1
A C E
EXAM PAPER
Student name: ______________________
2018 YEAR 12
YEARLY EXAMINATION
Mathematics Extension 2
General Instructions
� Working time - 180 minutes � Write using black pen � NESA approved calculators may be used � A reference sheet is provided at the back of this paper � In Questions 11-16, show relevant mathematical reasoning and/or
calculations
Total marks: 100
Section I –10 marks � Attempt Questions 1-10 � Allow about 15 minutes for this section Section II – 90 marks � Attempt Questions 11-16 � Allow about 165 minutes for this section
Year 12 Mathematics Extension 2
2
SectionI10marksAttemptquestions1-10Allowabout15minutesforthissectionUsethemultiple-choiceanswersheetforquestions1-101. Letarg(𝑧) = (
)foracertaincomplexnumberz.Whatisarg(𝑧*)?
(A) −7𝜋5
(B) −3𝜋5
(C) 2𝜋5
(D) 3𝜋5
2. If 1ln(tan56𝑥)1 + 𝑥:
√<
=𝑑𝑥, whichofthefollowingintegralsusesthecorrectsubstitution?
(A)1 ln𝑢𝑑u√<
=
(B)1
ln𝑢1 + tan:𝑢
𝑑𝑢(C
=
(C)1 ln𝑢𝑑𝑢(<
=
(D)1
ln𝑢1 + tan:𝑢
𝑑𝑢(<
=
3. Ellipse: (𝑥 − 1):
9+𝑦:
4= 1 Hyperbola: 𝑥: − 𝑦: = 4
Howmanypointsdothegraphsoftheaboveequationshaveincommon? (A) 0 (B) 1 (C) 2 (D) 34. Aforceofmagnitude4Nactsinthenorth-eastdirectionandanotherforceof3Nactsin
theeasterlydirection.Whatistheresultantmagnitude(inN)ofthesetwoforces? (A)
G25 − 12√2 (B)
G25 + 12√2 (C) 25 + 12√2 (D) 25 − 12√2
Year 12 Mathematics Extension 2
3
5. Whatistheeccentricityoftheellipse3𝑥: + 5𝑦: − 12𝑥 + 30𝑦 + 42 = 0? (A)
I25
(B)I35
(C)I52
(D)I53
6.
Thecurve𝑦 = 4𝑥: − 𝑥Jandthex-axisbetweenx=0andx=2isrotated2𝜋radiansaboutthey-axis.WhichofthefollowingisanexpressionforthevolumeVofthesolidformed?
(A)𝜋1 K4 − 𝑦
J
=𝑑𝑦
(B)2𝜋1 K4 − 𝑦
J
=𝑑𝑦
(C)4𝜋1 K4 − 𝑦
J
=𝑑𝑦
(D)8𝜋1 K4 − 𝑦
J
=𝑑𝑦
7. Aparticleofmass4kgmovesinacircularmotiononasmoothfrictionlesstableataspeed
of3m/s.Itisattachedtoafixedpointinthemiddleofthetablebyalight,inelasticstringoflength2metres.Whatisthetensioninthestring?
(A) 6N (B) 12N (C) 18N (D) 36N8. Theequation𝑥J + 𝑝𝑥 + 𝑞 = 0 where𝑝 ≠ 0 and𝑞 ≠ 0hasroots𝛼, 𝛽, 𝛾and𝛿.Whatis𝛼4 + 𝛽4 + 𝛾4 + 𝛿4?
(A) –4q (B) 𝑝2 − 2𝑞 (C) 𝑝4 − 2𝑞 (D) 𝑝4
Year 12 Mathematics Extension 2
4
9. Thepoint𝑃 V𝑐𝑝, XYZliesontherectangularhyperbola𝑥𝑦 = 𝑐:.
WhatistheequationofthenormaltothehyperbolaatP?
(A) 𝑝𝑦 − 𝑐 = 𝑝<(𝑥 − 𝑐𝑝)
(B) 𝑝𝑥 −1𝑝𝑦 = 𝑐𝑝: [1 −
1𝑝:\
(C) 𝑥 + 𝑝𝑞𝑦 = 𝑐(𝑝 + 𝑞)
(D) 𝑥 + 𝑝:𝑦 = 2𝑐𝑝 10. Aparticleisprojectedwithaspeedof20m/sandpassesthroughapointPwhose
horizontaldistancefromthepointofprojectionis30mandwhoseverticalheightabovethepointofprojectionis8.75m.Whatistheangleofprojection?
