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MATHEMATICAL METHODS (CAS)

Written examination 2Thursday 8 November 2012

Reading time: 11.45 am to 12.00 noon (15 minutes) Writing time: 12.00 noon to 2.00 pm (2 hours)

QUESTION AND ANSWER BOOK

Structure of bookSection Number of

questionsNumber of questions

to be answeredNumber of

marks

12

225

225

2258

Total 80

• Studentsarepermittedtobringintotheexaminationroom:pens,pencils,highlighters,erasers,sharpeners,rulers,aprotractor,set-squares,aidsforcurvesketching,oneboundreference,oneapprovedCAScalculator(memoryDOESNOTneedtobecleared)and,ifdesired,onescientificcalculator.Forapprovedcomputer-basedCAS,theirfullfunctionalitymaybeused.

• StudentsareNOTpermittedtobringintotheexaminationroom:blanksheetsofpaperand/orwhiteoutliquid/tape.

Materials supplied• Questionandanswerbookof24pageswithadetachablesheetofmiscellaneousformulasinthe

centrefold.• Answersheetformultiple-choicequestions.

Instructions• Detachtheformulasheetfromthecentreofthisbookduringreadingtime.• Writeyourstudent numberinthespaceprovidedaboveonthispage.• Checkthatyournameandstudent numberasprintedonyouranswersheetformultiple-choice

questionsarecorrect,andsignyournameinthespaceprovidedtoverifythis.

• AllwrittenresponsesmustbeinEnglish.

At the end of the examination• Placetheanswersheetformultiple-choicequestionsinsidethefrontcoverofthisbook.

Students are NOT permitted to bring mobile phones and/or any other unauthorised electronic devices into the examination room.

©VICTORIANCURRICULUMANDASSESSMENTAUTHORITY2012

SUPERVISOR TO ATTACH PROCESSING LABEL HEREVictorian Certificate of Education 2012

2012MATHMETH(CAS)EXAM2 2

SECTION 1–continued

Question 1Thefunctionwithrulef (x)=–3sin π x

5

hasperiod

A. 3

B. 5

C. 10

D. π5

E. π10

Question 2Forthefunctionwithrulef (x)=x3–4x,theaveragerateofchangeoff (x)withrespecttoxontheinterval [1,3]isA. 1B. 3C. 5D. 6E. 9

Question 3Therangeofthefunctionf :[–2,3)→ R, f (x)=x2 – 2x–8isA. RB. (−9,–5]C. (–5,0)D. [−9,0]E. [–9,–5)

SECTION 1

Instructions for Section 1Answerallquestionsinpencilontheanswersheetprovidedformultiple-choicequestions.Choosetheresponsethatiscorrect forthequestion.Acorrectanswerscores1,anincorrectanswerscores0.Markswillnotbedeductedforincorrectanswers.Nomarkswillbegivenifmorethanoneansweriscompletedforanyquestion.

3 2012MATHMETH(CAS)EXAM2

SECTION 1–continuedTURN OVER

Question 4Giventhatgisadifferentiablefunctionandkisarealnumber,thederivativeofthecompositefunction g (ekx)isA. kg′(ekx)ekx

B. kg (ekx)

C. kekx g (ekx)

D. kekx g′(ex)

E. 1k

ekxg′(ekx)

Question 5Lettheruleforafunctiongbeg (x)=loge((x–2)

2).Forthefunctiong,theA. maximaldomain=R+andrange=RB. maximaldomain=R\{2}andrange=RC. maximaldomain=R\{2}andrange=(–2,∞)D. maximaldomain=[2,∞)andrange=(0,∞)E. maximaldomain=[2,∞)andrange=[0,∞)

Question 6Asectionofthegraphoff isshownbelow.

y

x

4π

2π3

4π−

2π−

4π− O

TheruleoffcouldbeA. f (x)=tan(x)

B. f (x)=tan x −

π4

C. f (x)=tan 24

x −

π

D. f (x)=tan 22

x −

π

E. f (x)=tan12 2

x −

π



Question 7Thetemperature,T°C,insideabuildingthoursaftermidnightisgivenbythefunction

f :[0,24]→ R,T(t)=22–10cosπ12

2t −( )

