2007 Maths Methods CAS 3 & 4 Exam 1 solutions · ©THE HEFFERNAN GROUP 2007 Maths Methods (CAS) 3 & 4 Trial Exam 1 solutions 5 To find the endpoints: Method 1 Because two complete

_____________________________________________________________________ ©THE HEFFERNAN GROUP 2007 Maths Methods(CAS) 3 & 4 Trial Exam 1 solutions

MATHS METHODS (CAS) 3 & 4 TRIAL EXAMINATION 1 SOLUTIONS 2007 Question 1 The function ( ) ( )2log −= xxf e is defined for

So ( )∞= ,2fd (1 mark)

Question 2 a. ( )( )xgf exists iff .

Now, { }0\ and RdRr fg == .

{ }fg drRR

⊄

⊄ ,0\ Since

so ( )( )xgf does not exist. (1 mark)

b. i.

( )( )

121

21

+=

⎟⎠

⎞⎜⎝

⎛=

x

xgxfg

(1 mark) ii.

{ }0\

))((

R

dd fxfg

=

=

(1 mark)

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©THE HEFFERNAN GROUP 2007 Maths Methods (CAS) 3 & 4 Trial Exam 1 solutions

2

Question 3

a. ( ) ( ) 21

3 ,,1: +−

=→∞x

xfRf

Let

Swap x and y

( )( )

( ) 12

3 So

12

32

31

3121

32

21

3

1 +−

=

+−

=

−=−

=−−

−=−

+−

=

−

xxf

xy

xy

yxy

x

yx

(1 mark)

b. ff rd =−1

Do a quick sketch of

( )( )∞=

∞=

− ,2 So

,2

1f

f

d

r

(1 mark)

(1 mark)

x

2

1

y

)(xfy =


3

Question 4 a. ( ) ( )xexf 2sin=

( )( )

( )

( )( )xx

x

x

ee

eudxdu

dudy

dxdy

ududy

uyey

22

2

2

cos2

2cos=

rule)(Chain

cos

sinsinLet

=

⋅

⋅=

=

=

=

(1 mark) – for knowing to use the chain rule and attempting to use it

(1 mark) – correct answer

b. ( )

( ) ( ) ( )

( )( ) ( )( )23

22

23

23

3

2

log232

2

log2312

2log

xx

xxxxx

xxx

xx

dxdy

xxx

y

e

e

e

+

+−+=

+

+−×+=

+=

(1 mark) – for knowing to use the quotient rule and attempting to use it

(1 mark) – correct answer

x

x

edxdu

eu

2

2

2

Let

=

=


4

Question 5

For ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛−=3

2sin2 πxy the period is (1 mark) correct period

This means that for [ ]ππ ,−∈x there will be two complete periods of the graph. x-intercepts Method 1

The graph is a sine graph with a period of π that has been translated 3π units to the right so

that the x-intercepts will occur at ⎟⎠

⎞⎜⎝

⎛ 0,3π , at ⎟

⎠

⎞⎜⎝

⎛=⎟

⎠

⎞⎜⎝

⎛+ 0,

650,

32πππ , at

⎟⎠

⎞⎜⎝

⎛ −=⎟

⎠

⎞⎜⎝

⎛+− 0,

60,

32πππ and at ⎟

⎠

⎞⎜⎝

⎛ −=⎟

⎠

⎞⎜⎝

⎛+− 0,

320,

3ππ

π . (1 mark) for x- intercepts

Method 2 The x- intercepts occurs when 0=y .

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛−=3

2sin2 πxy

becomes

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛−=3

2sin20 πx

So, 03

2sin =⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛−πx

...,2,,0,,2,3...3

2 ππππππ

−−−=⎟⎠

⎞⎜⎝

⎛−x

...,,2,0,

2,,

23...

3π

πππ

ππ −−

−=−x

...,34,

65,

3,

6,

32,

67... ππππππ −−−

=x

ππππππ

≤≤−−−

= xx since6

5,3

,6

,32

(1 mark) for x- intercepts y - intercept )0( =x

Angle measured clockwise because it is negative.

y-intercept occurs at ( )3,0 − (1 mark)

S

T

A

C

32π


5

To find the endpoints: Method 1 Because two complete periods occur for [ ]ππ ,−∈x the right endpoint and left endpoint will have the same y-coordinate and it will be the same as the y-intercept. So left endpoint is ( )3,−−π and the right endpoint is ( )3,−π . (1 mark) Method 2

( ) ( )

3232

232

3sin2

3sin2

32sin2

34sin2

38sin2

322sin2

342sin2

32sin2:endpointRight

32sin2 :endpointLeft

−=−×=

−×=⎟⎠

⎞⎜⎝

⎛−×=

⎟⎠

⎞⎜⎝

⎛−×=⎟

⎠

⎞⎜⎝

⎛ −=

⎟⎠

⎞⎜⎝

⎛=⎟

⎠

⎞⎜⎝

⎛ −=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛ −=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛−−=−

π

ππ

ππ

ππ

πππ

πππ ff

3−= Left endpoint is ( )3,−−π . Right endpoint is ( )3,−π .

