Download pdf - 2018 Mathematical Methods-nht Written examination 2 · 3 2018 MATHMETH EXAM 2 (NHT) SECTION A – continued TURN OVER D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I

MATHEMATICAL METHODSWritten examination 2

Monday 4 June 2018 Reading time: 2.00 pm to 2.15 pm (15 minutes) Writing time: 2.15 pm to 4.15 pm (2 hours)

QUESTION AND ANSWER BOOK

Structure of bookSection Number of

questionsNumber of questions

to be answeredNumber of

marks

A 20 20 20B 4 4 60

Total 80

• Studentsarepermittedtobringintotheexaminationroom:pens,pencils,highlighters,erasers,sharpeners,rulers,aprotractor,setsquares,aidsforcurvesketching,oneboundreference,oneapprovedtechnology(calculatororsoftware)and,ifdesired,onescientificcalculator.CalculatormemoryDOESNOTneedtobecleared.Forapprovedcomputer-basedCAS,fullfunctionalitymaybeused.

• StudentsareNOTpermittedtobringintotheexaminationroom:blanksheetsofpaperand/orcorrectionfluid/tape.

Materials supplied• Questionandanswerbookof21pages• Formulasheet• Answersheetformultiple-choicequestions

Instructions• Writeyourstudent numberinthespaceprovidedaboveonthispage.• Checkthatyournameandstudent numberasprintedonyouranswersheetformultiple-choice

questionsarecorrect,andsignyournameinthespaceprovidedtoverifythis.• Unlessotherwiseindicated,thediagramsinthisbookarenot drawntoscale.• AllwrittenresponsesmustbeinEnglish.

At the end of the examination• Placetheanswersheetformultiple-choicequestionsinsidethefrontcoverofthisbook.• Youmaykeeptheformulasheet.

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

©VICTORIANCURRICULUMANDASSESSMENTAUTHORITY2018

SUPERVISOR TO ATTACH PROCESSING LABEL HEREVictorian Certificate of Education 2018

STUDENT NUMBER

Letter

2018MATHMETHEXAM2(NHT) 2

SECTION A – continued

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Question 1

Letf:R → R,f (x)=3–2cos π x4

.

TheperiodandrangeofthisfunctionarerespectivelyA. 4and[−2,2]

B. 8and[1,5]

C. 8πand[1,5]

D. 8πand[−2,2]

E. 12 and[−1,5]

SECTION A – Multiple-choice questions

Instructions for Section AAnswerallquestionsinpencilontheanswersheetprovidedformultiple-choicequestions.Choosetheresponsethatiscorrectforthequestion.Acorrectanswerscores1;anincorrectanswerscores0.Markswillnotbedeductedforincorrectanswers.Nomarkswillbegivenifmorethanoneansweriscompletedforanyquestion.Unlessotherwiseindicated,thediagramsinthisbookarenotdrawntoscale.

3 2018MATHMETHEXAM2(NHT)

SECTION A – continuedTURN OVER

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Question 2Thediagrambelowshowspartofthegraphofapolynomialfunction.

x

y

–5 –4 –3 –2 –1 O 1 2 3 4 5

ApossibleruleforthisfunctionisA. y=(x+2)(x–1)(x–3)B. y=(x+2)2(x–1)(x–3)C. y=(x+2)2(x–1)(3–x)D. y=−(x−2)2(x–1)(3–x)E. y=−(x+2)(x–1)(x–3)

Question 3Adiscreterandomvariablehasabinomialdistributionwithameanof3.6andavarianceof1.98Thevaluesofn(thenumberofindependenttrials)andp(theprobabilityofsuccessineachtrial)areA. n=4andp=0.9B. n=5andp=0.72C. n=6andp=0.6D. n=8andp=0.45E. n=12andp=0.3

Question 4IfAandBareeventsfromasamplespacesuchthatPr(A)=0.6,Pr(B)=0.3andPr(A ∪ B)=0.7,then Pr(A ∩ B′)isequaltoA. 0.12B. 0.18C. 0.2D. 0.3E. 0.4



