mathematics advanced 2u - ace paper 2...year 12 mathematics advanced 2 section i 10 marks attempt...

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A C E

EXAM PAPER

Student name: ______________________

PAPER 2 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

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SectionI10marksAttemptquestions1-10Allowabout15minutesforthissectionUsethemultiple-choiceanswersheetforquestions1-10

1. Whatisthevalueof! (6$% − 4))$?+

,

(A) –2

(B) –1

(C) 0

(D) 1

2. Thediagramshowsthegraphof- = /0($ − 1).

Howmanysolutionsaretheretotheequation/0($ − 1) = $% − 1? (A) 0

(B) 1

(C) 2

(D) 3

3. Aninfinitegeometricserieshasafirsttermof3andalimitingsumof1.8.Whatisthecommonratio?

(A) −0. 3̇ (B) −0. 6̇ (C) −1.5 (D) −3.75


3

4. Whatisthederivativeof/08? (A) 3$%/08 (B) 3$/08 (C) 3$%/90: (D) $9/08;+

5. Asampleof14peoplewereaskedtoindicatethetime(inhours)theyhadspentwatchingtelevisiononthepreviousnight.Theresultsaredisplayedinthedotplotbelow.

• • • • • • • • • • • • • •0 1 2 3 4 5

Whatisthemeanandsamplestandarddeviationofthesetimes?Giveyouranswerscorrecttoonedecimalplace.

(A) $< = 2.0ands=1.5 (B) $< = 2.1ands=1.5 (C) $< = 2.1ands=1.6 (D) $< = 2.6ands=1.2

6. Thenormaldistributionshowstheresultsofamathematicsassessmenttask.Ithasameanof70andastandarddeviationof10

Whatpercentageofresultslieintheshadedregion?

(A) 16%

(B) 32%

(C) 34%

(D) 68%

7. Theaccelerationofaparticlemovinginastraightlineisgivenbytheformula> = 12? + 6.Initiallytheparticleisatx=5metresandtheinitialvelocityoftheparticleis–36m/s.Whenistheparticleatrest?

(A) t=0 (B) t=1 (C) t=2 (D) t=3


4

8. Theequationofleast-squareslineofbestfitisgivenbyy=mx+cwhere A = B CDCEandF = -< −A$̅ Whatisthegradientoftheleast-squareslineofbestfitgivenr=0.561,H0 = 1.987andHK = 4.579?

(A) 0.24

(B) 1.29

(C) 7.13

(D) 16.21

9.

Apossibleequationforthegraphshownaboveis:

(A) - = cos O$ + π6Q

(B) - = sin O$ − π6Q

(C) - = sin O$ + π6Q

(D) - = −sin O$ − π6Q

10. Anareaisboundedbythecurve - = 23T9 − $% thecoordinateaxesandlinex=2.

Whatisanapproximationforthisareausingthetrapezoidalruleandthreefunctionvalues?

(A) 1.82

(B) 2.69

(C) 3.26

(D) 3.63


5

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

Simplify-

-% − 4 −2

- − 22

Question12(2marks)

(a) Express sinUcosU + cos9U

sinU asasingletrigonometricratio. 1

(b) Hencesolve sinUcosU + cos9U

sinU = 1 for0 ≤ U ≤ 2π. 1


6

Question13(3marks) Marks

Differentiate (a) tan5$ 1

(b)ln$$ 1

(c) $cos$ 1 Question14(3marks)

Thesecondtermofanarithmeticseriesis39andthesixthtermis19.Whatisthesumofthefirsttenterms?

3

Question15(2marks)

Findtheanti-derivativeof4 − $;9. 2


7


(a) Sketchthegraphsof- = 4 − $%and- = 3onthesamenumberplane. 2 (b) Thegraphof- = 3cutstheparabolaatPandQ.

WhatarethecoordinatesofPandQ?1

(c) Calculatetheareaboundedbythegraphsof- = 4 − $%and- = 3. 2 Question17(2marks)

Riley’sclassachieveda72%meanand8%standarddeviationfortheirprojectwork.WhatwasRiley’smarkifheachievedaz-scoreof–2.5?

