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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_
www.haeseandharris.com.auWeb:
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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
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.
USING THE INTERACTIVE STUDENT CD
NOTE ON ACCURACY
ERRATA
CD LINK
TI
C
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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
18 FUNCTIONS (Chapter 1)
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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
FUNCTIONS (Chapter 1) 19
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
20 FUNCTIONS (Chapter 1)
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)
FUNCTIONS (Chapter 1) 21
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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
22 FUNCTIONS (Chapter 1)
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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
FUNCTIONS (Chapter 1) 23
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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
24 FUNCTIONS (Chapter 1)
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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
FUNCTIONS (Chapter 1) 25
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26 FUNCTIONS (Chapter 1)
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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FUNCTIONS (Chapter 1) 27
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 >
28 FUNCTIONS (Chapter 1)
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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FUNCTIONS (Chapter 1) 29
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.
c Find f1(x) using variable interchange.
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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30 FUNCTIONS (Chapter 1)
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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FUNCTIONS (Chapter 1) 31
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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32 FUNCTIONS (Chapter 1)
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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FUNCTIONS (Chapter 1) 33
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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34 FUNCTIONS (Chapter 1)
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.
c Find f1(x) using variable interchange.
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
38 SEQUENCES AND SERIES (Chapter 2)
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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
SEQUENCES AND SERIES (Chapter 2) 39
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
40 SEQUENCES AND SERIES (Chapter 2)
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
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1_02\040IB102.CDR 22 August 2005 14:25:21 DAVID2
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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
SEQUENCES AND SERIES (Chapter 2) 41
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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
42 SEQUENCES AND SERIES (Chapter 2)
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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
SEQUENCES AND SERIES (Chapter 2) 43
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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
44 SEQUENCES AND SERIES (Chapter 2)
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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
SEQUENCES AND SERIES (Chapter 2) 45
(See chapter for exponent simplification.)3
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Y:\...\IBBK1_02\045IB102.CDRFri Apr 30 15:14:24 2004
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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
46 SEQUENCES AND SERIES (Chapter 2)
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Y:\...\IBBK1_02\046IB102.CDRWed Mar 10 16:10:18 2004
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SEQUENCES AND SERIES (Chapter 2) 47
) 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