IGCSE Additional Mathematics by Aarav Ranjan

SuePemberton Cambridge IGGSE®and O Level Additional Mathematics Coursebook Second edition il Cambridge UNIVERSITY PRESS

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4. ,4 A'/'gj, >*. M Contents Acknowledgements IntroductionHowtousethisbook1Functions 1.1 Mappings 1.2Definitionofafunction 1.3Compositefunctions 1.4Modulusfunctions 1.5 Graphsof))=|f(x)|wheref(x)islinear 1.6Inversefunctions 1.7 Thegraphofafunctionanditsinverse SummaryExaminationquestions Simnltaneousequationsandqnadratics 2.1 Simultaneousequations(onelinearandonenon-linear) 2.2 Maximumandminimumvaluesofaquadraticfunction 2.3 Graphsofy=|f(x)|wheref(x)isquadratic 2.4Quadraticinequalities2.5Rootsofquadraticequations 2.6Intersectionofalineandacurve SummaryExaminationquestions Indicesandsnrds 3.1 Simplifying expressionsinvolving indices 3.2 Solving equationsinvolving indices 3.3Surds 3.4 Multiplication,divisionandsimplificationofsurds 3.5 Rationalisingthedenominatorofafraction 3.6 Solvingequationsinvolvingsurds Summary FactorsExaminationquestionsandpolynomials 4.1 Adding,subtractingandmultiplyingpolynomials 4.2Divisionofpolynomials 4.3Thefactortheorem 4.4 Cubicexpressionsandequations 4.5Theremaindertheorem Summary ExaminationquestionsEquations,inequalitiesandgraphs 5.1 Solvingequationsofthetype\ax-b\=\cx-d\ 5.2Solvingmodulusinequalities 5.3 Sketchinggraphsofcubicpolynomialsandtheirmoduli 5.4 Solvingcubicinequalitiesgraphically 5.5 Solvingmorecomplexquadraticequations SummaryExaminationquestions Logarithmicandexponentialfunctions 6.1 6.2Logarithmstobase10Logarithmstobasea 6.3 Thelawsoflogarithms VI vii viii 1 2 3 5 7 10 12 15 2319182528 34 423739 44 46 5049 51 55 57 60 63 67 6770717375 8278 86 87 89 9490 98 103102 105 107 111 112115 118

CambridgeIGCSEand0LevelAdditionalMathematics6.4Solvinglogarithmicequations 120 6.5 Solvingexponentialequations 122 6.6 Changeofbaseoflogarithms 124 6.7 Naturallogarithms 126 6.8 Practicalapplicationsofexponentialequations 128 6.9 Thegraphsofsimplelogarithmicandexponentialfunctions 129 6.10 Thegraphsoiy=ke"''+aandy=k\n{ax+b)wheren,k,a andbareintegers 130 6.11 Theinverseoflogarithmicandexponentialfunctions 133 Summary 134 Examinationquestions 135 7 Straight-linegraphs 138 7.1 Problemsinvolvinglengthofalineandmidpoint 140 7.2 Parallelandperpendicularlines 143 7.3 Equationsofstraightlines 145 7.4 Areasofrectilinearfigures 148 7.5 Convertingfromanon-linearequationtolinearform 151 7.6 Convertingfrom linearform to a non-linear equation 155 7.7 Findingrelationshipsfromdata 159 Summary 165 Examinationquestions 165 8 Circularmeasure 170 8.1 Circularmeasure 171 8.2 Lengthofanarc 174 8.3 Areaofasector 177 Summary ISO Examinationquestions 181 9 Trigonometry 186 9.1 Anglesbetween0°and90° 187 9.2 Thegeneraldefinitionofanangle 190 9.3 Trigonometricratiosofgeneralangles 192 9.4 Graphsoftrigonometricfunctions 195 9.5 Graphs ofy=|f(v)|,where f(x) is a trigonometricfunction 205 9.6 Trigonometricequations 209 9.7 Trigonometricidentities 214 9.8 Furthertrigonometricequations 216 9.9 Furthertrigonometricidentities 218 Summai7 220 Examinationquestions 221 10Permutationsandcombinations 224 10.1 Factorialnotation 225 10.2 Arrangements 226 10.3 Permutations 229 10.4 Combinations 234 Summary 237 Examinationquestions 238 11Series 243 11.1 Pascal'striangle 244 11.2 Thebinomialtheorem 249 11.3 Arithmeticprogressions 252 11.4 Geometricprogressions 257 11.5 Infinitegeometricseries 262 11.6 Furtherarithmeticandgeometricseries 267 Summary 270 Examinationquestions 271

12Differentiation1 12.1 Thegradientfunction 12.2 Thechainrule 12.3 Theproductrule 12.4 Thequotientrule 12.5 Tangentsandnormals 12.6 Smallincrementsandapproximations 12.7 Ratesofchange 12.8 Secondderivatives 12.9 Summary12.10StationarypointsPracticalmaximumandminimumproblemsExaminationquestions 13Vectors13.1Furthervectornotation13.2Positionvectors 13.3 Vectorgeometry 13.4 SummaryConstantvelocityproblemsExaminationquestions 14Differentiation2 14.1 Derivativesofexponentialfunctions 14.2 Derivativesoflogarithmicfunctions 14.3 Derivativesoftrigonometricfunctions 14.4 15SummaryFurtherapplicationsofdifferentiationExaminationquestionsIntegration 15.1 Differentiationreversed 15.2 Indefiniteintegrals 15.3 Integrationoffunctionsoftheform{ax+b)n 15.4 Integrationofexponentialfunctions 15.5 Integrationofsineandcosinefunctions Integrationoffunctionsoftheform—and ° X FurtherFurtherindefiniteintegrationDefiniteintegrationdefiniteintegration Areaunderacurve 15.6 15.7 15.8 15.9 15.10 ax+ 15.11 b Areaofregionsboundedbyalineandacurve Summary Examinationquestions 16Kinematics 16.1 Applicationsofdifferentiationinkinematics 16.2 Applicationsofintegrationinkinematics SummaryExaminationquestions Answers Index 274275 280282285 287 291 298294 300305 315310311 317 319 323 327 336335331 337341345350356357361362365 367 372370368 375 383378 385 397392 398 402 404 412 419418 422 454

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Coverartwork;Shestakovych/ShutterstockChapter1Fanjianhua/Shutterstock;Chapter2zhudifeng/Shutterstock;Chapter3LACUNA Fig.DESIGN/GettyImages;Chapter4MichaelDechev/Shutterstock;Fig.4.1SteveBower/Shutterstock;4.2Laboko/Shutterstock;Fig.4.3irin-k/Shutterstock;Chapter5zentilia/Shutterstock;Chapter 6Peshkova/Shutterstock;Chapter7ittipon/Shutterstock;Chapter8ZhuQiu/FyeFm/GettyImages;Chapter9pauldowning/GettyImages;Fig.9.1aarrows/Shutterstock;Chapter10GinoSantaMaria/Shutterstock;Fig.10.1snake3d/Shutterstock;Fig.10.2KeithPublicover/Shutterstock;Fig.10.3 AleksandrKurganov/Shutterstock;Fig.10.4AfricaStudio/Shutterstock;Chapter11elfinadesign/Shutterstock;Chapter12AlenKadr/Shutterstock;Chapter13muratart/Shutterstock;Chapter14Neamov/Shutterstock;Chapter15AhuliLabutin/Shutterstock;Chapter16AlexLMX/Getty

CambridgeIGCSEand0LevelAdditionalMathematics Acknowledgements

Pastexamination paper questions throughoutare reproduced by permission ofCambridge Assessment InternationalEducation.

'Challenge'questions have been included at the end ofmostexercises to challenge and stretch highabilitystudents. Towards the end ofeach chapter,there is asummary ofthe key concepts to help students consolidate whatthey havejustlearnt.This is followed by a'Past paper'questionssection,which contains real questionstakenfrompastexaminationpapers.

A Practice Book is also available in the CambridgeIGCSL^ Additional Mathematicsseries, which offers studentsfurthertargetedpractice.Thisbookcloselyfollowsthechaptersandtopicsofthecoursebook offering additional exercises to help students to consolidate conceptslearntand to assess theirlearning aftereachchapter.ATeacher'sResource,tooffersupportandadvice,isalsoavailable.

Introduction

ThishighlyillustratedcoursebookcoverstheCambridgeIGCSl^andOLevelAdditionalMathematics syllabuses(0606 and 4037). The course is aimed atstudents who are currendy studying or have previously studied CambridgeICCS^ Mathematics(0580) or Cambridge O Level Mathematics(4024).

'Classstudies,Wherethecontentinonechapterincludestopicsthatshouldhavealreadybeencoveredinpreviousarecapsectionhasbeenprovidedsothatstudentscanbuildontheirpriorknowledge.discussion'sectionshavebeenincludedtoprovidestudentswiththeopportunitytodiscussandlearnnewmathematicalconceptswiththeirclassmates,withtheirclassteacheractingasthefacilitator.Theaimoftheseclassdiscussionsectionsistoimprovethestudent'sreasoningandofal communicationskills.

CambridgeIGCSEand0LevelAdditionalMathematics How to use this book Chapter-eachchapterbeginswithasetoflearningobjectivestoexplainwhatyouwilllearninthischapter. Chapter 2 Simultaneousequationsandquadratics Thissectionwillshowyouhowto: ■ solve simultaneous equations in two unknowns by elimination or substitution ■ findthemaximumandminimumvaluesofaquadraticfunction ■ sketch graphs of quadratic functions and find their range for a given domain ■ sketchgraphsofthefunctiony=|f{:c)twhereffx)isquadraticandsolveassociatedequations ■ determine the numberof roots of a quadratic equation and the related conditions for a line to intersect,beatangentornotintersectagivencurve ■ solvequadraticequationsforrealrootsandfindthesolutionsetforquadraticinequalities^ Recap-checkthatyouarefamiliarwiththeintroductoryskillsrequiredfor thechapter. Youshouldalreadyknowhowtosolvelinearinequalities.Twoexamplesareshownbelow. Solve 2(x -5) < 9 2* - 10 < 9 2x < 19 X < 9.5 expandbrackets add 10tobothsides dividebothsidesby2 Solve 5-3x^17 -3x 12 X « -4 subtract5fiom bothsides dividebothsidesby-3 ClassDiscussion-additionalactivitiestobedoneintheclassroomforenrichment. CLASS DISCUSSION Solveeachofthesethreepairsofsimultaneousequations. 8x + = 7 Sx+by=-9 2)1Sx+y=10=15-6x 102*+5=.3)1-6)=-4x Discussyouranswerswithyourclassmates. Discusswhatthegraphswouldbelikeforeachpairofequations.

Worked Example-detailed step-by-step approachesto help students solve problems. Findthevalueof; ®5! 11! Answers g 8!8x7x6x/x^x^x^)0 si ^x/x/x/yl =8x7x6 = 336 bill=HxlOx9xXx/x^x/xy^>0x^yl 813! /x/x^x.^x/x/xix/X3X2X1 _990 6 = 165 Note-quicksuggestionstoremindyouaboutkeyfactsandhighlightimportant points.

6Thisdesignismadefrom 1bluecircle,4orangecircles and16greencircles. Thecirclestoucheachother.Giventhattheradiusofeach greencircleis1unit,findthe exactradiusof atheorangecircles, bthebluecircle.

Summary-attheendofeachchaptertoreviewwhatyouhavelearnt. Summary Oneradian(b)isthesizeoftheanglesubtended atthecentreofacircle,radiusr,byanarcoflengthr.

Challenge-challengeyourselfwithtougherquestionsthatstretchyourskills

o Note: logic30canalsobewrittenasIg30orlog30.

When9ismeasuredinradians: ■ thelengthofarcAB=rO mtheareaofsectorAOB=-AO. 2

CHALLENGEQ

ACTIVITY Usegraphingsoftwaretoconfirmthat: • y=cosX+1isatranslationofy=cosxbythevector ^ fo)•y=cosX+2isatranslationofy=cosxbythevector • y =cosx-3 is a translation of y =cosx by the vector ^ and fo1 •y=tanx+1isatranslationofy=tanxbythevectory=tanx-2isatranslationofy=tanxbythevector -9

CambridgeIGCSEand0LevelAdditionalMathematics

Examinationquestions

Thefunctionfissuchthatf(x)=4x^- +ax+b,whereaandbareconstants.Itisgiventhat 2*-1isafactoroff(x)andthatwhenf(*)isdividedbyx+2theremainderis20.Findthe remainderwhenf(x)isdividedbyx-1.

Examinationquestions-exam-stylequestionsforyoutotestyourknowledgeand understandingattheendofeachchapter.

Workedexample

[6]

Cat?tbridgeIGCSEAddiiumalMathematics0606Paper11Q2Nov2011 Answer f(x)=4x'-8x^+ax+b If2x - 1 is afactor,then f j= a + 26 = 3 •d) Remainder=20whendividedbyx+2,meansthatf(-2)=20. 4(-2)'-8(-2f+a(-2)+6=20 -32- 32- 2a + A = 20 -2a+b=84 (2) From(1)a=3-26. Substitutingin(2),gives;-2(3-26)+6=84 -6+46+6=8456=906=18 Soa=—33,6=18. Remainderwhenf(x)=4x'-8x^-33x+18isdividedby(x-I)isf(I). Remainder=4(1)'-8(1)'^-33(1)+18 ==4-8-33+18-19 Activity-foryoutoapplyyourtheoreticalknowledgetoapracticaltask.

'■■■ Chapter 1 Functions I Thissectionwillshowyouhowto: understandandusetheterms:function,domain,range(imageset),one-onefunction,inverse function and composition offunctions usethenotationf(*)=2x^-f5,f:xi->5x-3,f~^(x) andf^(x) understandtherelationshipbetweeny=f(x)andy=|f(x)| solvegraphicallyoralgebraicallyequationsofthetype \ax+b\=c and \ax+b\=cx+d explaininwordswhyagivenfunctionisafunctionorwhyitdoesnothaveaninverse find the inverse of a one-one function and form composite functions use sketch graphs to show the relationship between a function and its inverse.

CambridgeIGCSEand0LevelAdditionalMathematics 1.1 Mappings Input Output 1 2 3 4 2 43 5 iscalledamappingdiagram. Theruleconnectingtheinputandoutputvaluescanbewrittenalgebraically as:X X+1. Thisisreadas'xismappedtox+T. x+1 TheThemappingcanberepresentedgraphicallybyplottingvaluesofx+1againstvaluesofx.diagramshowsthatforoneinputvaluethereisjustoneoutputvalue. TheItiscalledaone-onemapping.tablebelowshowsone-one,many-oneand one-manymappings. many-one X one-many :y/x o X Foroneinputvalue thereisjustoneoutput value. Fortwoinputvalues thereisoneoutput value. Foroneinputvaluetherearetwooutput values. Exercise1.1 Determinewhethereachofthesemappingsisone-one,many-oneor one-many. 2xi-^x^+5 xelR 4 Xi->2* xeR x>0 6 x1-^x^+1 xeR,x^O 1 X 1-^ X-1-I X€ 3 X l-> x^ xe 1 5 X XG X 12 7 X XG X xGR,x>0 8 XI—>±X xeR,XS:0

WORKEDEXAMPLE1 f(x)=2x-l x€lR,-l^x^3 a Writedownthedomainofthefunctionf. b Sketchthegraphofthefunctionf. cWritedowmtherangeofthefunctionf. Answers a Thedomainis —1^x^3. bThegraphofy=2x—1hasgradient2anday-interceptof-1. WhenX=-1,y=2(-l)-1=-3 Whenx=3,y=2(3)-1=5 f(x) (3,5) <u be Uc O (-1,-3)T" Domain c Therangeis—3^f(x)^5.

Chapter1:Functions

Definitionofafunction

Afunctionisarulethatmapseachxvaluetojustoneyvalueforadefinedset ofinputvalues. Thismeansthatmappingsthatareeither< arecalledfunctions. ^ I many-one Themappingxi->x-l-1wherexeIR,isaone-onefunction. , . ^f:x^-»x^-l xeIR Itcanbewrittenas < ^ l^f(x)=X+1 X€[R (f:Xi->X-I-1isreadas'thefunctionfissuchthatxismappedtox-I-T)f(x)representstheoutputvaluesforthefunctionf.

Sowhenf(x)=x-t-1,f(2)=2-1-1=3. Thesetofinputvaluesforafunctioniscalledthedomainofthefunction. Thesetofoutputvaluesforafunctioniscalledtherange(orimageset)of thefunction.

1.2

CambridgeIGCSEand0LevelAdditionalMathematics

WORKEDEXAMPLE2

4 Thefunction fis defined by f(x)= x^ -4 for xe R. Findtherangeoff.

6 The function fis defined by f(x)=(2x + 1)^-5 for x ^ Findtherangeoff.

Theminimumvalueofthe expressionis0+3=3andthis minimumoccurswhenx=2. So thefunction f(x)=(x-2)^ +3 thewillhaveaminimumpointatpoint(2,3).

7 Thefunctionfis defined by f:x i-> 10-(x-3)^ for 2^ Findtherangeoff. -301 X X 53 7.

5 Thefunctionfisdefined by f(x)=(x-1)^ +5 for x ^1. Findtherangeoff.

Domain (6,19) u be - C

Thefunctionfisdefinedbyf(x)=(x-2)^+3for0^x^6. FindSketchthegraphofthefunction.therangeoff. Answers f(x)=(x-2)^+3isapositivequadraticfunctionsothegraphwillbeoftheform\J {x-2f)+3

Exercise1.2

1 WhichofthemappingsinExercise1.1arefunctions? 2 Findtherangeforeachofthesefunctions. a f(x)-X-5, -2^X^7 b f(:>c)=Sx+2, c f(x)= 7-2x, -l^x^4 d f(x)= x^, 1f( f f(x)=2^ -3 x) 3 The function g is defined as g(x)= x^ + 2 for x ^ 0. Writedowntherangeofg.

Thispartoftheexpressionisasquaresoitwillalwaysbe>0. Thesmallestvalueitcanbeis0.Thisoccurswhenx=2.

WhenX=0,y=(0-2)^+3=7. WhenX=6,y=(6-2)^+3=19. Therangeis3^f(x)^^19.

WORKEDEXAMPLE3

8Thefunctionfisdefinedbyf(x)=3+yjx-2forx^2. Findtherangeoff. 1.3 Composite functions

Thefunctionfisdefinedbyf(x)=(x—2)^-3forx>—2. 2x+6 fThefunctiongisdefinedbyg(x) Findfg(7). X-2 or X>2. Answers fg('7) =f(4) =(4-2f-3 =1 g acts on 7firstand g(7) 2(7)+6 7-2 fisdrefunction'take2,squareanddientake3'

Chapter1:Functions

Whenonefunctionisfollowedbyanotherfunction,theresultingfunctionis calledacompositefunction. fg(x)meansthefunction g acts on x first,thenfacts on the result. Note: (x)meansff(x),soyouapplythefunctionftwice.

Mostfunctionsthatyoumeetarecombinationsoftwoormorefunctions. Forexample,thefunctionx\-^2x+bisthefunction'multiplyby2andthen add5'.Itisacombinationofthetwofunctionsgandfwhere: g:X1-^2x (thefunction'multiplyby2') f:Xi->X+5 (thefunction'add5') So,X 2x+5isthefunction'firstdogthendof.

g f

WORKEDEXAMPLE4

CambridgeIGCSEand0LevelAdditionalMathematics

f(x)=:2x-l for xeR g(x)= x^ +5 for X e R Find afg(x) bgf(x) Cf-(x). Answers afg(x) ==f(x^+5)2(x2+5)-1 =2x^+9 gactsonXfirstandg(x)=x-+5 fisthefunction'doubleandtake1' b gf(x) =g(2x-1) =(2x-1)2+5 factsonXfirstandf(x)=2x-1 gisthefunction'squareandadd5' * expandbrackets c f^(x) ===fr(x)f(2x-l)2(2x-l)-1 =4x—3 F(x)meansff(x) .factsonXfirstandf(x)=2x-1 fisthefunction'doubleandtake1'

=--4x^—4x+1+54x^-4x+6

Exercise1.3 1 f:Xi->2x+3forxeR g:Xi->x^-1forXeR Findfg(2). 2 f(x)=x^-1 for xeR g(x)=2x+3for XeR Findthevalueofgf(5). 3 f(x)=(x + 2)^ -1 for X e Findf2{3). 4 Thefunction fis defined by f(x)= 1+-Jx-2 for x ^ 2. Thefunctiongisdefinedbyg(x)=—-1for x>0. X Findgf(18). 5 The function fis defined by f(x)=(x -1)^ +3 for x > -1. +4 Thefunctiongisdefinedbyg(x)= for x>5. X—5 Findfg(7).

3x+5 x-2' X-1 X^2.

Chapter!:Functions

Itisimportanttonotethatthemodulusofanynumber(positiveornegative)is alwaysapositivenumber.

Thefunctionfisdefinedbyf:x 3x+1 for x€ R.

10Thefunctionfisdefined,forxGR,byf:xi-> Thefunctiongisdefined,forxgR,byg:x c Solvetheequationgf(x)=12. 11f(x)=(x+4)^+3forX>0 g(x)=— for X>0 X Solvetheequationfg(x)=39.

h:Xh-»X+2forx>Q k: i-»yfx for x>0 Expresseachofthefollowingintermsofhandk. a X h-> yfx +2 b X i-> \Jx +2

The function g is defined by g : x: ^ for x ^ 2.

12 The function g is defined by g(x)= x^ -1 for x ^ 0.

Themodulusofanumberisthemagnitudeofthenumberwithoutasign attached. Themodulusof4iswritten141. |4|=4and|-4|=4

Themodulusofanumberisalsocalledtheabsolutevalue.

Solvetheequationgf(x)=5. g(x)=x^+2forXeR h(x)=3x-5 for xeR Solvetheequationgh(x)=51. f(x)=x^-3for X>0 3 g(x)=— for X>0 X Solvetheequationfg(x)=13.

Expresseachofthefollowingasacompositefunction,usingonlyfand/org: a xi-»(x-l)" b xH^x^-l c xi->x-2

Thefunctionhisdefinedbyh(x)=2x-7for x^0. Solvetheequationgh(x)=0.

1.4Modulusfunctions d Xi->X"

13Thefunctionfisdefinedbyf:xi->x^forxgR. Thefunctiongisdefinedbyg:x x-1for xGR.

meansthat x= orx=-k.

ax+b=-k

CambridgeIGCSEand0LevelAdditionalMathematics Themodulusofx,writtenas]x|,isdefinedas: ^X ifX>0 I x|= < 0 if X =0 L-X ifX<0 CLASSDISCUSSION Alisaysthattheseareallrulesforabsolutevalues: l-Jc +^|=|xl lx;)/|=|x|+\^\X\/\ |x| I/I r|^|/=x^ Discusseachofthesestatementswithyourclassmatesanddecideiftheyare:Alwaystrue Sometimestrue Nevertrue Youmustjustifyyourdecisions.

Thestatement1x|=fe,wherek^0, Thispropertyisusedtosolveequationsthatinvolvemodulusfunctions.So,ifyouaresolvingequationsoftheform\ax+b\k,yousolve theequations ax+b=k and Ifyouaresolvingharderequationsoftheform\ax+b\=cx+d,yousolve theequationsax+b=cx+d and ax+b=-{cx+d). Whensolvingthesemorecomplicatedequationsyoumustalwayscheckyour answerstomakesurethattheysatisfytheoriginalequation.

Answers

WORKEDEXAMPLE5 Solve. a|2x+l|=5 b|4x-3| c -10 dIX-3|=2x a|2x+l|=5 2x+1=5 or 2x+1=-5 2x=4X=2 2x=-6 X=—3

3x4-2 =2 X4-1 e x4-|x-5|=8 14X4-12 =3 X4-4 f 9—11 — x|= 2x 3 Solve aIX-l|=3 d I x^ — 5x|= X g I 2x^ 4-1| = 3x 4- X I = 2-X x^ — 3j = X 4- 3 ^ 7x 4- 6|= 6X' b|x^4-l|=10 c e |x^-4|= x4-2 f h |2x^-3x|= 4-x i 4 Solveeachofthefollowingpairsofsimultaneousequationsa3i=x4-4b3i=xC3i=3x j' =|x^-16| 3! =|3x-2x^| )) =|2x^-5|

CHECK:!2x2+l|=5/and !2x-3+1|=.5/ Solutionis:x=-3or2. 14x-3l=X 4x-3=X or 4x-3=-X 3x=3 5x=3 X=1 X=0.6 CHECK:I4X0,6-3|=0.6/and |4X1-3|=1/ Solutionis:x=0.6or1. c|x^-10|=6 x^-10=6 or X .2 x^=16 10=-6 x2=4 X=±2X=±4 CHECK:I(-4f-10|=6/, |(-2f-10|= 6/,|(2)^-10|=6/ and|(4f-10|=6/ Solutionis:x=-4,-2,2or4. IX-3|=2x X-3=2x or X=-3 andCHECK:1-3-3|11-3!=2X1/ Solutionis:X=1. X—3=—2x 3x=3 X=1 2X-3X Exercise1.4 1 Solve a |3x:-2|=10 X-1 gd ^-5 4 2 Solve 2x-5 a =6=1=8 b 12x4-9]=5 2x4-7 1 c 16-5xI=2 7-2x =4 X4-1 2x 1 2 5 =4 i 12x-5|=X X4-3 d |3x-5|=x4-2

Chapter1:Functions

V II / 0 X

Youthenreflectinthex-axisthepartoftheline thatisbelowthex-axis. WORKEDEXAMPLE6 Sketchthegraphof3;= graphmeetstheaxes. Answers ,showingthecoordinatesofthepointswherethe FirstsketchthegraphofV~ ~ Thelinehasgradient—anda 31-intercept of-1.

X -x-1 2

y 0 X y ■■■ ■v-'iSv' > ,.---'2 X

Youthenreflectinthex-axisthepart ofthelinethatisbelowthex-axis.

CambridgeIGCSEand0LevelAdditionalMathematics 1.5 Graphs of y =|f(x)| where f(x)is linear Consider drawing the graph of 3)=|x|. Firstdrawthegraphof31=%.

InWorkedexample5yousawthatthereweretwo answers,x =—3or x =2,to the equation|2x+1| — 5. These can also be found graphically by finding the x-coordinates ofthe points ofintersection ofthe graphsof y =|2x+11 and y =5asshown. Chapter1:Functions y=2x+1 y=5

Exercise1.5 1 Sketch the graphsofeach ofthefollowingfunctionsshowing the coordinates ofthe points where the graph meets the axes. ay=|x+l| • .by=,l2x-3| cy=l5-xl 1 e y=110-2xI f yy= —X+32 6 3X 2 a Complete the tableofvaluesfor y=]x-21+3; X -2 -I 0 '1 2 3 4 y 6 4 b Drawthegraphofy=|x—2|+3for-2^x^4. Draw the graphs ofeach ofthe following functions, ay=|x|+l . by=|x|-3 cy=2-lx| d y=lx-3|+l e y=|2x+6|-3

Inthesameworkedexampleyoualsosawthattherewasonlyoneanswer,x=1,totheequation|x-31=2x. Thiscanalsobefoundgraphicallybyfindingthex-coordinatesofthepointsofintersectionofthe graphsof y =|x — 3| and y=2xasshown. ^U-3l

Giventhateachofthesefunctionsisdefinedforthedomain-3« findtherangeof a f:xi-»5-2x b g:xh-»l5-2xl c h;xt-^5-l2x|, 4,

c Solve the equation f(x)=2.

It is important to remember that not everyfunction hasaninverse.

Step 2: Interchange the xand y variables.Step 3: Rearrange to make y the subject. ► y = ■*-X=5y-2 X + 2 f-^x) X + 2

The inverse ofafunction f(x) is thefunction that undoeswhatf(x)hasdone.

6 a Sketch the graph of 3; =|2x + 41 for -6 < x < 2,showing the coordinatesofthe points where the graph meetsthe axes, b On the same diagram,sketch the graph of 31 = x+ 5. cSolvetheequation|2x+41=x+5.

The domain off~^(x)isthe range off(x).

7 A function fis defined by f(x)=|2x-6|-3, for -1 ^ x ^8. aSketchthegraphof31=f(x).bStatetherangeoff.

8aSketchthegraphof3;=13x-41for-2<x<5,showingthe coordinatesofthe pointswherethegraph meetstheaxes, bOnthesamediagram,sketchthegraphof3;=2x.

An inversefunction f-^(x)can existif,and onlyif,the function f(x)is a one-onemapping.

CSolvetheequation2x=13x-41.

Therangeoff~^(x)isthedomainoff(x).

The inverse ofthe function f(x) is written asf~Hx).

Thestepstofindtheinverseofthefunctionf(x)=5x-2are: Step 1: Write the function as 31 = >■ y = 5x - 2 f-Hx)

You should already know how to find the inverse function ofsome simple one-onemappings.

1.6Inversefunctions

9 a Sketch the graph of f(x)=|x +2 +1x-21. b Use yourgraph tosolvetheequation|x+2|+|x-2|= 6.

ICHALLENGEQ

CambridgeIGCSEand0LevelAdditionalMathematics 5 f:Xi->3-2x for -1 x 4 g:xi->(3-2x| for h:Xi-»3-12xI for -1=€x 4 Findtherangeofeachfunction.

CLASSDISCUSSION

Exercise1.6 1 f(x)=(x + 5)^ -7 for X 5= -5. Find an expression for f~'(x). 0 2 f(x)= y. ^2, ^ ^ 0- Find an expression for f~^(x).

Discussthefunction

f(x)= doeshaveaninverse?

3f(x)=(2x-3)+1forx^1—.Findanexpressionforf~^(x). 4 f(x)=8-Vv-3 for x^3. Find anexpressionforf~^(x).

(x+5)2-1=1(x+5)2=2 X+5=±^J2 X=—5+V2 »:=-5+\/2orX=—5—y/2

y=yjx+l-5 X=yjy+l-5 X+5=^Jy+I (x+5)2=3+1 3=(x+5)2-1 b f(35)=>/35Tr-5=l

Chapter!:Functions

f(x)= forxeR. Doesthefunctionfhaveaninverse? Explainyouranswer.Howcouldyouchangethedomainoffsothat

Step1:Writethefunctionas31= Step2:Interchangethexandyvariables. Step3;Rearrangetomakeythesubject. f-'(x)=(x+5)2-1

WORKEDEXAMPLE7 f(x)=Vx+1—5forX -1 a Findanexpressionforf~'(x). b Solvetheequationf~'(x)=f(35). Answers a f(x)=-v/x+1—5forX^-1

Therangeoffisf(x)^-5sothedomainoff~'isx&-5. Hencetheonlysolutionoff~^(x)=f(35)isx=-5+>/2.

CambridgeIGCSEand0LevelAdditionalMathematics

forx#2 2-X Express and intermsofx. 6 f : X -> (x + 2)^ -5 for X > -2 a Find an expression for f~^(x). b Solve the equation f ^(x) = 3. 7 f(x)=(x-4)^ +5 for x > 4 aFindanexpressionforf~^(x).

5f:xh->5x-3forx>0

bSolvetheequationf^(x)=f(0) ,, 2x+3 8 g(x)= for X>1 X—1 a Findan expressionforg~^(x). b Solve theequation g (x)=5. 9 f(x)= ^+ 2 for xeIR g(x)= x^ -2x for xeR a Findf"H^«)- ^ Solvefg(x)=f"H»:)10 f(x)= x^ + 2 for xeIR g(x)= 2x +3 for xe K Solve the equation gf(x)=g~^(l7). 2x+8 f11f:Xi-> orX^2 X-2 Solvetheequationf(x)=g~^(x).

g:Xi->-—-forX>-5 12 f(x)=3x-24 for x 5= 0.Write down the range off ^ 13f:xi-^x+6 forx>0 g;xi-^>/x forx>0 Expressx x^-6intermsoffandg. 14f:X 3-2xfor0<X^5 g:xi->l3-2xlfor0^x=s5 Stateh:xi-»3-|2x|for0^x=s5whichofthefunctionsf,gandhhasaninverse. 15f(x)=x^+2forX^0 g(x)=5x-4for x^0 a Write down the domainoff~^ b Write down therangeofg ^ 16Thefunctionsfandgaredefined,forxeIR,by f:X Sx-k,whereAisapositiveconstant 5x-14 g;X wherex -1. X+1 a Find expressionsforf~^ and g"^ b Findthevalueofkforwhichf"'(5)=6. c Simplify g"^g('c)-

Theinversefunctionisf X+1 ThedomainoffMs-3^x^5andtherangeoffMs-1 f'(x)^3.-1 Drawingfandf^onthesamegraphgives; (3,5); b=; (-1,-3) Q Note; Thegraphsoffandf"^arereflectionsofeachotherintheliney=X. Thisistrueforallone-onefunctionsandtheirinversefunctions. Thisisbecause;ff"^(x)=x=f-^f(x).

1.7 The graph of a function and its inverse InWorkedexample1youconsideredthefunction f(x)=2x-1xgR, —1^X^3. Thedomainoffwas-1^x<3andtherangeoffwas-3^f(x)^5.

Somefunctionsarecalledself-inversefunctions becausefanditsinversef ^ arethesame. Iff(x)=—forX5!:0,thenf~'(^)=~forx#0. X X So f(x)= — for X *0is an example ofa self-inverse function. X

Whenafunctionfisself-inverse,thegraphoffwillbesymmetricalaboutthe liney=X.

17f:Xh-» for xgR g:xi-^x-8 for xgIR Expresseach ofthefollowingasacompositefunction,usingonlyf,g,f~^ and/org"^; ^ a X (x-8)3 b xi-^x^+8 c xh-»x3—8 I d Xi->(x+8)3

Chapter1:Functions

WORKEDEXAMPLE8 f(x)=(x-2)^,2 X^5. Onthesameaxes,sketchthegraphsof31=f(x)and3=f'(x),showingclearlythe pointswherethecurvesmeetthecoordinateaxes. Answers y^Ux-tf WhenX=5,jy=9. Thispartoftheexpressionisasquaresoitwillalways be>0.Thesmallestvalueitcanbeis0.Thisoccurs whenX=2. 10-

CambridgeIGCSEand0LevelAdditionalMathematics

y CL y^^f(X) v =^ 0 4 5 ^ r'(x)

CLASSDISCUSSION

R /// Reflectfiny=/ 9 / y r 0 2 'a 5 8 ]'qJ )y 10} 1-81 : 1 i 1 j\I 1-6 ■ 4 L h / -i 1/!,-! f-i h'l h 1 '-i/ j 1/ ! |o 1 2! i! 5 10^

inverse

Sundeepsaysthatthediagramshowsthegraphof thefunction f(x:)= x" for x> 0,together with its functiony=f~^(x). IsSundeepcorrect?Explainyouranswer.

Chapter1:Functions

Onacopyofthegrid,drawthegraph oftheinverseofthefunctionf. o ■ 04 f A fi ( > 01 ( f X Q 4 -fi-

Exercise1.7

Sketch,on asingle diagram,the graphsof y =g(x) and y =g~\x),showing thecoordinatesofanypointswherethecurvesmeetthecoordinateaxes. f(x)=4x-2 for -1^X^3. Sketch,on a single diagram,the graphs of y =f(x) and y =f~Hx),showing thecoordinatesofanypointswherethelinesmeetthecoordinateaxes. Thefunctionfis defined by f: x i-» 3-(x + 1)^ for x ^-1. a Explainwhyfhasaninverse, b Find an expression for f~' in terms of x. C On the same axes,sketch the graphs of y =f(x) and y =f~^(x),showing thecoordinatesofanypointswherethecurvesmeetthecoordinateaxes.

Onacopyofthegrid,drawthegraph oftheinverseofthefunctiong.

y (j 4 g O h C) o > f1^ -4 -fif(x)= +3,X^0.

On thesame axes,sketch the graphsof y=f(x) and y =f~'(x),showing the coordinates ofany points where the curves meet the coordinate axes. g(x)=2* for X€K On thesameaxes,sketch the graphsof y=g(x) and y=g"^(x),showing the coordinates ofany points where the curves meet the coordinate axes. g(x)=x^—1forX^0.

Theinverseofafunctionf(x)isthefunctionthatundoeswhatf(x)hasdone. Theinverseofthefunctionf(x)iswrittenasf"^(x). The domain off~^(x) is the range off(x). Therangeoff^x)isthedomainoff(x).

An inverse functionf~'(x)can existif,and onlyif,thefunction f(x)is a one-one mapping. The graphsoffand f~^ are reflections ofeach otherin the line y = x. I

Thesetofinputvaluesforafunctioniscalledthedomainofthefunction.

Themodulusofx,writtenas|x|,isdefinedas: X ifx>0 0 ifX=0 -X ifX<0 Compositefunctions

Summary Functions A function is a rule that maps each x-value tojust one y-value for a defined set ofinput values. MappingsthatareeitherJ arecalledfunctions. Imany-one

Thesetofoutputvaluesforafunctioniscalledtherange(orimageset)ofthefunction.

Cambridge IGCSE and0LevelAdditional Mathematics CHALLENGEQ 8 2%+7 I:Xi-> —forX^z X—2 aFind intermsofx. b Explain what this implies about the symmetry ofthe graph ofy=f(x).

Modulusfunction

fg(x)meansthefunctiongactsonxfirst,thenfactsontheresult, f^(x)meansff(x). Inversefunctions

Examinationquestions Workedexample Thefunctionsfandgaredefinedby 9V f(x)= forX>0, X+1 g(x)=Vx+1forX>-1. a Findfg(8). [2] ax b Find an expression forf (x),giving your answer in the form where a, b and care ix+cintegerstobefound. " [3] C Find an expression for g~'(x),stating its domain and range. [4] CambridgeIGCSEAdditionalMathematics0606Paper21Q12i,ii,iiifun2014 Answers a g(8)= =3 fg(8)=f(3) 2(3) =3+11.5 b f^C^v)=ff(x) =f 2x X+1 2x X+1 2x X+14x _X+1 3x+1X+14x +1 substitute3forxin c2x+1 2x 2x substitute forxin X+1 X+1 simplifymultiplynumeratoranddenominatorbyx+1 3x+1 a=4,&=3andc=1

y=y/x+l X = yjy + I X"=)!+1 y=x'^ g 1 Therangeofgisg(x)>0sothedomainofgMsx>0. The domain ofg is x > -1 so the range ofg~' is x > -1.

CambridgeIGCSEAdditionalMathematics0606Paper11Qli,iiJun2012

CambridgeIGCSEAdditionalMathematics0606Paper11Q12ai,iiNov2013

CambridgeIGCSEAdditionalMathematics0606Paper11Qli,iiNov2012

g(x:)=yjx+lforX>-1 Step1:Writethefunctionas})= Step2:Interchangethexandyvariables. Step3:Rearrangetomakeythesubject.

CambridgeIGCSEAdditionalMathematics0606Paper21QINov2011

Exercise1.8 Exam Exercise 1 Solve the equation 14x-5[=21. [3]

aSketchthegraphofy=12x-51,showingthecoordinatesofthepointswherethegraph meetsthecoordinateaxes. [2] y 0 X bSolve12x-51=3. [2]

4 A function fis such thatf(x)=3x^-1 for-10^ x 8. a Findtherangeoff. [3] b Writedownasuitabledomainforfforwhichf~'exists. [1]

CambridgeIGCSEand0LevelAdditionalMathematics

aSketchthegraphof=13+5x|,showingthecoordinatesofthepointswhereyourgraph meetsthecoordinateaxes. [2] b Solve the equation 13+5x|=2. [2]

Chapter1:Functions 5 Thefunctionsfandgaredefinedforrealvaluesofxby f(x)=y]x-\-3forX>1, x-2 g(x)= 2x—3 forX>2. a Findgf(37). b Findanexpressionforf"'(x). C Find an expression for g"'(x). 1 6Afunctiongissuchthatg(»;)= for1 2x-1 a Findtherangeofg. b Findg"'^(x). C Writedownthedomainofg"^^)d Solve g^(x)= 3. [2] [2] [2] CambridgeIGCSEAdditionalMathematics0606Paper21Q4Noxi2014 X^3. [2][1][1] [3] CambridgeIGCSEAdditionalMathematics0606PaperIIQ9i-ivNov2012 7 a Thefunctionsfandgaredefined,forxeR,by f:XH->2x+3 g:X x'-^-1. Findfg(4). [2] b Thefunctionshandkaredefined,forx>0,by h:Xi->X+4 k:Xh-> yfx. Expresseachofthefollowingintermsofhandk. [2][1][1] CambridgeIGCSEAdditionalMathematics0606Paper21Q5a,bi,ii,iiiNov2011 X Vx+4 X X+8 ii X x^-4 8Thefunctionfisdefinedbyf(x)=2-\lxA-bfor-5^x<0. i Writedowntherangeoff. ii Findf~n^)andstateitsdomainandrange. 4 Thefunctiongisdefinedby g(x)=— for -5^x<-1. iii Solve fg(x)=0. [2] [3][4] CamimdgeIGCSEAdditionalMathematics0606Paper11Q6Jun2016

10i Ontheaxesbelow,sketchthegraphsofy=2-xandy=13+2x|. [4] 9-y 6 O 6^ D ii Solve 13 +2x|=2 [3]

CambridgeIGCSEAdditionalMathematics0606Paper12Q4Mar2016

CambridgeIGCSEand0LevelAdditionalMathematics [2][1]

Thefunctionfissuchthatf(x)=2x^-8x+5. i Showthatf(x)=2{x+a)^+b,whereaandbaretobefound. ii Hence,orotherwise,writedownasuitabledomainforfsothatf"^exists, bThefunctionsgandharedefinedrespectivelyby g(x)= +4,X^0, h(x)=4x-25,X^0, i Writedowntherangeofgandofh"^ [2] ii On a copy ofthe axes below sketch the graphs ofy = g(x) and ofy = g~^(x),showing thecoordinatesofanypointswherethecurvesmeetthecoordinateaxes. [3] }'O iii Findthevalueofxforwhichgh(x)=85. [4]

CambridgeIGCSEAdditionalMathematics0606PaperIIQ1lai,ii,bi,ii,iiiJun2011

9 a

Chapter 2 Simultaneous equations and quadratics Thissectionwillshowyouhowto: m solvesimultaneousequationsintwounknownsbyeliminationorsubstitution ■ findthemaximumandminimumvaluesofaquadraticfunction m sketchgraphsofquadraticfunctionsandfindtheirrangeforagivendomain ■ sketchgraphsofthefunctiony==|f(jc)|wheref(x)isquadraticandsolveassociatedequations ■ intersect,determinethenumberofrootsofaquadraticequationandtherelatedconditionsforalinetobeatangentornotintersectagivencurve ■ solvequadraticequationsforrealrootsandfindthesolutionsetforquadraticinequalities.

CambridgeIGCSEand0LevelAdditionalMathematics RECAP(simultaneousequations) Youshoukialieadyknowhowlouseagraphicalmethodoranalgebraicmethodtosoh'esimultaneousequations wherebothequationsarelinear. Tosolvesiniultaneouslinearequationsalgebraicallyyoushouldknowboththeeliminationmethodandthe substitutionmethod. -(2)-(1) Eliminationmethod Solve AddMultiply5x+2y=252x-y=l(2)by2.5x+2y=254*-2y=2theequationstoeliminatey. 9x=27 *=3 Substitutefor*inequation(2).6-y=1y=5 SolutionisX=3,y=5. -(1) "(2) Substitutionmethod Solve 5*:+2y=252*-y=1 Makeythesubjectofequation(2). y=2ac—1 Substituteforyinequation(1). 5x+2(2x-l)=25 9x-2=25 9x=27 X=3 Substitutefora:inequation(2). 6-y=1y=5 Solutionis*=3,y=5. CLASSDISCUSSION Solveeachofthesethreepairsofsimultaneousequations.Sx+3y=7 3x+5y=-9 2ySx+y=10=15-6a; 102x+5=3y-6y=-4a: Discussyouranswerswithyourclassmates. Discuss whatthe graphs would be like for each pair ofequations.

« RECAP(quadraticequations)

CLASSDISCUSSION

Solve each ofthese quadratic equations. -8x+15=0 +4x+4=0 x^+2x+4=0 Discuss your answeis with your classmates. Discuss what the graphs would be like for each of the functions y = - 8x+15,y ^ + 4x+4 and y = + 2x + 4. 2.1 Simultaneous equations(one linear andonenon-linear)

Inthissectionyouwilllearnhowtosolve simultaneous equations where one equation is linearandthesecondequationisnotlinear. The diagram shows the graphs of y = x +1 andy=x^-5. Thecoordinatesofthepointsofintersection ofthe two graphs are(-2,-1)and (3,4). (-2,-1) x+1 y= x^+5

Factorisationmethod Solve -4x-12=0. Factorise: (5C-6)(x+2)=0 x-6=0or*:+2=0 Solutionis x=6or*=-2. Completingthesquaremethod Solve x^-4x-12=0. Completethesquare. (*:-2f-4-12=0 (*-2)^=16 Squarerootbothsides. X-2=±4 x—2=4or*:-2=—4 Solutionis X=6orX=-2. Quadraticformulamethod Solve -4x-12=0. Identifya,bandc a=\,b=—4andc=-12. Usetheformula: -b±4b^ •4ac X 2a Substitutefora,bandc. 4±V(-4f-4X1X(-12) 2x1 4± 2 4±8 X 4+82 4-8 X = or X = 2 2 Solutionisx=6orx=-2.

You should already know how to use a graphical method or an algebraic method to solve quadratic equations. To solve quadratic equadons algebraically you should know the factorisation metlK)d,the quadraticformula methodandthecompletingthesquaremethod.

Chapter2:Simultaneousequationsandquadratics

Wesaythat x = —2,y = —1 and x =3, y =4 are thesolutionsofthe simultaneousequations y= x+1 and y = x^-5. Thesolutionscanalsobefoundalgebraically: y = X+1 (1) y=x^-5 (2) Substituteforyfrom(1)into(2): x+l=x^-5 rearrange -X-6=0 factorise (x+2)(x-3)=0 X=-2orX=3 Substitutingx=-2into(1)givesy=-2+1=-1.Substitutingx=3into(1)givesy=3+1=4. The solutions are: x =-2,y =-1 and x =3,y =4. WORKED EXAMPLE 1 J Solvethesimultaneousequations. 2x+2y=7 X Oh or II 00 Answers 2x+2y =7 ~(l) —4y^=8 7-2yFrom , X= ^ jSubstituteforxin expandbrackets 49_28y+4y^4/-8 ! 4 mulliphliolhsidesby4 149-28y+4y^-16y^=32 rearrange' 12y2+28y-17=0 (63>+17)(23i-1)=0 faciorise 1v=—2—ory= 1 5 Substituting y=-2—into givesX=D—. 1.Substituting 7 ~ g givesX=3. Thesolutionsare:x=6 3 5 1 '^ ^ 6 ^"^* = ^'7=2'

CambridgeIGCSEand0LevelAdditionalMathematics

25 The straightline y = 2x-2 intersects the curve x^ - y = 5 at the points AandB. GiventhatAliesbelowthex-axisandthepointPliesonABsuchthat AP:PB=3:I,findthecoordinatesofP.

19Calculatethecoordinatesofthepointswheretheliney—1—2xcutsthe curve =2. 20Thesumoftwonumbersxand))is11. The productofthe two numbers is 21.25. a Writedowntwoequationsinxandy. bSolveyourequationstofindthepossiblevaluesofxandy. 21 Thesum ofthe areasoftwosquaresis818cm^. Thesumoftheperimetersis160cm.Findthelengthsofthesidesofthesquares.

22 The line y = 2-2x cuts the curve - y^ = 5 at the points A and B. FindthelengthofthelineAB.

23 Theline 2x+5y =1 meetsthecurve x^+5xy-4y^+10=0 atthe FindpointsAandB.thecoordinatesofthemidpointofAB.

24Theline y=x-10intersectsthecurve x^+ +4x+6y-40=0 at thepointsAandB. FindthelengthofthelineAB.

26Theline x-2y=2 intersectsthecurve x+y^=10attwopointsAandB. Find the equation ofthe perpendicular bisector ofthe line Ai5.

Chapter 2: Simultaneous equations and quadratics Exercise2.1 Solvethefollowingsimultaneousequations. 1 9 y= 2 y=X—6 3 y=X—1 4 II y=X+6 00 II + C^l x2+y2=25 y=2x+2 5 x^—xy=0 6 3y=4x-5 7 2x+y=7 8 X-y=2 X+y=1 x2+3xy=10 xy=6 2x2_3^2 9 X+2y=7 10 y=2x 11 xy=2 12 y2=4x x^+y^=10 OC II + M( X+y=3 2x+y=4 13 o II oe + 14 X+y=4 15 y=3x 16 X-2y 2x2+3^=1 x2+y2=10 II CM M( 4y2-3x2_j 17 3+X+xy=0 18 xy=12 2x+5y=8 (x-l)(y+2)= 15

Ifa>0,thecurvehasaminimumpointwhich occursatthelowestpointofthecurve.

Thegeneralequationofaquadraticfunctionisf(x)=ax^+bx+c,where«,bandcareconstantsandai^Q. Thegraphofthefunctiony=asc+bx+cis,calledaparabola.Theorientationoftheparaboladependsonthe ^ rr* • _r 2 r rvalueofa,thecoefficientofxl r

The maximum and minimum points can also be called turning points orstationarypoints. Every parabola has a line ofsymmetrythat passes through the maximum or minimum point.

-l\ 0 k X -T

2.2Naximumandminimumvaluesofaquadraticfunction

CambridgeIGCSEand0LevelAdditionalMathematics

f(x)=x^-3x-4 xeIR Findtheaxiscrossingpointsforthegraphofy-f(x). Sketchthegraphofy=f(x)andusethesymmetryofthecurvetofindthe coordinatesoftheminimumpoint. Statetherangeofthefunctionf(x). Answers a y=x^-3x-4 WhenX=0,y=-4. Wheny=0, X?-3x-4 =0 (x+1)(x-4)=0 X=-IorX =4 Axescrossingpointsare:(0,-4),(-1,0)and(4,0).

X 1.5

,If«<0,thecurvehasamaximumpointwhichoccursatthehighestpointofthecurve.

WORKEDEXAMPLE2

bThelineofsymmetrycutsthex-axis midwaybetween-Iand4. Sothelineofsymmetryisx=1.5 WhenX=1.5,y=(1.5)2-3(1.5)-4. y=-6.25. Minimumpoint=(1.5,-6.25). c Therangeisf(x)^-6.25. y=x^-3x-4

Express2*2-4x+5intheformpix-q)^+r,wherep,^andrareconstantsto Answersbefound. 2*? — 4x+5=p{x— q)^+r Expandingthebracketsandsimplifyinggives: 2»?-4x+5=p}?-2pqx+pq^+r Comparingcoefficientsofx?,coefficientsofxandtheconstantgives:2=p(1)-4=-2pq(2)5=pf+r(3)

Substitutingjb==2inequation(2)givesq=I.Substitutingp=2and^=1inequation(3)givesr=3.

Rearrangingthesegivethefollowingimportantresults: x^ + 2dx =(x + d)^ - d^ — 2dx =(x- d)^ - d^

Tocompletethesquareforx^+lOx-3: 10-2=5 x2+lOx-3=(x+5f-52-3 x2 +lOx-3=(x + 5)^ -28

Chapter2:Simultaneousequationsandquadratics

Completingthesquare

ToThisisknownascompletingthesquare.completethesquareforx^+8x: 8-2=4 /\ x^ +8x =(x +4)^ -4^ x^ +8x =(x +4)^ -16

Tocompletethesquarefor2x2_ ^24 mustfirsttakeafactorof2outof theexpression: 2x2-8x+14^2[x2-4x+7] 4-2=2 /\ x2-4x +7 =(x- 2)^ -22 +7 x2-4x +3=(x- 2)^ +3 So, 2x2 -8x +6 = 2[(^ _ 2)2 + 3]= 2(x - 2)^ +6 Youcanalsouseanalgebraicmethodforcompletingthesquare,asshowninthefollowingexample.

WORKEDEXAMPLE3

So2a2-4x+5=2(x-1)2+3.

Ifyou expand the expressions (x+d)^ and(x-df you obtain the results: (x + d)^ = x^ + 2<ix + d^ and (x- d)^ = -2dx + d^

CambridgeIGCSEand0LevelAdditionalMathematics

Completingthesquareforaquadraticexpressionorfunctionenablesyouto: writedownthemaximumorminimumvalueoftheexpression writedownthecoordinatesofthemaximumorminimumpointof thefunction • sketchthegraphofthefunction

•

Thispartoftheexpressionisasquareso itwillalwaysbe>0.Thesmallestvalueit canbeis0.Thisoccurswhenx=I.

• writedownthelineofsymmetryofthefunction • statetherangeofthefunction.

Theminimumvalueoftheexpressionis 2x0+3=3and thisminimumoccurswhenx=1. Sothefunction y=2x--4x+5 willhaveaminimum pointatthepoint(1,3). WhenX=0,y=5. Thegraphofy-2x--4x+5 cannowbesketched: y=2x^-4x+5 Thelineofsymmetryisx=1. Therangeisy^3. Thegeneralruleis: Foraquadraticfunctionf(x)=ax^+bx+cthatis written in theform f(x)= a(x- h)^+ k, I ifa>0,theminimumpointis{h,k) " ifa<0,themaximumpointis{h,k).

•

cSketchthegraphof31=f(x),showingthecoordinatesofthepointswherethe graphintersectsthexand31-axes.

Chapter2:Simultaneousequationsandquadratics

f(x)=2+8x—2x^ x:eR a Findthevalueofa,thevalueof6andthevalueofcforwhich[(x)=a-l>(x+cf. b Writedownthecoordinatesofthemaximumpointonthecurve y=f(*).

2+8x-2x^=a-6(x^+2cx+ )

Substituting6=2inequation(2)givesc=—2.

Themaximumvalueoftheexpressionisl0-2x0=10andthismaximumoccurs whenX=2. Sothefunctiony=2+8x-2x^willhavemaximumpointatthepoint(2,10). c y 2+8x-2x^

WORKEDEXAMPLE4

dStatetherangeofthefunctionf(x). Answers a 2+8x-2x^=a-b{x+c)^

dTherangeisf(x)10. 2+yj5

Thispartoftheexpressionisasquaresoitwillalways be^0.Thesmallestvalueitcanbeis0.Thisoccurs ^^•henX=2.

2+8x-2x^=a—bx^-2bcx-bc^

Substitutingb=2andc=-2inequation(3)givesa=10. Soa=10,6=2andc=-2. v=10-2(x-2)

y=2+8x-2x'

Comparingcoefficientsofx^,coefficientsofxandtheconstantgives: -2=-b (1) 8=-2bc (2) 2=a-bc^ (3)

WhenWhenX=0,y=2.y=0, 10-2(x-2)^=0 2(x-2)^=10(x-2)^=5 X-2=+V5 X=2±yfb X=2—y/5orX=2+Vs (x=-0.236orX=4.24to3sf). Axes crossing points are:(0,2),(2- y/E, O) and (2+ y/E, o).

8aExpress4x2+2x+5intheforma{x+b)+c,wherea,bandc areconstants. bDoesthefunctiony=4x2^2x+5meetthex-axis? Explainyouranswer. 9 f(x)=2x2-8x+1

ax^—8x+15 b x^—lOx—5 cx^-fix+2 d x^—3x+4 ex^+fix+5 f x^+fix+9 gx^+4x-17 h x^+5x+fi

Exercise2.2

3Expresseachofthefollowingintheform(x-m)^+n.

1 Usethesymmetryofeachquadraticfunctiontofindthemaximumor Sketchminimumpoints.eachgraph,showingallaxiscrossingpoints, ay= -bx-& by= -x-20 cy=x^+4x-21 dy=x^+3x-28 ey= +4x+1 fy=15+2x

CambridgeIGCSEand0LevelAdditionalMathematics

2Expresseachofthefollowingintheform{x—m)^+n.

ax^—8x b x^—lOx cx^—5x d x^—3x ex^+4x f x^+7x gx^+9x h x^+3x

4 Express each of the following in the form a(x - p)^ + q. a 2x2_ +3 b 2x2-12x+1 ^ 3x2_i2x+5 d 2x2-3x+2 e2x2+4j(;+1f 2x2^ _3 g2x2_3^^5 3j(.2_^0 5 Express each of the following in the form m-{x- nf. a fix—x2 b lOx—x2 c 3x—x2 d \26E 8x—x2 xpresseachofthefollowingintheforma-{x+b). a 5-2x-x2 b 8-4x-x2 c 10-5x-x2 d

7 Express each ofthefollowingin theform a-pi^x+ . a 9-fix-2x2 y_4j(;_2x2 ^ 7 gx-2x2 7+3x-x22+5x-3x2

a Express 2x2 _ 4. y form a(x + b^ + c, where a and b areintegers. bFindthecoordinatesofthestationarypointonthegraphofy=f(x). 10 f(x)= x2-X-5 for x6 R a Findthesmallestvalueoff(x)andthecorrespondingvalueofx.

b Hence write down a suitable domain forf(x)in order that f~H^) exists. 11f(x)=5-7x-2x2y-Qj.g(R

a Write f(x) in the form p-2{x- q^,where p and q are constants to befound. b Writedowntherangeofthefunctionf(x).

Chapter2:Simultaneousequationsandquadratics

aExpress7+5x-x^intheforma-(x+ ,wherea,andb areconstants.

forxGIR a Express 14 + 6x - 2x^ in the form a + b{x + c)^, where a, b and c areconstants.

15f(x)=4x^+6x-8wherex^m

b Writedownthecoordinatesofthestationarypointonthegraph ofy=f(x). c Sketchthegraphofy=f(x).

a Write f(x) in the form 2(x +of + b, where a,and ^are constants to befound.

b Find the coordinates ofthe turning pointofthe function f(x),stating whetheritisamaximumorminimumpoint.

c Explainwhyf(x)hasaninverseandfindanexpressionforf~'(x)intermsofx.

a Express 1+4x-x^ in theform a-{x-\-bf,where aand b are constants tobefound. b Findthecoordinatesoftheturningpointofthefunctionf(x),stating whetheritisamaximumorminimumpoint.

16f(x)=1+4x-x^forX^2

c Findtherangeoff.

13f(x)=7+5x- for0^X^7

Findthesmallestvalueofmforwhichfhasaninverse.

12f(x:)=14+6jc-

14 Thefunction fis such that f(x)= 2x^-8x+3.

bWritedownasuitabledomainforfsothatf exists.

d State,giving a reason,whether or notfhasan inverse.

CambridgeIGCSEand0LevelAdditionalMathematics 2.3 Graphs ofy=|fM|where f(x)is quadratic

•reflectinthex-axisthepartofthecurvey-ax^+bx+cthatisbelow thex-axis.

Sketchthegraphofy=|x^-2x-3| Answers Firstsketchthegraphofy=x"-2x -3. WhenX=0,y=-3. Sothey-interceptis-3. Wheny=0, x^-2x-3=0 (x+l)(x-3)=0 X=-1 or X=3. Sothex-interceptsare-1and3. -1+3The x-coordinate ofthe minimum point = ^ =1. They-coordinateoftheminimumpoint=(l)^-2(l)-3=-4. Theminimumpointis(1,-4). \ ^ OG CM tT( II -i\o -3 (1.-4)

WORKEDEXAMPLE5

Nowreflectinthex-axisthepartofthecurvey=x^-2x-3thatisbelowthex-axis. \ y=\x^--2x-3| (1,4) / \ ^ -1\O _/3

Tosketch the graph ofthe modulusfunction y =|ax^ +bx+c\, you must: • firstsketch the graph of y = ax"^ + bx + c

Chapter2:Simultaneousequationsandquadratics Asketchofthefunction)>=|«:^+4x-12|isshownbelow. 7 + II \ (-2,16) / \ / Y -6 O Nowconsiderusingthisgraphtofindthenumberofsolutionsoftheequation I +4x-I2I=Awhere 0. y=\x^+4:x-12\ J= 20 IX^+4x- 121 = 20 has2solutions ><=16|x^+4x-12|=16has3solutions (-2,16) -----X 12 -6 O y=l |x^+4x-12|=7has4 solutions |x^+4x-12|=0 has 2solutions X Theconditionsforthenumberofsolutionsoftheequation|x^+4x-12|=^ are: ValueofA ^=0 0<k<16 k=l6 /s>16 Numberofsolutions , 2 .■ ti ^ 4 3 2 Equationsinvolving|f(x)|,wheref(x)isquadratic,canbesolvedalgebraically: To solve x^ + 4x - 12i = 16 : x^+4x-12=16 x^+4x-28=0 X = X = ± V4^ - 4 X 1 X (-^ 2x1 -4± or or x^+4x-12=-16 or x^+4x+4=0 or (x + 2)(x + 2) = 0 X=-2 X = -2 ± 4V2 (x = 3.66 or X = -7.66 to 3 sf) The exact solutions are x = -2 - 4V2 or x = -2 or x = -2 + 4V2.

bFindthevalueofkforwhichl(x-8)(x-3)l=Ahasthreesolutions. Solvetheseequations. |x2-6|=10 b|x2-2|=2 1x2+2x1=24 e a d c IX -5x|=6 x--5x+l|=3 f|x2+3x-1|=3 g x2+2x-4|= 5 h |2x2-3|= 2x 1 |x2-4x + 7l = 4

b Sketch the graph of3=|(x-7)(x+l)|.

Cambridge IGCSE and0LevelAdditional Mathematics Q Note:

a Findthecoordinatesofthestationarypointonthecurve 3=|(x-7)(x+l)|.

a Writef(x)in theform a-{x+b^,where aand bareconstants, b Sketchthegraphof3=f(x).

Thegraphofj;=|+4x-121issketchedbelowshowingthese threesolutions. j +4x- 12| -2,16 -2+4^/2-6 2+4^/2 Exercise2.3

c Find the set ofvalues ofkfor which|(x-7)(x +l)|= k has foursolutions.

a Findthecoordinatesofthestationarypointonthecurve 3=|(x + 5)(x + 1)|.

cSketchthegraphof3=|f(x)|. f(x)= 2x^ + X-3 a Writef(x)in theform a(x+b)'^+c, where a,band care constants, b Sketch the graph of 3 =|f(x)|.

b Find the set ofvalues of kfor which |(x + 5)(x +l)|= k has twosolutions. a Findthecoordinatesofthestationarypointonthecurve 3 =|(x -8)(x -3)|.

a b y =\x^ -2x-3\ 5x+4 e = I 2x^ -llx;-6| f 3; = 1 + 5x - 2| ^ — 4x+3| d3=I -2x-8| f(x)=1-4x-x^

1Sketchthegraphsofeachofthefollowingfunctions.

Chapter2:Simultaneousequationsandquadratics CHALLENGEQ 8 Solvethesesimultaneousequations.a3;=x+lb2y-X+4: 3)=I -2x-3 :y= 1 X-3 C y=2x y 1 2x--41 2.4Quadraticinequalities RECAP Youshouldalreadyknowhowtosolvelinearinequalities. Twoexamplesareshownbelow. Solve2{x-5)<9 2x-10<92x<19 X<9.5 expandbrackets add10tobothsidesdividebothside>ib\2 Solve5-3x^17 -3.V^12 .V«-4 subtract5frombothsidesdi\idebothsidesbv-3 Q Note: Itisveryimportantthat\()uremembertherulethatwhenyoumultiply ordividebothsidestjfanineqtialit\b\anegativenumberthenthe inequalitysign mustbereversed.Thisisillustratedin thesecond exampleabo\e,whenbothsidesoftheinequalityweredividedbv-3. CLASSDISCUSSION 7x+12Robertis asked to solve the inequality ^3. Hewrites: 7X+/7^3X 4X^-12 Sox^-3 Annacheckshisanswerusingthenumber-4. Shewrites: Wheu.,x=—4, {y-X(-4)+12)^(-4)={~±&)-i-(-4)=4 t-f&MtK=—4tsavalueofxthatsatisfiestheoriQliA.alitvequaLLty SoRobert'saia-swertukstbeLtvcorrecti DiscussRobert'sworkingoutwithyourclassmatesandexplainRobert'serror, 7x+12Nowsolvetheinequality ^^3correctly.

Exercise2.4 1 Solve. a (x+3)(x-4)>0 b d x(x—5)<0 e (x-5)(x-l)^0 c (x-3)(x+7)^0 (2x+l)(x-4)<0f (3-x)(x+l)^0 g (2x+3)(x-5)<0 h {x-bf^Q i (x-3)^ Solve. ax'"^+5x-14<0 bx^+X-6^0 c d x2+2x-48>0 e 2x2-X-15^0 f x2-9x 0 +20^0 5x2^ 4.4> Q

Thesolutionis x<-1 and x>4. Solve2x2 15 Answers y Rearranging:2x2^j(._i5^q Sketchthegraphof3=2x2^_15 \ When3=0,2x2-1-X-15=0 y=2x^+X-15 /+ (2x-5)(x+3)=0 -3\ 0 X=2.5 or X=-3. \

QuadraticCambridgeIGCSEand0LevelAdditionalMathematicsinequalitiescanbesolvedbysketchingagraphandconsidering whenthegraphisaboveorbelowthex-axis.

Thesolutionis-3^x^2.5. /2.5 X

Sketchthegraphof31=x^-3x—4. / When3=0,x^—3x-4=0 (x+l)(x-4)=0 X=-1 or X=4.

Sothex-axiscrossingpointsare-1and4. +\ /+ -i\o X Forx^—3x-4>0youneedtofind therangeofvaluesofxforwhichthecurveispositive(abovethex-axis).

Sothex-axiscrossingpointsare-3and2.5. For2x2^ _15^QyQyneedtofind curvetherangeofvaluesofxforwhichtheiseitherzeroornegative(belowthex-axis).

WORKEDEXAMPLE6 Solve -3x-4>0. Answers \ ^ ^I II

Chapter2: Simultaneousequations and quadratics 3Solve. a <18-3x c x(3-2x)^1 e (x+3)(1-x)< X-1 12x<x2+35 x^ +4x < 3(x +2) (4x+3)(3x — 1)< 2x(x+3) 4 Findthesetofvaluesofxfor which ax^-llx+24<0 and 2x+3<13 bx^-4x^12 and 4x-3>1 c x(2x-1)< 1 and 7-2x<6 d x2-3x-10<0 and x^-lOx+21<0 ex^+x-2>0 and x^-2x-3^0. 5Solve. a1x^+2x-2|<13 b 1 x^ -8x+61 <6 C CHALLENGEQ 6 Find the range ofvalues of xfor which 3x^-2x-8 <0. 2.5 Roots of quadratic equations The answers to an equation are called the roots ofthe equation. Consider solving the following three quadratic equations using the -b±yjb'^—4ac formulax= 2a x^+2x-8=0 x^+6x+9=0 x^+2x+6=0 _-2± -4X1X(-8) 2x1 -2±-v/36 X = 2 X=2orX=-4 -6± -4X1X9 X 2x1 -2±>/0 X 2 X=-1orX=-1 -2±V22-4X1X6 X 2x1 -2±V-20 X 2 nosolution 2distinctroots 2equalroots 0roots The partofthe quadraticformula underneath the square rootsign is called thediscriminant. discriminant=b^—4ac

Cambridge IGCSE and0LevelAdditional Mathematics

-4ac

Findthevaluesofkforwhich —Sx+6—k{x—2)hastwoequalroots. Answers -3x+6=k{x-2) x^—3x+6—kx+2k=0 x^-{3+k)x+6+2k=0 Fortwoequalroots -4ac=0. (3+kf-4x\x{6+2k)=0 +9-24-8ft=0 ft2-2ft-15=0 (ft+3)(ft-5)=0 Soft=-3 or ft=5.

6^-4ac Natureofroots >0 2realdistinctroots =0 2realequalroots <0 0realroots

0realroots

Shapeofcurvey=ax^+bx+c >0 2realdistinct

WORKEDEXAMPLE8

Thesign(positive,zeroornegative)ofthediscriminanttellsyouhowmanyrootsthereareforaparticularquadraticequation. Thereisaconnectionbetweentherootsofthequadraticequation ax^+bx+c=Qandthecorrespondingcurvey=ax"^+bx+c. Natureofrootsof ax^+6*+c=0 roots X Thecurvecutsthe x-axisat2distinctpoints. =0 2realequalroots or Thecurvetouchesthex-axisat1 point. <0 or Thecurveisentirelyaboveorentirelybelow thex-axis.

6 Find the values of k for which kx^ + 2{k +S)x + k = 0 has two distinctroots.

4 Findthevaluesofkforwhich3x^+2x+^=0hasnorealroots.

1 Statewhethertheseequationshavetwodistinctroots,twoequalrootsor noroots. a +4iX+4 b x^+4x-21 c +9x+1 e -6x + 2 f 4x^ + 20x + 25 g 3x^ + 2x + 7 32 d X-3x+15 h 5x^-2x—9

8 Find the values of kfor which 4x^ -{k-2)x +9=0 has two equal roots.

WORKEDEXAMPLE9

5 Find the valuesofkfor which(A+l)x^+^x-2^=0 hastwoequal roots.

11Showthattherootsoftheequationkx^+5x-2k=0arerealanddistinct forallrealvaluesofk.

Chapter2:Simultaneousequationsandquadratics

7 Findthevaluesofkforwhich 3x^-4x+5-^=0hastwodistinctroots.

Findthevaluesofkforwhichx +^x+9=0hastwoequalroots.

10Showthattherootsoftheequation +{k-2)x-2k=0arerealand distinctforallrealvaluesofk.

FindthevaluesofAforwhich +{k—2)x+4=0hastwodistinctroots. Answers

Findthevaluesofkforwhich ftx^-4x+8=0 hastwodistinctroots.

SoCriticalvaluesare-2and6.ft<-2orft>6.

x^+{k-2)x+4=0 \ / r

9 Find the values of kfor which 4x^ +4{k- 2)x + k -0 has two equal roots.

Fortwodistinctroots 4ac>0 \ / (ft-2f-4x1x4>0 \ /+ ft2-4ft+4 -16>0 \ / ft^-4ft -i2>o (ft+2)(ft--6)>0

Exercise2.5

Situation1 Situation2 Situation3 2pointsofintersection 1pointofintersection 0pointsofintersection

Whenconsideringtheintersectionofastraightlineandaparabola,thereare threepossiblesituations.

Thenumberofpointsofintersectionwilldependonthevalueofb'-4ac.Thedifferentsituationsaregiveninthetablebelow. =4ac Natureofroots Lineandcurve >0 2realdistinctroots 2distinctpointsofintersection =0 2realequalroots 1pointofintersection(lineisatangent) <0 0realroots nopointsofintersection

CambridgeIGCSEand0LevelAdditionalMathematics

Thelinedoesnotintersectthecurve. Youhavealreadylearntthattofindthepointsofintersectionoftheline 9 y-x-^withtheparabola y=x"-3x-4 yousolvethetwoequations simultaneously. Thiswouldgivex~-3x-4=x-6 x~-4x+2=0. Theresultingquadraticequationcanthenbesolvedusingthe quadraticformula; -b ± ylb~ — 4ac X 2a

Thelinecutsthecurveattwo distinctpoints. Thelinetouchesthecurveat onepoint.Thismeansthat thelineisatangenttothe curve.

Theconditionforaquadraticequationtohaverealrootsisb-4ac^0.

2.6 Intersection ofa line and a curve

Sincethelineisatangenttothecurve, -4ac=0. (-6)^-4xlx(4-^)=0 36-16+4^=0k=-b

WORKEDEXAMPLE12

Findthevalueofkforwhichy=2x+kisatangenttothecurvey- -4x+4. Answers x^-4x+4=2x+k x^-6x+(4-k)—0

Findtherangeofvaluesofkforwhichy=x-5intersectsthecurvey=kx'-6attwo distinctpoints. Answers kx^kx^—6=X-5-X-1=0

Criticalvaluesare-6and2. So-6<A<2.

WORKEDEXAMPLE11

Sincethelineintersectsthecurveattwodistinctpoints, b'^—4ac>0. (-if-4xkx(-1)>0 1+4A>0 1

WORKEDEXAMPLE10

Findthevaluesofkforwhichy=kx-3doesnotintersectthecurvey=x'^—2x+1. Answers x^-2x+1=A;x-3 x^-x(2+^)+4=0

Sincethelineandcurvedonotintersect. b'^-4ac<0. \ / (2+ -4X1X4<0 \ / +4fe+4-16<0+4;i-12<0 +\ /+ ik+6)ik-2)<0 -6\ /2 '^

Chapter2:Simultaneousequationsandquadratics

2 Findthevalueofkforwhichthex-axisisatangenttothecurve = x^ + -k)x-{4k +3). 3 Findthevaluesoftheconstantcforwhichtheliney=x+cisatangentto 2 thecurvey-Sx+ X 4

9 The line 3; = mx +6 is a tangent to the curve 3= x^ -4x + 7. Findthepossiblevaluesofm.. Summary

CambridgeIGCSEand0LevelAdditionalMathematics

Completingthesquare

For a quadraticfunction f(x)=ax^+ bx+ cthatis written in theform f(x)= a{x- Jif+k, 1ifa>0,theminimumpointis{h,k) iiifa<0,themaximumpointis{h,k).

Findthesetofvaluesofkforwhichtheline y=3x+l cutsthecurve y=x^+kx+2intwodistinctpoints. 5 Theliney=2x+kisatangenttothecurve +2x)i+20=0. a Findthepossiblevaluesofk. b ofForeachofthesevaluesofk,findthecoordinatesofthepointofcontactthetangentwiththecurve.

Findthesetofvaluesofmforwhichtheliney=mx-5doesnotmeetthe curvey=x^-5x+4.

Findthesetofvaluesofkforwhichtheliney=k-xcutsthecurve y=x^-7x+4intwodistinctpoints.

Findthevaluesofkforwhichy=kx+1isa.tangenttothecurve y=2x^+x+S.

Exercise2.6

7 Findthevaluesofkforwhichtheline3;=fex-10meetsthecurve x^+y^=lOx. 8

Chapter2:Simultaneousequationsandquadratics Quadraticequation{ax^+bx+c=0)and correspondingcurve(y =ax^+bx+c) >0 Natureofrootsof ax^+bx+c-0 2realdistinctroots Shapeofcurvey-ax^+ +c ■X or Thecurvecutsthex-axisat2distinctpoints. =0 2realequalroots or Thecursetouchesthex-axisat1point. <0 0realroots or The curve is entirelv above or entirelv below the x-axis. Quadratic curve and straight line Situation1 Situation 2 Situation3 2pointsofintersection 1pointofintersection 0pointsofintersection Thelinecutsthecurveattwo distinctpoints. The line touches the curve at onepoint.Thismeansthatthe lineisatangenttothecurve. Thelinedoesnotintersectthecurve. Solving simultaneously the equation of the curve with the equation of the line will give a quadratic equation ofthe form ax^ + bx + c = 0. The discriminant - 4ac, gives information about the rootsoftheequationandalsoabouttheintersectionofthecurvewiththeline. -4ac Natureofroots Lineandcurve >0 2 real distinct roots 2distinctpointsofintersection =0 2realequalroots 1pointofintersection(lineisatangent) <0 norealroots nopointsofintersection The condition for a quadratic equation to have real roots is b^ — 4ac ^ 0.

CambridgeIGCSEAdditionalMathematics0606Paper21Q8Jun2013

Answers y=2x-8 - .(1) 2x^+y'^-5xy+32=0 (2) Substitutefor))from(1)in(2): 2x^+i2x-8)^-5x{2x-8)+32=0 2x^+4x2_32x+64-lOx^+40x+32=0 -4x2+ ^90_Q x2-2x-24=0 (x-6)(x+4)=0 X=6orX=-4 When X 6,))= 2(6)-8= 4. WhenX=-4,y=2(-4)-8=-16. ThepointsofintersectionareA(6,4)andB(-4,-16). UsingPythagoras;AB'^=20'^+102 AB-=500AB= AB=22.4to3sf expandbrackets collectliketerms divideby-4 factorise A(6,4) 5(-4,-16) Exercise2.7 ExamExercise 1 Findthesetofvaluesofkforwhichtheliney=^(4x-3)doesnotintersectthecurve y = 4x2 g^ _ g_ [5] CambridgeIGCSEAdditionalMathematics0606Paper11Q4Jun2014 2 Findthesetofvaluesofxforwhichx(x+2)<x. [3] CambridgeIGCSEAdditionalMathematics0606Paper21Q1Jun2014 3 a Express 2x2 _ +5 form p[x- +r, wherep,qand rareconstantstobefound.[3] b Flencestatetheleastvalueof2x2_ +5 valueofxatwhichthisoccurs. [2] CambridgeIGCSEAdditionalMathematics0606Paper21Q5i,iiJun2014

CambridgeIGCSEand0LevelAdditionalMathematics Examination questions Workedexample Theliney=2x-8cutsthecurve2x^+ -5x)!+32=0atthepointsAandB. FindthelengthofthelineAB. m

9Findthesetofvaluesofkforwhichtheliney-2x-5cutsthecurvey=x^+kx+\\in twodistinctpoints. [6]

CambridgeIGCSEAdditionalMathematics0606Paper21Q1Nov2013

CambridgeIGCSEAdditionalMathematics0606Paper21Q4Jun2011

4Findthesetofvaluesofkforwhichthecurve)»=(A+l)x:^-3x+(^+1)liesbelowthex-axis. [4] CambridgeIGCSEAdditionalMathematics0606Paper11Q2Nov2013

CambridgeIGCSEAdditionalMathematics0606Paper21Q4i,iiNov2012

a Findthevalueofaandofbforwhichf(x)=a-{x+b)^andhencewritedownthe coordinatesofthestationarypointofthecurvey=f(x). bOntheaxesbelow,sketchthegraphofy=f(x),showing thecoordinatesofthepointwhereyourgraphintersects they-axis. [3] O [2]

Itisgiventhatf(x)=4+8x-x^.

CambridgeIGCSEAdditionalMathematics0606Paper21Q2Nov2011

8Giventhatthestraightliney=3x+cisatangenttothecurvey=x^+9x+k,expresskin termsofc. [4]

Chapter2:Simultaneousequationsandquadratics

Findthevaluesofkforwhichtheline3;=A-6xisatangenttothecurvey=x{2x+k). [4]

10 The equation ofa curve is given by y = 2x^ +ax+14, where ais a constant. Given that this equation can also be written as y = 2{x- + b, where is a constant,find a thevalueofaandofb, [2] btheminimumvalueofy. [1]

CambridgeIGCSEAdditionalMathematics0606Paper21Q5Nov2010

11Findthesetofvaluesofmforwhichtheliney=mx-2cutsthecurvey=x^+8x+7intwo distinctpoints. [6]

CambridgeIGCSEAdditionalMathematics0606Paper11Q2Nov2012

CambridgeIGCSEAdditionalMathematics0606Paper11Qli,iiNov2010

Findthesetofvaluesofxforwhich <6-5x. [3]

CambridgeIGCSEAdditionalMathematics0606Paper11Q1Nov2015 15 a FindthesetofvaluesofXforwhich 4x?+19x-5 0. [3]

b i Express x? +8x-9 in the form (xAd)^ + b. where a and & are integers. [2] ii Use youranswerto partitofind thegreatestvalue of 9-8x-x? andthe value ofxat whichthisoccurs. [2] iii Sketchthegraphofy=9-8x-x^,indicatingthecoordinatesofanypointsof intersectionwiththecoordinateaxes. [2]

CambridgeIGCSEAdditionalMathematics0606Paper21Q9Jun2015

1 1 1 1 1

1 1 1 * 1 2 3 4^ [4]

CambridgeIGCSEAdditionalMathematics0606Paper11Q8i,ii,iiiNov2013

CambridgeIGCSEAdditionalMathematics0606Paper11Q1Jun2016

CambridgeIGCSEandpLevelAdditionalMathematics12aOnacopyofthegridtotheright,sketchthe graphofy=I(X-2)(X+3)|for-5^x^4, andstatethecoordinatesofthepointswhere thecurvemeetsthecoordinateaxes. -5-4-3-2-10

c Given that ^is a positive constant,state the set ofvalues of kfor which l(x-2)(x + 3)|= k has2solutionsonly. [3]

13Findthevalueofkforwhichthecurvey=2x^-3x+ Ipassesthroughthepoint(4,-7), [1] ilmeetsthex-axisatonepointonly. [2]

b Findthecoordinatesofthestationarypointonthecurvey=|(x-2)(x+3)|. [3]

14Findtherangeofvaluesofkforwhichtheequation =8x-2xk has2realdistinct roots. [4]

Chapter 3 Indices and Surds Thissectionwilishowyouhowto: ■ performsimpleoperationswithindicesandsurds,includingrationalising thedenominator.

