2014-mg-maths-ext-1

2014HSCMathematicsExtension1MarkingGuidelinesSectionIMultiple-choiceAnswerKeyQuestionAnswer1D2A3C4D5B6B7A8D9C10C–1–BOSTES2014HSCMathematicsExtension1MarkingGuidelinesSectionIIQuestion11(a)CriteriaMarks•Providesacorrectsolution3•Obtainsaquadraticinx,orequivalentmerit2•Attemptstosolvequadraticinx+2x,orequivalentmerit1Sampleanswer:2Letn=x+xn2
6n+9=0(n
3)2=0n=32Now,x+=3xx2
3x+2=0(x
2)(x
1)=0x=2orx=1Question11(b)CriteriaMarks•Providesacorrectsolution2•Findsacorrectexpressionfortheprobabilityofrainon1or2days,orequivalentmerit1Sampleanswer:P(rain)=0.1(
Prain)=1
0.1=0.9Probabilitythatitrainsonfewerthan3daysis30)30+3030(0.9×0.929×0.1+0.928(0.1)2012–2–BOSTES2014HSCMathematicsExtension1MarkingGuidelinesQuestion11(c)CriteriaMarks•Drawsacorrectsketch2•Drawsagraphshowingthecorrectrangeorshape,orequivalentmerit1Sampleanswer:
yy=6tan
1x262
3y3–3– 5BOSTES2014HSCMathematicsExtension1MarkingGuidelinesQuestion11(d)CriteriaMarks•Providesacorrectsolution3•Obtainsacorrectprimitive,orequivalentmerit2•Attemptsthegivensubstitution,orequivalentmerit1xx=u2+1|x=2u=1dx=2udu|x=5u=2uu2+1122udu=2u33+u12=283+2
13
1=210320=3x
1dx2Sampleanswer:–4–BOSTES2014HSCMathematicsExtension1MarkingGuidelinesQuestion11(e)CriteriaMarks•Providesacorrectsolution3•Obtainsonecorrectinterval,orequivalentmerit2•Obtainsaquadraticinequality,orequivalentmerit1Sampleanswer:Multiplybyx2:2(2+5)xx26×xx2iexx(2+5)6xxx2+520()
6xxx2
6x)0(+5xx
5
1)0()(x
0x1orx5–5–BOSTES2014HSCMathematicsExtension1MarkingGuidelinesQuestion11(f)CriteriaMarks•Findsthecorrectderivative2•Usesanappropriatedifferentiationrule,orequivalentmerit1Sampleanswer:exlnxy=xx1xe.+exlnx
exlnx.1dyx=2dxxxe
ex+exlnxxlnxx=2xQuestion12(a)Question12(a)(i)CriteriaMarks•Providesacorrectanswer1Sampleanswer:Distancetravelledis(2+2)m,ie4m.–6–BOSTES2014HSCMathematicsExtension1MarkingGuidelinesQuestion12(a)(ii)CriteriaMarks•Providesacorrectsolution2•Identifieswhentheparticleisfirstatrest,orequivalentmerit1Sampleanswer:Whenatrest,x
=0ie6cos3t=0
cos3t=03t=2t=6
Firstatrestwhent=6Now,x=2sin3tx
=6cos3t

x=
18sin3twhent=6

x=
18sin2=
18×1acceleration=
18m/s2–7– BOSTES2014HSCMathematicsExtension1MarkingGuidelinesQuestion12(b)CriteriaMarks•Providesacorrectsolution3•Obtainscorrectintegralexpressionforvolumeintermsofcos8x,orequivalentmerit2•Obtainsintegralexpressionforvolumeoftheformkcos24xdx,0
8orequivalentmerit1Sampleanswer:y=cos4xV=8y2dx08=cos24xdx081+cos8x=dx20sin8x8=x+280=+0
0
0282volume=unit316–8–BOSTES2014HSCMathematicsExtension1MarkingGuidelinesQuestion12(c)CriteriaMarks•Providesacorrectsolution3•Obtainsexpressionforv2possiblyinvolvinganundeterminedconstant2•Attemptstousea=12dv2()dxorequivalentmerit1Sampleanswer:
xd1v22=x

=2
edx2
x12v=2
e2dx2
x2=2x
e+c1
2
x2=2x+2e+cWhenx=0,v=41()16=0+2+c2c=6
x12v=2x+2e2+62
x22v=4x+4e+12–9–BOSTES2014HSCMathematicsExtension1MarkingGuidelinesQuestion12(d)CriteriaMarks•Providesacorrectsolution2•Attemptstousethebinomialtheorem.1Sampleanswer:Consider(1+x)n.Fromthebinomialtheorem:
n
n
n
nn(1+x)n=2+……+0+1x+2xnxLetx=–1
n
n
n
nnn(1+–1=0=–1+()2+……+())()–1–10+12nnnnnn0=
+
……+()012
1nQuestion12(e)CriteriaMarks•Providesavalidexplanation1Sampleanswer:yxOx1x2aThetangentatx=xisdrawn,anditintersectsthexaxisatx2.1Asshowninthediagram,xisfurtherawayfrom
thanx1.2–10– BOSTES2014HSCMathematicsExtension1MarkingGuidelinesQuestion12(f)CriteriaMarks•Providesacorrectsolution3•ObtainsAandB,orequivalentmerit2•ShowsA–B=2,orequivalentmerit1Sampleanswer:A
Be
0.03tT=Ast,T23A=23T=23
Be
0.03tt=0T=22=23
BB=21
0.03tT=23
21eNow,whenT=0
0.03t10=23
21e
0.03t21e=1313
0.03te=2113
0.03t=ln2113ln21t=
0.03=15.985……!t16minutes–11–BOSTES2014HSCMathematicsExtension1MarkingGuidelinesQuestion13(a)CriteriaMarks•Providesacorrectproof3•Usesinductiveassumption,orequivalentmerit2•Establishescaseforn=1,orequivalentmerit1Sampleanswer:2n+(––1)n+1n=1,21+(––1)2=2+1=3whichisdivisibleby3Assumetrueforn=kieAssume2k+(––1)k+1isdivisibleby3ie2k+(––1)k+1=3MwhereMisaninteger.Weneedtoprovetrueforn=k+1ieThat2k+1+(–1)k+2isdivisibleby3k+2k+22k+()()()1+
1=22k+
1k+2k+2=23M+()
1
1+()k+2=32M+()
1Whichisamultipleof3sinceMandkareintegers.ieif2k+(–1)k+1isdivisibleby3,then2k+1+(–1)k+1isdivisibleby3
Bytheprincipleofmathematicalinduction2n+(–1)n+1isdivisibleby3foralln
1.2k()k+1=3M

1()k+2=3M+
1–12–BOSTES2014HSCMathematicsExtension1MarkingGuidelinesQuestion13(b)(i)CriteriaMarks•Providesacorrectsolution2•UsesPythagoras’Theoremandattemptstodifferentiate,orequivalentmerit1Sampleanswer:L=402+x2dLdx=12.1.2xx=402+x2x=L=cos
Question13(b)(ii)402+x2CriteriaMarks•Providesacorrectsolution1Sampleanswer:dL dLdx= dt dxdtdL= 3dx=3cos
(frompart(i)).–13–BOSTES2014HSCMathematicsExtension1MarkingGuidelinesQuestion13(c)(i)CriteriaMarks•Providesacorrectsolution2•Establishesonecoordinate,orequivalentmerit1Sampleanswer:P(2at,at2)S(0,a)t2:10×t2+2ata×t2+1×at2Q=t2+1,t2+12at2at2Q=t2+1,t2+1Qu