2014 Specialist Mathematics Written examination 22014 SPECMATH EXAM 2 4 SECTION 1 – continued...
Transcript of 2014 Specialist Mathematics Written examination 22014 SPECMATH EXAM 2 4 SECTION 1 – continued...
SPECIALIST MATHEMATICSWritten examination 2
Monday 10 November 2014 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
1 22 22 222 5 5 58
Total 80
• Studentsarepermittedtobringintotheexaminationroom:pens,pencils,highlighters,erasers,sharpeners,rulers,aprotractor,set-squares,aidsforcurvesketching,oneboundreference,oneapprovedCAScalculatororCASsoftwareand,ifdesired,onescientificcalculator.CalculatormemoryDOESNOTneedtobecleared.
• StudentsareNOTpermittedtobringintotheexaminationroom:blanksheetsofpaperand/orwhiteoutliquid/tape.
Materials supplied• Questionandanswerbookof23pageswithadetachablesheetofmiscellaneousformulasinthe
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.
©VICTORIANCURRICULUMANDASSESSMENTAUTHORITY2014
SUPERVISOR TO ATTACH PROCESSING LABEL HEREVictorian Certificate of Education 2014
STUDENT NUMBER
Letter
2014SPECMATHEXAM2 2
SECTION 1–continued
Question 1Theasymptotesofthehyperbolagivenby
x y−( )− =
39 4
12 2
intersectthecoordinateaxesat
A. 0 4 5 0 4 5 3 0, . , , . , ,−( ) ( ) −( )
B. 0 2 0 2 3 0, , , , ,( ) −( ) ( )
C. 0 2 0 2 3 0, , , , ,( ) −( ) −( )
D. 0 4 5 0 4 5 3 0, . , , . , ,−( ) ( ) ( )
E. 2 0 2 0 0 3, , , , ,( ) −( ) −( )
Question 2Theellipsegivenbyx2 – 6x + 2y2 + 8y+16=0hascentre,lengthofhorizontalsemi-axisandlengthofverticalsemi-axisrespectivelyof
A. −( )3 2 1 2, , ,
B. −( )2 3 1 12
, , ,
C. 3 2 12
1, , ,−( )
D. −( )3 2 12
1, , ,
E. 3 2 1 12
, , ,−( )
Question 3Thefeaturesofthegraphofthefunctionwithrule f x
x xx x
( ) = − +− −
2
24 3
6include
A. asymptotesatx=1andx = –2B. asymptotesatx=3andx = –2C. anasymptoteatx=1andapointofdiscontinuityatx=3D. anasymptoteatx=–2andapointofdiscontinuityat x=3E. anasymptoteatx=3andapointofdiscontinuityatx = –2
SECTION 1
Instructions for Section 1Answerallquestionsinpencilontheanswersheetprovidedformultiple-choicequestions.Choosetheresponsethatiscorrect forthequestion.Acorrectanswerscores1,anincorrectanswerscores0.Markswillnotbedeductedforincorrectanswers.Nomarkswillbegivenifmorethanoneansweriscompletedforanyquestion.Taketheacceleration due to gravitytohavemagnitudegm/s2,whereg=9.8.
3 2014SPECMATHEXAM2
SECTION 1–continuedTURN OVER
Question 4Thedomainof arcsin 2 1x −( ) is
A. [ , ]−1 1
B. [ , ]−1 0
C. 0 1,[ ]
D. −
12
12
,
E. 0 12
,
Question 5Ifthecomplexnumberzhasmodulus 2 2 andargument
34π,thenz2isequalto
A. –8i
B. 4i
C. −2 2i
D. 2 2i
E. –4i
Question 6Giventhati n = pandi 2=–1,theni 2n+3intermsofpisequaltoA. p2 – iB. p2 + iC. –p2
D. –ip2
E. ip2
Question 7Thesumoftherootsofz3 – 5z2 + 11z–7=0,wherez C∈ ,is
A. 1 2 3+ i
B. 5i
C. 4 2 3− i
D. 2 3i
E. 5
2014SPECMATHEXAM2 4
SECTION 1–continued
Question 8Theprincipalargumentof
− −+
3 2 62 2
ii is
A. −13
12π
B. 712π
C. 1112π
D. 1312π
E. −11
12π
Question 9Thecircle z i− − =3 2 2 isintersectedexactlytwicebythelinegivenby
A. z i z− = +1
B. z i z− − = −3 2 5
C. z i z i− − = −3 2 10
D. Im(z) = 0
E. Re(z) = 5
Question 10Alargetankinitiallyholds1500Lofwaterinwhich100kgofsaltisdissolved.Asolutioncontaining2kgofsaltperlitreflowsintothetankatarateof8Lperminute.Themixtureisstirredcontinuouslyandflowsoutofthetankthroughaholeatarateof10Lperminute.ThedifferentialequationforQ,thenumberofkilogramsofsaltinthetankaftert minutes,isgivenby
A. dQdt
Qt
= −−
16 5750
B. dQdt
Qt
= −+
16 5750
C. dQdt
Qt
= +−
16 5750
D. dQdt
Qt
=−
100750
E. dQdt
Qt
= −−
81500 2
5 2014SPECMATHEXAM2
SECTION 1–continuedTURN OVER
Question 11
Let dydx
x xy y x= − = =3 2 1and when .
