MATHEMATICAL METHODS (CAS)

Written examination 2Wednesday 9 November 2011

Reading time: 11.45 am to 12.00 noon (15 minutes) Writing time: 12.00 noon to 2.00 pm (2 hours)

QUESTION AND ANSWER BOOK

Structure of bookSection Number of

questionsNumber of questions

to be answeredNumber of

marks

12

224

224

2258

Total 80

Materials supplied

Instructions

student numbername student number

and

At the end of the examination

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

SUPERVISOR TO ATTACH PROCESSING LABEL HERE

Figures

Words

STUDENT NUMBER

Letter

2011

SECTION 1

Question 1d

A. d2

52

, −⎛⎝⎜

⎞⎠⎟

B.

C. d −⎛

⎝⎜

⎞⎠⎟

52

0,

D. 0 52

, −⎛⎝⎜

⎞⎠⎟

d

E. 5

20+⎛

⎝⎜

⎞⎠⎟

d ,

Question 2perpendicular

A. 12

B. –2

C. 12

D. 4

E. 2

Question 3x + a x3 – 13x2 – ax a a

A. – 4B. –3C. – 1D. 1E. 2

SECTION 1

Instructions for Section 1all

correct

not

SECTION 1TURN OVER

Question 4

e f x x

A. f xf x

( )( )

B. 2 f xf x

( )( )

C. f xf x( )( )2

D. e f x

E. e f x

Question 5g R g x 2 4x

A. g–1 R g–1 xx2 4

2

B. g–1 R g–1 x x 2

C. g–1 R g–1 xx2

4

D. g–1 R g–1 xx2 4

2

E. g–1 R R g–1 xx2 4

2

Question 6X

f xx x e

elsewheree( )

log ( )=

≤ ≤⎧⎨⎩

10

X E XA.B.C. 1D.E.

SECTION 1

Question 7

f x 3 x f h f hf 3 8 5.A.B.C.D.E.

Question 8f R R f x x x

g 32

5,⎡⎣⎢

⎞⎠⎟ R g x x

h f + g hA.

B. −⎡⎣⎢

⎤⎦⎥

34

10,

C. −⎛⎝⎜

⎤⎦⎥

34

154

,

D. 34

13,⎡⎣⎢

⎞⎠⎟

E. 32

13,⎡⎣⎢

⎞⎠⎟

SECTION 1TURN OVER

Question 9y f x

4

–5 –3 O

y

x

y f x

–3–5 O

4

A.

–3 O

B.

–3 O

D.

–3–5 O

4

E.

–3–5 O

C.

x x x

x x

y y

y y

y

SECTION 1

Question 10

A. 45

B. 25

C. 23

D. 13

E. 34

Question 11f x e x

A. e

B. 13

6log ( )e

C. loge3125

43⎛

⎝⎜

⎞⎠⎟ −

D. 13

31254

3loge⎛⎝⎜

⎞⎠⎟ −

E. 5 5 2 2 3

3log ( ) log ( )e e

Question 12X

a X > aa

A.B.C.D.E.

SECTION 1TURN OVER

Question 13

A.B.C.D.E.

Question 14

y

xO 1

1

e 2

y = e 2x

i. 1

2

0

xe dx

ii. e e dxx2 2

0

1

iii. 2

2

1

eye dy

iv. 2

1

log ( )2

ee x

dx

A. ii. B. ii. iii.C. i. ii. iii. iv.D. ii. iv.E. i. iv.

SECTION 1

Question 15

3

–16

23

y

xO

A. y x= +⎛⎝⎜

⎞⎠⎟ +2

61cos π

B. y x= −⎛⎝⎜

⎞⎠⎟ +2 4

61cos π

C. y x= −⎛⎝⎜

⎞⎠⎟ −4 2

121sin π

D. y x= +⎛⎝⎜

⎞⎠⎟ −3 2

61cos π

E. y x= +⎛⎝⎜

⎞⎠⎟ −2 4 2

31sin π

Question 16not

f R R f x |x2

A. fB. fC. f x x.D. f x 2x x E. f x 2x x

Question 17

y x x= +32

A. 4x y

B. 4y + x

C. y x= +

41

D. x – 4y –5

E. 4y + 4x 20

SECTION 1TURN OVER

Question 18x3 x2 + 15x + w x

A. w < 25B. wC. wD. w w > 25E. w > 1

Question 19f R R f x x2

y

y

x

x1 2 3 4 5 6

1

O

O 2 3 4 5 6

p p

A. 10B. 15C. 20D. 25E. 30

2011 MATHMETH(CAS) EXAM 2 10

SECTION 1 – continued

Question 20A part of the graph of g: R R, g(x) = x2 – 4 is shown below.

A

B

y

xaO

The area of the region marked A is the same as the area of the region marked B.The exact value of a isA. 0B. 6

C. 6 D. 12E. 2 3

Question 21For two events, P and Q, Pr(P Q) = Pr(P Q).P and Q will be independent events exactly whenA. Pr(P ) = Pr(Q)

B. Pr(P Q ) = Pr(P Q)

C. Pr(P Q) = Pr(P) + Pr(Q)

D. Pr(P Q ) = Pr(P Q)

E. Pr(P) = 12

11

END OF SECTION 1TURN OVER

Question 22

c a a b b c

A. 1 1 1

log ( ) log ( ) log ( )c a ba b c

B. 1 1 1

log ( ) log ( ) log ( )a b cc a b

C. 1 1 1

log ( ) log ( ) log ( )a b cb c a

D. 1 1 1

log ( ) log ( ) log ( )a b ca b c

E. 1 1 1

log ( ) log ( ) log ( )c b aab ac cb

2011 MATHMETH(CAS) EXAM 2 12

SECTION 2 – Question 1 – continued

Question 1Two ships, the Elsa and the Violet, have collided. Fuel immediately starts leaking from the Elsa into the sea.The captain of the Elsa estimates that at the time of the collision his ship has 6075 litres of fuel on board and

he also forecasts that it will leak into the sea at a rate of t2

5 litres per minute, where t is the number of minutes

that have elapsed since the collision.a. At this rate how long, in minutes, will it take for all the fuel from the Elsa to leak into the sea?

