mathematics advanced 2u - ace paper 1...year 12 mathematics advanced 3 3. a factory produces bags of...

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A C E

EXAM PAPER

Student name: ______________________

PAPER 1 YEAR 12

YEARLY EXAMINATION

Mathematics Advanced

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 2 hours and 45 minutes for this section

Year 12 Mathematics Advanced

2

SectionI10marksAttemptquestions1-10Allowabout15minutesforthissectionUsethemultiple-choiceanswersheetforquestions1-101. Whatisthesolutiontotheequation2cos%& − 1 = 0inthedomain0 ≤ & ≤ 2π?

(A) & =π6,11π6

(B) & =π4,7π4

(C) & =π4,5π4,7π4,11π4

(D) & =π4,3π4,5π4,7π4

2.

Whichofthefollowingpropertiesmatchestheabovegraph?

(A) 3′(&) > 0and3′′(&) < 0

(B) 3′(&) > 0and3′′(&) > 0

(C) 3′(&) < 0and3′′(&) < 0

(D) 3′(&) > 0and3′′(&) > 0


3

3. Afactoryproducesbagsofcashews.Theweightsofthebagsarenormallydistributed,

withameanof900gandastandarddeviationof50g.Whatisthebestapproximationforthepercentageofbagsthatweighmorethan1000g?

(A) 0% (B) 2.5% (C) 5% (D) 16%

4. Whatisthevalueof= (>?@ + 1)B&C

D?

(A) >?

(B) 13>?

(C)13(>? + 1)

(D)13(>? + 2)

5. WhatisthegradienttothecurveF = (& − G)(&% − 1)atthepointwhenx=–2? (A) −3G − 6 (B) −5G − 1 (C) 4G + 11 (D) 5G + 46.

Whatisthecorrelationbetweenthevariablesinthisscatterplot? (A) Weaknegative (B) WeakPositive (C) Moderatenegative (D) Moderatepositive


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7. AsectionofthegraphF = 3(&)isshownbelow.

Whichofthefollowingisthecorrectfunctionfortheabovegraph?

(A) 3(&) = tanI12J& −

π4KL

(B) 3(&) = tanI2J& −π4KL

(C) 3(&) = tanI12J& −

π2KL

(D) 3(&) = tanI2J& −π2KL

8. Thegraphofthederivativefunctionisshownbelow.

WhereisthefunctionF = 3(&)increasing? (A) {& ∶ & > 0} (B) {& ∶ & > 2} (C) {& ∶ −3 < & < 2} (D) {& ∶ & < −3}or{& ∶ & > 2}


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9. Thetablebelowshowsthepresentvalueofa$1annuity.

Presentvalueof$1

Endofyear 3% 4% 5% 6%

5 4.5797 4.4518 4.3295 4.2124

6 5.4172 5.2421 5.0757 4.9173

7 6.2303 6.0021 5.7864 5.5824

8 7.0197 6.7327 6.4632 6.2098

Whatisthepresentvalueofanannuitywhere$12,000iscontributedeachyearforsixyearsintoanaccountearning3%perannumcompoundinterest?

(A) $15183.83 (B) $54956.40 (C) $65006.40 (D) $72000.0010. Whichofthefollowinggraphscouldnotrepresentaprobabilitydensityfunctionf(x)? (A)

(B)

(C)

(D)


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SectionII90marksAttemptquestions11-16Allowabout2hoursand45minutesforthissectionAnswereachquestioninthespacesprovided.Yourresponsesshouldincluderelevantmathematicalreasoningand/orcalculations.Question11(2marks) Marks

Differentiatethefollowingfunctionswithrespecttox. (a) 3(&) = sin& + &% 1 (b) 3(&) = ln(&% + 1)? 1 Question12(3marks)

Forthearithmeticsequence4,9,14,19,…. (a) Writetheruletodescribethenthterm. 1 (b) Whatisthe25thterm? 1 (c) Findthesumofthefirst100terms. 1


7

Question13(4marks) Marks

AcontinuousrandomvariableXhasafunctionfgivenby

3(&) = R|3 − &| 2 ≤ & ≤ 4

0 otherwise

(a) FindX(2 ≤ Y ≤ 3.5) 2 (b) FindX(2 ≤ Y ≤ 2.5) 2 Question14(4marks)

Differentiate (a) 2>@cos& 2

(b)tan&& 2

Question15(1mark)

Find=(2& + 3)CD B& 1


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Tran’sindustrialunitproducesaluminiumrods.Inthepastweektheindustrialunithasproducedaluminiumrodswithameanweightof12.5kilogramsandastandarddeviationof0.5kilograms.