CambridgeIGCSEand0LevelAdditionalMathematics Youshouldalreadyknowthemeaningofthewordsindex,power,exponentandbase: index,exponentorpower 2x2x2x2=2^ t base Thepluralofthewordindexisindices. (TheseYoushouldalsoknowandbeabletoapplythefollowingrulesofindices:rulesarealwaystrueifthehasesaandharepositiverealnumbersandtheindicesmandnarerealnumbers.) RULE1: RULE2 RULE3 RULE4 RULE5 m^,n in+n a X a = a m n in-ii d in-u a a =■ a or —77= 0. / m\ n inn{a ) =a a" X b" = {ab) " a" _ /rt\" RULE 6: RULE7: RULE 8: RULE 9: " 1rt = 1 a = 1 a" = 4a a" =C-te)'"= ■0 y y-v: IfthebasesaandbarenegativethentheindicesmandnmustheintegersforRules1-5tobealwaystrue.Rules6and7arealsotrueifbaseaisnegative. 3.1 Simplifying expressions involving indices When simplifying expressions involving indices you often need to use more than one of the rules for indices. In Worked Example 1 and Exercise 3.1 you may assume that all bases are positiverealnumbers. WORKEDEXAMPLE1 .i Simplify ,^ (x'y) (4x^y)^ X ^]9x''y^ (4)^(:«^)'(y)'x(9)5 (/)^ / 3,-2 2,-2(x ) (y ) x~ 48 y 6-4 y 3 3c~^y"T = 48x'^ /

Chapters:IndicesandSurds Exercise3.1 1 Simplifyeachofthefollowing. a g b xx^ n8 j yj25x^ y e k CX-i-X f ix-')^ .2...e4.3i Sx yx5x y I ^j9x^y~'^ X ^jSx^ y~^ 2Simplifyeachofthefollowing.^X\x^ -2 (?/3c)^ X V/ X3 X Vj? (3xy)^X (2x^yl 3Giventhat 4Giventhat (36x^).4^2 =2"3^x\evaluatea,6andc. 8xX3x a 5Giventhat 4_2 a^b3 a_i25^3 =x"y,findthevalueofaandthevalueofb. = a iP,find the value of xand the value ofy. 6Giventhat _2.4X ,5-* „ =a b,findthevalueofxandthevalueofy. 3 1 7 Simplify(1 + x)^-(1+x) 3.2 Solving equations involving indices Considertheequation2"=64. Theequationhasanunknownexponent(orindex)andiscalledanexponentialequation.Youhavealreadylearnthowtosolvesimpleexponential equationssuch as 2*=64.In this section,you will learn how to solve more complicatedexponentialequations.

CambridgeIGCSEand0LevelAdditionalMathematics CLASSDISCUSSION Thediagramshowsthegraphof3)=2*for-3^x^3. Discuss thefollowing questions with your classmates: • Howdoesthegraphbehavewhenx>3? •Howdoesthegraphbehavewhenx<-3? •Howdoesthegraphofy=3*comparewiththe graphofy=2*? •Howdoesthegraphofy=4*comparewiththe graphsofy=2*andy=3*? -3 -1 3^1 -& -XNowanswerthefollowingquestion: •Ifa>0,isitpossibletofindavalueofxforwhicha*iseither0ornegative? v=2' 3 WORKEDEXAMPLE2 Solvetheequation2'^*^'=8. Answers 23,:+1 change8to2® 23*+1_2^ equatetheindices 3x+1 =3 soiveforx 3x=2 2 X— WORKED3EXAMPLE3 Solvetheequation3^*'x9*'=243. Answers 32*-i^9*-i^243 changetobase3 32*-1^(32).-!^35 3^*"^X3^*"^=3® addtheindicesontheiefthandside 34*-3^35 equatetheindices 4x-3=5 solveforX H II 00 X=2

Chapters:IndicesandSurds WORKEDEXAMPLE4 A Solvethesimultaneousequations.9'(27^)=1 4''+(V2f=128 Answers 9*(27^)=1 changetobase3 4'=-f-(V2)^=128 changetobase2 32*^33,^30 2^* 22^-2' addiudiresonthelefthandside subtractindicesonthelefthandside 32*+3j_gO 2^"-^^2' equatingtheindicesgives2x+Sj=0 equating the indices gives 2x — ■^3'=7 2x+ 3^1 =0 subli act the two equations 2x--y =7 2 3^))=-7,so3;=-2. Substituting31=-2into2a:+ =0givesx=3. The solution is x = 3,31 = -2.

WORKEDEXAMPLE5 JH 2 a SolvetheequationAy +?>y -1=0. b Use your answer to part a to solve the equation 4(2*)^ + 3(2*) -1=0. Answers a 43^ + 3y- 1 = 0 factorise (43-1)(3+1) =0 43-1 = 0 or 3+1 = 0 1 1 3=—or3=—1 b4(2*)^+3(2*)-1=0 2 comparingwith4y +3y—1=0givesy=2* 2'' =i-or2* =-l replace — with2"^ 4 ' 4 2*=2"^or2*=-l X 2Solving2=2 ,givesx= 2. Thereisnosolutionto2*= -1, since 2* > 0 for all real values of x. Thesolutionisx=-2.

CambridgeIGCSEand0LevelAdditionalMathematics Exercise3.2 1 Solveeachofthefollowingequations, a 52>:=57x-i b 4^*+1=43^-2 d =3,9x+5

2Solveeachofthefollowingequations. b 4^"-256 in-1 ~4 a 2"^^=32 d3 2n+ 1 =27 1 e2' c 2"'"^ =128 g5n+1 f2 3n+ 2 128 125 h 5^'-16=1 3 Solveeachofthefollowingequations. 4.3 3^*-1=27* .3x+4 o4x+12 a 2*=4d 3*=9*^^ e4 =8^ j 5x^+3= 252^ h4 5-3x 1 ,x+1 k 3*'"^=27' fc53^-7=252^252^^+1=1253^^+2 i I 2'^*"~^-8*=0 4 Solveeachofthefollowingequations. a 2^*x4*^'=64 b 2^*""^x8*'^=128 c(2^"*)(4^*"^^)=8 5 Solveeachofthefollowingequations. d 3*^'x9^-*= 27 ,2x+ 1272x gO-X gx+3 ix+ 4 8~ 64 4I* db 4* ^Sx 2?-X H 00 272x 32*+1 36-x- 9X+3 b 2^*X5*=8000 c 52^x4*= 10001 6 Solveeachofthefollowingequations. a 3^*x2*=— b 18 7 Solveeachofthefollowingpairsofsimultaneousequations. a 4*4-21"=16 b 27*=9(3^) c 125*4-51'=25 32*x9>=27 2*4-81'=1 8 a Solvetheequation -7y-4=0. bUseyouranswertopartatosolvetheequation2(2*)2-7(2*)-4=0. 2 9 a Solvetheequation4y =15+7y. bUseyouranswertopartatosolvetheequation4(9*)=15+7(3*). 310 a Solve the equation 3y = 8+ y. ^ ^ b Useyouranswertopartatosolvetheequation3x2=8+3x2.

Chapters:IndicesandSurds 33Surds Asurdisanirrationalnumberoftheform wherenisapositiveinteger thatisnotaperfectsquare. ,^/5 and are all surds. ^/9 isnotasurd because V9=V?=3. I— I— I— 3Other examples ofsurds are 2+ \5, V7 - V2 and —5~~Whenananswerisgivenusingasurd,itisanexactanswer. CLASSDISCUSSION ontoThefrogcanonlyhoplilypadsthat Itcontainsurds.isallowedtomove alongarow(westor east)oracolumn (northorsouth)but isnotallowedtomove diagonally. Findtheroutethatthe frogmusttaketocatch thefly. , ll'+V25^!v..I .■ » \ „.^1 ♦ You can collect like terms together. 6^/^^+ 3Vir= 9-Jn and 5>/7 - 2^7 = 3>/7 WORKEDEXAMPLE6 Simplify4(5 - ^/3) - 2(5^3 - l)Answers 4(5-V3)-2(5V3-1) =20-4a/3-10\/3+2 = 22 -14V3 expandthebrackets collectliketerms

CHALLENGEQ

5 Thenumberintherectangle onthesideofthetriangleis thesumofthenumbersatthe adjacentvertices. Findthevalueofx,thevalueof yandthevalueofz. 17-10a/2

CHALLENGEQ

6Thisdesignismadefrom and1bluecircle,4orangecircles16greencircles. Thecirclestoucheachother.Giventhattheradiusofeach greencircleis1unit,findthe exactradiusof atheorangecircles, bthebluecircle.

CambridgeIGCSEand0LevelAdditionalMathematics Exercise3.3 1 Simplify. a 3>/5+7V5 2 A 3>/5+7V3 Simplify. 3A+B b 3VT0+2VT0 c 8^^TT+^/TT B2>/5-3Nf3 C 2>/3->/5 bA-C c2A+3B d 6^3-V3 d 5A+2B-C 3 Thefirst4termsofasequenceare 2+3V7 2+5^/7 2+7V7 2+9^/7. a Writedownthe6thtermofthissequence, b Find the sum ofthe first5terms ofthis sequence. C Writedownanexpressionforthenthtermofthissequence. 4 a Find the exactlength ofAB. ^ b Findtheexactperimeterofthetriangle.

3.4Multiplication,divisionandsimplificatlonofsurds

Youcanmultiplysurdsusingtherule;\[ax"lb=4ab

Simplify. a VSxVb b (V8)' c 2V5X3V3 Answers aV3X>/5=V3^=VT5 b(Vs)^=VSXVS=VM=8 Note: c 2a/5X3V3=6Vl5 Expandandsimplify. a (4-V3)^ b (V3+5>/2)(V2+V3) Answers a (4-V3)' =(4-V3)(4-V3) ==16-4V3-4V3+319-8V3 b(V3+5V2)(V2+V3) =V6+3+10+5V6=13+6a/6

scpiarsrrxeansmultiplybyitself Ixpindthebrackets I lifeetmtis expandthebraGkets eelleetEketerms V98 can be simplified using the multiplication rule. -M=V49x2=V49XV2=7V2

WORKEDEXAMPLE7

CambridgeIGCSEand0LevelAdditionalMathematics WORKEDEXAMPLE91 Simplify -Vl2 Answers V75->/l2 75=25X3and12=4x3 II X X =5anda/4=2 =5x^/3-2xa/3 collectliketerms =3a/3 Youcandividesurdsusingtherule: •\/a a b WORKEDEXAMPLE10 SimplifyNfn" Answers 4(5-^/3)-2(5V3-1) =20-4>/3-10a/3+2 Exercise3.4 Simplify. at 00| X b a/2Xa/^ c a/5Xa/6 d e (VII)" f g 3a/2x5a/3 h 7a/5x2V7 Simplify. a/112 U a/^ a/T2 a/T7 "V28 D c a/3 U A[i2 f a/T5 g Vm a/4 ® a/108 f V3 a/6 n a/^ . V5 a/^ 1# 9a/^ 1 a/120 a/^ J 2a/TT 3a/5 1 a/^ Simplify. aa/8 b a/12 c d a/^ e f g a/T8 h a/32 1 a/^ j a/90 k I a/99 ma/44 n a/125 0 a/117 P a/200

Chapters:IndicesandSurds q " 5 yV^x^/5 Simplify. r VSOOO V ^!w X^flO w 2 ^/8xVS 3 •VSX^/6 a loo + b a/12+a/3 c a/^+3a/5 d V^+2V3 e a/^-2V8 f a/125+a/^ g V45-a/5 h a/^-5a/5 i a/175-a/^+a/^ j |o + joK |oo k a/200-2VI8+ a/72 I 5a/^-3a/63-a/7 m V^+2V^+4a/45 n 5a/T2-3^/48+ 0 a/^+a/8-a/98+

5 Expandandsimplify. a ^/2(3+^/2) d^/3(^/^+5) g(>/2+l)(V2-l) j(3->/2)(3+V2) m(4+2^/3)(4-2V3) b>/3(2V3+a/T2) e ^/3(^/3-l) h k(V3+5)(^/3-1)(4+V3)(4->f3) n(V7+^/5)(^/7+2^/5) c a/2(5-2V2) f^/5(2V5+^/^) i (2+ a/5)(2V5 +1) I (1 + a/5)(1- Vs) o(3+2^/2)(5+2^/2)

8a FindthevalueofAC'^. bFindthevalueoftanx. Writeyouranswerintheform whereaandbareintegers, c Findtheareaofthetriangle. Writeyouranswerintheform 5V2 wherepandqareintegers. 9 Acuboidhasasquarebase. The sides ofthe square are oflength(l +V2)cm. Theheightofthecuboidis(S-V2)cm. Findthevolumeofthecuboid. Expressyouranswerin theform a+b^}2,where aand bareintegers.

6 Expandandsimplify. (2+V5)' b(5-^/3)' (4+5V3)^ d (^/2+^/3)^

7 Arectangle hassidesoflength(2+^/8)cm and(7->/2)cm. Findtheareaoftherectangle. Expressyouranswerin theform a+b^I2,where aand bareintegers.

Khadeejasaysthat: '7%e aj-UvX)irroAunuU tvumJxtYS irrxUunuU.' Discussthisstatementwithyourclassmatesanddecideifthisstatementis: Alwaystrue Sometimestrue Nevertrue Youmustjustifyyourdecision. Yourationalisethedenominatorofafractionwhenitisasurd.Torationalisethedenominatorofafractionmeanstoturnanirrational denominatorintoarationalnumber. InExercise3.4,question5m youfoundthat This(4+2V3)(4-2^)=(4)2-8V3+8V3-(2^3)^=16-12=4isanexampleoftheproductoftwoirrationalnumbers(4+2^3)and (4-2V3)giving a rational number (4). The4+2^/3and4-2^/3arecalledconjugatesurds.productoftwoconjugatesurdsalwaysgivesarationalnumber.

Forfractionsoftheforma-bdc.Forfractionsoftheform a+ftVc. 1 a+6Vc 1 a ,multiplynumeratoranddenominatorby,multiplynumeratoranddenominatorby

CHALLENGEQ 11 a Given that a/6'-4^ = Va- Vfe, find the value of a and the value of b. b Given that "^23-6^6-4^2 = Vc + Vrf, find the value of <~ and the value of d. 3.5

Theproductofconjugatesurds a+bdc and a— b4cisarationalnumber. •Soyoucanrationalisethedenominatorofafractionusingtheserules:Forfractionsoftheformmultiplynumeratoranddenominatorby-id.

CambridgeIGCSEand0LevelAdditionalMathematics

Rationalising the denominator of a fraction

CLASSDISCUSSION

%CHALLENGEQ 10Findtheexactvalueofy2+V3-^2-Vs. (Hint:letx=V2+VS-V2-VSandthensquarebothsidesoftheequation.)

Chapters:IndicesandSurds Note: To find products such as(4+ 2V^4- 2'\/3), it is quicker to use the algebraicidentity: 2 2{x+y){x-y)=x -y So(4+2^X4-2yf3)=4^-(2^13f=16-12=4 WORKEDEXAMPLE11 Rationalisethedenominatorsof , 2 . 5 . V7+3V2 V5 Answers 2+Vs V?-V2 ■ multiplynumeratoranddenominatorby^ Vs2V5V5 22V5+VF ^ 5(2-VF) (2+VF)(2-VF) =10-5^/F =10-5VF V7+3V2 V7-V2 =(^^+3V2)(V7+V2) (V7-V2)(^/7+V2) =7+VI4+3V14+6 (V7)^-(V2)^ _I3+4VI4 multiplynumeratoranddenominatorby(2->/F) use{jf+ t)=-r^—/toexjrandthedenominator multiplynumeratoranddenominatorby(^^7+^/2) use(a'+ 1;)=x^—yttoexpandthedenominator Exercise3.5 1Rationalisethedenominators. m 1 h 3 9 V2 <5 □ V2 C V3 a a/6 4 V5 f 12 V3 g 4 VT2 h 10 V8 3 j V2 If V3 I a/T2 <8 V32 <15 a/156 5 7 1+a/5 a/3-1 2V2 2V3 0 <5 P a/3 3-V2 r 14-V7 <7

Writeasasinglefraction. a — +"7=^^— a/3+1 a/3-1 :+ A/7+V2 a/7-a/2 :+ 1 4-a/3 4+a/3 5 The area ofarectangle is(8+a/To)cm^. The length ofone side is(a/5 + a/2)cm. Find thelength ofthe otherside in theform oa/K+6a/2,where aand bare integers.

7 A right circular cylinder has a volume of(2B+ 14^/3)7^cm^and a base radius of(2+a/s)cm. Find its heightin theform{a+ b^)cm,where aand bareintegers. !+&a/2

8aFindthevalueoftanx. Writeyouranswerintheform wherea,bandcareintegers. bFindtheareaofthetriangle. ^ P+^a/2 Writeyouranswerintheform if—, wherep,9andrareintegers.

CambridgeIGCSEand0LevelAdditionalMathematics 2Rationalisethedenominatorsandsimplify. 1 b 1 1+a/2 3+a/b B f a/7 2+VB 2-a/7 1 j 8 2V3-A/2 a/7-a/B 1 g 3+>/7 2 2-^/3 d h 45 2V3-3 3 Rationalisethedenominatorsandsimplify. 2-V3 .I+V2 .V2+I 2+V3^/5+l3-VS 3-^/2Vl7-^^lT Viv+VTT g 2V2-1 ^/3+VV V7-V2 V7+a/2 ^/^-a/23a/^+a/^ 4

9Findthevalueofcos6. Writeyouranswerintheform wherea,bandcareintegers. d+ b<l 1+2a/2 (2+a/7)cm Bern

6 A cuboid has a square base oflength(2+ a/b)cm. Thevolumeofthecuboidis(I6+7a/5)cm^. Findtheheightofthecuboid. Expressyouranswerintheforma+ ,whereaandbareintegers.

Chapters:IndicesandSurds CHALLENGEQ 10Thebluecirclehasradius2andthegreen circlehasradius1.A6isacommontangent andallthreecirclestoucheachother. Findtheradiusofthesmallercircle. 141' > A 3 B 3.6Solvingequationsinvolvingsurds WORKED EXAMPLE 12JHH Solve>/5X=^/^X+Vy. Answers <Ex=V2x+V7 collecta'sononeside II X X factorise dividebothsides -V2) V7 V5-V2 multiplynumeratoranddenominatorby(V5+ V7(V5+V2) 2 2 use(.*+)')(a—y)=.v—_vtoe.xpandthedenominator(V5-V2)(V5+V2) V35+Vl4 V^+Vl4 5-2 Vl4 ^ 3

CambridgeIGCSEarid0LevelAdditionalMathematics WORKEDEXAMPLE13 Solvethesimultaneousequations. 3x+by=20 V2X-5V2}'=8 Answers 3x+531=20 3V2X+5V23)V2X-5V2y=8=20V2V2X-b^y-8 4V2x=20V2+820V2+8 X = 4V2 (20V2+8)xV2 404V2xvf2+8^/2 8 a: multiplythefirstequationbyV2 addthetwoequationstoeliminatey dividebothsideshy4^ multiplynumeratoranddenominatorby\/2 X=5+V2 Substitutingx=5+V2inthefirstequationgives 3(5+ v/2)+5y =20 I5+3V2+5y=20 5y=5-3V2 5-3V2 y=Sothesolutionisx=5+s/2, ■3V2 CLASSDISCUSSION Haroonwrites: X ^2 X +2 X^ +¥x X^ + ix " 12 ^ 0 (X + f/x; "2)^0 f&x="S X=J Discussthiswithyourclassmates. Explainwhyhenowhastwovaluesforx. In the class discussion you found that ifyou square both sides ofan equation and then solve the resulting equation, you sometimes find that you have an answer that is not a valid solution of the original equation. Hence, it is always important to check your answers by substituting the answers back into the original equation.

CheckX=4intheoriginalequation: V3X!+4-V2X +1=Vl6-^/9=l,sox=0isalsoasolutionoftheoriginal equation. Sothefinalanswerisx=0orx=4.

Chapters:IndicesandSurds

WORKEDSothefinalanswerisx=4.EXAMPLE15

CheckX=4intheoriginalequation: VT=2and2x1-6=2,sox=4isasolutionoftheoriginalequation.

SolveV3x+4-V2x+1=1 Answers V3x+4-V2x+1=1 V3x+4=1+V2x+1 3x+4 =(l+V2x+l)(l+V2x+l) 3x+4=1+2V2x+1+2x+1x+2=2V2x+1 (x+2)^=4(2x+l) x^+4x+4=8x+4x^-4x=0 x(x-4)=0 X=0orX=4 fetsiaieamfyftternfimmmMs expwdm-immhrnkskfesftetemtes

CheckX=0intheoriginalequation: V3xO+4-V2xO+1=V4—Vr=l,sox=Oisasolutionoftheoriginalequation.

Solveyfx=2x—6. Answers Vx=2x-6 X=(2x-6)(2x-6) 4x^-25x+36X=4x^—24x+36=0 (4x-9)(x-4)=0 X =^or X =4 4 QCheck X =^in the original equation: squarebothskteexjMii-dthehracte-S c'olktttliketerms faetortse =Iand 2 X•-6=-1,so X =Iis not a valid solution ofthe original equation.

WORKEDEXAMPLE14

iite mpmddM'toeteisi eoitegiJfe* tmmM'

V36+6xc V2x+ 3 — X=0 f Vx+4 +2= X i y[x = 2x- 6

6 The roots ofthe equation x^ - 2^6 x+ 5 = 0 are pand q, where p> q. Write ^intheform wherea, bandcareintegers.

7 Find the positive root ofthe equation (4 - ^/2) x^ - (l + 2V2) x- 1 = 0. Write your answer in the form where a, b and c are integers.

Exercise3.6 1 Solvetheseequations, a VT2x-•n/5x=V3 b ^/ToX=V5x+V2

3Solvetheseequations, a10^/x—4=7Vx+6 d V2x-1-3=0 g 3+V5x+6=12 4 Solvetheseequations, a V2x-1=X dVlO-2x+X=1 g Vx+5+1=X j v/3x+l-X-1=0 5 Solvetheseequations, a V2x+7=Vx+3+1 cV3x+4-V2x+1=1 e ■Vx+ 1 + V2x+3=5 b x + y = b V6x+2)! = 12 e x+V2)i = 5 + 4^/2 x+^=8 b 6Vx + 4 = 9Vx + 3 e V2x-3 =6 h V5x^ — 8 = 2x ViyX = 2^/3 x+ Vh c3x+))=6 4x+ 3}i = 8 — SVB c 7Vx + 9 = 10Vx+ 10 f V5x- 1 = Vx+7 i 12- Vx + 5 = 7 b ■VX+6=X e Vx+ 15 = x+ 3 h v/4x-3 +2x- 1 =0 k 2x+3- V20x+9 =0 b Vx = Vx- 5 + 1 d "Vx+ 1 - Vx= 1 f Vl6-2x=2+

CambridgeIGCSEand0LevelAdditionalMathematics

2 Solvethesesimultaneousequations. a 3x-y =5^2 2x+y=5 d 2x+y=11 5x-3)i=ll>/7

Chapters:IndicesandSurds Summary Rulesofindices RULE1 m. RULE2 RULESRULE4RULE5RULE6 RULE7 RULESRULE9 , n m + n a X a = a m n Im-n) a ^a =a or a"xb^={ab)'' n ri a_ia\" b"" U/ 0 1 a=\ _ 1 ^ ~ n 1 a"=^ an=mr= = a Rulesofsurds RULE1: W^=VaxVi RULE 2: = ^ RULES: >/ax^/a=a Theproductofconjugatesurdsa+ and a-iVcisarationalnumber. mM Examination questions Worked example 2+V3 Withoutusingacalculator,findthevalueofcos6,givingyouranswer rs. CambridgeIGCSEAdditionalMathematics0606Paper21Q3Nov2011 intheform ,wherea,bandcareintegers.

CambridgeIGCSEand0LevelAdditionalMathematics

squarebracketsinthenumerator (2 + n/3)^ = 4 + 2V3 + 2>/3 + 3 = 7 + 4V3)

Answers

collectliketermsinthenumerator multiply numerator and denominatoi'by(2 - VS) (2 + 4>/3)(2- V3) = 4-2V3 + 8\f3-12 = -8 + 6V3 (2+>i3)(2-Nl3)=2'-(V3)2=l dividenumeratoranddenominatorby2 1 Solve 2'^'"''=^. 64 2Solvethesimultaneousequations, 4* ^ = 1024 256> [4]

Cambridge IGCSEAdditional Mathematics 0606Paper 21 Q9 Nov 2014 5 Withoutusing acalculator, express 0(l + "n/s)"^ in the form a+ 6^3, where aand bare integers tobefound. [4]

CambridgeIGCSEAdditionalMathematics0606Paper21QllaJun2014 32^ X 9^ = 243 [5] 3Solvetheequation 362y-5^ 02^-1 03, 210^^®' Cambridge IGCSEAdditional Mathematics 0606 Paper 21 Q5 Nov 2013 [4]

ExamExercise3.7exercise

CambridgeIGCSEAdditionalMathematics0606Paper21Q5bJun2012 4 Integers a and bare such that [a+ + a- ftVs = 51. Find the possible values of aand the corresponding values of b. [0]

CambridgeIGCSEAdditionalMathematics 0606Paper21 Q2Jun 2014

Rearrangetomakecos6thesubject: (2+^/3)^+2^-3^cos6= cos6= cos6= 2X(2+V3)X2 7+4^3+4-9 4(2+ 2+4^/3 4(2+^/3) ._(2+4V3)(2-VS) 4(2+V3)(2-V^ ■8+6^/3 cos6=cos6 4 -4+3V3

Usingthecosinerule: 3^=(2+V3)^+2^-2X(2+^/3)X2Xcos0

a i Show that 3V5-2^/2 isasquare rootof 53-12Vl^. [1] [4][1]

10Donotuseacalculatorinthisquestion. 1+3V3 6+2V3 XI 5-V3

CambridgeIGCSEAdditionalMathematics0606Paper11Q7a,hi,iiNov2012

Calculatorsmustnotbeusedinthisquestion. B V5-2 V5-1 The diagram shows a triangle ABCin which angle A=90°. SidesABand ACare V5-2and V5+1 respectively. Find tan Bin theform a+&V5,where aand bare integers. [3]

9 Given that ^ =p"qC,findthevaluesofa,b,andc. [3] p^q^r

ii State the other square root of 53- 12^10. b Express in theform a+Z>>/6,where aand bare integerstobefound.

The diagram shows two parallelograms that are similar. The base and height,in centimetres,of each parallelogram isshown. Given that x,the heightofthesmaller parallelogram, icP+ IS [5]^findthevalueofeachoftheintegerspandq. CambridgeIGCSEAdditionalMathematics0606Paper22Q6Mar2016

CambridgeIGCSEAdditionalMathematics0606Paper11Q7iJun2013

CambridgeICCSEAdditionalMathematics0606Paper21Q2Nov2013

CambridgeIGCSEAdditionalMathematics0606Paper12Q2Mar2016

Donotuseacalculatorinanypartofthisquestion.

Donotuseacalculatorinthisquestion. (4V5_2)'^. rExpress— intheformpMb+q,wherej&and^areintegers. [4]

Chapter 4 Factors and polynomials Thissectionwillshowyouhowto: ■ usetheremainderandfactortheorems ■ Mfindfactorsofpolynomialssolvecubicequations.

WORKED EXAMPLE 1 IfP(x)= -5 andQ(x)=x®+2x-1,findanexpressionfor aP(x)+Q(x), bP(x)-Q(x), c2Q(x), dP(x)Q(x). Answers a P(x) + Q(x) = 2x® - 6x^ - 5 + x^ + 2x -1 collect like terms = 3x® — 6x^ + 2x — 6 b P(x) - Q(x) = (2x' - 6x2 _ 5) _ + 2x - l) remove brackets = 2x® - 6x2 _ 5 _ ^3 _ 2x + 1 collect like terms = X® - 6x2 - 2x - 4 c 2Q(x) = 2(x3 + 2x-1) =2x^+4x-2 d P(x)Q(x) = (2x® — 6x2 _ s){x^ + 2x — l) = 2x'^ (x® + 2x — l) — 6x2 ^ 2x — l) - 5(x^ + 2x - l) = 2x® + 4x^ - 2x' — 6x^ — 12x® + 6x2 _ 5^3 _ ^ g =2x®-6x®+4x^-19x®+6x2_jq^^5

Degree Name ax+b, a^0 1 linear ax^+bx+c, a^0 2 quadratic ax^-1-bx'^+CX+d, a^0 3 cubic ax^+bx^+cx^ ■¥dx+e, a^d 4 quartic

The next example is a recap on how to add, subtract and multiply polynomials.

Chapter4:Factorsandpolynomials 4.1Adding,subtractingandmultiplyingpolynomials

Thehighestpowerofxinthepolynomialiscalledthedegreeof thepolynomial. Youalreadyknowthespecialnamesforpolynomialsofdegree1,2and3. These are shown in the table below together with the special name for a polynomialofdegree4.Polynomialexpression

•thecoefficients areconstants • is called the leading coefficientand 0 • Uq iscalledtheconstantterm.

•Xisavariable • nisanon-negativeinteger

Apolynomialisaanexpressionoftheform +... +Cq where:

Exercise4.1

CLASSDISCUSSION

3Simplifyeachofthefollowing,a(2x-3)(x+2)+(x+l)(x-1) b (3x + l)(x^ + 5x + 2)-(x^ -4x + 2)(x + 3) c (2x3 + ^ + 3^ _ 4)_ + 2)(x3 - x2 + 5x + 2)

4Iff(x)=2x^-X-4andg(x)=x^+5x+2,findanexpressionfor af(x)+»g(x) b [f(x)]^ c f^(x) d gf(x). b (x3+2x'^-l)(3x +2) d (x+2)'^(3x3^^ f {Sx-lf

CambridgeIGCSEand0LevelAdditionalMathematics

P(x) is a polynomial ofdegree p and Q(x) is a polynomial ofdegree q, where p> q. Discuss with your classmates what the degree ofeach of the following polynomials is; P(^)+Q(x) 2P(x) Q(x)+5 -3Q(x) P2{x) [Q(^)]' P(x)Q(x) QPW Q(x)-P(x)

1 If P(x)= 3x'^ + -1 and Q(x)= 2x^ + +1,find an expressionfor a P(x)+Q(x) b 3P(x)+Q(x) c P(x)-2Q(x) d P(x)Q(x). 2 Find the following products, a (2x-l)(4x^+X+2) c (3x^ + 2x - 5)(x^ + + 4) e(x^-5x+2)^

Step3:Repeattheprocess x^-3x+2 X-2)x^-5x^+8x-4 x3-2x2-3x2+8^—3x2 2x-4 2x-4 divide2xbyx;2x-rx=2multiply(.X—2)by2,2(x—2)=2x—4 subtract,(2x—4)—(2x—4)=0 So(x®-5x2gjc-4)-5-(x 2)=x2-3x+2.

Chapter4:Factorsandpolynomials 4.2 Division of polynomials To be able to divide a polynomial by another polynomial you first need to rememberhowtodolongdivisionwithnumbers. The stepsfor calculating 5508 17 are: 324 17)5508 Divide55by17 51 i 3 X 17-51 40 55-51=4,bringdownthe0fromthenextcolumn 3_4 Divide40by17,2x17=34 68 40-34=6,bringdownthe8fromthenextcolumn 68 Divide68by17,4x17=68 0 68-68=0 So 5508-17=324 dividend divisor quotient Thesameprocesscanbeappliedtothedivisionofpolynomials. WORKED EXAMPLE 2 1 Dividex'-5x^+8x-4by x-2. Answers Step1: X? X-2)-5x^+8x-4 x^-2x^ i —Sx?+8x dividethefirsttermofthepolynomialhjx,x^^x multiply(x-2)byx2,x^{x-2)=x®-2x2 subtract,(x®—5x2)_^^3_^x'^)=-3x2 bringdownthe8xfromthenextcolumn Step2:Repeattheprocess x^-3x X-2) -5x^+8x-4 x^-2x^ -3x^+8x —3x^+6x ,, 2x-4 divide-3x2 ^—3^2^x=—3x multiply(x—2)by—3x,—3x(x—2)=—3x2^ subtract,(-3x2+ _(—3x2+ _2x bringdownthe-4fromthenextcolumn

Step2;Repeattheprocess 2x2_0^ x+3)2x3+0x2- x+51 2x3+6x2 -6x2_ ^ divide-6x2 —6x2 ^=—6x —6x2-18x' multiply(x+3)by-6x,-6x(x+3)=-6x2 I7x+51 subtract,(-6x2-x)-(-6x2_jg^^=17x bringdownthe51fromthenextcolumn

Step3:Repeattheprocess x2- 6x+17 X+3)2x3+0^2_ 52 2x3^ H DL —6x2_Jgjj. 17x+51 divideI7xby.x,I7x^x=17 17x+51 multiply(x+3)by17,17(x+3)=17x+51 0 subtract,(17x+51)-(I7x+51)=0

Divide2x''-x+51byx+3. Therearenox2termsin 2x^ -x+51sowewriteitas2x3 q^2_+5J Answers Step1: 2x2 x +3)2x3+0x2- 52 dividethefirsttermofthepolynomialbyx,2x3 ^=2x2 2r+6x2 1 multiply(x+3)by2x2,2x2(x+3)=2x3 g^2 -6x2_^ subtract,(2x3^ ^5^2^=-6x2 bringdownthe-xfromthenextcolumn

CambridgeIGCSEand0LevelAdditionalMathematics J

WORKEDEXAMPLE3r

fdb (x^ — x2-3x+2)-5-(x — 2) (x3(x3-3x-2)-J-(x-2)+2x2_9^_ig)^(;c+2) b d (6x3+11x2-3x-2)-5-(3x+1)(3x3_21^2+4^_28)-(x-7)

So(2x3_j(.+5j^^ 3)_2x^-6x+17. Exercise4.2 Simplifyeachofthefollowing, a (x'"® +3x^-46x-48)^(x+1) c (x^-20x2 +loOx-125)-(x-5) e (x'^ — 3x2 _ 32^ 33^ _ *7^ Simplifyeachofthefollowing. a (3x^ +8x2^3^ _ 2) (x +2) c (3x3-11x2+20)-(x-2)

Therefore2x-Iisafactorof4x3-2x2+gjj,_4

InWorkedexample2youfoundthatx-2dividedexactlyinto (x3-5x2 _4j (x3-5x2+g^ _ 4^^ _ 2)= x2-3x+2

4.3Thefactortheorem

Chapter4:Factorsandpolynomials 3 Simplify. -Sx^—4x+4 a X-1 5x^—50%+8 '3x2+i2x-2 4aDividex^—1by(x+1). ^ 2x3+9^2+25 b X+5 ^ x3 -14x -15 x2—3x—5 bDividex^-8by(x-2).

Thiscanalsobewrittenas: (x3-5x2^gj^ _ 4^ _ _ 2)(x2-3x+2)

IfapolynomialP(x)isdividedexactlybyalinearfactor x-ctogivethe polynomialQ(x),then P(x)=(x-c)Q(x).

Substituting x = cinto thisformula gives P(c)= 0. Hence: IfforapolynomialP(x),P(c)=0then x-cisafactorofP(x).

Thefactortheoremcanbeextendedto: IfforapolynomialP(x), =0 then ax-bisafactorofP(x).

Forexample,when x =2, 4x3 -8x2 -X + 2 = 4(2)3 _ g(2)2 -2+2- 32-32- 2+2= 0. Thereforex-2isafactorof4x3_ ^^2.

Thisisknownasthefactortheorem.

Forexample,whenx=—, 4x3-2x2+8x-4 = 4|^ij

CLASSDISCUSSION

CambridgeIGCSEand0LevelAdditionalMathematics

Discusswith your classmates which ofthefollowing expressions are exactlydivisibleby*-2. -X-2 2x^+5x^-4x-3 X®-4x^+8x-8 2x^- +Sx^-2x-5 x^-8 3x^-8x-8 dx^-lOx^-18 -4x-4 x^+X+10 WORKEDEXAMPLE4 Showthat X-3isafactorof —6x^+llx—6 by aalgebraicdivision b thefactortheorem. Answers aDivide -6x^+llx ■6 byX-3. — 3x+ 2 X - 3)x^ - 6x^ + llx- 6 X? — 3x^ -3x2 + 11^ -3x2 ^ 2x- 6 2x- 6 0 The remainder = 0, so x - 3 is a factor of x^ - 6x2 + - 6 b Letf(x)=x^—6x2+ _5 iff(3)=0,thenx—3isafactor. f(3) = (3)®-6(3^ + 11(3)-6 =27-54+33-6 = 0 SoX-3isafactorof x^-6x2 +llx-6.

2Findthevalueofaineachofthefollowing, a X+1isafactorof6x^+27x^+ax+8. bX+7isafactorofx^-5x^-6x+a. c 2x+3isafactorof4x^+ax^+29x+30.

3 X-2isafactorofx^+ax^+bx-4. Expressbintermsofa.

aUsethefactortheoremtoshow:AC-4isafactorof-6x+8

d 3ac+1isafactorof +1 -3x-2.

Chapter4:Factorsandpolynomials

bAC+1isafactorof -Sx-2 c x-2isafactorof5ac^-17ac^+28

4 Find the value ofaand the value ofbin each ofthefollowing, a x^+3x-10 isafactorofx^+ax^+6x+30. b 2x^-1Ix+5 isafactorof ax^-17x^+bx-15. c 4x^-4x-15 isafactorof4x^+ax^+6x+30.

2x^+x—\isafactorof2x^-x^+ax+b. Findthevalueofaandthevalueofb.

"(1) -(2) Exercise4.3 1

Usingthefactortheoremf =0andf(-l)=0. f(|)=0 gives 2(i) =0 1 1 «.ev +-+i=0 4 4 2 a=-2b f(-l)=0gives2(-if-(-if+a(-l)+b=0 -2-l-a+b=0 a=b-3 (2)=(1)givesb-3=-2b 5b=3b=l Substitutingin(9)givesa=-2. Soa=-2,6=1.

WORKEDEXAMPLE5

Answers

Letf(x:)= —x^+ax+b. If2x^+jc-1=(2*-l)(x:+1)isafactoroff(x),then2x-1andAS+1arealsofactors off(*).

Factorise —3x^—13x+15completely. Answers Letf(x)= —3x^—13x+15.

f(l)=(if-3X(if-13X(1)+15=0 SoX-1isafactoroff(x).

bExplainwhyx+3isalsoafactoroftheexpression.

Findthepossiblevalueofa.

Method1(bytrialanderror) f(x)=x^-3x^-13x+15

-5x+6andx^-6*^+llx+ahaveacommonfactor.

CambridgeIGCSE and0LevelAdditional Mathematics

5 Itisgiventhat

8X+aisafactorofx^+8x^+Aax-3a. aShowthat -4a^+3a=0. b Findthepossiblevaluesofa.

WORKEDEXAMPLE6

Thepositiveandnegativefactorsof15are ±1,+3,±5and±15.

Hence ifyou know one factor ofa cubic expression it is possible to then factorisetheexpressioncompletely.Thenextexampleillustratesthree differentmethodsfordoingthis.

f(-3)=(-3)^-3X(-3)^-13X(-3)+15=0 SoX+3isafactoroff(x).

f(5)=(5)^-3X(5)^-13X(5)+15=0 SoX-5isafactoroff(x).

Hencef(x)=(x-l)(x-5)(x+3)

Considerfactorisingx^-5x^+8x-4completely.

InWorked example 2youfound that (x^-5x^ +8x-4)-s-(x-2)= x^-3x+2. Thiscan be rev/ritten as: x^ — 5x^ +8x-4=(x-2)(x^ — 3x +2).

6 X-2 is acommon factor of 3x^-{a-b)x-8 and -(a+ b)x +30.

Factorisingcompletelygives: x^-5x^+8x-4=(x-2)(x-2)(x-1).

f(1)=(1)® -3X(1)'-13X(1)+15=0 SoX-1isafactoroff(x).

Findthevalueofaandthevalueofb.

73c-3and2jc-1arefactorsof2*^-px^-2qx+q. a Findthevalueofpandthevalueofq.

Theotherfactorscanbefoundbyanyofthefollowingmethods.

4.4 Cubic expressions and equations

2x3_3^2_igjj_g=2x3+ +4)jj2+(2fc_4)x_8

Equating coefficientsofx^: 6-1=-3 b=-2 f(x)=(x-l)(x^-2x-15) =(x-l)(x-5)(x+3)

Solve2x'-3x2-I8x-8=0. Answers

2x^-3x2_jgj|._g_(jj+2)(2x2+ _4^ expandandcollectliketerms

WORKEDEXAMPLE7

SoX+2isafactoroff(x).

Chapter4: Factorsand polynomials

f(x)=(x-l)(x^-2x-15) =(x-1)(x—5){x+3)

Method2(bylongdivision) )?—2x-15 X -l)x^-3x2- 13^+15 —2}P— ISx -2}P+2x —15x+15—15x+15 0

Letf(x)=2x®-3x2_ _g

Thepositiveandnegativefactorsof8are±1,±2,±4and±8. f(-2)=2(-2)®-3X(-2)^-18X(-2)-8=0

Method3(byequatingcoefficients)Sincex—1isafactor,x^—3x^—13x+15canbewrittenas: x^ — 5x^-13x +15=(x-l)(ax^ +bx+c) coefficientof is1,soa since1x1=1 constanttermis-15,soc=-15 since-1X-15=15 x^-5x^-13x+15=(x-l)(x^+bx-15) expandandcollectliketerms x^-3x^-13x+15=x^+(6-l)x^+{-b-15)x+15

2x^-3x2 _ jgjj _ g _ +2)(ax2+6x+c) coefficientof is2,so«=2 since1x2=2 constanttermis-8,soc=-4 since2X-4=-8

f(-l)=2(-l)®+7X(-1)2-2X(-1)-1 0

TheLetf(x)=2x^+7x2_2x-1.positiveandnegativefactorsof-1are+1.

Byinspection,^ So2x-1isafactorof: 2x®+7x2-2x-1=(2x-l)(ax2+6x+c) 1=0. coefficientof is2,so«=1 constanttermis-1,soc=1 since2x1=2 since-1X 1=-1 2x'+7x2_2x-1=(2x-l)(x2+6x+l)

Equating2x®+7x2-2x-1=2x'+{2b-l)x2+(2-6)x-1coefficientsofx^:2b-1=7 b=4

Equatingcoefficientsof 6+4= b f(x)=(x+2)(2x2_7;,;_4) =(x+2)(2x+l)(x-4) Hence(x+2)(2x+l)(x-4)=0. 1 SoX=-2orX=-^orX=4. 2' -3 -7

Solve2x^+7x2-2x-1=0. Answers

f(1)=2(1)^+7X(1)2-2X(1)-1 0

WORKEDEXAMPLE8

SoX-1andx+1arenotfactorsoff(x).

So2x^+7x2_2x-1=(2x-l)(x2+4x+1). 1 -4±V42-4X1X1 X = — or X 2 2x1 1 -4±273 X = — or X 2 2 x =^orx = -2+ >/3 orx = -2->/3

CambridgeIGCSEand0LevelAdditionalMathematics

Exercise4.4 1 a Showthatx-1 isafactorof2x®-x^-2x+l. b Hence factorise 2x® - x^ - 2x + 1 completely.

b d f h x3 - 6x2 + llx - 6 = 0 2x3 + 3;c2 -17x + 12 = 0 2x3 ^ — 5x - 4 = 0 2x3 _ 3^2 _ 29x + 60 = 0 ca +5x2 +2x2 4x 7x 2 = 0 2 = 0 x-' + 8x2 ^ y2x - 9bx3 d 2x3 ^ 3^2 0 17x+12=0 Solve the equation 2x3 ^ g^2 - 14^ - 9 = 0. Express roots in the form a ± ftVc , where necessary.

Equatingcoefficientsofx^:ft-3=1 ft = 4 So x^ + x^ — 36 = (x — 3)(x^ + 4x + 12)

SoX-3isafactoroff(x). x^+x^-36=(x-3)(ax^+bx ■¥ c) coefficient of is 1, so ct = 1 since 1x1 = 1 constant term is —36, so - 12 since-3X12=-36 x^ + x^ - 36 = (x — 3)(x^ + bx + 12) x^ + x^-36 = x^+{b- 3)x2 + (12 - 3b)x - 36

(Note: x^ + 4x + 12 cannot be factorised into two further linear factors, since the discriminant<0.)

2 Factorise these cubic expressions completely. a x^+2x^-3x-10 b c 2x-'' -9x^ -18x d e 2x^-13x2+I7x+12 f e 4x® -8x2 _^ 2 h x^ + 4x2 x^ — 8x2 j4 3x® +2x2 _ Y^x+6 2x^ +3x2 _32^+ 15 3Solvethefollowingequations. aX®—3x2_33^+35=0 c 3x3+17x2+18X-8 = 0 e 2x3-3x2-llx + 6 = 0 g 4x3 + 12x2 + 5x - 6 = 0 Solvethefollowingequations. Express roots in the form a ± b-Jc , where necessary.

Notallcubicexpressionscan befactorisedinto3linearfactors.

Considerthecubicexpressionx®+ -36. Let f(x)= x^ +x^ -36. f(3)=(3)'+(3)'-36=0

Chapter4: Factorsand polynomials

8 f(x) is a cubic polynomial where the coefficient of is 1. Findf(x)whentherootsoff(x)=0are a-2,1and5 b-5,-2and4 c-3,0and2.

10 f(x) is a cubic polynomial where the coefficient of x^ is 1. Therootsoff(x)=0are-3,1+^2and1-V2.

Substituting x =3in the polynomial gives f(3)= 2(3)'^ -4(3)^ + 7(3)-37 = 2. When 2x^ -4x^ + 7x-37 is divided by x — 3,there is a remainder. 2x^+ 2x+13 X-3)2x^-4x2+ 7^_37 2x^-6x^2x^+7x2x^-fix 13x-3913x-37 Theremainderis2.Thisisthesamevalueasf(3). f(x)=2x''—4x^+7x—3fi,canbewrittenas f(x)=(x-3)(2x2+2x+13)+2.

bHenceshowthatx^-x^-x-2=0hasonlyonerealrootandstate Solvetheequation +8x^+12x=9. Writeyouranswerscorrectto2decimalplaceswherenecessary, a Showthat X-2isafactorof x^-x^-x-2. X thevalueofthisroot.

9 f(x)isacubicpolynomialwherethecoefficientofx''is2.Findf(x)whentherootsoff(x)=0are a—0.5,2and4 b0.5,1and2 c—1.5,1and5.

Cambridge IGCSE and0Level Additional Mathematics

Express f(x) as a cubic polynomial in x with integer coefficients. 12 2x +3 is afactor of 2x^ +[a^ +l)x^-3x^ +(l-a^)x +3. a Showthat4a^-9<2^+4=0. b Findthepossiblevaluesofa. 4.5 The remainder theorem Considerf(x)=2x^-4x^+7x-37.

Express f(x) as a cubic polynomial in x with integer coefficients.

11f(x)isacubicpolynomialwherethecoefficientofx'^is2. The roots off(x)=0 are ^,2+ y/S and 2-^/3.

Ingeneral: IfapolynomialP(x)isdividedbyx-ctogivethepolynomial Q(x)andaremainderR,thenP(x)=(x-c)Q(x)+R.

=-189+54+120+17 =2

7x^+15x+5

■

WORKEDEXAMPLE9

Chapter4:Factorsandpolynomials

Findtheremainderwhen7x'+6x^-40x+17isdividedby(x+3)byusing a algebraicdivision b thefactortheorem. Answers a Divide7x^+6x^-40x+17by(x+3). x+3j7r'+6x2-40x+17 Theremainderis2. b Letf(x)=7x^+6x^-40x+17. Remainder-f(-3) =7(-3f+6(-3)^-40(-3)+17

IfapolynomialP(x)isdividedbyx-c,theremainderisP(c).

7x^+21x^ -15x^—15x^—40x-45x 5x+17 5x+152

Substituting x = c into thisformula gives P(c)= R. Thisleadstotheremaindertheorem:

TheRemainderTheoremcanbeextendedto: IfapolynomialP(x)isdividedbyax-b,theremainderis

2(lf+fl(lf-9(1)+6=1 2+a—9+6=1 a+6=8 (1)

Whenf(x)=2*®+ax^-9x+bf{x)isdividedbyx-1,theremainderis1.

Findthevalueofb. c When2x®+x^+cx-10 isdividedby2x-1,theremainderis-4.

Whenf(jc)isdividedbyx+2,theremainderis19meansthat:f(-2)=19.

WORKEDEXAMPLE10

a=3and6=5 Exercise4.5 1Findtheremainderwhen +2x^-X+3 isdividedby x-1 x^-6x^+llx-7 isdividedby x-2 x^-3x^-33x+30 isdividedby x+2 2x^-x^-18x+11 isdividedby2x-1. a Whenx^+x^+ax-2 isdividedby x-1,theremainderis5.

f(x)hasafactorofx-1andleavesaremainderof3whendividedbyx+2. Findthevalueofaandofb. f(x)=x^+ax^+llx+6

f(x)hasafactorofx-2andleavesaremainderof24whendividedbyx-5. Findthevalueofaandofb.

2{-2f+a{-2f-9(-2)+6=19 -16+4a+18+6=19 4a+6=l7 (2) (2)-(1)gives3a=9 a=3

Findthevalueofc. f(x)=X®+ax^+6x-5

Substitudnga=3inequation(2)gives6=5.

Findthevalueofaandofb. Answers

Whenf(x)=2x'+ax^—%x+bf(x)isdividedbyx-1,theremainderis1meansthat:f(l)=1.

Whenf(x)isdividedbyx+2,theremainderis19.

CambridgeIGCSEand0LevelAdditionalMathematics

Findthevalueofa. b When2x^-6x^+7x+6 isdividedby x+2,theremainderis3.

b Solve the equation in parta completely.

7 f(x)= 2x^ +(a+l)x^ -ax +ft

When f(x) is divided by x-1,the remainder is 6.

Chapter4:Factorsandpolynomials

9f(x)=x^-5x^+ax+ftf(x)hasafactorofx-2.

Therootsoff(x)=0are2,3,andk.

a Expressftintermsofa.

10f(x)=x^+ax^+ftx+c

b Findtheremainderwhenf(x)isdividedbyx+1.

Whenf(x)isdividedby x-1,theremainderis5. Whenf(x)isdividedby x-2,theremainderis14. Showthata=-4andfindthevalueofft. 8 f(x)=ax^+ftx^+5x-2

f(x)hasafactorof 2x-1 and leavesaremainderof21 when divided by x-2.

Findthevalueofaandofft

aFindthevalueofaandofft.

12f(x)=x^-8x^+ftx-20 When f(x) is divided by x — 1,the remainder is R.

Whenf(x)isdividedbyx-1,theremainderis—8.

b Findtheremainderwhenf(x)isdividedbyx-1.

aFindthevalueofk.

5 f(x)= +ax+b f(x)hasafactorofx-3andleavesaremainderof15whendividedby x+2. aFindthevalueofaandofb. b Solvetheequationf(x)=0. 6 f(x)=4x^+8x^+ax+b f(x)hasafactorof2x-1 andleavesaremainderof48whendividedby x-2. aFindthevalueofaandofb.

11f(x)=4x^+ax^+13x+ft

WhenWhenf(x)isdividedbyx+a,theremainderisR.f(x)isdividedbyx—a,theremainderis2R.

bWhenf(x)isdividedbyx+1,theremainderis—9.

Findthevalueofk.

b Find theremainderwhen the expression isdivided by x+1.

13f(x)^x^+2x^-6x+9

a Showthat3a^-2a^-18a-9=0.

Whenf(x)isdividedbyx-2,theremainderis4i2.

Whenf(x)isdividedby2x+1,theremainderis-6. Findthevalueofaandofft

Therootsoff(x)=0are1,k,andk+I. Whenf(x)isdividedby x-2,theremainderis20. aShowthat -3A-18=0. b HencefindthepossiblevaluesofA.

Thefactortheorem:

If,forapolynomialP(x),P(c)=0thenx-cisafactorofP(x).If,forapolynomialP(x),=0thenax-bisafactorofP(x).

CHALLENGEQ 16 f(x)= x^ + ax'^ + bx + c

CambridgeIGCSE and0LevelAdditional Mathematics 14f(x)= +Ax-15

Summary

WhenWhenf(x)isdividedbyx-1,theremainderisR.f(x)isdividedbyx+4,theremainderis-R. aFindthevalueofk. b Hencefindtheremainderwhentheexpressionisdividedbyx+2. 15 P(x)=5(x-l)(x-2)(x-3)+a(x-l)(x-2)+A(x-l)+c ItisgiventhatwhenP(x)isdividedbyeachofx-1,x-2andx-3theremaindersare7,2and1 respectively.Findthevaluesofa,b,andc.

Theremaindertheorem:

IfapolynomialP(x)isdividedbyx-c,theremainderisP(c).IfapolynomialP(x)isdividedbyax-b,theremainderis •

Workedexample aand bareconstants.Itisgiven that 2x-lisafactoroff(x)and thatwhenf(x)isdivided byx+2theremainderis20.Find the remainderwhenf(x)isdividedbyx-1. f(x)= 4x^ — 8x^ +ax + b If2x-1isafactor,thenf v2 1 -2+ a+2b=S (1) Remainder=20whendividedbyx+2,meansthatf(-2)=20. 4(-2)^ -8(-2f + a(-2)+ b = 20 -52-32-2a+b=20 -2a+b=84 (2) From(1)a=5-2b. Substituting in(2),gives: -2(3-2b)+ b = 84 —6+4b+6=845b=90b=18 So a=-33,b=18. Remainderwhenf(x)=4x®-8x^-33x+18isdividedby(x-1)isf(l). Remainder = 4(1)^ -8(1)^ -33(1)+18 ==4-8-33+18-19

Thefunctionfissuch thatf(x)=4x®-Sx^+ax+b, where

Chapter4:Factorsandpolynomials Examination questions

CambridgeIGCSEAdditionalMathematics0606Paper11Q2Nov2011 Answer

CambridgeIGCSEand0LevelAdditionalMathematics Exercise4.6 Exam Exercise

CambridgeIGCSEAdditionalMathematics0606Paper12Q7Mar2016

CambridgeIGCSEAdditionalMathematics0606Paper21Q2Nov2010

Cambrid^IGCSEAdditionalMathematics0606Paper21Q5i,iiJun2011

7 The polynomial f(x)= ax^+7x^ -9xA-b is divisible by 2x- 1. The remainder when f(x) is dividedbyx-2is5timestheremainderwhenf(x)isdividedbyx+1. i Showthata=6andfindthevalueofb. [4] ii Using the values from part a,show that f(x)=(2x-l)(cx^ A-dxA-e^, where c, d and e are integerstobefound, ill Hence factorise f(x) completely. [2] [2]

3aTheremainderwhentheexpressionx®+9x^+bxA-cisdividedbyx-2istwicethe remainderwhentheexpressionisdividedbyx-1.Showthatc=24. [5] bGiventhatx+8isafactorofx^^+9x-A-bxA-24,showthattheequationx^+9x^+Z'x+24=0hasonlyonerealroot. [4]

4Theexpressionx^+8x^A-px-25leavesaremainderof/?whendividedbyx-1anda remainderof-J?whendividedbyX+2. a Findthevalueofp. [4] b Hence find the remainder when the expression is divided by x +3. [2]

CambridgeIGCSEAdditionalMathematics0606Paper21Q12i,iiNov2012

CambridgeIGCSEAdditionalMathematics0606Paper21Q10i,iiNov2012

6Theexpressionx'^+ax'^-15x+bhasafactorofx-2aridleavesaremainder75whendivided byX+3.Findthevalueofaandofb. [5]

Thefunctionf(x)=ax^+4x^+bx-2,whereaandbareconstants,issuchthat2x-1isa factor. Given that the remainder when f(x) is divided by x-2is twice the remainder when f(x)isdividedbyx+1,findthevalueofaandofb. [6]

CambridgeIGCSEAdditionalMathematics0606Paper11Q6Nov2013

1aShowthatX-2isafactorof -14x^+32. [1] b HencefactorlseSx^—14x^+32completely. [4]

5Factorisecompletelytheexpression2x^-llx^-20x-7. [5]

CambridgeIGCSEAdditionalMathematics0606Paper11Q4Nov2010

m Equations,Inequalities and graphs Thissectionwiiishowyouhowto: ■ solvegraphicallyoralgebraicallyequationsofthetype|aa:+6|=\cx+d\ ■ solvegraphicallyoralgebraicallyinequalitiesofthetype\ax-^b\>c(c^O), |ax+Z»|^c(c>0)and|ax+6|=s|cx+d| ■ solve cubic inequalities in theform k{x-a){x-b){x—c)^d graphically ■ sketchthegraphsofcubicpolynomialsandtheirmoduli,whengiveninfactorisedform ■ usesubstitutiontoformandsolvequadraticequations.

p'-?'=(|p|-l?l)(l/'l+l?l)

• l/'hkl •\p\<\q\

Thenextworkedexampleshowsyouhowtosolveequationsoftheform \ax+b\-\cx+d\ using algebra. To solve this type ofequation you can use the techniques that you learntin Chapter 1 you use the rule:

and|=q^youcansaythat: p^-q^=\pf-\qf

Usingthedifferenceoftwosquaresthengives:

CambridgeIGCSEand0LevelAdditionalMathematics 5.1 Solving equations ofthe type \ ax+ b\= \cx+ d\

can

|p|=|<?| P^ =q^ WORKEDEXAMPLE1 Solvetheequation\x-b\=|x+1|usinganalgebraicmethod. Answers Method1 |x-5|=I*+1| X-5=X+l0=6 or or x-5=-(x+1) 2x=4x=2 CHECK:|2-5|=|2+1|/ MethodThesolutionisx=22 |x-5|=!*: ■+1| (x-5)^=(x+1)^ x^-lOx+25= +2x+1 12x=24x=2 0-6isfalse use \p\ ^1^1 expandsimplify

CLASSDISCUSSION

Usingthestatementabove,explainhowthesethreeimportantresultscanbeobtained: (ThesymbolOmeans'isequivalentto'.) <q^,ifq*0

Usingthefactthat\p\=

Chapter5:Equations,inequalitiesandgraphs Theequation|3c-5|=|3c+l|couldalsohavebeensolvedgraphically. _V=|.\-+ll j=|x-51 _8 -6 -4 -2 O 2 4 6 8 1;0 12^ Thesolutionisthex-coordinatewherethetwographsintersect. WORKEDEXAMPLE2 Solvetheequation12x+1|=|x-3|usinganalgebraicmethod. Answers Method1 |2x+l|=|x-3| 2x+1=X—3 or 2x+1=—(x—3) orx=-4 3x=22 X= 3 CHECK:|2(-4)+l|=|-4-SK, Solutionis X=— or x=-4 3 Method2 |2x+l|-|x-3| (2x+lf=(x-3)''^ 3x^4x^+4x+1=x^—6x+9+lOx-8=0 (3x-2)(x+4)=0 2 X=—orX=-4 3 2 2 2x-+1 --3 3 3 / use IpI =|<7| « expand simplify factorise

CambridgeIGCSEand0LevelAdditionalMathematics The equation 12x+1|=|x-3| could also be solved graphically. -6 -4 -2 O y=\2x+1| >'= W 1 ;/ ' J IXI / 4—J-A. \ f\ y 1XK / 1 x f' 1 ^1 i \ \ \ \4V" \ /! / ! i 1 ! / i Y 1 f 1 : V1 ■ V \/ 1 r \ 1 1 1 1 1 1 1 6 10 12 X Fromthegraph,oneoftheanswersisclearlyx=-4.Theexactvalueoftheotheranswerisnotsoobviousandalgebraisneededtofindthisexactvalue. The graph of y=|x-3| can be written as: y-x-3 if y=—(x-3) if x<3 Thegraphofy=12x+1|canbewrittenas: y=2x+1 if x^—— y=—{2x+1) if x<1 Thesecondanswerisfoundbyfindingthexvalueatthepointwhere y=~{x—5)andy=2x+1 intersect.2x+1=—X—32x+1=—x+33x=2 2 X = Hencethesolutionisx=-4orx=— 3

litiesandgraphs WORKEDEXAMPLE3 Solve|x+41+|x-5|=11. Answers |x+4|+|x-5|=11 subtractjx-5|frombothsides |x+4|=11—|x—5| splittheequationintotwoparts x+4=ll-|x-5| -(1) x+4=|x-5|-ll -(2) Usingequation(1)|x-5|=7-x .splitthise{[uaiionintotwo[)ar[s x-5=7-x or x-5=-(7--x) 2x=12 or 0=-2 0=-2isfalse x=6 Usingequation(2)1X-5|=X+15 splitthisequationintotwoparts x-5=x+i5 or x-5=-(x+15) o II M( OU o 1! O 0=20isfalse x=-5 CHECK:|6+4|+!6-5i=11«^,1-5+41+1-5-51=11/ Thesolutionisx=6orx=-5. Exercise5.1 1Solve. a |2x-l|=|x| d |5x+l|=|l-3x| g |3x-2|=|2x+5| b |x+5|=|x-4| e |l-4x|=|2-x| h |2x-l|=2|3-x| 2 Solvethesimultaneousequations |x-5|and))=18-x|. 3 Solve the equation 6|x+2|^+7|x+2|-3=0. 4a Solvetheequationx^-6|x|+8=0. b Usegraphingsoftwaretodrawthegraphoff(x)=x^-6|x|+8, C Use your graph in partb to find the range ofthe function f. CHALLENGEQ 5 SoI\ {■ the equation |x + l|+| 2x-3| = 8. c|2x-3|=|4-x| X 1-- :|3x+2| i |2-x| = 5 1 , —x+12

Thegraphsof31=[2x-1|and31=3intersectatthepointsAandB. |2x-ll= 2x-l if X -(2x-1) if X< 2 y=|2x-1 AtA,theline3;=-(2x-1)intersectstheline3=3. ! -(2x-l)=3 I -2x+l=3 2x=-2x=-l AtB,theline3=2x-1intersectstheline3=3. 2x-l=3 2x=4x=2 Tosolvetheinequality12x-1|<3youmustfindwherethegraphofthefunction 3=12x-1|isbelowthegraphof3=3. Hence,thesolutionis -1<x<2.

Thefollowingexamplesillustrate thedifferentmethodsthatcanbeusedwhensolvingmodulusinequalities.

Inequalities

CambridgeIGCSEand0LevelAdditionalMathematics

5.2 Solving modulus

Twousefulpropertiesthatcanbeusedwhensolvingmodulusinequalitiesare: -q^p^q and \p\^q p^-q or p^q I

CHALLENGEQ Solve the simultaneous equations y=|x —51 and 3;=13—2x|+2. CHALLENGEQ 7 Solve the equation 2[3x +4;;-21 +3^25-5x + 2:>^=0.

WORKEDEXAMPLE4

Solve12x-1|<3. Answers Method1(usingalgebra) 12x-l|<3 use|p <q <=> -q < p < q -3<2*-1<3 -2<2x<4-1<X<2 Method2(usingagraph)

Chapter5:Fquations,inequalitiesandgraphs WORKEDEXAMPLE5 Solve12x+3|>4. Answers Method1(usingalgebra) |2x+3|>4 use\p\>q p<-q or p> 2x+3<-4 2x<-77 x< 2 or 2*:+3>4 or 2x>1 1 or X > Method2(usingagraph) Thegraphsof31=12X+3|and31=4intersect atthepointsAandB. |2x+3|= 2x+3 if i—(2*+3)if X<-1 a y 4At4,the line 31=-(2x+3)intersects the line3=4. _p _] O -(2x+3)=4 -2x-3=4 2x=-7 7 X= 2 AtB,theline3=2x+3intersectstheline3=4. 2x+3=42x=11 X= 2 y=\2x+3| 3=4 /B X Tosolvetheinequality12x+3|>4youmustfindwherethegraphofthefunction3=12x+3|isabovethegraphof3=4. Hence,thesolutionisx<-—orx>—. 2 2

CambridgeIGCSEand0LevelAdditionalMathematics WORKEDEXAMPLE6 Solvetheinequality|2x+ll>\5-x\ Answers Method1(usingalgebra) |2x+l|>|3-x| (2x+l)^ 4:X^+4x+1>9-6x+ Sx^+lOx-8^0 (3x-2)(x+4)^0 IpI \q\ ^ factorise Criticalvaluesare — and-4. 3 y=Sx^ + lOx — 8 Hence,x<-4orx3=— 3 Method2(usingagraph) Thegraphsofy=12x+1|and y=13-x|intersectatthepointsAandB. 1. 2 1 2 X-3 if X3:3 -(x-3) if X<3 AtA,theliney=-(2x+1)intersectstheliney=-(x-3). 2x+1=X-3 X=-4 AtB,theliney=2x+1intersectstheliney=-(x-3). 2x+1=-(x-3) 3x=2 2 X= 3 2x+l if -(2x+1) if X< 3—x|=|x—3|=

Chapter5:Fquations,inequalitiesandgraphs I2x+ll -7-6-5-4-3-2-1O 1 23 56 Tosolvetheinequality|2^?+1|^13-x|youmustfindwherethegraphofthe functiony=\2x+1\isahovethegraphofj;=13—x|. 2 Hence,x^-4orx^ Exercise5.2 y=|x-2| 1 2 3 4^ 67 8 9 Ip^ Thegraphsofy=|x-2|and3)=12x-10|areshownonthegrid. Writedownthesetofvaluesofxthatsatisfytheinequality|x-2|>|2x-10|. 2a Onthesameaxessketchthegraphsofy=13x-6|andy=14-x|. b Solve the inequality 13x-6|^|4-x|. 3 Solve. a |2x-3|>5 d |2x-7|>3 4Solve. a |2x-3|^x-l 5Solve. a |2x-l|^|3x| d |4x+3|>|x| b |4-5x|^9 e |3x+ll>8 b |5+x|>7-2x b |x+l|>|x| e |x+3|3=|2x| c |8-3x|<2 f|5-2x|^7 c |x-2|-3x^l c |x|>|3x-2| f 12x|<IX—3|

WORKEDEXAMPLE7 a Sketchthegraphofthefunctiony=(2x-1)(2-x)(x+1). b

In thissection youwillleam howtosketch graphsoffunctionsoftheform y=k(x-a)(x-b)(x-c) and theirmoduli. intercepts. help find the generalshape ofthe curve you need to consider what happens to yasXtendstopositiveinfinity(i.e.asx—>+0°) yasXtendstonegativeinfinity(i.e.asx—>-0°) Hencesketchthegraphofy=|(2x-1)(2-x)(x+1)| Whenx=0,y=-lx2x1=-2..-.Thecurveinterceptsthey-axisat(0,-2).Wheny=0,(2x-1)(2—x)(x+1)=0

•

Answers a

Whensketchinggraphsofthisformyoushouldshowthegeneralshapeofthecurveandalloftheaxis

CambridgeIGCSEand0LevelAdditionalMathematics b|a:-2|^|*+5| e |jc+2|<—x-5 2 6Solve, a|x+l|>|x-4| d |2x+3l^l*-3| 7 Solve. a 2|x-3l>|3«+l| b 3|x-l|<|2x+ll 8Solvetheinequality\x+2k\^\x-3k\whereAisapositiveconstant.9Solvetheinequality\x+Sk\<4\x-k\whereAisapositiveconstant. c|x+l|^|3^i:+5| f|3x-2|^|x+4| c|2x-5l^3l2x+l| CHALLENGEQ 10Solve|3x+2|+|3x-2|^8. 5.3 Sketching graphs of cubic polynomials and their moduli

2x-l=0 2-x=0 x+l=0 1 x= 2 X=2 X=—1 .*.Thecurveinterceptsthex-axisat (2,0)and(-1,0).

Chapter5:Fquations,inequalitiesandgraphs Asx—>+00,y—AsX—>y—>+00 Thegraphofthefunctiony =(2x-l)(2-x)(x+l)is: \ ^ y=(2x-l)(2-x)(x+1) -1\ 0 /l 2) X /2 \ b Tosketch the curve 3i =|(23c-l)(2-*:)(x+l)| you reflectin the x-axis the parts ofthecurvey={2x-1)(2-x)(x+1)thatarebelowthese-axis.y=|(2x-l)(2-«)(x+1)1 WORKEDEXAMPLE8 aSketchthegraphofthefunctiony={x-if(x+1). b Hence sketch the graph of y =|(x-l)^{x+l)|. Answers Whenx=0,y=(-1)^x1=1. WhenyThecurveinterceptsthey-axisat(0,1).=0,(x-1)(x-1)(x+1)=0 X—1=0 X—1=0 x+l=0 X=1(repeatedroot) x=-1 /. Thecurveinterceptsthe»-axisat(1,0)and(-1,0). Asx—>+00,y—>+00Asx—^y—>—oo

y=2(x+l)-(7-2x)

Cambridge IGCSE and0LevelAdditional Mathematics

A\Jc

FindthecoordinatesofthepointsA,BandCwherethecurveinterceptsthex-axisandthepointDwherethecurveinterceptsthe positivey-axis. y=(x-2)(x+1)(x-3) Sketch each ofthese curves and indicate clearly the axis intercepts, a y=(x-2)(x-4)(x+3) b y=(x+2)(x+l)(3-x) c y=(2x+l)(x+2)(x-2) d y=(3-2x)(x-1)(x+2)

Thegraphofthefunctiony=(x-1)^(*+1)is: y=(x~\)-{x+1) bTosketchthecurvey=|(a:-l)^(x+l)|youreflectinthese-axisthepartofthe curvey={x—1)^(x+1)thatisbelowthex-axis. y jy=|(x-l)2(x+1)1 -1 0 1 a; In worked examples 7 and 8,you considered the value ofy as x —> + <» to determine the shape of the cubiccurve.Itis often easier torememberthatforacurve with equation y=k{x-a){x-b)(x-c)the shape ofthe graph is: ifftispositive ifftisnegative Exercise5.3

FindthecoordinatesofthepointAandthepointB,whereAisthepointwherethecurveinterceptsthe interceptspositivex-axisandBisthepointwherethecurvethepositivey-axis.

Chapter5:Fquations,inequalitiesandgraphs

ay-x^{^x+2^ C ))=(x+l)^(x:-2) Sketcheachofthesecurvesandindicateclearlytheaxisintercepts. b y=x^{5-2x) d y=(x-2)'(10-3x)

CHALLENGEQ 9

Thediagramshowsthegraph of y=k{x-aY(x-b). Findthevaluesofa,bandk. -1 -4J X

CHALLENGEQ 10Thediagramshowsthegraphof y=|A(x-a)(x-6)(x-c)| wherea<b<c. Findthevaluesofa,b, candk.

Factoriseeachofthesefunctionsandthensketchthegraphofeachfunctionindicatingclearlythe axisintercepts. a c 7 a y=9x 2x^+x^-25x+12 b y= +4x^+X-6 d y=2x^+3x^-29x 60 b a On thesameaxessketch thegraphsof y=x(x-5)(x-7)and y=x(7-x),showingclearlythe pointsatwhichthecurvesmeetthecoordinateaxes.

Use algebra to find the coordinates ofall the points where the graphsintersect. On the same axes sketch the graphs of y=(2x-l)(x+2)(x+1)and y=(x+1)(4-x),showing clearly the points atwhich the curves meet the coordinate axes, b Use algebra to find the coordinates ofall the points where the graphs intersect.

Exercise5.4

1Thediagramshowspartofthegraphofy=x{x-2)(ie+1). Usethegraphtosolveeachofthefollowinginequalities a x(x-2)(5c+1)^0, b x(x—2)(x+l)^l, c x(x-2)(x+1)^-2. y=x(x-2)(x+1)

5.4Solvingcubicinequalitiesgraphically

Inthissectionyouwilllearnhowtouseagraphicalmethodtosolveinequalitiesofthe formk{x-a){x-b){x-c) d.

CambridgeIGCSEand0LevelAdditionalMathematics

WORKEDEXAMPLE9

Thediagramshowspartofthegraphofy=—(x-3)(x-2)(x+2). 6 inequalityUsethegraphtosolvethe{x—5){x-2){x+2)^6. y=-(x-3)(x-2)(x+2) 6 Answers -(x-3)(x-2)(x(x-3)(x-2)(x+2)«6+2)«1 You6needtofindwherethe curvey=—(x-3)(x-2)(x+2) 6 isbelowtheliney=1. Theredsectionsofthegraphrepresentwherethecurvey=—(x-3)(x-2)(x+2) 6 istheliney=1. Thesolutionisx^-1.6or1^x^3.6. dividebothsidesby6 y=-(x-3)(x-2)(x+2) 6 aj'=1

Use the graph to solve each ofthefollowing inequalities a (l-x)(x-2)(x+1)^-3, b (l-x)(x-2)(x+l)^0, c (1—x)(x-2)(x+l)>—1. (ffl^(x-2)(x+l)

5.5 Solving more complex quadratic equations Youmaybeaskedtosolveanequationthatisquadraticinsomefunctionofx WORKEDEXAMPLE10 Solvetheequation4x'* 17*2+4=0. Answers Method1(substitutionmethod) 4x^-17x^+4:=0 usethesubstitutionv= 4y^—\7y+4:=0 factorise (43i-1)(>-4):=0 solve li 0 or ^-4:=0 II afH . *H-1 or y =4 substitutex~fory 2 1 X= 4 2 or X =4 x=±2 or X =±2

Use the graph to solve each ofthefollowing inequalities a (jc+lf(2-a:)^0, b (jc+l)^(2-x)^4, C (x+lf(2-a:)^3. 1)2(2-X)

Chapters:Fquations,inequalitiesandgraphs

2 The diagram shows part of the graph of 3)=(»:+l)^(2-x).

Thediagramshowspartofthegraph of3)=(l-a:)(x-2)(x+l).

CambridgeIGCSEand0LevelAdditionalMathematics Method2(factorisedirectly) 4x:^-17x^+4=0 (43c^-1)(x^-4)=:0 4x^-1=0 or x^-4=0 1 or 1 x=±— or 2 x=±2-4 WORKEDEXAMPLE11 i 1 i Usethesubstitution y=x^ tosolvetheequation3x^-5x^+2=0. Answers 2 1 1 3x3-5x3+2=0 let y=x3 3/-5y+2=0 factorise (3y-2)(y-l)=0 solve 3y-2=0 2 or y-1=0 or 7=1 1 substitutex3fory x3=3 1 or x3=1 cubebothsides x= u. \3 or x=(1)3 II or x=1 WORKEDEXAMPLE12 So\\etheequation x-4>/x-21=0. Answers x-4^/x—21=0 usethesubstitutionv=^/x /-4y-21=0 (7-7)(7+3)=0 O II OC + 0U o II 1> OC Ou l> II substituteVxfory OC II u O x> 11 \/x=-3hasnorealsolutions H

Chapter5:Fquations,inequalitiesandgraphs WORKEDEXAMPLE13 Solve the equation 8^4* 33^2*^+4=0. Answers 8(4^)-33(2*)+4=0 8(2')^-33(2=')+4=0 8y^—33y+4=0 (83)-1)(ji-4)=0 1 -v=— or v=4 ■^ 8 2"=- or 2'=4 8 x=-3 or x=2 ..\24'-'canbewrittenas (2^) ~(^ ) letv=2^ substitute 2* for y i = 2"® and 4 = 2^ Exercise 5.5 1 Findtherealvaluesofxsatisfyingthefollowingequations. ax^-5x'^+4=0 bx^+x^-6=0 c 20x^+64=0 d x' + 2x'-8 = 0 e x'-4x'-21 = 0 f 2x'-17x'-9 = 0 A 1 4 X X 4x''+6=llx' ' 6 3 ^ X X 2 WriteUsethequadraticformulatosolvetheseequations.youranswerscorrectto3significantfigures. a x"-8x^+1=0 d 2x®-3x^-8=0 3 Solve. a x-7\/x + 10 = 0 d ^/^(2+^/^)=35 g 2x+4=9^/x b x^-5x^-2=0 e 3x®-5x^-2=0 b x-y/x-12 = 0 c 2x^+x^-5=0 f 2x®-7x''-3=0 c X+ 5y/x — 24 = 0 8x-18>/x+9 = 0 f 6x+ll^/^-35 = 0 h Syfx H —16 ■sjx 2yfx H—j== 9 Vx 4Solvetheequation2x^—7x^+6=0. 5 The curve y= \fx and the line 5y= x+4 intersect at the points Pand Q. a Writedownanequationsatisfiedbythex-coordinatesofPandQ. b SolveyourequationinpartaandhencefindthecoordinatesofPandQ. 6 Solve. a2^"-6(2")+8=0 b3^'^-10(3^)+9=0c2(2^")-9(2")+4=0 c3^'^'-28(3'')+9=0d2^'^^-33(2'')+8=0e3^''^^+3(3'')-2=0

.CambridgeIGCSEand0LevelAdditionalMathematics 7 = and g(x)= +7x.Solve gf(x)=2. 8 f(x)=x^-2 and g(x)=x"-5x. Solve gf(x)=6. 9 f(x)=x'^+3x and g(x)=x^-4x. Solve gf(x)=0. Summary Solving modulusequations Tosolvemodulusequationsyoucanusetheproperty: \a\=\b\ =b'^ SolvingmodulusinequalitiesTosolvemodulusinequalitiesyoucanusetheproperties: Ia1^6 -b a b \a\'^b a^—bora^b \a\>\b\ <=i' > b'^ \a\ <\b\ <b'^, b Q The graph of j;=^(jc-a)(x-6)(A:-c) Thex-axisinterceptsare(a,0),{b,0)and(c,0).Theshapeofthegraphis if^is positive ^ negative The graph of >>=|A(ji:-a)(x:-A)(jc-c)| Tosketch thecurve y=|/t(x-a)(x-i)(x-c)|you reflectin the x-axisthe partsof thecurvey=k{x-a){x—b){x-c)thatarebelowthex-axis. i

Answers a The graph of)!= x^(x-2); When x=0,y=0^ X (-2)=0.

Thecurveinterceptsthex-axisat(0,0)and(2,0).

Worked example a On thesameaxessketch the graphsoiy= x^{x-2)and31=x(6-x),showingclearly thepointsatwhichthecurvesmeetthecoordinateaxes. [5]

Thecurveinterceptsthex-axisat(0,0)and(6,0). AsAsX—>+00,y—>-00Xy—>—°a y j y=x'^{x-2) 6\ y=x(x-6)\

Examination questions

Examinationstylequestion

Wheny=0,x^(x-2)=0 x=0 x=0 x-2=0 X=0 (repeatedroot) x=2

bUsealgebratofindthecoordinatesofthepointswherethegraphsintersect. [6]

Asx^-00,AsX—>+00,y—>+00y-»-ooThegraphofy=x(6-x):Whenx=0,y=0x6=0.Thecurveinterceptsthey-axisat(0,0).

Thecurveinterceptsthey-axisat(0,0).

Wheny=0,x(6-x)=0 x=0 6-x=0 x=0 x=6

Chapter5: Fquations,inequalitiesand graphs

CambridgeIGCSEand0LevelAdditionalMathematicsbAtthepointsofintersection: x^{x-2)=x(6-x) x^{x-2)-3c(6-x)=0 x[x(x-2)-(6-x)]=0 x(x?-X-6)=0 x(x-3)(x+2)=0 X=0 or X=3 or x=-2 x=0 => y=0 x=3 ^ y—9 x=-2 => }i=-16 Thepointsofintersectionare(-2,-16),(0,0)and(3,9). Exercise5.6 1ExamStyleExerciseSolvetheequation12x-3|=13x-5|. 32Solvetheinequality12x-1|>7.Solvetheinequality17-5x|<3. 4 Solvetheinequality|x|>13x-2|. 5 Solve the inequality|x-1| <|x + 2|. 16 Solvetheinequality|x+2|< -X-1 7 Solvetheinequality|x+2^|>|x-^|where isapositiveconstant. 8 a Solve the equation|x-13|= 14. b Hence solve the equation \y^ -13|= 14. [3] Examinationstylequestion [3] Examinationstylequestion [3] Examinationstylequestion [4] Examinationstylequestion [4] Examinationstylequestion [4] Examinationstylequestion [4] Examinationstylequestion [1][3] Examinationstylequestion 9 Sketch the graph of y=x(3-2x)(x-4),showing clearly the points atwhich the curve meets the coordinateaxes. [3] Examinationstylequestion

10 Sketch the graph of 3!=2(2x-l)(x-3)(x+l),showing clearly the points at which the curve meetsthecoordinateaxes. [4] Examinationstylequestion 11 l)(x-l)(2-x)

Examinationstylequestion 14 a On thesame axessketch the graphsof y=(x-3)(x+1)^ and y=—,showingclearlythe pointsatwhichthecurvesmeetthecoordinateaxes, b Hence state the number ofreal roots ofthe equation (x-3)(x+l)2 6 [1][5] 15 a Factorise completely 2x®+x^-25x+12.

The diagram shows part ofthe graph of y=2(x+l)(x-l)(2-x). Usethegraphtosolvetheinequality(x+l)(x-l)(2-x)>-l. [3]

Chapter5:Fquations,inequalitiesandgraphs

Examinationstylequestion [5]

Examinationstylequestion 13 a Factorise completely X®+x^-6x. [3] b Hencesketch thecurve with equation y=x®+x^-6x,showingclearlythe pointsatwhich thecurvemeetsthecoordinateaxes. [3]

Examinationstylequestion 12 Sketchthegraphofy=(x-4)(x-l)(x+2),showingclearlythepointsatwhichthecurve meetsthecoordinateaxes. [4] Hencesketchthecurvey=|(x-4)(x-l)(x+2)|. [1]

b Hence sketch the curve with equation y=2x^+x^-25x+12,showing clearly the points at whichthecurvemeetsthecoordinateaxes.