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Question 5Asetofthreenumbersthatcouldbethesolutionsofx3 + ax2+16x+84=0isA. {3,4,7}B. {–4,–3,7}C. {–2,–1,21}D. {–2,6,7}E. {2,6,7}

Question 6Thesumofthesolutionstotheequation 3 sin(2x)=–3cos(2x)forx ∈[0,2π]isequalto

A. π3

B. 76π

C. 113π

D. 133π

E. 143π

Question 7Sixballsnumberedfrom1to6areplacedinajar.Aballistakenrandomlyfromthejaranditsnumberisrecorded.Thisballisreturnedtothejar,andasecondballisthentakenrandomlyanditsnumberisrecorded.Thesumofthetworecordednumbersisthencalculated.Theprobabilitythatthesumofthetworecordednumbersis7,giventhatthefirstrecordednumberisodd,isequalto

A. 13

B. 14

C. 16

D. 112

E. 19



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Question 8Partofthegraphofy=f(x)isshownbelow.

x

y

y = f (x)

–3 –2 –1 O 1 2 3 4

–3

–2

–1

1

Thegraphofy=f ′(x)isbestrepresentedby

x

x = 1

y = –1

y

O

x

x = 1y

Ox

x = 1y

O

x

x = 1

y = –1

y = 0

y = 0

y = 0

y

O

x

x = 1y

O

A. B.

C. D.

E.



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Question 9AcontinuousrandomvariableXhasanormaldistributionwithameanof40andastandarddeviationof5.ThecontinuousrandomvariableZhasthestandardnormaldistribution.Pr(–2<Z<1)isequaltoA. Pr(40<X<55)B. Pr(35<X<50)C. Pr(30<X<50)D. Pr(10<X<30)E. Pr(X>30)–Pr(X<45)

Question 10

Therangeofthefunction f R f x x x: , ,−

→ ( ) = − +

12

2 2 3 43 is

A. 4 2 4 2−( + ),

B. −

12

2,

C. 4 2 4 2−( + ,

D. −

12

2,

E. 4 2 4 2− + ,

Question 11Themaximaldomainofthefunctiong,whereg(x)=loge(–2x),isA. RB. R –

C. R+

D. [0,∞)E. (–∞,0]

Question 12

Theaveragevalueoff(x)=x2–2xovertheinterval[1,a]is133.

ThevalueofaisA. 2

B. 3

C. 103

D. 5

E. 163



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Question 13Thefunctionfhasthepropertyf(2x)=(f(x))2–2forallrealnumbersx.Apossibleruleforthefunctionf(x)is

A. 1

42x +B. cos(x)

C. 2loge(x2+1)

D. ex + e–x

E. x2

Question 14

Thegraphofthefunctionf isobtainedfromthegraphofthefunctiongwithrule g x x( ) cos= −

3

6π

bya

dilationofafactorof12fromthex-axis,areflectioninthey-axis,atranslationof

π6unitsinthe

negativexdirectionandatranslationof4unitsinthenegativeydirection,inthatorder.Theruleoff is

A. f x x( ) cos= − −

−

32 3

4π

B. f x x( ) cos( )= − −32

4

C. f x x( ) cos( )= − −32

4

D. f x x( ) cos= − −

−3

2 34π

E. f x x( ) cos= − +

−

32 3

4π

Question 15

If f x dx( )−∫ = −

3

28 and f x dx( ) =∫ 10

2

3,thevalueof f x dx( )

−∫ 3

3is

A. 2B. –2C. –18D. 18E. 0

Question 16Letf:R+→ R,f (x)=–loge(x)andg :R → R,g(x)=x2+1.Thedomainandrangeoff(g(x))arerespectivelyA. RandR+ ∪{0}B. RandR –

C. [1,∞)andR+ ∪{0}D. R+andR+ ∪{0}E. RandR – ∪{0}



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Question 17IfF(x)isanantiderivativeoff (x)andF(4)=–6,thenF(8)isequaltoA. f ′(8)+6