(1 mark)

The graph of ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛−=3

2sin2 πxy is shown below.

(1 mark) correct shape including amplitude

x

y

2

-2

32π

−3π

−32π

3π

3,−−π 3,−π3−

6π

−65π


6

Question 6 a.

( ) ( )

( )(given) 160

1Prsymmetryby 5Pr9Pr

⋅=

−<=

<=>

ZXX

(1 mark)

b. ( )97Pr << XX (conditional probability) ( )

( )( )( )

42258450

84.05.0

a.part from 1601

509Pr1

7Pr9Pr

97Pr

=

=

=

⋅−

⋅=

>−

<=

<

<∩<=

XXX

XX

(1 mark)

(1 mark)

(1 mark)

1 3 75 9 11 1310 2 3-3 -2 -1Z

X


7

Question 7 a. Since ( )xf represents a probability density function then

( )∫ =+3

1

11 dxax

required as 41

14

122

8

112

32

9

12

3

1

2

−=

−=

=+

=⎭⎬⎫

⎩⎨⎧

⎟⎠

⎞⎜⎝

⎛+−⎟

⎠

⎞⎜⎝

⎛+

=⎥⎦

⎤⎢⎣

⎡+

a

a

a

aa

xax

(1 mark) b.

( )

85

183

1812

84

8

241

1412Pr

2

1

2

2

1

2

2

1

=

+−=

⎭⎬⎫

⎩⎨⎧

⎟⎠

⎞⎜⎝

⎛+−−⎟

⎠

⎞⎜⎝

⎛+−=

⎥⎦

⎤⎢⎣

⎡+−=

⎥⎦

⎤⎢⎣

⎡+×−=

⎟⎠

⎞⎜⎝

⎛+−=< ∫

xx

xx

dxxX

(1 mark)

(1 mark)

(1 mark)


8

Question 8 a.

Note that at the points ( ) ( ) ( )5,0 and 0,3,0,1 the graph of ( )xfy = has cusps; i.e. “pointy bits” not smooth curves since the graph of ( )xfy = is being reflected in the x-axis.

(1 mark)

b. ( ) ( )∫ ∫−=3

1

5

3

dxxfdxxfA

(1 mark)

x

y

1 3 5

)(xfy =

)(xfy =


9

Question 9 a.

( ) ( )

required as 10

2

2

51

21

51 where

2121

heightbase21area

aa

aa

xxfafa

ba

−=

⎟⎠

⎞⎜⎝

⎛−××=

−=××=

××=

××=

b.

(1 mark)

Maximum occurs when

Note: we know we have a maximum because the graph of the function 102

2aaA −= is

an inverted parabola.

We have a maximum when 25

=a . (1 mark)

units square 85

units square 625.04025

625.025.14025

4050

1025.625.1

4025

45

105.2

22.5 area ,5.2or when

102area ,

25When

22

=

==

−=−=

−=−=

−==−== aaaa

(1 mark)

102

21

102Let

2

adadA

aaA

−=

−=

(1 mark)


10

Question 10

The gradient of a normal to x

axy61 is 3 2 −+= .

(1 mark)

Also the gradient of the normal .

21

6331

61When

−=

=−

=−

x

xx

(1 mark)

The x-coordinate of the point where the normal hits the curve is .

The curve and the normal both pass through the point .

Substituting this point into (1 mark)

12112

91043

65 So

413

65

213

65 gives

32

2

=

−=

−=

+×=

+⎟⎠

⎞⎜⎝

⎛−×=

+=

a

a

a

axy

(1 mark)


11

Question 11

At the point of intersection of the graphs,

( )( ) 02306

Let

06

6

2

2

2

=−+

=−+

=

=−+

−=

mmmm

me

ee

ee

x

xx

xx

( )2log solution no2 or 3 So

e

xx

xee=

=−=

The x-coordinate of the point of intersection is ( )2loge . (1 mark)

b. a

(1 mark)

Total 40 marks

(1 mark)

( ) ( ){ }( )

( )( )

( )

( ) ( )( )

( )( )

( )

( ) units square 2564log

211

2422log

211

222log

20

22log6

26

6

6Area

6

2log6

00

2log22log

2log

0

2

2log

0

2

2log

0

2

2

−=

+−−=

++−−=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−⎟

⎟⎠

⎞⎜⎜⎝

⎛−−=

⎥⎦

⎤⎢⎣

⎡−−=

−−=

−−=

∫

∫

e

e

e

e

xx

xx

xx

e

ee

e

e

e

e

eeee

eex

dxee

dxee

(1 mark)

(1 mark)

(1 mark)

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2007 Maths Methods CAS 3 & 4 Exam 1 solutions · ©THE HEFFERNAN GROUP 2007 Maths Methods (CAS) 3 & 4 Trial Exam 1 solutions 5 To find the endpoints: Method 1 Because two complete