haese harris publications - islandmaths12 - home standard level... · haese & harris publications...

for the international studentMathematics SL

Mathematics

John Owen

Robert Haese

Sandra Haese

Mark Bruce

Specialists in mathematics publishing

HAESE HARRIS PUBLICATIONS&

InternationalBaccalaureate

DiplomaProgramme

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MATHEMATICS FOR THE INTERNATIONAL STUDENTInternational Baccalaureate Mathematics SL Course

This book is copyright

Copying for educational purposes

Acknowledgements

Disclaimer

John Owen B.Sc., Dip.T.

Robert Haese B.Sc.

Sandra Haese B.Sc.

Mark Bruce B.Ed.

Haese & Harris Publications

3 Frank Collopy Court, Adelaide Airport, SA 5950, AUSTRALIA

Telephone: +61 8 8355 9444, Fax: + 61 8 8355 9471

Email:

National Library of Australia Card Number & ISBN 978-1-876543-03-7

Haese & Harris Publications 2004

Published by Raksar Nominees Pty Ltd

3 Frank Collopy Court, Adelaide Airport, SA 5950, AUSTRALIA

First Edition 2004

2005 three times , 2006, 2007, 2008

Cartoon artwork by John Martin. Artwork by Piotr Poturaj and David Purton

Cover design by Piotr Poturaj

Computer software by David Purton

Typeset in Australia by Susan Haese (Raksar Nominees). Typeset in Times Roman 10 /11

The textbook and its accompanying CD have been developed independently of the International

Baccalaureate Organization (IBO). The textbook and CD are in no way connected with, or endorsed

by, the IBO.

. Except as permitted by the CopyrightAct (any fair dealing for the purposes of

private study, research, criticism or review), no part of this publication may be reproduced, stored in 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 be made to Haese &

Harris Publications.

: Where copies of part or the whole of the book are made under

Part VB of the Copyright Act, the law requires that the educational institution or the body that

administers it has given a remuneration notice to Copyright Agency Limited (CAL). For information,

contact the CopyrightAgency Limited.

: While every attempt has been made to trace and acknowledge copyright, the

authors and publishers apologise for any accidental infringement where copyright has proved

untraceable. They would be pleased to come to a suitable agreement with the rightful owner.

: All the internet addresses (URLs) given in this book were valid at the time of printing.

While the authors and publisher regret any inconvenience that changes of address may cause readers,

no responsibility for any such changes can be accepted by either the authors or the publisher.

Reprinted (with minor corrections)

\Qw_ \Qw_

[email protected]

www.haeseandharris.com.auWeb:

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1_00\002IB100.CDR Wednesday, 12 March 2008 11:50:59 AM PETERDELL

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

Mathematics for the International Student: Mathematics SL has been written to embracethe syllabus for the new two-year Mathematics SL Course, which is one of the courses ofstudy in the International Baccalaureate Diploma Programme. It is not our intention to definethe course. Teachers are encouraged to use other resources. We have developed the book inde-pendently of the International Baccalaureate Organization (IBO) in consultation with manyexperienced teachers of IB Mathematics. The text is not endorsed by the IBO.

This package is language rich and technology rich. The combination of textbook and interac-tive Student CD will foster the mathematical development of students in a stimulating way.Frequent use of the interactive features on the CD is certain to nurture a much deeper under-standing and appreciation of mathematical concepts.

The book contains many problems from the basic to the advanced, to cater for a wide rangeof student abilities and interests. While some of the exercises are simply designed to buildskills, every effort has been made to contextualise problems, so that students can see every-day uses and practical applications of the mathematics they are studying, and appreciate theuniversality of mathematics.

Emphasis is placed on the gradual development of concepts with appropriate worked exam-ples, but we have also provided extension material for those who wish to go beyond thescope of the syllabus. Some proofs have been included for completeness and interest al-though they will not be examined.

For students who may not have a good understanding of the necessary background knowl-edge for this course, we have provided printable pages of information, examples, exercisesand answers on the Student CD. To access these pages, simply click on the Backgroundknowledge icon when running the CD.

It is not our intention that each chapter be worked through in full. Time constraints will notallow for this. Teachers must select exercises carefully, according to the abilities and priorknowledge of their students, to make the most efficient use of time and give as thorough cov-erage of work as possible. Investigations throughout the book will add to the discovery aspectof the course and enhance student understanding and learning. Many Investigations are suit-able for portfolio assignments and have been highlighted in the table of contents. Reviewsets appear at the end of each chapter and a suggested order for teaching the two-year courseis given at the end of this Foreword.

The extensive use of graphics calculators and computer packages throughout the book en-ables students to realise the importance, application and appropriate use of technology. Nosingle aspect of technology has been favoured. It is as important that students work with apen and paper as it is that they use their calculator or graphics calculator, or use a spreadsheetor graphing package on computer.

The interactive features of the CD allow immediate access to our own specially designedgeometry packages, graphing packages and more. Teachers are provided with a quick andeasy way to demonstrate concepts, and students can discover for themselves and re-visitwhen necessary.

FOREWORD


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Instructions appropriate to each graphic calculator problem are on the CD and can be printedfor students. These instructions are written for Texas Instruments and Casio calculators.

In this changing world of mathematics education, we believe that the contextual approachshown in this book, with the associated use of technology, will enhance the students under-standing, knowledge and appreciation of mathematics, and its universal application.

We welcome your feedback.

Email:

Web: JTO RCH

SHH MFB

[email protected]

www.haeseandharris.com.au

Thank you

The authors and publishers would like to thank all those teachers who offered advice andencouragement. Many of them read the page proofs and offered constructive comments andsuggestions. These teachers include: Marjut Menp, Cameron Hall, Paul Urban, FranOConnor, Glenn Smith, Anne Walker, Malcolm Coad, Ian Hilditch, Phil Moore, JulieWilson, David Martin, Kerrie Clements, Margie Karbassioun, Brian Johnson, Carolyn Farr,Rupert de Smidt, Terry Swain, Marie-Therese Filippi, Nigel Wheeler, Sarah Locke, RemaGeorge.

For the first year, it is suggested that students work progressively from Chapter 1 throughto Chapter 16, although some teachers may prefer to leave Chapter 16 Vectors in 3dimensions until the second year.

Descriptive statistics and Probability (Chapters 18 and 19) could possibly be taught thefirst year. Alternatively, calculus could be introduced (Chapters 20-22), but most teacherswill probably prefer to leave calculus until the second year and have students workprogressively from Chapter 20 though to Chapter 29.

We invite teachers who have their preferred order, to email their suggestions to us. We canput these suggestions on our website to be shared with other teachers.

TEACHING THE TWO-YEAR COURSE A SUGGESTED ORDER


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The CD is ideal for independent study. Frequent use will nurture a deeper understanding ofMathematics. Students can revisit concepts taught in class and undertake their own revisionand practice. The CD also has the text of the book, allowing students to leave the textbook atschool and keep the CD at home.

The icon denotes an active link on the CD. Simply click the icon to access a range ofinteractive features:

spreadsheets and worksheets

video clips

graphing and geometry software

graphics calculator instructions

computer demonstrations and simulations

background knowledge

For those who want to make sure they have the prerequisite levels of understanding for thisnew course, printable pages of background information, examples, exercises and answers areprovided on the CD. Click the Background knowledge icon.

Graphics calculators: Instructions for using graphics calculators are also givenon the CD and can be printed. Instructions are given for Texas Instruments andCasio calculators. Click on the relevant symbol (TI or C) to access printableinstructions.

Students are reminded that in assessment tasks, including examination papers, unless other-wise stated in the question, all numerical answers must be given exactly or to three significantfigures.

If you find an error in this book please notify us by emailing .

As a help to other teachers and students, we will include the correction on our website andcorrect the book at the first reprint opportunity.

