Question 4
(a)
2 log7 x + 3 = 2 log x 7 ⇒ 2 log7 x + 3 =
⇒
2
log7 x
2(log7 x)2 + 3 log7 x − 2 = 0
⇒ (2 log7 x − 1)(log7 x + 2) = 0
⇒ log7 x =
log7 x =
or
log7 x = −2
√
1
⇒ x = 71/2 = 7 2
log7 x = − 2 ⇒ x = 7 −2 =
(b)
1
2
1
49
!
!
!
1 1
1 1
+
+
=0
α β
α β
!
1
α+β
2
+
=0
x −
αβ
αβ
Correct equation is: x 2 −
⇒
Student’s equation
x2
!
1
1
−
+
= 0 is correct IF:
α+β
αβ
α+β
1
=
αβ
α+β
⇒ (α + β)2 = αβ
⇒ α 2 + αβ + β 2 = 0
[1]
Using ‘completing the square’ method:
1
1
α 2 + αβ + β 2 = α 2 + αβ + β 2 + β 2 − β 2
4
4
!2
3
1
= α + β + β 2 > 0 for all α, β ∈ <
2
4
But from [ 1 ]: α 2 + αβ + β 2 = 0
∴ There exist NO real values of α and β which will give the correct answer. 5
AM 27 / MAY 2014 / 1
Question 2
n1
 −→



OA







−→



OB



Let 


−→



OC







−→


 OD
C
A
Π1
B
(a)
= a = 4i + 2j + 3k
= b = 9i + 2j + 2k
= c = 6i − j + 2k
= d = 5i − 3j + 14k
Let n1 denote vector normal to plane Π1 .
−→
−→
AB × AC = (b − a) × (c − a)
= (5i − k) × (2i − 3j − k)
i
= 5
2
j
0
−3
k − 1 −1
= i(0 − 3) − j(−5 + 2) + k(−15 − 0)
= −3i + 3j − 15k
= −3(i − j + 5k)
∴ n1 = i − j + 5k
Vector equation of plane Π1 containing points A, B and C is:
.
.
.
.
⇒ r.(i − j + 5k) = 4 − 2 + 15
⇒ r.(i − j + 5k) = 17 or x − y + 5z = 17 r n1 = a n1 ⇒ r (i − j + 5k) = (4i + 2j + 3k) (i − j + 5k)
(b)
Ax + By + Cz + D Using d = √
:
A2 + B2 + C 2 5(1) + (−3)(−1) + 14(5) − 17 d =
√
1 + 1 + 25
5 + 3 + 70 − 17 =
√
27
D(5, −3, 14)
61
√
3 3
√
61 3
=
units 9
d
=
Π¹ : x – y + 5z – 17 = 0
17
AM 27 / MAY 2014 / 2
Question 6
ln 2
Z
(b)
Given:
(e x + e − x )n dx
0
3 5
To prove: nIn = 4(n − 1)In−2 +
2 2
Proof:
! n−1
Using result in (a):
ln 2
Z
n
(e x + e − x )n dx = 4(n − 1)
ln 2
Z
0
(e x + e − x )n−2 dx
0
h
i ln 2
+ (e x − e − x )(e x + e − x )n−1
0
1
⇒ nIn = 4(n − 1)In−2 + 2 −
2
⇒ nIn = 4(n − 1)In−2 +
(c)
Volume generated = π
3 5
2 2
ln 2
Z
y 2 dx
0
=π
ln 2
Z
(e x + e − x )4 dx
0
= πI4
Using reduction formula in (b) with:
3 5
8 2
!3
3 5
n = 2 −→ I2 = 2 × 1 × I0 +
4 2
!1
n = 4 −→ I4 = 1 × 3 × I2 +
where I0 =
ln 2
Z
0
2
dx = x ln
0
= ln 2
∴ Volume generated = πI4
!
!#
"
3 125
15
+
= π 3 2 ln 2 +
8
8 8
= 49.145 cubic units AM 27 / MAY 2014 / 2
28
!
1
2+
2
! n−1
! n−1
Question 3
(a)

 a
P2 = 
b
P =
2
(P2 )T

0 


c
 
0   a 2
 = 
 
c
ab + bc

 a

b


⇒ 
a2
ab + bc

0 


c2
 
0   a2
 = 
 
c2
0

ab + bc 


c2
Equating corresponding elements:
ab + bc = 0 ⇒ b(a + c) = 0
a+c = 0
⇒
(b)
(since bc , 0)
∴ |P| , 0 

y = 2x has coordinates of the form (k , 2k)



Any point on the line 


 y = x has coordinates of the form (K , K)
So, under transformation matrix P:
   

 k   K 
 a
P   =   ⇒ 
2k
K
b


⇒ 

0 


c
   
 k   K 
  =  
   
2k
K
  
  K 
 =  
  
bk + 2ck
K
ak


ak = K



⇒ 


 (b + 2c)k = K
⇒ a = b + 2c
From (a): c = −a ⇒ b = 3a
∴ a ∈ <, a , 0, b = 3a, c = − a AM 27 / SEP 2014 / 1
(inadmissible, since bc , 0)
0 = ac
c
= − c2 < 0
(c)
b=0
a = −c ⇒
a
|P| = b
or
40