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STUDENT NUMBER Letter

SPECIALIST MATHEMATICSWritten examination 2

Monday 12 November 2012 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

12

225

225

2258

Total 80

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

• StudentsareNOTpermittedtobringintotheexaminationroom:blanksheetsofpaperand/orwhiteoutliquid/tape.

Materials supplied• Questionandanswerbookof25pageswithadetachablesheetofmiscellaneousformulasinthe

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.

©VICTORIANCURRICULUMANDASSESSMENTAUTHORITY2012

SUPERVISOR TO ATTACH PROCESSING LABEL HEREVictorian Certificate of Education 2012

2012SPECMATHEXAM2 2

SECTION 1–continued

Question 1Thegraphwithequation y

x x=

− −1

2 62hasasymptotesgivenby

A. x = − 32, x =2andy = 1

B. x = − 32and x =2only

C. x =32, x =–2andy = 0

D. x = − 32, x =2andy = 0

E. x =32and x =–2only

Question 2Arectangleisdrawnsothatitssideslieonthelineswithequationsx =–2,x=4,y =–1andy =7.Anellipseisdrawninsidetherectanglesothatitjusttoucheseachsideoftherectangle.Theequationoftheellipsecouldbe

A. x y2 2

9 161+ =

B. ( ) ( )x y+

++

=1

93

161

2 2

C. ( ) ( )x y−

+−

=1

93

161

2 2

D. ( ) ( )x y+

++

=1 1

2 2

363

64

E. ( ) ( )x y−

+−

=1

363

641

2 2

SECTION 1

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

3 2012 SPECMATH EXAM 2

SECTION 1 – continuedTURN OVER

Question 3The graph of y = f (x) is shown below.

y

x

2

1

–1

O

–2

–2 –1 1 2

All of the axes below have the same scale as the axes in the diagram above.

The graph of yf x

=1( )

is best represented by

A.

y

xO

B.

y

xO

C.

y

xO

D.

xO

y

E.

y

xO

2012SPECMATHEXAM2 4

SECTION 1–continued

Question 4Thedomainandrangeofthefunctionwithrule f x x( ) ( )= − +arccos 2 1

2π arerespectively

A. [–2,0]and[0,π ]

B. [–2,0]and π π2

32

,

C. [0,1]and[0,π ]

D. [0,1]and π π2

32

,

E. [0,π ]and[0,1]

Question 5

If z = 2 45

cis _ π andw = z9,then

A. w =16 2 365

cis π

B. w = −16 25

cis π

C. w =16 2 45

cis π

D. w = −9 25

cis π

E. w = 9 2 45

cis π

Question 6Foranycomplexnumberz,thelocationonanArganddiagramofthecomplexnumberu i z= 3 canbefoundby

A. rotatingzthrough 32π inananticlockwisedirectionabouttheorigin

B. reflectingzaboutthex-axisandthenreflectingaboutthey-axis

C. reflectingzaboutthey-axisandthenrotatinganticlockwisethrough π2 abouttheorigin

D. reflectingzaboutthex-axisandthenrotatinganticlockwisethroughπ2 abouttheorigin

E. rotatingzthrough 32π inaclockwisedirectionabouttheorigin

5 2012SPECMATHEXAM2

SECTION 1–continuedTURN OVER

Question 7Thesetofpointsinthecomplexplanedefinedby| z + 2i | = | z |correspondstoA. thepointgivenbyz = –iB. thelineIm(z) = –1C. thelineIm(z) = –iD. thelineRe(z) = –1E. thecirclewithcentre–2iandradius1

Question 8Ifz = a + bi,wherebothaandbarenon-zerorealnumbersandz ∈ C,whichofthefollowingdoesnot representarealnumber?A. z z+

B. | z |C. z z

D. z2 – 2abi

E. z z z z−( ) +( )

Question 9Euler’sformulaisusedtofindy2,where

dydx

x= cos( ),x0=0,y0=1andh=0.1

Thevalueofy2 correcttofourdecimalplacesisA. 1.1000andthisisanunderestimateofy (0.2)B. 1.1995andthisisanoverestimateofy (0.2)C. 1.1995andthisisanunderestimateofy (0.2)D. 1.2975andthisisanunderestimateofy (0.2)E. 1.2975andthisisanoverestimateofy (0.2)

2012 SPECMATH EXAM 2 6

SECTION 1 – continued

Question 10The diagram that best represents the direction field of the differential equation dy

dxxy= is

A.

