mathematics sl - isp math - ispmath - home · 2010-01-16mathematics sl - isp math - ispmath - home
TRANSCRIPT
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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 Publications3 Frank Collopy Court, Adelaide Airport, SA 5950, AUSTRALIATelephone: +61 8 8355 9444, Fax: + 61 8 8355 9471Email:
National Library of Australia Card Number & ISBN 1 876543 03 5
Haese & Harris Publications 2004
Published by Raksar Nominees Pty Ltd3 Frank Collopy Court, Adelaide Airport, SA 5950, AUSTRALIA
First Edition 2004
Cartoon artwork by John Martin. Artwork by Piotr Poturaj and David PurtonCover design by Piotr PoturajComputer 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 InternationalBaccalaureate Organization (IBO). The textbook and CD are in no way connected with, orendorsed by, the IBO.
. Except as permitted by the Copyright Act (any fair dealing for thepurposes of private study, research, criticism or review), no part of this publication may bereproduced, 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 madeunder Part VB of the Copyright Act, the law requires that the educational institution or the bodythat administers it has given a remuneration notice to Copyright Agency Limited (CAL). Forinformation, contact the Copyright Agency Limited.
: While every attempt has been made to trace and acknowledge copyright, theauthors and publishers apologise for any accidental infringement where copyright has proveduntraceable. They would be pleased to come to a suitable agreement with the rightful owner.
: All the internet addresses (URLs) given in this book were valid at the time ofprinting. While the authors and publisher regret any inconvenience that changes of address maycause readers, no responsibility for any such changes can be accepted by either the authors or thepublisher.
\Qw_ \Qw_
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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
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 worksheetsvideo clipsgraphing and geometry softwaregraphics calculator instructionscomputer demonstrations and simulationsbackground 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 KNOWLEDGE11
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 notationD Algebraic simplificationE Linear equations and inequalitiesF Absolute value (modulus)G Product expansionH Factorisation
: Another factorisationtechnique
I Formula rearrangementJ Adding and subtracting algebraic fractionsK Congruence and similarity
ANSWERSSummary of circle propertiesSummary of measurement facts
A Relations and functionsB Interval notation, domain and rangeC Function notation
: Fluid filling functionsD Composite functions,
E The reciprocal function
F Inverse functionsG The identity function
Review set 1AReview set 1B
A Number patternsB Sequences of numbersC Arithmetic sequencesD Geometric sequencesE SeriesF Sigma notation
: Von Kochs Snowflake curveReview set 2AReview set 2BReview set 2C
A Index notationB Negative basesC Index lawsD Rational indices
10
CDCDCDCDCDCDCDCD
CDCDCDCDCD1112
1619222426
27
28313233
36384144515758595960
62636571
Investigation
Investigation
Investigation
E Algebraic expansionF Exponential equationsG Graphs of exponential functions
: Exponential graphsH GrowthI Decay
Review set 3AReview set 3BReview set 3CReview set 3D
A IntroductionB Logarithms in base 10
: Discovering the laws oflogarithms
C Laws of logarithmsD Exponential equations (using logarithms)E Growth and decay revisitedF Compound interest revisitedG The change of base rule
Review set 4AReview set 4B
A Introduction 104: occurs naturally 104: Continuous compound
interestB Natural logarithms
: The laws of naturallogarithms
C Laws of natural logarithmsD Exponential equations involvingE Growth and decay revisitedF Inverse functions revisited
Review set 5AReview set 5B
A Families of functions: Function families
B Key features of functionsC Transformations of graphsD Functional transformations
Review set 6
A Assumed knowledge: Finding where lines
meet using technologyB Equations of linesC Distance between two pointsD Midpoints and perpendicular bisectors
Review set 7A
73747576798184848586
8890
9293969799