[email protected]

USING THE INTERACTIVE STUDENT CD

NOTE ON ACCURACY

ERRATA

CD LINK

TI

C


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mailto:[email protected]

6 TABLE OF CONTENTS

TABLE OF CONTENTS

BACKGROUND KNOWLEDGE

11

1 FUNCTIONS 15

2 SEQUENCES AND SERIES 35

3 EXPONENTS 61

to access, click active icon on CD

* Portfolio Assignments

*

denotes ideas for possible

Abbreviations used in this book

A Operations with surds (radicals)

B Standard form (scientific notation)

C Number systems and set notation

D Algebraic simplification

E Linear equations and inequalities

F Absolute value (modulus)

G Product expansion

H Factorisation

: Another factorisationtechnique

I Formula rearrangement

J Adding and subtracting algebraic fractions

K Congruence and similarity

ANSWERS

Summary of circle properties

Summary of measurement facts

A Relations and functions

B Interval notation, domain and range

C Function notation

: Fluid filling functions

D Composite functions,

E The reciprocal function

F Inverse functions

G The identity function

Review set 1A

Review set 1B

A Number patterns

B Sequences of numbers

C Arithmetic sequences

D Geometric sequences

E Series

F Sigma notation

: Von Kochs Snowflake curve

Review set 2A

Review set 2B

Review set 2C

A Index notation

B Negative bases

C Index laws

D Rational indices

10

CD

CD

CD

CD

CD

CD

CD

CD

CD

CD

CD

CD

CD

11

12

16

19

22

24

26

27

28

31

32

33

36

38

41

44

51

57

58

59

59

60

62

63

65

71

Investigation

Investigation

Investigation

E Algebraic expansion

F Exponential equations

G Graphs of exponential functions

: Exponential graphs

H Growth

I Decay

Review set 3A

Review set 3B

Review set 3C

Review set 3D

A Introduction

B Logarithms in base 10

: Discovering the laws oflogarithms

C Laws of logarithms

D Exponential equations (using logarithms)

E Growth and decay revisited

F Compound interest revisited

G The change of base rule

Review set 4A

Review set 4B

A Introduction 104

: occurs naturally 104

: Continuous compoundinterest

B Natural logarithms

: The laws of naturallogarithms

C Laws of natural logarithms

D Exponential equations involving

E Growth and decay revisited

F Inverse functions revisited

Review set 5A

Review set 5B

A Families of functions

: Function families

B Key features of functions

C Transformations of graphs

D Functional transformations

Review set 6

A Assumed knowledge

: Finding where linesmeet using technology

B Equations of lines

C Distance between two points

D Midpoints and perpendicular bisectors

Review set 7A

73

74

75

76

79

81

84

84

85

86

88

90

92

93

96

97

99

100

101

102

105

108

109

110

112

112

114

116

117

120

120

122

123

126

127

131

135

138

141

144

146

*

*

*

Investigation

Investigation 1

Investigation 2

Investigation 3

Investigation

Investigation

Investigation

e

e

4 LOGARITHMS 87

5 NATURAL LOGARITHMS 103

6 GRAPHING AND TRANS-

FORMING FUNCTIONS 119

7 COORDINATE GEOMETRY 129

f gx

1x

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TABLE OF CONTENTS 7

Review set 11

A Areas of triangles

B Sectors and segments

C The cosine rule

D The sine rule

: The ambiguous case

E Using the sine and cosine rules

Review set 12A

Review set 12B

A Observing periodic behaviour

B Radian measure and periodic propertiesof circles

C The unit circle (revisited)

D The sine function

: The family

: The family

: The families

E Modelling using sine functions

F Equations involving sine

: The area under an arch of284

G The cosine function 285

H Solving cosine equations 287

I Trigonometric relationships 289

J Double angle formulae 292

: Double angle formulae 292

K The tangent function 295

L Tangent equations 298

M Other equations involving 301

Review set 13A 302

Review set 13B 302

Review set 13C 303

Review set 13D 304

Review set 13E 305

A Introduction

B Addition and subtraction of matrices

C Multiples of matrices

D Matrix algebra for addition

E Matrix multiplication

F Using technology

G Some properties of matrix multiplication

H The inverse of a matrix

I Solving a pair of linear equations

J The determinant

K The inverse of a matrix

L systems with unique solutions

233

236239241244245248251252

257

260265270271

271

273275278

308311314316317321325328331334337337

Investigation

Investigation 5

*

*

*

*

Investigation 1

Investigation 2

Investigation 3

Investigation 4

12 NON RIGHT ANGLED TRIANGLE

TRIGONOMETRY 235

13 PERIODIC PHENOMENA 255

14 MATRICES 307

and

tanx

Review set 7B

Review set 7C

A Function notation

B Graphs of quadratic functions

: Graphing

: Graphing

C Completing the square

D Quadratic equations

E The quadratic formula

F Solving quadratic equations withtechnology

G Problem solving with quadratics

H Quadratic graphs (review)

I The discriminant,

J Determining the quadratic from a graph

K Where functions meet

L Quadratic modelling

Review set 8A

Review set 8B

Review set 8C

Review set 8D

Review set 8E

A Binomial expansions

: The binomial expansionsof

: values

B The general binomial expansion

Review set 9

A Pythagoras rule (review) 205

B Pythagoras rule in 3-D problems 207

: Shortest distance 208

C Right angled triangle trigonometry 209

D Finding sides and angles 211

E Problem solving using trigonometry 217

F The slope of a straight line 221

Review set 10A 222

Review set 10B 223

Review set 10C 223

A The unit quarter circle

B Obtuse angles

C The unit circle

: Parametric equations

147148

153154

155

155160162168

170171174178182185186190191191192193

196

196199199202

226228231232

*

*

Investigation 1

Investigation 2

Investigation 1

Investigation 2

Investigation

Investigation

8 QUADRATIC EQUATIONS AND

FUNCTIONS 149

9 THE BINOMIAL THEOREM 195

10

203

11 THE UNIT CIRCLE 225

PRACTICAL TRIGONOMETRY

WITH RIGHT ANGLED

TRIANGLES

or CnrCn r

(a+ b)n ; n > 4

y = a (x h)2 + k

y = a (x ) (x )

f : ! ax2 + bx+ c

y = sin

y = sin (xC) y = sinx+D

y = Bsin x; B > 0

y = A sinx

3 3

3 33 3

2 2


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8 TABLE OF CONTENTS

* Investigation: Using matrices incryptography

Review set 14A

Review set 14B

Review set 14C

Review set 14D

Review set 14E

A Vectors

B Operations with vectors

C Vectors in component form

D Vector equations

E Vectors in coordinate geometry

F Parallelism

G Unit vectors

H Angles and scalar product

Review set 15A

Review set 15B

Review set 15C

Review set 15D

A 3-dimensional coordinates

B 3-dimensional vectors

C Algebraic operations with 3-D vectors

D Parallelism

E Unit vectors

F Collinear points and ratio of division extension

G The scalar product of 3-D vectors

Review set 16A

Review set 16B

A Vector and parametric form of a line in2-dimensional geometry

B The velocity vector of a moving object

C Constant velocity problems

: The two yachts problem

D The closest distance

E Geometric applications of

F Lines in space

G Line classification

Review set 17A

Review set 17B

A Statistical enquiries

: Statistics from the internet

B Populations and samples

341342343344345345

348352360365366368369371376376377378

380383386389390

391392395396

399401403405406409411414416417

CDCDCD

Investigation

Investigation

15 VECTORS IN 2-DIMENSIONS 347

16 VECTORS IN 3-DIMENSIONS 379

17 LINES IN THE PLANE AND

IN SPACE 397

18 DESCRIPTIVE STATISTICS 419

BACKGROUND KNOWLEDGE IN

STATISTICS

420

to access, click

active icon on CD

C Presenting and interpreting data CD

ANSWERS CD

A Continuous numerical data and histograms 421

B Measuring the centre of data 425

: Merits of the meanand median 427

C Cumulative data 442

D Measuring the spread of data 445

E Statistics using technology 453

F Variance and standard deviation 456

G The significance of standard deviation 461

Review set 18A 463

Review set 18B 465

A Experimental probability

: Tossing drawing pins

: Coin tossing experiments

: Dice rolling experiments

B Sample space

C Theoretical probability

D Using grids to find probabilities

E Compound events

: Probabilities ofcompound events

: Revisiting drawing pins

F Using tree diagrams

G Sampling with and without replacement

: Sampling simulation

: How many should I plant?

H Pascals triangle revisited

I Sets and Venn diagrams

J Laws of probability

K Independent events revisited

Review set 19A

Review set 19B

: The speed of fallingobjects

A Rate of change

B Instantaneous rates of change

: Instantaneous speed

Review set 20

A The idea of a limit

: The slope of a tangent

B Derivatives at a given -value

C The derivative function

: Finding slopes offunctions with technology

D Simple rules of differentiation

: Simple rules ofdifferentiation

E The chain rule

Investigation

Investigation 1

Investigation 2

Investigation 3

Investigation 4

Investigation 5

Investigation 7

Investigation 1

Investigation 2

Investigation 1

Investigation 2

Investigation 3

470470471472474475479480

481481485487490491492494499503505505

508509513513519

522522525530

531533

534537

* Investigation 6

x

19 PROBABILITY 467

20 INTRODUCTION TO CALCULUS 507

21 DIFFERENTIAL CALCULUS 521

r = a + tb


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TABLE OF CONTENTS 9

26 INTEGRATION 649

27 TRIGONOMETRICINTEGRATION 689

28 VOLUMES OF REVOLUTION 701

29 STATISTICAL DISTRIBUTIONS 709

ANSWERS AND INDEX 745

A Reviewing the definite integral

B The area function

: The area function

C Antidifferentiation

D The fundamental theorem of calculus

E Integration

F Integrating and

G Integrating by substitution

H Distance from velocity

I Definite integrals

: and areas

J Finding areas

K Problem solving by integration

Review set 26A

Review set 26B

Review set 26C

A Basic trigonometric integrals

B Integrals of trigonometric functionsof the form

C Definite integrals

D Area determination

Review set 27A

Review set 27B

A Solids of revolution

B Volumes for two defining functions

Review set 28

A Discrete random variables

B Discrete probability distributions

C Expectation

: Concealed number tickets

D The mean and standard deviation of adiscrete random variable

E The binomial distribution

F Mean and standard deviation of abinomial random variable

: The mean and standarddeviation of a binomial random variable

G Normal distributions

: Standard deviationsignificance

: Mean and standard

deviation of

H The standard normal distribution( -distribution)

