Bill Pender
David Sadler, Derek Ward
Brian Dorofaeff, William McArthur SECOND
ShaftesburyRoad,CambridgeCB28EA,UnitedKingdom OneLibertyPlaza,20thFloor,NewYork,NY10006,USA 477WilliamstownRoad,PortMelbourne,VIC3207,Australia 314–321,3rdFloor,Plot3,SplendorForum,JasolaDistrictCentre,NewDelhi–110025,India 103PenangRoad,#05–06/07,VisioncrestCommercial,Singapore238467
CambridgeUniversityPress&AssessmentisadepartmentoftheUniversityofCambridge. WesharetheUniversity’smissiontocontributetosocietythroughthepursuitof education,learningandresearchatthehighestinternationallevelsofexcellence. www.cambridge.org
© BillPender,DavidSadler,DerekWard,BrianDorofaeff andJuliaShea2019
© BillPender,DavidSadler,DerekWard,BrianDorofaeff andWilliamMcArthur2025
Thispublicationisincopyright.Subjecttostatutoryexception andtotheprovisionsofrelevantcollectivelicensingagreements, noreproductionofanypartmaytakeplacewithoutthewritten permissionofCambridgeUniversityPress&Assessment.
Firstpublished2019
SecondEdition2025 2019181716151413121110987654321
CoverdesignedbySardineDesign
TextdesignedbydiacriTech
TypesetbydiacriTech
PrintedinChinabyC&COffsetPrintingCo.Ltd.
AcataloguerecordforthisbookisavailablefromtheNationalLibraryofAustraliaat www.nla.gov.au
ISBN978-1-009-65480-7
Additionalresourcesforthispublicationatwww.cambridge.edu.au/GO
ReproductionandCommunicationforeducationalpurposes
TheAustralian CopyrightAct1968 (theAct)allowsamaximumofonechapteror10%ofthepagesofthis publication,whicheveristhegreater,tobereproducedand/orcommunicatedbyanyeducationalinstitution foritseducationalpurposesprovidedthattheeducationalinstitution(orthebodythatadministersit)hasgiven aremunerationnoticetoCopyrightAgencyLimited(CAL)undertheAct. FordetailsoftheCALlicenceforeducationalinstitutionscontact:
CopyrightAgencyLimited Level12,66GoulburnStreet
SydneyNSW2000
Telephone:(02)93947600
Facsimile:(02)93947601
Email:memberservices@copyright.com.au
ReproductionandCommunicationforotherpurposes
ExceptaspermittedundertheAct(forexampleafairdealingforthepurposesofstudy,research,criticism orreview)nopartofthispublicationmaybereproduced,storedinaretrievalsystem,communicatedor transmittedinanyformorbyanymeanswithoutpriorwrittenpermission.Allinquiriesshouldbemadeto thepublisherattheaddressabove.
CambridgeUniversityPress&AssessmenthasnoresponsibilityforthepersistenceoraccuracyofURLsfor externalorthird-partyinternetwebsitesreferredtointhispublicationanddoesnotguaranteethatanycontenton suchwebsitesis,orwillremain,accurateorappropriate.Informationregardingprices,traveltimetablesandother factualinformationgiveninthisworkiscorrectatthetimeoffirstprintingbutCambridgeUniversityPress& Assessmentdoesnotguaranteetheaccuracyofsuchinformationthereafter.
CambridgeUniversityPress & AssessmentacknowledgestheAboriginalandTorresStraitIslanderpeoplesofthis nation.Weacknowledgethetraditionalcustodiansofthelandsonwhichourcompanyislocatedandwherewe conductourbusiness.WepayourrespectstoancestorsandElders,pastandpresent.CambridgeUniversityPress & AssessmentiscommittedtohonouringAboriginalandTorresStraitIslanderpeoples’uniqueculturaland spiritualrelationshipstotheland,watersandseasandtheirrichcontributiontosociety.
Introductionandoverview vii
Acknowledgements ix
Abouttheauthors x
1 Algebrareview
1A Expandingbrackets ...............................
1B Factoring
1C Algebraicfractions ...............................
1D Solvingquadraticequations ..........................
1E Solvingsimultaneousequations
ReviewofChapter1 ...............................
2 Numbersandsurds
2A Realnumbersandintervals
2B Surdsandtheirarithmetic ...........................
2C Furthersimplificationofsurds
2D Rationalisingthedenominator
ReviewofChapter2 ...............................
3 Functionsandgraphs
3A Functionsandfunctionnotation ........................
3B Functions,relations,andgraphs ........................
3C Reviewoflineargraphs
3D Quadraticfunctions—factoringandthegraph ................
3E Completingthesquareandthegraph
3F Thequadraticformulaeandthegraph
3G Powers,cubics,andcircles ...........................
3H Twographsthathaveasymptotes
3I Directandinversevariation ..........................
4 Equationsandinequations
4A Linearequationsandinequations
4B Quadraticequationsandinequations
4C Thediscriminant ................................
4D Quadraticidentities
ReviewofChapter4 ...............................
5 Transformationsandsymmetry
5A Translationsofknowngraphs
5B Reflectioninthe y-axisand x-axis .......................
5C Evenandoddsymmetry
5D Horizontalandverticaldilations
5E Theabsolutevaluefunction ..........................
5F Compositefunctions
5G Combiningtransformations ..........................
5H Continuityandpiecewise-definedfunctions ..................
ReviewofChapter5
6 FurthergraphsEXTENSION
6A Solvingtwoparticularinequations
6B Thesignofafunction
6C Sketchingreciprocalfunctions .........................
6D Sketchingsumsanddifferences
6E Modifyingafunctionusingabsolutevalue ...................
6F Inverserelationsandfunctions ........................
6G Inversefunctionnotation
6H Definingfunctionsandrelationsparametrically ...............
ReviewofChapter7
7 Trigonometry
7A Trigonometrywithright-angledtriangles
7B Problemsinvolvingright-angledtriangles ...................
7C Trigonometricfunctionsofageneralangle ..................
7D Quadrant,sign,andrelatedacuteangle
7E Givenonetrigonometricfunction,findanother ................
7F Trigonometricidentities
7G Trigonometricequations
7H Thesineruleandtheareaformula ......................
7I Thecosinerule
7J Problemsinvolvinggeneraltriangles .....................
ReviewofChapter7 ...............................
8
Linesinthecoordinateplane
8A Lengthsandmidpointsoflinesegments ....................
8B Gradientsoflinesegmentsandlines
8C Equationsoflines
8D Furtherequationsoflines ...........................
8E Usingpronumeralsinplaceofnumbers
ReviewofChapter8 ...............................
9 Exponentialandlogarithmicfunctions
9A Indices
9B Fractionalindices ................................
9C Logarithms
9D Thelawsforlogarithms
9E Equationsinvolvinglogarithmsandindices ..................
9F Exponentialandlogarithmicgraphs
9G Applicationsofthesefunctions .........................
ReviewofChapter9 ...............................
10 Differentiation
10A Tangentsandthederivative
10B Thederivativeasalimit
10C Arulefordifferentiatingpowersof x .....................
10D Thenotation dy/dx forthederivative
10E Thechainrule ..................................
10F Differentiatingpowerswithnegativeindices .................
10G Differentiatingpowerswithfractionalindices
10H Theproductrule ................................
10I Thequotientrule
10J Ratesofchange
10K Averagevelocityandaveragespeed ......................
10L Instantaneousvelocityandspeed
ReviewofChapter10 ..............................
11 PolynomialsEXTENSION
11A Thelanguageofpolynomials
11B Graphsofpolynomialfunctions ........................
11C Divisionofpolynomials
11D Theremainderandfactortheorems
11E Consequencesofthefactortheorem .....................
11F Sumsandproductsofzeroes
11G Geometryusingpolynomialtechniques ....................
ReviewofChapter11 ..............................
12 Euler’snumber
12A Theexponentialfunctionbase e ........................
12B Transformationsofexponentialfunctions
12C Thelogarithmicfunctionbase e
ReviewofChapter12 ..............................
13 Radianmeasureofangles
13A Radianmeasureofanglesize
13B Solvingtrigonometricequations
13C Arcsandsectorsofcircles ...........................
13D Trigonometricgraphsinradians
ReviewofChapter13 ..............................
14 Probability
14A SetsandVenndiagrams
14B Probabilityandsamplespaces .........................
14C Samplespacegraphsandtreediagrams
14D Venndiagramsandtheadditiontheorem
14E Multi-stageexperimentsandtheproductrule ................
14F Probabilitytreediagrams
14G Conditionalprobability .............................
ReviewofChapter14 ..............................
15 Dataandprobability
15A Randomvariablesandfrequencytables ....................
15B Cumulativefrequency
15C Groupeddata
ReviewofChapter15 ..............................
16 FurthertrigonometryEXTENSION
16A Three-dimensionaltrigonometry
16B Trigonometricfunctionsofcompoundangles .................
16C Thedouble-angleformulae
16D TrigonometricEquations ............................
16E Thesumofsineandcosinefunctions
ReviewofChapter16
17 CombinatoricsEXTENSION
17A Factorialnotation
17B Orderedselectionswithandwithoutrepetition
17C Orderedselections—threemoreprinciples .................
17D Orderedselectionswithidenticalelements
17E Countingunorderedselections ........................
17F Usingcountinginprobability ..........................
17G Arrangementsinacircle
ReviewofChapter17 ..............................
18 ThebinomialtheoremEXTENSION
18A BinomialexpansionsandPascal’striangle
18B Binomialexpansionswithseveralvariables ..................
18C Thebinomialtheorem
18D Usingthegeneralterm .............................
18E IdentitiesinPascal’striangle ..........................
18F FurtheridentitiesinPascal’striangle
ReviewofChapter18 ..............................
Answers
Thischaptercontainsvariouslooselyrelateddiscussionsaboutfunctionsandtheirgraphs.
▶ Sections6A–6Bdealwithsometrickiertypesofinequations,usingalgebraicandgraphical–geometric methods.
▶ Sections6C–6Eintroducefurthertransformationsofthegraphsoffunctions.Thegraphofthereciprocalof agraphedfunctionisdrawn—aprocedurethatrequiresalittlemorediscussionofasymptotes.Thesumand differenceoftwographedfunctionsisdrawn.Andthecompositionsofafunctionwiththeabsolutevalue functionaredrawnusingreflectionsinthe x-axisand y-axis.
▶ Section6F–6Gintroduceinverserelationsandfunctionsgraphically,withtheircorrespondingreflectionsin thediagonalline y = x.Thisisyetanothertypeoftransformationofaknowngraph.Section6Gdevelops theformalnotationforinversefunctions.
▶ ThelastSection6Hintroducesparameterssothatequationsoffunctions,andrelationsingeneral,canbe expressedandgraphedintermsoffunctions x and y ofasingleparameter.
Asalways,computersketchingofcurvesisveryusefulindemonstratinghowthefeaturesofagraphare relatedtothealgebraicpropertiesofitsequation,andingainingfamiliaritywiththevarietyofgraphsandtheir transformations.
SomequestionsinSections6Cand6HusethetrigonometryofthegeneralangleandPythagoreanidentities, andthesequestionscouldbedelayeduntilaftertrigonometryisreviewedandextendedinChapter7.
Thechapterisconceptuallydemanding,particularlySection6D.ReadersmayprefertoleaveSection6Duntil laterintheyear—anappropriateplacecouldbebeforeChapter16:Furthertrigonometry,whenthesum a sin x + b cos x oftwotrigonometricgraphsneedstobesketched.
Learningintentions
• Solveabsolutevalueinequationsoftheform |ax + b| < k inthreeways.
• Solveinequationswith x inthedenominatorbymultiplyingthroughbyitssquare.
Thissectionisdevotedtotwoparticulartypesofinequations:
• Absolutevalueinequationsoftheform |ax + b| < k
• Inequationswhere x isinthedenominator,suchas 5 x 4 ≥ 1.
ThenextSection6Bwillsolveinequationsusingatableoftestpoints.
Sofarwehavesolvedlinearandquadraticinequations:
• Wesolvedlinearinequationslikelinearequations,exceptthatwhenmultiplyingordividingbyanonzero number,theinequalityisreversed.
• Wesolvedquadraticinequationsusingthegraph,andalsousingtestpointsafterfindingthezeroes.
Rememberthatwe solveinequations suchas x + 2 < 3,andwe proveinequalities suchas(x 2)2 ≥ 0,but thattheword‘inequality’isoftenusedforbothmeanings.Donotbesurprisedifyouareaskedto‘solvean inequality’.
Ourtaskistosolveinequations |ax + b| < k,where a 0, b,and k areconstants (andtheinequalitysignis < ,or ≤ ,or > ,or ≥ ).
Drawingagraphisprobablytheclearestmethodofsolution.Takingasastandardexampletheinequation |2x + 1| > 3:
1Solvinganabsolutevalueinequationsuchas |2 x + 1| > 3 graphically
• Solvethe equation |2x + 1| = 3usingthemethodsofSection5E,Box19.
• Hencedrawthegraphsof y = |2x + 1| and y = 3ontheonesetofaxes.
• Readthesolutionoff thegraph.
Example1 Solvinganabsolutevalueinequationgraphically
Solve |2x + 1| > 3graphically.
Solution
Firstsolvetheequation |2x + 1| = 3 2x + 1 = 3or2x + 1 = 3 x = 1or 2.
Thegraphof y = |2x + 1| isVupwards,with x-intercept x = 1 2
Fromthesketchof |2x + 1| and y = 3together,thesolutionoftheinequalityis x < 2or x > 1. 2 x y
Method2forsolvinganabsolutevalueinequation—Usedistanceonthe numberline
Wecanusedistanceonthenumberlinetosolvetheinequality.Butwemustbecarefulindealingwiththe coefficientof x,particularlywhenitisnegative.
First,wereviewformulaeforabsolutevalueusingdistanceonthenumberline:
2Absolutevalueanddistance
• |x| = distancefrom x tozeroonthenumberline • |x a| = distancefrom x to a onthenumberline
Theeasycase,wherethecoefficientof x is1
Whenthecoefficientof x is1,wecanimmediatelyusethesecondformulaabove.
Example2 Thecoefficientof x is1
Solvetheseinequationsonthenumberline.
Thegeneralcase,wherethecoefficientof x isnot1
Therearetwoinitialstepstogetitintotheformwherethecoefficientof x is1.
3Solving | ax + b| < k,foranynon-zerovalueof a
Step1: Force a tobepositivebywritingsay |− 2x + 3| as |2x 3|
Step2: Dividethroughbythenowpositivenumber a
Step3: Solveusingdistance,asintheeasycaseabove.
Example3 Thecoefficientof x isnot1
Solvetheseinequationsusingdistanceonthenumberline.
Method3forsolvinganabsolutevalueinequation—Algebraicapproach
WesawinSection5Ehowanabsolutevalueequationoftheform |
| = k canbesolvedalgebraicallyby rewritingtheequation.
Rewriteanequation |
Asimilarapproachcanbetakentosolvinganinequationsuchas |
4Solvinganabsolutevalueequationorinequationalgebraically
• Rewriteanequation | f (x)| = k as f (x) = k or f (x) = k
• Rewriteaninequation | f (x)| < k as k < f (x) < k
• Rewriteaninequation | f (x)| > k as f (
) < k or f (x) > k
Example4 Solvinganabsolutevalueinequationalgebraically
Solution
a AsinSection5E,thefirststepinsolvingtheequation |9 2x| = 5is 9 2x = 5or9 2x = 5, givingtwosolutions, x = 2or x = 7.
Usingtheseconddotpoint, |9 2x| < 5
< 9 2x < 5 9 14 < 2x < 4 ÷ ( 2) 7 > x > 2 thatis,2 < x < 7. b
Usingthethirddotpoint,
2x ≤−14or 2x ≥−4 ÷ ( 2) x ≥ 7or x ≤ 2 thatis, x ≤ 2or x ≥ 7. c
Inequationsthathavenosolutions,orareinequalities
Theabsolutevalue | f (x)| cannotbenegative.Thusif k isnegative:
• | f (x)| = k and | f (x)| < k havenosolutions,and
• | f (x)|≥ k istrueforallvaluesof x inthedomainof f (x).
Example5 Inequationsthathavenosolutions,orareinequalities
Solvetheseabsolutevalueinequationsalgebraically: |x 7| < 5 a |6 x|≥−1 b
Solution
Theabsolutevalueisnevernegative,sothereare nosolutions. a Absolutevalueisalwaysatleastzero,soevery realnumberisasolution.Thustheinequationisan inequality.
Solvinginequationswith x inthedenominator—multiplythroughbyits square
Whenwetrytosolvethisinequation,weimmediatelyrunintoaproblem: 5 x 4 ≥ 1.
Thedenominator x 4issometimespositiveandsometimesnegative.Thusifweweretomultiplybothsidesby thedenominator x 4,theinequalitysymbolwouldreversesometimesandnotothertimes.
Toavoidusingcases,themoststraightforwardapproachistomultiplythroughinsteadbythe squareofthe denominator,whichisalwayspositiveorzero.
5Solvinganinequationwiththevariableinthedenominator
• Multiplythroughbythesquareofthedenominator
▷ Becarefultoexcludethezeroesofthedenominatorfromthesolutions.
▷ Oncethefractionshavebeencleared,therewillusuallybecommonfactorsonbothsides.These should not bemultipliedout,becausethefactoringwillbeeasieriftheyareleftunexpanded.
Analternativeapproachusingatableofsignsispresentedinthenextsection.
Example6 Solvinganinequationbymultiplyingthroughbythesquare
Solve 5 x 4 ≥ 1.
Solution
First, x 4,becausetheLHSisundefinedwhen x = 4.
Thekeystepistomultiplybothsidesby (x 4)2
× (x 4)2 5(x 4) ≥ (x 4)2,and x 4, Don’texpandthebracketshere!Youwillonlyhavetore-factor.
(x 4)2 5(x 4) ≤ 0,and x 4, (x 4)(x 4 5) ≤ 0,and x 4, (x 4)(x 9) ≤ 0,and x 4.
Fromthediagram,4 < x ≤ 9. x y 49
Noticethat x = 4isnotasolutionoftheoriginalinequationbecausesubstituting x = 4intotheLHSofthe originalinequationgives 5 4 4 ,whichisundefined.
1 Solvefor x usingdistanceonthenumberline.
3 Solve:
3 Multiplybothsidesby x2 andhencesolve:
4 Collectliketermswherenecessarythenmultiplyby x2 tosolve:
5a Usingdistanceonthenumberline,solve:
b Next,solvetheinequationsinparts(a)graphically.
c Finally,solvetheinequationsinpart(a)algebraically.
6 Multiplybothsidesofeachinequationbythesquareofthedenominatorandhencesolveit.Donotmultiply outanycommonfactor.
7 Ineachcase,firstsimplifytheinequation,thenusemultiplicationbythesquareofthedenominatortosolve for x.
8a Solvebyanysuitablemethod,andgraphthesolutiononanumberline.
b Solvetheinequationsinparts(ii),(iv),and(vi)bytheothertwoofthethreemethodsthatyoudid not useinpart(a).
9 Firstsimplifyeachinequation,thensolvefor x.
x + 1| + 2 < 3 a |x 2|− 1 > 3 b |x 3|− 1 ≥ 4 c
10 Solvefor x: x 1 x + 2 > 2
11a Showthatthedoubleinequation2 ≤|x|≤ 6hassolution2 ≤ x ≤ 6or 6 ≤ x ≤−2.
b Similarlysolve:
12 Saywhethereachstatementistrueorfalse.Ifitisfalse,giveacounterexample.
13 Solvetheseequationsandinequations.
14 Considertheinequation x + 1 x < 2x
Explainwhy x mustbepositive. a Hencesolvetheinequation. b
15 Solvetheinequation1 + 2x x2 ≥ 2 x
16 Considertheinequation |x a| + |x b| < c,where a < b.
a If a ≤ x ≤ b,show,usingdistancesonanumberline,thattherecanonlybeasolutionif b a < c
b If b < x,show,usingdistancesonanumberline,that x < a
c If x < a,show,usingdistancesonanumberline,that x > a + b c 2
d Henceshowthateither x a + b 2 < c 2 orthereisnosolutiontotheoriginalproblem.
e Hencefindthesolutionof |x + 2| + |x 6| < 10.
Learningintentions
• Constructatableoftestpointsdodgingaroundzeroesanddiscontinuities.
• Usethetableoftestpointstosolvevarioustypesofinequation.
• Usethetableoftestpointsasafirststepincurve-sketching.
Thissectioncompletesthestoryofaverygeneralmethodofusingtestpointstoanalysethesignofanyfunction, providedonlythatwecanfinditszeroesanddiscontinuities.Thepatternofsignsproducedbythesetestpoints canbeusedtosolveinequations,whicharetheimmediatepointofthissection.
Butthepatternalsoprovidesthebasisforsketchingcurvesofeverykind,andwillbecentralparticularlytocurve sketchingusingcalculusinYear12.
SketchingfactoredcubicsinSection3G,andsolvingquadraticinequationsinSection4B,usedatableoftest valuesdodgingaroundthezeroestoseewherethefunctionchangedsigns.Thefunctionsinthischapter,however, mayalsohavebreaksor discontinuities,andweneedtododgearoundthemaswell,asthediagrambelow demonstrates:
6Wherecanafunctionchangesign?
Theonlyplaceswhereafunctionmaypossiblychangesignarezeroesanddiscontinuities.
• Thegraphabovehasdiscontinuitiesat x = c, x = d, x = e,and x = f ,andhaszeroesat x = a and x = b
• Thefunctionchangessignatthezero x = a andatthediscontinuities x = c and x = e,andnowhereelse.
• Noticethatitdoesnotchangesignatthezero x = b oratthediscontinuities x = d and x = f .
ThestatementinBox6goestotheheartofwhattherealnumbersareandwhatcontinuitymeans.Inthiscourse, thesketchaboveissufficientjustification.
Asaconsequence,wecanexaminethesignofafunctionusingatableoftestvaluesdodgingaroundanyzeroes anddiscontinuities.Weaddathirdrowforthesign,sothatthetablebecomesa tableofsigns
Toexaminethesignofafunction,drawupaa tableofsigns usingtestvaluesthatdodgearoundallthe zeroesanddiscontinuities.
Findingthezeroesofafunctionhasbeenaconstantconcerneversincequadraticswereintroducedinearlier years.Tofinddiscontinuities,wewillbeusingourstandardassumptionthat allthefunctionsinthecourseare continuouseverywhereintheirdomains,exceptwherethereisanobviousproblem.
Solvingpolynomialinequationsusingatableofsigns
AsimplerformofthistableofsignswasintroducedinSection3Gtosketchcubics.Cubics,andmoregenerally polynomials,donothaveanydiscontinuities,sothetestvaluesonlyneedtododgearoundtheirzeroes—ifwe canfindthem.
Section3G’sconcernwastosketchthefunction,whereasheretheconcernistosolveinequations.Thegraphsare notnecessaryhere,andoursketcheswillbeincomplete,butthesketchesallowustoseethewholesituation.
Example7
Sketchingafactoredpolynomialandsolvinganinequality
a Drawupatableofsignsofthefunction y = (x 1)(x 3)(x 5).
b Statewherethefunctionispositiveandwhereitisnegative.
c Solvetheinequation(x 1)(x 3)(x 5) ≤ 0.
d Confirmtheanswersbysketchingwhatisnowknownaboutthegraph.
Solution
a Therearezeroesat1,3and5,andnodiscontinuities.
x 0123456
y 15030 3015
sign 0 + 0 0 +
b Hence y ispositivefor1 < x < 3or x > 5,and negativefor x < 1or3 < x < 5.
c x ≤ 1or3 ≤ x ≤ 5.
Example8 Factoringapolynomial,solvinganinequality,andsketching thecurve
Solve x3 + 1 ≤ x2 + x usingatableofsigns.Thenconfirmwithasketch.
Solution
Movealltermstotheleft,thenfactorbygrouping,
x 3 x 2 x + 1 ≤ 0
x 2(x 1) (x 1) ≤ 0 (x 2 1)(x 1) ≤ 0 (x + 1)(x 1)2 ≤ 0.
TheLHShaszeroesat1and 1,andnodiscontinuities.
x 2 1012
y 90103
sign 0 + 0 +
Hence x ≤−1or x = 1.
Solvinginequationsinvolvingdiscontinuities
Whenthefunctionhasdiscontinuities,themethodisthesame,exceptthatthetestvaluesnowneedtododge arounddiscontinuitiesaswellasthezeroes.
Again,asketchisusefulforunderstanding.BecausehorizontalasymptotesareinYear12,however,aclueis neededbeforethegraphreallytellsthestory.
a Findthezeroesanddiscontinuitiesof y = x 1 x 4 ,andexamineitssign.
bi Write x 1 = (x 4) + 3,hencefindthehorizontalasymptote.
ii Identifytheverticaldiscontinuitiesofthefunction,andsketchit.
Solution
a Thereisazeroat x = 1,andadiscontinuityat x = 4. x 01245
y 1 4 0 1 2 ∗ 4 sign + 0 −∗ +
Hence y ispositivefor x < 1or x > 4,andnegativefor1 < x < 4. x y 4 1 1
bi y = (x 4) + 3 x 4 = 1 + 3 x 4
Hence y → 1as x →∞ andas x →−∞,so y = 1isahorizontalasymptoteinbothdirections.
ii As x → 4 , y →−∞,andas x → 4+ , y →∞,so x = 4isaverticalasymptote.
Findthezeroesanddiscontinuitiesof y = 1 1 + x2 ,andexamineitssign.
Solution
Thedenominator1 + x2 isalwaysatleast1.Hencethefunctionisdefinedforallreal x,andis alwayspositive.
Atableofsignsisnowunnecessary,butwecandrawoneup.Therearenodiscontinuitiesand nozeroes,soonetestvalue f (0) = 1confirmsthatthefunctionisalwayspositive. x 0 y 1 sign +
Wenowhavetwowaystosolveaninequationwith x inthedenominator.Forcomparison,hereisaninequality with x inthedenominatorsolvedbothways.Firstitissolvedbymultiplyingbothsidesbythesquareofthe denominator.Thenitissolvedusingatableofsigns.Comparethetwoquitedifferentapproaches.
Solve 3 x + 2 ≤ x usingatableofsigns.
Solution
Collectingeverythingontheleft, 3 x + 2 x ≤ 0, usingacommondenominator, 3 x2 2x x + 2 ≤ 0, andfactoring, (3 + x)(1 x) x + 2 ≤ 0.
TheLHShaszeroesat x = 3and x = 1,andadiscontinuityat x = 2.
Sothesolutionis x ≥ 1or 3 ≤ x < 2.
(Nosketchhere—relyjustonthetableofsigns.)
Solve 3 x + 2 ≤ x byfirstmultiplyingthoughbythesquareofthedenominator.
Solution
Noticethatthecommonfactor(x + 2)isnevermultipliedout.Thatwouldcauseaseriousproblem,becauseof theeffortrequiredtore-factortheexpandedcubic.
x + 2
× (x + 2)2 3(x + 2) ≤ x(x + 2)2,and x 2, x(x + 2)2 3(x + 2) ≥ 0,and x 2, (x + 2)(x 2 + 2x 3) ≥ 0,and x 2, (x + 2)(x + 3)(x 1) ≥ 0,and x 2.
TheLHScannowbesketchedusingatableofsigns.
Fromthegraph,
Thepurposeofthisexerciseistosolveinequationsusingatableofsigns.Inthecaseofpolynomials,thistable alsoallowsthesketchtobedrawn,asinSection3G.Asketchmakesthesituationclearer,butthesketchisnot necessaryforobtainingthesolution.
1 UsethegivengraphoftheLHStohelpsolveeachinequation. x y
(x 1)(x 2) ≤ 0 a
2a Explainwhythezeroesof y =
Thencopyandcompletethetableofsigns.
b Usethetableofsignstosolve(x + 1)2(1 x) ≥ 0.
c Sketchthegraphtoconfirmthesolutioninpart(b).
3 Usethethreestepsofthepreviousquestiontosolveeachinequation.
4 Firstfactoreachpolynomialcompletely,thenusethemethodsofQuestions2and3tosketchitsgraph. (Hint:takeoutanycommonfactorsfirst.)
5 Fromthegraphsinthepreviousquestion,orfromthetablesofsignsusedtoconstructthem,solvethe followinginequations.Beginbygettingalltermsontotheoneside.
6 Here isQuestion10fromExercise6A.Thistime,collectalltermsontheLHSasasimplifiedsingle fraction.Findthezeroesanddiscontinuitiestouseinatableofsigns,andhencesolveeachinequation.
7a Findthezeroesanddiscontinuitiesof y = x2 x 3 and constructatableofsigns.
b Hencesolve x2 x 3 < 0.
8 Ifnecessary,collectalltermsontheLHSandfactor.Findanyzeroesanddiscontinuities,thendrawupa tableofsignswithinterlacingvaluesinordertosolvetheinequation.
9a Factoreachequationcompletely,andhencefindthe x-interceptsofthegraph.Factorparts(ii)and(iii)by groupinginpairs.
b Foreachfunctioninthepreviousquestion,examinethesignofthefunctionaroundeachzero,andhence drawagraphofthefunction.
10 Findallzeroesofthesefunctions,andanyvaluesof x wherethefunctionisdiscontinuous.Thenanalysethe signofthefunctionbytakingtestpointsaroundthesezeroesanddiscontinuities.
11 Multiplythroughbythesquareofthedenominator,collectalltermsononesideandthenfactortoobtain afactoredcubic.Sketchthiscubicbyexaminingtheinterceptsandthesign.Hencesolvetheoriginal inequation.
12 Solvetheinequationsinthepreviousquestionbythealternativemethodofcollectingeverythingonthe LHS,findingacommondenominator,identifyingzeroesanddiscontinuities,anddrawingupatableof signs.
13a Provethat f (x) = 1 + x + x2 ispositiveforall x.
b Provethat f (x) = 1 + x + x2 + x3 + x4 ispositiveforall x.Considerseparatelythethreecases x ≥ 0, 1 < x < 0and x ≤−1.Groupthefivetermsintopairsindifferentwayswiththesecondandthirdcases. c Usesimilarmethodstoprovethatforallintegers n ≥ 0, f (x) = 1 + x + x 2 + ··· + x 2n 1 + x 2n ispositiveforall x.
d Provethat x = 1istheonlyzeroof f (x) = 1 + x + x2 + + x2n 1,forallpositiveintegers n
14 Let f (x) = a 2 1 x + a 1 x a with a > 1.
Forwhatvaluesof k does f (x) = k haveasolution? a Solve f (x) < 2usinganyappropriatemethod. b
Nowchooseasuitablevaluefor a anduseshiftingtosolve 1 x 1 1 x 7 < 4 3 . c
Learningintentions
• Sketchthereciprocalofafunctiongiventhegraphofthefunction.
• Determineitsverticalandhorizontalasymptotes,anditsdomainandrange.
Section6C–6Epresentfurtherwaystotransformorcombinegraphs.Ineachcase,theemphasiswillbemoreon therelationshipbetweenthegraphsthanontheequationsofthevariousfunctions.
Thissectionsketchesthereciprocalfunction g(x) = 1 f (x) ofagraphedfunction f (x).Readingfromthegraphof y = f (x),wedeveloppropertiesofthisreciprocalfunction y = g(x),andusethesepropertiestosketch y = g(x). Thereciprocaltrigfunctions—cosec,sec,andcot—aredevelopedinChapter7.
Therearemanysmalldetailsinvolvedinsketchingreciprocals,soitisbettertogiveseveralexamples,then followthemwithsomegeneralstatements.
Thesketchshows y = 2x.Let g(x) = 1/ f (x)bethereciprocalfunctionof f (x).
Reasoningfromthesketch:
a Whatsigndothevaluesof y = g(x)have?
b Whatisthe y-interceptof y = g(x)?
c Whathappensto y = g(x)as x →∞?
d Whathappensto y = g(x)as x →−∞?
e Hencesketch y = g(x).
f Whatarethedomainandrangeof f (x)and g(x)?
Solution
a f (x)isalwayspositive,soitsreciprocalisalwayspositive.
b f (0) = 1,andthereciprocalof1is1,so g(0) = 1.
c As x → +∞, f (x) →∞,so g(x) → 0+ ,sothe x-axisisahorizontalasymptote ontheright.
d As x →−∞, f (x) → 0+ ,so g(x) → +∞
f Theybothhavedomainallreal x,andrange y > 0.
Note: g(x) = 2 x is,ofcourse,thereflectionof y = f (x)inthe y-axis.Butthe intentionhereistoargueusingreciprocals. e y
8Graphingthereciprocal—signandhorizontalasymptotes
Let f (x)beagraphedfunction,andlet g(x) = 1/ f (x)beitsreciprocalfunction.
• When f (x)issmallandpositive, g(x)islargeandpositive.
▷ When f (x)islargeandpositive, g(x)issmallandpositive.
• When f (x) = 1, g(x) = 1,andviceversa.
• If f (x) → +∞ as x →∞,then g(x) → 0+ as x →∞,
▷ andthe x-axisisahorizontalasymptoteontheright.
• If f (x) → 0+ as x →∞,then g(x) → +∞ as x →∞.
Thelasttwodotpointshaveequivalentbehaviourontheleft. Allfourdotpointshaveequivalentbehaviourwhen f (x)isnegative.
Again,aninitialexampleshouldmakesthingsclear.
Analysingzeroesandverticalasymptotes.
Sketch f (x) = x 2,thenwith g(x) = 1/ f (x),andreasoningfromthesketch:
a Whereis g(x)undefined?
b Does y = g(x)haveanyzeroes?
c Whereis y = g(x)aboveandbelowthe x-axis?
d Identifyanyverticalasymptotes.
e Identifyanyhorizontalasymptotes.
f Whatisthe y-interceptof y = g(x)?
g Wheredoes y = g(x)intersectwith y = f (x)?
h Hencesketch y = g(x)onthesamesetofaxes.
i Whatarethedomainandrangeof f (x)and g(x)?
Solution
a Zerohasnoreciprocal,so g(x)isundefinedatthezero x = 2of f (x).
b Thenumberzeroisnotthereciprocalofanynumber,so g(x)isneverzero.
c g(x)ispositivewhen f (x)ispositive,andnegativewhen f (x)isnegative.
d As x → 2+ , f (x) → 0+ ,so g(x) → +∞ As x → 2 , f (x) → 0 ,so g(x) →−∞. Hence x = 2isaverticalasymptote.
e As x → +∞, f (x) →∞,so g(x) → 0+ . As x →−∞, f (x) →−∞ ,so g(x) → 0 Hencethe x-axisisahorizontalasymptoteleftandright.
f When x = 0, f (0) = 2,so g( 2) = 1 2
g Theonlynumbersthataretheirownreciprocalare1and 1. Hence y = f (x)and y = g(x)intersectat(1, 1)and(3,1).
i Domain:For f (x),allreal x.For g(x), x 2.
Range:For f (x),allreal y.For g(x), y 0.
y
9Graphingthereciprocal—zeroesandverticalasymptotes
Let f (x)beagraphedfunction,andlet g(x) = 1/ f (x)beitsreciprocalfunction.
• Zeroistheonlynumberthathasnoreciprocal.
▷ Hence g(x)isundefinedateveryzeroof f (x).
• Zeroistheonlynumberthatisnotareciprocalofanynumber.
▷ Hence g(x)isneverzero.
• When f (x)hasazero, y = g(x)normallyhasaverticalasymptote(butcheck,becausetherearesome ratherbizarreexceptions).
Thesedeserveparticularattention.
10Graphingthereciprocal—domainandrange
Let f (x)beagraphedfunction,andlet g(x) = 1/ f (x)beitsreciprocalfunction.
• Thedomainof g(x)isthedomainof f (x)withallthezeroesof f (x)removed.
• g(x)isneverzero,sotherangeof g(x)doesnotcontainzero.
• Infinityisnotanumber: Don’teverbetemptedtosaythatif f (x)hasanasymptoteat x = a,thenthe function g(x) = 1/ f (x)iszeroat x = a.
Thesymbols ∞ and −∞ arepartofmathematics,buttheyarenotnumbers,andtheycertainlydonothave reciprocals.
• Thereciprocalofthereciprocalmaynotbetheoriginalfunction: Referringtoourpreviousexample,startinsteadwithreciprocalfunction,thatis, let f (x) = 1 x 2 .Then f (x)isundefinedat x = 2,soitsreciprocal g(x) = 1/ f (x) isalsoundefinedat x = 2. Thatis, g(x) = x 2,where x 2.
Thisisnottheoriginalline y = x 2oftheexample,becauseitszero x = 2hasnow beenremoved.
Sketchingthereciprocalfromthegraph,withnoequation
Inthepreviousexamples,theequationofthefunctionwasthereforreference.Butinthetwoexamplesbelow, onlythesketchisgiven.
Giveneachsketchof y = f (x)below,sketch g(x) = 1/ f (x),thenstatethedomainandrangeof g(x).
a x y b x y
Solution
a Thepoint( 3,2)iscalleda localmaximum becauseforasmall regionaround x = 3, f (x)isgreatestat x = 3.Similarly,(3, 2)is calleda localminimum
Corresponding, g(x)hasalocalminimumat( 3, 1 2 )andalocal maximumat(3, 1 2 ).
Thedomainof g(x)is x 4,0or4(removethezeroes),andthe rangeof g(x)is y 0.
b As x →∞, f (x) → 2,so g(x) → 1 2 ,so y = 1 2 isahorizontalasymptoteonthe right.
Similarly,as x →−∞,
isahorizontal asymptoteontheleft.
Domainof g(x): x 0.Range: y <
Part(a)introducedtheideaof localmaximaandminima,andtheirrelationshipwiththereciprocalfunction.
Part(b)showedhowtodealwithhorizontalasymptotesotherthanthe x-axis.
Note: Question9and13concernthegraphsofthethreetrigonometricfunctionssin θ,cos θ,andtan θ.The graphsaredrawninthequestions,butreadersunfamiliarwiththemmayliketodelayattemptingthese questionsuntilafterstudyingChapter7.
1a Sketch y = 1 x 1 aftercarryingoutthefollowingsteps.
i Statethenaturaldomain,andfindthe y-intercept.
ii Findanypointswhere y = 1or y = 1.
iii Explainwhy y = 0isahorizontalasymptote.
iv Drawupatableofvaluesandexaminethesign.
v Identifyanyverticalasymptotes,andusethetableofsignstowritedownitsbehaviournearany verticalasymptotes.
b Repeatfor y = 2 3 x c Repeatfor y = 2 x + 2 d Repeatfor y = 5
2 Followsteps(i)–(v)ofQuestion1(a)toinvestigatethefunction y = 2 (x 1)2 andhencesketchitsgraph.
Showanypointswhere y = 1.
3 Investigatethedomain,zeroes,signandasymptotesofthefunction y = 1 (x 2)2 andhencesketchits graph.Showanypointswhere y = 1.
4a Let f (x) = x + 1.
Graph y = f (x)showingtheinterceptswiththeaxes. i Alsoshowthepointswhere f (x) = 1and f (x) = 1. ii
Henceonthesamenumberplanesketch y = 1 f (x) iii
b Followsimilarstepsforthefunction g(x) = 2 x.
5 Let y = f (x)where f (x) = 1 3 (x + 1)(x 3).
a Showthat y = 1at x = 1 √7and x = 1 + √7.Plotthesepoints.
b Completethegraphof y = f (x)showingthevertex,theinterceptswiththeaxesandthepointswhere f (x) = 1.
c Whatistherangeof y = f (x)?
d Hencesketch y = 1 f (x) onthesamenumberplane.
e Whatistherangeof y = 1 f (x) ?
6a Followsimilarstepstoquestion5tosketch y = 1 4 (4 x2)and y = 4 4 x2
b Whatistherangeof y = 1 4 (4 x2)?
c Whatistherangeof y = 4 4 x2 ?
7a Sketch y = 1 2 (x2 + 1)showingthepointswhere y = 1.
b Whatistheminimumvalueof 1 2 (x2 + 1)?
c Sketch y = 2 x2 + 1 onthesamenumberplane.
d Explainwhythe x-axisisanasymptoteto y = 2 x2 + 1 .
e Whatisthemaximumvalueof 2 x2 + 1 ?
8 Let f (x) = x2 + 2x 3.
Whatisthemaximumof f (x)? a
Sketch y = f (x)and y = 1 f (x) onthesamenumberplane. b
Explainwhythe x-axisisanasymptotefor y = 1 f (x)
c Whatistheminimumof 1 f (x) ? d
9 Thediagramsshowfirst y = cos θ then y = sin θ for0◦ ≤ θ ≤ 360◦ . y 180° 360° y
ai Copythesketchof y = cos θ,andaddtoitthesketchof y = 1 cos θ
ii Whatarethedomainandrangeof y = 1 cos θ inthisinterval?
bi Copythesketchof y = sin θ,andaddtoitthesketchof y = 1 sin θ .
ii Whatarethedomainandrangeof y = 1 sin θ inthisinterval?
10 Provethatthesymmetryofafunctionispreservedwhentakingreciprocals.Thatis,provethatthe reciprocalofanevenfunctionisanevenfunction,andprovethatthereciprocalofanoddfunctionisanodd function.
11a Graph y = 2 + x x byfirstnotingthat y = 1 + 2 x
b Hencegraph y = x 2 + x
12 Sketchthereciprocalofeachfunctionshown,showingallsignificantfeatures.
13 Thesketchshows y = tan θ for0◦ ≤ θ ≤ 360◦ .
a Whatisthedomainof y = tan θ inthisinterval?
b Whereistan θ = 0inthisinterval?
c Hencestatethedomainof y = 1 tan θ inthisinterval.
d Whatislim θ→90◦ 1 tan θ ?
e Copythegraphof y = tan θ,andaddtoitthegraphof y = 1 tan θ
f Whatistherangeofthereciprocalfunction? y
14 Ineachcasethegraphof y = f (x)isgiven.Sketchthegraphof y = 1 f (x) ,payingcarefulattentiontothe domain,anyasymptotes,andanyrelevantlimits.
y x a y x b
15 TheargumentsinthesolutiontoExample15(a)seemstorelyonthefollowingassertionaboutthegraphsof afunction y = f (x)anditsreciprocalfunction y = ( f (x)) 1: ‘Whenonecurvehasalocalmaximum,theotherhasalocalminimum.’ Thisstatementisnotstrictlytrue.Statethequalificationthatneedstobemadeinthisstatement,andgivean examplewherethequalificationisnecessary.
16 Let y = 1 x 2 .Writedownpreciselytheequationofthereciprocalfunction.
• Sketchthesumanddifferenceoftwographedfunctions.
Theproblemaddressedinthissectionistotakethesketchesoftwofunctions f (x)and g(x),andworkingjust fromthosesketches,sketchtheirsumanddifference:
s(x) = f (x) + g(x)and d(x) = f (x) g(x).
Thepronumerals s(x)and d(x)usedherearenotanymathematicalconvention,theyarejustconvenientnotation inthissection.
Theword‘ordinate’
The ordinate ofapointisthe y-coordinate,ormoregenerally,thecoordinateontheverticalaxis.Theoperations inthissectionandthenextroutinelyactonthe y-coordinatesofpoints,sothisshorterwordisuseful.The coordinateonthehorizontalaxisissometimesreferredtoasthe‘abcissa’,plural‘abcissae’,butthatwordisnot necessaryinthecourse.
Thegraphsoftwofunctions f (x)and g(x)aresketchedtotheright,andwewantto sketchthesum s(x) = f (x) + g(x).Theequationsofthefunctionswillusuallynotbe given,butitisconvenienttostatethemherewhiledevelopingthemethod.Theyare:
f (x) = x 2 36and g(x) = 5x , andhereisatablesofvalues:
Eachordinateof s(x)isthesumoftheordinatesof f (x)and g(x).Thisisthekey ideathatleadstoeverythingelse.Thesecondsketchadds y = s(x)totheothertwo graphs.
Nowletusconsiderhowwecouldhavesketched s(x)ifwehadnothadthe equations. x y ygx() ysx() yfx()
0 Addtheordinateswherepossible—thekeyidea.Thishasbeendoneinthetopdiagram.
1 Ifonecurve,say f (x),hasazeroat x = a,then
s(a) = 0 + g(a) = g(a), so s(x)hasthesameordinateas g(x)at x = a,andthecurvesmeetthere. Thishappensatboththezeroes x = 6and x = 6of f (x) = x2 36,where s(x)meets g(x).Italsohappensat thezero x = 0of g(x) = 5x,where s(x)meets f (x).
2 Iftheordinatesareoppositesat x = a,then
s(a) = f (a) + g(a) = 0,
so s(x)hasazeroat x = a.Thishappensat x = 9andat x = 4.
3 Ifthetwocurvesmeetat x = a,sothattheirordinatesareequal,then
s(a) = f (a) + g(a) = 2 f (a)(or2g(a)) andtheordinateof s(x)istwicetheordinateof f (x)(orof g(x)).Thishappensat x = 4andat x = 9.
Wecannotfindfromthegraphsalonetheprecisepositionoftheminimum,sointheabsenceoftheequations, thisisnotrequiredinthesketch.Butonceweknowtheequations,theminimumisthevertexoftheparabola
s(x) = (x 2 36) + 5x = x 2 + 5x 36 = (x + 9)(x 4), wheretheaverageofthezeroesis x = 1 2 (4 9) = 2 1 2 ,and s( 2 1 2 ) = 42 1 4 .
Hereisanexampleinwhichonecurvehasahorizontalandavertical asymptote.Asuccessionofstepsallowustosketchthesum s(x) = f (x) + g(x).
1 Thecurvesintersectat(2,2)andat( 1, 1),so
s(2) = f (2) + g(2) = 2 + 2 = 4,
s( 1) = f ( 1) + g( 1) = 1 + ( 1) = 2.
2 Thecurve y = f (x)hasazeroat x = 2,so
s( 2) = f ( 2) + g( 2) = 0 + ( 2) = 2.
3 Addtheordinatesat x = 1,so
s(1) = f (1) + g(1) = 3 + 1 = 4.
4 Therearenovaluesof x whereordinatesareopposites,so s(x)hasno zeroes.
5 Becausetherearenozeroesfor s(x),itcanonlychangesignatthe asymptote x = 0(nextstep).Hence s(x)isnegativefor x < 0,andpositive for x > 0.
Dealingwiththehorizontalandverticalasymptotesof f (x):
6 Vertically,onbothsidesofthe y-axis:
As x → 0+ , f (x) → +∞ and g(x) → 0,so s(x) → +∞.
As x → 0 , f (x) →−∞ and g(x) → 0,so s(x) →−∞. Thusthe y-axisisaverticalasymptoteto y = s(x).
7 Horizontally,ontherightandtheleft:
As x →∞, f (x) → 1and g(x) → +∞,so s(x) → +∞. As x →−∞, f (x) → 1and g(x) →−∞,so s(x) →−∞.
Note: Thereisafinaldetailthatseemsbeyondthecourse,buthasbeenaddedtothediagramforcompleteness. Youwillseethatathirdasymptote,anobliqueasymptote,hasbeendrawn.Hereistheargumentforit. Forlargevaluesof x,positiveornegative, f (x)isalmost1,so y = s(x)isalmostthesamegraphas y = 1 + g(x).
Hence y = s(x)eventuallylookslikealineparallelto y = g(x).
Havingnowarguedfromthegraphsalone,wewillnowrevealtheequationsofthetwofunctions—theyare f (x) = 2 x + 1and(obviously) g(x) = x.Theirsumis
s(x) = x + 1 + 2 x = x2 + x + 2 x . Thishasnozeroesbecausethediscriminant ∆= 1 8 = 7isnegative,andclearlythegraphhasavertical asymptoteat x = 0.
11Sketchingthesumoftwosketchedfunctions
• Tosketchthesum s(x) = f (x) + g(x)oftwosketchedfunctions:
0 Addtheordinateswhereverpossible.Thisisthekeyidea.
• Somesystematicapproaches:
1 Ifonecurvehasazero,then s(x)meetstheothercurveatthatvalueof x.
2 Ifthecurvesmeet,thentheordinateof s(x)isdoubletheordinateof f (x).
3 Iftheordinatesof f (x)and g(x)areopposites,then s(x)hasazero.
4 Thesignof s(x)everywherewillusuallybeclearnow.
• Toclarifyanyverticalorhorizontalasymptoticbehaviour:
5 If f (x) →∞ or g(x) →∞,thensoalsodoes s(x).
If f (x) →−∞ or g(x) →−∞,thensoalsodoes s(x).
If,however, f (x)and g(x)goinoppositedirections,wemaynotbeabledeterminewhathappens withthesum.
AsdiscussedinSection5A,atranslationof y = f (x)up5is y = f (x) + 5.Thisis f (x) + g(x)where g(x) = 5isa constantfunction,sotranslationsupanddownarespecialcasesoftheconstructioninthissection.
Nowletussketchthedifference d(x) = f (x) g(x)ofthesametwofunctions f (x) = x 2 36and g(x) = 5x , butthistimewewillworkfromthesketchesaloneandthenconfirmthesketchusing atableofvalues.Thedifferenceisthesumof f (x)and g(x),soitcouldbedoneby reflecting g(x)inthe x-axisandthenaddingthegraphs,butitiseasiertosketchitin onestep. x y ygx() yfx()
0 Subtracttheordinateswherepossible—thekeyidea.
1 If f (x)hasazeroat x = a,then d(a) = 0 g(a) = g(a),sotheordinateof d(x)is theoppositeoftheordinateof f (x).Thishappensat x = 6andat x = 6. If g(x)hasazeroat x = a,then d(a) = f (a) 0 = f (a),sotheordinateof d(x)isthe sameastheordinateof f (x),andthecurve d(x)meetsthecurve f (x).Thishappens at x = 0.
2 Iftheordinatesof f (x)and g(x)areopposites,thentheordinateof d(x)isdoublethe ordinateof f (x).Thishappensat x = 9andat x = 4.
3 Ifthetwocurvesmeetat x = a,thentheyhaveequalordinatesthere,so d(x)hasa zeroat x = a.Thishappensat x = 4andat x = 9. x y ygx() yfx() ydx()
Hereisatableofvaluestoconfirmthesearguments: x 9 6 4
f (x) 450 20 35 36 35 20045
g(x) 45 30 20 505203045
d(x) 90300 30 36 40 40 300
Again,wecannotfindfromthegraphsalonethepreciselocationoftheminimum,soitisnotneededinthe sketch.Itisthevertexoftheparabola y = d(x).Readersshouldcompletethesquarefor d(x)andshowthatthe vertexis(2 1 2 , 42 1 4 ).
Theprocedureshereareverysimilartothepreviousexampleofaddinggraphswithasymptotes,sothereisno needforanotherexample—andkeepinmindthatsubtractinggraphsmeansaddingtheoppositeofthesecond graph.
ThestructuredQuestion10inExercise4Dbelowpresentsthestepsinsubtractinggraphs,andtheEnrichment Question15dealswiththeobliqueasymptotes.
12Sketchingthedifferenceoftwosketchedfunctions
• Tosketchthedifference d(x) = f (x) g(x)oftwosketchedfunctions.