(a) Qualitycontrolrequiresanyaluminiumrodwithaz-scorelessthan–1toberejected.Whatistheminimumweightthatwillbeaccepted?

1

(b) Aluminiumrodswithaz-scoregreaterthan2arealsorejected.Whatisthemaximumweightthatwillbeaccepted?

1

Question17(2marks)

WhatistheareaenclosedbetweenthecurvesF = &% + 1andF = 3& + 1? 2


9


FindthevalueofkifF = >\@sin&andBFB&

− 3F = >\@cos&. 3

Question19(3marks)

Thediagrambelowshowsanativegarden.Allmeasurementsareinmetres.

(a) UsetheTrapezoidalRulewith4intervalstofindanapproximatevaluefortheareaofthenativegarden.

2

(b) If25millimetresofrainfellovernight,howmanylitresofrainfellonthe

nativegarden?Assume1m? = 1000L.1


10


ConsiderthefunctionsF = &%andF = &% − 3& + 2. (a) Sketchthetwofunctionsonthesameaxes. 2 (b) Henceorotherwisefindthevaluesofxsuchthat&% > (& − 1)(& − 2). 1 Question21(2marks)

Statetheamplitudeandperiodofthefunction3(&) = 4 + 3cos Jπ&2K 2


11


Thenormaldistributionshowstheresultsofamathematicsassessmenttask.Ithasameanof60andastandarddeviationof10.

(a) Whatisthemathematicsassessmentresultwithaz-scoreof–2? 1

(b) Whatisthez-scoreofamathematicsassessmentresultof65? 1 Question23(2marks)

Find= (sec%2&)B&_`

D 2

Question24(2marks)

Howmanysolutionsdoestheequation|cos(2&)| = 1havefor0 ≤ & ≤ 2π? 2


12


Afunction3(&)isdefinedby3(&) = &%(3 − &). (a) FindthestationarypointsforthecurveF = 3(&)anddeterminetheirnature. 2 (b) SketchthegraphofF = 3(&)showingthestationarypointsandx-intercepts. 2 (c) FindtheequationofthetangenttothecurveatthepointX(1,2). 1 Question26(2marks)

ConstructarecurrencerelationintheformabcC = ab × (1 + e) − ftomodelthebalanceofaloanof$58000borrowedat6%perannum,compoundingmonthly,withpaymentsof$810permonth.

2


13


Tenkilogramsofchlorineisplacedinwaterandbeginstodissolve.AfterthourstheamountAkgofundissolvedchlorineisgivenbyg = 10>h\i

(a) CalculatethevalueofkgiventhatA=3.6whent=5.Answercorrecttothreedecimalplaces.

2

(b) Afterhowmanyhoursdoesonekilogramofchlorineremainundissolved?

Answercorrecttoonedecimalplace.2

Question28(2marks)

Thethirdandseventhtermsofageometricseriesare1.25and20respectively.Whatisthefirstterm?

2


14


Thetablebelowshowsforearmlengthandhandlength.

Forearm(incm) 25.0 25.6 26.0 26.6 27.0 27.4 28.0 28.6 29.0 29.2Hand(incm) 17.2 17.6 18.2 18.4 19.0 19.0 19.8 19.8 20.4 20.6

(a) Drawascatterplotusingtheabovetable. 1

(b) Drawalineofbestfitonthescatterplot. 1

(c) Charlottehasaforearmwhoselengthis27.8cm.Whatisherexpectedhandlength?

1

(d) CalculatethevalueofthePearson’scorrelationcoefficient.Answercorrecttofourdecimalplaces.

2


15


Florenceleft$1000inherwillforWorldVision.Herinstructionswerethatthismoneybeinvestedat5%interest,compoundedannually.