I Applications of the normal distribution

Review set 29A

Review set 29B

Review set 29C

650655656656658662668670673675677677682685686687

690

692695698700700

702705708

710712714717

717721

724

726727

730

733

734739741742743

Investigation 1

Investigation 2

Investigation 1

Investigation 2

Investigation 3

Investigation 4

z

f (ax+ b)

f (u)u0 (x)

R baf (x) dx

eax+b (ax+ b)n

Investigation 4

Investigation

Investigation 1

Investigation 2

Investigation 3

Investigation 1

Investigation 2

: Differentiating composites

F Product and quotient rules

G Tangents and normals

H The second derivative

Review set 21A

Review set 21B

Review set 21C

A Functions of time

B Time rate of change

C General rates of change

D Motion in a straight line

: Displacement, velocityand acceleration graphs

E Curve properties

F Rational functions

G Inflections and shape type

H Optimisation

I Economic models

Review set 22A

Review set 22B

A Derivatives of exponential functions

: The derivative of

: Finding when

and

B Using natural logarithms

C Derivatives of logarithmic functions

: The derivative of

D Applications

Review set 23A

Review set 23B

A The derivative of , ,

B Maxima/minima with trigonometry

Review set 24

A Areas where boundaries are curved

: Finding areas usingrectangles

B Definite integrals

:

Review set 25

538541545550552553554

556558560564

568571579583587597600602

606606

607

611615615618622623

627632634

638

640642

646

648

a

x

x x x

ln

sin cos tan

22 APPLICATIONS OF

DIFFERENTIAL CALCULUS 555

23 DERIVATIVES OF

EXPONENTIAL AND

LOGARITHMIC FUNCTIONS 605

24 DERIVATIVES OF

TRIGONOMETRIC FUNCTIONS 625

25 AREAS WITHIN CURVED

BOUNDARIES 637

,

dy

dx= ax

y = axy = ax

R 10

1 x2dxR 1

0

x2 1 dx

z = x xs


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SYMBOLS AND NOTATION USED IN THIS BOOK

This notation is based on that indicated by the International Organisation of Standardisation.

N the set of all natural numbers

f0, 1, 2, 3, 4, 5, ..... gZ the set of all integers

f0, 1, 2, 3, 4, 5, .....gZ+ the set of all positive integers

f1, 2, 3, 4, 5, .....gQ the set of all rational numbers

Q+ the set of all positive rational

numbers

R the set of all real numbers

R+ the set of all positive real numbers

n(S) the number of elements in set S

2 is an element of=2 is not an element of? the empty set, or null set

U the universal set

[ union\ intersectionjxj the modulus of x,

or the absolute value of x

j x j = x if x > 0 or x if x 6 0

dy

dxthe derivative of y with respect to x

f 0(x) the derivative of f(x) withrespect to xZ

y dx the indefinite integral of y with

respect to xZ ba

y dx the definite integral of y withrespect to x from x = a tox = b

ex the exponential function

logax the logarithm of x, in base a

ln x the natural logarithm of x, loge x

sinx, cosx the circular functionsand tanx

P(x, y) point P with coordinates x and y

]A the angle at A

]PQR the angle between QP and QR

PQR the triangle with vertices P, Q and Rv vector v!AB the vector from point A to point B

un the nth term of a sequence

d the common difference of anarithmetic sequence

r the common ratio of a geometricsequence

Sn u1 + u2 + u3+ ..... +unthe sum of the first n terms of asequence

S1 the sum to infinity of asequence

nXi=1

ui u1 + u2 + u3+ ..... +un

n

r

the binomial coefficient of the(r + 1)th term in the expansionof (a+ b)n

f : x 7! y f is the function where x goes to yf(x) the image of x operated on by f

f1(x) the inverse function of f(x)f g the composite function of f and glimx!a

f(x) the limit of f(x) as x tends to a

a the position vector of A,!OA

i, ,j k unit vectors in the direction of the

yx --, and z-axis respectively

jaj the magnitude of aa b the scalar product of a and bA1 the inverse of matrix Adet A the determinant of matrix A

I the identity matrix under P(A) the probability of event A

P(A0) the probability of event not AP(A/B) the probability of A occurring

given that B has occurred

N(, 2) the normal distribution withmean and variance 2

population mean

2 population variance

population standard deviation

x sample mean

s 2n sample variance

sn sample standard deviation


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SUMMARY OF CIRCLE PROPERTIES

BACKGROUND KNOWLEDGE

A circle is a set of points which are equidistant froma fixed point, which is called its centre.

The circumference is the distance around the entirecircle boundary.

An arc of a circle is any continuous part of the circle.

A chord of a circle is a line segment joining any twopoints of a circle.

arcchord

centre

circle

A semi-circle is a half of a circle. A diameter of a circle is any chord passing

A radius of a circle is any line segment

A tangent to a circle is any line which

diameter

radius

tangent

point of contact

through its centre.

joining its centre to any point on the circle.

touches the circle in exactly one point.

BACKGROUND KNOWLEDGE AND GEOMETRIC FACTS 11

Before starting this course you can make sure that youhave a good understanding of the necessary backgroundknowledge.

Click on the icon alongside to obtain a printable set ofexercises and answers on this background knowledge.

BACKGROUND

KNOWLEDGE

Below is a summary of well known results called theorems. Click on the appropriate icon to

revisit them.

Name of theorem Statement Diagram

Angle in a

semi-circle

The angle in a semi-

circle is a right angle.If then ]ABC = 90o.

OA C

B

GEOMETRY

PACKAGE


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12 BACKGROUND KNOWLEDGE AND GEOMETRIC FACTS

Name of theorem Statement Diagram

Chords of a

circle

The perpendicular

from the centre of

a circle to a chord

bisects the chord.

If then AM = BM.

A

MB

O GEOMETRYPACKAGE

Radius-tangent The tangent to a cir-

cle is perpendicular

to the radius at the

point of contact.

If then ]OAT = 90o.

Tangents from

an external

point

Tangents from an ex-

ternal point are equal

in length.

If then AP = BP.

Angle at the

centre

The angle at the centre

of a circle is twice the

angle on the circle sub-

tended by the same arc.

If then ]AOB = 2]ACB.

Angles

subtended

by the

same arc

Angles subtended by an

arc on the circle are

equal in size.

If then ]ADB = ]ACB.

Angle between

a tangent and

a chord

The angle between a tan-

gent and a chord at the

point of contact is equal

to the angle subtended

by the chord in the al-

ternate segment.

If then ]BAS = ]BCA.

A

B

O P

AB

DC

A

B

S

C

T

GEOMETRY

PACKAGE

GEOMETRY

PACKAGE

GEOMETRY

PACKAGE

GEOMETRY

PACKAGE

GEOMETRY

PACKAGE

A

T

O

SUMMARY OF MEASUREMENT FACTS

PERIMETER FORMULAE

The distance around a closed figure is its perimeter.

A B

C

O


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BACKGROUND KNOWLEDGE AND GEOMETRIC FACTS 13

P =4 l P =2(l+w) P =a+ b+ c l=( )360 2rorC=2rC=d

triangle

b

c

a

square

l

l

w

rectangle circle

r d

arc

r

AREA FORMULAE

Shape Figure Formula

Rectangle Area = length width

Triangle Area = 12base height

Parallelogram Area = base height

T

Trapezoidor

rapezium Area =(

(

a+ b

2h

Circle Area = r2

SectorArea =

360r2

length

width

base base

height

r

r

height

base

a

b

h)

)

The length ofan arc is a

fraction of thecircumference

of a circle.

RECTANGULAR PRISM

A = 2(ab+bc+ac)

SURFACE AREA FORMULAE

ab

c

For some shapes we can derive formula for perimeter The formulae

for the most common shapes are given below:

a .


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14 BACKGROUND KNOWLEDGE AND GEOMETRIC FACTS

CYLINDER CONE

Object Outer surface area

Hollow cylinder A=2rh

(no ends)

Open can A=2rh+r2

(one end)

Solid cylinder A=2rh+2r2

(two ends)

r

h

hollow

hollow

r

h

hollow

solid

r

h

solid

solid

Object Outer surface area

Open cone A=rs(no base)

Solid cone A=rs+r2

(solid)

r

s

r

s

SPHERE

Area,

A = 4r2

VOLUME FORMULAE

r

Object

Solids of

uniform

cross-section

Volume of uniform solid

= area of end length

Pyramids

and

cones

Volume of a pyramid

or cone

= 13

(area of base height)

SpheresVolume of a sphere

= 43r3

r

height

endheight

end

base

height

base

h

height

VolumeF eigur


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FunctionsFunctions

11Chapter

A

B

C

D

E

F

G

Relations and functions

Interval notation, domain andrange

Function notation

: Fluid fillingfunctions

Composite functions,

The reciprocal function

Inverse functions

The identity function

Review set 1A

Review set 1B

Investigation

f g

Contents:

x ! 1x


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The charges for parking a car in a short-term car park at an

Airport are given in the table shown alongside.