0 Subtracttheordinateswhereverpossible.Thisisthekeyidea.
• Somesystematicapproaches:
1 If f (x)hasazeroat x = a,then d(a) = g(a).
If g(x)hasazeroat x = a,then d(a) = f (a).
2 Ifthecurvesmeetat x = a,then d(a)hasazerothere.
3 Iftheordinatesareoppositesat x = a,then d(a) = 2 f (a) = 2g(a).
4 Thesignof d(x)everywherewillusuallybeclearnow.
• Toclarifyanyverticalorhorizontalasymptoticbehaviour:
5 If f (x) →∞ or g(x) →−∞,then d(x) →∞
If f (x) →−∞ or g(x) →∞,then d(x) →−∞
If,however, f (x)and g(x)gointhesamedirection,wemaynotbeabledeterminewhathappens withthedifference.
Theseneedattention,butthereisnorealdifficultyinvolved.
13Domainandrangeofthesumanddifference
Let s(x)and d(x)bethesumanddifferenceoftwofunctions f (x)and g(x).
• s(x)and d(x)havedomaintheintersectionofthedomainsof f (x)and g(x).
• Sortouttherangesof s(x)and g(x)fromthegraphs,ortheequations.
1 Eachdiagrambelowshowsthegraphoftwofunctions, y = f (x)and y = g(x).Copyeachdiagramtoyour bookanddrawthegraphof y = f (x) + g(x),byaddingordinates.Trytodistinguishtheoriginalgraphsfrom thegraphofthesum—usedifferentcolours,ordottheoriginalgraphs.
2 CopyeachdiagraminQuestion1toafreshnumberplane.Subtractordinatestosketchthegraphsof y = f (x) g(x).
3 Thediagramtotherightshowsthegraphsof y = f (x),where f (x) = x4,andof y = g(x),where g(x) = x2
a Copythediagramtoyourbook.
b Onthesamesetofaxesandinadifferentcolour,sketch y = f (x) g(x)by subtractingordinates.Paycarefulattentiontopointswherethegraphscross, andtothezeroesof f (x)and g(x). x y
4 Thediagramtotherightshowsthegraphsof y = f (x),where f (x) = x2,andof y = g(x),where g(x) = x.
a Copythediagramtoyourbook.
b Onthesamesetofaxesandinadifferentcolour,sketch y = f (x) + g(x) byaddingordinates.Paycarefulattentiontopointswhere g(x) = f (x), becauseatthosepoints f (x) + g(x) = 0.Noticealsothezeroesof f (x) because f (x) + g(x) = f (x)atthosepoints,andthezeroesof g(x),because f (x) + g(x) = g(x)atthosepoints.
5a Plot y = x3 and y = x onthesamenumberplane,notinganypointsofintersection.
b Hencesketchthegraphofthedifference, y = x3 x.
6 Sketch y = x4 and y = x(2 x),thensketchthedifference y = x4 x(2 x).
7 Whensketchingthesumsofabsolutevaluegraphsinthisquestion,itishelpfultorememberthatthesum oftwolinearfunctionsisitselfalinearfunction.Thusthefollowinggraphswillbemadeupofstraight-line sections.
a Graph f (x) = |x + 1| and g(x) = |x 1|,thengraph: y = f (x) + g(x) i y = f (x) g(x) ii
b Graph f (x) = |2x| and g(x) = |x 1|,thengraph:
y = f (x) + g(x) i y = f (x) g(x) ii
8 Graphsthefunctions y = x2 and y = |x| onthesamesetofaxes.
ai Explainwhythegraphof y = x2 + |x| mustlieabovebothoriginalgraphs.
ii Hencesketchthesum y = x2 + |x|.Takecarewiththeshapeat x = 0.
bi Explainwhythegraphof y = x2 −|x| isonorbelowthe x-axisfor 1 ≤ x ≤ 1.
ii Hencesketchthedifference y = x2 −|x|.Takecarewiththeshapeat x = 0.
9a Sketch y = √x and y = x,payingcarefulattentiontothedomainandtothepointswherethesegraphs intersect.
b Hencesketch y = x √x.
10 Thegraphontherightisarepeatoftheoneusedinthetheorytoinvestigate thesumoffunctionsthatinvolvesanasymptote.Inthisquestionyouwill investigatetheirdifference
d(x) = f (x) g(x).
a Thegivencurvesintersectat x = 1and x = 2.Whatfeatureofthegraph of y = d(x)occursatthesevalues?
b Writedownthezeroof f (x)andhenceevaluate d(x)there.
c Evaluate d(1)byusingtheordinatesof f (x)and g(x).
d Copyandcompleteeachstatementrelatingtotheverticalasymptoteof y = f (x).
ii
As x → 0+ , f (x) → ... and g(x) → ... so d(x) → ... i As x → 0 , f (x) → ... and g(x) → ... so d(x)
e Copyandcompleteeachstatementrelatingtothehorizontalasymptoteof y = f (x).
As x → +∞, f (x) → and g(x) → so d(x) → i
As x →−∞, f (x) → and g(x) → so d(x) → ii
f Brieflyexplainwhy d(x)approaches y = 1 x forlargevaluesof x.
g Sketch y = d(x)showingallthesefeatures.
11 Foreachpairoffunctions,firstgraph y = f (x)and y = g(x)onthesamenumberplane.Theninadifferent colourgraphthesum y = f (x) + g(x).
f (x) = 1 x and g(x) = x a f (x) = 1 x and g(x) = 1 x + 2 b f (x) = 2x and g(x) = x 1 c f (x) = 2x and g(x) = 2 x d
12 Foreachpairoffunctionsinthepreviousquestion,sketchthedifference y = f (x) g(x).
13 Itmaybethatif y = f (x)and y = g(x)exhibitevenoroddsymmetrythenweknowthesymmetryof s(x) = f (x) + g(x)andof d(x) = f (x) g(x).
a Trytocompletethefollowingtable,predictingthesymmetryoftheresultingfunctions,andtestyour claimsontheexamplesinthisexercise.
f (x)even, g(x)even f (x)odd, g(x)odd f (x)even, g(x)odd s(x)
d(x)
b Proveformallythatif f (x)and g(x)arebothoddthen s(x)isalsoodd.Thatis,provethat s( x) = s(x) forall x initsdomain.
14 Considerthetwofunctions f (x) = 1 2 x 3and g(x) = 1 + 1 x 1 .Thepurposeofthisquestionistosketch thegraph y = f (x) + g(x).
a Findthe x-coordinatesofthepointswhere f (x) = g(x).
b Plottheline y = f (x)andthehyperbola y = g(x)onthesamenumberplane,andmarkthepointsfoundin part(a).
c Showthat f (x) + g(x) ( 1 2 x 2) → 0as x → +∞ andas x →−∞.Addtheline y = 1 2 x 3toyourgraph. Thisisan obliqueasymptote tothecurve y = f (x) + g(x).
d Completeyoursketchofthesum y = f (x) + g(x).
15 Inthetext,thesum s(x) = f (x) + g(x)wasdrawnforthefunctions f (x) = 2 x + 1and g(x) = x,andthe difference d(x) = f (x) g(x)wasdrawninQuestion10.Theobliqueasymptotesof s(x)and d(x)werealso drawn.Thisquestionnowfindstheequationsoftheobliqueasymptotestothetwocurvesbythemethod usedinthepreviousquestion.
a Simplify s(x) = f (x) + g(x),andshowthat s(x) (x + 1) = 2 x .Henceexplainwhy y = x + 1isanoblique asymptotetothecurve y = s(x).
b Similarlysimplify d(x) = f (x) g(x),andsofinditsobliqueasymptote.
Learningintentions
• Sketch y = | f (x)| and y = f (|x|),giventhegraphof y = f (x).
• Recognisethesefunctionsascompositesof f (x)withtheabsolutevaluefunction.
• Associatetheresultinggraphswithreflectionsintheaxes.
Theprobleminthissectionistotakesomegraph y = f (x),andfromitsketch y = | f (x)| and y = f (|x|).
Theseoperationscanbedoneverysimplyusingreflectionsinthe x-axisand y-axis.
Identifyingthecomposites
Let g(x) = |x| betheabsolutevaluefunction.Then
| f (x)| = g f (x) isthecompositeformedby f (x)followedby g(x),and
f (|x|) = f g(x) isthecompositeformedby g(x)followedby f (x).
Thusthesetwomodifiedfunctions y = | f (x)| and y = f (|x|)arecompositesof f (x)andtheabsolutevaluefunction inthetwopossibleorders.
Areviewoftheabsolutevaluefunction
Ourgraphswillrelyonthealgebraicdefinitionof |x| usingcases:
x,for x ≥ 0(forexample, |5| = 5and |0| = 0) x,for x < 0(forexample, |− 5| = 5).
Expressedinwords:
• Theabsolutevalueofapositivenumberorzeroisunchanged.
• Theabsolutevalueofanegativenumberistheopposite,soispositive. Thusanabsolutevaluecanneverbenegative.
Sketching y = | f ( x)| fromthesketchof y = f ( x)
Eachabsolutevaluetransformationwillbeillustratedinturnusingthegraphtothe right,whichistheparabola
f (x) = (x + 1)(x 3).
Graphing y = | f (x)| requirestwoarguments:
• Whenthegraphisaboveoronthe x-axis, f (x)ispositiveorzero.Hence | f (x)| = f (x),sothegraphisunchanged. y x
• Whenthegraphisbelowthe x-axis, f (x)isnegative.Hence | f (x)| = f (x)isthe oppositeof f (x).
■ Thusreplaceeverypartofthegraphbelowthe x-axisbyitsreflectionbackabove the x-axis.
Atableofvalueshelpstounderstandthesituation:
)
14Tosketch y = | f ( x)| fromthesketchof y = f ( x)
• Everythingaboveandonthe x-axisstaysthesame.
• Replaceeverythingbelowthe x-axisbyitsreflectionbackabovethe x-axis.
Sketching y = f (| x|) fromthesketchof y = f ( x)
Theproceduretograph y = f (|x|)againhastwoarguments:
• Totherightoforonthe y-axis, x ispositiveorzero. Hence |x| = x,sothegraphisunchanged.
• Totheleftof x-axis, x isisnegative,so |x| = x,and f (|x|) = f ( x).
■ Thusreplaceeverypartofthegraphleftofthe y-axisbyitsreflectionbackacross the y-axis.
Thetableofvaluesillustratingthishasapreliminaryrowfor |x|: x 4 3 2 101234
15Tosketch y = f (| x|) fromthesketchof y = f ( x)
• Everythingtotherightofthe y-axisstaysthesame.
• Replaceeverythingleftofthe y-axisbyitsreflectionbackacrossthe y-axis.
Theresultingfunctioniseven,thatis,ithaslinesymmetryinthe y-axis.Toillustratethis,thetableofvaluesis clearlysymmetricabout x = 0.
Combiningthesetransformationstosketch y = f (| x|)
Thesetwoabsolutevaluetransformationcanbecombined,asinthetwoexamplesbelow.
Earlierinthissection,wesketched f (x) = (x + 1)(x 3)anditstransformations y = | f (x)| and y = f (|x|).Use thesegraphstosketch y = f (|x|) :
Solution
Startwiththeearliersketchofeither | f (x)| or f (|x|). y x
Usingthegraphof y = f (x)sketchedtotheright,sketch:
y = | f (x)|
Solution
y
y = f (|x|)
y = f (|x|) y x
1 Thegraphof y = f (x)issketchedtotheright.Todraweachtransformation,copy thegraphanddrawthetransformedgraphinadifferentcolouronthesameaxes. Replaceanypartofthegraphbelowthe x-axisbyitsreflectionabovethe x-axis.Thiswillgiveyouthegraphof y = | f (x)|.
a
b x y
Graphonlythepartsof y = f (x)thataretotherightofthe y-axis.Thenaddthe reflectionofthesepartsinthe y-axis.Thiswillgiveyouthegraphof y = f (|x|).
Noticethesymmetryinthe y-axis.
c
Usingthegraphof y = f (|x|),replaceanypartofthegraphbelowthe x-axisbyitsrefectionabovethe x-axis.Thisgivesthegraphof y = f (|x|)
d
Usingthegraphof y = | f (x)|,graphonlythepartsthataretotherightofthe y-axis.Thenaddthe reflectionofthesepartsinthe y-axis.Whatdoyounoticeabouttheresultofthisandtheanswertopart (c)?
2 Ineachcase,followthestepsofQuestion1andusethegivengraphof y = f (x)tosketch:
3 Sketcheachgivenfunctionanduseittothensketch:
4a Explainwhythegraphsof y = 2x and y = |2x| arethesame.
b Write y = 2|x| usingcases,thensketchitsgraph. DEVELOPMENT
5a Sketch y = f (x)where f (x) = (x + 1)(x 2),showingall x-interceptsandthevertex.
b Hencesketch:
6 Repeatthestepsofthepreviousquestionforthefunction f (x) = x 1 1.
7a Graph y = f (x),where f (x) = 1 x 1 + 1.Becarefultoidentifyanyinterceptswiththeaxesandany asymptotes.
b Hencesketch:
8a Let f (x) = x(x 2).
Sketch y = f (x). i
Userepeatedtransformationstosketch y = | f (|x|)| ii
b Repeatpart(a)forthefunction f (x) = (x + 1)(3 x).
9a Showthat y = |2 x| isneitherevennoroddandgraphit.
b Showthat y = 2 −|x| isevenandusepart(a)tographthisnewfunction.
c Hencegraph y = 1 2 −|x|
d Finallygraph y = 1 2 −|x| .
10a Let f (x)beanyfunction.Explainwhy g(x) = f (|x|)iseven.
b Let f (x)beanoddfunction.Showthat h(x) = | f (x)| iseven.
11a Whenwillthegraphsof y = f (x)and y = | f (x)| bethesame?
b Whenwillthegraphsof y = f (x)and y = | f (x)| besymmetricinthe x-axis?
CHALLENGE
12a ReadcarefullytheinstructioninQuestion1(b)forgraphing y = f (|x|).Writeasimilarinstructionfor graphing |y| = f (x).
b Testyourinstructionbygraphing |y| = f (x)foreachofthefunctionsinQuestion2.
13 Sketch |y| = |x|
14a Describethegraphof y = f (−|x|)intermsofthegraphof y = f (x).
b Whattypeofsymmetrymust y = f (−|x|)possess?
15 Let f (x)beanyfunction,andlet g(x) = f (|x|)and h(x) = 1 2 ( f (x) + f ( x)).
Provethatboth g(x)and h(x)areevenfunctions. a Are g(x)and h(x)alwaysthesamefunction?Ifsothenproveit,otherwisegiveacounter-example. b
16a Investigatethegraphsofthesequenceoffunctions
b Showthatthe2nd,4th,8th,...functionsinthissequencecanbesimplifiedto
Learningintentions
• Constructtheformulaoftheinverseofarelationorfunction.
• Usethehorizontallinetesttoseewhethertheinverserelationisafunction.
Mathematicsisfullofinverseprocesses:
• Theinverseofmultiplyingby7isdividingby7—andtheinverseofdividingby7ismultiplyingby7.
• Theinverseofshiftingup3isshiftingdown3—andtheinverseofshiftingdown3isshiftingup3.
• Theinverseofreflectinginthe y-axisisreflectinginthe y-axis—thisprocessisitsowninverse.
Thissectionusesgeometricandgraphicalmethodstoobtaintheinverseofarelation,andtofindasimple geometricconditionfortheinversetobeafunction.
Cubethenumber5andget125.Theinverseprocessistakingthecuberoot,whichsends125backto5.We candothiswithanynumber,positive,negativeorzero,sothecubingfunction y = x3 hasawell-defined inverse function y = 3 √x thatsendsanyoutputbacktoitsoriginalinput.Whenthesetwofunctionmachinesareput togetherinachain,ineitherorder,the composition ofthetwofunctionsis theidentityfunction thatmapsevery numbertoitself:
Theexchangingofinputandoutputcanalsobeseeninthetwotablesofvalues,wherethetworowsare interchanged:
Thisexchangingofinputandoutputmeansthatthecoordinatesofeachorderedpairareexchanged.Rememberingthatarelationwasdefinedsimplyasasetoforderedpairs,weareledtoadefinitionofinversethatcanbe appliedtoanyrelation,whetheritisafunctionornot:
• The inverserelation ofanyrelationisobtainedbyreversingeachorderedpair.
• Theinverserelationoftheinverserelationistheoriginalrelation.
Thesecondstatementfollowsfromthefirstbecauseeachorderedpairreturnstoitsoriginalstatewhenreverseda secondtime.Forexample,thepair(2,8)intheoriginalbecomesthepair(8,2)intheinverse,andreversedgoes backto(2,8).
Reversinganorderedpairmeansthattheoriginalfirstcoordinateisreadoff thevertical axis,andtheoriginalsecondcoordinateisreadoff thehorizontalaxis.
Geometrically,thisexchangingofthetwocoordinatescanbedonebyreflectingthe pointinthediagonalline y = x.Thiscanbeseenbycomparingthegraphsof y = x3 and y = 3 √x,whicharedrawnhereonthesamepairofaxes.
17Thegraphoftheinverserelation
Thegraphoftheinverserelationisobtainedbyreflectingtheoriginalgraphinthediagonalline y = x
Domainandrangeoftheinverserelation
Exchanging x-and y-coordinatesmeansthedomainandrangeareexchanged:
18Domainandrangeoftheinverserelation
• Thedomainoftheinverseistherangeoftherelation.
• Therangeoftheinverseisthedomainoftherelation.
Whenthecoordinatesareexchanged,the x-variablebecomesthe y-variableandthe y-variablebecomesthe x-variable.Hencethemethodforfindingtheequationandrestrictionsoftheinverseis:
19Theequationoftheinverserelation
• Tofindtheequationandrestrictionsoftheinverserelation,write x for y and y for x everytimeeach variableoccurs.
• Thisprocesscanbeappliedtoanyrelationwhoseequationandrestrictionsareknown,whetherornotit isafunction.
Forexample,theinverseofthefunction y = x3 is x = y3.Thisparticularequationcanthenbesolvedfor y togive y = 3 √x,confirmingthatinthisparticularcase,theinverserelationisagainafunction.
Example18 Graphingfunctionandinversetogether
a Writedowntheinverserelationofthefunction y = x2 .
b Graphbothrelationsonthesamenumberplane,showingthereflectionline.
c Writedownthedomainandrangeofbothrelations.
d Istheinverserelationafunction?
Solution
a Writing x for y and y for x,theinverseis x = y 2 .
c Fortheoriginal,domain:allreal x,range: y ≥ 0. Fortheinverse,domain: x ≥ 0,range:allreal y. Noticehowthedomainandtherangehavebeen swapped.
d Theinverseisnotafunction—itfailsthevertical linetest. b x y
Example19 Graphingfunctionandinversetogether
Repeatthepreviousquestionsforthefunction y = x3 + 2.
Solution
a Writing x for y and y for x,theinverseis x = y 3 + 2, whichis y = 3 √x 2.
c Forboth,domainandrangeareallrealnumbers.
d Theinverseisagainafunction,bytheverticalline test.
Whenthereareanyrestrictions,then x and y mustbeswappedintheseaswell,asinthenextexample.
Example20 Formingtheequationoftheinversewithrestrictions
Considerthefunction y = 2x 3,where x > 1.
a Writedowntheequationandrestrictionoftheinverserelation.
b Rewritetheinverseasafunctionwith y asthesubject,changingtherestrictiontoarestrictionon x
Solution
a Thefunctionis y = 2x 3,where x > 1, sotheinverserelationis x = 2y 3,where y > 1.
b Solvingfor y, y = 1 2 (x + 3),where y > 1. Therestriction y > 1is 1 2 (x + 3) > 1 x
Ingeneral,theinverseofarelationisnotafunction.Forexample,thesketchesbelowshowthegraphsofanother function,witharestriction,anditsinverse:
Thefirstgraphisafunction,passingthe verticallinetest.Butwhenwereaditbackwards,thevalue y = 3gives twoanswers, x = 1and x = 1.Accordingly,whenwesketchtheinverserelation,weseethatitfailsthevertical linetest,with x = 3crossingthegraphtwice.
Butreflectionin y = x exchangesverticalandhorizontallines!Nowwecanseeimmediatelyfromthefirstgraph thatitsinverseisnotafunction,becauseitfailsthe horizontallinetest —theline y = 3crossesitmorethanonce. Wedidn’tneedtodrawthesecondgraphtoknowthattheinverseisnotafunction.Allweneedtoknowisthat thefirstgraphfailsthehorizontallinetest.
20Horizontallinetestforwhethertheinverseisafunction
• Geometrically,theinverserelationofagivenrelationisafunctionifandonlyifnohorizontalline crossestheoriginalgraphmorethanonce.
Wecouldalsohavedoneittheslowway—solvetheequation x = 4 y2 oftheinverserelationtogive y = √4 x or y = √4 x ,showingagainthatformanyvaluesof x,thereismorethanonevalueof y,sothe inverseisnotafunction.
1 Drawtheinverserelationofeachrelationbyreflectingintheline y = x
2 Usetheverticalandhorizontallineteststodeterminewhichrelationsandwhichinverserelationsdrawnin Question1arealsofunctions.
3 Determineeachinversealgebraicallybyswapping x and y andthenmaking y thesubject. y = 3x 2 a
4 Foreachfunctioninthepreviousquestion,drawagraphofthefunctionanditsinverseonthesamenumber planetoverifythereflectionproperty.Drawaseparatenumberplaneforeachpart.
5a Findeachinversealgebraicallybyswapping x and y andthenmaking y thesubject. y = 1 x + 1 i y = 1 x + 1 ii y = x + 2 x 2 iii y = 3x x + 2 iv
b Forparts(i)and(iv)above,findthedomainandrangeofthefunction,andthedomainandrangeofthe inversefunction.
6 Swap x and y andsolvefor y tofindtheinverseofeachfunction.Whatdoyounotice,andwhatisthe geometricsignificanceofthis?
7a Grapheachrelationanditsinverserelation,thenfindtheequationoftheinverserelation.Whichofthe fouroriginalrelationsarefunctions,andwhichinverserelationsarefunctions?
= x2 4 iii
b Forparts(i)and(iv)above,findthedomainandrangeoftherelation,andthedomainandrangeofthe inverserelation.
8 Writedowntheinverseofeachfunction,solvingfor y ifitisafunction.Sketchthefunctionandtheinverse onthesamegraphandobservethesymmetryintheline y = x.
9 Eachfunctionbelowhasarestriction.Writedownitsinverserelation.Thenattempttosolveitfor y.Ifthe inverseisafunction,rewritetherestrictionasarestrictionon x.Iftheinverseisnotafunction,giveavalue of x thatcorrespondstotwoormorevaluesof y. y = 3x 10,where x < 2
= x3 + 2,where x < 3
10a Factorise f (x) = x2 2x 3inordertoshowthatthegraphof y = f (x)failsthehorizontallinetest.
b Let g(x) = x2 2x 3for x ≥ 1.Explainwhythisfunctionhasaninverseandfinditsequation.
11a Showthattheinversefunctionof y = ax + b x + c is y = b cx x a .
b Henceshowthat y = ax + b x + c isitsowninverseifandonlyif a + c = 0.
12a Showthattheinverseof y = 2x + 2 x 2 isnotafunction.
b Showthattheinverseof y = 2x 2 x 2 isafunction.
13a Showthatifthedomainofanevenfunctioncontainsanon-zeronumber,thenitsinverseisnota function.
b Istheinverseofanoddfunctionalwaysafunction?Ifnot,giveacounter-example.
14a ThepolynomialinQuestion10(a)hastwolinearfactorsandfailsthehorizontallinetest,soitsinverseis notafunction.Explainwhyallpolynomialswithatleasttwolinearfactorsmustautomaticallyfailthe horizontallinetest.
b Doanypolynomialshaveaninversewhichisafunction?
Learningintentions
• Usethealternativedefinitionofinversefunctionintermsofcomposition.
• Usethestandardnotationfortheinversefunction.
• Restrictafunctionorrelationappropriatelysothatitsinverseisafunction.
Thissectionignoresrelations,anddealsonlywithfunctionsandinversefunctions.
Afunction f (x)mustbydefinitionpassthe verticallinetest,meaningthateveryvalueof x inthedomain correspondstoexactlyone y-value.Fortheinverseof f (x)tobeafunctionalso, f (x)mustpassthe horizontal linetest,meaningthateveryvalueof y intherangecorrespondstoexactlyone x-value.
Thustheconditionfortheinverseof f (x)tobeafunctionisthat f (x)isa one-to-onecorrespondence betweenthe elementsofthedomainandtheelementsoftherange.Suchafunctioniscalledsimply one-to-one.
• Afunction f (x)iscalled one-to-one,ora one-to-onecorrespondence,ifforeveryvalueof y inthe range,thereisexactlyoneelement x inthedomainsothat f (x) = y
• Theinverserelationofafunction f (x)isalsoafunction:
▷ ifandonlyif f (x)passesthehorizontallinetest,and
▷ ifandonlyif f (x)isone-to-one,and
▷ ifandonlyiftheequationoftheinversecanbesolveduniquelyfor y.
Thecompositeofafunctionanditsinversefunctionistheidentity
Supposethat f (x)isaone-to-onefunctionwithinversefunction g(x).Then g(x)isalsoaone-to-onecorrespondence,withthesamepairingofthedomainandrangeasprovidedby f (x),butwiththecorrespondingpairs reversed.Whenweapply f (x)then g(x),itisasifnothinghashappened,thatis,
g f (x) = x,forall x inthedomainof f (x).
Andbecause g(x)sendseachnumberbackwhereitcamefrom,itsinverseis f (x),andtheothercomposite f g(x) isalsoanidentityfunction,
f g(x) = x,forall x inthedomainof g(x).
Whenwerestrictdiscussiontofunctions,thesetwoconditionscanbetakenasanalternativedefinitionofan inversefunction:
• Thefunction g(x)istheinversefunctionofafunction f (x)ifandonlyif: g f (x) = x,forall x inthedomainof f (x),and f g(x) = x,forall x inthedomainof g(x).
Thatis,ifandonlyif g f (x) and f g(x) arebothidentityfunctions.
• An identityfunction isafunction I(x)suchthat:
I(x) = x,forall x inthedomainof I(x).
Supposethat f (x)isaone-to-onefunction,thatis,itsinverseisalsoafunction.Thenthatinversefunctionis writtenas f 1(x).Theindex 1usedheremeans‘inversefunction’andmustnotbeconfusedwithitsmore commonuseforthereciprocalofanumber.Toreturntotheoriginalexampleinthelastsection, If f (x) = x 3,then f 1(x) = 3 √x
Becareful: f (x) 1 = 1 f (x)
Aswehaveseen,theinversefunction f 1(x)isalsoone-to-one,withinverse f (x),andifthefunctionandthe inversefunctionareappliedsuccessivelyineitherorder,theresultistheoriginalnumber.Usingtheexample above,
23Inversefunctionnotation
• Ifafunction f (x)isone-to-one,thenitsinverserelationisalsoaone-to-onefunction,andiswrittenas f 1(x).
• Thecompositeofthefunctionanditsinverse,ineitherorder,sendseverynumberforwhichitis definedbacktoitself:
f 1 f (x) = x,forall x inthedomainof f (x) f f 1(x) = x,forall x inthedomainof f 1(x).
• Tofindtheformulafor f 1(x)fromtheformulafor f (x):
▷ Convertto y = notationtogeneratetheinverserelation.
▷ Write y for x and x for y inalltheequationsandrestrictions.
▷ Solvefor y,andconverttothenotation f 1(x) =
Examples21–22demonstratethemethoddescribedinthefinaldotpoint.
Example21 Verifyingtheformaldefinitionofaninversefunction
Findtheequationof f 1(x)foreachfunction,thenverifythat f 1 f (x) = x and f f 1(x) = x f (x) = x3 + 2 a f (x) = 6 2x,where x > 0 b
Solution
a Let y = x 3 + 2.
Thentheinversehasequation x = y 3 + 2(thekeystep) andsolvingfor y, y = 3 √x 2.
Hence f 1(x) = 3 √x 2.
Verifying, f 1 f (x) = 3 (x3 + 2) 2 = 3 √x3 = x and f f 1(x) = 3 √x 2 3 + 2 = (x 2) + 2 = x
b Let y = 6 2x,where x > 0.
Thentheinversehasequation x = 6 2y,where y > 0(thekeystep) y = 3 1 2 x,where y > 0.
Therestriction y > 0means3 1 2 x > 0 x < 6,
so f 1(x) = 3 1 2 x,where x < 6.
Verifying, f 1 f (x) = f 1(6 2x) = 3 1 2 (6 2x) = 3 3 + x = x and f f 1(x) = f (3 1 2 x) = 6 2(3 1 2 x) = 6 6 + x = x
Findtheinverserelationofeachfunction.Iftheinverseisafunction,findanexpressionfor f 1(x),andverify that f 1 f (x) = x and f f 1(x) = x f (x) = 1 x 1 + x a f (x) = x2 9 b
Whatissurprisingabouttheresultofpart(a)?
Solution
a Let y = 1 x 1 + x .
Thentheinversehasequation x = 1 y 1 + y (thekeystep) × (1 + y) x + xy
(now y occursonlyonce)
Noticethatthisfunction f (x)anditsinverse f 1(x)areidentical,sothatifthefunction f (x)isapplied twice,eachnumberissentbacktoitself.
Thus f f (2) = f 1 3 = 1 1 3 ÷ 2 3 = 2 andingeneral, f f (x) = 1 1 x 1+x 1 + 1 x 1
= (1 + x) (1 x) (1 + x) + (1 x) = x
b Thefunction f (x) = x2 9failsthehorizontallinetest.Forexample, f (3) = f ( 3) = 0,whichmeansthat the x-axismeetsthegraphtwice.Hencetheinverserelationof f (x)isnotafunction. Alternatively,theinverserelationis x = y2 9,whichonsolvingfor y gives y = √x + 9or √x + 9, whichisnotunique,sotheinverserelationisnotafunction.
Whenafunctionisnotone–to-one,thatis,itsinverseisnotafunction,itisoftenconvenienttorestrictthe domainofthefunctionsothatthisnewrestrictedfunctionhasaninversefunction.
Theclearestexampleisthesquaringfunction f (x) = x2,whoseinverserelationisnotafunctionbecause,for example,49hastwosquareroots,7and 7.
If,however,werestrictthedomainof f (x) = x2 to x ≥ 0anddefineanew restrictedfunction
g(x) = x 2,where x ≥ 0,
thenthenewrestrictedfunction g(x)isone-to-one,andthushasaninverse function.Thisinversefunctionhasequation g 1(x) = √x,whereasexplained earlier,thesymbol √ means‘takethepositivesquareroot(orzero)’.
Totherightarethegraphsoftherestrictedfunctionanditsinversefunction,with theunrestrictedfunctionanditsinverserelationshowndotted.
TheseideaswillbedevelopedagreatdealfurtherinYear12,wheninverseofthetirgonometricfunctionsare developed.
1 Let f (x) = 2x 8and g(x) = 1 2 x + 4.
a Verifybysubstitutionthat: g f (5) = 5 i f
b Whatdoyouconcludeaboutthefunctions f (x)and g(x)?
2 Eachpairoffunctions f (x)and g(x)areknowntobemutualinverses.Showineachcasethat f g(2) = 2 and g f (2) = 2,andthat f g(x) = x and g f (x) = x.
f (x) = x + 13and g(x) = x 13 a
(x) = 7x and g(x) = 1 7 x b f (x) = 2x + 6and g(x) = 1 2 (x 6) c f (x) = x3 6and g(x) = 3 √x + 6 d
3a Findtheinversefunction f 1(x)of f (x) = 2x + 5.Begin‘Let y = 2x + 5’,thenswap x and y tofindthe inverse,thensolvefor y,thenwritedowntheequationof f 1(x).
b Checkyouranswerbycalculating f 1 f (x) and f f 1(x)
c Similarlyfindtheinversefunctionsofeachfunction,andcheckeachanswer.
f (x) = 4 3x i f (x) = x3 2 ii
4 Explainwhethertheinverserelationisafunctionbytestingwhetheritisone-to-one.Ifitisafunction,find f 1(x),specifyingitsdomain.Thenverifythetwoidentities f 1 f (x) = x and f f 1(x) = x
f (x) = x2 a f (x) = √x b f (x) = x4 c f (x) = x3 + 1
(x) = x + 1 x 1 l
5 Let f (x)betherestrictedfunction f (x) = 3x 2,where1 ≤ x ≤ 4.
a Findtheinversefunction f 1(x),beingcarefultoadditsrestriction.
b Showthat f 1 f (x) and f f 1(x) arebothidentityfunctions,andfindtheirrespectivedomains.A sketchmaymakethesituationclearer.
6a Whatisthegradientoftheline y = ax + b?
b Writedowntheinverserelationof y = ax + b
c Whataretheconditionsforthisinverserelationtobeafunction?
d Whentheinverseisafunction,solveitfor y,finditsgradient,andexplainwhythegradientsofthe functionanditsinversebothhavethesamesign.
e Giveanargumentusingreflectionintheline y = x foryouranswersinpart(c).
7 Sketchonseparategraphs:
y = x2 a y = x2,for x ≥ 0 b
Drawtheinverseofeachonthesamegraph,thencommentonthesimilaritiesanddifferencesbetweenparts (a)and(b).
8 Theparabola y = f (x),where f (x) = (x 3)2 + 1,hasitsvertexat(3,1).Theinverseof f (x)isnota function.
ai Janinesaysthatwhensheappliestherestriction x ≥ a to f (x)theinverseisafunction.Whatisthe leastvalueof a?
ii Findtheequationoftheinverseinthiscase.
bi Jacobsaysthatwhenheappliestherestriction x ≤ b to f (x)theinverseisafunction.Whatisthe greatestvalueof b?
ii Findtheequationoftheinverseinthiscase.
9a Let f (x) = x2 2x 3withtherestriction x ≥ a.Itisknownthattheinverseof f (x)isafunction.Using thepreviousquestionasaguide,findtheleastvalueof a andfind f 1(x)inthatcase.
b Dothesameforthefunction f (x) = 5 4x x2 withtherestriction x ≤ a
10a Let f (x)and g(x)beone-to-onefunctions.Thatis,bothpassthehorizontallinetest.Let h(x) =
Showthattheinversefunctionof
bi Findtheinversefunctionof h(x) = 1 x 3 .
ii Express h(x)asthecompositionofthereciprocalfunctionandalinearfunction,andhenceuse part(a)tofinditsinversefunction.
11 Suggestrestrictionsonthedomainsofeachfunctiontoproduceanewfunctionwhoseinverseisalsoa function(theremaybemorethanoneanswer).Drawtherestrictedfunctionanditsinverse.
y = √4 x2 a y = 1 x2 b y = x3 x c y = √x2 d
)
CHALLENGE
12a Let f (x) = ax + b and g(x) = αx + β.Find g f (x) ,andhenceprovethattheconditionfor f (x)and g(x) tobemutuallyinversefunctionsis
α = a 1 and β = a 1b .
b Findthreelinearfunctions f (x), g(x)and h(x),noneofwhosegraphspassthroughtheorigin,withno twographsparallel,suchthat h g f (x) istheidentityfunction.
13 Doestheemptyfunctionhaveaninversefunction?Ifsothenwhatisit?
Learningintentions
• Dealwithafunctionorrelationdefinedparametrically.
• EliminatetheparametertoobtaintheCartesianequationofthecurve.
Thereisaningeniouswayofhandlingcurvesbymakingeachcoordinateafunctionofasinglevariable,calleda parameter.Eachpointonthecurveisthenspecifiedbyasinglenumber,ratherthanbyapairofcoordinates.
ThesectionusessometrigonometrythatisonlyreviewedinChapter7.Anglesofanymagnitudeandthe Pythagoreanidentitiesareneeded.Readersmayprefertodelaystudyingtheseexamplesandquestionsuntil Chapter7iscompleted.
SDBhitsasixattheSydneyCricketGround.
• Acameramaninadistantstand,exactlybehindthepathoftheball,seestheballriseandfall,withitsheight y inmetresgivenby y = 5t2 + 25t,where t istimeinsecondsafterthestrike.
• AdronefilmingtheshotfromhighabovetheCricketGroundseestheballmoveacrosstheground,with distance x fromthebatsmangivenby x = 16t
Thecameramanandthedronetogetherhaveacompleterecordoftheball’sflight.(Ignoreparallaxhere,and regardbothobserversas‘distant’).Fromtheirresults,wecandrawupatableofvaluesofthe(x, y)positionof theballateachtime t:
t 0122 1 2 345
x 0163240486480
y 0203031 1 4 30200
Theresulting(x, y)-graphshowsthepathoftheball,andeachpointonthe graphcanbelabelledwiththecorrespondingvalueoftime t
Thisvariable t iscalleda parameter,andtheequations
x = 16t, y = 5t2 + 25t arecalled parametricequationsofthecurve
Itispossibleto eliminatetheparameter t fromthesetwoequations. Solvingthefirstequationfor t, t = x 16 , thensubstitutingintothesecond, y = 5 256 x 2 + 25 16 x , y = 5x2 + 400x 256
andfactoringdisplaysthezeroes, y = 5x(80 x) 256
• Acurveinthe(x, y)-planemaybe parametricallydefined,meaningthat x and y aregivenasfunctions ofathirdvariable t calleda parameter.
▷ Thesetwoequationsfor x and y intermsof t arecalled parametricequations ofthecurve.
• Inmanysituations,theparameter t maybe eliminated togiveasingleequationin x and y forthecurve.
▷ Thesingleequationin x and y iscalledthe Cartesianequation ofthecurve.
Theletter t isoftenusedfortheparameterbecauseitstandsfor‘time’.Otherlettersareoftenused,however, particularly θ and φ foranangle,and p and q
Wecanreversetheprocessofeliminatingtheparameter,and parametrise familiarcurves.Somestraightforward parametrisationsaregivenbelowofaparabola,acircle,andarectangularhyperbola.
Theparabola x2 = 4y canbeparametrisedbythepairofequations
x = 2t and y = t2
becauseeliminationof t gives x2 = 4y.Thevariablepoint(2t, t2)nowrunsalong thewholecurveastheparameter t takesalltherealnumbersasitsvalues:
Thesketchshowsthecurvewiththesevenplottedpointslabelledbytheirparameter.Thecurvecanberegarded asa‘bentandstretchednumberline.’
Thecircle x2 + y2 = r2 canbeparametrisedusingtrigonometricfunctionsby
x = r cos θ and y = r sin θ
ThisparametrisationusesthePythagoreanidentity,becausesquaringthetwo equationsandaddingthem:
,becausecos
Noticefromthetableofvaluesbelowthatwiththeseequations,eachparameter correspondstojustonepoint,buteachpointcorrespondstoinfinitelymany differentvaluesoftheparameter,alldifferingbymultiplesof360◦:
Inourpreviousexamples,themapfromparameterstopointswasalwaysone-to-one,butinthisparametrisation ofthecircle,themapismany-to-one.
Thecircleisarelation,butnotafunction.Thegreatadvantageofparametrisingthecircleisthatitisnow describedbyapairof functions,whichinmanysituationsareeasiertohandlethantheoriginalrelation.
Aparametrisationoftherectangularhyperbola
Therectangularhyperbola xy = 1canbeparametrisedalgebraicallyby
x = t and y = 1
Thereisaone-to-onecorrespondencebetweenthepointsonthecurveandthe realnumbers,withtheoneexceptionthat t = 0doesnotcorrespondtoanypoint:
Example23 UsingtrigidentitiestofindtheCartesianequaton
FindtheCartesianequationsofthecurvesdefinedbytheparametricequations:
Describepart(a)geometrically.
Solution
Fromthefirst, p = 1 4 x,andsubstitutingintothe second, y = 1 16 x 2 + 1, whichisaparabolawithvertex(0,1)andconcave up. a Squaring, x 2 = sec2 θ, and y 2 = sin2 θ = 1 cos2 θ
Exercise6H
Note: SomequestionsinthisexerciseusetrigonometryreviewedinChapter7,inparticular,anglesofany magnitudeandthePythagoreanidentities:
1 Considertheparametricequations x = t 2and y = 2t 1.
a Completethetabletotheright.
b Explainfromthetablewhythegraphisaline.
c Fromthetable,findthe y-interceptandthegradient.
d Eliminate t tofindtheCartesianequation,andcheckitfrompart(c). t 2 1012 x y
2a Theparametricequations x = 2t 3and y = t + 1representaline. Findthepoints A and B withparameters t = 0and t = 1,andhencefindthegradientoftheline. i Findthevalueof t thatmakes x = 0,andhencefindthe y-intercept. ii Checkyouranswersbyeliminating t toformaCartesianequation. iii
b Repeatthestepsinpart(a)fortheselines: x = 2t 3and y = 6t 5 i x = 2t 3and y = 3t 2 ii
3a Completethetablebelowforthecurve x = 4t, y = 2t2 andsketchitsgraph. t 6 4 2 101246 x y
b EliminatetheparametertofindtheCartesianequationofthecurve.
c Thecurveisaparabola.Whatvalueof t givesthecoordinatesofthevertex?
4 Repeatthepreviousquestionforthecurve
5a Showthatthepoint cp, c p liesonthehyperbola xy = c2,where c isaconstant.
b Completethetableofvaluesbelowfor x = 2 p, y = 2 p andsketchthegraph. p 3 2 1 1 2 1 4 0 1 4 1 2 123 x y
c Explainwhathappensas p →∞, p →−∞, p → 0+ and p → 0 .
6a Showthatthepoint(a cos θ, b sin θ)liesontheellipse
bi Completeatableofvaluesforthecurve x = 4cos θ, y = 3sin θ,takingthevalues θ = 0◦,30◦,60
, 90◦,120◦ , ... ,360◦ . ii SketchthecurveandstateitsCartesianequation. DEVELOPMENT
7 EliminatetheparameterandhencefindtheCartesianequationofthecurve. x = 3 p, y = 2
+ 1 p , y = p2 + 1 p2 c
8a Showthat x = a + r cos θ and y = b + r sin θ defineacirclewithcentre(a, b)andradius r. b Hencesketchagraphofthecurve x = 1 + 2cos θ, y = 3 + 2sin θ.
9 Showbyeliminationthat x = t2 1 t2 + 1 and y = 2t t2 + 1 almostrepresenttheunitcircle x2 + y2 = 1.Whatpoint ismissing?
10 [ParametersandCurveOrientation]Let A = (1,2)and B = (2,1).
ai Showthat x = 1 + t, y = 2 t,0 ≤ t ≤ 1parameterisesthelinesegment AB. ii Describehowthepoint P(1 + t,2 t)movesas t increasesfrom0to1.
bi Showthat x = 2 u, y = 1 + u,0 ≤ u ≤ 1alsoparameterisesthelinesegment AB. ii Describehowthepoint P(2 u,1 + u)movesas u increasesfrom0to1.
c Thepoints A and B alsolieonthecirclewithcentre(1,1)andradius1.Considerthepoints P(1 + cos t,1 + sin t)and Q(1 + sin t,1 + cos t),where0◦ ≤ t ≤ 90◦ inbothcases.Explainthedifference betweenthecurvestracedoutbythepoints P and Q
11
DifferentparametricrepresentationsmayresultinthesameCartesianequation.Thegraphicalrepresentation,however,maydifferduetorestrictionsinthedomainorrange.
a
b
FindtheCartesianequationofthecurve x = 2 t, y = t 1andsketchitsgraph.
FindtheCartesianequationofthecurve(sin2 t,cos2 t).Explainwhy0 ≤ x ≤ 1and0 ≤ y ≤ 1andsketch agraphofthecurve.
c
FindtheCartesianequationofthecurve x = 4 t2 , y = t2 3.Explainwhy x ≤ 4and y ≥−3andsketch agraphofthecurve.
12 Acertaingraphhasparametricequations x = at + b and y = ct + d,wheretheconstants a, b, c and d arereal numbers.
a
Eliminate t fromtheseequationsandhenceshowthattheyrepresentastraightline.Expressyouranswer ingradient-interceptform.
b
c
Arethereanyrestrictionsthatneedtobeplacedontheanswertopart(a)?
Howcouldtheanswertopart(a)bewritteninordertoavoidtheserestrictions?Explainyouranswer.
13a Showthattheparametricequations x = c(sec θ tan θ)and y = c(sec θ + tan θ)with 90◦ < θ < 90◦ representstheportionofthehyperbola xy = c2 inthefirstquadrant.
b Whatrestrictionon θ isneededtogettheportioninthethirdquadrant?
14 FindtheCartesianequationofthecurve x = 3 + r cos θ, y = 2 + r sin θ,anddescribeitgeometricallyif: r isconstantand θ isvariable, a θ isconstantand r isvariable. b
15 [ParametersandTransformations]
a Thecirclewithcentretheoriginandradius2hasparametricequations x = 2cos θ and y = 2sin θ with 180◦ < θ ≤ 180◦ .
Byconsideringtranslations,writedowntheparametricequationswhenthiscircleisshiftedright1 andup3. i
Byconsideringdilations,writedowntheparametricequationswhentheoriginalcircleisstretched horizontallyby2andverticallyby 1 2 . ii
b Acertaincurvehasparametricequations x = f (t)and y = g(t).Ignoringanypossiblerestrictionson t, answerthefollowing.
Describethecurvegeneratedby x = f (t) + h and y = g(t) + k. i
Describethecurvegeneratedby x = af (t)and y = bg(t). ii
16 Showbyeliminationthat x = 2
Whatpointismissing?
17a Explainwhytheparametricequations x = cos t, y = sin t, z = t describeaspiral.
18a Showthatthepoint(a sec θ, b tan θ)liesonthecurve x2 a2 y2 b2 = 1.
b Completeatableofvaluesforthecurve x = 4sec θ, y = 3tan θ,where0◦ ≤ θ ≤ 360◦.Whathappens when θ = 90◦ and θ = 270◦?
c Sketchthecurve(ithastwoasymptotes)andstateitsCartesianequation.
19 AfterfindingtheCartesianequation,sketchthecurvewhoseparametricequationsare x = 1 2 (2t + 2 t )and y = 1 2 (2t 2 t ).
20 Arelationisdefinedparametricallyby x = f (t)and y = g(t).
a Whattransformationoftherelationoccurswhen t isreplacedby t if:
f (t)and g(t)arebotheven, i
f (t)isevenand g(t)isodd, iii
f (t)and g(t)arebothodd, ii
f (t)isoddand g(t)iseven. iv
b Whatistherelationshipbetweenthisrelationandtherelationdefinedby x = g(t)and y = f (t)?
c Whereisthegraphoftherelation x = | f (t)| and y = |g(t)| located?
d Whereisthegraphoftherelation x = f (t)and y = f (t)located?
• Createyourownsummaryofthischapteronpaperorinadigitaldocument.
• Thisautomatically-markedquizisaccessedintheInteractiveTextbook.AprintablePDFWorksheetversionis alsoavailablethere.
• Checklist AvailableintheInteractiveTextbook,usethechecklisttotrackyourunderstandingofthelearningintentions. PrintablePDFandworddocumentversionsarealsoavailablethere.
Note: Graphingsoftwarecouldbeveryhelpfulinthisexercise.
1 Solveeachinequation.
2 Solveeachinequationbymultiplyingbothsidesbythesquareofthedenominator.
3 Considerthefunction f (x) = x(x + 2)(x 3).
Writedownthezeroesofthefunction,anddrawupatableofsigns. a Copyandcomplete:‘ f (x)ispositivefor...,andnegativefor...’ b Writedownthesolutionoftheinequation x(x + 2)(x 3) ≤ 0. c Sketchthegraphofthefunctiontoconfirmtheseresults. d
4 Considerthefunction y = (1 x)(x 3)2
Writedownthezeroesofthefunctionanddrawupatableofsigns. a Hencesolvetheinequation(1 x)(x 3)2 ≥ 0. b Confirmthesolutionbysketchingagraphofthefunction. c
5 Solveeachinequationinquestion4usingthetable-of-signsmethod.Firstmoveeverythingtotheleft-hand side,thenmaketheLHSintoasinglefraction,thenidentifyitszeroesanddiscontinuities,thendrawupa tableofsigns,thenreadthesolutionfromthetable.
6 Considerthelinearfunction f (x) = x 2.
Sketch y = f (x),clearlyindicatingthe x-and y-intercepts. a Alsoshowonyoursketchthepointswhere y = 1and y = 1. b Hencesketchthegraphof y = 1 f (x) onthesamenumberplane. c
Writedowntheequationoftheverticalasymptoteof y = 1 f (x) ,thencopyandcompletethesentence,‘As x → 2 , y → ,andas x → 2+ , y → ’ d
7 Considerthequadraticfunction f (x) = 3 x2 .
Sketch y = f (x),clearlyindicatingthe x-and y-intercepts. a
b
c
Alsoshowonyoursketchthepointswhere y = 1and y = 1.
Hencesketchthegraphof y = 1 f (x) onthesamenumberplane.
8
Sketchthereciprocalofeachfunctiongraphedbelow,showingalltheimportantfeatures.
9a
Findtheequationsoftheverticalasymptotesofeachfunction. y = 2 x + 1 i y = 2x + 1 x 2 ii y = 4x x2 25 iii
b Inpart(iii)above,identifythezeroesanddiscontinuitiesanddrawupatableofsigns.Thendescribethe behaviourofthecurveneareachverticalasymptotebycopyingandcompleting,‘As x → 5 , y → ... , andas x → 5+ , y → ... ’(andsimilarlyfor 5).
10 Considerthefunction y = 2x x2 1 .
a
Showthatitisanoddfunction.
b Identifyanyverticalandhorizontalasymptotes. c Hencesketchthegraph. d
Findthezeroesanddiscontinuities,anddrawupatableofsigns.
11 Thegraphsof y = f (x)and y = g(x)aresketchedtotheright.Onseparatenumber planes,sketch:
a y = f (x) + g(x)
b y = f (x) g(x)
12 Thegraphsof y = f (x)and y = g(x)aresketchedtotheright.
a Onseparatenumberplanes,sketch:
i y = f (x) + g(x)
ii y = f (x) g(x)
b Whatsymmetryshouldbeevidentinyoursketches?
13 Thegraphof y = f (x)issketchedtotheright.Onthreeseparatenumberplanes, sketch:
a y = | f (x)|
b y = f (|x|)
c y = | f (|x|)| x y
14 Copyeachdiagrambelow,thensketchtheinverserelationofthefunction.Alsostatewhetherornotthe inverserelationisafunction.
15 Findtheequationoftheinversefunctionforeachfunction.
16 Findtheinversefunction f 1(x)ofeachfunction,andthenconfirmalgebraicallythat
f 1(x) = x and f 1 f (x) = x .
