MATHEMATICAL METHODSWritten examination 1

Wednesday 7 November 2018 Reading time: 9.00 am to 9.15 am (15 minutes) Writing time: 9.15 am to 10.15 am (1 hour)

QUESTION AND ANSWER BOOK

Structure of bookNumber of questions

Number of questions to be answered

Number of marks

9 9 40

• Studentsarepermittedtobringintotheexaminationroom:pens,pencils,highlighters,erasers,sharpenersandrulers.

• StudentsareNOTpermittedtobringintotheexaminationroom:anytechnology(calculatorsorsoftware),notesofanykind,blanksheetsofpaperand/orcorrectionfluid/tape.

Materials supplied• Questionandanswerbookof14pages• Formulasheet• Workingspaceisprovidedthroughoutthebook.

Instructions• Writeyourstudent numberinthespaceprovidedaboveonthispage.• Unlessotherwiseindicated,thediagramsinthisbookarenotdrawntoscale.• AllwrittenresponsesmustbeinEnglish.

At the end of the examination• Youmaykeeptheformulasheet.

Students are NOT permitted to bring mobile phones and/or any other unauthorised electronic devices into the examination room.

©VICTORIANCURRICULUMANDASSESSMENTAUTHORITY2018

SUPERVISOR TO ATTACH PROCESSING LABEL HEREVictorian Certificate of Education 2018

STUDENT NUMBER

Letter

2018MATHMETHEXAM1 2

THIS PAGE IS BLANK

3 2018MATHMETHEXAM1

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Question 1 (3marks)

a. If y x x dydx

= − + −( ) , .3 643 2 3 find 1mark

b. Let f xex

x( ) =

cos( ).

Evaluate f ′(π). 2marks

InstructionsAnswerallquestionsinthespacesprovided.Inallquestionswhereanumericalanswerisrequired,anexactvaluemustbegiven,unlessotherwisespecified.Inquestionswheremorethanonemarkisavailable,appropriateworkingmustbeshown.Unlessotherwiseindicated,thediagramsinthisbookarenotdrawntoscale.

2018MATHMETHEXAM1 4

Question 3–continued

Question 2 (3marks)

Thederivativewithrespecttoxofthefunction f:(1,∞)→Rhastherule ′ = −−( )

f xx

( ) 12

12 2

Giventhat f(2)=0,find f(x)intermsofx.

Question 3 (5marks)Let f :[0,2π]→R, f (x)=2cos(x)+1.

a. Solvetheequation2cos(x)+1=0for0≤x≤2π. 2marks

5 2018MATHMETHEXAM1

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b. Sketchthegraphofthefunction fontheaxesbelow.Labeltheendpointsandlocalminimumpointwiththeircoordinates. 3marks

y

x

4

3

2

1

0

–1

–2

53π4

3ππ

323π

π 2π

2018MATHMETHEXAM1 6

Question 4 (2marks)LetXbeanormallydistributedrandomvariablewithameanof6andavarianceof4.LetZbearandomvariablewiththestandardnormaldistribution.

a. FindPr(X>6). 1mark

b. FindbsuchthatPr(X>7)=Pr(Z<b). 1mark

Question 5 (3marks)

Let f:(2,∞)→R,where f xx

( )( )

.=−12 2

Statetheruleanddomainof f–1.

7 2018MATHMETHEXAM1

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Question 6 (4marks)Twoboxeseachcontainfourstonesthatdifferonlyincolour.Box1containsfourblackstones.Box2containstwoblackstonesandtwowhitestones.Aboxischosenrandomlyandonestoneisdrawnrandomlyfromit.Eachboxisequallylikelytobechosen,asiseachstone.

a. Whatistheprobabilitythattherandomlydrawnstoneisblack? 2marks

b. Itisnotknownfromwhichboxthestonehasbeendrawn.

Giventhatthestonethatisdrawnisblack,whatistheprobabilitythatitwasdrawnfromBox1? 2marks

2018MATHMETHEXAM1 8

Question 7 (5marks)LetPbeapointonthestraightliney=2x–4suchthatthelengthofOP,thelinesegmentfromtheoriginOtoP,isaminimum.

a. FindthecoordinatesofP. 3marks

b. FindthedistanceOP.Expressyouranswerintheformb

a b,whereaandbarepositive

integers. 2marks

9 2018MATHMETHEXAM1

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CONTINUES OVER PAGE

2018MATHMETHEXAM1 10

Question 8–continued

Question 8 (7marks)Let f:R→R, f(x)=x2ekx,wherekisapositiverealconstant.

a. Showthat f ′ (x)=xekx(kx+2). 1mark

b. Findthevalueofkforwhichthegraphsofy=f (x)andy=f ′ (x)haveexactlyonepointofintersection. 2marks

11 2018 MATHMETH EXAM 1

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Let g x xek

kx( ) .= −

2 The diagram below shows sections of the graphs of f and g for x ≥ 0.

0

x = 2

2

f

y

g

x

Let A be the area of the region bounded by the curves y = f (x), y = g (x) and the line x = 2.

c. Write down a definite integral that gives the value of A. 1 mark

d. Using your result from part a., or otherwise, find the value of k such that Ak

=16 . 3 marks

2018MATHMETHEXAM1 12

Question 9 –continued

Question 9 (8marks)Considerapartofthegraphofy=x sin(x),asshownbelow.

