2015 specialist mathematics written examination 1 · pdf filespecialist mathematics written...

SPECIALIST MATHEMATICSWritten examination 1

Friday 6 November 2015 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,sharpenersandrulers.

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

Materials supplied• Questionandanswerbookof10pageswithadetachablesheetofmiscellaneousformulasinthe

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.

©VICTORIANCURRICULUMANDASSESSMENTAUTHORITY2015

SUPERVISOR TO ATTACH PROCESSING LABEL HEREVictorian Certificate of Education 2015

STUDENT NUMBER

Letter

2015SPECMATHEXAM1 2

THIS PAGE IS BLANK

3 2015SPECMATHEXAM1

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

Question 1 (3marks)

ConsidertherhombusOABCshownbelow,whereOA a→=�i andOC

→= + +� � �i j k ,andaisapositivereal

constant.

C

O

B

A

a. Finda. 1mark

b. ShowthatthediagonalsoftherhombusOABCareperpendicular. 2marks

2015SPECMATHEXAM1 4

Question 2 (4marks)A20kgparcelsitsonthefloorofalift.

a. Theliftisacceleratingupwardsat1.2ms–2.

Findthereactionforceoftheliftfloorontheparcelinnewtons. 2marks

b. Findtheaccelerationoftheliftdownwardsinms–2sothatthereactionoftheliftfloorontheparcelis166N. 2marks

Question 3 (4marks)Thevelocityofaparticleattimetsecondsisgivenbyr i j k( ) ( )t t t= − + −4 3 2 5 ,wherecomponentsaremeasuredinmetrespersecond.

Findthedistanceoftheparticlefromtheorigininmetreswhent=2,giventhat� � �r i k( )0 2= − .

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Question 4 (4marks)

a. Findallsolutionsof z i z C3 8= ∈, incartesianform. 3marks

b. Findallsolutionsof( ) ,z i i z C− = ∈2 83 incartesianform. 1mark

Question 5 (3marks)Findthevolumegeneratedwhentheregionboundedbythegraphofy=2x2–3,theliney=5andthey-axisisrotatedaboutthey-axis.

2015SPECMATHEXAM1 6

Question 6 (4marks)Theaccelerationams–2ofabodymovinginastraightlineintermsofthevelocityvms–1isgivenby a = 4v2.

Giventhatv = ewhenx=1,where xisthedisplacementofthebodyinmetres,findthevelocityofthebodywhenx=2.

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Question 7 (5marks)a. Solvesin( ) sin( ), [ , ]2 0 2x x x= ∈ π . 3marks

b. Find x x x x: ) ), , ,cosec( cosec(2 02 2

< ∈

∪

π ππ . 2marks

2015SPECMATHEXAM1 8

Question 8–continued

Question 8 (7marks)

a. Showthat tan( ) log sec2 12

2x dx x ce∫ = ( ) + . 2marks

Thegraphof f x x( ) ( )=12arctan isshownbelow.

O

y

x

b. i. Writedowntheequationsoftheasymptotes. 1mark

ii. Ontheaxesabove,sketchthegraphof f –1,labellinganyasymptoteswiththeirequations. 1mark

9 2015SPECMATHEXAM1

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c. Find f 3( ). 1mark

d. Findtheareaenclosedbythegraphof f,thex-axisandtheline x = 3 . 2marks

2015SPECMATHEXAM1 10

END OF QUESTION AND ANSWER BOOK

Question 9 (6marks)

Considerthecurverepresentedby x xy y2 232

9− + = .

a. Findthegradientofthecurveatanypoint(x,y). 2marks

b. Findtheequationofthetangenttothecurveatthepoint(3,0)andfindtheequation ofthetangenttothecurveatthepoint 0 6,( ) .

Writeeachequationintheformy = ax + b. 2marks

c. Findtheacuteanglebetweenthetangenttothecurveatthepoint(3,0)andthetangenttothecurveatthepoint 0 6,( ) .

Giveyouranswerintheformkπ,wherekisarealconstant. 2marks

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