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43. (a, d)
44. (a, b) : Let A(x 1 , y 1 ), B(x 2 , y 2 )) 2
Since, cos P = 1 3
be the points of intersection.
On solving, x 2
⇒ 3[( 4λ) + ( 4λ−2) − ( 4λ+ 2)] = 2⋅4λ( 4λ
− 2)
= a(2x + 1)
⇒ x 2 – 2ax – a = 0
⇒ 3{16λ 2 – 32λ} = 8λ(4λ – 2) ⇒ 16λ 2 = 80λ
∴ x 1 + x 2 = 2a, x 1 x 2 = –a
∴ λ = 5
Now, AB = 40 ( x2 − x1) + ( y2 − y1)
The sides are 18, 20 and 22.
= 40
⇒ ( x2 x1) 2
1 1 1 1 0 0
− + {( 2 x2 − x1)}
= 40
⇒ 5{(x 2 – x 1 ) 2 } = 40 ⇒ (x 1 + x 2 ) 2 – 4x 1 x 2 = 8
49. (b, c) : D = α β γ = α β−α γ −α
⇒ 4a 2 + 4a = 8 ⇒ a 2 + a – 2 = 0 ⇒ a = 1, –2
2
α
2
β
2
γ
2
α
2 2
β −α 2 2
γ −α
45. (a, d) : dy
(C 2 → C 2 – C 1 , C 3 → C 3 – C 1 )
− y tanx=
2xsec
x
1 1
dx
= ( β−α)( γ − α) = ( β−α)( γ −α)( γ −β)
= (α – β)(β – γ)(γ – α)
It is a linear differential equation. β + α γ + α
I.F. = e –∫ tan x dx = e –In(sec x) D = 0 ⇒ trivial as well as non-trivial solution
= cos x
The solution is y cos x = 2x sec x cos x dx = x 2 and so the number of solutions will be infinite.
+ c
We have y(0) = 0 c = 0 y = x 2 D ≠ 0 ⇒ system has only trivial solution.
secx
y π π 2
2
π 2
⎛ ⎞
⎡
7 7
⎝
⎜ ⎟
4⎠
16
8 2
sec x tan x
1 ⎛ 1+ 3x+
1⎞
⎛ 1− 3x+
1⎞
⎤
50. (3) : ⎢⎜
( 3x
+ 1)
⎝ 2 ⎠
⎟ −
⎝
⎜ ⎟ ⎥
⎢
2 ⎠ ⎥
y ′ ⎛ 2 2
⎣
⎦
⎝ ⎜ π⎞
⎠
⎟ = ⋅ π ⋅ + π ⋅ = π + π
2 2
3 3 9 2 3 2
1 ⎡
7 7⎤
43 3 3
=
⎢( 1+ 3x+
1) −( 1− 3x + 1)
+ ⎣
⎦
⎥
...(i)
7
2 ( 3x
1)
46. (b, d) : In any series of (2n – 1) terms, the middle
7 7
term is t n . According to problem, t n of A.P., G.P. and Now, ( 1+ 3x+
1) −( 1− 3x+
1)
H.P. are a, b, c respectively. Hence, a, b, c are A.M.,
G.M. and H.M. respectively.
⎡ 7
= (
7 3
2⎣
C 1 3x+
1)+ C 3
( 3x+
1)
A.M. ≥ G.M. ≥ H.M. ⇒ a ≥ b ≥ c
Further, (G.M.) 2 7 5
= (A.M.) × (H.M.)
+ ( + )
7 7
C 5 3x 1 + C 7
( 3x+
1) ⎦
b 2 = ac ⇒ ac – b 2 = 0
2 3
= 2 3x+ 1× [ 7+ 35( 3x+ 1) + 21( 3x+ 1) + ( 3x+
1)]
47. (a, b) : By geometrical condition, line L is parallel Now, putting above value in (i), so the given
to the line of intersection of P 1 and P 2 .
A vector along L is ( i+ 2 ̂ j− k ) × ( 2 ̂ i− j+
k expression becomes
) 1
= ̂i−3̂j−
5k̂
2 6 [42 + 105x + 21(3x + 1)2 + (3x + 1) 3 ]
Any point on L is A(λ, –3λ, –5λ)
So, degree of given expression is 3.
The foot of perpendicular from A to plane P 1 is 51. (5) : A has rank 3
α− λ β+
3λ γ λ λ λ λ
= = + 5
=− − 6 + 5 +
∴
1 1
|A| = 0 ⇒ α = 5
=−
5x
4x
1 2 −1
1+ 4+
1 6
e − e
52. (1) : lim
The foot of perpendicular is
x→0
x
⎛ 1 1 1⎞
⎛
2 ⎞
λ− − λ− − λ +
⎝
⎜ , 3 , 5
⎠
⎟
⎛
2 ⎞
( 4x)
( 5x)
6 3 6
+ + + ∞
⎝
⎜
⎠
⎟ − ⎜1+ 4x
+
1 5x
...
2
⎟
2
P
⎝
⎜ + ... ∞⎠
⎟
48. (b, d) : Let PN = 2λ – 2,
= lim
2 l– 2
x→0
QL = 2λ and MR = 2λ + 2
M
x
N
So PQ = 4λ – 2,
2 ⎛ 25 16⎞
2l+ 2
x+ x −
⎝
⎜ ⎟ + ... ∞
⎠
QR = 4λ + 2, RP = 4λ
= lim
2 2
= 1
x→0
Q 2l L
R
x
–2
CHEMISTRY TODAY | APRIL ‘17 83