(a) HowmuchmoneywouldbegiventoWorldVisionafter100years?Giveyouranswertothenearestdollar.

1

(b) Florencehasrequestedherfamilyinvestafurther$1000atthebeginningof

eachsubsequentyearatthesameinterestrate.HowmuchmoneywouldbegiventoWorldVisionafter100yearsifherfamilyfollowedFlorence’sinstructions?Giveyouranswertothenearestdollar.

2

Question31(3marks)

Evaluatethefollowingdefiniteintegrals.

(a) = &%%

hC+ 1B& 1

(b) = √3& + 4B&k

hC 2


16


Thetablebelowshowsthefuturevalueofa$1annuity.

Futurevalueof$1

Endofyear 4% 6% 8% 10%

1 1.00 1.00 1.00 1.00

2 2.04 2.06 2.08 2.10

3 3.12 3.18 3.25 3.31

4 4.25 4.37 4.51 4.64

(a) Whatwouldbethefuturevalueofa$32000peryearannuityat8%perannumfor4years,withinterestcompoundingannually?

1

(b) Anannuityof$6300isinvestedeverysixmonthsat8%perannum,compoundedbiannuallyfor2years.Whatisthefuturevalueoftheannuity?

1

Question33(4marks)

Considerthefunction3(&) =1

1 + &%

(a) Findthevalueof3′(&). 2 (b) FindthecoordinatesofthepointonthecurveF = 3(&)atwhichthetangent

isparalleltothex-axis.2


17


Anobjectismovinginastraightlineanditsvelocityisgivenby;

o = 1 − 2sin2pforp ≥ 0

wherevismeasuredinmetrespersecondandtinseconds.Initiallytheobjectisattheorigin.

(a) Findthedisplacementx,asafunctionoft. 2 (b) Whatisthepositionoftheobjectwhenp = _

?? 1

(c) Findtheaccelerationa,asafunctionoft. 1 (d) Sketchthegraphofa,asafunctionoft,for0 ≤ p ≤ π. 2 (e) Whatisthemaximumaccelerationoftheobject? 1


18


Sketchthefollowgraphsonthesamenumberplane. 3

F = √&, F = √& − 1, F = √& − 1

Question36(2marks)

ClassAhas24studentsandachievedameanonanassessmenttaskof75.5%.ClassBhas28studentsandachievedameanonthesameassessmenttaskof80.5%.Whatwasthemeanmarkforbothclasses.Answercorrecttoonedecimalplaces.

2


19


AV8supercarsracetrackconsistsoftwosemi-circularcurvesandtwostraights.Thedimensionsoftheracetrackareshownbelow.Thetotallengthoftheracetrackis4.8km.

(a) Letxkmrepresentthelengthofthestraightandykmrepresentthediameterofthesmallersemicircle.Showthat:

2

F =9.6 − 4&3π

(b) TheaveragespeedofaV8supercaronthisracetrackisdependentonthe

lengthofthestraight.Itisgivenby:3

s = 200 − I

&?

27+π6FL

Whatisthelengthofthestraightthatmaximizesthespeed?


20


(a) SketchthegraphF = |2& − 4|. 2 (b) Usingthegraphfrompart(a),orotherwise,findallvaluesofmforwhichthe

equation|2& − 4| = t& + 1hasexactlyonesolution.2

Question39(2marks)

ThePearson’scorrelationcoefficientbetweenstudentsassessmentresultandtheirheightwas0.12.Whatisthemeaningofthiscorrelation?

2

Question40(2marks)

Theheightsofagroupoffriendsarenormallydistributedwithameanof167cmandastandarddeviationof12cm.Whatpercentageofthegrouparemorethan179cmtall?

2

Endofpaper


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1

ACEExaminationPaper1Year12MathematicsAdvancedYearlyExaminationWorkedsolutionsandMarkingguidelinesSectionI Solution Criteria1. 2cos%𝑥 − 1 = 0

cos%𝑥 =12orcos𝑥 = ±

1√2

𝑥 =π4,3π4,5π4,7π4

1Mark:D

2. 𝑓′(𝑥) > 0(increasing)𝑓′′(𝑥) < 0(concavedown)

1Mark:A

3. 𝑧 =𝑥 − �̅�𝑠

=1000 − 900

50

= 295%ofscoreshaveaz-scorebetween–2and2\2.5%haveaz-scoregreaterthan2.