2


8


Thetablebelowshowsthepresentvalueinterestfactorsforsomemonthlyinterestratesandloanperiodsinmonths.

Presentvalueof$1

Period 0.0060 0.0065 0.0070 0.0075

46 40.09350 39.64965 39.21263 38.78231

47 40.84841 40.38714 39.93310 39.48617

48 41.59882 41.11986 40.64856 40.18478

49 42.34475 41.84785 41.35905 40.87820

Jessicaborrows$16000foracar.Shearrangestorepaytheloanwithmonthlyrepaymentsover4years.Sheischarged8.4%perannuminterest.

(a) FindJessica’smonthlyrepayment.Answertothenearestcent. 2

(b) CalculatetheamountofinterestJessicawillpayoverthetermoftheloan.Answertothenearestdollar.

2

Question19(2marks)

Evaluate ! ($% + sin2$))$.Z[

, Writeyouranswercorrecttothreedecimalplaces. 2


9


Sketchthegraphofe($) = 3 − 2$ + 1. 3Labelallaxisintercepts.Labeleachasymptotewithitsequation. Question21(2marks)

Thevariablesprofitmadeandamountspentonadvertisingarestronglycorrelatedwithacorrelationcoefficientr=0.9.Whatconclusionscanyoudrawfromthisinformation?

2


10


Themeanforaclasstestis64%andthestandarddeviationis12.5. (a) InaclasstestMarcushasaz-scoreof2.Whatdoesthatmean? 1

(b) Fletcherhasamarkof51.5%.Whatishisz-score? 1

(c) Aylasaidshehadaz-scoreof3butHannahisunconvinced.Why? 1 Question23(3marks)

Thedisplacementofanobjectattime(t)secondsisgivenby: 3

$ = 3/;%f + 10/;f + 4?

Findthetimetheobjectcomestorest.


11


Laraborrows$50,000topurchasefurnitureforhersmallbusiness.Theinterestiscalculatedmonthlyatarateof2%permonth.Sheintendstorepaytheloanwithinterestintwoannualinstalmentsof$Mattheendofthefirstandsecondyears.

(a) WriteanexpressioninvolvingMforthetotalamountowedbyLaraafter12months,justafterthefirstinstalmentof$Mhasbeenpaid.

1

(b) Showthath = $50000 × 1.02%k

1.02+% + 1 2

(c) Whatwillbethetotalamountofinterestpaidonthisloan? 2 Question25(2marks)

TheprobabilitydensityfunctionforthecontinuousrandomvariableXis: 2

e($) = l$9 0 ≤ $ ≤ 2

0 otherwise

WhatisthevalueofE(X)?


12


(a) Onthesamesetofaxes,sketchthegraphsof- = sin$and- = 1 − cos$overthedomain0 ≤ U ≤ π.

2

(b) Findthevaluesofxforwhichsin$ = 1 − cos$inthedomain0 ≤ U ≤ π. 1 (c) Findtheareabetween- = sin$and- = 1 − cos$overthedomain0 ≤ U ≤ π. 2 Question27(2marks)

Whatistheequationofthenormaltothecurve- = $% − 4$atthepoint(1,–3)? 2


13


Afunctione($)isdefinedbye($) = 7 + 4$9 − 3$k. (a) Findthecoordinatesofthestationarypointsforthecurve- = e($). 2 (b) Findallvaluesofxforwhiche′′($) = 0. 1 (c) Determinethenatureofthestationarypoints. 2 (d) Sketchthegraphof- = e($)forthedomain−1 ≤ $ ≤ 2. 2


14


Thescatterplotbelowshowstherelationshipbetweenageandfitnesslevel.

(a) Drawalineofbestfitonthescatterplot.Findthegradientofthisline. 2

(b) Lachlanis30yearsold.Whatishisexpectedfitnesslevel? 1

(c) CalculatethevalueofthePearson’scorrelationcoefficient.Answercorrecttotwodecimalplaces.