Examinationstylequestion

[4]

Examinationstyleqiiestion

b Usealgebratofindthecoordinatesofthepointswherethegraphsintersect.

[6]

16a Onthesameaxessketchthegraphsofy=(6-x)andy=4x(4-x),showingthepoints atwhichthecurvesmeetthecoordinateaxes. [5]

CambridgeIGCSEand0LevelAdditionalMathematics

-v <-v\ % A v^'I m /5^>' V^ y, ^•KP" iiW V r SviKf Ay'l\i rfY/- ■ ■'^-/ « Chapters Logarithmic and exponential functions Thissectionwillshowyouhowto: m usesimplepropertiesofthelogarithmicandexponentialfunctionsincludingInxande* ■ usegraphsofthelogarithmicandexponentialfunctionsincludingInxande'^andgraphsof/(e"*+o andkln(ox+b)wheren,k,aandbareintegers ■ usethelawsoflogarithms,includingchangeofbaseoflogarithms ■ solve equations of the form o'' = £i. « M ■1

Considertheexponentialfunctionf(x)=10*. Tosolve 10*=30. Youcansay10'=10and10'-'=100. So 1<X<2. Thegraphofy=10*couldbeusedtogivea moreaccuratevalueforxwhen10*=30. Fromthegraph,x=1.48. Thereisafunctionthatgivesthevalueofxdirectly. If10*=30thenx=logiQ30. isreadas'log30tobase10'. logisshortforlogarithm.

CambridgeIGCSEand0LevelAdditionalMathematics

6.1Logarithmstobase10

On your calculator,for logs to the base 10, you use the or(^1^ key. Soif10*=30 then X=log1030 X=1.477to4sf. Hencetheruleforbase10is: Ify=10*thenx=logjoy. Thisrulecanbedescribedinwordsas: logioyisthepowerthat10mustberaisedtoinordertoobtainy. For example, logiglOO = 2 since 100 = 10^. y=10* and y=logjQX areinversefunctions. y 11 1 0 /i X y=log,„x

:y 100- y =10* 80r 60-40i i ^i: j ! 1 ?^ f 1 1 ! f^ 1 1 i i ii.4^ X Q Note: logio30can also be written asIg30orlog30.

Chapter6:Logarithmicandexponentialfunctions CLASSDISCUSSION Discusswithyourclassmateswhyeachofthesefourstatementsistrue. logio1=0 [logio10*=XforX€rlogio10=1 X= for X >0^ WORKEDEXAMPLE1 aConvert10*=45tologarithmicform, bSolve10*=45givingyouranswercorrectto3sf. Answers a Method110*=45 Step1:Identifythebaseandindex;Ihebaseis10.TheindexisX.Step2:Starttowriteinlogform:Inlogformtheindexalwaysgoesonits Step3:Completethelogform: SoX=logio45. Method2 10*=45 logiolO*=logio45X=logio45 b 10*=45 X=logio45 X=1.65 ownandthebasegoesatthebaseofthe logarithm.SoX=logjo?

Fillinthelastnumber.X=logjo45Takelogstobase10ofbothsides.logiolO*=X

CambridgeIGCSEand0LevelAdditionalMathematics WORKED EXAMPLE2 J^H|||||^H|||H

Step2:Starttowriteinexponentialform: Writethebaseandtheindexfirst. So102-9=p

a Convertlogio*=2.9toexponentialform. b Solvelogiox=2.9givingyouranswercorrectto3sf. Answers a Method1 logiox=2.9

Step1:Identifythebaseandindex: Thebaseis10.Theindexis2.9. (Inlogformtheindexisalwayson itsown.)

Step3:Completetheexponentialform: O SoX=10^'® Method2 logiox=2.9 IQlogioA:^102.9 lO^ogto'^=X X=lO^-^ b logiox=2.9 X=lO^-^ x=794 WORKEDEXAMPLE3 Findthevalueof a logiolOO000 b lOgigO.OOl C lOgiglOO-v/TO. Answers a logiglOO000=logiglO^ =5 write100000asapowerof10,100000=10® b logioO.OOl=logiolO-^ =-3 write0.001asapowerof10,0.001=10"® c logiol00>/10 = logiolO" write100-JToasapowerof10 -2.5 100VT0=102X10«-5=10"

Chapter6:Logarithmicandexponentialfunctions Exercise6.1 1Convertfromexponentialformtologarithmicform. a 10^=1000 b lO'-^=100 c 10®=1000000 dlO""=2 e10^=15 f10"^=0.06 2Solveeachoftheseequations,givingyouranswerscorrectto3sf. a 10^=75 b 10^=300 c 10^=720 d10^=15.6 elO'^=0.02 f10^=0.005 3Convertfromlogarithmicformtoexponentialform. a Ig100000=5 d Ig^=7.5 b lglO=l e Ig X = 1.7 1 =-3c Ig^1000 f IgX=-0.8 VTo J' 4 Solve each ofthese equations,giving your answers correct to 3sf. aIgX=5.1 blgx=3.16 cIgx=2.16 dIgX=-0.3 elgx=-1.5 fIgX=-2.84 5 Withoutusingacalculator,findthevalueof a Ig 10000 b IgO.Ol c IgVlO d Ig(^) e Ig(loVlo) f Ig*"^^^® 6.2 Logarithms to base a Inthelastsectionyoulearntaboutlogarithmstothebase of10. The same principles can be applied to define logarithms inotherbases. Ify=ofthenx=leggy. lOggffl=1 lOggl=0 loggfl*=X X= Theconditionsforlog^x tobedefinedare: • a>0anda^1 • X>0 y y=a" II 1 // y=log^^ 0 /I X

WORKEDEXAMPLE4

Step1:Identifythebaseandindex: Step2:Starttowriteinlogform: Step3:Completethelogform: So4=log2l6.

Convert2''=16tologarithmicform. Answers Method1 2'*=16

Step 2: Start to write in exponential form: Write the base and the index first. So 7^ = ? Step 3: Complete the exponendal form: II OC So 72 = 49 Method2 log749=2 ylog,49 _ y2 7log,49^4g 49 = 72

CambridgeIGCSEand0LevelAdditionalMathematics

Convertlog749=2toexponentialform. AnswersMethod1 log749=2 Step1:Idendfythebaseandindex: Thebaseis7.Theindexis2. (Inlogformtheindexisalwaysonitsown.)

Takelogstobase2ofbothsides,log22''=4, WORKED EXAMPLE5 ■jjjjjjjjj^HH

Thebaseis2.Theindexis4. Inlogformtheindexalwaysgoesonits ownandthebasegoesatthebaseofthe logarithm.So4=logg?Fillinthelastnumber.4=log^lb Method2 2^=16 Iog22^=log2l6 4=log2l6

WORKEDEXAMPLE6Findthevalueof b logs128 c log4^alogs81 Answers a logs81=logs3^ =4 write81asapowerof3,81=3^ b log2l28=logs2^ =7 C log4 =log44-2 write128asapowerof2,128=2^ 1 1 ._2write—asapowerof4,—-4 16 ' 16

Chapter6:Logarithmicandexponentialfunctions

Exercise6.2 1 Convertfromexponentialformtologarithmicform, a43=64 b2^=32 c5^=125 d 02=36 g e 2--^= h X]_ 32 f 3-'= 81 a = c 2 Convertfromlogarithmicformtoexponentialform, alog24=2 blog264=6 clogjl=0 e log366 = -^d log,9=2 f log82=i g log^l=0 3 Solve. a logjX=4 d log3X = i g log2(x-1)= 4 4 Findthevalueof a log4l6 e logs243 i log648 5 Simplify. alog^x^ e log. 6 Solve. alogs(log2 =1 h log,8=)) b loggX=2 e log,144=2 h log3(2x +1)= 2 i log„^=c c log^x=4 f log,27=3 ' log5(2-3x)=3 b logjSl f log2(8V2) • 11 log7 c log464 glog5(25V5) k log5^ b log,^ c log,(x^/x) d log20.25 .1 h logs <1 log.:^ X\/x X /f log,(V^) g log,j^^j h log b log2(log5x)= 2 Vn/x^

CambridgeIGCSEand0LevelAdditionalMathematics 6.3 The laws of logarithms If X and y are both positive and a > 0and a * I,then the following laws of logarithmscanbeused; Multiplicationlaw (^)=log^x+log„3; Proofs: log« =log„ X = logJa'°g''^ + '°s„>) =\og„x+log„y log«log« Divisionlaw =log„x-log„3; ) logflV=log« ==log«log„x-log„ylog„-v-lognv"! Using the power law, log„ j= log„x' =-log«^ Thisgivesanotherusefulruletoremember:-log^x Powerlaw logaC^)"logaW"=mlog^x = log„(a'"'°g«'^) = m log^x WORKEDEXAMPLE71 aUsethelawsoflogarithmstosimplifytheseexpressions.Ig8+lg2 blog4l5 log45 C21og34+51og32 Answers a Ig8+lg2 blog4l5-rlog45 C 2log34+5log32 =lg(8X2) =log342+log32® =lgl6 =log43 =log3(16x32) =log3512

Giventhat\og^p=xand\og5q=y,expressintermsofxand/or a log5/'+logg?'' b logs/)^-loggT? c logs I Answers a ==logs^+logslogs/'+3logs?x+Sy blogs/*^-logsV? =2l0g5p-il0g5? „ 1 -2x--y WORKEDEXAMPLE8o Chapter6:Logarithmicandexponentialfunctions Exercise6.3 1 Writeasasinglelogarithm. aloggS+loggS blog3l2-log32 d 2 log74-31og72 e ^log325+ log34 c 31og52+log58 f 2logy1-1+logy8 g l+log43 h lg5-2 3-log4l0 2 Write as a single logarithm,then simplify your answer, a log256-log27 b log6l2+log63 c -l d log3l5--^log325 e log440-ilog4l25 Z 6 1 3 Simplify, a 21og53--|log54+log58 4 a Express 16and 0.25 as powers of2. log3l6 b Hence,simplify log og236-log23 f ^log3l6-2 log36 b 2+^log249-log221 5 Simplify. logy4 30.25 logy27 logs64 d logslOQ logy2 logyS log30.25 loggO.Ol 6Giventhatu=logs*,find,insimplestformintermsofu a X b logj 25 logs d logs Xy/x 125 7Giventhatlog^p=xandlog4?=y,expressintermsofxand/ory a log4(4/)) b c log4/'+log4 d pq. 8 Giventhatlog^x=5andlog^y=8,find b loga log„ ^1^ log« d log«(xy). 9 Giventhatlog^x=12andlog„y=4,findthevalueof a log„ r \ X \j/ b log« VX'/ log«(x^) d log,,

WORKEDEXAMPLE9 Solve. a 2logf,(x+2)=logs{2x+19) b 4log,2-log,4=8 Answers a 2logs(x+2)=logs{2x+19) usethepowerlaw logs +2)^=logs(2*+19) useequalityoflogarithms (x+2)^=2x+19 expandbrackets x^+4x+4=2x+19 x^+2x-15=0 (x-3)(x+5)=0 X=3orX=-5 Checkwhenx=3:

You have already learnt how to solve simple logarithmic equations. In this section you will learn how to solve more complicated equations. Itisessential,whensolvingequationsinvolvinglogs,thatallrootsarechecked intheoriginalequation.

2logg(x+2)=2loggS=logg25isdefined ,logg(2x+19)=logg25 isdefined

CambridgeIGCSEand0LevelAdditionalMathematics

6.4Solvinglogarithmicequations

SoX—3isasolution,sincebothsidesoftheequationaredefinedandequivalent invalue. Checkwlien.r=-5: 2logg(x+2)=2logg(-3)isnotdefined SoX=-5isnotasolutionoftheoriginalequation. Hence,thesolutionisx=3. b 41og,2-log,4=2 usethepowerlaw log,2'-log,22=2 usethedivisionlaw log,2^-2=2log,2^=2 log,4=2 converttoexponentialform x2_4 X=±2 Sincelogarithmsonlyexistforpositivebases,x=-2isnotasolution. Checkwhenx=2: 4log22-log24=4-2log22 ==4-22

SoX=2satisfiestheoriginalequation. Hence,thesolutionisx=2.

CHALLENGEQ 7 a Show that Ig(x^)!)=18 can be written as 21gx+Ig))=18. b Ig(x^)i)=18 and Ig =2. 17^ FindthevalueofIgxandIg^i.

4Solve. a log^64-log,4=1 c log^4-21og^3=2

5 Solve. a (log5^)^ - Slogs(x)+ 2 = 0 c (logs^f-log5(x3)= 18 b log42x-log45=log43 d logg(x+3)= 2l0gg4+logg5 blg(5x)-lg(x-4)=1 dlogg(2x+3)=2+logg(2x-5) f lg(4x +5)+ 21g2 = l +lg(2x-l)

CHALLENGEQ 6

2 Solve. a loggX+loggS=2 clog2(x+4)=1+logs(x-3)elogs(10^+3)-logs(2^-1)=2

3Solve. alog5(x+8)+log5(x+2)=log520x: c 2log4 X-log4(x +3)=1 eloggX+logg(2x-5)=1 g lg(x+5)+lg2x=2

Chapter6: Logarithmic and exponentialfunctions

1Solve. a Iog2X+log24=log220 cIog4(x-5)+log45=21og4l0

aSolvethesimultaneousequations,x^;=64logx7=2 c log4(x+3))= 21og4X log4;y=log43+log4X e logioa=2logjo^* logio(2«-^)= 1 b 2"=4^ 21g3!=Igx+Ig5 d xy=640 21ogiox-logio^=2 f 4*?'= log2:y-log2x=1

b loggX+logg(X-2)=lOgglS d Igx+lg(x+1)=lg20 f 3+2log2X=log2(14x-3) h lgx+lg(x-15)=2 b log^ie+log^4=3 d logJ5=2+log^5 b (loggxf-log5(x2)= 15 d 2(log2x)^ + 51og2(x^)= 72

Exercise6.4

CambridgeIGCSEand0LevelAdditionalMathematics 6.5 Solving exponential equations In Chapter 3 you learnt how to solve exponential equations whose terms could be converted to the same base.In this section you will learn how to solve exponentialequationswhosetermscannotbeconvertedtothesamebase. WORKEDEXAMPLE10 Solve,givingyouranswerscorrectto3sf. a3"^=40 b5-'^+'=200 Answers3^^=40 XIgS''=lg40Ig3=Ig40 ,.igio lg3 X~3.36 52a:+i 200 lg52x+)^ig200 (2x+l)lg5=lg200 2x-tl=lS^ lg5 2x+l=3.292.. 2x=2.292.. X=1.15 usetakelogsofbothsidesthepowerrule dividebothsidesbyIg3 usetakelogsofbothsidesthepowerrale dividebothsidesbylg5 WORKEDEXAMPLE11 Solve3(2^^)-2^-''-8=0. Answers 3(22x)_2>t+i-8=0 3(22^)-2(2'')-8=0 3/-2y-8=0 {y-2)(3y+4)=0 Wheny=2 2=2" X=14 Wheny= 3 -1= 2'' 3 replace2-^+'with2(2*)usethesubstitutiony=2* factorise therearenosolutionstothisequationsince2*isalways positive Hence,thesolutionisx=1.

4 Use thesubstitution y=3* tosolve theequation 3^*+2= 5(3*).

Chapter6:Logarithmicandexponentialfunctions

Solve,givingyouranswerscorrectto3sf. a2^^-70 b 3^=20 c 5''=4 e 3^+'=55 f =20 g 1"-'"=40 i5^+1=3^+2 j 4X-1^5^+1 1^32X+3^53x+] d h I 2^"=150—gx+4 24--5x_^X+4: 2 a Show that 2*"^^ -2*"'=15 can be written as 2(2*)-—(2*)= 15. bHencefindthevalueof2*. C Findthevalueofx. 3 Solve,givingyouranswerscorrectto3sf. a 2*+2-2*=4 b 2*+*-2*"'-8=0 c 3*+i-8(3*-^)-5=0 d 2*+2-2*-^ =12 e 5*-5*+2+125=0

Exercise6.5 1

5 Solve,givingyouranswerscorrectto3sf. a 32^-6X3*+5=0 b 4^^-6x4*-7=0 c 22'=-2*-20=0 d 5^'=-2(5*)-3=0 6 Use the substitution u = 5" to solve the equation 5^*-2(5*"^')+21 = 0. 7 Solve,givingyouranswerscorrectto3sf. a 2^*+2*+'-15=0 b 6^*-6*+'+7=0 c32==-2(3*+')+8=0 d42'=+!=17(4*)-15 8 Solve,givingyouranswerscorrectto3sf. a 4*-3(2*)-10=0 b 16*+2(4*)-35=0 c 9* - 2(3*+')+8 = 0 d 25*+ 20 = 12(5*) CHALLENGEQ 9 32*+!X5*-i=27*X52* Findthevalueofa15* X. CHALLENGEQ 10Solvetheequations,givingyouranswerscorrectto3significant figures. a |3*+2|=|3*-10| b |2"^'+3|=|2*+10| 3^4 =5(3N)+24 d41*1=5(2'*')+14

6.6Changeofbaseoflogarithms Yousometimesneedtochangethebaseofalogarithm. A logarithm in base b can be written with a different base cusing the change ofbaserule. \ia,b,c>Qandb,c then: logc« Proof: Ifx=log^a,thenb"=a takelogsofbothsides log^Z)^=log^fl usethepowerrule X log,6 = log^rt divide both sides by logc b log,a X = log,,a log,^log,.blog,^

Ifc=ainthechangeofbaserule,thentherulegives: log^a = loga^

CambridgeIGCSEand0LevelAdditionalMathematics CHALLENGEQ 11 Solve the inequality 12*^'-l|< 12*-Sl giving your answer in exact form.

WORKEDEXAMPLE12 Changelogg?tobase10.Henceevaluateloga?correctto3sf. Answers log,7=^=2.81 lg2 Q Note: Some calculators have a f log,D 1 key. Thiscanbeusedtoevaluatelogg7direcdy.

Hence,thesolutionisx=3. d logy0.0025

2Giventhatu=log4x:,find,insimplestformintermsofu. alog^4, blogJ6, clog^2, dlog^S. 3 Giventhatloggji=x,expressintermsofx. a log,,9, b log9(9:^), c logj^ d log3(81)i).

7 a Expresslog4XintermsofloggX. b Using your answer ofpart a,and the substitution u = loggX, solve the equationlog4X+loggX=12.

SoCheckwheny=-2:logs(~2)isnotdefinedX=-2isnotasolutionoftheoriginalequation.

SoX=3isasolution,sincebothsidesoftheequationaredefinedandequivalent invalue.

Chapter6:Logarithmicandexponentialfunctions

Thechangeofbaserulecanbeusedtosolveequationsinvolvinglogarithmswithdifferentbases.

Exercise6.6 tt 1 111 Usetherulelog^a= toevaluatethesecorrectto3sf. logio^ a logalO b loggSS c loggS

4 a Given that log^x = 20 and log^)i=5, find log,x. bGiventhatlog^X=15andlog^F=6,findthevalueoflog^F. 5 Evaluate log^^2 x loggjf?. 6 Solve. a loggS+log9(x+4)=log525 b 21og42+logy(2x+3)=log327

Solvelog,*=logg(x+6). Answers log,X=loggfx+6) changelogg(x+6)tobase3 log,(x+6)log3X=7'logs9 logs9=log,3-=2multiplybothsidesby2

2loggX=log,(x+6) usethepowerrule loggX^=log,(x+6) useequalityoflogs x^=X+6 x^-X-6=0 (x-3)(x+2)=0 X=3or X=-2 Checkwhen.y=3: logs3isdefinedandisequalto1 lOgg(3+6)=logg9isdefinedandisequalto1

WORKEDEXAMPLE13

CHALLENGEQ 12 Solvethesimultaneousequations

Ify=e* then x=Iny. y=InX isthereflectionof y=e^ intheliney=x. y=InX andy=e* areinversefunctions.

d 4 lOggX = lOggX+2

Alltherulesoflogarithmsthatyouhavelearntso faralsoapplyfornaturallogarithms. y=X y=Inx;

8Solve, a loggX+5log4X=14 c 5loggx-log4X=3 9 a Expresslog^3intermsofalogarithmtobase3. b Usingyouranswerofparta,andthesubstitutionu=loggX,solvethe equationloggX=3-21og^3.

log5^+log„5=2log4X+6log^4-5=0 logsy=4-4log^5

2loggy=log5l25+log3X V= 6.7Naturanogarithms

10Solve. a loggx=9log^3 b c log4X-4log^4+3=0 d e loggx-9log^2=8 f 11a Expresslog4XintermsofloggX. b Expresslogsyintermsofloggy. c Hencesolve,thesimultaneousequations

6log4X+3loggy=16 loggX-2log4y=4.

Thereisanothertypeoflogarithmtoaspecialbasecallede. Thenumbereisanirrationalnumberande=2.718. Thenumbereisaveryimportantnumberinmathematicsasithasveryspecial properties.You will learn aboutthese special properties in Chapters 15 and 16. Logarithmstothebaseofearecallednaturallogarithms. In X isusedtorepresentlog^x.

CambridgeIGCSEand0LevelAdditionalMathematicsblOggX+2lOggX=4

Chapter6:Logarithmicandexponentialfunctions ^0.2 Exercise6.7 1Useacalculatortoevaluatecorrectto3sf. ae^ b e^-^ ce" 2Useacalculatortoevaluatecorrectto3sf. aIn4 b In2.1 cIn0.7 3 Withoutusingacalculatorfindthevalueof d e"^ d In0.39 ae 4Solve In5 iln64 b e2 c 3e'"2 ae'"*=7 b lne''=2.5 ce^'"*=36 dd -ln| —p 2 -In* =20 5 Solve,givingyouranswerscorrectto3sf. ae''=70 .2* =28 ce *+i =16 .2*-! =5 6 Solve,givingyouranswersintermsofnaturallogarithms. ae"=7 b 2e'=+l=7 ce 2*-5 =3 »3*-l =4 7 Solve,givingyouranswerscorrectto3sf. aInX=3 bInX=-2 cIn(x+1)=7 d In(2x-5)=3 8 Solve,givingyouranswerscorrectto3sf. aInx®+Inx=5 b =2e*~' cln(x+5)-Inx=3 9 Solve,givingyouranswersinexactform.aln(x-3)=2 d e*=2e-* c e^*-4e*=0b =7 e e^*-9e*+20=0f e*+6e~*=5 10Solve,givingyouranswerscorrectto3sf. a -2e''-24=0b e^''-5e*+4=0 c e*+2e-*=80 11Solvethesimultaneousequations,givingyouranswersinexactform, a Inx=21ny b e®*"^=3e®* Iny-Inx=1 e^*=Se*"^^ 12 Solve 5ln(7-e^*)=3,giving your answer correct to 3significant figures. 13Solveex-xe®''"'=0. CHALLENGEQ 14 Solve 5x^-x^e^*+2e^* =10 giving your answers in exactform.

1 Atthestartofanexperimentthenumberofbacteriawas100. Thisnumberincreasessothataftertminutesthenumberofbacteria,N,is givenbytheformula A=100X2'.

b Estimatetheyearinwhichthepopulationwouldbeexpectedtohave firstdecreasedto5000.

2 Atthebeginningof2015,thepopulationofaspeciesofanimalswas estimatedat50000.

aWhenf=0, T=75e-^'^^^+20 =75e"+20 usee =95 Temperatureofthedrinkwhenfirstmadeis95°C.

WORKEDEXAMPLE14

bWhen«=6, T=75e-°°2^®+20 =86.5 Temperatureofthedrinkwhent=6is86.5°C. 0 c WhenT=65, 65=75e-®"2'+20 45=75e-°02' In0.6=0.6=-0.02t _In0.6 '~ -0.02 t=25.5to3sf subtract20frombothsides dividebothsidesby75 takeInofbothsides dividebothsidesby-0.02

aEstimatethenumberofbacteriaafter12minutes, bEstimatethetime,inminutes,ittakesforthenumberofbacteriato exceed10000000.

CambridgeIGCSEand0LevelAdditionalMathematics

Thisnumberdecreasedsothat,afteraperiodofnyears,thepopulationwas 50000e'-^''3". a Estimatethepopulationatthebeginningof2020.

c FindthevalueoftwhenT-65. Answers

Thetemperature,T°C,ofahotdrink,tminutesafteritismade,isgivenby r=75e-«"2'H-20. a Findthetemperatureofthedrinkwhenitwasmade, b Findthetemperatureofthedrinkwhent-6.

6.8 Practical applications of exponential equations

Inthissectionyouwillseehowexponentialequationscanbeappliedto real-lifesituations.

Exercise6.8

a Findtheinitialpopulationofthesefish, bEstimatethenumberofthesefishafter6weeks, cEstimatethenumberofweeksittakesforthenumberofthesefishto havefallento—ofthenumberintroduced.

aFindthevalueofb.

Thepopulation,N,ofthisspeciesoffishaftertweeksisgivenbytheformula N=500e-o-^'.

• Thegraphissaidtobeasymptotictothe negativex-axis.

The volume ofwater in a container, V cm^,at time t minutes,is given by theformula y=2000e-*'. Wheny=1000,t=15. aFindthevalueofk.

Chapter6:Logarithmicandexponentialfunctions

Thearea,ndaysafterthemeasurementsstarted,isgivenbytheformula A=A^b".

b Find the value of A^ and explain what Aq represents, c Estimatethenumberofdaysfortheareaofthispatchofmouldto exceed7cm^.

Thegraphofy=e^

The value,$1^ ofa house n years after it was builtis given by the formula y=250000e"".

6.9 The graphs of simple logarithmic and exponentialfunctions

Youshouldalreadyknowthepropertiesofthe graphsy=e*andy=Inx.

• y=e*interceptsthey-axisat(0,1).

• WhenXgetscloserto-oo,thenygetscloserto0.

• Also,asX—>-l-oo,theny—>+0°.

b FindthevalueofFwhen t=22. Aspeciesoffishisintroducedtoalake.

• e*>0forallvaluesofx.

Whenn=2,A=1.8andwhenn=5,A=2.4.

c Estimatethenumberofyearsforthishousetodoubleinvalue. The area, Acm^,ofa patch of mould is measured daily.

y=X y=Inx

• Thiscanbewrittenas:Asx—>-<»,theny—>0.

Whenn=3,y=350000. aFindtheinitialvalueofthishouse, bFindthevalueofa.

TheCambridgeIGCSEand0LevelAdditionalMathematicsgraphofy=lnx •))=Inxinterceptsthex-axisat(1,0). • InXonlyexistsforpositivevaluesofx. • AsX->0,theny—>-«>. •Thegraphisasymptotictothenegativey-axis. • AsX—>+°o,theny—>+0°. y =e* and y=Inx areinversefunctions,sotheyare mirrorimagesofeach otherintheliney=x. Exercise6.9 1 Useagraphingsoftwarepackagetoploteachofthefollowingfamilyof curvesfork=S,2,1,-1,-2and-3. ay=e^ by=he" Cy=e"+k Describethepropertiesofeachfamilyofcurves. 2 Useagraphingsoftwarepackagetoploteachofthefollowingfamilyof curvesforft=3,2,1,-1,-2and-3. a y = In ftx b y = ft In X c y = In(x + ft) Describe the properties ofeach family ofcurves. 6.10Thegraphsofy=/ce"*+oandy=k{n(ax+b) where n,k,a and b are integers CLASSDISCUSSION Consider the function y =2e^*+ 1. Discussthefollowinganddecideonthemissinganswers: 1 When X =0,y =... 4 As x y 2They-interceptis(...,...) 5Asx-»-«>,y—>... 3Wheny=0,Xis... 6Theliney=...isan... (RememberNowsketchthegraphofy=2e+1andcompareyouranswerwithyourclassmates.toshowanyaxiscrossingpointsandasymptotesonyoursketchgraph.)

Chapter6:Logarithmicandexponentialfunctions WORKEDEXAMPLE15 Sketchthegraphof =3e 5. Answers WhenX=0, y=3e°—5 ==3-5-2 hencethe3hinterceptis(0,-2) When31=0,0=3e~^*-5 -=e-2^ 3 In If1 ■ X=-0.255 hencethex-interceptis(-0.255,0) AsX +°o,e~^*—»0so31-» -5 hencetheasymptoteis3=-5 AsX—>-00,e"^''—>°oso31—> 00 Thesketchgraphof3=3e"^* -5is: 3=3e-2^-5\ 3' -0.255\ 0 X -2 Y——5 -5 y ^

Sketchthegraphof31=4In(Sac+5). Answers Whenx=0,31=4In5, =6.44 When3=0,0=4In(2*+5) e"=2x+5 x1=2*:+5=-2 hencethe3-interceptis(0,41n5) hencethex4nterceptis(-2,0)

Exercise 6.10 1 Sketch the graphs ofeach ofthe following exponential functions. [Remember to show the axis crossing points and the asymptotes.] a y = 2e* - 4 63 = Se"" + 6 ^ 3 = 5e* + 2 d 3 = 2e"* +6 63 = 3e~* - 1 f 3 = -2e~* + 4 g 3 = 4e^* +1 h 3 = 2^^" +8 I 3 = -e^' + 2 2 Sketch the graphs of each of the following logarithmic functions. [Remember to show the axis crossing points and the asymptotes.] a 3 = ln(2x + 4) b 3 = ln(3x-6) c 3 = ln(8-2x) d 3 = 21n(2x + 2) e 3 = 41n(2x-4) f 3 = -31n(6x-9)

CambridgeIGCSEand0LevelAdditionalMathematics WORKEDEXAMPLE16

0=ln(2»:+5)

InXonlyexistsforpositivevaluesofx. So41n(2x+5)onlyexistsfor2x+5>0 X>-2.5. AsX—¥+00, 3—>00 As X—>—2.5,3-»-00 hencetheasymptoteisx=-2.5 Thesketchgraphof3=41n(2x+5)is: ;=-2.5 / ^n 5 r 0 X ■3 = 41n(2x + 5)

=-^y forX>3

WORKEDEXAMPLE17Findtheinverseofeachfunctionandstateitsdomain, af(x)=2e"^*+3for3C€R affx)=20^=^+3for*GIRStep1:Writethefunctionasy=

2 y=31n(2x-4) X=31n(2y-4) f = ln(2y-4) * es=2y-4 * 2y=e®+4 1^ y=-e®+2 2 Exercise6.11 1

StepStep1:Writethefunctionasy=2:Interchangethexandyvariables."Step3:Rearrangetomakeythesubject. f~Tx)=—e3+2forxG ThefollowingfunctionsareeachdefinedforxGR. Find f~^(x)for each function and state its domain, a f(x)=e*-I-4 b f(x)=e*-2 c f(x)=5e*-1 d f(x)=3e^*+1 e f(x)=5e^*-1-3 f f(x)=4e~^*4-5 g f(x)=2-e* h f(x)=5-2e~^*

Chapter6: Logarithmicand exponentialfunctions 6.11

Step2:Interchangethexandyvariables.Step3:Rearrangetomakeythesubject. f-\x) f(x)=31n(2x-4),x>2

The inverse of logarithmic and exponential functions

bf(x)=3In(2x-4)forx>2 Answers

y=2e-^'+3 *=2e-^>+3 X-3= ^=e-'>

In Chapter 1 you learnthow to find the inverse ofa one-one function. Thissectionshowsyouhowtofindtheinverseofexponentialandlogarithmic functions.

2Findf^(x)foreachfunction, a f(jc)=ln(»:+1),x>-1 c f(x)=21n(x+2),X>-2 e f(x)= 31n(2x-5), X > -^ b fW=ln(x-3),X>3 d f(x)= 21n(2x +1),X >-^ f f(x)= -51n(3x-l),x>i g(x)=In(2x+1)forx>3 f(x)=e^*+1forxeR a Statetherangeoff(x). cStatethedomainoff~^(x). 4 f(x)= e* for x6 R g(x)= ln5x for x > 0 a Find i fg(x) ii gf(x). bSolveg(5c)=3f"'(x). 5 f(x)= e®* for xe R g(*:)= Inx for x > 0 a Find ifg(x)ii gf(*). bSolvef(x)=2g"H*)6 f(x)=e^*forxe R a Findfg(x). bSolvef(x)=8g"'(x). Summary Therulesoflogarithms \iylogafl■=a"thenx=log^);.=1log.l=0log^a*=XX= log^ = log^x + =log^x-log^)) log^ (x)" = b Find f ^(x). d Find f~^f(x).

Product rule:

CambridgeIGCSEand0LevelAdditionalMathematics

Power rule:

log^j; Divisionrule:log„

y=x y = log^x Changeofbase:log^a= j^specialcase:log^ ^=-log^xj Uia.case:log..=^1 L iog«^J

m log^x

Chapter6:Logarithmicandexponentialfunctions LogarithmsNaturallogarithmstothebaseofearecalled Innaturallogarithms.Xisusedtorepresentloggx.Ifj=e*thenjc=In;y.Alltherulesoflogarithmsapplyfor naturallogarithms. y=Q'' \ \ \ \ 1 / // / / ✓ / ✓ / / / ✓ / ✓ / ✓ ✓ ^01^ ✓ / / / > 0/ /l X * / ✓ ✓ ✓ X / =InX Examination questions Worked example By changing the base of log2a4,express (log2a4)(l + log^2) as a single logarithm to base a. [4] CambridgeIGCSEAdditionalMathematics0606Paper21QllbJun2014 Answer log2a4= log«2log«2atoga4+log„a 1log«4+log„2 (log2«4)(l + log^2)= loga4 (l + log,2)l+log„2 ^ (log«4)(l + log„2) l+log„2 =loga4 usetheproduetruleonthedenominator rememberthatlog^a=1 divide numerator and denominator by (l + log^2)

CambridgeIGCSEAdditionalMathematics0606Paper11Q4Nov2012

CambridgeIGCSEAdditionalMathematics0606Paper11Q4i,iiJun2013 [2] CambridgeIGCSEAdditionalMathematics0606Paper21QpaJun2012 [3]

CambridgeIGCSEAdditionalMathematics0606Paper11Q2Jun2011 Usingthesubstitutionu=5",orotherwise,solve 52X+1 ^7(5x)_2. [5]

CambridgeIGCSEAdditionalMathematics0606Paper11Q4i,iiNov2014 Solvethefollowingsimultaneousequations. logg(x +3)= 2+loggji logg(jc+)))=3 [5]

ExamExercise

CambridgeIGCSEAdditionalMathematics0606Paper21Q3Nov2014 Functionsgandharedefinedby g(x)= 4e*-2 for x € IR, h(x)=In5xforx>0. a Find [2] b Solvegh(x)=18. [3]

CambridgeIGCSEand0LevelAdditionalMathematics

6Solvetheequation3^*=1000,givingyouranswerto2decimalplaces.

CambridgeIGCSEAdditionalMathematics0606Paper21Q4i,ii,iiiNov2013 a Giventhatlog4X= findthevalueofx. Solve 21og4y-log4(5y-12)= ^. [4][1]

CambridgeIGCSEAdditionalMathematics0606Paper11Q12bi,iiNov2013 Giventhatlog^X=5andlog^F=2,find a log.X^, [1] b log^ [1] C logxyA [2]

1 a Using the substitution y -5",show thatthe equation -5*"^'+2= 2(5*) can be writtenintheformay^+by+2=0,whereaandbareconstantstobefound. [2] Hence solve the equation 5^*^'- 5*^' + 2 = 2(5*). [4]

7ExpressIga+3Ig6-3asasinglelogarithm.

Exercise6.12

Thetemperature,T°Celsius,ofan object,tminutesafteritisremovedfrom aheatsource,is givenby T=55e-oi'+15. a Find the temperature ofthe objectattheinstantitisremovedfrom the heatsource. [1] b Find the temperature ofthe objectwhen f=8. [1] c FindthevalueoftwhenT-25. [3]

Chapter6:Logarithmicandexponentialfunctions

12 a i Sketch the graph of y=e''-5,showing the exact coordinates ofany points where the graph meets the coordinate axes. [3] ii Findtherangeofvaluesofkforwhichtheequatione"—5=khasnosolutions. [1] b Simplify log„^/2+log„8+log„ j,giving your answer in theform plog^2,where p ISaconstant, c Solvetheequationloggx-logg4x:=1. [4][2]

CambridgeIGCSEAdditionalMathematics0606Paper22QIOMar2015 13a Solvethefollowingequationstofindpandq. g,-l^22/-+!=4'' 9^-^X3'=81 [4] b Solvetheequationlg(3x-2)+lg(x+l)=2-Ig2. [5]

CambridgeIGCSEAdditionalMathematics0606Paper21Q5Nov2015

CambridgeIGCSEAdditionalMathematics0606Paper21Q3Jun2016

CambridgeIGCSEAdditionalMathematics0606Paper21Q8i,ii,iiijun2011

11Donotuseacalculatorinthisquestion. i Find the value of -log^ p^. ii Ill Show that =(igy)^, where y is a constant to be found,logs10 [1][1][2] iv Solve log^2x+log^3x:=log^600. [2]

CambridgeIGCSEAdditionalMathematics0606Paper21Q1Jun2015

10 a Write loggv x: as alogarithm to base 3. [2] b Given that log„y=3(log„15-log„3)+1,express yin termsofa. [3]

Chapter 7 Straight-line graphs Thissectionwillshowyouhowto: ■ solvequestionsinvolvingmidpointandlengthofaline ■ usetheconditionfortwolinestobeparallelorperpendicular ■ interprettheequationofastraight-linegraphintheformy=mx+c ■ transformgivenrelationships,includingy-ax"and31=0/?*,tostraight-lineformandhence determineunknownconstantsbycalculatingthegradientorinterceptofthetransformedgraph. .JliliiP'

Chapter?:Straight-linegraphs Youshouldalreadybefamiliarwiththefollowingcoordinategeometrywork:Lengthofaline,gradientandmidpoint Pisthepoint(Xj,y^)andQisthepoint yg)MisthemidpointofthelinePQ. The length of the line PQ — )^ +(y2 ~ 3*1 T ThegradientofthelinePQ*2—*1 The coordinates ofMare ^ ^ ^ ^ j. ¥* ■ .Q(*2'?2) M GradientsGradientsofparallellinesIftwolinesareparallelthentheirgradientsareequal.ofperpendicularlines gradientm gradient-i Ifalinehasagradientofm,alineperpendiculartoit hasagradientof-—. m equalgradients Thisrulecanalsobewrittenas: Ifthegradientsofthetwoperpendicularlinesarem\ andm^,thentojXm^=-\. Theequationofastraightline Theequationofastraightline\.sy=mx+cwherem=thegradientandc=they-intercept.

a FindthelengthofAB. b FindthemidpointofAB. Answers a(-3,7) tt (*i.yi) (6,-2) ix2,yz) decidewhichvaluestouseforxi,vi, >'2 AB = a/{3C2 - + U - hf =^(6--3)'+(-2-7f =9V2 b Midpoint = ^ f-3+67+-2 ~l 2 ' 2 =(1.5,2.5)

7.1 Problems involving length of a line and midpoint

You need to know how to apply the formulae for the midpointand the length ofalinesegmenttosolveproblems.

WORKEDEXAMPLE2

CambridgeIGCSEand0LevelAdditionalMathematics

WORKEDEXAMPLE1 Aisthepoint(-3,7)andBisthepoint(6,-2).

ThedistancebetweentwopointsP(7,a)andQ(a+1,9)is15. Findthetwopossiblevaluesofa. Answers (7,a) (a+1,9) tt t t (^i.yi) (j*^.3^) decidewhichvaluestouseforX|,y[,X9,j'l

Using Pd = >/(*2 - +(>2 -hf and PQ = 16. ^(a+1-if+(9-af =15 V(a-6)^ +(9-af =15 square both sides (a-6)^+(9-af=225 -12a+36+81-18a+ =225 collecttermsononeside 2a^-30a-108=0 dividebothsidesby2 a^-15a-54=0 factorise (a-18)(a+3)=0 solve a—18=0ora+3=0 Hencea=18ora=-3.

WORKEDEXAMPLE4

Chapter?:Straight-linegraphs

TheKEDIXAMPLE3coordinatesofthemidpointofthelinesegmentjoiningA{-5,11)andB(p,q), are(2.5,-6). Findthevalueofpandthevalueofq. Answers (-5,11) {p,q) ft ft decidewhichvaluestouseforSC],y^,x^, —_—^^—J midpoint=(2.5,-6). —5*4"p Equatingthe^coordinatesgives:—^=2.5 -b+p=bjb=10 Equatingthe3K;oordinatesgives: ^ =-6 114-9=-129=-23 Hence/>=10andq=-23.

ThreeoftheverticesofaparallelogramABCDareA(-10,1),5(6,-2)andC(14,4). a FindthemidpointofAC. b FindthecoordinatesofD. Answers a Midpointof AC = ~ ^•^)b LetthecoordinatesofZ)be{m,n).

SinceABCDisaparallelogram,themidpointofBDisthesameasthemidpointof AC. MidpointofBD 6+m -2+n]=(2,2.5) Equating the ^coordinates gives: ^^ = 2 —26+m=4m=-2-{-fiEquating the ^coordinates gives: —^— = 2.5 —2-t-w=5n=7

ShowthattriangleABCisaright-angledisoscelestriangle.

Thistrianglehassidesoflength5>/3cm,2-v/6cmand cm. Priyasaysthatthetriangleisright-angled. Discusswhethersheiscorrect. Explainyourreasoning.

CambridgeIGCSEand0LevelAdditionalMathematics

ThecoordinatesofDare(-2,7).D(-2.7)

CLASSDISCUSSION

1 Find the length ofthe line segmentjoining a(2,0)and(5,0) b(-7,4)and(-7,8) c(2,1)and(8,9) d(-3,1)and(2,13)e(5,-2)and(2,-6) f(4,4)and(-20,-3) g (6,-5)and(1,2) h (-3,-2)and(-1,-5) i (-7,7)and(5,-5).

C(14,4) A(-10,1) 5(6,-2)

4 The distance between two points P(10,2b) and -5)is 5-s/IO. Findthetwopossiblevaluesofb.

2CalculatethelengthsofthesidesofthetrianglePQRUseyouranswersto determine whetherornotthe triangle isright-angled. a P(3,11),Q(5,7),/?(ll,10) b P(-7,8),Q(-l,4),i?(5,12) c P(-8,-3),Q(-4,5),i?(-2,-6) 3 A(-l,0),fi(l,6)andC(7,4).

5ThedistancebetweentwopointsP(6,-2)andQ^{2a,a)is5. Findthetwopossiblevaluesofa.

Exercise7.1

6Findthecoordinatesofthemidpointofthelinesegmentjoining a (5,2)and(7,6) b (4,3)and(9,11) c(8,6)and(-2,10) d (-1,7)and (2,-4) e (-7,-8)and (-2,3) f (2a,-Sb)and (4a,5b).

WORKEDEXAMPLE5

Chapter?: Straight-line graphs

Thecoordinatesof3pointsareA{8 ft,2),B{-2,ft)andC(-8,2ft). FindthepossiblevaluesofkifA,BandCarecollinear. Answers IfA,BandCarecollinearthentheylieonthesameline.gradientofAB=gradientofBC k-2 2k-k -2-(8-k) -8-(-2) k-2 k crossmultiply k-\0 -6 -6(ft-2)=A(A-IO) expandbrackets -6k+I2=k^-lOA collecttennsononeside k^-4k-12=0 factorise (k-6){k+2)=0 solve ft—6=0or^+2=0 Henceft=6orft=-2.

9 The point P{2k, k) is equidistantfrom A(-2,4)and B{7,-5). Findthevalueofk.

10IntriangleABC,themidpointsofthesidesAB,BCandACareP(2,3),<2(3, 5)andR(-4,4)respectively.FindthecoordinatesofA,BandC. 7.2 Parallel and perpendicular lines Youneedtoknowhowtoapplytherulesforgradientstosolveproblems involvingparallelandperpendicularlines.

7ThecoordinatesofthemidpointofthelinesegmentjoiningP(-8,2)and Q{a,b),are(5,-3). Findthevalueofaandthevalueofb.

8 ThreeoftheverticesofaparallelogramABCDareA{-7,6), 8) andC(7,3). a FindthemidpointofAC. bFindthecoordinatesofD.

CambridgeIGCSEand0LevelAdditionalMathematics WORKEDEXAMPLES TheverticesoftriangleABCdneA(-4,2),B(5,-5)andC{k,k+2).FindthepossiblevaluesofkifangleACBis90°. Answers SinceangleACBis90°, gradientofACxgradientofBC=-1. (fe+2)-2 ^^{k+2)-{-5) k-(-4) k-5 k k+7 =-1 =-1 k+4 k-5 k{k+7)=-{k+4){k-5) k^+7k=-{k^-k-20) +7k=-k^+k+20 2k^+ -20=0 +3*-10=0 {k+5){k-2)=Q A+5=0orft-2=0 Hencek--5ork-2. Thetwopossiblesituationsare: A{-4,2) C{k,k+2)B{5,-5) y A(-4,2) y A(-4,2)^ C(2,4) X \ \ X C(-5^^3r^ 5(5,-5) 5(5,-5) Exercise7.2 1FindthegradientofthelineABforeachofthefollowingpairsofpoints, a A(1,2) 5(3,-2) b A(4,3) 5(5,0) c A(-4,4) 5(7,4) d A(1,-9) 5(4,1) e A(-4,-3) 5(5,0) f A(6,-7) 5(2,-4) 2Writedownthegradientoflinesperpendiculartoalinewithgradient d 1 1 -2i 2 as b-i c? 2 5 4 TwoverticesofarectangleABCDareA(3,-5)and5(6,-3). a FindthegradientofCD. b FindthegradientofBC.

You should already know that the equation ofa straight line is y=mx+c wherem=thegradientandc=they-intercept.

Chapter?:Straight-linegraphs

A(-1,-5),5(5,-2)and C(l,1). ABCDisatrapezium. ABis parallel to DCand angle BAD is 90°. FindthecoordinatesofD. The midpoint of the line segmentjoining P(-2,3) and (2(4,-1) is M. ThepointChascoordinates(-1,-2). ShowthatCMisperpendiculartoPQ. A(-2,2),5(3,-1)andC(9,-4).

Thereisanalternativeformulathatcanbeusedwhenyouknowthegradient ofthestraightlineandapointontheline.

CHALLENGEQ 9 A is the point(-2,0)and Bis the point(2,6). Find the pointCon the x-axissuch thatangle ABCis90°.

7.3 Equations of straight lines

Thecoordinatesof3pointsare A(-4,4),B{k,—2)and C{2k +1,—6). FindthevalueofkifA,BandCarecollinear. The verticesoftriangle ABCare A{—k,—2),B{k,—4)and C(4,k — 2). Find the possiblevaluesofkifangle ABCis90°.

Consideraline,with gradient m,which passesthrough the known point A(xj, yi) and whose general point is P{x, y). *P{x,y) y^) o

a Find the gradientofABand the gradientofBC. bUseyouranswertopartatodecidewhetherornotthepointsA,BandC arecollinear.

CambridgeIGCSEand0LevelAdditionalMathematics GradientofAP=m,hence———= m X — X. multiplybothsidesby{x-xi) y- yi = m{x-Xj). The equation of a straight line, with gradient m,which passes through the point(xj,yj)is: WORKEDEXAMPLE7 Findtheequationofthestraightline awithgradient2andpassingthroughthepoint(4,7) b passingthroughthepoints(-5,8)and(1,-4). Answers aUsingy-yi=m(x-Xi)withot=2, =4and =7 y-7y-7=2(x-4)=2*-8 y=2x-l b (-5,8) (1,-4) tt f t l*!.)'!) (X2,y2) decidewhichvaluestousefor Xi,31,x^,yz Gradient=m 2-x, l-(-5) =-2 Usingy-yi=m{x-*1)withm=-2,xi=-5and3;!=8. 31-8=-2{x+5) y-8=-2x-10y=-2x-2

a parallel to theline y = 2x+4, passing through the point(6,2) bparalleltothelinex+2y=5,passingthroughthepoint(2,-5) c perpendicular to the line 2x+Sy = 12, passing through the point(7,3) dperpendiculartotheline4x-y=&,passingthroughthepoint(4,-1).

MidpointofAB= =(5,6). So the perpendicular bisectoris the line passing through the point(5,6)with gradient _1 2" 1

2 Find the equation oftheline passing through a (3,2)and (5,7) b (-1,6)and (5,-3) c (5,-2)and (-7,4).

WORKEDEXAMPLE8

Findtheequationofthelinewith a gradient3andpassingthroughthepoint(6,5) b gradient-4and passing through the point(2,-1) c gradient -- and passing through the point(8,-3).

5 Find the equation of the perpendicular bisector of the line segment joining the points a (1,3)and(-3,1) b (-1,-5)and(5,3) c (0,-9)and(5,-2).

Gradientoftheperpendicularis—~•

Exercise7.3

Findtheequationoftheline

Chapter7:Straight-linegraphs

Find the equation ofthe perpendicular bisector ofthelinejoining A{S,2)and 5(7,10). yz-husing usingmjXw?2=-1 using ^1+^2 h+yz Answers GradientofAB- —-=—=2. 7-3 4

Usingy-y^=m(x-Xj)withXj=5,31,=6and m= 3-6= -|(x-5) 31=——X+8.5 2 x+231=17 multiply both sides by 2and rearrange

4 Pisthepoint(2,5)andQisthepoint(6,0). A line Iis drawn through Pperpendicular to PQto meet the 3!-axis at the pointR. a Find the equation ofthe line I. b Find thecoordinatesofthe pointR. c Find theareaoftriangle OPPwhere Oisthe origin.

bFindthecoordinatesofP. c FindthelengthsofACandBP.

b Find the equation oftheline through A which is perpendicularto theline /j.

Discusswithyourclassmates,howyoucanfindthearea oftriangleABC. Trytofindasmanydifferentmethodsaspossible. Compare the ease ofuse ofeach ofthese methods. y 9 8 7 6-54321 0012345678910^

The line li has equation Sx +2y = 12. Theline hasequation3*=2x-1.

Pisthefootoftheperpendicularfrom5toAC. a FindtheequationofBP.

The perpendicular bisector ofthe linejoining A(-l,4)and 5(2,2) intersectsthex-axisatPandthe3)-axisatQ. a FindthecoordinatesofPandofQ. b FindthelengthofPQ.

CHALLENGEQ 10

Thelinesliand^intersectatthepointA.

d Use your answers to partcto find the area oftriangle ABC. Q

Thecoordinatesofthree pointsare A(l,5),5(9,7)and C{k,-6).

a FindthecoordinatesofA.

CLASSDISCUSSION

MisthemidpointofA5andMCisperpendiculartoA5. a FindthecoordinatesofM.

ThecoordinatesoftrianglePQRare P(-3,-2),Q{5,10)and 5(11,-2).

a Find the equation ofthe perpendicular bisectorsof i PQ ii QR. b Find the coordinates ofthe pointwhich is equidistantfrom P,Q and 5. Note: Thepoint iswherethe perpendicular bisectorsofthe sidesintersect. 7.4 Areas of rectilinear figures

bFindthevalueofk. The coordinates of triangle ABCare A(2,-1), 7) and C(14, 5).

c Find the area oftriangle OPQwhere Ois the origin.

CambridgeIGCSEand0LevelAdditionalMathematics

aAnswers1

Thereisamethodthatyoumightnothaveseenbefore.Itisoftenreferredto as the'shoestring'or'shoelace' method.You do not have to know this method fortheexamination,butyoumayfinditusefultoknow. If the vertices of triangle A5Care A(xj,y,), 5(x2.y2) and C(x3,3)3), then: AreaoftriangleABC=-^^1^2+ -Hx^y-^-Xgyi-x^y^-*1^3 Thiscomplicatedformulacanbewrittenas; ji >3 y\ The productsin the direction are given positive signsand the productsin the direction are given negative signs. 3^1 ^8 -^3 Forthetriangleintheclassdiscussion 1 A(l,3),J5(9,1)and C(3,8): ^ area oftriangle ABC = ^|1x1+9x8+3x3-3x9-1x3-8x1 = -^11 +72-^9-27-3-81 =-|44| 2 =22units^ This method can be extended for use with polygons with more than3sides.

WORKEDEXAMPLE9

Chapter?:Straight-linegraphs

TheverticesofapentagonABCDEzreA(0,-1),5(5,1),C(3,4),5(-l,6)andC(-3,2). a Find the area ofthe pentagon using the'shoestring'method, b Find the area ofthe pentagon using the'boxingin' method. Areaofpentagon =-|0+20+18+(-2)+3-(-5)-3-(-4)-(-18)-0 =31.5units® Q Note: Ifyoutakethe verticesinan anticlockwise direction aroundashape, thentheinside ofthemodulus signwillbepositive.Ifyou intaketheverticesaclockwise direction,then theinsideofthe modulussign willbenegative.

3 Triangle PQPwhere P(l, 4) , Q{-S, -4) and R{7, k) is right-angled at Q, a Find the value of fe. b Find the area of triangle PQR. 4 Aisthepoint(-4,0)andBisthepoint(2,3). Mis the midpoint of the line AB. PointCissuchthatMC= • a FindthecoordinatesofMand C. b Show that CMis perpendicular to AB. c Find the area of triangle ABC. 5 Angle ABCis 90° and Mis the midpoint ofthe line AB. The point Clies on the 31-axis. a Findthe coordinatesofBand C. bFindtheareaoftriangleABC. A(8,3)

1Findtheareaofthesetriangles, a A(-2,3),B(0,-4),C(5,6)

Cambridge IGCSE and 0Level Additional Mathematics b Forthe'boxingin'method,youdrawarectangle aroundtheoutsideofthepentagon. .-7: ■ 4 / r 6 /I ^/ 5- /■ 1 r r 1 ^ ■1 /—! / 41 / / I 1 i\ 5 !|V 3- \i 14 2- \ i L\: —!\114.5 \ 1 1 X yO4 -5 -2 -IX p. 1 ^ 3^4 5 f Area ofpentagon = area ofrectangle - sum ofthe outside areas = (8x7)-(4+6+5+5+4.5) =56-24.5 =31.5units^ b P(-3,l), Q(5,-3), i?(2,4)

Exercise 7.4

2 Find the area of these quadrilaterals. a A(1,8),B(-4, 5), C(-2, -3), D{4, -2) b P{2, 7), Q{-5, 6), P(-3, -4), S{7, 2)

Aisthepoint(-2,0)andCisthepoint(6,4).ACandBDarediagonalsofthesquare,whichintersectatM.

TheperpendiculartothelineABatthepointAcrossesthe^j-axisatthe pointC. aFindthecoordinatesofC. bFindtheareaoftriangleABC. ABisparalleltoDCandBCisperpendiculartoAB. aFindthecoordinatesofC. bFindtheareaoftrapeziumABCD.

a FindthecoordinatesofM,BandD. bFindtheareaofABCD. Thecoordinatesof3ofthe verticesofaparallelogram ABCDare A(—4,3), 5(5,-5)andC(15,-1). a Find thecoordinatesofthe pointsofintersection ofthe diagonals, b Find the coordinates ofthe point D. CFindtheareaofparallelogramABCD. 7.5Convertingfromanon-linearequationto linearform Somesituationsin therealworld can be modelled using anequation. Consideran experimentwhere asimple pendulum,oflength Lcm,travels fromAto5andbacktoAinatimeofTseconds.Thetableshowsthetime taken,T,fordifferentlengthsL. 30

5(9,3) A(l,l)

D{S.5,8)

Chapter?: Straight-line graphs

L 5 10 15 20 25

ABCDisasquare.

T 0.45 0.63 0.78 0.90 1.00 1.10 B

Aisthepoint(-4,5)andBisthepoint(5,8).

Theydonotlieonastraightline.

The fact that points belief that

Sohowcanyoufindtheequationofthecurve?

the

IfyouaretoldthattheruleconnectingthevariablesTandLisbelievedtobe graph

the ruleconnectingthevariablesTandLisT=ayfL.

of Tagainst yfT.To do this,you must firstmakeatableofvaluesforTand yfZ. 41 2.24 3.16 3.87 4.47 5.00 5.48 T 0.45 0.63 0.78 0.90 1.00 1.10 T-—1 T T r 1 2 3 4 5 6 VI

30 L

T = a-JL,you should draw the

now lie on a straight line confirms the

Cambridge IGCSE and0LevelAdditional Mathematics WhenthegraphofTagainstLisdrawn,thepointslieonacurve.

Thenon-linearequationy=ax+—becomesthelinearequation: Method2 Y=mX+c,whereY=xy,X=x^,m=a and c=b b y=ax+

Intheexampleaboveyouhaveconvertedfromanon-lineargraphtoa lineargraph. Before you can model more complicated relationships,you mustfirstlearn howtochoosesuitablevariablesforYandXtoconvertfromanon-linear equationintothelinearformY=mX cwheremisthegradientofthe straightlineandcistheF-intercept.

Chapter?:Straight-linegraphs

Multiplyingbothsidesoftheequationbyxgives: xy=ax^+b Nowcomparexy=ax^+bwith Y-mX+c: xy= aIx^1+ b ;= i ^ t + c

WORKEDEXAMPLE10

The gradient ofthe straightline tells you the value ofa. TT• , , 1.1-0.45 Usingthetwoendpoints,gradient=—————=0.2. 5.48—2.24 Hence the approximate rule connecting Tand Lis T=O.2VI

ItisimportanttonotethatthevariablesXandFinF=mX-fcmustcontain onlytheoriginalvariablesxandy.Theymustnotcontaintheunknown constantsaandb.

Converty-ax+—^whereaandbareconstants,intotheformY-mX+c. X Answers Method1 y=ax+ X

Dividingbothsidesoftheequationbyxgives: y b ^=a+ y b Nowcompare-=a+—withY=mX+c\ X X t t t t Y= ,nX+ c

Thenon-linearequationy=ax+—becomesthelinearequation: X "V 1 Y=mX+c,whereY=^,X=—,m=bandc=a

WORKEDEXAMPLE11 whereaandbareconstants,intotheformY=mX+c. Answers -bx y-ae Takingnaturallogarithmsofbothsidesgives Iny=\na-bxIny=-bx+Ina

Iny=In(ae"*^In31=Ina+Ine"**

SimilarlyCambridgeIGCSEand0LevelAdditionalMathematicstheconstantsmandcmustcontainonlytheoriginalunknownconstantsaandb.Theymustnotcontainthevariablesxandy.

Converty=ae

NowcompareIny=—bx+InawithY=mX+c: (1^=-b(^+Ina t t T t Y = m X + c The non-linear equation y =ae"'* becomes the linear equation: Y=mX+c,whereY=\ny,X=x,m=-bandc=Ina Exercise7.5 1Converteachofthesenon-linearequationsintotheformY=mX-I-c,where aand b are constants.State clearly whatthe variables Xand Yand the constantsmand crepresent. (Note:theremaybemorethanonewaytodothis.) bb y=ax^—ay=ax^-fb c y=ax -bx d y{a-x)=bx g X=axy+by /— b e y=ayjx+—j= yjx 1 r- b h —a\JX I— y vx fy=^+b Converteach ofthese non-linear equations into the form Y= mX+ c, where a and b are constants.State clearly what the variables Xand Yand theconstantsmandcrepresent. (Note:theremaybemorethanonewaytodothis.) a y=10'"*+' b y=e°*"' d y=a¥ g a = e*^+^ ex^y'=e^hy=oe'* cy=ax' f xa^=b

Chapter?:Straight-linegraphs 7.6 Convertingfrom linearform to a non-linear equation EXAMPLET2 Findyintermsofx. a y b xy ^^^T6,8) ^1,7) > 2) ~d 0 X Answers aThelinearequationisF=mX+c,whereY =yand X=x^. r A- . 8-4 2Gradient=m= = 6-0 3 F-intercept=c-4 C) HenceF=-X+4. 3 2 Thenon-linearequationisy=—x^+4. bThelinearequationisF=mX+c,whereF 2-7 Gradient =m= 6-1=-1 Using F=mX+c,m——\,X=&andF=2 =xyandX=x. 2=-1X6+c ii CO Hence Y=-X+8. Thenon-linearequationis =-x+8,8 =—1H X

Nowcomparelg)i=xIg6+IgavrithY=mX+c: )=lg/50+lg« 1 ft t Y X + 13-5 Gradient=m= =2 6-2 Igb=2 b=lO^ Using Y=mX+c,m=2,X=2andF= 5=2X2+t: c=1 Iga=1 a=10^ Hence a=10andb=100.

(6,13

Variablesxandyaresuchthaty=aXb",where aandbareconstants. ThediagramshowsthegraphofIg)!againstx, passingthroughthepoints(2,5)and(6,13). Findthevalueofaandthevalueofb. Answers y=ax b" Takinglogarithmsofbothsidesgives: lg3i=lgalg3i=Ig(aX6^)+lgi'Ig^i=xlgZ)+Iga

Cambridge IGCSE and0LevelAdditional Mathematics

WORKEDEXAMPLE13

Chapter?: Straight-line graphs Exercise7.6 1Thegraphsshowpartofastraightlineobtainedbyplottingyagainstsome functionofx. Foreachgraph,expressyintermsofx. O (1,5) (5,3) (4,-1)2 2 Foreachofthefollowingrelations i expressyintermsofx O (5,6) O ii findthevalueofywhenx=2. c xy (6,15) (1,5) O f .(2,6) fy \i2,9) X+y /ao,8) ylx -^5,6) y(6,1) / ^1.2)0 / x^ /(2,-2) 0 0 X

a Express Ig y in terms of x.

Variablesxandyarerelatedsothat,whenIgyisplottedontheverticalaxisandxisplottedonthehorizontalaxis,astraight-linegraphpassing throughthepoints(6,2)and(10,8)isobtained.

CambridgeIGCSEand0LevelAdditionalMathematics

aExpressy'^intermsof2*.

b Find the value of xand the value ofysuch that — = 3. X

6 Variablesxandyarerelatedsothat,whene''isplottedonthevertical axisandx^isplottedonthehorizontalaxis,astraight-linegraphpassing throughthepoints(3,4)and(8,9)isobtained.

aExpressyintermsofx,givingyouranswerintheformy=axx^. b Find the value ofX when y =51.2.

Variablesxandyarerelatedsothat,whenInyisplottedontheverticalaxisandInxisplottedonthehorizontalaxis,astraight-linegraphpassing throughthepoints(1,2)and(4,11)isobtained.aExpressInyintermsofx.

bExpressyintermsofx,givingyouranswerintheformy=ax10**.

10Variablesxandyaresuchthat,when Iny isplottedontheverticalaxisand Inx isplottedonthehorizontalaxis,astraight-linegraphpassingthrough thepoints(2.5,7.7)and(3.7,5.3)isobtained.

b FindthevalueofXwhen)!=11.

Variablesxandyarerelatedsothat,when—isplottedonthevertical X axisandxisplottedonthehorizontalaxis,astraight-linegraphpassing throughthepoints(2,4)and(5,-2)isobtained. a Expressyintermsofx.

b Expressyintermsofx.

a FindthevalueofIny when Inx=0. bGiventhaty=axx*,findthevalueofaandthevalueofb.

8 Variablesxandyarerelatedsothat,whenIgyisplottedonthevertical axisandIgxisplottedonthehorizontalaxis,astraight-linegraphpassing throughthepoints(4,8)and(8,14)isobtained.

3 Variables xand y are related so that,when is plotted on the vertical axis X and x^is plotted on the horizontalaxis,astraight-line graph passing through(2,12)and(6,4)isobtained. Expressyintermsofx.

4Variablesxandyarerelatedsothat,wheny'^isplottedontheverticalaxis and 2* is plotted on the horizontal axis,a straight-line graph which passes throughthepoint(8,49)withgradient3isobtained.

a Expresse^ intermsofx. b Expressyintermsofx.

Answers a takelogsofbothsides usethemultiplicationlaw usethepowerlawlgy=lg{ky=kxx"Xx")lgy=lgk+\gx''\gy=nlgx+Ig^ NowcompareIgy=nlgx+lgkwithY=mX+c\ (g)= n(^+ Ig k t f t t Y = m X + c •HencethegraphofIgyagainstIgxneedstobedrawnwheregradient=n•interceptonverticalaxis=Igk.

anon-linearequation. In this section,you will learn how to apply what you havejustlearntin sections 7.5and7.6tofindthenon-linearequationconnectingtwovariables. WORKEDEXAMPLE14 X 5 10 20 40 80 y 2593 1596 983 605 372 Thetableshowsexperimentalvaluesofthevariablesxandy.

Chapter?:Straight-linegraphs 7.7 Finding relationshipsfrom data

is easy to establish theconnectionusingtheequationy-mx c.

Ifthe

Whenexperimentaldataiscollectedfortwovariables,itisusefulifyoucanthenestablishthemathematicalrelationshipconnectingthetwovariables. dataforms straight-line when graph is plotted,it Itismoreusual,however,forthedatatolieonacurveandtobeconnectedby aByplottingasuitablestraight-linegraph,showthatxandyarerelatedbythe equationy=kXx",wherekandnareconstants, bUseyourgraphtoestimatethevalueofkandthevalueofn.

CambridgeIGCSEand0LevelAdditionalMathematics Tableofvaluesis Igx 0.699 1.000 1.301 1.602 1.903 Igy 3.414 3.203 2.993 2.782 2.571 b Thepointsformanapproximatestraightline,soxandyarerelatedhythe equationy=kXx". n=gradient _ 2.571-3.414 ~1.903-0.699 «-0.7 Ig lgA=interceptonverticalaxis=3.9 k=10^-9 k=7943

Chapter?: Straight-line graphs WORKEDEXAMPLE15 X 2 4 6 8 10 y 4.75 2.19 1.42 1.05 0.83 whereaandh Thetableshowsexperimentalvaluesofthevariablesxandv. a+bx Thevariablesareknowntoberelatedbytheequationy——— ■. are constants. a Drawthegraphof againstx. b Useyourgraphtoestimatethevalueofaandthevalueofb. Analternatemethodforobtainingastraight-linegraphfortheequation a+bx 1 y= —istoplotxyontheverticalaxisand—onthehorizontalaxis. c Withoutdrawingasecondgraph,estimatethegradientandtheinterceptonthe vertical axis of this graph. Answers a First make a table ofvalues for of x^y and x: X 2 46 8 10 2 x^y 19 35.0 51.1 67.2 83 The graph of x^y against xis; x-v 90 80 70 60 50 40 30 20 10 O > / 1 2 3 4 6 7 8 9 10 b Using y = a+bx x^y = bx + a Now compare x^y = bx-¥ a with Y = mX + c: (^x^= b (^+ (I t t t t F = X 4- c 83-19Hence, b=gradient= =8 anda=y-intercept=3 10 0=3and6=8 C Using -2 a + bx x2 a + bx y xy a xy=—+ X aNowcompare xy =—-i-6 with Y=mX+c: X(J)= «X0+ 6 t t t t Y=m X-1- c Gradient=o=3andinterceptonverticalaxis=6=8.

CambridgeIGCSEand0LevelAdditionalMathematics Exercise7.7 1 0.5 1.0 1.5 2.0 2.5 y 1.00 3.00 3.67 4.00 4.20

0.189 0.200

The table shows experimental values ofthe variables xand y. aDrawthegraphofxyagainstx^. b Useyourgraphtoexpressyintermsofx. c Findthevalueofxandthevalueofyforwhichxy=12.288.

0.4 0.5

Thetableshowsexperimentalvaluesofthevariablesxandy. a Copyandcompletethefollowingtable. X y1 b Draw the graph of- against —. y X c Expressyintermsofx. Findthevalueofxwheny=0.16. X 1 2 3 4 5 y 12.8 7.6 6.4 6.2 6.4

Thetableshowsexperimentalvaluesofthevariablesxandy.aCopyandcompletethefollowingtable. X xy b Drawthegraphofxyagainstx. c Expressyintermsofx. Findthevalueofxandthevalueofyforwhichxy=2. X 0.1 0.2 0.3 y 0.111 0.154 0.176

5 X 10 100 1000 10000 y 15 75 378 1893

The table shows experimental values ofthe variables xand y.

Thetableshowsexperimentalvaluesofthevariablesxandy.

bUseyourgraphtoestimatethevalueofWqandk. cEstimatethevalueofmwhent=27.

Thetablebelowshowsexperimentalvaluesoftandm. t 10 20 30 40 50 m 40.9 33.5 27.4 22.5 18.4 a DrawthegraphofInmagainstt.

Thevariablesareknowntoberelatedbytheequationy=axb"where aandbareconstants, a DrawthegraphofIgyagainstx. b Useyourgraphtoestimatethevalueofaandthevalueofb. 7 X 2 4 6 8 10 y 4.9 13.3 36.2 98.3 267.1

Thetableshowsexperimentalvaluesofthevariablesxandy. Thevariablesareknowntoberelatedbytheequationy=axe""wherea andnareconstants, a DrawthegraphofInyagainstx.

b Useyourgraphtoestimatethevalueofaandthevalueofn.

Chapter?:Straight-linegraphs

Thevariablesareknowntoberelatedbytheequationy=kx"wherekand nareconstants. a DrawthegraphofIgyagainstIgx. b Useyourgraphtoestimatethevalueofkandn. 6 X 2 4 6 8 10 y 12.8 32.8 83.9 214.7 549.8

Themass,mgrams,ofaradioactivesubstanceisgivenbytheformulariQe"'",wheretisthetimeindaysafterthemasswasfirstrecordedand m mo andkareconstants.

b Use your graph to estimate the value ofaand the value of b. C Estimatethevalueofxwheny=50. X 2 4 6 8 10 0.10 0.33 1.08 3.48 11.29

10 X 0.2 0.4 0.5 0.7 0.9 y 36 12 9 6 4.5

Thetableshowsexperimentalvaluesofthevariablesxandy.Thevariablesareknowntoberelatedbytheequationy-10°xb",where aandbareconstants. a DrawthegraphofIgyagainstx.

b Use your graph to estimate the value ofaand the value of b. c Estimatethevalueofxwheny=5.

Thetableshowsexperimentalvaluesofthevariablesxandy.Thevariablesareknowntoberelatedbytheequationy= X+b ,wherea andbareconstants, a Drawthegraphofyagainstxy. b

Thetableshowsexperimentalvaluesofthevariablesxandy. a DrawthegraphofIny againstInx. b Express yin terms ofx. Analternatemethodforobtainingtherelationshipbetweenxandyisto plot Igy ontheverticalaxisand Igx onthehorizontalaxis. C Withoutdrawingasecondgraph,findthegradientandtheinterceptontheverticalaxisofthisgraph.

CambridgeIGCSEand0LevelAdditionalMathematics X 2 4 6 8 10 y 30.0 44.7 66.7 99.5 148.4

The table shows experimental values ofthe variables x and y. Thevariablesareknowntoberelatedbytheequationy= wherea andbareconstants. a DrawthegraphofInyagainstx.

AnUseyourgraphtoestimatethevalueofaandthevalueofb.alternatemethodforobtainingastraight-linegraphfortheequation y= C Withoutdrawingasecondgraph,estimatethegradientandthe a 1 IStoplotXontheverticalaxisand-onthehorizontalaxis. X+b y interceptontheverticalaxisofthisgraph. 11 X 2 5 15 25 60 y 11.5 5.54 2.30 1.53 0.76

Ifalinehasagradientofm,alineperpendiculartoithasagradientof

SolutionsWorkedexampletothisquestionbyaccuratedrawingwillnotbeaccepted.

O m,5) A(l,4)

ThediagramshowsaquadrilateralABCDinwhichAisthepoint(1,4),andBisthepoint(6,5).AngleABCisarightangleandthepointCliesonthex-axis.ThelineADisparalleltotheyandthelineCDisparalleltoBA.Find axis a theequationoftheline CD, btheareaofthequadrilateralABCD.

TheIfthegradientsofthetwoperpendicularlinesare?Wjandm^,thenwijxmg=-I.equationofastraightline y=mx+cwherem=thegradientandc=they-intercept. 3)-jij = »w(x-JCj) where m =the gradientand (xj.jii) isaknown pointon theline.

. m

[5] [4] CambridgeIGCSEAdditionalMathematics0606Paper21Q10i,iiNov2010

Non-linearequations

To converta non-linear equation involving xand yinto a linear equation,express the equation in theformY=mX-I-c,whereXandYareexpressionsinxand/ory. Examination questions

Chapter?:Straight-linegraphs Summary Lengthofalinesegment,gradientandmidpoint Length ofPQ = yj(x2" +(y^")'i)^ 3*2~yiGradientofPQ X QiXry^^ i M MidpointofPQ= X\+*2 3*1 3*2 P(Xi,yi)2 2 Paralleland perpendicularlines Iftwolinesareparallelthentheirgradientsareequal. 1

CambridgeIGCSEand0LevelAdditionalMathematics Answers a GradientofAB 5-46-1 GradientofEC=-5 EquationofEC: y 1_ 5 When3>=0, Cisthepoint(7,0) EquationofCD: EquationofCDis — y\ = m(x-Xj) -5=-5(x-6) 3)-5=30-5x3=35-5x 0=35-5x X=7 31-3)]= m(x-Xj) ,-0 = i(*-7) j= ^(«-7) usem-{m2 1tofindgradientofEC usem=-5,Xj=6andyi=5 use m = —, =7and >>1 = 0 5 b ADis parallel to the j-axis,hence the x-coordinate ofDis 1. Substitutingx=1intotheequation3'=—(x-7)givesy=-1.2. 5 Use the 'shoestring' method with A(l,4), E(6,5), C(7,0)and D(l,-1.2). AreaofAECD= 2 1 X 5 X 0 x.,,x 4 =-|5+0+(-8.4)+4-24-35-0-(-1.2)1-1-57.21 2' ' =28.6units^ Exercise7.8 Examexercise 1 ThepointPliesonthelinejoiningA(-2,3)and5(10,19)suchthatAP-.PE-1:3. a Showthatthex-coordinateofPis1andfindthe3)-coordinateofP. b TheFindtheequationofthelinethroughPwhichisperpendiculartoAE.linethroughPwhichisperpendiculartoAEmeetsthe31-axisatthepointQ. C FindtheareaofthetriangleAQB. [2][3][3] CambridgeIGCSEAdditionalMathematics0606Paper11Q8Nov2014

CambridgeIGCSEAdditionalMathematics0606Paper11Q10i,ii,iiiNov2013

FindthevalueofAandofb. [5] CambridgeIGCSEAdditionalMathematics0606Paper11Q2Jun2013

CambridgeIGCSEAdditionalMathematics0606Paper11Q8i,ii,iiiJun2014

aByplottingasuitablestraightlinegraph,showthatTandparerelatedbytheequation p=kV",where/Eandnareconstants. [4] Useyourgraphtofind b thevalueofn, [2] cthevalueofpwhenV=35. [2]

Solutionstothisquestionbyaccuratedrawingwillnotbeaccepted. The points A(-3,2) and B{1,4) are vertices ofan isosceles triangle ABC,where angle B = 90°. a FindthelengthofthelineAB. [1] b FindtheequationofthelineBC. [3] c FindthecoordinatesofeachofthetwopossiblepositionsofC. [6]

Variablesxandyaresuchthaty=Ab",whereAandbareconstants.ThediagramshowsthegraphofInyagainstx,passingthroughthepoints(2,4)and(8,10). Iny O (8,10) (2,4)^

Chapter?:Straight-linegraphs 2 ThetableshowsvaluesofvariablesVandp. V 10 50 100 200 p 95.0 8.5 3.0 1.1

CambridgeIGCSEand0LevelAdditionalMathematics5Solutionstothisquestionbyaccuratedrawingwillnotbeaccepted. 5(7,7) A(ll,4)

CambridgeIGCSEAdditionalMathematics0606Paper21QIOJun2012 Thetableshowsexperimentalvaluesoftwovariablesxandy. X 1 2 3 4 5 y 3.40 2.92 2.93 3.10 3.34

7 ThepointsAandBhavecoordinates(-2,15)and(3,5)respectively. TheperpendiculartothelineABatthepointA{-2,15)crossesthey-axisatthepointC. FindtheareaoftriangleABC. [6]

Itisknownthatxandyarerelatedbytheequationy= +ox,whereaandbareconstants, a Completethefollowingtable. y4x [1] b On a grid plot y4x against xyfx and draw astraightline graph. [2] c Useyourgraphtoestimatethevalueofaandofb. [3] d Estimatethevalueofywhenxis1.5. [1] CambridgeIGCSEAdditionalMathematics0606Paper21Q9i-ivNov2011

CambridgeIGCSEAdditionalMathematics0606Paper11Q7Jun2011 V.

The diagram shows a trapezium A5CZ)with vertices A(11,4), 5(7,7), C(-3, 2) and D. ThesideADisparalleltoBCandthesideCDisperpendiculartoEC. FindtheareaofthetrapeziumABCD. [9]

Solutionstothisquestionbyaccuratedrawingwillnotbeaccepted. ThepointsAandBhavecoordinates(2,-1)and(6,5)respectively.

Solutionstothisquestionbyaccuratedrawingwillnotbeaccepted. Two points A and JShave coordinates(-3,2)and(9,8)respectively. FindthecoordinatesofC,thepointwherethelineABcutsthey-axis.FindthecoordinatesofD,themid-pointofAB.FindtheequationoftheperpendicularbisectorofAB.

CambridgeIGCSEAdditionalMathematics0606Paper21Q8Nov2015

ThepointChascoordinates(10,-2).

CambridgeIGCSEAdditionalMathematics0606Paper11Q8i,iiNov2010

1 Find the equation ofthe perpendicular bisector ofAB,giving your answer in the form ax+by=c,wherea,bandcareintegers.

Chapter?: Straight-line graphs

8 Variables x and y are such that,when In3; is plotted against Inx, a straightline graph passing throughthepoints(2,5.8)and(6,3.8)isobtained. (6,3.8) (2,5.8)

CambridgeIGCSEAdditionalMathematics0606Paper22Q8Mar2015

ii Find theequation ofthelinethrough Cwhichis parallelto AB. [2]

[3][1][2][1] [3]

iii Calculate the length ofBC. [2] iv Show that triangle ABCis isosceles. [1]

CambridgeIGCSEAdditionalMathematics0606Paper11Q5Jun2015

V ShowthattheareaoftriangleABEisfourtimestheareaoftriangleECD.

10 The curve y = xy + s? — 4: intersects the line y =^x—\ at the points A and B. Find the equation ofthe perpendicular bisector ofthe line AB. [8]

The perpendicular bisector ofABcuts the y-axis at the pointE. iv FindthecoordinatesofE.

a b [4] Inx Find thevalueofIn3* when Inx =0. [2] Given that y= Ax'',find the value ofAand ofb. [5]

! Chapters Circular measure Thissectionwillshowyouhowto: m useradianmeasure M solveproblemsInvolvingthearclengthandsectorareaofacircle.

Itfollowsthatthecircumference(anarcoflength27tr)subtendsanangleof 2kradiansatthecentre.

27Cradians=360°nradians=180° WhenanangleiswrittenintermsofK,theradiansymbolisusuallyomitted. Hence,K=180°.

Haveyoueverwonderedwhythereare360degreesinonecomplete revolution?

• ancientastronomersclaimedthattheSunadvancedinitspathbyone degreeeachdayandthatasolaryearconsistedof360days

Chapters:Circularmeasure 8.1Circularmeasure

Theoriginalreasonforchoosingthedegreeasaunitofangularmeasureisactuallyunknownbutthereareanumberofdifferenttheories:

• theancientBabyloniansdividedthecircleinto6equilateraltrianglesandthensubdividedeachangleatOinto60furtherparts,resultingin 360divisionsinonecompleterevolution

•Degrees360hasmanyfactorswhichmakesdivisionofthecirclesomucheasier.arenottheonlywayinwhichyoucanmeasureangles.Inthissection youwilllearnhowtouseradian

measure.Thisissometimesreferredtoas thenaturalunitofangularmeasureanditisusedextensivelyinmathematicsbecauseitcansimplifymanyformulaeandcalculations. Inthediagrambelow,themagnitudeofangleAOBisIradian(1radianis writtenasIradorP).

AnarcequalinlengthtotheradiusofacirclesubtendsanangleofIradianatthecentre.

2 4

CambridgeIGCSEand0LevelAdditionalMathematics

Convertingfromdegreestoradians

Anglesthatarenotsimplefractionsof180°canbeconvertedusingthe followingrule:Tochangefromdegreestoradians,multiplyby Since7C=180°,--30°,—=18°etc.

71 Convertingfromradianstodegrees

Since180°=7C,then90°=—,45°=—etc.