There is an obvious relationship between time spent and the

cost. The cost is dependent on the length of time the car is

parked.

Looking at this table we might ask: How much would be

charged for exactly one hour? Would it be $5 or $9?

To make the situation clear, and to avoid confusion, we

could adjust the table and draw a graph. We need to indicate

that 2-3 hours really means for time over 2 hours up to andincluding 3 hours i.e., 2 < t 6 3.

Car park charges

Period (h) Charge

0 - 1 hours $5:001 - 2 hours $9:002 - 3 hours $11:003 - 6 hours $13:006 - 9 hours $18:009 - 12 hours $22:0012 - 24 hours $28:00

So, we

now haveCar park charges

Period Charge

0 < t 6 1 hours $5:001 < t 6 2 hours $9:002 < t 6 3 hours $11:003 < t 6 6 hours $13:006 < t 6 9 hours $18:009 < t 6 12 hours $22:0012 < t 6 24 hours $28:00

The parking charges example is clearly the latter as any real value of time ( t hours) in theinterval 0 < t 6 24 is represented.

For example: ft : 0 < t 6 24g is the domain for the car park relation f2, 1, 4g is the domain of f(1, 5), (2, 3), (4, 3), (1, 6)g.

RELATIONS AND FUNCTIONSA

exclusion

inclusion

charge ($)

time ( )t

3 6 9

10

20

30

12 15 18 21 24

In mathematical terms, because we have

relationship between two variables, time and

cost, the schedule of charges is an example

of

relation may consist of finite number of

ordered pairs, such as ), ),

), or an infinite number of or-

dered pairs.

a

a .

A a

( , ( ,

( , ( , )

relation

f g

1 5 2 34 3 1 6

16 FUNCTIONS (Chapter 1)

The set of possible values of the variable on the horizontal axis is called the of the

relation.

domain


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The set which describes the possible y-values is called the range of the relation.

For example: the range of the car park relation is f5, 9, 11, 13, 18, 22, 28g the range of f(1, 5), (2, 3), (4, 3), (1, 6)g is f3, 5, 6g.

We will now look at relations and functions more formally.

A relation is any set of points on the Cartesian plane.

A relation is often expressed in the form of an equation connecting the variables x and y.

For example y = x+ 3 and x = y2 are the equations of two relations.

These equations generate sets of ordered pairs.

Their graphs are:

However, a relation may not be able to be defined by an equation. Below are two examples

which show this:

RELATIONS

y y

x

x3

x y= 2

y x= + 32

4

(1) (2)

FUNCTIONS

A function is a relation in which no two different ordered pairs have

the same x-coordinate (first member).

We can see from the above definition that a function is a special type of relation.

y y

x x

All points in thefirst quadrantare a relation.

> 0, > 0x y

These 13 pointsform a relation.

TESTING FOR FUNCTIONS

Algebraic Test:

If a relation is given as an equation, and the substitution of any value

for x results in one and only one value of y, we have a function.

FUNCTIONS (Chapter 1) 17


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1 Which of the following sets of ordered pairs are functions? Give reasons.

a (1, 3), (2, 4), (3, 5), (4, 6) b (1, 3), (3, 2), (1, 7), (1, 4)c (2, 1), (2, 0), (2, 3), (2, 11) d (7, 6), (5, 6), (3, 6), (4, 6)e (0, 0), (1, 0), (3, 0), (5, 0) f (0, 0), (0, 2), (0, 2), (0, 4)

For example: y = 3x 1 is a function, as for any value of x there is only one valueof y

x = y2 is not a function since if x = 4, say, then y = 2.

Geometric Test (Vertical Line Test):

If we draw all possible vertical lines on the graph of a relation,

the relation:

is a function if each line cuts the graph no more than once is not a function if one line cuts the graph more than once.

DEMO

Which of the following relations are functions?

a b c

a b c

Example 1

y

x

y

x

y

x

y

x

y

x

a function a function

y

x

not a function

GRAPHICAL NOTE

If a graph contains a small open circle end point such as , the end point isnot included.

If a graph contains a small filled-in circle end point such as , the end pointis included.

If a graph contains an arrow head at an end such as then the graph continuesindefinitely in that general direction, or the shape may repeat as it has done previously.

EXERCISE 1A



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2 Use the vertical line test to determine which of the following relations are functions:

a b c

d e f

g h i

3 Will the graph of a straight line always be a function? Give evidence.

4 Give algebraic evidence to show that the relation x2 + y2 = 9 is not a function.

y y y

x xx

y y y

x

x

x

y y y

x

xx

INTERVAL NOTATION, DOMAIN AND RANGEB


DOMAIN AND RANGE

The domain of a relation is the set of permissible values that x may have.

The range of a relation is the set of permissible values that y may have.

For example:

(1)All values of x > 1 are permissible.So, the domain is fx: x > 1g.All values of y > 3 are permissible.So, the range is fy: y > 3g.

(2) x can take any value.

So, the domain is fx: x is in Rg.y cannot be > 1

) range is fy: y 6 1g.

y

x

y

x

(2, 1)


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x

x

x

x

a b

a b

(3) x can take all values except x = 2:

So, the domain is fx: x 6= 2g.Likewise, the range is fy: y 6= 1g.

y

x

y = 1

x = 2


Intervals have corresponding graphs.

For example:

fx: x > 3g or [3, 1[ is read the set of all x such that x is greater than or equalto 3 and has number line graph

fx: x < 2g or ]1, 2[ has number line graph

fx: 2 < x 6 1g or ]2, 1] has number line graph

fx: x 6 0 or x > 4gi.e., ]1, 0] or ]4, 1[ has number line graph

Note: for numbers between a and b we write a < x < b or ]a, b[.

for numbers outside a and b we write x < a or x > b

i.e., ]1, a[ or ]b, 1[.

The domain and range of a relation are best described where appropriate using interval

notation.

For example: The domain consists of all real

x such that x > 3 and wewrite this as

Likewise the range would be fy: y > 2g.

For this profit function:

the domain is fx: x > 0g the range is fy: y 6 100g.

the set of all such that

2

3

x

y

(3, 2)

range

domain

100

10

items made ( )x

profit ($)

range

domain

fx : x > 3g


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1 For each of the following graphs find the domain and range:

a b c

d e f

g h i

2 Use a graphics calculator to help sketch carefully the graphs of the following functions

and find the domain and range of each:

a f(x) =px b f(x) =

1

x2c f(x) =

p4 x

d y = x2 7x+ 10 e y = 5x 3x2 f y = x+ 1x

EXERCISE 1B

y

x

(

y

x

( 1, 1 (5, 3)

y

x

y =

x = 2

For each of the following graphs state the domain and range:

a b

a Domain is fx: x 6 8g.Range is fy: y > 2g.

b Domain is fx: x is in Rg.Range is fy: y > 1g.

Example 2

y

x

(4, 3)

y

x

y

x

(0, 2)

y

x

(1, 1)

y

x

(Qw_ Qr_), 6

y

x

y =

y

x

x = x =

1

y

x

(2, 2)

( 1, 2)

( 4, 3)



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g y =x+ 4

x 2 h y = x3 3x2 9x+ 10 i y = 3x 9

x2 x 2j y = x2 + x2 k y = x3 +

1

x3l y = x4 + 4x3 16x+ 3

Function machines are sometimes used to illustrate how functions behave.

For example:

So, if 4 is fed into the machine,2(4) + 3 = 11 comes out.

The above machine has been programmed to perform a particular function.

If f is used to represent that particular function we can write:

f is the function that will convert x into 2x+ 3.

So, f would convert 2 into 2(2) + 3 = 7 and

4 into 2(4) + 3 = 5.

This function can be written as:

f : x ! 2x+ 3

function f such that x is converted into 2x+ 3

Two other equivalent forms we use are: f(x) = 2x+ 3 or y = 2x+ 3

So, f(x) is the value of y for a given value of x, i.e., y = f(x).

Notice that for f(x) = 2x+ 3, f(2) = 2(2) + 3 = 7 and

f(4) = 2(4) + 3 = 5:Consequently, f(2) = 7 indicates that the point

(2, 7) lies on the graph of the function.

Likewise f(4) = 5 indicates that thepoint (4, 5) also lies on the graph.