Theaveragetemperatureinsidethebuildingbetween2amand2pmisA. 10°CB. 12°CC. 20°CD. 22°CE. 32°C

Question 8Thefunctionf :R→R,f (x)=ax3 + bx2 + cx,whereaisanegativerealnumberandbandcarerealnumbers.Fortherealnumbersp < m < 0 < n < q,wehavef (p)=f (q)=0andf ′(m)=f ′(n)=0.Thegradientofthegraphofy=f (x)isnegativeforA. (–∞,m)∪(n,∞)B. (m,n)C. (p,0)∪(q,∞)D. (p,m)∪(0,q)E. (p,q)

Question 9Thenormaltothegraphof y b x= − 2 hasagradientof3whenx =1.Thevalueofbis

A. −109

B. 109

C. 4

D. 10

E. 11

Question 10Theaveragevalueofthefunctionf :[0,2π]→ R,f (x)=sin2(x)overtheinterval[0,a]is0.4.Thevalueofa,tothreedecimalplaces,isA. 0.850B. 1.164C. 1.298D. 1.339E. 4.046



Question 11Theweightsofbagsofflourarenormallydistributedwithmean252gandstandarddeviation12g.Themanufacturersaysthat40%ofbagsweighmorethanxg.ThemaximumpossiblevalueofxisclosesttoA. 249.0B. 251.5C. 253.5D. 254.5E. 255.0

Question 12Demelzaisabadmintonplayer.Ifshewinsagame,theprobabilitythatshewillwinthenextgameis0.7.Ifshelosesagame,theprobabilitythatshewilllosethenextgameis0.6.Demelzahasjustwonagame.TheprobabilitythatshewillwinexactlyoneofhernexttwogamesisA. 0.33B. 0.35C. 0.42D. 0.49E. 0.82

Question 13AandBareeventsofasamplespaceS.

Pr(A ∩ B)= 25andPr(A ∩ B′ )= 3

7.

Pr(B′ |A)isequalto

A. 635

B. 1529

C. 1435

D. 2935

E. 23



Question 14

Thegraphof f :R+∪{0} → R, f (x)= x isshownbelow.Inordertofindanapproximationtotheareaoftheregionboundedbythegraphoff,they-axisandtheliney=4,Zoedrawsfourrectangles,asshown,andcalculatestheirtotalarea.

4

3

2

1

O

y x=

x

y

Zoe’sapproximationtotheareaoftheregionisA. 14

B. 21

C. 29

D. 30

E. 643



Question 15If f ′(x)=3x2–4,whichoneofthefollowinggraphscouldrepresentthegraphofy=f (x)?

A. y

x

B. y

x

C. y

x

D. y

x

E. y

x

Question 16Thegraphofacubicfunctionfhasalocalmaximumat(a,–3)andalocalminimumat(b,–8).Thevaluesofc,suchthattheequationf (x)+c=0hasexactlyonesolution,areA. 3<c < 8B. c >–3orc < –8C. –8 < c <–3D. c <3orc > 8E. c < –8



Question 17Asystemofsimultaneouslinearequationsisrepresentedbythematrixequation

mm

xy m

31 2

1+

=

.

Thesystemofequationswillhaveno solutionwhenA. m=1B. m=–3C. m ∈{1,–3}D. m ∈ R\{1}E. m ∈{1,3}

Question 18Thetangenttothegraphofy=loge(x)atthepoint(a,loge(a))crossesthex-axisatthepoint(b,0),whereb <0.Whichofthefollowingisfalse?A. 1 < a < e

B. Thegradientofthetangentispositive

C. a > e

D. Thegradientofthetangentis 1a

E. a > 0

Question 19Afunction f hasthefollowingtwopropertiesforallrealvaluesofθ .

f (π – θ )=–f (θ )and f (π – θ )=–f (–θ )

Apossiblerulefor f isA. f (x)=sin(x)

