MATHEMATICAL METHODSWritten examination 1

Wednesday 6 November 2019 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

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

sharpenersandrulers.• StudentsareNOTpermittedtobringintotheexaminationroom:anytechnology(calculatorsor

software),notesofanykind,blanksheetsofpaperand/orcorrectionfluid/tape.

Materials supplied• Questionandanswerbookof13pages• 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.

©VICTORIANCURRICULUMANDASSESSMENTAUTHORITY2019

SUPERVISOR TO ATTACH PROCESSING LABEL HEREVictorian Certificate of Education 2019

STUDENT NUMBER

Letter

2019MATHMETHEXAM1 2

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THIS pAgE IS BLANK

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Question 1 (4marks)

Let f R f xx

: 13

13 1

, , ( )∞

→ =

−.

a. i. Findf′(x). 1mark

ii. Findanantiderivativeoff (x). 1mark

b. Letg R R g x xx

: \ { } , ( ) sin( ) .− → =+

11π

Evaluateg′(1). 2marks

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

2019MATHMETHEXAM1 4

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Question 2 (4marks)

a. Let f R R f xx

: \ , ( ) .13

13 1{ }→ =−

Findtheruleoff –1. 2marks

b. Statethedomainoff –1. 1mark

c. LetgbethefunctionobtainedbyapplyingthetransformationTtothefunctionf,where

T xy

xy

cd

=

+

andc,d∈R.

Findthevaluesofcanddgiventhatg=f–1. 1mark

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Question 3 (3marks)Theonlypossibleoutcomeswhenacoinistossedareaheadoratail.Whenanunbiasedcoinistossed,theprobabilityoftossingaheadisthesameastheprobabilityoftossingatail.Johasthreecoinsinherpocket;twoareunbiasedandoneisbiased.Whenthebiasedcoinistossed,the

probabilityoftossingaheadis 13.

Jorandomlyselectsacoinfromherpocketandtossesit.

a. Findtheprobabilitythatshetossesahead. 2marks

b. Findtheprobabilitythatsheselectedanunbiasedcoin,giventhatshetossedahead. 1mark

2019MATHMETHEXAM1 6

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Question 4 (4marks)

a. Solve12 2

2−

=

∈ −[ ]cos cos .x x x for , π π 2marks

b. Thefunction f R f x x: −[ ]→ =

2

2π π, , ( ) cos isshownontheaxesbelow.

y

x�−�−2�

2

O

−1

−2

1

Letg R g x f x: , , ( ) ( ).−[ ]→ = −2 1π π

Sketchthegraphofgontheaxesabove.Labelallpointsofintersectionofthegraphsoffandg,andtheendpointsofg,withtheircoordinates. 2marks

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Question 5 (5marks)

Let f R R f xx

: \ { } , ( )( )

.1 21

12→ =−

+

a. i. Evaluatef (–1). 1mark

ii. Sketchthegraphoffontheaxesbelow,labellingallasymptoteswiththeirequations. 2marks

y

x

6

5

4

3

2

1

O 1 2 3 4 5−1−1

−2

−2

−3

−3

−4

−4

−5

b. Findtheareaboundedbythegraphoff,thex-axis,thelinex=–1andthelinex=0. 2marks

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Question 6 (3marks)Fredownsacompanythatproducesthousandsofpegseachday.Herandomlyselects41pegsthatareproducedononedayandfindseightfaultypegs.

a. Whatistheproportionoffaultypegsinthissample? 1mark

b. Pegsarepackedeachdayinboxes.Eachboxholds12pegs.LetP̂ betherandomvariablethatrepresentstheproportionoffaultypegsinabox.

Theactualproportionoffaultypegsproducedbythecompanyeachdayis 16.

FindPr6

<

1P̂ .Expressyouranswerintheforma(b)n,whereaandbarepositiverationalnumbers

andnisapositiveinteger. 2marks

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Question 7 (4marks)

Thegraphoftherelation y x= −1 2 isshownontheaxesbelow.Pisapointonthegraphofthisrelation,Aisthepoint(–1,0)andBisthepoint(x,0).

y

xBA

P (x, y)21y x= −

(x, 0)(–1, 0)

a. FindanexpressionforthelengthPBintermsofxonly. 1mark

b. FindthemaximumareaofthetriangleABP. 3marks

2019MATHMETHEXAM1 10

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Question 8 –continued

Question 8 (4marks)Thefunctionf :R→R,f (x)isapolynomialfunctionofdegree4.Partofthegraphoff isshownbelow.Thegraphoff touchesthex-axisattheorigin.

1 ,12

−

1 ,12

(–1, 0) (1, 0)O

y

x

a. Findtheruleoff. 1mark

Letgbeafunctionwiththesameruleasf.Let h D R h x g x x xe e: ( ) log ( ) log→ = ( ) − +( ), 3 2 ,whereDisthemaximaldomainofh.

b. StateD. 1mark

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c. Statetherangeofh. 2marks

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Question 9–continued

Question 9 (9marks)

Considerthefunctions f R R f x x x g R R g x ex: and → = + − → =, ( ) : , ( )3 2 2 .

a. Statetheruleofg f x( )( ). 1mark

b. Findthevaluesofxforwhichthederivativeof g f x( )( ) isnegative. 2marks

c. Statetheruleof f g x( )( ). 1mark

d. Solve f g x( )( ) = 0. 2marks

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e. Findthecoordinatesofthestationarypointofthegraphof f g x( )( ). 2marks

f. Statethenumberofsolutionstog f x f g x( ) ( )( ) + ( ) = 0. 1mark

END OF QUESTION AND ANSWER BOOK

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 2019

© VICTORIAN CURRICULUM AND ASSESSMENT AUTHORITY 2019

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