1Mark:B

4. D (𝑒FG + 1)𝑑𝑥 = J

13𝑒FG + 𝑥K

L

MM

L

= N13𝑒F + 1O −

13

=13(𝑒F + 2)

1Mark:D

5. 𝑦 = (𝑥 − 𝑎)(𝑥% − 1)= 𝑥F − 𝑎𝑥% − 𝑥 + 𝑎

𝑑𝑦𝑑𝑥

= 3𝑥% − 2𝑎𝑥 − 1Gradientatthepointwhenx=–2𝑚 = 3 × (−2)% − 2𝑎 × (−2) − 1= 4𝑎 + 11

1Mark:C

6. Correlationbetween0.5and0.74.\Moderatepositive.

1Mark:D

7. Period =π𝑛=π2

∴ 𝑛 = 2AlsoatranslationofX

Yinthepositivex-direction.

𝑓(𝑥) = tan[2\𝑥 −π4]^

1Mark:B

8. Increasingfunction𝑓′(𝑥) > 0{𝑥 ∶ −3 < 𝑥 < 2}

1Mark:C

9. Intersectionvalueis5.4172(3%and6years)𝑃𝑉 = 5.4172 × 12000

= $65006.40

1Mark:C

10. Fundamentalpropertyofaprobabilitydensityisthatforanyvalueofx,thevalueoff(x)isnon-negative.\Graph(A)has𝑓(𝑥) < 0

1Mark:A


2

SectionII 11(a) 𝑓(𝑥) = (sin𝑥 + 𝑥%)

𝑓′(𝑥) = cos𝑥 + 2𝑥1Mark:Correctanswer.

11(b) 𝑓(𝑥) = ln(𝑥% + 1)

𝑓′(𝑥) =2𝑥

𝑥% + 1

1Mark:Correctanswer.

12(a) a=4andd=5for4,9,14,19,….𝑇l = 𝑎 + (𝑛 − 1)𝑑= 4 + (𝑛 − 1) × 5= 5𝑛 − 1


12(b) 𝑇%m = 5 × 25 − 1= 124


12(c) 𝑆l =𝑛2[2𝑎 + (𝑛 − 1)𝑑]

=1002[2 × 4 + (100 − 1) × 5]

= 25150


13(a)

𝑃(2 ≤ 𝑋 ≤ 3.5) =12× 1 × 1 +

12× 0.5 × 0.5

= 0.625

2Marks:Correctanswer.1Mark:Showssomeunderstanding.

13(b) 𝑃(2 ≤ 𝑋 ≤ 2.5) =12× 0.5 × (1 + 0.5)

= 0.375

2Marks:Correctanswer.1Mark:Showsunderstanding.

14(a) 𝑑𝑑𝑥(2𝑒Gcos𝑥) = 2𝑒G(−sin𝑥) + cos𝑥2𝑒G

= 2𝑒G(cos𝑥 − sin𝑥)

2Marks:Correctanswer.1Mark:Appliestheproductrule.

14(b) 𝑑𝑑𝑥 N

tan𝑥𝑥 O =

𝑥 × sec%𝑥 − tan𝑥 × 1𝑥%

=𝑥sec%𝑥 − tan𝑥

𝑥%

2Marks:Correctanswer.1Mark:Appliesthequotientrule.

15D(2𝑥 + 3)ML 𝑑𝑥 =

(2𝑥 + 3)MM

11 × 2+ 𝐶

=(2𝑥 + 3)MM

22+ 𝐶



3

16(a)𝑧 =

𝑥 − �̅�𝑠

−1 =𝑥 − 12.50.5

𝑥 = (−1 × 0.5) + 12.5= 12

\Minimumweighttobeacceptedis12kg.


16(b)𝑧 =

𝑥 − �̅�𝑠

2 =𝑥 − 12.50.5

𝑥 = (2 × 0.5) + 12.5= 13.5

\Maximumweighttobeacceptedis13.5kg.