(A) tan−1 [
23\
(B) tan−1 [
32\
(C) tan−1 [
34\
(D) tan−1 [
43\
Year 12 Mathematics Extension 2
5
SectionII90marksAttemptquestions11-16Allowabout165minutesforthissectionAnswereachquestionintheappropriatewritingbooklet.Yourresponsesshouldincluderelevantmathematicalreasoningand/orcalculations.Question11(15marks) Marks (a) For𝑧 = 1 − 𝑖,𝑤 = 3 − 2𝑖, find: (i) |𝑧 + 𝑤| 1 (ii) 𝑧: − 𝑤: 2 (b) Findtheexactvalueof:
(i) 1𝑥 + 1
√𝑥: + 2𝑥 + 5𝑑𝑥
<
:
2
(ii) 1 K4− 𝑥:𝑑𝑥
√:
=
3
(c) Findtheequationofthetangenttothecurve𝑥2 − 𝑥𝑦 + 𝑦3 = 1atthepoint
P(1,1)tothecurve.3
(d) Find 1𝑑𝑥
9𝑥: + 6𝑥 + 5 3
(e) Theequation2𝑥< − 3𝑥: − 5𝑥 − 1 = 0hasroots𝛼, 𝛽and𝛾. 1
Findthevalueof1
𝛼<𝛽<𝛾<
Year 12 Mathematics Extension 2
6
Question12(15marks) Marks (a) Theregionunderthecurve𝑦 = 𝑒5bc andabovethex-axisfor−𝑎 ≤ 𝑥 ≤ 𝑎 is
rotatedaboutthey-axistoformasolid.
(i) Calculatethevolumeofthesolidusingthemethodofcylindricalshells. 3 (ii) Whatisthelimitingvalueofthevolumeofthesolidasaapproaches
infinity?1
(b) 𝑧1 = 1 + 𝑖 and𝑧2 = √3 − 𝑖 Find𝑧1 ÷ 𝑧2intheform𝑎 + 𝑖𝑏 whereaandbarereal. 1 Write𝑧6and𝑧:inmodulus-argumentform. 2 Writecos )(
6:asasurdbyequatingequivalentexpressionsfor𝑧1 ÷ 𝑧2. 1
(c) FindthevaluesofA,B,CandDsuchthat: 2 5𝑥3 − 3𝑥2 + 2𝑥 − 1
𝑥4 + 𝑥2=𝐴𝑥+𝐵𝑥2+𝐶𝑥 + 𝐷𝑥2 + 1
Henceevaluate 1
5𝑥< − 3𝑥: + 2𝑥 − 1𝑥J + 𝑥:
𝑑𝑥 2
(d) Solvethepolynomialequation𝑥4 − 6𝑥3 + 9𝑥2 + 4𝑥 − 12 = 0,giventhatthe
equationhasadoubleroot.3
Year 12 Mathematics Extension 2
7
Question13(15marks) Marks (a) IntheArganddiagram,ABCDisasquare,andOEandOFareparallelandequal
inlengthtoABandADrespectively.TheverticesAandBcorrespondtothecomplexnumbersz1andz2respectively.
(i) ExplainwhythepointEcorrespondsto𝑧: − 𝑧6. 1 (ii) WhatcomplexnumbercorrespondstopointF? 1 (iii) WhatcomplexnumbercorrespondstovertexD? 1
(b) Evaluatetheintegral 1ln𝑥𝑥:
𝑑𝑥 2
(c) Amotorbiketravelsaroundacircularbendthatisbankedatanangleof𝛼˚to
thehorizontal.Thebiketravelsatalineontheroadwheretheradiusofthecurveisrmetres,ataconstantspeed,sothereisnosidewaysfrictionalforceactingonthebike.
Findexpressionsfor𝑁sin𝛼and𝑁cos𝛼byresolvingforces. 2 Deriveanexpressionforthevelocity(v)ofthebike. 1 Findthevalueof𝛼˚iftheradiusofthecurveis120metresandtheroad
isbankedtoallowvehiclestotravelat90km/h.(Useg=10m/s)2
(d) Asolidisformedbyrotatingaboutthelinex=2theregionboundedbytheparabolay=x2,thex-axis,x=0andx=2.Findthevolumeofthissolidusingthemethodofslicing.
3
(e) Amotelhasfourvacantrooms.Eachroomcanaccommodateamaximumoffourpeople.Inhowmanydifferentwayscansixpeoplebeaccommodatedinthefourrooms?
2
Year 12 Mathematics Extension 2
8
Question14(15marks) Marks (a) (i) Considerthepolynomialequation:𝑎𝑥< + 𝑏𝑥: + 𝑐𝑥 + 𝑑 = 0.
Whataretherelationsbetweentherootsa,b,𝛾oftheequation,intermsofthecoefficientsa,b,candd?
