2015 Specialist Mathematics Written examination 2 · 2019-03-14 · SPECIALIST MATHEMATICS Written examination 2 Monday 9 November 2015 Reading time: 3.00 pm to 3.15 pm (15 minutes)

SPECIALIST MATHEMATICSWritten examination 2

Monday 9 November 2015 Reading time: 3.00 pm to 3.15 pm (15 minutes) Writing time: 3.15 pm to 5.15 pm (2 hours)

QUESTION AND ANSWER BOOK

Structure of bookSection Number of

questionsNumber of questions

to be answeredNumber of

marks

1 22 22 222 5 5 58

Total 80

• Studentsarepermittedtobringintotheexaminationroom:pens,pencils,highlighters,erasers,sharpeners,rulers,aprotractor,setsquares,aidsforcurvesketching,oneboundreference,oneapprovedCAScalculatororCASsoftwareand,ifdesired,onescientificcalculator.CalculatormemoryDOESNOTneedtobecleared.

• StudentsareNOTpermittedtobringintotheexaminationroom:blanksheetsofpaperand/orcorrectionfluid/tape.

Materials supplied• Questionandanswerbookof23pageswithadetachablesheetofmiscellaneousformulasinthe

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.

©VICTORIANCURRICULUMANDASSESSMENTAUTHORITY2015

SUPERVISOR TO ATTACH PROCESSING LABEL HEREVictorian Certificate of Education 2015

STUDENT NUMBER

Letter

2015SPECMATHEXAM2 2

SECTION 1 – continued

Question 1

Theellipsex y−( )

+−( )

=29

34

12 2

canbeexpressedinparametricformas

A. x=2+3t and y t= + +3 2 1 2

B. x=2+3sec(t)andy=3+2tan(t)C. x=2+9cos(t)andy=3+4sin(t)D. x=3+2cos(t)andy=2+3sin(t)E. x=2+3cos(t)andy=3+2sin(t)

Question 2

Therangeofthefunctionwithrule f x x x( ) = −( ) −

2

21arcsin is

A. −[ ]π , 0

B. −

π π2 2,

C. −−( ) −( )

22

22

x xπ π,

D. 0 4,[ ]

E. 0, π[ ]

Question 3Ifbothaandcarenon-zerorealnumbers,therelationa2x2+(1–a2)y2 = c2 cannotrepresentA. acircle.B. anellipse.C. ahyperbola.D. asinglestraightline.E. apairofstraightlines.

SECTION 1

Instructions for Section 1Answerallquestionsinpencilontheanswersheetprovidedformultiple-choicequestions.Choosetheresponsethatiscorrect forthequestion.Acorrectanswerscores1,anincorrectanswerscores0.Markswillnotbedeductedforincorrectanswers.Nomarkswillbegivenifmorethanoneansweriscompletedforanyquestion.Taketheacceleration due to gravitytohavemagnitudegm/s2,whereg=9.8.

3 2015SPECMATHEXAM2

SECTION 1 – continuedTURN OVER

Question 4Thetwoasymptotesofaparticularhyperbolahavegradients

23and −

23respectivelyandintersectatthe

point(2,1).Onebranchofthehyperbolapassesthroughthepoint(5,5).Theequationofthehyperbolais

A. x y−( )

−−( )

=24

19

12 2

B. x y−( )

−−( )

=24

19

1736

2 2

C. y x−( )

−−( )

=19

24

1736

2 2

D. y x−( )−

−( )=

14

29

32 2

E. x y−( )

−−( )

=29

14

32 2

Question 5

Given z ii

=++

1 31

,themodulusandargumentofthecomplexnumberz5arerespectively

A. 2 2 56

and

B. 4 2 512

and

C. 4 2 712

and

D. 2 2 512

and

E. 4 212

and − π

Question 6Whichoneofthefollowingrelationshasagraphthatpassesthroughthepoint1+2iinthecomplexplane?