2

Question30(2marks)

Statethedomainandrangeofe($) = √1 − $%. 2


15


TheprobabilitydensityfunctionforthecontinuousrandomvariableXis: 2

e($) = o19 (4$ − $

%) 0 < $ < 3

0 otherwise

Findvalueofq(r ≤ 2). Question32(3marks)

Determinetheequationofacurvegivenby)%-)$% = 12$ + 6 and(1,–2)asa

3

stationarypointonthecurve. Question33(2marks)

Simplifylim0→,sin6$$ 2


16


Acableoflength3metresistobebenttoformthehypotenuseandbaseofaright-angledtriangleABC.LetthelengthofthebaseBCisxmetres.

(a) WhatisthelengthofthehypotenuseACintermsofx? 1 (b) ShowthattheareaofthetriangleABCis0.5$√9 − 6$. 2 (c) Whatvalueofxgivesthemaximumpossibleareaofthetriangle? 3 (d) Findthemaximumpossibleareaofthetriangle. 1


17


Howmanysolutionsexistforxoftheequation/0 + $ + 2 = 0?Hint:Drawgraphs.

3

Question36(3marks)

Thenumberofstudentsabsentfromyear12forthepastninedayswasasfollows:15,18,14,17,18,13,12,29,20

(a) Whatisthemean?Answercorrecttoonedecimalplace. 1 (b) Findtheinterquartilerange? 1 (c) Is29anoutlierforthissetofdata?Justifyyouranswerwithcalculations. 1


18


Considerthefunctione($) = (2$ − 3)k. (a) Findthevalueofe′(1). 2 (b) Findequationofthetangentatthepoint(1,1)tothecurve- = (2$ − 3)k. 1 (b) Theequationofthetangentcutsthex-axisatAandthey-axisatB.

Findtheareaof∆wxy,whereOistheorigin.3

Endofpaper


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ACEExaminationPaper2Year12MathematicsAdvancedYearlyExaminationWorkedsolutionsandmarkingguidelinesSectionI Solution Criteria1. ! (6𝑥% − 4)𝑑𝑥

*

+= 2! (3𝑥% − 2)𝑑𝑥

*

+= 2[𝑥0 − 2𝑥]+*

= 2[(10 − 2 × 1) − (00 − 2 × 0)] = −2

1Mark:A

2.

Therearetwopointsofintersectionofthetwographs.

1Mark:C

3. a=3andS=1.8

𝑆 =𝑎

1 − 𝑟

1.8 =3

1 − 𝑟

1.8 − 1.8𝑟 = 31.8𝑟 = −1.2𝑟 = −0. 6̇

1Mark:B

4. 𝑑𝑑𝑥𝑒=> = 3𝑥%𝑒=> 1Mark:A

5. Results:{0,0,0,1,1,2,2,2,2,3,3,4,4,5}Calculator:𝑥? = 2.1ands=1.6

1Mark:C

6. Regionisoutsideonestandarddeviation100%–68%=32%

1Mark:B

7. 𝑎 = 12𝑡 + 6𝑣 = 6𝑡% + 6𝑡 + 𝐶Whent=0thenv=–36−36 = 6 × 0% + 6 × 0 + 𝐶or𝐶 = −36𝑣 = 6𝑡% + 6𝑡 − 36 = 6(𝑡 + 3)(𝑡 − 2)

\Particleatrest(v=0)whent=2

1Mark:C

8. 𝑚 = 𝑟𝑠H𝑠== 0.561 ×

4.5791.987

= 1.29 1Mark:B

9. TestequationswithpointsLMN, 0PandL%M

0, 1Ponthecurve.

(B)𝑦 = sin Lπ6−π6P = 0and𝑦 = sin Y

2π3−π6Z

= 1Correct

1Mark:B

10. 𝑦 =23_9 − 𝑥% =

ℎ2[𝑦+ + 𝑦% + 2 × 𝑦*]

=12a2 +

2√83

+ 2 ×2√53c = 3.6309. . . ≈ 3.63

1Mark:D


2

SectionII 11 𝑦

𝑦% − 4−

2𝑦 − 2

=𝑦

(𝑦 + 2)(𝑦 − 2)−

2𝑦 − 2

=𝑦 − 2(𝑦 + 2)(𝑦 + 2)(𝑦 − 2)

=−𝑦 − 4𝑦% − 4

2Marks:Correctanswer.1Mark:Findsacommondenominatororshowssomeunderstanding.