6 10 followingAnglesthatarenotsimplefractionsof7tcanbeconvertedusingtherule: Tochangefromradianstodegrees,multiplyby 180 n 180 (Itisusefultorememberthat1radian=1x =57°.) Tt WORKEDEXAMPLE1 aChange60°toradians,givingyouranswerintermsof7t. 371 bChange—radianstodegrees. Answers a Method1:180°=7tradiansradians flBOy _ 71 I 3 J "3 60°=— radians 3 b Method1: 7tradians=180° 7t radians=36° — radians=108° 5 Method2: 60 =f60x—1 I 180; 60°=— radians 3 radians Method2:37C —radians 5 371radians=108°

Chapter8:Circularmeasure Exercise8.1 1 Changetheseanglestoradians,intermsofTt. a 10° b20° c o O d o O e 15° f 120° g135° h 225° i 360° j 720° k80° I300° m 9° n 75° 0 210° Changetheseanglestodegrees. Tt 'i b ^ 6 c Tt 12 d Tt 9 e 2Tt3 ^ 47T 5 In ^ w h 5Tt Ti" i 3Tt20 j 9Tt To , fijt k 5 \ 37t m In T n 8Tt T 0 9Tt Y Writeeachoftheseanglesin radianscorrectto3sf. a32° b55° c 84° d 123° e 247° Writeeachoftheseanglesindegreescorrectto1decimalplace, al.Srad b2.5rad c 1.02rad dl.SSrad e0.58rad Copyandcompletethetables,givingyouranswersintermsof71. a Degrees 0 45 90 135 180 225 270 315 360 Radians 0 7t 27C Degrees 0 30 60 90 120 150 180 210 240 270 300 330 360 Radians 0 71 271 6 Useyourcalculatortofind asinl.Srad b tanO.Srad 7t sm Tt COS C sinl.2rad , Tt f tan—. 4 Note: Youdonotneedtochangetheangletodegrees.Youshouldsettheanglemodeonyour calculatortoradians. CHALLENGEQ 7 AnnaistoldthesizeofangleBACindegreesandsheisthenaskedtocalculatethelengthofthelineEC.She useshercalculatorbutforgetsthathercalculatorisin radianmode.Luckilyshestillmanagestoobtainthe correctanswer.GiventhatangleBACisbetween10° ofand15°,usegraphingsoftwaretohelpyoufindthesizeangleBACindegreescorrectto2decimalplaces.

WORKEDEXAMPLE21 —radiansatthecentreofacirclewithradius8cm.

CLASSDISCUSSION

Ifyouknowtheradius,rcm,andtheangle9(indegrees)atthecentre •ofthecircle,describehowyouwouldfind:arclength•areaofsector • perimeterofsector • lengthofchord • perimeterofsegment • areaofsegment. 8.2 Length

Youshouldalreadybefamiliarwiththefollowingmathematicalwordsthatareusedincirclequestions. arc sector minorsegmentandmajorsegment. of atthecentreofthecircleisoflengthr.Hence,ifanarcsubtendsanangleof6radiansatthecentre,thelengthofthearcisrd.

Findthelengthofthearcin 0 termsofit. Answers Arclength=rd =8x-6 4jt =—cm 3

an arc Fromthedefinitionofaradian,thearcthatsubtendsanangleof1radian

Arclength=rd

CambridgeIGCSEand0LevelAdditionalMathematics

chord

segment Discussandexplain,withtheaidofdiagrams,themeaningofeachofthesewordsExplainwhatismeantby; • minorarcandmajorarc • minorsectorandmzyorsector •

Anarcsubtendsanangleof 7t

Asectorhasanangleof2radiansandanarclengthof9.6cm. Findtheradiusofthesector. Answers Arclength=rO 9.6=rx2 r=4.8cm 12cm Thecirclehasradius5cmand centreO. PQisatangenttothecircleatthe pointP. QROisastraightline.Find a anglePOQinradians 5cm c theperimeteroftheshadedregion. Answers triangleQPOisright-angledsincePQisatangenttanPOQremembertohaveyourcalculatorinradianmodeAnglePOQ=tan =1.176... =1.18radians b 0Q2=122+52 0Q2=169 OQ=lS HenceQR=8cm. usingPythagoras usearePR=rOc Perimeter=PQ+QR+arcPR =12-h8+(5X1.176...) =25.9cm 175 WORKEDEXAMPLE3 WORKEDEXAMPLE4 Chapters: Circularrtieasure Exercise8.2 1 Find,in terms ofn,the arc length ofa sector of a radius6cm and angle ^ b radius 5cm and angle ^ Sn c radius10cmandangle 5k dradius18cmandangle—. 6

4cm 5 ABCDis a rectangle with AJ5=6cm and BC=16cm. OisthemidpointofBC. OAEDisasectorofacircle,centreO.Find a AO b angleADD,inradians ctheperimeteroftheshadedregion. Find a thelengthofarcAB b thelengthofchordAB ctheperimeteroftheshadedsegment. TriangleEFGisisosceleswithEG=EG=16cm. OHisanarcofacircle,centreE,withangleHEG=0.85 radians.Find a thelengthofarcGH b thelengthofEE c the perimeter ofthe shaded region. 3rad 16cm 16cm

Findthearclengthofasectorof a radius 8cm and angle 1.2 radians b radius 2.5cm and angle 0.8 radians. Find,inradians,theangleofasectorof a radius 4cm and arc length 5cm b radius 9cm and arc length 13.5cm. Findtheperimeterofeachofthesesectors.

CambridgeIGCSEand0LevelAdditionalMathematics

Chapters: Circularmeasure 8.3 Area ofasector Tofindtheformulafortheareaofasectoryouusetheratio: area ofsector angle in the sector area ofcircle complete angle atthe centre When6ismeasuredinradians,theratiobecomes areaofsector 2% e %r areaofsector= 2n Xjcr^ areaofsector=~r^d 2 WORKEDEXAMPLE5 271 Findtheareaofasectorofacirclewithradius6cmandangle radians. Giveyouranswerintermsof7t. Answers Areaofsector ■r^e 1 -2 2Tt -X6''X 2 3 =123tcm'' WORKED 6 i 2rad \Calculatetheareaoftheshadedsegmentcorrect to3sf. 1 8 cimx^ cm \ V ° J Answers -t Area oftriangle A0J5 = - x 8 x 8 x sin(2'^) using area oftriangle - - a/^sinC ==29.0975...29.0975... 1 1Area ofsector AOB = — x8x8x2 using area ofsector = „ ^ = 64 Areaofsegment=areaofsectorAOB-areaoftriangleAOB =64-29.0975... = 34.9cm2

CambridgeIGCSEand0LevelAdditionalMathematics

Exercise8.3

7Thecirclehasradius12cmandcentreO. PQisatangenttothecircleatthepointP.

QROisastraightline.Find a anglePOQ,inradians b theareaofsectorPGR c theareaoftheshadedregion. 8 AOBisthesectorofacircle,centreO,radius8cm. ACis a tangent to the circle atthe point A. CBOis a straightline and the area ofsector AOBis 32cm^. Find a angleAOB,inradians btheareaoftriangleAOC c theareaoftheshadedregion. 35cm

OAEDisasectorofacircle,centreO.Find a AO b angleADD,inradians c theareaoftheshadedregion.

4 POQisthesectorofacircle,centreO,radius10cm. The length ofarc PQis8cm.Find a anglePOQ,in radians b the areaofthesectorPOQ. 5 Asectorofacircle,radiusrcm,hasaperimeterof150cm. Findanexpression,intermsofr,fortheareaofthesector. 6

2Findtheareaofasectorof a radius4cm and angle 1.3radians b radius3.8cm and angle 0.6radians.

3Find,inradians,theangleofasectorof a radius 3cm and area5cm^ b radius 7cm and area 30cm^.

ABCDisarectanglewithAJ5=9cmandPC=18cm.OisthemidpointofBC.

Find,intermsof7t,theareaofasectorof a radius6cm and angle ^ b radius 15cm and angle ^ 5 C radius 10cm and angle 771 10 Sxtd radius9cmandangle—. ^ 6

9 TriangleEFGisisosceleswithEG=EG=9cm. GEIisanarcofacircle,centreE,withangle aHEG=0.6radians.FindtheareaofsectorofHEG btheareaoftriangleEEG c the area ofthe shaded region.

12AOBisasectorofacircle,centreO,withradius9cm.

FOG=1.2radians. EOHisasectorofacircle,centreO,withradius5cm. The shaded region hasan area of71.4cm^. Find the perimeteroftheshaded region. O ■ 97ccm 9Dcm 5cm

10Thediagramshowsacircle,centreO,radius12cm.

Theshadedregionhasaperimeterof18cm.Find a angleAOD,inradians b the area ofthe shaded region.

Angle GOD=0.5 radiansand angle ODGis arightangle. OC=5cm.Find a OD bCD c the perimeter ofthe shaded region d the area ofthe shaded region.

13EOGisasectorofacircle,centre O,with angle

11AODisasectorofacircle,centreO,radius4cm. BOGisasectorofacircle,centreO,radius10cm.

Chapters:Circularmeasure

AngleAOB=0radians. ArcAB-97tcm. aShowthat0= 371 b Find the area ofthe shaded region.

14Thediagramshowsasemi-circle,centreO,radius10cm. FHisthearcofacircle, centreE. Findtheareaof a triangleEOF cbsectorFOGsectorFEE[ d theshadedregion. HG CHALLENGEQ

15Thediagramshowsacircleinscribedinsideasquare ofsidelength10cm.Aquartercircleofradius10cm isdrawnwiththevertexofthesquareasitscentre. Findtheshadedarea. SummaryOneradian(l'^)is the size ofthe anglesubtended atthe centre ofa circle,radius r, by an arc oflength r.

When6ismeasuredinradians: thelengthofarcAB=rd theareaofsectorAOB=—r^O

CambridgeIGCSEand0LevelAdditionalMathematics

CHALLENGEQ

[2] [6][4]

Examination questions

Answer

cos0.9 OA-12=7.3047. 7.3to1dp Perimeter=arc CDE+EB+BA+AC arcCDE:arcCDE=id = 12X(271

EB=OB-0E=19.3047-12=7.3047...

AC= 22 OA 12

The diagram shows an isosceles triangle A05and a sector OCDE'Oofa circle with centre O. ThelineABisatangenttothecircle. AngleAOB=1.8radiansandtheradiusofthecircleis12cm. a Show thatthe distance AC=7.3cm to 1 decimalplace, b Find the perimeter oftheshaded region, c Findtheareaoftheshadedregion. CambridgeIGCSEAdditionalMathematics0606Paper21Q11Jun2011 a cos0.9=0A= 19.3047... -1.8) cos0.9=0B= henceneedtofindarcCDE,EBandBA where Ois the angle in the major sector = 2n- 1.8 using half ofthe isosceles triangle =19.3047. cos0.9 BA: BA = 2x(0Ax sin0.9)= 30.24378... =53.798...+7.3047...+30.24378...+7.3047. =98.7cm

Workedexample

Chapters:Circularmeasure

=53.798... 22 OB 12 EB:

Perimeter=arcCDE+EB+BA+AC

CambridgeIGCSEand0LevelAdditionalMathematics c areaofshadedregion=areaoftriangleOAB+areaofm^orsectorCDE =—absinC+—r^6 2 2 = i 19.3072 X sin1.8 +-X12^ x (2Tt -1.8) 2 2 =504cm^ Exercise8.4ExamExercise 1

CambridgeIGCSEAdditionalMathematics0606Paper21 QlOi,ii Nov 2013 V.

O R xcm-

CamlmdgeIGCSEAdditionalMathematics0606Paper21Qlla,bNov2014

The diagram shows a sector OPQofa circle with centre O and radius x cm. Angle POQis 0.8 radians. The point Slies on OQsuch that OS =5cm. The point Plies on OPsuch that angle OPSis a right angle. Given that the area of triangle OPSis one-fifth of the area ofsector OPQj,find a the area ofsector OPQin terms of x and hence show that the value of x is 8.837 correct to4significantfigures, btheperimeterofPQSR, c theareaofPQSR.

The diagram shows a circle with centre O and a chord AB. Theradiusofthecircleis12cmandangleAGEis1.4radians, a Find the perimeter ofthe shaded region. b Findtheareaoftheshadedregion.

CambridgeIGCSEAdditionalMathematics0606Paper11Q8i,ii,iiiJun2013 12cm

Chapters:Circularmeasure 6cm 16cm I6cm

ThediagramshowsasquareABCDofside16cm.Misthemid-pointofAB.ThepointsEandFareonADandBCrespectivelysuchthatAE=BE=6cm. FFisanarcofthecirclecentreM,suchthatangleEMEis6radians. a Showthat0=1.855radians,correctto3decimalplaces. [2] b Calculate the perimeter ofthe shaded region. [4] c Calculatetheareaoftheshadedregion. [3]

bFindtheperimeteroftheshadedregion. [5] C Findtheareaoftheshadedregion. [4]

radiusThediagramshowsaright-angledtriangleABCandasectorCBDCofacirclewithcentreCand12cm.AngleACB=1radianandACDisastraightline.aShowthatthelengthofABisapproximately10.1cm.[1]

CambridgeIGCSEAdditionalMathematics0606Paper21Qlli,ii,iiiJun2012

CambridgeIGCSEAdditionalMathematics0606Paper11QIONov2015

CambridgeIGCSEand0LevelAdditionalMathematics

The diagram shows two circles,centres A and B,each ofradius 10cm.The pointBlies on the circumferenceofthecirclewithcentreA.ThetwocirclesintersectatthepointsCandD.The pointElies on the circumference ofthe circle centre Bsuch that ABEis a diameter. i ExplainwhytriangleA5Cisequilateral. [1] ii Write down,in terms oftt,angle CBE. [1] Findtheperimeteroftheshadedregion. [5] Find the area ofthe shaded region. [3]

III iv

Thediagramshowsacircle,centreO,radius8cm.ThepointsPandQheonthecircle. 3n The linesPTand QTare tangents to the circle and angle POQ- — radians. I FindthelengthofPT. [2] ii Findtheareaoftheshadedregion. [3] iiiFindtheperimeteroftheshadedregion. [2] CambridgeIGCSEAdditionalMathematics0606Paper21Q4Jun2015 v_

III IV

The diagram shows a circle,centre O,radius rcm.Points A,Band Care such that A and B lieonthecircleandthetangentsatAandBmeetatC.AngleAOB=0radians. i GiventhattheareaofthemajorsectorAOBis7timestheareaoftheminorsectorAOB, findthevalueof0. [2] ii GivenalsothattheperimeteroftheminorsectorAOBis20cm,showthatthevalue of r, correct to 2 decimal places,is 7.18. [2]

Usingthevaluesof0andrfrompartsiandii,findtheperimeteroftheshadedregion ABC. [3] FindtheareaoftheshadedregionABC. [3] CambridgeIGCSEAdditionalMathematics0606Paper12Q9Mar2016

Chapters: Circularmeasure

Chapter 9 Trigonometry Thissectionwillshowyouhowto: m findthetrigonometricratiosofanglesofanymagnitude ■ determinetheamplitudeandperiodoftrigonometricfunctions m describetherelationshipbetweentrigonometricgraphs u sketchgraphsof31=|f(x)|,wheref(x)isatrigonometricfunction ■ drawandusethegraphsof3)=asinto+c, y=acosbx+c, 31=atanto+r whereaisapositive integer.6isasimplefractionorintegerandcisaninteger ■ usetrigonometricrelationships ■ solvesimpletrigonometricequations ■ provesimpletrigonometricidentities.

RECAP Chapters:Trigonometry Youshouldalreadyknowthefollowingtrigonometricrauos: smf= opposite hypotenuse cos6 adjacent hypotenuse sin y COS0= r tan^= tan( opposite adjacent y. X 9.1 Angles between 0° and 90^ WORKEDEXAMPLE1 Giventhatsin0=-7=andthat6isacute,findtheexactvaluesof sin6 a sin^0 b cos0 c tan0 tan6—cos0 Answers a sin^6=sin9xsin9 -A A. ~_45 b sin0=—;= V5 Theright-angledtriangletorepresent9is: UsingPythagoras'theorem,x X=1 1 >/5 H sin^9means(sin9)^ ence,cos9= c tan0=-=2 1 Vs 5 ■ V5 tan0- cos0 2 ^ sin0 V5 ~2S-1 2{2S+1) ~(2V5 -l)(2s/5 + 1) 2(275 +1) ~ 19 2-t475 19 multiplynumeratoranddenominatorby75 multiply numeratorand denominatorby(275 + l)

CambridgeIGCSEand0LevelAdditionalMathematics 7C 7C 7C Thesine,cosineandtangentof30°,45°and60°(or—,—and—)canbe obtainedexactlyfromthefollowingtwotriangles: ® ^ ^ Consideraright-angledisoscelestrianglewhosetwoequalsidesareoflength 1unit. The third side isfound using Pythagoras'theorem: Vl^ +1^ = y/2 sin45°= V2 tancos45°45°=1J2I 2y 2 n 1 sm—=-7= 4 yf2 n 1 cos—=—pr 4 ^/2 tan—=14 2 TheConsideranequilateraltrianglewhosesidesareoflength2units.perpendicularbisectortothebasesplitstheequilateraltriangleintotwocongruentright-angledtriangles. The heightofthe triangle can befound using Pythagoras'theorem: V22-1^ = V3 2 sin60° n sm— 3 2 Ao^ 2/ \ /60\ r \ cos60°=2 tan60°=y/S 1 22 sin30°= cos30°= 71 1 cos— 3 2 71 tan—= 3K 1 sm7t y/S COS— = 6 2 tan30°= V3 V3" 7C tan— 6 _1_ yJS Q Note: Youdonotneedtolearnthesetrianglesandratiosfortheexamination, butyou mayfind it useful to know them.

Chapter9:Trigonometry WORKEDEXAMPLE2Findtheexactvalueof a sin45°sin60° b cos^45° Answers 1 Js a sin45°sin60°=-4= X ^/2 2 2yj2s/3xV22^/2xV2 = A 4 b cos^45°=cos45°xcos45° _J_ J_ =1 ~2 7C .71 /3', 1 tan—+sin— vo+-7= 3 4 ^ 1-cos 71 1-1 =2>/3+V2. x2 71 n tan—t-sin 3 4 1 " 1-cos 3 rationalisethedenominator cos^45°means(cos45°) expandthedenominatorsimplifiestobrackets rationalisethedenominator 1 Exercise9.1 21 Given that tan^ ^ that 6is acute, find the exact values of asin0 bcos6 csin^6 dsin^6+cos^6 2+sin0 3-cos6 2 Giventhatsin9= andthat6isacute,findtheexactvaluesof a COS0 b tan0 c 1-sin^0b tan0 d sin0+COS0 cos0-sin0 e tan0 3 Giventhatcos9-—andthat0isacute,findtheexactvaluesof a sin0 b tan0 c tan0cos0 d sin^0+cos^0 cos0-tan0 1-cos^0

CambridgeIGCSEand0LevelAdditionalMathematics 4 Findtheexactvalueofeachofthefollowing, atan45°cos60° btan^60° c d sin45°+cos30° cos230° cos45°+cos60° 5 Findtheexactvalueofeachofthefollowing. f K n a sin—COS 4 3 Tt 5-tan 3 K Sin 3 u 2^b sm^ 1 1 .n n sin— cos 6 4 tan30°cos30° tan45°-sin30° 1+sin260° K tan 6 K COS 4 Tt tan 4 Tt -Sin 4 K tan 6 Tt Sin 6 9.2 The general definition of an angle Youneedtobeabletousethethreebasictrigonometricfunctionsfor anyangle. Todothisyouneedageneraldefinitionforanangle: SECOND FIRST QUADRANT QUADRANT P 0 X THIRD FOURTH QUADRANT QUADRANT AnangleisameasureoftherotationofalineOFaboutafixedpointO.Theangleismeasuredfromthepositivex-direction.Ananticlockwiserotationistakenaspositiveandaclockwiserotationistaken asnegative. saidTheCartesianplaneisdividedintofourquadrantsandtheangle6istobeinthequadrantwhereOPlies.Inthediagramabove,6isinthe firstquadrant.

Chapter9:Trigonometry WORKEDEXAMPLES DrawadiagramshowingthequadrantinwhichtherotatinglineOPliesforeachofthefollowingangles.IneachcasefindtheacuteanglethatthelineOPmakeswiththe »-axis. 2n a 240° b -70° c 490° d Answers a 240°isananticlockwiserotation b-70°isaclockwiserotation y. k \60°/ a X p oy,\70°/ * o >JO p acuteangle=60° acuteangle=70° 2Jt C 490°isananticlockwiserotation d —isananticlockwiserotation 490°=360°+130° y\ P. V ■^^90° 50A Q j X y\p 2n 7T\ 3 3 \ a X acute angle = 50° acute angle Exercise 9.2 1 Draw a diagram showing the quadrant in which the rotating line QPlies for each ofthe following angles. In each question indicate clearly the direction ofrotation and find the acute angle that the line QPmakes with the x-axis. a 110° b -60° c 220° d -135° e -300° 371 * T g 771 T 57t h 3 1371 1 9 57C 2 State the quadrant that OPlies in when the angle that OPmakes with the positivex-axisis a110° b300° c-160° d245° e-500° 71 * 4 g llTt h 57t 1371 J 971 T

y X T

Byconsideringthesignofyinthesecond,thirdandfourthquadrants,determinethesignsofthesineratioineachofthese quadrants. Byconsideringthesignofxinthesecond,thirdandfourth quadrants,determine the signs ofthe cosine ratio in each of thesequadrants. Byconsideringthesignofxandyinthesecond,thirdandfourth quadrants,determinethesignsofthetangentratioineachof thesequadrants.Whathappenstothetangentratiowhenx=0?

CambridgeIGCSEand0LevelAdditionalMathematics 9.3

CLASSDISCUSSION

Trigonometric ratios of general angles

Onacopyofthediagram,recordwhichratiosarepositiveineach quadrant. Thefirstquadranthasbeencompletedforyou. (Allthreeratiosarepositiveinthefirstquadrant.) sin6 1-1^5 II cos6 Xr tan0 =1 X y\ sincos tan

Ingeneral,trigonometricratiosofanyangle9inanyquadrant aredefinedas: sin0=—,COS0=—,tan0=—,x^0 wherexandyarethecoordinatesofthepointPandristhelengthofQPandr= Youneedtoknowthesignsofthethreetrigonometricratiosinthe fourquadrants. Inthefirstquadrantsin,cosandtanarepositive.(Sincex,yandrareallpositive.)Byconsideringthesignofxandyyoucanfindthesignofeachofthethree trigonometricratiosintheotherthreequadrants.

Expressintermsoftrigonometricratiosofacuteangles, acos(-110°) bsin125° Answers aTheacuteanglemadewiththe*-axisis70°. Inthethirdquadrantonlytanispositive, socosisnegative. cos(-110°)=^os70° b Theacuteanglemadewiththe*-axisis55°. Inthesecondquadrantsinispositive, sin125°=sin55°

7 S A ' ' j > O O x T / c y ■ S t A A Xl25° / 55°\ 0 T ' ► X c

Theresultsoftheclassdiscussion,canbesummarisedas: You can memorise this diagram using a mnemonicsuch as 'AllStudentsTrusttambridge'. 180° Chapter9:Trigonometry 90° 7' Sin All X Tan Cos 0°,360° 270°

WORKEDEXAMPLE4

CambridgeIGCSEand0LevelAdditionalMathematics WORKEDEXAMPLE5 Giventhatsin0=— andthat180°«0«270°,findthevalueoftan0andthevalue ofcos0. ^ Answers 0isinthethirdquadrant. tanispositiveandcosisnegativeinthisquadrant. +(-3)^=5'^ ,.2 =25-9=16 Sincex<0,*=-4 tan0 -3 3 . _ -4 4 —=—andCOS0=—= -4 4 5 5 7' S A xf LI I ^ O X-3;Jy iX^ T C Exercise9.3 1 Express the following as trigonometric ratios ofacute angles. a sin220° b cos325° c tan140° d cos(-25°) 7k h (etan600°27t Itan 471 f sin5 Otc I sm * 4 gtan- cos -") 2 Giventhatcos0=-andthat270°<0^360°,findthevalueof 5 a tan0 b sin0. 3 Giventhattan0=-VSandthat90°^0^180°,findthevalueof asin0 bcos0. 4 Giventhatsin0=—andthat0isobtuse,findthevalueof 13 aCOS0 btan0. 2 5 Giventhattan0=-andthat0isreflexfindthevalueof 3 asin0 bCOS0. 4 1 6 Given that tan A = — and cosB = —1=,where A and B are in the same 3 -v3 quadrant,find the value of asinA b cosA csinjB d tanjB. 12 3 7 GiventhatsinA= andcosB=-,whereAandBareinthesame 13 5 quadrant,findthevalueof acosA b tanA c sinB d tan5.

CLASSDISCUSSION

Consider taking a ride on a Ferris wheel,with radius 50 metres, whichrotatesataconstantspeed. You enter the ride from a platform that is level with the centre of thewheelandthewheelturnsinananticlockwisedirection.

9.4 Graphs of trigonometric functions Chapter9:Trigonometry

Sketchthefollowingtwographsanddiscusstheirproperties: • agraph ofyourverticaldisplacementfromthecentre ofthe wheelplottedagainstangleturnedthrough • agraph ofyourhorizontaldisplacementfromthecentreof thewheelplottedagainstangleturnedthrough. Thegraphsofy=sinxandy=cosx SupposethatOPmakesanangleofxwiththepositive horizontalaxisandthatPmovesaroundtheunitcircle, throughonecompleterevolution.ThecoordinatesofPwill be(cosx,sinx).

Thegraphofcosx againstxfor0°^x^360°is:

P(cosx,sinx) = sin X y=cosX I'X 70 360

The heightofPabove the horizontal axis changes from0—>l-^0->-l->0. Thegraphof sinx againstxfor0°^x 360°is: ThedistanceofPfromtheverticalaxischanges from1—>0—»-l—>0—>1.

CambridgeIGCSEand0LevelAdditionalMathematics

The red dashed lines at *= ±90°, x = 270° and x = 450° are asymptotes.The branches ofthe graph get closer and closer to the asymptotes without ever reachingthem.

Thegraphsof31 = sinx and 3=cos* can becontinued beyond 0° ^360°: y 10 90 18\) 270 450 540 ^

TheThetangentfunctionbehavesverydifferentlytothesineandcosinefunctions.tangentfunctionrepeatsitscycleevery180°soitsperiodis180°.

Thetangentfunctiondoesnothaveanamplitude.

Thegraphsofy=as\nbx+c,y=acosbx+candy=atanbx+c

You have already learnt about the graphs of y = sin x, y = cos* and y = tan*. Inthissectionyouwilllearnhowtosketchthegraphsofy=asinbx+c,y-acosbx+candy=aXznbx+cwhereaandbarepositive Youintegersandcisaninteger.canusegraphingsoftwaretoobservehowthevaluesofa,bandcaffectthetrigonometricfunctions.

^^1Thesine and cosinefunctionsare called periodicfunctionsbecause they repeatthemselvesoverandoveragain.

Thegraphofy=tanx 7 y 0 0 0009^ 180 ^0 360 4^ 540 ^ -1y=tanx 0/ISO270/3604^0/40 ^-180 -^0/6

The period ofa periodicfunction is defined asthelength ofone repetition orcycle. The basic sine and cosine functions repeat every 360°. We say they have a period of360°.(Period = 271,ifworking in radians.)

Theamplitudeofaperiodicfunctionisdefinedasthedistancebetweena Themaximum(orminimum)pointandtheprincipalaxis.basicsineandcosinefunctionshaveamplitude1.

ACTIVITY

Usegraphingsoftwaretoconfirmthat:

Usinggraphingsoftware,thegraphsofy=sinxandy=sin2xare: -l-" y=sin2x y=sinx Thegraphofy=sin2xisastretchofthegraphofy=sinx.

Chapters:Trigonometry

• y =2cosX is a stretch of y =cosx from the x-axis with stretch factor 2 « y=3cosXisastretchofy=cosxfromthex-axiswithstretchfactor3 and • y=2tanxisastretchofy=tanxfromthex-axiswithstretchfactor2 • y=3tanxisastretchofy=tanxfromthex-axiswithstretchfactor3. The graph of y =sin bx

Thegraphofy=osin*

Usinggraphingsoftware,thegraphsof))=sinxand))=2sinxare: t--y=sinX: Iy=2sinx

The graph of )i =2sin x is a stretch ofthe graph of )> =sin x. Ithasbeenstretchedfrom the x-axiswithastretchfactorof2. Theamplitudeof3)=2sinxis2andtheperiodis360°. Similarly,it can be shown that graph of y =3sin x is a stretch of y =sin x fromthex-axiswithstretchfactor3.Theamplitudeofy=3sinxis3andthe periodis360°.

Usinggraphingsoftware,thegraphsofy=sinxandy=sinx+1are: -I-' ^ ^ y=sinX . y=sinX+1 X 90 18bv 270 /^60

ACTIVITY

Usegraphingsoftwaretoconfirmthat:

•y=cos2xisastretchofy=cosxfromthey-axiswithstretchfactor 1 2 • y =cos3x is a stretch of y = cosx from the y-axis with stretch factor ^ and •y=tan2xisastretchofy=tanxfromthey-axiswithstretchfactor • y = tanSx is a stretch of y = tan x from the y-axis with stretch factor ^ The graph of y = sin x + c

CambridgeIGCSEand0LevelAdditionalMathematics Ithasbeenstretchedfromthey-axiswithastretchfactorof—. 2 Theamplitudeofy=sin2xis1andtheperiodis180°. Similarly,the graph of y =sinSx is a stretch,from the y-axis,of y =sin x withstretchfactor—. 3

Ithasbeentranslatedbythevector

Thegraphofy=sinx+1isatranslationofthegraphofy=sinx.

Theamplitudeofy=sinSxis1andtheperiodis120°.

Theamplitudeofy=sinx+1is1andtheperiodis360°.Similarly,thegraphofy=sinx+2isatranslationofy=sinxbythe fo'i vector .

Theamplitudeofy=sinx+2is1andtheperiodis360°.

Chapters:Trigonometry ACTIVITY Usegraphingsoftwaretoconfirmthat: • 3)=cosX+1 isatranslationof3)=cosxbythevector • 31=cosX+2isatranslationof31=cosxbythevector • y=cosX-3isatranslationof3;=cosxbythevector and • y=tanx+1 isatranslationofy=tanxbythevector ^0^ vl/ ^0^ V: ■ y=tanx-2isatranslationof3;=tanxbythevector ^ 0^ v-2y

Inconclusion, affects amplitude (amplitude=a) y=asinbx+c affects period . ^ 360° 2Jt' penod= or ^ b b, verticaltranslation translation ='^ y=acosbx+c affects (amplitudeamplitude=a) affects period ( . 360° 27c'| Ipenod=—^—or—I verticaltranslation translation ='^ y=atanbx+c stretchesthegraph affects Note:there is no period amplitude for a tangent ( 180° function periods—or f) verticaltranslation translation '^

Step3:Sketchthegraphofy=2cos3x: Stretch y=cos3xfromthex-axis withstretchfactor2. 360° P 120°eriod= Amplitude=2 yficos3x

Q^AO Period=—=120° 3 Amplitude=1 y=cos3x

CambridgeIGCSEand0LevelAdditionalMathematics

Sketchingtrigonometricfunctions

Step4:Sketchthegraphofy=2cos3x-1: 'O^iTranslate31=2cos3xby Q^AO Period= =120° Amplitude=2 -1, cos3x-1

The sketch graph ofa trigonometricfunction,such as y = 2cos3*-1 for 0°^ x ^ 360°, can be builtup in steps.

Step1:Startwithasketchof))=cosx: Period=360° Amplitude = 1 ^ -1 -2y=cosX j X 9 !0^^^70 3^0

Step2:Sketchthegraphof31=cos3x: Stretchy=cosxfromtheji-axis withstretchfactor—.3

-3y=3sin2x / \1 \ / \/ \ / \ X 94 \i10 2' 0 i\ /r i v/l y =3sin

is a

Chapters:Trigonometry WORKEDEXAMPLE6

f(x:)=3sin2xfor0°^ ^360° a Writedowntheperiodoff. Writedowntheamplitudeoff. Writedownthecoordinatesofthemaximumandminimumpointsonthecurve 3)=f(3c). Sketchthegraphof3=f(x). Useyouranswertopartdtosketchthegraphofy=3sin2x+1. Answers Perioda 360° b c =180° y=sinXhasitsmaximumandminimumpointsat: (90°,1),(270°,-1),(450°,1)and(630°,-1) Hence,f(x)=3sin2xhasitsmaximumandminimumpointsat: (45°,3),(135°,-3),(225°,3)and(315°,-3) 2x+1 translation of

the graph y =3sin 2x by the vector ^ ^ :3sin2x+1

Amplitude=3

CambridgeIGCSEand0LevelAdditionalMathematics WORKEDEXAMPLE7 a Onthesamegrid,sketchthegraphsofy=sin2xand3)=1+cos2xfor 0°«x«360°. b Statethenumberofrootsoftheequationsin2x=1+cos2xfor0°^ 360°. Answers a 3 1 0 -1 y=1+(:os2x //\ A X ( 80 2 50 iin2x b Thegraphsof3;=sin2x and 3)=1+cos2x intersecteachotherat4pointsin theinterval. Hence,thenumberofrootsofsin2*=1+cos2xis4. Exercise9.4 1 a Thefollowingfunctionsaredefinedfor0°^ ^360°. coordinatesForeachfunction,writedowntheamplitude,theperiodandtheofthemaximumandminimumpoints.if(x)=7cosxiif(x)=2sin2xiiif(x)=2cos3x iv f(x)=3sin—X 2 V f(j:)=4cosx+1 vi f(x)=5sin2x-2 b Sketch the graph ofeach function in parta and use graphing software tocheckyouranswers. a Thefollowingfunctionsaredefinedfor0 x=52)1. Foreachfunction,writedowntheamplitude,theperiodandthecoordinatesofthemaximumandminimumpoints. i f(x)=4sinx iv f(x)=3cos—x 2 II f(x)=cosSx III f(x)=2sin3x V f(x)=sin2x+3 vl f(x)=4cos2x-1 b Sketchthegraphofeachfunctioninpartaandusegraphingsoftware tocheckyouranswers.

Foreachfunction,writedowntheperiodandtheequationsofthe asymptotes. 1 f(«:)= tan2j!: ii f(a:)= 3tan^x Hi f(x:)= 2tan3x +1

Thegraphof3*=<2+bcoscx,for0°^ ^360°,isshownabove. Writedownthevalueofa,thevalueofbandthevalueofc. a Thefollowingfunctionsaredefinedfor0°^jc^360°.

bSketchthegraphofeachfunctionandusegraphingsoftwaretocheck youranswers.

Chapters:Trigonometry

Partofthegraphof31=osinftx+cisshownabove. Writedownthevalueofa,thevalueofbandthevalueofc. ] \ j j J X J I- I—

Thegraphof 3)=a+fisincx,for0^ ^7i,isshownabove. Writedownthevalueofa,thevalueofbandthevalueofc. y 43 2 1 O / \ / \ \ / 45 90 135 180 X

12a Onthesamegrid,sketchthegraphsofy=sinxandy=1+sin2xfor 0°^X<360°. b Statethenumberofrootsoftheequationsin2x—sinx+1=0for 0°^X^360°.

Themaximumvalueoffis5andtheperiodis72°. aWritedownthevalueofAandthevalueofB. b Writedowntheamplitudeoff. c Sketchthegraphoff.

FindthevalueofA,thevalueofBandthevalueofC. 9 f(x)=a+ftsincx

11f(x)=A+BsinCx for 0°^x^360°

CambridgeIGCSEand0LevelAdditionalMathematics a

Themaximumvalueoffis13,theminimumvalueoffis5andtheperiod is60°. Findthevalueofa,thevalueofbandthevalueofc.

b Sketch the graph ofeachfunction and use graphing software to check youranswers. 8 o 7t / 71 371 / 2n X 2 / '1

Thefollowingfunctionsaredefinedfor0 ^271.

Theamplitudeoffis3,theperiodis90°andtheminimumvalueoffis-2. aWritedownthevalueofA,thevalueofBandthevalueofC. b Sketchthegraphoff.

Foreachfunction,writedowntheperiodandtheequationsof theasymptotes. i f(x:)=tan4x: ii f(x)=2tanSx iii f(x)=5tan2x-3

10f(x)=A+ScosBxfor0°^x^360°

Partofthegraphofy=AtanBx+Cisshownabove.Thegraphpassesthroughthepoint

b Statethenumberofrootsoftheequationsinx=1+cos2xfor 0°^X^360°.

Chapters:Trigonometry

9.5 Graphs of y =|f(x)|,where f(x) is a trigonometric function You have already learnt how to sketch graphs of 3 =|f(x)| where f(x) is either linearorquadratic. In this section you will learn how to sketch graphs of 3 =|f(x)| where f(x) is a trigonometricfunction.

Answers a Step1:Sketchthegraphof3=sinx rl-y 0^f(x)^1 J'=Isin sinX 1Step2:Reflectinthex-axisthepartofthecurve3=sinxthatisbelowthex-axis.

b Statetherangeofthefunctionf.

WORKEDEXAMPLE8 a Sketchthegraphoff(x)=|sinx|for0^x^2jt.

13 a On the same grid,sketch the graphs of 3;= sinx and 31 = 1+cos2x for 0°^X^360°.

bTherangeofthefunctionfis0^f(x)^l.

14 a On the same grid,sketch the graphs of 3* = 3cos2x and 3;= 2+sinx for0°^X^360°.

b Statethenumberofrootsoftheequation3cos2x=2+sinxfor 0°^X^360°.

Exercise9.5

WORKEDEXAMPLE9

CambridgeIGCSEand0LevelAdditionalMathematics

bStatethenumberofrootsoftheequation|0.5+sinx|=cosxfor 0°^X^360°.

Theequation13cosx-2\=k,has2rootsfortheinterval0^x^27t. Findthepossiblevaluesofk.

bSketchthegraphof))=|2+5cosx|for0° x^180°

bStatethenumberofrootsoftheequation|tanx|=cosxfor 0°^X^360°.

7aOnthesamegrid,sketchthegraphsof)>=10.5+sinx|andy=cosx for0°^X^360°.

8 a On thesame grid,sketch the graphs of =11+4 cosx| and y=2+cosx for0°^x^360°.

bStatethenumberofrootsoftheequation11+4cosx|=2+cosxfor 0°^X^360°.

4 a Sketchthegraphofy=2+Scosxfor0°^x^180°.

The diagram shows the graph of f(x)=|a + 6coscx|, where a, 6and c arepositiveintegers.

Chapters:Trigonometry

Writedownthenumberofsolutionsoftheequation|2+5cosx|=1for 0°^X^180°.

a Sketchthegraphofy=2+5cosxfor0°^ ^180°.

bStatethenumberofrootsoftheequation|sin2x|=tanxfor 0^X^271.

Findthevalueofa,thevalueofbandthevalueofc.

bSketchthegraphof))=|2+3cosx|for0°^x^180°.

c Writedownthenumberofsolutionsoftheequation12+3sin2x|=1 for0°^X^180°.

6 a On the same grid,sketch the graphs of =|sin2x|and y = tanx for 0^X^27C.

CHALLENGEQ 10 360X

a On the same grid,sketch the graphs of )i =|tanx| and y - cosx for 0°^X^360°.

1oOl II

Considertheright-angledtriangle: ^

Thesketchgraphshowstherearefourvaluesofx,between-360°and360°,for whichsinx=0.5.

Thesetwoimportantruleswillbeneededtosolvesometrigonometric equationslaterinthissection. Considersolvingtheequation: sinx=0.5X=sin'(0.5)

X II OOo o

y sin0=— cosO=— tan0= r r X tan6 dividenumeratoranddenominatorbyr

CambridgeIGCSEand0LevelAdditionalMathematics 9.6 Trigonometric equations

Thefollowingrulescanbefoundfromthistriangle:

Acalculatorwillgivetheanswer: Thereare,however,manymorevaluesofxforwhich sinx=0.5. Considerthegraphofy=sinxfor -360°^X^360°: -360 -270 -180 -90 ^ 360^ -330 -210

30 150 -1- y=sinX

-I X 1 y ^ n use—=sin0and—=cosa r r r _r_ X tanO sin0 cost dividebothsidesby X use—=COS0and-=sin0 cos^6+sin^9=1

90 180 270

y X

Chapter9:Trigonometry

Thesketchgraphshowstherearetwovaluesofx,between0°and360°,forwhich cosX=-0.4. Usingthesymmetryofthecurve,thesecondvalueis(360°-113.6°)=246.4°. Hencethesolutionofcosx=-0.4for0° x 360°is x=113.6° or 246.4°.

WORKEDEXAMPLE10 Solve cosx=-0.4for 0°^x^360°. Answers

TX=-0.4=113.6° 1O

Youcanusethecalculatorvalueofx=30°,togetherv^iththesymmetryofthe curvetofindtheremaininganswers. Hencethesolutionofsinx=0.5for-360°^x^360°is; X=-330°,-210°,30°or150° cosx -H useacalculatortofindcos"(-0.4)to1decimalplace y=cos 113.6 9a^-----180 360-• y=-0.4

CambridgeIGCSEand0LevelAdditionalMathematics WORKEDEXAMPLE11 Solvetan2A=-1.8for0° /A /A 00o o Answers tan2A=-1.8 let x=2A tanx=-1.8 useacalculatortofindtan'(-1.8) =-60.95° I 1 / y=tanx -60.95 y y -90\/^ 90i/^80 2701^-^60'' ( 00 l-H IIUsingthesymmetryofthe 1 curve: X=-60.95 X=(-60.95+180) x=(119.05+180) =119.05 =299.05 UsingX=2A, 2A=-60.95 2A=119.05 2A=299.05 A=-30.5 A=59.5 A=149.5 Hencethesolutionoftan2A=-1.8for0°^ «180°is A=59.5°orA=149.5°.

Chapters:Trigonometry WORKEDEXAMPLE12 Solve sin^2A ~^j~ 0^ A ^ 7t. Answers sin sinx=0.6X=0.6435radians letX=2v4 3 useacalculatortofindsin"'0.6 0.6435ji 2 y=sinX -1Usingthesymmetryofthecurve: x=0.6435 x=jc-0.6435 =2.498 Using X=2A3 2A--=0.64353 2A--=2.498 3 A =i(^0.6435+1j A =i(^2.498+^jA: A-0.845 A=1.77 Hencethesolutionofsin^2A^J~ 0<A^7tis A=0.845or1.77radians.

WORKEDEXAMPLE13

Solvesin^x-2sinxcosx=0for0°^x^360°. Answers sin^X-2sinXcosx=0 factorise sinX(sinx-2cosx)=0 sinX=0 orsinx—2cosx=0 X=0°,180°,360° sinx=2cosx tanx=2X=63.4or180+63.4X=63.4°or243.4°

CambridgeIGCSEand0LevelAdditionalMathematics

WORKEDEXAMPLE14

Thesolutionofsin^x-2sinxcosx=0for0°^x^360°is X=0°,63.4°,180°,243.4°or360°.

Solve 1+cosx=3sin^x for 0°^x^360°. Answers 1+cosX=3sin^x 1+cosx=3(l-cos^x) 3cos^X+cosX-2 =0 (3cosX-2)(cosX+1)=0 3cosx-2=0 or 2 cosx= 3 X=48.2°or360-48.2 x=48.2°or311.8° cosx+1=0 cosx=-lX=180° usesin^x=1-cos^x expandbracketsandcollectterms factorise y=cosX

Thesolutionof1+cosx=3sin^xfor0°<x^360°is X=48.2°,180°or311.8°.

1 Solve each ofthese equationsfor 0°^ x ^ 360°. asinx=0.3 b cosx=0.2 ctanx=2 d e tanx=-1.4 f sinx=-0.8 g 4sinx-3=0 h

Solve4sin(2x-0.4)-5cos(2x-0.4)=0for 0^x^7t. Solveeachoftheseequationsfor0°^x^360°. a sinxtan(x-30°)=0 b 5tan2x-4tanx=0 c 3cos^x=cosx d sin^x+sinxcosx=0 e 5sinXcosX=cosx f sinxtanx=sinx 'X 360°.8 Solveeachoftheseequationsfor0° a 4sin^X=1 9 Solveeachoftheseequationsfor 0° a tan^X+2tanx-3=0 c 3cos^X-2cosX—1=0 e 3cos^X-3=sinx g 2cos^X-sin^x-2sinx-l=0 I 3cosx=8tanx K 10 f(x)=sinx for 0^x^ 2 Solvegf(x)=0.5. b 25tan^x=9 X^360°. db f h 22sin^X+sinx-1=0sin^X-cosx-1=0

Exercise9.6

sinX+5=6cos^x 1+tanXcosx=2cos^x g(x)=2x-1for X€IR

Solveeachoftheseequationsfor0°^x^360°. a 4sinx=cosx b 3sinx+4cosx=0 c5sinx-3cosx=0 d5cos2x-4sin2x=0

sinX=-0.6 2cosX+1=0 tanx=-3 5sinX+2=0sin2x=-0.62-3sin2x=0 6 7 sin(2x-3)=0.6 for 0<x <7t radians. (X TC I sin-+—=1for0<x<47Cradians. 12 6)

3Solveeachoftheseequationsfor0°^x 180°. asin2x=0.8 b cos2x=-0.6ctan2x=2 d e 5cos2x=4 f 7sin2x=-2 g l+3tan2x=0h

Chapters:Trigonometry

2 Solve each ofthe these equationsfor 0^ x ^ 271. acosX=0.5 b tanx=0.2 csinx=2 d esinx=-0.75 f cosx=-0.55 g 4sinx=l h

4Solveeachoftheseequationsforthegivendomains, a cos(x-30°)=-0.5 for 0°^x^360°. b 6sin(2x+45°)=-5for 0° x^180°. «540°. cos! X +^]= -0.5 for 0 < x < 271. 2cos| y1+ >/3 = 0 for 0°

WORKEDEXAMPLE15 theidentity (1+sinx)^ LHS=(1+sinx)^+(1-sinx)^+2cos^x

CambridgeIGCSEand0LevelAdditionalMathematics 9.7 Trigonometric identities sin^X+cos^X=1iscalledatrigonometricidentitybecauseitistrueforall valuesofx. Inthissectionyouwilllearnhowtoprovemorecomplicatedidentities involvingsinx,cosxand tanx. Whenprovinganidentity,itisusualtostartwiththemorecomplicatedside oftheidentityandprovethatitsimplifiestothelesscomplicatedside.Thisis illustratedinthenextexample. Note: LHSmeansleft-handsideandRHSmeansright-handside.

=(1+sinx)(l+sinx)+(1-sinx)(l-sinx)+2cos^x expandbrackets =1+2sinX+sin^x+1—2sinx+sin^x+2cos^x collectliketerms =2+2sin^X+2cos^x =2+2(sin^X+cos^x) usesin"x+cos^x=1 =2+2x1 ==4RHS WORKEDEXAMPLE16 Provetheidentity cosx=sinxtanx. cosx Answers LHS= 1 cosx 1 ■ - COS X cosx cosx 1-COS^X cosx cosx sinx =sinx X cosx =sinxtanx=RHS use1-cos^X=sin^ smx use tanx cosx

+(1-sinx)^ +2cos^x =4. Answers

Prove

Chapters:Trigonometry CLASSDISCUSSION Equivalenttrigonometricexpressions (YourDiscusswhyeachofthetrigonometricexpressionsintheblueboxessimplifytosinx.Createasmanytrigonometricexpressionsofyourownwhichsimplifytotanx.expressionsmustcontainatleasttwodifferenttrigonometricratios.)Compareyouranswerswithyourclassmates. sm X sinX sin''X+sinXcos''x sin'X tanXcosx tanXcosx 1-sinX tan'^x cos'^x sinX cosx Exercise9.7 1Proveeachoftheseidentities. sinx a tanx=cosx 1-sin^X c =cosx cosx / • \2 e(smX+cosx)=1+2sinxcosx cosXsmX tanx =1-sin^X cos^X-sin^X cosx+sinx +sinX=cosX f tan^X-sin^x=tan^ Xsin^X 2Proveeachoftheseidentities, acos^X-sin^x=2cos^x-1 c cos^X+sin^Xcos^x=cos^x e 2-(sin x + cosx)^ =(sin x-cosx)^ 3Proveeachoftheseidentities. cos^X—sin^X =cosX+smX cosx - smx cos"*X-sin^X cos^X =1-tan^X bcos^X-sin^X=1-2sin^x d 2(1+cosx)-(1+ cosx)^ = sin^x f cos^X+sin^X=sinix+cos^x db smx 1+cosx + 1+cosx sin^x (l — cos^x) = smx 2, smx cos^x(l- sin^x) tan^X

CambridgeIGCSEand0LevelAdditionalMathematics 9.8 Further trigonometric equations Thecosecant,secantandcotangentratiosThereareatotalofsixtrigonometricratios.Youhavealreadymettheratios sine,cosine and tangent.In this section you will learn aboutthe other three ratios,whicharecosecant(cosec),secant(sec)andcotangent(cot).These threeratiosaredefinedas:cosec0= sin01 sec9 1 cot9= cost 1 { cos6*^ I tan0 V sin0)f Considertheright-angledtriangle: sin0= r T cosecd=y ^ X ^ V cos6=— tan9= r X sec6= X cot$= X Thefollowingrulescanbefoundfromthistriangle: dividebothsidesby-I1-1- y T use—=tan0and—=sec0 x^ II CM + ^2 /^2 /\ X +1= r yy) yy) 1+tan^9=sec^9 dividebothsidesby X y y cot^9+\=cosec^9 r use—=cot0and—=cosec9 Theseimportantidentitieswillbeneededtosolvetrigonometricequationsin thissection.

Solve2cosec^x+cotx-8=0for0°^^360°. Answers 2cosec^x+cot*-8=0 2(l+cot^jc)+cotjc-8=0 2cot^X+cotX—6=0 (2cotx:-3)(cotx+2)=0 2cota:-3=0 orcotx+2=0 use1+cot-X=cosec^x expandbracketsandcollectterms factorise =tan X cotx=-2cotx = -26.6 y=—0.5tanx=tanx = X=33.7or33.7+180x=-26.6or-26.6+180or-26.6+360 X=33.7°or213.7° x=153.4°or333.4°(since-26.6°isoutofrange)

Thesolutionof2cosec^x+cotx—8=0for0°^x^360°is X=33.7°,153.4°,213.7°or333.4°.

Exercise9.8 1 Solve each ofthese equationsfor 0°^ x ^360°. 360° 217 acotx=0.3 b secx=4 c cosecx=—2 2 Solveeachofthetheseequationsfor0^x^27t. a cosecx=5 b cotx=0.8 c secx=—4 3 Solveeachoftheseequationsfor0°^x<180°. asec2x=1.6 b cosec2x=5 ccot2x=-1 4 Solveeachoftheseequationsforthegivendomains,asec(x-30°)=3for0°x360°bcosec(2x+45°)=-5for0°^x180° c cotl^x -t- = 2 for 0 < X < 271 d 3sec(2x+3)=4 for 0<x<7t 5 Solveeachoftheseequationsfor0°<x^360°. d 3secX-5=0 d 2cotx+3=0 d 5cosec2x=-7 a sec^X=4 b 9cot^X=4 c16cot^—X=9 2 Solveeachoftheseequationsfor0° x 360°. a3tan^X-secx-1=0 b 4tan^x+8secx=1 c 2sec^X=5tanx+5 d 2cot^x-5cosecx-1=0 e6cosx+6secx=13 f cotx+6sinx—2cosx=3 g 3cotx=2sinx h 12secx-lOcosx-9tanx=0

WORKEDEXAMPLE17 Chapters:Trigonometry

Cambridge IGCSE and0LevelAdditional Mathematics 9:9Furthertrigonometricidentities In this section you will learn how to prove trigonometric identities that involve anyofthesixtrigonometricratios. CLASSDISCUSSION Oddoneout cosx cotx cosec X — cotx cos x sin^X+cos^X cosecX simx cosec^x cotx sinx tanx Findthetrigonometricexpressionthatdoesnotmatchtheothersixexpressions. Createasmanyexpressionsofyourowntomatchtheoddoneout'. (Yourexpressionsmustcontainatleasttwo differenttrigonometricratios.) Compareyouranswerswithyourclassmates. smx cosec^X-cor cos X secX -tan^x WORKEDEXAMPLE18 Provetheidentity tan1+secXX+sinX cosecx. AnswersLHS1+secX tanX+sinX 1 sinX usesecx= andtanx= cosx cosx cosx sinx 1-sinX cosx cosx+1 sinx+sinxcosx cosx+1 sinx(cosx+1) 1 sinx = cosecx =RHS multiplynumeratoranddenominatorbycosx factorisethedenominator dividenumeratoranddenominatorby(cosx+1) 1 use cosecx sinx

4Proveeachoftheseidentities.sinXcosX a

1-cos^X sec^x =sec^X 1 =1-sin^X smx 1-cos^X1-sec^X1-cosec^x cosxcosxcotx+cotx 1 = = cosecX tan'*X =secx-tanx 1 1-sinx 1+sinx cosx sinx=2tanxsecx 1-tanx 1-cotx smx+cosx cosx + cosx cosecx+1 cosecx-1=2tanx 65 Show that(3+2sinx)^+(3-2sinxf+8cos^x has a constant value for all xand state this value, a Express 5sin^x-2cos^x in theform a+isin^x. b State the range ofthefunction f(x)=5sin^x-2cos^x for0^x^27t. CHALLENGEQ 7 a Expresssin^9+4cos0+2intheforma-(cos9-bf. b Hencestatethemaximumandminimumvaluesofsin^9+4cos9+2.

1Proveeachoftheseidentities. atanx+cotx=secxcosecx

Exercise9.9

2CcosecX-sina:=cosxcotxProveeachoftheseidentities. a (1+secx)(cosecx-cotx)=tanx b (1+secx)(l-cosx)= sinxtanx Ctan^X-sec^x+2=cosec^x-cot^x d (cotx + tan x)(cotX-tanx)= cosec^x-sec^ x 1 bs b sin^X inX+cosXcotx=cosecx dsecXcosecX-cotx=tanx =sinXcosX tanx+cotx sin^xcosx+cos^x sinx 1+tan^X =cotx = sec X cosec X tanx smxtan'^x 1+ tan^X 1+sinX =sin^X 1-sinx =(tanx + secx) 1 =secxcosecX cosx smx 1 1 + 1+cosx 1-cosx =2cosec^x cosX cosX e ^ 1- ^—=2secx 1+sinx 1-sinx d f h j db f +cos^X cos^X

3Proveeachoftheseidentities.

Chapters:Trigonometry

CambridgeIGCSEand0LevelAdditionalMathematics Summary Positiveandnegativeangles Angles measured anticlockwisefromthe positive -direction are positive. Angles measured clockwise from the positive -direction are negative. Diagram showing where sin,cos and tan are positive 90° 180°Sin > All X Tan Cos 0°,360° 270° Usefulmnemonic:AllStudentsTrustCambridge' Cosecant,secantandcotangent 1 „ 1 cosec u sin0 Trigonometricidentities sinx tanx cosx 11sin^X+cos^X=1-I-tan^X=sec^x+cot^X=cosec^X secy= COS0 coty= tan0

Examination questions

Statethevalueofa,ofhandofc. [3] CambridgeIGCSEAdditionalMathematics0606Paper11Q1Nov2013 period=7U= 27C ^0^ vly hencec=1. y=asin(te)+c a — amplitude a=3 b=2

Chapters:Trigonometry

Worked example The diagram showsthe graph of 31 = asin(6x)+c for 0^ x ^27t,where a,^>and care positive integers. y

The graph of 3;= 3sin2x has been translated by the vector a=3,^»=2,c=l Exercise9.10ExamExercise 1aSketchthecurvey=3cos2x-1for0°^x=€180°. [3] biStatetheamplitudeofl-4sin2x. [1] iiStatetheperiodof5tan3x+l. [1] CambridgeIGCSEAdditionalMathematics0606Paper11Q2Nov2014 2 a Solve2cos3x=cot3xfor0°<x 90°. [5] Tt'b Solvesecjy+~|=~2for0 ^7tradians. [4]

CambridgeIGCSEAdditionalMathematics0606Paper11Q11a,bNov2014 a Provethatsecxcosecx-cotx=tanx. [4] b Usetheresultfrompartatosolvetheequationsecxcosecx=3cotxfor0°<x<360°. [4] CambridgeIGCSEAdditionalMathematics0606Paper21Q10i,iiNov2014

CambridgeIGCSEand0LevelAdditionalMathematics ii cv, ^ l+sin0 cos64Showthat + =9sec6. cosd 1+sin0 5 a Solvetheequation2cosecx+ [4] CambridgeIGCSEAdditionalMathematics0606Paper21Q12a,bNov2013 COS* 0for0°^ <360°. [4] b Solve the equation 7sin(23)-l)= 5 for 0 < 31 ^ 5 radians. [5] CambridgeIGCSEAdditionalMathematics0606Paper11Qli,ii,iiiJun2013 6 On acopyofthe axes below sketch,for 0 ^2Jt,the graph of [2][2]a y=COS*-1 b 3;=sin2*. 3^1 2-1-27|C I It 37t T 27t * CStatethenumberofsolutionsoftheequationcos*-sin2*=1,for0^*^27C. [1] CambridgeIGCSEAdditionalMathematics0606Paper11Qli,ii,HiJun2013 7Provethat 1+sin0cos6 + -sin0 V cosQ =2+4tan^0. [4] CambridgeIGCSEAdditionalMathematics0606Paper21Q1Jun2013 8 a Given that 15cos^6+2sin^6= 7, show that tan^6= —. 14] 5 b Solve IScos^6+ 2sin26 = 7 for 0^6 radians. [3] CambridgeIGCSEAdditionalMathematics0606Paper11Q6i,iiJun2012

Chapter9:Trigonometry [3] 180°* a Thediagramshowsasketchofthecurvey=asin{bx)+cfor0°^ ^180°. Findthevaluesofa,bandc. b Giventhatf(x)=5cos3x+1,forallx,state i theperiodoff, [1] ii theamplitudeoff. [1] CambridgeIGCSEAdditionalMathematics0606Paper21Q3a,bi,ii/I'w2012 10 a Solve 4sinx=cosecx for0°^x^360°. [3] 180°b Solve tan^3)i-2sec33)-2=0 for 0°^ c Solve tan|^z-^j=^/3 for 0^z^2ti radians. [3][6] CambridgeIGCSEAdditionalMathematics0606Paper11QIOJun2015 11ShowthatVsec^0-1+Vcosec^O-1=secOcosecO. [5] CambridgeIGCSEAdditionalMathematics0606Paper11Q3Nov2015 12Solvethefollowingequations. 4sin2x+5cos2x=0for0°^x^l80° i cot^ +3cosec3»=3for 0° ;360° II cos ^j=-^ for 0 z^2ti radians,giving each answer as a multiple of n [4][5][3] CambridgeIGCSEAdditionalMathematics0606Paper21Q9Nov2015

i' i-i I'lM hapterlO Permutations and combinations Thissectionwillshowyouhowto: ■ findthenumberofarrangementsofndistinctitems ■ findthenumberofpermutationsofritemsfromndistinctitems ■ findthenumberofcombinationsofritemsfromndistinctitems ■ solveproblemsusingpermutationsandcombinations. 'W:

Chapter10:Permutationsandcombinations 10.1Factorialnotation 5x4x3x2x1iscalled5factorialandiswrittenas5! Itisusefultorememberthatn!=nx(n-1)1 Forexample5!=5x41 WORKEDEXAMPLE1 Findthevalueof8! 5! 11! Answers 8!_8X7X6 5! Fx /x/x/x/l' =8x7x6 =336 ^J4!__11x1Qx9x/x/x^x/x/x/x/x/ 813! Fx/x/xi^x/x/x^x/x3x2xI =9906165 Exercise10.1 1 Without using a calculator,find the value ofeach ofthe following. Usethexlkeyonyourcalculatortocheckyouranswers. a71 . 41 2121 11 21 61 3121 71 c 31 f 61 (31)11 51 51 71 — X 31 41 2 Rewriteeachofthefollowingusingfactorialnotation. a2x1 b6x5x4x3x2x1 c5x4x3 d17x16x15x14 10x9x8 3x2x1 12x11X10x9x8 4x3x2x1 CHALLENGEQ 3 Rewriteeachofthefollowingusingfactorialnotation. a n(n—1)(n-2)(n-3) b n(n-1)(n-2)(n-3)(n-4)(n-5) n{n—1)(n-2) 5x4x3x2x1 n{n-l)(n-2)(n-3)(n-4) 3x2x1

ThebookscouldbearrangedintheorderOGBR. Find the number ofdifferent ways that the 4 books can be arranged in aline. Youwillneedtobesystematic.Howmanywaysarethereofarrangingfivedifferentbooksinaline?

CLASSDISCUSSION

The first space can be filled in 3 ways with either A or B or C. Foreachofthese3waysoffillingthefirstspacethereare2waysoffillingthesecondspace. There are 3x2 ways of filling the first and second spaces. For each of the ways offilling the first and second spaces there isjust 1 way offilling the third space. Thereare3x2x1waysoffillingthethreespaces. The number of arrangements = 3x2x1 = 6. 3 X 2 X 1 is called 3 factorial and can be written as 3! ofIntheclassdiscussion,youshouldhavefoundthattherewere24differentwaysarrangingthe4books. 4! = 4x3x2x1 = 24 The number ofways ofarranging n distinct items in a line = n\. WORKEDEXAMPLE2 00000 \ J L ) L J) I -JJ a Findthenumberofdifferentarrangementsoftheseninecardsifthereareno restrictions. b FindthenumberofarrangementsthatbeginwithGRAD. C FindthenumberofarrangementsthatbeginwithGandendwithS.

Thesebooksarearrangedintheorder

CambridgeIGCSEand0LevelAdditionalMathematics 10.2 Arrangements

Theseare:ABC ACB BAG BCA CABandCBA. Thereare6differentarrangements. Method2 Consider filling 3spaces.nilmgd □□

TofindthenumberofwaysofarrangingthelettersA,BandCinalineyoucanusetwomethods. Method1 Listallthepossiblearrangements.

G.(Blue,Red, Orange,Green).

cThefirstandlastlettersarefixed,sotherearenow7letterstoarrangebetween theGandtheS. numberofarrangements=7!=5040

AnotherThereare4!X3!waysofhavingthesquaresatthestartandthecirclesattheend.possiblearrangementis: OOOiio I 1 Thereare3!x4!waysofhavingthecirclesatthestartandthesquaresattheend.totalnumberofarrangements=4!x3!+3!x4!=144+144=288

□□iOO

Answers a Thereare9differentcards. numberofarrangements=9!=362880

ThefirstfourlettersareGRAD,sotherearenowonly5letterslefttobearranged. numberofarrangements=5!=120

Thereare4!differentwaysofarrangingthefoursquares. Thereare31differentwaysofarrangingthethreecircles. Sothetotalnumberofpossiblearrangements=4!x3!=24X6=144. c Ifthesquaresarealltogether,apossiblearrangementis: oO'D □ □io c:

□□□□ooo

Findthenumberofdifferentarrangementsofthesesevenobjectsifthereare norestrictions. Findthenumberofarrangementswherethesquaresandcirclesalternate. Findthenumberofarrangementswhereallthesquaresaretogether. Findthenumberofarrangementswherethesquaresaretogetherandthecircles are together. Answers a Thereare7differentobjects.numberofarrangements=7! =5040

Thenumberofwaysofarrangingthe1blockoffoursquaresandthe3 circles = 4! Thereare4!waysofarrangingthefoursquareswithintheblockofsquares. Sothetotalnumberofpossiblearrangements=4!x4!=24x24=576. d Ifthesquaresaretogetherandthecirclesaretogether,apossiblearrangementis: □ □ Q

WORKEDEXAMPLE3

Chapter 10: Permutation',and combination-i

Ifthesquaresandcirclesalternate,apossiblearrangementis: □onoooD

2 Findthenumberofdifferentarrangementsoflettersineachofthe followingwords. aTIGER bOLYMPICS cPAINTBRUSH

4Ashelfholds7differentbooks. 4ofthebooksarecookerybooksand3ofthebooksarehistorybooks. a Findthenumberofwaysthebookscanbearrangedifthereare norestrictions. b Findthenumberofwaysthebookscanbearrangedifthe4cookerybooksarekepttogether.

3 a Findthenumberofdifferentfour-digitnumbersthatcanbeformed using the digits 3,5,7and 8 withoutrepetition. b Howmanyofthesefour-digitnumbersare i even li greaterthan8000?

CambridgeIGCSEand0LevelAdditionalMathematics

aEachFive-digitnumbersaretobeformedusingthedigits2,3,4,5and6.digitmayonlybeusedonceinanynumber.Eindhowmanydifferentfive-digitnumberscanbeformed. Howmanyofthesefive-digitnumbersare b even c greaterthan40000 d evenandgreaterthan40000?

7 a Eindthenumberofdifferentarrangementsofthelettersinthe wordORANGE. Findthenumberofthesearrangementsthat b beginwiththeletterO chavetheletterOatoneendandtheletterEattheotherend.

Threegirlsandtwoboysaretobeseatedinarow. Eindthenumberofdifferentwaysthatthiscanbedoneif a thegirlsandboyssitalternately b agirlsitsateachendoftherow c thegirlssittogetherandtheboyssittogether.

Findthenumberofdifferentarrangements a 4peoplesittinginarowonabench b 7differentbooksonashelf.

Exercise10.2

8 a Eindthenumberofdifferentsix-digitnumberswhichcanbemadeusing thedigits0,1,2,3,4and5withoutrepetition.Assumethatanumber cannotbeginwith0.

b Howmanyofthesix-digitnumbersinpartaareeven?

the number ofpermutations ofr items from n nldistinctitemsis"P^= (n-r)!

8x7x6x5x4x3x2x1 8! 8! So^Pg 8' -=8^=336. 5x4x3x2x1 5! (8-3)1 (8-3)1 51 The

10.3Permutations

For each of these 8 ways offilling the first space there are 7 ways offilling the secondspace. There are 8x7 ways of filling the first and second spaces. Foreachofthewaysoffillingthefirstandsecondspacesthereare6waysoffillingthethirdspace. There are 8x7x6 ways of filling the three spaces.

•

Chapter10:Permutationsandcombinations

96girlsand2boysaretobeseatedinarow.Findthenumberofwaysthatthiscanbedoneifthe2boysmusthaveexactly4girlsseatedbetweenthem.

•

Now consider having 8 letters A, B, C, D, E, F, G, H and 3 spaces to fill. Thefirstspacecanbefilledin8ways.

Inthelastsection,youlearntthatifyouhadthreelettersA,BandCand3spacestofill,thenthenumberofwaysoffillingthespaceswas3x2x1=3!

Thenumberofdifferentwaysofarrangingthreeletterschosenfromeight letters=8x7x6=336. The different arrangements of the letters are called permutations. Thenotationisusedtorepresentthenumberofpermutationsof3items chosenfrom8items. Notethat8x7x6canalsobewrittenas general rule for finding Note: Inpermutations,ordermatters. Bydefinition,0!=1.

□□□

CambridgeIGCSEand0LevelAdditionalMathematicsToexplainwhy0!=1,considerfindingthenumberofpermutationsof5 letterstakenfrom5letters. // t \\ 5 4 3 2 1 choices choices choices choices choice

Thefirstthreespacesmustbefilledwiththreeofthesixletters.Thereisachoiceof6forthefirstspace,5forthesecondspaceand4forthethird space.

Thereare6lettersand5digitstoselectfrom.Numberofarrangementsof3lettersfrom6letters=Numberofarrangementsof2digitsfrom5digits= Sonumberofpossiblesecuritycodes=®P3x^P2=120x20=2400. Method2 Thereare6lettersand5digitstoselectfrom.

choices choices choices choices choices

Thelasttwospacesmustbefilledwithtwoofthe5digits.Thereisachoiceof5forthefirstspaceand4forthesecondspace.Sonumberofpossiblesecuritycodes=6x5x4x5x4=2400.

230 WORKEDEXAMPLE4

(5 120 0! 120

Thenumberofwaysoffillingthe5spaceswiththe5letters=5x4x3x2x1=120. But,5P5= -5)! So =120.Hence0!mustbeequalto1. ^ Asecuritycodeconsistsof3lettersselectedfromA,B,C,D,E,Ffollowedby2digits selectedfrom5,6,7,8,9. Findthenumberofpossiblesecuritycodesifnoletterornumbercanberepeated. Answers Method1

Answers Method1 Thefirstnumbermustbea3andthelastnumbermustbea6.

WORKEDEXAMPLE5

Considerthenumberofchoicesforfillingeachofthefourspaces. / t \ \ 1 4 3 1 choice choices choices choice

Exercise10.3

Therearenowtwospacestofillusingtwooftheremainingfourdigits1,5,7and9.

4 Thereare8competitorsinalongjumpcompetition. Inhowmanydifferentwayscanthefirst,secondandthirdprizes beawarded?

Numberofwaysoffillingtheremainingtwospaces=P2= Thereare12differentnumbersthatsatisfytheconditions. 4!4! (4-2)! 2! 12.

Chapter10:Permutationsandcombinations

2 Findthenumberofdifferentwaysthat4books,chosenfrom6booksbe arrangedonashelf.

Method2

Findhowmanyevennumbersbetween3000and4000canbeformedusingthedigits 1,3,5,6,7and9ifnonumbercanberepeated.

Thereare12differentnumbersthatsatisfytheconditions. Method3

Numberofwaysoffillingthefourspaces=1x4x3x1

Inthisexampleitisnotimpracticaltolistallthepossiblepermutations. Theseare:3156 3516 3176 3716 3196 3916 3576 3756 3196 3916 3796 3976 Thereare12differentnumbersthatsatisfytheconditions.

3Howmanydifferentfive-digitnumberscanbeformedfromthedigits1,2, 3,4,5,6,7,8,9ifnodigitcanberepeated?

1 Withoutusingacalculator,findthevalueofeachofthefollowing. Usethe keyonyoucalculatortocheckyouranswers. a% b c"Pg d%

5CambridgeIGCSEand0LevelAdditionalMathematicsFindhowmanydifferentfour-digitnumbersgreaterthan4000thatcanheformedusingthedigits1,2,3,4,5,6and7ifnodigitcanbeusedmore thanonce.

I/) Z o Acombinationisaselectionofsomeitemswheretheorderofselectiondoes notmatter. Soforcombinationsorderdoesnotmatter. Considerthesetof5crayons. fromTofindthenumberofdifferentwaysofchoosing3crayonsthesetof5crayonsyoucanusetwomethods. MethodI Listallthepossibleselections. Ill no III HQ 111 101 HQ 101HI101 So there are 10differentways ofchoosing 3crayonsfrom 5.

7 Afour-digit numberisformed usingfour ofthe eight digits 1,2,3,4,5,6,7 and8.Nodigitcanbeusedmorethanonce.

You have alreadyinvestigated the number ofways ofarranging 4differentbooksinaline. Youarenowgoingtoconsiderthenumberofwaysyoucanselect3 booksfromthe4bookswheretheorderofselectiondoesnotmatter. Iftheorderdoesnotmatter,thentheselectionBROisthesameas OBR. Find the number ofdifferent ways ofselecting 3booksfrom the 4books.

6 Findhowmanyevennumbersbetween5000and6000canbeformedfrom thedigits2,4,5,7,8,ifnodigitcanbeusedmorethanonce.

9 Findhowmanydifferentevenfour-digitnumbersgreaterthan2000can heformed using the digits 1,2,3,4,5,6,7,8ifno digit may be used more thanonce.

Findhowmanydifferentfour-digitnumberscanbeformedif atherearenorestrictions bthenumberisodd c thenumberisgreaterthan6000 d thenumberisoddandgreaterthan6000.

8 Numbersareformedusingthedigits3,5,6,8and9. Nodigitcanbeusedmorethanonce. Findhowmanydifferent a three-digitnumberscanbeformed b numbersusingthreeormoredigitscanbeformed.

CLASSDISCUSSION

Chapter10: Permutations and combinations Method2 Thenumberofcombinationsof3from5= 10.4 Combinations 5! 5x4 3!2! 2! =10. The general ruleforfinding the number ofcombinationsofritemsfrom ndistinctitemsis: f nl rl(n—r)! Note: •Ine©mbinati©n§,©rdtrdotsnotmatttn • isrtadas'nehoostr\ So ®G|is read as"B choose S\ • canalsobewrittenas(^ • isthesameas Sothenumberof ofehoosittfSfrom10isthesameasthenumber ofwaysofehoosinf7from10-.Thisisbecausewhenyouchooseagroup ofSfrom 10,you are automatically left with a group of7. WORKEDEXAMPLE6 Ateamof6swimmersistobeselectedfromagroupof20swimmers. Findthenumberofdifferentwaysinwhichtheteamcanbeselected. Answers 20! Numberofwaysofselectingtheteam=

4boys 1 girl number ofways=^€4X =15 x4=60 3boys 2girls numberofways=®Cgx^€2=20x6=120

A B

WORKEDEXAMPLE9

Aquizteamof5studentsistobeselectedfrom6boysand4girls.

Iftherearemoreboysthangirlstherecouldbe:

Numberofwaysofchoosing3peoplefrom7fortentA= =35.

5boys 0girls numberofways= x^Cq=6x1=6

So,thenumberofpossibleselections=^Cgx^C2=84x21=1764.

Findthenumberofdifferentselectionsthatcanbemade. Answers

WORKEDEXAMPLE8

CambridgeIGCSEand0LevelAdditionalMathematics WORKEDEXAMPLE7

3coatsand2dressesaretobeselectedfrom9coatsand7dresses.

Findthenumberofpossibleteamsthatcanbeselectedinwhichtherearemoreboys thangirls. Answers

Numberofwaysofchoosing2dressesfrom7dresses=^C2.

Thediagramshows2differenttentsAandB.TentAholds3peopleandtentBholds4people.Findthenumberofwaysinwhich7peoplecanbeassignedtothetwotents. Answers

So,thenumberofwaysofassigningthe7peopletothetwotents=35.

Numberofwaysofchoosing3coatsfrom9coats=^Cg.

So,thetotalnumberofpossibleteams=6+60+120=186.

Chapter10:Permutationsandcombinations

Numberofwaysofchoosing1from2piecesbyBach=^Ci So,numberofpossibleselections='C2x^C2x^Ci=21x6x2=252. C Ifthereisatleastonepiecebyeachcomposertherecouldbe; 3Chopin 1 Mozart 1 Bach number ofways='C3X'^Ci x ^Ci =35 x 4 x 2=280 1 Chopin 3 Mozart 1 Bach number ofways='Cjx ^Cjx ^Cj=7x4x2=56 2 Chopin 2 Mozart 1 Bach number ofways=''C2 x ^Cg x ^Cj=21 x 6X 2=252 1 Mozart 2Bach number ofways='C2x'^Cj x ^C2=21 x 4x 1=84 Chopin 2 Mozart 2Bach number ofways='Ci X'*C2 x ^C2=7x6x1=42

WORKEDEXAMPLE10

Sofiahastoplay5piecesofmusicforhermusicexamination. Shehas13piecesofmusictochoosefrom. Thereare7pieceswrittenbyChopin,4writtenbyMozartand2writtenbyBach.

2Chopin

Findthenumberofwaysthe5piecescanbechosenif a therearenorestrictions b there mustbe 2pieces by Chopin,2pieces by Mozartand 1 piece by Bach c theremustbeatleastonepiecebyeachcomposer.

Answers aNumberofwaysofchoosing5from13=^^Cs=1287 b Numberofwaysofchoosing2from7piecesbyChopin= Number ofways ofchoosing 2from 4 pieces by Mozart=^Cj

So,totalnumberofways=280+56+252+84+42=714. 0 Exercise10.4 1 Withoutusingacalculatorfindthevalueofeachofthefollowing,andthen use the"C^key on your calculator to check your answers. a ^Ci b ^C3 c 4C4 d (^) e (I) f 2Showthat =^Cs. 3 Howmanydifferentwaysarethereofselectinga3photographsfrom10photographs b5booksfrom7books cateamof11footballersfrom14footballers? 4 How many differentcombinations of3letters can be chosenfrom the lettersP,Q,R,S,T?

9Anathleticsclubhas10longdistancerunners,8sprintersand5jumpers. Ateamof3longdistancerunners,5sprintersand2jumpersistobeselected. Findthenumberofwaysinwhichtheteamcanbeselected.

10Ateamof5membersistobechosenfrom5menand3women. Findthenumberofdifferentteamsthatcanbechosen aiftherearenorestrictions b thatconsistof3menand2women cthatconsistofnomorethan1woman.

11Acommitteeof5peopleistobechosenfrom6womenand7men.

Findthenumberofdifferentcommitteesthatcanbechosen aiftherearenorestrictionsbiftherearemorementhanwomen.

6Findthenumberofwaysinwhichthe8toyscanbeplacedintheboxessothat5toysareinboxAand3toysareinboxB.4pencilsand3pensaretobeselectedfromacollectionof8pencilsand 5pens. Findthenumberofdifferentselectionsthatcanbemade.

CambridgeIGCSEand0LevelAdditionalMathematics

A B

Thediagramshows2differentboxes,AandB. 8 different toys are to be placed in the boxes.

Findthenumberofdifferentwaysinwhichitispossibletoanswer a exacdy10questionscorrectly b exactly25questionscorrectly.

12Acommitteeof6peopleistobechosenfrom6menand7women. Thecommitteemustcontainatleast1man.Findthenumberofdifferentcommitteesthatcanbeformed.

13 A school committee of5 people is to be selected from a group of4teachers and7students.

FourofthelettersofthewordPAINTBRUSHareselectedatrandom.Findthenumberofdifferentcombinationsif athereisnorestrictiononthelettersselectedbtheletterTmustbeselected.

Atestconsistsof30questions. Eachansweriseithercorrectorincorrect.

Findthenumberofdifferentwaysthatthecommitteecanbeselectedif atherearenorestrictionsbtheremustbeatleast1teacherandtheremustbemorestudents thanteachers.

Summary Arrangementsinaline Thenumberofwaysofarrangingndistinctitemsinalineis nx(n-1)X(n-2)x...x3x2x1=n! Permutations Thenumberofpermutationsofritemsfromndistinctitemsis np ^(n—r)\ Inpermutations,ordermatters. TheCombinationsnumberofcombinationsofritemsfromndistinctitemsis n\"c,=(;!)=•r\{n—r)\ Incombinations,orderdoesnotmatter. si .i

Findthenumberofwaysthatthe6entertainerscanbeselected.

Chapter10: Permutationsand combinations

4oftheCDsareclassical,3arejazzand3arerock. Heselects5oftheCDsfromhiscollection.

16 In a group of15entertainers,there are6singers,5guitarists and 4comedians. Ashowistobegivenby6oftheseentertainers.

14Atestconsistsof10differentquestions.

4ofthequestionsareontrigonometryand6questionsareonalgebra. Studentsareaskedtoanswer8questions. a Find the number ofwaysin which students can select8questionsifthere arenorestrictions, b Findthenumberoftbeseselectionswhichcontainatleast4algebra questions.

Findthenumberofwayshecanmakehisselectionif a therearenorestrictions b hisselectionmustcontainhisfavouritejazzCD chisselectionmustcontainatleast3classicalCDs.

15Rafiuhasacollectionof10CDs.

Intheshowtheremustbeatleast1guitaristand1comedian. Theremustalsobemoresingersthanguitarists.

CambridgeIGCSEand0LevelAdditionalMathematics Examination questions

1 a i Findhowmanydifferent4-digitnumberscanbeformedusingthedigits1,2,3,4,5 and6ifnodigitisrepeated. [1]

Workedexample

iiHowmanyofthe4-digitnumbersfoundinpartiaregreaterthan6000? [1] ill How many ofthe 4-digit numbersfound in parti are greater than 6000and areodd? [1]

bAquizteamof10playersistobechosenfromaclassof8boysand12girls.

I Findthenumberofdifferentteamsthatcanbechoseniftheteamhastohaveequal numbersofgirlsandboys. [3]

II Findthenumberofdifferentteamsthatcanbechoseniftheteamhastoincludethe youngestandoldestboyandtheyoungestandoldestgirl. [2]

CambridgeIGCSEAdditionalMathematics0606Paper11Q1Oai,ii.Hi,hi,iiNov2014

Answers 1 a I Number of4-digit numbers= ®P4= 360. II Method1

Thefirstnumbermustbea6. There are now three spaces to fill using three ofthe remaining five digits 1,2,3,4 and 5. 5' 5p-Number ofways offilling the remaining three spaces= P3 Thereare60differentnumbersthatsatisfytheconditions. 5! (5-3) -=—=60. 2! Method2

Considerthenumberofchoicesforfillingeachofthefourspaces. / f \ \ 1 choice 5 choices 4 3 choiceschoices Numberofwaysoffillingthefourspaces=1x5x4x3. Thereare60differentnumbersthatsatisfytheconditions.

Whenthefirstandlastspaceshavebeenfilledtherewillbefournumbersleftto choosefrom. / f \ \ 1 4 3 3 choice choices choices choices

Themiddletwospacesmustthenbefilledusingtwooftheremainingfournumbers.Numberofwaysoffillingthefourspaces=1xx3=1x 41 41 ——x3=3x—=36. (4-2)! 2!

Numberofwaysoffillingthefourspaces=1x4x3x3.

Thelastnumbermustbea1,3or5.

Numberofpossibleteams=^€5x =56x792=44352.

iiTheteamincludestheyoungestandoldestboyandtheyoungestandoldestgirl.

Chapter10:Permutationsandcombinations

Thereare36differentnumbersthatsatisfytheconditions. Method2 Considerthenumberofchoicesforfillingeachofthefourspaces.

Numberofwaysofchoosing6from16= =8008.