A. zz = 5

B. Arg( )z =π3

C. z z i− = −1 2

D. Re(z)=2Im(z)

E. z z+ = 2

2015SPECMATHEXAM2 4


Question 7

If z i= +3 3 ,thenz63isA. realandnegativeB. equaltoanegativerealmultipleofiC. realandpositiveD. equaltoapositiverealmultipleofiE. apositiverealmultipleof1 3+ i

Question 8Arelationthatdoes notrepresentacircleinthecomplexplaneisA. zz = 4B. z i z i+ = −3 2C. z i z− = + 2D. z i− + =1 4E. z z+ =2 4

Question 9Letz1 = r1cis(θ1)andz2 = r2cis(θ2),wherez1andz1z2areshownintheArganddiagrambelow;θ1andθ2areacuteangles.

Im(z)

Re(z)

z1z2

z1

O

AstatementthatisnecessarilytrueisA. r2 > 1

B. θ1 < θ2

C. zz

r1

21>

D. θ1 = θ2

E. r1 > 1

5 2015SPECMATHEXAM2


Question 10

Usingasuitablesubstitution,thedefiniteintegral x x dx2

0

1

3 1+( )∫ isequivalentto

A. 19

252

32

12

0

1

u u u du− +

∫

B. 127

252

32

12

1

4

u u u du− +

∫

C. 19

252

32

12

1

4

u u u du− +

∫

D. 127

252

32

12

0

1

u u u du− +

∫

E. 13

252

32

12

1

4

u u u du− +

∫

Question 11Thevelocity–timegraphforabodymovingalongastraightlineisshownbelow.

0

–0.5

–1

0.5

1

1 2 3 4 5 6

v

t

ThebodyfirstreturnstoitsinitialpositionwithinthetimeintervalA. (0,0.5)B. (0.5,1.5)C. (1.5,2.5)D. (2.5,3.5)E. (3.5,5)

2015SPECMATHEXAM2 6


Question 12

Givendydx

y= −1

3 and y=4 when x=2,then

A. y ex

= −− −( )23 3

B. y ex

= +− −( )23 3

C. y ex

=− −

42

3( )

D. y ey x

=− −4 23

( )

E. y ex

= +−( )23 3

Question 13

y

x–3

–1

–2

1

2

–2 –1 O 1 2 3

Thedirectionfieldforacertaindifferentialequationisshownabove.Thesolutioncurvetothedifferentialequationthatpassesthroughthepoint(–2.5,1.5)couldalsopassthroughA. (0,2)B. (1,2)C. (3,1)D. (3,–0.5)E. (–0.5,2)

7 2015SPECMATHEXAM2


Question 14Adifferentialequationthathasy = x sin(x)asasolutionis

A. d ydx

y2

2 0+ =

B. x d ydx

y2

2 0+ =

C. d ydx

y x2

2 + = −sin( )

D. d ydx

y x2

2 2+ = − cos( )

E. d ydx

y x2

2 2+ = cos( )

Question 15

Thecomponentoftheforce� � �F i j= +a b ,whereaandbarenon-zerorealconstants,inthedirectionofthe

vector� � �w i j= + ,is

A. a b+

2 �

w

B. �F

a b+

C. a ba b

++

2 2 �

F

D. a b+( )�w

E. a b+

2 �

w

2015SPECMATHEXAM2 8


Question 16

60° 30°

T~1

T~ 2

W~

Thediagramaboveshowsamasssuspendedinequilibriumbytwolightstringsthatmakeanglesof60°and30°withaceiling.Thetensionsinthestringsare

�T1and �

T2,andtheweightforceactingonthemassisW∼ .Thecorrectstatementrelatingthegivenforcesis

A. � � � �T T W1 2+ + = 0

B. � � � �T T W1 2+ − = 0

C. � � �T T1 2× + × =

12

32

0

D. � � �T T W1 2× + × =

32

12

E. � � �T T W1 2× + × =

12

32

Question 17

PointsA,BandChavepositionvectors� � �a i j= +2 ,

� � � �b i j k= − +3 and

� � �c j k= − +3 respectively.