3 marks

SECTION 2

Instructions for Section 2Answer all questions in the spaces provided.In all questions where a numerical answer is required an exact value must be given unless otherwise

In questions where more than one mark is available, appropriate working must be shown.Unless otherwise indicated, the diagrams in this book are not drawn to scale.

13

SECTION 2 – Question 1TURN OVER

is not

t r

b.103

SECTION 2 – Question 1

g R g xπx80

⎛⎝⎜

⎞⎠⎟

y (m)

CO

x (m)

h

h R h x − ⎛⎝⎜

⎞⎠⎟60

40sin π x

c. y h x

15 2011 MATHMETH(CAS) EXAM 2

SECTION 2 – continuedTURN OVER

d.

i. Find sin π x dx40

0

40⎛⎝⎜

⎞⎠⎟∫

ii.

1 + 2 = 3 marks

Total 11 marks

SECTION 2 – Question 2

Question 2A

BX A,

Y B

f y

yy y

e yy

( )

. . ( )

=

<

≤ ≤

>

⎧

⎨⎪⎪

⎩⎪⎪ − −

0 0

160 4

0 25 40 5 4

a. i. X

ii. Y

b. Y

SECTION 2 – Question 2TURN OVER

c. i. Y

ii. a Y a

d. Pr( )Y ≤ =3 932

produced by machine B

2011 MATHMETH(CAS) EXAM 2 18

SECTION 2 – continued

All of the chocolates produced by machine A and machine B are stored in a large bin. There is an equal number of chocolates from each machine in the bin.It is found that if a chocolate, produced by either machine, takes longer than 3 seconds to produce then it can

e. A chocolate is selected at random from the bin. It is found to have taken longer than 3 seconds to produce. Find, correct to four decimal places, the probability that it was produced by machine A.

3 marks

Total 15 marks

SECTION 2TURN OVER

CONTINUES OVER PAGE

SECTION 2 – Question 3

Question 3a. f R R f x x3 + 5x i. f x

ii. f x x

b. p p R R p x ax3 + bx2 + cx + k a b c k i. p m m

ii. p m

c. q q R R q x x3

i. q–1 x

21

SECTION 2TURN OVER

ii. y q x y q–1 x

d. g g R R g x x3 + 2x2 + cx + k c k i. g c

ii. y g x y g–1 x k

2011 MATHMETH(CAS) EXAM 2 22

SECTION 2 – Question 4 – continued

Question 4Deep in the South American jungle, Tasmania Jones has been working to help the Quetzacotl tribe to get drinking water from the very salty water of the Parabolic River. The river follows the curve with equation y = x2 – 1, xTasmania has his camp site at (0, 0) and the Quetzacotl tribe’s village is at (0, 1). Tasmania builds a desalination plant, which is connected to the village by a straight pipeline.

y

x1

Parabolic River

village (0, 1)

desalination plant (m, n)

camp(0, 0)

(–1, 0)

pipeline

a. If the desalination plant is at the point (m, n) show that the length, L kilometres, of the straight pipeline that carries the water from the desalination plant to the village is given by

L m m= − +4 23 4.

3 marks

23

SECTION 2 – Question 4TURN OVER

b.

i.dLdm

ii.

SECTION 2 – Question 4

72

34

,⎛

⎝⎜⎜

⎞

⎠⎟⎟

x y x 7

2 .

y

c.

T x x k x= − + + −( )12

1 14

7 44 2 2 k

k

d. k1

2 13

i. dTdx

25

ii.

e. kk

f. k

END OF QUESTION AND ANSWER BOOK

MATHEMATICAL METHODS (CAS)

Written examinations 1 and 2

FORMULA SHEET

Directions to students

Detach this formula sheet during reading time.

This formula sheet is provided for your reference.

© VICTORIAN CURRICULUM AND ASSESSMENT AUTHORITY 2011

MATH METH (CAS) 2

This page is blank

3 MATH METH (CAS)

END OF FORMULA SHEET

Mathematical Methods (CAS)Formulas

Mensuration

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

13Ah

curved surface area of a cylinder: 2 rh volume of a sphere: 43

3r

volume of a cylinder: r 2h area of a triangle: 12bc Asin

volume of a cone: 13

2r h

Calculusddx

x nxn n( )= -1

x dx

nx c nn n=

++ ≠ −+∫

11

11 ,

ddxe aeax ax( )= e dx a e cax ax= +∫

1

ddx

x xelog ( )( )= 1

1x dx x ce= +∫ 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 rule: ddxuv u dv

dxv dudx

( )= + quotient rule: ddx

uv

v dudx

u dvdx

v⎛⎝⎜

⎞⎠⎟ =

−

2

chain rule: dydx

dydududx

= approximation: f x h f x h f x+( ) ≈ ( ) + ′( )

ProbabilityPr(A) = 1 – Pr(A ) Pr(A B) = Pr(A) + Pr(B) – Pr(A B)

Pr(A|B) = Pr

PrA BB∩( )( ) transition matrices: Sn = Tn S0

mean: μ = E(X) variance: var(X) = 2 = E((X – μ)2) = E(X2) – μ2

probability distribution mean variance

discrete Pr(X = x) = p(x) μ = x p(x) 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