1
(ii) Theequation36𝑥< − 12𝑥: − 11𝑥 + 2 = 0hasrootsa,bandc.Findaif𝛼 = 𝛽 + 𝛾.
1
(iii) Theequation𝑥< + 𝑏𝑥: + 𝑐𝑥 + 𝑑 = 0hasrootsa,bandc.Showthat𝑏< − 4𝑏𝑐 + 8𝑑 = 0if𝛼 = 𝛽 + 𝛾.
2
(b)
Thegraphof𝑓(𝑥) = 𝑒5b − 2isshownabove.Drawseparateone-thirdpagesketchesofthesefunctions.Indicateclearlyanyasymptotesandinterceptswiththeaxes.
𝑦 = |𝑓(𝑥)| 1 𝑦 = {𝑓(𝑥)}: 1
𝑦 =1
𝑓(𝑥) 2
𝑦 = ln𝑓(𝑥) 1 (c) AlightinextensiblestringOPisfixedattheendOandisattachedattheother
endPtoaparticleofmassm(inkg)whichismovinguniformlyinahorizontalcirclewhosecentreisverticallybelowanddistantx(inmetres)fromO.Letgbetheaccelerationduetogravity.
Showthattheperiodofmotionisgivenbytheformula: 3
𝑇 = 2𝜋I𝑥𝑔
Whatistheeffectonthemotionoftheparticleifthemassisdoubled? 1 Ifthenumberofrevolutionspersecondisincreasedfrom2to3,find
thechangeinx.Answercorrecttothreedecimalplacesanduse10ms-2forgravity.
2
Year 12 Mathematics Extension 2
9
Question15(15marks) Marks (a) ThecirclesXPYSandXYRQintersectatthepointsXandY.PXQ,PYR,QSY,PST
andQTRarestraightlines.
Explainwhy∠𝑆𝑇𝑄 = ∠𝑌𝑅𝑄 + ∠𝑌𝑃𝑆. 1 Showthat∠𝑌𝑅𝑄 + ∠𝑌𝑃𝑆 + ∠𝑆𝑋𝑄 = 𝜋 2 ProvethatSTQXisacyclicquadrilateral. 1 Let∠𝑄𝑃𝑌 = 𝛼and∠𝑃𝑄𝑌 = 𝛽.Showthat∠𝑆𝑇𝑄 = 𝛼 + 𝛽 3 (b) 𝑃(acos𝜃, 𝑏sin𝜃)and𝑃(acos𝜑, 𝑏sin𝜑)aretheendpointsofadiameterofthe
ellipseshownbelow.
TangentstotheellipseatP,Qcutthex-axisatX,Urespectively,andthey-axisatY,Vrespectively.
ShowthatthetangenttotheellipseatPisthefollowingequation. 2 𝑥cos𝜃
𝑎+𝑦sin𝜃𝑏
= 1
Showthat𝜑 = 𝜃 ± 𝜋 2 WhatarethecoordinatesofX,Y,U,Vintermsofa,band𝜃? 2
ShowthattheareaofXYUVis4𝑎𝑏|sin2𝜃| 2
Year 12 Mathematics Extension 2
10
Question16(15marks) Marks
(a) 𝐼� = 11
(1 + 𝑥:)�6
=𝑑𝑥𝑛 = 1,2,3, . ..
Showthat 𝐼𝑛+1 =
2𝑛 − 12𝑛
𝐼𝑛 +1
𝑛 × 2𝑛+1𝑛 = 1,2,3, … 3
Henceevaluate1
1(1 + 𝑥:)<
6
=𝑑𝑥 2
(b) Provetheidentity: cos3𝐴 −34cos𝐴 =
14cos3𝐴 2
Showthat𝑥 = 2√2cosAsatisfiesthecubicequation𝑥< − 6𝑥 + 2 = 0giventhatcos3𝐴 = − 6
:√:
2
Whatarethethreerootsoftheequation𝑥< − 6𝑥 + 2 = 0?Answercorrecttofourdecimalplaces.
1
(c) Given 𝑦 =𝑥<
𝑥: − 4
Findthecoordinatesofallthestationarypoints. 2 Whataretheequationsoftheasymptotesofthecurve? 2
Hencesketchthecurve 𝑦 =𝑥<
𝑥: − 4. 2
Endofpaper
Year 12 Mathematics Extension 2
11
Year 12 Mathematics Extension 2
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Year 12 Mathematics Extension 2
13
Year 12 Mathematics Extension 2
1
ACEExamination2018Year12MathematicsExtension2YearlyExaminationWorkedsolutionsandmarkingguidelinesSectionI Solution Criteria
1Arg(%&) = 7Arg(%)
=7+5= −
3+5
1Mark:B
2
Let/ = tan3456/65
=1
1 + 59
Whenx=0,u=0and5 = √3, / =<=
>ln(tan345)1 + 59
√=
@65 = > ln/6/
<=
@
1Mark:C
3
Thereare3pointsincommon.