17 Solvingthetwoequationssimultaneously.𝑥% + 1 = 3𝑥 + 1𝑥% − 3𝑥 = 0𝑥(𝑥 − 3) = 0\Pointofintersectionoccurswhenx=0andx=3.

𝐴 = D (3𝑥 + 1) − (𝑥% + 1)𝑑𝑥F

L

= D (3𝑥 − 𝑥%)𝑑𝑥F

L= u

3𝑥%

2−𝑥F

3vL

F

= u[3 × 3%

2−3F

3^ − [

3 × 0%

2−0F

2^v

=92squareunits

2Marks:Correctanswer.1Mark:Findsthepointsofintersectionorshowssomeunderstandingoftheproblem.

18 𝑦 = 𝑒xGsin𝑥𝑑𝑦𝑑𝑥

= 𝑒xG × cos𝑥 + sin𝑥 × 𝑘𝑒xG

= 𝑒xG(cos𝑥 + 𝑘sin𝑥)𝑑𝑦𝑑𝑥

− 3𝑦 = 𝑒xGcos𝑥

𝑒xG(cos𝑥 + 𝑘sin𝑥) − 3𝑒xGsin𝑥 = 𝑒xGcos𝑥𝑘𝑒xGsin𝑥 − 3𝑒xGsin𝑥 = 0

𝑒xGsin𝑥(𝑘 − 3) = 0𝑘 = 3

3Marks:Correctanswer.2Marks:Makessignificantprogresstowardsthesolution.1Mark:Findsthederivative.

19(a) 𝐴 =ℎ2[𝑦L + 𝑦Y + 2(𝑦M + 𝑦% + 𝑦F)]

=1.52[2 + 0 + 2(4.5 + 5.1 + 3.6)]

= 21.3m%\Areaofthenativegardenisapproximately21.3m2.

2Marks:Correctanswer.1Mark:Usestrapezoidalrule.

19(b) Now25mm = 0.025m𝑉 = 𝐴ℎ= 21.3 × 0.025= 0.5325mF = 532.5L

\532.5Lofwaterfellinthenativegarden.



4

20(a) 𝑦 = 𝑥% − 3𝑥 + 2 = (𝑥 − 1)(𝑥 − 2)

2Marks:Correctanswer.1Mark:Onegraphdrawncorrectly.

20(b) Solvesimultaneouslytofindthepointofintersection𝑥2 = 𝑥2 − 3𝑥 + 23𝑥 = 2𝑥 =

23

Therefore𝑥% > 𝑥% − 3𝑥 + 2when𝑥 >23


21 Amplitude=3

Period =2ππ2= 4

2Marks:Correctanswer.1Mark:Findseitheramplitudeortheperiod.

22(a) Studentswithaz-scoreof–2istwostandarddeviationsbelowthemean(60 − (2 × 10) = 40.\Ascoreof40hasaz-scoreof–2.


22(b) z-scorefor65

𝑧 =𝑥 − �̅�𝑠

=65 − 6010

= 0.5

\z-scoreis0.5


23D (sec%2𝑥)𝑑𝑥 = J

12tan2𝑥K

L

X�

X�

L

=12\tan

π4− tan0]

=12

2Marks:Correctanswer.1Mark:Findstheprimitivefunctionorshowssomeunderstanding.

24 Drawthegraphs:𝑦 = |cos(2𝑥)|and𝑦 = 1

\Thereare5solutions.



5

25(a) 𝑓(𝑥) = 𝑥%(3 − 𝑥) = 3𝑥% − 𝑥FStationarypoints𝑓′(𝑥) = 0𝑓′(𝑥) = 6𝑥 − 3𝑥%3𝑥(2 − 𝑥) = 0𝑥 = 0, 𝑥 = 2\Stationarypointsare(0,0)and(2,4)𝑓′′(𝑥) = 6 − 6𝑥At(0, 0), 𝑓′′(0) = 6 > 0MinimaAt(2, 4), 𝑓��(%) = −6 < 0Maxima

2Marks:Correctanswer.1Mark:Findsoneofthestationarypointsorrecognises6𝑥 − 3𝑥% = 0.

25(b) x-intercepts(y=0)𝑥2(3 − 𝑥) = 0𝑥 = 0, 𝑥 = 3

2Marks:Correctanswer.1Mark:Makessomeprogresstowardssketchingthecurve.