FUNCTION NOTATIONC

y

x

(2, 7)

( ) = 2 + 3x x

3

3

x

2 + 3x

I double the

input and

then add 3



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Note: f(x) is read as f of x and is the value of the function at any value of x. If (x, y) is any point on the graph then y = f(x). f is the function which converts x into f(x), i.e., f : x ! f(x). f(x) is sometimes called the image of x.

1 If f : x ! 3x+ 2, find the value of:a f(0) b f(2) c f(1) d f(5) e f(13)

2 If g : x ! x 4x

, find the value of:

a g(1) b g(4) c g(1) d g(4) e g(12)

3 If f : x ! 3x x2 + 2, find the value of:a f(0) b f(3) c f(3) d f(7) e f(32 )

4 If f(x) = 7 3x, find in simplest form:a f(a) b f(a) c f(a+ 3) d f(b 1) e f(x+ 2)

If f : x ! 2x2 3x, find the value of: a f(5) b f(4)

f(x) = 2x2 3xa f(5) = 2(5)2 3(5) freplacing x by (5)g

= 2 25 15= 35

b f(4) = 2(4)2 3(4) freplacing x by (4)g= 2(16) + 12

= 44

Example 3

EXERCISE 1C

If f(x) = 5 x x2, find in simplest form: a f(x) b f(x+ 2)

a f(x) = 5 (x) (x)2 freplacing x by (x)g= 5 + x x2

b f(x+ 2) = 5 (x+ 2) (x+ 2)2 freplacing x by (x+ 2)g= 5 x 2 [x2 + 4x+ 4]= 3 x x2 4x 4= x2 5x 1

Example 4



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INVESTIGATION FLUID FILLING FUNCTIONS

5 If F (x) = 2x2 + 3x 1, find in simplest form:a F (x+ 4) b F (2 x) c F (x) d F (x2) e F (x2 1)

6 If G(x) =2x+ 3

x 4 :

a evaluate i G(2) ii G(0) iii G(12 )b find a value of x where G(x) does not exist

c find G(x+ 2) in simplest form

d find x if G(x) = 3:

7 f represents a function. What is the difference in meaning between f and f(x)?

8 If f(x) = 2x, show that f(a)f(b) = f(a+ b).

9 Given f(x) = x2 find in simplest form:

af(x) f(3)x 3 b

f(2 + h) f(2)h

10 If the value of a photocopier t years after purchase isgiven by V (t) = 9650 860t dollars:

a find V (4) and state what V (4) means

b find t when V (t) = 5780 and explain what thisrepresents

c find the original purchase price of the photocopier.

11 On the same set of axes draw the graphs of three different functions f(x) such thatf(2) = 1 and f(5) = 3:

12 Find f(x) = ax+ b, a linear function, in which f(2) = 1 and f(3) = 11.

13 Find constants a and b where f(x) = ax+b

xand f(1) = 1, f(2) = 5.

14 Given T (x) = ax2 + bx+ c, find a, b and c if T (0) = 4, T (1) = 2 andT (2) = 6:

When water is added at to cylindrical container the depth

of water in the container is function of time.

This is because the volume of water

added is directly proportional to the time

taken to add it. If water was not added at

constant rate the direct proportionality

would not exist.

The depth-time graph for the case of

cylinder would be as shown alongside.

a a

a

a

a

constant rate

water

depth

time

depth

DEMO



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1 For each of the following containers, draw a depth v time graph as water is added:

a b c d

e f g h

a b c

d e f

time

depth

DEMO

The question arises: What changes in appearance of the graph occur for different shaped

containers? Consider vase of conical shape.a

What to do:

2 1

3

4 1

Use the water filling demonstration to check your answers to question

rite brief report on the connection between the shape of vessel and the corre-

sponding shape of its depth-time graph. ou may wish to discuss this in parts. For

example, first examine cylindrical containers, then conical, then other shapes. Slopes

of curves must be included in your report.

Draw possible containers as in question which have the following depth time

graphs:

.

W a a

Y

v

depth

time

depth

time

depth

time

depth

time

depth

time

depth

time



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Consider f : x ! x4 and g : x ! 2x+ 3.f g means that g converts x to 2x+ 3 and then

f converts (2x+ 3) to (2x+ 3)4.

This is illustrated by the two function machines below.

Algebraically, if f(x) = x4 and g(x) = 2x+ 3, then

(f g)(x) = f(g(x))= f(2x+ 3) fg operates on x firstg= (2x+ 3)4 ff operates on g(x) nextg

Likewise, (g f)(x) = g(f(x))= g(x4) ff operates on x firstg= 2(x4) + 3 fg operates on f(x) nextg= 2x4 + 3

So, in general, f(g(x)) 6= g(f(x)).

The ability to break down functions into composite functions is useful in differential

COMPOSITE FUNCTIONS, f gD

I doubleand then

add 3

I raise anumber to

the power 4

x

2 3x

2 3x

(2!\+\3)V

g-function machine

f-function machine

calculus.

Given f : x ! 2x+ 1 and g : x ! 3 4x find in simplest form:a (f g)(x) b (g f)(x)

Example 5

DEMO

Given f : x ! f(x) and g : x ! g(x), then the composite function of f andg will convert x into f(g(x)).

f g is used to represent the composite function of f and g.i.e., ff gg :means following and ,f g f g x f g x( ( = ( ( ) ) )) x ! f(g(x)).


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f(x) = 2x+ 1 and g(x) = 3 4xa ) (f g)(x) = f(g(x))

= f(3 4x)= 2(3 4x) + 1= 6 8x+ 1= 7 8x

b (g f)(x) = g(f(x))= g(2x+ 1)

= 3 4(2x+ 1)= 3 8x 4= 8x 1

Note: If f(x) = 2x+ 1 then f() = 2() + 1

f() = 2() + 1and f(3 4x) = 2(3 4x) + 1

1 Given f : x ! 2x+ 3 and g : x ! 1 x, find in simplest form:a (f g)(x) b (g f)(x) c (f g)(3)

2 Given f : x ! x2 and g : x ! 2 x find (f g)(x) and (g f)(x).

3 Given f : x ! x2 + 1 and g : x ! 3 x, find in simplest form:a (f g)(x) b (g f)(x) c x if (g f)(x) = f(x)

4 a If ax+ b = cx+ d for all values of x, show that a = c and b = d.

(Hint: If it is true for all x, it is true for x = 0 and x = 1.)

b Given f(x) = 2x+ 3 and g(x) = ax+ b and that (f g)(x) = x for allvalues of x, deduce that a = 12 and b = 32 .

c Is the result in b true if (g f)(x) = x for all x?

THE RECIPROCAL FUNCTIONE x 1xx ! 1

x, i.e., f(x) =

1

xis defined as the reciprocal function.

It

The two branches of

has graph: Notice that:

f(x) = 1x

is meaningless when x = 0

The graph of f(x) = 1x

exists in the first

and third quadrants only.

f(x

y

) =

=

1

1

x

x

is symmetric about y = x and

y = x

EXERCISE 1D

y

x

y xy x

GRAPHING

PACKAGE


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y

x

y x

y x

y x

y

x

1

1

y x

y x

y x x >

y x

y x y >


f(x) = 1x

is asymptotic (approaches)

to the x-axis and to the y-axis.

as x ! 1, f(x) ! 0 (above)as x ! 1, f(x) ! 0 (below)as y ! 1, x ! 0 (right)as y ! 1, x ! 0 (left)! reads approaches or tends to

1 Sketch the graph of f(x) =1

x, g(x) =

2

x, h(x) =

4

xon the same set of axes.

Comment on any similarities and differences.

2 Sketch the graphs of f(x) = 1x

, g(x) = 2x

, h(x) = 4x

on the same set of axes.

Comment on any similarities and differences.

A function y = f(x) may or may not have an inverse function.

If y = f(x) has an inverse function, this new function

must indeed be a function, i.e., satisfy the vertical line test and it must be the reflection of y = f(x) in the line y = x.

The inverse function of y = f(x) is denoted by y = f1(x).

If (x, y) lies on f , then (y, x) lies on f1. So reflecting the function in y = x has thealgebraic effect of interchanging x and y,

e.g., f : y = 5x+ 2 becomes f1 : x = 5y + 2.

For example, y = f1(x) is the inverse ofy = f(x) as

it is also a function it is the reflection of y = f(x)

in the oblique line y = x.

This is the reflection of y = f(x)in y = x, but it is not the inversefunction of y = f(x) as it fails thevertical line test.

We say that the function y = f(x)does not have an inverse.

Note: y = f(x) subject to x > 0

does have an inverse function.

Also, y = f(x) subject to

x 6 0 does have an inversefunction.

INVERSE FUNCTIONSF

EXERCISE 1E


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Note: If f includes point (a, b) then f1 includes point (b, a),i.e., the point obtained by interchanging the coordinates.

1 Consider f : x ! 3x+ 1.a On the same axes graph y = x, f and f1.

b Find f1(x) using coordinate geometry and a.

c Find f1(x) using variable interchange.

2 Consider f : x ! x+ 24

.

a On the same set of axes graph y = x, f and f1.

b Find f1(x) using coordinate geometry and a.


