2017 specialist mathematics written examination 2 · pdf filesection a – ot turn over 9...

SPECIALIST MATHEMATICSWritten examination 2

Friday 9 June 2017 Reading time: 10.00 am to 10.15 am (15 minutes) Writing time: 10.15 am to 12.15 pm (2 hours)

QUESTION AND ANSWER BOOK

Structure of bookSection Number of

questionsNumber of questions

to be answeredNumber of

marks

A 20 20 20B 6 6 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• Questionandanswerbookof23pages.• Formulasheet.• Answersheetformultiple-choicequestions.

Instructions• Writeyourstudent numberinthespaceprovidedaboveonthispage.• Checkthatyournameandstudent numberasprintedonyouranswersheetformultiple-choice

questionsarecorrect,andsignyournameinthespaceprovidedtoverifythis.• Unlessotherwiseindicated,thediagramsinthisbookarenotdrawntoscale.• 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.

©VICTORIANCURRICULUMANDASSESSMENTAUTHORITY2017

SUPERVISOR TO ATTACH PROCESSING LABEL HEREVictorian Certificate of Education 2017

STUDENT NUMBER

Letter

SECTION A – continued

2017SPECMATHEXAM2(NHT) 2

Question 1Thenumberofasymptotesofthegraphofthefunctionwithrule f x

x xx x

( ) = − ++ −

3

27 53 4

isA. 0B. 1C. 2D. 3E. 4

Question 2Theequationx2 + y2 + 2ky+4=0,wherekisarealconstant,willrepresentacircleonlyifA. k > 2B. k < –2C. k≠±2D. k<–2ork > 2E. –2 < k < 2

Question 3Forthefunctionf:R → R,f(x)=karctan(ax – b)+c,wherek>0,c>0anda,b ∈ R,f(x)>0if

A. c k<

π2

B. c k≥

π2

C. x ba

>

D. c k+ >π2

E. c ≥ π2

SECTION A – Multiple-choice questions

Instructions for Section AAnswerallquestionsinpencilontheanswersheetprovidedformultiple-choicequestions.Choosetheresponsethatiscorrectforthequestion.Acorrectanswerscores1;anincorrectanswerscores0.Markswillnotbedeductedforincorrectanswers.Nomarkswillbegivenifmorethanoneansweriscompletedforanyquestion.Unlessotherwiseindicated,thediagramsinthisbookarenot drawntoscale.Taketheacceleration due to gravitytohavemagnitudegms–2,whereg=9.8

SECTION A – continuedTURN OVER

3 2017SPECMATHEXAM2(NHT)

Question 4Ifsin(θ + ϕ)=aandsin(θ – ϕ)=b,thensin(θ)cos(ϕ)isequaltoA. ab

B. a b2 2+

C. ab

D. a b2 2−

E. a b+

2

Question 5GiventhatA,B,CandDarenon-zerorationalnumbers,theexpression 3 1

2 2x

x x+−( )

canberepresentedinpartialfractionformas

A. Ax

Bx

+−( )2

B. Ax

Bx

+−( )2 2

C. Ax

Bx

Cx

+−

+−( ) ( )2 2 2

D. Ax

Bx

Cx

+ +−2 2( )

E. Ax

Bxx

Cx Dx

+−

++−( ) ( )2 2 2



Question 6

Re(z)

Im(z)

–3 –2 –1 0 1 2 3

3

2

1

–1

–2

–3

S

TherelationthatdefinesthelineSaboveisA. z z i+ = +2 2

B. Arg( )z =34π

C. z z i− = +2 2

D. Im( )z =

+ −

Arg Arg3

4 4π π

E. z z i− = −2 2



Question 7

sin ( )cos ( )3 2

4

3

x x dx( )∫π

π

isequivalentto

A. u u du u x4 2

12

12

−( ) =∫ where cos( )

B. − −( ) =∫ u u du u x2 4

12

12

where cos( )

C. − −( ) =∫ u u du u x2 4

4

3

π

π

where sin( )

D. u u du u x2 4

4

3

−( ) =∫π

π

where sin( )

E. − −( ) =∫ u u du u x2 4

12

12

where sin( )



Question 8

x

y

2

1

–1

–1 10

Thedifferentialequationthatbestrepresentsthedirectionfieldaboveis

A. dydx

x y= − 2

B. dydx

y x= −

C. dydx

y x= −2 2

D. dydx

y x= −2

E. dydx

y x= +



Question 9ThegradientofthetangenttoacurveatanypointP(x,y)ishalfthegradientofthelinesegmentjoining PandthepointQ(–1,1).Thecoordinatesofpointsonthecurvesatisfythedifferentialequation