Numberofpossibleteams=8008. Exercise10,5 Exam Exercise 1aHowmanyevennumberslessthan500canbeformedusingthedigits1,2,3,4and5?

HiMethod1

Therearenow6placeslefttofilland16peoplelefttochoosefrom.

Thereare36differentnumbersthatsatisfytheconditions, bi Theremustbe5boysand5girls. Number ofways ofchoosing 5 boysfrom 8=^Cs. Number ofways ofchoosing 5 girls from 12= ^^€5.

Eachdigitmaybeusedonlyonceinanynumber. [4]

Thefirstnumbermustbea6. Thelastnumbermustbea1,3or5.

Thefirstnumbermustbea6.

CambridgeIGCSEand0LevelAdditionalMathematics b A committee of8 people is to be chosenfrom 7 men and 5 women.

i Calculatethenumberofdifferentcommitteesthatcanbeselected. [3]

Ofthesepaintings,5arebyPicasso,4byMonetand1byTurner.1Findthenumberofdifferentwaysthepaintingscanbedisplayedifthereare norestrictions. [1] ii Findthenumberofdifferentwaysthepaintingscanbedisplayedifthepaintingshyeachoftheartistsarekepttogether. [3]

I Find the number ofwaysin which this can be done. [1]

CambridgeIGCSEAdditionalMathematics0606Paper11Q10a,bi,iiJun2014

CambridgeIGCSEAdditionalMathematics0606Paper21Q9ai,ii,bi,iiNov2012

Oneofthe6seniorstudentsisacousinofoneofthe5juniorstudents. Ii Calculatethenumberofdifferentcommitteeswhichcanbeselectedifatmostoneof thesecousinsisincluded. [3]

Ticketsforaconcertaregivenoutrandomlytoaclasscontaining20students. Nostudentisgivenmorethanoneticket.Thereare15tickets.

Anartgallerydisplays10paintingsinarow.

b Acommitteeof4seniorstudentsand2juniorstudentsistobeselectedfromagroupof6 seniorstudentsand5Juniorstudents.

3 a Arrangementscontaining5differentlettersfromthewordAMPLITUDEaretobemade. Find 1 the number of5-letter arrangements ifthere are no restrictions, [1] Ii thenumberof5-letterarrangementswhichstartwiththeletterAandendwiththe letterE. [1]

CambridgeIGCSEAdditionalMathematics0606Paper11Q4ai,ii,bi,ii,iiiJun2012

Findthenumberofdifferentcommitteesthatcouldbeselectedif I thecommitteecontainsatleast3menandatleast3women, [4] if theoldestmanortheoldestwoman,butnotboth,mustbeincludedinthecommittee.[2]

Thereare12boysand8girlsintheclass.Findthenumberofdifferentwaysinwhich II 10 boys and 5girls get tickets, [3] IIIalltheboysgettickets. [1]

bThesixdigitsofthenumber738925arerearrangedsothattheresultingsix-digit numberiseven.Findthenumberofdifferentwaysinwhichthiscanbedone. [2] CambridgeIGCSEAdditionalMathematics0606Paper22Q2Mar2015

CambridgeIGCSEAdditionalMathematics0606Paper21Q6i,ii,iiiNov2010 a Jeanhasninedifferentflags.

Chapter10:Permutationsandcombinations

b 3DVDsand 2videotapes are to be selected from a collection of7DVDsand 5videotapes. Calculatethenumberofdifferentselectionsthatcouldbemade. [3]

Eachdigitmayonlybeusedonceinanynumber,aFindhowmanydifferentsix-digitnumberscanbeformed.

[1][1][1][3]

Ashelfcontains8differenttravelbooks,ofwhich5areaboutEuropeand3are aboutAfrica.

Six-digitnumbersare tobeformed usingthedigits3,4,5,6,7and9.

[2] ii Find the number ofdifferent ways the books can be arranged ifthe 5 books about Europearekepttogether. [2]

Nodigitcanbeusedmorethanonceinanynumber. Findhowmanydifferent4-digitnumberscanbeformedif a b c thetherearenorestrictions,numberislessthan4000, thenumberisevenandlessthan4000.

CambridgeIGCSEAdditionalMathematics0606Paper11Q4i-ivNew2011

[2][2][2]

i FindthenumberofdifferentwaysinwhichJeancanchoosethreeflagsfromher nineflags. [1]

Findhowmanyofthesesix-digitnumbersare b c d 5 a even, evengreaterthan500000,andgreaterthan500000.

iiJeanhasfiveflagpolesinarow.Sheputsoneofhernineflagsoneachflagpole. Calculatethenumberofdifferentfive-flagarrangementsshecanmake. [1]

I Findthenumberofdifferentwaysthebookscanbearrangediftherearenorestrictions.

CambridgeIGCSEAdditionalMathematics0606Paper21Q6ai,ii,bJun2011

A 4-digit number isformed by usingfour ofthe seven digits 1,3,4,5,7,8and 9.

Ateamof6tennisplayersistobechosenfrom10tennisplayersconsistingof7menand3 women.Findthenumberofdifferentteamsthatcouldbechoseniftheteammustinclude atleast1woman. [3]

An examination consists ofa section A,containing 10shortquestions,and a section B, containing5longquestions.Candidatesarerequiredtoanswer6questionsfrom sectionAand3questionsfromsectionB.Findthenumberofdifferentselections ofquestionsthatcanbemadeif i therearenofurtherrestrictions, iicandidatesmustanswerthefirst2questionsinsectionAandthefirstquestion insectionB. [2][2]

CambridgeIGCSEAdditionalMathematics0606Paper11Q4Nov2015

CambridgeIGCSEand0LevelAdditionalMathematics 8 a 9 a 10a A lock can be opened using only the number 4351.State whether this is a permutationoracombinationofdigits,givingareasonforyouranswer. [1] There are twenty numbered balls in a bag.Two ofthe balls are numbered 0,six are numbered1,fivearenumbered2andsevenarenumbered3,asshowninthetablebelow. Numberonball Frequency 52 3 7

CambridgeIGCSEAdditionalMathematics0606Paper21Q5Jun2015

[1][1][2]

A6-characterpasswordistobechosenfromthefollowing9characters, letters A B E F numbers 5 8 9 symbols * $

Eachcharactermaybeusedonlyonceinanypassword.Findthenumberofdifferent6-characterpasswordsthatmaybechosenifitherearenorestrictions, [1] ii thepasswordmustconsistof2letters,2numbersand2symbols,inthatorder, [2] iiithepasswordmuststartandfinishwithasymbol. [2]

6booksaretobechosenatrandomfrom8differentbooks. I Findthenumberofdifferentselectionsof6booksthatcouldbemade. Aclockistobedisplayedonashelfwith3ofthe8differentbooksoneachsideofit. Findthenumberofwaysthiscanbedoneif iitherearenorestrictionsonthechoiceofbooks, iii 3ofthe8booksaremusicbookswhichhavetobekepttogether.

Fouroftheseballsarechosenatrandom,withoutreplacement.Calculatethenumberofwaysthiscanbedonesothat i thefourballsallhavethesamenumber, [2] ilthefourballsallhavedifferentnumbers, [2] Hi thefourballshavenumbersthattotal3. [3]

CambridgeIGCSEAdditionalMathematics0606Paper12Q5Mar2016

Chapter 11 Series Mi. Thissectionwillshowyouhowto: usethebinomialtheoremforexpansionof{a+b)"forpositiveintegral« / \ ^a"~''b'^forabinomialexpansion V^J usethegeneralterm ■ recognisearithmeticandgeometricprogressions ■ usetheformulaforthenthtermandforthesumofthefirstntermstosolveproblemsinvolving arithmeticandgeometricprogressions ■ usetheconditionfortheconvergenceofageometricprogression,andtheformulaforthesumto infinityofaconvergentgeometricprogression.

•Eachnumberisthesumofthetwonumbersintherowaboveit.Thenextrowwouldbe:15101051 This row can then be used to write down the expansion of(a+ b)^. {a+bf=la®+5a^6+lOa^ft^

•Allofthetermshaveatotalindexof4(a^,a^6,a^6^,aJ?andb^).

Thereisasimilarpatternintheotherexpansions.ThecoefficientsalsoformapatternthatisknownasPascal'striangle. 11 '5' +ii Note:

The expansion of(a+ bf'can be used to expand(a+ b)^. {a+b)^={a+b){a+b)^ ={a+b)(<2^+2ab+b^) =c?+2c?b+alP^+c?b+2al^+I? = +Sa^b+3ab^+b^

CambridgeIGCSEand0LevelAdditionalMathematics 11.1 Pascal's triangle

Theword'binomial'means'twoterms'.

Ifyoulookattheexpansion of(a+b)^,youshould notice thatthe powersofa andbformapattern.

Similarlyitcanbeshownthat{a+b)'^=a^+Aa^b+ +Aab^+6^.

• Eachrowalwaysstartsandfinisheswitha1.

Writingtheexpansionsof(a+6)"outinorder: {a-¥b)^= la + 16 (a+b)2 la^ + 2ab + 16^ (a+6)^= la^+3c^6+3a6^+16^ (a+6)^=la^+4a^6+6a^6^+Aab^+16"^

Thewordisusedinalgebraforexpressionssuchasx+5and2x-Sy. Youshouldalreadyknowthat(a+b)^={a+b){a+b)= +2ab+b^.

•Thefirsttermis andthenthepowerofadecreasesby1whilstthepower ofbincreasesby1ineachsuccessiveterm.

Thesenumbersarecalledtetrahedralnumbers. WhatWhichothernumberpatternscanyoufindinPascal'striangle?doyounoticeifyoufindthetotalofeachrowinPascal'striangle?

Chapter11:Series

CLASSDISCUSSION

ForTherearemanynumberpatternsinPascal'striangle.example,thenumbers1,4,10and20havebeenhighlighted.

WORKEDEXAMPLE1

O 1 4 10

UsePascal'striangletofindtheexpansionof a (2+5x)3 b (2x-3)4 Answers a(2+5x)^Theindex=3sousethethirdrowinPascal'striangle.ThethirdrowofPascal'striangleis1,3,3and1. (2+5x)^=1(2)^+3(2)^(5x)+3(2)(5x)^+i Usetheexpansionof(n+6)^. =8+60x+150x?+125j(? b(2x-3)^ Theindex=4sousethefourthrowinPascal'striangle. ThefourthrowofPascal'striangleis1,4,6,4and1. (2jc-3)'*=1(2*:)^+4(2x)®(-3)+6(2x)^{—3)^ Usetheexpansionof(a+£>)''. +4(2x)(-3)3+1(-3)^ =Ifix'*-96*?+216x^-216x^+81

Sothecoefficientofx^is-40+240=200.

1X(-40X®)=-40x®and3xx80 =240x®

WritedownthesixthandseventhrowsofPascal'striangle.

Theterminx^comesfromtheproducts: (1+3x)(32—80x+80x~-40X®+lOx'*-x^)

ThefifthrowofPascal'striangleis1,5,10,10,5and1. (2-x)®=1(2)5+5(2)''(-x)+10(2)^(-x)2+io(2)2(-x)^+5(2)(-x)^+1(-x)® =32-80x+80x2_40^3+jqx^-x®

CambridgeIGCSEand0LevelAdditionalMathematics

a Expand(2-x)®. b Findthecoefficientof intheexpansionof(1+3x)(2-x)®. Answers a (2-x)®Theindex=5sousethefifthrowinPascal'striangle.

WORKEDEXAMPLE2

b (1+3x)(2-x)5=(1+3x)(32-80x+80x2_40^3+jqx^-x^)

Expand(1+2x)(l+3x)^.Thecoefficientofxinthe expansion of(2+axf is96. Findthevalueoftheconstanta. a Expand(3+x)^. bUseyouranswertopartatoexpress(3+Vs)intheforma+b-Jb. 2x^ (3+ 2x)^ 3--X 2

Exercise11.1 21

UsePascal'striangletofindtheexpansionsof (1 + xf b (1- xY f 9 e (x + f (3)+ 4)^ i (x - 23)^ j (3x - 4)^ Find the coefficient of x^ in the expansions of c (p+^r g {a-bf k 2 XH X 5 6 d Ihd h (2+ x)^ (2x + 3)^ 2 1 a (x + 4)^ b (1 + x)® c (3- x)^ e (x-2)^ f (2x+5)^ g (4x-3)® (4 + x)^ +(4- x)^ = A +5x2 4. FindthevalueofA,thevalueofBandthevalueofC.

Chapter11:Series 8 a Expand (1+jc)^. b Useyouranswertopartatoexpress i (l + y/E'f in the form a + b-yjs, ii (l- Vs) in the form c + rf-v/S. cUseyouranswerstopartbtosimplify(l+Vs)+(l-Vs). 9 a Expand(2 bFindthecoefficientofx®intheexpansionof(l+3x^)(2-x^)^. 10Findthecoefficientofxintheexpansion ( 1 Y11 Find the term independent of x in the expansion of I + —J .

CHALLENGEQ 12a Findthefirstthreeterms,inascendingpowersofy,inthe expansion of(2+y). b Byreplacingywith -4x^,findthecoefficientofx^inthe expansion of(2+ 3x -4x^)^. CHALLENGEQ 13 Thecoefficientofx^in the expansion of(3+ax)^ is 12timesthe / \4coefficient of x^ in the expansion of M + —j. Find the value of a. CHALLENGEQ 14a Given that ^x^^ ~ )~ ^~ value ofaand thevalueofb. b Hence,withoutusingacalculator,findtheexactvalueof 2+ CHALLENGEQ 15Giventhaty=x+—,express X 3 1 ■a X +-J mtermsofy b X® +-^ in terms ofy.

CambridgeIGCSEand0LevelAdditionalMathematics CLASSDISCUSSION

What do you have to do to find the number of routes to anystone? How manyroutes are there from the STARTstone to the FINISH stone? In the class discussion you should have found that the number of routes from the START stone to stone Qis10.

Hencethenumberofroutesisthesameasthenumberofdifferentcombinationsof3F'sand2S's. Thecombinationsare: the number of routes is 10. This is the same as ^€3 (or ^€2).

Thediagramshowsthereare3routesfromtheSTARTstonetostoneG.

D H I N M R n rv C G L Q u F K • 1 p| T w J 0 S V FINISH

FFESS FFSFS FFSSE ESFSE ESEES ESSFF SSFFF SESEE SEESE SEEES So

Thesteppingstonegame Therulesarethatyoucanmove East—► or Southj from anystone.

a A ii B b C ii D iii E c F ii G iii H iv I

1FindthenumberofroutesfromtheSTARTstonetoeachofthefoliowingstones.

Whatdo younoticeaboutyouranswers to parts a, band c?

There are 6 routes from the START to stone L. How couldyouhave calculated that there are 6 routeswithout drawing orvisualisingthem?

TomovefromSTARTtoQyoumustmoveFast(F)3andSouth(S)2,inanyorder.

Answers (3 + 4x)5 = 3^ + 5Ci 3^4x) + ®C2 5^(4xf +^Cg 32(4x)3 + 3(4*)^+ {4xy =243+1620X+4320*2+5760*®+3840*^+1024*®

So theexpansionof{a+hf is: (a + bf = ^Co + ®Ci a%+ +^C4 b^ Thiscanbewrittenmoregenerallyas: {a+by="Coa"+"Cia""^6+"Cg +"Cg0""^^+...+ ''6''+...+"C„6" But"Co=1and"C„=l,sotheformulasimplifiesto:(«+^>)"=a"+"Cia"-i6+"C2+ + "C,a""'■6'"+... +y) or n(a+6)"=a"+ I j f \ n v2y i) K a"-^6'"+...+6" -Si J

- *)2® = 22® + 25c, 22^ (-*) + 2®C2 22® (-*)2 + ... + 2® (-*)2» + ... + (-x)25 Thetermcontaining*2°is25C2oX2®x(-*)2''. = 53130 X 32 X *20 = 1700160*2" Sothecoefficientof*2"is1700160.

Find the coefficient of *2" in the expansion of (2 — *)2®. Answers

Usingacalculator: ^Co= These numbers are the same as numbersin fifth row ofPascal's triangle.

the

Pascal's11.2Thebinomialtheoremtrianglecanbeusedtoexpand{a+&)"foranypositiveintegern,butifnislargeitcantakea longtime.Combinationscanbeusedtohelpexpandbinomialexpressionsmorequickly.

Chapter11:Series

the

Theformulaeaboveareknownasthebinomialtheorem. WORKEDEXAMPLE3 Usethebinomialtheoremtoexpand(3+4x)®.

WORKEDEXAMPLE4 (2

1 ^Ci =5 ^C2= 10 = 10 ^€4=5 ^Cs= 1

CambridgeIGCSEand0LevelAdditionalMathematics Usingthebinomialtheorem, (1 + x)^= V +'Ci 1® X +'C2 1® 1^ X^+^C4 l^x^+... =1+ X+''C2x^+'C3x^+'C4x^+... But'Cj,'C2,'Cg and'C4 can also be written as: ''Ci=—=7 1!6! ^C2= 7! 7x6 2!5! 2! ^Cg7! 7x6x5 314! 3! 7! 7x6x5x4 413! 4! So,(1+ —1+7x+ 7x6 x^+ 7x6x5 X®+ 7x6x5x4 2! 3! 4! Thisleadstoanalternativeformulaforbinomialexpansions: x^+... (1+x)"=1+nx+ n{n -1)^^ n{n -l)(w - 2)^^ n{n- l)(n - 2)(n -3) ^4 21 31 41 + Thefollowingexampleillustrateshowthisalternativeformulacanbeapplied. WORKEDEXAMPLE5 Findthefirstfourtermsofthebinomialexpansionto a (l+S;;)' b (2-31)® Answers a (i+33/=H-7(33)+^(3>.f + 7x6x5 3! (37)'+b (2-3^ =1+213+189/+9453^+... 6 2|1 =2®|1-1 1 +61-11+.^f-lT + 6x5x47 3? ^ * 3! ■ 2® 2; 21V2 =2®^ L '4' -64-1923+24O32-I6O3®+... 1-33+^ Replacexby33;andwby7intheformula Theformulaisfor(1+x)"sotakeoutafactor of2. Replacexby^ ^^and n by6in the formula. Multiplytermsinbraeketsby2®.

Chapter11:Series

5 a Write down,in ascending powers of x, the first 4 terms in the expansion of(1 + 2x)®.

b Find the coefficient of x^ in the expansions of 1-- j(l + 2x)®. ^ f x 6 a Write down,in ascending powers of x, the first4 terms in the expansion of 1+ f xV®b Find the coefficient of x^ in the expansions of(1 + 3x)l 1+ —j .

b Find the coefficient of x^ in the expansions of(1- 4x)(l- 3x)^''. 8aFind,inascendingpowersofx,thefirst3termsoftheexpansionof(1+2x)'.bHencefindthecoefficientofx^intheexpansionof(1+2x)'(l-3x+5x^).

2 Use the binomial theorem to find the expansions of a (1+ e (x + 3))^ b (1-x)^ f (2-x)^ j (1c (l + 2x)' g {a-nf .2 V lOy V X,

Exercise11.2

4 Usethebinomialtheoremtofindthefirstthreetermsineachoftheseexpansions. b (l+2x)® c (l-3x)^ d (3+ 2x)®a (1 + x)'° e (3-x)® f 1^2+^xj g (5-x2)® h (4x-53!f''.

7 a Write down,in ascending powersofx,the first4termsin the expansion of(1-3x)"^

9 a Find,in ascending powers of x, the first4 terms of the expansion of(1 + x)'. b Hence find the coefficient of in the expansion of(1 + 31-312)'. f 3V10FindthecoefficientofxinthebinomialexpansionofIx J. 11 Find the term independent of x in the binomial expansion of|^x + • CHALLENGEQ 12 When(1+ ax)" is expanded the coefficients of and are equal. Findaintermsofn.

Findthetermin foreachofthefollowingexpansions, a(2+xf b(5+xf c(l+2x)® e (l-xf f {2-xf g (10-3x) d (3+ xf h (2x + 33)f I x^+ 1 2x2 d (3+2xf h (4-bxf^.

1WritethefollowingrowsofPascal'striangleusingcombinationnotation, arow3 brow4 c row5

AtIGCSElevel you learntthata numbersequence is an ordered setof numbers thatsatisfy a rule and thatthe numbersin thesequence are called the terms of the sequence. A number sequence is also called a progression. Thesequence5,8,11,14,17,...

iscalled anarithmetic progression.Each term differsfromthetermbeforebyaconstant.Thisconstantiscalledthecommon difference.

Thefirstfivetermsofanarithmeticprogressionwhosefirsttermisaand whosecommondifferenceisdare: a a+d term1 term2 Thisleadstotheformula: a+2(i term3 u+3d term4 a+4d term5 1 nthterm= +(w—l)d 1

WORKEDEXAMPLE7

Firstterm=1.2,commondifference=0.8.

inthearithmeticprogression-17,-14,-11,-8, ,58. Answers nthterm=a+{n-\)d usefl=-17,d=3andnthterm=58 solve n-1=25n=26

CambridgeIGCSEand0LevelAdditionalMathematics 11.3 Arithmetic progressions

FindWORKEDEXAMPLE6thenumberofterms

58=-17+3(n-l)

Thenotationusedforarithmeticprogressionsis: a=firstterm d=commondifference I=lastterm

Thefifthtermofanarithmeticprogressionis4.4andtheninthtermis7.6.Findthe firsttermandthecommondifference. Answers fifthterm=4.4 =>a+4d=4.4 (1) ninthterm=7.6 =>a+8d=7.6 (2) (2)-(1),gives4d=3.2 d=0.8 Substitutingin(1)givesa+3.2=4.4 a=1.2

CLASSDISCUSSION

Thenthtermofanarithmeticprogressionis11-3n.Findthefirsttermandthe commondifference. Answers Firstterm=11-3(1)=8 substituten=1 intonthterm=11—3w secondterm=11-3(2)= 5 substitute m=2 into nthterm=11—3n Commondifference=secondterm-firstterm=-3. Thesumofanarithmeticprogression Whenthetermsinasequenceareaddedtogethertheresultingsumiscalledaseries.

2UseGauss'smethodtofindthesumof a 2+4+6+8+ +394+396+398+400 b 3+6+9+12+...+441+444+447+450 c 17+24+31+38+ +339+346+353+360. 3 UseGauss'smethodtofindanexpression,intermsofn,forthesum l+2+3+4+...+(n-3)+(n-2)+(TO-l)+TO. Itcanheshownthatthesumofanarithmeticprogression, canbewrittenas: +0 or S„ =-|[2a+(n-l)d] Proof: Reversing: ci +(fl+d)+(<z+2^i)+ +(/—2(1)+(^l—(f)+/ I +(/—(i)+(/—2cO+...+((2+2d)+{a+d)+a Adding: 2S„=n{a+I)+{a+I)+{a+I) +...+{a+l) +ia+I)+{a+I) 2S„=n{a+I) n =-(«+/)

Itis said thatatthe age ofeight,thefamous mathematician Carl Gauss was asked to find the sum ofthenumbersfrom1to100.Histeacherexpectedthistasktokeephimoccupiedforsometime butGausssurprisedhisteacherbywritingdownthecorrectanswerafterjustacoupleofseconds. 1Hismethodinvolvedaddingthenumbersinpairs:1+100=101,2+99=101,3+98=101,...Canyoucompletehismethodtofindtheanswer?

WORKEDEXAMPLE8 A

l+2+3+4+...+97+98+99+100=?

use«thterm=-38when«=19anda=25 solve usenthterm=-87whena=25andd=-3.5 solve Sn=^{a+l) usea=25,1=-87andn=33 33 %=y(25-87) =-1023

The12thterminanarithmeticprogressionis8andthesumofthefirst13termsis78. Findthefirsttermoftheprogressionandthecommondifference. Answers nthterm =a+(n-l)d usewthterm=8whenn=12 8~a+lid -(1) S„ = ■|[2a+(n-l)d] use w= 13 and =78 1378 = Y(2a+12d) simplify 6=a+fid (2) (1) - (2) gives5d=2 d = 0.4 Substituting d=0.4 inequation (1) givesa=3.fi. Firstterm=3.fi,commondifference=0.4.

Using 1= a+{n-\)d, gives =^[2a+(n-l)d]

It is useful to remember the following rule that applies for all progressions: nthterm=S„-S,•n-l

Cambridge IGCSE and0LevelAdditional Mathematics

WORKEDEXAMPLE9

WORKEDEXAMPLE10

Inanarithmeticprogression,thefirsttermis25,the19thtermis-38andthelast termis-87.Findthesumofallthetermsintheprogression. Answers nthterm=a+(n-l)d -38=25+18dd=-3.5 nthterm=a+{n-l)d -87=25-3.5(n-l) n-l=32n=33

Exercise11.3 1

Thefirstterminanarithmeticprogressionisaandthecommondifference isd. Writedownexpressions,intermsofaandd,forthefifthtermandthe 14thterm. 2Findthesumofeachofthesearithmeticseries. a 2+9+16+... (15terms) b 20+11+2+... (20terms) c 8.5+10+11.5+... (30terms) d -2x-5x-8x-... (40terms)

Chapter11:Series WORKEDEXAMPLE11

Thesumofthefirstnterms,5„,ofaparticulararithmeticprogressionisgivenby S„=brr-3n. a Findthefirsttermandthecommondifference. b Findanexpressionforthenthterm. Answers a Si=5(1)2-3(1)=2 => firstterm=2 .%=5(2)2-3(2)=14 => secondfirstterm+secondterm=14term=14-2=12 Firstterm=2,commondifference=10. b Method1: nthterm=a+(n-l)d usea=2,d=10 =2+10(n-1) =lOn-8 Method2: nthterm=-Sn- j= 5n2-3n-[5(n-l)2-3(n-l)] =5n2-3n-(Sn^-lOn+5-3n+3) =10n-8

4 Thefirsttermofanarithmeticprogressionis2andthesumofthefirst12 termsis618. Findthecommondifference.

3 Findthenumberoftermsandthesumofeachofthesearithmeticseries, a23+27+31...+159 b28+11-6-...-210

5 Inanarithmeticprogression,thefirsttermis-13,the20thtermis82and thelasttermis112.aFindthecommondifferenceandthenumberofterms, b Findthesumofthetermsinthisprogression.

6 The first two termsin an arithmetic progression are 57and 46.The last termis-207.Findthesumofallthetermsinthisprogression.

7Thefirsttwotermsinanarithmeticprogressionare-2and5.Thelasttermintheprogressionistheonlynumberintheprogressionthatisgreater than 200.Find the sum ofall the terms in the progression.

b Findthe55thtermintermsofa.

Findthefirsttermandthecommondifference, b Giventhatthenthtermofthisprogressionis-97,findthevalueofn.

10Thefirsttermofanarithmeticprogressionis7andtheeleventhtermis32.Thesumofallthetermsintheprogressionis2790.Findthenumberof termsintheprogression.

Tlgivenby =—(4n+5).Findanexpressionforthenthterm.

18 The eighth term in an arithmetic progression is three times the third term. Showthatthesumofthefirsteighttermsisfourtimesthesumofthefirst fourterms. CHALLENGEQ 19Thefirsttermofanarithmeticprogressioniscos^xandthesecond termis1. a Writedownanexpression,intermsofcosx,fortheseventhtermof thisprogression.

8 Thefirsttermofanarithmeticprogressionis8andthelasttermis34.The sum ofthe firstsix terms is 58.Find the number ofterms in this progression.

9FindthesumofalltheintegersbetweenlOOand400thataremultiplesof6.

14Thesumofthefirstnterms,S„,ofaparticulararithmeticprogressionis givenby =-Sn^-2n.Findthefirsttermandthecommondifference.

15Thesumofthefirst nterms, ofaparticulararithmeticprogressionis

11Rafiubuysaboatfor$15500.Hepaysforthisboatbymakingmonthly paymentsthatareinarithmeticprogression.Thefirstpaymentthathemakes is$140andthedebtisfullyrepaidafter31payments.Findthefifthpayment.

fci:

16Acircleisdmdedintotwelvesectors.Thesizesoftheanglesofthesectors

The eighth term ofan arithmetic progression is -10 and the sum ofthe firsttwentytermsis-350. a

CambridgeIGCSEand0LevelAdditionalMathematics I

13Thesumofthefirstnterms,S,„ofaparticulararithmeticprogressionis givenbyS„=4n-+2n.Findthefirsttermandthecommondifference.

Thesumofthefirst25termsis15timesthesumofthefirst4terms,aFindaintermsofd.

b Show that the sum ofthe first twenty terms ofthis progression is 20+170sin2x.

17areinarithmeticprogression.Theangleofthelargestsectoris6.5timestheangleofthesmallestsector.Findtheangleofthesmallestsector.Anarithmeticsequencehasfirsttermaandcommondifferenced.

Chapter11:Series CHALLENGEQ 20Thesumofthedigitsinthenumber56is11.(5+6=11)aShowthatthesumofthedigitsoftheintegersfrom15to18is30.bFindthesumofthedigitsoftheintegersfrom1to100. 11.4Geometricprogressions Thesequence7,14,28,56,112,...iscalledageometricprogression.Each termisdoubletheprecedingterm.Theconstantmultipleiscalledthe commonratio. Otherexamplesofgeometricprogressionsare:Progression Commonratio 1,-2,4,-8,16,-32,... -2 81,54,36,24,16,10-,... 3 32 -8,4,-2,l.-i'T'24 1 2 Thenotationusedforageometricprogressionis: a=firstterm r=commonratio The firstfive terms ofa geometric progression whose first term is aand whose commonratioisrare: a ar term 1 term2 Thisleadstotheformula: ar^ term3 3 ar term4 ar term5 nthterm=ar

WORKEDEXAMPLE12

Thethirdtermofageometricprogressionis144andthecommonratiois—. Findtheseventhtermandanexpressionforthenthterm. ^ Answersnthterm=ar"' usenthterm=144whenn=3andr= -(I) seventh144a=64term=64nthterm=arn-l 64 729 (11

CambridgeIGCSEand0LevelAdditionalMathematics WORKEDEXAMPLE13 Thesecondandfourthtermsinageometricprogressionare108and48respectively.Giventhatallthetermsarepositive,findthefirsttermandthecommonratio.Hence,writedownanexpressionforthenthterm. Answers108=ar (2)-(l) (1) gives ar^ ar AS=a^- (2) 9410848 3 ^2 2, alltermsarepositive=>r>0 2 Substitutingr=—intoequation(1)givesa=162. 3 2 (2Firstterm=162,common ratio = —, nth term =1®2|— -1 258 ■ WORKEDEXAMPLE14 f 1YThe nth term ofa geometric progression is ~ g J ' common ratio Answers firstterm=30|-—|=-15 iV secondterm = 30(-lJ = W Commonratio= 2nd term _ 7.5 _ 1^ 1stterm -15 2 1 Firstterm=-15,commonratio--

Chapter11:Series WORKEDEXAMPLE15 Inthegeometricsequence2,6,18,54,. whichisthefirsttermtoexceed1000000? Answers nthterm usea=2andr=3 2x3"-^>1000000 divideby2andtakelogs logioS""^>logio500000 usethepowerruleforlogs (n-1)logio3>logio500000 dividebothsidesbylogio3 logio500000 n-1> logio3 n-1>11.94...n>12.94... The13thtermisthefirsttoexceed1000000. CLASSDISCUSSION 1Inthisclassdiscussionyouarenotallowedtouseacalculator.Considerthesumofthefirst10terms,5io,ofageometricprogressionwitho=1andr=5. Sio = 1 +5+52+5^... +5'+ 5® +5^ a Multiplybothsidesoftheequation abovebythecommon ratio,5,and completethe followingstatement. 5Sio-5+52+5-+5-+...+5-+5-+5b Whathappenswhen you subtractthe equationfor Sjofrom the equationfor5Sio? c Canyoufindanalternativewayofexpressingthesum5io? 2 Use the method from question 1 to find an alternative way ofexpressing each ofthe following a 3+3x2+3x2^+3x2^+... (12terms) b 32+32x-+32x(2 I2 c 27-18+12-8+... +32X + (15terms) (20terms) It can be shown that the sum ofa geometric progression,5„,can be written as: 1-r or S. Q 1 Note: Forthese formulae,r+=1 Eitherformulacanbeusedbutitisusuallyeasierto •usethefirstformulawhen-1<r<1. • usethesecondformulawhenr>1orwhenr<-l.

Thesecondtermofageometricprogressionis9lessthanthefirstterm. Thesum ofthe second and third terms is 30. Given that all the terms in the progression are positive,findthefirstterm. Answers secondterm=firstterm-9 ar=a-9 Q rearrangetomakeathesubject «= (1)1-r secondterm+thirdterm=30 ar+ =30 factorise ar(l +r)-30 (2) ar(l + r) 30(1 - r)(2) - (1 gives = ° a 9 simplify 3»2+13r-10=0 factoriseandsolve (3r-2)(r+5)=0 o 1. r=— or r=-5 23 aUtermsareposidve=>r>0 r= 2 ^ Substituting r=— into (1) givesa=27. Firsttermis27.

CambridgeIGCSEand0LevelAdditionalMathematics

Proof: rX (!)■ Sn = a + ar+ai^ + ... + ar" + ar""^ + ar"~' ar + ai^ + ... + ar^~^ + + ar"~^+ ar"rS„ = (2)(1) (2) - (1): rS„-Sn =ar"-a (r-l)5„=«(r"-l) a(r"-1) Sn= r-1 Multiplying numerator and denominator by -1 gives the alternative formula a(l-r") 1-r

Findthesumofthefirsttentermsofthe geometricseries2+6+18+54+... Answers oa(r"-1) r-1 use a = 2, r= 3 and w = 10 o 2(31"-1) 3-1 simplify =59048

WORKEDEXAMPLE17

WORKEDEXAMPLE16

a1,2,4,6,... b-1,4,-16,64,... .2 3 5 8 c81,27,9,3,... e 2,0.4,0.08,0.16,. 1111 11 11 f-5,5,-5,5,... 435

Thesecondtermofageometricprogressionis12andthefourthtermis27.Giventhatallthetermsarepositive,findthecommonratioandthe firstterm. 5 6

Exercise11.4 1

Thefirsttermofageometricprogressionis75andthethirdtermis27. Findthetwopossiblevaluesforthefourthterm.

11Findthesumofthefirsteighttermsofeachofthesegeometricseries, a 4+8+16+32+... b 729+243+81+27+... C 2-6+18-54+... d -5000+1000-200+40-

8Threeconsecutivetermsofageometricprogressionarex,x+6andx+9. Findthevalueofx.

9Inthegeometricsequence—,—,1,2,4,...whichisthefirsttermtoexceed 42 500000?

Thefirstterminageometricprogressionisaandthecommonratioisr.Writedownexpressions,intermsofaandr,fortheninthtermandthe 20thterm.

Identifywhetherthefollowingseqtiencesaregeometric.Iftheyaregeometric,writedownthecommonratioandtheeighthterm.

12Thefirstfourtermsofageometricprogressionare1,3,9and27.Findthe smallestnumberoftermsthatwillgiveasumgreaterthan2000000.

Thesixthand13thtermsofageometricprogressionare—and320 respectively.Findthecommonratio,thefirsttermandthe10thtermofthisprogression. 7

Thethirdtermofageometricprogressionis108andthesixthtermis-32. Findthecommonratioandthefirstterm.

Thesumofthesecondandthirdtermsinageometricprogressionis30. Thesecondtermis9lessthanthefirstterm.Giventhatallthetermsinthe progressionarepositive,findthefirstterm.

10Inthegeometricsequence 256,128,64,32,... whichisthefirsttermthat islessthan0.001?

13Aballisthrownverticallyupwardsfromtheground.Theballrisestoa^ heightof10mandthenfallsandbounces.Aftereachbounceitrisesto of the height of the previous bounce. ^ a Writedownanexpression,intermsofn,fortheheightthattheballrisesafterthenthimpactwiththeground.

14Thethirdtermofageometricprogressionisninetimesthefirstterm.The

Giveyouranswercorrecttothenearestsecond. 16Ageometricprogressionhasfirstterma,commonratiorandsumto nterms,S„. Showthat 2n_^2n CHALLENGEQ 171,1,3.19.127,i81, Show that the sum of the first 2n terms of this sequence is —(S"-3'"+2). CHALLENGEQ 18S„=6+66+666+6666+66666+... Findthesumofthefirstntermsofthissequence. 11.5 Infinite geometric series Aninfiniteseriesisaserieswhosetermscontinueforever. The geometric series where a = 2 and r = — is 2+1+^+—+^+... 24 8 Forthisseriesitcanbeshownthat Si = 2,5^= 3,S3=3i,54=3|,S5=3^,.... This suggests that the sum to infinity approachesthenumber4. Thediagramofthe2by2squareisavisualrepresentationofthisseries.Ifthepatternofrectanglesinsidethesquareiscontinuedthetotalareasoftheinsiderectangles Thisapproachesthevalue4.confirmsthatthesumtoinfinityof • c 1 1 1theseries2+lH 1 1 h...is4. 2 4 8 1 2 i 4

CambridgeIGCSEand0LevelAdditionalMathematicsbFindthetotaldistancethattheballtravelsfromthefirstthrowtothe impactwith the ground. sumofthefirstfourtermsisktimesthefirstterm.Findthepossiblevalues ofk. in1.05minutes.Hereduceshisspeedinsuchawaythateachkilometretakeshimtimesthetimetakenfortheprecedingkilometre.Findthetotaltime,minutesandseconds,Johntakestocompletethe10kmrace.

15Johncompetesina10kmrace.Hecompletesthefirstkilometrein4

fifth

Chapter11:Series Thisisanexampleofaconvergentseriesbecausethesumtoinfinityconverges onafinitenumber. CLASSDISCUSSION convergeUseaspreadsheettoinvestigatewhetherthesumofeachoftheseinfinitegeometricseriesordiverge.Iftheyconverge,statetheirsumtoinfinity. fl= r=2 5 a=-3,r= 1 "2 a=5,r-- 1 K a=—,r=-5 2 32 Findotherconvergentgeometricseriesofyourown.Ineachcasefindthesumtoinfinity.Canyoufindaconditionforrforwhichageometricseriesisconvergent? Consider the geometric series a+ ar+ar^+ar^+ ... + ar"'. a(\ - r") Thesum, ,isgivenbytheformula =—j . If-1<r<1,thenasngetslargerandlarger,r"getscloserandcloserto0. Wesaythatas n^oo, r"—>0. a(l-r") a(l-O) a Hence,asn—><», — = . I-r 1-r 1-r Thisgivestheresult 5„= 1—r providedthat-1<r<1 Q Note: whenThisisnottruer>1orwhenr=-1 WORKEDEXAMPLET8 Thefirstthreetermsofageometricprogressionare25,15and9. a Writedownthecommonratio, b Findthesumtoinfinity. Answers a Commonratio b S„= secondterm 15 3 firstterm 25 5 1-r25 1_! usea=25andr= =62.5

CambridgeIGCSEand0LevelAdditionalMathematics WORKEDEXAMPLE19 Ageometricprogressionhasacommonratioof—andthesumofthefirstfourterms is164. ^ a Findthefirsttermoftheprogression, b Findthesumtoinfinity. Answers a S,= 164= a(l-r"*) 1-r1- 4 ma 41164= 125a a=500 b S„= a 1-r 500 1 7 =2779 use54=164andr=simplify solveusea=500andr= 5 Exercuse11.5 1 Findthesumtoinfinityofeachofthefollowinggeometricseries. o 1 1 1 a 3+1+-+-+...39 O 8 8 8 C 84 1 1 h 5 25 125 w 1 1 1 1 1b 1 1 ... 2 48 16 d -162+108-72+48-... 2 Thefirsttermofageometricprogressionis10andthesecondtermis8. Findthesum toinfinity. 3 Thefirsttermofageometricprogressionis300andthefourthtermis 2 -2—.Findthecommonratioandthesumtoinfinity. 4 The first four terms ofa geometric progression are 1,0.8^,0.8^ and 0.8®. Findthesumtoinfinity. 5 a Write the recurring decimal 0.42 as the sum ofa geometric progression. b Use your answer to part a to show that 0.42 can be written as 11 33" 6 Thefirsttermofageometricprogressionis-120andthesumtoinfinityis -72.Findthecommonratioandthesumofthefirstthreeterms.

12Thefourthtermofageometricprogressionis48andthesumtoinfinityis threetimesthefirstterm.Findthefirstterm.

Chapter11:Series

Thesecondtermofageometricprogressionis6.5andthesumtoinfinity is26.Findthecommonratioandthefirstterm.

11Thefirstthreetermsofageometricprogressionare k+15,kand k-l2 respectively,find a thevalueofk b thesumtoinfinity.

10Thesecondtermofageometricprogressionis18andthefourthtermis 1.62.Giventhatthecommonratioispositive,find a thecommonratioandthefirstterm b the sum to infinity.

14Thefirsttermofageometricprogressionis1andthesecondtermis TC 7t 2sin?cwhere—<x<—.Findthesetofvaluesofxforwhichthis 2 2 progressionisconvergent.

13Ageometricprogressionhasfirsttermaandcommonratior.Thesumofthefirstthreetermsis62andthesumtoinfinityis62.5.Findthevalueofa andthevalueofr.

The first three terms ofa geometric progression are 175,k and 63.Given thatallthetermsintheprogressionarepositive,find a thevalueofk b the sum to infinity.

15Aballisdroppedfromaheightof12m.Aftereachbounceitrisesto—oftheheightofthepreviousbounce.Findthetotalverticaldistancethatthe balltravels.

8 Thesecondtermofageometricprogressionis-96andthefifthtermis 40-. 2 a Findthecommonratioandthefirstterm, b Findthesumtoinfinity.

Step 3: Remove the line segments that were used as the base ofthe equilateraltrianglesinstep2.

Pattern1 Pattern2 Pattern3 Pattern4

CambridgeIGCSEand0LevelAdditionalMathematics CHALLENGEQ 16

Thesethreestepsarethenrepeatedtoproducethenextpattern.

Starting with an equilateral triangle,a Koch snowflake pattern can be constructed using the following steps:

Step1: Divideeachlinesegmentintothreeequalsegments.

Step2: Drawanequilateraltriangle,pointingoutwards,whichhas the middle segmentfrom step 1 as its base.

Youaregiventhatthetriangleinpattern1hassidelengthxunits.aFind,intermsofx,expressionsfortheperimeterofeachofpatterns1,2,3and4andexplainwhythisprogressionfortheperimeterof thesnowflakedivergestoinfinity.

Pattern Area %/3x^n/3x^ +3 \/3x^ +3—^^^+1273/ ^ f Ml ^+3—5^+12—^+48 4 4 4 Henceshowthattheprogressionfortheareaofthesnowflake g convergesto—timestheareaoftheoriginaltriangle. 5

bShowthattheareaofeachofpatterns1,2,3and4canbewrittenas

Chapter11:Series CHALLENGEQ 17Acircleofradius1unitisdrawntouchingthethreeedgesofanequilateraltriangle. Threesmallercirclesarethendrawnateachcornerto Thistouchtheoriginalcircleandtwoedgesofthetriangle.processisthenrepeatedaninfinitenumberoftimes, aFindthesumofthecircumferencesofallthecircles, b Findthesumoftheareasofallthecircles. 11.6Furtherarithmeticandgeometricseries Some problems may involve more than one progression. CLASSDISCUSSION Hi a,b,c,... 21Giventhata,bandcareinarithmeticprogression,findanequationconnectinga,bandc.Giventhata,bandcareingeometricprogression,findanequationconnectinga,bandc. WORKEDEXAMPLE20 Thefirst,secondandthirdtermsofanarithmeticseriesarex,yand Thefirst, secondandthirdtermsofageometricseriesare X, andy.Giventhatx<0,find: a thevalueofxandthevalueofy b thesumtoinfinit)'ofthegeometricseries C thesumofthefirst20termsofthearithmeticseries. Answers a Arithmeticseriesis:x+y+ +... usecommondifferences y-X= -y 2y-x'^+X ---(1) Geometricseriesis:x+»?+y+... 2 y 9 X usecommonratios y (2) (1)and(2)give2x^= +x dividehyx(sincex 0)andrearrange 2x^-X-1=0 factoriseandsolve (2x+l)(x-l)=0 1 x=—orx=l 2 x^lsincex<0 Hence,x=--and y=--. 2 -^ 8

CambridgeIGCSEand0LevelAdditionalMathematics b 1—r 1 2 _1 C S„=-[2a+{n-\)d\ 5^0= 2 =61.25 rs' 1+19 V8. usea=— and r= 2 2 usen=20, .-,d=y-x= Exercise11.6

4 Ageometricprogressionhassixterms.Thefirsttermis486andthe 2 commonratiois—.Anarithmeticprogressionhas35termsandcommon 3 ^ equaldifference—.Thesumofallthetermsinthegeometricprogressionistothesumofallthetermsinthearithmeticprogression.Findthe first term and the last term ofthe arithmetic progression. 5 fifthThefirst,secondandthirdtermsofageometricprogressionarethefirst,andeighthtermsrespectivelyofanarithmeticprogression.Given thatthefirsttermineachprogressionis200andthecommonratioofthe geometricprogressionisr,where 1,find a thevalueofr b thefourthtermofeachprogression c thesumtoinfinityofthegeometricprogression.

The first,second and third terms ofa geometric progression are the first, fifthand11thtermsrespectivelyofanarithmeticprogression.Giventhatthefirsttermineachprogressionis48andthecommonratioofthe geometricprogressionisr,wherer*1,find a thevalueofr b thesixthtermofeachprogression.

Thefirsttermofaprogressionis25andthesecondtermis20. a Given that the progression is geometric,find the sum to infinity. bGiventhattheprogressionisarithmetic,findthenumberoftermsin theprogressionifthesumofallthetermsis-1550.

Thefirsttermofaprogressionis8andthesecondtermis12.Findthesumofthefirstsixtermsgiventhattheprogressionisaarithmeticbgeometric.

Thefirsttermofanarithmeticprogressionis12andthesumofthefirst16 termsis282.

TheThefirsttwotermsofageometricprogressionare80and64respectively.firstthreetermsofthisgeometricprogressionarealsothefirst,11thandnthtermsrespectivelyofanarithmeticprogression.Findthevalueofn.

a Findthecommondifferenceofthisprogression. Thefirst,fifthand nth termofthisarithmeticprogressionarethefirst, secondandthirdtermrespectivelyofageometricprogression. b Findthecommonratioofthegeometricprogressionandthevalueofn.

Thefirsttwotermsofaprogressionare5xand respectively. a Forthecasewheretheprogressionisarithmeticwithacommon differenceof24,findthetwopossiblevaluesofxandthecorresponding valuesofthethirdterm, b Forthecasewheretheprogressionisgeometricwithathirdterm g of-—,findthecommonratioandthesumtoinfinity.

Summary Binomialexpansions Ifnisapositiveintegerthen{a+b)"canbeexpandedusingtheformula (a+b)"=a"+"Ci + + + + or = a"+r L-ift+r ar-^b^+ a"-'"6''+...+6" andwhere"C, n n! rj {n-r)\r\ Inparticular, o , ..\n 1 . , n{n-\) 9 n{n-\){n-t) 3, n(n-l)(n-2)(n-3) 4(1+xj=I+nx+—— )C+ ^XT+— — — x^+...+x". 2! 3! 4! Arithmeticseries Foranarithmeticprogressionwithfirstterma,commondifferencedandnterms: • theAthterm=a+(A-I)<i • thelastterm= a+(n-l)rf •thesumoftheterms=S„=—(a+/)=—[2a+(n-l)d]. 2 2

CambridgeIGCSEand0LevelAdditionalMathematics Geometricseries Forageometricprogressionwithfirstterma,commonratiorandnterms: • thekthterm=or*"^ • thelastterm=ar"~^ 1-r r-1 Theconditionforageometricseriestoconvergeis-1<r<1. thesumoftheterms= = ■ Whenageometricseriesconverges, = a 1-r Examination questions Workedexample a Find thefirst4termsin theexpansion of(2+ in ascending powersof 3Vb Fi X. nd the term independent of x in the expansion of(2 + 11 1 . [3][3]

CambridgeIGCSEAdditionalMathematics0606Paper11Q3i,iiJun2015 Answer a Expanding(2+x^) using the binomial theorem gives 2® + ®Ci 2^x2 + ®C2 2^(x^)^ + 2^[x^f = 64 + 192x2 + 240x4 + 160x®.. b (2+ x2)6 l-^ =(64 +192x2+ 240x4+160x^..) 1-^+^ \ XJ V X X Term independentof x =(64 x 1)+|^192x2 x +^240x4 x =64-1152+2160 =1072 Exercise11.7 ExamExercise 1 a Find the firstfour termsin the expansion of(2+ x)^ in ascending powers ofx. b Hence find the coefficient of x^ in the expansion of (1 + 3x)(l - x)(2 + x)®. [3] [4] CambridgeIGCSEAdditionalMathematics0606Paper21Q7i,iiJun2013 ( 2 V 2aFindthefirst3terms,indescendingpowersofx,intheexpansionofIxH—-\. [3] b Hence find the term independent of x in the expansion of|^2 s "i"j "

CambridgeIGCSEAdditionalMathematics0606Paper11Q6i,iiNov2012

5 a Find,in ascending powersofx,thefirst3termsin the expansion of(2-5x)®, giving your answer in the form a + bx + cx^, where a, b and care integers. [3] \10 [3]b Find the coefficient ofxin the expansion of(2 — 5x)^^1+ —^ CambridgeIGCSEAdditionalMathematics0606Paper11Q6i,iiNov2010

/ XY 33 The coefficient of in the expansion of[^1 + gj > where n is a positive integer is -. aFindthevalueofn. b Usingthisvalueofn,findthetermindependentofxintheexpansionof \2 [4][4]

4 a Find the coefficient of in the expansion of ~^j • b Find the coefficient of in the expansion of(1+4x)^l- . CambndgeIGCSEAdditionalMathematics0606Paper11Q7i,iiNov2011 12 [3][2] CambridgeIGCSEAdditionalMathematics0606Paper21Q2i,iiJun2011

7 i Find thefirst4termsin the expansion of(2+ in ascending powersofx. [3] ii Find theterm independentofxin theexpansion of + . [3] CambridgeIGCSEAdditionalMathematics0606Paper11Q3Jun2015

CambridgeIGCSEAdditionalMathematics0606Paper12Q4Mar2015

6 i Write down,in ascendingpowersofx,thefirst3termsin theexpansion of(3+2x)®. Give each term in its simplestform. [3] ii Hencefind the coefficientofx^in the expansion of (2-x)(3+2x)®. [2]

8 a i Use the Binomial Theorem to expand (fl + 6)^, giving each term in its simplestform. [2] ii Hence find the term independentofxin the expansion of ^2x+—j. [2] b The coefficient of x^ in the expansion of[1+-] equals positiveinteger«, ^ 5n 1^ Findthevalueofthe [3] CambridgeIGCSEAdditionalMathematics0606Paper21Q8Jun2016

b Given thatthe nth term ofthe progressionis-22,find thevalueofn. [2]

CambridgeIGCSEand0LevelAdditionalMathematics

The second term ofa geometric progression is-576 and the fifth term is 243. Find a b c 14a thecommonratio thefirstterm thesumtoinfinity. [3] [1][2]

[3]

Examinationstylequestion

a Find thefirstterm ofthe progression and thecommon difference. [4]

Examinationstylequestion

An arithmetic progression hasfirstterm aandcommon difference d. Give thatthesum ofthe first100termsis25timesthesumofthefirst20terms. [2][3]

Ageometricprogressionhasfirstterm8andcommonratior.Asecondgeometric progression has first term 10 and common ratio —r.The two progressions have the samesum to infinity,S.Find the values ofrand the value ofS. [3]

The 15th term ofan arithmetic progression is 3and the sum ofthe first8 terms is 194.

15a Thesixthtermofanarithmeticprogressionis35andthesumofthefirsttenterms is 335.Find the eighth term. [4]

9Thefirsttermofageometricprogressionis35andthesecondtermis-14. a Findthefourthterm. [3] b Findthesumtoinfinity. [2]

b Writedownanexpression,intermsofa,forthe50thterm.

b Find thesum toinfinity. [2]

Examinationstylequestion

10Thefirstthreetermsofageometricprogressionare2k+6,k+12andkrespectively. Allthetermsintheprogressionarepositive.aFindvalueof

Examinationstylequestion

Examinationstylequestion a Finddintermsofa.

Examinationstylequestion

The 10th term ofan arithmetic progressionis4andthesum ofthefirst7termsis-28. Findthefirsttermandthecommondifference. [4] Thefirstterm ofa geometric progression is40and thefourth term is 5.Find thesum toinfinityoftheprogression.

Ageometricprogressionhasfirstterma,commonratiorandsumtoinfinity5. Asecondgeometricprogressionhasfirstterm3a,commonratio2randsumtoinfinity4S.Findthevalueofr.

Chapter11:Series 16a

b An arithmetic progression has first term -24.The nth term is-13.8 and the(2n)th termis-3.Findthevalueofn.

[4][3] Examinationstylequestion

Chapter 12 Differentiation 1 Thissectionwillshowyouhowto: usethenotationsf'(x),f"(x), 6y d^y d I dy dx'dx^' dxi^dx^ ■ usethederivativeofx"(foranyrationaln),togetherwithconstantmultiples,sumsandcomposite functionsofthese ■ differentiateproductsandquotientsoffunctions ■ applydifferentiationtogradients,tangentsandnormals,stationarypoints,connectedratesof change,smallincrementsandapproximationsandpracticalmaximaandminimaproblems ■ usethefirstandsecondderivativeteststodiscriminatebetweenmaximaand minima.

Youhaveleamthowtofindthegradientofastraightlinejoiningthetwopoints(xj,3)1)and (x^,y^)usingtheformula: y—y Gradient=— ^ Youshouldalsohaveleamthowtoestimatethegradientofacurve ataparticularpoint. Forexample,tofindthegradientofthecurvey=x^atthepoint(1,1)Step1:Drawthegraphof3=x^. Step2:Drawatangenttothecurveatthepoint(1,1). y—y Step3:Findthegradientofthetangentusing^ x^-x^ \ ^ /}' ^ \X .y=xinaccuracyThismethodonlygivesanapproximateanswer(becauseoftheofdrawingthetangent)anditisalsoverytimeconsuming. Inthischapteryouwillleamanaccuratemethodforfindingthegradientofacurve,whichdoesnotinvolvedrawing thecurvefiirst.Thisaccuratemethodiscalleddifferentiation.

Chapter12;Differentiation1 RECAP

12.1 The gradient function y y=X' X -V.2 ThediagramshowsapointA(x,y)onthecurvey=x^andapoint5whichis closetothepointA. ThecoordinatesofBare (x:+bx,y+8y) where 5x isasmallincreasein the value ofXand 5y is asmallincrease in the value ofy. The coordinates ofA and Bcan also be written as (x:, and +5x,(x +5x)^j. gradientofchordAB= X2—*1 _ (x:+ 6x:)^ - x^ (x +5*)- X _x^+2x8x+(Sjc)^-x^ dx _ 2x8x +(5x:)^ 5x: =2x:+5x:

Therulefordifferentiatingpowerfunctionsis: If))=x",then—-nx"'. dx Itiseasiertorememberthisruleas: 'multiplybythepowernandthensubtractonefromthepower' Sofortheearlierexamplewherey=x^, ^= 2 X x2-i dx =2x1 =2x.

Differentiation of powerfunctions

Notation

Hence,gradientofcurve=2x. differentiationThisprocessoffindingthegradientofacurveatanypointiscalledfromfirstprinciples.

Therearethreedifferentnotationsthatcanbeusedtodescribetherule above. dy1 Ify=x^,then——2x. dx 2 If f(x)= x^, then fX''^)= 2x. 3 —(x2)= 2x. dx dy —IScalledthederivativeofywithrespecttox. dx f'Cx) is called the gradientfunction ofthe curve y =f(3c). —(x^)= 2x means'ifyou differentiate x^ with respectto x,the resultis 2x'. dx Youdonotneedtobeabletodifferentiatefromfirstprinciplesforthe examinationbutyouareexpectedtoknowtherulesofdifferentiation.

CambridgeIGCSEand0LevelAdditionalMathematics As B A,bx ^0 and the gradient ofthe chord AB —> the gradient ofthe curveatA.

Chapter12:Differentiation1 WORKEDEXAMPLE1 Differentiatewithrespecttox. a x® b-7 cyfx x' d 3 Answers a —(x®)= 5x®-' b Afl dxVx® CO II =bx'^ =-3x-®-i =-3x-^ 3 x^ c = Apldx^ ' dxl^ J d ;^(3) dx H0 OC II 1 =—x2 2 II 0 H0 =0 2 1 2>/x Youneedtoknowandbeabletousethefollowingtworules:Scalarmultiplerule: I Addition/subtractionrule: £[f(x)± g(x)]= £[f(«)]±:^[g(^)] dx

CambridgeIGCSEand0LevelAdditionalMathematics WORKEDEXAMPLE2 1 5Differentiate -Sx^+—+ withrespecttox Answers —(2x^-3x'+ ,dxV 2x^ X^ yfxJ dx 2x^-3x^+—X^+5x2 2 = 2—(x®)- + — —(»: +5— 1 X 2 dx da: 2da: da: = 2(5a:'')-3(2x)+^(-3x-^)+5j^-ia:"2j ^ =lOx'*-6a: x'^ a:2 WORKEDEXAMPLE3 Findthegradientofthecurve31=(3a:+2)(2x-1)atthepoint(-1,3). Answers y=(3x+2)(2a:-1) y=Ga:^+a:-2 da: When a:=-l,^= 12(-1)+ 1 da:=-11 Gradientofcurveat(-1,3)is-11. Exercise12.1 1 Differentiatewithrespecttox. a X 4 f 1 X'^ k p X b g yfx qx^XX cx~ h yf}7 m x dx"® i x5 3 n *2 X Tx 1 e X1 j *3 O xTx

6Findthegradientofthecurvey=x^—2x^+5x-3atthepointwherethe curvecrossesthe))-axis.

4Findthecoordinatesofthepointonthecurvey=2x^-x-1atwhichthe gradientis7. X—45 Find the gradientofthe curve y = at the pointwhere the curve crossesthex-axis. *

Chapter12:Differentiation1 2 Differentiatewithrespecttox. a -5x+4 d Sx^ +--\ X x^ g -3 b 8x^-5x^-2 e2*:----4= X yjx , 5x^ — yfx j 5*^(x +1) m (3x +lf k X ^{2x-5) n (l c 7-2x^+4x ^ X+5 * ~7r -*-1 ' 7^ I -(«'-2) X o (2x-l)(3x+4) dy3 Findthevalueof—atthegivenpointonthecurve, dx ay=3x^-4by=4-2x^ c 3, 2.5 X atthepoint(1,-1) atatthepoint(-1,2)thepoint(-2,-2) dy=5x^-2x^-3atthepoint(0,-3) X+5 e 3= atthepoint(5,2) f y= X X—3 yfx atthepoint(9,2)

7 Thecurvey=2x'^+'7x-4andtheline31=5meetatthepointsPandQ. Find the gradientofthe curve at the pointPand at the point Q.

8 Thecurvey=ax^+hxhasgradient8when x=2andhasgradient-10 whenX=-1. Findthevalueofaandthevalueofb.

To differentiate y =(2x +3)^,you could expand the bracketsand then differentiate each term separately.This would take a long time to do.There is amoreefficientmethodavailablethatallowsyoutofindthederivativewithout expanding: Let M = 2x+3, then y -(2x+3)® becomes y = m®. The derivative ofthe composite function y =(2x +3)^ can then befound usingthechainrule; dy_dy^Au dx dw dx

Thegradientofthecurvey=ax+— atthepoint(-1,-3)is-7. X Findthevalueofaandthevalueofb. 10 Findthecoordinatesofthepointsonthiscurve y= 1-6x-1 wherethegradientis2.

CambridgeIGCSEand0LevelAdditionalMathematics 9

11 Thecurve "v=— -2x^-8x+5 andtheline y=x+5 meetatthe -^ 3 ^ apointsA,BandC.FindthecoordinatesofthepointsA,BandC. b Find the gradient ofthe curve at the points A,Band C. 12y—4x^+3x^—6x-1 d)i aFind dx d}'b Findtherangeofvaluesofxforwhich—3==0. dx Yiy—x^+x"^—16x—16 a Finddx b Findtherangeofvaluesofxforwhich dy dx 0. CHALLENGEQ 14Acurvehasequadony=x^—5x^+25x-+145x+10.Showthatthe gradientofthecurveisnevernegative. 12.2Thechainrule

Withpracticeyouwillfindthatyoucandothismentally: Consider the 'inside'of{2x +3)^ to be 2x +3. To differentiate(2x +3)®: Step1:Differentiatethe'outside'first:8(2x+3)^

Chapter12: Differentiation1

Alternatively,todifferentiatetheexpressionmentallv: 9 Write-— ^as2(5x^-1)-^. (5x^-1) Step 1: Differentiate the'outside'first:-8(5x^-1)~®

Letu=5x^-1 dudx Using =lOx so and y=2u-^ dy du -8u~ dj) _ d^i ^dw dx du dx =-8m-5XlOx =-8(5x^-1)-®XlOx —80x (Sx^ -1)^

Step2:Thendifferentiatethe'inside':2 Step 3:Multiply these two expressions: 16(2x +3)^

Findthederivativeofy=(2x+3)®. Answers y=(2x+3)® LetM=2x+3so y dw_9 j dy—-2 and dx du =8m' . dy dy dw Using X dx dw dx =8m^X2 =8(2x+3)^x2 =16(2x+3f

Findthederivativeofy Answers (5*:^ y= (5x^ -1)^

Step2:Thendifferentiatethe'inside':lOx Step3: Multiplythetwoexpressions: —80x(5x^—l)~^ = —80x (5x^ -1)®

WORKEDEXAMPLE4

WORKEDEXAMPLE5

8 The curve y = ^ passes through the point(1,4) and has gradient3 V^x+1 -—atthispoint. Findthevalueofaandthevalueofb. 12.3 The product rule Consider thefunction 3;= x^(x^+l). 31=x'+x^ —=7x®+2x dx

CambridgeIGCSEand0LevelAdditionalMathematics

6 Find the gradient ofthe curve y = x + at the points where the curve crossesthex-axis.

Exercise12.2 1 Differentiatewithrespecttox. a (x+2)^ b (3x-1)^ e ^ f 2(x-4)® 3 i +2)^ j(l-2x^y 2 Differentiatewithrespecttox. 1 3 a b { (2x+4) 4 x-1) 1 (x^ - 2x) (x - 1)® 3 Differentiatewithrespecttox. a \jx +2 b yj5x -1 e^3^^ f 4V2x-1 c (1-5x)^ g 6(5-x)^ k (x^ - 3x)^ (2-3x) 2 ^ (5x + 1)^ cV2x2_3 1 g \JSx—1 ix-7 h -(2x+5)^ 2 XH X dh 16 (2x^ -5) 1 2(3x - 2)^ d x/x^T^ 3 ^2-5x

4 Find the gradient of the curve y =(2x - 5)'' at the point (3,1).

5 Find the gradient of the curve y = at the point where the curve (xcrossesthe3)-axis. -2)'

7 Find the coordinates of the point on the curve y = yj{x^ -6x + 13) where thegradientis0.

Thefunctiony-x^(x^+l)canalsobeconsideredastheproductoftwo separate functions y = uv where « = x^ and n = x^ +1. To differentiate the productoftwofunctions you can use the productrule d. . dv du —{uv)=u 1-V—— dx dx dx

Chapter12:Differentiation1 Itiseasiertorememberthisruleas '(firstfunction X derivative ofsecondfunction)+(secondfunction x derivative offirstfunction) Sofor)i= +l), =(x^)—(«:^+1)+(x^+1) ,(^) firstdiffereiitiate + second differentiate second first =(x^)(5x'^)+(x^+l)(2x) =5x®+2x®+2x =7x®+2x WORKEDEXAMPLE6 Findthederivativeofy=(5x+1)v6x-1. Answers y=(5x+1)y/Qx—1 =(5x+l)(6x-1)2 d3i dx =(5x+1)^|^(6x-1)2j +|^(6x-1)2 fost^(liffpientiate + second differentiate second first =(5x +l)|^|(6x-1)"2(6)j+(V6x-l)(5) V. use the chain rule ,w±a^5V6;rrT y/6x-1 3(5x+l)+5(6x-l) y/6x-1 _45x-2 yJ&X—1 -(5x+1) writeasasinglefraction simplifythenumerator

Cambridge IGCSE and0LevelAdditional Mathematics EXAMPLE7

^= 0 when 2(x + 2)(2x-5)2(5x+1)=0 x+2=0 2x-5=0 5x+l=0 X=-2 X=2.5 X=-0.2

Findthe^coordinateofthepointsonthecurvey= gradientis0. X+2)^(2x-5)^ wherethe Answers y={x+2f{2x-5f first differentiate + second differeiitiate second first =(x+2)2[3(2x-5)2(2)]+(2x-5)3[2(x+2)1(1)] ^ ^ usethechainrule usethechainrule =6(x+2)2(2x-5)2+2(x+2)(2x-5)^ factorise =2(x+2)(2x-5)2[3(x+2)+(2x-5)] simplify =2(x+2)(2x-5)2(5x+1)

1 Use the productrule to differentiate each ofthefollowing with respectto x: a x(x + 4) b 2x(3x + 5) C a:(x + 2)^ d e Xyjx-5 f {x + 2)yfx g hV^(3-x2f i(2x+1)(x2+5) j (x + 4)(x-3f k (x-lf(x + 2f I (2x + lf(x-3)^ Find the gradient ofthe curve y = x^^x +2 at the point(2,8). Find the gradientofthe curve y ={x-1)^(x +3)^ atthe pointwhere x =2. Find the gradientofthe curve ))=(x+2)(x-5)^ atthe points where the curvemeetsthex-axis. Find the x-coordinate ofthe points on the curve 3; =(2x -3)^(x + 2)'' wherethegradientiszero.

Findthex-coordinateofthepointonthecurve3=(x-I-3)V4-xwhere thegradientiszero. 2 3 4

Exercise12.3

Thefunctiony= (2x-3) canbedifferentiatedbywritingthefunctionin theform y-{x^+1){2x-3)'andthenbyapplyingtheproductrule. X^"i"1 separateAlternatively,y=functions: u canbeconsideredasthedivision(quotient)oftwo y=—whereu=x^+\andu=2x-3. V Todifferentiatethequotientoftwofunctionsyoucanusethequotientrule: dwdu dxdfuVu dx ^ Itiseasiertorememberthisruleas: (denominatorxderivativeofnumerator)-(numeratorxderivativeofdenominator) (denominator)^

WORKEDEXAMPLE8

Usethequotientruletofindthederivativeof ■» = • (2x-3) Answers + 1 ^ " (2x -3) denominator differentiate numerator numerator differentiate denoimnator (2x-3)x^(x''+l) - {x'+l)x-^i2x-S) dx (2x-3)'^ j denominatorsquared 1 (2x-3)(3x2)-(x^-t■ 1)(2) (2x-3)'-^ 6x^ - 9x^ - 2x^ - 2 (2x - 3)4x® - 9x'^ - 2 (2x-3)^

Chapter12:Differentiation1 12.4 The quotient rule +1

V u

CambridgeIGCSEand0LevelAdditionalMathematics CLASSDISCUSSION dy Analternativemethodforfinding—inworkedexample8istoexpressyintheform dx j; =(x3+l)(2x-3ri andtothendifferentiateusingtheproductrule. Trythismethodandthendiscusswithyourclassmateswhichmethodyouprefer. WORKEDEXAMPLE9 Findthederivativeof)>= Answers (i+lf yJx+2 •Jx+2 denominator differentiate numerator numerator differentiate denominator dy dx Vx+2X—[(x+1)^1-(x+ifx-^fVx+21 L J r1V^ dx' denominatorsquared (^/;^T2)[2(x + 1)'(1)]-(x + 1)'[i(x + 2)4(1)] X+2 x+2 _ 4(x + l)(x + 2)-(x + 1)^ ~ 2(x + 2)>/jr+2 _ (x + 1)[4(x + 2)-(x + 1)] 2(x + 2)l =(x+l)(3x+7) 2(x + 2)l bymultiplynumeratoranddenominator2Vx+2 factorisethenumerator

CHALLENGEQ 8 The line 5x-5y = 2 intersects the curve x^y -5x + y + 2 = 0 at three points. aFindthecoordinatesofthepointsofintersection, b Find thegradientofthecurve ateach ofthe pointsofintersection. 12.5 Tangents and normals y=f(x) tangent normal

Theline perpendiculartothe tangentatthe pointA iscalled thenormalatA.

Chapter12: Differentiation1 Exercise12.4

1Usethequotientruletodifferentiateeachofthefollowingwithrespecttox: 1+2x 5—X 5x-2 fb 3x+2X+4 X X' 1 3x3xX-1+45—1 d h 5x-23-8xX+4 x2-2 X+32 Find the gradient of the curve y = ^ at the point (2,5).

7Findthecoordinatesofthepointonthecurvey=—, wherethe p, vx+1 gradientis—.

Findthecoordinatesofthepointsonthecurve)'= where —=0. ^ 2x-1 dx 7x-2 2x+3 4 Findthegradientofthecurvey= crossesthe))-axis. 5 Differentiate with respect to x: Vx . X a b , 2x+1 yjl-2x 6 Find the gradient ofthe curve y = ? vx+5 •J atthepointwherethecurve x-+2 byfx 3+X atthepoint(-4,-6). 2(x-5)

Cambridge IGCSE and0LevelAdditional Mathematics Ifthe value of — atthe point(x,,jij) is m,then the equation ofthe tangentis , dxgivenby: y~y\~ ~*i) The normalatthe point(xp y,) is perpendicular to the tangent,so the gradientofthenormalis-—andtheequationofthenormalisgivenby: m y — yi = (x — Xj), m ^0 m WORKEDEXAMPLE10 a1 ° u Findtheequationofthetangentandthenormaltothecurvej-ox--x+-atthe pointwherex=2. Answers y=3x^—X+8x~' —=6x-1-8x"^ dx Whenx=2,y=3(2)"-(2)+8(2)"'=14 and ^-6{2)-l-8(2)-''=9 d;c Tangent:passesthroughthepoint(2,14)andgradient=9 y-14=9(x-2) y=9x—4 1 Normal:passesthrough the point(2,14) and gradient-~g y-14=-i(x-2) 9y+X=128

NormalatJS:

Thenormalstothecurvey= -5x^+3x+1,atthepointsA(4,-3)and5(1,0), meetatthepointC. FindthecoordinatesofC. x^4,^ = -10(4)+3= =3(1)^-10(1)+3=-4 passesthroughthepoint(4,-3)andgradient= ^-(-3)=--(x-4) (1) passesthroughthepoint(1,0)andgradient (2) Addingequations(1)and(2)gives\by=-30 Wheny=-2,11(-2)=-x-29 Hence,Cisthepoint(—7,-2). AreaoftriangleABC= -3-^0'^-2^-3

b FindtheareaoftriangleABC. Answers a ^=3x^-lOx+3 dx When

4viv-7v4

11 dx WhenX=1,

7 5(1,0) ^\ X A(4,-3)

Chapter12:Differentiation1 WORKEDEXAMPLE11

11)1=-X-29

>=-2

dx NormalatA:

X=-7

31-0= 4y=X—1-

3(4f

=-|0+(-2)+21-(-3)-0-(-8)| = i|30| 2 =15units'^

10 A curve has equation 3 =(2- Vx) . ThenormalatthepointP(l,1)andthenormalatthepoint<2(9,1) intersectatthepointR. aFindthecoordinatesofR. b FindtheareaoftrianglePQR.

-^CambridgeIGCSEand0LevelAdditionalMathematics Exercise12.5

6Thetangentstothecurve3)=x^-5x+4,atthepoints(1,0)and(3,-2), meetatthepointQ. FindthecoordinatesofQ.

Findtheequationofthenormaltothecurveatthegivenvalueofx. a))=x^+5xatX=-1 b)>=3x^-4x+1atx=2 Sx''-7x^+2x atX=-1 =2x(x-3)^ atX=2 d y=A f:y X' X+2 X-2 atX=-2 atX=6 O3 Findtheequationofthetangentandthenormaltothecurve3;=5x at the point where x = 1. ^

4Thenormaltothecurvey=x^-2x+1atthepoint(2,5)intersectsthe 3!-axisatthepointP. FindthecoordinatesofP. X—1

5 Findtheequationofthetangentandthenormaltothecurvey= , \X Aatthepointwherethecurveintersectsthe}i-axis.

7Thetangenttothecurvey=3x^-1Ox-8atthepointPisparalleltothe line31=2x-5.FindtheequationofthetangentatP.

8 Acurvehasequation3=x^-x+6. a FindtheequationofthetangenttothiscurveatthepointP(-l,6). ThetangentatthepointQisparalleltothetangentatP. b FindthecoordinatesofQ. c FindtheequationofthenormalatQ.

9 A curve has equation 3= 4+(x -1)"*. ThenormalatthepointP(l,4)andthenormalatthepointQ(2,5) intersectatthepointR. FindthecoordinatesofR.

Find the equation ofthe tangent to the curve at the given value of x. a y = x^-3atx = l b y = x'^ +3x +2 at X =-2 cy=2x^+5x^-1atX=1 e ))=(x-3)(2x-1)^ at X=2 d )i=5h— atx=-2 X f y= at X=-3 ^ x+1

The diagram showsthe tangentto the curve y =f(x) atthe point P(x,3;).

IfPandQaresufficientlyclosethen: 831 dy 8x dx

FindtheareaofthetriangleMNO,whereOistheorigin. X—3

The gradientofthe chord PQis 83) 8x

13Theequationofacurveis3= .

Chapter12:Differentiation1

Thecurveintersectsthex-axisatthepointP.ThenormaltothecurveatPmeetsthe^j-axisatthepointQ.

ThegradientofthetangentatthepointPis—. dx The point Q[x+8x,y+831) is a pointon the curve close to the pointP.

12Theequationofacurveis31= X+2

11 A curve hasequation y- yfx{x -2)^. The tangent at the point P{3, Vs) and the normal at the point Q{9, 1) intersectatthepointR. a Show that the equation ofthe tangent at the point P(s, VS]is 19S l7ylS y= X -^ 6 2 b Find the equation ofthe normalat the point Q(l,-1).

TheThetangenttothecurveatthepointwherex-Smeetsthe31-axisatM.normaltothecurveatthepointwherex--3meetsthex-axisatN.

FindtheareaofthetrianglePOQ,whereOistheorigin.

12.6 Small increments and approximations y=fW Q(x+5x,3)+831) 5y tangent

Thevolume,Vcm®,ofaspherewithradiusrcmisV=—Tcr®. 3 Find,intermsofp,theapproximatechangeinVasrincreasesfrom10to10+ where pissmall. Answers 4 dVFdr -Jtr-' 3 4jcr^ Whenr=10,8F Using =dVdVdrdr400jt 4jt(10f = 40071 W8r P 8y»4007C/»

CambridgeIGCSEand0LevelAdditionalMathematics

Variablesxandyareconnectedbytheequationy= Findtheapproximateincreaseinyasxincreasesfrom2to2.05. Answers y=X®+x^ —=3x^+2x dx When X=2, =3(2)^ +2(2)=16 dx Using 5y _ dy =16 8x dx 8y 8y=16X0.05 8y=0.8

WORKEDEXAMPLE12

WORKEDEXAMPLE13

1 Variables xand yare connected bythe equation y - 2x^-Sx.

Find the approximate change in y as xincreasesfrom 2 to 2.01. r 2 8

Find,intermsofp,theapproximatechangeinyasxincreasesfrom4to 4+/?,wherepissmall.

Findtheapproximatechangeinyasxincreasesfrom1to1.02.

3 Variables xand yare connected by the equation x'^y = 400.

Variablesxandyareconnectedbytheequationy—|—x Find,intermsofp,theapproximatechangeinyasxincreasesfrom9to 9+/>,wherepissmall.

2 Variables x and y are connected by the equation y — 5x ~^•

Find,intermsofp,theapproximatechangeinyasxincreasesfrom2to 2+p,wherepissmall.

The periodic time, Tseconds,for a pendulum oflength Lcm is T = 271,/^. Vio Find the approximate increase in Tas Lincreasesfrom 40 to 41. 8 ycm '^xcm 2xcm

Find,in terms ofp,the approximate change in y as x increasesfrom 10 to 10+p,wherepissmall. ^

Chapter12:Differentiation1

Thevolumeofthesolidcuboidis360cm"^andthesurfaceareaisAcm^. a Expressyintermsofx.

u cu t. . 2 1080bShowthatA-4x'^H . X C Find,intermsofp,theapproximatechangeinAasxincreasesfrom2to2+j!?,wherepissmall.Statewhetherthechangeisanincrease ordecrease.

A curve has equation y =(x +1)(2x-S)"*.

A curve has equation y =(x -2)V2x + 1.

Exercise12.6

2 Discuss how the height ofwater in container B changes with time.

CLASSDISCUSSION

3 Discuss how the height ofwater in container C changes with time.

1 Discuss how the height ofwater in container A changes with time.

Cambridge IGCSE and0LevelAdditional Mathematics 12.7

h h

■im

Consider pouringwater at a constant rate oflOcm^s'^ into each ofthese three large containers.

Rates of change

WORKED EXAMPLE 14 Variables Vand iareconnectedbytheequation V= -8t+3. FindtherateofchangeofVwithrespectto twhen t=4. Answersy= -8t+3 dV =m-8 dt When t = 4, dt 10(4)-8 = 32

On copies of the axes below, sketch graphs to show how the height {h cm) varies with time {tseconds)foreachcontainer. h dvYou already know that — represents the rate of change ofy with respect to x. dx There are many situations where the rate ofchange of one quantity depends onthechangingvalueofasecondquantity. In the class discussion, the rate of change of the height ofwater at a particular time, t, can be found by finding the value of — at time t. (The gradient of thetangentattimet.)

Chapter12:Differentiation1 Connectedratesofchange Whentwovariablesxandybothvarywithathirdvariablet,thethree variablescanbeconnectedusingthechainrule: dt dac dt You may also need to use the rule that: dx_1 dy ~ W dx WORKEDEXAMPLE15

Variablesxandyareconnectedbytheequationy=x^-5x^+15. Answers

Giventhatxincreasesatarateof0.1unitspersecond,findtherateofchangeofy whenX=4.

dxy=x^—5x^+15and —=0.1 dt -^=3x2-lox dx WhenX=4, dy dx 3(4f 8 10(4) dy dy dx Usingthechainrule,—=—X ® dt dx dt =8X0.1=0.8 Rateofchangeofyis0.8unitspersecond.

Theverticalcross-sectionisanequilateraltriangle. Waterispouredintothecontaineratarateof24cm^s~'. a Showthatthevolumeofwaterinthecontainer,Vcm'',isgivenbyT=40y/3hr, wherehcmistheheightofthewaterinthecontainer, b Findtherateofchangeofhwhenh=\2. Answers a Lengthofsideoftriangle= h sin60° 2Sh Areaofaia„g.= =ix?|5x» =:^xi2o 3 =40V3 dT r- dV b —=80V3/tand—=24 dh dt dV r-Whenh=U, =80>/3(12) dh =960V3 TT• u u ■ ,d/t dh dVUsingthechainrule,—= x ® d< dV dt 1 960>/3 A 120 x24 y/S Rateofchangeofhis cmpersecond.

V=areaoftrianglex120

Thediagramshowsawatercontainerintheshapeofatriangularprismoflength 120cm.

CambridgeIGCSEand0LevelAdditionalMathematics WORKEDEXAMPLE16 120cm

Giventhatxincreasesatarateof2unitspersecond,findtherateof increaseofywhenx=1.

Variablesxandyareconnectedbytheequationy 2x-5 X-1

10AspherehasradiusrcmandvolumeVcm^.Theradiusisincreasingatarateof—cms~'.

Findtherateofincreaseofthevolumewhen V=97271.

Exercise12.7

2 Variables xand yare connected by the equation y = x + yjx-5.

4 Variables x and y are connected by the equation y = 2^-1'

11Asolidmetalcuboidhasdimensionsxcmbyxcmby5xcm.

The radius ofthe base is increasing at a rate of0.1 cms~^

The volume is increasing at a rate of2cm^s"'.

Giventhatxincreasesatarateof0.01unitspersecond,findtherateof changeofywhenx=8.

1 1 2 7 Variables x and y are connected by the equation ~ ~ g ~ ~•

5 Variablesxandyareconnectedbytheequationy= 2x x^+3

Asquare hassidesoflength xcm and area Acm^. The areaisincreasing ata constantrate of0.2cm^s"^

3 Variables x and y are connected by the equation y =(x - 3)(x + 5)"^.

Giventhatyincreasesatarateof0.1unitspersecond,findtherateof changeofxwhenx=-2.

Findtherateofincreaseofxwhenx=4.

1 Variables xand y are connected by the equation y = -5x.

12Aconehasbaseradiusrcmandafixedheight18cm.

Chapter12:Differentiation1

Giventhatxincreasesatarateof0.2unitspersecond,findtherateof changeofywhenx=-4.

Giventhatxincreasesatarateof0.05unitspersecond,findtherateof changeofywhen x-4.