3 For each of the following functions f

i find f1(x) ii sketch y = f(x), y = f1(x) and y = x on the same axes:

a f : x ! 2x+ 5 b f : x ! 3 2x4

c f : x ! x+ 3

Consider f : x ! 2x+ 3.a On the same axes, graph f and its inverse function f1.

b Find f1(x) using i coordinate geometry and the slope of f1(x) from aii variable interchange.

a f(x) = 2x+ 3 passes through (0, 3) and (2, 7).

) f1(x) passes through (3, 0) and (7, 2).

b i This line has slope2 07 3 =

1

2.

So, its equation is

y 0x 3 =

1

2

i.e., y =x 3

2

i.e., f1(x) =x 3

2

ii f is y = 2x+ 3, so f1 is x = 2y + 3) x 3 = 2y)

x 32

= y i.e., f1(x) =x 3

2

Example 6

EXERCISE 1F

y

x

(0, 3)

(3, 0)

(7, 2)

(2, 7)

y x( )

y x ( )

y x


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y

x-31

y

x

-2

5

y

x

2

2

y

x

5

(3, 6)

y

x1

y

x

4

4 Copy the graphs of the following functions and in each case include the graphs of

y = x and y = f1(x).

a b c

d e f

5 a Sketch the graph of f : x ! x2 4 and reflect it in the line y = x.b Does f have an inverse function?

c Does f where x > 0 have an inverse function?

6 Sketch the graph of f : x ! x3 and its inverse function f1(x).

7 The horizontal line test says that:

for a function to have an inverse function, no horizontal line can cut it more than once.

a Explain why this is a valid test for the existence of an inverse function.

b Which of the following functions have an inverse function?

i ii iii

Consider f : x ! x2 where x > 0.a Find f1(x).

b Sketch y = f(x), y = x and y = f1(x) on the same set of axes.

a f is defined by y = x2, x > 0

) f1 is defined by x = y2, y > 0

) y = px, y > 0i.e., y =

px

fas px is 6 0gSo, f1(x) =

px

b

y

x-11

y

x

2

y

x

Example 7

y

x

y x

@=!X' ! 0>

@=~!


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8 Consider f : x ! x2 where x 6 0.a Find f1(x).

b Sketch y = f(x), y = x and y = f1(x) on the same set of axes.

9 a Explain why f : x ! x2 4x+ 3 is a function but does not have an inversefunction.

b Explain why f for x > 2 has an inverse function.

c Show that the inverse function of the function in b is f1(x) = 2 +p

1 + x.

d If the domain of f is restricted to x > 2, state the domain and range of

i f ii f1.

10 Consider f(x) = 12x 1.a Find f1(x).

b Find i (f f1)(x) ii (f1 f)(x).

11 Given f : x ! (x+ 1)2 + 3 where x > 1,a find the defining equation of f1

b sketch, using technology, the graphs of y = f(x), y = x and y = f1(x)

c state the domain and range of i f ii f1.

12 Consider the functions f : x ! 2x+ 5 and g : x ! 8 x2

.

a Find g1(1). b Solve for x the equation (f g1)(x) = 9.

13 Given f : x ! 5x and g : x ! px,a find i f(2) ii g1(4)

b solve the equation (g1 f)(x) = 25.

14 Given f : x ! 2x and g : x ! 4x 3 show that(f1 g1)(x) = (g f)1(x).

15 Which of these functions are their own inverses, that is f1(x) = f(x)?

a f(x) = 2x b f(x) = x c f(x) = x d f(x) = 1x

e f(x) = 6x

In question 10 of the previous exercise we considered f(x) = 12x 1.We found that f1(x) = 2x+ 2 and that (f f1)(x) = x and (f1 f)(x) = x.

THE IDENTITY FUNCTIONG

e(x) = x is called the identity function of function

It is the unique solution of (f f1)(x) = (f1 f)(x) = e(x).y = f(x)


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1 For f(x) = 3x+ 1, find f1(x) and show that (f f1)(x) = (f1 f)(x) = x.

2 For f(x) =x+ 3

4, find f1(x) and show that (f f1)(x) = (f1 f)(x) = x.

3 For f(x) =px, find f1(x) and show that (f f1)(x) = (f1 f)(x) = x.

4 a B is the image of A under a reflection in the

line y = x.If A is (x, f(x)), what are the coordinates ofB under the reflection?

b Substitute your result from a into y = f1(x).What result do you obtain?

c Explain how to establish that f(f1(x)) = xalso.

1 Draw a graph to show what happens in the following jar-water-golf ball situation:

Water is added to an empty jar at a constant rate for two minutes and then one golf ball

is added. After one minute another golf ball is added. Two minutes later both golf balls

are removed. Half the water is then removed at a constant rate over a two minute period.

2 If f(x) = 2x x2 find: a f(2) b f(3) c f(12)3 For the following graphs determine:

i the range and domain ii the x and y-intercepts iii whether it is a function.

a b

4 For each of the following graphs find the domain and range:

a b

5 If h(x) = 7 3x:a find in simplest form h(2x 1) b find x if h(2x 1) = 2

6 If f(x) = ax + b where a and b are constants, find a and b for f(1) = 7 andf(3) = 5:

EXERCISE 1G

y

x

B

A

)(xfy

)(1 xfy

REVIEW SET 1A

y

x

y

x

y

x

x

y

1

-\Wl_T_(2, 5)

5


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7 Find a, b and c if f(0) = 5, f(2) = 21 and f(3) = 4for f(x) = ax2 + bx+ c.

8 For each of the following

containers draw a depth

v time graph as water is

added.

a b

9 Consider f(x) =1

x2.

a For what value of x is f(x) meaningless?

b Sketch the graph of this function using technology.

c State the domain and range of the function.

10 If f(x) = 2x 3 and g(x) = x2 + 2, find in simplest form:a f(g(x)) b g(f(x))

11 If f(x) = 1 2x and g(x) = px, find in simplest form:a (f g)(x) b (g f)(x)

12 Find an f and a g function given that:

a f(g(x)) =p

1 x2 b g(f(x)) =x 2x+ 1

2

1 If f(x) = 5 2x, find a f(0) b f(5) c f(3) d f(12)

2 If g(x) = x2 3x, find in simplest form a g(x+ 1) b g(x2 2)

3 For each of the following functions f(x) find f1(x) :

a f(x) = 7 4x b f(x) = 3 + 2x5

4 For each of the following graphs, find the domain and range.

a b

x

y

y x x = ( 1)( 5)

x

y

x2

(1, 1)

REVIEW SET 1B


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5 Copy the following graphs and draw the graph of each inverse function:

a b

6 Find f1(x) given that f(x) is: a 4x+ 2 b3 5x

4

7 Copy the following graphs and draw the graph of each inverse function:

a b

8 Given f(x) = 2x+ 11 and g(x) = x2, find (g f1)(3).9 Consider x ! 2x 7.

a On the same set of axes graph y = x, f and f1.

b Find f1(x) using coordinate geometry.


10 a Sketch the graph of g : x ! x2 + 6x+ 7.b Explain why g for x 6 3 has an inverse function g1.c Find algebraically, the equation of g1.

d Sketch the graph of g1.

11 Given h : x ! (x 4)2 + 3 where x > 4, find the defining equation of h1.

12 Given f : x ! 3x+ 6 and h : x ! x3

, show that

(f1 h1)(x) = (h f)1(x).

y

x

2

5

y

x

y

x

2

y

x

22


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22Chapter

Contents:

Sequences and seriesSequences and series

A

B

C

D

E

F

Number patterns

Sequences of numbers

Arithmetic sequences

Geometric sequences

Series

Sigma notation

: Von Kochssnowflake curve

Review set 2A

Review set 2B

Review set 2C

Investigation


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An important skill in mathematics is to be able to recognise patterns in sets of numbers, describe the patterns in words, and continue the patterns.

A list of numbers where there is a pattern is called a number sequence.

The members (numbers) of a sequence are said to be its terms.

For example, 3, 7, 11, 15, ..... form a number sequence.

The first term is 3, the second term is 7, the third term is 11, etc.

We describe this pattern in words:

The sequence starts at 3 and each term is 4 more than the previous one.

Thus, the fifth term is 19, and the sixth term is 23, etc.

1 Write down the first four terms of the sequences described by the following:

a Start with 4 and add 9 each time.

b Start with 45 and subtract 6 each time.

c Start with 2 and multiply by 3 each time.

d Start with 96 and divide by 2 each time.

2

a 8, 16, 24, 32, .... b 2, 5, 8, 11, .... c 36, 31, 26, 21, ....

d 96, 89, 82, 75, .... e 1, 4, 16, 64, .... f 2, 6, 18, 54, ....

3 Find the next two terms of:

a 95, 91, 87, 83, .... b 5, 20, 80, 320, .... c 45, 54, 63, 72, ....

