Chemistry_Today_April_2017_vk_com_stopthepress

Given that, at equilibrium,
⎛ n
⎜
⎝ n
A
B
⎞ nB
⎟ = ⎛ ⎞
⎠ ⎝ ⎜ n ⎠
⎟
eq.
A
initial
2V
+
4
RT
x V
=
RT
4V
−
2
RT x V
RT
x = 2 V
RT
( V
)
8
4
K
RT
c = 8
= 2
8 = 2
y
2V
( RT )
y = 8
*
36. (8) :
HC
H * *
HC 3
* *
* *
HO *
H
Cholesterol
3 H
37. (b): x + y + z = 0, x ≥ – 5, y ≥ – 5, z ≥ – 5
Let x = α – 5 , α ≥ 0, y = β – 5, β ≥ 0, z = γ – 5, γ ≥ 0
Now, (α – 5 + β – 5 + γ – 5) = 0 α + β + γ = 15
No. of integral solution = 15 + 3 – 1 C 3 – 1 = 17 C 2 = 136.
38. (c) : Let S = smoker, S′ = Non-smoker, D = death by
cancer
Using conditional probability, we can write
P(D) = P(S) P(D | S) or P(S′) P(D | S′)
20
80
1 4 x
0. 006 = ⋅ PDS ( | ) + ⋅ P( DS | ′)
= ⋅ x + ⋅
100 100
5 5 10
[Let P(D | S) = x and given P(D | S) = 10·P(D | S′)]
⇒ x = 3
140
⎡d1
0 0 0 ⎤
⎢
d
39. (d): Let A = ⎢
0
⎥
2 0 0
⎥
⎢0 0 d3
0 ⎥
⎢
⎥
⎣⎢
0 0 0 d4
⎦⎥
⎡d
1 2 0 0 0 ⎤
⎢
⎥
⎢
2
0 d2 2 0 0 ⎥
A = A.
A = ⎢
0 0 d3 2 ⎥
⎢
0 ⎥
⎢
⎣⎢
0 0 0 d4 2 ⎥
⎦⎥
Given, A 2 = A
2
∴ d i = d i (i = 1, 2, 3, 4) or d i (d i – 1) = 0
⇒ d i = 0 or 1 for i = 1, 2, 3, 4
∴ Each diagonal elements can be chosen in 2 ways
(either 0 or 1). As there are 4 diagonal elements.
∴ No. of ways = 2 × 2 × 2 × 2 = 16
∴ No. of non-zero diagonal matrices
= 16 – 1 = 15 [ One of them is zero matrix]
40. (a) : sin x + cosec x = 2 (given) ...(i)
Squaring both sides, we get
sin 2 x + cosec 2 x + 2 = 4 or sin 2 x + cosec 2 x = 2
∴ For n = 2, sin n x + cosec n x = 2
On cubing, equation (i) gives
sin 3 x + cosec 3 x + 3(2) = 8
or sin 3 x + cosec 3 x = 8 – 6 = 2
∴ For n = 3, sin n x + cosec n x = 2
For n = 4, (sinx + cosecx) 4 = 16
(sin 2 x + cosec 2 x + 2) 2 = 16
sin 4 x + cosec 4 x + 4 + 2 + 4(2) = 16
sin 4 x + cosec 4 x = 16 – 14 = 2
Proceeding in the same way, we find that
sin n x + cosec n x = 2 ∀ n ∈ N.
41. (b): Since, g(x) g(y) = g(x) g(y) + g(xy) – 2
Now, at x = 0, y = 2, we get g(0) g(2) = g(0) g(2)
g(0) – 2
5g(0) = 5 2g(0) –2 3g(0) = 3 g(0) = 1
g(x) is given in a polynomial and by the given
relation g(x) can not be linear.
Let g(x) = x 2 k g(x) = x 2 1 [g(0) = 1]
(x 2 1) (y 2 1) = x 2 1 y 2 1 x 2 y 2 1 – 2
lim g(x) = g(3) = 3 2 1 = 10
x→3
42. (b, d) : We have, 4x 2 + 9y 2 = 1 ...(i) & 8x = 9y ...(ii)
Differentiating (i) w.r.t. x, we get
4
8x
18y dy dy x
+ = 0 ⇒ =−
dx dx 9y
⇒ slope of tangent = −4 x
9y .
Also, slope of line (ii) = 8 9
Since line (ii) is parallel to the tangent.
−4x
8
= ⇒ x = –2y
9y
9
2 2 2 1 1
From (i), 44 ( y ) + 9y = 1⇒ y = ⇒ y=±
25 5
1 2
1 2
When y= , x=− ; when y=− , x=
5 5
5 5
∴ Points are ⎛ 2 1⎞
⎛ 2 1⎞
−
⎝
⎜ , ⎟ and −
⎠
⎜ ,
⎝ ⎠
⎟
5 5 5 5
82 CHEMISTRY TODAY | APRIL ‘17