B. f (x)=cos(x)

C. f (x)=tan(x)

D. f (x)=sin x2

E. f (x)=tan(2x)

Question 20AdiscreterandomvariableXhastheprobabilityfunctionPr(X=k)=(1–p)kp, wherekisanon-negativeinteger.Pr(X >1)isequaltoA. 1 – p + p2

B. 1 – p2

C. p – p2

D. 2p – p2

E. (1–p)2


END OF SECTION 1TURN OVER

Question 21

Thevolume,Vcm3,ofwaterinacontainerisgivenbyV=13πh3wherehcmisthedepthofwaterinthe

containerattimetminutes.Waterisdrainingfromthecontainerataconstantrateof300cm3/min.Therateofdecreaseofh,incm/min,whenh=5is

A. 12π

B. 4π

C. 25π

D. 60π

E. 30π

Question 22Thegraphofadifferentiablefunctionfhasalocalmaximumat(a,b),wherea <0andb >0,andalocalminimumat(c,d ),wherec>0andd <0.Thegraphofy=–|f (x–2)|hasA. alocalminimumat(a–2,–b)andalocalmaximumat(c–2,d )B. localminimaat(a+2,–b)and(c+2,d )C. localmaximaat(a+2,b)and(c+2,–d )D. alocalminimumat(a–2,–b)andalocalmaximumat(a–2,–d )E. localminimaat(c+2,–d )and(a+2,–b)


SECTION 2 – Question 1–continued

Question 1Asolidblockintheshapeofarectangularprismhasabaseofwidthxcm.Thelengthofthebaseis two-and-a-halftimesthewidthofthebase.

52x cm

x cm

h cm

Theblockhasatotalsurfaceareaof6480sqcm.

a. Showthatiftheheightoftheblockishcm,h= 6480 57

2− xx

.

2marks

SECTION 2

Instructions for Section 2Answerallquestionsinthespacesprovided.Inallquestionswhereanumericalanswerisrequiredanexactvaluemustbegivenunlessotherwisespecified.Inquestionswheremorethanonemarkisavailable,appropriateworkingmust beshown.Unlessotherwiseindicated,thediagramsinthisbookarenotdrawntoscale.



b. Thevolume,Vcm3,oftheblockisgivenbyV(x)= 5 6480 514

2x x( ).−

GiventhatV(x)>0andx >0,findthepossiblevaluesofx.

2marks

c. Find dVdx

,expressingyouranswerintheform dVdx

=ax2 + b,whereaandbarerealnumbers.

3marks

d. Findtheexactvaluesofxandhiftheblockistohavemaximumvolume.

2marks



Question 2

Letf :R\{2} → R, f (x)= 12 4x −

+3.

a. Sketchthegraphofy=f (x)onthesetofaxesbelow.Labeltheaxesinterceptswiththeircoordinatesandlabeleachoftheasymptoteswithitsequation.

x

y

O

3marks

b. i. Findf ′(x).

ii. Statetherangeoff ′.

iii. Usingtheresultof part ii. explainwhyfhasnostationarypoints.

1+1+1=3marks


SECTION 2 – Question 2–continuedTURN OVER

c. If(p,q)isanypointonthegraphofy=f (x),showthattheequationofthetangenttoy=f (x)atthispointcanbewrittenas(2p–4)2(y–3)=–2x+4p–4.

2marks



d. Findthecoordinatesofthepointsonthegraphofy=f (x)suchthatthetangentstothegraphatthese

pointsintersectat −

1, 7

2.

4marks

15 2012 MATHMETH(CAS) EXAM 2

SECTION 2 – continuedTURN OVER

e. A transformation T: R2 → R2 that maps the graph of f to the graph of the function

g: R\{0} → R, g (x) = 1x

has rule Txy

a xy

cd

= +0

0 1, where a, c and d are non-zero real numbers.

Find the values of a, c and d.