–5π –4π –3π –2π –π π 2π 3π 4π 5π0

y

x

a. i. Giventhat∫(x sin (x))dx = sin (x) – x cos (x) + c, evaluate (x sin (x))dxn

n( )

π

π+

∫1

whennis

apositiveevenintegeror0.Giveyouranswerinsimplestform. 2marks

ii. Giventhat∫(x sin (x))dx = sin (x) – x cos (x) + c, evaluate (x sin (x))dxn

n( )

π

π+

∫1

whennis

apositiveoddinteger.Giveyouranswerinsimplestform. 1mark

13 2018MATHMETHEXAM1

Question 9 –continuedTURN OVER

b. Findtheequationofthetangenttoy=x sin(x)atthepoint −

52

52

π π, . 2marks

c. ThetranslationTmapsthegraphofy=x sin(x)ontothegraphofy=(3π–x)sin(x),where

T R R Txy

xy

a: ,2 2

0→

=

+

andaisarealconstant.

Statethevalueofa. 1mark

2018 MATHMETH EXAM 1 14

END OF QUESTION AND ANSWER BOOK

d. Let f : [0, 3π] → R, f (x) = (3π – x) sin (x) and g : [0, 3π] → R, g (x) = (x – 3π) sin (x).

The line l1 is the tangent to the graph of f at the point π π2

52

,

and the line l2 is the tangent

to the graph of g at π π2

52

, ,−

as shown in the diagram below.

y

x0 π 2π 3π

l1

l2

Find the total area of the shaded regions shown in the diagram above. 2 marks

MATHEMATICAL METHODS

Written examination 1

FORMULA SHEET

Instructions

This formula sheet is provided for your reference.A question and answer book is provided with this formula sheet.

Students are NOT permitted to bring mobile phones and/or any other unauthorised electronic devices into the examination room.

Victorian Certificate of Education 2018

© VICTORIAN CURRICULUM AND ASSESSMENT AUTHORITY 2018

MATHMETH EXAM 2

Mathematical Methods formulas

Mensuration

area of a trapezium 12a b h+( ) volume of a pyramid 1

3Ah

curved surface area of a cylinder 2π  rh volume of a sphere

43

3π r

volume of a cylinder π r 2h area of a triangle12bc Asin ( )

volume of a cone13

2π r h

Calculus

ddx

x nxn n( ) = −1 x dxn

x c nn n=+

+ ≠ −+∫ 11

11 ,

ddx

ax b an ax bn n( )+( ) = +( ) −1 ( )( )

( ) ,ax b dxa n

ax b c nn n+ =+

+ + ≠ −+∫ 11

11

ddxe aeax ax( ) = e dx a e cax ax= +∫ 1

ddx

x xelog ( )( ) = 11 0x dx x c xe= + >∫ log ( ) ,

ddx

ax a axsin ( ) cos( )( ) = sin ( ) cos( )ax dx a ax c= − +∫ 1

ddx

ax a axcos( )( ) −= sin ( ) cos( ) sin ( )ax dx a ax c= +∫ 1

ddx

ax aax

a axtan ( )( )

( ) ==cos

sec ( )22

product ruleddxuv u dv

dxv dudx

( ) = + quotient ruleddx

uv

v dudx

u dvdx

v

=

−

2

chain ruledydx

dydududx

=

3 MATHMETH EXAM

END OF FORMULA SHEET

Probability

Pr(A) = 1 – Pr(A′) Pr(A ∪ B) = Pr(A) + Pr(B) – Pr(A ∩ B)

Pr(A|B) = Pr

PrA BB∩( )( )

mean µ = E(X) variance var(X) = σ 2 = E((X – µ)2) = E(X 2) – µ2

Probability distribution Mean Variance

discrete Pr(X = x) = p(x) µ = ∑ x p(x) σ 2 = ∑ (x – µ)2 p(x)

continuous Pr( ) ( )a X b f x dxa

b< < = ∫ µ =

−∞

∞

∫ x f x dx( ) σ µ2 2= −−∞

∞

∫ ( ) ( )x f x dx

Sample proportions

P Xn

=̂ mean E(P̂ ) = p

standard deviation

sd P p pn

(ˆ ) ( )=

−1 approximate confidence interval

,p zp p

np z

p pn

−−( )

+−( )

1 1ˆ ˆ ˆˆˆ ˆ