2015 Mathematical Methods (CAS) Trial Exam 1 Solutions
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Q3a
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Q1a Let 2(x − a )2 + b = − x 2 , expand and collect like terms,
(
)
3 x 2 − 4ax + 2a 2 + b = 0
To have one point of contact, ∆ = 0 , .: 2a 2 + 3b = 0
Pick a value for a , say a = 3 , then b = −6
2
Q1b b = − a 2
3
Q3b
Q3c
1
1
1
+1 → x =
+1 → x + 2 =
+1
x +1
y +1
y +1
1
→ x+2=
+ 1 , simplify and write y as the subject of
y − 2 +1
1
the equation, y =
+1
x +1
Q2a y =
1
1
+ 1 ≥ 0 , x ≠ −1 and
≥ −1
x +1
x +1
If x + 1 > 0 , x > −1 and 1 ≥ − x − 1 , i.e. x ≥ −2 , .: x > −1
If x + 1 < 0 , x < −1 and 1 ≤ − x − 1 , i.e. x ≤ −2 , .: x ≤ −2
.: D is (− ∞, − 2] ∪ (− 1, ∞ )
Q2b
 5π 
Q2c  a,
 is a continuous interval,
 6 
.: the range of g is also a continuous interval
h o g is defined if the range of g ⊆ D
.: the range of g ⊆ (− 1, ∞ )
.: 2 sin a = −1 , a = −
π
6
2015 Mathematical Methods (CAS) Trial Exam 1 Solutions
Q4a Let 2 x + 2 − 2 = x − 1 ,
and 2 x + 2 > 0 , i.e. x ≥ 0
(
) = ( x + 1) , 2x + 2 = x + 2
= (2 x ) , x + 2 x + 1 = 4 x , x
2x + 2
(x + 1)2
2 x + 2 = x + 1 where x ≥ 0
2
2
2
2
x + 1, x + 1 = 2 x
2
− 2 x + 1 = 0 , ( x − 1) = 0
2
.: x = 1 and y = 0 , the intersection is (1, 0) .
Q4b y = 2 x + 2 − 2 ,
dy
=
dx
1
2x + 2
dy
1
y = x − 1,
=
dx 2 x
The gradient of the common tangent is −
1
, .: 2 x + 2 = 4 x , x = 1 and y = 0
2 x
x y
.: the common tangent + = 1 is at (1, 0 ) and has a gradient
a b
b 1
of − =
a 2
1 0
1
.: + = 1 , a = 1 and b = −
a b
2
.: −
b
=
a
1
b
.
a
=
2x + 2
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Q5a 2 x − 1 > 0 and x + 1 > 0 , .: x >
1
1
and x > −1 , .: x >
2
2
1 
The domain is  , ∞  .
2 
x
Pr ( X = x )
Q5b As x → 0.5 + , the value of f (x ) → −∞ , .: x =
1
is an
2
= 0 , .:
1
5
32
2
10
32
3
10
32
4
5
32
13
, .: n = 2
16
1
5 10 1
Pr ( X ≤ 2 ) =
+
+
=
32 32 32 2
Q5c Let 2 log10 (2 x − 1) − log10 (x + 1) = 0 .
(2 x − 1)2
0
1
32
1
2
Pr ( X ≥ n ) =
asymptote of y = f (x ) . It is the only one.
(2 x − 1)2
=1
x +1
x +1
Expand and simplify to 4 x 2 − 5 x = 0 , x(4 x − 5) = 0
1
5
Since x > , .: x = and y = 0 .
2
4
5 
The only x-intercept is  , 0  .
4 
.: log10
Q9 Binomial distribution, N = 5 , p =
Q10a f (1) = k +
1
1
1
, f (2 ) = , f (5) = 3k +
5
5
5
π 
π
Q6 sin 46° = sin (45° + 1° ) = sin +

 4 180 
π
π
π
1
π
1
1 
π 
≈ sin +
× cos =
+
×
=
1 +

4 180
4
2 180
2
2  180 
dy
= e x (cos x + sin x ) + e x (sin x − cos x ) = 2e x sin x
dx
Q7a
Area under graph =
π
dy
= 2e x sin x , .:
dx
Q7b
3
∫ 2e
x
[
]
sin x dx = e x (sin x − cos x )
π
3
10 k
0
0
2
π
3
1
.: ∫ e sin x dx = e x (sin x − cos x )
2
0
[
x
=
]
+
1
1 1 3
1 1
 k + +  + 3 k + +  = 1
2
5 5 2
5 5
4
1
=1, k =
5
25
π
3
Q10b By inspection of the graph, the median m ∈ [2, 5]
0
 1 π
1  π3  3 1  0
e
−
− e (0 − 1) = e 3
 4
2   2 2 

( 3 − 1)+ 12
Q8a Equation of the inverse: ( y − 1)2 + 1 = x , ( y − 1)2 = x − 1 ,
y = 1± x −1
Q8b It is the same area as the region bounded by y = (x − 1) + 1
2
and y = 2 . When y = 2 , 2 = ( x − 1)2 + 1 , x = 0, 2
2
( [
])
2
(
)
Area = ∫ 2 − ( x − 1) + 1 dx = ∫ 1 − ( x − 1) dx
2
0
2
0
2

(x − 1)3  = 4
= x −

3  0 3

1
(x − 2 ) + 1 = x + 3
25
5
25
m+3
.: f (m ) =
25
Area under the graph from x = m to x = 5 :
1m+3 8 
1
+ (5 − m ) = , m 2 + 6m − 30 = 0 and m > 0

2  25
25 
2
f (x ) =
.: m = −3 + 39
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2015 Mathematical Methods (CAS) Trial Exam 1 Solutions
© itute.com 2015
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