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Page 1: MATHEMATICAL METHODS (CAS) - vcaa.vic.edu.au · 2013 Mathematical Methods (CAS) Written examination 1 Author: Victorian Curriculum and Assessment Authority \(VCAA\) Subject: 2013

MATHEMATICAL METHODS (CAS)Written examination 1

Wednesday 6 November 2013 Reading time: 9.00 am to 9.15 am (15 minutes) Writing time: 9.15 am to 10.15 am (1 hour)

QUESTION AND ANSWER BOOK

Structure of bookNumber of questions

Number of questions to be answered

Number of marks

10 10 40

• Studentsarepermittedtobringintotheexaminationroom:pens,pencils,highlighters,erasers,sharpeners,rulers.

• StudentsareNOTpermittedtobringintotheexaminationroom:notesofanykind,blanksheetsofpaper,whiteoutliquid/tapeoracalculatorofanytype.

Materials supplied• Questionandanswerbookof14pages,withadetachablesheetofmiscellaneousformulasinthe

centrefold.• Workingspaceisprovidedthroughoutthebook.

Instructions• Detachtheformulasheetfromthecentreofthisbookduringreadingtime.• Writeyourstudent numberinthespaceprovidedaboveonthispage.

• AllwrittenresponsesmustbeinEnglish.

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

VICTORIANCURRICULUMANDASSESSMENTAUTHORITY2013

Figures

Words

STUDENT NUMBER Letter

SUPERVISOR TO ATTACH PROCESSING LABEL HEREVictorian Certificate of Education 2013

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2013MATHMETH(CAS)EXAM1 2

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Question 1 (5marks)

a. Ify=x2loge(x),finddydx . 2marks

b. Letf (x)=ex2.

Findf ' (3). 3marks

InstructionsAnswerallquestionsinthespacesprovided.Inallquestionswhereanumericalanswerisrequired,anexactvaluemustbegivenunlessotherwisespecified.Inquestionswheremorethanonemarkisavailable,appropriateworkingmustbeshown.Unlessotherwiseindicated,thediagramsinthisbookarenotdrawntoscale.

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Question 2 (2marks)Findananti-derivativeof(4–2x)–5withrespecttox.

Question 3 (2marks)Thefunctionwithrule g(x)hasderivativeg′(x)=sin(2πx).

Giventhatg(1)=1π,findg(x).

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Question 4 (2marks)

Solvetheequation sin x2

12

= − forx [2π,4π].

Question 5 (4marks)a. Solvetheequation2log3(5)–log3(2)+log3(x)=2forx. 2marks

b. Solvetheequation3–4x =96–xforx. 2marks

Page 6: MATHEMATICAL METHODS (CAS) - vcaa.vic.edu.au · 2013 Mathematical Methods (CAS) Written examination 1 Author: Victorian Curriculum and Assessment Authority \(VCAA\) Subject: 2013

2013MATHMETH(CAS)EXAM1 6

Question 7–continued

Question 6 (3marks)Letg:R → R, g(x)=(a–x)2,whereaisarealconstant.

Theaveragevalueofgontheinterval[–1,1]is3112 .

Findallpossiblevaluesofa.

Question 7 (6marks)Theprobabilitydistributionofadiscreterandomvariable,X,isgivenbythetablebelow.

x 0 1 2 3 4

Pr(X=x) 0.2 0.6p2 0.1 1- p 0.1

a. Showthatp=23 orp=1. 3marks

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b. Letp=23 .

i. CalculateE(X). 2marks

ii. FindPr(X ≥E(X)). 1mark

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Question 8 (3marks)Acontinuousrandomvariable,X,hasaprobabilitydensityfunction

f xx x

( ) =

∈[ ]

π π4 4

0 2

0

cos ,if

otherwise

Giventhatddx

x x x x xsin cos sinπ π π π4 4 4 4

=

+

,findE(X).

Page 9: MATHEMATICAL METHODS (CAS) - vcaa.vic.edu.au · 2013 Mathematical Methods (CAS) Written examination 1 Author: Victorian Curriculum and Assessment Authority \(VCAA\) Subject: 2013

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CONTINUES OVER PAGE

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2013MATHMETH(CAS)EXAM1 10

Question 9–continued

Question 9 (6marks)Thegraphoff (x)=(x -1)2-2,x [-2,2],isshownbelow.Thegraphintersectsthex-axiswherex=a.

7

8

9

6

5

4

3

2

1

O

–1

–2

–3

–3 –2 –1 1 2 3

y

xa

a. Findthevalueofa. 1mark

b. Ontheaxesabove,sketchthegraphofg(x)=| f (x)|+1,forx [-2,2].Labeltheendpointswiththeircoordinates. 2marks

Page 11: MATHEMATICAL METHODS (CAS) - vcaa.vic.edu.au · 2013 Mathematical Methods (CAS) Written examination 1 Author: Victorian Curriculum and Assessment Authority \(VCAA\) Subject: 2013

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c. Thefollowingsequenceoftransformationsisappliedtothegraphofthefunction g:[–2,2]→ R, g(x)=| f (x)|+1.

• atranslationofoneunitinthenegativedirectionofthex-axis

• atranslationofoneunitinthenegativedirectionofthey-axis

• adilationfromthex-axisoffactor13

Find i. theruleoftheimageofg afterthesequenceoftransformationshasbeenapplied 2marks

ii. thedomainoftheimageofgafterthesequenceoftransformationshasbeenapplied. 1mark

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Question 10 –continued

Question 10 (7marks)

Letf:[0,∞)→R,f(x)=2 5ex- .

Aright-angledtriangleOQPhasvertexOattheorigin,vertexQonthex-axisandvertexPonthegraphoff,asshown.ThecoordinatesofPare(x,f(x)).

y = f(x)

P(x, f(x))

y

O Qx

a. Findthearea,A,ofthetriangleOQPintermsofx. 1mark

Page 13: MATHEMATICAL METHODS (CAS) - vcaa.vic.edu.au · 2013 Mathematical Methods (CAS) Written examination 1 Author: Victorian Curriculum and Assessment Authority \(VCAA\) Subject: 2013

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Question 10 –continuedTURN OVER

b. FindthemaximumareaoftriangleOQPandthevalueofxforwhichthemaximumoccurs. 3marks

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END OF QUESTION AND ANSWER BOOK

c. LetSbethepointonthegraphoffonthey-axisandletTbethepointonthegraphoffwith

they-coordinate12.

FindtheareaoftheregionboundedbythegraphoffandthelinesegmentST. 3marks

y = f(x)

S

y

xO

12

T

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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 2013

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MATHMETH (CAS) 2

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Page 17: MATHEMATICAL METHODS (CAS) - vcaa.vic.edu.au · 2013 Mathematical Methods (CAS) Written examination 1 Author: Victorian Curriculum and Assessment Authority \(VCAA\) Subject: 2013

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(X 2) – 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


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