A. dydx

yx

=+−1

2 1( )

B. dydx

yx

=−+

2 11

( )

C. dydx

xy

=−+1

2 1( )

D. dydx

xy

=−+

2 11

( )

E. dydx

yx

=−+1

2 1( )

Question 10

Asolutiontothedifferentialequationdydx

x y x yex y=

+ − −+

cos( ) cos( )canbeobtainedfrom

A. eydy x

edx

y

xsin ( )sin ( )∫ ∫= −2

B. eydy

edx

y

xcos( )∫ ∫= 2

C. eydy x

edx

y

xcos( )cos( )∫ ∫= −2

D. eydy e x dx

yx

−−∫ ∫=sin ( )sin ( )2

E. eydy x

edx

y

xcos( )sin ( )∫ ∫= 2

Question 11

Twoparticleshavepositionsgivenby

r i j123 4= −( ) + +( )t t b and

r i j ,22 25 1= + −( )t t wheret≥0andbis

arealconstant.Theparticleswillcollideifthevalueofbis

A. 2 33

−

B. 3 1−

C. 2 33

+

D. − −2 33

E. − −3 1



Question 12If

u i j k= + −3 6 2 and

v i j k ,= + −2 2 thenthevectorresoluteof

u inthedirectionof

v is

A. 2073 6 2

i j k+ −( )

B. 2092 2

i j k+ −( )

C. 20493 6 2

i j k+ −( )

D. 2032 2

i j k+ −( )

E. 373 6 2

i j k+ −( )

Question 13

D

A B

C

O

LetOABCDbearightsquarepyramidwhere� � � �a b c and d= = = =OA OB OC OD, , .

AnequationcorrectlyrelatingthesevectorsisA.

a c b d+ = +

B.

a c . d c−( ) −( ) = 0C.

a d b c+ = +

D.

a d . c b−( ) −( ) = 0E.

a b c d+ = +

Question 14Giventhatthevectors

a i j k b i j k and c i j k= + − = − + = − +, 2 2 4λ arelinearly dependent,thevalue ofλisA. –10B. –8C. 2D. 4E. 8



Question 15 Aparticleofmass2kghasaninitialvelocityof

i j− 6 ms–1.Afterachangeofmomentumof6i j

− 2 kgms–1,theparticle’svelocity,inms–1,isA. 3i j

−

B. 2i j

−12

C. 4i j

− 7

D. 2i j

+ 5

E. 11i j

+ 2

Question 16ApersonofmassMkgcarryingabagofmassmkgisstandinginaliftthatisacceleratingdownwards at ams–2.TheforceoftheliftflooractingonthepersonhasmagnitudeA. Mg + mgB. Mg+(M + m)aC. Mg–(M + m)aD. (M + m)(g + a)E. (M + m)(g – a)

Question 17Theacceleration,ams–2,ofaparticlemovinginastraightlineisgivenbya = v2+1,wherevisthevelocityoftheparticleatanytimet.TheinitialvelocityoftheparticlewhenatoriginOis2ms–1.ThedisplacementoftheparticlefromOwhenitsvelocityis3ms–1isA. loge(2)

B. 12

103

loge

C. 12

2log ( )e

D. 12

52

loge

E. loge45


END OF SECTION A

Question 18Xisarandomvariablewithameanof5andastandarddeviationof4,andYisarandomvariablewithameanof3andastandarddeviationof2.IfXandYareindependentrandomvariablesandZ = X – 2Y,thenZwillhavemeanμandstandarddeviationσgivenbyA. μ=–1,σ = 0

B. μ=–1,σ = 4 2

C. μ=2,σ = 8

D. μ=2,σ = 4 2

E. μ=–1,σ = 2 6

Question 19Thepetrolconsumptionofaparticularmodelofcarisnormallydistributedwithameanof 12L/100kmandastandarddeviationof2L/100km.Theprobabilitythattheaveragepetrolconsumptionof16suchcarsexceeds13L/100kmisclosesttoA. 0.0104B. 0.0193C. 0.0228D. 0.3085E. 0.3648

