The Mathematical Association of Victoria Trial Examination ...physicsservello.com.au/files/2011 MAV...

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2011 MAV MATHS METHODS CAS EXAM 1 - SOLUTIONS © The Mathematical Association of Victoria, 2011 Page 1 The Mathematical Association of Victoria Trial Examination 2011 Maths Methods CAS Examination 1 - SOLUTIONS Question 1 a. ( ) ) tan( x x dx d Using the product rule 1M ) ( sec ) tan( 2 x x x + = 1A b. i. ( ) x e dx d x 2 2 + 2 2 2 + = x e 1A ii. dx x e e x x + + 2 ) 1 ( 4 2 2 + + = dx x e e x x 2 2 2 2 2 2 ( ) x e x e 2 log 2 2 + = 1A Question 2 a. Shape and closed circles for endpoints 1A All coordinates correct 1A b. dx x + = 2 0 3 2 2 ) 1 ( 0 2 1 value Average 1A dx x + = 1 0 3 2 2 ) 1 ( due to symmetry 1 0 3 5 2 ) 1 ( 5 3 + = x x 1A + = 0 5 3 ) 2 0 ( 6 . 2 = 1A (1, 2) ( 1 , 2 4 3 + ) (3, 3 4 +2) (0, 3)

Transcript of The Mathematical Association of Victoria Trial Examination ...physicsservello.com.au/files/2011 MAV...

Page 1: The Mathematical Association of Victoria Trial Examination ...physicsservello.com.au/files/2011 MAV Maths Methods CAS...Trial Examination 2011 Maths Methods CAS Examination 1 - SOLUTIONS

2011 MAV MATHS METHODS CAS EXAM 1 - SOLUTIONS

© The Mathematical Association of Victoria, 2011 Page 1

The Mathematical Association of Victoria Trial Examination 2011

Maths Methods CAS Examination 1 - SOLUTIONS Question 1

a. ( ))tan(xxdxd

Using the product rule 1M )(sec)tan( 2 xxx += 1A

b. i. ( )xedxd x 22 +

22 2 += xe 1A

ii. dxxe

ex

x

∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛++2)1(4

2

2

∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛++= dxxe

ex

x

2222 2

2

( )xe xe 2log2 2 += 1A

Question 2 a.

Shape and closed circles for endpoints 1A All coordinates correct 1A

b. dxx∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−=

2

0

32

2)1(02

1 valueAverage 1A

dxx∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

1

0

32

2)1( due to symmetry

1

0

35

2)1(53

⎥⎥⎦

⎢⎢⎣

⎡+−= xx 1A

⎟⎠⎞⎜

⎝⎛ −−−+= 053)20(

6.2= 1A

(1, 2)

( 1− , 243 + ) (3, 3 4 +2)

(0, 3)

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2011 MAV MATHS METHODS CAS EXAM 1 - SOLUTIONS

© The Mathematical Association of Victoria, 2011 Page 2

Question 3 62 =+ ykx

6)1(3 =−+ ykx For no solution the lines are parallel

0132

=−k

k or using ratios

213 −= k

k or using gradients

13

2 −−=−k

k 1M

06)1( =−−kk

062 =−− kk 0)2)(3( =+− kk

3=k or 2−=k 1A Check to make sure they are not the same line

Using ratios 663 ≠

k or

66

21 ≠−k or using the y-intercept

16

26

−≠k

When 3=k , 166

33 == , hence the same line

When 2−=k , 66

23 ≠−

, hence parallel lines, no solution 1A

Question 4

0)1(log)1(log2 22 =++− xx

0)1(log)1(log 22

2 =++− xx

0)1()1(log 22 =+− xx 1M

1)1()1( 2 =+− xx 1A

01)1)(12( 2 =−++− xxx

023 =−− xxx 1A 0)1( 2 =−− xxx

0≠x , 251−≠x

251+=x 1A

Question 5 a. xexf −−=1)(

Let xey −−=1 Inverse: swap x and y

yex −−=1 1M xe y −=− 1

)1(log xy e −−= 1A The domain is )1 ,(−∞ 1A OR OR

)1(log)( where,)1 ,(: 11 xxfRf e −−=→−∞ −− 2A b. (0, 0) 1A

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2011 MAV MATHS METHODS CAS EXAM 1 - SOLUTIONS

© The Mathematical Association of Victoria, 2011 Page 3

Question 6 ( ) 213)()( 33 +=−+== xxxfgxh 1A

⎪⎩

⎪⎨⎧

−<−−−≥+=

33

33

2 ,22 ,2)(

xxxxxh 1A

⎪⎩

⎪⎨⎧

−<−−>=′

32

32

2 ,32 ,3)(

xxxxxh 1A

Question 7

a. ⎟⎟⎠

⎞⎜⎜⎝

⎛⎥⎦

⎤⎢⎣

⎡−

+⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−=⎥

⎤⎢⎣

⎡′′

11

3002

yx

yx

)1(2 +−=′ xx , )1(3 −=′ yy 1M

12−′

−= xx , 13+′

= yy

Substitute into 112 −+

=x

y

1413

−′

−=+′

xy 1H

612 −′

−=′x

y

The equation of the image is 612 −−=x

y 1A

b. Translation one unit to the right 1A Reflection in the y-axis 1A Dilation by a factor of six from the x-axis 1A OR OR There are other solutions such as Reflect in the x-axis 1A Dilate by a factor of 6 from the x-axis 1A Translate by 1 unit in the positive x direction 1A both translations correct Translate by 12 units in the negative y-direction

The order must be correct Question 8 a. Using sum of probabilities is 1 1M

96.04101.003.03

=+=++++

qpqpp

Since

16.096.06

96.0242

==

=+⇒=

pp

pppq 1A

b.

Pr X < 2 | X < 3( ) = Pr X < 2! X < 3( )Pr X < 3( )

=Pr X < 2( )Pr X < 3( ) =

0.640.96

= 23

1A

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2011 MAV MATHS METHODS CAS EXAM 1 - SOLUTIONS

© The Mathematical Association of Victoria, 2011 Page 4

Question 9 a.

k2

cos !3t!

"#$%& +1

!"#

$%&dt

0

3

' =1

k23!sin !

3t!

"#$%& + t

()*

+,-0

3

=1

k2

3!sin !( )+ 3(

)*+,-. 3

!sin 0( )+ 0(

)*+,-

/01

234=1

k25 3 =1 6 k = 2

3

b.

Pr T > 2( ) = 13

f t( ) dt2

3

!

= 13

3!sin !( )+ 3"

#$%&'( 3

!sin 2!

3)*+

,-. + 2

"#$

%&'

/01

234

= 133( 3

!5 32

( 2/016

2346

= 131( 3 3

2!)*+

,-.

OR 13( 32!

Question 10

( ) 0)2(cos1 3 =+ xdxd

Using the chain rule 1M

( ) 0)2(cos12

)2sin()2(cos63

2

=+

x

xx

The denominator cannot equal zero Hence solve 0)2sin()2(cos2 =xx 1M

0)2cos( =x or 0)2sin( =x 1M

...23,

22 ππ=x or ...2,,02 ππ=x

ππππ= ,43,

2,4,0x but

2π≠x because 1)2(cos3 −≠x

πππ= ,43,

4,0x 1A

END OF SOLUTIONS

1M

1A

1A

1A