1Mark:D
4Force = E39 + 49 − 2 × 3 × 4 × cos135˚
= K25 + 12√2
1Mark:B
5
359 + 5L9 − 125 + 30L + 42 = 0359 − 125 + 5L9 + 30L = −42
3(59 − 45 + 4) + 5(L9 + 6L + 9) = −42 + 12 + 453(5 − 2)9 + 5(L + 3)9 = 15(5 − 2)9
5+(L + 3)9
3= 1
\ P = √5, Q = √3
R9 = 1 −Q9
P9= 1 −
S√3T9
S√5T9 =
25orR = U
25
1Mark:A
6
Areaofthesliceisanannulus.459 − 5V = L
5V − 459 + 4 = 4 − L(59 − 2)9 = 4 − L59 − 2 = ±E4 − L
59 = 2 ± E4 − LXY = +(Z9 − [9)XL
= +\S2 + E4 − LT − S2 − E4 − LT]XL = 2+E4 − LXL
Y = lim`a→@
c 2+V
ad@
E4 − LXL = 2+> E4 − LV
@6L
1Mark:B
Year 12 Mathematics Extension 2
2
7e =
fg9
[
=4 × 39
2
= 18N
1Mark:C
8
Sumoftheroots
(j + k + l + X) = −Q
P=0
1= 0
jV + mj + n = 0kV + mk + n = 0lV + ml + n = 0XV + ∂e + n = 0 jV + kV + lV + XV + m(j + k + l + X) + 4n = 0
jV + kV + lV + XV + m × 0 + 4n = 0jV + kV + lV + XV = −4n
1Mark:A
9
5L = p9
L + 56L
65= 0
6L
65= −
L
5
6L
65= −
p
m÷ pm = −
p
pm9= −
1
m9
\Gradientofthenormalism9L −
p
m= m9(5 − pm)
mL − p = m=(5 − pm)
1Mark:A
10
Equationsofprojectilemotion5 = Yrcosj30 = 20rcosj
r =3
2cosj①
L = −1
2tr9 + Yrsinj
8.75 = −1
2× 10 × r9 + 20rsinj
35= −20r9 + 80rsinj② Substitutingequation(1)intoequation(2)
35 = −20 × w3
2cosjx9
+ 80 × w3
2cosjx × sinj
35 = −45sec9j + 120tanj7 = −9(tan9j + 1) + 24tanj
9tan9j − 24tanj + 16 = 0(3tanj − 4)9 = 0
3tanj = 4
tanj =4
3
j = tan34 w4
3x
1Mark:D
Year 12 Mathematics Extension 2
3
SectionII Solution Criteria
11(a)
(i)
|% + z| = |4 − 3{|= 5
1mark:Correct
answer.
11(a)
(ii)
%9 − z9 = (1 − {)9 − (3 − 2{)9= −2{ − (9 − 12{ + 4{9)= −2{ − 5 + 12{= −5 + 10{
2marks:
Correctanswer.
1mark:Shows
some
understanding.
11(b)
(i)
Usethesubstitution/ = 59 + 25 + 56/65
= 25 + 2
0.56/ = (5 + 1)65Whenx=2thenu=13andwhenx=3thenu=20
>5 + 1
√59 + 25 + 565 = >
0.56/
/49
9@
4=
=
9
= |/49}4=
9@
= √20 − √13
2marks:
Correctanswer.
1mark:Finds
theprimitive
functionorsets
upthe
integration
using
substitution.
11(b)
(ii)
Usethesubstitution5 = 2sin~656~
= 2cos~
65 = 2cos~6~Whenx=0then~ = 0andwhen5 = √2then~ =
<
V
> E4− 5965 = > 2cos~ × 2cos~6~
<V
@
√9
@
= 4> w12+12cos2~x 6~
<V
@
= 4 |12~ +
14sin2~}
@
<V
= 4 w+8+14x
=+2+ 1
3marks:
Correctanswer.
2marks:Finds
theprimitive
function.
1mark:
Correctly
expressesthe
integralin
termsofq
11(c) 59 − 5L + L= = 1
25 − wL + 56L65x + 3L9
6L65
= 0
(3L9 − 5)6L65
= L − 25 AtP(1,1)6L65
=L − 253L9 − 5
= −12
\Equationofthetangent
L − 1 = −12(5 − 1)
5 + 2L − 3 = 0
3marks:
Correctanswer.