B. –6+f ′(4)

C. ∫48

f x dx( )

D. − +( )∫ 64

8f x dx( )

E. − + ∫64

8f x dx( )

Question 18Considerthegraphsoffandgbelow,whichhavethesamescale.

x

x = 1

y = 2

y

O

y = f (x)

y = g(x)

x

x = –2

y = –2

y

O

IfTtransformsthegraphoffontothegraphofg,then

A. T R Rxy

xy

T: ,2 2 1 00 1

34

→

=

+

−−

B. T R Rxy

xy

T: ,2 2 1 00 1

34

→

=

−

+

−−

C. T R Rxy

xy

T: ,2 2 1 00 1

30

→

= −

+−

D. T R Rxy

xy

T: ,2 2 2 00 1

→

=

−−

E. T R Rxy

xy

T: ,2 2 1 00 2

→

=

−−


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END OF SECTION ATURN OVER

Question 19Aboxcontains20000marblesthatareeitherblueorred.Therearemorebluemarblesthanredmarbles.Randomsamplesof100marblesaretakenfromthebox.Eachrandomsampleisobtainedbysamplingwithreplacement.Ifthestandarddeviationofthesamplingdistributionfortheproportionofbluemarblesis0.03,thenthenumberofbluemarblesintheboxisA. 11000B. 16000C. 17000D. 18000E. 19000

Question 20Letfbeaone-to-onedifferentiablefunctionsuchthatf(3)=7,f(7)=8,f ′(3)=2andf ′(7)=3.Thefunctiongisdifferentiableandg(x)=f –1(x)forallx.g′(7)isequalto

A. 12

B. 2

C. 16

D. 18

E. 13


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SECTION B – Question 1–continued

Question 1 (9marks)Letf:R → R,f (x)=x4–4x–8.

a. Givenf (x)=(x–2)(x3 + ax2 + bx + c),finda,bandc. 1mark

b. Findtwoconsecutiveintegersmandnsuchthatasolutiontof (x)=0isintheinterval(m,n),wherem<n<0. 2marks

SECTION B

Instructions for Section BAnswerallquestionsinthespacesprovided.Inallquestionswhereanumericalanswerisrequired,anexactvaluemustbegivenunlessotherwisespecified.Inquestionswheremorethanonemarkisavailable,appropriateworkingmust beshown.Unlessotherwiseindicated,thediagramsinthisbookarenotdrawntoscale.


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SECTION B – continuedTURN OVER

Thediagrambelowshowspartofthegraphoffandastraightlinedrawnthroughthepoints(0,–8)and(2,0).AsecondstraightlineisdrawnparalleltothehorizontalaxisandittouchesthegraphoffatthepointQ.ThetwostraightlinesintersectatthepointP.

x

y

O

P Q

y = f (x)

2

–8

c. i. Findtheequationofthelinethrough(0,–8)and(2,0). 1mark

ii. StatetheequationofthelinethroughthepointsPandQ. 1mark

iii. StatethecoordinatesofthepointsPandQ. 2marks

d. AtransformationT R R Txy

xy

d: ,2 2

0→

=

+

isappliedtothegraphoff.

i. FindthevalueofdforwhichPistheimageofQ. 1mark

ii. Let(m′,0)and(n′,0)betheimagesof(m,0)and(n,0)respectively,underthetransformationT,wheremandnaredefinedinpart b.

Findthevaluesofm′andn′. 1mark


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Question 2 (18marks)Rebecca’sRoboticsmanufacturesthreetypesofcomponentsforrobots:sensors,motorsandcontrollers.Themanufacturingprocessesforeachtypeofcomponentareindependent.Itisknownthat8%ofallofthesensorsmanufacturedaredefective.

a. Arandomsampleoffivesensorsisselected.