100101102
105108
109110112112114116117
120120122123126127
131
135138141144146
*
*
*
Investigation
Investigation
Investigation
Investigation
Investigation 1
Investigation 2
Investigation 3
e
e
4 LOGARTIHMS 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 trianglesB Sectors and segmentsC The cosine ruleD The sine rule
: The ambiguous caseE Using the sine and cosine rules
Review set 12AReview set 12B
A Observing periodic behaviourB Radian measure and periodic properties
of circlesC The unit circle (revisited)D The sine function
: The family: The family
: The families
E Modelling using sine functionsF Equations involving sine
: The area under an arch of284
G The cosine function 285H Solving cosine equations 287I Trigonometric relationships 289J Double angle formulae 292
: Double angle formulae 292K The tangent function 295L Tangent equations 298M Other equations involving 301
Review set 13A 302Review set 13B 302Review set 13C 303Review set 13D 304Review set 13E 305
A IntroductionB Addition and subtraction of matricesC Multiples of matricesD Matrix algebra for additionE Matrix multiplicationF Using technologyG Some properties of matrix multiplicationH The inverse of a matrixI Solving a pair of linear equationsJ The determinantK The inverse of a matrixL 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 TRIANGLETRIGONOMETRY 235
13 PERIODIC PHENOMENA 255
14 MATRICES 307
and
tanx
Review set 7BReview set 7C
A Function notationB Graphs of quadratic functions
: Graphing
: Graphing
C Completing the squareD Quadratic equationsE The quadratic formulaF Solving quadratic equations with
technologyG Problem solving with quadraticsH Quadratic graphs (review)I The discriminant,J Determining the quadratic from a graphK Where functions meetL Quadratic modelling
Review set 8AReview set 8BReview set 8CReview set 8DReview set 8E
A Binomial expansions: The binomial expansions
of: values
B The general binomial expansionReview set 9
A Pythagoras rule (review) 205B Pythagoras rule in 3-D problems 207
: Shortest distance 208C Right angled triangle trigonometry 209D Finding sides and angles 211E Problem solving using trigonometry 217F The slope of a straight line 221
Review set 10A 222Review set 10B 223Review set 10C 223
A The unit quarter circleB Obtuse anglesC 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 ANDFUNCTIONS 149
9 THE BINOMIAL THEOREM 195
10
203
11 THE UNIT CIRCLE 225
PRACTICAL TRIGONOMETRYWITH RIGHT ANGLEDTRIANGLES
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 incryptographyReview set 14AReview set 14BReview set 14CReview set 14DReview set 14E
A VectorsB Operations with vectorsC Vectors in component formD Vector equationsE Vectors in coordinate geometryF ParallelismG Unit vectorsH Angles and scalar product
Review set 15AReview set 15BReview set 15CReview set 15D
A 3-dimensional coordinatesB 3-dimensional vectorsC Algebraic operations with 3-D vectorsD ParallelismE Unit vectorsF Collinear points and ratio of division
extensionG The scalar product of 3-D vectors
Review set 16AReview set 16B
A Vector and parametric form of a line in2-dimensional geometry
B The velocity vector of a moving objectC Constant velocity problems
: The two yachts problemD The closest distanceE Geometric applications ofF Lines in spaceG Line classification
Review set 17AReview 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 ANDIN SPACE 397
18 DESCRIPTIVE STATISTICS 419
BACKGROUND KNOWLEDGE INSTATISTICS
420 to access, click
active icon on CD
C Presenting and interpreting data CDANSWERS CD
A Continuous numerical data and histograms 421B Measuring the centre of data 425
: Merits of the meanand median 427
C Cumulative data 442D Measuring the spread of data 445E Statistics using technology 453F Variance and standard deviation 456G The significance of standard deviation 461
Review set 18A 463Review set 18B 465
A Experimental probability: Tossing drawing pins: Coin tossing experiments: Dice rolling experiments
B Sample spaceC Theoretical probabilityD Using grids to find probabilitiesE Compound events
: Probabilities ofcompound events
: Revisiting drawing pinsF Using tree diagramsG Sampling with and without replacement
: Sampling simulation: How many should I plant?