(x) = 1 2 x + 4 a
17 Alineisdefinedparametricallybytheequations x = t + 2and y = 2t + 6.
Copyandcompletethetablebelow. t 5 4 3 2 101 x y a
Usethetabletosketchtheline,markingeachpointwithits t-value. b Eliminatetheparameter t tofindtheCartesianequationoftheline. c
18 Aparabolaisdefinedparametricallybytheequations x = 1 2 t and y = 1 4 t2 .
Copyandcompletethetablebelow.
t 6 4 2 101246 x y a
Usethetabletosketchtheparabola,markingeachpointwithits t-value. b Eliminatetheparameter t tofindtheCartesianequationoftheparabola. c
19 Acurveisdefinedparametricallybytheequations x = cos θ 1and y = sin θ + 1.
Usetheidentitycos2 θ + sin2 θ = 1tofindtheCartesianequationofthecurve. a Describethecurve,andthensketchit. b
20 Acurveisrepresentedparametricallybytheequations x = 2t and y = 1 t + 1 . FindtheCartesianequationofthecurveandhencesketchit.
Oncethevariable x hasbeenintroducedintoarithmetic,polynomialexpressionssuchas4x3 7x + 5arise naturally.Indeedeveryexpressionthatcanbeformedusingjustthethreeoperationsofaddition,subtractionand multiplicationcanbewrittenasapolynomial—simplyexpandbrackets,andcollectliketerms.
Thecoursehasalreadydiscussedlinearandquadraticfunctionsindetail,andthischapterbeginsthesystematic studyofpolynomialsofhigherdegree.
PolynomialswerementionedbrieflyinSection3G,andsomereaderswillalreadyhavefamiliaritywiththeir graphs,withlongdivisionofpolynomials,andwiththeremainderandfactortheorems.Thelatersectionsinthis chapterdeveloptherelationshipsbetweenthecoefficientsandthezeroes.
Theproblemoffactoringagivenpolynomialisaconstanttheme,andthefinalSection11Gappliesthemethods ofthechaptertogeometricproblemsaboutpolynomialcurves,circles,andrectangularhyperbolas.
Learningintentions
• Definepolynomial,leadingtermandcoefficient,degree,monic,constant.
• Classifyzero,linear,quadratic,cubicandquarticpolynomials.
• Add,subtract,andmultiplypolynomials,andobservetheresultingdegrees.
• Defineidenticallyequalpolynomials,andusetherelationshiptothecoefficients.
Polynomialsareexpressionssuchasthequadratic x2 5x + 6orthequartic3x4 2 3 x3 + 4x + 7.Theyhave occurredalreadyinthecourse,butnowourlanguageandnotationneedstobemoreprecise.
1Polynomials
• A polynomial isanexpressionoftheform
P(x) = an
n 1 xn 1 + ···
+ a0, where a0, a1,..., an arerealnumbers,and n isawholenumber.
• A polynomialfunction isafunctionthatcanbewrittenasapolynomial.
Theterm a0 iscalledthe constantterm.Thisisthevalueofthepolynomialat x = 0,so a0 isthe y-interceptofits graph.
Thetermofhighestindexwithnon-zerocoefficientiscalledthe leadingterm ofthepolynomial.Itscoefficientis calledthe leadingcoefficient,anditsindexiscalledthe degree.Forexample,thepolynomial
P(x) = 5x 6 3x 4 + 2x 3 + x 2 x + 9 hasleadingterm 5x6 andleadingcoefficient 5,andhasdegree6,writtenas deg P(x) = 6.
Forconvenience,theconstantterm a0 isregardedinthiscontextasbeingatermofindex0.Thisisbecause a0 canbewrittenas a0 x0 providedthat x 0.Butneveractuallywritetheterm a0 as a0 x0,because00 isundefined.
A monicpolynomial isapolynomialwhoseleadingcoefficientis1.Forexample, P(x) = x3 2x2 3x + 4is monic.Everynon-zeropolynomialisauniquemultipleofamonicpolynomialbecause an 0,so
Thezeropolynomial,andpolynomialsoflowdegree,havestandardnames.
• The zeropolynomial Z(x) = 0isaspecialcase.Ithasaconstantterm0.Butithasnotermwithanon-zero coefficient.Henceithasnoleadingterm,noleadingcoefficient,andmostimportantly,nodegree.Itsgraphis the x-axis—meaningthateveryrealnumberisazeroofthezeropolynomial.
• A constantpolynomial isapolynomialwhoseonlytermistheconstantterm: P(x) = 4, Q(x) = 3 5 , R(x) = π, Z(x) = 0.
Apartfromthezeropolynomial,allconstantpolynomialshavedegree0,havenozeroes,andareequaltotheir leadingtermandtotheirleadingcoefficient.
• A linearpolynomial isapolynomialofdegree1:
P(x) = x 3, Q(x) = 4x + 7, R(x) = 1 2 x
Warning: Aconstantfunctionisalinearfunction.ButaconstantpolynomialisNOTalinearpolynomial, becauseitdoesnothavedegree1—ithasdegree0oristhezeropolynomial.
• Apolynomialofdegree2iscalleda quadraticpolynomial:
P(x) = 3x 2 + 4x 1, Q(x) = 1 2 x x 2 , R(x) = 9 x 2
Noticethatthecoefficientof x2 mustbenon-zeroforthedegreetobe2.
• Polynomialsofhigherdegreearecalled: cubics (degree3), quartics (degree4), quintics (degree5),....
2Thedegreeofapolynomial P(x))
• The leadingterm of P(x)isthetermofhighestindexwithnon-zerocoefficient.
▷ The degree of P(x)istheindexoftheleadingterm.
▷ The leadingcoefficient of P(x)isthecoefficientoftheleadingterm.
▷ A monicpolynomial isapolynomialwhoseleadingcoefficientis1.
• The zeropolynomial Z(x) = 0hasnoleadingterm,andsohasnodegree.
• A constantpolynomial P(x) = a0 isapolynomialwithonlyaconstantterm.
▷ If a0 0,ithasdegree0.Butif a0 = 0,itisthezeropolynomial Z(x) = 0.
• A linearpolynomial P(x) = a1 x + a0 isapolynomialofdegree1,so a1 0.
• Polynomialsofhigherdegreearecalled quadratic,cubic,quartic,quintic,...
Whentwopolynomialsareaddedorsubtracted,theresultsareagainpolynomials:
(5x 3 4x + 3) + (3x 2 3x 2) =
Thezeropolynomial Z(x) = 0isthe zeroforaddition,intheusualsensethat P(x) + Z(x) = P(x),forall polynomials P(x).The oppositepolynomial P(x)ofanypolynomial P(x)isobtainedbytakingtheoppositeof everycoefficient.Thenthesumof P(x)and P(x)isthezeropolynomial.Forexample, (4x 4 2x 2 + 3x 7) + (
Thedegreeofthesumordifferenceisusuallythemaximumofthedegreesofthetwopolynomials,asinthetwo examplesabove,wherethepolynomialshavedegrees2and3andtheirsumhasdegree3.If,however,thetwo polynomialshavethesamedegree,theleadingtermsmaycanceloutanddisappear,forexample, (x 2 3x + 2) + (9 + 4x x 2) = x + 11,whichhasdegree1, orthetwopolynomialsmaybeopposites,inwhichcaseeverythingcancelsoutsothattheirsumiszeroandthus hasnodegree.
3Degreeofthesumanddifference
Let P(x)and Q(x)benon-zeropolynomialsofdegree n and m respectively.
• If n m,thendeg P(x) + Q(x) = maximumof m and n.
• If n = m,thendeg P(x) + Q(x) ≤ n or P(x) + Q(x) = 0.
Anytwopolynomialscanbemultiplied,givinganotherpolynomial:
Theconstantpolynomial I(x) = 1istheidentityformultiplication,inthesensethat P(x) × I(x) = P(x),forall polynomials P(x).Multiplicationbythezeropolynomial,ontheotherhand,alwaysgivesthezeropolynomial.
Iftwopolynomialsarenon-zero,thenthedegreeoftheirproductisthesumoftheirdegrees,becausetheleading termoftheproductisalwaystheproductofthetwoleadingterms.
4Degreeoftheproduct
If P(x)and Q(x)arenon-zeropolynomials,then deg P(x) × Q(x) = deg P(x) + deg Q(x).
Identicallyequalpolynomials
Weneedtobeclearwhatismeantbysayingthattwopolynomialsarethesame.
5Identicallyequalpolynomials
• Twopolynomials P(x)and Q(x)arecalled identicallyequal iftheyareequalforallvaluesof x: P(x) = Q(x),forall x,oftenwrittenas P(x) ≡ Q(x).
• Iftwopolynomialsareidenticallyequal,thenthecorrespondingcoefficientsofthetwopolynomialsare equal.
Theseconddotpointneedsproof.ThisisdevelopedintheEnrichmentsectionofExercise11B.Inthemeantime, hereisanexamplethatusestheresult.
Example1 Findingcoefficientsinidenticallyequalpolynomials
Find a, b, c, d,and e if ax4 +
. Solution
Expanding,(x2 3)2 = x4 6x2 + 9.
Nowcomparingcoe
Themostsignificantproblemofthischapteristhefactoringofagivenpolynomial.Forexample,
isaroutineexpansionofafactoredpolynomial,butitisnotatallclearhowtomoveintheotherdirectionfrom theexpandedformbacktothefactoredform.
If P(x)isapolynomial,thentheequationformedbyputting P(x) = 0isa polynomialequation.Forexample, usingthepolynomialinthepreviousparagraph,wecanformthepolynomialequation
x
+
= 0.
Solvingpolynomialequationsandfactoringpolynomialfunctionsarecloselyrelated.Usingthefactoringofthe previousparagraph,
x(x + 2)2(x 2)2(x 2 + x + 1) = 0,
sothesolutionsare x = 0, x = 2(doubleroot)and x = 2(doubleroot),wherethequadraticfactor x2 + x + 1has nozeroes,because ∆= 3.Thusfactoringapolynomialintolinearandirreduciblequadraticfactorssolvesthe correspondingpolynomialequation.
Thesolutionsofapolynomialequationarecalled roots,whereasthe zeroes ofapolynomialfunctionarethe valuesof x wherethevalueofthepolynomialiszero.Thedistinctionbetweenthetwowordsisnotalways strictlyobserved.
1 Statewhetherornoteachexpressionisapolynomial.
2 Foreachpolynomial,state:
thedegree, i theleadingcoefficient, ii theleadingterm, iii theconstantterm, iv whetherornotthepolynomialismonic. v Expandthepolynomialfirstwherenecessary.
+
3 If P(x) = 5x + 2and Q(x) = x2 3x + 1,find:
(x) P(x) d
4 If P(x) = 5x + 2, Q(x) = x2 3x + 1and R(x) = 2x2 3,show,bysimplifyingtheLHSandRHSseparately, that:
(P(x) + Q(x)) + R(x) = P(x) + (Q(x) + R(x)) a P(x) (Q(x) + R(x)) = P(x)Q(x) + P(x)R(x) b
(P(x)Q(x)) R(x) = P(x) (Q(x)R(x)) c
5 Factoreachpolynomialcompletely,andwritedownallitszeroes.
x3 8x2 20x a 2x4 x3 x2 b x4 81 c x4 5x2 36 d
6 Foreachpolynomial,determine:
thedegree, i theleadingcoefficient, ii theconstantterm. iii
(2x3 3)3 a (2x2 + 1)(3x3 2)(4x4 + 3)(5x5 4) b
7a Thepolynomials P(x)and Q(x)havedegrees p and q respectively,and p q.Whatisthedegreeof:
P(x)Q(x), i
P(x) + Q(x)? ii
b Whatdifferenceswoulditmakeif P(x)and Q(x)bothhadthesamedegree p?
c Giveanexampleoftwopolynomials,bothofdegree2,whichhaveasumofdegree0.
8 Writedownthemonicpolynomialwhosedegree,leadingcoefficient,andconstanttermareallequal.
9 Find a, b and c,giventhatthefollowingpairsofpolynomialsareidenticallyequal.
ax2 + bx + c = 3x2 4x + 1,forall x a
(a b)x2 + (2a + b)x = 7x x2,forall x b
a(x 1)2 + b(x 1) + c = x2,forall x c
a(x + 2)2 + b(x + 3)2 + c(x + 4)2 = 2x2 + 8x + 6,forall x. d
10a Supposethat P(x) = ax4 + bx3 + cx2 + dx + e iseven,sothat P( x) = P(x).
Showthat b = d = 0.
b Supposethat Q(x) = ax5 + bx4 + cx3 + dx2 + ex + f isodd,sothat Q( x) = Q(x).
Showthat b = d = f = 0.
c Giveageneralstatementofthesituationinparts(a)and(b).
11 Supposethat P(x), Q(x), R(x)and S (x)arepolynomials.Indicatewhetherthefollowingstatementsaretrue orfalse.Provideacounter-exampleforanyfalsestatements.
If P(x)iseven,then P′(x)isodd. a If Q′(x)iseven,then Q(x)isodd. b
If R(x)isodd,then R′(x)iseven.
c If S ′(x)isodd,then S (x)iseven. d
12a Find a and b sothat x4 + 1 = (x2 + ax + 1)(x2 + bx + 1).
b Find a and b sothat x4 + x2 + 1 = (x2 + ax + 1)(x2 + bx + 1).
c Find a and b sothat x4 x2 + 1 = (x2 + ax + 1)(x2 + bx + 1).
d Showthatallthequadraticfactorsinparts(a),(b)and(c)areirreducible.
13a Whatisthecoefficientof x inthepolynomial B(x) = (1 + x)n ?
b Whatisthecoefficientof xn inthepolynomial G(x) = (1 + x + x2 + ··· + xn)2 ?
Learningintentions
• Sketchpolynomials,usingbehaviouras x →∞ andas x →−∞
• Definemultiple(orrepeated)zeroes,andsimple(sometimescalledsingle)zeroes.
• Takeaccountofsimpleandmultiplezeroesinsketches.
Wehaveassumedthatthegraphofapolynomialfunctioniscontinuousforallvaluesof x.Weassumealsothatit is smooth,withoutsharpcornersanywhere.
Thissectionwillconcentrateontwomainconcerns.
• Howdoesthegraphbehaveforlargepositiveandnegativevaluesof x?
• Giventhefullfactoringofthepolynomial,howdoesthegraphbehavenearitsvarious x-intercepts?
InYear12,wewillpursuefurtherquestionsaboutstationarypointsandinflectionsthatarenotalsozeroesofthe polynomial.
Itshouldbeintuitivelyobviousthatforlargepositiveandnegativevaluesof x,thebehaviourofthecurveis governedentirelybythesignofitsleadingterm.
Forexample,thecubicgraphsketchedontherightbelowis
P(x) = x 3 4x = x(x 2)(x + 2).
Forlargepositivevaluesof x,theleadingterm x3 completelyswampstheother term 4x.Hence P(x) →∞ as x →∞.
Similarly,forlargenegativevaluesof x,theterm 4x isnegligiblecomparedwiththe farbiggernegativevaluesoftheleadingterm x3.Hence P(x) →−∞ as x →−∞. x y −22
Everypolynomialofodddegreehasagraphthatsimilarlydisappearsoff diagonallyoppositecorners.Thecurve iscontinuous,soitmustbezerosomewhereatleastonce.Hereisthegeneralsituation.
6Behaviourofpolynomialsforlarge x
Supposethat P(x)isapolynomialofdegreeatleast1withleadingterm an xn .
• As x →∞, P(x) →∞ if an ispositive,and P(x) →−∞ if an isnegative.
• As x →−∞, P(x)behavesthesameaswhen x →∞ ifthedegreeiseven,and P(x)behavesinthe oppositewayifthedegreeisodd.
▷ Itfollowsthateverypolynomialofodddegreehasatleastonezero.
SeetheEnrichmentsectionforaformalproof.
Ifthepolynomialcanbecompletelyfactored,thenitszeroescanbequicklyreadoff, andwecanconstructatableoftestvaluestodecideitssign.Here,forexample,isthe tableoftestvaluesandthesketchof
P(x) = (x + 2)3 x 2(x 2).
x 3 2 10123
y 450 30 2701125 x y −22
Thefunctionchangessignaround x = 2and x = 2,wheretheassociatedfactors(x + 2)3 and(x 2)haveodd degrees,butnotaround x = 0,wherethefactor x2 hasevendegree.
Becausethecurveissmooth—withoutsharpcorners—at x = 0,weknowthatthecurvewillbeincreasing ontheleftof x = 0,decreasingontherightof x = 0,andstationaryat x = 0.Thisproducesa turningpoint at theorigin—the x-axisisatangentthere,andthecurve turnsover smoothlyfromincreasingtostationaryto decreasingwithoutcrossingthe x-axis.
At x = 2,ourtableofvaluestellsusthatthecurvecrossesthe x-axis.WeshallseeinYear12thatthecurveis momentarilyflatthere,witha horizontalinflection onthe x-axisat x = 2—the x-axisisatangenttothecurve thatactuallycrossesthecurvethere.Thiscorrespondstothefactor(x + 2)3 havingodddegreegreaterthan1. Provingallthisrequirescalculus,buttheresultisobviousbycomparisonwiththeknowngraphofthevery simplepolynomialfunction y = x3 thatwefirstdrewinSection3G.
Somelanguageisneededhere.Takethepolynomial P(x) = (x + 2)3 x2(x 2).
• Thezero x = 2iscalleda triplezero.
• Thezero x = 0iscalleda doublezero (aswithquadraticsinChapter3).
• Thezero x = 2iscalleda simplezero.
Thetriplezero x = 2andthedoublezero x = 0arecalled multiple or repeatedzeroes.
• Supposethat x α isafactorofapolynomial P(x),and P(x) = (x α)m Q(x),where Q(x)isnotdivisibleby x α
Then x = α iscalleda zeroofmultiplicity m
• Azeroofmultiplicity1iscalleda simplezero.
• Azeroofmultiplicity2orgreateriscalleda multiplezero ora repeatedzero.
• Forloworders,weusetheterms doublezero, triplezero,and quadruplezero,....
Note: Theterm singlezero issometimesusedinplaceof simplezero,butthattermisbestavoidedbecauseof theconstantverbalconfusionthatitcauses.
Hereisthestatementofhowapolynomialgraphbehaveswhenitcrossesthe x-axisatazero,asjustifiedaboveas farasispossibleinYear11.
8Multiplerootsandtheshapeofthecurve
Supposethat x = α isazeroofapolynomial P(x).
• If x = α isasimplezero(sometimescalledasinglezero),thenthecurvecrossesthe x-axisat x = α at anangle,andisnottangenttothe x-axisthere.
• If x = α hasevenmultiplicity,thenthecurveistangenttothe x-axisat x = α,anddoesnotcrossthe x-axisthere.Thepoint(α,0)isa turningpoint
• If x = α hasoddmultiplicityatleast3,thenthecurveistangenttothe x-axisat x = α,butcrossesthe x-axis.Thepoint(α,0)isa horizontalinflection
Note: WhenweweredealingonlywithquadraticsinChapter3,weoftenusedtheterms‘equalroots’and ‘distinctroots’.Neverusethosetermsinthecontextofpolynomials,becauseoftheverbalambiguities thatinevitablyarise.
Example2 Sketchingfactoredpolynomialsnearsimpleandmultiplezeroes
Ineachpart,namethezeroesandtheirmultiplicity,andthensketch,showingthebehaviournearany x-intercepts:
a P(x) = (x 1)2(x 2)
b Q(x) = x3(x + 2)4(x2 + x + 1)
c R(x) = 2(x 2)2(x + 1)5(x 1)
Solution
Inpart(b), x2 + x + 1hasnozeroes,because ∆= 1 4 < 0.
1isadoublezero, 2isasimplezero.
2isadoublezero, 0isatriplezero. b x y 8 −112
1isaquintuplezero, 1isasimplezero, 2isadoublezero.
Giveanexampleofapolynomialfactoredintolinearfactorsthat:
a hastwomultiple(orrepeated)zeroes,andthreesimplezeroes(sometimescalled‘singlezeroes’). b hasdegree9,andhastwotriplezeroesandonedoublezero,andisnon-monic.
Solution (x + 5)2(x 1)(x 2)(x 3)x7 a 12(x + 8)3(x + 7)3(x + 5)2(x + 3) b
1 Writedownthezeroesofeachpolynomial,andclassifyeachzeroasasimplezerooramultiplezero.
Note: Readersshouldbeawarethattheterm‘repeatedzero’isoftenusedfor‘multiplezero’,andthatthe term‘singlezero’issometimesusedfor‘simplezero’.
2 Writedownthezeroesofeachpolynomial,andstatethemultiplicityofeachzero.
3 Writedowntherootsofeachpolynomialequation,andidentifyeachrootasasimple,double,tripleor quadrupleroot.
4 Sketchthegraphsoftheselinearpolynomials,clearlyindicatingallinterceptswiththeaxes.
5 Sketchthegraphsofthesequadraticpolynomials,clearlyindicatingallinterceptswiththeaxes.
6 Sketchthegraphsofthesecubicpolynomials,clearlyindicatingallinterceptswiththeaxes. y = x3 a y = x3 + 2 b
= (x 4)3 c y = (x 1)(x + 2)(x 3) d y = x(2x + 1)(x 5) e y = (1 x)(1 + x)(2 + x) f y = (2x + 1)2(x 4) g y = x2(1 x) h
= (2 x)2(5 x) i
7 Sketchthegraphsofthesequarticpolynomials,clearlyindicatingallinterceptswiththeaxes.Identifyall doublezeroes,triplezeroes,andquadruplezeroes,
F(x) = x4 a F(x) = (x + 2)4 b
F(x) = x(3x + 2)(x 3)(x + 2) c F(x) = (1 x)(x + 5)(x 7)(x + 3) d F(x) = x2(x + 4)(x 3) e F(x) = (x + 2)3(x 5) f
F(x) = (2x 3)2(x + 1)
DEVELOPMENT
8 Thesepolynomialsarenotfactored,butthepositionsoftheirzeroescanbefoundbytrialanderror.Copy andcompleteeachtableofvalues,thensketchthegraph,andstatehowmanyzeroesthereare,andbetween whichintegerstheylie.
9 Sketcheachpolynomialfunction,clearlyindicatingallinterceptswiththeaxes.
(x) = x(2x + 3)3(1 x)4 c
(
) = (x + 2)2(3 x)3
(x) = (x + 1)(4 x2)(x2 3x 10) d
10 Usethegraphsdrawninthepreviousquestiontosolvetheseinequations.
(x 2)
11 Factoreachpolynomial,thensketchit,showingallinterceptswiththeaxes.
12 Considerthepolynomial P(x) = x4 5x2 + 4x + 13.
Showthat P(x)canbeexpressedintheform(x2 a)2 + (x b)2 . a
Howmany x-interceptsdoesthegraphof P(x)have?Explainyouranswer. b
13 Atwhatpointsdothegraphsofthepolynomials f (x) = (x + 1)n and g(x) = (x + 1)m intersect,where m n? (Hint:Considerthecaseswhere m and n areoddandeven.)
14 Toprove: Let P(x) = an xn + an 1
+ a0 beapolynomialofdegreeatleast1.Thenthe leadingtermdominatesthebehaviourof P(x) forlargepositivevaluesof x,andforlargenegativevalues of x.
Writedown P(x) xn . a
b
c
Takethelimitofthisexpressionas x →∞
Drawaconclusionaboutthebehaviourof P(x)as x →∞
Drawaconclusionaboutthebehaviourof P(x)as x →−∞ d
e
Whatconclusion,ifany,canyoudrawaboutthezeroesofapolynomialofodddegree,andaboutthe zeroesofapolynomialofevendegree?
15 Toprove: Let P(x) beapolynomialsuchthat P(x) = 0 forall x.Then P(x) isthezeropolynomial Z(x) = 0 thathasnotermwithanon-zerocoefficient.
Usethepreviousquestiontoprovethat P(x)cannothavedegree ≥ 1,andhenceisaconstantpolynomial P(x) = a0 a
Explainwhytheconstant a0 cannotbenon-zero. b
16 Toprove: If P(x) = Q(x) forall x,then P(x) and Q(x) havethesamecoefficients.
Usethepreviousquestion,andthepolynomial A(x) = P(x) Q(x),toprovethisresult.
Learningintentions
• Applythedivisionalgorithmforpolynomials,andcompareitwithintegerdivision.
Section11Ahadexamplesofadding,subtracting,andmultiplyingpolynomials,operationsthatarequite straightforward.Thedivisionofonepolynomialbyanother,however,ismoreelaborate.
Itcanhappenthatthequotientoftwopolynomialsisagainapolynomial:
Butusually,divisionresultsinrationalfunctions,notpolynomials:
Thereisacloseanalogyherebetweentheset Z ofallintegersandthesetofallpolynomials.Inbothcases, everythingworksnicelyforaddition,subtraction,andmultiplication,buttheresultsofdivisiondonotusually liewithintheset.Forexample,although20 ÷ 5 = 4isaninteger,thedivisionoftwointegersusuallyresultsina fractionratherthananinteger,asin23 ÷ 5 = 4 3 5
Inbothcases,thebestwaytohandledivisionistouseremainders.
Ontherightisanexampleofthewell-knownlongdivisionalgorithmfor integers,appliedhereto197 ÷ 12.Thenumber12iscalledthe divisor, 197iscalledthe dividend,16iscalledthe quotient,and5iscalledthe remainder
Theresultofthedivisioncanbewrittenas 197 12 = 16 5 12 ,butwecanavoid fractionscompletelybywritingtheresultas:
197 = 12 × 16 + 5,dividend = divisor × quotient + remainder.
Theremainder5mustbelessthan12,otherwisethedivisionprocesscouldbecontinued.Thusthegeneralresult fordivisionofintegerscanbeexpressedas:
• Let p (the dividend )and d (the divisor)beintegers,with d > 0.Thenthereareuniqueintegers q (the quotient)and r (the remainder)suchthat p = dq + r and0 ≤ r < d
• Whentheremainder r iszero, d isa divisorof p,andtheinteger p factors as p = d × q
Themethodofdividingonepolynomialbyanotherissimilartothemethodofdividingintegers.
• Ateachstep,dividetheleadingtermoftheremainderbytheleadingtermofthedivisor.Continuethe processforaslongaspossible.
• Unlessotherwisespecified,expressthefinalanswerintheform dividend = divisor × quotient + remainder.
Noteonmissingterms: Intheexamplebelow,itisvitaltonoticethatthereisnotermin x2 inthedividend. Therearetwoapproachestodealwiththis:
• Leaveagapforthe x2 column.Thisiswhathasbeendonebelow.
• Ifyouprefernottohaveanygaps,writeinthemissingterm0x2 . Whynottrybothmethods?Bothareverysatisfactory.
Divide3x4 4x3 + 4x 8by:
x 2 a x2 2 b
Giveresultsfirstinthestandardmanner,thenusingrationalfunctions.
Solution
Ineachpart,thestepshavebeenannotatedtoexplainthemethod.
a 3x3 + 2x2 + 4x + 12
x 2 3x4 4x3 + 4x 8(divide x into3x4,givingthe3x3 above) 3x4 6x3 (multiply x 2by3x3 andthensubtract)
2x3 + 4x 8(divide x into2x3,givingthe2x2 above) 2x3 4x2 (multiply x 2by2x2 andthensubtract)
4x2 + 4x 8(divide x into4x2,givingthe4x above) 4x2 8x (multiply x 2by4x andthensubtract) 12x 8(divide x into12x,givingthe12above) 12x 24(multiply x 2by12andthensubtract) 16(thisisthefinalremainder)
Hence3x4 4x3 + 4x 8 = (x 2)(3x3 + 2x2 + 4x + 12) + 16, or,writingtheresultusingrationalfunctions,
3x4 4x3 + 4x 8 x 2 = 3x 3 + 2x 2 + 4x + 12 + 16 x 2 .
b 3x2 4x + 6
x2 2 3x4 4x3 + 4x 8(divide x2 into3x4,givingthe3x2 above)
3x4 6x2 (multiply x2 2by3x2 andthensubtract)
4x3 + 6x2 + 4x 8(divide x2 into 4x3,givingthe 4x above)
4x3 + 8x (multiply x2 2by 4x andthensubtract)
6x2 4x 8(divide x2 into6x2,givingthe6above)
6x2 12(multiply x2 2by6andthensubtract) 4x + 4(thisisthefinalremainder)
Hence3x 4 4x 3 + 4x 8 = (x 2 2)(3x 2 4x + 6) + ( 4x + 4), or 3x4 4x3 + 4x 8 x2 2 = 3x 2 4x + 6 + 4x + 4 x2 2 .
Thedivisiontheorem
Thedivisionprocessillustratedabovecanbecontinueduntiltheremainderiszeroorhasdegreelessthanthe degreeofthedivisor.Thusthegeneralresultforpolynomialdivisionis:
11Divisionofpolynomials
• Supposethat P(x)(the dividend)and D(x)(the divisor)arepolynomialswith D(x) 0.Thenthereare uniquepolynomials Q(x)(the quotient)and R(x)(the remainder)suchthat:
1 P(x) = D(x)Q(x) + R(x),
2 eitherdeg R(x) < deg D(x),or R(x) = 0.
• Whentheremainder R(x)iszero,then D(x)iscalleda divisor of P(x),andthepolynomial P(x) factors intotheproduct P(x) = D(x) × Q(x).
Forexample,inthetwolongdivisionsinExample4above:
• Inpart(a),theremainderafterdivisionbythedegree1polynomial x 2isthepolynomial16ofdegree0.
• Inpart(b),theremainderafterdivisionbythedegree2polynomial x2 2isthelinearpolynomial 4x + 4of degree1.
TheuniquenessofthequotientandremainderinthefirstdotpointaboveisprovenintheEnrichmentsectionof thefollowingexercise.
1 Performeachintegerdivision,andwritetheresultintheform p = dq + r,where0 ≤ r < d.Forexample, 30 ÷ 7 = 4,remainder2,so30 = 4 × 7 + 2.
63 ÷ 5
2 Uselongdivisiontoperformeachdivision.Expresseachresultintheform P(x) = D(x) Q(x) + R(x). (x2 4x + 1) ÷ (x + 1) a (x2 6x + 5) ÷ (x 5) b
(x3 x2 17x + 24) ÷ (x 4) c (2x3 10x2 + 15x 14) ÷ (x 3) d
(4x3 4x2 + 7x + 14) ÷ (2x + 1) e (x4 + x3 x2 5x 3) ÷ (x 1) f (6x4 5x3 + 9x2 8x + 2) ÷ (2x 1) g (10x4 x3 + 3x2 3x 2) ÷ (5x + 2) h
3 Expresstheanswerstoparts(a)–(d)ofthepreviousquestioninrationalform,thatis,as P(x) D(x) = Q(x) + R(x)
D(x) .
4 Uselongdivisiontoperformeachdivision.Expresseachresultinthestandardform P(x) = D(x) Q(x) + R(x).
(x3 + x2 7x + 6) ÷ (x2 + 3x 1) a (x3 4x2 2x + 3) ÷ (x2 5x + 3) b
(x4 3x3 + x2 7x + 3) ÷ (x2 4x + 2) c (2x5 5x4 + 12x3 10x2 + 7x + 9) ÷ (x2 x + 2) d
5a Ifthedivisorofapolynomialhasdegree3,whatarethepossibledegreesoftheremainder?
b Ondivisionby D(x),apolynomialhasremainder R(x)ofdegree2.Whatarethepossibledegreesof D(x)?
6 Uselongdivisiontoperformeachdivision.Takecaretoensurethatthecolumnslineupcorrectly.Express eachresultintheform P(
). (x3 5x + 3) ÷ (x 2) a
(x3 3x2 + 5x 4) ÷ (x2 + 2)
x3 3) ÷ (2x 4) e
x3 + x2 11) ÷ (x + 1)
Writetheanswerstoparts(c)and(f)aboveinrationalform,thatis,intheform
7a Uselongdivisiontoshowthat P(x) = x3 + 2x2 11x 12isdivisibleby x 3,andhenceexpress P(x)as theproductofthreelinearfactors.
b Findthevaluesof x forwhich P(x) > 0.
8a Uselongdivisiontoshowthat F(x) = 2x4 + 3x3 12x2 7x + 6isdivisibleby x2 x 2,andhence express F(x)astheproductoffourlinearfactors.
b Findthevaluesof x forwhich F(x) ≤ 0.
9a Findthequotientandremainderwhen x4 2x3 + x2 5x + 7isdividedby x2 + x 1.
b Hencefind a and b sothat x4 2x3 + x2 + ax + b isexactlydivisibleby x2 + x 1.
10a Uselongdivisiontodividethepolynomial f (x) = x4 x3 + x2 x + 1bythepolynomial d(x) = x2 + 4. Expressyouranswerintheform f (x) = d(x)q(x) + r(x).
b Hencefindthevaluesof c and d suchthat x4 x3 + x2 + cx + d isdivisibleby x2 + 4.
11 If x4 2x3 20x2 + mx + n isexactlydivisibleby x2 5x + 2,find m and n.
12 Supposethat P(x) = x4 + x3 5x2 22x + 5and D(x) = x2 + 3x + 5.
a
Findthepolynomials Q(x)and R(x),where R(x)isoflowerdegreethan D(x),sothat P(x) = D(x) Q(x) + R(x).
Henceexplainwhy P(x)and D(x)cannothaveacommonzero. b
13 Considerthecubicequation x3 kx + (k + 11) = 0.
Uselongdivisiontoshowthat k = x2 + x + 1 + 12 x 1 a
Hencefindalltheintegervaluesof k forwhichtheequationhasatleastone positive integersolutionfor x. b
14 Toprove: Let P(x) and D(x) bepolynomialswith D(x) 0.Thenthereexistsuniquepolynomials Q(x) and R(x) suchthat P(x) = D(x)Q(x) + R(x) andeither R(x) = 0 or deg R(x) < deg D(x)
Existencefollowsfromthelongdivisionalgorithm,twoexamplesofwhichhavebeengiven.Toprove uniqueness,let Q1(x)and R1(x)bepolynomialssuchthat P(x) = D(x)Q1(x) + R1(x),andeither R1(x) = 0ordeg R1(x) < deg D(x).
Showthat D(x) Q(x) Q1(x) = R1(x) R(x). a UsethepossibledegreesofLHSandRHStoprovetheresult. b
Learningintentions
• Proveandusetheremaindertheoremtofindremainders.
• Proveandusethefactortheoremtofindzeroes.
• Identifypossiblezeroesbyfactoringtheconstantterm.
• Combinethefactortheoremandlongdivisiontofactorapolynomial.
Longdivisionofpolynomialsisacumbersomeprocess.Itisthereforeusefultohavetwotheorems,calledthe remaindertheorem andthe factortheorem,thatprovideinformationabouttheresultsofadivisionwithoutthe divisionactuallybeingcarriedout.Inparticular,thefactortheoremgivesasimpletestwhetheraparticularlinear polynomialisafactor.
Theremaindertheoremisaremarkableresultwhich,inthecaseoflineardivisors,allowstheremaindertobe foundwithoutthelongdivisioneverbeingperformed.
12Theremaindertheorem
Supposethat P(x)isapolynomialand α isaconstant. Thentheremainderafterdivisionof P(x)by x α is P(α).
Proof Because x α isapolynomialofdegree1,thedivisiontheoremtellsusthatthereareuniquepolynomials Q(x)and R(x)suchthat P(x) = (x α)Q(x) + R(x), whereeither R(x) = 0ordeg R(x) = 0.
Hence R(x)iszerooranon-zeroconstant,whichwecanwritemoresimplyas r, sothat
Substituting x = α gives P(α) = (α α)Q(α) + r andrearranging, r = P(α),asrequired.
Findtheremainderwhen3x4 4x3 + 4x 8isdividedby x 2: bylongdivision, a bytheremaindertheorem. b
Solution
a In Example4ofSection11C,performingthedivisionshowedthat 3x 4 4x 3 + 4x 8 = (x 2)(3x 3 + 2x 2 + 4x + 12) + 16, thatis,theremainderis16.
b Alternatively,substituting x = 2into P(x),
remainder = P(2)(Thisistheremaindertheorem.)
= 48 32 + 8 8
= 16,asexpected.
Example6 Findingcoefficientsusingtheremaindertheorem
Thepolynomial P(x) = x4 2x3 + ax + b hasremainder3afterdivisionby x 1,andhasremainder 5after divisionby x + 1.Find a and b
Solution
Applyingtheremaindertheoremforeachdivisor, P(1) = 3
1 2 + a + b = 3
a + b = 4. (1)
Also P( 1) = 5
1 + 2 a + b = 5
a + b = 8.(2)
Adding(1)and(2), 2b = 4, andsubtractingthem, 2a = 12.
Hence a = 6and b = 2.
Thefactortheorem
Theremaindertheoremtellsusthatthenumber P(α)istheremainderafterdivisionby x α.But x α isafactor ifandonlyiftheremainderafterdivisionby x α iszero,so:
13Thefactortheorem
Supposethat P(x)isapolynomialand α isaconstant. Then x α isafactorof P(x)ifandonlyif P(α) = 0.
Thisisaquickandeasywaytotestwhether x α isafactorof P(x).
Example7 Usingthefactortheoremandlongdivisiontofactorapolynomial
a Showthat x 3isafactorof P(x) = x3 2x2 + x 12,and x + 1isnot.
b Thenuselongdivisiontofactorthepolynomialcompletely.
Solution
a P(3) = 27 18 + 3 12 = 0,so x 3isafactor.
P( 1) = 1 2 1 12 = 16 0,so x + 1isnotafactor.
b Longdivisionof P(x) = x3 2x2 + x 12by x 3(whichweomit)gives
P(x) = (x 3)(x 2 + x + 4), andbecause ∆= 1 16 = 15 < 0forthequadratic,thisfactoringiscomplete.
Thefactortheoremgivesusthebeginningsofanapproachtofactoringpolynomials.Thisapproachwillbe furtherrefinedinthenexttwosections.
14Factoringpolynomials—theinitialapproach
• Usetrialanderrortofindanintegerzero x = α of P(x).
• Thenuselongdivisiontofactor P(x)intheform P(x) = (x α)Q(x).
Ifthecoefficientsof P(x)areallintegers,thenalltheintegerzeroesof P(x)aredivisorsoftheconstant term.
Proof Wemustprovetheclaimthatifthecoefficientsof P(x)areintegers,theneveryintegerzeroof P(x)isa divisoroftheconstantterm.
Let
wherethecoefficients an, an 1,...
Substitutinginto P(α) = 0gives
1, a0 areallintegers,andlet x = α beanintegerzeroof P(x).
so a0 isanintegermultipleof α
Example8
Factoringtheconstanttermtofindpossiblezeroes
Factor P(x) = x4 + x3 9x2 + 11x 4completely.
Solution
Becauseallthecoefficientsareintegers,anyintegerzeroisadivisoroftheconstantterm 4.Thuswetest1,2, 4, 1, 2and 4. P(1) = 1 + 1 9 + 11 4 = 0,so x 1isafactor.
Afterlongdivision(omitted), P(x) = (x 1)(x 3 + 2x 2 7x + 4).
Let Q(x) = x 3 + 2x 2 7x + 4,then Q(1) = 1 + 2 7 + 4 = 0,so x 1isafactor.
Againafterlongdivision(omitted), P(x) = (x 1)(x 1)(x 2 + 3x 4).
Factoringthequadratic, P(x) = (x 1)3(x + 4).
Note: Inthenexttwosections,andagaininYear12,wewilldevelopmethodsthatwilloftenallowlongdivision tobeavoided.
1 Withoutdivision,findtheremainderwhen P(x) = x3 x2 + 2x + 1isdividedby: x 1 a x 3 b x + 2 c x + 1 d x 5 e
2 Withoutdivision,findwhichofthefollowingarefactorsof F(x) = x3 + 4x2 + x 6.
x 1 a x + 1 b x 2 c x + 2 d x 3 e x + 3 f
3a Find k,if x 1isafactorof P(x) = x3 3x2 + kx 2.
b Find m,if 2isazeroofthefunction F(x) = x3 + mx2 3x + 4.
c Whenthepolynomial P(x) = 2x3 x2 + px 1isdividedby x 3,theremainderis2.Find p.
d Forwhatvalueof a is3x4 + ax2 2divisibleby x + 1?
4 Factoreachpolynomialandsketchitsgraph,indicatingallinterceptswiththeaxes.Rememberthatany integerzeroesaredivisorsoftheconstantterm.
5 SolvetheseequationsbyfirstfactoringtheLHS.
+
6a Showthat P(x) = x3 8x2 + 9x + 18isdivisiblebyboth x 3and x + 1.
b Byconsideringtheleadingtermandconstantterm,express P(x)asaproductofthreelinearfactors.
7a Showthat P(x) = 2x3 x2 13x 6isdivisiblebyboth x 3and2x + 1.
b Byconsideringtheleadingtermandconstantterm,express P(x)asaproductofthreelinearfactors.
8a Whenthepolynomial P(x) = 2x3 x2 + ax + b isdividedby x 1theremainderis16,andwhenitis dividedby x + 2theremainderis 17.Find a and b.
b Thepolynomial P(x)isgivenby P(x) = x3 + ax2 + bx 18.Find a and b giventhat x + 2isafactorof P(x),and 24istheremainderwhen P(x)isdividedby x 1.
9 Withoutdivision,findtheremainderwhen P(x) = x3 + 2x2 4x + 5isdividedby: 2x 1 a 2x + 3 b 3x 2 c
a
10 If P(x) = 2x3 + x2 13x + 6,evaluate P 1 2 .Henceuselongdivisiontoexpress P(x)infullyfactored form.
b
Given P(x) = 6x3 + x2 5x 2,evaluate P 2 3 thenexpress P(x)infactoredform.
11 Iseither x + 1or x 1afactorof xn + 1,where n isapositiveinteger? (Hint:Considerthecaseswhere n isevenoroddseparately.)
12 P(x)isanoddpolynomialofdegree3.Ithas x + 4asafactor,andwhenitisdividedby x 3the remainderis21.Find P(x).
a Find p sothat x p isafactorof4x3 (10p 1)x2 + (6p2 5)x + 6. b
13 Whenthepolynomial P(x)isdividedby(x 1)(x + 3),thequotientis Q(x)andtheremainderis2x + 5. Writedownadivisionidentitybasedonthisinformation. a Hence,byevaluating P(1),findtheremainderwhen P(x)isdividedby x 1. b Whatistheremainderwhen P(x)isdividedby x + 3? c
14 Thepolynomial P(x)isdividedby(x 1)(x + 2).Supposethatthequotientis Q(x)andtheremainderis R(x).
Explainwhythegeneralformof R(x)is ax + b,where a and b areconstants. a If P(1) = 2and P( 2) = 5,find a and b.(Hint:Usethedivisionidentity.) b
15 Thepolynomial P(x)isdividedby(x + 4)(x 3).If P( 4) = 11and P(3) = 3,usethesameapproachas thepreviousquestiontofindtheremainder.
16a Whenapolynomialisdividedby(2x + 1)(x 3),theremainderis3x 1.Whatistheremainderwhen thepolynomialisdividedby2x + 1?
b When x5 + 3x3 + ax + b isdividedby x2 1,theremainderis2x 7.Find a and b
c Whenapolynomial P(x)isdividedby x2 5,theremainderis x + 4.Findtheremainderwhen P(x) + P( x)isdividedby x2 5.(Hint:Writedownthedivisionidentity.)
17 Whenthepolynomial P(x)isdividedby x2 k2,theremainderis ax + b
Showthat a = 1 2k P (k)
Giventhat P(x) = 8x5 4x4 + 6x3 11x2 2x + 3and k = 1 2 ,find a and b,andhencefactor P(x)fully. b
18a Usethefactortheoremtoprovethat a + b + c isafactorof a3 + b3 + c3 3abc.Thenfindtheotherfactor. (Hint:Regarditasapolynomialin a.)
b Factor ab3 ac3 + bc3 ba3 + ca3 cb3 .
19a Ifallthecoefficientsofamonicpolynomialareintegers,provethatalltherationalzeroesareintegers. (Hint:LookcarefullyattheproofunderBox14.)
b Ifallthecoefficientsofapolynomialareintegers,provethatthedenominatorsofalltherationalzeroes (inlowestterms)aredivisorsoftheleadingcoefficient.
Learningintentions
• Developconsequencesofthefactortheoremtohelpfactorapolynomial.
• Developboundsonthenumberofzeroesofapolynomialofacertaindegree.
• Developatestfortwopolynomialstobeidenticallyequal.
• Developgeometricconsequencesofthefactortheorem.
Thefactortheoremhasanumberofstraightforwardbutveryusefulconsequences.Theyarepresentedhereassix successivetheorems.
A.Severaldistinctzeroes
Supposethatseveraldistinctzeroesofapolynomialhavebeenfound,probablyusingtestsubstitutionsintothe polynomial.
15Distinctzeroes
Supposethat α1, α2,... αs aredistinctzeroesofapolynomial P(x).Then(x α1)(x α2) ··· (x αs)isa factorof P(x).
Proof Because α1 isazero, x α1 isafactor,and P(x) = (x α1)P1(x).
Because P(α2) = 0but α2 α1 0, P1(α2)mustbezero. Hence x α2 isafactorof P1(x),and P1(x) = (x α2)P2(x),so P(x) = (x α1)(x α1)P2(x).
Continuingsimilarlyfor s steps,(x α1)(x α2) (x αs)isafactorof P(x).
B.Alldistinctzeroes
If n distinctzeroesofapolynomialofdegree n canbefound,thenthefactoringiscomplete,andthepolynomial istheproductofdistinctlinearfactors.
16Alldistinctzeroes
If α1, α2,..., αn are n distinctzeroesofapolynomial P(x)ofdegree n,then P(x) = a(x α1)(x α2) (x αn), where a istheleadingcoefficientof P(x).
Proof Bytheprevioustheorem,(x α1)(x α2) (x αn)isafactorof P(x), so P(x) = (x α1)(x α2) ··· (x αn)Q(x),forsomepolynomial Q(x).
But P(x)and(x α1)(x α2) ··· (x αn)bothhavedegree n,so Q(x)isaconstant.
Equatingcoefficientsof xn,theconstant Q(x)istheleadingcoefficient.
Factoringpolynomials—findingseveralzeroesfirst
Ifwecanfindseveralzeroesofapolynomial,thenwehaveaquadraticorcubicfactor,andthelongdivisions requiredcanbereduced,orevenavoidedcompletely.
17Factoringpolynomials—findingseveralzeroesfirst
• Usetrialanderrortofindasmanyintegerzeroesof P(x)aspossible.
• Usinglongdivision,divide P(x)bytheproductoftheknownfactors.
Ifthecoefficientsof P(x)areallintegers,thenanyintegerzeroof P(x)mustbeoneofthedivisorsofthe constantterm.
Whenthisprocedureisappliedtothepolynomialfactoredintheprevioussection,oneratherthantwolong divisionsisrequired.
Factor P(x) = x4 + x3 9x2 + 11x 4completely(doneinExample8).
Solution
Asbefore,allthecoefficientsareintegers,soanyintegerzeroisadivisoroftheconstantterm 4.Thatis,we test1,2,4, 1, 2and 4.
P(1) = 1 + 1 9 + 11 4 = 0,so x 1isafactor.
P( 4) = 256 64 144 44 4 = 0,so x + 4isafactor.
Afterlongdivisionby(x 1)(x + 4) = x2 + 3x 4(omitted),
P(x) = (x 2 + 3x 4)(x 2 2x + 1).
Factoringbothquadratics, P(x) = (x 1)(x + 4) × (x 1)2 = (x 1)3(x + 4).
Note: Themethodsofthenextsectionwillallowthisparticularfactoringtobedonewithnolongdivisions.
Thenextexampleinvolvesapolynomialthatfactorsintodistinctlinearfactors—nothingmorethanthefactor theoremisrequiredtocompletethetask.
Example10 Findingallthezeroesbythefactortheorem
Factor P(x) = x4 x3 7x2 + x + 6completely.
Solution
Thedivisorsoftheconstantterm6are1,2,3,6, 1, 2, 3and 6.
P(1) = 1 1 7 + 1 + 6 = 0,so x 1isafactor.
P( 1) = 1 + 1 7 1 + 6 = 0,so x + 1isafactor.
P(2) = 16 8 28 + 2 + 6 = 12 0,so x 2isnotafactor.
P( 2) = 16 + 8 28 2 + 6 = 0,so x + 2isafactor.
P(3) = 81 27 63 + 3 + 6 = 0,so x 3isafactor.
Wenowhavefourdistinctzeroesofapolynomialofdegree4.
Hence P(x) = (x 1)(x + 1)(x + 2)(x 3).(Noticethat P(x)ismonic.)
C.Themaximumnumberofzeroes
Ifapolynomialofdegree n had n + 1zeroes,thenbythefirsttheoremabove(Box15),itwouldbedivisiblebya polynomialofdegree n + 1,whichisimpossible.
18Maximumnumberofzeroes
Apolynomialofdegree n hasatmost n zeroes.
Theprevioustheoremtranslateseasilyintoaconditionforapolynomialtobethezeropolynomial.
19Avanishingcondition
• Supposethat P(x)isapolynomialthathasnotermofdegreemorethan n,yetiszeroforatleast n + 1 distinctvaluesof x.Then P(x)isthezeropolynomial.
• Inparticular,theonlypolynomialthatiszeroforallvaluesof x isthezeropolynomial(aswasproven inExercise11BEnrichment).
Proof Supposethat P(x)hadadegree.Thisdegreemustbeatmost n becausethereisnotermofdegree morethan n.Butthedegreemustalsobeatleast n + 1becausethereare n + 1distinctzeroes.Thisisa contradiction,so P(x)hasnodegree,andisthereforethezeropolynomial.
Note: Thisagainhighlightsthefactthatthezeropolynomial Z(x) = 0isquitedifferentinnaturefromallother polynomials.Itistheonlypolynomialwithaninfinitenumberofzeroes—infacteveryrealnumberisa zeroof Z(x).Associatedwiththisisthefactthat x α isafactorof Z(x)forallrealvaluesof α,because Z(x) = (x α)Z(x)(whichistriviallytrue,becausebothsidesarezeroforall x).Nowonderthenthatthe zeropolynomialdoesnothaveadegree!
Amostimportantconsequenceofthislasttheoremisaconditionfortwopolynomials P(x)and Q(x)tobe identicallyequal.
20Anidenticallyequalcondition
• Supposethat P(x)and Q(x)aredegree n polynomialsthathavethesamevaluesforatleast n + 1values of x.
• Thenthepolynomials P(x)and Q(x)areidenticallyequal(meaningthattheyareequalforallvalues of x),andtheircoefficientsareequal.
• Inparticular:
▷ Alinearpolynomialisdeterminedbytwovalues.
▷ Aquadraticpolynomialisdeterminedbythreevalues.
▷ Acubicpolynomialisdeterminedbyfourvalues.
▷ Aquarticpolynomialisdeterminedbyfivevalues.
Proof Let F(x) = P(x) Q(x).Because F(x)iszerowhenever P(x)and Q(x)havethesamevalue,itfollows that F(x)iszeroforatleast n + 1valuesof x,sobytheprevioustheorem, F(x)isthezeropolynomial,so P(x) = Q(x)forallvaluesof x
AndwestatedbeforeinBox5ofSection11A—proveninExercise11BEnrichment—thatitnow followsthatthecoefficientsareequal.