Question 20Themassofsuspendedmatterintheairinaparticularlocalityisnormallydistributedwithameanof μmicrogramspercubicmetreandastandarddeviationofσ=8microgramspercubicmetre.Themeanof 100randomlyselectedairsampleswasfoundtobe40microgramspercubicmetre.Basedonthis,a90%confidenceintervalforμ,correcttotwodecimalplaces,isA. (38.68,41.32)B. (26.84,53.16)C. (38.43,41.57)D. (24.32,55.68)E. (37.93,42.06)


SECTION B – Question 1–continuedTURN OVER

Question 1 (12marks)

a. i. Useanappropriatedoubleangleformulawith t =

tan 5

12π todeduceaquadratic

equationoftheformt2 + bt + c=0,wherebandcarerealvalues. 2marks

ii. Henceshowthat tan .512

2 3π

= + 1mark

SECTION B

Instructions for Section BAnswerallquestionsinthespacesprovided.Unlessotherwisespecified,anexactanswerisrequiredtoaquestion.Inquestionswheremorethanonemarkisavailable,appropriateworkingmust beshown.Unlessotherwiseindicated,thediagramsinthisbookarenotdrawntoscale.Taketheacceleration due to gravitytohavemagnitudegms–2,whereg=9.8


SECTION B – Question 1–continued

Consider f R f x x: , , ( ) arctan3 6 3 33 6

+

→ =

−

π .

b. Sketchthegraphoffontheaxesbelow,labellingtheendpointswiththeircoordinates. 3marks

1

0.5

–0.5

–10 –5 5 10

y

xO

c. Theregionbetweenthegraphoffandthey-axisisrotatedaboutthey-axistoformasolidofrevolution.

i. Writedownadefiniteintegralintermsofythatgivesthevolumeofthesolidformed. 2marks

ii. Findthevolumeofthesolid,correcttothenearestinteger. 1mark


SECTION B – continuedTURN OVER

d. Afishpondthathasashapeapproximatelylikethatofthesolidofrevolutioninpart c.isbeingfilledwithwater.Whenthedepthishmetres,thevolume,Vm3,ofwaterinthepondisgivenby

V h h= +

− −tan π6

33

Ifwaterisflowingintothepondatarateof0.03m3perminute,findtherateatwhichthedepthisincreasingwhenthedepthis0.6m.Giveyouranswerinmetresperminute,correcttothreedecimalplaces. 3marks



Question 2 (11marks)Onerootofaquadraticequationwithrealcoefficientsis 3 + i.

a. i. Writedowntheotherrootofthequadraticequation. 1mark

ii. Hencedeterminethequadraticequation,writingitintheformz2 + bz + c=0. 2marks

b. Plotandlabeltherootsof z z z3 22 3 4 0− + = ontheArganddiagrambelow. 3marks

Im(z)

Re(z)

3

2

1

–1

–2

–3

O–3 –2 –1 1 2 3



c. Findtheequationofthelinethatistheperpendicularbisectorofthelinesegmentjoiningtheoriginandthepoint 3 + i.Expressyouranswerintheformy = mx + c. 2marks

d. Thethreerootsplottedinpart b.lieonacircle.

Findtheequationofthiscircle,expressingitintheform z − =α β,whereα,β ∈ R. 3marks



Question 3 (12marks)BacteriaarespreadingoveraPetridishataratemodelledbythedifferentialequation

dPdt

P P= −( )21 ,0<P<1

wherePistheproportionofthedishcoveredafterthours.

a. i. Express21P P−( ) inpartialfractionform. 1mark

ii. Henceshowbyintegrationthat t c PPe

−=

−

2 1

log , wherecisaconstantofintegration. 2marks

iii. IfhalfofthePetridishiscoveredbythebacteriaatt=0,expressPintermsoft. 2marks



Afteronehour,atoxinisaddedtothePetridish,whichharmsthebacteriaandreducestheirrateofgrowth.Thedifferentialequationthatmodelstherateofgrowthisnow

dPdt

P P P= −( ) −21

20 for t≥1

b. FindthelimitingvalueofP,whichisthemaximumpossibleproportionofthePetridishthatcannowbecoveredbythebacteria.Giveyouranswercorrecttothreedecimalplaces. 2marks

c. Thetotaltime,Thours,measuredfromtimet=0,neededforthebacteriatocover80%ofthePetridishisgivenby

T P P P dP s

q

r

= −( ) −

+

1

21

20

whereq,rands ∈ R.

