is an integer, so 761 is not a member of the sequence.

2 a 83¡ 87 = ¡479¡ 83 = ¡475¡ 79 = ¡4

So, assuming that the pattern continues, consecutive terms

differ by ¡4.) the sequence is arithmetic with u1 = 87, d = ¡4.

b un = u1 + (n¡ 1)d= 87 + (n¡ 1)(¡4)= 87¡ 4n+ 4

i.e., un = 91¡ 4n

c u40 = 91¡ 4(40)= 91¡ 160= ¡69

d Let un = ¡143 = 91¡ 4n) 4n = 234

) n = 58 12

but n is an integer, so

¡143 is not a memberof the sequence.

3 a un = 3n¡ 2 u1 = 3(1)¡ 2 = 1 un+1 = 3(n+ 1)¡ 2 = 3n+ 1un+1 ¡ un = (3n+ 1)¡ (3n¡ 2)

= 3, a constant

So, assuming that the pattern continues,

consecutive terms differ by 3.

) the sequence is arithmetic with u1 = 1

and d = 3.

b u1 = 1, d = 3 c u57 = 3(57)¡ 2 = 169d Let un = 450 = 3n¡ 2 i.e., 3n = 452 and so n = 150 23

So, try the two values on either side of n = 150 23

, i.e., for n = 150 and n = 151 :

u150 = 3(150)¡ 2 and u151 = 3(151)¡ 2= 448 = 451

i.e., u151 = 451 is the least term which is greater than 450.

4 a un =71¡ 7n2

= 35 12¡ 7

2n u1 =

71¡ 7(1)2

= 32

un+1 =71¡ 7(n+ 1)

2=71¡ 7n¡ 7

2=64¡ 7n2

= 32¡ 72n

un+1 ¡ un = (32¡ 72n)¡ (35 12 ¡ 72n) = ¡ 72 a constant

So, assuming that the pattern continues, consecutive terms differ by ¡ 72

.

) the sequence is arithmetic with u1 = 32, d = ¡ 72 .

b u1 = 32, d = ¡ 72 c u75 =71¡ 7(75)

2= ¡227

IBHL_WS

50 Mathematics HL, Chapter 2 – SEQUENCES AND SERIES

d Let un = ¡200 = 71¡ 7n2

i.e., ¡400 = 71¡ 7n ) 7n = 471) n = 67 2

7

So, try the two values on either side of n = 67 27 , i.e., for n = 67 and n = 68 :

u67 =71¡ 7(67)

2= ¡199 and u68 = 71¡ 7(68)

2= ¡202 1

2

i.e., terms of the sequence are less than ¡200 for n > 68.

5 a The terms are consecutive,

) k ¡ 32 = 3¡ kfequating common differencesg

) 2k = 35 and so k = 17 12

c The ter

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