Usingtheconditionthatpolynomialsareidenticallyequal
Find a, b, c and d,if x3 x = a(x 2)3 + b(x 2)2 + c(x 2) + d foratleastfourvaluesof x.
Solution
Becausethecubicsareequalforfourvaluesof x,theyareidenticallyequal.
Substituting x = 2, 6 = d.
Equatingcoefficientsof x 3,1 = a.
Hencetheidentityisnow
Substituting x = 0, 0 = 8 + 4b 2c + 6 2b c = 1. (1)
Substituting x = 1, 0 = 1 + b c + 6 b c = 5. (2)
Solving(1)and(2)simultaneously, b = 6and c = 11.
Herearesomeofthegeometricversionsofthefactortheorem—theytranslatetheconsequencesaboveintothe languageofcoordinategeometry.Youwillalreadyhaveseentheminoperationwhendealingwithgraphsof quadratics.
21Geometricimplicationsofthefactortheorem
1 Thegraphofapolynomialfunctionofdegree n iscompletelydeterminedbyany n + 1pointsonthe curve.
2 Thegraphsoftwodistinctpolynomialfunctionscannotintersectinmorepointsthanthemaximumof thetwodegrees.
3 Alinecannotintersectthegraphofapolynomialofdegree n inmorethan n points.
Example12 Examiningintersectionsofcurvesusingthefactortheorem
Byfactoringthedifference F(x) = P(x) Q(x),describetheintersectionsbetweenthecurves P(x) = x4 + 4x3 + 2and Q(x) = x4 + 3x3 + 3x,andfindwhere P(x)isabove Q(x).
Solution
Subtracting, F(x) = x 3 3x + 2.
Substituting, F(1) = 1 3 + 2 = 0,so x 1isafactor. F( 2) = 8 + 6 + 2 = 0,so x + 2isafactor.
Afterlongdivisionby(x 1)(x + 2) = x2 + x 2, F(x) = (x 1)2(x + 2).
Hence y = P(x)and y = Q(x)aretangentat x = 1,butdonotcrossthere,andalsointersectalsoat x = 2, wheretheycrossatanangle—seeBox22belowaboutthisstep.
Because F(x)ispositivefor 2 < x < 1or x > 1,andnegativefor x < 2, P(x)isabove Q(x)for 2 < x < 1 or x > 1,andbelowitfor x < 2.
Simpleandmultiplerootswhenfindingintersectionsofpolynomials
Whentwopolynomialgraphscross,asintheworkedexampleabove,solvingthemsimultaneouslyresultsina polynomialequation,whichmayhavemultipleroots.Thebehaviouratamultiplerootisexactlyanalogoustothe behaviourofasinglepolynomialatamultiplezero,asdescribedinBox8inSection11B:
22Thebehaviouratanintersdectionoftwopolynomialgraphs
Supposethat x = åisarootofthepolynomialequationobtainedwhentwopolynomialgraphs y = P(x) and y = Q(x)aresolvedsimultaneously.
• If x = α isasimpleroot(sometimescalledasingleroot),thenthecurvescrosseachotheratanangle, andarenottangenttoeachotherthere.
• If x = α hasevenmultiplicity,thenthecurvesaretangenttoeachother,anddonotcross.
• If x = α hasoddmultiplicityatleast3,thenthecurvesaretangenttoeachother,andcrossatthepoint ofintersection.
Again,theproofreliesonYear12calculus.Butthisresultdoesfollowimmediatelyfromtheearlierstatementin Box8,ifoneconsidersinsteadthe difference F(x) = P(
),aswasdoneintheworkedexampleabove.
The fundamentaltheoremofalgebra cannotbeprovenintheExtension2course,butthetheoremhelpsusto understandtheimportanceofcomplexnumbersforpolynomials.Ittellsusthateverypolynomialequationof degree n ≥ 1hasexactly n roots,providedfirstthatrootsarecountedaccordingtotheirmultiplicity,andsecondly thatcomplexrootsarealsocounted.Forexample:
• x3 = 0hasoneroot x = 0,butthisroothasmultiplicity3.
• x3 1 = 0,whichfactorsto(x 1)(x2 + x + 1) = 0,hasroot x = 1,butalsohasthetwocomplexrootsof x2 + x + 1 = 0.
Thismeansthatthegraphofapolynomialofdegree n ≥ 2intersectseverylineinexactly n points,providedfirst thatpointswherethelineisatangentarecountedaccordingtotheirmultiplicity,andsecondlythatcomplex pointsofintersectionarealsocounted.Thistheoremprovidesthefundamentallinkbetweenthealgebraof polynomialsandthegeometryoftheirgraphs,andallowsthedegreeofapolynomialtobedefinedalgebraically asthehighestindex,orgeometricallyasthenumberoftimeseverylinecrossesit.
1 Usethefactortheoremtowritedowninfactoredform: amoniccubicpolynomialwithzeroes 1,3and4, a amonicquarticpolynomialwithzeroes0, 2,3and1, b acubicpolynomialwithleadingcoefficient6andzeroesat 1 3 , 1 2 and1. c
2a Showthat2and5arezeroesof P(x) = x4 3x3 15x2 + 19x + 30.
b Henceexplainwhy(x 2)(x 5)isafactorof P(x).
c Divide P(x)by(x 2)(x 5)andhenceexpress P(x)astheproductoffourlinearfactors.
3 Usetrialanderrortofindasmanyintegerzeroesof P(x)aspossible.Uselongdivisiontodivide P(x)bythe productoftheknownfactorsandhenceexpress P(x)infactoredform.
4 RefertoBox19toanswerparts(a)and(b).
Thepolynomial(a 2)x2 + (1 3b)x + (5 2c)hasthreezeroes.Whatarethevaluesof a, b and c? a
b
Thepolynomial(a + 1)x3 + (b 3)x2 + (2c 1)x + (5 4d)hasfourzeroes.Whatarethevaluesof a, b, c and d?
5a If3x2 4x + 7 = a(x + 2)2 + b(x + 2) + c forall x,find a, b and c. b If2x3 8x2 + 3x 4 =
, b, c and d.
c Usesimilarmethodstoexpress x3 + 2x2 3x + 1asapolynomialin(x + 1).
d Ifthepolynomials2x2 + 4x + 4and a(x + 1)2 + b(x + 2)2 + c(x + 3)2 areequalforthreevaluesof x,find a, b and c.
6a Apolynomialofdegree3hasadoublezeroat2.When x = 1ittakesthevalue6andwhen x = 3ittakes thevalue8.Findthepolynomial.
b Twozeroesofapolynomialofdegree3are1and 3.When x = 2ittakesthevalue 15andwhen x = 1ittakesthevalue36.Findthepolynomial.
7 Showthat x2 3x + 2isafactorof P(x) = xn(2m 1) + xm(1 2n) + (2n 2m),where m and n arepositive integers.
8 Iftwopolynomialshavedegrees m and n,where m > n,whatisthemaximumnumberofintersectionpoints oftheirgraphs?
9 Explainwhythegraphofacubicpolynomialwiththreedistinctzeroesmusthavetwoturningpoints.
10 Theline y = k meetsthecurve y = ax3 + bx2 + cx + d fourtimes.Findthevaluesof a, b, c and d interms of k.
11 Find,inexpandedform,themonicdegreefivepolynomialwhosezeroesare0, 1,1,2 √2and2 + √2.
12 Byfactoringthedifference F(x) = P(x) Q(x),describetheintersectionsbetweenthecurves P(x)and Q(x).
P(x) = 2x3 4x2 + 3x + 1, Q(x) = x3 + x2 8 a
P(x) = x4 + x3 + 10x 4, Q(x) = x4 + 7x2 6x + 8 b
P(x) = 2x3 + 3x2 25, Q(x) = 3x3 x2 + 11x + 5 c
P(x) = x4 3x2 2, Q(x) = x3 5x d
P(x) = x4 + 4x3 x + 5, Q(x) = x3 3x2 2x + 5 e
13 If a and b arenon-zero,and a + b = 0,provethatthepolynomials A(x) = x3 + ax2 x + b and B(x) = x3 + bx2 x + a haveacommonfactorofdegree2butarenotidenticalpolynomials.Whatisthecommon factor?
14a Factor xn 1.
b Supposethattherootsoftheequation xn 1 = 0are x = 1, α1, α2, , αn 1.Showthat(1 α1)(1 α2) (1 αn 1) = n
15 Supposethat P(x) = x5 + x2 + 1and Q(x) = x2 2.If r1, r2, r3, r4 and r5 arethefivezeroesof P(x),findthe valueof Q(r1) × Q(r2) × Q(r3) × Q(r4) × Q(r5).
16 Supposethat P(x)isapolynomialof odddegree n.Itisknownthat P(k) = k k + 1 for k = 0,1,2, ..., n.
Writedownthezeroesofthepolynomial(x + 1) P(x) x a
b Find P(n + 1). c
Let A betheleadingcoefficientofthepolynomial(x + 1) P(x) x.Factorthepolynomial,andhence showthat A = 1 1 × 2 × 3 ×···× n × (n + 1) = 1 (n + 1)!
UNCORRECTED
Learningintentions
• Developandusethesum-and-product-of-rootsformulaefordegrees2,3,and4.
• Usetheseformulaetohelpfactorpolynomials,andforotherpurposes.
Whenexpandingamonicpolynomialwithonlylinearfactors,suchas
P(x) = (x 2)(x 3)(x 5)(x 7) = x 4 17x 3 + 101x 2
247x + 210, thefourzeroes2,3,5and7ofthepolynomialclearlydeterminethefivecoefficients1, 17,101, 247and210. Butwhataretheformulaeforthecoefficients?
Ifwestudythecoefficient 17ofthetermin x3,andtheconstantterm210,itdoesnottakelongtorealisethat
17 = (2 + 3 + 5 + 7) = (sumofthezeroes)
210 =+2 × 3 × 5 × 7 =+(productofthezeroes).
Itisnotatallclear,however,howthecoefficients101of x2,and 247of x,arerelatedtothezeroes.(Weinvite readerstotryworkingitoutbeforecontinuing.)
Thissectionanswersthesequestionsforquadratic,cubic,andquarticpolynomials.Knowingtheseformulaeisa helpfultechniquewhentryingtofactorapolynomial,andhasothermoregeneralapplications.
Thezeroesofaquadratic
Let P(x) = ax2 + bx + c beaquadraticwithzeroes α and β andleadingcoefficient a.Then P(x) = a(x α)(x β) bythefactortheorem.
Expanding, P(x) = a(x α)(x β) = ax2 a(α + β)x + aαβ
Hencethecoefficientof x is a timesthesumofthezeroes,andtheconstantis +a timestheproductofthe zeroes, b = a(α + β)and c =+
Thisisjustwhathappenedwiththequarticabove,exceptformultiplicationbytheleadingcoefficient a.Solving forthesumandproductofthezeroes:
23Sumandproductofzeroesofaquadratic
Let P(x) = ax2 + bx + c havezeroes α and β.Then α + β = b a and αβ =+ c a .
Example13 Usingthesumandproductofzeroesofaquadratic
ai Showbysubstitutionthat x = 5isazeroof P(x) = 3x2 30x + 75.
ii Usesum-of-zeroestofindtheotherzero.
iii Useproduct-of-zeroestofindtheotherzero.
b Findthesumofthezeroesof Q(x) = x2 + 7x 11,andhencefinditsaxisofsymmetry.
Solution
ai P(5) = 3 × 25 30 × 5 + 75 = 0.
ii α + β =+ 30 3 α + 5 = 10 α = 5,so5isadoublezero.
b α + β = 7 1 = 7.
iii αβ = 75 3 α × 5 = 25 α = 5,soagain,5isadoublezero.
Theaxisofsymmetryismidwaybetweenzeroes,soitis x = 3 1 2
(Thiscalculationisexactlythesameastheformula x = b 2a fortheaxis.)
Thezeroesofacubic
Let P(x) = ax3 + bx2 + cx + d beacubicwithzeroes α, β and γ andleadingcoefficient a.Then P(x) = a(x α)(x β)(x γ)bythefactortheorem.
Expanding, P(x) = a(x α)(x β)(x γ) = ax 3 a(α + β + γ)x 2 + a(αβ + βγ
soaswesawbeforewiththequarticandthequadratic,
b = a(α + β + γ)and d = aαβγ , wherethenegativesignoftheproductcomesfromthecubeof 1. Butthenewphenomenonhereisthecoefficientof x, c = a(αβ + βγ + γα) =+(sumofproductsofpairsofzeroes).
Solvingforthesumsandproductsofthezeroes:
24Sumsandproductsofzeroesofacubic
Let P(x) = ax3 + bx2 + cx + d havezeroes α, β and γ.Then α + β + γ = b a (sumofthezeroes) αβ + βγ + γα =+ c a (sumofproductsofpairsofzeroes) αβγ = d a (productofthezeroes)
Example14
Usingthesumandproductofzeroesofacubic
a Showthat 6isazeroof P(x) = x3 4x2 39x + 126.
b Usesumandproductofzeroestofindtheothertwozeroes.
c Checktheformulaforthesumofproductsofpairsofzeroes.
Solution
a P( 6) = 216 144 + 234 + 126 = 0.
b α + β 6 = 4 1
α + β = 10 and αβ × ( 6) = 126 1 αβ = 21. Hencebyinspection, α and β are3and7. (Youhavebeenusingthis‘byinspection’approachforyearstosolvequadratics.)
c αβ + βγ + γα = 3 × 7 + 7 × ( 6) + ( 6) × 3 = 39,whichisthecoefficientof x.
Supposethatthefourzeroesofthequarticpolynomial
P(x) = ax 4 + bx3 + cx 2 + dx + e
are α, β, γ and δ.Bythefactortheorem(seeBox16inSection11E), P(x)isamultipleoftheproduct (x α)(x β)(x γ)(x δ):
P(x) = a(x α)(x β)(x γ)(x δ) = ax 4 a(α + β + γ + δ)x 3 + a(αβ + αγ + αδ
Equatingcoefficientsoftermsin x3 , x2 , x andconstantsnowgives:
25Zeroesandcoefficientsofaquartic
(sumofthezeroes)
(sumofproductsofpairsofzeroes)
(sumofproductsoftriplesofzeroes)
(productofthezeroes)
Thesecondformulagives thesumoftheproductsofpairsofzeroes,andthethirdformulagives thesumofthe productsoftriplesofzeroes.
Themethodisthesameforalldegrees.Notationisunfortunatelyamajordifficultyhere,andtheresultsarebetter writteninwords.Supposethat α1, α2,... αn arethe n zeroesofthedegree n polynomial
P(x) = an xn + an 1 xn 1 + ··· + a1 x + a0
26Zeroesandcoefficientsofapolynomial
sumofthezeroes = α1 + α2 + + αn = an 1 an
sumofproductsofpairsofzeroes =+ an 2 an
sumofproductsoftriplesofzeroes = an 3 an
productofthezeroes =
Noticethealternatingsignsofthesuccessiveresults.Itisunlikelythatanythingapartfromthefirstandlast formulaewouldberequired.
Example15 Usingthesumandproductofzeroesofacubic
Let α, β and γ betherootsofthecubicequation x3 3x + 2 = 0.Usetheformulaeabovetofind:
Checktheresultwiththefactoring x3 3x + 2 = (x 1)2(x + 2)obtainedinthesolutionofExample12in Section11E.
Solution
Factoringpolynomialsusingthefactortheoremandthesumand productofzeroes
Longdivisioncanbeavoidedinmanysituationsbyapplyingthesumandproductofzeroesformulaeafteroneor morezeroeshavebeenfound.Herearethestepsthatwehavedevelopedsofarforfactoringpolynomials:
27Factoringpolynomials—thestepssofar
• Usetrialanderrortofindasmanyintegerzeroesof P(x)aspossible.
• Usesumandproductofzeroestofindtheotherzeroes.
• Alternatively,uselongdivisionof P(x)bytheproductoftheknownfactors.
Ifthecoefficientsof P(x)areallintegers,thenanyintegerzeroof P(x)mustbeoneofthedivisorsofthe constantterm.
Inthenextexample,wefactorapolynomialfactoredtwicealready,butthistimethereisnoneedforanylong divisionatall.
Usingthesumandproductofzeroesofaquartic
Factor F(x) = x4 + x3 9x2 + 11x 4completely.
Solution
Asbefore, F(1) = 1 + 1 9 + 11 4 = 0, and F( 4) = 256 64 144 44 4 = 0.
Letthezeroesbe1, 4, α and β
Then α + β + 1 4 = 1
α + β = 2. (1)
Also αβ × 1 × ( 4) = 4
αβ = 1. (2)
From(1)and(2), α = β = 1,so F(x) = (x 1)3(x + 4).
Example17 Usingthesumandproductofzeroesofacubic
Factorcompletelythecubic G(x) = x3 x2 4.
Solution
First, G(2) = 8 4 4 = 0.
Letthezeroesbe2, α and β
Then2 + α + β = 1
α + β = 1,(1) and2 × α × β = 4
αβ = 2. (2)
Substituting(1)into(2), α( 1 α) = 2
α2 + α + 2 = 0
Thisisanirreduciblequadratic,because ∆= 7,sothecompletefactoringis G(x) = (x 2)(x2 + x + 2).
Note: Thisprocedure—developingtheirreduciblequadraticfactorfromthesumandproductofzeroes—is reallylittleeasierthanthelongdivisionitavoids.
Ifsomeinformationcanbegainedabouttherootsofapolynomialequation,itmaybepossibletoforman identitywiththecoefficientsofthepolynomial.
Ifonezeroofthecubic f (x) = ax3 + bx2 + cx + d istheoppositeofanother,provethat ad = bc.
Solution
Weknowthatoneofthezeroesistheoppositeofanother,sowecanbeginwiththefollowingsentence,which expressesthisfactsymbolically:
Letthethreezeroesofthecubicbe α, α and β
Nowwecanusetheformulaforthesumoftheroots,
Thenusingtheformulafortheproductoftheroots,
Therearethreeproductsofpairsofroots,namely α2 and αβ and βα,sousingthesumofproductsofpairs ofroots,
Nowwemusteliminatetheroots α and β fromequations(1),(2),and(3).
Substituting(3)into(2), cβ = d, (4) anddividing(1)by(4), ac = b d , thatis, ad = bc
1 If α and β aretherootsofthequadraticequation x2 4x + 2 = 0,find:
2 If α, β and γ aretherootsoftheequation x3 + 2x2 11x 12 = 0,find:
+ β + γ a
Nowfindtherootsoftheequation x3 + 2x2 11x 12 = 0byfactoringtheLHS.Hencecheckyouranswers fortheexpressionsinparts(a)–(i).
3 If α, β, γ and δ aretherootsoftheequation x
= 0,find:
4 If α and β aretherootsoftheequation2x2 + 5x 4 = 0,find:
5a Thepolynomial P(x) = 2x3 5x2 14x + 8haszeroesat 2and4.Usethesumofthezeroestofindthe otherzero.
b Supposethat x 3and x + 1arefactorsof P(x) = x3 6x2 + 5x + 12.Usetheproductofthezeroesto findtheotherfactorof P(x).
6 Considerthepolynomial P(x) = x3 x2 x + 10.
Showthat 2isazeroof P(x). a
Supposethatthezeroesof P(x)are 2, α and β.Showthat α + β = 3and αβ = 5. b
Bysolvingthetwoequationsinpart(b)simultaneously,showthat α2 3α + 5 = 0. c Henceshowthattherearenosuchrealnumbers α and β. d Hencestatehowmanytimesthegraphofthecubiccrossesthe x-axis. e
7 Showthat x = 1and x = 2arezeroesof P(x),andusethesumandproductofzeroestofindtheotherone ortwozeroes.Noteanymultiplezeroes.
P(x) = x3 2x2 5x + 6 a P(x) = 2x3 + 3x2 3x 2 b
P(x) =
8a Findthevaluesof a and b forwhich x3 + ax2 10x + b isexactlydivisibleby x2 + x 12,andthenfactor thecubic.
b Findthevaluesof a and b forwhich x2 x 20isafactorof x4 + ax3 23x2 + bx + 60,andthenfindall thezeroes.
9a Ifoneoftherootsoftheequation x2 + bx + c = 0istwicetheotherroot,showthat2b2 = 9c.(Hint:Let therootsbe α and2α.)
b Ifoneoftherootsoftheequation x2 + px + q = 0isonemorethantheotherroot,showthat p2 = 4q + 1. (Hint:Lettherootsbe α and α + 1.)
10a Findtherootsoftheequation3x3 x2 48x + 16 = 0giventhatthesumoftwoofthemiszero.(Hint: Lettherootsbe α, α and β.)
b Findtherootsoftheequation2x3 5x2 46x + 24 = 0giventhattheproductoftwoofthemis3.(Hint: Lettherootsbe α, 3 α and β.)
c Findtherootsoftheequation x3 27x 54 = 0giventhattwoofthemareequal.(Hint:Lettheroots be α, α and β.)
d Findthezeroesofthepolynomial P(x) = 2x3 13x2 + 22x 8giventhatoneofthemistheproductof theothertwo.(Hint:Letthezeroesbe α, β and αβ.)
11a Twooftherootsoftheequation x3 + 3x2 4x + a = 0areopposites.Findthevalueof a andthethree roots.
b Twooftherootsoftheequation4x3 + ax2 47x + 12 = 0arereciprocals.Findthevalueof a andthe threeroots.
12 Findthezeroesofthepolynomial x4 3x3 8x2 + 12x + 16iftheyare α,2α, β and2β.
13a Ifonerootoftheequation x3 bx2 + cx d = 0isequaltotheproductoftheothertwo,showthat (c + d)2 = d(b + 1)2 .
b Thepolynomial P(x) = x4 + ax3 + bx2 + cx + d hastwozeroesthatareoppositesandanothertwozeroes thatarereciprocals.Showthat: b = 1 + d i c = ad ii
14 Thecubicequation x3 Ax2 + 3A = 0,where A > 0,hasroots α, β and α + β
Usethesumoftherootstoshowthat α + β = 1 2 A a
b
Usethesumoftheproductsofpairsofrootstoshowthat
Showthat A = 2√6. c
15 Thepolynomial P(x) = x3 Lx2 + Lx M haszeroes α, 1 α and β
a Showthat: α + 1 α + β = L i 1 + αβ +
= L ii β = M iii
b Showthateither M = 1or M = L 1.
16 Supposethatthepolynomial P(x) = ax3 + bx2 + cx + d haszeroes α, β and γ
Showthat α2 + β2 + γ2 = b2 2ac a2 . a
Henceexplainwhythepolynomial2x3 3x2 + 5x 8cannothavethreerealzeroes. b
17 If α, β and γ aretherootsoftheequation x3 + 5x 4 = 0,evaluate α3 + β3 + γ3 . (Hint: x = α, x = β and x = γ satisfythegivenequation.)
18 Supposethattheequation x2 + bx + c = 0hasroots α and β.
a Showthat b and c aretherootsoftheequation x2 + (α + β αβ)x αβ(α + β) = 0.
b Findthenon-zerovaluesof b and c forwhichtherootsoftheequationinpart(a)are: α and β i α2 and β2 ii
Learningintentions
• Usethemethodsofpolynomialstosolvegeometricproblems.
Thisfinalsectionaddsthemethodsoftheprecedingsections,particularlythesumandproductofzeroes,tothe availabletechniquesforstudyingthegeometryofvariouscurves.Thestandardtechniqueistoexaminetheroots oftheequationformedintheprocessofsolvingtwocurvessimultaneously. Thissectionisratherdemanding,andcouldallberegardedasEnrichment.
Whentwocurvesintersect,wecanformtheequationwhosesolutionsarethe x-or y-coordinatesofpointsof intersectionofthetwocurves.Themidpointoftwopointsofintersectioncanthenbefoundusingtheaverageof theroots.Tangentscanbeidentifiedascorrespondingtodoubleroots.
Thenextexamplecouldalsobedoneusingquadraticequations,butitisaclearexampleoftheuseofsumand productofroots.
Theline y = 2x meetstheparabola y = x2 2x 8atthetwopoints A(α,2α)and B(β,2β).
a Showthat α and β arerootsof x2 4x 8 = 0,andhencefindthecoordinatesofthemidpoint M of AB.
b Usetheidentity(α β)2 = (α + β)2 4αβ tofindthehorizontaldistance |α β| from A to B.Thenuse Pythagoras’theoremandthegradientofthelinetofindthelengthof AB.
c Findthevalueof b forwhich y = 2x + b isatangenttotheparabola,andfindthepoint T ofcontact.
Solution
a Solvingthelineandtheparabolasimultaneously,
x 2 2x 8 = 2x x 2 4x 8 = 0.
Hence α + β = 4, and αβ = 8.
Averagingtheroots, M has x-coordinate x = 2,andsubstitutingintotheline, M = (2,4).
b Weknowthat(α β)2 = (α + β)2 4αβ = 16 + 32 = 48
so |α β| = 4√3. Becausethelinehasgradient2,theverticaldistanceis8√3, sousingPythagoras, AB2 = (4√3)2 + (8√3)2
= 16 × 15
AB = 4√15.
c Solving y = x2 2x 8and y = 2x + b simultaneously, x 2 4x (8 + b) = 0
Becausethelineisatangent,lettherootsbe θ and θ Thenusingthesumofroots, θ + θ = 4, so θ = 2,
Usingtheproductofroots, θ2 = 8 b andbecause θ = 2, b = 12.
Sotheline y = 2x 12isatangentat T (2, 8).
Geometricproblemsusingsumandproductofroots
Thesumandproductofrootscanmakesomeinterestinggeometricproblemsquitestraightforward.
Alinewithgradient m throughthepoint P( 1,0)onthecubic y = x3 x crossesthecurveattwofurther points A and B with x-coordinates α and β respectively.
a Sketchthesituation.
b Showthat α and β satisfythecubicequation x3 (m + 1)x m = 0.
c Showthatthemidpoint M of AB liesontheverticalline x = 1 2 .
d Findthelinethrough P tangenttothecubicatapointdistinctfrom P
Solution
a Thecubicfactorsas y = x(x 1)(x + 1), sothezeroesare x = 1, x = 0and x = 1.
b Thelinethrough P( 1,0)hasequation y = m(x + 1).
Solvingthelinesimultaneouslywiththecubic,
c Thecubichasroots α, β and1. Usingthesumofroots, α + β + ( 1) = 0(thereisnotermin x 2) so α + β = 1.
Hencethemidpoint M of AB has x-coordinate 1 2 (α + β) = 1 2 ,so M liesontheline x = 1 2
d Theline PM isatangentatanotherpointonthecurvewhenthepoints A, M and B coincide,thatis,when α = β, andbecause α + β = 1,thismeansthat α = β = 1 2
Usingtheproductofroots, α × β × ( 1) = m (theconstanttermis m) 1 2 × 1 2 × ( 1) = m m = 1 4 ,
sothetangentthrough P is y = 1 4 (x + 1).
Note: Sketchesshouldbedrawninallthesequestionstomakethesituationclear.
1a Showthatthe x-coordinatesofthepointsofintersectionoftheparabola y = x2 6x andtheline y = 2x 16satisfytheequation x2 8x + 16 = 0.
b Solvethisequation,andhenceshowthatthelineisatangenttotheparabola.Findthepoint T ofcontact.
2a Showthatthe x-coordinatesofthepointsofintersectionoftheline y = 2x + b andtheparabola y = x2 6x satisfythequadraticequation x2 4x b = 0.
b Supposethatthelineisatangenttotheparabola,sothattherootsofthequadraticequationareequal.Let theserootsbe α and α.
Usingthesumofroots,showthat α = 2. i
Usingtheproductofroots,showthat α2 = b,andhencefind b ii
Findtheequationofthetangentanditspoint T ofcontact. iii
3 Theline y = x + 1meetstheparabola y = x2 3x at A and B
a Find α + β,andhencefindthecoordinatesofthemidpoint M of AB. b
Showthatthe x-coordinates α and β of A and B satisfytheequation x2 4x 1 = 0.
4a Showthatthe x-coordinatesofthepointsofintersectionoftheline y = 3 x andthecubic y = x3 5x2 + 6x satisfytheequation x3 5x2 + 7x 3 = 0.
b Showthat x = 1and x = 3arerootsoftheequation,andusethesumoftherootstofindthethirdroot.
c Explainwhythelineisatangenttothecubic,thenfindthepointofcontactandtheotherpointof intersection.
5a Showthatthe x-coordinatesofthepointsofintersectionoftheline y = mx andthecubic y = x3 5x2 + 6x satisfytheequation x3 5x2 + (6 m)x = 0.
b Supposenowthatthelineisatangenttothecubicatapointotherthantheorigin,sothattherootsofthe equationare0, α and α.
Usingthesumofroots,showthat α = 5 2 i
Usingtheproductofpairsofroots,showthat α2 = 6 m,andhencefind m ii
Findtheequationofthetangentanditspoint T ofcontact. iii
6 Theline y = x 2meetsthecubic y = x3 5x2 + 6x at F(2,0),andalsoat A and B.
a
Showthatthe x-coordinates α and β of A and B satisfy x3 5x2 + 5x + 2 = 0.
Find α + β,andhencefindthecoordinatesofthemidpoint M of AB b
c
Showthat αβ = 1,thenusetheidentity(α β)2 = (α + β)2 4αβ toshowthatthehorizontaldistance |α β| between A and B is √13.
HenceusePythagoras’theoremtofindthelength AB d
7 Alinepassesthroughthepoint A( 1, 7)onthecurve y = x3 3x2 + 4x + 1.Supposethatthelinehas gradient m andistangenttothecurveatanotherpoint P onthecurvewhose x-coordinateis α
a
b
c
Showthatthelinehasequation y = mx + (m 7).
Showthatthecubicequationwhoserootsarethe x-coordinatesofthepointsofintersectionoftheline andthecurveis x3 3x2 + (4 m)x + (8 m) = 0.
Explainwhytherootsofthisequationare 1, α and α,andhencefindthepoint P andthevalueof m
8 Thepoint P(p, p3)liesonthecurve y = x3.Alinethrough P withgradient m intersectsthecurveagainat A and B.
Findtheequationofthelinethrough P a
Showthatthe x-coordinatesof A and B satisfytheequation x3 mx + mp p3 = 0.
c
b Hencefindthe x-coordinateofthemidpoint M of AB,andshowthatforfixed p, M alwaysliesonaline thatisparalleltothe y-axis.
9a Theequation x3 (m + 1)x + (6 2m) = 0hasarootat x = 2andadouble rootat x = α.Find α and m.
b Writedowntheequationoftheline ℓ passingthroughthepoint P( 2, 3)with gradient m.
c Thediagramshowsthecurve y = x3 x + 3andthepoint P( 2, 3)onthe curve.Theline ℓ cutsthecurveat P,andistangenttothecurveatanother point A onthecurve.Findtheequationoftheline ℓ P() −2−3 , x y 3
10a Usethefactortheoremtofactorthepolynomial y = x4 4x3 9x2 + 16x + 20,giventhattherearefour distinctzeroes,thensketchthecurve.
b Theline ℓ: y = mx + b touchesthequartic y = x4 4x3 9x2 + 16x + 20attwodistinctpoints A and B. Explainwhythe x-coordinates α and β of A and B aredoublerootsoftheequation x4 4x3 9x2 + (16 m)x + (20 b) = 0.
c Usethetheoryofthesumandproductofrootstowritedownfourequationsinvolving α, β, m and b
d Hencefind m and b,andwritedowntheequationof ℓ
Note: Iftwocurvestoucheachotherat P,thentheyaretangenttoeachotherat P
11a Find k andthepointsofcontactiftheparabola y = x2 k touchesthequartic y = x4 attwopoints.
b Find k andthepoint T ofcontactiftheparabola y = x2 k touchesthecubic y = x3
c Find k andthepointsofcontactiftheparabola y = x2 k touchesthecircle x2 + y2 = 1attwopoints.
12 Acirclepassingthroughtheorigin O istangenttothehyperbola xy = 1at A,andintersectsthehyperbola againattwodistinctpoints B and C.Provethat OA ⊥ BC.
13 Thediagramtotherightshowsthecircle x2 + y2 = 1andtheparabola y = (λx 1)(x 1),where λ isaconstant.Thecircleandparabolameetinthe fourpoints
P(1,0), Q(0,1), A(α, φ), B(β, ψ).
Thepoint M isthemidpointofthechord AB.
a Showthatthe x-coordinatesofthepointsofintersectionofthetwocurves satisfytheequation λ2 x 4 2λ(1 + λ)x 3 + (λ2 + 4λ + 2)x 2 2(1 + λ)x = 0.
b Usetheformulaforthesumoftherootstoshowthatthe x-coordinateof M is λ + 2 2λ .
c Useasimilarmethodtofindthe y-coordinateof M,andhenceshowthat M liesonthelinethroughthe origin O parallelto PQ.
d Forwhatvaluesof λ istheparabolatangenttothecircleinthefourthquadrant?
e Forwhatvaluesof λ arethefourpoints P, Q, A and B distinct,withrealnumbersascoordinates.
• Createyourownsummaryofthischapteronpaperorinadigitaldocument.
• Thisautomatically-markedquizisaccessedintheInteractiveTextbook.AprintablePDFWorksheetversionis alsoavailablethere.
• Checklist AvailableintheInteractiveTextbook,usethechecklisttotrackyourunderstandingofthelearningintentions. PrintablePDFandworddocumentversionsarealsoavailablethere.
1 Considerthepolynomial P(x) = 2x3 5x2 6x 11.State: thedegreeof P(x), a theleadingcoefficientof P(x), b theleadingtermof P(x), c theconstanttermof P(x). d
2 Thepolynomial P(x)hasdegree3.Writedownthedegreeofthepolynomial: 3 P(x) a (P(x))3 b
3 Findthecoefficientof x2 inthepolynomial P(x) = (x2 3x 7)(2x2 + 4x 9).
4a Sketchthegraphofthepolynomialfunction y = (x + 2)2(x 1)(x 3),showingallinterceptswiththe coordinateaxes. b Hencefindthevaluesof x forwhich(x + 2)2(x 1)(x 3) < 0.
5 Sketchthegraphofthepolynomial P(x) = x3 x5 .
6 Supposethatthepolynomial P(x) = 2x3 + 7x2 4x + 5isdividedby D(x) = x 3. Findthequotient Q(x)andtheremainder R(x). a Writedownadivisionidentityusingtheinformationabove. b
7 Withoutlongdivision,findtheremainderwhen P(x) = x3 5x2 + 1isdividedby: x 3 a x + 2 b
8a Usethefactortheoremtoshowthat x 2isafactorof P(x) = x3 19x + 30. b Hencefactor P(x)fully.
9 Findthevalueof k giventhat x + 3isafactorof P(x) = x3 + 4x2 + kx 12.
10 Findthevaluesof b and c giventhat x + 1isafactorof P(x) = x3 + bx2 + cx 7,andtheremainderis 12 when P(x)isdividedby x 5.
11 Findthevaluesof h and k giventhat x + 2isafactorof Q(x) = (x + h)2 + k,andtheremainderis16when Q(x)isdividedby x.
12 Thepolynomial P(x)isdividedby(x + 1)(x 2).Supposethatthequotientis Q(x)andtheremainderis R(x).
Explainwhythegeneralformof R(x)is ax + b,where a and b areconstants. a When P(x)isdividedby x + 1theremainderis10,andwhen P(x)isdividedby x 2theremainderis 8.Find a and b.(Hint:Usethedivisionidentity.) b
13 Supposethatthepolynomial Q(x) = x2 6x 4haszeroes α and β.Withoutfindingthezeroes,findthe valueof:
14 If α, β and γ aretherootsoftheequation x3 + 10x2 + 5x 20 = 0,find:
15 Theequation x3 + 5x2 + cx + d = 0hasroots 3,7and α.
Usethesumoftherootstofind α a
Usetheproductoftherootstofind d b
Usethesumoftherootsinpairstofind c c
16 Theequation6x3 17x2 5x + 6 = 0hasroots α, β and γ,where αβ = 2.
Usetheproductoftherootstofind γ. a
Usethesumoftherootstofindtheothertworoots. b
17 Onerootoftheequation ax2 + 2bx + c = 0isthereciprocalofthesquareoftheotherroot.Showthat a3 + c3 + 2abc = 0.
18 Solvetheequation9x3 27x2 + 11x + 7 = 0giventhattherootsare α β, α and α + β.
19 Findthezeroesofthepolynomial P(x) = 8x3 14x2 + 7x 1giventhattheyare α β , α and αβ.
20 Theline y = 9x + 5isthetangenttothecurve y = x3 3x2 atthepoint A( 1, 4).Thelineintersectsthe curveatanotherpoint B.Supposethatthe x-coordinateof B is α
Writedownthecubicequationwhoserootsarethe x-coordinatesof A and B. a
Explainwhytherootsofthisequationare 1, 1and α. b Hencefindthepoint B. c
Thefirstsectionofthischapterappliestrigonometrytothree-dimensionalproblems.Threenewissuesarise immediatelywhenmovingfrom2to3dimensions.
▶ Visualisationisthefirstnewissuehere,andacleardiagramdisplayingallthetrianglesisrequired—atask thatmayinvolveseveralattempts.
▶ Thesecondnewissueistoidentifyallrightangles,andtounderstandwhatitmeansforaplaneandaline,or twoplanes,tobeperpendicular.
▶ Thethirdnewissueistounderstandhowtoidentifymoregenerallytheanglebetweenalineandaplane, andbetweentwoplanes.
Three-dimensionalworkwillcontinueintheYear12studyof2and3-dimensionalvectors,wherethesesame threeissueswillarise.
Thenexttwosectionsdevelopaseriesoftrigidentitiesinvolvingcompoundanglesanddoubleangles.These arerequiredinYear12whenthetrigfunctionsaredifferentiated,which,andasalwayswithnewfunctions,will beginwithanappealtogeometry.Theseidentitieswillalsoberequiredwhensketchingfunctionsinvolvingthe trigfunctions,andinmodellingvariousperiodicphenomena.
Thenewidentities,togetherwiththosedevelopedpreviously,areappliedinSection16Dtothesolutionofa widevarietyoftrigonometricequationsthatariselaterwhenmodellingwithcalculus.
ThefinalSection16Edealswithfunctionsoftheform a sin x + b cos x,whichsurprisinglyturnouttobesine orcosinefunctions—shiftedhorizontally,anddilatedverticallytogiveadifferentamplitude.Theseresults andmethodsrequiretransformationoftrigonometricfunctionspresentedintheYear12Advancedcourse,soa preliminarysubheadingquicklyappliesthemethodsofChapter5tothetwotransformationsthatareneeded.
ThemethodsofthislastsectionarethefirststepinFourieranalysis,whichisthebasisofavastnumberofwave phenomenainscience,engineeringandmedicine—ourearssplitasoundwaveintoindividualfrequencies, asisclearwhenwelistentomusic,andtheanalysisofgravitywavesisattheveryedgeofphysicsrightnow. JosephFourierwasaFrenchscientistdeeplyinvolvedwithNapoleonBonaparteandtheFrenchRevolution. Aswellashisground-breakingworkwiththeanalysisofperiodicphenomena,hewasanexpertinEgyptian archaeology.
Learningintentions
• Interpretdatagivenindiagramorwrittenformabouta3Dtrigproblem.
• Applytrigonometrytosolvea3Dproblem.
Trigonometryisbasedontriangles,whicharetwo-dimensionalobjects.Whentrigonometryisappliedtoathreedimensionalproblem,thediagrammustbebrokenupintoacollectionoftrianglesinspace,andtrigonometryis thenappliedtoeachtriangleinturn.Acarefullydrawndiagramisalwaysessential.
Twonewideasaboutanglesareneeded—theanglebetweenalineandaplane,andtheanglebetweentwo planes.Pythagoras’theoremremainsfundamental.
TrigonometryandPythagoras’theoreminthreedimensions
Herearethestepsinasuccessfulapproachtoathree-dimensionalproblem.
1TrigonometryandPythagoras’theoreminthreedimensions
1 Drawacarefulsketchofthesituation.
2 Notecarefullyallthetrianglesinthefigure.
3 Mark,ornote,allrightanglesinthesetriangles.
4 Alwaysnamethetriangleyouareworkingwith.
Example1 UsingPythagoras’theoremina3Dfigure
Therectangularprismsketchedbelowhasdimensions:
= 4cmand
a UsePythagoras’theoremin △CFG tofindthelengthofthediagonal FC.
b Similarlyfindthelengthsofthediagonals AC and AF
c UsePythagoras’theoremin △ACG tofindthelengthofthespace diagonal AG.
d Usetrigonometryin △BAF tofind ∠BAF (nearestminute).
e Usetrigonometryin △GAF tofind ∠GAF (nearestminute).
Solution
a In △CFG, FC2 = 32 + 42 , FC = 5cm. usingPythagoras,
b In △ABC, AC2 = 52 + 42 (Pythagoras)
AC = √41cm.
In △ABF, AF2 = 52 + 32 (Pythagoras)
AF = √34cm.
c In △ACG,theangle ∠ACG isarightangle,and AC = √41and CG = 3.
Hence AG2 = AC2 + CG2 ,
= 41 + 32
AG = √50
= 5√2cm. usingPythagoras,
d In △BAF,theangle ∠ABF isarightangle,and AB = 5and BF = 3.
Hencetan ∠BAF = BF AB = 3 5
∠BAF ≑ 30◦58′
e In △GAF,theangle ∠AFG isarightangle,and AF = √34and FG = 4.
Hencetan ∠GAF = FG AF = 4 √34
∠GAF ≑ 34◦27′
In3Dspace,aplane P andaline ℓ canberelatedinthreedifferentways:
• Inthefirstdiagram,thelinelieswhollywithintheplane.
• Intheseconddiagram,thelinenevermeetstheplane.Wesaythatthelineandtheplaneare parallel.
• Inthethirddiagram,thelineintersectstheplaneinasinglepoint P.
Whentheline ℓ meetstheplane P inthesinglepoint P,itcandosointwodistinct ways.
Intheupperdiagram,theline ℓ isperpendiculartoeverylineintheplanethrough P. Wesaythatthelineis perpendicular totheplane.
Inthelowerdiagram,theline ℓ isnotperpendicularto P.Toconstructtheangle θ betweenthelineandtheplane:
• Chooseanotherpoint A ontheline ℓ
• Constructthepoint M intheplane P sothat AM ⊥P
Then ∠APM istheanglebetweenthelineandtheplane.
Example2 Findingtheanglebetweenalineandaplane
Findtheanglebetweenaslantedgeandthebaseinasquarepyramidofheight8metreswhosebasehasside length12metres(nearestminute).
Solution
UsingPythagoras’theoreminthebase ABCD,
AC2 = 122 + 122
AC = 12√2metres.
Theperpendicularfromthevertex V tothebasemeetsthebaseatthemidpoint M ofthediagonal AC
In △MAV,tan ∠MAV = MV MA = 8 6√2
∠MAV ≑ 43◦19′ , andthisistheanglebetweentheslantedge AV andthebase.
In3Dspace,twoplanesthatarenotparallelintersectinaline ℓ.Toconstructthe anglebetweentheplanes:
• Takeanypoint P onthislineofintersection.
• Constructtheline p through P perpendicularto ℓ lyingintheplane P
• Constructtheline q through P perpendicularto ℓ lyingintheplane Q
Theanglebetweentheplanes P and Q isdefinedtobetheangle θ betweenthese twolines p and q.
Example3 Findingtheanglebetweentwoplanes
Inthepyramidofthepreviousexample,findtheanglebetweenanobliquefaceofthepyramidandthebase.
Solution
Let P bethemidpointoftheedge BC
Then VP ⊥ BC and MP ⊥ BC, so ∠VPM istheanglebetweentheobliquefaceandthebase.
In △VPM,tan ∠VPM = VM PM =
Three-dimensionalproblemsinwhichnotrianglecanbesolved
Inthefollowingclassicproblem,therearefourtrianglesformingatetrahedron,butnotrianglecanbesolved, becausenomorethantwomeasurementsareknowninanyoneofthesetriangles.
Themethodistointroduceapronumeral h = TF fortheheight,thenworkaroundthefigureuntil four measurementsareknownintermsof h inthebasetriangle—atthispointanequationin h canbeformedandsolved.
Example4 Solvingaprobleminwhichnotrianglecanbesolved
Amotoristdrivingonlevelgroundsees,duenorthofher,atowerwhoseangleofelevationis10◦.After driving3kmfurtherinastraightline,thetowerisinthedirectionduewest,withangleofelevation12◦ Howhighisthetower? a Inwhatdirectionisshedriving? b
Solution
Letthetowerbe TF,andletthemotoristbedrivingfrom A to B
a Therearefourtriangles,noneofwhichcanbesolved.
Let h betheheightofthetower.
In △TAF, AF = h cot10◦ .
In △TBF, BF = h cot12◦ .
Wenowhaveexpressionsforfourmeasurementsin △ABF,sowecanusePythagoras’theoremtoforman equationin h
In △ABF, AF2 + BF2 = AB2 h2 cot2 10◦ + h2 cot2 12◦ = 32 h2(cot2 10◦ + cot2 12◦) = 9
sothetowerisabout407metreshigh.
b Let θ = ∠FAB,thenin △AFB,sin θ
soherdirectionisaboutN64◦E.
Hereisasummaryofwhathasbeensaidabout3Dproblems:
2Three-dimensionaltrigonometry
1 Understandanglesbetweenalineandaplane,andbetweentwoplanes.
2 Drawacarefuldiagramofthesituation,markingallrightangles.
3 Aplandiagram,lookingdown,isusuallyagreathelp.
4 Identifyeverytriangleinthediagram,toseewhetheritcanbesolved.
5 Ifonetrianglecanbesolved,thenworkfromitaroundthediagramuntiltheproblemissolved.
6 Ifnotrianglecanbesolved,assignapronumeraltowhatistobefound,thenworkaroundthediagram untilanequationinthatpronumeralcanbeformedandsolved.
1 Thediagramtotherightshowsarectangularprism.
a UsePythagoras’theoremtofindthelengthofthebasediagonal BE
b Hencefindthelengthoftheprismdiagonal BH
c Find,correcttothenearestdegree,theangle α that BH makeswiththe baseoftheprism.
2 Thediagramtotherightshowsacube.
a Writedownthesizeof:
b UsePythagoras’theoremtofindtheexactlengthof:
AF ii AG
c Hencefind,correcttothenearestdegree:
∠GAF
∠AGB
3 Thediagramtotherightshowsatriangularprism.
a Findtheexactlengthof:
i AC ii AF
b Whatisthesizeof ∠ACF?
c Find ∠AFC,correcttothenearestdegree.
4 Thediagramtotherightshowsasquarepyramid.Thepoint C isthecentreofthebase,and TC is perpendiculartothebase.
a Writedownthesizeof: i ∠CMQ ii ∠TCM iii ∠TCQ
b Findthelengthof: i CM ii CQ
c Find,correcttothenearestdegree: i theanglebetweenasidefaceandthebase, ii theanglebetweenaslantedgeandthebase.
5 Thediagramtotherightshowsarectangularprism.
a Writedownthesizeof:
i ∠ABF ii ∠DBF
b Find,correcttothenearestdegree,theanglethatthediagonalplane DBFH makeswiththebaseoftheprism.
6 Thediagramtotherightshowsasquareprism.Theplane ABC isinside theprism,and M isthemidpointofthebasediagonal BC.
a Findtheexactlengthof MD
b Hencefind,correcttothenearestdegree,theanglethattheplane ABC makeswiththebaseoftheprism.
7 Twolandmarks P and Q onlevelgroundareobservedfromthetop T ofa verticaltower BT ofheight30m.Landmark P isduesouthofthetower,while landmark Q isdueeastofthetower.Theanglesofelevationof T from P and Q are15◦ and18◦ respectively.
a Showthat BP = 30tan75◦ andfindasimilarexpressionfor BQ.
b Find,correcttothenearestmetre,thedistancebetweenthetwolandmarks.
8 Atree BT isduenorthofanobserverat P andduewestofanobserverat Q Thetwoobserversare50mapartandthebearingof Q from P is36◦.The angleofelevationof T from Q is28◦ .
a Showthat BQ = 50sin36◦
b Hencefindtheheight h ofthetreecorrecttothenearestmetre.
c Find,correcttothenearestdegree,theangleofelevationof T from P
9 Twomonuments A and B are400mapartonahorizontalplane.Theangleof depressionof A fromthetop T ofatallbuildingis18◦.Also, ∠TAB = 52◦ and ∠TBA = 38◦
a Showthat TA = 400cos52◦ .
b Findtheheight h ofthebuilding,correcttothenearestmetre.
c Find,correcttothenearestdegree,theangleofdepressionof B from T .
10 Thediagramshowsacubeofside2cm,withdiagonals AG and CE intersectingat P.Thepoint M isthemidpointofthefacediagonal EG. Let α betheacuteanglebetweenthediagonals AG and CE
a Whatisthelengthof PM?
b Findtheexactlengthof EM.
c Writedowntheexactvalueoftan ∠EPM
d Hencefind α,correcttothenearestminute.
11 Thediagramshowsarectangularpyramid. X and Y arethemidpointsof AD and BC respectivelyand T isdirectlyabove Z. TX = 15cm, TY = 20cm, AB = 25cmand BC = 10cm.
a Showthat ∠XTY = 90◦ .
b UsingeithersimilartrianglesorPythagoras’theorem,showthat TZ = 12cm.
c Hencefind,correcttothenearestminute,theanglethatthefrontface DCT makeswiththebase.
12 Twoobserversat A and B onhorizontalgroundare300mapart.From A,the angleofelevationofthetop C ofatallbuilding DC is32◦.Itisalsoknownthat
∠DAB = 59◦ and ∠ADB = 78◦
a Showthat AD = 300sin43◦ sin78◦
b Hencefindtheheightofthebuilding,correcttothenearestmetre.
13
Aballoon B isduenorthofanobserver P anditsangleofelevationis62◦.Fromanotherobserver Q 100metresfrom P,theballoonisduewestanditsangleofelevationis55◦.Lettheheightoftheballoonbe h metresandlet C bethepointonthelevelgroundverticallybelow B
a Showthat PC = h cot62◦,andwritedownasimilarexpressionfor QC
b Explainwhy ∠PCQ = 90◦
c UsePythagoras’theoremin △CPQ toshowthat
h2 = 1002 cot2 62◦ + cot2 55◦ .
d Hencefind h,correcttothenearestmetre.
14 Fromapoint P duesouthofaverticaltower,theangleofelevationofthetopofthetoweris20◦.Froma point Q situated40metresfrom P anddueeastofthetower,theangleofelevationis35◦.Let h metresbe theheightofthetower.
Drawadiagramtorepresentthesituation.
b
a Showthat h = 40 √tan2 70◦ + tan2 55◦ ,andevaluate h,correcttothenearestmetre.
