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SPECIALIST MATHEMATICSWritten examination 1

Friday 8 November 2013 Reading time: 9.00 am to 9.15 am (15 minutes) Writing time: 9.15 am to 10.15 am (1 hour)

QUESTION AND ANSWER BOOK

Structure of bookNumber of questions

Number of questions to be answered

Number of marks

9 9 40

• Studentsarepermittedtobringintotheexaminationroom:pens,pencils,highlighters,erasers,sharpeners andrulers.

• Studentsarenotpermittedtobringintotheexaminationroom:notesofanykind,acalculatorofanytype,blanksheetsofpaperand/orwhiteoutliquid/tape.

Materials supplied• Questionandanswerbookof11pageswithadetachablesheetofmiscellaneousformulasinthe

centrefold.• Workingspaceisprovidedthroughoutthebook.

Instructions• Detachtheformulasheetfromthecentreofthisbookduringreadingtime.• Writeyourstudent numberinthespaceprovidedaboveonthispage.

• AllwrittenresponsesmustbeinEnglish.

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

©VICTORIANCURRICULUMANDASSESSMENTAUTHORITY2013

SUPERVISOR TO ATTACH PROCESSING LABEL HEREVictorian Certificate of Education 2013

2013SPECMATHEXAM1 2

THIS PAGE IS BLANK

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InstructionsAnswerallquestionsinthespacesprovided.Unlessotherwisespecified,anexactanswerisrequiredtoaquestion.Inquestionswheremorethanonemarkisavailable,appropriateworkingmust beshown.Unlessotherwiseindicated,thediagramsinthisbookarenotdrawntoscale.Taketheacceleration due to gravitytohavemagnitudegm/s2,whereg=9.8.

Question 1 (3marks)Abodyofmass10kgisheldinplaceonasmoothplaneinclinedat30°tothehorizontalbyatensionforce,Tnewtons,actingparalleltotheplane.a. Onthediagrambelow,showallotherforcesactingonthebodyandlabelthem. 1mark

T

30°

b. FindthevalueofT. 2marks

2013SPECMATHEXAM1 4

Question 2 (4marks)

Evaluatex

x xdx

−− +∫

55 62

0

1

.

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Question 3 (4marks)ThecoordinatesofthreepointsareA(–1,2,4),B(1,0,5)andC(3,5,2).

a. FindAB. 1mark

b. ThepointsA,BandCaretheverticesofatriangle. Provethatthetrianglehasarightangleat A. 2marks

c. Findthelengthofthehypotenuseofthetriangle. 1mark

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Question 4 (6marks)a. Statethemaximaldomainandtherangeofy=arccos(1–2x). 2marks

b. Sketchthegraphofy=arccos(1–2x)overitsmaximaldomain.Labeltheendpointswiththeircoordinates. 2marks

y

xO

c. Findthegradientofthetangenttothegraphofy=arccos(1–2x)at x = 14. 2marks

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Question 5 (5marks)Acontainerofwaterisheatedtoboilingpoint(100°C)andthenplacedinaroomthathasaconstanttemperatureof20°C.Afterfiveminutesthetemperatureofthewateris80°C.

a. UseNewton’slawofcooling dTdt

k T= − −( )20 ,whereT °Cisthetemperatureofthewater

attime tminutesafterthewaterisplacedintheroom,toshowthate k− =5 34. 2marks

b. Findthetemperatureofthewater10minutesafteritisplacedintheroom. 3marks

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Question 6 (4marks)Findthevalueofc,wherec R,suchthatthecurvedefinedby

y ex

cx

2132

+−

=−( )

hasagradientof2wherex =1.

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Question 7 (6marks)Thepositionvectorr( )~ t ofaparticlemovingrelativetoanoriginOattimetsecondsisgivenby

r i j~( ) sec( ) tan( ) , ,~ ~t t t t= + ∈

4 2 0

2π

wherethecomponentsaremeasuredinmetres.

a. Showthatthecartesianequationofthepathoftheparticleis x y2 2

16 41− = . 2marks

b. Sketchthepathoftheparticleontheaxesbelow,labellinganyasymptoteswiththeirequations. 2marks

x

y

O

c. Findthespeedoftheparticle,inms–1,when t = π4. 2marks

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Question 8 (4marks)Findallsolutionsofz4–2z2+4=0,z Cincartesianform.

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END OF QUESTION AND ANSWER BOOK

Question 9 (4marks)

Theshadedregionbelowisenclosedbythegraphofy =sin(x)andthelinesy =3xandx = π3 .

Thisregionisrotatedaboutthex-axis.

1

2

3

x =y = 3x

y = sin(x)

O

y

x

π3

π3

Findthevolumeoftheresultingsolidofrevolution.

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 2013

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