12(a)sin𝜃cos𝜃 +

cos0𝜃sin𝜃

=cos𝜃sin𝜃

(sin%𝜃 + cos%𝜃)

= cot𝜃

1Mark:Correctanswer.

12(b) cot𝜃 = 1

𝜃 =π4or

5π4


13(a) 𝑑𝑑𝑥(tan5𝑥) = sec25𝑥 ×

𝑑𝑑𝑥(5𝑥)

= 5sec%5𝑥


13(b)𝑑𝑑𝑥 Y

ln𝑥𝑥 Z

=𝑥 × 1𝑥 − ln𝑥 × 1

𝑥%

=1 − ln𝑥𝑥%


13(c) 𝑑𝑑𝑥(𝑥cos𝑥) = −𝑥sin𝑥 + cos𝑥


14 𝑇h = 𝑎 + (𝑛 − 1)𝑑𝑇% = 𝑎 + 𝑑 = 39①𝑇N = 𝑎 + 5𝑑 = 19②Equation②−①4𝑑 = −20𝑑 = −5Substitute𝑑 = −5intoequation①𝑎 − 5 = 39

𝑎 = 44 𝑆h =

𝑛2[2𝑎 + (𝑛 − 1)𝑑]

=102[2 × 44 + (10 − 1) × (−5)]

= 215

3Marks:Correctanswer.2Marks:Findsthefirsttermandthecommondifference.1Mark:FindstwoequationsusingthenthtermofaAPorshowssomeunderstanding.

15 !4 − 𝑥l0 𝑑𝑥 = 4𝑥 +12𝑥l% + 𝐶

= 4𝑥 +12𝑥%

+ 𝐶

2Marks:Correctanswer.1Mark:Findsoneoftheterms.


3

16(a)

2Marks:Correctanswer.1Mark:Drawsoneofthegraphscorrectly.

16(b) 𝑦 = 4 − 𝑥%①𝑦 = 3②Substitute3foryintoequation①3 = 4 − 𝑥%𝑥% = 1𝑥 = ±1\CoordinatesareP(–1,3)andQ(1,3)


16(c)𝐴 = 2! [(4 − 𝑥%) − 3]𝑑𝑥

*

+

= 2! −𝑥% + 1*

+𝑑𝑥

= 2 a−𝑥0

3+ 𝑥c

+

*

= 2 ao−10

3p + 1c

=43squareunits

2Marks:Correctanswer.1Mark:Correctlysetsuptheintegral.

17𝑧 =

𝑥 − �̅�𝑠

−2.5 =𝑥 − 728

𝑥 = (−2.5 × 8) + 72= 52%

\Riley’smarkwas52%.

2Marks:Correctanswer.1Mark:Usesthez-scoreformulawithatleastonecorrectvale.

18(a) 𝑟 = +.+vw*%

= 0.0070,𝑛 = 4 × 12 = 48Intersectionvalueis40.64856Letthemonthlyrepaymentbex.

𝑃𝑉 = 40.64856 × 𝑥16000 = 40.64856 × 𝑥

𝑥 =1600040.64856

= 393.6178… ≈ $393.62\Jessica’smonthlyrepaymentis$393.62.

2Marks:Correctanswer.1Mark:Findstheintersectionvalueorshowssomeunderstanding.


4

18(b) Totalrepaid = 393.6178. . .× 48= 18893.6582. . . ≈ $18894

Interest = 18894 − 16000= $2894

\Jessica’sinterestontheloanis$2894.

2Marks:Correctanswer.1Mark:Findsthetotalamounttoberepaid.

19! (𝑥% + sin2𝑥)𝑑𝑥 = a

𝑥0

3−12cos2𝑥c

+

MN

MN

+

= Lπ6P

0

3−12cos L2 ×

π6P − a

00

3−12cos2 × 0c

= aπ0

648−14c +

12

= 0.2978. . .≈ 0.298

2Marks:Correctanswer.1Mark:Findstheprimitivefunctionorshowssomeunderstanding.