H Pascals triangle revisitedI Sets and Venn diagramsJ Laws of probabilityK Independent events revisited
Review set 19AReview set 19B
: The speed of fallingobjects
A Rate of changeB Instantaneous rates of change
: Instantaneous speedReview set 20
A The idea of a limit: The slope of a tangent
B Derivatives at a given -valueC The derivative function
: Finding slopes offunctions with technology
D Simple rules of differentiation: Simple rules of
differentiationE 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 integralB The area function
: The area functionC AntidifferentiationD The fundamental theorem of calculusE IntegrationF Integrating andG Integrating by substitutionH Distance from velocityI Definite integrals
: and areasJ Finding areasK Problem solving by integration
Review set 26AReview set 26BReview set 26C
A Basic trigonometric integralsB Integrals of trigonometric functions
of the formC Definite integralsD Area determination
Review set 27AReview set 27B
A Solids of revolutionB Volumes for two defining functions
Review set 28
A Discrete random variablesB Discrete probability distributionsC Expectation
: Concealed number ticketsD The mean and standard deviation of a
discrete random variableE The binomial distributionF Mean and standard deviation of a
binomial random variable: The mean and standard
deviation of a binomial random variableG Normal distributions
: Standard deviationsignificance
: Mean and standarddeviation of
H The standard normal distribution( -distribution)
I Applications of the normal distributionReview set 29AReview set 29BReview 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 compositesF Product and quotient rulesG Tangents and normalsH The second derivative
Review set 21AReview set 21BReview set 21C
A Functions of timeB Time rate of changeC General rates of changeD Motion in a straight line
: Displacement, velocityand acceleration graphs
E Curve propertiesF Rational functionsG Inflections and shape typeH OptimisationI Economic models
Review set 22AReview set 22B
A Derivatives of exponential functions: The derivative of: Finding when
and
B Using natural logarithmsC Derivatives of logarithmic functions
: The derivative ofD Applications
Review set 23AReview set 23B
A The derivative of , ,B Maxima/minima with trigonometry
Review set 24
A Areas where boundaries are curved: Finding areas using
rectanglesB 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 OFDIFFERENTIAL CALCULUS 555
23 DERIVATIVES OFEXPONENTIAL ANDLOGARITHMIC FUNCTIONS 605
24 DERIVATIVES OFTRIGONOMETRIC FUNCTIONS 625
25 AREAS WITHIN CURVEDBOUNDARIES 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 numbersf0, 1, 2, 3, 4, 5, ..... g
Z the set of all integersf0, 1, 2, 3, 4, 5, .....g
Z+ the set of all positive integersf1, 2, 3, 4, 5, .....g
Q the set of all rational numbersQ+ the set of all positive rational
numbersR the set of all real numbersR+ the set of all positive real numbers
n(S) the number of elements in set S2 is an element of=2 is not an element of? the empty set, or null setU the universal set[ union\ intersectionjxj the modulus of x,
or the absolute value of xj 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 withrespect to xZ b
a
y dx the definite integral of y withrespect to x from x = a tox = b
ex the exponential functionlogax the logarithm of x, in base aln x the natural logarithm of x, loge xsinx, cosx the circular functionsand tanxP(x, y) point P with coordinates x and y]A the angle at A]PQR the angle between QP and QRPQR 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 sequenced the common difference of an
arithmetic sequencer the common ratio of a geometric
sequenceSn u1 + u2 + u3+ ..... +un
the 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 ff1(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 theyx --, 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 AI the identity matrix under P(A) the probability of event AP(A0) the probability of event not AP(A/B) the probability of A occurring
given that B has occurredN(, 2) the normal distribution with
mean and variance 2
population mean2 population variance population standard deviationx sample means 2n sample variancesn 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.
BACKGROUNDKNOWLEDGE
Below is a summary of well known results called theorems. Click on the appropriate icon torevisit 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
GEOMETRYPACKAGE
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12 BACKGROUND KNOWLEDGE AND GEOMETRIC FACTS
Name of theorem Statement Diagram
Chords of a
circle
The perpendicularfrom the centre ofa circle to a chordbisects the chord.
If then AM = BM.A
MB
O GEOMETRYPACKAGE
Radius-tangent The tangent to a cir-cle is perpendicularto the radius at thepoint of contact.
If then ]OAT = 90o.
Tangents from
an external
point
Tangents from an ex-ternal point are equalin length.
If then AP = BP.
Angle at the
centre
The angle at the centreof a circle is twice theangle on the circle sub-tended by the same arc.
If then ]AOB = 2]ACB.
Angles
subtended
by the
same arc
Angles subtended by anarc on the circle areequal in size.
If then ]ADB = ]ACB.
Angle between
a tangent and
a chord
The angle between a tan-gent and a chord at thepoint of contact is equalto the angle subtendedby the chord in the al-ternate segment.
If then ]BAS = ]BCA.
A
B
O P
A B
D C
A
B
S
C
T
GEOMETRYPACKAGE
GEOMETRYPACKAGE
GEOMETRYPACKAGE
GEOMETRYPACKAGE
GEOMETRYPACKAGE
AT
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
basea
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 formulaefor 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 anAirport are given in the table shown alongside.
There is an obvious relationship between time spent and thecost. The cost is dependent on the length of time the car isparked.
Looking at this table we might ask: How much would becharged for exactly one hour? Would it be $5 or $9?
