案例数学试卷

CAMBRIDGEINTERNATIONALEXAMINATIONS

GCEAdvancedSubsidiaryLevelandGCEAdvancedLevel

MARKSCHEMEfortheOctober/November2012series

9709MATHEMATICS

9709/42

Paper4,

umrawmark50

Thismarkschemeispublishedasanaidtoteachersandcandidates,toindicatetherequirementsoftheexamination.ItshowsthebasisonwhichExaminerswereinstructedtoawardmarks.ItdoesnotindicatethedetailsofthediscussionsthattookceatanExaminers’meetingbeforemarkingbegan,whichwouldhaveconsideredtheacceptabilityofalternativeanswers.

MarkschemesshouldbereadinconjunctionwiththequestionpaperandthePrincipalExaminerReportforTeachers.

Cambridgewillnotenterintodiscussionsaboutthesemarkschemes.

CambridgeispublishingthemarkschemesfortheOctober/November2012seriesformostIGCSE,GCEAdvancedLevelandAdvancedSubsidiaryLevelcomponentsandsomeOrdinaryLevelcomponents.

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MarkScheme

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MarkSchemeNotes

Marksareofthefollowingthreetypes:

MMethodmark,awardedforavalidmethodappliedtotheproblem.Methodmarksarenotlostfornumericalerrors,algebraicslipsorerrorsinunits.However,itisnotusuallysufficientforacandidatejusttoindicateanintentionofusingsomemethodorjusttoquoteaformula;theformulaorideamustbeappliedtothespecificprobleminhand,e.g.bysubstitutingtherelevanttiesintotheformula.CorrectapplicationofaformulawithouttheformulabeingquotedobviouslyearnstheMmarkandinsomecasesanMmarkcanbeimpliedfromacorrectanswer.

Accuracymark,awardedforacorrectanswerorintermediatestepcorrectlyobtained.Accuracymarkscannotbegivenunlesstheassociatedmethodmarkisearned(orimplied).

Markforacorrectresultorstatementindependentofmethodmarks.

Whenapartofaquestionhastwoormore“method”steps,theMmarksaregenerallyindependentunlesstheschemespecificallysaysotherwise;andsimilarlywhenthereareseveralBmarksallocated.ThenotationDMorDB(ordep*)isusedtoindicatethataparticularMorBmarkisdependentonanearlierMorB(asterisked)markinthescheme.Whentwoormorestepsareruntogetherbythecandidate,theearliermarksareimpliedandfullcreditisgiven.

ThesymbolimpliesthattheAorBmarkindicatedisallowedforworkcorrectlyfollowingonfrompreviouslyincorrectresults.Otherwise,AorBmarksaregivenforcorrectworkonly.AandBmarksarenotgivenforfortuitously“correct”answersorresultsobtainedfromincorrectworking.

Note: B2orA2meansthatthecandidatecanearn2or0.

B2/1/0meansthatthecandidatecanearnanythingfrom0to2.

Themarksindicatedintheschememaynotbesubdivided.Ifthereisgenuinedoubtwhetheracandidatehasearnedamark,allowthecandidatethebenefitofthedoubt.Unlessotherwiseindicated,marksoncegainedcannotsubsequentlybelost,e.g.wrongworkingfollowingacorrectformofanswerisignored.

Wrongormissingunitsinananswershouldnotleadtothelossofamarkunlesstheschemespecificallyindicatesotherwise.

Foranumericalanswer,allowtheAorBmarkifavalueisobtainedwhichiscorrectto3s.f.,orwhichwouldbecorrectto3s.f.ifrounded(1d.p.inthecaseofanangle).Asstatedabove,anAorBmarkisnotgivenifacorrectnumericalanswerarisesfortuitouslyfromincorrectworking.ForMechanicsquestions,allowAorBmarksforcorrectanswerswhicharisefromtakinggequalto9.8or9.81insteadof10.

Thefollowingabbreviationsmaybeusedinamarkschemeorusedonthescripts:AEF AnyEquivalentForm(ofanswerisequallyacceptable)

AG AnswerGivenonthequestionpaper(soextracheckingisneededtoensurethatthedetailedworkingleadingtotheresultisvalid)

BODBenefitofDoubt(allowedwhenthevalidityofasolutionmaynotbeabsoluyclear)

CAOCorrectAnswerOnly(emphasisingthatno“followthrough”fromapreviouserrorisallowed)

CWO CorrectWorkingOnly–oftenwrittenbya‘fortuitous’answerISW IgnoreSubsequentWorking

MR Misread

PA PrematureApproximation(resultinginbasicallycorrectworkthatisinsufficientlyaccurate)

SOS SeeOtherSolution(thecandidatemakesabetttemptatthesamequestion)

SR SpecialRuling(detailingthemarktobegivenforaspecificwrongsolution,oracasewheresomestandardmarkingpracticeistobevariedinthelightofaparticularcircumstance)

Penalties

MR–1ApenaltyofMR–1isdeductedfromAorBmarkswhenthedataofaquestionorpartquestionaregenuinelymisreadandtheobjectanddifficultyofthequestionremainunaltered.InthiscaseallAandBmarksthen e“followthrough”marks.MRisnotappliedwhenthecandidatemisreadshisownfigures–thisisregardedasanerrorinaccuracy.AnMR–2penaltymaybeappliedinparticularcasesifagreedatthecoordinationmeeting.