25(c) 𝑓′(𝑥) = 6𝑥 − 3𝑥%Gradientofthetangentatthepoint𝑃(1,2)𝑚 = 6 × 1 − 3 × 1% = 3𝑦 − 𝑦M = 𝑚(𝑥 − 𝑥M)𝑦 − 2 = 3(𝑥 − 1)

𝑦 = 3𝑥 − 1or3𝑥 − 𝑦 − 1 = 0


26 𝑟 =0.0612

= 0.005

𝐷 = 810and𝑉L = 58000

Recurrencerelation𝑉l�M = 𝑉l × (1 + 𝑟) − 𝐷

= 𝑉l × 1.005 − 810

2Marks:Correctanswer.1Mark:Substitutesonecorrectvalueintotherecurrencerelation.

27(a) 𝐴 = 10𝑒�x�3.6 = 10𝑒�x×m

𝑒�mx = 0.36−5𝑘ln𝑒 = ln0.36

𝑘 =ln0.36−5

= 0.2043. . . .≈ 0.204

2Marks:Correctanswer.1Mark:Makessomeprogresstowardsthesolution


6

27(b) 𝐴 = 10𝑒�x�1 = 10𝑒�L·%LY...×�

𝑒�L·%LY...×� = 0.1−0.204 × 𝑡 × ln𝑒 = ln0.1

𝑡 =ln0.1

−0.204. . .

= 11.2689. . .≈ 11.3hours

\Onekilogramofchlorinedissolvesafter11.3hours.

2Marks:Correctanswer.1Mark:Makessomeprogresstowardsthesolution

28 𝑇l = 𝑎𝑟l�M𝑇F = 𝑎𝑟% = 1.25①𝑇� = 𝑎𝑟� = 20②Dividingthetwoequations𝑎𝑟�

𝑎𝑟%=

201.25

𝑟Y = 16𝑟 = ±2

𝑇� = 𝑎 × (±2)� = 20

𝑎 =2064

=516

\Firsttermis mM�

2Marks:Correctanswer.1Mark:FindstwoequationsusingthenthtermofaGPorshowssomeunderstanding.

29(a)


29(b) Seelineofbestfitontheabovescatterplot. 1Mark:Correctanswer.

29(c) Whenforearmlength=27.8thenhandlength=19.4cm(fromthescatterplot)\Charlotte’shandlengthshouldbe19.4cm.


29(d) UsethecalculatortofindPearson’scorrelationcoefficient.𝑟 = 0.990691…≈ 0.9907

2Marks:Correctanswer.1Mark:Findsavalueofrcloseto0.99.


7

30(a) 𝐹𝑉 = 𝑃𝑉(1 + 𝑟)l

= 1000(1 + 0.05)MLL

= 131501.257. . .

≈ $131501

\Worldvisionwillreceive$131501


30(b) 𝐴MLL = 1000(1.05)MLL + 1000(1.05)�� + ⋯+ 1000(1.05)M

GPwith𝑎 = 1000(1.05),r=1.05andn=100

𝐴MLL =1000(1.05)[1.05MLL − 1]

1.05 − 1

= 2740526.41. . .

≈ $2740526

\Worldvisionwillreceive$2740526after100years.

2Marks:Correctanswer.1Mark:IdentifiesaG.P.with100terms.

31(a)D 𝑥%%

�M+ 1𝑑𝑥 = u

𝑥F

3+ 𝑥v

�M

%

= �[2F

3+ 2^ − �

−1F

3+ (−1)��

= 6


31(b)D √3𝑥 + 4𝑑𝑥 = J

29(3𝑥 + 4)

F%K�M

YY

�M

=29× JN(3 × 4 + 4)

F%O − N(3 × (−1) + 4)

F%OK

= 14

2Marks:Correctanswer.1Mark:Findstheprimitivefunction.

