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
2019MATHMETHEXAM1 8
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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
2019MATHMETHEXAM1 12
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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