2 marks

2012 MATHMETH(CAS) EXAM 2 16

SECTION 2 – Question 3 – continued

Question 3Steve, Katerina and Jess are three students who have agreed to take part in a psychology experiment. Each student is to answer several sets of multiple-choice questions. Each set has the same number of questions, n, where n is a number greater than 20. For each question there are four possible options (A, B, C or D), of which only one is correct.a. Steve decides to guess the answer to every question, so that for each question he chooses A, B, C or D

at random. Let the random variable X be the number of questions that Steve answers correctly in a particular set.

i. WhatistheprobabilitythatStevewillanswerthefirstthreequestionsofthissetcorrectly?

ii. Find,tofourdecimalplaces,theprobabilitythatStevewillansweratleast10ofthefirst20 questions of this set correctly.

iii. Use the fact that the variance of X is 7516

to show that the value of n is 25.

1 + 2 + 1 = 4 marks



IfKaterinaanswersaquestioncorrectly,theprobabilitythatshewillanswerthenextquestioncorrectly is 3

4 . Ifsheanswersaquestionincorrectly,theprobabilitythatshewillanswerthenextquestion incorrectlyis 2

3 .

Inaparticularset,KaterinaanswersQuestion1incorrectly.b. i. CalculatetheprobabilitythatKaterinawillanswerQuestions3,4and5correctly.

ii. FindtheprobabilitythatKaterinawillanswerQuestion25correctly.Giveyouranswercorrecttofourdecimalplaces.

3+2=5marks



c. TheprobabilitythatJesswillansweranyquestioncorrectly,independentlyofheranswertoanyotherquestion,isp (p > 0). LettherandomvariableYbethenumberofquestionsthatJessanswerscorrectlyinanysetof25.

IfPr(Y >23)=6Pr(Y =25),showthatthevalueofp is 56

.

2marks

d. Fromthesesetsof25questionsbeingcompletedbymanystudents,ithasbeenfoundthatthetime,inminutes,thatanystudenttakestoanswereachsetof25questionsisanotherrandomvariable,W,whichisnormally distributedwithmeana andstandarddeviationb.

Itturnsoutthat,forJess,Pr(Y≥18)=Pr(W≥20)andalsoPr(Y≥22)=Pr(W≥25). Calculatethevaluesofaandb,correcttothreedecimalplaces.

4marks



Question 4TasmaniaJonesisinthejungle,searchingfortheQuetzalotltribe’svaluableemeraldthathasbeenstolenandhiddenbyaneighbouringtribe.Tasmaniahasheardthattheemeraldhasbeenhiddeninatankshapedlikeaninvertedcone,withaheightof10metresandadiameterof4metres(asshownbelow).Theemeraldisonashelf.Thetankhasapoisonousliquidinit.

r m

4 m

tap

shelf

10 m

h m

2 m

a. Ifthedepthoftheliquidinthetankishmetres i. findtheradius,rmetres,ofthesurfaceoftheliquidintermsofh

ii. showthatthevolumeoftheliquidinthetankisπh3

75m3.

1+1=2marks



Thetankhasatapatitsbasethatallowstheliquidtorunoutofit.Thetankisinitiallyfull.Whenthetapisturnedon,theliquidflowsoutofthetankatsucharatethatthedepth,hmetres,oftheliquidinthetankisgivenby

h t t= + −10 11600

12003( ),

wheret minutesisthelengthoftimeafterthetapisturnedonuntilthetankisempty.b. Showthatthetankisemptywhent=20.

1mark

c. Whent =5minutes,find i. thedepthoftheliquidinthetank

ii. therateatwhichthevolumeoftheliquidisdecreasing,correcttoonedecimalplace.

1+3=4marks



d. Theshelfonwhichtheemeraldisplacedis2metresabovethevertexofthecone. Fromthemomenttheliquidstartstoflowfromthetank,findhowlong,inminutes,ittakesuntilh=2.

(Giveyouranswercorrecttoonedecimalplace.)

2marks

e. Assoonasthetankisempty,thetapturnsitselfoffandpoisonousliquidstartstoflowintothetankatarateof0.2m3/minute.

Howlong,inminutes,afterthetankisfirstemptywilltheliquidonceagainreachadepthof2metres?

2marks

f. Inordertoobtaintheemerald,TasmaniaJonesentersthetankusingavinetoclimbdownthewallofthetankassoonasthedepthoftheliquidisfirst2metres.Hemustleavethetankbeforethedepthisagaingreaterthan2metres.