To make the situation clear, and to avoid confusion, wecould adjust the table and draw a graph. We need to indicatethat 2-3 hours really means for time over 2 hours up to andincluding 3 hours i.e., 2 < t 6 3.
Car park chargesPeriod (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, wenow have Car 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 $28:00
The parking charges example is clearly the latter as any real value of time ( t hours) in theinterval 0 < t 6 24 is represented.
For example: ft : 0 < t 6 24g is the domain for the car park relation f2, 1, 4g is the domain of f(1, 5), (2, 3), (4, 3), (1, 6)g.
RELATIONS AND FUNCTIONSA
exclusioninclusion
charge ($)
time ( )t3 6 9
10
20
30
12 15 18 21 24
In mathematical terms, because we haverelationship between two variables, time andcost, the schedule of charges is an exampleof
relation may consist of finite number ofordered 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 therelation.
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 exampleswhich 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 havethe 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 valuefor 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 > bi.e., ]1, a[ or ]b, 1[.
The domain and range of a relation are best described where appropriate using intervalnotation.
For example: The domain consists of all realx 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 functionsand 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 and4 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 andf(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 x3
3
x
2 + 3x
I double theinput andthen 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 existc find G(x+ 2) in simplest formd 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) meansb find t when V (t) = 5780 and explain what this
representsc 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 depthof water in the container is function of time.
This is because the volume of wateradded is directly proportional to the timetaken to add it. If water was not added at
constant rate the direct proportionalitywould not exist.The depth-time graph for the case ofcylinder would be as shown alongside.
a aa
a
a
constant rate
water
depth
time
depthDEMO
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 shapedcontainers? Consider vase of conical shape.a
What to do:
2 1
3
4 1
Use the water filling demonstration to check your answers to questionrite 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. Forexample, first examine cylindrical containers, then conical, then other shapes. Slopesof curves must be included in your report.Draw possible containers as in question which have the following depth timegraphs:
.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() + 1f() = 2() + 1
and 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
GRAPHINGPACKAGE
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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 tox 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 fi 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 a
ii 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 isy 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)
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5
y
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y
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5
(3, 6)
y
x1
y
x
4
4 Copy the graphs of the following functions and in each case include the graphs ofy = 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 +
p1 + x.
d If the domain of f is restricted to x > 2, state the domain and range ofi 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 theline 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 ballis added. After one minute another golf ball is added. Two minutes later both golf ballsare 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
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x
y
x
y
x
x
y
1
-\Wl_T_(2, 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 followingcontainers draw a depthv time graph as water isadded.
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
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5
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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 twoterms:
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 ValueStep 3: In cell A2,
type the number 7Step 4: In cell A3,
type the formula =A2 + 4
Step 5: Press ENTER. Your spreadsheetshould look like this:
Step 6: Highlight cell A3 and positionyour cursor on the right handbottom corner until it changesto a . Click the left mousekey and drag the cursordown to Row 10.This is called filling down.
Step 7: You should have generated thefirst nine members of the numbersequence as shown:
To form a number pattern with a spread-sheet like start with and add eachtime follow the given steps.
7 4
A spreadsheet consists of a series ofrectangles called and each cell
has a position according to theand 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.
SPREADSHEETA a
a
spreadsheet is computer software program that enablesyou to do calculations, write messages, draw graphs and dowhat if calculations.This exercise will get you using spreadsheet to constructand investigate number patterns.
SEQUENCES AND SERIES (Chapter 2) 37
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Consider the illustratedtower of bricks. The toprow, or first row, has threebricks. The second row hasfour bricks. The third rowhas five, etc.
If un represents the number of bricks in row n (from the top) thenu1 = 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 begraphed. Each term is represented by adot.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 asthose above.
graphics calculator sequences
SEQUENCES OF NUMBERSB1 rowst
2 rownd
3 rowrd
0 112345
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 toget 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 the
amounts still owing are: $1000, $925, $850, $775, ......
Instead of adding (or subtracting) a fixed number to get the next number in a sequence wesometimes 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 sequencesand 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
OPENING PROBLEM
A circular stadium consists of sections as illustrated. The diagram shows thetiers of concrete steps for the final section, . Each seat is to be
m wide. AB, the arc for the front of the first row is m long.For you to consider:
How wide is each concrete step?What is the length of the arc of the backof 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 thestadium?What is the radius of the playing surface?