2marks:
Evaluatesthe
derivativeatP
tofindthe
gradientofthe
tangent.
1mark:
Differentiates
implicitly.
Year 12 Mathematics Extension 2
4
11(d) Completethesquare959 + 65 + 5 = 9w59 +
2
35x + 5
= 9 w59 +2
35 +
1
9x − 1 + 5
= 9 w5 +1
3x
9
+ 4
= (35 + 1)9 + 29
Therefore>
65
959 + 65 + 5= >
65
(35 + 1)9 + 29
=1
6tan34 w
35 + 1
2x + �
3marks:Correctanswer.2marks:Completesthesquareandsetsupintegration.1mark:Showssomeunderstandingoftheproblem.
11(e)jkl = −
6
P=1
2
1
j=k=l==
1
(jkl)=
=1
0.5== 8
1mark:correctanswer
12(a)(i)
CylindricalshellswithradiusofxandheightR3ÄÅ
Y = lim`Ä→@
c2+5R3ÄÅX5
Ç
Äd@
= +> 25R3ÄÅ65
Ç
@
= −+ÉR3ÄÅÑ@
Ç
= +S1 − R3ÇÅT
2marks:Correctanswer.1mark:Findstheradiusandheightofthecylindricalshell
12(a)(ii)
limÇ→Ö
+S1 − R3ÇÅT = +
1mark:Correctanswer.
12(b)(i)
%4
%9=
1 + {
√3 − {×√3 + {
√3 + {
=√3 − 1
4+ {
√3 + 1
4
1mark:Correctanswer.
12(b)(ii) %4 = √2(cos
+
4+ {sin
+
4)
2marks:Correctanswer.1mark:Findsoneofthepointsinmodulus-argumentform.%9 = 2 Ücos á−
+
6à + {siná−
+
6àâ
Year 12 Mathematics Extension 2
5
12(b)(iii)
%4%9=√22Ücos á
+4—+6à + {siná
+4—+6àâ
=√3 − 14
+ {√3 + 14
Equatingtherealparts1
√2cos w
5+12x =
√3 − 14
\cos w5+12x =
√6 − √24
1mark:Correctanswer.
12(c)(i)
55= − 359 + 25 − 1 = ã5(59 + 1) + å(59 + 1) + (�5 + ç)59= ã5= + ã5 + å59 + å + �5= + ç59
Therefore(ã + �)5= = 55=①(å + ç)59 = −359②
ã5 = 25③å = −1④
HenceA=2,B=–1,C=3andD=–2
2marks:Correctanswer.1mark:Findstwoofthepronumeralsorshowssomeunderstanding.
12(c)(ii) >
55= − 359 + 25 − 15V + 59
65 = >w25−159+35 − 259 + 1
x65
= >w25−159+
3559 + 1
−2
59 + 1x65
= 2ln5 +15+32ln(59 + 1) − 2tan345 + �
2marks:Correctanswer.1mark:Correctlyfindsoneoftheintegrals
12(d) ê(5) = 5V − 65= + 959 + 45 − 12ê′(5) = 45= − 1859 + 185 + 4 Bytrialanderror(x–2)isadoubleroot(ê(2) = ê′(2) = 0) Dividing5V − 65= + 959 + 45 − 12by59 − 45 + 4gives59 − 25 − 3 Thereforeê(5) = 5V − 659 + 959 + 45 − 12
= (5 − 2)9(5 − 3)(5 + 1) \x=–1,2and3
3marks:Correctanswer.2marks:Findsthedoublerootormakessignificantprogress.1mark:Usesthederivativeofthefunction.
13(a)(i)
OE//ABandOE=AB\OABEisaparallelogramLetwbethevectorthatcorrespondtopointE.z + %4 = %9\z = %9 − %4
1mark:Correctanswer.
Year 12 Mathematics Extension 2
6
13(a)(ii)
íe ⊥ íîandOF=OE\Fcorrespondsto{(%9 − %4)(multiplyingacomplexnumberbyicorrespondstoananticlockwiserotationabouttheoriginthrough90˚)
1mark:Correctanswer.
13(a)(iii)
SinceAD//OFandAD=OFPointDcorrespondstothecomplexnumber:%4 + {(%9 − %4) = %4(1 − {) + {%9
1mark:Correctanswer.
13(b) Integrationbyparts
> ln559 65 = > ln5 × 665 w−
15x65
= − ln55 −> 665 ln5 ×−
15 65
= − ln55 − 15 + �
= − ln5 + 15 + �
2marks:Correctanswer.1mark:Setsupintegrationbyparts.