Findtherateofchangeofthevolumewhenr=10.

Thecuboidisheatedandthevolumeincreasesatarateof0.5cm'^s"^

9 Acube hassidesoflength xcm and volume Vcm^.

FindtherateofincreaseofxwhenV=512.

FindtherateofincreaseofxwhenA=16.

Giventhatxincreasesatarateof0.02unitspersecond,findtherateof changeofywheny=1.

Giventhat increasesatarateof0.1unitspersecond,findtherateof changeofywhen*=9.

Ifyoudifferentiateywithrespecttoxyouobtaindy. dy dx —iscalledthefirstderivativeofywithrespecttox. dy d

13Waterispouredintotheconicalcontaineratarateof5cm®s^

dx^ 6x-14

iscalledthesecondderivativeofywithrespecttox.f(x)=X®-7x^+2x+1 or dy = dx 3x^-14x+2 = oror

CambridgeIGCSEand0LevelAdditionalMathematics

Ifyoudifferentiate—withrespecttoxyouobtain (jSy dx dx writtenas—^. dx2 ydx) whichcanalsobe d^y. dx^ Sofory=X®-7x^+2x+1

14 Water is poured into the hemispherical bowl at a rate of47tcm®s \

After iseconds,the volume ofwater in the bowl, Fcm®,is given by V = 87i/i^ -i7t/i® ,where h cm is the height of the water in the bowl, a Findtherateofchangeofhwhen =2.

After tseconds,the volume ofwater in the container, Vcm®,is given by ^~12 ^ heightofthewaterinthecontainer. a Findtherateofchangeofhwhenh=b. bFindtherateofchangeofhwhenh=10.

bFindtherateofchangeofhwhenh=4. 12.8Secondderivatives

d^y

f'(x)= 3x2-14^+2 f"(x)=6x-14

WORKEDEXAMPLE17 Giventhaty=3x^-—,find . Answers y=3x^-Sx"' —=6x+5x"^ dx

d^y dx^ 6-lOx-® = 6-1? x'

Exercise12.8

2 Find —^ for each ofthe followingfunctions. dx^ / ..xs Ax-1 a y=x(x-A) b y= c x+2 d :y=7^ e y= y= f y= —1 X-5

3 Giventhatf(x)=x^-7x^+2x+1,find af(l) bf'(l) cf"(l).

Chapter12:Differentiation1

5Acurvehasequationy=2x^-15x^+24x+fi. dy d^yCopyand complete the table to show whether — and —^ are positive (+), dx dx"^ negative(-)orzero(0)forthegivenvaluesofx. X 0 1 2 3 4 5 dy; dx dx^ dy 6Acurvehasequationy=2x^+3x^-3fix+5.Findtherangeofvaluesofxforwhichboth—and d2y —jarebothpositive, dx"^ d^ d 7Giventhaty=x^-2x+5,showthat4—j+(x-1)—=2y. dx'' dx

4Acurvehasequationy=Ax^+3x^-fix-1. dii —-0whenx=-1andwhenx=0.5.aShowthat XX+1—3 2x+5 3x—1 dx bFindthevalueof dx^ whenX=-1andwhenx=0.5.

iCHALLENGEQ d^'v dyfi 8 Given that y = SVx,show that 4x^ —^+ 4x — = y.6 dx

1 Find dx^ foreachofthefollowingfunctions. ay=5x^-7x+5 by=2x^+3^:^-1 c31=4 x^ . / 4 f y=dy={Ax+1) ey d^y -Jx+3

CambridgeIGCSEand0LevelAdditionalMathematics 12.9

Theredsectionsofthecurveshowwherethegradientisnegative(wheref(x) isadecreasingfunction)andthebluesectionsshowwherethegradientis positive(where f(x) is an increasingfunction).The gradient ofthe curve is zeroatthepointsP,QandR.

Ataminimumpoint: .■i=o dx •thegradientisnegativetotheleftoftheminimumandpositivetotheright

Stationary points

Atamaximumpoint: . ■4:=o dx • the gradient is positive to the left of the maximum and negativetotheright Minimum points The stationary points Pand R are calledminimum points.

Considerthegraphofthefunctiony=f(x)shownbelow. <0 dx

y=f(x) O X

Apointwherethegradientiszeroiscalledastationarypointora turningpoint. Maximum points ThestationarypointQiscalledamaximumpointbecausethevalueofyat thispointisgreaterthanthevalueofyatotherpointsclosetoQ.

Chapter12:Differentiation1 StationarypointsofinflexionThereisathirdtypeofstationarypoint(turningpoint)calleda pointofinflexion. Atastationarypointofinflexion: . =o dx jfrom positive tozero and then to positive again • thegradientchanges <or l^from negative tozero and then to negative again Youdonotneedtoknowahoutpointsofinflexionfortheexamination.Theyhavebeenincludedhereforcompleteness. WORKEDEXAMPLE18 Findthecoordinatesofthestationarypointsonthecurvey=x''-3x+1and determinethenatureofthesepoints.Sketchthegraphofy= -3x+1. Answers y=x^-3x+1 -3 dx Forstationarypoints ^=0. dx 3x2-3=0 x2-1=0 (x+l)(x-1)=0 X=-1orX=1 WhenX=:-l,y=(-1)^-3(-l)+1=3 Whenx=l,y=(l)^-3(l)+l=-l Thestationarypointsare(-1,3)and(1,-1). ^-r -0-9 dy dx 3(-l.l)2-3=positive 0 3(—0.9)2_g_fjegajiye directionof tangent shapeof curve 0.9 1 dy dx 3(0.9)2-3=negative 0 3(1.1)2—3=positive directionoftangent shapeof curve

CambridgeIGCSEand0LevelAdditionalMathematics So(-1,3)isamaximumpointand(1,-1)isaminimumpoint. Thesketchgraphof -3a:+1is: (-1,3) y= -3x+I

Secondderivativesandstationarypoints

Consider movingfrom left to rightalong a curve,passing through a maximumpoint: Thegradient,—,startsasapositivevalue,decreasestozeroatthemaximum dx pointand then decreases to a negative value. df rd-V.

Since—decreasesasxincreases,thentherateofchangeof—isnegative. di) d(d))^ d^y[Therateofchangeof iswrittenas -J Thisleadstotherule: dv d^'v If^=0and—f<0,thenthepointisamaximumpoint. dx dx^ Now,considermovingfromlefttorightalongacurve,passingthrougha minimumpoint:

Thestationarypointsare(1,17)and(4,-10). 0= 12.-30 d2y WhenX=1,—-r=-18whichis<0 dx'^ d2y WhenX=4,—5-=18whichis>0 dx"' So(1,17)isamaximumpointand(4,-10)isaminimumpoint. Thesketchgraphofy=2x®-15x2+24x+6is: y V=2x''-15x-+24x+6 i t \ / X 1 (4,-10)

Findthecoordinatesofthestationarypointsonthecurvey= -I5x^+24x+6and determinethenatureofthesepoints.Sketchthegraphofy=2x^-15x^+24x+6. Answers y=2x^—\bx^+24x+6 dy dx =6x^-30x+24 For stationary points,-^ = 0. ^ dx 6x^-30x+24=0 x^-5x+4=0 (x-l)(x-4)=0 X=1orX=4 WhenX=1,y=2(1)®-15(1)2+24(1)+6=17 WhenX=4,y=2(4)®-15(4)2+24(4)+6=-10

Chapter12:Differentiation1 d)! Thegradient,—,startsasanegativevalue,increasestozeroattheminimum dx pointandthenincreasestoapositivevalue. dy d)>Since increasesasxincreases,thentherateofchangeof—ispositive. dx d;* Thisleadstotherule: dy d^yIf — =0 and —^> 0,then the pointis a minimum point. dx dac'^

WORKEDEXAMPLE19

dFindthecoordinatesofthepointonthecurvewherethegradientis minimumandstatethevalueofthe minimumgradient.

Thecurvey=x^+ax+bhasastationarypointat(1,3).

aFindthevalueofaandthevalueofb.

Findthecoordinatesofthestationarypointsoneachofthefollowing curvesanddeterminethenatureofeachofthestationarypoints. ay= -12x+8 by=(5+x)(l-x) cy= -I2x+2 dy=x^+x^-16x-16 e y = x(3—4x-x^) f y =(x-1)(x^ —6x+2)

Find X+1 _2x+5 X+1 y = —I- yfx X x^-5x+3 x+1 —andhenceexplainwhythecurvehasnoturningpoints, dx Thecurvey=2x^+ax^-12x+7hasamaximumpointatx=-2.

aFindthevalueofaandthevalueofb.

b Determinethenatureofthestationarypoint(1,-1).

The curve y = ax +-^ has a stationary point at(-1,-12).

b Determinethenatureofthestationarypoint(-1,-12).

CHALLENGEQ

b Determinethenatureofthestationarypoint(1,3).

CambridgeIGCSEand0LevelAdditionalMathematics

aFindthevalueofaandthevalueofb.

b Findthecoordinatesofthesecondstationarypointonthecurve. c Determinethenatureofthetwostationarypoints.

2 Findthecoordinatesofthestationarypointsoneachofthefollowing curvesanddeterminethenatureofeachofthestationarypoints. a y —yfx+yfx by=x^ X d y= 2x x^+9 e y= 3

Theequationofacurveisydy

Thecurvey=x^+—+bhasastationarypointat(1,-1). X

8 The curve y = 2x^ - 3x'^ + ax + b has a stationary point at the point (3,-77). a Findthevalueofaandthevalueofb.

Exercise12.9

c Find the coordinates ofthe other stationary pointon the curve and determinethenatureofthisstationarypoint.

Findthevalueofa.

The dimensionsofthe box mustbe positiveso x=3. Whenx=3,V=4(3)^-78(3)^+360(3)=486.

ThestationaryvalueofTis486andoccurswhenx=3.d^V„ d^y WhenX=3,—=24(3)-156=-84whichis<0 Thestationaryvalueisamaximumvalue.

Therearemanyproblemsforwhichyouneedtofindthemaximumor minimumvalueofafunction,suchasthemaximumareathatcanheenclosed withinashapeortheminimumamountofmaterialthatcanbeusedtomake acontainer.WORKEDEXAMPLE20 xcm 15cm 24cm

Thediagramshowsa24cmby15cmsheetofmetalwithasquareofsidexcm removed from each comer.The metal is then folded to make an open rectangular boxofdepthxcmandvolumeVcm^.

a ShowthatV=4x^-78*:^+360x

b FindthestationaryvalueofTandthevalueof*forwhichthisoccurs, c Determinethenatureofthisstationaryvalue. Answers a V=lengthxbreadthxheight =(24-2x)(15-2x)x =(360 — 78x +4x^^x =4x^-78*2+360x dV dx =12*2-I56x+360 Stationaryvaluesoccurwhen-^=0. dx 12a?-156X+360=05£^-13*:+30=0 (x-10)(x-3)=0 X=10orX=3

Chapter12:Differentiation1

12.10Practicalmaximumandminimumproblems

CambridgeIGCSEand0LevelAdditionalMathematics WORKEDEXAMPLE21 T p yva Q PQSTApieceofwire,oflength2m,isbenttoformtheshapePQRST.isarectangleandQRSisasemi-circlewithdiameterSQ. FT=x:mandPQ^=ST= m. ThetotalareaenclosedbytheshapeisAm^. a Expressyintermsofx. b ShowthatA=x 2 o dA d^A c Find—and—-r. dx dx'' d FindthevalueforxforwhichthereisastationaryvalueofA. e Determinethemagnitudeandnatureofthisstationaryvalue. Answers a Perimeter=PQ,+arcQRS+ST+TP 2= y +1 X + y + Jc 2=2y+ijtx+X 2y = 2-X -ijix ,1 1 b A=areaofrectangle-t-areaofsemi-circle JlX^ 19 1 ? 1 2 =X—x^—Jtx'^+—nx'^ 2 4 8 12 1 2 =X x^ JIX"' 2 8 =1-X 7CX dx 4 d^A , 1 =-1 It dx^ 4

Thediagramshowsarectangulargarden withafenceon threeof isitssidesandawallonitsfourthside.Thetotallengthofthefence100mandtheareaenclosedisAm^. a Show that A = ix(100- x). bFindthemaximumareaofthegardenenclosedandthevalueofxfor whichthisoccurs.

ThestationaryvalueofAis (4+7l) anditisamaximumvalue. Thesumoftwonumbersxandyis8. a Expressyintermsofx. i Given thatP= x)), write down an expression for Pin terms of x. ii FindthemaximumvalueofP. ci Giventhat5= ,writedownanexpressionforS,intermsofx. ii FindtheminimumvalueofS.

Chapter 12: Differentiation 1 d Stationaryvaluesoccurwhen dx =0. 1 1—* %x=0 4 4—43c-Jcx=0 x:(4+Ji)=4 Stationaryvalueoccurswhenx e Whenx= (4+7t) A= 4 (4+7l) 1 WhenX= (4+7t)' (4+n) 2L(4+it) 4(4+7i)-8-271 (4+ itf 8+2k (4+ Ji)^ 2(4+71) (4+ 7C)^ 2 (4+7l) 1 1 u-V,•=-1 71whichIS<0 4 1 7C 8 (4+7l),

Exercise12.10 1

3Thevolumeofthesolidcuboidis576cm^andthesurfaceareaisAcm^. a Expressyintermsofx. 1728 bShowthatA=4x^H . X C FindthemaximumvalueofAandstatethedimensionsofthe cuboidforwhichthisoccurs. ycm xcm 2xcm

d Find the value for rfor which there is a stationary value of A. e Determine the magnitude and nature ofthis stationary value. The diagram shows a window madefrom arectangle with base 2rm and height h m and asemicircle ofradius rm.The perimeter ofthe windowis6mandthesurfaceareaisAm^.

e Determine the magnitude and nature ofthisstationary value. ABCDisarectanglewithbaselength2punits,and areaAunits^.

dA d^A cfind—and dr dr d FindthevalueforrforwhichthereisastationaryvalueofA. eDeterminethemagnitudeandnatureofthisstationaryvalue. hm 2rm > 2punits

Expresshintermsofr. b ShowthatA=6r-2r^-—Ttr'^. 2 ^ ,dA ^d^A cFind—and—-.drdr^

A cuboid has a total surface area of400cm^and a volume of Fcm^. The dimensions of the cuboid are 4x cm by x cm by h cm. a ExpresshintermsofTandx.

b Show that A = 2p{4:-p'^). c Find the value ofpfor which A has a stationary value

dFindthisstationaryvalueanddetermineitsnature. Asolidcylinderhasradiusrcmandheighthcm. Thevolumeofthiscylinderis25071cm^andthesurfaceareaisAcm a Expresshintermsofr.

A piece of wire,oflength 60cm,is bent to form a sector ofa circle with radiusrcmandsectorangleQradians.Thetotalareaenclosedbythe shapeisAcm^. a Express6intermsofr.

dX X[ / 1 / 1 / 1 / 1 / 1 / 1 / 1 / 1 / 1 II r Kl / A< *B \ X

bShowthatV=160x5 C FindthevalueofxwhenTisamaximum.

The points A and B lie on the x-axis and the points CandDlieonthecurvey=4—x^. aExpressBCintermsofp.

d Find the value for rfor which there is astationary value ofA.

bShowthatA=27tr^+

bShowthatA=30r-r^. ,p.,dA ^d^A cFind—and—-.drdr

CambridgeIGCSE and0LevelAdditional Mathematics

Thediagramshowsasolidformedhyjoiningahemisphereofradiusrcmtoacylinderofradiusrcmandheighthcm.Thesurface areaofthesolidis2887tcm^andthevolumeisFcm^.aExpresshintermsofr. 5

d Determinethenatureofthisstationaryvalue.

10Apieceofwire,oflength50cm,iscutintotwopieces.Onepieceisbenttomakeasquareofsidexcmandtheotherishentto

bShowthatV=1447lr TCr^. 6

makeacircleofradiusrcm.Thetotalareaenclosedbythetwoshapesis Acm^. a Expressrintermsofx. (71+4)x2-lOOx+625 bShowthatA= 71 c FindthestationaryvalueofAandthevalueofxforwhichthisoccurs. Giveyouranswerscorrectto3sf.

Chapter12:Differentiation1

11Thediagramshowsasolidcylinderofradiusrcmandheight2hcmcutfromasolidsphereofradius5cm.Thevolumeofthecylinder isFcm^. a Expressrintermsofh.

CFindtheexactvalueofrsuchthatFisamaximum.

b Show that V = 27t/i(25-h'^). c FindthevalueforhforwhichthereisastationaryvalueofV.

CHALLENGEQ

12Thediagramshowsahollowconewithbaseradius12cmandheight24cm. Asolidcylinderstandsonthebaseoftheconeandtheupperedge touchestheinsideofthecone. The cylinder has base radius rcm,height h cm and volume Fcm^. a Expresshintermsofr.

b ShowthatV=27ir^(12-r).

C Findthevolumeofthelargestcylinderwhichcanstandinside thecone. 24cm 12cm

CambridgeIGCSEand0LevelAdditionalMathematics CHALLENGEQ 13Thediagramshowsarightcircularconeofbaseradiusrcmandheighthcmcutfromasolidsphereofradius10cm.Thevolumeofthe coneisVcm^. a Expressrintermsofh. b Showthat y= (20-/1). C FindthevalueforhforwhichthereisastationaryvalueofV. d Determinethemagnitudeandnatureofthisstationaryvalue. dy Ify=x",then =nx"' dx £[fM±gW]= £[fW]±£[gW] dy_dy^dtt dx du dx d- , dw dtt —(ttu)=u hV dx dx dx Summary Rulesofdifferentiation Powerrule: Scalarmultiplerule: Addition/subtractionrule: Chainrule: Productrule: Quotientrule: Tangentsandnormals Ifthe value of — atthe point(xj,yj) is m,then: dx the equation ofthe tangentisgiven by y-yj= to(x-Xj) theequation ofthe normalisgiven by y — yj=-—(x-Xj) m Smallincrementsandapproximations„6y dyIf5xand5yaresufficientlysmallthen ~—. ox dx Stationarypoints Stationary points(turning points)ofafunction y =f(x) occur when = 0. Ax d/u dxvw du dv V u dx dx dy

b Given that rcan vary,show thatthestationaryvalue ofPcan be written in theform ky/2, where ^is a constant to be found. [4]

CambridgeIGCSEAdditionalMathematics0606Paper11QlO(part)Nov2011

c Determine the nature ofthis stationary value and find the value of0for which it occurs. [2]

Chapter12:Differentiation1 Firstderivativetestformaximumandminimumpoints Atamaximum point: dy u =z0 I dx a thegradientispositivetotheleftofthemaximumand negativetotheright Ataminimumpoint: . ^= 0 dx ® thegradientisnegativetotheleftoftheminimumandpositivetotheright Secondderivativetestformaximumand minimum points fd'"^)! <0 I If -^ = 0 and < dx 'I m dx^dx^ ,thenthepointisamaximumpoint >0,thenthepointisaminimumpoint Examination questions Worked example r cm rcm

a By first expressing 0in terms ofr,show thatthe perimeter,Pcm,ofthe shaded region is givenbyP=4r+—. [6]

The figure showsasector ABCofa circle centre C,radius 2rcm,where angle ACEis30radians. The pointsD,E,Tand Glie on an arc ofa circle centre C,radius rcm.The pointsDand Gare the mid-pointsofCAandCBrespectively.AnglesDCEandEGGareeach0radians.Theareaofthe shadedregionis5cm^.

CambridgeIGCSEand0LevelAdditionalMathematics Answers a AreaofsectorABC-(areaofsectorDCE+areaofsectorFCG)=5 -r^e=b 5r^6=5 0=4 P=CE+arcED+m+arcAB+5G+arcGP+PC =r+rO+r+(2r)(30)+r+r0+r =4r+8r0 = 4r- + 8r X^ , 8 =4r+ r —P=4r+8r"^=4-8r-2 dr dP Stationaryvaluesoccurwhen—=0. dr 4-4= 0 r2=2 r=^ Whenr=yl2,P=4 =8>/2 Stationaryvalue ofPis 8^2. cWhenr=y/2,6= 1 1 2 When r = yj2, dr d^P dr^ d^P =4-8r-2=16r-3 16 dr^ (V2) 3"=positive The stationary value is a minimum and occurs when ^ ~ ^ •

CambridgeIGCSEAdditionalMathematics0606Paper21Q8iNov2014

CambridgeIGCSEAdditionalMathematics0606Paper21Q7i,ii,iiijun2014

CambridgeIGCSEAdditionalMathematics0606Paper11Q5i,iiJun2012

4Asectorofacircleofradiusrcmhasanangledradians,where6<%.Theperimeterofthe sectoris30cm. aShowthatthearea,Acm^,ofthesectorisgivenbyA=15r-r^. [3] b Given that rcan vary,find the maximum area ofthe sector. [3]

1 Find the coordinates ofthe stationary pointon the curve y = x'^ + 16 [4]

7aFindtheequationofthetangenttothecurvey=x^+2x^-3x+4atthepointwherethecurvecrossesthey-axis.[4]

2 Giventhat y= Ay kx ,showthat—= .9 2+ Ax (2+ xf ,whereAisaconstanttobefound. [3]

CambridgeIGCSEAdditionalMathematics0606Paper21Q8i,iiJun2014

Chapter12:Differentiation1

CambridgeIGCSEAdditionalMathematics0606Paper21Q6Jun2013

8Thenormaltothecurvey=x^+6x^-34x+44atthepointP(2,8)cutsthex-axisatAand they-axisatB.Showthatthemid-pointofthelineABliesontheline 4y=x-I-9. [8]

CambridgeIGCSEAdditionalMathematics0606Paper21Q6Nov2012

bFindthecoordinatesofthepointwherethetangentmeetsthecurveagain. [3]

Exercise12.11 ExamExercise

CambridgeIGCSEAdditionalMathematics0606Paper11QlNov2014

3 Given thata curve has equation y = — +2-s/x,where x>0,find a ^ [2] dx' b [21 Hencedx^orotherwise,find C thecoordinatesandnatureofthestationarypointonthecurve. [4]

CambridgeIGCSEAdditionalMathematics0606Paper21Q3i,iiNov2013 X^+8 6 Findtheequationofthenormaltothecurvey X-2 atthepointonthecurvewherex=4. [6]

5 a Given that y = f— x - ,find —. [2] U J dx b Hence find the approximate change in y as x increasesfrom 12 to 12+ p,where pis small.[2]

facesThediagramshowsanemptycontainerintheformofanopentriangularprism.Thetriangularareequilateralwithasideofxcmandthelengthofeachrectangularfaceisycm.Thecontainerismadefromthinsheetmetal.Whenfull,thecontainerholds200^/3cm^. i Show that Acm^,the total area ofthe thin sheet metal used,is given by , ^/3^ 1600 A= + 2 ^ [5] Giventhatxandycanvary,findthestationaryvalueofAanddetermineitsnature. [6]

CambridgeIGCSEAdditionalMathematics0606Paper21Q7i,iiNov2010 11 X cmxcm xcm

CambridgeIGCSEAdditionalMathematics0606Paper21Q7Jun2012 10 xcm 45cm xcmxcm 60cm A rectangular sheet of metal measures60cm by 45cm.A scoop is made by cutting outsquares, ofsideXcm,fromtwocornersofthesheetandfoldingtheremainderasshown. a Show that the volume, Vcm^ ofthe scoop is given by 7=2700x-165x2+2x\ [2] bGiventhatxcanvary,findthevalueofxforwhichVhasastationaryvalue. [4]

CambridgeIGCSEAdditionalMathematics0606Paper12Q9Mar2015

CambridgeIGCSEand0LevelAdditionalMathematics 6489 Given that f(^c)= ,find the value ofxfor which f"(x)= 0. [6]

y Chapter 13 Vectors Thissectionwillshowyouhowto:usevectorsinanyform,e.g. a AB,p,at+b'i ■ usepositionvectorsandunitvectors ■ findthemagnitudeofavector;addandsubtractvectorsandmultiplyvectorsbyscalars ■ composeandresolvevelocities.

Twovectorsaresaidtobeequaliftheyarethesamelength andareinthesamedirection.

Youshouldalreadybefamiliarwiththefollowingvectorwork; Quantitiesthathavebothmagnitudeanddirectionarecalledvectors.Quantitiesthathaveonlymagnitudearecalledscalars.ABmeansthedisplacementfromthepointAtothepointB. 41-

For example,if a =| ^ |and b thena+banda-bcanbeshownonavectordiagramas: a+banda-bcanalsobefoundasfollows: '44 f2+ 4^ lu U+1 a+b 2-43-1 -2 Multiplicationbyascalar

Forthediagram,AB The'magnitude'ofthevectorABmeansthe'length'ofthevectorABandisdenotedby ABI. IABIiscalledthemodulusofthevectorAB.

Thevectora+bmeansthevectorafollowedbythevectorb. Thevectora-bmeansthevectorafollowedhythevector-b. Theresultantvectorisoftenshovrawithadoublearrow. Itisdrawnfromthestartingpointtothefinishingpoint.

I"ini u a + b Start Finish 7 -bSt/ art

a-b

'-6 12

UsingPythagorasforthediagram,I AB I-V(4)^+(-3)^ =5.

Thevectora+acanbewrittenas2a. 2a, Thevectorsaand2aareexamplesofparallelvectors. Twovectorsareparallelifonevectorcanbewrittenasamultipleoftheothervector. Forexample: 0 and(s)areparallelandinthesamedirectionbecause1.^1=5'^ 2 -4 -6andI Iareparallelandinoppositedirectionsbecause IfCollinearpointsAB=MCthenthepointsA,BandCarecollinear. (ThisisbecausethelinesABandACmustbeparallelandthepointAliesonbothlines.)

Thevector-aisthesamelengthasthevectorabutisintheoppositedirection. Additionandsubtractionofvectors

CambridgeIGCSEand0LevelAdditionalMathematics

Chapter13:Vectors 13.1

WORKEDEXAMPLE1 Find I^|. aPQ^=41-2j bUsingPythagoras,|PQ\= Youcouldbeaskedtofindtheunitvectorinthedirectionofagivenvector.Themethodisoutlineinthefollowingexample.

+(-2)^=-JW=2-\/5.

WORKEDEXAMPLE2 =41+3j FindtheunitvectorinthedirectionofthevectorEF. Answers FirstfindthelengthofthevectorEF: EF^=4^+3^ usingFvthagoras HencetheunitvectorinthedirectionofEFis: 1 -(4i+3j)

ThevectorABinthediagramcanbewrittenincomponentformas ABcanalsobewrittenas41+3j,where: iis a vector oflength 1 unitin the positive x-direction and jis a vector oflength 1 unitin the positive))-direction. Note: Avectoroflength1unitiscalledaunitvector.

Answers P Q

^4^v3y B y/i J JL j ik j A i

Furthervectornotation

aWritePQintheformai+b\. b

CambridgeIGCSEand0LevelAdditionalMathematics WORKEDEXAMPLE3 a=-2i+3j, b=4i-j and c=-221+18j. FindAand suchthatla+/tb=c. Answers la+;Ub c l(-2i+3j)+^(41-j)=-22i+18j Equatingthei'sgives -21+4fl=-22 -l+2/x=-ll (1) Equatingthej'sgives 31-ju=18 61-2/t=36 (2) Addingequations(i)and(2)gives 51=251=5 Substitutingfor1inequation(1)gives -5+2jU=-11 2fi=-6 /t=-3 So1=5,/i=-3. Exercise13.1 1 Writeeachvectorintheformai+ . a AB b AC C aB d AE e BE f Be gEA hDB iBe 2 Findthemagnitudeofeachofthesevectors. a-2i b4i+3j c5i-12j d-81-6j e7i+24j f 151-8j g-41+4j h 51-lOj 3 ThevectorABhasamagnitudeof20unitsandisparalleltothe vector41+3j. FindAB. 4 ThevectorPQhasamagnitudeof39unitsandisparalleltothe vector121-5jFindPQ. 5 Findtheunitvectorinthedirectionofeachofthesevectors. a61+8j b51+12j c-41-3j d81-15j e31+3j 6 p=81-6j,q=-21+3jandr=101 Find A E D 'c b' a2q b 2p+q 1 p-3r 1-r-p-q.

OP= ^3^ v2. or OP=31+2j NowconsidertwopointsAandBwithpositionvectorsaandb. A B i b o ABmeansthepositionvectorofBrelativetoA. AB=Ad+OB Hence:=-0A+OB Ab=OB-OA or AB=b-a

13.2Positionvectors ForO,meansthedisplacementofthepointPfromO.thisdiagram,thepositionvectorofPis

Chapter13:Vectors 7 p=9i+12j,q=3i-3jand r=7i+j Find a|p+q| b|p+q+r|. 8 p=7i-2j and q=i+jUj FindAand suchthat Ap+q=36i-13j. 9 a=5i-6j,b=-i+2j and c=-131+18j. FindAand/XsuchthatAa+/ib=c.

ThepositionvectorofapointPrelativetoanorigin,

CambridgeIGCSEand0LevelAdditionalMathematics WORKEDEXAMPLE4

1 ThepointPliesonPQsuchthatPR-—PQ. bFindthepositionvectorofP Answers a P(l=OQ-OP =(10i-3j)-(4i+5j) PQ-6i-8j PR =-PQ 4^ = i(6i-8j) 4 =1.5i-2j OR=OP+m OR=(4i+5j)+(L5i-2j)=5.5i+3j OQ=101-3jandOP=41+5j collectI'sandj's WORKEDEXAMPLES RelativetoanoriginO,thepositionvectorsofpointsA,BandCare-21+5j,101-j

AC=OC-OA

OB-OA collectI'sandj's Answers

RelativetoanoriginO,thepositionvectorofPis4i+5jandthepositionvectorofQislOi-3j.aFindPQ. and A(21+j) respectively.GiventhatCliesonthelineAB,findthevalueofA. AB +5j =(101-j)-(-21+5j) =121-6j IfCliesonthelineAB,thenAC=kAB. =A(21+j)-(-21+5j) =(2A+2)l-(5-A)j (2A+2)1-(5-A)J=12^:1-6ftj =101-jand

kAB=1^(121-6j) =12^1-6ki Hence,

OA =-21 21+j)andOA--21+5j collectI'sandj's -d) multiplybothsidesby2 -(2)

EquatingtheI'sgives: 2A+2=I2k Equatingthej'sgives: 5-X=6k 10-2X=12k Usingequation(1)andequation(2)gives 2A+2=10-2A 4A=8 A=2 OC=A,( OB

1 Find AB,in theform a\+ b], for each ofthe following, aA(4,7)andB(3,4) bA(0,6)andB(2,-4) cA(3,-3)andB(6,-2) dA(7,0)andB(-5,3) e A(-4,-2)andB(-3,5) 2 a f A(5,-6)andB(-1,-7).

Chapter13:Vectors

ThepointCliesonABsuchthatAC=SCB. aFindAP. bFindtheunitvectorinthedirectionofAB. C FindthepositionvectorofC.

Exercise13.2

ThevectorRelativetoanoriginO,thepositionvectorofAis4i-2jandthepositionofPisAi+2j.unitvectorinthedirectionofABis0.3i+0.4j.FindthevalueofA.

7 RelativetoanoriginO,thepositionvectorofAis ^ 10^ -11 -10 10. andthe positionvectorsofPis

aFindAB. ThepointsA,PandClieonastraightlinesuchthatAC=2AP.bFindthepositionvectorofthepointC.

4RelativetoanoriginO,thepositionvectorofAis-7i-7jandthe positionvectorofPis9i+5j.

3O is the origin,Pis the point (1,5) and PQ ~ I ^ j• Find OQ. -2 3' -5, bOistheorigin,Eisthepoint(-3,4)andEF FindthepositionvectorofF. cOistheorigin,Misthepoint(4,-2)andNM= FindthepositionvectorofN.

3ThevectorOAhasamagnitudeof25unitsandisparalleltothe vector-3i+4j. The vector OB has a magnitude of26 units and is parallel to the vector12i+5j. Find: aOA b'OB cAB dIabI.

5 RelativetoanoriginO,thepositionvectorofPis-2i-4jandtheposition vectorofQis8i+20j.aFindPQ. bFindI^|. C FindtheunitvectorinthedirectionofPQ. d FindthepositionvectorofM,themidpointofPQ.

10RelativetoanoriginO,thepositionvectorofAis61+6jandtheposition vectorof is121-2j. aFindAB. The point Clies on ABsuch that AC =^AB. bFindthepositionvectorofC.

ThepointsA,BandCaresuchthatBC=2AB.FindthepositionvectorofC.

CambridgeIGCSEand0LevelAdditionalMathematics

14RelativetoanoriginO,thepositionvectorofAis—61-I-4jandthe positionvectorof5is181-I-6j.Cliesonthey-axisandOC=OA-I-XOB. FindOC.

15 Relative to an origin O,the position vector ofPis 81 -t- 3j and the position vector of Qis -121-7j.Plieson the x-axis and OR = OP+flOQ. FindOR.

11RelativetoanoriginO,thepositionvectorofAis1 1andtheposition (5^vectorofBis

positionvectorsofBis aFind: i 1"^! ii iil Ilil. The points A,Band Clie on a straightline such that AC= CB. bFindthepositionvectorofthepointC.

(3^

9 Relative to an origin O,the position vector ofA is 31-2j and the position vectorofBis151+7j. a FindAB. The point Clies on ABsuch that AC =^ AB. b FindthepositionvectorofC.

GiventhatCliesonthelineAB,findthevalueofX.

13RelativetoanoriginO,thepositionvectorsofA,BandCare-21+7j, 21-j and 61+ Ajrespectively.

8 Relative to an origin O,the position vector ofA is |j the ^24^ 18

12RelativetoanoriginO,thepositionvectorsofpointsA,BandCare -51-llj,231-4jandA(1-3j)respectively.

aFindthevalueofAwhenAC=17. b FindthevalueofAwhenABCisastraightline. c FindthevalueofAwhenABCisaright-angle.

Chapter13:Vectors CHALLENGEQ 16 Relative to an origin O,the position vectors ofpoints P, Qand i?are -6i+8j,-4i+2j and 5i+5j respectively. a Findthemagnitudeof: I ^ li ^ iii b ShowthatanglePQRis90°. cIfOP=AOQ+liOR,findthevalueofAandthevalueoffl. 13.3 Vector geometry WORKEDEXAMPLE6 OA=ti,OB=h,BX=-=BAandOY=-OA. 5 4 a Findintermsofaandb: \ M W WC iii OX iv^ bGiventhatOP=AOX,findOPintermsofA,aandb. CGiventhatBP=jlBY,findOPintermsofjl,aandb. d FindthevalueofAandthevalueofII. AnswersBA=OA-0B==:SL-h i BX=-H=-(a-b) 5 5 ii^=0B+H=b+-(a-b)=-a+-b 5 5 5 V RY=10+0Y=-b+-'dA=-a-b 4 4 b OP=AOX =A—a+—b\55 3A 2A^ =—a+—b 5 5

CambridgeIGCSEand0LevelAdditionalMathematics c OP=OB+BP = b +/i(|a-b) 3/i a+(1-jii)b dEquatingthecoefficientsofaforOPgives: 3A_3^ T~T A_£ 4A"S~4=5jU (1) dividebothsidesby3 multiplybothsidesby20 2A , —=1-u 5 ^ EquatingthecoefficientsofbforOPgives:multiplybothsidesby5 2A=5-5/x (2) Addingequation(1)andequation(2)gives: 6A=5 A =^ 5 2 SubstitutingA=—inequation(1)givesp= —■ 6 3 5 2 Hence, A = — and U = —■ 6 3 WORKEDEXAMPLE7 OA =3a, OB —4bandMisthemidpointofOB. OP: PA=4:3and^ = XBP. a Findintermsofaandb: B i AB ii AM. b Find in terms of A, a and b: M^ u i BX ii JiX. ^ 1 C IfM, Xand A are collinear, find the value ofA. ^ p A Answers a i AB =A0+7)B = -3a + 4b W Im=MO+OA =-2b+3a=3a-2b

Chapter13:Vectors b \ BX =Xbp =XiW+'op) = Xf^-Ab+^'OA I2A . = a-4Ab 7 li MX =MB+H OU 12-^=2b+ a-4Ab 7 useOA=3a -TTS 12?^ useBX=-—a-47.b 7 collecta'sandb's 12A a+(2-4A)b

IfM,XandAarecollinear,thenMX=kMA. 19^ -^a +(2-4A)b = A(3a-2b)

Usingequation(1)andequation(2)gives: 2A -1 = ^ 7 14A-7=4AlOA=7A=0.7 dividebothsidesby-2

Equatingthecoefficientsofbgives: -2k=2-4A k=2A-1 (2)

Exercise13.3 1 OA=a,OB=b. RisthemidpointofOAandOP=SOB. AQ = XaB and^=ju^. a FindOQintermsofA,aandb. b FindOQintermsofn,aandb. c FindthevalueofAandthevalueofji. 2 AB=5a,^=3aandH=b. AX=AACandDX=j^iDB. aFindintermsofaandb, iAD,iiDB. b FindintermsofA,jU,aand/orb, iAX, iiDXc FindthevalueofAandthevalueoffx.

Equatingthecoefficientsofagives: 3.=1M 7 dividebothsidesby3 7 -d)

3CambridgeIGCSEand0LevelAdditionalMathematicsOP=a,PY=2b,andOQ=3b.

5 OA^a,OB^b.

M is the midpoint ofAB and OY —^ OA. 'ox=XOMandH aFindintermsofaandb, iAB iiOM. cbFindOXintermsofA,aandb.FindOXintermsofjl,aandb. d FindthevalueofAandthevalueoffl. OA=a,OB=h,BX=-BAandOY^-OA. 5 7 'op = XOX and^=[lEY.

BisthemidpointofODandAC=— QA. 'ax = AAD and H =iiBC. ^ a FindOXintermsofA,aandb. bFindOXintermsof aandb. c FindthevalueofAandthevalueoffl.

aFindOPintermsofA,aandb. b FindOPintermsoffl,aandb. cOAFindthevalueofAandthevalueoffl.=a,OB=handOistheorigin. 'ox-XOA and OY=flOB. aI FindBXintermsofA,aandb. ii FindAFintermsoffl,aandb. b 5BP=2^andAF=W. i FindOPintermsofA,aandb. ii FindOPintermsoffi,aandb. iii FindthevalueofAandthevalueoffl.

OX = WY and =^jL^. a FindOXintermsofA,aandb, b FindOXintermsofji,aandb, C Find the value ofA and the value offl.

4 OA=a,OB=b.

Chapter13:Vectors 8 O,A,BandCarefourpointssuchthat 'OA=7a-5b,'OB=2a+5band^=-2a+13b. aFind iAC iiAB. b Useyouranswerstopartatoexplainw yBliesonthelineAC. CHALLENGEQ 9 OA=aandOB=b. OA:A£=1:3 and A5:5C=1:2. OB=BD a Fin^in termsofaand/orb^ ^ i OE ii OD iii OC. bFind,intermsofaand/orb, iC£ iiCO iiiDE. CUseyouranswerstopartbtoexplainwhyC,DandEarecollinear. dFindtheratioCD:DE. 13.4 Constant velocity problems Ifan objectmoveswith aconstantvelocity,v,where v =(41-2j)ms"\ thevelocitycanberepresentedonadiagramas: Velocityisaquantitythathasbothmagnitudeanddirection. Themagnitudeofthevelocityisthespeed. IfV=(41-2j)ms-ithen, speed=yJ{Af+{-2f =2>/5ms""^ Youshouldalreadyknowtheformulaforanobjectmovingwithconstantspeed:speed= distancetravelled timetaken Similarly,theformulaforanobjectmovingwithconstantvelocityis: ,. displacementraocity= timetaken

The velocity ofa particle travelling north-east at 4>/2 m s"^ can be written in the form (ai -t- ms"^: b cos45°= and sin45°4>/2 - 4V2 a=4>/2Xcos45° b=4>^xsin45 a=4 6=4 Hencethevelocityvectoris(4!+4j)ms"^

Splitting a velocity into its components

The velocity ofa particle travelling on a bearing of120° at20ms canbewrittenintheform(xi+yj)ms"^: sin60°=— and cos60°= 20 20 a:=20Xsin60° y=20xcos60° jc=lOyfE y=10 Hence the velocity vector is (lO-v/S^i-lOj) ms"^ □

AnobjecttravelsataconstantvelocityfrompointAtopointB. AB = (321-24j)mandthetimetakenis4s.Find: a thevelocity bthespeed. Answers ,. displacement 321-241 ,avelocity=— = =(8i-6j)ms" timetaken 4 b speed = yjiSf + (-6)^ = 10 ms"^ :ity 2^kmh

Thepositionat2pm=Thepositionat3pm= '.a-l-f 6^ 1113 13 \ /<g^ v9/ Theposition at4pm ~ I jgJ ^ M-(" 13y 14 12 10 8 46 2 km relative O s 111, % 3 pm % 4 P»n 2 4 6 8 1012 14 16 18 20 22

Consider a boat sailing with velocity At 12 00 hours the boat is at the point A with position vector to an origin O. Thediagramshowsthepositionsoftheboat at 12 00 hours, 1 pm, 2 pm, 3 pm, 4 pm The position at 1 pm =

WORKEDEXAMPLE8

CambridgeIGCSEand0LevelAdditionalMathematics

1 a Displacement =(21i+54j)m,time taken =6seconds.Find the velocity. b Velocity =(5i-6j)ms"\ time taken =6seconds.Find the displacement. c Velocity =(-4i+4j)kmh"\ displacement =(-50i+50j)km. Findthetimetaken.

3 A helicopterfliesfrom a pointPwith position vector(50i+100j)km to a pointQ. The helicopter flies with a constantvelocity of(30i-40j)kmh"^ and takes 2.5hourstocompletethejourney.FindthepositionvectorofthepointQ.

2 Acar travelsfrom a pointA with position vector(60i-40j)km toa pointB with position vector (—50i+18j)km. The car travels with constantvelocity and takes5 hours to complete Findthejourney.thevelocityvector.

b A boatsailson a bearingof030°with aspeed of20kmh"^ Findthevelocityvectoroftheboat, c A plane flieson a bearing of240°with aspeed of100ms"^ Findthevelocityvectoroftheplane.

5 A particle startsata pointPwith position vector(-80i+60j)m relative to anoriginO. The particle travelswith velocity(I2i-16j)ms~^. a Findthespeedoftheparticle, b Findthepositionvectoroftheparticleafter i1second ii2seconds iii3seconds, c FindthepositionvectoroftheparticletsecondsafterleavingP.

Chapter13:Vectors

Exercise13.4

Hence the position vector,r,ofthe boat t hours after 1200 hours is given by theexpression: r 3 ,13 -2, Thisleadstothegeneralrule:

4 a Acar travels north-eastwith aspeed of18>/2 kmh"K Findthevelocityvectorofthecar.

Ifan object has initial position a and moves with a constant velocity v,the positionvectorr,attimet,isgivenbytheformula:r=a+/v.

\ '5' -{-t J 112J

11 Attime t=0,boatPleavesthe origin and travelswith velocity(31-I-4j)kmh"^

Also at time t =0,boat Qleaves the pointwith position vector (-101 -I- 17j)km and travels with velocity (51 -I- 2j)kmh"^

a WritedownthepositionvectorsofboatsAandBafter2hours, b Find the distance between boatsPand Qwhen t=2.

8At1200hours,aboatsailsfromapointP.

Thepositionvector,rkm,oftheboatrelativetoanoriginO,thoursafter 1200isgivenbyr= a Write down the position vector ofthe pointP. b Write down the velocity vector ofthe boat, c Findthespeedoftheboat.

CambridgeIGCSEand0LevelAdditionalMathematics

6 At 1200 hours,a ship leaves a point Qwith position vector (lOi+38j)km relativetoanoriginO.Theshiptravelswithvelocity(6i-8j)kmh"^ a Find the speed ofthe ship, b Findthepositionvectoroftheshipat3pm. CFindthepositionvectoroftheshipthoursafterleavingQ. d Findthetimewhentheshipisatthepointwithpositionvector (61i-30j)km.

d FindthedistanceoftheboatfromPafter4hours.

7 At 1200 hours,a tanker sails from a pointPwith position vector (5i+12j) kmrelativetoanoriginO.Thetankersailssouth-eastwithaspeedof I2V2kmh-^ a Findthevelocityvectorofthetanker, b Findthepositionvectorofthetankerat i1400hours ii 1245hours. CFindthepositionvectorofthetankerthoursafterleavingP.

9At1500hours,asubmarinedepartsfrompointAandtravelsadistanceof 120kmtoapointB.

Thepositionvector,rkm,ofthesubmarinerelativetoanoriginO,thours after 15 00 is given by r = ^ . ^20-H a Write down the position vector ofthe point A. b Write down the velocity vector ofthe submarine. C FindthepositionvectorofthepointB.

10 At 1200 hours two boats,A and B,have position vectors (-101+40j)km and (701+lOj)km and are moving with velocities(201+lOj)kmh"^ and (-101 -I- 30j)kmh"^ respectively.

a FindthepositionvectorsofAandBat1500hours. bFindthedistancebetweenAandBat1500hours.

Chapter13:Vectors Summary Positionvectors ABmeansthepositionvectorofBrelativetoA. AB=OB-OA or AB=b-a Ifanobjecthasinitialpositionaandmoveswithaconstantvelocityv, the position vector r,at time t, is given by the formula:r = a+iv. Velocity ,,, . displacementVelocity=—' timetaken Examinationquestions Workedexample In this question| ^ | is a unit vector due eastand| ^ At12:00acoastguard,atpointO,observesashipwithpositionvector Theshipis movingatasteadyspeedoflOkmh"'onabearingof330°. isaunitvectorduenorth. kmrelativetoO. ^16^ vl2. a Findthevalueofpsuchthat kmh representsthevelocityoftheship. [2] b Write down,in termsoft, the position vector ofthe ship,relative to O,thours after 1200. [2] c FindthetimewhentheshipisduenorthofO. [2] d Find the distance ofthe shipfrom Oat this time. [2] CambridgeIGCSEAdditionalMathematics0606Paper21Q7i-ivNov2012 Answers a cos60°= 10 X=10Xcos60° X=5 and sin60°= 10 y=10Xsin60° y=5-\/3 Thevelocityvectoroftheshipis Hence, p = 5>/3. ms

CambridgeIGCSEand0LevelAdditionalMathematics flfil '16-5t' r ,5>/3, 12+ 5y/St^ c r = 16-5f 12+5>/3r V y ShipisduenorthofOwhen 16- =0 t=3.2 Time =1512 '16-5x3.2' ^ 0 ^ r J2+5^x 3.2^ J2+16>/3^d Whent=3.2,r= Distance ofship from 0=12+16^3 = 39.7km to 3sf. ExamExercise13.5Exercise 1RelativetoanoriginO,thepositionvectorsofthepointsAandBare21-3jand111+42j arespectively.WritedownanexpressionforAB. The point Clies on ABsuch that AC =^AB. [2] [4]b TheFindthelengthofOC.pointDliesonOAsuchthatDCisparalleltoOB. CFindthepositionvectorofD. [2] CambridgeIGCSEAdditionalMathematics0606Paper21Q8i,ii,iiiJun2012 InthediagramOA=a,OB=bandAP=—AB. aGiventhatOX=pOP,wherepisa.constant,expressOXintermsofp,aandb. [3] b GivenalsothatAX=XOB,whereX,isaconstant,useavectormethodtofindthevalue of p and ofX. [5] CambridgeIGCSEAdditionalMathematics0606Paper21Q8i,iiNov2011

RelativetoanoriginO,pointsA,BandChavepositionvectorsrespectively.Alldistancesaremeasuredinkilometres.Amandrivesataconstantspeeddirectly fromAtoBin20minutes.

5 a ThefourpointsO,A,BandCaresuchthatOA=5a,OB=15b,OC=24b-3a. ShowthatBliesonthelineAC. [3] b RelativetoanoriginO,thepositionvectorofthepointPis1-4jandthepositionvectorof thepointQis31+7j.Find i 1^I, [2] 11 the unitvectorin the direction PQ_, [1] IN the position vector ofM,the mid-pointofPQ. [2] CambridgeIGCSEAdditionalMathematics0606Paper21Q7Jun2015 (5] '-lo' and UJ I I2J I—100

CambridgeIGCSEAdditionalMathematics0606Paper21Q3Jun2011

3RelativetoanoriginO,thepositionvectorsofthepointsAandBarei-4jand71+20j respectively.The point Clies on ABand issuch thatAC=— AB.Find the position vector of C andthemagnitudeofthisvector. [5]

Chapter13:Vectors

4ThepositionvectorsofthepointsAandBrelativetoanoriginOare-21+I7jand61+2j respectively. FindthevectorAB. [1] FindtheunitvectorinthedirectionofAB. [2]

ThepositionvectorofthepointCrelativetotheoriginOissuchthatOC=OA+mOB, wherewisaconstant.GiventhatCliesonthex-axis,findthevectorOC. [3]

1 Calculatethespeedinkmh-iatwhichthemandrivesfromAtoB. HenowdrivesdirectlyfromBtoCatthesamespeed. [3] 11 FindhowlongittakeshimtodrivefromBtoC. [3] CambridgeIGCSEAdditionalMathematics0606Paper21Q3Nov2015

CambridgeIGCSEAdditionalMathematics0606Paper22Q5Mar2015

ThefigureshowspointsA,BandCwithpositionvectorsa,bandcrespectively,relative toanoriginO.ThepointPliesonABsuchthatAP:AB=3:4.ThepointQliesonOCsuch thatOQ:QC=2:3. ExpressAPintermsofaandbandhenceshowthatOP=-(a+3b). [3] Find PQ in terms ofa,b and c. [3] ii Given that5PQ = 6PC,find c in terms ofa and b. [2] CambridgeIGCSEAdditionalMathematics0606PaperIIQ9Jun2013

CambridgeIGCSEand0LevelAdditionalMathematicsThepositionvectorsofpointsAandBrelativetoanorigin0areaandbrespectively.The pointPissuchthatOP=jlOA.ThepointQissuchthatOQ=^OB.ThelinesAQandBP intersectatthepointK I ExpressAQ_intermsofA,aandb. ii ExpressBPintermsofH,aandb. ItisgiventhatSAP=AQand9>BR=IBP. iii Express OR in terms ofA,a and b. ExpressORintermsofn,aandb. HencefindthevalueofflandofA. IV V [1] [1][2][2][3]

CambridgeIGCSEAdditionalMathematics0606PaperIIQ12Nov2014

The points Sand Thave position vectorss and t with respectto an origin O. ThepointsO,SandTdonotlieinastraightline.Giventhatthevector2s+ is paralleltothevector(/t+3)s+9twhere//isapositiveconstant,findthevalueoffl.

ThepointsAandBhavepositionvectorsaandbwithrespecttoanoriginO. ThepointCliesonABandissuchthatAB:ACis1:X.Findanexpression forOCintermsofa,bandX.

[3][2][3][3]

Chapter13:Vectors 9 a The vectors p and q are such thatp = Hi-24jand q= 21+ aj. I ii Findthevalueofeachoftheconstantsaandpsuchthatp+2q=(«+y9)i-20j. Usingthevaluesofaand/Sfoundinparti,findtheunitvectorinthedirection p+2q.

CambridgeIGCSEAdditionalMathematics0606Paper22Q10Mar2016

■- ■" --v. i=S>-4-' K- :.:<•. ..; ■■• Thissectionwillshowyouhowto: H differentiatesinX,cosx,tanx,e'^and Inxtogetherwithconstantmultiples,sumsandcomposite functionsofthese.

Chapter13:Vectors dv In Chapter 12you learnt thatif y =x^,then — =2x Thiscanalsobewrittenas:iff(A:)= thenf'(x)=2x. Graphingsoftwarecanbeusedtoshowthefunctionf(x)=x^ anditsgradient(derived)function("'(x). i Iv ■" 3" \ \ \ \ 1y = i // ✓ / /✓y ^/ ✓ / ✓ /✓ / 1 1 -2 -1 XI- 0 -[ 2 ^ -3-2✓ -4\ \ \ 14.1 Derivatives of exponentialfunctions CLASSDISCUSSION Graphingsoftwarehasbeenusedtodrawthegraphsofy=2"and3)=3*togetherwiththeir gradient(derived)functions. 10-3 86431=2' 2- gradient function -2 -1 O gradient function Discusswithyourclassmateswhatconclusionscanbemadefromthesetwographs.

CambridgeIGCSEand0LevelAdditionalMathematics InChapter6youlearntabouttheexponentialfunctiony=e"where e«2.718. Thisfunctionhastheveryspecialpropertythatthegradientfunctionis identicaltotheoriginalfunction.Thisleadstotherule: —(e*)= e* dx^' Thederivativeofe^'^^ Considerthefunctiony=e^^*\ Let y=e" where m=f(x) du = e" -=f'W dx ^ ^ Usingthechainrule:—=—x dx dw dx =e"Xf'(x) =f'(x)Xe^^*' = f'(x)X e^^*^ Inparticular: —fe®*"'"*]=axe'""'"* dx^ -■

Chapter14: Differentiation2 WORKEDEXAMPLE1 Differentiatewithrespecttox. a e" b Answers a —(e5^)= 5 xe®^=5e®^ c xê e2x d X differentiateoriginai index funttion dx =(2x-3)X =(2x-3)e"' -3x differentiate oiigituil index runction —(xê =x^X—(e®*)+e X—(x^) dx^ ' dx^' dx^' =x^X(-3e~^'')+ x2x =—3xê"®*+2xe~®* ==X6-3"(2-3x) dx /„2xN X^(x)^ * dx dx XX2e^''-e^""X1 e^''(2x-1) quotientproductrnlerule WORKEDEXAMPLE2 Acurvehasequationy=(e^''+ dy Findthevalueof—whenx=0. dx Answers y = + e®*)^ —=5(e^''+e®')^X(26^"+3e®'') dx WhenX=0,—=5(e®+e°)^x(2e®+3e°) dx =5X2^X5 =400 chainsnle

CambridgeIGCSEand0LevelAdditionalMathematics CLASSDISCUSSION Bywriting2as findanexpressionfor—(2*). d dDiscuss with yourclassmateswhetheryoucan findsimilarexpressionsfor —(3*)and —(4*). dx djc Exercise14.1 1 Differentiatewithrespecttox. b e^*>7* a e d2e"^* ge" j 2(3-e^'^) e 6e2 h 5x—3e^ e*+e"* c f 2 Differentiatewithrespecttox. axe" b x^e^" e* e X d yjx g e*e^+1-1 h xe^" ^2* 3 Findtheequationofthetangentto 5 a y= .2x +3atX=0 3e^^ ^3*+l C f 2+^ 5^x^ + e*^j 3xe* e2* yfx x^e"-5 e*+1 by= +1atx=In5 c y = x^(l+e*) at x =1. 4Acurvehasequationy=5e^*-4x-3. The tangent to the curve atthe point(0,2)meets the x-axis at the point A. FindthecoordinatesofA. 5 Acurvehasequationy=xe*.aFind,intermsofe,thecoordinatesofthestationarypointonthiscurveanddetermineits nature. b Find,in terms ofe,the equation ofthe normal to the curve at the pointP(l,e). c The normal atPmeets the x-axis at A and the y-axis at B.Find,in terms ofe,the area of triangleOAB,whereOistheorigin.

Chapter14:Differentiation2 14.2 Derivatives of logarithmic functions

CLASSDISCUSSION

Graphingsoftwarehasbeenusedtodrawthe graph of =In X together with its gradient (derived)function. Thegradientfunctionpassesthroughthe points 1^1 sj, (l.l),(^2, and [^5, . Discusswithyourclassmateswhatconclusionsyoucanmakeaboutthegradientfunction. y-\nx J' 41 31 1t 1 2\ y=InX 1 gradient fonction -i O -1 /f 1 1 1 1' /I 2 3 4 5 6 7^ -2 / -4--3Fromtheclassdiscussionyoushouldhaveconcludedthat: —(Inx)= dx X LetThederivativeofInf(x)Considerthefunctiony=Inf(x).y=\nuwherem=f(x). dy _ 1 6.U u dw —=f(x) dx Usingthechainrule:—= x dx du dx =—Xf'(x) u f'(x)f(x) dx DnfW]=^

CambridgeIGCSEand0LevelAdditionaiMathematics Inparticular: —[in(ax+fe)]= dx ax+b WORKEDEXAMPLE3 Differentiatewithrespecttox. a InSx b ln(5x-7) Answers a —(InSx)= dx 8x •< c ln(2x- +5) -'inside'differentiated -'inside' d Xn-Jx—10 1 X — rIn(2x^ + 5)1 = dx 5x—7 4x 2x^-I-5 -'inside'differentiated 'inside' -'inside'differentiated -'inside' 1-(X-10)-2(1)^ d Method1:—fInVx-101= dx Vx-10 1 -'inside'differentiated -'inside' 2(x-10) Method2:usingtherulesoflogarithmsbeforedifferentiating. dx [^In Vx - 10]= dxdx ln(x-10)2—ln(x-10) useIna'"=m\na =•ix^[ln(x-10)] 2 dx 1 1 = - X ■ 2 X-10<1 2(x-10) •'inside'differentiated ■ 'inside'

Chapter14:Differentiation2 InxSx WORKEDEXAMPLE4 Differentiatewithrespecttox a5x®In2x Answers a—(5x®In2x)=5x®x—(In2x)+InSxx (5x®) dx^ ' dx dx^' 9 =5x®X—+In2xX30x® 2x productrule =5x®+30x®In2x —(Inx)-lnx: dx dx_ :l3xjdxl (3x)^ 1 3xX—InxX3 X 9x2 1—Inx 3x2 quotientrule WORKEDEXAMPLE5 % Acurvehasequation31=3xsinx+—. a Show that 3= ^In(2x -1)-In(x2 + l). dti bHence,findthevalueof—whenx=1. Answers a 3=In x2+1 =ln(2x-l)i-ln(x2+l) = iln(2x -1)-ln(x2 + l) use uselog„{xT=mlog^x dy d^c —riln(2xdxL2 ^ 1) i X;i[ln(2« -1)]- +1)] 2 dx' 1 =-X 2x 2 2x-1 x2+1 1 2x 2x-1 x2+1 x2+1-2x(2x-1) (2x-l)(x2+1) _ —3x2+2x+1 ~(2x-l)(x2+l) WhenX=1,—=0. dx

CambridgeIGCSEand0LevelAdditionalMathematics WORKEDEXAMPLE6 Acurvehasequationy=log,,[bx-2). a Show that y = ■j-^[ln(5x-2)]. dv b Hence, find the value of — when x = 2. ax Answers a 31=log2(5x-2)_ln(5x-2) use log;,a = logc« log,b In2 1 In2 [ln(5x-2)] b ^ = J_x-l-[,n(5x-2)]dx ln2 dx^ 1 5 X In2 5x-2 5 (5x-2)In2 dj_ 5 When X = 2, dx 8In2 Exercise14.2 1Differentiatewithrespecttox. B Inbx b lnl2x d 2 + ln(l-x^) e ln(3x + l) f4 g ln(2-5x)'^ h 2x + ln cln(2x+3) f In + 2 i 5 - In ■ (2 j In(In3c) kIn(yfx+l) 2Differentiatewithrespecttox. a xlnx b 2x^lnx dSxlnx^ 4 g e x^ln(lnx) . ln(2x + l) Inx -3x) ( ln(x^ + Inx) c (x - 1)In X ^ ln2x X ln(x^ -l) ' 2x + 3 3 A curve has equation = x^ In3x. Find the value of and —j at the point where x = 2. dx dx*^

Chapter14:Differentiation2 4 Use the lawsoflogarithms to help differentiate these expressions with respectto ^ C lnj^x(x- S)"*]a InvSx+T b In- ,(2x "2.x+1^ d In X-1 -5) g In 2x+3 hIn (x-5)(x -1-1) (x-1-3)(x—1) f In In X(x-I-1) X-f2 (x -fXl)(2x-3)(x-1) 5 Find — foreachofthefollowing. dx a y=loggX cb3)=log2x^;y=log4(5x-1) d'v 6 Find—foreachofthefollowing. dx ae^=4x^-1be^=5x^-2x C e''=(x-I-3)(x-4) 7 A curve has equation x = — -I-1J. d'v ^ Findthevalueof—whenx=1. dx 14.3Derivativesoftrigonometricfunctions N CLASSDISCUSSION ote: Use change of base oflogarithms before differentiating. Q Note: Takethenaturallogarithmofbothsides oftheequationbeforedifferentiating. Graphing software has been used to draw the graphs of j;= sinx and y =cosx together with theirgradient(derived)functions. y=sinx y=COSX 1gradient function 1 \ functiongradient/ y=cosx 0 / \ \ / 0 \ / / \ 1, TV, 2\ A /37t \ ■' ^ /ZTC /y=sin:r \ A 2\ /tt pn /^ in ^ -1- -1- ''vy Discuss withyour classmateswhatconclusionsyou can make from these two graphs.

CambridgeIGCSEand0LevelAdditionalMathematics From the class discussion youshould haveconcluded thatif is measured m —(sinx)=cosx dx dx (cosx)=—sin X The derivative of tan x can be found using these two results together with the quotientrule. dx (tanx)= dxVcosxy d d cosxX—(sinx)-sinxx—(cosx) dx dx (cosx)^ cosxXcosx-sinxx(-sinx) cos^X cos^X+sin^xcos^X =cos^xsec^X usethequotientrule usecos^X+sin^x=1 1 use = secx cosx —(tanx)=sec^x dx

Chapter14: Differentiation2 1 WORKED EXAMPLE 7 Differentiatewithrespecttox. 0 3tanx a7cosX b sinX c d X (5-3cosx)® Answers a -^(7cosx)= dx 7—(cosx) dx -7sinX b-^(x^sinx)= dx X -^(sinx)+ sinx x —(x^) dx' dx^^ productrule x^cosX+2xsinx ^ d f3tanx^ X X -^(3tanx)- 3tanx x —fx) dx dx'^ quotientruledxl X ) x2 XX3sec^X-3tanxX1 x^ 3xsec^X-3tanx dr,d ^[(5-3cos x2 x)®]= 8(5 -Bcosxf X 3sinx = 24sinX(5-3cosx)^ chainrule Derivatives of s'm{ax+b),cos{ax + b) and tan(oA^+/}) Consider the function y = sin{ax + b) where x is measured in radians. Let )>=sinM where u=ax+b. d)) du = COSM du dx: Using the chain rule:^ ^ ^ dx du dx =cosuxa =acos{ax+b) dx [sin(ax+i)]= a cos(ax+b)

Similarly,itcanbeshownthat: d[ dx dx cos(ax+6)]= —a sin(ax+ b) [tan(ax+b)\= a sec^(ax+b)

CambridgeIGCSEand0LevelAdditionalMathematics Q Note: Itisimportanttorememberthat,incalculus,allanglesaremeasuredin radiansunlessaquestiontellsyouotherwise. WORKEDEXAMPLE8 Differentiatewithrespecttox. a 5sin|^^-2x^ b xcosSx X sin(2x + d (1+3tan2*)^ Answers dx 5sin = 5 X -2xjX(-2) =-10cos b—(xcosSx)=XX—(cosSx)+cosSxX-^(x) dx dx dx productrule =XX(-3sin3x)+cosSxx(1) =cos3x-SxsinSx ^dx sinf2x + jc'i I 4J. [sm[2x + 5]] ^2x+ ^(2*)- X 1^2 cos^2x +^ quotient rule sin^^2x + 2xsin^2x+ ~ 2x^cos|^2x + sin^^2x + d —r(l+3tan2xfl= 5(l+3tan2x)^ x6sec22x dx•- ■' chainrule = 30sec^ 2x(1 + 3tan2x)^ Exercise14.3 1 Differentiatewithrespecttox.a2+sinxb2sinx+3cosx d 3sin2x e 4tan5x g tan(3x + 2) h sin(^2x+|j c 2cosx-tanx f 2 cos 3x - sin 2x i 2cos(

Chapter14:Differentiation2 2 Differentiatewithrespecttox. sin''X d (3-cosx)'^ b 5cos^(3x) e2sin3(2x+ 3 Differentiatewithrespecttox. a sin X d Xtan' tanx 1 sin-^2x b2sin2xcos3x 5 e h k cos3xsin^c 2+cosx 3x sin2x Differentiate with respect to x. -| ^COSX ^g(sinx+cosx) g e*(cosX+sinx) j x^ln(cosx) gCosS* e e*sinx h x^e™" , sin3x sin^X-2cosx 3cos'^x+2tan^I2x K X tanx X cosx sinx 3x—1sinX+cosX sinX—cosX C e'' 1 e'cos—X 2 In(sinx) xsinx 5Findthegradientofthetangentto K a y—2xcos3xwhenx= 3 1 d6aBywritingsecxas ,find—(secx). cosx dx 1 db By writing cosecx as ^ ,find—(cosecx). sinx dx COSX dCBywritingcotxas^ ,find—(cotx). sinx dx d)!7 Find—foreachofthefollowing. a =sin3x b e>=3cos2x . 2-cosX , 7t0y= whenX3tanx 4 Note: Takethenaturallogarithmofbothsides ofthe equation before differentiating. CHALLENGEQ 8Acurvehasequationy=Asinx+Bsin2x. The curve passes through the point p[-,3]and has a gradient of 3^/2 ^ 71 ; whenX=— 2 4 FindthevalueofAandthevalueofB.

CambridgeIGCSEand0LevelAdditionalMathematics CHALLENGEQ 9 Acurvehasequationy=Asinx+Bcos2x. Thecurve hasagradientof 5-y/3 when *~ ^gradientof 6+2\f2when ^. 4 FindthevalueofAandthevalueofB. 14.4Furtherapplicationsofdifferentiation You need to be able to answer questions that involve the differentiation of exponential,logarithmic and trigonometric functions. WORKEDEXAMPLE9 Acurvehasequation =3xsinx+—. The curve passes through the point '''j• aFindthevalueofa. b FindtheequationofthenormaltothecurveatP. Answers Tt a WhenX=—,y=3x n fn) n — X sin — + 2 I2J 6 _3ji n 571 J'-y+g- 3 Hence,a= 5k o b y=33csinx+ 6 —=3xcosX+3sinX dx Normal: passes through the point , j and gradient 5n l( 5k 1 K y —=——X+ 3 3 6 1 llJi 3 6

Acurvehasequation3;= Inx. Findtheapproximateincreasein31asxincreasesfrometoe+p,wherepissmall. Answers y=x^Inx ■^ = X - + Inx X 2x dx X =x+2xlnx product rule WhenX=e. =e+2elne =3e Using dx Sy 3ep InX Variablesxandyareconnectedbytheequationy=^—. Giventhat31increasesatarateof0.1unitspersecond,findtherateofchangeofx whenX=2. Answers and =0.1 quotient rule (2x + 5)^ (2x+5)-2xlnx dx x(2x+5f d3i_(4+5)-4In2 dx ~ 2(4 + 5f 9-41n2 WhenX 162 Usingthechainrule, = — X 162 xO.l 9-41n2 Rateofchangeofxis2.60unitspersecondcorrectto3sf. 351 WORKEDEXAMPLE10 WORKEDEXAMPLE11 Chapter i4: Differentiation 2

Acurvehasequationy-e*(2sin23i:-3cos2x)for0<x<—radians. Findthe^coordinateofthestationarypointonthecurveanddeterminethenatureof thispoint. Answers y=e"*(2sin2*:-3cos2*) dx =e"*(4cos2*+6sin2x)-e*(2sin2*-3cos2*) =e"*(7cos2x+4sin2x) dy Stationarypointsoccurwhen—=0. dx e"*(7cos2x+4sin2x)=0 7cos2x+4sin2x=0 2x=2.0899 tan2x=—buttheyareoutsidethe d^y —i=e"*(—14sin2x+8cos2x)—e~*(7cos2x+4sin2x) dx"^ =e~*(cos2x-18sin2x) WhenX=1.045, ^i<0. dx^ Hencethestationarypointisamaximumpoint.

or 7 tan2x= 4

4 % range0<^ ~•

X=1.045 noe"'=0solution Thereareothervaluesof xforwhich 7

productrule

CambridgeIGCSEand0LevelAdditionalMathematics WORKEDEXAMPLE12

ThediagramshowsanisoscelestrapeziumPQ/iSwithareaAcm^. AngleSPQ=anglePQR—6radians. PS=QR=10cmandSR=16cm. a ShowthatA=160sin0+100sin0cos0. cbFindthevalueof0forwhichAhasastationaryvalue,Determinethenatureofthisstationaryvalue. Answers a A = ^(a+ b)h i(PQ+SP)xlOsin0 use h=10sin0

Chapter14:Differentiation2 WORKEDEXAMPLE1310cm 16cm —* ]()cm P Q

fl24 When0=1.146,^<0. dx^ Hencethestationaryvalueisamaximumvalue.

usePQ_=10cos0+16+10cos0=16+20cos0 = —[(16+20cos0)+16]X 10sin0 =5sin0(32+20cos0) =160sin0+100sin0cos0 dA b——=16Ocos0+[lOOsin0(—sin0)+lOOcos^0]usetheproductruleon100sin0 djc =160cos0-100sin^0+lOOcos^0 =160cos0-100(l-cos^0)+100cos^0 =200cos20+160cos0-100

dA Stationaryvaluesoccurwhen—=0. dx: 200cos20+160cos0-100=0 10cos^0+8cos0—5=0 cos0=0.412 orcos0=-1.212 0=1.146radians nosolution d^A usesin^0=1usethequadraticformula =-400cos0sin0-160sin0

5 Acurvehasequationy=xe".

ThetangenttothecurveatthepointP(l,e)meetsthe^(-axisatthepointA.

Cambridge IGCSE and0LevelAdditional Mathematics Exercise14.4

8 Variablesxandyareconnectedbytheequationy=— . X"t"3

The normal to the curve atPmeets the x-axis atthe pointB.

Findtheapproximatechangeinyasxincreasesfrom1to1+p,where pissmall.

7Variablesxandyareconnectedbytheequationy=3+ln(2x-5)

9 Variablesxandyareconnectedbytheequationy=3+2x-5e"*.

bThenormaltothecurveatBmeetsthex-axisatthepointC.

FindthecoordinatesofC.

ThenormaltothecurveatPmeetsthex-axisatQ. FindthecoordinatesofQ.

FindtheapproximatechangeinyasxincreasesfromIn2toln2+/>, wherepissmall. ln(x^-2)

Find theequation ofthenormaltothecurveatthe pointonthecurvewhere x = —.

FindtheareaoftriangleOAB,whereOistheorigin.

10Acurvehasequationy= ^—-—. X Find the approximate changein yas xincreasesfrom yfS to yfS +p,where pissmall.

Find the approximate change in y as xincreasesfrom 4to 4+jb, where pissmall. InX

Find the approximate increase in y as xincreasesfrom — to — + p,where pissmall. ^ ^

6 Variablesxandyareconnectedbytheequationy=sin2x.

2Acurvehasequation3)=xsin2xfor0^^7tradians.

The curve crosses the x-axis atA and the ^i-axis atB. aFindthecoordinatesofAandB.

3 Acurvehasequationy= +1.

I 7C Ia Find the equation ofthe normal to the curve at the point P —,— .

4Acurvehasequationy=5-e^*.

b The normal atPintersects the a:-axis at Qand the 31-axis at R. Find the coordinates of Qand R.

Thecurvecrossesthe^^-axisatP.

cFindtheareaoftriangleOQRwhereOistheorigin. 1

1 A curve has equation y = 3sin|^2jc+^j.

Thecurvecrossesthex-axisatthepointAandhasaminimumpointatB. FindthecoordinatesofAandthecoordinatesofB.

ThecurvehasaminimumpointatPandamaximumpointatQ. aFindthecoordinatesofPandthecoordinatesofQ.

13Acurvehasequationy=Ae^'+£e~^'. The gradientofthe tangentatthe point(0,10)is-12.

11Findthecoordinatesofthestationarypointsonthesecurvesanddeterminetheirnature. X ay=xe^ by= c)i=e*-7x+2 d 3;= 5e^*-lOx-1 e 31 = -8)e"* f 3= x^lnx: Inx g y= x^ ln{x^ +1)

12Findthecoordinatesofthestationarypointsonthesecurvesand determinetheirnature. n a 31 =4sinX+3cosx for 0^ a:^ X X b 3»=6cos-+8sin- for 0=Sx 27t c 31=5sin d y= (-§) n 5k for ^X< 6 6 for0<X<K sinx e 3!=2sinxcosx+2cosx for 0^ ^7t

bFindthecoordinatesoftheturningpointonthecurveanddetermine itsnature. 14 A curve has equation 3)= xlnx.

15Acurvehasequationy=x^e".

b The tangent to the curve at the point A(1,e) meets the 3)-axis at the pointB. ThenormaltothecurveatthepointA(1,e)meetsthe3-axisatthe pointC. FindthecoordinatesofBandthecoordinatesofC. c Find the area oftriangle ABC.

aFindthevalueofAandthevalueofB.

Chapter14:Differentiation2

12cm Summary Exponentialfunctions A(e-)= e' djc [e°*"^*]= aeax+b d* Logarithmicfunctions —(Inx)= djc X Trigonometricfunctions ^[ln(ajc + ft)]= ° djc ax+b dx (sinx)=cos* d* [sin {ax + ft)]= <2 cos(ax 4- ft) —(cos*)=-sin* d* ^ —[cos(a*-t- ft)]=-a sin(a*+ft) d*^ -■ d* (tan*) = sec^ * d* [tan(fl* -f- ft)] = a sec^(ax + ft)

CHALLENGEQ inscribedinacircle,centreO,radiusrcm. ORP=0radians. TrianglePQRhasanareaofAcm^. sin20+r-sin20cos20. stationaryvalueanddeterminethenatureofthisstationaryvalue.

16Thediagramshowsasemi-circlewithdiameterEFoflength12cm. AngleGEF=6radiansandtheshadedregionhasanarea ofAcm^. a Showthat A=360+18sin20. bGiventhat0isincreasingatarateof0.05radiansper 71second,findtherateofchangeofAwhen^ radians.

17ThediagramshowsanisoscelestrianglePQR

aShowthatA=

Cambridge IGCSE and0LevelAdditional Mathematics

PR=QRandangle

bFindthevalueof0forwhichAhasa

Chapter14: Differentiation2 Examinationquestions Workedexample ThefigureshowsasectorOABofacircle,centreO,radius10cm.AngleAOB=20radianswhere 71 0<0<—.AcirclecentreC,radiusrcm,touchesthearcABatthepointD.ThelinesOAandOB aaretangentstothecircleatthepointsEandTrespectively.Writedown,intermsofr,thelengthofOC. [1] b Henceshowthatr= 10sin0 [2] 1+sin0 C Giventhat0canvary,find—whenr=—. ^ d0 3 TTd Given thatrisincreasing at2cms"',find the rate atwhich 0isincreasing when 0= —. 6 [6][3] CambridgeIGCSEAdditionalMathematics0606Paper11Q10(part)Nov2011 Answers a OC=OD-CD OC=10-r b Using triangle OCE: T sin0=OCsin0= 10-r r=10sin0-rsin0 r+rsin0=10sin0 r(l+sin0)= lOsin0 10sin0 useOC=10-r multiply both sides by(10-r) collecttermsinvolvingr factorise divide both sides by(1 +sin0) r 1+sin0

CambridgeIGCSEand0LevelAdditionalMathematics 10sin0 r 1+sin0 dr_(1+sin0)lOcos0-10sin0cos0 d0 (1+sin0)^ dr 10cos0 (1)d0 (1 + usethequotientrule sin0)^ 10 10 10sin0 Whenr=—, —= 3 3 1+sin0 30sin0=10+10sin0 20sin0=10 sin0= 1 y/S Ifsin0=—,thencos0=—. 2 2 Substitutinginequation(1)gives: dr d0 10X 2 20^/3 ^ ^ 1 ^ ^dr 20>/3d When0=—,sin0=-,cos0=— and —= 6 2 2 d0 9 TT• U U • , d0 d0 drUsingthechainrule: —=—x dt dr df 9 X2 20^/3 9 ~10^/3 _Sy/S 10 Exercise14.5ExamExercise 1aFindtheequationofthetangenttothecurve —In*atthepointonthecurve wherex=1. bShowthatthetangentbisectsthelinejoiningthepoints(-2,16)and(12,2). [4] [2] dv 2Find—when dx a y b )!= cos2xsin tanx 1+Inx 3/ CambridgeIGCSEAdditionalMathematics0606Paper11Q5i,iiNov2014 [4][4] CambridgeIGCSEAdditionalMathematics0606Paper21Q10i,lljun2014

.aFind,intermsofIn2,thecoordinatesofA,B,CandD.,AreaoftriangleBPD1 b Giventhat ;———=--,expresskintermsofIn2. AreaoftnangleAPC k

CambridgeIGCSEAdditionalMathematics0606Paper11Q5i,iiNov2012

CambridgeIGCSEAdditionalMathematics0606Paper11Q1Oi,ii,HiJun2012 P=(l,ln2) 3i=m(x+l)-lnx [8][3]

3Variablesxandyarerelatedbytheequationy=10-4sin^x,where0^x^.

5Variablesxandyaresuchthaty=e^*+e

A curve has equation y = 2xsinx -I- —.The curve passes through the point p{—,a . 3 V2 / a Find,intermsof%,thevalueofa. [1]

changeGiventhatxisincreasingatarateof0.2radianspersecond,findthecorrespondingrateofofywheny=8.[6]

a Finddx dy bByusingthesubstitutionu=e^",findthevalueofywhen—=3. dx [2][4]

bUsingyourvalueofa,findtheequationofthenormaltothecurveatP. [5]

Chapter14; Differentiation2 K

c Given that x is decreasing at the rate of0.5 units s \find the corresponding rate of changeofywhenx=1. [3]

CambridgeIGCSEAdditionalMathematics0606Paper21Q12(part)Ncrv2011

4 Giventhaty= dy cos4xfind

Thediagramshowspartofthecurvey=ln(x+1)-Inx.Thetangenttothecurveat thepointP(l,In2)meetsthex-axisatAandthey-axisatB.Thenormaltothecurve atPmeetsthex-axisatCandthey-axisatD.

CambridgeIGCSEAdditionalMathematics0606Paper21Q3Jun2013 [3]

CambridgeIGCSEAdditionalMathematics0606Paper11Q5i,iiNov2010

K 71 b the approximate change in y when xincreasesfrom — to — + p,where pis small. [2]

CambridgeIGCSEAdditionalMathematics0606Paper12Q6Mar2015 In "hsl

CambridgeIGCSEAdditionalMathematics0606Paper11Q5Nov2015 [5]i Findthecoordinatesofthestationarypointsonthecurve. ii Showthat ^ wherepandqareintegerstobefound,anddeterminedx^ (x^+1)3 the nature ofthe stationary points ofthe curve. [5]

12 The diagram shows a circle, centre O,radius rcm.The points A and Blie on the circle such that angleAOB=26radians. iFind,intermsofrand6,anexpressionforthelengthofthechordAB. 10 [1] 6+sin0iiGiventhattheperimeteroftheshadedregionis20cm,showthatr= dr It lilGiventhatrand6canvary,findthevalueof—when6 da 6 ivGiventhatrisincreasingattherateofIScms-^findthecorrespondingrateof rt changeof0when6=—. o [4][2][3]

CambridgeIGCSEand0LevelAdditionalMathematics tan2x dv 8I Giventhaty= ,find . X dx ii Hencefindtheequationofthenormaltothecurvey wherex=—. 8 tan2x atthepoint [3][3]

CambridgeIGCSEAdditionalMathematics0606Paper11Q7Jun2015

10Variablesxandyaresuchthat=(x-3)In(2x^+l). dy i Findthevalueof—whenx=2. dx iiHencefindtheapproximatechangeinywhenxchangesfrom2to2.03. [4][2]

ThepointA,wherex=0,liesonthecurvey=— .ThenormaltothecurveatA X Xmeetsthex-axisatthepointB. i Find theequation ofthisnormal. [7] ii FindtheareaofthetriangleAOB,whereOistheorigin. [2]

CambridgeIGCSEAdditionalMathematics0606Paper22Q11Mar2016

CambridgeIGCSEAdditionalMathematics0606Paper11QllJun2016

11Acurvehasequationy= x^+1

^iiiiiigr*' Chapter 15 Integration Thissectionwillshowyouhowto: ■ useintegrationasthereverseprocessofdifferentiation 1 integratesumsoftermsinpowersofx,including-and X ax+b integratefunctionsoftheform(ax+b)", e'"■'^sin(ox+fj),cos{ax+b) evaluatedefiniteintegralsandapplyintegrationtotheevaluationofplaneareas.

