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SPECIALIST MATHEMATICSWritten examination 1
Tuesday 5 June 2018 Reading time: 2.00 pm to 2.15 pm (15 minutes) Writing time: 2.15 pm to 3.15 pm (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:anytechnology(calculatorsorsoftware),notesofanykind,blanksheetsofpaperand/orcorrectionfluid/tape.
Materials supplied• Questionandanswerbookof11pages• Formulasheet• Workingspaceisprovidedthroughoutthebook.
Instructions• Writeyourstudent numberinthespaceprovidedaboveonthispage.• Unlessotherwiseindicated,thediagramsinthisbookarenotdrawntoscale.• AllwrittenresponsesmustbeinEnglish.
At the end of the examination• Youmaykeeptheformulasheet.
Students are NOT permitted to bring mobile phones and/or any other unauthorised electronic devices into the examination room.
©VICTORIANCURRICULUMANDASSESSMENTAUTHORITY2018
SUPERVISOR TO ATTACH PROCESSING LABEL HEREVictorian Certificate of Education 2018
STUDENT NUMBER
Letter
2018SPECMATHEXAM1(NHT) 2
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THIS PAGE IS BLANK
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InstructionsAnswerallquestionsinthespacesprovided.Unlessotherwisespecified,anexactanswerisrequiredtoaquestion.Inquestionswheremorethanonemarkisavailable,appropriateworkingmust beshown.Unlessotherwiseindicated,thediagramsinthisbookarenotdrawntoscale.Taketheacceleration due to gravitytohavemagnitudegms–2,whereg=9.8
Question 1 (3marks)Alightinextensiblestringhangsoverafrictionlesspulleyconnectingmassesof3kgand7kg, asshownbelow.
3 kg
7 kg
a. Drawalloftheforcesactingonthetwomassesonthediagramabove. 1mark
b. Calculatethetensioninthestring. 2marks
2018SPECMATHEXAM1(NHT) 4
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Question 2 (3marks)Let
a i j k= − +3 2 m and
b i j k ,= − +2 3 wherem ∈ R.
Findthevalue(s)ofmsuchthatthemagnitudeofthevectorresoluteof
a parallelto
b isequalto 14.
Question 3 (3marks)
Findsin(t)giventhat t =
+
arccos arctan .12
1334
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Question 4 (4marks)Throughoutthisquestion,useanintegermultipleofstandarddeviationsincalculations.Thestandarddeviationofallscoresonaparticulartestis21.0
a. Fromtheresultsofarandomsampleofnstudents,a95%confidenceintervalforthemeanscoreforallstudentswascalculatedtobe(44.7,51.7).
Calculatethemeanscoreandthesizeofthisrandomsample. 2marks
b. Determinethesizeofanotherrandomsampleforwhichtheendpointsofthe95%confidenceintervalforthepopulationmeanoftheparticulartestwouldbe1.0eithersideofthesamplemean. 2marks
2018SPECMATHEXAM1(NHT) 6
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Question 5 (4marks)
Evaluate 12 52
1
2 3 1
x xdx
+ +
−
∫ .
Question 6 (4marks)Giventhaty=(x–1)e2xisasolutiontothedifferentialequationa d y
dxb dydx
y2
2 + = ,findthevaluesofaandb,whereaandbarerealconstants.
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Question 7 (4marks)
a. Find ddx
x( ) .1 212−
2marks
b. Hence,findthelengthofthecurvespecifiedby y x= −1 2 from x = 12to x = 3
2. Give
youranswerintheformkπ,k ∈ R. 2marks
2018SPECMATHEXAM1(NHT) 8
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Question 8–continued
Question 8 (6marks)Acircleinthecomplexplaneisgivenbytherelation|z–1–i |=2,z ∈ C.