x

y

–4 –3 –2 –1 O 1 2 3 4

–3–2–1

1234

–4

B.

x

y

–4 –3 –2 –1 O 1 2 3 4

–4

–3

–2

–1

1

2

3

4

C.

y

–4 –3 –2 –1 O 1 2 3 4

–4

–3

–2

–1

1

2

3

4

x

D.

x–4 –3 –2 –1 O 1 2 3 4

–4

–3

–2

–1

1

2

3

4

y

E.

x–4 –3 –2 –1 O 1 2 3 4

–4–3–2–1

1234

y

7 2012SPECMATHEXAM2

SECTION 1–continuedTURN OVER

Question 11

If d ydx

x x2

22= − and dy

dx= 0 at x=0,thenthegraphofywillhave

A. alocalminimumat x =12

B. alocalmaximumatx =0andalocalminimumatx = 1

C. stationarypointsofinflectionatx =0andx=1,andalocalminimumatx = 32

D. astationarypointofinflectionatx =0,nootherpointsofinflectionandalocalminimumatx = 32

E. astationarypointofinflectionatx =0,anon-stationarypointofinflectionatx =1andalocalminimum

at x = 32

Question 12Thevolumeofthesolidofrevolutionformedbyrotatingthegraphofy = 9 1 2− −( )x aboutthex-axisisgivenby

A. 4 3 2π ( )

B. π 9 1 2

3

3− −( )

−∫ ( )x dx

C. π 9 1 2

2

4− −( )

−∫ ( )x dx

D. π 9 1 2 2

2

4− −( )

−∫ ( )x dx

E. π 9 1 2

4

2

− −( )−∫ ( )x dx

2012SPECMATHEXAM2 8

SECTION 1–continued

Question 13

Usingasuitablesubstitution, sin ( )cos ( )3 4

0

3x x dx

π

∫ canbeexpressedintermsofuas

A. u u du6 4

0

3−( )∫

π

B. u u du6 4

1

12

−( )∫

C. u u du6 4

12

1

−( )∫

D. u u du6 4

0

32

−( )∫

E. u u du4 6

0

32

−( )∫

Question 14Aparticleisactedonbytwoforces,oneof6newtonsactingduesouth,theotherof4newtonsactinginthedirectionN60°W.Themagnitudeoftheresultantforce,innewtons,actingontheparticleisA. 10

B. 2 7

C. 2 19

D. 52 24 3−

E. 52 24 3+

9 2012 SPECMATH EXAM 2

SECTION 1 – continuedTURN OVER

Question 15The vectors a i j k~ ~ ~ ~= + −2 3m and b i j k~ ~ ~ ~= − +m2 are perpendicular for

A. m = −23

and m = 1

B. m = −32

and m = 1

C. m =23

and m = –1

D. m =32

and m = –1

E. m = 3 and m = –1

Question 16The distance between the points P (–2, 4, 3) and Q (1, –2, 1) isA. 7

B. 21

C. 31

D. 11

E. 49

Question 17If u~ ~ ~ ~= − +2 2i j k and v~ ~ ~ ~= − +3 6 2i j k, the vector resolute of v~ in the direction of u~ is

A. 2049

3 6 2( )~ ~i j k~

− +

B. 203

2 2( ~ ~ ~i j k )− +

C. 207

3 6 2( )~ ~ ~i j k− +

D. 209

2 2( )~ ~ ~i j k− +

E. 19

2 2( )~ ~ ~− + −i j k

2012SPECMATHEXAM2 10

SECTION 1–continued

Question 18Thevelocity–timegraphforthefirst2secondsofthemotionofaparticlethatismovinginastraightlinewithrespecttoafixedpointisshownbelow.

v

t1 2

20

–20

–40

Theparticle’svelocityvismeasuredincm/s.Initiallytheparticleisx0cmfromthefixedpoint.Thedistancetravelledbytheparticleinthefirst2secondsofitsmotionisgivenby

A. vdt0

2

∫

B. vdt x0

2

0∫ +

C. vdt vdt1

2

0

1

∫ ∫−

D. | |vdt0

2

∫

E. vdt vdt x1

2

0

1

0∫ ∫− +

Question 19Abodyismovinginastraightlineand,aftertseconds,itisxmetresfromtheoriginandtravellingatvms–1.Giventhatv = x,andthatt =3wherex =–1,theequationforx intermsoft isA. x = et–3

B. x = – e3–t

C. x t= −2 5

D. x t= − −2 5

E. x = – et–3

11 2012SPECMATHEXAM2

SECTION 1–continuedTURN OVER

Question 20Particlesofmass3kgandmkgareattachedtotheendsofalightinextensiblestringthatpassesoverasmoothpulley,asshown.