CambridgeIGCSEand0LevelAdditionalMathematics InChapter12,youleamtabouttheprocessofobtaining—whenyouknowy. Thisprocesswascalleddifferentiation. dx Youlearnttherulefordifferentiatingpowerfunctions: Ify=x",then =nx""'. dx Applyingthisruletofunctionsoftheformy=x^+cyouobtain: y = x^ + 5.3 ^ y = x2+2 I dy =2x y=x2 - v2 y=x^—3 dx Inthischapteryouwillleamaboutthereverseprocessofobtainingywhenyouknowreverseprocessiscalledintegration. dy dx anNote:Thereareinfinitenumberoffunctionsthatwhendifferentiated give2x. 15.1Differentiationreversed CLASSDISCUSSION dyFind foreachofthefollowingfunctions, dx , y=—x^+3 4 y=i3C^-0.8 1 3 ^ y=—X+4 3 1 2 1y=—X +1 ^ -2 1 x2-7 dy Discussyourresultswithyourclassmatesandtrytofindaruleforobtainingyif—=x^ Describeyourruleinwords. ijS# Fromtheclassdiscussionyoushouldhaveconcludedthat: If-^=x"theny=——x""^^+c,wherecisanarbitraryconstantandn^-1. dx n+1 Itiseasiertorememberthisruleas: 'increasethepowernby1togetthenewpower,thendividebythenewpower'.

Chapter15:Integration WORKEDEXAMPLE1 Findyintermsofxforeachofthefollowing. Answersdjc a d*y= 41+1 =-x®+c b^=^ dx . d^i b —dx 1 i+i X2 +C -x^+c23 =-y^x'+C d^)_1 dx X® dx 1 y=-3+1 J_ -2 1 -x-®+^+c 2x2 + c WORKEDEXAMPLE2 Findyintermsofxforeachofthefollowing.dy = a 6x2 +4x dx x2 Answers —=6x2_5ju-2+ dx _63 ^ S'' '(-l)- X =2x®+5x~'+2x2^^=2x®+—+2x2+^ X ■^ = 8x® - -X-® - 2x» dx 3 b ■^ = 8x®--^-2 dx 3x d;y_(x-2)(x+5) 84 y=-x* 3(-4) x"^ --x' +c = 2x^ +-x^ -2x + c 3 = 2x^ +—^—2x+c 3x^ dy x2+3x-10 dx yfx 3 1 j[ =X2 +3x2 _10x~2 1 — % — 10 9 5 3 1 =-x2+2x2-20x2+C 5 dx yfx write in index form readyfor integration write in index form ready for integration writeinindexformreadyforintegration

CambridgeIGCSEand0LevelAdditionalMathematics WORKEDEXAMPLE3 dvAcurveissuchthat —=(1-a:)(3A:-2)and(2,8)isapointonthecurve, dx Findtheequationofthecurve. Answers Ay (l dx -x)(3x-2) =-3x2+5x-2 =-3x2+5^1_2^0 5 ■y=-X®+—x2-2x+c 2 WhenX=2, =8 8=-(2f+|(2)'-2(2)+c 8=-8+10-4+cc=10 5 Theequationofthecurveis;y=-x*+—x2-2x+10. expand brackets writeinindexformreadyforintegration Exercise15.1 1 Find y in terms of xfor each of the following. a-^=12x4 dx d^=dx X® b^= dx e ^ = -L dx 2x^ ^ = 7x3d}i dx d)) dx 3^ yfx 2Findyintermsofxforeachofthefollowing. a = 7x® +2x'^ +3 dx dy 3 15 c — = —7 7+x dx X x2 3Findyintermsofxforeachofthefollowing. a -^ = 3x(x-2)dx b — = 2x^ -3x^ +5x dx dx X x' d)i dx = x2 (4x2 _ 3^ g ^ = (x+2V^fdx ' dy _ — 3x dx 2x^ dy_x^—4x2^2 dx 2x2 d); dx d ^= f h d); dx d); dx x(x-3)(x+4) (2x - 3)(x - 1) x^ (3x + 5)(x - 2) -v/x

Acurveissuchthat ^—. dx Given that the curve passes through the point(-1,10),find the equation of thecurve. , .2 dy (2- ) 7

dv 4Acurveissuchthat—=3*^-4x+1. dx Giventhatthecurvepassesthroughthepoint(0,5)findtheequationof thecurve. cJ-v 5 Acurveissuchthat—=fix(x-1). dx Giventhatthecurvepassesthroughthepoint(1,-5)findtheequationof thecurve. . . u u 2x^+fi 6

Indefinite integrals

The special symbol J is used to denote integration.

Acurveissuchthat—= p • dx vx Giventhatthecurvepassesthroughthepoint(9,14),findtheequationof thecurve. dv 8 Acurveissuchthat—=kx^-2xwherekisaconstant. dx Giventhatthecurvepassesthroughthepoints(1,fi)and(-2,-15)findthe equationofthecurve. d^V9

When you need to integrate x^,for example,you write I' x 1 ^dx=—X +c. Jx^ dx is called the indefinite integral of x^ with respect to x. Itiscalled'indefinite'becauseithasinfinitelymanysolutions.Usingthisnotationtheruleforintegratingpowersofxcanbewrittenas: Jx"dx= n+1 x""^^+c,wherecisaconstantandn^-1

15.2

Acurveissuchthat—^=12x-12. dx2 Thegradientofthecurveatthepoint(2,9)is8. a Expressyintermsofx. b Showthatthegradientofthecurveisneverlessthan2. dy 10Acurveissuchthat—^=kx-5whereAisaconstant. dx j Thegradientofthenormaltothecurveatthepoint(2,-1)is—. 3 Findtheequationofthecurve.

Chapter15:Integration

CambridgeIGCSEand0LevelAdditionalMathematics Thissectionprovidesyouwithpracticeatusingthisnewnotationtogether withthefollowingrules; JA:f(x) d* = Ajf(x) dx,where ft is a constant J[f(»:)±gW]d*= Jf(a;)dx ± Jg(x)dx WORKEDEXAMPLE4 Find: a [x^(lOx^ -8x + 3)dx b f dx J J xvx Answers a Jx^(lOx^ -8x +3)dx = J(lOx^ -8x® +3x^)dx 10x5 8x^ 3x5 + +c 5 4 3 =2x5 x^-2 dx /[ 3 5 3\ = *2-2X 2 dx X2--——X2+c © H) 2^ -i =—x2+4x 2+C 72o(— 4 =—x'vx+—^+c 7 Vx Exercise15.2 1 Findeachofthefollowing. a J4x^ dx b J12x^ dx 4 I d dx x 2 Findeachofthefollowing. a J(x + 2)(x + 5)dx b J(x — l)(2x +3)dx d I (Jx +sfdx e |x(x-l)^dxJ + 3)' Jx(x -1)^ J 2x ^ dx 6 J d ■y/x x J(x-5)^dx J^(x-4)dx

Chapter15:Integration 3 Findeachofthefollowing. I J dx -SVic dx e J J x^-8 2x3 x^+5 X^Vx dxdx (x +1)^ Sx^J l(^ dx\2 + dx 15.3Integrationoffunctionsoftheform(ax+b)" InChapter12youlearntthat: dr 1 (2 (2 dx12X8 x+5)'' x+5)' Hence f(2x + 5)'dx = ——(2x + 5)^ + c. J 2x8 Thisleadstothegeneralrule: f(ax + b)"dx = —"(ax+ 6)"^^ + c, n ^ -1 and a ^ 0 J a(n+1) FindWORKEDEXAMPLE5 Answers a J(3x-8)'3dx= -(3x-8)^'+c3(5+1)^ ' =—{3x-8f+c 18 8 (4x+ly dx = 8j(4x +ir3dx (4x++c 8 4(-3+l)' -(4x+l)~^+c 1 + c (4x+1)' 19 1 1 ■(2x-7)"r'+c <4-) =12V2x-7+c

CambridgeIGCSEand0LevelAdditionalMathematics Exercise15.3 1 Find: g J(x + 2)® d j3(2-3x) I a/x+1 dx J(2x -5) [(7x + 2) 1(— J Ux+3 (4x +1)^. 3dxdx dx c J2(3x + 2)^dx f J^(3x-if dx i 2(3-2xr d)!

3Acurveissuchthat—=V2x^n. dx Giventhatthecurvepassesthroughthepoint(4,11)findtheequationof thecurve. dx 1 4 Acurveissuchthat—-= .^ . dx vlO-X Giventhatthecurvepassesthroughthepoint(6,1),findtheequationof thecurve. dx « 5 Acurveissuchthat —=k{2x-3) where ft isaconstant.dx J The gradientofthe normal to the curve atthe point(2,2)is —. 8 Findtheequationofthecurve.

InChapter15,youlearntthefollowingrulesfordifferentiatingexponential functions: —(e*)=e* —(e®*"*"*)=ae'^"''* dx^' dx^ ' Sinceintegrationisthereverseprocessofdifferentiation,therulesfor integratingexponentialfunctionsare: Je*dx=e*+c J' 1 dx=-e'"+*+c I

2Acurveissuchthat—^= dx Giventhatthecurvepassesthroughthepoint(0,-1.95)findtheequation ofthecurve.

6 A curve is such that^= 2(ftx-l)^ where ft is a constant. Given that the curve passes through the points(0,1) and (1,8)find the equationofthecurve. 15.4 Integration of exponentialfunctions

Chapter15:Integration WORKEDEXAMPLE6 Find dx J'dx bIe dx cIe®"®dx. J' Answers /■a I = — e^* + c 2 1b e *dx -e~^*+c -7 1 i. 7 c I e®* ® dx = —e®* ® + c 6 Exercise 15.4 1 Find: aIe®*dx /■ dx d e dxJ !■g I dx 2 Find a Je*(5J' f4e^ fe=-= dxdx dx b J +l)' dx I +4 dx 1 5e®* 2e* dx f 3 Find: a J|^2e*+ dx b J(x^- dx c J Je2 dx J2e^*dx J dx J(3e*+e"*)^dx f (e''^ -2e^f J e®^' 3x^e^* — 4e dx 12x^e* ci'v 4 A curve is such that — = 2e-'' + e"*. dx Given that the curve passes through the point (0, 4), find the equation of thecurve. dy 9_ — - ke'^ " +4x, where Aisaconstant.5 Acurveissuchthat dx At the point (2, 10) the gradient of the curve is 1. a Findthevalueofk. b Find the equation of the curve. d^v6 Acurve issuch that —^ = 8e d -2x x^ dv Giventhat—=2whenx=0andthatthecurvepassesthroughthepoint f, 2 \ I 1> , find the equation of the curve.

CambridgeIGCSEand0LevelAdditionalMathematics 7 The point P(—,51 lies on the curve for which — = 2e^ ^ U y d;c ThepointQ(l,k)alsoliesonthecurve, aFindthevalueofk. ThenormalstothecurveatthepointsPand<2intersectatthepointR. bFindthecoordinatesofR. 15.5 Integration of sine and cosine functions InChapter14,youlearnthowtodifferentiatesineandcosinefunctions: d d —(sinx)=cosX —I dx dx d d —(cos5c)=-SinX dx dx integratingSinceintegrationisthereverseprocessofdifferentiation,therulesforsineandcosinefunctionsare: -[sin(ax + 6)]= acos(ax + b) [cos(ax + 6)]= -asin(ax + b) I J Q [[cos(ax + 6)] dx = — sin(ax + i)+ r J a f[sin(ax+&)]dx=-—cos(ax++c J a cosXdx=sinX+csinXdx=-cosx+c Note: Itisimportanttorememberthattheformulaefordifferentiatingand integratingthesetrigonometricfunctionsonlyapplywhenxismeasured inradians. WORKEDEXAMPLE7Find: J sin 2x dx b J cos5x dx c fssinJ 2 dx Answers a jsin 1 cosi2xdx=—cos2x+c2s5xdx=—sin5x+c 5 c X r X I3sin—dx=3Isin—dx J 2 J 2 = 3X[^-2cos 2 I X =-6cos—+c2

Find: +2cos(5x-1)]dxcos2x+5sin5x)dx Answers cos2x +5sin3*:) dx = 3J cos2x dx +5J sin3x dx = 3 X f— sin 2x1+ 5 x( — — cos =—sin2x cos3x+c x^ +2cos(5x -1)] dx = J x^ dx +2J cos(5x — 1)dx 1s « =-x^+2X +c =—X®+—sin(5x-1)+c WORKEDEXAMPLE8 Chapter15:Integration Exercise15.5 1 Findsin4xdx a d g 2 Fine a 2cos2xdx 1 jesi cos2xdx sinSxdx 5sin(2-3x)dx h j2cos(2x-7) dx i (1-sinx)dx 5x 3cos2x-TUsin— dx 2) e J(e^* -5sin2x) dx 3 Acurveissuchthat—=cosx-sinxdx dx b fd X sin—dx 3cos(2x+1)dx4sin(l-5x)dx (>/x - 2cos3x] dx 1 3x\ , —-cos— dx sin—Idx2. GiventhatthecurvepassesthroughthepointI—,31,findtheequationof thecurve d}i 4 Acurveissuchthat—=1-4cos2x dx dx Given that the curve passes through the point I —,1j,find the equation of thecurve dy 5 Acurveissuchthat—=4x-6sin2x. dx Giventhatthecurvepassesthroughthepoint(0,-2),findtheequationof thecurve.

Sinceintegrationisthereverseprocessofdifferentiation,therulesforintegrationare: J —dx=Inx+c,X>0 X J —^—dx=—ln(ax+b)+c,ax+b>Q ax+b a

ThetangentstothecurveatthepointsPandQintersectatthepointR. bFindthecoordinatesofR. 15.6 Integration offunctions ofthe form - and —^ X ax+b InChapter14,youlearntthefollowingrulesfordifferentiatinglogarithmicfunctions: -(Inx) X>0 —[in{ax+^)]= dx ax+b ,ax+b>0 dx X ItisimportanttorememberthatInxisonlydefinedforx>0.

8 The point[ —,5 lies on the curve for which — = 4sinf2xl2 j dx { 2) a Findtheequationofthecurve. b Find the equation ofthe normal to the curve at the pointwhere x = —. n9 The point P|—,3 1 lies on the curve for which — = 3cos|3x dx 71 Thepoint thecurve, aFindthevalueofk.

CambridgeIGCSEand0LevelAdditionalMathematics

6 A curve is such that —^= 45cos3x +2sin x. , dx'^ dy pointGiventhat—=-2whenx=0andthatthecurvepassesthroughthe(ti,-1),findtheequationofthecurve, dy7 Acurveissuchthat—=kcos3x—4,whereAisaconstant. dx Atthepoint(tc,2)thegradientofthecurveis-10. aFindthevalueofk. b Findtheequationofthecurve.

Chapter15:Integration WORKEDEXAMPLE9 Findeachoftheseintegralsandstatethevaluesofxforwhichtheintegralisvalid. 1 ' 1 6 2x+5 -dx I 8 3-4x -dx Answers a f^dx=5fi JX JX dx =51nx+c,x>0 -dx=6f—^—dx 2 b f--A_dx =6f J2x+5 J x+5 6|^|ln(2x+5)+c =3In(2x+5)+c,X> f —^dx=sf —^dx J3—4x J3—4x = 8(^^jln(3-4x)+f =-2In(3-4x)+c,*< validfor2x+5>0validfor3-4x>0 CLASSDISCUSSION Nicola is asked to find Jx 'dx. Shetriestousetheformula dx= Nicolaisalsoaskedtofind I" J(5x +2) n+1dx. -x""^'4-ctoobtainheranswer. Shetriestousetheformulaf{ax+b)"dx= —-(ox+ +ctoobtainheranswer. J o(n+l) a{n+1) DiscusswithyourclassmateswhyNicola'smethodsdonotwork.

CambridgeIGCSEand0LevelAdditionalMathematics CLASSDISCUSSION RajuandSaraareaskedtofind J Raju writes: f —7- -r dx = f -— ^ J2(3x-l) J6x2(3x-l) -dx dx 1 =—ln(6x-2)+c Sara^ r 1 . 1 r 1 writes: I— rox=—|J 2(3x-l) 2J 3x-l dx = ^In(3x -1)+ c Decide:whoiscorrectanddiscussthereasonsforyourdecisionwithyourclassmates. Exercise15.6 1 Fine a X dg j J 3xdx 7 2x-l dx J j +2 4 2—3x dxdx 1 I 8xdx5 dx 2(5x-l) dx 2 Find: a K7x+1+— I dx X, d f^dx J X (■ 5x + 2y/x J ^^2g dx 1(1+?] d« J3-5x3 I 2x^ dx X^ — Xyfxy/x dx dji 5 3 Acurveissuchthat—= 7 dx 2x-1 forx>0.5. dx 21(3'-1) lOxe* Given that the curve passes through the point (1, 3), find the equation of the curve. dv 5 4 Acurveissuchthat—=2xH—forx>0. dx X Given that the curve passes through the point (e, e^), find the equation ofthe curve. 5 Acurveissuchthat—=—— forx>-e. dx X-I-e Given that the curve passes through the point (e, 2 +ln2), find the equation of the curve.

Chapter15:Integration CHALLENGEQ dy 2 6 ThepointP(l,-2)liesonthecurveforwhich—^=3 . dx X Thepoint(2(2,k)alsoliesonthecurve, a Findthevalueofk. ThetangentstothecurveatthepointsPand(2intersectatthepointR b FindthecoordinatesofR 15.7 Further indefinite integration Thissectionusestheconceptthatintegrationisthereverseprocessof differentiationtohelpintegratecomplicatedexpressions. If — rF(x)l = f(x), then ff(x)dx = F(x) dx J +c ShowWORKEDEXAMPLE10that +1^ 2(3x^—3x—l) dxW4x—3 a/(4x - 3f I 3x''^—3x—1 Hencefindj V(4x -3)' Answers Lety x^+l V4x-3 dj) dx (>/4x — 3)2x —(x^+1) i(4x - 3) (4) 2xV4x-3 4x—3 2(x^+ 1) V4x-3 J 3x^-3x-1 ■\]{4x—3/ 4x - 3 2x(4x - 3) - 2(x^ +1) (4x -3)V4x -3 2(4x^ - 3x - x^ — 1) V(4x - 3)^ 2(3x2-3x-l) V(4x-3)^ 1f2(3x2-3x-l),dx = - I \ — ' dx2J V(4x-3f x2+1 multiplyquotientrulenumeratorand denominatorby-Jix-3 2V4x-3 + c

CambridgeIGCSEand0LevelAdditionalMathematics WORKEDEXAMPLE11 Differentiatexsinxwithrespecttox. Hence find J,cos« dx Answers Letdv=Xsinx — =(x)(cosx)+(sinx)(l) productrule =XcosX+sinX Hence J(x cosx + sin x)dx = x sin x JX cosX dx + J sin x dx = xsin x Jxcosx dx = xsinx - J sinx dx JXcosX dx = Xsin x +cosx + c WORKEDEXAMPLE12 Differentiate x^yj2x—1 withrespecttox. I V2x-1y 5 Hencefindj —dx Answers Lety=x®-v/2x-1 dy dx (x^) -X2X(2x-1)"2 2 +3x^V2x—1 + V2x-1 X(3x2) ^J2x^ _x^+3x2^2x-1) ~ yj2x-1 7x»-3x2 I V2x-1 7x3_3;x2+5 V2x-1 dx -I —,7x3_3^2dx+I rV2x-1 J 7x3_3^2 f 5 JJ2x-l = X^yj2x -1 H ^-r-(2x -1)2+1 2Xdx 2dx =x^y/2x-1+5y/2x-1+C =(x3+5)■J2x — 1 + c productrule

Exercise15.7 x+5 . , d)i x-6 1 a Giventhaty= , ,showthat V2 dx bHencefind I x-1 X—6 yl{2x-lf ■ dx. V(2x-1)^ 2aDifferentiate(3x^-l)^withrespecttox.bHencefind-1)'dx. 3 a Differentiate xInx withrespectto x. b Hence find J In x dx. dAnx^ 1-Inx 4 a Showthat— = ^—. dxVXy X f AnX^b Hence find J I—g-I dx. 5 a Giventhaty=xVx^-4,find bHencefind 1 bHencefind J x^-2 Vx^—43(x+ (x—3) d)i dx dx. 9fx 3) 6 a Giventhaty=3(x+1)Vx-5,showthat—=^/ . dx 2vx-5 Vx-5 dx. 7 a Find-^fxe^*-— dxl 2 2* b Hencefind Ixe^*dxI' df sinx ^ k 8aShowthat—|:: |canbewrittenintheform 7,andstate dxVI-cosx thevalueofk. cosx-1 bHencefind JcosX-1 dx. d^ /cjc*9 a Giventhatv=(x+8)yjx-4,showthat—= , ,andstatethe , J,, dx V:c-4 valueofk. X kx b H dence find J 10a Giventhaty= ■v/x—4 1 x^—7 r 4x b Hencefind I — to dx.J(x2-7f ,show x. that kxd)! dx - 7)' -,andstatethevalueofk. 11aFind—(2xAnx). dx^ ^ b Hence find Jx^Inx dx.

Chapter15:Integration

Notethatthec'scancelout,sotheprocesscanbesimplifiedto: l4 1 x^dx i: =21 x^dxiscalledthedefiniteintegralofx^withrespecttox. Itiscalled'definite'becausethereisonlyonesolution. Hence,theevaluationofadefiniteintegralcanbewrittenas: [*f(x)dx =[f(x)]* = F(6)- F(a) Ja

- CambridgeIGCSEand0LevelAdditionalMathematics 12a Differentiatexcosxwithrespecttox. b Hence find J.s.n.d.. dy13aGiventhaty= (sin2x+cos2x),showthat—=4e^*cos2x. /• dx b Hencefind e^*cos2xdx. CHALLENGEQ 14 a Find ^(x^yl2x-7 ). I 5x^—14x+3 bHencefindI 1 —dx. yl2x-7 15.8 Definite integration Youhavelearntaboutindefiniteintegralssuchas fx^ dx = —x^ + c J 3 wherecisanarbitraryconstant. Jx^ dx is called the indefinite integral of x^ with respect to x.

X4^+

Itiscalled'indefinite'becauseithasinfinitelymanysolutions. Youcanintegrateafunctionbetweentwodefinedlimits.Theintegralofthefunctionx"withrespecttoxbetweenthelimitsx=1and X =4 is written as: x* dx Themethodforevaluatingthisintegralis c 2. ri s 1'Xdx= —X+c Ji [3 J, = XH+c =21

Chapter15:Integration Thefollowingrulesfordefiniteintegralsmayalsobeused. j kf{x) dx = A j f(x) dx,where ^ is a constant Ja Vfl fi b fi b [f(x)±g(x)]dx=If(x)dx±Ig(x)dx Ja Ja ^a WORKEDEXAMPLE13 Evaluate Answers •2 f^ J, 1 X®+3 dx b J VSx +1 dx c J -1 10 (3-2x) ITdx t^dx=f(x"''+3x~^)dxJ] Jl = r-x^+—x-'l' [4 -1 Jl =(^i(2r-3(2)-^]-[i(ir-3(ir' =51 [V3x+1dx=f(3x+1)2dx Jo Jo 1 3 -(3x+l)5 (3) '2 3 -(3x+1)2.9^ '. xl62 ■Xl2 ■(?)-(!) 14 f' 10 f' 0(3-2,)-'d. [P2 10 )(-l) ,3 — 2x -1 - fjJ u (3 - 2xr

CambridgeIGCSEand0LevelAdditionalMathematics WORKEDEXAMPLE14 Evaluate: (■2 a J dx: b J ^(3cos2x+5)dx Answers • 2 J, dx _2»:-3 (2e')-(2e-') 2e-e 2e2-2 J^(5+3cos2x)d*=j^5x+^sin2xj* (5n 3 n\ 3 = 1-—sin— - 0+—smOU 2 2j I 2 j |]-(OfO)4 571+6 In Section 15.10 you will learn why the modulus signs are included in the following integration formulae: J—dx=ln|x|+c J ——dx=—ln|ax+b\+c ax+b a Note: It is normal practice to only include the modulus sign when finding definite integrals. The next two worked examples show how these formulae are used in definite integrals. WORKEDEXAMPLE15 Findthevalueof 8 J2 2x +1 dx. Answers 8i.22x+1 dx= —ln2x+l L2 ' = (41n|7| )-(41n|5|) =4(ln7-ln5) = 41n5 substitutelimits simplify

Chapter15:Integration WORKEDEXAMPLE16 Eindthevaluenf f^ ^ J32—3x Answers r ^ dx= J32—3x În|2-3x|j substitutelimits =(-31n|-13l)-(-31n|-7|) simplify -31nl3+31n7 =3(ln7-lnl3) 7 =31n13 Exercise15.8 1 Evaluate. a J dx fS (x^+2)dx Jo j J \/x(x +3)dx gd ^-3 Idx JlX J (5x^ — 3x)dx 1V X 2Evaluate. •2J (2x + l)' dx X+4dx '-1(x +4) 3Evaluate. i: r g f (e^ + e^^)^ dx Jo a I e^* dx 'o I d I dx '0 Jo |.i 2 J-i(3x+5)^ b Iê''*dxJ: dx i: It 2*-l dx 3e*—Î dx e" c J (5x + 2)dx i f (2x -l)(3x -5)dx J-3 'ri2Vx—Îdx vx. yjix+if r" 6 J-4 -\/4 — 3x dx dx 2 c I Se"^"" dx 'o J. Jo dx 'I ^3+86^^^ ) 2e^ dx

CambridgeIGCSEand0LevelAdditionalMathematics Evaluate. •71 a I sinXdxJ Jo n Jj(2cosX -sin2x)d X e f^(3+cos2x)dx Jo K ^(2x-sin2x)dx Jo 5Evaluate. » 5 a I ^ dx r 2 Ji3x+1 f— J-22x+1 j: 1 dx 2x+3 4 2 ■dx f Ji 1- 3xdx 6 Evaluate. 3x—1 'k f" n 1 X 2x 1 dx J, 7t 3 7t c I "sin 2x dx )o I 6, 7C Jj(sin3x-cos2x)dx p 3 Ja2x-1 1-33-2x dxdx 2x 2x dx 2 2 GiventhatI dx=--In7,findthevalueofk.Ji3x-l 3 4:X A 8 a Giventhat:: r=2+ ^,findthevalueoftheconstantA. 2 2x+3 b Henceshowthat x + 3 4:X 5 dx = 2-31n-. 0 2x -f 3 3J' Jo 9 a Find the quotient and remainder when 4x^ + 4x is divided by 2x+l. J' Jo *^ 4" 1 b Hence show that I — — dx = 2 - — In 3. lo 2x+l 2 CHALLENGE Q 6x^—8x 10 Find the value of I dx. Ji 2x-l

Chapter15:Integration 15.9 Further definite integration Thissectionusestheconceptthatintegrationisthereverseprocessofdifferentiationtohelpevaluatecomplicateddefiniteintegrals. WORKEDEXAMPLE17 Sx dv Giventhaty=,„ ,find—. yjx+5 dx f- 3 Henceevaluate , dx. Jo ^(x- + bf Answers 3x dy dx ^ yjx^+5 (a/x^ + 5)(3)-(3x) i(x2 + 5)4(2x) 3a/x2+5 x^+5 3x^ a/x^+5 X+5 -3(x^+5)-3x^ (x^+ 5)a/x^ + 5 15 r a/(x^ + 5)^ yj{x^ + 5)"^ dx 5Jo 15 +5)' 3x dx 5[_Vx^+5 Ia/TTsJ a/O+5 multiplyquotientrulenumeratoranddenominator byVx"+5 WORKEDEXAMPLE18 Giventhaty=xcos2x,find dx Henceevaluate 2xsin2xdx.1:

CambridgeIGCSEand0LevelAdditionalMathematics Answers Let31=xcos2x—=(x)X(-2sin2x)+(cos2x)X(1) dx productrule =cos2x-2xsin2x 7 J:(cos2x-2*:sin2x)dx=[xcos2x](' I ®cos2x dx- I 2xsin2xdx Jo J( 7C [-sin2x I - I ^ 2xsin2x dx = — 2 Jo Jo 12 12 Xcos—I-(0Xcos0) 6 3; I 2xsin2xdx —sin2x2 6 7C 0 ~i2 (I.itWl.nl ^ =-sm—--sin0 V2 3) V2 )12 y/S _Jt_ T'u 3j3-n 12 Exercise15.9 1 a Giventhat3)=(x+1)y/2x-1,find b Henceevaluate I , dx. Jiyj2x-1 dx 2 a Given that y = xv3x^ +4,find b Henceevaluate I J( dx ^3x^+2 3V3x^~+^ d}' dx 3 a Giventhat y= —^,find ^ X+5 dx bHenceevaluate fJifx (x 4x ^ +5)' dx. X~l~2 dv 4aGiventhaty= ,find—. V3 dx+4 x f fix+4 b HenceevaluateI =dx. J( '0 V(3x+4)' 5 a Showthat d X cosx+xsinx xVcosx cos X f4cosX+XsmX bHenceevaluate dx. Jo 5cosX

Chapter15: Integration CHALLENGEQ 6 a Find—(xsinx). dx bHenceevaluateIxcosxdx Jo CHALLENGEQ 7 a Find—(x^In x). dx^ ' b Hence evaluate I 4xIn x dxr CHALLENGEQ 8 a Giventhat))=xsin3x,find dy dx bHenceevaluate1xcos3xdx Jo 15.10Areaunderacurve ^\v = f(x) A(x) 0 X Ifyou define the area under the curve y = f(x) to the left ofx as A(x),then as xincreases then A(x) alsoincreases. Now consider a small increase in x,say 6x,which results in a small increase, 5A, in area. // II A(x) M 0 XX+5x 6A=areatotheleftof(x+5x)-areatotheleftofx5A=A(x+6x)-A(x) Nowconsiderthearea5Awhichisapproximatelyarectangle. 5A=y5x y 5A so —-y ox 5x

Cambridge IGCSE and0LevelAdditional Mathematics i A X A U dA dAAs Ox -> 0,then > —,hence — - y. ox dx dx dA TrdA , r If—=V,thenA= a;dx. dx J II / 0 a b X andHencetheareaoftheregionboundedbythecurvey=f(x),thelinesx-aX=bandthex-axisisgivenbythedefiniteintegral. Area=If(x)dx,wheref(x)^0 w a WORKEDEXAMPLE19 Findtheareaoftheshadedregion. Answers Area=j dx l3 <f]-© =6—units^ 3 O f: >'=X^ 2 3 WORKEDEXAMPLE20 J' YinAtheareaoftheshadedregion. 1t 1 11 1 1 3 X

Chapter15:Integration Answers Area fsr 1^ X+yC^ X+2e2 d* 3+2e2 \ 3 I =2+2e2-2e2 =2[l+>/;(e-l)] =7.67units^ l+2e2

WORKEDEXAMPLE21

IfIntheexamplessofar,therequiredareahasbeenabovethex-axis.therequiredareabetweeny=f(x)andthex-axisliesbelowthex-axis,then

Findtheareaoftheshadedregion. Answers •3 J, 1 =-4.5 Areais4.5units^. The required region could consistofa section above the x-axis and a section belowthex-axis. If this happens you must evaluate each area separately.

f(x)dxwillbeanegativevalue. Ja IHence,for a region that lies below the x-axis, the area is given as I f(x)dx

Cambridge IGCSE and0LevelAdditional Mathematics Thisisillustratedinthefollowingexample. WORKEDEXAMPLE22 Findthetotalareaoftheshadedregions. O / \y=2sinx / \ 371 7t\ i f2nX Answers 9Km f" jSsinxdx=[-2cosx]q Jo =(—2cosn)-(-2cos0)=(2)-(-2) =4 J. 3jt 2 j.. _ r o 1 2 2sinxdx=[-2cosx]^^ =^-2cos^j-(-2cos7t) =(0)-(2) =-2 Hence,thetotalareaoftheshadedregions=4+2=6units^. X+c.In Section 15.8 you were given the formula J — dx = ln| The use ofthe modus symbols in this formula can be explained by considering thesymmetrypropertiesofthegraphof>'=—• >' 1 2- \ \ 1 1-8 -i2 -i 0 1 2 3 * -1--2-

Chapter15:Integration 1 f"M Theshadedareasthatrepresenttheintegrals —dxandI —dxareequali magnitude. ^ ^ ^ However,oneoftheareasis f 1 f^1 belowthex-axis,whichsuggeststhat I —dx=-I —dx. J-3 X J2 X f®1Evaluating I —dx gives: J2X (•.Si QJ — dx =[In X = In3-In 2 = In f 1 3 2 Thisimpliesthat I —dx=-In—=In—. ^ J-3 X 2 3 1 Ifyoutryusingintegrationtofindthevalueof I —dx youobtain J-3X (•-21 9 I — dx =[In X ]l! = In(-2)- In(-3)= In J-3 X 3 Thereis,however,aproblemwiththiscalculationinthatInxisonlydefinedfor x>0 soIn(-2)andIn(-3)donotactuallyexist. Hence for x < 0, we say that f —dx = ln|x|+ c. JX Exercise15.10 1 Findtheareaofeachshadedregion.y=X®-6x^+9x y-5yjx-X =3x+ O 1 X

CambridgeIGCSEand0LevelAdditionalMathematics O y=2sinx+Scosx E X 2 Findtheareaofeachshadedregion.y=x{x-2) C ^ \ O y=x+2 b vi 3;=1+sin2x K n 3! = x-5cos| 2x+3 :y /-2 -l\ 0 A X y={x+2)(x+l)(x-1) Findthetotalshadedregion. 4Sketchthefollowingcurvesandfindtheareaofthefiniteregionor regions bounded by the curves and the x-axis. a y = x(x + l) b y =(x + 2)(3-x) C y = x(x2-4) d y = x(x-2)(x+4)e y = x(x-l)(x-5) f y = x^(4-x)

* b Thelinex-pdividestheregioninpartaintotwoequalparts. Findthevalueofp.

Chapter15:Integratipri'

CHALLENGEQ 8

6aFindtheareaoftheregionenclosedbythecurvey=—,thex-axisand thelinesx=1andx=4.

7aShowthat—fxe*-e*)=xe*. dx ^ b Use your resultfrom parta to evaluate the area ofthe shaded region. y=xe"

CHALLENGEQ 9 a Show that —(xinx)= 1+Inx. dx b Use your resultfrom part a to evaluate the area of the shaded region. y=\nx

Giventhattheshaded

The diagram shows part ofthe curve y = ^ regionhasarea8,findtheexactvalueofk.

5 Find the area enclosed by the curve y = -j=,the x-axis and the lines x =4 andX=9. 12

The following example shows possible method area enclosed curve straightline. curve y = line atthepoint(4,4).Findtheareaoftheshaded regionboundedhythecurveand theline.Answers

CambridgeIGCSEand0LevelAdditionalMathematics CHALLENGEQ 10a Showthat—(xcosx)=cosx-xsinx. dx b Use your resultfrom parta to evaluate the area ofthe shaded region. y = xsin X n X 15.11Areaofregionsboundedbyalineandacurve

The

27x intersects the

by a

Thereisanalternativemethodforfindingtheshadedareaintheprevious

Area=areaundercurve-areaoftriangle dx X4X4 -8 392 WORKEDEXAMPLE23

and a

for finding the

example.

y = x

Chapter15:Integration \ y= A / /0a ^ h X ( y=gix) Iftwofunctionsf(x)andg(x)intersectatx=aandx=b, thenthearea,A,enclosedbetweenthetwocurvesisgivenby: A = f f(x)dx - r g(x)dx •a ^a Sofortheareaenclosed by y= 2>/x and y=x: 3*1 y=2Vx O 4 using f(x)=2%/x and g(x)= x area = f f(x)dx- f g(x)dx Jo Jo I 2Vx dx - I X dx Jo Jo |»4 = J {^yfx — x)dx ^X2-—X^ \ 2 0 ^4 ^ 1 —X42—X4^ 3 2 =2— units^ 3 ^ 3 xO43 1 -X02--X02 This alternative method is the easiest method to use in this nextexample.

X=±2

CambridgeIGCSEand0LevelAdditionalMathematics

WORKEDEXAMPLE24

Thecurvey= -Aintersectsthexsaxis lineatthepointsAandBandintersectsthey=X-2atthepointsBandC. aFindthecoordinatesofA,BandC. bFindtheareaoftheshadedregion boundedbythecurveandtheline. Answers a Wheny=0,x^-4=0 Aisthepoint(-2,0)andBisthepoint(2,0) Forintersectionofcurveandline: x^—4=X—2 x^-X—2=0 {x+\){x-2)=0 Whenx=-I,y=-S Cisthepoint(-1,-3) Area = (x -2)dx - (x^-4)dx =J(x-2-x^+4)dx=J(x+2-x^) =[—x^+2x-—x^ = 4.5 units^

•(■••-D-V-l) \ ^ / ^ \a BIX \ ^ X A = X- -4 y = X -2 dx

CLASSDISCUSSION

X=-1or3c=2

Discuss tvith your classmates howyou could find the shaded area, A, enclosed by the curve y = x{8- x) the line y = 5x and the x-axis. Can you find more than one method? Calculate the area using each ofyour different methods. Discuss with your classmates which methodyou preferred =3x \ y = x(8 - x)

A 0 8\ X

<•2

Chapter15: Integration Exercise15.11

Findtheareaoftheregionboundedbythecurve y=2+3x- thelinex=2andtheliney=2.

Sketchthefollowingcurvesandlinesandfindtheareaenclosedbetween theirgraphs. a b c d e y = x"^ +1 and y =5 y=x'^-2x+3and x+y=9 y = ^/x and 1 y=4x-x^and2x+y=0 y-{x-l)(x-5)and y=x-1

4 Findtheareaoftheshadedregion.

=1+cosx y=2+3x-X =3x'^+2 y=2sin2x+3cosx

Findtheareaoftheregionenclosedbythecurvey=1+cosxandtheliney=1.

Findtheareaoftheregionboundedbythecurve y=3x^+2,theliney=14andthey-axis.

The curve y = V2x + 1 meets the y-axis at the point P. Thetangentatthepoint<2(12,5)tothiscurvemeets they-axisatthepointR.

X-I-3 andy=4 x+3

CambridgeIGCSEand0LevelAdditionalMathematics

10FindtheareaoftheshadedregionPQR.Thediagramshowsthegraphsofy=2-fcos2xand y=1-I-2cos2xfor0 x n. Findtheareaoftheshadedregion. (2,8)/^ \y=fix<p 0 fi X R P y=2+cos2x y=1+2cos2x

Thediagramshowsthegraphsofy= Findtheareaoftheshadedregion.

CHALLENGEQ 11

Sketchthefollowingpairsofcurvesandfindtheareaenclosedbetween atheirgraphsforx3=0.y=andy=x{2-x) b))= and31=4x-3x^ Find the shaded area enclosed by the curve y = SVx the ^ line y = 10- X and the x-axis. ^ y-S^fx y=10-X

8Thetangenttothecurvey=fix-x^atthepoint(2,8) cutsthex-axisatthepointP. aFindthecoordinatesofP. bFindtheareaoftheshadedregion.

The area. A, bounded by the curve y = f(x), the x-axis and the lines x = a and x = b is given by the formula A = f f(x) dx if f(x) ^ 0. Ja V = f(^)

Chapter15: Integration Summary

dx*- J Integrationformulae 1

Areaunderacurve

integrationasthereverseofdifferentiation — rF(x:)l = f(x), then ff(jc) dx = F(3£:)+ c. dJ x=n+\ +c,wherecisaconstantandn^-\ f(ax + by dx = ^—"(ex + 6)"^^ + c, n -1 and a ^0 J a(a+l) I J dx=— +c|e' f[cos(ax + ^»)] dx = — sin(ax + b)+ c V 0/ sinx dx = -cosx + c J[sin(ax +6)]dx = —cos(ax ■¥b)+c e*dx=e*+c cosx dx = sinx + c J 1 dx = Inx+c, X > 0 1 ——dx = —ln(ax + ft) + c, ax + 6 > 0 ax-^b a f 1 f 1 1I —dx = ln|x| + c I dx = —ln| ax + 6| + c J X J ax+b a Rules for indefinite integration Ikfix)dx=kjf(x)dx,where/eisaconstant J[f(x)+g(x)] dx= Jf(x)dx±Jg(x)dx Rules for definite integration If Jf(x)dx= F(x)+c,then J f(x)dx=[F(x)]* = F{b)-F(a). IAf(x)dx=Af(x)dx,whereAisaconstant Ja Ja f^b I [f(x) ± g(x)l dx = j f(x) dx ± I g(x) dx Ja Ja

Workedexample

Thediagramshowspartofacurvesuchthat—=3x^-6x-9. dx PointsAandBarestationarypointsofthecurveandlinesfromAandBaredrawnperpendicular tothex-axis.Giventhatthecurvepassesthroughthepoint(0,30),find a theequationofthecurve, b thex-coordinatesofAandB, c theareaoftheshadedregion. Answers a —=3x^-6x-9 dx y=x^-3x^-9x+c

UsingX=0,y=30givesc=30. Theequationofthecurveisy=x^-3x^-9x+30. b Stationarypointsoccurwhen —=0 3x^-6x-9=0 x^-2x—3=0 (x+l)(x-3)=0 X=-1orX=3

d\'

[3][4][4] CambridgeIGCSEAdditionalMathematics0606Paper11Q12(part)Nov2012

Thex-coordinateofAis-1andthex-coordinateof

Cambridge IGCSE and0LevelAdditional Mathematics Areaboundedbythegraphsoftwofunctions Iftwofunctions f(5£:) and g(x)intersect at x = a and x = b,then thearea,A,enclosedbetweenthetwocurvesisgivenbytheformula A = j f(x)dx - I g(x)dx. Ja Ja y=g(-^) Examination questions

is3.

Chapter15:Integration c Area = J y dx r3 — J -3x^ -9x + 30)dx Qy^ —-x^-—+30x (3)' ,.,3 9(3)^-(3)^ +30(3) (-1)^ ( 1)3 9(-l)^ +30(-l) =76 Areaofshadedregionis76units^. Exercise15.12 Exam Exercise 1 a Given that y = e*^,find —. [2] dx f a bUseyouranswertopartatofindIxe*dx. [2] f2 2 *' c Henceevaluate I xe*dx. [2] Jo CambridgeIGCSEAdditionalMathematics0606Paper11Q5i,ii,iiiJun2014 dy 1 f1 5^2Acurveissuchthat—=4x+ ^forx>0.Thecurvepassesthroughthepoint ~• cix +ij \^2 6J a Findtheequationofthecurve. [4] b Find the equation ofthe normal to the curve at the point where x = 1. [4] CambridgeIGCSEAdditionalMathematics0606Paper11Q7i,iiJun2014 3 a Find J [2] ^ f^V 6Hence find the value ofthe positive constant k for which J 11 -jdx = 2. [4]

CambridgeIGCSEAdditionalMathematics0606Paper11Q5i,iiJun2013 The diagram shows partofthe curve of y =9x^— x^,which meetsthe x-axis atthe origin O and at the point A.The line y — 2x +18 = 0 passes through A and meets the y-axis at the pointB. NOTTO SCALE y=9x^-X' y-2x+18=0

b 4

CambridgeIGCSEand0LevelAdditionalMathematics a Showthat,forx^0,9x^-x^^108. [4] bFindtheareaoftheshadedregionboundedbythecurve,thelineABandthey-axis. [6] CambridgeIGCSEAdditionalMathematics0606Paper11Qll(part)Jun2012 5 Thediagramshowspartofthecurvey—2sinSx.Thenormaltothecurvey=2sinSxatthe Kpoint where ^ ~ ^ meets the y-axis at the point P. y=2sin3x O n NOTTO SCALE a FindthecoordinatesofP. [5] b Findtheareaoftheshadedregionboundedbythecurve,thenormalandthey-axis. [5] CambridgeIGCSEAdditionalMathematics0606Paper11Qll(part)Jun2012 6 Findtheexactvalueof 5+ aandbareintegers. 8 4x—3 0^ 0 ^ 7 aFindthevalueoftheconstantAsuchthat =3+ f^6x —5 7 bHenceshowthat dx=6+2In—. Js2x—3 3 8aDifferentiate4x^ln(2x+1)withrespecttox. 2x b 1 Gi dy dx,giving your answerin theform ln((3e^),where [5] Examinationstylequestion [2][5] Examinationstylequestion [3] 2x-3 2x-3 venthaty= ii Hencefind J X+4 , ,showthat = -rdx j 5x+20 d {yjx + 2) x. iii Henceevaluate ydx.(^/^)I '5x+20 [4][2][2] CambridgeIGCSEAdditionalMathematics0606Paper11Q9Nov2013

Chapter15:integration 9 Donotuseacalculatorinthisquestion. df ^ i Showthat— =pxe'^",wherepisanintegertobefound.dxV4 ) /•In 2 iiHencefindtheexactvalueof xe"'*dx,givingyouranswerintheform Jo [4] 10 [4]aIn2+-,wherea,bandcareintegerstobefound. c CambridgeIGCSEAdditionalMathematics0606Paper11Q5Jun2016 Find J ^3x-x2 dx. [2] Thediagramshowspartofthecurvey=3x-x^andthelinesy=3xand2y=27-3x. Thecurveandtheliney=3xmeetthex-axisatOandthecurveandtheline2y=27-3x meetthex-axisatA. 2y=Tl-3x O A X ii FindthecoordinatesofA. [1] iii Verify that the coordinates ofBare(3,9). [1] iv Find the area ofthe shaded region. [4] CambridgeIGCSEAdditionalMathematics0606Paper21QllJun2016

Thissectionwillshowyouhowto: ■ ofapplydifferentiationtokinematicsproblemsthatinvolvedisplacement,velocityandaccelerationaparticlemovinginastraightline\withvariableorconstantacceleration,andtheuseofx-tand u-rgraphs.

•Youshouldalreadyknowaboutdistance-timegraphsandspeed-timegraphs.Distance-timegraphscanbeusedtofindouthowthedistancechangeswithtime.Thegradientofadistance-timegraphrepresentsthespeed. speed.Thesteeperalineis,thegreaterthe

Distanceandspeedareexamplesofscalarquantities. (Scalarquantitieshavemagnitudebutnodirection.)Inthischapteryouwilllearnaboutdisplacement,velocityandacceleration. Displacement,velocityandaccelerationareexamplesofvectorquantities. (Vectorquantitieshavemagnitudeanddirection.)

distance Astraightlineshowsthatthespeedis constant(steady). Ahorizontallineshows> thatthespeediszero.

speedAlinewithnegativegradientmeanstheisdecreasing.(Itisdecelerating.)

Chapter16:Kinematics

O time

Speed-timegraphscanbeusedtofindouthowthespeedchangeswithtime. •Thegradientofaspeed-timegraphrepresentstheacceleration.•Theareaunderaspeed-timegraphrepresentsthedistancetravelled. speed speedgradientmeanstheis,thegreatertheisconstant.acceleration.

Alinewithpositive gradientmeansthe speedisincreasingr(Itisaccelerating.) Alinewithzero Thesteepertheline

time

Ifa particle movesin a straightline,with s{t),then rateofchangeofdisplacementwithrespecttotime,—,isthevelocity,v,of theparticleattimet.

CambridgeIGCSEand0LevelAdditionalMathematics 16.1Applicationsofdifferentiationinkinematics Displacement displacement,Consideraparticle,P,travellingalongastraightlinesuchthatitssmetres,fromafixedpointO,tsecondsafterpassingthrough O,isgivenby5=4^ Thegraphofsagainsttfor0^i^5is: 5 5 4 3 2 1 -1 -2 -3 -4 <) 5 K ^ 5=4t-t^\ -5-*

the d5

Thiscanberepresentedonamotiondiagramas: Attimet=S,thedisplacementis3mandthetotaldistancetravelledis 4+(4-3)=5m.

ds dt

Attimef=4,thedisplacementisOmandthetotaldistancetravelledis 4+4=8m. Attime^=5,thedisplacementis-5mandthetotaldistancetravelledis 4+9=13m. Velocityandacceleration

displacementfunction

t-5 t=A t=3 Jt=2 t=0 t=l I I I -I 1 1 r -5-4-3-2-1 0 1 2 3 4 5^ Attimet=2,thedisplacementoftheparticlefromOis4m.Whent=2theparticlestopsinstantaneouslyandreversesitsdirectionof motion.

Ifthe velocityfunction is v{t),then the rate ofchange ofvelocity with respect totime,—,istheacceleration,a,oftheparticleattimet.

Itisimportantthatyouareabletointerpretthesignsfordisplacements, velocitiesandaccelerations. TheseinterpretationsforaparticlePrelativetoafixedpointOare summarisedinthefollowingtables.

Chapter16:Kinematics

Earlierinthischapteryouconsideredthisdisplacement-timegraphforthefirst5secondsofthemotionofaparticleP.Thedisplacementoftheparticle,smetres,fromafixedpointO,tseconds after passing through O,was given by s = 4t- t^.

dt a du dt dt^

CLASSDISCUSSION

Workinginsmallgroups,completethefollowingtasks. Find an expression,in terms of t, for the velocity, y ms~\ ofthis particle. Drawthegraphofvagainsttfor0 a positive b zero c negative. Whatcanyousayaboutthedirectionofmotionoftheparticlewhenvisapositivebzerocnegative?

f 5. 3Statethevaluesoftforwhichvis

I ) 5 W r s=4t-P'\ .)t

5 Find the acceleration, a ms"^,ofthis particle and interpret your answer. Now,discussyourconclusionswiththewholeclass.

CambridgeIGCSEand0LevelAdditionalMathematics Signsfordisplacements 5<0 5=0 5>0 Pistotheleftof0 Pisat0 Pistotherightof0 Signsforvelocityv V<0 i;=0 v>0 Pismovingtotheleft Pisinstantaneouslyat rest Pismovingtotheright Signsforaccelerationa a<0 a=0 a>0 velocityisdecreasing velocitycouldbe maximumorminimum orconstant velocityisincreasing

WORKEDEXAMPLE1

—\2t^+Abt.

Thevelocityis9ms~^whent=2. b v=5t^-24t+45 a=—=6t-24 dt TheWhent=4,a=6(4)-24=0accelerationis0ms"^whent=4.

Aparticlemovesinastraightlinesothat,tsecondsafterpassingafixedpointO,its displacement,smetres,fromOisgivenby5= aFindthevelocitywhent=2. b Findtheaccelerationwhen<=4. c Findthevaluesoftwhentheparticleisinstantaneouslyatrest, dFindthedistanceoftheparticlefromOwhent=7. e Findthetotaldistancetravelledbytheparticleduringthefirst7seconds. Answers a s=t^-\2t^+45t v=^= -24<+45 d/ Whent=2,v=5{2f-24(2)+45=9

cFindthedistancetravelledduringthethirdsecond, d Findtheaccelerationoftheparticlewhent=2.5. Answers aWhen<=0,^=In[2(0)+5]=ln5

WORKEDEXAMPLE2

Aparticlemovesinastraightlinesothatthedisplacement,smetres,fromafixedpointO,isgivenbyi=In{2t+5),wheretisthetimeinsecondsafterpassingapoint aXontheline.FindOX b Findthevelocitywhent-10.

OX=ln5==1.61m b s=In(2f+5) ds 2 theparticleisatXattimet=0 V (1/ 2<+5 Whent=10,v= 2(10)+5 Thevelocityis0.08ms"'. 0.08

Whent=3,s={Sf-12(3)^+45(3)=54.Whent=5,s=(5)^-12(5)2+45(5)=50. sC t=7 X X t=0 yt=3 0 50 54 70^

Totaldistancetravelled=54+(54-50)+(70-50)=78m.

Theparticleis70mfromOwhent=7.

t=?>ort=5

Theparticleisatinstantaneousrestwhenn=0. 3^2-24t+45=0 <2-8t+15=0 (t-3)(t-5)=0

Theparticleisatinstantaneousrestwheni=3and<=5. Whent=1,s=(7)^-12(7)2+45(7)=70.

Chapter16: Kinematics

Whent=0,s=0. Criticalvaluesarewhent=5andt=5.

Theaccelerationis-0.04ms"^. --0.04

Cambridge IGCSE and0LevelAdditional Mathematics

u=0when-2^-^'+3e-^'=0 2e-2'=3e-S' e'=2 ,3 f=In Itisinstantaneouslyatrestwhen«=In—=0.405s. b When t=1,s=20[e-2(')-e"^')]=1.711. DisplacementfromOis1.71m.

i=0 t=2 t=S X > X X In5 In9In11^ 11

Distancetravelledduringthethirdsecond=In11-In9=In—=0.201m. d V 2t+5dv =2(2t+5)-' a=—=-2{2t+5)'^x2= ^ dt {2t+5f Whent=2.5,a= [2(2.5)+5]'

Aparticlemovesinastraightlinesuchthatitsdisplacement,imetres,fromafixed pointOonthelineattimeIsecondsisgivenbyj=20[e"^'-e"^']. a Findvalueoftwhentheparticleisinstantaneouslyatrest, b FindthedisplacementoftheparticlefromOwhent=\. C Findthetotaldistancetravelledduringthefirstsecondofitsmotion. Answers as=20[e~^'-e~®']t;=—=20[-2e-2'+3e-^']

WORKEDEXAMPLE3

Sincei;>0forallvaluesoft,thereisnochangeinthedirectionofmotionoftheparticle. Thethirdsecondisfromt=2x.ot-?>. When<=2,s=In[2(2)+5]-In9.Whenf=3,sIn[2(3)+5]=In11.

Chapter16:Kinematics cWheni=0,s=0. 3 Criticalvalueiswhent=In— 2 Whent=In—,5=20 2 2.963. t=l 3i=InI t=0 0 1.711 2.963 Totaldistancetravelled=2.963+(2.963—1.711)=4.21m. WORKEDEXAMPLE4 Aparticletravelsinastraightlinesothat,tsecondsafterpassingthroughafixed point O,its velocity w ms"\ is given hy v = 5cos j. aFindthevalueof(whentheparticlefirstcomestoinstantaneousrest, b Find the acceleration ofthe particle when ^ ^• Answers a Theparticleisatrestwhenv=0. 5cos cos| 2'~ ^ t_K 2~2 t = 7t Itisatinstantaneousrestwhent=n. b V=5cos dwd< 5.fA =—sm 2 Uj When t — —,a = — — sinf—1 = -1.25 3 2 U; Whent=—theaccelerationis-1.25ms

Aparticlemovesinastraightlinesothat,tsecondsafterpassingthrougha fixedpointO,itsdisplacement,smetres,fromOisgivenby5=In(1-I-2t).

Aparticle,movinginastraightline,passesthroughafixedpointO. Itsvelocityums"',<secondsafterpassingthroughO,isgivenby V=cos3<-I-sin3b

b Findtheinitialaccelerationoftheparticle.

c Showthattheparticleisdeceleratingforallvaluesoft.

a Findthevalueoftwhenthevelocityoftheparticleis0.4ms~'.

Aparticle,movinginastraightline,passesthroughafixedpointO. Its velocity u m s"', tseconds after passing through O,is given by v a Findtheinitialvelocityoftheparticle.

50 {2,t+2)' 6e2'-2t.

a Find the value ofiwhen the velocityofthe particlefirstequals4ms"'.

d Sketchthevelocity-timegraphforthemotionoftheparticle.

b Findthedistancetravelledbytheparticleduringthethirdsecond, c Findtheaccelerationoftheparticlewhen ^=1.5.

Aparticle,movinginastraightline,passesthroughafixedpointO. Its velocity ums~', iseconds after passing through O,is given by u = a FindthevelocityoftheparticleasitpassesthroughO. bFindthevalueoftwhenthevelocityis0.125ms"'. c Findtheaccelerationoftheparticlewhent=\.

Exercise16.1

Aparticlemovesinastraightlinesuchthatitsdisplacement,5metres,from a fixed point Oon the line at time tseconds is given by 5 = 9[ln(3i -I- 2)]. a Findthevalueof<whenthedisplacementoftheparticlefromOis36m.

b Findtheaccelerationoftheparticlewhen t=5.

a Findthevalueof<whentheparticleisfirstinstantaneouslyatrest, b Findtheaccelerationoftheparticlewhent=K.

Aparticlestartsfromrestandmovesinastraightlinesothattseconds afterpassingthroughafixedpointO,itsvelocityums~',isgivenby V = 5(l-e-'). a Findthevelocityoftheparticlewhen t=In100. b Statethevaluewhichvapproachesastbecomesverylarge. C Findtheaccelerationoftheparticlewhenu=4.

b Findthevelocityoftheparticlewhen/=1.

Aparticletravelsinastraightlinesothat,tsecondsafterpassingthrougha fixed point O,its velocity y ms~',is given by u = 8cos|^^j.

CambridgeIGCSEand0LevelAdditionalMathematics

bFindanexpressionforthevelocityandtheaccelerationoftheparticle intermsoft.

Chapter16:Kinematics

c Findthetotaldistancetravelledduringthefirst2secondsofitsmotion.

b Find the displacementofthe particle from Owhen t =2.

b Find the time when the particle firstcomes to rest and its distance from Oatthisinstant.

a Findthetimewhentheparticleisinstantaneouslyatrest.

10 A particle movesin a straightline so that its displacementfrom a fixed pointO,isgivenby5=2t+2cos2fiwheretisthetimeinsecondsafterthe motionbegins. a Findtheinitialpositionoftheparticle.

C Find the time when the particle first comes to rest and its distance from Oatthisinstant.

aFindanexpressionforthevelocityandtheaccelerationoftheparticle intermsoft.

8 A particle startsfrom restand movesin a straightline so that, tseconds afterleavingafixedpointO,itsdisplacement,smetres,isgivenby 5=2-2cos2t.

9Aparticlemovesinastraightlinesuchthatitsdisplacement,smetres,from a fixed point Oon the line at time tseconds is given by 5 = 50[e"^' - e""^'].

11 A particle, moving in a straightline,passes through a fixed point O. Itsvelocityums"^tsecondsafterpassingthroughO,isgivenby V-kcosAt,wherekisa.positiveconstant.

a Findthevalueoftwhentheparticleisfirstinstantaneouslyatrest, b Findanexpressionfortheacceleration,a,oftheparticletsecondsafter passingthrough O. ^TC cGiventhattheaccelerationoftheparticleis10ms~^whent=—,find 24 thevalueofk.

d Findthetimewhentheaccelerationoftheparticleiszeroforthefirst timeanditsdistancefromOatthisinstant.

12Aparticlemovesinastraightlinesuchthatitsdisplacement,smetres, fromafixedpointOatatimetseconds,isgivenby 5=2^for0^f 4, s=8+2In(f-3)forf>4.

aFindthedistanceOX. CbFindthevalueoftwhentheparticleisfirstatrest.Findthevaluesoftforwhichthevelocityispositive,

Applications of integration in kinematics

a Find the initial velocity ofthe particle. bFindthevelocityoftheparticlewhen it=2 lii=6.

d Find the values of tfor which the velocity is negative. 16.2

d Sketch the displacement-time graph for the motion ofthe particle, eFindthedistancetravelledbytheparticleinthe8thsecond.

In the last section you learnt that when a particle moves in a straightline wherethedisplacementfromafixedpointOonthelineiss,then:

13Aparticlemovesinastraightlinesothatthedisplacement,5metres, from a fixed point O,is given by s = 2t^ -I7t^ +40t-2,where tis the timeinsecondsafterpassingapointXontheline.

di V = dt _dw_d^5 dt dt^ and Conversely,ifaparticlemovesinastraightlinewheretheaccelerationofthe particleisa,then:

c Find the acceleration ofthe particle when I 1 =2 ii i =6.

"= Jadf ands=\vdt Inthissectionyouwillsolveproblemsthatinvolvebothdifferentiation andintegration.

CHALLENGEQ

CambridgeIGCSEand0LevelAdditionalMathematics CHALLENGEQ

Chapter16; Kinematics Thefollowingdiagramshouldhelpyourememberwhentodifferentiateand whentointegrate. displacement(s) V = d5 I,v= r/dt dv d''5 d< d^^ velocity{v) acceleration{a) V=a(it WORKEDEXAMPLES AparticlemovinginastraightlinepassesafixedpointOwithvelocity8ms Itsaccelerationams"^,tsecondsafterpassingthroughOisgivenbya= +1. aFindthevelocitywhent=3. b FindthedisplacementfromOwhent=3. Answers a fl!= +1 V=Iad< 1 = J(2t+1) = +t + c dt UsingV=8whent=0,givesc=8 V= -¥t+S TheWhent=S,v=(3)^+(3)+8=20particle'svelocitywhent=3is20ms"'. b t;=i^+f+8 5 = Jv dt = J(<^ + < +8)dt I s 1 9 o —~t H—t +8t+c 3 2 Usings=0whent=0,givesc=0 I s 1 9 o s=-t^+-{2+8t 3 2 When t = 3, s = -(3)'+ -^(3)^ + 8(3)= 37.5 3 2 Itsdisplacementwhen^=3is37.5m.

Aparticle,movinginastraightline,passesthroughafixedpointO. Itsvelocityums"',tsecondsafterpassingthroughO,isgivenbyu=6e^'+2t. a Findtheaccelerationoftheparticlewhent=\.

CambridgeIGCSEand0LevelAdditionalMathematics

WORKEDEXAMPLE6

Alternativemethod

cSinceu>0forallvaluesoft,thereisnochangeinthedirectionofmotionof theparticle. Whent=0,s=0 Whent=2,s=2e^<2)+(2)^-2=2e®+2Distancetravelledduringthefirst2seconds=2e®+2=809m.

bItsaccelerationwhent=1is364ms"^.V=6e'"+2/ s=Iud< J = J(6e^'+ 2t)dt =2e®'+ +c Usings=0whent=0,givesc=-2

b FindanexpressionforthedisplacementoftheparticlefromO. C Findthetotaldistancetravelledbytheparticleinthefirst2secondsofitsmotion. Answers a aV=6e®'+2t=—=ISe®'+2 dt When1=1,a= +2=364

Sincethereisnochangeindirectionofmotion: s = Ju dt Note:f (6e3'+ 2t)dt Jo =(2e®+4)-(2e«+O) ■ [2e^'+ inThisalternativemethodcanonlybeusedwhentherehasbeennochangedirectionofmotionduringthe relevanttimeinterval. =2e®+2 =809m

Thedisplacement,smetres,isgivenbys=2e''+t^-2.

Chapter16:Kinematics

Aparticlestartsfromrestandmovesinastraightlinesothat,/secondsafterleavinga fixedpointO,itsvelocity,wms"',isgivenbyw=4+8cos2t. Findtherangeofvaluesofthevelocity, Findtherangeofvaluesoftheacceleration. Tt Answers a u„i„-4+8(-l)=-4 Hence,-4^u^12. b u=4+8cos2t a=—=-16sin2tdt and -1^cos2f^1 and =4+8(1)=12 a„i„=-16(l)=-16 Hence,-16^a^16.WhenV=0,4+8cos2<=0 cos2t-and -1^sin2<^1 and =-16(-1)=16 2t=^ 21 2t^ 3 _^ "3 particle first comes to rest when ^ ^ ■ d 5 -I'dt = J(4+8cos2t)dt =4t+4sin2<+c Usings=0whent=0,givesc=0 Particlechangesdirectionwhent=—. Whent= When t ;= 4(^|j+4sin(^Y]" =4 j+4sin(7t)= 2j:= 6.2832 6.283 7.653 Totaldistancetravelled=7.653+(7.653-6.283)=9.02m.

cFindthevalueoftwhentheparticlefirstcomestoinstantaneousrest.

KThe

i=4<+4sin2t n

I y 0

dFindthedistancetravelledduringthetimeinterval0^t .

V=4+8cos2t

WORKEDEXAMPLE7

2 A particle, moving in a straightline,passes through a fixed point O. Its velocity ums~\ tsecondsafter passing through O,is given by = a Findtheaccelerationoftheparticlewhent=2.

CambridgeIGCSEand0LevelAdditionalMathematics

c Find the distance travelled by the particle in the 3rd second.

bFindanexpressionforthedisplacementoftheparticlefromO.

c Find the total distance travelled by the particle in the first2seconds of itsmotion.Giveyouranswercorrecttothenearestmetre.

4Aparticle,movinginastraightline,passesthroughafixedpointO. Its velocity ums"^ tsecondsafter passing through O,isgiven by V = t+2cos|—). 37t

b Find the time taken before the particle returns to O. c Findthedistancetravelledbytheparticleinthe2ndsecond, dFindthedistancetravelledbytheparticlebeforeitcomestoinstantaneousrest, e Findthedistancetravelledbytheparticleinthefirst12seconds.

32 it+

bFindanexpressionforthedisplacementoftheparticlefromO.

3Aparticle,movinginastraightline,passesthroughafixedpointO. Its velocity ums"\ tsecondsafter passing through O,isgiven by f=4e^'+2t. a Findtheaccelerationoftheparticlewheni=1.

FindthedisplacementoftheparticlefromOwhent=—andits acceleration at this instant. ^

1Aparticle,movinginastraightline,passesthroughafixedpointO. Its velocity ums~\ tsecondsafter passing through O,is given by f= 10^ aFindthevelocityoftheparticlewhentheaccelerationisfims"^.

C Find,to the nearest metre,the displacementofthe particle from O whent=2.

5Aparticle,movinginastraightline,passesthroughafixedpointO. Its velocity ums~\ tsecondsafter passing through O,isgiven hy v =4e^'+6e~^'. a Showthatthevelocityisneverzero, bFindtheaccelerationwhent=In2.

Exercise16.2

10AparticlestartsfromrestatafixedpointOandmovesinastraightline towards a point A.The velocity, w ms~\ ofthe particle, tseconds after leavingO,isgivenbyw=8-8e"^'.

Aparticlemovesinastraightline,sothat,tsecondsafterleavingafixed point O,its velocity, t;ms"\ is given hy v = pt^ + qt -\2,where p and qareconstants. When t =2 the acceleration ofthe particle is 18m s~^. When?=4thedisplacementoftheparticlefromOis32m. Findthevalueofpandthevalueofq.

8 A particle movingin astraightline passesa fixed point Owith velocity 18ms"^ Itsaccelerationoms"^,isecondsafterpassingthroughOisgivenbya=3<-12. a Findthevaluesoftwhentheparticleisinstantaneouslyatrest. b Findthedistancetheparticletravelsinthe4thsecond, cFindthetotaldistancetravelledinthefirst10secondsofitsmotion.

a Findthevalueoftwhentheparticlefirstcomestoinstantaneousrestat thepointP.

11Aparticletravelsinastraightlinesothat,tsecondsafterpassingthrougha fixed point O,its speed, w ms~\ is given by v = 2cos|^^j-1.

b GiventhattheparticlereachesAwhent=2findthedistanceOA. Giveyouranswercorrectto3sf.

a Findtheaccelerationoftheparticlewheni=In5.

b Findthedistancetravelledbytheparticlebeforeitfirstcomesto instantaneousrest.Giveyouranswercorrectto3sf.

Chapter16:Kinematics

A particle movingin a straightline passes a fixed point Owith velocity 10ms"^ Itsaccelerationams"^,tsecondsafterpassingthroughOisgiven bya=3-2t.

a Findtherangeofvaluesfortheacceleration.

bFindthetotaldistancetravelledfromt=0tot=2n.

a Findthevalueoftwhentheparticleisinstantaneouslyatrest. b Sketchthevelocity-timegraphforthemotionoftheparticle. cFindthetotaldistancetravelledinthefirst7secondsofitsmotion.Give youranswercorrectto3sf.

9 Aparticlestartsfromrestandmovesinastraightlinesothat,tseconds afterleaving afixed point O,its velocity,ums"\is given by u=3-I-6cos2<.

Cambridge IGCSE and0LevelAdditional Mathematics CHALLENGEQ

CHALLENGEQ

b Findanexpression,intermsoft,forthedisplacementoftheparticle. C Findthesecondvalueoftforwhichtheparticleisatinstantaneousrest, d Findthedistancetravelledduringthe4thsecond.

Summary

Therelationshipsbetweendisplacement,velocityandaccelerationare: displacement(s) _dw_d^s At d<^ velocity(v) acceleration(a)

Aparticleisatinstantaneousrestwhenv=0.

12Aparticlemovesinastraightlinesothattsecondsafterpassingthrougha fixed point O,its acceleration,a ms~^,is given hy a-pt+ q, where pand q areconstants.

13AparticlestartsfromapointOandmovesinastraightlinesothatits Kt displacement,icm,fromOattime/secondsisgivenby5=2/sin—. 3 a Findexpressionsforthevelocity,v,andtheacceleration,a,ofthe particleattimetseconds, b Showthat18(y/-s)=t^{9a+K^s).

TheparticlepassesthroughOwithvelocity3ms~^andacceleration-2ms~^. Theparticlefirstcomestoinstantaneousrestwhent=2.aFindthevalueofpandthevalueofq.

a Find the velocity ofthe particle as it passes through O. [1]

Aparticletravelsinastraightlinesothat,tsafterpassingthroughafixedpointO,itsvelocity, wcms"^ is given by f = 4e^'-24^.

Answer a When t^O,v = 4e(2xo)-24(0)= 4. Thevelocityofthe particleasitpassesthrough Ois4cms"\ b Thethirdsecondisfromt=2tot=S. f3=0for2^t^S,hencethereisnochangeindirectionofmotioninthistimeinterval.

Chapter16:Kinematics

CambridgeIGCSEAdditionalMathematics0606Paper21Q11(part)Nov2012

Examination questions

b Findthedistancetravelledbytheparticleinthethirdsecond. [4]

c Find an expression for the acceleration ofthe particle and hence find the stationary value of the velocity. [5]

Worked example

Distancetravelled (.3 = J (4e2'- 24t)dt =[2e^'-I2t^]=(2e®108)-(26" 48) =2e®-2e''-60 Distancetravelledinthirdsecond=638m. Acceleration=—(v) dr ^ =-(4e dt 2( 24t) =8e^'-24 Accelerationis(8e^^-24)cms"^. du Stationaryvaluesofvoccurwhen —=0. 8e2<-24=0 =3 2t=ln3 t=-ln3 2 1 When t=-ln3,v=4e'"3-121n3=12-121n3 2 Stationary value ofvelocity is 12(1-In 3)cms"^

a WritedownthevelocityofthebodyasitpassesthroughO.