a. SketchthecircleontheArganddiagrambelow. 1mark
4
3
2
1
O
–1
–2
–3
–4
–4 –3 –2 –1 1 2 3 4
Im(z)
Re(z)
b. i. Writetheequationofthecircleintheform(x – a)2+(y – b)2 = candshowthatthe
gradientofatangenttothecirclecanbeexpressedas dydx
xy
=−−
11
. 2marks
ii. Findthegradientofthetangenttothecirclewherex=2inthefirstquadrantofthecomplexplane. 1mark
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c. FindtheequationsofallraysthatareperpendiculartothecircleintheformArg(z)=α. 2marks
2018SPECMATHEXAM1(NHT) 10
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Question 9–continued
Question 9 (9marks)a. i. Giventhatcot(2θ)=a,showthattan2(θ)+2atan(θ)–1=0. 2marks
ii. Showthattan( ) .θ = − ± +a a2 1 1mark
iii. Hence,showthattan ,π12
2 3
= − giventhatcot( ) ,2 3θ = whereθ ∈(0,π). 1mark
b. Findthegradientofthetangenttothecurve y=tan(θ) atθ π=
12. 2marks
11 2018SPECMATHEXAM1(NHT)
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END OF QUESTION AND ANSWER BOOK
c. Asolidofrevolutionisformedbyrotatingtheregionbetweenthegraphofy=tan(θ),
the horizontalaxis,andthelinesθ π=
12andθ π
=3aboutthehorizontalaxis.
Findthevolumeofthesolidofrevolution. 3marks
SPECIALIST MATHEMATICS
Written examination 1
FORMULA SHEET
Instructions
This formula sheet is provided for your reference.A question and answer book is provided with this formula sheet.
Students are NOT permitted to bring mobile phones and/or any other unauthorised electronic devices into the examination room.
Victorian Certificate of Education 2018
© VICTORIAN CURRICULUM AND ASSESSMENT AUTHORITY 2018
SPECMATH EXAM 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 43π r3
area of a triangle 12 bc Asin ( )
sine ruleaA
bB
cCsin ( ) sin ( ) sin ( )
= =
cosine rule c2 = a2 + b2 – 2ab cos (C )
Circular functions
cos2 (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
+ =+
−1tan ( ) tan ( ) tan ( )
tan ( ) tan ( )x y x y
x 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=
−
3 SPECMATH EXAM
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Circular functions – continued
Function sin–1 or arcsin cos–1 or arccos tan–1 or arctan
Domain [–1, 1] [–1, 1] R
Range −
π π2 2
, [0, �] −
π π2 2
,
Algebra (complex numbers)
z x iy r i r= + = +( ) =cos( ) sin ( ) ( )θ θ θcis
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)
Probability and statistics
for random variables X and YE(aX + b) = aE(X) + bE(aX + bY ) = aE(X ) + bE(Y )var(aX + b) = a2var(X )
for independent random variables X and Y var(aX + bY ) = a2var(X ) + b2var(Y )
approximate confidence interval for μ x z snx z s
n− +
,
distribution of sample mean Xmean E X( ) = µvariance var X
n( ) = σ2
SPECMATH EXAM 4
END OF FORMULA SHEET
Calculus
ddx
x nxn n( ) = −1 x dxn
x c nn n=+
+ ≠ −+∫ 11
11 ,
ddxe aeax ax( ) = e dx
ae cax ax= +∫ 1
ddx
xxelog ( )( ) = 1 1
xdx 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 dxa
ax c= +∫ddx
xx
sin−( ) =−
12
1
1( ) 1 0
2 21
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
( )( )
( ) ,ax b dxa n
ax b c nn n+ =+
+ + ≠ −+∫ 11
11
( ) logax b dxa
ax b ce+ = + +−∫ 1 1
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
arc length 1 2 2 2
1
2
1
2
+ ′( ) ′( ) + ′( )∫ ∫f x dx x t y t dtx
x
t
t( ) ( ) ( )or
Vectors in two and three dimensions
r = i + j + kx y z
r = + + =x y z r2 2 2
� � � � �ir r i j k= = + +ddt
dxdt
dydt
dzdt
r r1 2. cos( )= = + +r r x x y y z z1 2 1 2 1 2 1 2θ
Mechanics
momentum
p v= m
equation of motion
R a= m