13(c)(i)
Resolvingforces
ecosj = fg9[ − ïsinj
0 = fg9[ − ïsinj
\ïsinj = fg9[
Alsoesinj = ïcosj −mg
\ïcosj = mg
2marks:Correctanswer.1mark:Findseitherïcosjorïsinj
13(c)(ii) ïsinj
ïcosj =fg9[ft
tanj = g9[t
g9 = [ttanjg = E[ttanj
1mark:Correctanswer.
13(c)(iii)
Frictionalforceis0
g = 90kmh =90 × 100060 × 60
= 25m/s
tanj = g9[t
= 259120 × 10
= 2548
j = 27˚31′
2marks:Correctanswer.1mark:Usestheresultsfortanjwithonecorrectvalue.
Year 12 Mathematics Extension 2
7
13(d) SamevolumeasL = (5 − 2)9rotatedaboutthey-axisAreaofthesliceisacirclewitharadiusofx–2andheighty.XY = +(5 − 2)9XL
= +SEL − 2T9XL
Y = lim`a→@
c+
V
ad@
SEL − 2T9XL
Y = +> L − 4EL + 46LV
@
= + |1
2L9 −
8
3L=9 + 4L}
@
V
= + w8 −8
3× 8 + 16x
=8+
3cubicunits
3marks:Correctanswer.2marks:Correctintegralforthevolumeofthesolid.1mark:Setsuptheareaoftheslice
13(e) 6peoplein4roomswithnorestrictions:46ways6peoplein4roomswitharoomof6: C4V 6peoplein4roomswitharoomof5: C4V × C4
= × Cúù ways
\6peoplein4roomswitharoomof4:4ù– C4
V – C4V × C4
= × Cúù = 4020
2marks:Correctanswer.1mark:Showssomeunderstanding
14(a)(i) j + k + l = −
Q
P
jk + kl + lj =p
P
jkl = −6
P
1mark:Correctanswer.
14(a)(ii) j + k + l = −
12
36
j + j =1
3
j =1
6
1mark:Correctanswer.
14(a)(iii) j + k + l = −
Q
1
j + j = −Q
j = −Q
2
w−Q
2x=
+ Q × w−Q
2x9
+ p × w−Q
2x + 6 = 0
−Q=
8+Q=
4−Qp
2+ 6 = 0
−Q= + 2Q= − 4Qp + 86 = 0Q= − 4Qp + 86 = 0
2marks:Correctanswer.1mark:Findsaorshowssomeunderstanding.
14(b)(i)
1mark:Correctanswer.
Year 12 Mathematics Extension 2
8
14(b)(ii)
1mark:Correctanswer.
14(b)(iii)
2marks:Correctanswer.1mark:Showsoneasymptoteonly.
14(b)(iv)
1mark:Correctanswer.
14(c)(i)
LetwbetheangularvelocityoftheparticlePaboutC.TheforcesactingontheparticleareitsweightmgandthetensionT1inthestring.ResolvingtheforcesatP:Horizontallyf[ü9 = †4cos á
+2 − ~à = †4sin~
3marks:Correctanswer.2marks:Derivingwormakingsignificantprogress.1mark:Resolvestheforcesverticallyandhorizontally
Verticallyft = †4cos~Dividingtheabovetwoequationsf[ü9ft = †4sin~
†4cos~
[ü9t = tan~buttan~ = [
5
\[ü9t = [
5 orü = Kt5
Year 12 Mathematics Extension 2
9
Thetimeforonecompleterevolution(ortheperiodofmotion)
† =2+
ü
=2+
Kt5 = 2+U
5
t
14(c)(ii)
Inthedivisionofthetwoequationsofmotionthemassiscancelledoutandhencethereisnoeffectonthemotionifthemassisdoubled.
1mark:Correctanswer.
14(c)(iii)
Makingxthesubjectoftheaboveequation5 =
t
ü9 Letw1andw2betheangularvelocitiesofPintwosituationsintheproblem.ü4 = 2revolutionspersec = 4+radianspersec ü9 = 3revolutionspersec = 6+radianspersec Usingtheaboveequationforx
54 =t
(ü4)9=
t
16+9
59 =t
(ü9)9=
t
36+9
54 − 59 =t
+9w1
16−1
36x
=5t
+9
≈ 0.035m\Theparticlerisesbyabout0.035metres
2marks:Correctanswer.1mark:Showingsomeunderstandingoftheproblem.
15(a)(i)
In∆†¶Z∠®†© = ∠™Z© + ∠™¶®(Exteriorangleofatriangleisequaltothesumofthetwointerioroppositeangles)
1mark:Correctanswer.