ThecosineofangleABCisequalto

A. 56 10

B. 76 13

C. −16 13

D. −721 6

E. −26 13

9 2015SPECMATHEXAM2


Question 18Thepositionvectorsoftwomovingparticlesaregivenby

� � �r i j1 t t t( ) = +( ) + +( )2 4 3 22 and

� � �r i j2 6 4t t t( ) = ( ) + +( ) ,wheret≥0.

TheparticleswillcollideatA. 3 3 5

� �i j+ .

B. 6 5� �i j+

C. 3 4 5� �i j+ .

D. 0 5.� �i j+

E. 5 6� �i j+

Question 19Alightinextensiblestring passesoverasmoothpulley,asshownbelow,withparticlesofmass1kgandmkgattachedtotheendsofthestring.

1 kg

m kg

Iftheaccelerationofthe1kgparticleis4.9ms–2 upwards,thenmisequaltoA. 1B. 2C. 3D. 4E. 5

2015SPECMATHEXAM2 10

END OF SECTION 1

Question 20Anobjectismovinginastraightline,initiallyat5ms–1.Sixteensecondslater,itismovingat11ms–1intheoppositedirectiontoitsinitialvelocity.Assumingthattheaccelerationoftheobjectisconstant,after16secondsthedistance,inmetres,oftheobjectfromitsstartingpointisA. 24B. 48C. 73D. 96E. 128

Question 21AblockofmassMkgisonaroughhorizontalplane.AconstantforceofFnewtonsisappliedtotheblockatanangleofθ tothehorizontal,asshownbelow.Theblockhasaccelerationams–2andthecoefficientoffrictionbetweentheblockandtheplaneisμ.

θ

F

TheequationofmotionoftheblockinthehorizontaldirectionisA. F–μMg = MaB. Fcos(θ)–μMg = MaC. Fsin(θ)–μ(Mg–Fcos(θ )) = MaD. Fcos(θ )–μ(Fsin(θ )–Mg) = MaE. Fcos(θ )–μ(Mg–Fsin(θ )) = Ma

Question 22Aballisthrownverticallyupwithaninitialvelocityof7 6 ms–1,andissubjecttogravityandairresistance.

Theaccelerationoftheballisgivenby��x v= − +( )9 8 0 1 2. . ,wherexmetresisitsverticaldisplacement,andvms–1isitsvelocityattimetseconds.Thetimetakenfortheballtoreachitsmaximumheightis

A. 3

B. 5

21 2

C. loge(4)

D. 1021 2

E. 10loge(4)

11 2015SPECMATHEXAM2

SECTION 2 – Question 1–continuedTURN OVER

Question 1 (12marks)

Consider y x= −2 2sin ( ).

a. Usetherelationy2=2–sin2(x)tofinddydx intermsofxandy. 1mark

b. i. Writedownthevaluesofywherex=0andwhere x = π2. 1mark

ii. Writedownthevaluesof dydx

wherex=0andwhere x = π2. 1mark

SECTION 2

Instructions for Section 2Answerallquestionsinthespacesprovided.Unlessotherwisespecified,anexactanswerisrequiredtoaquestion.Inquestionswheremorethanonemarkisavailable,appropriateworkingmust beshown.Unlessotherwiseindicated,thediagramsinthisbookarenotdrawntoscale.Taketheacceleration due to gravitytohavemagnitudegm/s2,whereg=9.8.


SECTION 2 – Question 1–continued

Nowconsiderthefunction f withrule f x x( ) sin ( )= −2 2 for 02

≤ ≤x π .

c. Findtherulefortheinversefunction f –1,andstatethedomainandrangeof f –1. 3marks

d. Sketchandlabelthegraphsof f and f –1ontheaxesbelow. 2marks

y

x

1

0 1



e. Thegraphsof f and f–1 intersectatthepointP(a,a).