20 Interceptis(0,1).Asymptotesarex=–1andy=3.

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

21 Strongpositivecorrelationindicatesthatwhenonevariableincreasestheothervariableincreases.\Increasedspendingofadvertisingareassociatedwithincreasedprofits.

2Marks:Correctanswer.1Mark:Showssomeunderstanding.

22(a) Az-scoreof2istwostandarddeviationsabovethemean.Thatis,Marcusscored89%intheclasstest.


22(b)𝑧 =

𝑥 − �̅�𝑠

=51.5 − 6412.5

= −1\Fletcher’sz-scoreis–1



5

22(c)𝑧 =

𝑥 − �̅�𝑠

3 =𝑥 − 6412.5

𝑥 = (3 × 12.5) + 64= 101.5%

\Aylaneedstoscore101.5%inthetest(impossible).


23 Theobjectcomestorestwhen�̇� = 0𝑥 = 3𝑒l% + 10𝑒l + 4𝑡�̇� = −6𝑒l% − 10𝑒l + 4= −2(3𝑒l% + 5𝑒l − 2)

Let𝑚 = 𝑒l−2(3𝑚% + 5𝑚 − 2) = 0−2(3𝑚 − 1)(𝑚 + 2) = 0Hence3𝑚 − 1 = 0or𝑚 + 2 = 0(Nosolution:𝑒l ≠ −2)

𝑚 =13

𝑒l =13

𝑡 = −ln Y13Z

= ln3 \Objectcomestorestafterln3seconds.

3Marks:Correctanswer.2Marks:Findsandfactorisesthequadraticequation.1Mark:Correctlydifferentiatesx.

24(a) P=$50000,r=0.02permonth,n=12months𝐹𝑉 = 𝑃𝑉(1 + 𝑟)h𝐴*% = 50000 × (1 + 0.02)*% − 𝑀

= 50000 × 1.02*% − 𝑀


24(b) Amountowedattheendofthesecondyear.𝐴%w = (50000 × 1.02*% − 𝑀) × 1.02*% − 𝑀

= 50000 × 1.02%w − 𝑀 × 1.02*% − 𝑀= 50000 × 1.02%w − 𝑀(1.02*% + 1)

Now𝐴%w = 0whentheloanispaidoff.0 = 50000 × 1.02%w − 𝑀(1.02*% + 1)

𝑀(1.02*% + 1) = 50000 × 1.02%w

𝑀 =50000 × 1.02%w

1.02*% + 1

2Marks:Correctanswer.1Mark:Findsthecorrectexpressionfor𝐴%worshowssomeunderstandingoftheproblem.

24(c)𝑀 =

50000 × 1.02%w

1.02*% + 1

= $35455.5950…Totalpaid = $35455.5950…× 2

= $70911.1900…Interest = $70911.1900. . . −$50000

= $20911.1900. . .≈ $20911.19

\Totalamountofintertestpaidwas$20911.19

2Marks:Correctanswer.1Mark:Makessignificantprogress.


6

25 Tofindtheexpectedvalueormean

! 𝑥(𝑥0)𝑑𝑥%

+= ! 𝑥(𝑥0)

%

+𝑑𝑥

= ! 𝑥w𝑑𝑥%

+

= a𝑥

5c+

%

= 6.4


26(a)

2Marks:Correctanswer.1Mark:Drawsoneofthecurves.

26(b) 𝑥 = 0or𝑥 =π2(fromthegraph) 1Mark:Correct

answer.26(c)

𝐴 = ! [sin𝑥 − (1 − cos𝑥)]𝑑𝑥 +! [(1 − cos𝑥) − sin𝑥]𝑑𝑥M

M%

M%

+

= [−cos𝑥 − 𝑥 + sin𝑥]+M% + [𝑥 − sin𝑥 + cos𝑥]M

%

M

= L0 −π2+ 1 − (−1 − 0 + 0P + oπ − 0 − 1 − L

π2− 1 + 0Pp

= 2 −π2+π2

= 2squareunits


27 𝑦 = 𝑥% − 4𝑥𝑦′ = 2𝑥 − 4Atthepoint(1,–3))𝑦 = 2 × 1 − 4 = −2Gradientofthenormal𝑚*𝑚% = −1