Findthevaluesofq,rands,givingthevalueofqcorrecttotwodecimalplaces. 2marks

d. GiventhatP=0.75whent =3,useEuler’smethodwithastepsizeof0.5toestimatethevalueofPwhent=3.5.Giveyouranswercorrecttothreedecimalplaces. 3marks



Question 4 (8marks)Acricketerhitsaballattimet=0secondsfromanoriginOatgroundlevelacrossalevelplayingfield.Thepositionvector

r( )t ,fromO,oftheballaftertsecondsisgivenby

r i j( ) .t t t t= + −( )15 15 3 4 9 2 ,where

i isaunitvectorintheforwarddirection,

j isaunitvector

verticallyupanddisplacementcomponentsaremeasuredinmetres.

a. Findtheinitialvelocityoftheballandtheinitialangle,indegrees,ofitstrajectorytothehorizontal. 2marks

b. Findthemaximumheightreachedbytheball,givingyouranswerinmetres,correcttotwodecimalplaces. 2marks

c. Findthetimeofflightoftheball.Giveyouranswerinseconds,correcttothreedecimalplaces. 1mark



d. Findtherangeoftheballinmetres,correcttoonedecimalplace. 1mark

e. Afielder,morethan40mfromO,catchestheballataheightof2mabovetheground.

HowfarhorizontallyfromOisthefielderwhentheballiscaught?Giveyouranswer inmetres,correcttoonedecimalplace. 2marks



Question 5 (10marks)A5kgmassisinitiallyheldatrestonasmoothplanethatisinclinedat30°tothehorizontal.Themassisconnectedbyalightinextensiblestringpassingoverasmoothpulleytoa3kgmass,whichinturnisconnectedtoa2kgmass.The5kgmassisreleasedfromrestandallowedtoaccelerateuptheplane.Takeaccelerationtobepositiveinthedirectionsindicated.

positiveacceleration

5 kg

T1

T1

3 kg

2 kg

T230°

positive acceleration

a. Writedownanequationofmotion,inthedirectionofmotion,foreachmass. 3marks

b. Showthattheaccelerationofthe5kgmassisg4 ms–2. 1mark

21 2017 SPECMATH EXAM 2 (NHT)


c. Find the tensions T1 and T2 in the string in terms of g. 2 marks

d. Find the momentum of the 5 kg mass, in kg ms–1, after it has moved 2 m up the plane, giving your answer in terms of g. 2 marks

e. A resistance force R acting parallel to the inclined plane is added to hold the system in equilibrium, as shown in the diagram below.

5 kg

T1

T1

3 kg

2 kg

T230°

R

Find the magnitude of R in terms of g. 2 marks



Question 6 (7marks)Abankclaimsthattheamountitlendsforhousingisnormallydistributedwithameanof$400000andastandarddeviationof$30000.Aconsumerorganisationbelievesthattheaverageloanamountishigherthanthebankclaims.Tocheckthis,theconsumerorganisationexaminesarandomsampleof25loansandfindsthesamplemeantobe$412000.

a. Writedownthetwohypothesesthatwouldbeusedtoundertakeaone-sidedtest. 1mark

b. Writedownanexpressionforthepvalueforthistestandevaluateittofourdecimalplaces. 2marks

c. Statewithareasonwhetherthebank’sclaimshouldberejectedatthe5%levelofsignificance. 1mark

d. Whatisthelargestvalueofthesamplemeanthatcouldbeobservedbeforethebank’s claimwasrejectedatthe5%levelofsignificance?Giveyouranswercorrecttothenearest 10dollars. 1mark


e. Iftheaverageloanmadebythebankisactually$415000andnot$400000asoriginallyclaimed,whatistheprobabilitythatarandomselectionof25loanshasasamplemeanthatisatmost$410000?Giveyouranswercorrecttothreedecimalplaces. 2marks

END OF QUESTION AND ANSWER BOOK

SPECIALIST MATHEMATICS

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 2017

© VICTORIAN CURRICULUM AND ASSESSMENT AUTHORITY 2017

SPECMATH EXAM 2

Specialist Mathematics formulas

Mensuration

area of a trapezium 12 a b h+( )

curved surface area of a cylinder 2π rh

volume of a cylinder π r2h

volume of a cone 13π r2h

volume of a pyramid 13 Ah

volume of a sphere 43π r3

area of a triangle 12 bc Asin ( )

sine ruleaA

bB

cCsin ( ) sin ( ) sin ( )