Section K
0 45 14 4
1 2 3
12 3 13
: :
1
2
3
4
5
6
32
14.4 m
20.25 m
A B
C D
r
to centreof circularstadium
13 m
Secti no K
Row 1
NUMBER SEQUENCES
THE GENERAL TERM
EXERCISE 2B
40 SEQUENCES AND SERIES (Chapter 2)
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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)ng
3 List the first five terms of the sequence f15 (2)ng.
An arithmetic sequence is a sequence in which each term differs from theprevious 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 thenb 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 dthe 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 anIn 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, 13d 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 bymultiplying by the same non-zero constant.
For example: 2, 10, 50, 250, .... is a geometric sequence as2 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, 12then 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 r3 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 termshave 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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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 = u1r4 = 162 .... (2)
So,u1r
4
u1r=
162
6 f(2) (1)g
Example 8
46 SEQUENCES AND SERIES (Chapter 2)
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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, 4
p3, 12, 12
p3, .... 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 (p
2)n1:
Next we need to find n such that un > 1400 .Using a graphics calculator with Y ^1 = 6 ( (
p2) n 1), we view a
table of values:
So, the first term to exceed 1400 is u17 where u17 = 1536.
Note: Later we can solve problems like this one using logarithms.
Example 9
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$5000 is invested for 4 years at 7% p.a. compound interest.What will it amount to at the end of this period?
u5 = u1 r4 is the amount after 4 years= 5000 (1:07)4 ffor a 7% increase 100% becomes 107%g+ 6553:98 f5000 1:07 4 g
So, it amounts to $6553:98 .
Example 10
=
48 SEQUENCES AND SERIES (Chapter 2)
^
Consider the following: You invest $1000 in the bank.You leave the money in the bank for 3 years.You are paid an interest rate of 10% p.a.The interest is added to your investment each year.
An interest rate of 10% p.a. is paid, increasing the value of your investment yearly.
Your percentage increase each year is 10%, i.e., 100% + 10% = 110% of the value atthe start of the year, which corresponds to a multiplier of 1:1 .
After one year your investment is worth $1000 1:1 = $1100After two years it is worth $1100 1:1
= $1000 1:1 1:1= $1000 (1:1)2 = $1210
After three years it is worth $1210 1:1= $1000 (1:1)2 1:1= $1000 (1:1)3
This suggests that if the money is left in your account for n years it would amount to$1000 (1:1)n.Note: u1 = $1000 = initial investment
u2 = u1 1:1 = amount after 1 yearu3 = u1 (1:1)2 = amount after 2 yearsu4 = u1 (1:1)3 = amount after 3 years...u15 = u1 (1:1)14 = amount after 14 years
...un = u1 (1:1)n1 = amount after (n 1) years
un+1 = u1 (1:1)n = amount after n yearsIn general, un+1 = u1rn is used for compound growth u1 = initial investment
r = growth multipliern = number of years
un+1 = amount after n years
COMPOUND INTEREST
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9 a What will an investment of $3000 at 10% p.a. compound interest amount to after3 years?
b What part of this is interest?
10 How much compound interest is earned by investing 20 000 Euro at 12% p.a. if theinvestment is over a 4 year period?
11 a What will an investment of 30 000 yen at 10% p.a. compound interest amount toafter 4 years?
b What part of this is interest?
12 How much compound interest is earned by investing $80 000 at 9% p.a., if the investmentis over a 3 year period?
13 What will an investment of 100 000 yen amount to after 5 years if it earns 8% p.a.compounded semi-annually?
14 What will an investment of 45 000 amount to after 28 months if it earns 7:5% p.a.compounded quarterly?
15 How much money must be invested now if you require $20 000 for a holiday in 4 yearstime and the money can be invested at a fixed rate of 7:5% p.a. compounded annually?
16 What initial investment is required to produce a maturing amount of 15 000 in 60months time given that a fixed rate of 5:5% p.a. compounded annually is guaranteed?
17 How much should I invest now if I want a maturing amount of 25 000 Euro in 3 yearstime and the money can be invested at a fixed rate of 8% p.a. compounded quarterly?
18 What initial investment is required to produce a maturing amount of 40 000 yen in 8years time if your money can be invested at 9% p.a., compounded monthly?
How much should I invest now if I want the maturing value to be $10 000in 4 years time, if I am able to invest at 8:5% p.a. compounded annually?
u1 = ?, u5 = 10000, r = 1:085
u5 = u1 r4 fusing un+1 = u1 rng) 10 000 = u1