Find,correcttofourdecimalplaces,theprobabilitythat

i. exactlytwooftheseselectedsensorsaredefective 2marks

ii. exactlytwooftheseselectedsensorsaredefective,giventhatatmosttwosensorsinthesamplearedefective. 2marks

b. Arandomsampleof50sensorsisselectedanditisfoundthattheproportionofdefectivesensorsinthissampleis0.08

Determineanapproximate90%confidenceintervalfortheproportionofdefectivesensors,correcttofourdecimalplaces. 2marks

Aholeisdrilledintoeachmotor.Thedepthoftheholeisnormallydistributedwithameanof20mmandastandarddeviationof0.3mm.

c. Whatistheprobabilitythat,forarandomlyselectedmotor,thedepthoftheholeisgreaterthan20.6mm?Giveyouranswercorrecttofourdecimalplaces. 1mark


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SECTION B – Question 2–continuedTURN OVER

Thedepthoftheholedrilledintoamotormustbewithin0.5mmofthemean,otherwisethemotorisdefective.

d. Whatistheprobabilitythatamotorisdefective,correcttofourdecimalplaces? 2marks

e. Rebeccadeliversanorderforfivesensorsandfivemotors.

Whatistheprobabilitythattheordercontainsexactlytwodefectivecomponents?Giveyouranswercorrecttothreedecimalplaces. 3marks

f. Aknobisattachedtoeachcontroller.Theheightofaknobisnormallydistributedwithameanof30mm.Iftheknobonacontrollerhasaheightgreaterthan30.4mmorlessthan29.6mm,thenthecontrollerisdefective.

Rebeccawantstoensurethatlessthan2%ofallcontrollersmanufacturedaredefective.

Whatisthemaximumstandarddeviationoftheheightofaknob,inmillimetres,thatcanbeattachedtoacontrollersothatlessthan2%ofcontrollersaredefective?Giveyouranswercorrecttotwodecimalplaces. 2marks


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SECTION B – continued

Theweight,w,ingrams,ofcontrollersismodelledbythefollowingprobabilitydensityfunction.

C ww w w

( )( ) ( )

=− − ≤ ≤

3640000

330 290 290 330

0

2

elsewhere

g. Determinethemeanweight,ingrams,ofthecontrollers. 2marks

h. Determinetheprobabilitythatarandomlyselectedcontrollerweighslessthanthemeanweightofthecontrollers.Giveyouranswercorrecttofourdecimalplaces. 2marks


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CONTINUES OVER PAGE


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Question 3 (13marks)Thefrontofabuildinghasalengthof80mandaheightof20m.Onthefrontofthebuildingisaglasspanelthatliesbetweentwoboundarycurves,asshownbytheshadedregioninthediagrambelow.Theboundarycurvesoftheregionaredefinedovertheinterval[0,80]withtherules

y x

y x

1

2

52 10

15

254 10

10

=

+

=

+

sin

sin

wherexisthehorizontaldistance,inmetres,andyistheverticaldistance,inmetres,measuredrelativetoanorigin,O,atthebottomleftcornerofthefrontofthebuilding.

x

y

O

a. Findthetotalareaoftheglasspanel,insquaremetres,correcttotwodecimalplaces. 2marks

LetDbetheverticaldistancebetweentheupperandlowerboundarycurves.

b. FindtheminimumvalueofD,inmetres,andthevalue(s)ofxwherethisminimumoccurs. 3marks


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c. WhatistheaveragevalueofD,inmetres,correcttotwodecimalplaces? 2marks

Theboundarycurvesovertheinterval[0,80]aregeneralisedto

c x a x

c x a x

1

22

1015

1010

( ) sin

( ) sin

=

+

=

+

wherea ∈ R+.

d. Theboundarycurvesdonotintersectfora ∈(0,p).

Findthemaximalvalueofp. 3marks

e. Findthevalueofaforwhichtheareaoftheglasspanelisamaximum.Alsostatethemaximumarea,insquaremetres,correcttotwodecimalplaces. 3marks


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Question 4 (20marks)Letf:(0,∞)→ R,f (x)=x–xloge(x).Partofthegraphof f isshownbelow.