UsingEuler’smethodwithastepsizeof0.1,theapproximationtoy whenx =1.1isA. 0.9B. 1.0C. 1.1D. 1.9E. 2.1
Question 12
If dydx
x= +( )2 16 andy=5whenx=1,thenthevalueofy whenx =4isgivenby
A. ( )2 161
4x dx+ +( )∫ 5
B. ( )2 161
4x dx+∫
C. ( )2 161
4x dx+ +∫ 5
D. ( )2 161
4x dx+ −∫ 5
E. ( )2 161
4x dx+ −( )∫ 5
Question 13Usingthesubstitutionu x= +1 then
dxx x( )+ +∫ 2 10
2canbeexpressedas
A. 1121
3
u udu
( )+∫
B. 2120
2
udu
+∫
C. 111
3
u udu
( )+∫
D. 14
112 20
2
u udu
( )+∫
E. 2 1121
3
udu
+∫
2014SPECMATHEXAM2 6
SECTION 1–continued
Question 14
y
x
Thedifferentialequationthatisbestrepresentedbytheabovedirectionfieldis
A. dydx x y
=−1
B. dydx
y x= −
C. dydx y x
=−1
D. dydx
x y= −
E. dydx y x
=+1
7 2014SPECMATHEXAM2
SECTION 1–continuedTURN OVER
Question 15Ifθistheanglebetween a i j k~ ~ ~ ~
= + −3 4 and b i j k~ ~ ~ ~= − +4 3 ,then cos 2θ( ) is
A. −45
B. 725
C. − 725
D. 1425
E. − 2425
Question 16Twovectorsaregivenby a i j k~ ~ ~
= + −4 3~ m and b i j k~ ~ ~ ~= − + −2 n ,wherem n R, .∈ +
If a = 10~
and a~ isperpendicularto b~ ,thenmandnrespectivelyare
A. 5 3 33
,
B. 5 3 3,
C. −5 3 3,
D. 93 5 9393
,
E. 5,1
Question 17Theaccelerationvectorofaparticlethatstartsfromrestisgivenby
a ( i 0 ) j k , where~ ~ ~~ ) sin( ) cos( .t t t e tt= − + − ≥−4 2 2 2 20 02
Thevelocityvectoroftheparticle,v(~ ),t isgivenby
A. − − + −8 2 40 2 40 2cos( ) sin(t t e ti ) j k~ ~ ~
B. 2 2 1 2 10 2cos( ) sin(t t e ti 0 ) j k~ ~ ~+ + −
C. 8 8 2 40 2 40 402− ( )( ) − ( ) + ( − )−cos sint t e ti j k~ ~ ~
D. 2 2 2 1 2 10 102cos sint t e t( ) −( ) + ( ) + ( − )−i 0 j k~ ~ ~
E. 4 2 4 20 2 20 20 2cos sint t e t( ) −( ) + ( ) + ( − )−i j k~ ~ ~
2014SPECMATHEXAM2 8
SECTION 1–continued
Question 18Abodyonahorizontalsmoothplaneisacteduponbyfourforces, F F F and F~ ~ ~ ~, , ,1 2 3 4 asshown.
Theforce F~1actsinanortherlydirectionandtheforce F~4
actsinawesterlydirection.
60°
60°150°
N
EW
S
F4~
F1~F2~
F3~
Giventhat F = 1, F = 2, F = 4 and F = 5,~1 ~2 ~3 ~4 themotionofthebodyissuchthatit
