2014specmath1 w
DESCRIPTION
sqsTRANSCRIPT
SPECIALIST MATHEMATICSWritten examination 1
Friday 7 November 2014 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
8 8 40
• Studentsarepermittedtobringintotheexaminationroom:pens,pencils,highlighters,erasers,sharpenersandrulers.
• Studentsarenotpermittedtobringintotheexaminationroom:notesofanykind,acalculatorofanytype,blanksheetsofpaperand/orwhiteoutliquid/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.
©VICTORIANCURRICULUMANDASSESSMENTAUTHORITY2014
SUPERVISOR TO ATTACH PROCESSING LABEL HEREVictorian Certificate of Education 2014
STUDENT NUMBER
Letter
2014SPECMATHEXAM1 2
THIS PAGE IS BLANK
3 2014SPECMATHEXAM1
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InstructionsAnswerallquestionsinthespacesprovided.Unlessotherwisespecified,anexactanswerisrequiredtoaquestion.Inquestionswheremorethanonemarkisavailable,appropriateworkingmust beshown.Unlessotherwiseindicated,thediagramsinthisbookarenotdrawntoscale.Taketheacceleration due to gravitytohavemagnitudegm/s2,whereg=9.8.
Question 1 (5marks)Considerthevector a i j k
~= − −3 2~ ~ ~
, where i j and k~ ~ ~, areunitvectorsinthepositivedirections
ofthex,yandzaxesrespectively.
a. Findtheunitvectorinthedirectionof a~. 1mark
b. Findtheacuteanglethat a~makeswiththepositivedirectionofthex-axis. 2marks
c. Thevector b i j k~= + −2 3 5~ ~ ~
.m
Giventhat b~isperpendicularto a
~,findthevalueofm. 2marks
2014SPECMATHEXAM1 4
Question 2 (5marks)Thepositionvectorofaparticleattime t≥0 isgivenby
r i j~ ~ ~t t t t( ) = −( ) + − +( )2 4 12
a. Showthatthecartesianequationofthepathfollowedbytheparticleis y = x2–3. 1mark
b. Sketchthepathfollowedbytheparticleontheaxesbelow,labellingallimportantfeatures. 2marks
y
xO
c. Findthespeedoftheparticlewhen t=1. 2marks
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Question 3 (5marks)Let f beafunctionofacomplexvariable,definedbytherule f z z z z z( ) = − + − +4 3 24 7 4 6.
a. Giventhat z = i isasolutionof f z( ) = 0, writedownaquadraticfactorof f z( ) . 2marks
b. Giventhattheotherquadraticfactorof f z( ) hastheform z2 + bz + c,findallsolutions of z4 – 4z3 + 7z2 – 4z+6=0 incartesianform. 3marks
2014SPECMATHEXAM1 6
Question 4 (3marks)Findthegradientofthenormaltothecurvedefinedby y = –3e3xe y atthepoint(1,–3).
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Question 5 (5marks)a. Forthefunctionwithrule f x x x( ) = ( ) ( )96 3 3cos sin , findthevalueofasuchthat
f x a x( ) = ( )sin .6 1mark
b. Useanappropriatesubstitutionintheformu g x= ( ) tofindanequivalentdefiniteintegralfor96 3 3 62
36
12 cos sin cosx x x dx( ) ( ) ( )∫ ππ
intermsofu only. 3marks
c. Henceevaluate 96 3 3 6236
12 cos sin cosx x x dx( ) ( ) ( )∫ ππ
,givingyouranswerintheform
k k Z, ∈ . 1mark
2014SPECMATHEXAM1 8
Question 6 (5marks)
a. Verifythat aa a−
= +−4
1 44
. 1mark
Partofthegraphof y x
x=
−( )2 4isshownbelow.
54321
O–1
–1 1 2 3 4
y
x
b. Theregionenclosedbythegraphof yx
x=
−( )2 4andthelines y=0,x =3 and
x =4 isrotatedaboutthex-axis.
Findthevolumeoftheresultingsolidofrevolution. 4marks
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Question 7 (5marks)Consider f x x x( ) = ( )3 2arctan .
a. Writedowntherangeof f. 1mark
b. Showthat ′( ) = ( ) ++
f x x xx
3 2 61 4 2arctan . 1mark
c. Henceevaluatetheareaenclosedbythegraphofg x x( ) = ( )arctan ,2 thex-axisand
thelines x x= =12
32
and . 3marks
2014SPECMATHEXAM1 10
END OF QUESTION AND ANSWER BOOK
Question 8 (7marks)Abodyofmass5kgisheldinequilibriumbytwolightinextensiblestrings.OnestringisattachedtoaceilingatAandtheothertoawallatB.ThestringattachedtotheceilingisatanangleθtotheverticalandhastensionT1newtons,andtheotherstringishorizontalandhastensionT2newtons.Bothstringsaremadeofthesamematerial.
A
θ
ceiling
wall
B5 kg body
a. i. Resolvetheforcesonthebodyverticallyandhorizontally,andexpressT1intermsofθ. 2marks
ii. ExpressT2intermsofθ. 1mark
b. Showthat tan secθ θ( ) < ( ) for 0 2< <θ π . 1mark
c. Thetypeofstringusedwillbreakifitissubjectedtoatensionofmorethan98N. Findthemaximumallowablevalueofθsothatneitherstringwillbreak. 3marks
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 2014
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