3 kg

m kg

Iftheaccelerationofthe3kgmassis4.9m/s2upwards,thenA. m=4.5B. m=6.0C. m=9.0D. m=13.5E. m=18.0

Question 21Aparticleofmass3kgisactedonbyavariableforce,sothatitsvelocityvm/swhentheparticleisxmfromtheoriginisgivenbyv = x2.Theforceactingontheparticlewhenx =2,innewtons,isA. 4B. 12C. 16D. 36E. 48

2012SPECMATHEXAM2 12

END OF SECTION 1

Question 22A12kgmassissuspendedinequilibriumfromahorizontalceilingbytwoidenticallightstrings.Eachstringmakesanangleof60°withtheceiling,asshown.

60° 60°

12 kg

Themagnitude,innewtons,ofthetensionineachstringisequaltoA. 6gB. 12 gC. 24 gD. 4 3 gE. 8 3 g

13 2012SPECMATHEXAM2

SECTION 2 – Question 1–continuedTURN OVER

Question 1Acurveisdefinedbytheparametricequations

x=cosec(θ) + 1 y=2cot(θ)

a. Showthatthecurvecanbeexpressedinthecartesianform

( )x y− − =1

412

2

2marks

SECTION 2

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

2012SPECMATHEXAM2 14

SECTION 2 – Question 1–continued

b. Sketchthecurvedefinedbytheparametricequations

x =cosec(θ) + 1 y =2cot(θ)

labellinganyasymptoteswiththeirequations.

5

4

3

2

1

O

–1

–2

–3

–4

–5

–5 –4 –3 –2 –1 1 2 3 4 5

y

x

4marks

15 2012SPECMATHEXAM2

SECTION 2–continuedTURN OVER

c. Theregionboundedbythecurveandthelinex =3isrotatedaboutthex-axis. Findthevolumeofthesolidformed.

2marks

d. Findthegradientofthecurvewhereθ π=

76.

3marks

2012SPECMATHEXAM2 16

SECTION 2 – Question 2–continued

Question 2a. Giventhatcos ,π

123 22

=

+ showthatsin .π12

2 32

=

−

2marks

b. i. Expressz i13 22

2 32

=+

+− inpolarform.

ii. Writedownz14inpolarform.

1+1=2marks

c. OntheArganddiagrambelow,shadetheregiondefinedby

z z z: ) )Arg( Arg(z) Arg(1 14≤ ≤{ }∩{z:1≤| z |≤2}, z C∈ .

Im(z)

Re(z)O–1–2–3 321

3

2

1

–1

–2

–3

2marks

17 2012SPECMATHEXAM2

SECTION 2–continuedTURN OVER

d. Findtheareaoftheshadedregioninpart c.

2marks

e. i. Findthevalue(s)ofnsuchthatRe(z1n) =0,where z i1

3 22

2 32

=+

+− .

ii. Findz1nforthevalue(s)ofnfoundinpart i.

3+1=4marks

2012SPECMATHEXAM2 18

SECTION 2 – Question 3–continued

Question 3Acaracceleratesfromrest.ItsspeedafterTsecondsisVms–1,where

V T= −17

61tan ,π T≥0

a. Writedownthelimitingspeed ofthecarasT → ∞.

1mark

b. Calculate,correcttothenearest0.1ms–2,theaccelerationofthecarwhenT=10.

1mark

c. Calculate,correcttothenearestsecond,thetimeittakesforthecartoacceleratefromrestto25ms–1.

2marks

19 2012SPECMATHEXAM2

SECTION 2 – Question 3–continuedTURN OVER

Afteracceleratingto25ms–1,thecarstaysatthisspeedfor120secondsandthenbeginstodeceleratewhilebraking.Thespeedofthecartsecondsafterthebrakesarefirstappliedisvms–1where

dvdt

t= −− 1100

145 2( ),

untilthecarcomestorest.d. i. Findvintermsoft.

ii. Findthetime,inseconds,takenforthecartocometorestwhilebraking.

2+2=4marks

2012SPECMATHEXAM2 20

SECTION 2–continued

e. i. Writedowntheexpressionsforthedistancetravelledbythecarduringeachofthethreestagesofitsmotion.

ii. Findthetotaldistancetravelledfromwhenthecarstartstoacceleratetowhenitcomestorest. Giveyouranswerinmetrescorrecttothenearestmetre.

