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