PA–1ThisisdeductedfromAorBmarksinthecaseofprematureapproximation.ThePA–1penaltyisusuallydiscussedatthemeeting.

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1 M1 ForusingWD=Fdcosα

WD=45×25cos14o A1

Workdoneis1090J(1.09kJ) A1 3

2 (i) [0.6=0+0.3a] M1

Accelerationis2ms2 A1 2

(ii) [mg–T=2m,T–(1–m)g

=2(1–m)] M1

[m=T/8T–(10–1.25T)=2–0.25T

or

T=8m8m–(10–10m)=2–2m] M1

T+1.25T+0.25T=10+2

or

m=0.6andT=8m A1

m=0.6andtensionis4.8N A1 4

Forusingv=0+at

ForapplyingNewton’s2ndlawtoAortoB

Foreliminatingorevaluatingm

Alternativeforpart(ii)

[{m+(1–m)}×2={m–(1–m)}×g] M1 Forusing(mA+mB)a=(mA–mB)gm=0.6 A1

[mg–T=2morT–(1–m)g=2(1–m)] M1 ForapplyingNewton’s2ndlawtoA

ortoB,substitutingformandsolvingforT

Tensionis4.8N A1

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3 Forusings=ut+½at2forABorAC

M1

55=5u+12.5a A1

(55+65)=10u+50aor

65=5vB+12.5aandvB=u+5a A1

M1 Forsolvingforaoru

a=0.4(oru=10) A1

u=10(ora=0.4) A1ft 6

Alternative

vB=(55+65)÷(5+5) ForcalculatingthespeedatBasthemeanspeedforthemotionfromAto

M1 C.

vB=12ms1 A1

ForcalculatingthespeedatX,whereXisthepointwherethecarpasses2.5safterpassingthroughA,

as55÷5=11ms1 B1

[a=(12–11)÷2.5] M1 Forusinga=(vB–vX)÷2.5

a=0.4 A1

u=vX–a×2.5=11–0.4×2.5=10 B1

4 (i) [Y12=682–(–60)2,Y32=1002–962. M1Y1=68sin28.1o,Y3=100sin16.3o]

Forcorrectmagnitudes(32,75,28) A1

Componentsare–32,75and–28 A1ft 3

(ii) [R2=(–60+0+96)2+(–32+75–28)2] M1

Magnitudeis39N A1

[θ=tan1{(–32+75–28)÷(–60+0+

96)}] M1

Directionis22.6o(or0.395radc)

anticlockwisefrom+vex-axis. A1 4

ForusingY2=F2–X2

orforfindingtheangles(sayαandβ)betweenthesofmagnitudes68and100,respectively,andthex-axis.Thenfindthetworelevantmagnitudesfrom68sinαand100sinβ

CanbescoredbyimplicationifthefinalA1isscoredforthecorrectanswertopart(i)

ForusingR2=X2+Y2

Forusingθ=tan1(Y/X)

Acceptjust‘22.6fromx-axis’orjust‘θ=22.6’

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5

(i) Accelerationfort<0.8is4/0.8[5=10sinθ]

θ=30°

B1M1

A1

3

Forusinga=gsinθ

Alternativeforpart(i)

(i)

[mgh=½m42ands={(0+4)÷2}×0.8]

ForusingPEloss=KEgainands÷t

M1

=(u+v)÷2(AtoB)

sinθ=0.8/1.6

A1

θ=30o

A1

(ii) Accelerationfor0.8<t<4.8is–4/(4.8–0.8)

[mgsin30o–F=m(–1)]

mgsin30o+m

µ

mgcos30o

Coefficientis0.693

B1

M1

ForusingNewton’ssecondlaw

M1

Forusingµ F/R

ftfollowingawronganswerforθin

A1ft

part(i)

A1

5

Accept0.69

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(i)

[30000/v–1000–1250g×30/500=1250a]

vbottom=30000/(1250×4+1000+750)

and

vtop=30000/(1250×0.2+1000+750)[½1250(152–4.44….2)]

IncreaseinKEis128000J(128kJ)

M1M1A1M1

A1

5

ForusingDF=30000/vForusingNewton’s2ndlaw

ForusingKEgain=

½m(v2v 2)

top bottom

Alternativeforpart(i)

(i)

[F–1000–1250g×30/500=1250a]

M1

ForusingNewton’ssecondlawtofindthedrivingatthebottomandthetop

ForusingDF=30000/vtofind

vbottomandvtop

ForusingKEgain=

½m(v2v 2)

top bottom

Fbottom=1250×4+1000+750=6750and

Ftop=1250×0.2+1000+750=2000

A1

[vbottom=30000/6750andvtop=30000/2000]

M1

M1

[½1250(152–4.44….2)]

IncreaseinKEis128000J(128kJ)

A1

(ii) PEgain=1250g×30and

WDagainst=1000×500

B1

[WDcar=128000+375000+500000]

M1

ForusingWDbycar’sengine=

KEgain+PEgain+WDagainst

Workdoneis1000000J(1000kJ)

A1ft

3

ftincorrectanswerin(i)

SpecialRuli