15(a)(ii)
∠™¶® = ∠™´®(Anglesinthesamesegmentareequal)∠™Z© + ∠™´© = +(Oppositeanglesofacyclicquadrilateralaresupplementary)∠¶´™ + ∠™´© = +(Straightlinemeasures180˚)\∠™Z© = ∠¶´™\∠™Z© + ∠™¶® + ∠®´© = ∠¶´™ + ∠YXS + ∠®´© = +(Straightlinemeasures180˚)\∠™Z© + ∠™¶® + ∠®´© = +
2marks:Correctanswer.1mark:Makessomeprogresstowardsthesolution.
15(a)(iii)
∠®†© + ∠®´© = ∠™Z© + ∠™¶® + ∠®ã© = +(from(i)and(ii))\STQXisacyclicquadrilateralastheoppositeanglesaresupplementary.
2marks:Correctanswer.1mark:Makessomeprogresstowardsthesolution.
Year 12 Mathematics Extension 2
10
15(a)(iv)
∠´™® = ∠´¶®and∠®´™ = ∠®¶™(Anglesinthesamesegmentareequal)(Anglesinthesamesegmentareequal)j = ∠©¶™ = ∠´¶® + ∠®¶™
= ∠´™® + ∠®´™
= ∠´®©(Exteriorangleofatriangleisequaltothesumofthetwointerioroppositeangles)∠®†© = + − ∠®´©(OppositeanglesofcyclicquadrilateralSTQXaresupplementary)∠®†© = ∠¶´®(PXQisastraightline)∠®†© = ´©® + ´®©
= ∠¶©™ + ∠©¶™
= k + j (Exteriorangleofatriangleisequaltothesumofthetwointerioroppositeangles)
3marks:Correctanswer.2marks:Makessignificantprogresstowardsthesolution1mark:Appliesarelevantcircletheorem.
15(b)(i)
5 = acos~andL = Qsin~6L
65=6L
6~ ×
6~
65
= Qcos~ ×1
−asin~
= −Qcos~
asin~
Equationofthetangent
L − Qsin~ = −Qcos~
asin~(5 − acos~)
LPsin~ − PQs{Ø9~ = −5Qcos~ + PQcos
9~
5Qcos~ + LPsin~ = PQ(s{Ø9~ + cos
9~)
\5cos~
P+Lsin~
Q= 1
2marks:Correctanswer.1mark:Findsthegradientofthetangent.
15(b)(ii)
GradientsofOP.f =
L9 − L4
59 − 54
=Qsin~
acos~
=Q
Ptan~
Similarly,thegradientofOQis∞Çtan±
HenceO,P,Qarecollinear(gradientsareequal).tan~ = tan±since~ ≠ ±then± = ~ ± +
2marks:Correctanswer.1mark:FindsthegradientoftheOPandOQ
15(b)(iii)
5cos~
P +
Lsin~
Q = 1
Tangentcutsthex-axisatXá Ç
≥¥µ∂, 0àandy-axisatYá0, ∞
µ∑∏∂à
Similarly,sincecos± = −cos~andsin± = −sin~then Uá− Ç
≥¥µ∂, 0àandYá0,− ∞
µ∑∏∂à
2marks:Correctanswer.1mark:Findsthecoordinatesofonepoint.
Year 12 Mathematics Extension 2
11
15(b)(iv)
XYUVisarhombussincethediagonalsXUandYVbisecteachotheratrightanglesatO.ã = 1
2 5L =12 × πY × ™Y
= 12 × ∫
2Pcos~∫ × ∫
2Qsin~∫
= 4PQ|sin2~|
2marks:Correctanswer.1mark:DeducesthatXYUVisarhombus.
16(a)(i) ªº = > 1
(1 + 59)º4
@65Ø = 1,2,3, . . .
= [5(1 + 59)3º]@4 − > 5(−Ø)(1 + 59)3º344
@(25)65
= 23º + 2Ø> [(1 + 59) − 1](1 + 59)3º344
@65
= 23º + 2Ø> É(1 + 59)3º − (1 + 59)3(ºø4)Ñ4
@65
= 23º + 2تº − 2تºø42تºø4 = (2Ø − 1)ªº + 23ºªºø4 =
2Ø − 12Ø ªº +
1Ø × 2ºø4
3marks:Correctanswer.2marks:Makessignificantprogresstowardsthesolution1mark:Correctlyappliesintegrationbyparts.
16(a)(ii) ª= =
34 ª9 +
116
= 34 w12 ª4 +
14x +
116
= 38 ª4 +
14
ª4 = > 11 + 59 65
4
@
= [tan345]@4= +4
\> 1(1 + 59)= 65 =
3+ + 832
4
@
2marks:Correctanswer.1mark:AppliestherecurrencerelationtofindanexpressionforI3
16(b)(i)
cos=ã = cos(2ã + ã)= cos9ãcosã − sin9ãsinã= (2cos9ã − 1)cosã − 2cosãsinãsinã= 2cos=ã − cosã − 2cosãsin9ã= 2cos=ã − cosã − 2cosã(1 − cos9ã)= 4cos=ã − 3cosã
14 cos
=ã = cos=ã − 34 cosA
2marks:Correctanswer.1mark:Makessomeprogresstowardsthesolution.