CambridgeIGCSEAdditionalMathematics0606Paper11Q5i-ivNov2011

Abodymovesinastraightlinesothat,isafterpassingthroughafixedpointO,its displacementfromOissm.Thevelocityyms"^issuchthatv=5cos4i.

bFindthelengthoftimeforwhichtheparticleistravellingwithconstantvelocity, c Find the total distance travelled by the particle. [3][1][1]

3 Aparticlemovesinastraightlinesuchthatitsdisplacement,xm,fromafixedpointOattime ts,isgivenby x=3+sin2<,wheret^0. a Findthevelocityoftheparticlewhent=0. [2]

CambridgeIGCSEAdditionalMathematics0606Paper11Q9i,ii,iiJun2011

2 Aparticlemovesinastraightlinesothat,tsafterpassingthroughafixedpointO,itsvelocity, 0 wms~\isgiven by y=2f-11H .Find theacceleration ofthe particle whenitisat instantaneous rest. ^ ^ ^ [7]

Thevelocity-timegraphrepresentsthemotionofaparticlemovinginastraightline, a Findtheaccelerationduringthefirst5seconds.

b Find the value oftwhen the particle is first atrest. [2] c Findthedistancetravelledbytheparticlebeforeitfirstcomestorest. [2] d Find the acceleration ofthe particle when t- —. [2]

b Find the value ofiwhen the acceleration ofthe body is first equal to 10ms"^. CFindthevalueofswhent-b. [1][4][4]

CambridgeIGCSEAdditionalMathematics0606Paper21Q9Jun2012

CambridgeIGCSEand0LevelAdditionalMathematics Exercise16.3 ExamExercise 1 t»ms 20 24fs

CambridgeIGCSEAdditionalMathematics0606Paper21Q2i,ii,iiiJun2013

a AparticlePmovesinastraightline.Startingfromrest,Pmoveswithconstant accelerationfor30secondsafterwhichitmoveswithconstantvelocity,^ms-^,for 90seconds.Pthenmoveswithconstantdecelerationuntilitcomestorest;the magnitudeofthedecelerationistwicethemagnitudeoftheinitialacceleration, i Use the information to complete the velocity-time graph. [2] w ms" Ok I I I I ' I tseconds

CambridgeIGGSEAdditionalMathematics0606Paper22Q12Mar2016

CambridgeIGCSEAdditionalMathematics0606Paper21Q6Jun2015

I FindthetimeatwhichPfirstcomestoinstantaneousrest. [2]

CambridgeIGCSEAdditionalMathematics0606Paper21QIONov2015

Chapter16:Kinematics

5 A particle Pis projectedfrom the origin Oso that it movesin a straightline.At time tseconds after projection,the velocity ofthe particle, m s"\ is given by v = 2^^- 14i+ 12.

ii FindanexpressionforthedisplacementofPfromOattimetseconds. [3] lil FindtheaccelerationofPwhent=2). [2]

ii Given that the particle travels 450 metres while it is accelerating,find the value ofkandtheaccelerationoftheparticle. [4]

b A body Qmovesin astraightline such that,tsecondsafter passing afixed point,its acceleration,ams~2,is given hy a=3t^+6.When /=0,the velocity ofthe bodyis5ms~b Findthevelocitywhent=3. [5]

CambridgeIGCSEAdditionalMathematics0606Paper22Q11Mar2015

7 A particle is movingin astraightlinesuch thatitsvelocity,fms-^,tsecondsafter passing afixed pointOis1/=e^'-6e~^'-1. Find an expression for the displacement,5m,from Oofthe particle after t seconds. [3] Using the substitution u= e^',or otherwise,find the time when the particle is atrest. [3] Findtheaccelerationatthistime. [2]

8 A particlePisprojectedfrom the origin Oso thatit movesin astraightline.Attime iseconds afterprojection,the velocityofthe particle,vms^i,isgiven by i*=9t^-63t+90. i Show thatPfirstcomestoinstantaneousrestwhen t=2. [2] ii Find the acceleration ofPwhen t= 3.5. [2] iii Find an expressionfor the displacementofPfrom Oattime tseconds. [3] iv FindthedistancetravelledbyPainthefirst2seconds, [2] b in thefirst3seconds. [2]

CambridgeIGCSEand0LevelAdditionalMathematicsAnswersChapter1 Exercise1.1 1 one-one 2many-one 3 one-one 4 one-one 5 one-one 6one-one 7 one-one 8one-many Exercise1.2 1 1,2,3,4,5,6and7 2a-7^f(x) 2 b2^f(x)^17 c -1<f(x:)^9 d0^f(x)^9 e-^f(x)^8 8 f 5 3 g(a:)^2 4 f(x)^-4 5 f(x)^5 6f(x)^-5 7 f(x)«10 8f(x)^3 IExercise1.39251 36754 1 5 67 6 ahk bkh 7 X= 8 « =-L=,=4 3 9 « = ±i 4 10x=2.5 II x=5 12X=3orX=4 13afg bgf cg^ d Exercise1.4 1 a-2-,4 b-2,-7 3 c 0.8,1.6 d -23,25 e-5,-2 g16,24 i 1^ 5 3' f-0.5,7.5 h -5,3| a -4-,-1.9 b -0.8,0 6 d 0.75,3.5c-5.6,4 f x=33 fdb-3,30,4,6-2,-1,0,3h-1,2 e6.5 3 eca-2,2-3,-1,2-2,1,3 g0.5,1 I 0,2,6 4 a(-4,0),(3,7),(5,9) b (0,0),(1,1),(2,2) c(1,3),(2.5,7.5) Exercise1.5 l avshape,vertexat(-1,0),y-intercept1 b Vshape,vertexat(1.5,0), cy-intercept3Vshape,vertexat(5,0),y-intercept5 d Vshape,vertexat(-6,0), y-intercept3 eVshape,vertexat(5,0),y-intercept10 f Vshape,vertexat(18,0), y-intercept6 2 a X -2 -1 0 1 2 3 4 y 7 6 5 4 3 4 5 3 caVshape,vertexat(0,1)bVshape,vertexat(0,-3)Ashape,vertexat(0,2)dVshape,vertexat(3,1), y-intercept4 e Vshape,vertexat(-3,-3), y-intercept3 4 a-3«f(x) 11 b0^g(x)«11 c-3^h(x)^5 5-5^f(x)^5,0^g(x)^5, -5^h(x)^3 6 a Vshape,vertexat(-2,0), y-intercept4 b straightlinethrough(-5,0), (0,5) c-3,1 7 a Vshape,vertexat(3,-3), by-intercept3-3^f(x)«7 c0.5,5.5 8 a V shape,vertex at oj, y-intercept4 bstraightlinethrough(-2,-4), (5,10) c0.8,4

9 a b -3,3 21 3 54 6 r\x)=yJx+7-5 t-'(,)=^ f'(«:)= 3+y/x—I {'(x)=(8- +3 ft -1/X X+3S ag'(a:)= X-2 9 af-^(x)=2(x-2) bnosolution 100 11-2,3.5 12f-i(3c)^0 13f-ig-i 14f,h 15 a X^2 b g(x) 0 16 a r'(x)= ^ 3 5 b13 ijaf-ig'cgf-i c X -X b g-if d f-ig-i Exercise1.7 1 .44 3 7 10 2x-7 9 X 8 7 b20 65 b294 43 b3.25 0'2-1 0 1 2 34 56 78 910*^ M— h+^ 1 /j 7^1i 134'"*'^^'IIIIIIIIiii' 24"'"!"g'T8'ilOTlS* 4-_ Sttli 7 afisaone-onefunction b r'(x)=y/3^-l c ;4M12 1 t rI"' ,-2 4' :-444ri.. -10-'-8- \ \ ^12- rTO 8 a f"'(x)= 2x+7 X-2 b Thecurveissymmetrical abouttheliney=x Exercise1.8 1 6.5,-4 2aVshape,vertexat(-0.6,0), y-intercept3 b-0.2,-1 3aVshape,vertexat(2.5,0),y-intercept5 b1,4 4 a -1^f(x)«299 b X^0 5 a-3 bf-'(x)=(x+3)2+1 -1/\ 3x—2 '®<'>=1741 6a0.2^g(x)=s1 1%„-l/"X 1+X 2x c0.2«x^1 d1.25

CambridgeIGCSEand0LevelAdditionalMathematics 7 a33 bi kh ii h2 illh-'k-ior(kh)-i 8 i 2-V5<f(x)^2 11f-'(x)=(2- -5 domain2-V5<x^2 range-5^f (x)<0 iiiX=-4 9 a i 2(x-2)2-3 ii x^2 big(x)354,h~^(x)320 ii 8 10 12 14 iii8.5 10i =|3+2x| )i=2-X ii-i-5 Chapter2 Exercise2,1 1 X=%y=9andx=-2,31=4 2 X=-1,31=-7andx=4,31=-2 3 X=-S,y--4and x=4,3;=3 4 X=1,31=4and x=-2,31=-2 5 X=0.5,31- X=0,3=1 6 X=-1,3=-3andx=2,3=1 7 X=1.5,3~ X=2,3=3 8 X=3,3=1andx=9,y=7 9 X=1.8,3=2.6andx=1,3=3 10 X=-1,3=-2and x=1,3=2 11 X=1,3=2andx=2,3=1 12 X=1,3=2and x=4,y--4 1 . 1113x=l,3=--andx=--,3= 14 15X=3,3=1andx=1,3=3X=-1,3=-3andx=1,3=3 16 X=0,3=-0.5andx=-1, 3=-l 17 X =-1,3=2 and x =7^,3=-1.4 18 X=-1.5,3=-8and x=4,3=3 19(-0.2,1.4)and(1,-1) 20 a X+3=11,X3=21.25 b X=2.5,3=8.5and x=8.5, 3=2.5 2117cmand23cm 22 6^/5 or13.4to3sf 23(0.5,0) 245V2or7.07to3sf 25(2,2)263=-2x-1 Exercise2.2 1 amin(2.5,-12.25),axis crossingpoints(-1,0),(6,0), (0,-6) bmin(0.5,-20.25),axis crossingpoints(-4,0),(5,0), (0,-20) cmin(-2,-25),axiscrossing points(-7,0),(3,0),(0,-21) dmin(—1.5,—30.25),axis crossingpoints(-7,0),(4,0), (0,-28) emin(-2,-3),axiscrossing points (-2->/3,O), (-2+ V3,0),(0, 1) fmax(1,16),axiscrossing points(-3,0),(5,0),(0,15)

2a(x-4)'-16b(x-5)'-25 c(x-2.5)^-6.25 d(x-1.5)'-2.25 e (x +2)'-4 f(x+3.5)'-12.25g(x+4.5)'-20.25h(x+1.5)'-2.25 3 a (x-4)'-1 b(x-5)'-30 c (x-3)'-7 d(x-1.5)'+1.75 e (x+3)'-4 f (x+3)'+0 g(x+2)'-21 h(x+2.5)'-0.25 4 a 2(x-2)'-5 b 2(x-3)'-17 c 3(x-2)'-7 d 2(x-0.75)'+0.875 e2(x+1)'-1 f 2(x +1.75)'-9.125 g 2(x-0.75)'+3.875 11 (h3 -4)+5-12 a 9-(x-3)' b25-(x-5)' c2.25-(x-1.5)' d 16-(x-4)'

y=\x^-2x-2,\ -2 -1 0 1 2 3 4 5 * ji=|l-4*-x^ 3 a 2(x +0.25)'-3.125 y=\x^-5x+4\ Ij -1 0 1 2 3 4 5 6 * >=|2x®+*-3| ji=|*'-2*-8| -3-2 -1 0 1 2 3 4 5 * -3 -2 -1 0 1 2 3 * a (3,16) b 1 y \ i—\~\r\ \/v =|( -2-1 01 234567* 2 0 2 4 6 c0<A<16 yt|2*2-ii*-6| 5 a

b k>4: 6 a(5.5,6.25) b 6.25 7 a-4,4 b -2,0,2 c-1,2,3,6 d -6,4 5-y/^,^5+>/^e ,1,4, ^ f-4,-2,-1,1

a 6-(jc+1)^ b l2-ix+2f c 16.25-(x+2.5f d 9.25-(x +1.5)' a 13.5-2(x +1.5)' b 3-2(x+l)' c 15-2(x-2)' d 4—-sfx--! 12 I 6) 8 a 4 +4.75 bNo,since4.75>0 a2(x-2)'-7 b(2,-7) 10a-5.25,0.5 11 a 118 1 b b X>0.5 X<11 8 12 a 18.5-2(x-1.5)' b(1.5,18.5) cnshapedcurve,vertex =(1.5,18.5) 13 a 13.25-(x-2.5)' b(2.5,13.25),maximum c-7 f(x)«13.25 d No,itisnotaone-one function. 14a 2(x-2)'-5 h 15-0.75 16a 5-(x-2)' b(2,5)maximum cOne-onefunction, f~'(x)=2+V5^ Exercise2.3 j=|3c2_4*+3| 0 1 2 3 4 5 * >=|3*=+5*-2| 2a5-(x+2) bnshapedcurve,vertex=(-2,5) (-3,4)

CambridgeIGCSEand0LevelAdditionalMathematics g-1-VTo,-1,-1+VIo ^l+V? -1+n/7 h , 2 2 1 -1,3 8 a (-1,0),(2,3),(4,5) b (-2,1),(-1,1.5),(2,3), c(5,4.5)(1,2),(2,4) Exercise2.4 1 aX<-3,X>4 b1<X<5 c X^-7,X^3 d0<X<5 e-0.5<X<4 f —1^X^3 g-1.5<X<5 h—oo<X<oo i X—3 2 a-7<X<2 bX«-3,X^2 c4^X^5 dX<-8,X>6 e—2.5 X^3 fX<-1,X>-0.8 3 a-6<X<3 bX<5,X>7 cX^0.5,X5=1 d-3<X<2 eX<-4,X>1 f-0.5<x<0.6 4 a3<X<5 b1<X 6 c0.5<X<1 d3<X<5 eX<-2,X^3 5 a -5<X<3 b 0<x<2,6<x<8 c0<x<2,4<x<6 6 -1-<X<2 3 Exercise2.5 1 atwoequalroots btwodistinctroots ctwodistinctrootsdnorootsetwodistinctroots ftwoequalroots gnoroots htwodistinctroots 2 k=±6 3 ^<0.5 4 ^>3 5 0,-i 6 k>-1.5 7 k>S3 8 k=-10,ft=14 9 ft=1,ft=4 Exercise2.6 1 ft=—3,ft=5 2 ft=-7,ft=-3 3 c=±4 4 ft<1,ft>5 5 a ft=±10 b(-2,6),(2,-6) 6 ft>-5 7 ft^0.75 8 —11<wi<1 9 m=-2,m=-6 Exercise2.7 1 3<ft<4 2 -1<X<0 ( 1Y 47 X H I 4j 8 3 a2 k 47 , 1 b — when x= 8 4 4 .<-£ 2 5 —6<X<1 6 -2,-18 7 a a=20,ft=-4,(4,20) bnshapedcurve,vertex =(4,20),)i-intercept=(0,4) 8 k—c+9 9 ft<-6,ft>10 10aa=-12,ft=-4 b-4 11Tra<2,m>14 12 a j=|(x-2)(x+3)| \ >1\ : 10 \ ® I \/4 A / \/^ -1—1T1 y : -4-20 2 4* _1 2' 4 J 1. 25 c ft> 4 13i-27 9 II 8 14ft<215a-5 X« 1 bi (x+4)2-25 ii25,-4

Ill 1 d 7 6 a—1 b3 11 17 ax=—,y= 6 ■' 3 / ^ 3 1 II II H Q \ , 73 19 i X c =\ „ 1 ^ Chapters Exercise3.1 1 a x'® b x'® c X ® d X® e x~^ f X-'® g x^ h x^ i 15x®/ j 5x^3>~^ k 16x'y 1 6x®3|-® 2 a x"* b X® c x"* d 9 ^ -1-X 3) 3 a a=l, b = 3, c=5 4 a 7 « = —. 6=1 2 3 5 X=-,V=-1 5■' 6 *=5,31=8 1 7 ^<{1 + x)^ Exercise3.2 1 a b 3 c —3or2 d —or5 2 c 52 a 4 b 2 d 1 e -1 g-4 h ±4 3 a 6 b -1 d -10 e -4^ 3 g-3 h 4i 3 j 1 or3 k -1 or4 1 -0.5or2 4 a 0.8 b 1.5 d 8 5 a 0 b 3 5 16 2 9 a --or3 b 1 4 10 a -i or 3 b 9 6 Exercise3.3 1 a 10V5 b 5VT0 c qVTT d 5V3 2 a 5V5+4V3 b 4V5+5V3 c12V5+5V3 d 20V5+27V3 3 a 2+I3V7 bIO+35V7 c 2+(2n-l)V7 4 a b 11 +VM 5 x= 5-3V2, 3 = 8 + 4^2, z=12-7V2. 6 aH-V2 b3+2V2 Exercise3.4 1 a 6 b 12 c VM d 2 e 13 f 5V5 g isVe h 14V^ 2 a 2 b V2 c 2 d i e i f V5 2 V5h - if jV2 k6 I<5 3 a 2V2 b 2V3 c 2V5 d 2V7 e 5V2 g 3V2 h 4V2 j 3VT0 k 3V7 m2Vir n5^^5 p IOV2 q 5V3 s Vs t V3 f6V2 i 4V5 I 3VTr 0 3VI3 r lOVM u 2V5 V 10V2 w 2V10 X 4V3 y 35 b3V3 c5V5 e 0 f9V^ h -3V5 i 6V7 k 10V2 I 0 4 a 9^/3 d 7>f3 g 2V5 j 8V2 m20V5 n 8V3 o6V^ 5 a 2+3V2 b 12 c -4+5V2 d 9+5V3 e 3- V3 f 20 g 1 h -2 + 4V3 i I2 + 5V5 j 7 k 13 1 -4 m4 n 17+3V35 0 23+I6V2 6 a9+4V5 b28-lOVS c 91 +40^3 d 5+2V6 7(10+12V2)cm2 8 a 293 b a = 9, b= 10 cp=A5,q=2 9(11+7V2)cm^ 10 ^/2 11 a a = 4, ^» = 2b c = 9, d = 2 Exercise3.5 3V2 1 V5 a 5 VI 2 4V5 c 3^/3 f 4V3

CambridgeIGCSEand0LevelAdditionalMathematics_2V3,5V2.3V28^^I^ V5 Vl3 5 ' 13 m 5V2 _ 7V3 .5+ V5 A ® K 3-V3 -2+3V2 r-I+2V7 2 a-I+V2 3-V7 d3-V53+V5 e-10+5V5 f-7-2V7 g4+2V3 i 2V3+V2 . 10 3 a7-4V3 5+4V2 5+3V2 9-2V14 ' ^^ e2+V5 g -5+-in ,30+ 2 ^ —7— ^ . 3V7-V24 a V3 b g 12+VS 13 5 (2V5-V2)cm 6 (4-Vs)cm 7 7-2V3cm 1+5V2 ^ 15+14V2 ~~r~ " 28 a 23-4V7 ® 30 106-4V2 Exercise3.6 6+VT5 ^ VTO+2V51 a —^ b 5 V85+2V15 c ? a I+V2,3-2V2 b2+V6,3-Ve c2+V5,-3V5 d3+V7,5-2V7 e 3-V2,5+V2 100 c nosolution e 19.5 g15 1 20 fd52 h2V2 4 a1 d-3 g 4 i 4 b3 c3 el f5 h nosolution j 0or1 j0or2 doa-3or1b9e3 7+2V6 c0or4 f-4.5 5 7 2+V2 ' 2 1Exercise3.72or3 2 x=S,y=-0.5 3 -4.5 4 a=-3,2 6=-18,12 56-3V3 6 aI7V5+1145-I2VIO+8=53-I2VT0 11-3-i5+2V2 b-1+Ve 87+3V5 13 9 a= ,b-0,c=\ 6 10 p=-27,q=2S Chapter4 Exercise4.1 1 a3x'+2x^+Sx^ b9x'+2x^+7x^-2 c3/-4x'-3 d +3x®+4x'+5x*-2x' +x^-1 2a8/-4x'+2x:'+3x-2 bSx*+8x^+4x^-Sx-2 c 3x®+5/-3jc'+7x^+8x-20 fed3;c®+12x'+13x'+3x'-4—10%^+29x^—20x:+427x'-27x^+9x-1 3 a3x^+X—7 b2x^+17*^+21;c-4 c2x'+5x^-8x^-x^-19x 4 a x'+7x^+X-4 b 4x''-4x'-15x^+8x+16 c8x'-8x'-32x^+17x+32d4x''-4x'-5x^+3x-2 Exercise4.2 1 ax^+2x-48 b x^+X-1 cx^-15x+25 d x^+2x+1 ex^+4x—5 f x'-9 2a3x'+2x-lb2x'+3x-2 c 3x'-5x-10 d 3x^+4 3 a 3x^-4 b 2x^-X+5 cX-4dX+3

4 a X®- +X-1 7 b2 Exercise4.6 b +2x+4 8 ax'-4x^-7x+10 1 b(x—2)(x-4)(3x+4) b x'+3x^-ISx-40 2 a=-2,6=2.5 Exercise4.3 2 a29 c x'+x^-6x 3 b b^-4ac=-11 b546 9 a2x'-1Ix^+lOx+8 4 a-17 b71 c12 b2x'-7x'+7x-2 5 (x+l)(x-7)(2x+1) 3 -2-2a c2x'-9x^-8x+15 6 a=7,b=-8 4 aa=0,b=—19 10 x'+x^—7x—3 7 i a=6,6=2 b a=2,b=58 11 2x'-9x^+6x-1 ii(2x-l)(3x^+5x-2) ca=-l2,b=-7 12 b2, 8 iii(2x-l)(3x-l)(x+2) 5 -66,18 Exercise4.5 Chapters 6 a=13.5,b=5.5 1 a5 b-1 Exercise5.1 7 a p=\,q=9 c76 d2 1 a1,- b 3 2 8 b0,1,3 2 a5 b57 c -\,l d -1,0 Exercise4.4 c11 3 1 b(2x-l)(x+l)(x--1) 3 a=4,i=0 1 3 ^ 6 2 e--,- f —, 2 a (x-2)(x^+4x+5) 4 a=-6,b=-6 35 5 7 3 7 b(x+4)(x+2)(x- 2) 5 aa=-8,6=15 g--,7 h ® 5 4 c x(2x+3)(x-6) ^„-1+721-1-V^ 2 ' 2 14 6 d(x-7)(x+l)(x- 2) ' 3' 7 e(2x+l)(x-3)(x--4) 6 aa=-9,b=2 2 13 3 X=—,yf(x-2)(x+3)(3x-1) b5 2 ^ 2 7 5 g(2x-l)(2x+l)(x -2) 7 6=2 3 x = — orx = 3 3 h(2x—l)(x—3)(x+5) 8 a=6,6=—3 4 aX=±2,X=±4 3 a -5,1,7 b 1,2,3 9 a6=12-2a cf(x)^-l c -4,-2,1 d -4,1,1.5 ba=5,6=2 c X=-2,X= 3 3 -4,--,1 10 a5 b-72 e-2,0.5,3 f x=0,)i=5orx=2,31=32 11 aa=-8,b=—5 6 g-2,-1.5,0.5 h -4,2.5,3 b-30 7 5x^4,y= 4 a 1,-3+V7 12 32 b -3,--±->/37 13 ^„-7+713-7-713 Exercise5.2 b3, , 1— 4<X<82 2 6 6 1 c 2,-2±V3 14 a3 b-5 d 1.5,1,4 15 a=2,6=-5,c=7 5 -0.5,-2±Vl3 16 b-3,6 6 -5.54,-3,0.54

2a interceptsshapedcurve,axis(-3,0),(2,0), (4,0),(0,24) b shapedcurve,axis intercepts(-2,0),(-1,0), (3,0),(0,6) c shapedcurve,axis intercepts (-2,0), oj,(2,0),(0,-4)

3 A 5(0,14)

dV/%shapedcurve,axis intercepts (-2,0),(l,0),(^|,oj,(0,-6)

CambridgeIGCSEand0LevelAdditionalMathematics 2 a bx^lorx:^2 3 ax<-lorx>4 b-1«X^ 5 o 10c2<X< 3 dX<2orX>5 e X<-S or X> f-1^X^6 4 a-^X^2 3 u 2 ^1 bx>— cx> 3 4 5 ax=s-lorx& bX> 2 c -<X<1 2 d x<-lorx> 5 e-l^x=s3f -3<x<l 3 3 6 ax>— bx^ 2 2 cx^-2orX^ 2 d fe—6^X^0-14<X<2x^-—orx^32

4a shapedcurve, axisintercepts(-2,0)and (0,0)where(0,0)isa minimumpoint b shapedcurve,axis intercepts ojand (0,0)where (0,0)isaminimumpoint c interceptsshapedcurve,axis(-1,0),(2,0), (0,-2)where(-1,0)isa maximumpoint dKy\shapedcurve,axis intercepts(2,0), oj,(0,40) where (2,0) isaminimumpoint 5 a -1 0 1 2 3 * -1 O 1 2 *

6a interceptsshapedcurve,axis(-3,0),(0,0), (3,0) b/X/shapedcurve,axis intercepts(-3,0),(-2,0), (1,0),(0,-6) c/X/shapedcurve,axis intercepts(-4,0),|—,0j, (3,0),(0,12) ^ d/X-^shapedcurve,axis intercepts (-3,0),|-—,0j, (4,0),(0,-60) ^ ^ ^

7 a-7<X<1 K 2 b-<X<4 5 c X^-2orX^ 8 X> 2 k Ik 9 x<—orx> 5 3 ,^ 4 4 10 — X^ 3 3 Exercise5.3 1 £>(0,A(-1,0),B(2,0),C(3,0),6)

7 a b(0,0),(4,12),(7,0) 8 a >t bluecurveaxisintercepts: (-2,0),(-l,0),Q,oj,(0,-2) redcurveaxisintercepts:, (-1,0),(4,0),(0,4) b(-3,-14),(-1,0),(1,6) 9 a=\,b=2andk=-2 10a=-1,b=l,c=2,k=S 1Exercise5.4ax^-lor0^jc^2bx^2.1cX-1.4or1^X^1.4 2aX^2 bX>-2 c-1.9^X 0.3orX^1.5 3 aX>2.5 cb—l^x^lorx^2X<-0.8or0.6^X^2.2 Exercise5.5 1 ca±2,±1±4,±2 e ±77 b fd±3 g ±^,±f h ±1 a±0.356,12.81 b±2.32 c ±1.16 d -1.11,1.42 e1.26,-0.693 f ±1.40 a4,25 b 16 c 9 d25 9 9 A f 25 16'4 9 g 7,16 4 h —25 9'^ 10 shapedcurve,axis intercepts (-1,0)fi,0J,(3,0), (0,6) ; 11X<-0.8or0.55<x<2.25 12a interceptsshapedcurve,axis(-2,0),(1,0), (4,0),(0,8) b 1 i 16 27 4 —,8 8 5 a 5^=x+4 b (1,1),(16,4) 6 a1,2 b0,2 c -1,2 d -2,3 e -1 7 -2 8 1,2 9 -4,-3,0,1 Exercise5.6 g 1 x=—orx=2 5 2 X<-3orX>4 3 -<X<2 5 4 -<X<1 2 1 5 XS= 2 6 —6<X< 3 * 7 X> 2 8 a -1,27 a-1,3 9 shapedcurve,axis intercepts(0,®)' oj'(4,0) -3 -2 -1 O 1 2 3 4 5 " 13ax(x-2)(x+3) b shapedcurve,axis crossingpoints(-3,0),(0,0), (2,0) 14 a b2 15a(x+4)(2x-l)(x-3) b shapedcurve,axis intercepts(-4,0), > (3,0),(0,12) 16 a i-1,2 b(0,0),(2,16),(8,-128) Chapters Exercise6.1 1 aIg1000=3 blglOO=2

Cambridge IGCSE and0LevelAdditional Mathematics c dIg1000000=6X=Ig2eX=Ig15 fX=Ig0.06 a1.88 b2.48 c2.86 d1.19 e-1.70 f -2.30 a10®=10000b10'=1 c 10"®= dx=10'® f x=10^® a126000 ec1450.0316a4 c 0.5 i e X=10 b1450 d0.501 f0.00145b-2 d i 3 l.7 e1.5 Exercise6.2 1 a cblog464=3log232=5log5125=3dlogs36=2 1 f 2.5 elog2 32 1 =-5 flog,—=-4 S381 glog,6=2 hlog,4=yilog,c=b a 2'=4 b c 5"=1 d 1 e36^=6 f g x°=1 h i a'=b 2®=64 3^=9 1 8®=2x'=8 a 16 c625e12h4 a2d-2 jg2.5-0.5 1000 5 a2 d-2 2 3 a8 g b fdb e h kb ehb9V3 -43145 -1.5 311 3 -6 7 6 625 g17 fc33.5 iI0.5-0.5 c1.5 f3.5 Exercise6.3 1 alogj15 clogs64 elog,20 dblog,6log,2 f log,(i) 1 glogd12 hIg®20 i log46.4 2 a3 b2 c d1 e 1.5 f 3 a log,36b log.f 4 a 2',2-' b-2 5 a2 d-1 b 3 c 6 a5" b m-2 c 1+-« 2 d -m-33 2 7 a 1+X b 2-x cX+23 d 8 a -8 b-5.5 c13 d 34 9 a 8 b20 c14 d 0 c-3 1Exercise6.4a5b7.5 d77 2a12 b8 d3 e0.7 3 a 2,8 b 5 d4 e3 g5 h20 4 a16 b4 d73 5 a5,25c—,15625 125 c25 c10 f7.5 c 6 f 1 3 2 2 c 3 b—,3125 125512,16 6aX=4,31=16bX=20,31=10 c X=4,3=12 d X=40,3=16 ea=6.25,6=2.5 fx=1.25,3=2.5 7 bIgX=8,Ig3=2 Exercise6.5 1 a6.13 b 2.73 c 0.861 d2.41 e 2.65 f 1.66 g6.90 h 5.19 i 1.15 j-13.4 k0.641 10.262 2 b10 c 3.32 3 a0.415 b2.42 c 2.46 d1.63 e 1.03 4 5-0.751,1.38a0,1.46 b1.40 c2.32 6 0.683,1.21 7 a1.58 b 0.257,0.829 c0.631,1.26d0.792,0.161 d0.683

'."'Jfcrtitv~;L-'•" ;i"i,7-^ji<i6v Answers 8 a2.32 b1.16 c0.631,1.26 d 0.431,1.43 9 a0.6 b-0.189 10a1.26 b2.81 c±1.89d±2.81 11 X< logiologio32 1Exercise6.6a3.32b3.18 c1.29 1 a u d-3.08 ,2 45 6 1 c 2m 1 a X c2x a314 a23 1, a-log^x M d A 2u b1+* d4+2x b0.4 b23 b256 8 a 16 c1.59 » - 1 logs* 1 "a ^.27 c ,4 256 db91.87 b 3,9 b5 d 16,64 e-,512 1 f25 1 a ^ glogs^ cx=32,y=2 12X=6.75,>=13.5 Exercise6.7 1 a7.39 b4.48 c1.22 d0.0498 2a1.39 b0.742 c-0.357 d-0.942 3 a5 b8 c6 d-2 4 a7 b2.5 c6 d0.05 5 a4.25 b1.67 c1.77 d1.30 6 aln7 b ln3 ci(5+ln3)di(l+31n2) Z o 7a20.1 b0.135 c1100 d12.5 8 a3.49 b-2.15 c0.262 9a3+6^ bi(l+ln7) 1 d —In2 2 c21n2 e2In2,ln5 fIn2,ln3 10 a 1.79 b 0,1.39 c-3.69,4.38 1 111aX=-r,31= e e b X=lnO.6,y=In0.12 120.822 130,0.4 14 ±y/2, ^ln5 1Exercise6.8a409600 b16.61 2a43000 b2091 3 a0.0462 b724 4a500 b82.6 c2.31 5 a$250000 b0.112 c 6.18 6 a 1 3 u 81 ,b—, representsthearea ofthepatchatthestartof themeasurements c6.72 Exercise6.10 1 aasymptote:y--4, y-intercept:(0,-2)x-intercept:(In2,0) basymptote:y=6, y-intercept:(0,9) casymptote:y=2, y-intercept:(0,7) dasymptote:y=6, y-intercept:(0,8) easymptote:y=-1, y-intercept:(0,2) x-intercept:(In3,0) fasymptote:y=4, y-intercept:(0,2) x-intercept:(-in2,0) gasymptote:y=1, hy-intercept:(0,5)asymptote:y=8, y-intercept:(0,10) i asymptote:y=2, y-intercept:(0,1) x-intercept: ^^ln2,oj 2aasymptote:x=-2,x-intercept:(-1.5,0) y-intercept:(0,ln4) basymptote:x=2, x-intercept:|^2^,oj casymptote:x=4, x-intercept:(3.5,0) dy-intercept:(0,ln8)asymptote:x=-1, x-intercept:(-0.5,0) y-intercept:(0,2ln2)

"" Cambridge IGCSE and 0LevelAdditional Mathematics easymptote:x=2,jc-intercept:(2.5,0) fasymptote:x=1.5, x-intercept: ^1^,0^ Exercise6.11 1 abf"'(x)=ln(x-4),X>4f"'(x)=ln(x+2),X>-2 df >-l>1 ef '(«)= +5 ff-'(«)=i(e^'"+l) 3 af(x)>1 br'(x)=iln(x-l) cX>1 d X 4 ai5x b yfb 5 a I x^ b-In22 6a(2x+1)^ bX=ln2 iix+ln5. ii3x Exercise6.12 1 a 5/-7y+2=0 b 0,-0.569 2 X=5.8,y=2.2 3 a = f'(x)= j, X e f-'(x)=iln[^^^j, x>3 f rHx)=-iln(^^],x>5 g r'(x)=ln(2-x),X<2 h = x<5 2 abr'(x)=e'-lrHx)=e'^+3 c r^x)=e"®''-2 df-'(x)=i(e"®'-l) 4 5 6 7 bX=1 a10 a2 3.14 b-5 b4,6 Ig 1000 8 -0.569 9 a70 c 4S,SS,5yf5; right-angled 4 5and-5 5 1and3 6 a(6,4) b (6.5,7) c(3,8) d (0.5,1.5) fe(-4.5,-2.5){3a,b) 7 a=18,A=-8 8 a(0,4.5) b(1,1) 9 1 10A(-5,2),B(9,4),C(-3,6) b39.7 c17.0 10 a ilogjx O b y=125a 11i-2 11-n illy=5 ivX=10 1Exercise7.2a-2b-3 d331 a 3 12a1 1 4/ d -- d / ^ / 3 a 1 b 0 1/5 * ^ / 4 (-3,-1) 3 1 ® 11 k^-5 1 e b2 523 2 b |log„2 13a p=5,q=2 cX=36 bX=4 Chapter7 1Exercise7.1a3b4 c10 d13 e5 f25 g yfM h y/T3 \ I2V2 2 a 2V5,SS,-M; right-angled b 2>/l3,10,4>/l0;not right-angled 5' 2 bnotcollinear 7 5 8 1,2 9 (11,0) Exercise7.3 1 a y=3x—13 b y=-4x+7 1 cy=—X+1 ^ 2 2 a cb2y=5x-113x-1-2y=9X+2y=1 c0 f -i 4 5 c 2

3 a3'=2x-10 bX+231=-8 c23=3x-15 d X+43=0 4 a 4x-53=-17 b (0,3.4) c 3.4units^ 5 ay=-2x cb3x+43=25x+73=-26 6 » 5(0,2!) bMi 4 c 1.6875units^ 7 a(3,2) b33=2x+5 8a(5,6) bk=8 9 a3=-2x+13 b(6,1) c 6>/5,3V5 d45units^ 10 a i 2x+33=14, 1 II y=—X2 b(4,2) Exercise7.4 1 a27.5units^b22units^ 2 a54.5units^b76units^ 3 a^=-9 b 50units^ 4 a(-1,1.5),(2,-4.5) c 22.5units^ 5 a (4,5),(0,-3) b20units^ 6 a(0,-7) b 60units^ 7 a (7.5,9) b38.25unks^ 8 a (2,2),(4,-2),(0,6) b 40units^ 9 a(5.5,1) b (6,7) c116units^ Exercise7.5 1 a 3=ax^+d,y=3,X=X", m=a,c=b b 3X=ax^+b,Y=xy,X=x', m=a,c=b C——ax—b,Y=—,X=X, X X m=a,c=—b dX=~b—+a,Y-x,X=—, y y m-—b,c—a ey4x=ax+b,Y-yyfx, X=x,m=a,c=b fx^3=bx^+a,Y=x^y, X=x^,m=b,c=a g—=ax+b,Y=~, y y X=x,m-a,c=b u Vx Vx =ax-b,Y= , 3 3 X=x,m=a,c=-b 2 alg3=ax+6,y=lg3,X=X, m=a,c=b bIn3=ax-6,F=In3,X=X, m—a,c-—b c lg3=61gx+Iga,Y=lg3, X=IgX,wi=6,c=Iga dlg3=xlg6+Iga,Y=lg3,X-x,m=lg6,c=Iga eln3=-—Inx+—,F=ln3, b b X=lnx,»n=--,c-b b f Igx=-3lga+Igi,F=Igx, X=3,7W=-Iga,c=lg6 gx^=-by+Ina,F=x'^, X=y,m=—b,c=Ina h In3=6x+Ina,F=In3, X—x,m=b,c=\na Exercise7.6 1 a3=2x 1,b V=—X'+3 5 c 3=>/x+1 H 14 nd 3=—XH— •^ 2 2 e3=-2X2'+7 , 5, 27 t3=—InxH ^ 4 4 9 li 1 ii 1 0 II y= x^-1 3 bii y 2x^+3x ii 14 cii y S-, X ii 3 d11 y=(l3-2x'f ii 25 eii y= 52 9 —X 4 2 ii 32 fii y= 0.5 -0.3 X-X 3^/2 3 3=-2x'+16x' 4 a 3' = 3(2')+ 25 b 5 5 a 3=-2x'+8x bX=2.5,3=7.5 6 ae'=x^+1 b 3= ln(x^ + 1) a igy=-x-7 b3=10"'X10

CambridgeIGCSEand0LevelAdditionalMathematics 8 a31=100x2 b0.64 9 aln3=31nx-l h ^ 0y= e 10a12.7 b a=e'",b=-2 Exercise7.7 1 a X 0.5 1.0 1.5 2.0 2.5 0.5 3 5.51 8 10.5 c3= 5-? X dX= 0.8,3 =2.5 2 a 1 X 10 5 3.33 2.5 2 1 y 9.01 6.49 5.68 5.29 5 c V = 2x d0.22 1+8x 3 by=0.8x+ X cX=0.6,y=20.48 4 bnify-50,k-0.02 c29.1 5 b =3,fe=0.7 6 ba=5,6=1.6 7 ba=1.8,rz=0.5 8 ba=0.2,6=3 c4.56 9 ba=-1.5,6=1.8c8.61 10ba=3.6,b=-0.1 Cgradient=3.6,intercept=0.1 11 U .1 -"-S11b3=eXX cgradient=-0.8,intercept=3 Exercise7.8 1 a3=7 b 3x+43=31 c 12.5 45 6 ballow-1.4to-1.6 c allow13to16 a2\f5 b3=-2x+6 c(3,0),(-1,8) A= b=e 55 3 a0.559 b0.960 c1.47 d2.15 e4.31 4 a74.5° b143.2°c59° d104.9°e33.2° 5 a ca=3,6=0.4 d3.05 7 12.5 8 a 6.8 b A =898,b=-0.5 9 1 2x+33=14 11 3+2 =|(x-10) 10ill\f^X+33=2 11i(0,3.5) ill2x+3=11 Chapters Exercise8.1 ii(3,5) iv(0,11) Tt a 18 57t d 18 n 9 2n e g 371 . .o h — I 27c 4 j 47t 71 m 20 a 90° d20° g 126° j162° m315° I Tt 1257C447t9 571 T2 b 30° e 120° h 75° k 216° n 480° f ^ 3 571 T 7ik 0 6 c 15° f144° 1 27° I 540° o 810° xVx 1 2.83 5.20 8 11.18 ^Vx 3.40 4.13 5.07 6.20 7.47 Degrees 0 45 90 135 180 Radians 0 7t 4 7t 2 371 4 7t Degrees 225 270 315 360 Radians 57C 4 371 2 77t 4 271 b Degrees 0 30 60 90 120 Radians 0 71 7C 7C 27C 6 3 2 3 Degrees 150 180 210240 270 Radians 57t 6 n 7n 6 4713 37t2 Degrees 300330 360 Radians 57t 3 ll7t 6 27t 6 a0.964 b 1.03 c0.932 d1 e0.5 f1 7Exercise12.79°8.2 1 3711571a — cm 2 b 271cm cm d 157tcm 2 3 4 4 a9.6cm b2cm a1.25rad b 1.5rad a 12.4cm c 31cm a 10cm c 38.5cm a 23cm c41.3cm b32cm b 1.85rad b 18.3cm

7 a13.6cm b21.1cm c34.7cm Exercise8.3 1 =. K 2 u 135 21 a Sttcm^ b 2ncm^ ctf 9 j135 9 c 35ncm^ d ncm^ 4 2 a10.4cm^ b4.332cm^ 3 a1.11rad b1.22rad 4 a 0.8rad b 40cm^ 5 r(75-r) 6 a 9yj2cm b ^ 2 81 2 c —n cm 2 7 a 1.24rad b 89.3cm^ c121cm^ 8 a Irad b 49.8cm^ c 17.8cm2 9 a24.3cm2 b37.7cm^ c13.4cm2 10 b 547t-36>/2 11 a §rad b 18cm^ 7 12a4.39cm b2.40cm C 15.5cm d 15.0cm^ 1334.4cm 14a45.5cm^ b57.1cm^ c80.8cm2 d21.7cm2 1514.6cm^ Exercise8.4 1 aOAx^ b19.8or19.9 c24.95to25 2 a74.1 b422or423 3 b54.6or54.5or54.55 c115.25or115.3or115 4 a54.3 c 187 5 iallsidesareequaltotheradii ofthecircles,whicharealso equal 2n iii58 6 7 3 V 148 .2or58.3 19.3 ii79.1 iii57.5 iii11.6 4 V1.08^Area^1.11 Chapters Exercise9.1 2Vl3^3Vl3 1 a b 13 13 c — d 1 13 7+Vl3 2 a 9 23 V2+23 c — d 25 5 23V2-2V^ ® 10 3 a — b 4^/3 7 4>/3 d 1 7-196>/3 48 4 a b3 .y/3+j2 2 d e 2 7 5 a :/! 4 >/6 1 .-6+1073 c XJ 3 3 e 2->^f 273-76 Exercise9.2 l a 70°/ J10° '-60° ^220° -135° 2 dasecondbfourthcthirdthirdethirdffirst gfourthhthird ifirst jfirst Exercise9.3 1 a-sin40° bcos35° c-tan40° d cos25° e tan60° TT g-tan4 I -tan7t , n fsin h cos6 n jsin 4 2 3 a 7^ 5 73 b -i 2

I! amplitude=1,period (2".1)

3, C

Iii amplitude=2,period iii period= Sn57: ■4(f-4

2 a1 amplitude=4,period -2".[5.4(1.-4]

V amplitude=4,period =360°,(0,5),(180,-3),(360,5) vi amplitude=5,period =180°,(45,3),(135,-7),(225,3),(315,-7)

vi amplitude=4,period=7t, (0,3),(^|,-5j,(7t,3), (^y,-5j,(27C,3)

1, B

3 a = 2, 6 = 4, c = l 4 a = 3, 6 = 2, c = 4 5 a=3,6=2,c=3 6 a i period=90°,x=45°, x=135°,x=225°,x=315° iiperiod=360°,x=180° iii period=60°,x=30°, x=90°,x=150°,x=210°, x=270°, x=330° 71 7 ai period=—, 7t 371 571 X f X J X ~~ y 8 8 8 7n 971 lilt x = y, x-y, X-—, _ 13Jt _ 1^ ^ ~ ~ 8 ii period=—, 3 7t 7t 571 X = —, X = —, X = 6 2 6 7n Sn llTtx-—,x-Y'X- g 37t 571 X = —, X — —, X = —, 4 4 4 7n X iv amplitude=3,period=4Tt (0, 3), (271,-3) amplitude=1,period=7i, v=2'& = = =

11 amplitude=2,period =180°,(45,2),(135,-2), (225,2),(315,-2) iii amplitude=2,period =120°,(0,2),(60,-2),(120,2),(180,-2),(240,2),(300,-2),(360,2)

CambridgeIGCSEand0LevelAdditionalMathematics 12 4 a 13 b -A 12 5 a2VT3 13 3>/l3 13 6 ac 7 a 3 13 4 c 5 Exercise9.4 b 5 d V2 b -1? d -1 3 1 a1 amplitude=7, period=360°,(0,7), (180,-7),(360,7)

3 cos ox 11 a A

iv =amplitude=3,period720°,(180,3)

4 in4x 12 b 2 13 b 2 14 b 4 Exercise 9.5 1 a f(x) 0 b 0 « f(x) « 1 C 0 ^ f(x) ^ 3 d 0 ^ f(x) « 1 e 0 « f(x) « 2 f 0«f(x)«2 g 1^f(x)^3 h0«f(x)«6 i 0 ^ f(x) « 7

7t 77C '2'T e1.82,2.75 K 7k 6'T 5 a14.0°,194.0° b 126.9°,306.9° c31.0°,211.0° d25.7°,115.7°,205.7°,295.7° 6 0.648,2.22 7 a 0°,30°,180°,210°,360° b0°,38.7°,180°,218.7°,360° c70.5°,90°,270°,289.5° d0°,135°,180°,315°,360° e11.5°,90°,168.5°,270° f 0°,45°,180°,225°,360° 8 a30°,150°,210°,330° b31.0°,149.0°,211.0°,329.0° 9 a45°,108.4°,225°,288.4° b 30°,150°,270° c0°,109.5°,250.5°,360° d 60°,180°,300° fe0°,180°,199.5°,340.5°,360°19.5°,160.5°,210°,330° g 19.5°,160.5°,270° h30°,150°,270° i 19.5°,160.5° 100.848 Exercise9.8 1 a73.3°,253.3° b 75.5°,284.5° c210°,330° d 53.1°,306.9° 2 a0.201,2.94b0.896,4.04 c1.82,4.46 d2.55,5.70 3 a 25.7°,154.3° b 5.8°,84.2° c 67.5°,157.5° d 112.8°,157.2° 4 a100.5°,319.5° b 73.3°,151.7° d 2.56,5.70 e1.28,2.00 5 a 60°,120°,240°,300° b56.3°,123.7°,263.3°,303.7° c106.3°,253.7° 6 a41.4°,180°,318.6° cb113.6°,246.4°,71.6°,153.4°,251.6°,333.4° d 19.5°,160.5° e48.2°,311.8° f 18.4°,30°,150°,198.4° g60°,300° h23.6°,30°,150°,156.4° Exercise9.9 5 26 6 a-2+7sin^x b-2^f(x)«5 7a7-(cos0-2f b6,-2 Exercise9.10 1 bi 4 ii 60° 2 a10°,50° , n 57t b—, 6 6 3 b54.7°,125.3°,234.7°,305.3° 5 a164.1°,344.1° b0.898,1.67,4.04,4.81

2 c4 4 c 3 6 b5 3c2 75b2b2 8 b4 9 1<^<5 10 a-\,b-4:,c=2 Exercise9.6 1 a 17.5°,162.5° b41.4°,318.6° c63.4°,243.4° d 216.9°,323.1° e125.5°,305.5° f 233.1°,306.9° g48.6°,131.4° h120°,240° n 5n 2 a-, 3 3 b0.197,3.34 cnosolutions d 1.89,5.03 e3.99,5.44 f 2.15,4.13 g 0.253,2.89 h3.55,5.87 3 a26.6°,63.4° b63.4°,116.6° c31.7°,121.7° d108.4°,161.6° fe18.4°,161.6°98.3°,171.7° g 80.8°,170.8° h20.9°,69.1° 4 ca150°,270°b95.7°,129.3°225°,315°

CambridgeIGCSEand0LevelAdditionalMathematics 6 c3 8 b0.902,2.24 9 aa=S,b=8,c=7 b1 120° 11 5 10a30°,150°,210°,330°. b60°,180°,23.5°,96.5°,143.5° 271 57t ^ T'T 121 64.3°,154.3° li 90°,194.5°,345.5° 571 1371 III—, 12 12 Chapter10 Exercise10.1 1 a 5040 b 12 c 840 d 336 d 6 e 60 f 20 g 4200 2 a 2! b 6! c .5! 2: d 17! 13! e 10! 7!3: f 12: 7!4! nl (n-4)! n! ^(n-3)!5! Exercise10.2 b d {n—6)! nl (n-5)!3! 1Exercise10.3a6720 b360 c6652800 d5040 2 360 3 15120 4 336 5 480 6 18 7 a1680 b840 c630 d330 8 a60 b300 9Exercise72010.4 1 a5 b20 c1 d70 el f35 _8! 3!51 5!3! 3 a120 b21 c364 4 10 5 56 6 700 7 a210 b84 8 a30045015 b142506 9 67200 1 a24 b 5040 10 a 16 b 30 2 a120 b40320 11 a1287 b756 c3628800 12 1709 3 a24 b16 11 6 13 a 462 b 350 14 a45 b 45 4 a5040 b576 15 a 252 b 126 5 a 120 b 72 c 72 d42 16 2000 6 a12 b 36 c 24 Exercise10.5 7 a720 b 120 c 48 1 a 28 b1 420 11 8 a 600 b312 9 288 2 a1 362880011 3 ai 15120 11210 bi 15504 113696 ill56 4 a720 b240 c480 d168 5 ai40320 112880 b350 6 a840 b 240 c 80 7 ai118415120 b240 8 a permutationbecausethe ordermatters b155 11420 ill70 9 ai28 1!20160 Hi720 b203 10ai60480 11144 ill1680 b12100 11420 Chapter11 Exercise11.1 1 1,6,15,20,15,6,1 1,7,21,35,35,21,7,1 2 a1+3x+3x'+x'b1-4x+6x"-4x'+x** c p'^+ +^P^q'+'^pq^+q* d8+12x+Ox'^+x' e x°+5x^31+lOx^y*'+lOx^y +5xy+3° f +I2y-+4831+64 g -3a'b+3ab^-b^ h16x''+32x"''3+24x^3*'^+8x3' 1 x'-6x"3+12x3^-83' j 81x'-432x'+864x^-768x +256 kX'+6x+—H— X X

I 6 ^ 1^—4 ^ -9 IX xH—X X 3 2' 4 b 10a 16 d8 e40 g5760 h -i 2 8c-12 f160 4 5 67 A=2048, 1280,C=40 1+14x+74x'+216x'+297x' +162x' a=8 a81+108x+54x'+12x'+x' b 376+168n/5 8 a l+5x+10x'^+10x''+5x''+x'' bi 76+4473 ii76-4473 c152 9 a16-32x'+24x'-8x'+x" b64 10 90 11 i 4 12a32+80>'+80/ b400 13 5 14aa=24,b=128 b 8072 15 a -3y b 3)® — 5y^ +5y Exercise11.2 1 a^Co ^Ci ^C2 ^Ca b4/^ 4^4/^ 4r^ 4c v_^0 ^2 ^3^4 -5c 5/^ 5c 5/^C *...0 *^2 ^5 ^4 ^5 2 a1+4x+6x^+4x''*+x'' b l-5x+10x'-10x'+5x'-x' c 1+8x+24x^+32x'+ 16x^ d27+27x+9x-+x' ex^+4x""'3+6x"3"+4xy'^+y* f 32-80x+80x'-40x'+lOx' 32a6' 16 2 3 X V v't\ vV J 1 h ' 2 10 100 2000 100000 kX'-15x'+90x 270^405 243 X'' I x'"+3x*'+ 15 -X + 5 _1^ 2 16x' 3 1 + r+ 16x' 64x' a40x' b175000x'' c160x''d720x' e -20x'' f -5376x"' g -9450000x' h -954204160000X-' a 1+lOx+45x^ bl+16x+112x" cl-21x+189x' ed729+2916x+4860x'19683-59049x+78732x' f256+512x+448x^ g1953125-3515625x' +2812500x' h 1048576x"'-13107200x"3 +73728000xY a 1+12x+60x^+160x'' b140 ,13 392 1433 6 aiH xH X H X 2 2 4 g —Sa^h+24a'b^ +16// h 16x'' +96x7'+216x7" + 216x3''+81>'' • 1 4 3^ 27, I —X—X-I x"-54x+81 ^377 b 4 7 a1-30x+405x--3240x' b-4860 8 a1+14x+84x^ b47 9 a 1+7x+21x"+35x' 10b7-9452 3 11 12 a= n-2 Exercise11.3 1 a+4d,a+I3d 2 a765 b-1310 c907.5 d-2420x 3 a35,3185 b15,-1365 4 9 5 a5,26 b 1287 6 -1875 7 2985 8 40 9 12450 10 45 11 $236 12 Si OC Hr37 13 6,8 14 -5,-6 15 1t—'IInO 00 + »1— 16 8° 17 a a-6d b10a 19 a6—5cos^ X 20 b901

CambridgeIGCSEand'OLevelAdditionalMathematics Exercise11.4 1 anotgeometric b -4,16384 1 1 c 327 dnotgeometric enotgeometric f -1,5 32 45 ar®,ar'® ±16.2 ^,8 2 2,—,40 64 27 -12 22nd 10 19 11a1020 b 10933 c-3280 d-4166.656 12 14 13 a logj b67.232m 1440,-20 1550minutes19seconds 182(10"^'-10-9n) 27 Exercise11.5 2 3 45 a4.5 c 5010 -1.250 2I 9 d-97.2 42 42 0.42= + 100 10000 + 42 6 -93^ 3 3 7 0.5,13 8 a -0.75,128 b737 9 a105 b437.5 10a—,60 10 b857 11 a 60 b375 12162 13a=50,r= 5 14 <X< 6 6 15 84m 16a3x,4x,—X,—X 3 9 commonratio>1 17 a 571 b ilH Exercise11.6 1 a108 b166.25 2 a125 b31 3 a 1.5 b364.5,78 Exercise11.7 1 a 64+192x+240x^+160x® b64 b72 b-5 bb38.5-640 ax®+12x"^+60 a n=6 a-27.5 a64-960x+6000x' 729+2916x+4860x2 i6804 64+192x2+240x^+160x® i1072 8 ai +4:a^b+6a^b^ +4ab^+6"* 24 It 25 bn=6 9 a-2.24 b25 10a24 b162 11a =2a!b99a 12aa=38,d=-2.5 3 13 b768 bn=25 a 4 14 a 41 15a a 16 a 5 Chapter12 Exercise12.1 b r= d= 1 ,6 '7 2 c 438S=10b80 b18 1000000 +... 4 12.5,63.5 ^ a 4x'^ b 9x' c T OC 5 3 aT b84.375,162.5 d -6x~' e —X~2 f -5x~'' 4 1 4 —X 2 52 —X 2 1 c 800 g h i X 5 6 7 a0.75 19 b-,n=10 4 j -x"^ 3 k -X2 » 3 i 1 X 2 8 a m1 n 3i —X 0 5^ —X X=—3 or8,3rdterm=33or88 2 3 b -2 -7I P 6x' q 3x2 r 2x 5' 7 s 1 X 2 t 3-f X^ 2

2 a6x--5 cb40x^—6x-6x'^+4 d6x-2x''^+2x~^ 1 e2+x""^+-X 2 1-- 5 f—x'^ X 2 2 g1+3x"'^ 1 h5+-X 2 3- 1-i 1 i-X---X +-X 2 2 2 j15x-+lOx k-2x"-+lOx"' I 1+2x~^ m18x+6 n-6x^+6x'' 0 12x+5 3 a6 b4 c-2 2d0 e-0.2 f 9 4 (2,5) 5 0.25 6 5 7 -11,11 8 a=3,i=—4 9 a=-2,b=5 11 a (-3,2),(0,5),(9, 14) 12b13,-8,37a12x^+6x—6bX^-1andx 0.5 13a3x"+2x-16 b—^X^2 3 "^= 4(»'-2)'+(« +5)9>0 Exercise12.2 1 a 9(x+2)' b 21(3x-l)® c -30(1-5x)' d2(ix-7 U e 4(2x+lf f 12(x-4f g -30(5-x)' h 8(2x+5)' i 8x(x^ + 2)^ j -28x(l-2x-f k5(2x-3)(x'-3x)' 2 a 1 d e (x+iy 6 (2x - 1)^' 15 (2-3x)64x f (2x"-5)" 8(x-1) (x'-2x)" 5 gh (x-1)" 30 3 a (5x+1)^ 6 (3x-2)' 1 d 2Vx+2 5 2V5x-1 2x V2x'-3 3x-+2 e 2Vx''+2x 2 3(3—2x)3 g \l2x-1 3 2(3x-l)5 h—5-^ (2-5x).'' 4 8 5 2 6 0.75,-3 7 (3,2) 8 Exercisea=8,b=S12.3 1 a2x+4 b12x+10 c (x+2)"(4x+2) dx(x-1)^(5x-2) 3x-10 2n/x-5 ^ 3x+2 2yfx 5x''+12x gh 2 4 2Vx+3 (3- x'")^(l3x^ -3) 2%/x 1 2(3x^ + X+5) j (4x+9)(x-3)~ k 2(x-l)(x+2)(2x+l) I14(x-l)(x-3f(2x+1)^ 9 385 49,0 5-2,0,1.5 2 6 13 Exercise12.4 1 a 11 h 10 (5-x)"' (x+4)~ 7 H 1 (3x+Af 00 OC x(5x — 4) f 9 X*+1 (5x-2)" (x- - 1)'

Cambridge IGCSE and 0Level Additional Mathematics 15 +8x+2 ® ~(3x-l)' ^ 2 -4 3(0,0),(1,1) 4 ^ 9 5 a 1-2x 2\/x[2x+1)^ 1-X b ^ (1-2x)i x{x^+4) C ^ r (x"+2)2 ^ -5(x-3) 2-Jx(x+3)^ 6 4 7 (3,-2) 8 a(-2,-2.4),(0.4,0),(2,1.6) b 25'29'25 Exercise12.5 1 a cb3)=4x-6y=-X-2y=16x-10 2 e y——3x—31olfy=—X+2 4 4 1 ,1 2 a y=--x-4-■^ 3 3 1b y=--x+5c y=-x-3d31=2x+7.563=—O.lx—3.8 f 3=4x-22 1 17 3 3=8x-6, 3=--:x:+ 4 (0,5.2) 5 9 1 "V = X , 16 2 6 (2.-3) 7 3=2x-20 8 a 3 = X + 8 b(1,6) 1 C "V = —X+2 13 2 9 (1,5.25) 10 a (7,4) b 12units'^ 11 b 3=-0.4x -0.6 12 216units^ 13 22.5units^ 9 ■ 1 2 4 1 u 113 a —cms' b —cms 5 5n Exercise12.6 1 0.21 2 0.68 3 -0.8p 4 2p 6 3^5 25p 7 n 1808 a , = ^ c -254jb,decrease Exercise12.7 1 0.15unitspersecond 2 0.125unitspersecond 3-4unitspersecond 4 -0.25unitspersecond 5 0.5unitspersecond 6 unitspersecond 150 7 -0.08unitspersecond 8 0.025cms"' 9 —cms"^ 96 10 324cm®s~' 11 cms"^ 480 12 12Ttcm®s-^ -114 a - cms Exercise712.81a10 18 ^ -7 1 ?! K1 -1b —cms12 b 12x+6 d 320(4x + l)' f^ (2x+1)2 (x+3)2 2 a 12(x-4)(x-2) 8x - 6 8 (x-sy 2(x"^ + 6x^ + 3x + 2) 50 e 3 : f 3 (x - 5)^ 102 (3x -1)' 3 a -3 b -9 c -8 4 b -18, 18 X 0 1 2 3 4 5 dy dx + 0 0 + + + + 6 x>2 Exercise 12.9 1 a (6,-28) minimum b (-2, 9) maximum c (-2, 18)maximum, (2,-14)minimum^222^ d -2-, 14— maximum,V327j (2,-36) minimum fl 14 e (-3,-18) minimum, I ^ maximum

f 2 14 3'27 minimum a(4,4)minimum b(-1,3)minimum c(4,3)minimum 1 maxim 3 um,(4,-18)3a3)= ,—Iminimum, 3. (^■5) 3, — I maximum e(-2,-4)maximum,(0,0)minimum f (-4,-13) maximum, (2, -1) minimum dj) dx {x+1)Y,numeratorof dy IS never zero dx 4 «=3 5 a a=-3,b=b b minimum c (-1,7), maximum 6 a a=2,b=-4 b minimum 7 a a=8,b=-4 b maximum 8 a a = -36, b = 4 b (-2,48) c(-2,48)=maximum, (3, -77) = minimum d (0.5,-14.5),-37.5 Exercise 12.10 1 a 3=8-X b i P = 8x 11 16 c1S=x^+(8-x)^ ii 32 2 b A = 1250,X = 50 288 X' c A=432,12cmby6cmby V 8cm 4 a h = 4x' ^ 3 a e= r C 30-2r,-2 d 15 e 225,maximum a = 3--r(7t+2) 2 c 6—4r—nr,—4—7t 6 4+Tt18 , maximum 4+n 7 a 4-p2S c 3 32V3 d ,maximum 8 a h250 c 47ir 50071 , 4Tt + IOOOtt d 5 e 1507T,minimum 144 3 9 a h = r r 2 12VTo 10 a r 25-2x 71 c A=87.5,X=7.00 11 a r = V255V3 3 d maximum 12a h=24-2r c 51271 13 a r = ^J20h c 13-3 ^ 32000 d 7t, maximum 81 Exercise12.11 1 (2, 12) 2 k=4 1 13 a + X vx X 1 2x'-^ c (1, 3), minimum 4 b 56.25 cm^ 5 a 2 b -256/7 6 y-—X+10 2 7 ay- -3x+4 b (-2, 10) 9 x=9 10 b x=10 11i!A=246,minimum Chapter13 Exercise 13.1 1 a i-3j b 3i-2j c 41-j d 2i e 1 + 3j f -21 + j g-21h-31-2ji-1-j 2 a 2 b 5 c 13 d 10 e 25 f 17 g 4V2 h 5V5 3 161+12j 4 361- 15j 5 a ^(3i+4j) b ^(5i+12j)

CambridgeIGCSEand0LevelAdditionalMathematics c -^(4i+ 3j)5 d ^(8i-15j) a 4i b 14i-9j c-26i-3j d -i+3j a15 b /X--3,A=5 //=3,A=-2 Exercise13.2 1 a-i-3j c3i+j e i+7j 104 db2i-10j-121+3j f-61-j a-151+20j b241+lOj c 151+j d V226 al6i+12j b^(4i+3j) 5 c51+2ja10i+24j b26 c ^(5i+ 12j) d 31 +8j 15 6 7 8 -2120 a1 29 1130 ill38.1 ■ (■") 9 a 121+9j 10 a 61-8j 11 M vJ 12 3 b 71+j b7.51+4j 13 a 22 or -8 b -9 c 1 14 6j 20. " y 16 a i 2V1O I!^/^M ill S-M ,7 2 c A = -, u = 3 3 Exercise13.3 1 a (l-A)a + Ab b -/xa + (l-/x)b + 3/xb ,3 1 c A = -, u = 5 5 2 a 1 2a - b ii 3a + b b i A(5a—b) ii ^(3a+b) , 5 3 cA=-,u= 8 8 3 a A(a+2b) b /xa + (3-3/x)b , 3 3 c A= u= 5^ 5 4 a (l-A)a + 2Ab b ya +(l-/x)b D c A= =; 5 a i -a+b 1 1. II -a+-b 2 2 b -Aa + -Ab 2 2 c ^/xa+(l-A/)b Jib 4 d A--, /i = 3 2 a - Aa + - Ab 5 5 c A 2b ^ ^ 31' ^ ~ 31 7 a 1 Aa-b ii -a + ^b 2 3 b i -Aa + -Ab 5 5 .. 1 3 . II —a+-ub 4 4 5 4 IIIA=-,u= 8^ 5 8 a 1 -9a+18b ii -5a+10b b AC = — AB, so AB and AC 5 areparallelandAlieson bothlines 9 ai 4a ii 2b iii -2a+3b b i 6a - 3b ii 2a - b iii 4a-2b bCE=?>CD,soCEandCDare parallelandCliesonboth lines d 1:2 Exercise13.4 1 a (3.51 + 9j) ms"' b (301 —36j)m c 12.5hours 2 (-22i + 11.6j)kmh-^ 3 1251km 4 a (181 + 18j)kmh'"' b (101 + 10^/3j)kmh"' c (-50^/3i - 50j)ms ^ 5 a20ms"' b i (-68i + 44j)m ii (-561 + 28j) m

iii(-44i+12j)m c r alOkmh"' b(28i+14j)kin c r = d2030 7a(I2i-12j)kmh-ibi(29i-12j)kmii(l4i+3j)km c r = i ']^ti(l2j l^-12j 8a(lOi+6j)kmb(5i+12j)kmh-i c13kmh"' d52 9a(I5i+20j)kmb(8i+6j)kmh"' 10c(llli+92j)kma(50i+70j)km, (40i+100j)km b31.6km 11 a (6i+8j)km,21jkm b14.3km Exercise13.5 1 a 9i+45j b13 c-i-2j 3 •• 3 4 a /.(fa +fb) (5i+12j)km,13km i 8i-15j ii ^(8i-15j) iii-531 5 a sVs b ^~(2l+ llj) c2i+1.5j 6 i 51kmh-^ ii40minutes 7 i Ab-a iifxa-b 2 1 iii-a+-Ab 3 3 • Ik 7 IV-b+-ua 8 8 16, 3 Vn=—,A= 21 8 8 i AP=—(b-a) 4 —• 1 3 2 iiPQ=—-a--b+-c 4 4 5 iiic=—(9b-5a) 16 9 ai a=2,y3=13 ii ^(31-4j) 5 b^=(1-A)a + Ab c//=3 k ^ ^ I 10x:(l + e'') 2 axe'+e' b2xV+2xe" c-3xe"'+3e"'e' dV^e'+ g {4x-1)

3 2x'2e' "(e'-lf h2xe'' e' + 2xe'+ 2x +5) 3 a5x+83)=10b5x-6)1+15In5=0 cy—3ex+2x-2e-1 4 A 5 a h'-i)'minimum by-e (x-1) 2e ^ (2e^ +1)^ 4e Exercise14.2 1 a i b i Chapter14 c 1Exercise14.1a7e'' b 3e'' e c15e'' d K v' 00 X e-3e"2 f 3e3"i g g2xe''''' h 2^3e^ i i -3e-''' j -4e^' k 2x+3 3x6+1 20 5x—2 3 3x—2 1 2x x^-1 1 2x+4 1 h2-X J (Vx+l)Vx x(x^ + lnx) 2 a1+lnx b2x+4xlnx c1 l-lnx d10+51nx^ X +2xln(lnx) I 1 XInX 2x'-^+1 2-Jx nX

CambridgeIGCSEand0LevelAdditionalMathematics ,l-ln2x 4 f 5 gX' x(lnx)^ 2 21n(2x+l) x^(2x+1) X 3x^(2x+3)-2(x''-l)ln(x''-l) (2x +3f(x'-l) 3 2+41n6,3+21n6 3 4 a 6x+225-2x 5(x-1) x(x-5) -3 g h i 5 a b (2x+l)(x-1) x-4 x(2-x) x^+4x+2 x(x+l)(x+2) 2-6x-2x^ (x-5)(x+l)(2x+3) —3x-1 (x+3)(x—1) -x^+6x-3 x(x—l)(x+l)(2x-3) X1ln3 2 Xln2 6 a b (5x-1)ln4 8x 4x'-1 15x'-2 5x'-2x 2x—1 (x+3)(x—4) 7 0.2 Exercise14.3 1 a cosX b2cosx-3sinx c-2sinX-sec^x d6cos2x e20sec^5x f-6sin3x-2cos2x g 3sec^(3x+2) h 2cos| 2x + ^ -2 (sinx-cosx) 1 -6sin 2 a3sinxcosx b-30cos3xsin3x c2sinXcosx+2sinx d 4(3-cosxf sinx COS!2xH— 6 + e 12sin^^2x + f-12sinxcos'X 8tan|^2x-^jsec^^2x- —j 3 aXcosX+sinX b-6sin2xsin3x +4cos2xcos3x c x^sec^X+2xtanx 3 2' d —xtanÎ — Isec^ 3 X +tan'| fe15tan3xsec3xsecx+xtanxsecx g X sec X - tan x 1+2cosX (2+cosx) .(3x-l)cosx-3sinx ' (3x -1)' j-6cot2xcosec'2x k3cosec2x-6xcot2x cosec2x 4 a-sinXe b-5sin5xe"^' c sec'xe*^"' d (cosX-sinx) e e"(sinx+cosx) f —e*I2cos 2 1 . 1 ^ — X - sin — X 2 2) g20"cosx h xe''°''(2-xsinx) 1 cotx j x[21n(cosx)-xtanx] ,3cos3x-2sin3x I ,2x-I XCOSX+sinx-xsinx 5 a-2 ^3V2-8 6 6 a tanxsecx b-cotxcosecx c-cosec^X 7 a3cot3x b-2tan2x 8 A=%B=-5 9 A=4,B=-S Exercise14.4 1 4x-2431-n 2 a2x+23=n ba(f.o).«(o,=) 71^ . 2 c — units 8 3 Q(l,0) 4 a A(îln5,oj,fi(0,4) bC(-8,0) 5 ê(2e^+l)

6 V2/) 7 ^ \p ^ -P4^ 2 102^/3p 11 a (-2,-2e'')minimum b (-1, maximum, (0,0)minimum c(ln7,9-7ln7)minimum d(0,4)minimum e (4,^-2,-4e^)minimum,8e-^)maximum 1f |e g ' minimum maximum h (-^/e - 1, e"') maximum, (0,0)minimum, (Ve- 1, maximum 12 a (0.927,5)maximum b(1.85,10)maximum c(0,5)maximum, (5.-5)mi. d ,3.10jmilminimumminimum 5n SjS minimum .6' 2 13 a A=2,B=8 b ln4,8^ minimum 14 A(l,0),fi(e-',-e-') 15 a />(0,0),Q(-2,4e-^) b B(0,-2e),c(^0,e+^j c -("36 + —1 21 3ej 16 b 2.7cm^ persecond 17—maximum o Exercise14.5 1 a)i=2x-1 a—cos2xcos 3 -2sin2xsin (f) Kf) 1 43 a (sec^x)(l+Inx) (tanx) ^ (l+ lnx)'"" -0.8 2xcos4x+4x^sin4xcos^4x u ^ b p 2^ 5 a2e''-2e-'' cb2.5-7.25 6 a ^(l+21n2,0), B\0,-+ln2 c(l-iln2,o). Z)(0,-2+ln2) b ii-(In2)' 471 7 aah 9 1971b2^= X 6 8 1^=dji_2xsec^2x-tan2x dx x^ .2 .. 8

71^ / n\ II y = —X71 32(71-2)1 8/ 9i3)+ln3=-^xii—(ln3)2 ln3 2 101 8+ln9 ii0.0393 11i (-1,-0.5),(1,0.5) iip=2,q=-6,(-1,-0.5) minimum,(1,0.5)maximum 12 i 2rsin0 iii -17.8 iv-0.842 Chapter15 Exercise15.1 125 1 a 31=—X +c ^ 5 —X 9 +c 74 —X 4 + c x^ + c 1 2x +c 6-s/x + c x'+ 2x® 5+3x+c X® T~ 3x' 4 5x' + + c 2 1 x'' 15 + X x^ + — + c 2 2 X® 1 X® 2x+c c^= d 31== X 3 a 31=x^-3x^+c 4b 31=-x^-x'+c 5 8f x'„, c3=-x^+—+2x+c 5 3 4 % X X ^ d 3=—+ 6x +c 4 3 x' 3 ,ey-—+—+C62x251 f y= + X 2x X 3 6- 9x^ h y= 20Vx+c 5 3 4 3=x^-2x^+X+5 5 3=2x'-3x^-4 o 6 „6 3=x X+32S 7 3=8y/x—4xH—x^+8 3

CambridgeIGCSEand0LevelAdditionalMathematics ,2* 8 y=2x'- +5 9 a y='2.x —6x^+8x+1 10))=2x^-5x+1 Exercise15.2 1 d Hc 8 • -f c b2x®+c 4 d X e6\/x+c x' 7x' f +c cVx + c 2 a 1- 1-lOx+c 3 2 U 2x' x' ^ b i 3x+c 3 2 ^ - 2 OC c ox+25x+c 3 x'' d4x^^ 1-9x+c 2 x" 2x' x' e 1 1-c 4 3 2 3x' f —-3x-^+c7 3 a X-I he X ^ x' 2 4 X _J 1 1_ 3x 3x^ 9x d 2x^-6>yx+c + c 2x2 2 + c ^ x' 6 9 f r+c 2 X 4x Exercise15.3 1 a —(x+2)'"+c 10 b—(2x-5)'+e 14^ c —(3x+2)'°+c 15 d -^(2-3x)'+ c 3 1 e —(7x+2)3+c 28^ ' f—(3x-1)2+c 15 g12Vx+1+c h 5(5x+3)' 1 + c 2 31=4(3-2xf(4x+1)® 20 + c -2 3 3 4 3=5-2VIO-X 5 3=(2x—3)^+1 (3x-1)® +86 ^ = ^ -1 Exercise15.4 e®' 1 a —+c 5 _\ c2e2+c e4e*+c ,9* b —+c 9 +c g ^7i+4 T 18 + c + c *• f —+c 2 ^5-2* h ■ + c 2 a5e' + e •C 9v b he +X+c 9e'' e'"' 2 2 de"-4e-'+c +6x+c he 4 2 f —-2e''-4e-'+e 5 3a2e'+2>/x+e u 1 s Se''"-'b —X' he 3 2 1 c — + he 4 3x 4 3= -e"''+4 5 ak=-n b3=7e'-^+2x'-5 6 3= 20"^'+6x — 6 7 a 6-e '4e'-e-2 24e-4e'-21 2e-2' 4e-4 Exercise15.5 1 a-—cos4x+e 4 b—sin2x+c 2 X c-3cos—+c3dsin2x+e e-2cos3x+e f —sin(2x+l)+e g ^cos(2-3x)+ e h sin(2x-7)+e 4 i —cos(l—5x)+e 5 2 a x+cosx+e b—x^—sin3xI+e 31 3 2 c —sin2x+—71cos he 2 5 2 ^ 1 2 3x d sm he X 3 2

e +5cos2x)+ c f 4>/x-2cos—+c 2 3 31=sinX+cosx+2 4 ■y = x- 2sin2;x: + 34 5 y = 2x^ + 3cos2x - 5 6 3 = 5cos3x—2sinx+A 7 a ^=6 b31-2sin3x-4x+2+47t n'8 a 3) = 5-2cos 2x by--—X+—+5-\/S ^ 26 9 ak=2 '5n-2 ,3] Exercise 15.6 1 a81nx+c b 51nx+c c —InX+c2 d —InX+c 3 e — ln(3x + 2) + c f -—ln(l-8x)+c 7 g — ln(2x-l)+c h -—ln(2-3x)+c 1 ln(5x -1)+ c 7x^ 2 a l-x+21nx+c 2 b X 1-4Inx+c X 9 c 25x 301nx+c X 1 2x'' 51nx +c f 3x^ r-121nx + c 3x3 g5InX f=-+c vx u 2x2h Inx + c =.3.v Inx + c 3 3=—ln(2x—1)+3 4 3 = x^ + 51nx-5 5 3=1+ln(x+e) 6 a l-21n2 b(21n2,21n2-3) Exercise15.7 1 , X + 5 b +c d X+InX+c yj2x-l 2 a 30x(3x^-l)"' b -^(3x^ -1)° +c 30^ ' 3 a Inx+1 b xlnx—x+c ii 1 Inx4 b +c X X 2(x" - 2)5 a -4—=^ bVx^—4—x-\/x^-4+c 2 2 6 b—(x+l)>/x—5+ 3 7 a 2xe'' b -xe^" - + c 2 4 8 a ^ = 1 , 5sinX b +c 1—cosX 9 a k = b —(x+8)n/x-4+c 3 10 a k=-2 2 b x' -7 + c 11a 2x"+6xInx u 1 3, 1 3 b — X" inX x +c 3 9 12 a cosX-Xsinx b sinx—xcosx+c 13 b (sin2x+cos2x)+c 14 a 5x'-14x b (x' +3)V2x-7 Exercise15.8127b 1.5 c 24 15 e 10.5 f 7.5 5 12— h 12 52 81 i 75.5 26.4 k 3!^ 24 I 5! 3 68 b 18-3 c 12.4 0.5 e 0.25 f 8 -1) b i(e-l) 5(e<-l) 2e' d 5(e^-l) 2e 1f -(e^ + 4e-3) g ^(3e'+8e'+6e'-17) h 2 i 4e' 4 a2 " e' 2 3 5 2e' 2 371

CambridgeIGCSEand0LevelAdditionalMathematics n^-8 16 5 a-ln2 3 c —In5 2 d V3-| f i(3-72) b -Inll 2 d —In32 2,11 ,c 9 e--In— f21n 3 2 7 6 a 2+|lny b ln9-10 c2.iln^ 2 3 7 k=5 8 A=-6 9 a quotient=2jc+1, remainder=-1 102--ln3 4 Exercise15.9 Sx 1 a V2x-1 b 5 6x^+4 2 a , b4 V 3 a4 a 3x^+4 2x 1 (x'+5 f 9 3x+2 5 b 2^Ji8x+4f 71V2 b2 6 a sinX+Xcosx+c b-(71-2) 2 7 ax+2xlnx+c b 1+e' 8 a sin3x+3xcos3x b—(n-2) 18 Exercise15.10 1 a 6ec65 2 a1 1 b 132 d 1+e-e f ^ " 43 b5>/2-5 c41n2 dln5 _1 57T^ 72 12 1 a 6 c 8 5 b 206 d 49 1 f213 5 6 78 e324 12 a9 b 1+e^ k=-(e"+1) 2 b1.6 9 10b51n5-4b71-1 Exercise15.11 1 2 2 3-^ 3 16 4 5 371 5 a 10d36 6 a i 3 7 34 b 20 e 20 b 5 4 8 a(-2,0)b 429 4-15 c 110 71 35 11 61n6 2 Exercise15.12 1 a2xe' h b—e 2 c26.8 2 ay- 2x^- ■X1+1 b8x+343!=93 +1 3 a X-t he X 4 b628units^ 5 a (0,1.85) 6 In(25e^) 7 a A=4 8x3 8 bk=2 b0.292units^ a 2x+l +12x2ln(2x+l) b11 10 + c Vx+2 9 i -4xe'^* ii41n216 .32 2I 10I —X^ x2+C 2 5 ii (9,0) Chapter16 Exercise16.1 III 40 iv16.2 b 10ms2 a12.5ms"^ c-2.4ms 2 a 6ms a4.95ms b5 c 1ms 4 a -(e''-2) b 5.4ms

5 a2 c -0.25ms^ 471 6 a 3 7 a / bln|—Im b-1.90ms^ b -3ms~^ 8 a u=4sin2i,fit=8cos bi=—,5=4 2 1 9 ^t=-ln2 2 b0.899m c 24.1m 10a2mawayfromO bv=2-4sin2<,a=-8cos2< Tt Tt Clt——,S= 4 2 11 a < 71 8 ba=-4Asin4i c k=5 12 a 2ms-' b i 2ms-' II — ms 3 0ms 2 e21n—m 4 13a2m bi= c0^i<—andt>4 3 d-<t<4 3 Exercise16.2 1 alOms"' b15seconds c12—m d166—m 3 3 e189-m 3 2 a-Ims"^ b 5 c 1.6m 3 a 2(4e^+1)m bj=2e^'+ -2 C 111m mt t+2 ct=6 97t' 14I +6Ini,^ms^ 5 b 29:y5ms-^ c 109m 6 p=6,q--6 7 at=5 b c62.5m 8 at=2,t=6 b5.5m C 112m 9a-12«a^12b5.74m 10a0.64ms-^ b12.1m 11 27111 a t= b (4VS.f]m "a P = ^'q = -^ bs= +3t 12 d —m 12 27T/ nt ^ nt 13a77= cos h2sm—, 3 3 3 27t'''i . nt a = sm h 9 3 471 nt —cos 3 3 Exercise16.3 1 a3.2 b15 32 54 n6 a2 d4 a 5 4 777 24 c312 c 1 c 1.14 t=1 2 3' ii-2 i -t^-7t^+l2t 3 17 ms 120 135 ii A=30,a=1 b50m.s~' t-S.5 8 ii0 iii37®-—72+907 2 iva78b88.5

absolutevalueseemodulus acceleration405,412-13,418 signsfor 405-6 addition ofpolynomials71-2 ofvectors316 addition/subtractionrule,fordifferentiation277, 310 amplitude 196,198-9 angles between0°and90°187-90 approximationsgeneraldefinition190-1negative190,220positive190,220291-3,310 arcs174 length 174-6 areas boundedbygraphsoftwofunctions 393,398 rectilinearfigures 148-51 regionsboundedbylineandcurve 392-6 sector177-80 undercurves385-92,397 arithmetic progressions 252-7 commondifference252 formulafornthterm252 notation252 sum253-5 arithmeticseries253-5,267-9 arrangements226-9 baserulefor22650 changeof124-6 binomial244 binomialexpansions249-51,269 binomialtheorem249-51,269 boxinginmethod149,150Cartesianplane190 chainrule280-2,310 changeofbaserule 124-6 chords174 gradients of 275-6,291 circularmeasure170-80 collinearpoints316

Index

CambridgeIGCSEand0LevelAdditionalMathematics

combinations232-7 orderand232-3 ruleforfinding233 commondifference252 commonratio257 completingthesquaremethod 25,29-30 composite functions 5-7 conjugatesurds60 convergentseries263 cosecantfunction(cosec)216-17cosinefunction(cos)187,192 amplitude196,199 derivativeof345-50 integrationgraphsof195-6,199-200,202of370-2 period196,199 signof192-3,220 cotangentfunction(cot)216-17 cubicequations78-82 cubicinequalities102-3 cubicpolynomials71,78-82,98-101 curve areaofregionboundedbyline 392-6 areaunder385-92,397 intersectionoflineand42-4 degree,definiteintegrals378-85,397ofpolynomial71

degrees 171 convertingfrom radians to 172 convertingtoradiansfrom172 denominator,rationalising60-3 derivatives276 ofcos(ax+V) 347 ofe^^'''338-9 ofexponentialfunctions337-40,350-6 first298 oflnf(x)341-2 oflogarithmicfunctions341-5,350-6 second298-9 andstationarypoints302-3 ofsin(ax+ft)347 oftan(ax+ft)347 oftrigonometricfunctions345-56 seealsodifferentiation

differentiation 274-311,336-56 addition/subtraction rule 277,310 applicationsinkinematicsof 404-12 approximations 291-3,310 chainrule280-2,310 firstderivative298 fromfirstprinciples276 notation276 powerrule 276-80,310 productrule282-4,310 quotientrule 285-7,310 ratesofchange 294-8 connected295 reversed,integration as 362-5,372,375,397 scalar multiple rule 277,310 secondderivatives298-9 andstationarypoints302-3 seealsoderivatives;stationarypoints discriminant39—40 displacement316,404,412-13,418 signsfor 405-6 distance403 distance-timegraphs403 dividend73 division ofpolynomials73-5 ofsurds58 divisionlaw 118 divisor73 domain3,12 e126,338 equationseliminationmethod24 cubic78-82 involvingindices seeexponentialequations involving surds 63-6 logarithmic 120-1 modulus90-4 ofquadraticseequadraticequationsroots39type|ax+^>1=|cx+d|90-4

factortheorem75-8,86 factorialnotation225 factorisationmethod25 factors,ofpolynomials 75-86 Ferriswheel195 fractions,rationalisingdenominator60-3

'equivalentto'symbol90 exponentialequationspracticalapplications128-9 solving51-4,122-3 exponentialfunctions111-35 derivativesof337-40,350-6 graphsof 129-32 integrationof368-70 inverseof133-4 exponentsseeindices

geometricprogressions257-62 commonratio257 formulafornthterm257 notation257 sum259-60 geometricseries262-9,270 conditionforconvergence263,270 infinite262-7 gradientgeometry,ofvectors323-7 curve275-6 line139,275 parallellines139,143-5 perpendicularlines 139,143-5 gradientfunction275-80 graphs 89-106 ofcubicpolynomialsandtheirmoduli98-101 ofexponentialfunctions129-32 offunctionsandinverses15-18 oflogarithmicfunctions129-32 insolvingcubicinequalities102-3 in solving equations oftype \ax+ b\ = \cx= d\ 91,92 in solving modulus inequalities 94-8 straight-lines seestraight-line graphs oftrigonometricfunctions/ratios195-205 ofy=ofy=acosbx+c196,199asin+c196,199

functions 1-18 composite 5-7 definition3 domain3,12 exponentialseeexponentialfunctions gradient 275-80 inverse12-15 ofexponentialfunctions133-4 graphsof 15-18 oflogarithmicfunctions133-4 logarithmicseelogarithmicfunctions modulus7-9 periodic196 range 3,12 trigonometricseetrigonometricfunctions

of3)=asinX197oiy=atanbx-\rc196,199

of3;=cos X 195-6 of:y=e'=129 oiy=ke"'+a130-2 oiy=k\nax+b130-2of31=sinbx197-8 of31=sinX195-6 of31=sin x+c 198 of3=tan X 196 of3=lf(x)l wheref(x)islinear10-12 wheref(x)isquadratic34-7 where f(x) is trigonometricfunction 205-7 imagesetseerange indefinite integrals 365-7,375-8,397 indices49-54 rulesof50,67 simplif)dng expressions involving 50-1 solving equations involving 51-4,122-3 inequalities cubic102-3linear37modulus94-8 quadratic37-9 inflexion,pointsof301-2 instantaneousrest406,418 integration361-98 applicationsin kinematicsof 412-18 area ofregions bounded by line and curve 392-6 areaundercurve385-92,397 areasboundedbygraphsoftwofunctions393, 398 ofcosinefunctions370-2definite378-85,397 as differentiation reversed 362-5,372,375,397 ofexponentialfunctions 368-70 formulaefor397 1 372offunctionsofform {ax+b) -5,380-1 1 offunctionsofform 372-5,380,388-9 offunctionsofform(ax:+6)"367-8 indefinite365-7,375-8,397 ofpowersofx365 ofsinefunctions370-2 symbolfor365

inversefunctions12-15 ofexponentialfunctions133-4 graphsof15-18 oflogarithmicfunctions133-4 kinematics 402-18 applicationsofdifferentiation in 404-12 applicationsofintegrationin 412-18 lawsoflogarithms118-19,134 length linearc174—6139 problemsinvolving140-3 linearlinearinequalities37polynomials71 line(s) equations ofstraight 139,145-8 gradient 139,275 intersectionofcurveand42-4 length 139 mid-point 139 parallel139,143—5 perpendicular139,143-5 problemsinvolvinglengthandmidpoint140-3

Cambridge IGCSE and0Level Additional Mathematics

see alsostraight-line graphs logarithmic equations,solving 120-1 logarithmic functions 111-35 changeofbaseof124—6 derivativesof341-5,350-6 graphsof129-32 inverseof 133-4 laws/rulesoflogarithms118-19,134naturallogarithms126-7,135productrulefor135 tobase10 112-15 tobasea115-17 many-onemappings2 mappings2many-one2 one-many2 one-one2 maximumpoints28-33,300 firstderivativetestfor300,311 practicalproblems305-10 secondderivativetestfor302,311 mid-pointofline139 problemsinvolving 140-3

morecomplex 103-6 quadraticformulamethod25 rootsof39-41 quadraticformula 39 quadraticformulamethod 25 quadraticinequalities37-9 quadraticpolynomials71 quarticpolynomials71 quotient73 quotientrule 285-7,310 radians171 convertingfromdegreesto 172 convertingtodegreesfrom 172 range 3,12 ratesofchange 294-8 connected295 rectilinearfigures,areas 148-51 remaindertheorem82-6 rest,instantaneous406,418 roots ofequations39 ofquadraticequations39-41

multiplication 71-2 quadratic 71 quartic 71 remaindertheorem82-6 subtraction 71-2 positionvectors 319-23,331 powerlaw118 powerrule276-80,310 powersseeindices productrule fordifferentiation282-4,310 forlogarithms135 progressions 252 arithmeticseearithmeticprogressionsgeometricseegeometricprogressions

minimumpoints28-33,300 firstderivativetestfor300,311 practicalproblems305-10 secondderivativetestfor303,311 modulus ofnumber7-9 ofvector316 modulusequations90-4 modulusinequalities94-8 motiondiagrams404 multiplication ofpolynomials71-2 ofsurds57-8 ofvectorbyscalar316 naturalmultiplicationlaw118logarithms126-7,135 non-linearequations convertingfromlinearformto 155-8 convertingtolinearform 151-4 findingrelationshipsfromdata 159-64 normals287-91 equationsof 288,310 one-oneone-manymappings2mappings2 order andcombinations 232-3 andpermutations229-30 parabola28 intersectionoflineand42-4 Pascal'striangle 244-8 permutationsperiod196,199periodicfunctions196229-32 ordernotation229and229-30 ruleforfinding229 pointsofinflexion301-2 polynomials70-86 addition71-2 cubic71,78-82,98-101 definition71 degree 71 division73-5 factortheorem75-8,86 factorsof75-86linear71

ruleforall254 quadrants 190-1 quadraticequations25,28-37,39-41,103-6

completingthesquaremethod25,29-30 conditionforrealroots42,45 andcorrespondingcurves45 factorisation method 25 graphsofy=|f(x)|wheref(x)isquadratic34-7 maximumvalues28-33minimumvalues28-33

derivativeof346-50 graphsof196,199 period196,199 signof192-3,220 tangents 45,287-91 equationsof288,310 tetrahedralnumbers245 translation198-9 trigonometricequations208-13,216-17trigonometricfunctions/ratios187

derivativesof345-56 ofgeneralangles192-4graphsof195-205 graphsofy=|f(x)|wheref(x)istrigonometric function 205-7 signsof192-3,220 sketching200-2 seealsocosecantfunction;cosine function;cotangent function;secantfunction; sinefunction;tangent function trigonometricidentities214—15,218-19,220 trigonometry186-220 equivalenttrigonometricexpressions215 seealsoangles;trigonometric equations;trigonometric functions/ratios; turningtrigonometricidentitiespointsseestationarypoints vectors315-31 addition316 componentform317 constant velocity problems 327-30 definition316,403 equal316 i,geometry323-7jnotation317

surds55-67 conjugate 60 division58 multiplication 57-8 rationalisingdenominatoroffraction60-3 rulesof67 simplification 57-60 solvingequationsinvolving63-6 tangentfunction(tan)187,192

CambridgeIGCSEand0LevelAdditionalMathematics scalarmultiplerule,fordifferentiation277,310 scalars316,403 secantfunction(sec) 216-17 sectors174area177-80 segments174 series243-70 arithmetic253-5,267-9 convergent263 definition253 geometricseegeometricseries infinite262 seealsoprogressions shoestring/shoelacemethod 149 simultaneous equations 23-4,25-7 eliminationmethod24 onelinearandonenon-linear 25-7 substitutionmethod24 sinefunction(sin) 187,192 amplitude 196,198-9 derivativeof345-50 equivalenttrigonometricexpressionsfor215 integrationgraphsof195-9,201-2of370-2period196,199 signof192-3,220 smallincrements291-3,310 speed403 constant327 seealsovelocity speed-timegraphs403 stationary points(turning points) 28,300-4,310 secondofinflexion301-2derivativesand302-3 steppingseealsomaximumpoints;minimumpointsstonegame248straight-linegraphs138-65 areasofrectilinearfigures148-51 convertingfromlinearformtonon-linear equation155-8 convertingfromnon-linearequation tolinearform151-4 equations ofstraightlines 139,145-8 finding relationshipsfrom data 159-64 seealsoline(s) substitutionmethod24 subtraction ofpolynomials71-2 ofvectors316

velocity327,331,404-5,412-13,418 signsconstant,problems327-301-30for405-6

splittingintocomponents328-9

magnitude 316 modulus316 multiplication byscalar 316 notation316,317-19 position319-23,331 subtraction 316 unit317