Mathematics Extension 2 - Blueprint Education

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


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


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\


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<


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


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


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


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


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


11


12


13


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


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


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.


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


12(b)(ii) %4 = √2(cos

+

4+ {sin

+

4)

2marks:Correctanswer.1mark:Findsoneofthepointsinmodulus-argumentform.%9 = 2 Ücos á−

+

6à + {siná−

+

6àâ


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


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



6

13(a)(ii)

íe ⊥ íîandOF=OE\Fcorrespondsto{(%9 − %4)(multiplyingacomplexnumberbyicorrespondstoananticlockwiserotationabouttheoriginthrough90˚)


13(a)(iii)

SinceAD//OFandAD=OFPointDcorrespondstothecomplexnumber:%4 + {(%9 − %4) = %4(1 − {) + {%9


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


13(c)(iii)

Frictionalforceis0

g = 90kmh =90 × 100060 × 60

= 25m/s

tanj = g9[t

= 259120 × 10

= 2548

j = 27˚31′

2marks:Correctanswer.1mark:Usestheresultsfortanjwithonecorrectvalue.


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


14(a)(ii) j + k + l = −

12

36

j + j =1

3

j =1

6


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)



8

14(b)(ii)


14(b)(iii)

2marks:Correctanswer.1mark:Showsoneasymptoteonly.

14(b)(iv)


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


9

Thetimeforonecompleterevolution(ortheperiodofmotion)

† =2+

ü

=2+

Kt5 = 2+U

5

t

14(c)(ii)

Inthedivisionofthetwoequationsofmotionthemassiscancelledoutandhencethereisnoeffectonthemotionifthemassisdoubled.


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)


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.



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.


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 + 2Øªº − 2Øªºø42Øªºø4 = (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


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



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 + 2Øªº − 2Øªºø42Øªºø4 = (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


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


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



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.

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