15 Fromtwopoints P and Q onlevelground,theanglesofelevationofthetop T ofa38mtowerare26◦ and22◦ respectively.Point P isduesouthofthe tower,andthebearingof Q fromthetoweris100◦T.
a Showthat PB = 38tan64◦,andfindasimilarexpressionfor QB
b Usethecosineruletodetermine,correcttothenearestmetre,thedistance between P and Q
16 Inthediagram, TF representsaverticaltowerofheight x metresstandingon levelground.From P and Q atgroundlevel,theanglesofelevationof T are22◦ and27◦ respectively. PQ = 63metresand ∠PFQ = 51◦ .
a Showthat PF = x cot22◦ andwritedownasimilarexpressionfor QF
b Usethecosineruletoshowthat x2 = 632 cot2 22◦ + cot2 27◦ 2cot22◦ cot27◦ cos51◦ .
c Findthevalueof x tothenearestinteger.
17 Thediagramshowsatowerofheight h metresstandingonlevelground.Theangles ofelevationofthetop T ofthetowerfromtwopoints A and B onthegroundnearby are55◦ and40◦ respectively.Thedistance AB is50metresandtheinterval AB is perpendiculartotheinterval AF,where F isthefootofthetower.
a Find AT and BT intermsof h.
b Whatisthesizeof ∠BAT ?
c UsePythagoras’theoremin △BAT toshowthat h = 50sin55◦ sin40◦ sin2 55◦ sin2 40◦
d Hencefindtheheightofthetower,correcttothenearestmetre.
18 Inthediagramofatriangularpyramid, AQ = x, BQ = y, PQ = h, ∠APB = θ,
∠PAQ = α and ∠PBQ = β.Also,therearethreerightanglesat Q.
a Showthat x = h cot α andwritedownasimilarexpressionfor y
b UsePythagoras’theoremandthecosineruletoshowthat cos θ = h2 (x2 + h2)(y2 + h2) .
c Henceshowthatsin α sin β = cos θ.
19 Thediagramshowsatriangularpyramid,allofwhosefacesareequilateraltriangles— suchasolidiscalleda regulartetrahedron.Supposethattheslantedgesareinclinedat anangle θ tothebase.Showthatcos θ = 1 √3
20 Asquarepyramidhasperpendicularheightequaltothesidelengthofitsbase.Showthattheanglebetween aslantedgeandabaseedgeitmeetsistan 1 √5.
21 Threetourists T1, T2 and T3 atgroundlevelareobservingalandmarkwhosetopweshallcall L. T1 isdue northof L, T3 isdueeastof L,and T2 isonthelineofsightfrom T1 to T3 andbetweenthem.Theanglesof elevationto L from T1, T2 and T3 are25◦,32◦ and36◦ respectively.
a
Showthattan ∠LT1T2 = cot36◦ cot25◦ .
Usethesinerulein △LT1T2 tofind,correcttothenearestminute,thebearingof T2 from L. b
Learningintentions
• Developandapplythecompoundangletrigformulae.
ThederivativesofthetrigonometricfunctionswillbedevelopedinYear12.Beforethatcanbedone,however, variousformulaeinvolvingcompoundanglesneedtobeestablished,beginningwiththeexpansionofobjects suchassin(x + h),tan(x y)andcos2x.Suchtrigonometricidentitiesaremostimportantforallsortsofother reasonsaswell,andmustbethoroughlymemorised.Theirdevelopmentistheconcernofthissectionandthe next.
Aswithallfundamentalresultsintrigonometry,theseformulaerequireaninitialappealtogeometry,inthis casecircles,Pythagoras’theoremintheformofthedistanceformula,thecosinerule,andthePythagorean trigonometricidentity.
Theapproachgivenherebeginswiththeexpansionofcos(α β)andusesthatresulttoderivetheother expansions.Therearemanyalternativeapproaches.
Theexpansionof cos(α − β)
Weshallprovethatforallrealnumbers α and β, cos(α β) = cos α cos β
Proof Lettherayscorrespondingtotheangles α and β intersectthecircle x2 + y2 = r2 atthepoints A and B respectively.Thenbythedefinitionsof thetrigonometricfunctionsforgeneralangles, A = (r cos α , r sin α)and B = (r cos β , r sin β).
Nowwecanusethedistanceformulatofind AB2:
=
(cos
(cos
(1
Buttheangle ∠AOB is α β,sobythecosinerulein △AOB,
AB2 = r 2 + r 2 2r 2 cos(α β) = 2r 2 1 cos(α β)
Equatingthesetwoexpressionsfor AB2 , cos(α β) = cos α cos β + sin α sin β
Note: Itwasclaimedintheproofthat ∠AOB = α β.Thisisnotnecessarilythecase,becauseit’salsopossible that ∠AOB = β α,orthat ∠AOB differsfromeitherofthesetwovaluesbyamultipleof2π.But thecosinefunctioniseven,anditisperiodicwithperiod2π.Soitwillstillfollowineverycasethat cos ∠AOB = cos(α β),whichisallthatisrequiredintheproof.
Thesixcompound-angleformulae
Hereareallsixcompound-angleformulaeforthesine,cosineandtangentfunctions,followedbytheremaining fiveproofs.
3Thecompound-angleformulae
A sin(α + β) = sin α cos β + cos α sin β
B cos(α + β) = cos α cos β sin α sin β
C tan(α + β) = tan α + tan β 1 tan α tan β
D sin(α β) = sin α cos β cos α sin β
E cos(α β) = cos α cos β + sin α sin β
F tan(α β) = tan α tan β 1 + tan α tan β
Theseformulaeareofcoursetruewhethertheangleisgivenindegreesorradians.
Proof WeproceedfromformulaE,whichhasalreadybeenproven.
B Replacing β by β inE,whichistheexpansionofcos(α β), cos(α + β) = cos α ( β)
= cos α cos( β) + sin α sin( β) = cos α cos β sin α sin β,becausecosineisevenandsineisodd.
A Usingtheidentitysin θ = cos( π 2 θ), sin(α + β) = cos π 2 (α + β) = cos ( π 2 α) β)
= cos( π 2 α)cos β + sin( π 2 α)sin β
= sin α cos β + cos α sin β
D Replacing β by β,andnotingthatcosineisevenandsineisodd, sin(α β) = sin α cos β cos α sin β
C Becausetan θ istheratioofsin θ andcos θ, tan(α + β) = sin(α + β) cos(α + β) = sin α cos β + cos α sin β cos α cos β sin α sin β = tan α + tan β 1 tan α tan β ,afterdividingtopandbottombycos α cos β.
F Replacing β by β,andnotingthatthetangentfunctionisodd,
tan(α β) = tan α tan β 1 + tan α tan β .
Example5 Usingthecompoundangleformulae
Expresssin(x + 2π 3 )intheform a cos x + b sin x
Solution
sin(x + 2π 3 ) = sin x cos 2π 3 + cos x sin 2π 3
= 1 2 sin x + 1 2 √3cos x
Giventhatsin α = 1 3 andcos β = 4 5 ,where α isacuteand π 2 < β < 0,findsin(α β)andcos(α + β).
Solution
First,usingthediagramsontheright,
cos α = 2 3 √2andsin β = 3 5 .
Sosin(α β) = sin α cos β cos α sin β
= 4 15 + 6 15 √2
= 2 15 (2 + 3√2),
andcos(α + β) = cos α cos β sin α sin β
= 8 15 √2 + 3 15 = 1 15 (8√2 + 3).
Thevariouscompound-angleformulaecanbeusedtofindexpressionsinsurdsfortrigonometricfunctionsof manyanglesotherthanoneswhoserelatedanglesarethestandard30◦,45◦ and60
Example7 Findingexactvaluesofanglesotherthan30◦,45◦,and60◦
Findexactvaluesof:
Solution
Therearemanyalternativemethods.
◦ = sin(30
1 Usethecompound-angleformulaetoexpand:
2 Expressasasingletrigonometricfunction:
3 Usethecompound-angleformulaetoprove:
sin(90◦ + A) = cos A a
cos(90◦ A) = sin A b
tan(180◦ + A) = tan A d cos(270◦ A) = sin A e
4 Provetheidentities:
sin(A + 45◦) = 1 √2 (sin A + cos A) a
tan(360◦ A) = tan A c
sin(360◦ A) = sin A f
2cos(θ + π 3 ) = cos θ √3sin θ b
c sin(A 30◦) = 1 2 (√3sin A cos A) d
tan( π 4 A) = 1 tan A 1 + tan A
5a Iftan α = 1 2 andtan β = 1 3 ,findtan(α + β).
b Ifcos A = 4 5 andsin B = 12 13 ,where A and B arebothacute,findsin(A + B).
c Ifsin θ = 2 3 andcos φ = 1 4 ,where θ and φ arebothacute,findcos(θ φ).
6 Proveeachidentity.
sin(A + B) + sin(A B) = 2sin A cos B a
cos(x y) cos(x + y) = 2sin x sin y b
c
sin(x + y) + cos(x y) = (sin x + cos x)(sin y + cos y)
7 Findfromthediagram,usingappropriatecompound-angleresults:
a sin ∠BAC
b cos ∠BAC
8a Byexpressing15◦ as(45◦ 30◦),showthat: sin15◦ = √3 1 2√2 i
b Hencefindsurdexpressionsfor:
◦ i
◦ ii
9 Inthediagramopposite,itisknownthattan
a Showthattan(α +
b Writedownanalternativeexpressionfortan(α + β).
c Henceshowthat b =
Ifsin
11 Usecompound-angleformulaetofindtheexactvalueof:
Showthatsin(
Hencesimplifysin
13 Showthat:
◦ iii
14 Proveeachidentity: 2sin(x y)
cos(x + y) cos(x y) = cot x cot y a
sin(θ + φ)
b sin(α β)
c
cos(θ φ) = tan θ + tan φ 1 + tan θ tan φ
sin α sin β + sin(β γ) sin β sin γ + sin(γ α) sin γ sin α = 0
QP R T 3 cm2 cm 1 cm
a Writedowntheformulaforsin(α + β).
b Henceshowthattheangles α and β inthe diagramabovehavesum45◦ .
Thediagramshowstworectangles.Eachrectangleis 6cmlongand3cmwide,andtheyshareacommon diagonal PQ.Showthattan α = 3 4
17 Inthediagramopposite, P and Q arelandmarksthatare160metresand 70metresduenorthofpoints A and B respectively.Points A and B lie130 metresapartonawest–eastroad,and C isapointontheroadbetween A and B sothat ∠PCQ = 45◦.Find AC. (Let AC = x and ∠ACP = α.)
18a Provethetrigonometricidentity
b Henceshowthatinanytriangle
Learningintentions
• Developandapplythevariousdouble-angletrigformulae.
Substituting θ = α = β intothecompoundangleformulaegivesformulaefordouble-angleexpressionssuchas sin2θ
Inthecompound-angleformulasin(α + β) = sin α cos β + cos α sin β,replacebothangles α and β bytheone angle θ.Thisgives:
sin2θ = sin(θ + θ) = sin θ cos θ + cos θ sin θ = 2sin θ cos θ
Thesameproceduregivesexpansionsofcos2θ andtan2θ:
4Thedouble-angleformulae
sin2θ = 2sin θ cos θ
Theexpansionofcos2θ canthenbecombinedwiththePythagoreanidentitytogivetwootherformsofthe expansion,firstbyusingsin
,thenbyusingcos
5The cos2θ formulae
cos2θ = cos2 θ sin2 θ andrearrangingthelasttwoidentities:
Thelasttwoareonlyrearrangements—perhapsjustremembertheyarethere.
Example8 Provingfurtherdouble-angleidentities
Provethat(sin x + cos x)2 = 1 + sin2x.
Solution
LHS = (sin x + cos x)2 = sin2 x + cos2 x + 2sin x cos x = 1 + sin2x = RHS.
Example9 Provingfurtherdouble-angleidentities
Expresscot2x intermsofcot x.
Solution
cot2x = 1 tan2x = 1 tan2 x 2tan x = cot2 x 1 2cot x ,afterdivisionoftopandbottombytan2 x.
Furtherexactvaluesusingthedouble-angleformulae
Thedouble-angleformulaecanbeusedtofindexactvaluesofthetrigonometricfunctionsatanglessuchas22
Example10 Findingexactvaluesoffurtheracuteangles
Findtheexactvalueoftan22
Solution
1 Usethedouble-angleformulaetosimplify:
2 Provethat: (cos A sin A)(cos A + sin A) = cos2A a (sin α cos α)2 = 1 sin2α b sin2θ = 2sin θ sin( π 2 θ) c 1 1 tan θ 1 1 + tan θ = tan2θ d
3a Ifcos α = 4 5 ,findcos2α
b Ifsin x = 2 3 ,findcos2x
c Ifsin θ = 5 13 and θ isacute,findsin2θ
d Iftan A = 1 2 ,findtan2A.
4 Ifsin x = 3 4 and π 2 < x < π,findtheexactvalueofsin2x
5 Proveeachidentity.
cos4 α sin4 α = cos2α a cos2x + cos x = (cos x + 1)(2cos x 1) b sin2A 1 cos2A = cot A c cos θ sin θ sin2θ = cos θ cos2θ d tan( π 4 + α) tan( π 4 α) = 2tan2α e sin2θ + sin θ cos2θ + cos θ + 1 = tan θ f
6a Showthat 1 cos2x 1 + cos2x = tan2 x.
b Hencefindtheexactvalueoftan π 8
7 Eliminate θ fromeachpairofparametricequations. x = 2 + cos θ, y = cos2θ a x = tan θ + 1, y = tan2θ b
8 Points A, B, C,and W lieinthesameverticalplane.Abirdat A observesa wormat W atanangleofdepression θ.Afterflying20metreshorizontallyto B,theangleofdepressionofthewormis2θ.Ifthebirdflewanother10metres horizontallyitwouldbedirectlyabovetheworm.Let WC = h
Writetan2θ intermsoftan θ a
Usethetworight-angledtrianglestowritetwoequationsin h and θ. b
Useparts(a)and(b)toshowthat 1 10 = 60 900 h2 c
Henceshowthat h = 10√3metres. d
Howcould h havebeenfoundwithouttrigonometry? e
9 In △ABC, b c = 4 3 and B = 2C.Usethesineruletoshowthatcos C = 2 3
10a Bywriting3θ as2θ + θ andusingcompound-angleanddouble-angleresults,provethatcos3θ = 4cos3 θ 3cos θ. b Henceshowthatcos 2π 9 isarootoftheequation8x3 6x + 1 = 0.
11 Usedouble-angleformulaetoshowthat: 2sin 4π 5 cos π 5 = sin 2π 5 a cos2 4π 7 sin2 3π 7 = cos 6π 7 b
12 Proveeachidentity.
cot2α + tan α = cosec2α a
sin3A sin A + cos3A cos A = 4cos2A b
tan2x cot x = 1 + sec2x c sin2θ cos2θ + 1 sin2θ + cos2θ 1 = tan(θ + π 4 ) d
1 + sin2α 1 + cos2α = 1 2 (1 + tan α)2 e
13a Showthattan3θ = 3tan θ tan3 θ 1 3tan2 θ
cosec4A + cot4A = 1 2 (cot A tan A) f
b Atower AB hasheight h metres.Theangleofelevationofthetopofthetower atapoint C 20metresfromitsbaseisthreetimestheangleofelevationata point D 80metresfurtherawayfromitsbase.Usetheidentityinpart(a)to showthat h = 100 √7 metres.
14a Writedowntheexactvalueofcos45◦ . b Henceshowthat:
15a Showthat 8 4√3 = √6 √2. b Showthattan165◦ = √3 2.
c Henceshowthattan82
Learningintentions
• Applythecompoundangleformulaetothesolutionoftrigonometricequations.
• Decidewhichmethodormethodstouseforparticulartrigonometricequations.
Trigonometricequationsoccurwhenevertrigonometricfunctionsandtheirgraphsarebeinganalysed,andcareful studyofthemisessential.TheywereaddressedindegreesinChapter7,andinradiansinChapter13.The identitiesofthelasttwosectionsnowallowfurtherequationstobesolved.
Aswehaveseen,factoringisfundamentaleverywhereinourcourse.Thedifferenceofsquaresistheimmediately obviousapproachtothenextworkedexample.
Example11
Solvingatrigequationbyfactoring.
Solvesin2 x 4cos2 x = 0,for π ≤ x ≤ π,byfactoring(threedecimalplaces).
Solution
Theequationis sin2 x 4cos2 x = 0.
Usingdifferenceofsquares,(sin x + 2cos x)(sin x 2cos x) = 0
÷ cos2 x (tan x + 2)(tan x 2) = 0.
Hencetan x = 2ortan x = 2.
Therelatedangleis1.107148...(leaveinthecalculator) so x ≑ 2.034, 1.107,1.107,or2.034.
Equationsrequiringalgebraicsubstitutionsorfactoring
Thisisareviewofthemethodaspresentedearlier.
Example12
Solvingtrigequationsusingsubstitutionorfactoring
Solve2cos x = 1 + sec x,for0 ≤ x ≤ 2π: usingasubstitution, a usingidentitiesandfactoring. b
Solution
Let u = cos x.
Then2u = 1 + 1 u
2u 2 u 1 = 0 (2u + 1)(u 1) = 0 u = 1or u = 1 2 , so,cos x = 1orcos x = 1 2 . Hence x = 0,2π, 2π 3 ,or 4π 3 . a
Hence x = 0,2π, 2π 3 ,or 4π 3 . b
2cos x = 1 + sec x × cos x 2cos2 x = cos x + 1 2cos2 x cos x 1 = 0 (2cos x + 1)(cos x 1) = 0 cos x = 1orcos x = 1 2
Equationswithmorethanonetrigonometricfunction,butthesame angle
Thisiswheretrigonometricidentitiescomeintoplay.
6Equationswithmorethanonetrigonometricfunction
Trigonometricidentitiescanusuallybeusedtoproduceanequationinonlyonetrigonometricfunction.
Solvetheequation2tan θ = sec θ,for0◦ ≤ θ ≤ 360◦: usingtheratioidentities, a bysquaringbothsides. b
Solution
a 2tan θ = sec θ 2sin θ cos θ = 1 cos θ sin θ = 1 2 θ = 30◦ or150◦
b Squaring,4tan2 θ = sec2 θ 4sec2 θ 4 = sec2 θ sec2 θ = 4 3
Checkingeachsolution, θ = 30◦ or150◦ .
Method(b)inthepreviousexampleisintendedasawarning—avoidsquaringanequationifpossible,because squaringmayintroduceextrasolutions.
Ifanequationhastobesquared,eachsolutionmustbecheckedintheoriginalequationtoseewhetheritisa solutionornot.Herearetwoverysimpleequations,onealgebraicandonetrigonometric,wheretheeffectof squaringisobvious.
a Supposethat √x = 5.
Squaring, x = 25.
But √25 = 5,not 5.
Infact,therearenosolutions.
b Supposethatcos x = 1.
Squaring,cos2 x = 1
1 sin2 x = 1
sin2 x = 0
sin x = 0.
x = nπ,where n ∈ Z.
Butcos nπ = 1,foreveryoddmultipleof π
Whendifferentanglesareinvolvedinthesametrigonometricequation,theusualapproachistousecompoundangleidentitiestochangeallthetrigonometricfunctionstofunctionsoftheoneangle.
7Equationsinvolvingdifferentangles
Usecompound-angleidentitiestochangeallthetrigonometricfunctionstofunctionsoftheoneangle.
Frequentlysuchanequationcanbesolvedbymorethanonemethod.
Example14 Usingthedouble-angleformulae
Usethetan2θ formulatosolvetan4x = tan2x,for0 ≤ x ≤ π 2
Solution
tan4x = tan2x 2tan2x 1 tan2 2x = tan2x 2tan2x = tan2x + tan3 2x
tan3 2x 3tan2x = 0
tan2x(tan2 2x 3) = 0
tan2x = 0,or √3,or √3.
Therestrictionon2x is0 ≤ 2x ≤ π, sothesolutionsare 2x = 0,or π,or π 3 ,or 2π 3 x = 0,or π 2 ,or π 6 ,or π 3 .
Equationsinvolvingdifferentanglesandfunctions
Thesixtrigfunctionsareallverycloselyrelated.Thebestapproachisusually:
8Approachingtrigonometricequations
• First,trytogetalltheanglesthesame.
• Thentrytogetallthetrigonometricfunctionsthesame.
Example15 Solvingequationsinvolvingdifferentanglesandfunctions
Solvecos2x = 4sin2 x 14cos2 x,for0 ≤ x ≤ 2π: bychangingalltheanglesto x, a bychangingalltheanglesto2x b
Solution cos2x = 4sin2 x 14cos2 x cos2 x sin2 x = 4sin2 x 14cos2 x 15cos2 x = 5sin2 x tan x = √3or
Homogeneousequations
Equationshomogeneousinsin x andcos x areanimportantspecialcase.
9Homogeneousequations
• Anequationiscalled homogeneousin sin x and cos x ifthesumoftheindicesofsin x andcos x ineach termisthesame(andthesumiscalledthe degree).
• Tosolveanequationhomogeneousinsin x andcos x,dividethroughbyapowerofcos x toproducean equationintan x alone.
Theexpansionsofsin2x andcos2x arehomogeneousofdegree2insin x andcos x.Also,1 = sin2 x + cos2 x can beregardedasbeinghomogeneousofdegree2.
Solvesin2x + cos2x = sin2 x + 1,for0 ≤ x ≤ 2π.
Solution
Expanding,2sin x cos x + (cos2 x sin2 x) = sin2 x + (sin2 x + cos2 x) 3sin2 x 2sin x cos x = 0(thisishomogeneousinsin x andcos x.)
÷ cos2 x 3tan2 x 2tan x = 0 tan x(3tan x 2) = 0 tan x = 0ortan x = 2 3 .
Hence x = 0or π or2π,or x ≑ 0.588or3.730.
Atrigonometricequationcanoftenbesolvedinmorethanoneway
WorkedExample11abovewasaclassicdifference-of-squaresequation.Butitisalsoaclassicforahomogeneousequationapproach.
TrigonometricequationsoftenhavealternativeapproachesSolvesin2 x 4cos2 x = 0,for π ≤ x ≤ π,asa homogeneousequation.
Solution
Dividingthroughbycos2 x givestan2 x 4 = 0. Therestisobvious,andproceedsalongthesamelinesasworkedExample11.
TheEquations sin x = sin α, cos x = cos α,and tan x = tan α
Itispossibletosolvetheseequationsveryneatlyandquickly,providedthatyoukeepyoureyeontheAllStations ToCentralquadrantsdiagram.
• Onesolutionis α
• WeknowfromtheASTCquadrantsdiagramthattherearesolutionsinoneoftheotherthreequadrants—find oneexampleofthesesolutions.
• Addintegermultiplesof360◦ (orof2π ifworkinginradians)tothesetwosolutions,tofindallthesolutions satisfyingthegivenrestrictionon x
• Butif α isaboundaryangle,justreadthesolutionsoff thegraph
Example18 Solvingequationsoftheform sin x = sin α
a Solvesin x = sin25◦,for 180◦ ≤ x ≤ 180◦
b Solvetan x = tan0.2,for0 ≤ x ≤ 2π.
c Solvecos x = cos160◦,for 360◦ ≤ x ≤ 360◦ .
d Solvesin x = sin 3π 2 ,for 2π ≤ x ≤ 2π.
Solution
a sin x ispositiveinquadrants1and2,sothetwosolutionsare25◦ and155◦
b BytheASTCdiagram,tan x > 0inquadrants1and3. Hencethetwosolutionsin0 ≤ x ≤ 2π are0.2,and0.2 + π (Alternatively,thegraphof y = tan x hasperiod π ,andeachbranchbetweentwoasymptotesisincreasing ateverypoint.)
c Theangle160◦ hasrelatedangle20◦,andcos x < 0inquadrants2and3,sothetwosolutionsin
0◦ ≤ x ≤ 360◦ are160◦ and200◦ Hencethefoursolutionsin 360◦ ≤ x ≤ 360◦ are 200◦ , 160◦,160◦,and200◦
d Thisisaboundaryangle—fromthegraph,thetwosolutionsare 3π 2 and π 2
1 Considertheequationsin2x cos x = 0.
Byusingadouble-angleformulaandthenfactoring,showthatcos x = 0orsin x = 1 2 . a Hencesolvetheequationfor0 ≤ x ≤ 2π b
2 Considertheequationcos2x cos x = 0.
Byusingadouble-angleformulaandthenfactoring,showthatcos x = 1or 1 2 . a Hencesolvetheequationfor0 ≤ x ≤ 2π. b
3 Considertheequationsin(x + π 4 ) = 2cos(x π 4 ).
Usecompound-angleformulaetoshowthattan x = 1. a Hencesolvetheequationfor0 ≤ x ≤ 2π b
4 Usecompound-angleformulaetosolve,for0 ≤ θ ≤ 2π. sin(θ + π 6 ) = 2sin(θ π 6 ) a
(Hint:Inpart(d),writecos3θ ascos(2θ + θ).)
5 Usedouble-angleformulaetosolve,for0 ≤ x ≤ 2π
sin2x = sin x a sin2x + √3cos x = 0 b
3sin x + cos2x = 2 c
x + 3cos x + 2 = 0 d tan2x + tan x = 0 e
x = tan x f
6 Solve,for0◦ ≤ θ ≤ 360◦,givingsolutionscorrecttothenearestminutewherenecessary.
2sin2θ + cos θ = 0 a 2cos2 θ + cos2θ = 0 b 2cos2θ + 4cos θ = 1 c
+ sin θ = 1 e
θ + 13cos 1 2 θ = 5 g
7 Considertheequationtan( π 4 + θ) = 3tan( π 4 θ).
Showthattan2 θ 4tan θ + 1 = 0. a
2 θ f
Henceusethequadraticformulatosolvetheequationfor0 ≤ θ ≤ π b
8 Giventheequation2cos x 1 = 2cos2x:
Showthatcos x = 1 4 (1 + √5)orcos x = 1 4 (1 √5). a
Hencesolvetheequationfor0 ≤ x ≤ 2π b
9a Showthatsin(α + β)sin(α β) = sin2 α sin2 β.
b Hencesolvetheequationsin2 3θ sin2 θ = sin2θ,for0 ≤ θ ≤ π.
10a Showthatsin3x = 3sin x 4sin3 x.
b Hencesolvetheequationsin3x + sin2x = sin x,for0 ≤ x ≤ 2π.
11a Giventheequationsin(θ + π 6 ) = cos(θ π 4 ),showthattan θ = √6 √3 √2 + 2.
b Hencesolvetheequationfor0 ≤ θ ≤ 2π
12 Solve,for0 ≤ θ ≤ 2π,givingsolutionscorrecttotwodecimalplaceswherenecessary.
cos2 2θ = sin2 θ a cos2θ + 3 = 3sin2θ [Hint:Write3as3(cos2 θ + sin2 θ).] b
CHALLENGE
13a Usetheproduct-to-sumidentity2cos A cos B = cos(A + B) + cos(A B)toprovethesum-to-product identitycos P + cos Q = 2cos P + Q 2 cos P Q 2
b Hencesolvetheequationcos4x + cos x = 0,for0 ≤ x ≤ π.
14a Showthatcos3x = 4cos3 x 3cos x.
b Bysubstituting x = 2cos θ,showthattheequation x3 3x 1 = 0hasroots x = 2cos20◦ , 2sin10◦ or 2cos40◦
c Useasimilarapproachtofind,correcttothreedecimalplaces,thethreerealrootsoftheequation x3 12x = 8√3.
15a If t = tan x,showthattan4x = 4t(1 t2) 1 6t2 + t4 .
b Iftan4x tan x = 1,showthat5t4 10t2 + 1 = 0.
c Showthatsin A sin B = 1 2 (cos(A B) cos(A + B)) andthat cos A cos B = 1 2 (cos(A B) + cos(A + B))
d Henceshowthat π 10 and 3π 10 bothsatisfytan4x tan x = 1.
e Hencewritedown,intrigonometricform,thefourrealrootsofthepolynomial equation5x4 10x2 + 1 = 0.
16 Considertheequationcos2x = sin3x,for0◦ ≤ x ≤ 360◦
Solvetheequationovertherestricteddomainbywritingsin3x ascos(90◦ 3x). a
b Hencedeterminetheexactvalueofsin18◦ c
Alternatively,usetheidentitysin3x = 3sin x 4sin3 x andthefactorisation
4u3 2u2 3u + 1 = (u 1)(4u2 + 2u 1)toshowthatsin x = 1, 1 + √5 4 or 1 √5 4
Learningintentions
• Express a sin x + b cos x intheform R sin(x + α)orrelatedforms.
• Understandtheresultasadilatedandshiftedsineorcosinefunction.
• Solvetrigequationsoftheform a sin x + b cos x = c andrelatedinequations.
Thissectionanalyseswhathappenswhenthesineandcosinecurvesareadded,and,moregenerally,when multiplesofthetwocurvesareadded.Thesurprisingresultisthat y = a sin x + b cos x isstillasineorcosine wave,whateverthevaluesof a and b are,butshiftedsidewaysandstretchedvertically.
Theseformsfor a sin x + b cos x alsogiveasystematicmethodofsolvinganyequationoftheform a cos x + b sin x = c.
Preliminarynotesonverticaldilationsandhorizontalshiftsof y = sin x and y = cos x
TheboxedresultsbelowfollowimmediatelyfromthenotesontransformationsinChapter5.Theyalsoestablish theamplitudeofverticallydilatedsineorcosinefunctions.
10Verticaldilationsandhorizontalshiftsof y = sin x and y = cos x
• Whenshiftedright α:
)and
andtheresultingwavefunctionsstillhave amplitude1.
• Whendilatedverticallywithfactor R > 0: y = sin x −→ y = R sin x and y = cos x −→
andtheresultingwavefunctionsnowhave amplitude R.
),
• Whenshiftedright α anddilatedverticallywithfactor R > 0(ineitherorder): y = sin x −→ y = R sin(x α)and y = cos x −→ y = cos(x α), andtheresultingwavefunctionshave amplitude R.
Example19 Sketchingverticaldilationsandhorizontalshiftsof sin x and cos x
Sketch,over 360◦ ≤ x ≤ 360◦,onsinglepairsofaxes:
y = sin x and y = 2sin x a y = cos x and y = 2cos x b y = sin x and y = sin(x 120◦) c y = cos x and y = cos(x 120◦) d y = sin x and y = 2sin(x 120◦) e y = cos x and y = 2cos(x 120◦) f
y = sin x + cos x bygraphicalmethods
Wecannowreturntothesumof y = sin x and y = cos x.Chapter6preparedthegroundforthisbysketchingthe sumoftwogivensketchedgraphs.
Thediagramtotherightshowsthetwographsof y = sin x and y = cos x.Fromthesetwographs,thesumfunction y = sin x + cos x hasbeendrawnonthesamediagram—the crossesrepresentobviouspointstomarkonthegraphofthesum.
• Thenewgraphobviouslyhasthesameperiod2π as y = sin x and y = cos x.Itlookslikeawave,andwithin[0,2π]there arezeroesatthetwovalues x = 3π 4 and x = 7π 4 wheresin x and cos x takeoppositevalues.
• Thenewamplitudeisbiggerthan1.Thevalueat x =
2,soifthemaximumoccursthere, asseemslikely,theamplitudeis √2.
• Thissuggeststhattheresultingsumfunctionis y = √2sin(x + π 4 ),becauseitisthestretchedsinecurveshifted left π 4 .Wecancheckthisbyexpansion:
Thatisexactlywhatweexpectedfromthesketchesofthegraphs.
Thegeneralalgebraicapproach—theauxiliaryangle
Itistrueingeneralthatanyfunctionoftheform f (x) = a sin x + b cos x canbewrittenasasinglewavefunction. Therearefourpossiblestandardformsinwhichthewavecanbewritten,andtheprocedureisdonebyexpanding thestandardformandequatingcoefficientsofsin x andcos x.
• Anyfunctionoftheform f (x) = a sin x + b cos x,where a and b areconstants(notbothzero),canbe writteninanyoneofthefourforms:
y = R sin(x α) y = R cos(x α)
y = R sin(x + α) y = R cos(x + α)
where R > 0and0◦ ≤ α < 360◦.Theconstant R = √a2 + b2 isthesameforallforms,butthe auxiliary angle α dependsonwhichformischosen.
• Tobegintheprocess,expandthestandardformandequatecoefficientsofsin x andcos x
• Becarefultoidentifythequadrantinwhichtheauxiliaryangle α lies.
Thenextexamplecontinueswiththeexamplegivenatthestartofthesection,andshowsthesystematicalgorithm usedtoobtaintherequiredform.
Example20 Usingtheauxiliaryangletorewriteasumofsineandcosine
Express y = sin x + cos x inthetwoforms: R sin(x + α), a R cos(x + α), b where,ineachcase, R > 0and0 ≤ α < 2π.Thensketchthecurve,showingallinterceptsandmaximaand minimaintheinterval0 ≤ x ≤ 2π.
Solution
a Expanding,
soforall x,
Equatingcoefficientsofsin x, R cos α = 1, (1)
equatingcoefficientsofcos x, R sin α = 1. (2)
Squaringandadding, R2 = 2,
andbecause R > 0, R = √2.
From(1), cos α = 1 √2 , (1A) andfrom(2), sin α = 1 √2 , (2A)
sotherelatedangle α isinthe1stquadrant,withrelatedangle π 4
Hence α = π 4 ,andsin x + cos x = √2sin(x + π 4 ).
Thegraphis y = sin x shiftedleftby π 4 ,andstretchedverticallybya factorof √2.Thusthe x-interceptsare x = 3π 4 and x = 7π 4 ,thereisa maximumof √2when x = π 4 ,andaminimumof √2when x = 5π 4
b Expanding, R cos(x + α) = R cos x cos α R sin x sin α, soforall x, sin x + cos x = R cos x cos α R sin x sin α
Equatingcoefficientsofcos x, R cos α = 1, (1)
equatingcoefficientsofsin x, R sin α = 1. (2)
Squaringandadding, R2 = 2,
andbecause R > 0, R = √2.
From(1),cos α = 1 √2 ,(1A)
andfrom(2),sin α = 1 √2 ,(2A)
sotheauxiliaryangle α isinthe4thquadrant,withrelatedangle π 4 ˙
Hence α = 7π 4 ,sosin x + cos x = √2cos(x + 7π 4 ).
Thegraphabovecouldequallywellbeobtainedfromthis.
Itis y = cos x shiftedleftby 7π 4 andstretchedverticallybyafactorof √2. (Alternatively,shiftrightby π 4 ,becausetheperiodis2π.)
Unlessspecialanglesareinvolved,theauxiliaryangle α willneedtobeapproximatedonthecalculator.Degrees orradianmeasuremaybeused,butthenextexampleusesdegreestomaketheworkingalittlemoreintuitive.
Example21 Approximatingtheauxiliaryangleandskeetchingthecurve
a Express y = 3sin x 4cos x intheform y = R cos(x α),where R > 0and0◦ ≤ α < 360◦,giving α correct tothenearestdegree.
b Sketchthecurve,showing,correcttothenearestdegree,allinterceptsandturningpointsintheinterval 180◦ ≤ x ≤ 180◦ .
Solution
a Expanding, R cos(x
, soforall x, 3sin x 4cos
Equatingcoefficientsofcos x, R cos α = 4, (1) equatingcoefficientsofsin x, R sin α = 3. (2)
Squaringandadding, R2 = 25, andbecause R > 0, R = 5.
From(1), cos α = 4 5 ,(1A) andfrom(2),sin α = 3 5 ,(2A) so α isinthe2ndquadrant,withrelatedangleabout37◦ .
Hencetheauxiliaryangleis α ≑ 143◦,and3sin x 4cos x = 5cos(x α),
b Thegraphis y = cos x shiftedrightby α ≑ 143◦ andstretched verticallybyafactorof5.
Thusthe x-interceptsare x ≑ 53◦ and x ≑ 127◦,thereisa maximumof5when x ≑ 143◦,andaminimumof 5when x ≑ 37◦
(Findthe y-intercept 4bysubstituting x = 0intotheoriginal equation.)
Anoteonthecalculatoranditsapproximationsfortheauxiliaryangle
InworkedExample21,theexactvalueof α is α = 180◦ sin 1 3 5 ,because α isinthe2ndquadrant.The calculatorholdsthisvaluecorrecttomanydecimalplaces.
Whentherearesubsequentcalculationstodo,asinworkedExample22below,thisvalueshouldbestoredin memoryandusedwhenevertheauxiliaryangleisrequired.Re-entryofanapproximationmayleadtorounding errors.
Solvingequationsoftheform a sin x + b cos x = c,andassociated inequations
OncetheLHSisinoneofthefourstandardforms,thesolutionstothisequationareeasilyobtained.Worked Example22continuesfromworkedExample21.
Note: Itisalwaysimportanttokeeptrackofrestrictionsonthecompoundangle.
Usingtheauxiliaryangletosolveequationsandinequations
a Usingthepreviousworkedexample,anditscalculatorapproximation,solve3sin x 4cos x = 2,for 180◦ ≤ x ≤ 180◦,correcttothenearestdegree.
b Henceusethegraphtosolve3sin x 4cos x ≤−2,for 180◦ ≤ x ≤ 180◦
Solution
a Weknowthat3sin x 4cos x = 5cos(x α),where α ≑ 143◦ , sotheequationis5cos(x α) = 2, cos(x α) = 2 5 . Thus x α isinquadrant2or3,withrelatedangleabout66◦ Therestrictionon x is 180◦ ≤ x ≤ 180◦ thatis,about 323◦ ≤ x α ≤ 37◦ , so x α ≑ 114◦ or 246◦ x ≑ 30◦ or 103◦ .
Becarefultousethecalculator’smemoryhere,fortheangle α,andfortheangle66◦ above. Neverre-enterapproximationsofthoseangles.
b Thegraphtotherightshowsthepreviouslydrawngraphof y = 3sin x 4cos x withthehorizontalline y = 2added. Thisroughlyverifiestheanswersinpart(a).
Italsoshowsthatthesolutionoftheinequation
3sin x 4cos x ≤−2isabout 103◦ ≤ x ≤ 30◦
12Theauxiliary-anglemethodforequationsoftheform a sin x + b cos x = c
• GettheLHSintooneofthefourforms R sin(x + α)or R sin(x α)or R cos(x + α)or R cos(x α).
• Solvetheresultingequation,keepingapproximationsinthecalculatormemory.
• Usethegraphtosolveanyassociatedinequation.
1a Whattransformationmoves y = cos x to y = 4cos x?Drawbothcurvesononesetofaxes,andstatethe newamplitude.
b Whattransformationmoves y = sin x to y = sin(x π 4 )?Drawbothcurvesononesetofaxes.
2 Find R and α exactly,if R > 0and0 ≤ α < 2π,and: R sin α = √3and R cos α = 1, a R sin α = 3and R cos α = 3. b
3 Find R (exactly)and α (correcttothenearestminute),if R > 0and0◦ ≤ α < 360◦,and:
R sin α = 5and R cos α = 12, a R cos α = 2and R sin α = 4. b
4a Ifcos x sin x = A cos(x + α),showthat A cos α = 1and A sin α = 1.
b Findthepositivevalueof A bysquaringandadding.
c Find α,if0 ≤ α < 2π.
d Statethemaximumandminimumvaluesofcos x sin x,andthefirstpositivevaluesof x forwhichthey occur.
e Solvetheequationcos x sin x = 1,for0 ≤ x ≤ 2π.
f Writedowntheamplitudeandperiodofcos x sin x.Hencesketch y = cos x sin x,for0 ≤ x ≤ 2π. Indicateonyoursketchtheline y = 1andthesolutionstotheequationinpart(e).
5a If √3cos x sin x = B cos(x + θ),showthat B cos θ = √3and B sin θ = 1.
b Find B,if B > 0,bysquaringandadding.
c Find θ,if0 ≤ θ < 2π.
d Statethegreatestandleastpossiblevaluesof √3cos x sin x,andthevaluesof x closestto x = 0for whichtheyoccur.
e Solvetheequation √3cos x sin x = 1,for0 ≤ x ≤ 2π.
f Sketch y = √3cos x sin x,for0 ≤ x ≤ 2π.Onthesamediagram,sketchtheline y = 1.Indicateonyour diagramthesolutionstotheequationinpart(e).
6 Let4sin x 3cos x = A sin(x α),where A > 0and0◦ ≤ α < 360◦ .
Showthat A cos α = 4and A sin α = 3. a
Showthat A = 5and α = tan 1 3 4 b
c
Hencesolvetheequation4sin x 3cos x = 5,for0◦ ≤ x ≤ 360◦.Givethesolution(s)correcttothe nearestminute.
7 Considertheequation2cos x + sin x = 1.
Let2cos x + sin x = B cos(x θ),where B > 0and0◦ ≤ θ < 360◦.Showthat B = √5and θ = tan 1 1 2 . a
Hencefind,correcttothenearestminutewherenecessary,thesolutionsoftheequation,for0◦ ≤ x ≤ 360◦ b
8 Letcos x 3sin x = D cos(x + φ),where D > 0and0◦ ≤ φ < 360◦
Showthat D = √10and φ = tan 1 3. a
Hencesolvecos x 3sin x = 3,for0◦ ≤ x ≤ 360◦.Givethesolutionscorrecttothenearestminutewhere necessary. b
9 Considertheequation √5sin x + 2cos x = 2.
TransformtheLHSintotheform C sin(x + α),where C > 0and0◦ ≤ α < 360◦ a
Find,correcttothenearestminutewherenecessary,thesolutionsoftheequation,for0◦ ≤ x ≤ 360◦ b DEVELOPMENT
10 Solveeachequation,for0◦ ≤ x ≤ 360◦,bytransformingtheLHSintoasingle-termsineorcosinefunction.
Givesolutionscorrecttothenearestminute.
3sin x + 5cos x = 4 a 6sin x 5cos x = 7 b
7cos x 2sin x = 5 c 9cos x + 7sin x = 3 d
11 Find A and α exactly,if A > 0and0 ≤ α < 2π,and: A sin α = 1and A cos α = √3, a A
α = 5and A sin α = 5. b
12ai Express √3cos x + sin x intheform A cos(x + θ),where A > 0and0 < θ < 2π
ii Hencesolve √3cos x + sin x = 1,for0 ≤ x < 2π
bi Expresscos x sin x intheform B sin(x + α),where B > 0and0 < α < 2π ii Hencesolvecos x sin x = 1,for0 ≤ x < 2π
ci Expresssin x √3cos x intheform C sin(x + β),where C > 0and0 < β < 2π
ii Hencesolvesin x √3cos x = 1,for0 ≤ x < 2π
di Express cos x sin x intheform D cos(x φ),where D > 0and0 < φ < 2π.
ii Hencesolve cos x sin x = 1,for0 ≤ x < 2π.
13 Solve,for0◦ ≤ x ≤ 360◦,givingsolutionscorrecttothenearestminute:
2sec x 2tan x = 5 a 2cosec x + 5cot x = 3 b
14a Giventheequationsin θ + cos θ = cos2θ,showthattan θ = 1orcos θ sin θ = 1.
b Hencesolvesin θ + cos θ = cos2θ,for0 ≤ θ < 2π
15 Solve,for0 ≤ x ≤ 2π: sin x cos x = √1.5 a √3sin2x cos2x = 2 b sin4x + cos4x = 1 c
16a Showthat(√3 + 1)cos2x + (√3 1)sin2x = 2√2cos(2x π 12 ).
b Hencesolve(√3 + 1)cos2x + (√3 1)sin2x = 2,for π < x ≤ π
17 Thecurrent I inamperesinacircuitafter t secondsisgivenby I = 2sin(t π 3 ) cos(t + π 2 ).Findthe maximumcurrent,andtheearliesttimethatitoccurscorrecttothenearesttenthofasecond.
18ai Showthatsin x cos x = √2sin(x π 4 ).
ii Hencesketchthegraphof y = sin x cos x,for0 ≤ x ≤ 2π
iii Useyoursketchtodeterminethevaluesof x inthedomain0 ≤ x ≤ 2π forwhichsin x cos x > 1.
b Useasimilarapproachtopart(a)tosolve,for0 ≤ x ≤ 2π: sin x + √3cos x ≤ 1 i sin x √3cos x < 1 ii √3sin x + cos x < 1 iii cos x sin x ≥ 1 2 √2 iv
19 Considerthefunction y = 4sin(3x + π 6 ) + 2sin3x
Showthatthemaximumvalueofthefunctionis2 5 + 2√3. a Find,correcttoonedecimalplace,thefirstpositivevalueof x forwhichthemaximumvalueoccurs.
b
20a Showthatifcos(x α) = cos β,thentan x = tan(α + β)ortan x = tan(α β).
Hint:Ifcos A = cos B,then A = 2nπ + B or A = 2nπ B,where n isaninteger.
b Showthat2cos x + 11sin x = 5√5cos x tan 1 11 2
c Considertheequation2cos x + 11sin x = 10,for0 ≤ x < 2π
Bywritingtheequationintheformcos(x α) = cos β andusingpart(a),showthattan x = 4 3 or 24 7 i
Deducethattheequationhasrootstan 1 4 3 and π tan 1 24 7 ii Provethatoneoftherootsistwicetheother. iii
• Createyourownsummaryofthischapteronpaperorinadigitaldocument.
• Thisautomatically-markedquizisaccessedintheInteractiveTextbook.AprintablePDFWorksheetversionis alsoavailablethere.
• Checklist AvailableintheInteractiveTextbook,usethechecklisttotrackyourunderstandingofthelearningintentions. PrintablePDFandworddocumentversionsarealsoavailablethere.
1 Fromtwopoints P and Q onhorizontalground,theanglesofelevationofthetop T ofa10mmonumentare 16◦ and13◦ respectively.Itisalsoknownthat ∠PBQ = 70◦,where B isthebaseofthemonument.
Showthat PB = 10tan74◦,andwritedownasimilarexpressionfor QB. a Hencedetermine,correcttothenearestmetre,thedistancebetween P and Q b
2 Thepoints P, Q and B lieinahorizontalplane.From P,whichisduewestof B,the angleofelevationofthetopofatower AB ofheight h metresis42◦.From Q,which isonabearingof196◦ fromthetower,theangleofelevationofthetopofthetower is35◦.Thedistance PQ is200metres.
c
Showthat PB = h cot42◦,andwritedownasimilarexpressionfor QB. a Explainwhy ∠PBQ = 74◦ . b Usethecosineruletoshowthat
Hencefindtheheightofthetower,correcttothenearestmetre. d
3 Atriangularpyramid ABCD hasbase BCD andperpendicularheight AD
a Find BD and CD intermsof h.
b Usethecosineruletoshowthat2h2 = x2 √3 hx.
c Let u = h x .Writetheresultofthepreviouspartasaquadraticequationin u,and henceshowthat
4 Simplify,usingthecompound-angleresults:
α cos2α + cos4α sin2α e
5 Simplify,usingthedouble-angleresults: 2sin2θ cos2θ a
2 1 2 x sin2 1 2 x b 2cos2 3α 1 c 2tan35◦ 1 tan2 35◦ d 1 2sin2 25◦ e
6 Giventhattheangles A and B areacute,andthatsin A = 3 5 andcos B = 5 13 ,find:
A a
B d
7a Bywriting75◦ as45◦ + 30◦,showthat:
◦ = √3 + 1 2√2 i
b Henceshowthat:
8 Usethecompound-angleanddouble-angleresultstofindtheexactvalueof: 2sin15◦ cos15◦ a cos35◦ cos5◦
9 Proveeachidentity.
θ(tan θ + cot θ) = 2 c
1 + tan A = tan2A g tan2A(cot A tan A) = 2, (providedcot A tan A) h
10 Anoffice-workerislookingoutawindow W ofabuildingstandingonlevel ground.From W,acar C hasanangleofdepression α,whileaballoon B directly abovethecarhasanangleofelevation2α.Theheightoftheballoonabovethecar is x,andtheheightofthewindowabovethegroundis h
a Showthat tan α h = tan2α x h .
b Henceshowthat h x = 1 tan2 α 3 tan2 α . x h W
11 Solveeachequationfor0 ≤ x ≤ 2π.
sin2x + cos x = 0 a
x = sin x b
x + 8cos x + 5 = 0 c cos2x + 5sin x + 2 = 0 d
x π 6 ) = cos(x π 3 ) e
x = 3tan x f
12a Usecompoundanddouble-angleformulaetoprovethatcos3x = 4cos3 x 3cos x
b Hencesolvecos3x + sin2x + cos x = 0,for0 ≤ x ≤ 2π.
13a Expresssin x cos x intheform R sin(x α),where R > 0and0 < α < π 2 .
b Hencesolvesin x cos x = √2,for0 ≤ x ≤ 2π
14a Express √3cos x + sin x intheform A cos(x θ),where A > 0and0 < θ < π 2
b Hencesolve √3cos x + sin x = 1,for0 ≤ x ≤ 2π
15a Express2sin x + √5cos x intheform R sin(x + α),where R > 0and α isacute.
b Hencesolve2sin x + √5cos x = 3,for0◦ ≤ x ≤ 360◦,writingthesolutionindegreescorrecttoone decimalplace.
16a Express3cos x 2sin x intheform A cos(x + θ),where A > 0and θ isacute.
b Hencesolve3cos x 2sin x = 1,for0◦ ≤ x ≤ 360◦,writingthesolutionscorrecttothenearestminute.
Arithmeticbeginswithcounting,andthroughoutallbranchesandapplicationsofmathematics,counting continuestobeimportant.
▶ HowmanyNSWnumberplatesarepossiblewiththecurrentpattern: Letter + Letter + Letter + Digit + Digit + Letter?
▶ Howmanypossible8-bitbyteshavethreeonesandfivezeroes?
▶ Howmanywayscanfourchickensbechosenfromaflockof20chickens?Thischapterintroducesand developssomeofthestandardmethodsofcounting.
Afterintroducing factorialnotation,suchas5! = 5 × 4 × 3 × 2 × 1,threesuccessivebasiccountingmethodsare addressed:
▶ Countingorderedselectionsallowingrepetition.Thisisdoneusingpowers.
▶ Countingorderedselectionswithoutrepetition,called permutations,Thisrequiresthenumbers nPr ,which areevaluatedusingfactorials.
▶ Countingunorderedselections,whicharejustsubsets,andarealsocalled combinations.Thisrequiresthe numbers nCr ,alsoevaluatedusingfactorials.
Countingarrangementsinacircleoraroundaroundtableisalsoaddressed.
Countingisparticularlyimportantinprobability.Whatistheprobabilityofbeingdealttwoacesandtwojacks inaBridgehandof13cards?Thecountingmethodsdevelopedhereareappliedtoprobabilitytocountsample andeventspaces.
ThefinalChapter18examinesbinomialexpansions(x + y)n.Itturnsoutthatthebasiccountingmethodshere, andparticularlythenumbers nCr ,areessentialforunderstandingtheseexpansions.
Anoteontheorderofthechapters: Somebasicsettheoryisobviouslyhelpfulincombinatorics.In particular,thestandardcountingrule:
oftenmakesacomplicatedsituationmuchclearer,andcomplementarysetsoccurroutinely.Readerswho havenotyetstudiedtheAdvancedChapter14:Probabilityareadvisedtoworkattheveryleastthroughthe introductorysettheoryinSection14Abeforeembarkingonthischapter.Farbetterwouldbetoworkthrough thewholeofChapter14,becauseprobabilityiscentraltoSection17F.