32(a) Intersectionvalueis4.51(8%and4years)𝐹𝑉 = 4.51 × 32000

= $144320


32(b) Intersectionvalueis4.25(4%and4years)𝐹𝑉 = 4.25 × 6300

= $26775


33(a) 𝑓(𝑥) =1

1 + 𝑥%= (1 + 𝑥%)�M

𝑓′(𝑥) = −(1 + 𝑥%)�% × 2𝑥

=−2𝑥

(1 + 𝑥%)%



8

33(b) Thetangenthasthesamegradientasthex-axis(parallel)Thex-axishasagradientof0(horizontalline)

𝑓′(𝑥) =−2𝑥

(1 + 𝑥%)%= 0

−2𝑥 = 0𝑥 = 0

When𝑥 = 0then𝑦 =1

1 + 0%= 1

\Pointis(0,1)

2Marks:Correctanswer.1Mark:Findsthegradientofthetangentormakessomeprogress.

34(a) 𝑥 = D(1 − 2sin2𝑡)𝑑𝑡

= 𝑡 + cos2𝑡 + 𝐶Initiallyt=0andx=00 = 0 + cos(2 × 0) + 𝐶𝐶 = −1∴ 𝑥 = 𝑡 + cos2𝑡 − 1

2Marks:Correctanswer.1Mark:Integratesthevelocityfunction.

34(b) When𝑡 =π3then

𝑥 =π3+ cos \2 ×

π3] − 1

=π3−12− 1 =

π3−32


34(c)𝑎 =

𝑑𝑑𝑡(1 − 2sin2𝑡)

= −4cos2𝑡


34(d) a = −4cos2𝑡for0 ≤ 𝑡 ≤ π.

2Marks:Correctanswer.1Mark:Drawsthegeneralshapeofthecurve.

34(e) −1 ≤ cos2𝑡 ≤ 1−4 ≤ −4cos2𝑡 ≤ 4 (orfromthegraph)\Maximumaccelerationis4ms-2



9

35

3Marks:Correctanswer.2Marks:Drawstwoofthegraphscorrectly1Mark:Showssomeunderstanding.

36 ClassAtotalnumberofmarks75.5 × 24 = 1812.ClassBtotalnumberofmarks80.5 × 28 = 2254

Mean =1812 + 225424 + 28

= 78.1923…%

≈ 78.2%

\Meanmarkforbothclassesis78.2%

2Marks:Correctanswer.1Mark:Makesso

37(a) 𝑃 = 2𝑥 +12× π × 𝑦 +

12× π × 2𝑦

4.8 = 2𝑥 +12π × 3𝑦

9.6 = 4𝑥 + 3π𝑦

𝑦 =9.6 − 4𝑥3π

2Marks:Correctanswer.1Mark:Findsanexpressionfortheperimeter.

37(b) Expressthespeedintermsofx

𝑆 = 200 − [𝑥F

27+π6𝑦^

= 200 − [𝑥F

27+π6×9.6 − 4𝑥3π

^

= 200 −𝑥F

27−9.6 − 4𝑥18

𝑑𝑆𝑑𝑥

= −3𝑥%

27+418

Maximumlengthofthestraightoccurswhen𝑑𝑆𝑑𝑥

= 0

−3𝑥%

27+418

= 0

3𝑥% = 6𝑥 = √2kmCheck

When𝑥 = √2kmthen𝑑%𝑆𝑑𝑥%

= −6𝑥27

= −6 × √227

< 0(Maxima)

3Marks:Correctanswer.2Marks:Findsthelengthofthestraightformaximumspeed.1Mark:DifferentiatestheSformulawithrespecttox.


10

38(a)

2Marks:Correctanswer.1Mark:Drawsthegeneralshapeorshowssomeunderstanding.

38(b)

Fromthegraph

𝑚 < −2or𝑚 ≥ 2or𝑚 = −12

2Marks:Correctanswer.1Mark:Findsoneofthesolutions.

39 Assessmentresultsincreaseasheightincreases.Lowpositivecorrelation.Notastrongrelationship.

2Marks:Correctanswer.1Mark:Showsunderstanding

40 𝑧 =𝑥 − �̅�𝑠

=179 − 167

12

= 168%ofscoreshaveaz-scorebetween–1and1.\32%÷2=16%haveaz-scoregreaterthan1.

2Marks:Correctanswer.1Mark:Findsthez-score.

mathematics advanced 2u - ace paper 1...year 12 mathematics advanced 3 3. a factory produces bags of...

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