1O 2 3ex

y

a. Findthevaluesofxforwhich

i. –1<f ′(x)<– 12

2marks

ii. 12<f ′(x)<1 1mark


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SECTION B – Question 4–continuedTURN OVER

b. i. Findtheequationofthetangenttothegraphoffatthepoint(a,f (a))intheform y=mx + c. 1mark

ii. Findthecoordinatesofthepointofintersectionofthetangenttothegraphoff at x=a

andthetangenttothegraphoff at xa

=1. 2marks

iii. Hence,findthecoordinatesofthepointofintersectionofthetangentstothegraphoff at

x=eand xe

=1.Expresseachcoordinateintermsofe. 1mark

c. i. Foravalueofb > e,thetangenttofatthepoint(b,f(b))andthetangenttofatthepoint(2,f(2))intersectthex-axisatthesamepoint.

Findthevalueofb. 2marks

ii. Ifthetangenttofatthepoint(p,f(p)),where1<p<e,andthetangenttofatthe point(q,f(q)),whereq > e,intersectonthex-axis,showthatpq=q p. 2marks


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d. Findtheequationofthetangenttothegraphoffatthepointwhere x e=12. 1mark

e. Partofthegraphoff,withthetangenttothegraphatPwhere x e=12,isshownbelow.

Eisthepointcorrespondingtothex-axisinterceptofthistangent. Fisthepointonthistangentwherey=1. Gisthepointcorrespondingtothelocalmaximumofthegraphoff. Histhepoint(1,0). Qisthepoint(e,0).

1

O 1 eH

G

Q

F

E x

y

1 12 2( , ( ))P e f e

i. FindthecoordinatesofthepointsEandF. 2marks

ii. FindtheareaofthequadrilateralEFGH. 2marks

iii. FindtheareaofthetriangleQGH. 1mark


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END OF QUESTION AND ANSWER BOOK

iv. Findanapproximationfortheareaoftheshadedregionbycalculatingtheaverageoftheareasfoundinpart e.ii.andpart e.iii. 1mark

v. Findtheerroroftheapproximationobtainedinpart e.iv.asapercentageoftheactualarea.Giveyouranswercorrecttotwodecimalplaces. 2marks

MATHEMATICAL METHODS

Written examination 2

FORMULA SHEET

Instructions

This formula sheet is provided for your reference.A question and answer book is provided with this formula sheet.

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

Victorian Certificate of Education 2018

© VICTORIAN CURRICULUM AND ASSESSMENT AUTHORITY 2018

MATHMETH EXAM 2

Mathematical Methods formulas

Mensuration

area of a trapezium 12a b h+( ) volume of a pyramid 1

3Ah

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 triangle12bc Asin ( )

volume of a cone13

2π r h

Calculus

ddx

x nxn n( ) = −1 x dxn

x c nn n=+

+ ≠ −+∫ 11

11 ,

ddx

ax b an ax bn n( )+( ) = +( ) −1 ( )( )

( ) ,ax b dxa n

ax b c nn n+ =+

+ + ≠ −+∫ 11

11

ddxe aeax ax( ) = e dx a e cax ax= +∫ 1

ddx

x xelog ( )( ) = 11 0x dx x c xe= + >∫ 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 ruleddxuv u dv

dxv dudx

( ) = + quotient ruleddx

uv

v dudx

u dvdx

v

=

−

2

chain ruledydx

dydududx

=

3 MATHMETH EXAM

END OF FORMULA SHEET

Probability

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

Pr(A|B) = Pr

PrA BB∩( )( )

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

Sample proportions

P Xn

=̂ mean E(P̂ ) = p

standard deviation

sd P p pn

(ˆ ) ( )=

−1 approximate confidence interval

,p zp p

np z

p pn

−−( )

+−( )

1 1ˆ ˆ ˆˆˆ ˆ