A. isinequilibrium.B. movestothewest.C. movestothenorth.D. movesinthedirection30°southofwest.E. movestotheeast.
Question 19Thevelocityvectorofa5kgmassmovinginthecartesianplaneisgivenbyv i j~ ~ ~
sin cos ,t t t( ) = ( ) + ( )3 2 4 2 wherevelocitycomponentsaremeasuredinm/s.Duringitsmotion,themaximummagnitudeofthenetforce,innewtons,actingonthemassisA. 8B. 30C. 40D. 50E. 70
9 2014SPECMATHEXAM2
SECTION 1–continuedTURN OVER
Question 20
3 kg
5 kg
Particlesofmass3kgand5kgareattachedtotheendsofalightinextensiblestringthatpassesoverafixedsmoothpulley,asshownabove.Thesystemisreleasedfromrest.Assumingthesystemremainsconnected,thespeedofthe5kgmassaftertwosecondsisA. 4.0m/sB. 4.9m/sC. 9.8m/sD. 10.0m/sE. 19.6m/s
Question 21Theacceleration,inms–2,ofaparticlemovinginastraightlineisgivenby–4x,wherex metresisitsdisplacementfromafixedoriginO.Iftheparticleisatrestwherex=5,thespeedoftheparticle,inms–1,wherex=3isA. 8
B. 8 2
C. 12
D. 4 2
E. 2 34
2014SPECMATHEXAM2 10
END OF SECTION 1
Question 22Thevelocity–timegraphbelowshowsthemotionofabodytravellinginastraightline,wherevms–1isitsvelocityafter tseconds.
O
10987654321
–1–2–3–4–5–6
v
t1 2 3 4 5 6 7 8 9 10
Thevelocityofthebodyoverthetimeinterval t∈[ ]4 9, isgivenby v t t( ) = − −( ) +9
164 92 .
Thedistance,inmetres,travelledbythebodyoverninesecondsisclosesttoA. 45.6B. 47.5C. 48.6D. 51.0E. 53.4
11 2014SPECMATHEXAM2
TURN OVER
Workingspace
2014SPECMATHEXAM2 12
SECTION 2 – Question 1–continued
Question 1 (11marks)Considerthefunction f withrule ( )( )
9( )2 4
f xx x
=+ −
overitsmaximaldomain.
a. Findthecoordinatesofthestationarypoint(s). 3marks
b. Statetheequationsofallasymptotesofthegraphof f. 2marks
SECTION 2
Instructions for Section 2Answerallquestionsinthespacesprovided.Unlessotherwisespecified,anexactanswerisrequiredtoaquestion.Inquestionswheremorethanonemarkisavailable,appropriateworkingmust beshown.Unlessotherwiseindicated,thediagramsinthisbookarenotdrawntoscale.Taketheacceleration due to gravitytohavemagnitudegm/s2,whereg=9.8.
13 2014SPECMATHEXAM2
SECTION 2–continuedTURN OVER
c. Sketchthegraphof f for x∈ −[ ]6 6, ontheaxesbelow,showingasymptotes,thevaluesofthecoordinatesofanyinterceptswiththeaxes,andthestationarypoint(s). 3marks
–6 –4 –2 O 2 4 6
6
4
2
–2
–4
–6
y
x
Theregionboundedbythecoordinateaxes,thegraphof f andthelinex=3,isrotatedaboutthex-axistoformasolidofrevolution.
d. i. Writedownadefiniteintegralintermsofxthatgivesthevolumeofthissolidofrevolution. 2marks
ii. Findthevolumeofthissolid,correcttotwodecimalplaces. 1mark
2014SPECMATHEXAM2 14
SECTION 2 – Question 2–continued
Question 2 (13marks)Considerthecomplexnumber z i1 3 3= − .
a. i. Expressz1inpolarform. 2marks
ii. FindArg z14( ). 1mark
iii. Giventhat z i1 3 3= − isonerootoftheequation z3 24 3 0+ = , findtheothertworoots,expressingyouranswersincartesianform. 2marks
b. i. Findthevalueof z i z i1 12 2+( ) −( ), where z i1 3 3= − . 1mark
ii. Showthattherelation z i z i+( ) −( ) =2 2 4 canbeexpressedincartesianformas x y2 22 4+ +( ) = . 2marks
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SECTION 2–continuedTURN OVER
iii. Sketch{ : }z z i z i+( ) −( ) =2 2 4 ontheaxesbelow. 2marks
Im(z)
Re(z)–6 –4 –2 O 2 4 6
6
4
2
–2
–4
–6
c. Thelinejoiningthepointscorrespondingtok – 2iand − +( )2 k i ,wherek<0,istangentto thecurvegivenby{ : }z z i z i+( ) −( ) =2 2 4 .
Findthevalueofk. 3marks
2014SPECMATHEXAM2 16
SECTION 2 – Question 3–continued
Question 3 (10marks)Let a i j k and b i j k~ ~ ~ ~ ~ ~ ~ ~ .= + + = − −3 2 2 2
a. Express a~ asthesumoftwovectorresolutes,oneofwhichisparallelto b~ andtheotherofwhichisperpendicularto b~ .Identifyclearlytheparallelvectorresoluteandtheperpendicularvectorresolute. 5marks
17 2014SPECMATHEXAM2
SECTION 2–continuedTURN OVER
OABC isaparallelogramwhereDisthemidpointofCB .