Learningintentions
• Defineandevaluatefactorials.
• Manipulatefactorialsbyunrollingand/orcancelling.
Productssuchas5 × 4 × 3 × 2 × 1 = 120,called factorials.Theyoccurthroughoutmathematics,butareparticularly importantincombinatorics.Thischapterintroducesfactorialsandsomeoftheirelementaryarithmetic.
Howmanywayscanfivepeopleformaqueueatabusstop?Solutionsofsuchproblemswillbeformalisedinthe nextsectionasthe multiplicationprinciple,butthisparticularproblemcanbesolvedstraightforwardly,andthe boxesbelowhelptovisualisethesolution.Theresultbelowisa factorial:
• Choosethepersonattheheadofthequeueinfiveways.
• Fourpeopleremain,sothepersoninsecondplacecanbechoseninfourways.
• Threepeopleremain,sothepersoninthirdplacecanbechoseninthreeways.
• Twopeopleremain,sothepersoninfourthplacecanbechosenintwoways.
• Andthereisnowonlyonewaytochoosethepersoninthelastplace.
1stplace 2ndplace 3rdplace 4thplace 5thplace 5 4 3 2 1
Numberofpossiblequeues = 5 × 4 × 3 × 2 × 1 = 120.
Thisproductiscalled‘5factorial’,withsymbol5! = 5 × 4 × 3 × 2 × 1.
Manyapparentlyunrelatedsituationsinmathematicsalsogeneratefactorials.Forexample,whatisthefifth derivativeof y = x5 ?
y ′ = 5x 4
y ′′ = 5 × 4 × x 3
y ′′′ = 5 × 4 × 3 × x 2
y(4) = 5 × 4 × 3 × 2 × x
y(5) = 5 × 4 × 3 × 2 × 1 = 120.
Thedefinitionof n factorial
Hereisthegeneraldefinitionof n!,called‘n factorial’,statedtwoways:
1Thedefinitionof n factorial
• Foreachwholenumber n ≥ 1,thenumber n!(n factorial )istheproductofallwholenumbersfrom n downto1: n! = n × (n 1) × (n 2) ×···× 3 × 2 × 1, anddefine0! = 1.
• Thebetterdefinition,however,is recursive —firstdefine0! = 1,andthenforeachsuccessivevalue of n,sayexactlyhowtodefine n!intermsof(n 1)!.
0! = 1, n! = n × (n 1)!,forallwholenumbers n ≥ 1.
Therecursivedefinitionhasthreeadvantages.Itavoidsthedots ··· inthefirstdefinition,thevalue0! = 1isbuilt morefirmlyintothedefinition,andasweshallsoonsee,itgivesabetterinsightintohowtomanipulatefactorial notation
Thefactthat0!isdefinedtobe1needssomeexplanation.
• An emptyproduct isregardedasbeing1,becauseifnothinghasyetbeenmultiplied,theregisterremainsat1 whereitwasoriginallysetinpreparationforperformingmultiplication.
• Inasimilarway,an emptysum is0,becauseifnothinghasyetbeenadded,theregisterremainsat0whereit wasoriginallysetinpreparationforperformingaddition.
Usingtherecursivedefinitionthatdefineseachfactorialintermsofitspredecessor:
andsoon,increasingveryquicklyindeed.Ifyourcalculatorhasafactorialbuttonlabelled x! or n! ,useit straightawaytoseethatatleastthe calculator believesthat0! = 1.Noticealsotheerrormessageif n isnota wholenumber,becausethedomainofthefunction n!isthewholenumbers0,1,2,....
Therecursivedefinitionof n!givenandusedaboveisthekeyideainmanycalculations.Successiveapplications ofthedefinitioncanbethoughtofas unrolling thefactorialfurtherandfurther:
8! = 8 × 7!(unrollingonce)
= 8 × 7 × 6!(unrollingtwice)
= 8 × 7 × 6 × 5!(unrollingthreetimes) andsoon.Thisideaisvitalwhentherearefractionsinvolved.
Example1
Simplifyingafactorialexpressionbyunrolling
Simplifyeachexpressionusingunrollingtechniques:
= 10 × 9 × 8 × 7!
= 10 × 9 × 8 = 720
n! (n r)! = n(n 1)(n 2) ··· (n r + 1)(n r)! (n r)! = n(n 1)(n 2) (n r + 1) r factors
(n + 2)!
n 1)!
(n + 2)(n + 1)n(n 1)! (n 1)! = (n + 2)(n + 1)n
Example2 Simplifyingafactorialexpressionusingacommondenominator
Simplifyeachexpressionusingacommondenominator.
2Calculatingbyunrollingfactorials
• Afactorialcanbeunrolledonestepatatime: 7! = 7 × 6! = 7 × 6 × 5! = 7 × 6 × 5 × 4! =
• Cancelfractionswithfactorialsbyunrollinguntilthefactorialscancel:
4! ≑ 7 × 6 × 5 × 4! 4! = 7 × 6 × 5 = 210.
Examiningtheprimefactorsinafactorial
a Withoutusinganydeviceortheinternet,findhowmanyfinalzeroestherearewhen25!isevaluated.
b Canthisbecheckedonacalculatororotherdevice?
Solution
a Everythingdependsonthefactthat10 = 5 × 2,and5and2areprimes. Thereareplentyoffactorsof2in25!,sothenumberoffinalzeroesisthenumberoffactorsof5inthe product.Therearefactorsof5in5,10,15,20and25(whichis52),making6altogether,so25!has6final zeroes.
b Itdependsonthedeviceorwebpage.Withoutsomereasoning,itwillneedtogivetheanswercorrectto26 significantfigures!
1 Useargumentssimilartothatusedattheverystartofthissectiontoexpressthefollowingusingfactorial notation.
a Thenumberofwaystoformaqueuewith8people.
b Thenumberofwaystoarrange4distinctbooksonabookself.
c Thenumberofways3particularpeoplecanbeplacedfirst,secondandthirdinacompetition,assuming thateachisplacedinoneofthesethreepositions.
d Samdropshisportablekeyboardandall101keysfalloff.Howmanywayscanheputthekeysbackon thekeyboard,assumingthattheyareallinterchangable?
e Whentheteachercallsonstudentsinalphabeticalorder,AndrzejZywiecisupsetthatheisalwayscalled last.Inaclassof20,howmanywayscouldtherollbecalled,ifthealphabeticalorderingrestrictionwere dropped?
2 Usethedefinitionof n!andmethodsofunrollingfactorialstoevaluateeachexpression.Donotusea calculator.
3 Usethefactorialbutton n! onyourcalculatortoevaluateeachexpression.
4 If f (x) = x6,findexpressionsfor:
)
5 Simplifybyunrollingfactorialsappropriately:
! (n 1)!
+ 2)! n! e (n 2)!
6 Simplifybytakingoutacommonfactor:
7 Writeeachexpressionasasinglefraction. 1 n! + 1 (n 1)! a
8a If f (x) = xn,find:
′(x) i
b If f (x) = 1 x ,find:
9a Showthat k × k! = (k + 1)! k!
. iv
b Hencebyconsideringeachindividualtermasadifferenceoftwoterms,sumtheseries1 × 1! + 2 × 2! + 3 × 3! + ··· + n × n!
10a Findthelargestpowerof: 2, i 10,thatisadivisorof10! ii
b Findthelargestpowerof: 2, i 5, ii 7, iii 13,thatisa divisorof100! iv
11 [Arelationshipbetweenhigherderivativesofpolynomialsandfactorials]
a If f (x) = 11x3 + 7x2 + 5x + 3,showthat: f (0) = 3 × 0! i f ′(0) = 5 × 1! ii f ′′(0) = 7 × 2! iii f ′′′(0) = 11 × 3! iv f (k)(0) = 0,forall k ≥ 4. v
Henceexplainwhy f (x)canbewritten f (x) = f (0) 0! + f ′(0) x 1! + f ′′(0) x2 2! + f ′′′(0) x3 3! .
b Showthatif f (x) = an xn + an 1 xn 1 + + a1 x + a0 isanypolynomial,then f (k)(0) =
ak k!,for k = 0,1,2,..., n, 0,for k > n,
andhenceexplainwhy f (x) = f (0) 0! +
(0) x 1! +
Because f (x)isapolynomial,this powerseries hasonlyfinitelymanyterms.Therearewaysof generalisingthistosomeotherfunctionsthatarenotpolynomials.
12a Evaluate k (k + 1)! ,for k = 1,2,3,4and5.
b Evaluate 1 (1 + 1)! + 2 (2 + 1)! + ··· + n (n + 1)! ,for n = 1,2,3,4and5.
c Writetheexpressioninpart(b)as S n,toindicatethatitisasumwhosevaluedependson n.Thatis, S n = 1 (1 + 1)! + 2 (2 + 1)! + + n (n + 1)!
Makeareasonableguessaboutthevalueof S n.Hencefindlimn→∞ S n
d Provethat k (k + 1)! = 1 k! 1 (k + 1)! ,andhenceproduceaproofofpart(c)usingacollapsingsum.
13 Expressusingfactorialnotation:
14 [Stirling’sformula]Thefollowingformulaistoodifficulttoproveatthisstage,butitismostimportant becauseitprovidesacontinuousfunctionthatapproximates n!forintegervaluesof n:
Showthattheformulahasanerrorofapproximately2.73%for3!and0.83%for10!Findthepercentage errorfor60!
Learningintentions
• Developthemultiplicationprincipleofcounting,usingboxestoillustrate.
• Countorderedarrangementsallowingrepetition.
• Countorderedarrangementswithoutrepetition.
• Understandpermutations,usethenotation nPr ,andevaluate nPr usingfactorials.
• Developthecountingprinciple—dealwiththedifficultiesfirst.
Twoquestionsdominatediscussionofcountinginthenextfoursections.
• Aretheselectionswearecountingorderedorunordered?
• Iftheyareordered,isrepetitionallowedornot?
Thesequestionsgeneratethethreesituationsdescribedinthechapterintroduction.
Sections17B–17Ddevelopthetheoryofcountingorderedselections,withandwithoutrepetition,then Section17Edealswithunorderedselections.
Thismorning,Stefanchoseashirttowearfromthe12thatheowns.Thenwithoutworryingaboutamatch,he choseapairoftrousersfromthe6pairsheowns.Howmanypossibleoutfitscouldhehavewalkedoutofthe housewearing?
Theansweris12 × 6 = 72.Foreachchoiceoftrousers,hehas12choicesofshirt,making12possibleoutfits.But therewere6choicesoftrousers,so
numberofoutfits = 12 + 12 + 12 + 12 + 12 + 12 = 12 × 6.
Themultiplicationprinciplemaybeappliedtomorethantwosuccessivechoices.Stefanalsochoseapairof shoesfromhis10pairs,andatiefromhisextraordinarycollectionof100wonderfulties.Takingthesechoices intoaccountaswell,
numberofoutfits = 12 × 6 × 10 × 100 = 72000.
3Ageneralcountingprinciple—themultiplicationprinciple
• Supposethataselectionistobemadein r stages.Supposethatthefirststagecanbechosenin n1 ways, thesecondin n2 ways,...,the rthin nr ways.Then numberofwaysofchoosingthecompleteselection = n1 × n2 ×···× nr
• Thesuccessivechoicescanbevisualisedinaboxdiagram,asdescribedbelow,
Anorderedselectioncanusuallyberegardedasasequenceofchoicesmadeoneaftertheother.Aboxdiagram isanefficientsetting-outhere,andkeepstrackofthesesuccessivechoices.InthesituationabovewithStefan’s outfits:
shirt trousers shoes tie 12 6 10 100
Numberofoutfits = 12 × 6 × 10 × 100 = 72000.
Usingthemultiplicationprinciple
Howmanyfive-letterwordscanbeformedinwhichthesecondandfourthlettersarevowelsandtheother threelettersareconsonants?
Note: Unlessotherwiseindicated,alwaystaketheletter‘y’asaconsonant.
Solution
Thereare5vowelsand26 5 = 21consonants.Wecanselecteachletterinorder:
1stletter 2ndletter 3rdletter 4thletter 5thletter 21 5 21 5 21
Hencenumberofwords = 21 × 5 × 21 × 5 × 21 = 231525.
Thisisthefirstsituationlistedinthechapterintroduction.Supposethat r-letterwordsareformedfrom n distinct letters,whereanylettercanbeusedanynumberoftimes.Theneachsuccessiveletterinthewordcanbechosen in n ways:
1stletter 2ndletter 3rdletter ··· rthletter n n n ... n
giving nr distinctwordsaltogether.Theresultisthusasimplepower:
If r lettersarechosensuccessivelyfrom n distinctletters,allowingrepetitionofletters,andplacedin order,then:
numberofarrangements = n r
Example5
Countingorderedselectionsallowingrepetition
a Howmanysix-digitnumberscanbeformedentirelyfromodddigits?
b Howmanyofthesenumberscontainatleastoneseven?
Solution
a Therearefiveodddigits,sothenumberofsuchnumbersis56
b Wefirstcountthenumberofthesesix-digitnumbersnotcontaining7.Suchnumbersareformed fromthedigits1,3,5and9,sothereare46 ofthem.Subtractingthisfromtheanswertopart(a), numberofnumbers = 56 46 = 11529.
Thisisthesecondofthethreesituationslistedinthechapterintroduction.Countingorderedselectionswithout repetitiontypicallyinvolvesfactorials,becauseaseachstageiscompleted,thenumberofpossibleobjects diminishesby1.
Note: Thephrases‘with(orallowing)replacement’and‘with(orallowing)repetition’areusedalmost interchangeably.Similarly,thephrases‘withoutreplacement’and‘withoutrepetition’arealsoused almostinterchangeably.
Example6 Countingorderedselectionswithoutrepetition
Inhowmanywayscanaclassof16selecta4-personcommitteeconsistingofapresident,avice-president,a treasurer,andasecretary?
Selectinorderthepresident,thevice-president,thetreasurerandthesecretary.
president vice-president treasurer secretary 16 15 14 13
Hencethereare16 × 15 × 14 × 13 = 16! 12! possiblecommittees(thatis,43680).
A permutation isanarrangementofobjectschosenfromacertainfinitesetwithoutrepetition(thatis,without replacement).Forexample,thewordsABC,CED,EABandDBCaresomeofthemany3-letterpermutations takenfromthe5-memberset{A,B,C,D,E}.
Thesymbol nPr isusedtodenotethenumberofpermutationsof r letterschosenwithoutrepetitionfromasetof n distinctletters.Thepreviousexampleiseasilygeneralisedtoshowthatthereare n! (n r)! suchpermutations,so thisbecomestheformulafor nPr : 1stletter 2ndletter 3rdletter 4thletter rthletter n n 1 n 2 n 3 n r + 1
Hence nPr = n(n 1)(n 2)(n 3) (n r + 1) = n(n 1)(n 2)(n 3) ×···× 2 × 1 (n r)(n r 1) ×···× 2 × 1 = n! (n r)!
5Permutations—orderedselectionswithoutrepetition
• A permutation isanarrangementofobjectschosenfromacertainfinitesetwithoutrepetition(thatis, withoutreplacement).
• Weoftenimaginepermutationsas words formedbychoosing,withoutrepetition,lettersfromasetof distinctletters.
• Thenumberof r-letterpermutationschosenfromasetof n distinctletters,is
nPr = n! (n r)!
• Thenumberofpermutationsofall n distinctlettersis
nPn = n! 0! = n!.
Theterm‘distinctobjects’isoftennecessaryinthissituation.
Scientificcalculatorshaveabuttonlabelled n Pr thatwillfindvaluesof nPr .Forlowvaluesof n and r,the answersareexact,butforhighervaluestheyareonlyapproximations.Makeapracticeofevaluatingthese numberbyhandwhenitisreasonabletodoso,becausesuchcalculationsgreatlyhelptheintuition.
Example7 Countingpermutationstakenfroma5-memberset.
a Howmany3-letterwordscanbeformedwithoutrepetitionfromthe5-memberset{A,B,C,D,E}?
b Howmany5-letterwordswithoutrepetitioncanbeformedfromtheset?
Solution
a ByBox5,numberofpermutations =
b ByBox5,numberofpermutations = 5! = 120.
A permutationofaset isanarrangementofallthemembersofthesetinsomeorder.Box5establishedthatthe numberofsuchpermutationis nPn = n!.
Example8 Permutationsofallmembersandofallbutonemember
a Howmany8-digitnumbers,andhowmany9-digitnumbers,canbeformedfromthe9non-zerodigitsifno repetitionisallowed?
b Commentonthetwoanswersinpart(a).
Solution
a Numberof8-digitnumbers =
b Theresultsarethesame,becauseevery8-digitnumbercanbeextendedinoneandonlyonewaytoa 9-digitnumber,thatis,byaddingtheunuseddigit.
Example9 Usingthecalculatorforverylargenumbers
Inhowmanywayscan20peopleand40peopleformaqueue?Answercorrecttotwosignificantfigures.
Solution
Ageneralcountingprinciple—dealwiththerestrictionsfirst
Manyproblemshaverestrictionsinthewaythingscanbearranged.Theserestrictionsshouldbedealtwithfirst. Itisalsoimportanttokeepinmindthattheorderoftheboxesusuallyrepresentstheorderthechoicesaremade in,nottheorderingoftheobjects,andthattheycanbeusedinsurprisinglyflexibleways.
6Ageneralcountingprinciple—dealwiththedifficultiesfirst
• Whencountingorderedselections,dealwithanyrestrictionsfirst.
• Usuallyplacetheboxesintheorderinwhichtheselectionsaremade.
Eightpeopleformtwoqueues,eachwithfourpeople.Albertwillonlystandintheleft-handqueue,Bethonly intheright-handqueue,andCharlesandDianainsistonstandinginthesamequeue.Inhowmanywayscan thetwoqueuesbeformed?
Solution
PlaceAlbertinanyofthe4possiblepositionsintheleft-handqueue, ThenplaceBethinanyofthe4positionsintheright-handqueue. PlaceCharlesinanyoftheremaining6positions.
PlaceDianainoneofthe2remainingpositionsinthesamequeue. Thereremain4unfilledpositions,whichcanbefilledin4!ways:
Albert Beth Charles Diana lastfourpositions 4 4 6 2 4!
Hencenumberofways = 4 × 4 × 6 × 2 × 4! = 4608.
1 ListallthepermutationsofthelettersofthewordDOG.Howmanyarethere?
2 ListallthepermutationsofthelettersEFGHI,beginningwithF,takenthreeatatime.
3 FindhowmanyarrangementsofthelettersofthewordFRIENDarepossibleifthelettersaretaken: fouratatime, a sixatatime. b
4 Findhowmanyfour-digitnumberscanbeformedusingthedigits5,6,7,8and9if: nodigitistoberepeated, a anyofthedigitscanoccurmorethanonce. b
5 Howmanythree-digitnumberscanbeformedusingthedigits2,3,4,5and6ifnodigitcanberepeated? Howmanyofthesearegreaterthan400?
6 Inhowmanywayscansevenpeoplebeseatedinarowofsevendifferentchairs?
7 Thesymbol nPr isthenumberofwaysofarranging r objectsselectedwithoutreplacementfrom n objects. Usethisexpression,andthe n Pr buttononyourcalculator,toanswerthefollowingquestions.
a Fromagroupof10,threepeoplelineuptobuytickets.Howmanywayscanthishappen?
b Fivecardsareeachlabelleduniquelywithoneofthedigits1,2,3,4,5.Threeofthefivecardsareplaced downinarow.Howmanywayscanthecardsbearranged?
c Onehundredpeopleeachbuyoneticketinalottery.Howmanywayscanthefirstthreeplacesbe awarded?
8 Theexpression nr isthenumberofwaysofarranging r objectsselectedwithreplacementfrom n objects. Usethisexpressiontoanswerthefollowingquestions.
a Howmanythree-digitnumberscanbewrittendownfromthedigits1to9?Thedigitsneednotbe distinct.
b Onehundredpeopleeachbuyoneticketinalottery.Ticketsareselectedonebyoneatrandomfroma barrelandthenreplacedbetweenselections.Howmanywayscanthefirstthreeplacesbeawarded?
c Acomputersendsastringoftenbinarydigits,thatis,eachsymbolcanonlybe0or1.Howmanysuch ten-digitstringsarepossible?
9 Eightrunnersareparticipatingina400-metrerace.
a Inhowmanywayscantheyfinish?
b Inhowmanywayscanthegold,silverandbronzemedalsbeawarded?
10a Ifyoutossacoinandrolladie,howmanyoutcomesarepossible?
b Ifyoutosstwocoinsandrollthreedice,howmanyoutcomesarepossible?
11 Awomanhasfourhats,threeblouses,fiveskirts,twohandbagsandsixpairsofshoes.Inhowmanyways canshebeattired,assumingthatshewearsoneofeachitem?
12 JackhassixdifferentfootballcardsandMeghasanothereightdifferentfootballcards.Inhowmanyways canoneofJack’scardsbeexchangedforoneofMeg’scards?
13 InSydney,landlinephonenumbers,ignoringtheareacode,consistofeightdigits.Letusconsiderthose startingwiththedigit9.
a Howmanysuchphonenumbersarepossible?
b Howmanyoftheseendinanoddnumber?
c Howmanyconsistofodddigitsonly?
d Howmanyaretherethatdonotcontainazero,andinwhichtheconsecutivedigitsalternatebetweenodd andeven?
14 Usersofautomatictellermachinesarerequiredtoenterafour-digitPIN(personalidentificationnumber). FindhowmanyPINs: arepossible, a consistoffourdistinctdigits, b consistofodddigitsonly, c startandendwiththesamedigit. d
15a Ifrepetitionsarenotallowed,howmanyfour-digitnumberscanbeformedfromthedigits1,2,...8,9? Howmanyoftheseendin1? b Howmanyoftheseareeven? c Howmanyaredivisibleby5? d Howmanyaregreaterthan7000? e
16 Repeatthepreviousquestionifrepetitionsareallowed.
17 InTasmania,acarlicenceplateconsistsoftwolettersfollowedbyfourdigits.Findhowmanyoftheseare possible:
a iftherearenorestrictions,
b ifthereisnorepetitionoflettersordigits,
c ifthesecondletterisXandthethirddigitis3, d ifthelettersareDandQandthedigitsare3,6,7and9.
18a InhowmanywayscanthelettersofthewordNUMBERbearranged?
b HowmanybeginwithN?
c HowmanybeginwithNandendwithU?
d InhowmanyistheNsomewheretotheleftoftheU?
19 Inhowmanywayscanaboatcrewofeightwomenbearrangedifthreeofthewomencanonlyrowonthe bowsideandtwootherscanonlyrowonthestrokeside?
20 Amotorbikecancarrythreepeople:thedriver,onepassengerbehindthedriverandoneinthesidecar.If amongfivepeople,onlytwocandrive,inhowmanywayscanthebikebefilled?
21a Howmanyfive-digitnumberscanbeformedwithoutrepetitionfromthedigits2,3,4,5and6?
b Howmanyofthesenumbersaregreaterthan56432?
c Howmanyofthesenumbersarelessthan56432?
22a Integersareformedfromthedigits2,3,4and5,withrepetitionsnotallowed. Howmanysuchnumbersarethere? i Howmanyofthemareeven? ii
b Repeatthetwopartstothisquestionifrepetitionsareallowed.
23a Howmanyfive-digitnumberscanbeformedfromthedigits0,1,2,3and4ifrepetitionsarenot allowed?
b Howmanyoftheseareodd?
c Howmanyaredivisibleby5?
24a If 8Pr = 336,findthevalueof r
b If7 × 2
,find
c Usingtheresult nPr = n! (
r)! ,provethat:
25 Recallthatawholenumberisdivisibleby3ifandonlyifthesumofthedigitsisdivisibleby3.
a Howmany5-digitwholenumbersaredivisibleby3?
b Howmany5-digitwholenumberswithno0saredivisibleby3?
c Howmany5-digitwholenumberswithno0saredivisibleby2?
d Howmany5-digitwholenumberswithno0saredivisibleby6?
Learningintentions
• Developthegroupingprinciple—orderthegroups,thenordereachgroup.
• Developthecomplementaryprinciple—dealwiththecomplementarysituation.
• Developthecasesprinciple—splittheproblemupintocases.
Section17Bdevelopedthemultiplicationprinciple,andtheprincipleofdealingwiththedifficultiesfirst.This sectiondevelopsthreefurtherprinciples.
Insomeorderingproblems,particularmembersmustbegroupedtogether.Thisproducesa compoundordering, inwhichthevariousgroupsmustfirstbeordered,andthentheindividualsorderedwithineachgroup.
7Ageneralcountingprinciple—usegroupingincounting
• Firstorderthegroups,thenordertheindividualswithineachgroup.
• Agroupmayconsistofasingleindividual.
Fourboysandfourgirlsformaqueueatthebusstop.Onecouplewanttostandtogether.Theotherthreegirls wanttostandtogether,buttheotherthreeboysdon’tcarewheretheystand.Howmanyacceptablewaysare thereofformingthequeue?
Solution
Therearefivegroups—thecouple,thegroupofthreegirls,andthethreegroupseachconsistingofone individualboy.Firstorderthefivegroups,thenorderthecouple,thenorderthethreegirls. orderthe5groups orderthecouple orderthe3girls 5! 2! 3! Hencenumberofways = 5! × 2! × 3! = 1440.
ManyprobabilityquestionshavealreadybeensolvedinChapter14usingcomplementaryevents.Thesame principleappliestocounting—counttheunacceptableorderings,thensubtractthemfromthetotalnumberof orderings.Theword‘not’mayormaynotbetheretopromptyou.
Howmany7-letterwordscanbeformedfromthelettersA,B,C,D,E,F,Gifthetwovowelsmustbe separatedbyatleastoneconsonant?
Solution
Numberoforderingswithoutrestriction = 7!
NumberoforderingswithAandEtogether = 2! × 6! (OrdertheAandEandgluethemtogether,thenorderthe6groups.)
HencenumberoforderingswithAandEapart = 7! 2 × 6! = 3600.
8Ageneralcountingprinciple—dealwiththecomplementarysituation
Counttheunacceptableorderings,thensubtracttocounttheacceptableorderings.
Sometimesthefiddlyconditionsofaproblemmeanthatonesetofboxesisnotenoughtosolveit,andseparate casesneedtobeconsidered.
Inthissituation,thecasesshouldnotoverlap,oriftheydo,thenumberintheoverlapneedstobesubtracted.
9Ageneralcountingprinciple—usecases
• Thecasesshouldnotoverlap.
• Ifthecasesdooverlap,theoverlapneedstobesubtracted.
ThissameprinciplewasexpressedinSection14Aonsettheory,bytheformulainBox4ofthatchapter:
Usingcasesinadifficultcountingproblem
a Howmanywholenumberslessthan1000areoddandgreaterthan500,ormultiplesof5andlessthan200?
b Howmanywholenumberslessthan1000areoddandgreaterthan200,ormultiplesof5andlessthan500?
Solution
a Thereare250oddnumbersbetween500and1000. Thereare40multiplesof5lessthan200(includingzero). Thetwocasesdonotoverlap,so
Total = 250 + 40 = 290.
b Thereare400oddnumbersbetween200and1000. Thereare100multiplesof5lessthan500(includingzero). Butthetwocasesoverlapthistime—thereare30oddnumbersfrom200to500thataremultiplesof5. Hence
Total = 400 + 100 30 = 470.
1 Howmanyrearrangementsarethereofthelettersofeachword,ifthevowelsmustbetogether? BOARDS a RIO b QUIT c TROUNCE d
2 HowmanyarrangementsofthewordMATHSarepossible,if:
a theTandHmustremaintogether?
b theTHmustremaintogetherandinthisorder?
3 HowmanywayscanAndrew,Becky,Courtney,DionandElliesitinarow,ifAndrewandBeckysit togetherandDionandElliesittogether?
4 Inhowmanywayscanthreedifferentmathematicsbooks,sixdifferentsciencebooksandfourdifferent Englishbooksbeplacedonashelf,ifthebooksrelatingtoeachsubjectaretobekepttogether?
5 AclassisaskedtodeterminehowmanywaysthelettersofthewordSOLARcanbebearranged,ifthefirst twopositionscannotbothbevowels.
a Jackdecidestousecases(thewordeitherstartswithavoweloritdoesnot).UseJack’smethodto answerthequestion.
b Jilldecidestoconsiderthecomplementarysituation(bothfirstpositionsarevowels).UseJill’smethodto answerthequestion.
6 Afamilyoftwoparentsandtwochildrenaregoingonacartrip.Onlytheparentscandrive.Ifthefather drives,thenthemotherwillsitinthebackseatwhereshefeelssafer.Thefatheralwayssitsinafrontseatof thecar.Howmanyarrangementsarepossible,iftwositinthefrontandtwointheback?
7 Howmanynumberscanbewrittendownusingeachdigitof789atmostonce,iftheresultmustbeatleast 80?
8 FindhowmanyarrangementsofthelettersofthewordUNIFORMarepossible:
a ifthevowelsmustoccupythefirst,middleandlastpositions,
b ifthewordmuststartwithUandendwithM,
c ifalltheconsonantsmustbeinagroupattheendoftheword, d iftheMissomewheretotherightoftheU.
9 FindhowmanyarrangementsofthelettersofthewordBEHAVING: endinNG, a beginwiththreevowels, b havethreevowelsoccurring together. c
10 AMathstestistoconsistofsixquestions.Inhowmanywayscanitbearrangedsothat: a theshortestquestionisfirstandthelongestquestionislast, b theshortestandlongestquestionsarenexttooneanother?
11 InMorsecode,lettersareformedbyasequenceofdashesanddots.Howmanydifferentlettersisitpossible torepresentifamaximumoftensymbolsareused?
12 Fourboysandfourgirlsaretositinarow.Findhowmanywaysthiscanbedoneif: theboysandgirlsalternate, a theboysandgirlssitindistinctgroups. b
13 Five-letterwordsareformedwithoutrepetitionfromthelettersofPHYSICAL.
a Howmanyconsistonlyofconsonants?
b HowmanybeginwithPandendwithS?
c Howmanybeginwithavowel?
d HowmanycontaintheletterY?
e Howmanyhavethetwovowelsoccurringnexttooneanother?
f HowmanyhavetheletterAimmediatelyfollowingtheletterL?
14a Howmanyseven-letterwordscanbeformedwithoutrepetitionfromthelettersofthewordINCLUDE?
b HowmanyofthesedonotbeginwithI?
c HowmanyendinL?
d Howmanyhavethevowelsandconsonantsalternating?
e HowmanyhavetheCimmediatelyfollowingtheD?
f HowmanyhavethelettersNandDseparatedbyexactlytwoletters?
g HowmanyhavethelettersNandDseparatedbymorethantwoletters?
15 Repeatparts(a)–(d)ofthepreviousquestionifrepetitionisallowed.
16a Inhowmanywayscantenpeoplebearrangedinaline:
i withoutrestriction, ii ifoneparticularpersonmustsitateitherend, iii iftwoparticularpeoplemustsitnexttooneanother, iv ifneitheroftwoparticularpeoplecansitoneitherendoftherow?
b Inhowmanywayscan n peoplebeplacedinarowof n chairs: i ifoneparticularpersonmustbeoneitherendoftherow, ii iftwoparticularpeoplemustsitnexttooneanother, iii iftwoofthemarenotpermittedtositateitherend?
17 Fiveboysandfourgirlsformaqueueatthecinema.Therearetwobrotherswhowanttostandtogether,the remainingthreeboyswishtostandtogether,andthefourgirlsdon’tmindwheretheystand.Inhowmany wayscanthequeuebeformed?
18 Eightpeoplearetoformtwoqueuesoffour.Inhowmanywayscanthisbedoneif:
a therearenorestrictions,
b Jimwillonlystandintheleft-handqueue,
c SeanandLiammuststandinthesamequeue?
19 Thereareeightswimmersinarace.Inhowmanywayscantheyfinishiftherearenodeadheatsandthe swimmerinLane2finishes:
a immediatelyaftertheswimmerinLane5,
b anytimeaftertheswimmerinLane5?
20 Fivebackpackersarriveinacitywheretherearefiveyouthhostels.
a Howmanydifferentaccommodationarrangementsarethereiftherearenorestrictionsonwherethe backpackersstay?
b Howmanydifferentaccommodationarrangementsarethereifeachbackpackerstaysatadifferentyouth hostel?
c Supposethattwoofthebackpackersarebrotherandsisterandwishtostayinthesameyouthhostel. Howmanydifferentaccommodationarrangementsarethereiftheotherthreecangotoanyoftheother youthhostels?
21 Numberslessthan4000areformedfromthedigits1,3,5,8and9,withoutrepetition.
Howmanysuchnumbersarethere? a
Howmanyofthemaredivisibleby5? c
Howmanyofthemareodd? b
Howmanyofthemaredivisibleby3? d
22 [Derangements]A derangement of n distinctlettersisapermutationofthemsothatnoletterappearsinits originalposition.Forexample,DABCisaderangementofABCD,butDACBisnot.Denotethenumberof derangementsof n lettersby D(n).
a BylistingallthederangementsofA,AB,ABCandABCD,findthevaluesof D(1), D(2), D(3)and D(4).
b SupposethatwehaveformedaderangementofthefivelettersABCDE.Letthelastletterinthe derangementbeX,andexchangeXwithE—thisputsEbacktoitsoriginalposition.EitherXisnow alsoinitsoriginalpositionsothatthreelettersareawayfromtheiroriginalpositions,orXisnotinits originalpositionsothatfourlettersareawayfromtheiroriginalpositions.Henceexplainwhy
D(5) = 4 × D(4) + 4 × D(3).
c Usethisformulatoevaluate D(5).Thenapplythecorrespondingargumentsandformulaetoevaluate D(6), D(7)and D(8).
Learningintentions
• Countrearrangementsofawordwhoselettersarenotalldistinct.
• Usecasestocountmorecomplicatedsituations.
• Developthespecialcasesofwordswithonlytwodistinctletters.
Sofar,ourpermutationshavebeendrawinglettersfromawordwhoselettersarealldistinct.Thissectiondeals withrearrangementsof allthelettersofawordwhoselettersarenotalldistinct.
Findingthenumberofdifferentwordsformedusingallthelettersoftheword‘PRESSES’iscomplicatedbythe factthattherearethreeSsandtwoEs.Ifthesevenletterswerealldifferent,wewouldconcludethat numberofways = 7!
Butwehave overcountedbyafactorof 2! = 2,becausetheEscanbeinterchangedwithoutchangingtheword. Wehavealso overcountedbyafactorof 3! = 6,becausethethreeSscanbepermutedamongstthemselvesin 3!wayswithoutchangingtheword.Takingaccountofbothovercountings: numberofways = 7! 2! × 3! = 420.
Thismethodiseasilygeneralised.Continuinginthelanguageof‘words’:
Supposethatawordof n lettershas r1 alikeofonetype, r2 alikeofanothertype,..., rk alikeofafinal type.Thenthenumberofdistinctwordsthatcanbeformedbyrearrangingthelettersis numberofwords
If ri = 1,thatis,ifthereisonlyoneletterofthe ithtype,then ri! = 1,soitdoesn’tmatterwhetherweincludeitin theformulaornot.
Threeidenticalwineglassesandfiveidenticaltumblersaretobearrangedinarowacrossthefrontofa cupboard.
a Inhowmanywayscanthisbedone(countingonlythepatterns)?
b Howdoesthischangeifoneofthewineglassesbecomesclearlychipped,andtwotumblersbreakandare replacedbytwoidenticaltumblersdifferentfromtheotherthree?
a Thereare8glasses,3alikeofonetypeand2alikeofanother, sonumberofways = 8! 3! × 5! = 56.
b Therearenow2alikeofonetype,1ofanother,3ofanother,2ofanother, sonumberofways = 8! 2! × 1! × 3! × 2! = 1680(the1!canbeomitted).
Asmentionedintheprevioussection,manycountingproblemsaretoocomplicatedtobeanalysedcompletelyby asingleboxdiagramorasinglemethod.Insuchsituations,theuseofcasesisunavoidable.Attentionshouldbe given,however,tominimisingthenumberofdifferentcasesthatneedtobeconsidered.
Howmany6-letterwordscanbeformedbyusingthe7lettersoftheword‘PRESSES’?
Solution
WeomitinturneachofthefourlettersP,R,EandS. Thisleavessixletters,whichwemustthenarrangeinorder.
a IfanSisomitted,therearethen2Esand2Ss, sonumberofwords = 6! 2! × 2! = 180.
b IfanEisomitted,therearethen3Ss, sonumberofwords = 6! 3! = 120.
c IfPorRisomitted(2cases),therearethen2Esand3Ss, sonumberofwords = 6! 3! × 2! × 2(doublingforthetwocases) = 120.
Hencetotalnumberofwords = 180 + 120 + 120 = 420.
Thissituationinvolvestwocountingformulaethatwillbevitalinthenextsection.
ThesetwoformulaearespecialcasesofmethodsthathavealreadybeendevelopedinSections17Bandthis section,andarebestintroducedwithanexample.
Example16
Countingwordsconsistingofjusttwodifferentletters
Anopinionpollasks8independentquestions,eachtobeansweredYorN.
a Howmanypossibleanswersheetsarethere?
b Howmanypossibleanswersheetshave5Ysand3Ns?
Solution
AnsweringthepollgeneratesawordsuchasYNNYYNYYmadeupofYsandNs.
a ByBox4inSection17B,thenumberofwordsis28.Orusingboxes:
b ByBox10above,thenumberofwordsis 8! 5! × 3!
Bothformulaeforthegeneralcasenowfolloweasilyfromthisexample.
11Rearrangementsofwordscontainingonlytwodifferentletters
• Thenumberof n-letterwordsconsistingonlyofYsandNsis: numberofwords = 2n
• Thenumberof n-letterwordsconsistingof r Ysand n r Nsis: numberofwords = n! r! × (n r)! .
Afterthenextsection,wewillbeabletousetheconcisesymbol nCr forthisexpression n! r! × (n r)!
Example17
Arrangementsofobjectsofjusttwotypes
a Inaqueueoftenadults,howmanypatternsofmenandwomenarepossible?
b Sixwomenandfourmenformaqueueatthebusstop.Howmanypatternsofmenandwomenare possible?
Solution
a ByBox11Dotpoint1,thenumberofpossiblepatternsis210 = 1024.
b ByBox11Dotpoint2,thenumberofpossiblepatternsis 10! 6! × 4! = 210.
1 Findthenumberofpermutationsofthefollowingwordsifallthelettersareused. BOB a ALAN b SNEEZE c TASMANIA d BEGINNER e FOOTBALLS f EQUILATERAL g COMMITTEE h WOOLLOOMOOLOO i
2 Thesixdigits1,1,1,2,2,3areusedtoformasix-digitnumber.Howmanynumberscanbeformed?
3 Sixcoinsarelineduponatable.Findhowmanypatternsarepossibleifthereare: fivetailsandonehead, a fourheadsandtwotails, b threetailsandthreeheads. c
4 Eightballs,identicalexceptforcolour,arearrangedinaline.Findhowmanydifferentarrangementsare possibleif:
a allballsareofadifferentcolour, b therearesevenredballsandonewhiteball, c therearesixredballs,onewhiteballandoneblackball, d therearethreeredballs,threewhiteballsandtwoblackballs.
5 Fiveidenticalgreenchairsandthreeidenticalredchairsarearrangedinarow.Findhowmanyarrangements arepossible: iftherearenorestrictions, a iftheremustbeagreenchaironeitherend. b
6 Amotoristtravelsthrougheightsetsoftrafficlights,eachofwhichisredorgreen.Heisforcedtostopat threesetsoflights.
a Inhowmanywayscouldthishappen?
b Whatothernumberofredlightswouldgiveanidenticalanswertopart(a)?
7 InhowmanywayscanthelettersofthewordSOCKSbearrangedinaline: a withoutrestriction, b sothatthetwoSsaretogether, c sothatthetwoSsareseparatedbyatleastoneotherletter, d sothattheKissomewheretotheleftoftheC?
8a FindthenumberofarrangementsofthelettersinSLOOPSif: therearenorestrictions, i thetwoOsaretogether, ii thetwoOsaretobeseparated, iii theOsaretogetherandtheSsaretogether. iv
b InhowmanyarrangementsofthelettersinTATTOOarethetwoOsseparated?
9 InhowmanywayscanthelettersofthewordDECISIONSbearranged: a withoutrestriction, b sothatthevowelsandconsonantsalternate, c sothatthevowelscomefirstfollowedbytheconsonants, d sothattheNissomewheretotherightoftheD?
10 InhowmanywayscanthelettersofthewordPROPORTIONALITYbearrangedsothatthevowelsand consonantsstilloccupythesameplaces?
11 Aformhastenquestionsinorder,eachofwhichrequirestheanswer‘Yes’or‘No’.Findthenumberof waystheformcanbefilledin: withoutrestriction, a ifthefirstandlastanswersare‘Yes’, b iftwoare‘Yes’andeightare‘No’, c iffiveare‘Yes’andfiveare‘No’, d ifmorethansevenanswersare‘Yes’, e ifanoddnumberofanswersare‘Yes’, f ifexactlythreeanswersare‘Yes’,andtheyoccur together, g ifthefirstandlastanswersare‘Yes’andexactly fourmoreare‘Yes’.
12 Containersareidentifiedbyarowofcoloureddotsontheirlids.Thecoloursusedareyellow,greenand purple.Inanyarrangement,therearetobenomorethanthreeyellowdots,nomorethantwogreendotsand nomorethanonepurpledot.
a Ifsixdotsareused,whatisthenumberofpossiblecodes?
b Whatisthenumberofdifferentcodespossibleifonlyfivedotsareused?
13a Howmanyfive-letterwordscanbeformedbyusingthelettersofthewordSTRESS?
b Howmanyfive-letterwordscanbeformedbyusingthelettersofthewordBANANA?
14 FindhowmanyarrangementsofthelettersofthewordTRANSITIONarepossibleif: therearenorestrictions, a theIsaretogether, b theIsaretogether,andsoaretheNs,andsoare theTs, c theNsoccupytheendpositions, d anNoccupiesthefirstbutnotthelastposition, e theletterNisnotateitherend, f thevowelsaretogether. g
15 Tencolouredmarblesareplacedinarow.
a Iftheyareallofdifferentcolours,howmanyarrangementsarepossible?
b Whatistheminimumnumberofcoloursneededtoguaranteeatleast10000differentpatterns?(Thiswill needaguess-and-checkapproach.)
16 FindhowmanyarrangementsthereareofthelettersofthewordGUMTREEif:
a therearenorestrictions, b theEsaretogether,
c theEsareseparatedby:
one, i two, ii three, iii four, iv fiveletters, v d theGissomewherebetweenthetwoEs, e theMissomewheretotheleftofbothEsandtheUissomewherebetweenthem, f theGissomewheretotheleftoftheUandtheMissomewheretotherightoftheU.
17 IfthelettersofthewordGUMTREEandthelettersofthewordKOALAarecombinedandarrangedinto asingletwelve-letterword,inhowmanyofthesearrangementsdothelettersofKOALAappearintheir correctorder,butnotnecessarilytogether?
18 Bobisabouttohanghiseightshirtsinthewardrobe.Hehasfourdifferentstylesofshirt,withtwoidentical shirtsineachstyle.Howmanydifferentarrangementsarepossible ifnotwoidenticalshirtsarenexttooneanother?
19 [Derangementsandthenumber e]A derangement wasdefinedinquestion22ofExercise14Casa permutationthatleavesnoletterunmoved,and D(n)wasdefinedasthenumberofderangementsof n letters. Anotherapproachtofindingaformulafor D(n)usesthe inclusion–exclusionprinciple introducedinthe EnrichmentsectionofExercise12C.Consider,forexample,thederangementsofABCD.
a HowmanypermutationsarethereofABCD?
b Howmanyoftheseleaveinitsoriginalposition: A, i B, ii
Eachofthesenumberswillneedtobesubtractedfromtheanswertopart(a).
c Indoingpart(b),however,wehavesubtractedsomeofthepermutationstwice.Forexample,someof themwouldleavebothAandBunmoved.Thusweneedtoaddbackthenumberofpermutationsthat leaveanytwoparticularlettersunmoved.Howmanyofthesearethere?
d Nowyouwillneedtosubtractthenumberofpermutationsthatleavethreelettersunmoved,andsoon. Hencefindanexpressionfor D(4).
e Rearrangeyourexpressionintotheform D(4)
f Explainhowthiscanbegeneralisedto
g WeshallproveinanEnrichmentquestionoftheYear12volumethatthesequenceinthebracketsof part(f)aboveconvergesto1/e as n →∞.Givebothacombinatorialandaprobabilisticinterpretationof thisresultintermsoftheratioofthenumbersofpermutationsandderangementsof n letters.
• Countallsubsets,andall r-membersubsets,ofan n-memberset.
• Usethenotation nCr ,andprovethat nC0 + nC1 + nC2 + ··· + nCn = 2n
• Developtheformula nCn r = nCr ,andexplainitusingcomplements.
Thissectionturnsattentionawayfromtheorderedselectionsofthepreviousthreesectionstowardsthecounting ofunorderedselections.Anunorderedselectionofdistinctobjectschosenfromacertainsetisjust asubsetofthe set.Thusgivenan n-memberset S ,therearetwobasicquestionstoask:
• Howmanysubsetsdoes S havealtogether?
• Howmany r-membersubsetsdoes S have,for r = 0,1,..., n?
Hereisanexampletoillustratethesituation.Let S bethefive-memberset S = {A,B,C,D,E}.
Hereisthelistofallthesubsetsof S ,arrangedbythenumberofmembers: 10-membersubsets: ∅ (theemptyset)
51-membersubsets:{A},{B},{C},{D},{E} 102-membersubsets:{A,B},{A,C},{A,D},{A,E},{B,C}, {B,D},{B,E},{C,D},{C,E},{D,E} 103-membersubsets:{A,B,C},{A,B,D},{A,B,E},{A,C,D},{A,C,E}, {A,D,E},{B,C,D},{B,C,E},{B,D,E},{C,D,E} 54-membersubsets:{A,B,C,D},{A,B,C,E},{A,B,D,E}, {A,C,D,E},{B,C,D,E} 15-membersubset:{A,B,C,D,E}
making1 + 5 + 10 + 10 + 5 + 1 = 32 = 25 subsetsaltogether.
Choosingasubsetof S requireslookingateachmemberof S inturnanddecidingwhethertoincludeitinthe subsetornot.LetuswritethecodeYfor‘Yes’ifitisincluded,andNfor‘No’ifitisnotincluded.Usingthis code,eachsubsetof S correspondstoafive-letterwordmadeupofYsandNs.Forexample,
YYNYN ←→ {A,B,D}andNNNNY ←→ {E}.
Hencethetotalnumberofsubsetsof S isthenumberoffive-letterwords madeupofYsandNs.ByBox11ofSection17D,orusingtheboxtothe righttovisualisethechoices,thisis25,so
Numberofsubsetsof S = 25 = 32,asdemonstratedabove.
Thissameargumentappliestosetsofanyfinitesize,sothegeneralresultis:
Numberofsubsetsofan n-memberset = 2n .
Howmanythree-membersubsetsdoes S have?
UsingthesamecodeofYsandNs,thenumberof3-membersubsetsof S isthenumberof5-letterwordswith3 Ysand2Ns.ThusbyBox11,seconddotpoint:
Numberof3-membersubsetsof S = 5! 3! × 2! = 10,asdemonstratedabove.
Again,thisargumentappliestosetsofanyfinitesize,sothegeneralresultis:
Numberofr-membersubsetsofan n-memberset = n! r! × (n r)! .
Thenotation nCr
Thenotation nCr isaconvenientshorthand,andmeansthenumberof r-membersubsetsofan n-memberset. Thuswehavetheformula nCr = n! r! × (n r)!
Let S bean n-memberset,andlet0 ≤ r ≤ n
• Thesymbol nCr denotesthenumberof r-membersubsetsofan n-memberset.
• Totalnumberofallthesubsetsof S = 2n
• Numberof r-membersubsetsof S = nCr = n! r!(n r)!
• Addingupthesubsetswith0,1,2,... n membersgivestheformula: nC0 + nC1 + nC2 + ··· + nCn = 2n ,
becauseLHSandRHSarebothcountingthetotalnumberofsubsetsof S .
Thesymbol nCr isspokenas‘n choose r’,andhasanalternativenotation n r .
Thenotations nPr and nCr areintendedtobesimilar.TheletterCstandsfor‘combination’—anoldtermfor ‘unorderedselection’—justastheletterPstandsfor‘permutation’.Byaconvenient,butfalse,etymology, Calsostandsfor‘Choose’,hencethemorerecentconventionofsaying‘n choose r’for nCr
Scientificcalculatorshaveabuttonlabelled n Cr thatwillfindvaluesof nCr .Theanswersareexactforlowvalues of n and r,butaswith nPr ,theyareonlyapproximationsforhighervalues.Again,makeapracticeofevaluating thesenumbersbyhandwhenreasonable—suchcalculationsgreatlyhelptheintuition.
Thereisanotherstandardwaytoprovetheformulafor nCr .InSection17Bwesawthatthenumberofthreeletterwordsformed,withoutrepetition,fromtheset S = {A,B,C,D,E} offivelettersis 5P3 = 5 × 4 × 3 = 60. Whenweturntounorderedselections,however,therearesixdistinctwordsthatallcorrespond,forexample,to thethree-membersubset{B,C,E}:
BCE,BEC,CBE,CEB,EBC,ECB ←→ {B,C,E}
Thereasonforthisisthatthereare 3P3 = 3 × 2 × 1 = 6waysoforderingthesubset{B,C,E}.Thusthecorrespondencebetweenthree-letterwordsandthree-membersubsetsismany-to-one,withasix-foldovercounting. Hencethenumberofthree-membersubsetsis60 ÷ 6 = 10,asillustratedabove.
Ingeneral, nPr = n! (n r)! wordsof r letterscanbeformedwithoutrepetitionfromthemembersofan n-member set S .Butevery r-membersubsetcanbeorderedin r Pr = r!ways,sothecorrespondencebetweentheordered selectionsandtheunorderedselectionsismany-to-one,withovercountingbyafactorof r!.
Hencethenumberof(unordered) r-membersubsetsoftheset S is nPr ÷ r Pr = n! (n r)! ÷ r! = n! r! × (n r)! .
Calculationsof nCr
Herearesomeexamplesofusingtheformulatocalculate nCr forsomevaluesof n and r
Example18 Calculationsinvolvingtheformulafor nCr
a Usetheformulafor nCr toevaluate: 8C5 i
b Find 16C5,leavingyouranswerfactoredintoprimes.
c EvaluatethetermsontheLHStoprove 5C
Solution
ai
=
b 16C5 = 16 × 15 × 14 × 13 × 12 1 × 2 × 3 × 4 × 5
= 2 × 14 × 13 × 12
= 24 × 3 × 7 × 13(Checkthisonthecalculator.)
c LHS = 5! 0! × 5! + 5! 1! × 4! + 5! 2! × 3! + 5! 3! × 2! + 5! 4! ×
= 1 + 5 + 10 + 10 + 5 + 1
= 32 = RHS,asshownonthefirstpageofthissection.