2+1=3marks

21 2012 SPECMATH EXAM 2

SECTION 2 – Question 4 – continuedTURN OVER

Question 4The position vector of the International Space Station (S), when visible above the horizon from a radar tracking location (O) on the surface of Earth, is modelled by

r i j~ ~ ~( ) sin ( . . ) cos ( . . )t t t= −( ) + −( ) −( )6800 1 3 0 1 6800 1 3 0 1 6400π π ,,

for t∈[ ]0 0 154, . ,

where i~ is a unit vector relative to O as shown and j~ is a unit vector vertically up from point O. Time t is

measured in hours and displacement components are measured in kilometres.

i∼

j∼

P

S hy

xO

r(t)∼

a. Find the height, h km, of the space station above the surface of Earth when it is at point P, directly above point O.

Give your answer correct to the nearest km.

1 mark

b. Find the acceleration of the space station, and show that its acceleration is perpendicular to its velocity.

3 marks

2012 SPECMATH EXAM 2 22

SECTION 2 – continued

c. Find the speed of the space station in km/h. Give your answer correct to the nearest integer.

2 marks

d. Find the equation of the path followed by the space station in cartesian form.

2 marks

e. Find the times when the space station is at a distance of 1000 km from the radar tracking location O. Give your answers in hours, correct to two decimal places.

3 marks

23 2012SPECMATHEXAM2

SECTION 2 – Question 5–continuedTURN OVER

Question 5Atherfavouritefunpark,Katherine’sfirstactivityistoslidedowna10mlongstraightslide.Shestarts fromrestatthetopandacceleratesataconstantrate,untilshereachestheendoftheslidewithavelocityof 6ms–1.a. Howlong,inseconds,doesittakeKatherinetotraveldowntheslide?

1mark

Whenatthetopoftheslide,whichis6mabovetheground,Katherinethrowsachocolateverticallyupwards.Thechocolatetravelsupandthendescendspastthetopoftheslidetolandonthegroundbelow.Assumethatthechocolateissubjectonlytogravitationalaccelerationandthatairresistanceisnegligible.b. Iftheinitialspeedofthechocolateis10m/s,howlong,correcttothenearesttenthofasecond,doesit

takethechocolatetoreachtheground?

2marks

c. AssumethatittakesKatherinefoursecondstorunfromtheendoftheslidetowherethechocolatelands.

Atwhatvelocitywouldthechocolateneedtobepropelledupwards,ifKatherineweretoimmediatelyslidedowntheslideandruntoreachthechocolatejustasithitstheground?

Giveyouranswerinms–1,correcttoonedecimalplace.

2marks

2012SPECMATHEXAM2 24

SECTION 2 – Question 5–continued

Katherine’snextactivityistorideaminispeedboat.Tostopatthecorrectboatdock,sheneedstostoptheengineandallowtheboattobeslowedbyairandwaterresistance.Attimetsecondsaftertheenginehasbeenstopped,theaccelerationoftheboat,ams–2,isrelatedtoitsvelocity,vms–1,by

a v= − −1

10196 2 .

Katherinestopstheenginewhenthespeedboatistravellingat7m/s.d. i. Findanexpressionforvintermsoft.

ii. Findthetimeittakesthespeedboattocometorest. Giveyouranswerinsecondsintermsof π.

25 2012SPECMATHEXAM2

iii. Findthedistanceittakesthespeedboattocometorest,fromwhentheengineisstopped. Giveyouranswerinmetres,correcttoonedecimalplace.

3+2+3=8marks

END OF QUESTION AND ANSWER BOOK

SPECIALIST MATHEMATICS

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 2012

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 n1

111 ,

ddxe aeax ax( ) =

∫ = +e dx

ae cax ax1

ddx

xxelog ( )( )= 1

∫ = +1xdx x celog

ddx

ax a axsin( ) cos( )( )=

∫ = − +sin( ) cos( )ax dxa

ax c1

ddx

ax a axcos( ) sin( )( )= −

∫ = +cos( ) sin( )ax dxa

ax c1

ddx

ax a axtan( ) sec ( )( )= 2

∫ = +sec ( ) tan( )2 1ax dx

aax c

ddx

xx

sin−( ) =−

12

1

1( )

∫

−=

+ >−1 0

2 21

a xdx x

a c asin ,

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

1tan

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