OB and AD intersectatpointP.
LetOA→
= a~ andOC→
= c.~
C D B
O
P
A
b. i. GiventhatAP AD→ →
=α , writeanexpressionfor AP→
intermsofα, a~ and c.~ 2marks
ii. GiventhatOP OB→ →
= β ,writeanotherexpressionfor AP→
intermsofβ, a~ and c.~ 1mark
iii. Hencededucethevaluesofαandβ. 2marks
2014SPECMATHEXAM2 18
SECTION 2 – Question 4–continued
Question 4 (12marks)Atawaterfunpark,aconicaltankofradius0.5mandheight1misfillingwithwater.Atthesametime,somewaterflowsoutfromthevertex,wettingthoseunderneath.Whenthetankeventuallyfills,ittipsoverandthewaterfallsout,drenchingallthoseunderneath.Thetankthenreturnstoitsoriginalpositionandbeginstorefill.
1 m
0.5 m
h
Waterflowsinataconstantrateof0.02πm3/minandflowsoutatavariablerateof
0 01. π h m3/min,wherehmetresisthedepthofthewateratanyinstant.
a. Showthatthevolume, Vcubicmetres,ofwaterintheconewhenitisfilledtoadepthof h metresisgivenbyV h=
π12
3. 1mark
b. Findtherate,inm/min,atwhichthedepthofthewaterinthetankisincreasingwhenthedepthis0.25m. 4marks
19 2014SPECMATHEXAM2
SECTION 2 – Question 4–continuedTURN OVER
Thetankisemptyattimet=0minutes.
c. Byusinganappropriatedefiniteintegral, findthetimeittakesforthetanktofill.Giveyouranswerinminutes,correcttoonedecimalplace. 2marks
2014SPECMATHEXAM2 20
SECTION 2–continued
Anotherwatertank,shownbelow,hastheshapeofalargebucket(partofacone)withthedimensionsgiven.Waterfillsthetankatarateof0.05πm3/min,butnowaterleaksout.
0.75 m
0.5 m
x
1 m
Whenfilledtoadepthofxmetres,thevolumeofwater,V cubicmetres,inthetankisgivenby
V x x x= + +( )π48
6 123 2
d. Giventhatthetankisinitiallyempty, findthedepth,xmetres,asafunctionoftimet. 5marks
21 2014SPECMATHEXAM2
SECTION 2 – Question 5–continuedTURN OVER
Question 5 (12marks)Thediagrambelowshowsblocksofmass3kgand5kgonasmoothplaneofinclinationθdegreestothehorizontal,connectedbyatautlightinextensiblerope.The5kgblockisconnectedbyanotherlightinextensibleropeviaasmoothpulleyatthetopoftheinclinedplanetoablockofmass2kg,hangingvertically.The2kgblockhasaccelerationupwardsofams–2.Theforcesoneachofthethreeblocksareshownonthediagram.
N
RT1
T2T2
5g
3g
T1
2gθ
a. i. Writedownanequationofmotionforthe2kgblock. 1mark
ii. Byresolvingforcesactingparalleltotheplaneontheothertwoblocks,writedownanequationofmotionforeachofthe3kgand5kgblocks,usingthesymbolsdefinedintheabovediagram. 2marks
2014SPECMATHEXAM2 22
SECTION 2 – Question 5–continued
iii. Showthatag
=−( )4 1
5sin( )
.θ
1mark
iv. Findtheangleθforthesystemtobeinequilibrium.Giveyouranswerindegrees,correcttoonedecimalplace. 1mark
Nowconsideradifferent situation wherethe3kgand5kgblocksareplacedatrestonarough plane.Theplaneisinclinedat30°tothehorizontal,asshown,andtherearenostringsattachedtotheblocks.Thecoefficientoffrictionbetweenthe3kgblockandtheplaneis0.11andthecoefficientoffrictionbetweenthe5kgblockandtheplaneis0.01.Initially,thedistancebetweenthetwoclosestfacesoftheblocksis3m.
b. i. Onthediagrambelow,show andlabeltheforcesactingoneachofthetwoblocks. 2marks
3 m
3 kg
5 kg
30°
23 2014SPECMATHEXAM2
ii. Calculatetheaccelerationofeachblockdowntheplane.Giveyouranswersinm/s2, correcttothenearest0.01m/s2. 2marks
iii. Calculatethetimetakenforthe5kgblocktocollidewiththe3kgblock.Giveyouranswerinseconds,correcttothenearest0.01s. 3marks
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 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