Findthelengthoftime,inminutes,correcttoonedecimalplace,thatTasmaniaJoneshasfromthetimeheentersthetanktothetimeheleavesthetank.

1mark



Question 5TheshadedregioninthediagrambelowistheplanofaminesitefortheBlackPossumminingcompany. Alldistancesareinkilometres.Twooftheboundariesoftheminesiteareintheshapeofthegraphsofthefunctions

f :R → R,f (x)=ex and g:R+ → R,g(x)=loge(x).

y

x–3 –2 –1 1 2 3

–3

–2

–1

O

1

2

3

x = 1

y = f (x) y = g(x)

y = –2

a. i. Evaluate f x dx( ) .−∫2

0

ii. Hence,orotherwise,findtheareaoftheregionboundedbythegraphofg,thexandyaxes,andtheliney=–2.



iii. Findthetotalareaoftheshadedregion.

1+1+1=3marks

b. Theminingengineer,Victoria,decidesthatabettersiteforthemineistheregionboundedbythegraphofgandthatofanewfunctionk:(–∞,a)→ R,k(x)=–loge(a – x),whereaisapositiverealnumber.

i. Find,intermsofa,thex-coordinatesofthepointsofintersectionofthegraphsofgandk.

ii. Hence,findthesetofvaluesofa,forwhichthegraphsofgandkhavetwodistinctpointsofintersection.

2+1=3marks


c. Forthenewminesite,thegraphsofgandkintersectattwodistinctpoints,AandB.ItisproposedtostartminingoperationsalongthelinesegmentAB,whichjoinsthetwopointsofintersection.

Victoriadecidesthatthegraphofkwillbesuchthatthex-coordinateofthemidpointofABis 2 . Findthevalueofainthiscase.

2marks

END OF QUESTION AND ANSWER BOOK

MATHEMATICAL METHODS (CAS)

Written examinations 1 and 2

FORMULA SHEET

Directions to students

Detach this formula sheet during reading time.

This formula sheet is provided for your reference.

© VICTORIAN CURRICULUM AND ASSESSMENT AUTHORITY 2012

MATHMETH (CAS) 2

This page is blank

3 MATHMETH (CAS)

END OF FORMULA SHEET

Mathematical Methods (CAS)Formulas

Mensuration

area of a trapezium: 12a b h+( ) volume of a pyramid:

13Ah

curved surface area of a cylinder: 2π rh volume of a sphere: 43

3π r

volume of a cylinder: π r 2h area of a triangle: 12bc Asin

volume of a cone: 13

2π r h

Calculusddx

x nxn n( ) = −1

x dx

nx c nn n=

++ ≠ −+∫

11

11 ,

ddxe aeax ax( ) = e dx a e cax ax= +∫

1

ddx

x xelog ( )( ) = 1 1x dx x ce= +∫ log

ddx

ax a axsin( ) cos( )( ) = sin( ) cos( )ax dx a ax c= − +∫

1

ddx

ax a axcos( )( ) −= sin( ) cos( ) sin( )ax dx a ax c= +∫

1

ddx

ax aax

a axtan( )( )

( ) ==cos

sec ( )22

product rule: ddxuv u dv

dxv dudx

( ) = + quotient rule: ddx

uv

v dudx

u dvdx

v

=

−

2

chain rule: dydx

dydududx

= approximation: f x h f x h f x+( ) ≈ ( ) + ′( )

ProbabilityPr(A) = 1 – Pr(A′) Pr(A ∪ B) = Pr(A) + Pr(B) – Pr(A ∩ B)

Pr(A|B) = Pr

PrA BB∩( )( ) transition matrices: Sn = Tn × S0

mean: µ = E(X) variance: var(X) = σ 2 = E((X – µ)2) = E(X2) – µ2

Probability distribution Mean Variance

discrete Pr(X = x) = p(x) µ = ∑ x p(x) σ 2 = ∑ (x – µ)2 p(x)

continuous Pr(a < X < b) = f x dxa

b( )∫ µ =

−∞

∞∫ x f x dx( ) σ µ2 2= −

−∞

∞∫ ( ) ( )x f x dx