𝑚 × −2 = −1

𝑚 =12= 0.5

Equationofthenormal𝑦 − 𝑦* = 𝑚(𝑥 − 𝑥*)

𝑦 − (−3) = 0.5(𝑥 − 1)2𝑦 + 6 = 𝑥 − 1

𝑥 − 2𝑦 − 7 = 0

2Marks:Correctanswer.1Mark:Findsthegradientofthenormal.


7

28(a) 𝑓(𝑥) = 7 + 4𝑥0 − 3𝑥wStationarypoints𝑓′(𝑥) = 0𝑓′(𝑥) = 12𝑥% − 12𝑥012𝑥%(1 − 𝑥) = 0𝑥 = 0, 𝑥 = 1When𝑥 = 0, 𝑦 = 7 + 4 × 00 − 3 × 0w = 7When𝑥 = 1, 𝑦 = 7 + 4 × 10 − 3 × 1w = 8\Stationarypointsare(0,7)and(1,8)

2Marks:Correctanswer.1Mark:Findsthefirstderivativeandequatesittozero

28(b) 𝑓′′(𝑥) = 024𝑥 − 36𝑥% = 012𝑥(2 − 3𝑥) = 0

𝑥 = 0or𝑥 =23


28(c) 𝑓′′(𝑥) = 12𝑥(2 − 3𝑥)At(0, 7), 𝑓′′(0) = 0PossibleofinflexionAt(1, 8), 𝑓′′(1) = −12 < 0Maxima At(0,7)checkforchangeinconcavity𝑥 = −0.1, 𝑓′′(𝑥) = 12 × −0.1(2 − 3 × −0.1) = −2.76 < 0𝑥 = 0.1, 𝑓′′(𝑥) = 12 × 0.1(2 − 3 × 0.1) = 2.04 > 0\(0,7)isapointofinflexionand(1,8)isamaxima.

2Marks:Correctanswer.1Mark:Findsthenatureofoneofthepoints.

28(d)

2Marks:Correctanswer.1Mark:Obtainsthecorrectgeneralshapeofthecurveorshowssomeunderstanding.

29(a) 𝑚 =RiseRun

= −880

= −0.1

\Gradientis–0.1

2Marks:Correctanswer.1Mark:Findsthelineofbestfitorshowssomeunderstanding.


8

29(b) Whenage=30thenfitnesslevel=7(fromthescatterplot)\Lachlan’sfitnesslevelshouldbe7.


29(c) Data:(10,8)(10,9)(20,8)(30,6)(30,7)(40,6)(40,8)(50,5)(50,6)(60,4)(60,6)(70,2)(70,3)(80,2)𝑟 = −0.9115…≈ −0.91

2Marks:Correctanswer.1Mark:Findsavalueofrcloseto–0.9.

30

Domain: − 1 ≤ 𝑥 ≤ 1Range:0 ≤ 𝑦 ≤ 1

2Marks:Correctanswer.1Mark:Domainorrange

31!

19

%

+(4𝑥 − 𝑥%)𝑑𝑥 =

19a2𝑥% −

𝑥0

3c+

%

=19ao2 × 2% −

20

3p − o2 × 0% −

00

3pc

=1627


32 𝑑%𝑦𝑑𝑥%

= 12𝑥 + 6

𝑑𝑦𝑑𝑥

= 6𝑥% + 6𝑥 + 𝐶*

At(1, −2)𝑑𝑦𝑑𝑥

= 0

6 × 1% + 6 × 1 + 𝐶* = 0

𝐶* = −12𝑑𝑦𝑑𝑥

= 6𝑥% + 6𝑥 − 12

𝑦 = 2𝑥0 + 3𝑥% − 12𝑥 + 𝐶%

(1,–2)satisfiestheequationofthecurve

2 × 10 + 3 × 1% − 12 × 1 + 𝐶% = −2

𝐶% = 5

∴ 𝑦 = 2𝑥0 + 3𝑥% − 12𝑥 + 5

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

33 lim=→+

sin6𝑥𝑥

= 6 lim=→+

sin6𝑥6𝑥

= 6

2Marks:Correctanswer.1Mark:Showsunderstanding.


