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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(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)
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
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
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
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