16(b)(ii)
Substituting5 = 2√2cosAinto5= + 65 + 2 = 0S2√2cosãT= + 6 × S2√2cosãT + 2 = 0
16√2cos=ã − 12√2cosã = −2cos=ã − 34 cosã = − 2
16√2
= − 18√2
\14 cos3ã = − 1
8√2
cos3ã = − 12√2
2marks:Correctanswer.1mark:Makessomeprogresstowardsthesolution.
Year 12 Mathematics Extension 2
12
16(a)(i) ªº = >
1(1 + 59)º
4
@65Ø = 1,2,3, . . .
= [5(1 + 59)3º]@4 − > 5(−Ø)(1 + 59)3º344
@(25)65
= 23º + 2Ø> [(1 + 59) − 1](1 + 59)3º344
@65
= 23º + 2Ø> É(1 + 59)3º − (1 + 59)3(ºø4)Ñ4
@65
= 23º + 2تº − 2تºø42تºø4 = (2Ø − 1)ªº + 23º
ªºø4 =2Ø − 12Ø ªº +
1Ø × 2ºø4
3marks:Correctanswer.2marks:Makessignificantprogresstowardsthesolution1mark:Correctlyappliesintegrationbyparts.
16(a)(ii) ª= =
34 ª9 +
116
=34 w12 ª4 +
14x +
116
=38 ª4 +
14
ª4 = >1
1 + 59 654
@
= [tan345]@4=+4
\>1
(1 + 59)= 65 =3+ + 832
4
@
2marks:Correctanswer.1mark:AppliestherecurrencerelationtofindanexpressionforI3
16(b)(i)
cos3ã = cos(2ã + ã)= cos9ãcosã − sin9ãsinã= (2cos9ã − 1)cosã − 2cosãsinãsinã= 2cos=ã − cosã − 2cosãsin9ã= 2cos=ã − cosã − 2cosã(1 − cos9ã)= 4cos=ã − 3cosã
14cos3ã = cos3ã −
34cosA
2marks:Correctanswer.1mark:Makessomeprogresstowardsthesolution.
16(b)(ii)
Substituting5 = 2√2cosAinto5= − 65 + 2 = 0S2√2cosãT
=+ 6 × S2√2cosãT + 2 = 0
16√2cos=ã − 12√2cosã = −2
cos=ã −34 cosã = −
216√2
= −18√2
\14 cos3ã = −
18√2
cos3ã = −12√2
2marks:Correctanswer.1mark:Makessomeprogresstowardsthesolution.
16(b)(iii) cos3ã = −
12√2
3ã = ±cos34 w−12√2
x + 2Ø+
wherenisanintegerTakingn=0,1,2andthepositivebranchtoobtainthethreeroots5 = 2√2cosA= 2.2618,−2.6017, 0.3399
1mark:Correctanswer.
Year 12 Mathematics Extension 2
13
16(c)(i) ê′(5) = (59 − 4)(359) − (5=)(25)
(59 − 4)9
= 59(359 − 12 − 259)(59 − 4)9
= 59(59 − 12)(59 − 4)9
Stationarypointsf’(x)=059(59 − 12) = 0\5 = 0or5 = ±√12 = ±2√3When5 = 2√3then5 = 3√3When5 = −2√3then5 = −3√3 \Stationarypointsare(0,0),S2√3, 3√3T,S−2√3,−3√3T
2marks:Correctanswer.1mark:Findsonestationarypoint
16(c)(ii) ê(5) = 5=
59 − 4 =5=
(5 + 2)(5 − 2) \Verticalasymptotesare5 = ±2
ê(5) = 5=59 − 4 = 5 + 45
(5 + 2)(5 − 2)limÄ→Ö
ê(5) = 5 \Obliqueasymptotey=x
2marks:Correctanswer.1mark:Findstheverticalorobliqueasymptotes.
16(c)(iii)
Whenx=0theny=0\Pointofintersectionwithcoordinateaxesis(0,0)S2√3, 3√3Tisaminima(ê′(5)changessignfrom–to+)S−2√3,−3√3Tisamaxima(ê′(5)changessignfrom+to–)L = ê(5)isanoddfunction
2marks:Correctanswer.1mark:Showsmostofthefeaturesofthecurve.