Anatural(orcanonical)correspondence— nCr = nCn−r
Supposethattwopeoplearetobechosenfromfivetomakeafternoontea.Thistaskcanbedoneintwoways:
• Choosethetwopeopleoutoffivewhowill makethetea
• Choosethethreepeopleoutoffivewhowill notmakethetea
Thusforeverychoiceoftwopeopleoutoffive,theremainingthreepeopleisacorrespondingchoiceofthree peopleoutoffive.Thisconfirmsthat 5C2 = 5C3
Butitalsogivesaone-to-onecorrespondencebetweenthetwo-membersubsetsandthethree-membersubsetsof afive-memberset:
{A,B} ←→ {C,D,E}
{A,C} ←→ {B,D,E}
{A,D} ←→ {B,C,E}
{A,E} ←→ {B,C,D}
{B,C} ←→ {A,D,E}
{B,D} ←→ {A,C,E}
{B,E} ←→ {A,C,D}
{C,D} ←→ {A,B,E}
{C,E} ←→ {A,B,D}
{D,E} ←→ {A,B,C}
Inthis natural or canonicalcorrespondence betweenthe2-membersubsetsandthe3-membersubsets,every subset T ispairedwithitscomplement T : T ←→ T
Let n and r bewholenumberswith0 ≤ r ≤ n,andlet S bean n-memberset.
• nCr = nCn r
• The r-membersubsetsof S are naturally or canonically pairedupwiththe(n r)-membersubsetsof S bypairingeachsubsetwithitscomplement.
Thecorrespondenceisbynomeansrestrictedtomathematics.Innormallanguage,asituationcanoftenbe describedjustaswellbysayingwhatitisnotasbysayingwhatitis—weareallfamiliarwith‘dayswhenno rainfell’or‘unforgivableactions’or‘invisibleenemies’.
Illustratingtheformulaefor nCr involvingcomplements
Write 7C2 and 7C5 infactorialnotation,showingthattheyareequal.
Solution
Hereareanumberofexamplesofcountingproblems.Asalways,wordssuchas‘atleast’,‘atmost’,‘not’and ‘excluding’shouldalwaysberegardedaswarningsthattheproblemmaybestbesolvedbyconsideringthe complementarysituation.
Example20 Using nCr incountingproblems
Tenpeoplemeettoplaydoublestennis.
a Inhowmanywayscanfourpeoplebeselectedfromthisgrouptoplaythefirstgame?(Ignorethe subsequentorganisationintopairs.)
b HowmanyofthesewayswillincludeMariaandexcludeAlex?
c Iftherearefourwomenandsixmen,inhowmanywayscantwomenandtwowomenbechosenforthis game?
d Againwithfourwomenandsixmen,inhowmanywayswillwomenbeinthemajority?
Solution
a Numberofways = 10C4 = 210.
b BecauseMariaisincluded,threefurtherpeoplemustbechosen,andbecauseAlexisexcluded,thereare noweightpeopletochoosethesethreefrom. Hencenumberofways = 8C3 = 56.
c Numberofwaysofchoosingthetwowomen = 4C2 = 6, numberofwaysofchoosingthetwomen = 6C2 = 15, sonumberofwaysofchoosingallfour = 15 × 6 = 90.
d Numberofwayswithonemanandthreewomen = 6C1 × 4C3 = 6 × 4 = 24, numberofwayswithfourwomen = 1, sonumberofwayswithamajorityofwomen = 24 + 1 = 25.
Let S = {2,4,6,8,10,12} bethesetofthefirstsixpositiveevennumbers.
a Howmanysubsetsof S containatleasttwonumbers?
b Howmanysubsetswithatleasttwonumbersdonotcontain8?
c Howmanysubsetswithatleasttwonumbersdonotcontain8butdocontain10?
Solution
a Numberof1-memberand0-membersubsets = 6C1 + 6C0 = 6 + 1 = 7, sonumberwithatleast2members = 26 7 = 57.
b Weconsidernowthe5-memberset T = {2,4,6,10,12}
Numberof1-memberand0-membersubsets = 5C1 + 5C0 = 6, sonumberwithatleast2members = 25 6 = 26.
c Because10hasalreadybeenchosen,weneedtochooseasubset,withatleastonemember,fromthe four-memberset U = {2,4,6,12}.
Numberof0-membersubsets = 1(theemptyset), sonumberofsuchsubsetswithatleastonemember = 24 1 = 15.
Tenpoints A1, A2,..., A10 arearrangedequallyspacedaroundacircle.
a Howmanytrianglescanbedrawnwiththesepointsasvertices?
b Howmanypairsofsuchtrianglescanbedrawn,iftheverticesofthetwotriangles aredistinct?
c Inhowmanysuchpairswillthetriangles: i notoverlap, ii overlap?
Solution
a Toformatriangle,wemustchoose3pointsoutof10, sonumberoftriangles = 10C3 = 120.
b Toformtwotriangles,firstchoose6pointsoutof10,whichcanbedonein 10C6 = 210ways. Takeanyoneofthose6points,thenchoosetheother2pointsinitstriangle,whichcanbedonein 5C2 = 10 ways.
Thesecondtriangleisthendrawnusingtheotherthreepoints. Hencenumberofpairsoftriangles = 210 × 10 = 2100.
c Toformtwonon-overlappingtriangles,wefirstchoose6pointsoutof10,whichagaincanbedonein 10C6 = 210ways.
These6pointscanbemadeintotwonon-overlappingtrianglesin3ways,byarrangingthe6pointsin cyclicorder,andchoosing3adjacentpoints.
i Hencenumberofnon-overlappingpairs = 210 × 3 = 630.
ii Bysubtraction,numberofoverlappingpairs = 2100 630 = 1470.
1 TwopeoplearechosenfromagroupoffivepeoplecalledP,Q,R,SandT.Listallpossiblecombinations, andfindhowmanythereare.
2 Findinhowmanywaysyoucanformagroupof: twopeoplefromachoiceofseven, a threepeoplefromachoiceofseven, b twopeoplefromachoiceofsix, c fivepeoplefromachoiceofnine. d
3a Findhowmanypossiblecombinationsthereareif,fromagroupoftenpeople: twopeoplearechosen, i eightpeoplearechosen. ii
b Whyaretheanswersidentical?
4 Fromapartyoftwelvemenandeightwomen,findhowmanygroupscanbechosenconsistingof: fivemenandthreewomen, a fourwomenandfourmen. b
5 Fournumbersaretobeselectedfromthesetofthefirsteightpositiveintegers.Findhowmanypossible combinationsthereareif: therearenorestrictions, a therearetwooddnumbersandtwoevennumbers, b thereisexactlyoneoddnumber, c allthenumbersmustbeeven, d thereisatleastoneoddnumber. e
6 Fourballsaresimultaneouslydrawnfromabagcontainingthreeidenticalgreenballsandsixidenticalblue balls.Findhowmanywaysthereareofdrawingthefourballsif: theballsmaybeofanycolour, a thereareexactlytwogreenballs, b thereareatleasttwogreenballs, c therearemoreblueballsthangreenballs. d
7 Acommitteeoffiveistobechosenfromsixmenandeightwomen.Findhowmanycommitteesare possibleif: therearenorestrictions, a allmembersaretobefemale, b allmembersaretobemale, c thereareexactlytwomen, d therearefourwomenandoneman, e thereisamajorityofwomen, f aparticularmanmustbeincluded, g aparticularmanmustnotbeincluded. h
8a WhatisthenumberofcombinationsofthelettersofthewordEQUATIONtakenfouratatime(without repetition)?
b Howmanyofthefour-lettercombinationscontainfourvowels?
c Howmanyofthefour-lettercombinationscontaintheletterQ?
9 AteamofsevennetballersistobechosenfromasquadoftwelveplayersA,B,C,D,E,F,G,H,I,J,Kand L.Inhowmanywayscantheybechosen: withnorestrictions, a ifthecaptainCistobeincluded, b ifJandKarebothtobeexcluded, c ifAisincludedbutHisnot, d ifoneofFandListobeincludedandtheother excluded? e
10a Considerthedigits9,8,7,6,5,4,3,2,1and0.Findhowmanyfive-digitnumbersarepossibleifthe digitsaretobein: descendingorder, i ascendingorder. ii
b Whydothesetwoquestionsinvolveunorderedselections?
11 Twelvepeoplearriveatarestaurant.Thereisonetableforsix,onetableforfourandonetablefortwo.In howmanywayscantheybeassignedtoatable?
12 Twentystudents,tenmaleandtenfemale,aretotravelfromschooltothesportsground.Eightofthemgoin aminibus,sixofthemincars,fourofthemonbikesandtwowalk.
a Inhowmanywayscantheybedistributedforthetrip?
b Inhowmanywayscantheybedistributedifnoneoftheboyswalk?
13 Tenpoints P1, P2,..., P10 arechoseninaplane,nothreeofthepointsbeingcollinear.
a Howmanylinescanbedrawnthroughpairsofthepoints?
b Howmanytrianglescanbedrawnusingthegivenpointsasvertices?
c Howmanyofthesetriangleshave P1 asoneoftheirvertices?
d Howmanyofthesetriangleshave P1 and P2 asvertices?
14 Tenpointsarechoseninaplane.Fiveofthepointsarecollinear,butnoothersetofthreeofthepointsis collinear.
a Howmanysetsofthreepointscanbeselectedfromthefivethatarecollinear?
b Howmanytrianglescanbeformedusingthreeofthetenpointsasvertices?
15 Fromastandarddeckof52playingcards,findhowmanyfive-cardhandscanbedealt: consistingofblackcardsonly, a consistingofdiamondsonly, b containingallfourkings, c consistingofthreediamondsandtwoclubs, d consistingofthreetwosandanotherpair, e consistingofonepairandthreeofakind. f
16a Inhowmanywayscanagroupofsixpeoplebedividedinto: twounequalgroups(neithergroupbeing empty), i twoequalgroups? ii
b Repeatpart(a)forfourpeople.
c Repeatpart(a)foreightpeople.
17 Findhowmanydiagonalstherearein: aquadrilateral, a apentagon, b adecagon, c apolygonwith n sides. d
18 Twelvepointsarearrangedinorderaroundacircle.
a Howmanytrianglescanbedrawnwiththesepointsasvertices?
b Inhowmanypairsofsuchtrianglesaretheverticesofthetwotrianglesdistinct?
c Inhowmanysuchpairswillthetriangles: notoverlap, i overlap? ii
19 Let S = {1,3,5,7,9,11,13,15,17,19} bethesetofthefirsttenpositiveoddintegers.
a Howmanysubsetsdoes S have?
b Howmanysubsetsof S containatleastthreenumbers?
c Howmanysubsetswithatleastthreenumbersdonotcontain7?
d Howmanysubsetswithatleastthreenumbersdonotcontain7butdocontain19?
20 Inhowmanywayscantwonumbersbeselectedfrom1,2,...,8,9sothattheirsumis: even, a odd, b divisibleby3, c divisibleby5, d divisibleby6? e
21 Therearetenbasketballersinateam.Findinhowmanyways: thestartingfivecanbechosen, a theycanbesplitintotwoteamsoffive. b
22 Nineplayersaretobedividedintotwoteamsoffourandoneumpire.
a Inhowmanywayscantheteamsbeformed?
b Iftwoparticularpeoplecannotbeonthesameteam,howmanydifferentcombinationsarepossible?
23 Byconsideringtheirprimefactorisations,findthenumberofpositivedivisorsof:
24a Thesixfacesofanumberofidenticalcubesarepaintedinsixdistinctcolours.Howmanydifferentcubes canbeformed?
b Adiefitsperfectlyintoacubicalbox.Howmanywaysarethereofputtingthedieintothebox?
25 Thediagramshowsa6 × 4grid.Theaimistowalkfromthepoint A inthetop left-handcornertothepoint B inthebottomright-handcornerbywalkingalongthe blacklineseitherdownwardsortotheright.Asinglemoveisdefinedaswalkingalong onesideofasinglesmallsquare,thusittakesyoutenmovestogetfrom A to B
a Findhowmanydifferentroutesarepossible: i withoutrestriction, ii ifyoumustpassthrough C, iii ifyoucannotmovealongthetoplineofthegrid, iv ifyoucannotmovealongthesecondrowfromthetopofthegrid.
b Noticethateveryroutemustpassthroughoneandonlyoneofthefivecrossedpoints.Henceprovethat
c Drawanothersuitablediagonaland,usingamethodsimilartothatinpart(b),provethat
d Drawupasimilar6 × 6grid,thenusingthesameideaasthatusedinparts(b)and(c),provethat
CHALLENGE
26 Apieceofartreceivesanintegermarkfromzeroto100foreachofthecategoriesdesign,techniqueand originality.Inhowmanywaysisitpossibletoscoreatotalmarkof200?
27 Howmanydifferentcombinationsarethereofthreedifferentintegersbetweenoneandthirtyinclusivesuch thattheirsumisdivisiblebythree?
28a Howmanydoublestennisgamesarepossible,givenagroupoffourplayers?
b Inhowmanywayscantwogamesofdoublestennisbearranged,givenagroupofeightplayers?
c Sixmenandsixwomenaretoplayinthreegamesofdoublestennis.Findhowmanywaysthepairings canbearrangedif:
therearenorestrictions, i eachgameistobeagameofmixeddoubles. ii
29 ReferringtoQuestion18ofExercise17D,Bob’sinterestinshirtshasmaturedrecently—henowhas 2n shirts,withtwoidenticalshirtsineachof n distinctstyles.Hestillwantstohanghisshirtsinthe wardrobesothatnotwoidenticalshirtsarenexttooneanother.
a Showthatthenumberofallowedarrangementsis
b Evaluatethisnumberfor n = 1,2,3,4,5and6,andfindthecorrespondingratioofthetotalnumberof arrangementstothenumberofallowedarrangements.
30 Computingisbasedon binarystrings,whicharesequencesof0sand1s,suchas00101and1001110.The probleminthisquestionistofindthenumber N ofbinarystringsthatcontainexactly a 0sandatmost b 1s, where a and b arewholenumbers.Wewillcountsuchstringsintwodifferentways,andhenceprovean interestingidentity.
a Using nCr notation,findhowmanybinarystringswithexactly a 0scontain: no1s i one1 ii two1s iii three1s iv r 1s. v
b Henceprovethat
c Let S beabinarystringconsistingof a 0sand r 1s,where r isatmost b.Extend S toalongerbinary string S 01 ... 1byadding0ontheright,andthenadding b r 1s.Howmany0sand1sarethereinthe resultingstring,andwhatisitstotallength?
d Describetheinverseprocessbywhichastringof(a + 1)0sand b 1scanbeconvertedtoastringof a 0s andatmost b 1s,andshowthatthetwoprocessesareone-to-onecorrespondences.
e Henceprovethat N = a+b+1Cb
f Showthat aC
Learningintentions
• Applycountingprocedurestofindingsampleandeventspacesinprobability.
Thepurposeofthissectionistoapplythecountingproceduresofthelastfoursectionstoquestionsabout probability.Inthesemorecomplicatedquestions,countingproceduresarerequiredforcountingbothsample spaceandeventspace.
Asalwaysinthistopic,thetwoquestionsthatneedtobeaskedare:
• Aretheselectionswearecountingorderedorunordered?
• Iftheyareordered,isrepetitionallowedornot?
Ifthequestionscanbedonewithorderedselectionsorwithunorderedselections,itisusuallyeasiertouse unorderedselectionsbecausethenumbersaresmaller.
Threecardsaredealtfromapackof52.
a Findtheprobabilitythatoneclubandtwoheartsaredealt,inanyorder.
b Findtheprobabilitythatoneclubandtwoheartsaredealtinthatorder.
Solution
a Letthesamplespacebethesetofallunorderedselectionsof3cardsfrom52, sonumberofunorderedhands = 52C3 = 52 × 51 × 50 3 × 2 × 1 . Wecannowchoosethehandbychoosing1clubfrom13in 13C1 = 13ways,andchoosingthe2hearts from13in 13C2 = 78ways,sothehandcanbechosenin13 × 78ways.
Hence P(1cluband2hearts) = 13 × 78 × 3 × 2 × 1 52 × 51 × 50 = 39 850 .
b Letthesamplespacebethesetofallorderedselectionsof3cardsfrom52, so numberoforderedhands = 52P3 = 52 × 51 × 50, andnumberofsuchhandsintheorder ♣♡♡ = 13 × 13 × 12.
Hence P(♣♡♡) = 13 × 13 × 12 52 × 51 × 50 = 13 850
Note: Theanswertopart(b)mustbe 1 3 oftheanswertopart(a),becauseinahandwithoneclubandtwo hearts,theclubcanbeanyoneofthreepositions.Thisindicatesthatitwouldbequitereasonabletodo part(a)usingorderedselections,andtodopart(b)usingunorderedselections,althoughthemethods chosenabovearemorenaturaltothewayinwhicheachquestionwasworded.
Thesamplespacesinthenexttwoexamplesareeasilyfound,butavarietyofmethodsisneededtoestablishthe sizesofthevariouseventspaces.
Afive-digitnumberischosenatrandom.Findtheprobability:
a thatitisatleast60000,
b thatitconsistsonlyofevendigits,
c thatthedigitsaredistinct,
d thatthedigitsaredistinctandinincreasingorder.
Solution
Thefirstdigitofafive-digitnumbercannotbezero,givingninechoices,buttheotherdigitscanbeanyoneof thetendigits.
Hencethenumberoffive-digitnumbers = 9 × 10 × 10 × 10 × 10 = 90000.
a Tobeatleast60000,thefirstdigitmustbe6,7,8or9, sothenumberoffavourablenumbersis4 × 10 × 10 × 10 × 10 = 40000. Hence P(atleast60000) = 4 9
b Ifallthedigitsareeven,therearefourchoicesforthefirstdigit(itcannotbezero)andfivechoicesforeach oftheotherfour.
Hencenumberofsuchnumbers = 4 × 5 × 5 × 5 × 5 = 2500, and P(alldigitsareeven) = 2500 90000 = 1 36
c Thisiscountingwithoutreplacement: 1stdigit
and P(digitsaredistinct) =
d Every unordered five-membersubsetofthesetofninenon-zerodigitscanbearrangedinexactlyoneway intoafive-digitnumberwiththedigitsinincreasingorder.(Notethatthedigitzerocannotbeused,because anumbercan’tbeginwiththedigitzero.)
Hencenumberofsuchnumbers = numberofunordered5-membersubsetsof{1,2,3,4,5,6,7,8,9} = 9C5 = 126, so P(digitsaredistinctandinincreasingorder) = 126 90000 = 7 5000 .
Example25 Aharderexampleusingavarietyofmethods
ContinuewithExample24,whereafive-digitnumberischosenatrandom,givingasamplespaceofsize 90000.Findtheprobabilitythat
a thenumbercontainsatleastone4,
b thenumbercontainsatleastone4andatleastone5,
c thenumbercontainsexactlythree7s,
d thenumbercontainsatleastthree7s.
a Thiscanbeapproachedusingthecomplementaryevent: numberoffive-digitnumberswithouta4 = 8 × 9 × 9 × 9 × 9 = 52488, sonumberoffive-digitnumberswitha4 = 90000 52488 = 37512. Hence P(atleastone4) = 37512 90000 = 521 1250 .
b Thisisbestapproachedusingthecountingrule |
Let A bethesetoffive-digitnumberswithouta4,andlet B bethesetoffive-digitnumberswithouta5. Then A ∩ B isthesetoffive-digitnumberwithno4 and no5,and A ∪ B isthesetoffive-digitnumberwith no4 or no5.
Fromline2ofthesolutiontopart(a), |A| = 52488, andsimilarly, |B| = 52488, thenproceedingagainasinpart(a), |A ∩ B| = 7 × 8 × 8 × 8 × 8 = 8672.
Nowweapplythecountingrule,
|−|A ∩ B| = 52488 + 52488 28672 = 76304, sothereare76304five-digitnumberswithno4 or no5.
Wewantthesetofallfive-digitnumberswithatleastone4 and atleastone5, whichisthecomplementaryset A ∪ B,and A ∪ B = 90000 76304 = 13696. Hence P(atleastone4andatleastone5) = 13696 90000 , = 856 5625 .
c Countingthenumberoffive-digitnumberswithexactlythree7srequirescases.Firstwecountthefive-digit stringswithexactlythree7s,byfirstplacingthethree7s,andthenchoosingthefirstandsecondnon-7 digits:
positionofthethree7s choosefirstnon-7 choosesecondnon-7
5C3 = 10 9 9
giving810suchstrings.Secondly,wemustsubtractthenumberoffive-digitstringswithexactlythree7s andbeginningwithzero:
positionofthethree7s choosetheothernon-7
4C3 = 4 9 giving36suchstrings.Hencethereare810 36 = 774suchnumbers,and P(numberhasexactlythree7s) = 774 90000 = 43 5000 .
d Thenumber77777istheonlyfive-digitnumberwithfive7s. Anyfive-digitnumberwithexactlyfour7shasoneofthefiveforms
7777,7
, wherethe ∗ in ∗7777isanon-zerodigit.Thereareeightnumbersofthefirstform,andnineoftheother fourforms,giving44numbersaltogether.
Hencethenumberwithatleastthree7s = 774 + 44 + 1 = 819, and P(atleastthree7s) = 819 90000 = 91 10000 .
1 Acommitteeofthreeistobeselectedfromtheninemembersinaclub.
a Howmanydifferentcommitteescanbeformed?
b Iftherearefivemenintheclub,whatistheprobabilitythattheselectedcommitteeconsistsentirelyof males?
2 Theintegersfrom1to10inclusivearewrittenontenseparatepiecesofpaper.Fourpiecesofpaperare drawnatrandom.Findtheprobabilitythat:
a thefournumbersdrawnare1,2,3and6, b thenumber9isoneofthenumbersdrawn, c thenumber8isnotdrawn, d thenumber7isdrawnbutthenumber1isnot.
3 Abagcontainsthreered,sevenyellowandfiveblueballs.Ifthreeballsaredrawnfromthebagsimultaneously,findtheprobabilitythat: allthreeballsareyellow, a alltheballsareofthesamecolour, b therearetworedballsandoneblueball, c alltheballsareofdifferentcolours. d
4 AsportscommitteeoffivemembersistobechosenfromeightAFLfootballersandsevensoccerplayers. Findtheprobabilitythatthecommitteewillcontain: onlyAFLfootballers, a onlysoccerplayers, b threesoccerplayersandtwoAFLfootballers, c atleastonesoccerplayer, d atmostonesoccerplayer, e Ian,aparticularsoccerplayer. f
5 Fromastandardpackof52cards,threeareselectedatrandom.Findtheprobabilitythat: theyarethejackofspades,thetwoofclubsand thesevenofdiamonds, a allthreeareaces, b theyarealldiamonds, c theyareallofthesamesuit, d theyareallpicturecards, e twoareredandoneisblack, f oneisaseven,oneisaneightandoneisanine, g twoaresevensandoneisasix, h exactlyoneisadiamond, i atleasttwoofthemarediamonds. j
6 Repeatthepreviousquestionifthecardsareselectedfromthepackoneatatime,andeachcardisreplaced beforethenextoneisdrawn.
7 Threeboysandthreegirlsaretositinarow.Findtheprobabilitythat: a theboysandgirlsalternate, b theboyssittogetherandthegirlssittogether, c twospecificgirlssitnexttooneanother.
8 Afamilyoffiveareseatedinarowatthecinema.Findtheprobabilitythat: a theparentssitontheendandthethreechildrenareinthemiddle, b theparentssitnexttooneanother.
9 Sixpeople,ofwhomPatrickandJessicaaretwo,arrangethemselvesinarow.Findtheprobabilitythat:
a PatrickandJessicaoccupytheendpositions, b PatrickandJessicaarenotnexttoeachother.
10
ThelettersofPROMISEarearrangedrandomlyinarow.Findtheprobabilitythat:
a thewordstartswithRandendswithS, b thelettersPandRarenexttooneanother, c thelettersPandRareseparatedbyatleastthreeletters, d thevowelsandtheconsonantsalternate, e thevowelsaretogether.
11 Thedigits3,3,4,4,4and5areplacedinarowtoformasix-digitnumber.Ifoneofthesenumbersis selectedatrandom,findtheprobabilitythat:
a itiseven, b itendsin5, c the4soccurtogether, d thenumberstartswith5andthenthe4sand3salternate, e the3sareseparatedbyatleastoneothernumber.
12 ThelettersofthewordPRINTERarearrangedinarow.Findtheprobabilitythat:
a thewordstartswiththeletterE,
b thelettersIandParenexttooneanother,
c therearethreelettersbetweenNandT, d thereareatleastthreelettersbetweenNandT.
13 ThelettersofKETTLEarearrangedrandomlyinarow.Findtheprobabilitythat:
a thetwolettersEaretogether, b thetwolettersEarenottogether,
c thetwolettersEaretogetherandthetwolettersTaretogether, d theEsandTsaretogetherinonegroup.
14 ThelettersofENTERTAINMENTarearrangedinarow.Findtheprobabilitythat: thelettersEaretogether, a twoEsaretogetherandoneisapart, b allthelettersEareapart, c thewordstartsandendswithE. d
15 Atankcontains20taggedfishand80untaggedfish.Oneachday,fourfishareselectedatrandom,andafter notingwhethertheyaretaggedoruntagged,theyarereturnedtothetank.Answerthefollowingquestions, correcttothreesignificantfigures.
a Whatistheprobabilityofselectingnotaggedfishonagivenday?
b Whatistheprobabilityofselectingatleastonetaggedfishonagivenday?
c Calculatetheprobabilityofselectingnotaggedfishoneverydayforaweek.
d Whatistheprobabilityofselectingnotaggedfishonexactlythreeofthesevendaysduringtheweek?
16 Abagcontainssevenwhiteandfiveblackdiscs.Threediscsarechosenfromthebag.Findtheprobability thatallthreediscsareblack,ifthediscsarechosen: a withoutreplacement, b withreplacement, c sothataftereachdrawthediscisreplacedwithoneoftheoppositecolour.
17 Sixpeoplearetobedividedintotwogroups,eachwithatleastoneperson.Findtheprobabilitythat: a therewillbethreeineachgroup, b therewillbetwoinonegroupandfourintheother, c therewillbeonegroupoffiveandanindividual.
18 Athree-digitnumberisformedfromthedigits3,4,5,6and7(norepetitionsallowed).Findtheprobability that:
thenumberis473, a thenumberisodd, b thenumberisdivisibleby5, c thenumberisdivisibleby3, d thenumberstartswith4andendswith7, e thenumbercontainsthedigit3, f thenumbercontainsthedigits3and5, g thenumbercontainsthedigit3or5, h alldigitsinthenumberareodd, i thenumberisgreaterthan500. j
19 Thedigits1,2,3and4areusedtoformnumbersthatmayhave1,2,3or4digitsinthem.Ifoneofthe numbersisselectedatrandom,findtheprobabilitythat: ithasthreedigits, a itiseven, b itisgreaterthan200, c itisoddandgreaterthan3000, d itisdivisibleby3. e
20a AsenatecommitteeoffivemembersistobeselectedfromsixLaborandfiveLiberalsenators.Whatis theprobabilitythatLaborwillhaveamajorityonthecommittee?
b Thesenatecommitteeistobeselectedfrom N LaborandfiveLiberalsenators.Usetrialanderrorto findtheminimumvalueof N,giventhattheprobabilityofLaborhavingamajorityonthecommitteeis greaterthan 3 4 .
21 FourbasketballteamsA,B,CandDeachconsistoftenplayers,andineachteam,theplayersarenumbered 1,2,...9,10.Fiveplayersaretobeselectedatrandomfromthefourteams.Findtheprobabilitythatofthe fiveplayersselected:
a threearenumbered4andtwoarenumbered9, b atleastfourarefromthesameteam.
22 Apokerhandoffivecardsisdealtfromastandardpackof52.Findtheprobabilityofobtaining:
a onepair, b twopairs, c threeofakind, d fourofakind, e afullhouse(onepairandthreeofakind), f astraight(fivecardsinsequenceregardlessofsuit,acehighorlow), g aflush(fivecardsofthesamesuit), h aroyalflush(ten,jack,queen,kingandaceinasinglesuit).
23 Fouradultsarestandinginaroomthathasfiveexits.Eachadultisequallylikelytoleavetheroomthrough anyoneofthefiveexits.
a Whatistheprobabilitythatallfouradultsleavetheroomviathesameexit?
b Whatistheprobabilitythatthreeparticularadultsusethesameexitandthefourthadultusesadifferent exit?
c Whatistheprobabilitythatanythreeofthefouradultscomeoutthesameexit,andtheremainingadult comesoutadifferentexit?
d Whatistheprobabilitythatnomorethantwoadultscomeoutanyoneexit?
24a Fivedinersinarestaurantchooserandomlyfromamenufeaturingfivemaincourses.Findthe probabilitythatexactlyoneofthemaincoursesisnotchosenbyanyofthediners.
b Repeatthequestionifthereare n dinersandachoiceof n maincourses.
25 [Thebirthdayproblem]
a Assuminga365-dayyear,findtheprobabilitythatinagroupofthreepeopletherewillbeatleastone birthdayincommon.Answercorrecttotwosignificantfigures.
b Ifthereare n peopleinthegroup,findanexpressionfortheprobabilityofatleastonecommonbirthday.
c Bychoosinganumberofvaluesof n,plotagraphoftheprobabilityofatleastonecommonbirthday against n for n ≤ 50.
d Howmanypeopleneedtobeinthegroupbeforetheprobabilityexceeds0.5?
e Howmanypeopleneedtobeinthegroupbeforetheprobabilityexceeds90%?
26 Duringthesevengamesofthefootballseason,MaxandBertmusteachmissthreeconsecutivegames.The gamestobemissedbyeachplayerarerandomlyandindependentlyselected.
a Whatistheprobabilitythattheybothhavethefirstgameoff together?
b WhatistheprobabilitythatthesecondgameisthefirstonewhereMaxandBertarebothmissing?
c WhatistheprobabilitythatMaxandBertmissatleastoneofthesamegames?
27 Eightplayersmakethequarter-finalsatWimbledon.Thewinnerofeachofthequarter-finalsplaysa semi-finaltoseewhoentersthefinal.
a Assumingthatalleightplayersareequallylikelytowinamatch,showthattheprobabilitythatanytwo particularplayerswillplayeachotheris 1 4 .
b Whatistheprobabilitythattwoparticularpeoplewillplayeachotherifthetournamentstartswith16 players?
c Whatistheprobabilitythattwoparticularplayerswillmeetinasimilarknockouttournamentif2n playersenter?
Learningintentions
• Developprocedurestocountarrangementsinacircle.
• Countarrangementsinacircletocalculateprobabilities.
Arrangementsinacircle,oraroundaroundtable,arecomplicatedbecausetwoarrangementsareregardedas equivalentifonecanberotatedtoproducetheother.Forexample,allthefiveround-tableseatingsbelowofKing Arthur,QueenGuinevere,SirLancelot,SirBorsandSirPercivalaretoberegardedasthesame:
Themoststraightforwardwayofcountingarrangementsinacircleistoseatthepeopleinorder,dealingwiththe restrictionsfirstasalways,butreckoningthatthereisessentiallyonlyonewaytoseatthefirstpersonwhosits down,becauseuntilthattime,alltheseatsareidentical.
14CountingArrangementsinaCircle
Thereisessentiallyonlyonewaytoseatthefirstperson,becauseuntilthen,alltheseatsareidentical.
KingArthur,QueenGuinevere,SirLancelot,SirBorsandSirPercivalsitarounda roundtable.Findinhowmanywaysthiscanbedone:
a withoutrestriction,
b ifQueenGuineveresitsatKingArthur’sleftthand,
c ifQueenGuineveresitsbetweenSirLancelotandSirBors,
d ifKingArthurandSirLancelotdonotsittogether.
Solution
a
Numberofways = 24.
b
Numberofways = 6.
c
Numberofways = 4.
d
Numberofways = 12.
Whenarranginggroupsaroundacircle,theprincipleisthesameastheprincipleforcompoundorderings establishedinSection17C.
15Arranginggroupsaroundacircle
• Firstchooseanorderforeachgroup.
• Thenarrangethegroupsaroundthecircle,reckoningthatthereisessentiallyonlyonewaytoplacethe firstgroup.
Fiveboysandfivegirlsaretositaroundatable.Findinhowmanywaysthiscanbedone:
a withoutrestriction,
b iftheboysandgirlsalternate,
c iftherearefivecouples,allofwhomsittogether,
d iftheboyssittogetherandthegirlssittogether,
e iffourcouplessittogether,butWalterandMaudedonot.
Solution
Numberofways = 9! = 362880. b
Numberofways = 5! × 4! = 2880.
c Eachcouplecanbeorderedin2ways,giving25 orderingsofthefivecouples.Thenseatthefivecouples aroundthetable: 1stcouple 2ndcouple 3rdcouple 4thcouple 5thcouple 1 4 3 2 1
Numberofways = 25 × 4! = 768.
d Theboyscanbeorderedin5!ways,andthegirlsin5!waysalso.Thenseatthetwogroupsaroundthe table: groupofboys groupofgirls 1 1
Numberofways = 5! × 5! × 1 = 14400.
e Ordereachofthefourcouplesin2ways,giving16orderingsofthecouples.Therearenowfourcouples andtwoindividualstoseataroundthetable,withtherestrictionthatMaudedoesnotsitnexttoWalter: Walter Maude 1stcouple 2ndcouple 3rdcouple 4thcouple 1 3 4 3 2 1
Numberofways = 24 × 3 × 4! = 1152.
Asalways,countingallowsprobabilityproblemstobesolvedbycountingthesamplespaceandtheeventspace.
ThreeTasmanians,threeNewZealanders,andthreeQueenslandersareseatedatrandomaroundaroundtable. Whatistheprobabilitythatthethreegroupsareseatedtogether?
Solution
Usingthesameboxesasbefore,thereare1 × 8!possibleorderings.Tofindthe numberoffavourableorderings,firstordereachgroupin3! = 6ways,thenorderthe threegroupsaroundthetablein1 × 2 × 1 = 2ways,sothetotalnumberoffavourable orderingsis6 × 6 × 6 × 2.
Hence P(groupsaretogether) =
1a Inhowmanywayscanfivepeoplebearranged: inaline, i inacircle? ii
b Inhowmanywayscantenpeoplebearranged: inaline, i inacircle? ii
2 Eightpeoplearearrangedin: astraightline, a acircle. b Inhowmanywayscantheybearrangedsothattwoparticularpeoplesittogether?
3 Bob,Betty,Ben,BradandBelindaaretobeseatedataroundtable.Inhowmanywayscanthisbedone: iftherearenorestrictions, a ifBettysitsonBob’sright-handside, b ifBradistositbetweenBobandBen, c ifBelindaandBettysitapart, d ifBenandBelindasitapart,butBettysitsnextto Bob? e
4 Fourboysandfourgirlsarearrangedinacircle.Inhowmanywayscanthisbedone:
a iftherearenorestrictions,
b iftheboysandthegirlsalternate,
c iftheboysandgirlsareindistinctgroups, d ifaparticularboyandgirlwishtositnexttooneanother, e iftwoparticularboysdonotwishtositnexttooneanother, f ifoneparticularboywantstositbetweentwoparticulargirls?
5 ThelettersA,E,I,P,QandRarearrangedinacircle.Findtheprobabilitythat: thevowelsaretogether, a AisoppositeR, b thevowelsandconsonantsalternate, c atleasttwovowelsarenexttooneanother. d
6 Inhowmanywayscantheintegers1,2,3,4,5,6,7,8beplacedinacircleif: therearenorestrictions, a alltheevennumbersaretogether, b theoddandevennumbersalternate, c atleastthreeoddnumbersaretogether, d thenumbers1and7areadjacent, e thenumbers3and4areseparated? f
7 Acommitteeofthreewomenandsevenmenistobeseatedrandomlyataroundtable.
a Whatistheprobabilitythatthethreefemaleswillsittogether?
b Thecommitteeelectsapresidentandavice-president.Whatistheprobabilitythattheyaresitting oppositeoneanother?
8 Findhowmanyarrangementsof n peoplearoundacirclearepossibleif: therearenorestrictions, a twoparticularpeoplemustsittogether, b twoparticularpeoplesitapart, c threeparticularpeoplesittogether. d
9 Twelvemarblesaretobeplacedinacircle.Inhowmanywayscanthisbedoneif:
a allthemarblesareofdifferentcolours, b thereareeightred,threeblueandonegreenmarble?
10 Therearetworoundtables,oneoakandonemahogany,eachwithfiveseats.Inhowmanywaysmaya groupoftenpeoplebeseated?
11 Asportscommitteeconsistingoffourrowers,threebasketballersandtwocricketerssitsatacirculartable.
a Howmanydifferentarrangementsofthecommitteearepossibleiftherowersandbasketballersbothsit togetheringroups,butnorowersitsnexttoabasketballer?
b Onerowerandonecricketerarerelated.Iftheconditionsin(a)apply,whatistheprobabilitythatthese twomembersofthecommitteewillsitnexttooneanother?
CHALLENGE
12 Agroupof n menand n + 1womensitaroundacirculartable.Whatistheprobabilitythatnotwomensit nexttooneanother?
13a Consideranecklaceofsixdifferentlycolouredbeads.Becausethenecklacecanbeturnedover, clockwiseandanti-clockwisearrangementsofthebeadsdonotyielddifferentorders.Hencefindhow manydifferentarrangementsthereareofthesixbeadsonthenecklace.
b Inhowmanywayscantendifferentkeysbeplacedonakeyring?
c Inhowmanywayscanoneyellow,tworedandfourgreenbeadsbeplacedonabraceletifthebeadsare identicalapartfromcolour?(Thiswillrequirealistingofpatternstoseeiftheyareidenticalwhenturned over.)
Becauseofbranching,itisdifficulttocountthenumberofpossiblehydrocarbonswithsay20carbonatoms.For example,forbutaneC4H10 therearetwocarbonchains,andforpentaneC5H12 therearethreecarbonchains:
Thisisahardproblem.Ifyoucanwritecomputerprogrammes,youcouldwriteyourownprogrammeto investigatethenumberofpossiblecarbonchainsforhydrocarbonswithincreasingnumbersofcarbonatoms. Perhapsyourprogrammecandisplaydiagramsofallthechains,aswithbutaneandpentaneabove.
• Createyourownsummaryofthischapteronpaperorinadigitaldocument.
• Thisautomatically-markedquizisaccessedintheInteractiveTextbook.AprintablePDFWorksheetversionis alsoavailablethere.
• Checklist AvailableintheInteractiveTextbook,usethechecklisttotrackyourunderstandingofthelearningintentions. PrintablePDFandworddocumentversionsarealsoavailablethere.
1 Howmanywayscan8peoplelineupinaqueue?
2 Byunrollingthefactorial(seeBox2inSection14A)simplify:
3 Useyourcalculatortoevaluate:
4 Howmanyfour-letterwords,withnorepeatedletters,canbeformedfromthelettersofJACKSON?
5 Anumberplateinacertaincountryconsistsof3lettersA–Zfollowedby4digits0–9.Howmanynumber platesarepossible?
6 FindhowmanyarrangementsofthelettersofthewordFOUNDERarepossible:
a ifthevowelsandconsonantsmustalternate, b ifthewordmuststartwithNandendwithD,
c ifalltheconsonantsmustbeinagroupattheendoftheword, d iftheRissomewheretotherightoftheU.
7 Fiveboysandfivegirlsaretositinarow.Findhowmanywaysthiscanbedoneif:
a therearenorestrictions,
b theboysandgirlssitindistinctgroups,
c aparticularboyandgirlmustsittogether.
8 HowmanywayscanthelettersofREPORTERbearranged?
9 HowmanywordsofthreeorfourlettersmaybeformedusingthelettersofSAMUEL?
10 Aquizconsistsoftenquestions,eachtakingtheanswerYesorNo.Howmanywaysisitpossibletoget6 correctand4wronganswers?
11 Acommitteeofsevenistobechosenfromsixmenandtenwomen.Findhowmanycommitteesare possibleif: therearenorestrictions, a allmembersaretobefemale, b allmembersaretobemale, c therearetobeexactlytwomen, d therearetobefourwomenandthreemen, e thereistobeamajorityofwomen, f aparticularmanmustbeincluded, g aparticularmanmustnotbeincluded, h Mustafarefusestobeonacommittee withYingYue.
12 Eightpeoplearriveatarestaurant.Findhowmanywayscantheybeassignedto: a alargetableforfiveandasmallertableforthree, b twoquitedifferenttablesforfour, c twoindistinguishabletablesforfour.
13 Acommitteeofsixisformedatrandomfromfourmenandthreewomen.Whatistheprobabilitythatitwill havemorementhanwomen?
14 Fromastandardpackof52cards,threeareselectedatrandom.Findtheprobabilitythat: theyarethequeenofspades,thethree ofclubsandthenineofhearts, a allthreearekings, b theyareallclubs, c theyareallofthesamesuit, d oneisredandtwoareblack, e oneisathree,oneisafive,andoneisaneight, f twoarefivesandoneisaseven, g atleasttwoofthemarespades. h
15 Threeboysandthreegirlsarearrangedinacircle.Inhowmanywayscanthisbedone: a iftherearenorestrictions, b iftheboysandthegirlsalternate, c iftheboysandgirlsareindistinctgroups, d ifaparticularboyandgirlwishtositnexttooneanother, e iftwoparticularboysdonotwishtositnexttooneanother, f ifaparticularboywantstositoppositeaparticulargirl?
InChapter11wediscussedthefactoringofapolynomialintoirreduciblefactors,sothatitcouldbewrittenina formsuchas
P(x) = (x 4)2(x + 1)3(x 2 + x + 1).
Inthischapterwewillnowstudyinmoredetailtheindividualbinomialpowerfactorssuchas(x 4)2 and (x + 1)3 thatappearinsuchafactoring,andtheirexpansions.Forexample,wehavealreadyseenthat
(x + 1)3 = x 3 + 3x 2 + 3x + 1.
Thecoefficientsinthegeneralexpansionof(x + y)n willbeinvestigatedthroughthepatternstheyformwhen theyarewrittendownin Pascal’striangle.Itturnsoutthatbasiccountingmethodsofthepreviouschapterare essentialforunderstandingtheseexpansions.
Pascal’striangledisplaysinaclearvisualformtheinterrelationshipsbetweenthecountingmethodsof Chapter17andthebinomialexpansionsofthischapter.Thereareagreatnumberofsymmetriesandother patterns.Sections18Eand18Finvestigatethesepatternsusingvariouscombinatoricandalgebraicmethods.
BinomialexpansionswillbeusedintheYear12topicofbinomialdistributions
Learningintentions
• Investigatethebinomialexpansions(1 + x)n,forlowwholenumbers n
• UsetheseexpansionstoconstructthefirstfewrowsofPascal’striangle.
• Noticeandexplainpatterns,suchastheadditionproperty,inPascal’striangle.
• UsePascal’striangletosolveproblemsinvolvingabinomialexpansion.
A binomial isapolynomialwithtwoterms,suchas1 + x or3x4 1 2 x2 .
A binomialexpansion istheexpansionofapowerofabinomial,forexample
(1 + x)3 = 1 + 3x + 3x 2 + x 3
Thisfirstsectionisrestrictedtotheexpansionof(1 + x)n andtothevarioustechniquesarisingfromsuch expansions.ThetechniquesarebasedonPascal’striangleanditsthreemostbasicproperties.
Someexpansionsof (1 + x)n
Herearetheexpansionsof(1 + x)n forlowvaluesof n.Thecalculationshavebeencarriedoutusingtworows sothatliketermscanbewrittenaboveeachotherincolumns.Inthisway,theprocessbywhichthecoefficients buildupcanbefollowedbetter.
Examinethelastexpansioninparticular,andworkoutwhytheboxed4 isthesumoftheboxed1 and3
(1 + x)0 = 1
(1 + x)1 = 1 + x (1 + x)2 = 1(1 + x) + x(1 + x) = 1 + x + x + x 2 = 1 + 2x + x 2
Noticehowtheexpansionof(1 + x)2 has3terms,thatof(1 + x)3 has4terms,andsoon.Ingeneral,theexpansion of(1 + x)n has n + 1terms,fromtheconstanttermin x0 = 1tothetermin xn.Becareful—thisisinclusive counting—thereare n + 1numbersfrom0to n inclusive.
Note: Thepower00 isundefined,andasaconsequence,thereshouldbecontinualqualificationstoavoidit. Butasiscustomaryinthistopic,wewillomitthemall,leavingthereadertointerpretsituationswherea problemmayarise.
Pascal’striangleandtheadditionproperty
Whenthecoefficientsintheexpansionsof(1 + x)n arearrangedinatable,theresultisknownas Pascal’striangle Thetablebelowcontainsthefirstfiverowsofthetriangle,copiedfromtheexpansionsabove,plusthenextfour rows,obtainedbycontinuingthesecalculationsupto(1 + x)8 .
Coefficientof:
1 1 11
2 121 3 133 1 4 1464 1
5 15101051
6 1615201561
7 172135352171
8 18285670562881
Threebasicpropertiesofthistriangleshouldquicklybecomeobvious.Theywillbeusedinthissection,and provenformallylater.
1ThreebasicpropertiesofPascal’striangle
1 Eachrowstartsandendswith1.
2 Eachrowisreversible.Thatis,therowsaresymmetric.
3 [Theadditionproperty]Everynumberinthetriangle,apartfromthe1s,isthesumofthenumber directlyabove,andthenumberaboveandtotheleft.
Thefirsttwopropertiesshouldbereasonablyobviousafterlookingattheexpansionsatthestartofthesection. Thethirdproperty,calledthe additionproperty,however,needsattention.ThreenumbersinPascal’striangle abovehavebeenboxedasanexampleofthis—noticehow
3 + 1 = 4.
Thebinomialexpansionsonthelastpagewerewrittenwiththecolumnsaligned,andwiththeseparticular coefficientsboxed,tomakethispropertystandout.
• Thesum3 + 1 = 4arisesbecausethecoefficientof x3 intheexpansionof(1 + x)4 isthesumofthecoefficients of x3 and x2 intheexpansionof(1 + x)3 .
Pascal’strianglecanbeconstructedusingtheserules,andthefirstquestioninExercise18Aasksforthefirst thirteenrowstobecalculated.
Pascal’striangleisoftendrawninthelayouttotheright below,likeanequilateraltriangle.Thislayoutdisplaysthe left-rightsymmetrywell,butitislessusefulforourpurposes becausewewantthecolumnstolineupwiththepowersof x. Nevertheless,youshouldbeabletoworkfrombothlayoutsof thetriangle. 1 11
Pascaldidnotdiscoverthetriangle,althoughhedidwriteanimportanttreatiseonitin1653.Itwasknownto theancientIndianmathematicianPingalainthe2ndcenturyBC,andlatertomediaevalPersianandChinese mathematicians,butthefirstknownoccurrenceinEuropeisina1527bookonbusinesscalculationsbyPetrus Adrianus.BelowisPascal’sversionofthetrianglefromhistreatise,withyetanotherlayout.
UsingPascal’striangle
Examples1–5illustratevariouscalculationsinvolvingthecoefficientsof(1 + x)n forlowvaluesof n.
Example1 UsingPascal’striangletowriteoutexpansions
UsePascal’striangletowriteouttheexpansionsof:
b Beparticularlycarefulwiththealternatingsignsinthisexpansion.
c Thesignsalsoalternateinthisexpansion.
Example2 UsingPascal’striangletoapproximatepowers
Byexpandingthefirstfewtermsof(1 + 0.02)8,findanapproximationof1.028 correcttofivedecimalplaces.
Solution
(1 + 0.02)8 = 1 + 8 × 0.02 + 28 × (0.02)2 + 56 × (0.02)3 + 70 × (0.02)4 + = 1 + 0.16 + 0.0112 + 0.000448 + 0.00001120 + ≑ 1.17166.
Theremainingfourtermsaretoosmalltoaffectthefifthdecimalplace.
Example3 Examiningaproductofbinomialspowers
a Writeouttheexpansionof 1 + 5 x 2 ,thenwriteoutthefirstfourtermsintheexpansionof(1 x)8 .
b Hencefind,intheexpansionof 1 + 5 x 2 (1 x)8: thetermindependentof x, i thetermin x. ii
Solution
a 1 + 5 x 2 = 1 + 10x 1 + 25x 2 (1 x)8 = 1 8x + 28x 2 56x 3 +
b Henceintheexpansionof 1 + 5 x 2 (1 x)8: i constantterm = 1 × 1 + (10x 1) × ( 8x) + (25x 2) × (28x 2) = 1 80 + 700 = 621.
ii termin x = 1 × ( 8x) + (10x 1) × (28x 2) + (25x 2) × ( 56x 3) = 8x + 280x 1400x = 1128x.
Example4 Findingpronumeralsgivenaconditiononabinomialexpansion
a Writedownthetermsin x4 and x3 intheexpansionof(1 + 2kx)6
b Find k ifthesetermsin x4 and x3 havecoefficientsintheratio2:3.
Solution
a (1 + 2kx)6 = + 15(2kx)2 + 20(2kx)3 + 15(2kx)4 + = + 60k2 x 2 + 160k3 x 3 + 240k4 x 4 + (Alternatively,justwritedownthetwoterms.)
b Put 240k4 160k3 = 2 3 . Then 3 2 k = 2 3 k = 4 9
Example5 Aharderexample—expandingapowerofatrinomial
Expand(1 + x + x2)4 usingPascal’striangle,bywriting1
Solution
Note: Question1constructstherowsofPascal’striangleupto n = 12.Thisis‘yourcopy’ofPascal’striangle —keepitinaprominentplaceforconstantuseinthischapter.Yourcalculatormaygiveyouparticular values,butitdoesnotdisplaythepatterns.
1 CompletealltherowsofPascal’strianglefor n = 0,1,2,3,...,12.
2 UsingPascal’striangletoprovidethebinomialcoefficients,givetheexpansionsof: (1 + x)6 a (1 x)6 b (1 + x)9 c
3 Continuethecalculationsoftheexpansionsof(1 + x)n atthebeginningofthissection,expanding(1 + x)5 and(1 + x)6 inthesamemanner.Keepyourworkincolumns,sothattheadditionpropertyofPascal’s triangleisclear.
4 Findthespecifiedtermineachexpansion.
a For(1 + x)11 : i findthetermin x2 , ii findthetermin x8 .
c For(1 + 2x)6 : i findthetermin x4 , ii findthetermin x5 .
b For(1 x)7 : i findthetermin x3 , ii findthetermin x5 .
d For 1 3 x 4 : i findthetermin x 1 , ii findthetermin x 2 .