4 Describe the following number patterns and write down the next three terms:

a 1, 4, 9, 16, .... b 1, 8, 27, 64, .... c 2, 6, 12, 20, ....

[Hint: In c 2 = 1 2 and 6 = 2 3.]5 Find the next two terms of:

a 1, 16, 81, 256, .... b 1, 1, 2, 3, 5, 8, .... c 6, 8, 7, 9, 8, 10, ....

d 2, 3, 5, 7, 11, .... e 2, 4, 7, 11, .... f 3, 4, 6, 8, 12, ....

NUMBER PATTERNSA

Describe the sequence: 14, 17, 20, 23, ..... and write down the next two terms.

The sequence starts at 14 and each term is 3 more than the previous term.

The next two terms are 26 and 29.

Example 1

EXERCISE 2A

g 480, 240, 120, 60, .... h 243, 81, 27, 9, .... i 50000, 10000, 2000, 400, ....

For each of the following write down description of the sequence and find the next two

terms:

a

36 SEQUENCES AND SERIES (Chapter 2)


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SPREADSHEET NUMBER PATTERNS

1

What to do:

Step 1: Open a new spreadsheet.

Step 2: In cell A1,type the label Value

Step 3: In cell A2,type the number 7

Step 4: In cell A3,type the formula =A2 + 4

Step 5: Press ENTER. Your spreadsheet

should look like this:

Step 6: Highlight cell A3 and positionyour cursor on the right hand

bottom corner until it changes

to a . Click the left mouse

key and drag the cursor

down to Row 10.This is called filling down.

Step 7: You should have generated the

first nine members of the number

sequence as shown:

To form a number pattern with a spread-

sheet like start with and add each

time follow the given steps.

7 4

A spreadsheet consists of a series of

rectangles called and each cell

has a position according to the

and it is in. Cell is shaded.

All formulae start with

cells

column

row B2

=

Filling down copiesthe formula from A3

to A4 and so on.

SPREADSHEET

A a

a

spreadsheet is computer software program that enables

you to do calculations, write messages, draw graphs and do

what if calculations.

This exercise will get you using spreadsheet to construct

and investigate number patterns.

SEQUENCES AND SERIES (Chapter 2) 37


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Consider the illustrated

tower of bricks. The top

row, or first row, has three

bricks. The second row has

four bricks. The third row

has five, etc.

If un represents the number of bricks in row n (from the top) then

u1 = 3, u2 = 4, u3 = 5, u4 = 6, ......

The number pattern: 3, 4, 5, 6, ...... is called a sequence of numbers.

This sequence can be specified by:

Using words The top row has three bricks and each successiverow under it has one more brick.

Using an explicit formula un = n+ 2 is the general term (or nth term)formula for n = 1, 2, 3, 4, 5, ...... etc.

Check: u1 = 1 + 2 = 3 X

u2 = 2 + 2 = 4 X

u3 = 3 + 2 = 5 X etc.

Early members of a sequence can be

graphed. Each term is represented by a

dot.

The dots must not be joined.

Step 8: Use the fill down process to answer the following questions:

a What is the first member of the sequence greater than 100?

b Is 409 a member of the sequence?

2

3

Now that you are familiar with a basic spreadsheet, try to generate the first 20

a Start with 132 and subtract 6 each time. Type:

Value in B1, 132 in B2, =B2 6 in B3, then fill down to Row 21.b Start with 3 and multiply by 2 each time. Type:

Value in C1, 3 in C2, =C22 in C3, then fill down to Row 21.

c Start with 1 000 000 and divide by 5 each time. Type:

Value in D1, 1 000 000 in D2, =D2/5 in D3, then fill down to Row 21.

members of the following number patterns:

Find out how to use a to generate of numbers such as

those above.

graphics calculator sequences

SEQUENCES OF NUMBERSB

1 rowst

2 rownd

3 rowrd

0 1

12345

2 3 4 5 6 7 8 9

etc.

n

un



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ARITHMETIC SEQUENCES

Number patterns like the one above where we add (or subtract) the same fixed number to

get the next number are called arithmetic sequences.

Further examples where arithmetic sequence models apply are:

Simple interest accumulated amounts at the end of each period.For example: on a $1000 investment at 7% simple interest p.a. (per annum) the

value of the investment at the end of successive years is:

$1000, $1070, $1140, $1210, $1280, ......

The amount still owed to a friend when repaying a personal loan with fixedweekly repayments.

For example: if repaying $75 each week to repay a $1000 personal loan theamounts still owing are: $1000, $925, $850, $775, ......

Instead of adding (or subtracting) a fixed number to get the next number in a sequence we

sometimes multiply (or divide) by a fixed number.

When we do this we create geometric sequences.

Consider investing $6000 at a fixed rate of 7% p.a. compound interest over a lengthy period.The initial investment of $6000 is called the principal.

After 1 year, its value is $6000 1:07 fto increase by 7% we multiply to 107%gAfter 2 years, its value is ($6000 1:07) 1:07

= $6000 (1:07)2After 3 years, its value is $6000 (1:07)3, etc.

The amounts $6000, $6000 1:07, $6000 (1:07)2, $6000 (1:07)3, etc. form ageometric sequence where each term is multiplied by 1:07 which is called the common ratio.

Once again we can specify the sequence by:

Using words The initial value is $6000 and after each successiveyear the increase is 7%.

Using an explicit formula

Other examples where geometric models occur are:

Problems involving depreciation.For example: The value of a $12 000 photocopier may decrease by 20% p.a.

i.e., $12 000, $12 000 0:8, $12 000 (0:8)2, ..... etc. In fractals, as we shall see later in the chapter on page 58.

GEOMETRIC SEQUENCES


un = 6000 (1:07)n1 for n = 1, 2, 3, 4, ......Check: u1 = 6000 (1:07)0 = 6000 X

u2 = 6000 (1:07)1 Xu3 = 6000 (1:07)2 X etc.

Notice that un is the amount after n 1 years.


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To solve problems like the Opening Problem and many others, a detailed study of sequences

and their sums (called series) is required.

A number sequence is a set of numbers defined by a rule for positive integers.

Sequences may be defined in one of the following ways:

by using a formula which represents the general term (called the nth term) by giving a description in words by listing the first few terms and assuming that the pattern represented continues

indefinitely.

un, Tn, tn, An, etc. can all be used to represent the general term (or nth term) of asequence and are defined for n = 1, 2, 3, 4, 5, 6, ....

fung represents the sequence that can be generated by using un as the nth term.For example, f2n+ 1g generates the sequence 3, 5, 7, 9, 11, ....

1 List the first five terms of the sequence:

a f2ng b f2n+ 2g c f2n 1gd f2n 3g e f2n+ 3g f f2n+ 11gg f3n+ 1g h f4n 3g i f5n+ 4g

NUMBER SEQUENCES

THE GENERAL TERM

EXERCISE 2B


OPENING PROBLEM

0Acircular stadium consists of sections as illustrated, with aisles in between. Thediagram shows the tiers of concrete steps for the final section, . Seatsare to be placed along every step, with each seat being m wide.AB, the arc

at the front of the first row is m long, while CD, the arc at the back of the

back row is . m long.

For you to consider:

How wide is each concrete step?

What is the length of the arc of the back of

Row , Row , Row , etc?

How many seats are there in Row , Row ,

Row , ...... Row ?

How many sections are there in the stadium?

What is the total seating capacity of the stadium?

What is the radius of the playing surface?

Section K

0 45

14 420 25

1 2 3

1 23 13

:

:

1

2

3

4

5

6

14.4 m

20.25 m

A B

C D

r

to centreof circularstadium

13 m

ti nSec o K

Row 1

IB_01cyan black

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1_02\040IB102.CDR 22 August 2005 14:25:21 DAVID2

2 List the first five terms of the sequence:

a f2ng b f3 2ng c f6 ( 12 )ng d f(2)ng3 List the first five terms of the sequence f15 (2)ng.

An arithmetic sequence is a sequence in which each term differs from the

previous one by the same fixed number.

For example: 2, 5, 8, 11, 14, .... is arithmetic as 5 2 = 8 5 = 11 8 = 14 11, etc.Likewise, 31, 27, 23, 19, .... is arithmetic as 27 31 = 23 27 = 19 23, etc.

fung is arithmetic , un+1 un = d for all positive integers n where d isa constant (the common difference).

Note: , is read as if and only if If fung is arithmetic then un+1 un is a constant and

if un+1 un is a constant then fung is arithmetic.

If a, b and c are any consecutive terms of an arithmetic sequence then

b a = c b fequating common differencesg) 2b = a+ c

) b =a+ c

2

i.e., middle term = arithmetic mean (average) of terms on each side of it.

Hence the name arithmetic sequence.

Suppose the first term of an arithmetic sequence is u1 and the common difference is d.

Then u2 = u1 + d ) u3 = u1 + 2d ) u4 = u1 + 3d etc.

then, un = u1 + (n 1)d

The coefficient of d is one less than the subscript.