= =

cosine rule c2 = a2 + b2 – 2ab cos (C )

Circular functions

cos2 (x) + sin2 (x) = 1

1 + tan2 (x) = sec2 (x) cot2 (x) + 1 = cosec2 (x)

sin (x + y) = sin (x) cos (y) + cos (x) sin (y) sin (x – y) = sin (x) cos (y) – cos (x) sin (y)

cos (x + y) = cos (x) cos (y) – sin (x) sin (y) cos (x – y) = cos (x) cos (y) + sin (x) sin (y)

tan ( ) tan ( ) tan ( )tan ( ) tan ( )

x y x yx y

+ =+

−1tan ( ) tan ( ) tan ( )

tan ( ) tan ( )x y x y

x y− =

−+1

cos (2x) = cos2 (x) – sin2 (x) = 2 cos2 (x) – 1 = 1 – 2 sin2 (x)

sin (2x) = 2 sin (x) cos (x) tan ( ) tan ( )tan ( )

2 21 2x x

x=

−

3 SPECMATH EXAM

TURN OVER

Circular functions – continued

Function sin–1 or arcsin cos–1 or arccos tan–1 or arctan

Domain [–1, 1] [–1, 1] R

Range −

π π2 2, [0, �] −

π π2 2,

Algebra (complex numbers)

z x iy r i r= + = +( ) =cos( ) sin ( ) ( )θ θ θcis

z x y r= + =2 2 –π < Arg(z) ≤ π

z1z2 = r1r2 cis (θ1 + θ2)zz

rr

1

2

1

21 2= −( )cis θ θ

zn = rn cis (nθ) (de Moivre’s theorem)

Probability and statistics

for random variables X and YE(aX + b) = aE(X) + bE(aX + bY ) = aE(X ) + bE(Y )var(aX + b) = a2var(X )

for independent random variables X and Y var(aX + bY ) = a2var(X ) + b2var(Y )

approximate confidence interval for μ x z snx z s

n− +

,

distribution of sample mean Xmean E X( ) = µvariance var X

n( ) = σ2

SPECMATH EXAM 4

END OF FORMULA SHEET

Calculus

ddx

x nxn n( ) = −1 x dxn

x c nn n=+

+ ≠ −+∫ 11

11 ,

ddxe aeax ax( ) = e dx

ae cax ax= +∫ 1

ddx

xxelog ( )( ) = 1 1

xdx x ce= +∫ log

ddx

ax a axsin ( ) cos( )( ) = sin ( ) cos( )ax dxa

ax c= − +∫ 1

ddx

ax a axcos( ) sin ( )( ) = − cos( ) sin ( )ax dxa

ax c= +∫ 1

ddx

ax a axtan ( ) sec ( )( ) = 2 sec ( ) tan ( )2 1ax dxa

ax c= +∫ddx

xx

sin−( ) =−

12

1

1( ) 1 0

2 21

a xdx x

a c a−

=

+ >−∫ sin ,

ddx

xx

cos−( ) = −

−

12

1

1( ) −

−=

+ >−∫ 1 0

2 21

a xdx x

a c acos ,

ddx

xx

tan−( ) =+

12

11

( ) aa x

dx xa c2 2

1

+=

+

−∫ tan

( )( )

( ) ,ax b dxa n

ax b c nn n+ =+

+ + ≠ −+∫ 11

11

( ) logax b dxa

ax b ce+ = + +−∫ 1 1

product rule ddxuv u dv

dxv dudx

( ) = +

quotient rule ddx

uv

v dudx

u dvdx

v

=

−

2

chain rule dydx

dydududx

=

Euler’s method If dydx

f x= ( ), x0 = a and y0 = b, then xn + 1 = xn + h and yn + 1 = yn + h f (xn)

acceleration a d xdt

dvdt

v dvdx

ddx

v= = = =

2

221

2

arc length 1 2 2 2

1

2

1

2

+ ′( ) ′( ) + ′( )∫ ∫f x dx x t y t dtx

x

t

t( ) ( ) ( )or

Vectors in two and three dimensions

r = i + j + kx y z

r = + + =x y z r2 2 2

� � � � �ir r i j k= = + +ddt

dxdt

dydt

dzdt

r r1 2. cos( )= = + +r r x x y y z z1 2 1 2 1 2 1 2θ

Mechanics

momentum

p v= m

equation of motion

R a= m

2017 specialist mathematics written examination 2 · pdf filesection a – ot turn over 9...

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