9

34(a) 𝐴𝐶 = (3 − 𝑥)metres 1Mark:Correctanswer.

34(b) (3 − 𝑥)% = ℎ% + 𝑥%ℎ% = 9 − 6𝑥 + 𝑥% − 𝑥%ℎ = √9 − 6𝑥

(h>0ashisaheight)

𝐴 =12𝑏ℎ

= 0.5𝑥√9 − 6𝑥m%

2Marks:Correctanswer.1Mark:Findstheheightofthetriangleorshowssomeunderstanding.

34(c) Maximumoccurswhen𝑑𝐴𝑑𝑥

= 0

𝐴 = 0.5𝑥√9 − 6𝑥𝑑𝐴𝑑𝑥

= 0.5 ¡𝑥 ×12(9 − 6𝑥)−

12 × (−6) + (9 − 6𝑥)

12 × 1¢

=12(9 − 6𝑥)l

*%[−3𝑥 + (9 − 6𝑥)*]

=−9𝑥 + 92√9 − 6𝑥

=9(1 − 𝑥)2√9 − 6𝑥

Now𝑑𝐴𝑑𝑥

=9(1 − 𝑥)2√9 − 6𝑥

= 0

\x=1

𝑥 = 0.9,𝑑𝐴𝑑𝑥

=9(1 − 0.9)

2√9 − 6 × 0.9> 0

𝑥 = 1.1,𝑑𝐴𝑑𝑥

=9(1 − 1.1)

2√9 − 6 × 1.1< 0

\Maximumoccurswhenx=1.

3Marks:Correctanswer.2Marks:Findsx=11Mark:Calculatesthefirstderivativeorhassomeunderstandingoftheproblem.

34(d) 𝐴 = 0.5𝑥√9 − 6𝑥= 0.5 × 1 × √9 − 6 × 1= 0.5√3m%

\Maximumpossibleareais0.5√3m%


35 Drawthegraphsof𝑦 = 𝑒=and𝑦 = −𝑥 − 2

\Thereis1solutionfor𝑒= + 𝑥 + 2 = 0(pointofintersection)

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

36(a) Usingthestatisticmodeonthecalculator.�̅� = 17.3333. . . ≈ 17.3



10

36(b) Usingthestatisticmodeonthecalculator.𝑄* = 13.5and𝑄0 = 19IQR = 𝑄0 − 𝑄*

= 19 − 13.5 = 5.5


36(c) OutlierUpperlimit = 𝑄0 + 1.5 × IQR

= 19 + 1.5 × 5.5= 27.25

\29isanoutlierasitisabovetheupperlimitof27.25


37(a) 𝑓(𝑥) = (2𝑥 − 3)w𝑓′(𝑥) = 4(2𝑥 − 3)0 × 2

= 8(2𝑥 − 3)0𝑓′(1) = 8(2 × 1 − 3)0

= −8

2Marks:Correctanswer.1Mark:Finds𝑓′(𝑥).

37(b) Equationofthetangentat(1,1)withgradient–8.𝑦 − 𝑦* = 𝑚(𝑥 − 𝑥*)𝑦 − 1 = −8(𝑥 − 1)

𝑦 = −8𝑥 + 9or8𝑥 + 𝑦 − 9 = 0


37(c)

Tofindthex-intercept(y=0)𝑦 = −8𝑥 + 9= −8 × 0 + 9 = 9

∴ 𝐵(0, 9)Tofindthey-intercept(x=0)𝑦 = −8𝑥 + 90 = −8𝑥 + 9

𝑥 =98

∴ 𝐴(98, 0)

𝐴 =12𝑏ℎ =

12×98× 9

=8116squareunits

\Areaof∆𝑂𝐴𝐵isv**Nsquareunits.

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

mathematics advanced 2u - ace paper 2...year 12 mathematics advanced 2 section i 10 marks attempt...

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