So, for an arithmetic sequence with first term and common difference d

the general term (or nth term) is un = u

u

1

1

+ (n 1)d.

ARITHMETIC SEQUENCESC

ALGEBRAIC DEFINITION

THE ARITHMETIC NAME

THE GENERAL TERM FORMULA



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1 Consider the sequence 6, 17, 28, 39, 50, .....

a Show that the sequence is arithmetic. b Find the formula for its general term.

c Find its 50th term. d Is 325 a member?

e Is 761 a member?

2 Consider the sequence 87, 83, 79, 75, .....

a Show that the sequence is arithmetic. b Find the formula for the general term.

c Find the 40th term. d Is 143 a member?

3 A sequence is defined by un = 3n 2:a Prove that the sequence is arithmetic. (Hint: Find un+1 un:)b Find u1 and d.

c Find the 57th term.

d What is the least term of the sequence which is greater than 450?

4 A sequence is defined by un =71 7n

2:

a Prove that the sequence is arithmetic. b Find u1 and d. c Find u75:

d For what values of n are the terms of the sequence less than 200?

Consider the sequence 2, 9, 16, 23, 30, .....

a Show that the sequence is arithmetic.

b Find the formula for the general term un.

c Find the 100th term of the sequence.

d Is i 828 ii 2341 a member of the sequence?

a 9 2 = 716 9 = 7

23 16 = 730 23 = 7

So, assuming that the pattern continues,

consecutive terms differ by 7

) the sequence is arithmetic with = 2, d = 7.

b un = u

u

1

1

+ (n 1)d ) un = 2 + 7(n 1) i.e., un = 7n 5

c If n = 100, u100 = 7(100) 5 = 695:

d i Let un = 828

) 7n 5 = 828) 7n = 833

) n = 119

ii Let un = 2341

) 7n 5 = 2341) 7n = 2346

) n = 33517

) 828 is a term of the sequence. which is not possible as n is an

In fact it is the 119th term. integer. ) 2341 cannot be a term.

Example 2

EXERCISE 2C



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5 Find k given the consecutive arithmetic terms:

a 32, k, 3 b k, 7, 10 c k + 1, 2k + 1, 13

d k 1, 2k + 3, 7 k e k, k2, k2 + 6 f 5, k, k2 8

6 Find the general term un for an arithmetic sequence given that:

a u7 = 41 and u13 = 77 b u5 = 2 and u12 = 12 12

Find k given that 3k + 1, k and 3 are consecutive terms of an arithmeticsequence.

Since the terms are consecutive,

k (3k + 1) = 3 k fequating common differencesg) k 3k 1 = 3 k) 2k 1 = 3 k) 1 + 3 = k + 2k

) 2 = k

or k =(3k + 1) + (3)

2fmiddle term is average of other twog

) k =3k 2

2which when solved gives k = 2:

Example 3

Find the general term un for an arithmetic sequence given thatu3 = 8 and u8 = 17:

u3 = 8 ) u1 + 2d = 8 ::::(1) fun = u1 + (n 1)dgu8 = 17 ) u1 + 7d = 17 ::::(2)

We now solve (1) and (2) simultaneously

u1 2d = 8u1 + 7d = 17) 5d = 25 fadding the equationsg) d = 5

So in (1) u1 + 2(5) = 8) u1 10 = 8

) u1 = 18

Now un = u1 + (n 1)d) un = 18 5(n 1)) un = 18 5n+ 5) un = 23 5n

Check:

u3 = 23 5(3)= 23 15= 8 X

u8 = 23 5(8)= 23 40= 17 X

Example 4



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c the seventh term is 1 and the fifteenth term is 39d the eleventh and eighth terms are 16 and 1112 respectively.

7 a Insert three numbers between 5 and 10 so that all five numbers are in arithmeticsequence.

b Insert six numbers between 1 and 32 so that all eight numbers are in arithmeticsequence.

8 Consider the finite arithmetic sequence 36, 35 13 , 3423 , ...., 30.

a Find u1 and d. b How many terms does the sequence have?

9 An arithmetic sequence starts 23, 36, 49, 62, ..... What is the first term of the sequenceto exceed 100 000?

A sequence is geometric if each term can be obtained from the previous one by

multiplying by the same non-zero constant.

For example: 2, 10, 50, 250, .... is a geometric sequence as

2 5 = 10 and 10 5 = 50 and 50 5 = 250.Notice that 102 =

5010 =

25050 = 5, i.e., each term divided by the previous one is constant.

Algebraic definition:

Notice: 2, 10, 50, 250, .... is geometric with r = 5. 2, 10, 50, 250, .... is geometric with r = 5.

Insert four numbers between 3 and 12 so that all six numbers arein arithmetic sequence.

If the numbers are 3, 3 + d, 3 + 2d, 3 + 3d, 3 + 4d, 12

then 3 + 5d = 12

) 5d = 9

) d = 95 = 1:8

So we have 3, 4:8, 6:6, 8:4, 10:2, 12.

Example 5

GEOMETRIC SEQUENCESD

un is geometric ,un+1

un= r for all positive integers n

where r is a constant (the common ratio).

f g



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If a, b and c are any consecutive terms of a geometric sequence then

b

a=c

bfequating common ratiosg

) b2 = ac and so b = pac where pac is the geometric mean of a and c.

Suppose the first term of a geometric sequence is u1 and the common ratio is r.

Then u2 = u1 r ) u3 = u1 r2 ) u4 = u1 r

3 etc.

then un = u1 rn1

The power of r is one less than the subscript.

So, for a geometric sequence with first term u1 and common ratio r,

the general term (or nth term) is un = u1rn1.

THE GEOMETRIC NAME

THE GENERAL TERM

For the sequence 8, 4, 2, 1, 12 , ......

a Show that the sequence is geometric. b Find the general term un.

c Hence, find the 12th term as a fraction.

a4

8= 12

2

4= 12

1

2= 12

12

1= 12

So, assuming the pattern continues, consecutive terms

have a common ratio of 12 :

) the sequence is geometric with u1 = 8 and r =12 :

b un = u1rn1 ) un = 8

12

n1or un = 2

3 (21)n1= 23 2n+1= 23+(n+1)

= 24n

c u12 = 8 ( 12 )11

=8

211

= 1256

Example 6


(See chapter for exponent simplification.)3


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1 For the geometric sequence with first two terms given, find b and c:

a 2, 6, b, c, .... b 10, 5, b, c, ..... c 12, 6, b, c, .....

2 a Show that the sequence 5, 10, 20, 40, ..... is geometric.

b Find un and hence find the 15th term.

3 a Show that the sequence 12, 6, 3, 1:5, ..... is geometric.b Find un and hence find the 13th term (as a fraction).

4 Show that the sequence 8, 6, 4:5, 3:375, .... is geometric and hence find the 10thterm as a decimal.

5 Show that the sequence 8, 4p

2, 4, 2p

2, .... is geometric and hence find, in simplestform, the general term un.

6 Find k given that the following are consecutive terms of a geometric sequence:

a 7, k, 28 b k, 3k, 20 k c k, k + 8, 9k

EXERCISE 2D

k 1, 2k and 21 k are consecutive terms of a geometric sequence. Find k.

Since the terms are geometric,2k

k 1 =21 k

2kfequating rsg

) 4k2 = (21 k)(k 1)) 4k2 = 21k 21 k2 + k

) 5k2 22k + 21 = 0) (5k 7)(k 3) = 0

) k = 75 or 3

Check: If k = 75 terms are:25 ,

145 ,

985 : X fr = 7g

If k = 3 terms are: 2, 6, 18: X fr = 3g

Example 7

A geometric sequence has u2 = 6 and u5 = 162. Find its general term.

u2 = u1r = 6 .... (1) fusing un = u1rn1 with n = 2gand u5 = u1r

4 = 162 .... (2)

So,u1r

4

u1r=

162

6 f(2) (1)g

Example 8



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) r3 = 27) r = 3

p27) r = 3

and so in (1) u1(3) = 6) u1 = 2

Thus un = 2 (3)n1:Note: (3)n1 6= 3n1

as we do not know the value of n.

If n is odd, then (3)n1 = 3n1If n is even, then (3)n1 = 3n1

7 Find the general term un, of the geometric sequence which has:

a u4 = 24 and u7 = 192 b u3 = 8 and u6 = 1c u7 = 24 and u15 = 384 d u3 = 5 and u7 =

54

8 a Find the first term of the sequence 2, 6, 18, 54, .... which exceeds 10 000.

b Find the first term of the sequence 4, 4p

3, 12, 12p

3, .... which exceeds 4800.

c Find the first term of the sequence 12, 6, 3, 1:5, .... which is less than 0:0001 :

Find the first term of the geometric sequence 6, 6p

2, 12, 12p

2, ....which exceeds 1400.

First we find un :

Now u1 = 6 and r =p

2

so as un = u1rn1 then un = 6 (

p2)n1:

Next we need to find n such that un > 1400 .

Using a g

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