Finda,correcttothreedecimalplaces. 1mark

Theregionboundedbythegraphof f,thecoordinateaxesandthelinex=1isrotatedaboutthex-axistoformasolidofrevolution.

f. i. Writedownadefiniteintegralintermsofxthatgivesthevolumeofthissolidofrevolution. 2marks

ii. Findthevolumeofthissolid,correcttoonedecimalplace. 1mark



Question 2 (12marks)a. i. OntheArganddiagram below,plotandlabelthepoints0+0i and 1 3+ i . 2marks

–3 –2 –1

–1

1

2

3

–2

–3

2 31O

Im(z)

Re(z)

ii. OnthesameArganddiagramabove, sketchtheline z i z− +( ) =1 3 and thecircle z − =2 1 . 2marks

iii. Usethefactthattheline z i z− +( ) =1 3 passesthroughthepointz=2,orotherwise,tofindtheequationofthislineincartesianform. 1mark



iv. Findthepointsofintersectionofthelineandthecircle,expressingyouranswersintheforma + ib. 3marks

b. i. Considertheequationz2–4cos(α)z+4=0,whereαisarealconstantand02

< <απ .

Findtherootsz1andz2ofthisequation,intermsofα,expressingyouranswersinpolarform. 3marks

ii. Findthevalueofαforwhich Argzz1

2

56

=

π. 1mark



Question 3 (10marks)Amanufacturerofbowtieswishestodesignanadvertisinglogo,representedbelow,wheretheupperboundarycurveinthefirstandsecondquadrantsisgivenbytheparametricrelations

x=sin(t), y t t=12sin ( ) tan ( ) for t∈ −

π π3 3, .

Thelogoissymmetricalaboutthex-axis.

y

x

–1

1

–1 1O

a. Findanexpressionfor dydx

intermsoft. 2marks



b. Findtheslopeoftheupperboundarycurvewhere t = π6.Giveyouranswerintheform

a bc

,wherea,bandcarepositiveintegers.1mark

c. i. Verifythatthecartesianequationoftheupperboundarycurveis y x

x=

−

2

22 1.

1mark

ii. Statethedomainforxoftheupperboundarycurve. 1mark



d. Showthat ddx

x x

x

ddx

x xarcsin( )( ) =−

+ −( )2

11

2

22 bysimplifyingtheright-handsideofthis

equation. 2marks

e. Hencewritedownanantiderivative intermsofx,tobeevaluatedbetweentwoappropriateterminals,andfindtheareaoftheadvertisinglogo. 3marks



Question 4 (12marks)Thepositionvector

�r( )t ,fromoriginO,ofamodelhelicoptertsecondsafterleavingthegroundis

givenby

� � �r i( ) cos sint t t

= +

+ +

50 25

3050 25

30π π jj k+

25t�

where�i isaunitvectortotheeast,

�j isaunitvectortothenorthand

�k isaunitvectorvertically

up.Displacementcomponentsaremeasuredinmetres.

a. i. Findthetime,inseconds,requiredforthehelicoptertogainanaltitudeof60m. 1mark

ii. FindtheangleofelevationfromOofthehelicopterwhenitisatanaltitudeof60m.Giveyouranswerindegrees,correcttothenearestdegree. 2marks

b. Afterhowmanysecondswillthehelicopterfirstbedirectlyabovethepointoftake-off? 1mark



c. Showthatthevelocityofthehelicopterisperpendiculartoitsacceleration. 3marks

d. Findthespeedofthehelicopterinms–1,givingyouranswercorrecttotwodecimalplaces. 2marks

e. Atreetophaspositionvector� � � �r i j k= + +60 40 8 .

