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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
SECTION 1 – continued
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
SECTION 1 – continuedTURN OVER
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
SECTION 1 – continued
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
SECTION 1 – continuedTURN OVER
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
SECTION 1 – continued
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
SECTION 1 – continuedTURN OVER
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.
2015SPECMATHEXAM2 12
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
13 2015SPECMATHEXAM2
SECTION 2 – continuedTURN OVER
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
2015SPECMATHEXAM2 14
SECTION 2 – Question 2–continued
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
15 2015SPECMATHEXAM2
SECTION 2 – continuedTURN OVER
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
2015SPECMATHEXAM2 16
SECTION 2 – Question 3–continued
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
17 2015SPECMATHEXAM2
SECTION 2 – Question 3–continuedTURN OVER
b. Findtheslopeoftheupperboundarycurvewhere t = π6.Giveyouranswerintheform
a bc
,wherea,bandcarepositiveintegers.1mark
c. i. Verifythatthecartesianequationoftheupperboundarycurveis y x
x=
−
2
22 1.
1mark
ii. Statethedomainforxoftheupperboundarycurve. 1mark
2015SPECMATHEXAM2 18
SECTION 2 – continued
d. Showthat ddx
x x
x
ddx
x xarcsin( )( ) =−
+ −( )2
11
2
22 bysimplifyingtheright-handsideofthis
equation. 2marks
e. Hencewritedownanantiderivative intermsofx,tobeevaluatedbetweentwoappropriateterminals,andfindtheareaoftheadvertisinglogo. 3marks
19 2015SPECMATHEXAM2
SECTION 2 – Question 4–continuedTURN OVER
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
2015SPECMATHEXAM2 20
SECTION 2 – continued
c. Showthatthevelocityofthehelicopterisperpendiculartoitsacceleration. 3marks
d. Findthespeedofthehelicopterinms–1,givingyouranswercorrecttotwodecimalplaces. 2marks
e. Atreetophaspositionvector� � � �r i j k= + +60 40 8 .
Findthedistanceofthehelicopterfromthetreetopafterithasbeentravellingfor45seconds.Giveyouranswerinmetres,correcttoonedecimalplace. 3marks
21 2015SPECMATHEXAM2
SECTION 2 – Question 5–continuedTURN OVER
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
2015SPECMATHEXAM2 22
SECTION 2 – Question 5–continued
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
23 2015SPECMATHEXAM2
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