5 Expand(1 + x)9 and(1 + x)10,andshowthatthesumofthecoefficientsinthesecondexpansionistwicethe sumofthecoefficientsinthefirstexpansion.
6 Withoutexpanding,simplify:
7 Findthecoefficientof x4 intheexpansionof(1 x)4 + (1 x)5 + (1 x)6 .
8 Findintegers a and b suchthat: (1 + √3)5 = a +
9 Verifybydirectexpansion,andbytakingoutthecommonfactor,that:
10 Donotuseacalculatorinthisquestion.
a Expandthefirstfewtermsof(1 + x)6,henceevaluate1.0036 tofivedecimalplaces.
b Similarly,expand(1 4x)5,andhenceevaluate0.965 tofivedecimalplaces.
11ai Expand(1 + x)4 asfarasthetermin x2 .
ii Hencefindthecoefficientof x2 intheexpansionof(1 5x)(1 + x)4 .
bi Expand(1 + 2x)5 asfarasthetermin x3 .
ii Hencefindthecoefficientof x3 intheexpansionof(2 3x)(1 + 2x)5 .
ci Expand(1 3x)4 asfarasthetermin x3
ii Hencefindthecoefficientof x3 intheexpansionof(2 + x)2(1 3x)4
12a When(1 + 2x)5 isexpandedinincreasingpowersof x,thethirdandfourthtermsintheexpansionare equal.Findthevalueof x.
b When(1 + x)7 isexpandedinincreasingpowersof x,thesixthtermistheaverageofthefifthandseventh termsintheexpansion.Findthevalueof x.
13 Findthecoefficientof:
a x3 in(3 4x)(1 + x)4
b x in(1 + 3x + x2)(1 x)3
c x4 in(5 2x3)(1 + 2x)5
d x0 in 1 x 3 3 1 + 2 x 2
14 Determinethevalueofthetermindependentof x intheexpansionof: (1 + 2x)4 1 1 x2 6 a 1 x 3 5 1 + 2 x 3 b
15a Intheexpansionof(1 + x)6:
i findthetermin x2 , ii findthetermin x3 ,
iii findtheratioofthetermin x2 tothetermin x3 , iv findthevaluesof(i),(ii)and(iii)when x = 3.
b Intheexpansionof 1 + 2 3x 7 :
i findthetermin x 5 , ii findthetermin x 6 , iii findtheratioofthetermin x 5 tothetermin x 6 , iv findthevaluesof(i),(ii)and(iii)when x = 2.
16a Findthecoefficientsof x4 and x5 intheexpansionof(1 + kx)8.Hencefind k ifthesecoefficientsarein theratio1:4.
b Findthecoefficientsof x5 and x6 intheexpansionof(1 3 4 kx)9.Hencefind k ifthesecoefficientsare equal.
17 [PatternsinPascal’striangle]CheckthefollowingresultsusingthetriangleyouconstructedinQuestion1. (Someofthesewillbeprovenlater.)
a Thesumofthenumbersintherowbeginning1, n,...isequalto2n
b Ifthesecondmemberofarowisaprimenumber,allthenumbersinthatrowexcludingthe1sare divisiblebyit.
c [Thehockey-stickpattern]Startatthetop1ofanycolumnofPascal’striangle,andgoverticallydown anynumberofrows.Thenthesumofthesenumbersisthenumberinthenextrowdownandtotheright. Forexample,ifyoustartatthe1inthe x2 column,andadddownwards:
1 + 3 + 6 + 10 + 15 + 21 = 56,whichisonerowdowninthe x3 column.
d [Thepowersof11]Ifarowismadeintoasinglenumberbyusingeachelementasadigitofthenumber, thenumberisapowerof11(exceptthataftertherow1,4,6,4,1,thepatterngetsconfusedbycarrying).
e Findthediagonalandthecolumncontainingthetriangularnumbers,andshowthataddingadjacentpairs givesthesquarenumbers.
18 Bywriting(1 + x + 3x2)6 as(1 + A)6,where A = x + 3x2,expand(1 + x + 3x2)6 asfarasthetermin x3 Henceevaluate(1.0103)6 tofourdecimalplaces.
19 [ThePascalpyramid]Byconsideringtheexpansionof(1 + x + y)n,where0 ≤ n ≤ 4,calculatethefirstfive layersofthePascalpyramid.
Learningintentions
• Understandthatapolynomial,andabinomial,canhavemorethanonevariable.
• Understandthatabinomialexpansion(x + y)n ishomogeneousin x and y
• Understandthatthecoefficientsof(x + y)n alsocomefromPascal’striangle.
• Solveproblemswithgeneralbinomialexpansions.
Themoregeneralcaseoftheexpansionof(x + y)n issimilar.Because x and y arebothvariables,however,the symmetriesoftheexpansionwillbemoreobvious.
Aminorpointaboutlanguage—wearenowwideningtheterm polynomial toincludetermsmadeupofpowers ofanynumberofvariables.Thus5xy4 isamonomial, x + y isabinomial,and
are trinomials.
(x + y)n
Herearetheexpansionsof(x + y)n forlowvaluesof n.Again,thecalculationshavebeencarriedoutwithlike termswritteninthesamecolumnsothattheadditionpropertyisclear—seetheboxedcoefficients.
(x + y)0 = 1
(x + y)1 = x + y
(x + y)2 = x(x + y) + y(x + y) = x 2 + xy + xy + y 2 = x 2 +
First,thecoefficientsintheexpansionsareexactlythesameasbefore,soasbeforetheycanbetakenfrom Pascal’striangle.Thisiseasilyprovenbysubstituting y = 1—thisturnstheexpansionof(x + y)n intothe expansionof(x + 1)n
Secondly,thepatternfortheindicesof x and y isstraightforward.
• Theexpansionof(x + y)3 hasfourterms,andineachtermtheindicesof x and y arewholenumbersadding to3.
• Similarlytheexpansionof(x + y)4 hasfiveterms,andineachtermtheindicesof x and y arewholenumbers addingto4.
Moregenerally,ineachsuccessiveexpansion,beginningwith(x + y)0 = 1,thetermsofthepreviousexpansion aremultipliedfirstby x,andthenby y,sothesumofthetwoindicesgoesupby1.
2Theexpansionof (x + y)n ishomogeneousofdegree n in x and y together
• Theexpansionof(x + y)n has n + 1terms,andineachtermtheindicesof x and y arewholenumbers addingto n.
▷ Theexpansionof(x + y)n iscalled homogeneousofdegree n in x and y together,becauseineach term,thesumoftheindicesof x and y is n.
• Thecoefficientsintheexpansionof(x + y)n arethesameasthecoefficientsintheexpansionof(1 + x)n Proof:Substitute y = 1.
Usingthegeneralexpansion
Thegeneralexpansionof(x + y)n isappliedsimilarlytotheexpansionof(1 + x)n
Example6
Expandingageneralbinomial
UsePascal’striangletowriteouttheexpansionsof:
Example7 Findingparticulartermsinabinomialexpansion
UsePascal’striangletowriteouttheexpansionof(2x + x 2)6,leavingthetermsunsimplified.Hencefind: thetermindependentof x, a thetermin x 3 b
Solution
Constantterm = 15 × (2x)4 × (x 2)2 = 15 × 24 × x 4 × x 4 = 240 a Termin x 3 = 20 × (2x)3 × (x 2)3 = 20 × 23 × x 3 × x 6 = 160x 3 b
Example8 Findingaterminabinomialexpansion
Expand(2 3x)7 asfarasthetermin x2,andhencefindthetermin x2 intheexpansionof(5 + x)(2 3x)7 .
Solution
(2 3x)7 = 27 + 7 × 2
= 128 1344 x + 6048 x 2 .
)
Hencethetermin x2 intheexpansionof(5 + x)(2 3x)7 = 5 × 6048 x 2 x × 1344 x = 28896 x 2
Exercise18B FOUNDATION
1 UsePascal’striangletoexpand:
+
)4
2 UsePascal’striangletoexpand: (1 + x2)4 a (1 3x2)3 b (x2 + 2y3)6 c x 1 x 9 d (
3 Simplifywithoutexpandingthebrackets: a y5 + 5y4(x y) + 10y3(x y)2 + 10y2(x y)3 + 5y(x
c x3 + 3x2(2y x) + 3x(2y x)2 + (2y x)3
d (x + y)6 6(x + y)5(x y) + 15(x + y)4(x y)2 20(x + y)3(x y)3 + 15(x + y)2(x y)4
6(x + y)(x y)5 + (x y)6
4ai Expand(4 + x)5 asfarasthetermin x3
ii Hencefindthecoefficientof x3 intheexpansionof(3 x)(4 + x)5
bi Expand(1 2x)6 asfarasthetermin x4
ii Hencefindthecoefficientof x4 intheexpansionof(1 3x)(1 2x)6 .
ci Expand(3 y)7 asfarasthetermin y4 .
ii Hencefindthecoefficientof y4 intheexpansionof(1 y)2(3 y)7 .
5a Expandandsimplify(x + y)6 + (x y)6
b Hence(andwithoutacalculator)provethat
6 Findthecoefficientof: x3 in(2 5x)(x2 3)4 a x5 in(x2 3x + 11)(4 + x3)3 b x0 in(3 2x)2 x + 2 x 5 c x9 in(x + 2)3(x 2)7 d
DEVELOPMENT
7ai UsePascal’striangletoexpand(x + h)3 .
ii If f (x) = x3,simplify f (x + h) f (x).
iii Henceusethedefinition f ′(x) = limh→0 f (x + h) f (x) h todifferentiate x3
b Similarly,differentiate x5 fromfirstprinciples.
8a Showthat(3 + √5)6 + (3 √5)6 = 20608.
b Showthat(2 + √7)4 + (2 √7)4 isrational.
c If(√6 + √3)3 (√6 √3)3 = a√3,where a isaninteger,findthevalueof a
9 Showthat 1 (
ThensimplifytheexpressionusingPascal’striangle.
10 Expand(x + 2y)5 andhenceevaluate:
a (1.02)5 correcttotofivedecimalplaces,
b (0.98)5 correcttotofivedecimalplaces.
byputtingtheLHSoveracommondenominator.
11a Expand: x + 1 x 3 i x + 1 x 5 ii x + 1 x 7 iii
b Hence,if x + 1 x = 2,evaluate:
12 Findthecoefficientsof x and x 3 intheexpansionof 3x a x 5 .Hencefindthevaluesof a ifthese coefficientsareintheratio2:1.
13 Thecoefficientsofthetermsin a3 and a 3 intheexpansionof ma + n a2 6 areequal,where m and n are non-zerorealnumbers.Provethat m2 : n2 = 10:3.
14a Expand x + 1 x 6 .
b If U = x + 1 x ,express x6 + 1 x6 intheform U6 + AU4 + BU2 + C.Statethevaluesof A, B and C
15a Bystartingwith ((x + y) + z)3,expand(x + y + z)3 .
b Findthetermindependentof x intheexpansionof(x + 1 + x 1)4
16 [TheSierpinskitrianglefractal]
a Drawanequilateraltriangleofsidelength1unitonapieceofwhitepaper.Jointhemidpointsofthe sidesofthistriangletoformasmallertriangle.Colouritblack.Repeatthisprocessonallwhitetriangles thatremain.Whatdoyounotice?
b DrawupPascal’striangleintheshapeofanequilateraltriangle,thencolouralltheevennumbersblack andleavetheoddnumberswhite.Whatdoyounotice?Thispatternwillbemoreevidentifyoutakeat leastthefirst16rows—perhapsuseacomputerprogramtogenerate100rowsofPascal’striangle.
Learningintentions
• Provethebinomialtheoremusingcombinatorics.
• Applythebinomialtheoremtoproblemswithbinomialexpansions.
• Applythebinomialtheoremtoproblemswithbinomialexpansions.
ThereaderhasprobablyrealisedalreadythattheentriesinPascal’striangleareexactlythenumbers nCr that aroseinthepreviouschapterwhencounting r-membersubsetsofan n-memberset.Oncethishasbeenproven below,wecanwritethegeneralbinomialexpansionoutexplicitlyas
orusingthealternativenotation n r for‘n choose r’,
(
Thisresultisknownasthe binomialtheorem.Itgivesastrikingconnectionbetweenthecountingmethodsofthe previouschapterandthealgebraicstructuresinthischapter,showingyetagaintheunitybetweenalgebraand arithmetic.
Wehavealreadyestablishedthat (
where ∗ denotesthedifferentcoefficients.Whatremainstobeprovenisthatthecoefficientof xn r yr is nCr
Asusual,themoststraightforwardwaytoestablishthisisbylookingathowthingsworkinaspecialcase.For example,letusestablishwhythecoefficientof x2y2 intheexpansionof(x + y)4 is 4C2 = 6.
Expanding(x + y)4 = (x + y)(x + y)(x + y)(x + y)withoutcollectingliketerms,orevenapplyingthecommutative lawofmultiplication,gives24 = 16terms:
(x + y)4 = xxxx (thetermwith4 x s)
+ xxxy + xxyx + xyxx + yxxx (thetermswith3 x sand1 y)
+ xxyy + xyxy + yxxy + xyyx + yxyx + yyxx (2 x sand2 y s)
+ xyyy + yxyy + yyxy + yyyx (thetermswith1 x and3 y s)
+ yyyy (thetermwith4 y s)
whichthenbecomes x4 + 4x3y + 6x2y2 + 4xy3 + y4 aftercollectingliketerms.
Themiddlerowaboveshowsthatthecoefficientof x2y2 is6becausetherearesixtermswithtwo x sandtwo y s.
Whyaretheresixterms?Thereasonis:
• thesesixtermsareallthewordsformedwith4letters,2alikeofonekind,and2ofanother.
Weshowedinthelastchapterthatthenumberofsuchwordsis 4! 2! × 2! = 6.
Moregenerally,whenweexpand(x + y)n,thecoefficientof xr yn r isthenumberofwordsformedwith n letters, r alikeofonekindand n r alikeofanother.
Nowwecanbringinthetheoryofthelastchapter,whereweusedthesymbol nCr forthenumberof r-member subsetsofan n-memberset.
Otherwiseexpressed,thisisthenumberofunorderedselectionsof r objectschosenfrom n objects.Weshowed therethat
nCr = n! r! × (n r)! , andwealsoshowedthatthiswasequaltothenumberof n-letterwordswith r lettersalikeofonekindand n r lettersalikeofanother.Thismeansthatthecoefficientof xr yn r canbewrittenas nCr ,givingaconciseformof theexpansion:
3Thebinomialtheorem,andaformulafortheentriesinPascal’striangle
• Forallwholenumbers n, (x + y)
where nCr = n! r!(n r)! = numberof r-membersubsetsofan n-memberset,for r = 0,1,..., n.
• Thisgivesanotationandaformulaforeachbinomialexpansioncoefficient.
• WenowhavetwointerpretationsofPascal’striangle:
▷ Itisatableofcoefficientsinthebinomialexpansion.
▷ Itisatableofthenumberof r-membersubsetsofan n-memberset.
Wecanalsowrite nCr = n × (n 1) ×···× (n r + 1) 1 × 2 ×···× r .
Remembertoothatthenotations nCr and n r meanexactlythesamething.Butdon’tmixthetwonotationsupin thesameproblembecauseitlooksdreadful.
Wecalculated nCr inSection17E.Theseexamplesusetheformulafor nCr tocalculatecoefficientsinbinomial expansions.
Example9 Usingthebinomialtheoremtocalculatetermsandcoefficients
Usethebinomialtheoremtocalculate:
a thecoefficientof x8 in(1 + x)12 ,
b thetermin x3y4 in(x + y)7 ,
c thecoefficientof A5 B5 in(2A 3B)10 (factoredintoprimes),
d Thetermin an 4b4 in(a 2b)n
Solution
a (1 + x)12 = + 12C8 × 14 × x 8 + ,
socoefficientof x 8 = 12! 8! × 4! = 12 × 11 × 10 × 9 4 × 3 × 2 × 1 = 495.
b (x + y)7 = + 7 4 × x 3 × y 4 + (alternativenotationfor nCr ),
sotermin x 3 y 4 = 7! 4! × 3! x 3 y 4 = 7 × 6 × 5 3 × 2 × 1 x 3 y 4 = 35 x 3 y 4 .
c (2A 3B)10 =
socoefficientof A5
d (a 2b)n = + nC4 × an 4 × ( 2b)4 + ,
sotermin
Someimportantvaluesof nCr,andthesymmetryoftherowsofPascal’s triangle
Theformulaefor nCr for n = 0,1and2areimportantenoughtobememorised: nC0 = n! 0! n! = 1 nC1 = n
WesawinChapter17that nCr = nCn r for r = 0,1,2,..., n.Itfollowsthereforethat nCn = 1, nCn 1 = n and nCn 2 = 1 2 n(n 1).
4Someimportantvaluesof nCr ,andtherowsymmetryofPascal’striangle
• Forallwholenumbers n, nC0 = nCn = 1, nC1 = nCn 1 = n,
Inparticular,alltherowsofPascal’strianglebeginandendwith1.
• Forallwholenumbers n,andfor r = 0,1,2,..., n, nCr = nCn r
Thatis,therowsarereversible.
CheckallthisonyourcopyofPascal’striangle.
Findthevalueof n if:
nC2 = 55 a
Solution
= 29 b
Weknowthat nC0 = 1and nC1 = n and nC2 = 1 2 n(n 1).
1 2 n(n 1) = 55 n 2 n 110 = 0
(n 11)(n + 10) = 0
Because n ≥ 0, n = 11.
n + 2n + 2 =
2 + n 56 = 0 (n 7)(n + 8) = 0
Because n ≥ 0, n = 7.
1 Usetheresult nCr = n! r!(n r)! toevaluatethefollowing.Donotuseacalculator—youwillneedtounroll thefactorialsymbol.CheckyouranswersagainstPascal’striangle.
2 Repeatthepreviousquestionforthesebinomialcoefficients.Rememberthealternativenotation
Cr .
3 Useyourcalculatorbutton n Cr toevaluate:
4a Expand(1 + x)4,andhencewritedownthevaluesof
b Hencefind:
5 Usethevaluesof nCr fromPascal’striangletofind:
6a Evaluate: 8 3 and 8 5 , i 7 4 and 7 3 ii
b If n 3 = n 2 ,findthevalueof n.
c If 12 4 = 12 n ,findthevalueof n
7a ByevaluatingtheLHSandRHS,verifythefollowingresultfor n = 8and r = 3: n r 1 + n r = n + 1 r
b Usethisidentitytosolveeachequationfor n 5 3 + 5 4 = n 4 i n 7 + n 8 = 11 8 ii
8 Findthespecifiedtermorcoefficientineachexpansion.
a For(1 + x)18,findthetermin x5
b For(1 + 6x)12,findthetermin x9
c For(1 7x)8,findthetermin x3
d For(1 + 3x)7,findthecoefficientof x2
9 Findthespecifiedtermsineachexpansion.
a For(2 + x)7,findthetermin x2
b For(x + 1 2 y)14,findthetermin x9y5
c For( 1 2 x 3y2)11,findthetermin x10y2
d For(a b 1 2 )20,findthetermin a2b9
10a Find x 0ifthetermsin x10 and x11 intheexpansionof(5 + 2x)15 areequal.
b Find x 0ifthetermsin x13 and x14 intheexpansionof(2 3x)17 areequal.
11a Intheexpansionof(1 + x)16,findtheratioofthetermin x13 tothetermin x11
b Findtheratioofthecoefficientsof x14 and x5 intheexpansionof(1 + x)20
c Intheexpansionof(2 + x)18,findtheratioofthecoefficientsof x10 and x16
12a Usethebinomialtheoremtoobtainformulaefor: nC0 i nC1 ii
b Hencesolveeachequationfor n
iv
C2 + nC1 = 22
c Usetheformulafor nC2 toshowthat n
,andverifytheresultonthethirdcolumnof Pascal’striangle.
13 Theexpression(1 + ax)n isexpandedinincreasingpowersof x.Findthevaluesof a and n ifthefirstthree termsare:
14a Intheexpansionof(2 + 3x)n,thecoefficientsof x5 and x6 areintheratio4:9.Findthevalueof n
b Intheexpansionof(1 + 3x)n,thecoefficientsof x8 and x10 areintheratio1:2.Findthevalueof n
15 Intheexpansionof x + 1 x 40 x 1 x 40 ,findthetermindependentof x.Giveyouranswerintheform nCr , andalsocorrecttofoursignificantfigures.
16 [Divisibilityproblems]
a Usethebinomialtheoremtoshowthat7n + 2isdivisibleby3,where n isapositiveinteger.(Hint: Write7 = 6 + 1.)
b Usethebinomialtheoremtoshowthat5n + 3isdivisibleby4,where n isapositiveinteger.
c Supposethat b, c and n arepositiveintegers,and a = b + c.Usethebinomialexpansionof(b + c)n to showthat an bn 1(b + cn)isdivisibleby c2.Henceshowthat542 248 isdivisibleby9.
17a Usethebinomialtheoremtoexpand(x + h)n
b Henceusethedefinition f ′(x) = lim
h todifferentiate xn fromfirstprinciples.
18 Usetheformula nCr = n! r!(n r)! toprovethatifthesecondmemberofarowisaprimenumber,allthe numbersinthatrow,excludingthe1s,aredivisiblebyit.
19 [ThesegeometricalresultsshouldberelatedtothenumbersinPascal’striangle.]
a Placethreepointsonthecircumferenceofacircle.Howmanylinesegmentsandtrianglescanbeformed usingthesethreepoints?
b Placefourpointsonthecircumferenceofacircle.Howmanysegments,trianglesandquadrilateralscan beformedusingthesefourpoints?
c Whathappensiffivepointsareplacedonthecircle?
d Howmanypentagonscouldyouformifyouplacedsevenpointsonthecircumferenceofacircle?
20 [Amoregeneralformofthebinomialtheorem]
a Showthatthebinomialexpansioncanbewrittenas
b Inthisform,itcanbeshownthattheexpansionistruefornegativeorfractionalvaluesof n,provided thattheRHSisregardedasthelimitofaninfinitesumofpowersof x.Thisiscalledthe powerseries expansion of(1 + x)n.Assumingthis,generatethebinomialexpansionsof:
c Verify,usingyourexpansionsinparts(b)(i)and(ii),that x 1 1 x = 1 (1 x)2 .(Thisassumesthata powerseriescanbedifferentiatedterm-by-term.)
Learningintentions
• Identifyandusethegeneraltermofabinomialexpansion.
Thissectioninvolveswritingdownandusingthegeneraltermofabinomialexpansion.Thismakesiteasierto identifyparticularterms,andtocomparetwoexpansions,bothofwhicharerequiredinthefinaltwosections.
The generalterm ofabinomialexpansionisanalgebraicexpressionthatisafunctionof r,sothateveryterm isobtainedbysubstitutingtheappropriatevalueof r.Forexample,thegeneraltermofthestandardbinomial expansion
,orwecanchoosetouse
5Thegeneraltermofabinomialexpansion
a Findthegeneraltermintheexpansionof(x2 + 3x)7 .
b Intheanswerstothesethreeparts,giveeachcoefficientfactoredintoprimes:
Findthetermin x12 i Findthetermin x10 ii
Solution
a Generalterm = 7Cr × (x 2)7 r × (3x)r = 7Cr × x 14 2r × 3r × x r = 7Cr × 3r × x 14 r .
bi Toobtainthetermin x12 , put14 r = 12
r = 2.
Hencethetermin x12 = 7C2 × 32 × x 12 = 7 × 6 1 × 2 × 32 × x 12 = 33 × 7 × x 12 .
iii Tofindthetermin x6,put14 r = 6
r = 8.
Findthetermin x6 iii
ii Toobtainthetermin x10 , put14 r = 10
r = 4.
Hencethetermin x10 = 7C4 × 34 × x 10 = 7 × 6 × 5 1 × 2 × 3 × 34 × x 10 = 34 × 5 × 7 × x 10 .
Thisisimpossible,becauseevaluating 7Cr requires0 ≤ r ≤ 7,sothereisnosuchterm(oralternatively,the termin x6 iszero).
a Findthegeneraltermintheexpansionof(2x2 x 1)9 .
b Findthetermin x12,givingthecoefficientfactoredintoprimes.
c Findtheconstantterm.
d Findthetermoflowestindex(withnon-zerocoefficient).
Solution
a Generalterm =
b Toobtainthetermin x12,put18
(Itisoftenbesttoleavelargenumbersfactoredintoprimes,asinparts(b)and(c).)
c Toobtaintheconstantterm,put18
d Thelowestpossibleindexiswhen r = 9(lookat 9Cr ),andthistermis ( 1)9 × 9C9 × 20 × x 18 27 = x 9 , whichisobviousbysimplylookingattheoriginalexpansion.
1 Writedownthegeneraltermforeachbinomialexpansion.
2 Considertheexpansion x2 + 1 x 9
a Showthateachtermintheexpansionof x2 + 1 x 9 canbewrittenas 9Ci x18 3i .
b Hencefindthecoefficientsof: x3 i x 3 ii x0 iii
3 Intheexpansionof(2x3 + 3x 2)10,thegeneraltermis 10Ck 2x3 10 k 3x 2 k .
a Showthatthisgeneraltermcanbewrittenas 10Ck 210 k 3k x30 5k
b Hencefindthecoefficientsoftheseterms,givingyouranswersfactoredintoprimes: x10 i x 5 ii x0 iii
4a Showthatthegeneraltermintheexpansionof x 2 5 x 15 canbewrittenas 15C j ( 1) j 5 j 2 j 15 x 15 2 j
b Hencefind,withoutsimplifying,thecoefficientsof: x11 i x ii x 5 iii
5 Findthetermindependentof x ineachexpansion. x + 3 x 8 a 2x3 1 x 12 b
6 Findthecoefficientofthepowerof x specifiedineachexpansion. x15 in x3 2 x 9 a
theconstanttermin 1 2x3 + x 20 c
2x 8 d x 1 in(x 1 + 2 7 x)5 e
7
)
5x2 +
7 Determinethecoefficientsofthespecifiedtermineachexpansion.
a For(3 + x)(1 x)15,findthecoefficientofthetermin x4
b For(2 5x + x2)(1 + x)11,findthecoefficientofthetermin x9
c For(x 3)(x + 2)15,findthecoefficientofthetermin x12
d For(1 2x 4x2) 1 3 x 9 ,findthecoefficientofthetermin x0 .
8a Findthecoefficientof x intheexpansionof x + 1 x 5 x 1 x 4 .
b Findthecoefficientof x2 intheexpansionof x 1 x 9 x + 1 x 5 .
c Findthecoefficientof y 3 intheexpansionof y + 1 y 10 y 1 y 7 .
9a Intheexpansionof(2 + ax + bx2)(1 + x)13,thecoefficientsof x0 , x1 and x2 areallequalto2.Findthe valuesof a and b
b Intheexpansionof(1 + x)n,thecoefficientof x5 is1287.Findthevalueof n bytrialanderror,andhence findthecoefficientof x10 .
10 Showthattherewillalwaysbeatermindependentof x intheexpansionof x p + 1 x2 p 3n ,where n isa positiveinteger,andfindthatterm.
11a Writedownthetermin xr intheexpansionof(a bx)12
b Intheexpansionof(1 + x)(a bx)12,thecoefficientof x8 iszero.Findthevalueoftheratio a b insimplest form.
12a Bywritingitas (1 x) + x2 4 ,expand(1 x + x2)4 inascendingpowersof x asfarastheterm containing x4 .
b Intheexpansionof(1 + 3x + ax2)n,where n isapositiveinteger,thecoefficientof x2 is0.Find,interms of n:
thevalueof a, i thecoefficientof x3 . ii
Learningintentions
• ProveaPascal’striangleidentitybysubstitutingintoabinomialexpansion.
• ProveaPascal’striangleidentitybyequatingcoefficientsinabinomialexpansion.
ThereareagreatnumberofpatternsinPascal’striangle.Somearequitestraightforwardtorecogniseandto prove,othersaremorecomplicated.Theyareveryimportantinallapplicationsofthebinomialtheorem,andin particular,somewillbeimportantinYear12binomialdistributions.
Note: HereiswhereyouabsolutelyneedtoreferconstantlytothePascaltriangleupto n = 12thatyou calculatedinthefirstquestionofExercise18A.Thissectionandthenextareallaboutsearchingfor patternsinthetriangle.
EachpatterninPascal’striangleisdescribedbyanidentityonthebinomialcoefficients nCr .Theseidentities sometimeshavearatherforbiddingappearance,anditisimportanttotakethetimetointerpreteachidentityas somesortofpatterninPascal’striangle.Usesmallvaluesof n suchas n = 3, n = 4,and n = 5toworkoutwhat theidentityissaying.
HereagainisthefirstpartofPascal’striangleformedbythenumbers nCr .Eachidentitythatisobtainedshould beinterpretedasapatterninthetriangleandverifiedthere,eitherbeforeoraftertheproofiscompleted.
Thereareanumberofapproachestoprovingidentities,andmanypatternscanbeprovenbytwoormoreofthese approaches.Thissectioncoverstwoapproaches(plusathirdintheEnrichmentsection)—theseareallbasedon thebinomialexpansion.Thefinalsectioncoversanotherthree,basedmoreoncombinatorics.
Amethodofproof—Substitutingavalueintoabinomialexpansion
Substituteintosomeformofthebinomialexpansion
Example13 Provingidentitiesbysubstitutingintoabinomialexpansion
Obtainidentitiesbysubstitutingintothebasicformofthebinomialexpansionaswrittenoutabove:
ThenexplainwhatpatterneachidentitydescribesinselectedrowsofPascal’striangletriangle.
Solution
a Substituting x = 1and y = 1,andswappingsides,gives n
whichwasproveninSection17Ebycombinatorics.InPascal’striangle,thismeansthatthesumofevery rowis2n.Forexample,1
b Substituting x = 1and y = 1,andswappingsides,gives
Thismeansthatthealternatingsumofeveryrowiszero. Forodd n,thisistrivial:1 5 + 10 10
c Substituting x = 1and y = 2,andswappingsides,gives
Takingasanexampletherow1,4,6,4,1,
Amethodofproof—Equatingcoefficients
Thismethodinvolvestakingtwoequalexpansionsandequatingcoefficients. Thefollowingidentityisquitedramatic,becauseitshowshowsquaringthetermsrelateseachrowtotherow twiceasfardownthetriangle.
Example14 Provingidentitiesbyequatingcoefficients
a Taking
andexpandingandequatingconstants,provethat
b TheninterprettheidentityinPascal’striangle.
Solution
a OntheRHS,theconstanttermis
OntheLHS,thefirstfactor(x + x 1)n canbeexpandedas
sotheconstanttermsintheproductontheLHSarisefromproductssuchas
ThustheconstanttermontheLHSisthesumoftheproducts
andbecause nCn k = nCk ,bythesymmetryoftherow, theconstanttermontheLHSis(
Equatingthetwoconstanttermsat(1)and(2)givestherequiredidentity:
b Thismeansthatiftheentriesofanyrowaresquaredandadded,thesumisthemiddleentryintherow twiceasfardown.LookatthePascaltriangleonthefirstpageofthissection,andpickouttherowsindexed by3and6.
3 1331 6 1615201561
Thesumofthesquaresoftherowindexedby3is1
= 20,whichisthemiddleterm 6C3 of therowindexedby6.
Addinginsteadthesquaresoftheentriesintherowindexedby6,
whichis 12C6,hopefullycalculatedintheveryfirstquestionofExercise18A.
1 [Substitution]Beginwiththebinomialexpansionfor n = 4,
anduseittoprovethefollowingresults.Explaineachresultintermsoftherow14641indexedby n = 4inPascal’striangle.
a Bysubstituting x = 1,showthat
b Bysubstituting x = 1,showthat
c Hence,byusingtheresultofpart(a),showthat
2 [Substitution]Usethegeneralbinomialexpansion, (1 + x)n = nC0 + nC
toprovethefollowingresults.Explaineachresultintermsoftherow15101051indexedby n = 5inPascal’striangle.
a Bysubstituting x = 1,showthat
b Bysubstituting x = 1,showthat
c Hence,byusingtheresultofpart(a),showthat
3 [Substitution]ThisquestionfollowsthesamestepsasQuestion2,andusestheexpansion
a Showthat
b Showthat
c CheckbothresultsinPascal’striangle,using n = 3and n = 4.
4 [Comparingcoefficients]
a Usethebinomialexpansionof(1 + x)n towriteanexpansionof(1 + x)(1 +
b Byequatingcoefficientsonbothsidesintheidentity(1 + x)(1 + x)n = (1 + x)n+1 deduceanidentity involving:
c Whatidentityhaveyouwrittendown?
5 [Comparingcoefficients]
a Showthat nCk = nCn k .
bi Bycomparingcoefficientsof x10 onbothsidesof(1 + x)10(1 + x)10 = (1 + x)20,writedownanidentity involving
ii Writedownthevalueof(
c Bycomparingcoefficientsof xn onbothsidesoftheidentity(1 + x)
(1 +
)n = (1 + x)2n,showthat (nC0)2 + (nC1)2 + + (nCn)2 = 2nCn
d CheckthisidentityonthePascaltrianglebyaddingthesquaresoftherowsindexedby n = 1,2,3,4,5 and6.
6a Bycomparingcoefficientsof xn+1 onbothsidesof(1 + x)n(1
,showthat
(2n)! (n 1)!(n + 1)! .
b Checkthisidentityontherowsindexedby n = 3,4,5and6ofthePascaltriangle.
7 [Comparingcoefficients]
a Byequatingcoefficientsof x4 intheexpansionof (1 + x)4 (1 x)4 = 1 x2 4 ,provethat 4C0 2 4C1
b Generalisethisresult,andproveit,byconsideringtheexpansionof (1
c CheckyouridentityonthePascaltriangle,for n = 4,5and6.
8 [Comparingcoefficients]
a Byexpandingbothsidesoftheidentity(1 + x)n+4 = (1 + x)n(1 + x)4,showthat n + 4
andstatethenecessaryrestrictionon r
b CheckthisidentityinPascal’striangle,using n = r = 5andusing n = 6and r = 4.
9 [Substitution]Usetheexpansionof(1 + x)2n,toprovethat 2nC0 + 2nC
10 [Comparingcoefficients]
a Findthecoefficientof xn+r intheexpansionof(1 + x)3n .
b Bywriting(1 + x)3
,provethatfor0
c CheckthisidentityinPascal’striangle,using n = 4and r = 3.
11 [Comparingcoefficients]
a Showthat xn(1 + x)n 1 + 1 x n =
b Henceprovethat1
CHALLENGE
ThenextquestionfollowsthesubstitutionmethodofQuestion1–2,butappliescalculustothebinomial expansionfirst.Ifyouhaven’tyetdonedifferentiation,youwillneedtoskipthisquestionforthemoment. Thisapproachofdifferentiation-then-substitutionisnotinthecourse,butitpresentsnoproblemsonceyou havedonedifferentiation.WehaveincludedithereinEnrichmentbecausesomeoftheresultingidentities areneededinYear12binomialdistributions.
12 [Substitutionanddifferentiation]Usethegeneralbinomialexpansion, (1 + x)n = nC0 +
toprovethefollowingresults.Explaineachresultintermsoftherow15101051indexedby n = 5inPascal’striangle.
a Differentiatebothsidesofthebinomialexpansion(∗)above.
b Bysubstituting x = 1,showthat1 × nC1 + 2 × nC2 + + n × nCn = n2n 1
c Bysubstituting x = 1,showthat1 × nC1 2 × nC2 + + ( 1)n 1n × nCn = 0.
13a Usetheresultsofthepreviousquestiontoshowthat
b CheckthisidentityinPascal’striangle,using n = 4,5and6.
14a Byconsideringtheidentity(1
b Whatisthecorrespondingresultif n isodd?
15 If(1 +
Learningintentions
• ProveaPascal’striangleidentityusingcombinatoricsassociatedwith nCr
• ProveaPascal’striangleidentityusingthefactorialexpressionof nCr
• ProveaPascal’striangleidentityusingpreviouslyprovenidentities.
ThisfinalsectionprovesPascal’striangleidentitiesusingthreemethodsthatareclosertothecombinatoricsand factorialsofChapter17.
Thereaderisthusinvitedtohaveadoubleviewofbinomialexpansions.Ontheonehand,theyarealgebraic expansions,withanappearancecomplicatedbythemanytermsintheexpansion.Ontheotherhand,theyare veryclosetothecombinatorialsituationpresentedinChapter17,becauseeachcoefficient nCr isthenumberof r-membersubsetsofan n-memberset.
Thefollowingexampledisplaysthecombinatoricsapproachwell,butitcanalsobedonebyequatingcoefficients. Thisallowsacomparisonofmethods.
a Ihave4twenty-centpiecesand6five-centpiecesinmypocket.Howmanysubsetsof3coinsaretherein mypocket?
i Ignoringthetwotypesofcoins,writetheresultas nCr .
ii Countseparatelythesubsetswith0,1,2and3five-centcoins.
iii WhatPascal’striangleidentityhaveyouproven?
b Provetheidentitybyequatingappropriatecoefficientsin (1 + x)4(1 + x)6 = (1 + x)10 .
c UsethevaluesfromPascal’striangletoverifytheidentity.
d IdentifyvisuallyalltogethertheentriesinthePascalTriangle,identifythepattern,andtrytoreproduceit elsewhereonthetriangle.
e Suggestageneralisationoftheidentity.
ai Numberofsubsets = 10C3.
ii Each3-coinsubsethas3twenty-centsand0five-cents,or2twenty-centsand1five-cent,or1 twenty-centand2five-cents,or0twenty-centsand3five-cents.Hence
Numberofsubsets = 4
b Thetermin x3 ontherightis 10C3 x3 .
Thetermin x3 ontheleftisthesumofthefourproducts:
Thisprovesthesameidentity.
c SubstitutingintotheidentityfromPascal’striangle(yourcopyorSection18E),
LHS = 4 × 1 + 6 × 6 + 4 × 15 + 1 × 20 = 4 + 36 + 60 + 20 = 120 = RHS.
d Visualisationislefttothereader.
e Hereisoneofmanypossiblegeneralisations.For n ≥ m ≥ 3,
TestitoutonthePascaltrianglefor n = 5and m = 4. Thenperhapsproveitbyanalogywithparts(a)or(b).
Amethodofproof—Writing nCr intermsoffactorials
Thismethodisnotparticularlyeasytofollow,butthefollowingexampleisthebest-knownapplicationofthe method.Again,theresulthasotherproofs,anditisinterestingtocomparethem.
Example16 Provingidentitiesbywriting nCr intermsoffactorials
a BywritingLHSandRHSintermsoffactorials,provethe additionproperty: nCr = n 1Cr 1 + n 1Cr ,for
1and1
b Giveacombinatorialproofoftheadditionproperty.
c Giveaproofcomparingcoefficients.
Solution
a RHS = (n 1)! (r 1)! × (n r)! + (n 1)! r!
= (n 1)! (r 1)! × (n r 1)! × 1 1
(
) + 1 r × 1 = (n 1)! (r 1)! × (n r 1)! × r r × (n r) + n r
= (n 1)! (r 1)! × (n r 1)! × n r × (n r) = n! r! × (n r)! = LHS
b Represent nCr asthenumberof r-personsubsetsofacrowdof n people. ThereisblokecalledGeoffreyinthecrowd, andevery r-personsubseteithercontainsGeoffrey,ordoesnotcontainGeoffrey.
Numberof r-membersubsetsthatcontainGeoffrey = n 1Cr 1 , (1) becauseweneedtochooseanother r 1peopleoutof n 1people.
Numberof r-membersubsetsnotcontainingGeoffrey = n 1Cr , (2) becauseweneedtochooseall r peopleoutof n 1people.
Addingthecountsin(1)and(2)gives nCr = n 1Cr 1 + n 1Cr
c Considerthetermsin xn intheexpansionof(1 + x)n = (1 + x)n 1 × (1 + x).
Termin xr ontheLHS = nCr x r
Termin xr ontheRHS = n
Equatingthecoefficientsof
Amethodofproof—Usingpreviouslyprovenidentities
Occasionallyapreviouslyprovenidentitycanbeusedinaproof.
Example17 Provingidentitiesbyusingpreviouslyprovenidentities
a Usetheadditionpropertytoprovethat
n+1C2 nC2 = n,forall n ≥ 2.
b Henceshowthat n
c Interpretpart(b)onthePascaltriangle.
Solution
a Bytheadditionproperty,
b Thisproofuses recursion,whichwillbecome mathematicalinduction inYear12. First,wecanrewritetheresulttopart(a)as
Thenapplyingthisformularepeatedly,
(aftermanymoresteps)
c LookatthethirdcolumnofthePascaltriangle(yourcopyorSection18E):
Note: Thenumbers1,3,6,10,15,...arecalledthe triangularnumbers becausetheyare thesuccessivetotalnumbersofdotsin1,2,3,4,5,...rowsofatriangulararray. WewillmeetthemagainwithsequencesandseriesinYear12.
Remarksonthevariousmethodsofproof
Thereaderhasprobablygainedsomeimpressionsaboutthesevariousmethods.
• Substitutionisverysimple,butdoesnotdevelopmanynewideas.
• Equatingcoefficientsandcombinatoricapproachesarebothexcellentmethods,andgenerateinterestingnew identities.
• Expandingfactorialscanbeveryfiddly.
• Usingpreviouslyprovenresultsmaybeusefulinsomesituations,asinallmathematics.
6SomemethodsofprovingidentitiesinPascal’striangle
• Substitutevaluesintoabinomialexpansion.
• Equatecoefficientsintwoequalbinomialexpansions.
• Usecombinatoricarguments.
• Expandthefactorialsintheexpressionsfor nCr
• Usepreviouslyprovenidentities.
• Differentiation-then-substitution(seeExercise18EEnrichment)isnotinthecourse,butitisstraightforward, andgeneratesmanyinterestingidentities,someofwhichwillbeneededinYear12binomialdistributions.
Forthenextthreequestions,youshouldusethedefinitionofthesymbol nCr asthenumberofwaysofchoosing an r-membersubsetfroman n-memberset.
1 [Combinatorics]Considertheset S = {A,B,C,D,E}
a Considerthesubset {A,B}.Whatlettershavebeenomittedfromthesubsetthatareinthewholeset?
b Listallthesubsetsof {A,B,C,D,E} containing2letters.
c Alongsideeachofthesetsin(a)writedownthesetthatdoesNOTcontainthosetwoletters.
d Usethecombinatorialdefinitionof nCr andparts(a)-(c)toexplainwhy 5C2 = 5C3.
e Explainwhy nCr = nCn r foranywholenumbers n and r with r ≤ n.
2 [Combinatorics]Considertheset S = {A,B,C,D}
a Writedownallthesubsetsof S ,thenexplainwhythereareexactly24 ofthem.
b Henceexplainwhy 4 0 + 4 1 + 4 2 + 4
c Explainwhy n 0 + n
= 2n,foranywholenumber n.
3 [Acombinatoricsproofoftheadditionproperty]Considerthesets
S = {A,B,C,D,E}and U = {A,B,C,D}.
ai Writedownallthe3-lettersubsetsof S thatdonotnotcontain E
ii Explainwhythisisalistofallthe3-lettersubsetsof U
bi Writedownallthe3-lettersubsetsof S thatcontain E
ii Explainhowtopairthemupwithallthe2-lettersubsetsof U
c Henceexplainwhy 4C2 + 4C3 = 5C3.
d Explainwhy nCr 1 + nCr = n+1Cr ,foranywholenumbers n and r with1 ≤ r ≤ n.
4 [Theadditionpropertyandtheformulafor nCr ]Inthisquestion,youwillprovethatif a, b, c and d areany fourconsecutivetermsinanyrowofPascal’striangle,then a a + b + c c + d = 2b b + c
a Considertherow1,7,21,35,35,21,7,1indexedby n = 7.Showthattheidentityholdsforeach sequence a, b, c, d offourconsecutivetermsfromthisrow.
b Choosefourconsecutivetermsfromanyotherrowandshowthattheidentityholds.
c Provetheidentitybyletting a =
and d = nCr+2.Youwillneedtousethe additionproperty,thentheformula nC
5 [Twocombinatoricsproofs]Wehaveseeninquestion2(a)thatsubstituting x = 1and x = 1intothe binomialexpansionproves
Herearecombinatoricsproofsoftheseresults.
a Let S bean n-memberset,andinterpreteach nCr asthenumberof r-membersubsetsof S .Henceprove thefirstidentity(1).
b Toprovethesecondidentity(2),chooseafixedelementAintheset S .Pairupeachsubset U not containingAwiththeuniquesubset U ∪{A} containingA.
i Explainwhytheprocedurearrangesallthesubsetsof S uniquelyintopairs.
ii Explainwhyonememberofeachpairhasanevennumberofmembers,andtheotherhasanodd numberofmembers.
iii Henceprovethat nC
6 [Theformulafor nCr ]Thisquestioninvolvestheformula
a Provethat
b Whatresulthaveyouproven?
7 [Aninequalityprovenusingtheformulafor nCr ]
a Provethat 2nCr < 2nCr+1,forallwholenumbers n and r with r < n
b Provethat 2n+1Cr < 2n+1Cr+1,forallwholenumbers n and r with r
DEVELOPMENT
8 [Thehockey-stickidentity]Lookatthecolumnindexedby r = 2inPascal’striangle:
Thegeneralformofthiswell-known hockey-stickidentity is
where n and r arewholenumberswith0 ≤ r ≤ n.Hereareproofsofthisidentityusingtheaddition property,andusingacombinatorialproof.
a [Aproofusingtheadditionproperty]
i Toprovetheparticularresult(∗),startwiththeright-handsideandwrite
thenkeepexpandingthetermwith r = 3.Tocompletetheproof,youwillneedtousethefactthat 3C3 = 2C2 = 1.
ii Generalisethistoaproofoftheidentity(∗∗).
b [Acombinatoricsproof]
i Thenumber 7C3 isthenumberof3-membersubsetsofa7-memberset S .Tochoosea3-member subset U of S = {1,2,3,4,5,6,7},makethechoiceinthefollowingway:
• Firstchoosefrom S thegreatestnumber k thatwillbeinthesubset U.Thismustbeoneofthe numbers3,4,5,6or7,because U istohavethreemembers,soitwillhavetwonumberssmaller than k.
• Thenchoosetheremainingtwonumbersin U fromthe k 1possiblenumbers1,2,..., k 1.
Explainwhythismethodofchoosingthesubsetyieldstheidentity(∗).
ii Generalisethistoaproofoftheidentity(∗∗).
9 [Acombinatorialapproachtothesumofsquaresinarow]Inthisquestionweconsiderbinarywords consistingonlyofthelettersAandB.
a Considerabinarywordconsistingof a + b letters,withAoccuring a timesandBoccuring b times.Show thatthereare a+bCa = a+bCb permutationsofsuchaword.
b Howmanypossiblepermutationsarethereofabinarywordwith2n letters,ifAandBbothoccur n times?
c Awordwith2n lettersmaybesplitdownthemiddleintotwowordsof n letters.Considertheexample wheretwoAsfallinthefirst n-letterword.
i Howmanyarrangementsarethereofthefirst n-letterbinarywordwithtwoAs?
ii Howmanyarrangementsarethereofthesecond n-letterbinarywordwith n 2oftheAsand2Bs?
iii Howmanyarrangementsarethereofaten-letterbinarywordwithtwoAsinthefirsthalfandtwoBs inthesecondhalf?
d Henceprovethat
10a Whentheentriesoftherow1,5,10,10,5,1indexedby n = 5inPascal’strianglearemultipliedby0, 1,2,3,4,5respectively,theresultsare0,5,20,30,20,5.Ignoringthezero,thisisfivetimestherow1, 4,6,4,1.Formulatethisresultalgebraically,for n = 5andthenforgenerally n,andproveitusingthe binomialtheorem.
b Whentheentriesoftherow1,5,10,10,5,1aredividedby1,2,3,4,5and6respectively,theresultis1, 2 1 2 ,3 1 2 ,2 1 2 ,1, 1 6 .Ifyouadd 1 6 atthestart,thisis 1 6 thoftherow1,6,15,20,15,6,1.Formulatethisresult algebraically,for n = 5andthenforgeneral n,andproveitusingthebinomialtheorem.
11a Findthevalueof
b Evaluate
+ 2
c Provethefollowingidentity,andverifyitusingtherowindexedby n = 4:
12ai If r > p + 1,showthat
ii Hencededucethatfor n
iii WhatisthesignificanceofthisresultinthePascal’striangle?
• Createyourownsummaryofthischapteronpaperorinadigitaldocument.
• Thisautomatically-markedquizisaccessedintheInteractiveTextbook.AprintablePDFWorksheetversionis alsoavailablethere.
• Checklist AvailableintheInteractiveTextbook,usethechecklisttotrackyourunderstandingofthelearningintentions. PrintablePDFandworddocumentversionsarealsoavailablethere.
1 WriteoutthefirstfiverowsofPascal’striangle.
2 Useyouranswertothepreviousquestiontoexpand:
+ x)5
3a Expand(1 + 7x)5 asfarasthetermin x2
b Hencefindthecoefficientof x2 intheexpansionof(1
4a Expand(1 + x)7 .
b Hencefindthefirstdecimalplaceof1.027 .
5 Expand:
6 Usetheresult n
toevaluateeachbinomialcoefficient.Donotuseacalculator—youwill needtounrollthefactorialsymbol.CheckyouranswersagainstthecopyofPascal’strianglethatyou developedinExercise15A.
7 Useyourcalculatortoevaluate:
8 UseyourunderstandingofthepatternsinPascal’striangletosimplify:
9 Solve nC2 = 28.
10 Findthecoefficientof x8 intheexpansionof(1 3x)(1 + 5x)14
11a Writeoutthefirstfewtermsintheexpansionof(1 + x)n .
b Byanappropriatesubstitution,provethat: n 0 + 2 × n 1 + 4 × n 2 + 8 × n 3 + ··· = 3n
c Verifythisresultfortherowindexedby n = 4inPascal’striangle.
12a Provethat nC0 + nC1 + nC2 + nC3 + ··· = 2n .
b WhatisthesignificanceofthisresultforPascal’striangle?
c Howmaythisresultbeinterpretedasasumofcombinations?
13a Showthatthegeneraltermintheexpansion(2x + 1 x2 )18 canbewritten: 18Cr 218 r x 18 3r
b Hencefindinthisexpansion:
i thetermindependentof x
ii thecoefficientof x9
iii thetermin x9
14 Inthisquestionweshallprovetheidentity
a Checkthisidentityinthecase n = 4and r = 0.
b Prove(∗)usingtheadditionproperty nCr + nCr+1 = n+1Cr+1.
c Prove(∗)byequatingcoefficientsof xr+2 intheidentity(1 +
d Acommitteeof8istobeformedfrom12people.
i Howmanycommitteesof8canbeformed? Thegroupof12includesBradandJanet.
ii HowmanycontainneitherBradnorJanet?
iii HowmanycontainoneofeitherBradorJanet?
iv Howmanycontainboth?
e Usepart(d)toprove(∗)inthecase n = 10and r = 6.Generaliseyourargument.