Findthedistanceofthehelicopterfromthetreetopafterithasbeentravellingfor45seconds.Giveyouranswerinmetres,correcttoonedecimalplace. 3marks



Question 5 (12marks)Aboatrampattheedgeofadeeplakeisinclinedatanangleof10°tothehorizontal.A250kg boattrailerontherampisunhitchedfromacarandamanattemptstolowerthetrailerdowntherampusingaropeparalleltotheramp,asshowninthediagrambelow.

lake 10°

boat trailer

rope

F~

Assumenegligiblefrictionforcesinthissituation.

a. Calculatetheconstantforce,Fnewtons,thatwouldberequiredtopreventthetrailerfrommovingdowntheramp.Giveyouranswercorrecttothenearestnewton. 1mark

b. Ifthemanexertsaforceof200Nviatherope,findtheaccelerationofthetrailerdowntheramp,assumingnegligiblefrictionforcesandairresistance.Giveyouranswerinms–2,correcttothreedecimalplaces. 2marks

c. Usingyourresultforaccelerationfrom part b.,findthespeedofthetrailerinms–1,correcttotwodecimalplaces,afterithasmoved30mdowntheramp,havingstartedfromrest. 2marks



Whenthetrailerrollsintothewater,itstops,thensinksverticallyfromrestsothatitsdepth xmetresaftertsecondsisgivenbythedifferentialequation

d xdt

dxdt

2

2 1 4 7= −

.

d. i. Showthattheabovedifferentialequationcanbewrittenas

1 4 1 77

. dxdv v

= − +−

, where v dxdt

= . 2marks

ii. Hence,showbyintegrationthat1.4x=–v–7loge(7–v)+7loge(7). 1mark

WhenthetrailerhassunktoadepthofDmetres,itisdescendingatarateof5ms–1.

iii. FindD,correcttoonedecimalplace. 1mark


iv. Writedownadefiniteintegralforthetime,inseconds,takenforthetrailertosinktothedepthofDmetresandevaluatethisintegralcorrecttoonedecimalplace. 3marks

END OF QUESTION AND ANSWER BOOK

SPECIALIST MATHEMATICS

Written examinations 1 and 2

FORMULA SHEET

Instructions

Detach this formula sheet during reading time.

This formula sheet is provided for your reference.

© VICTORIAN CURRICULUM AND ASSESSMENT AUTHORITY 2015

SPECMATH 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: 43 π r

3

area of a triangle: 12 bc Asin

sine rule: aA

bB

cCsin sin sin

= =

cosine rule: c2 = a2 + b2 – 2ab cos C

Coordinate geometry

ellipse: x ha

y kb

−( )+

−( )=

2

2

2

2 1 hyperbola: x ha

y kb

−( )−

−( )=

2

2

2

2 1

Circular (trigonometric) functionscos2(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

+ =+

−1 tan( ) tan( ) tan( )tan( ) tan( )

x y x yx 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=

−

function sin–1 cos–1 tan–1

domain [–1, 1] [–1, 1] R

range −

π π2 2, [0, �] −

π π2 2,

3 SPECMATH

Algebra (complex numbers)z = x + yi = r(cos θ + i sin θ) = r cis θ

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)

Calculusddx

x nxn n( ) = −1

x dx

nx c nn n=

++ ≠ −+∫ 1

111 ,

ddxe aeax ax( ) =

e dx

ae cax ax= +∫ 1

ddx

xxelog ( )( ) = 1

1xdx 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 dx

aax c= +∫

ddx

xx

sin−( ) =−

12

1

1( )

1 02 2

1

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

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

constant (uniform) acceleration: v = u + at s = ut +12

at2 v2 = u2 + 2as s = 12

(u + v)t

TURN OVER

SPECMATH 4

END OF FORMULA SHEET

Vectors in two and three dimensions

r i j k~ ~ ~ ~= + +x y z

| r~ | = x y z r2 2 2+ + = r~ 1. r~ 2 = r1r2 cos θ = x1x2 + y1y2 + z1z2

Mechanics

momentum: p v~ ~= m

equation of motion: R a~ ~= m

friction: F ≤ µN

Documents

2015 Specialist Mathematics Written examination 2 · 2019-03-14 · SPECIALIST MATHEMATICS Written examination 2 Monday 9 November 2015 Reading time: 3.00 pm to 3.15 pm (15 minutes)