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Again, c 2 a 2 b 2
= + + 2a⋅b
− −
⇒ ⋅ 48 144 48
a b=
=−72
2
| a× b+ c × a|
= | a × b + a × b | = 2 | a × b
|
2 2 2
= 2 ab −( a ⋅b
) = 2
2 12 ⋅ 48 − − 2
( 72)
= 48 3
44. (a, d) : Let centroid of the triangle PTN is (α, β)
at 2 + ( − at 2 ) + 2a+
at 2 2at
⇒ α=
and β =
3
3
Eliminating ‘t’ we get,
⎡ 2
9β
⎤
3α
= a ⎢ + 2
2
⎥
⎣⎢
4a
⎦⎥
The locus of (α, β) is
2
9y
2 4a ⎛ 2a⎞
3x
= + 2a
⇒ y = x −
4a
3 ⎝
⎜ ⎟
3 ⎠
∴
45. (a, d)
⎛ 2a
⎞
vertex
⎝
⎜ ,
⎠
⎟ focus
3 0 , ( a, 0 )
46. (a, b) : sinx + cos x = 1+
sin2x
So, 1< sinx + cos x ≤ 2.
y
y = [|sinx| + |cosx|] = 1. (–3, 1)
(3, 1) y = 1
x 2 + y 2 = 10
x
O
⇒ 2x+ 2y dy = 0
dx
So, angle is either tan –1 (–3) or tan –1 (3).
47. (a, c) : x 2 + y 2 – 10x + 21 = 0
x 2 – 10x + (y 2 + 21) = 0
It has real roots if D 0 100 – 4(y 2 + 21) 0
y 2 + 21 25 y 2 4 –2 y 2
Also, y 2 + (x 2 – 10x + 21) = 0 will have real roots if
D 0 0 – 4(x 2 – 10x + 21) 0 (x – 3)(x – 7) 0
3 x 7
48. (b): We have, lim| | [cos x
x
]
...(i)
x →0
When x 0, then 0 cos x 1 [cos x] = 0 when
x 0
0
From (i), we have lim| x | = lim 1 = 1
x→0
x→0 49. (a, c) : PA ( ∪B)
≥ 3 4 and 1 8
Let P(A) + P(B) be x.
3
x−P( A∩B)
≥
4
≤PA
( ∩B)
≤
3
8
3
1
⇒ x− ≥P( A∩B) ≥ ⇒ x ≥ 7 P(A ∪ B) ≤ 1
4
8 8
⇒ x – P(A ∩ B) ≤ 1
3 11
⇒ x−1≤P( A∩B)
≤ ⇒ x ≤
8 8
50. (a, b, d)
51. (c) : Area of ∆PQR = 1 × 4 2× 8= 16 2 sq units
2
Area of ∆PQS = 1 ×
2 2 × 4 2 = 4 2 sq units
ar ∆ PQS
ar ∆PQR = 1 4
52. (b): Equation of perpendicular bisector of SR is x = 4
...(i)
Equation of perpendicular bisector of PS is
y− = − 1
2 ( x−
0) or 2y+ x= 2 ...(ii)
2
Circumcentre is point of intersection of (i) and (ii),
x= 4, y=−
2 ∴ C( 4, − 2)
2 2
∴ radius = PC = () 3 + ( 3 2) = 3 3 units
53. (b): P(X ≥ 3) = 1 – P(X ≤ 2)
= 1 – {P(X = 1) + P(X = 2)}
= 1 – {P(6) + P(6′ 6)} = 1 – P(6) – P(6′)P(6)
1 5 1
− −
= 1 − − × = 1 − 1 5 36 6 5 25
− = =
6 6 6 6 36 36 36
54. (d): Required probability
⎛
P
X ≥ 6⎞
PX
PX
⎝
⎜
X > ⎠
⎟ = ( ≥ 6)
1
= − ( ≤ 5)
3 1−PX
( ≤3)
1−PX
( ≤3)
1
= − { PX ( = 1) + PX ( = 2) + ..... + PX ( = 5)}
1− { PX ( = 1) + PX ( = 2) + PX ( = 3)}
⎧
− + ⋅ + ⎛ ⎝ ⎜ ⎞ 2
⎠ ⎟ ⋅ + + ⎛ ⎝ ⎜ ⎞ 4
⎪1
5 1 5 1 5 ⎫
⎨
⎠ ⎟ 1⎪
1
... ⋅ ⎬
6 6 6 6 6 6 6
=
⎩⎪
⎭⎪
⎧1
5
1 − + ⋅ + ⎛ 6 6 ⎝ ⎜ ⎞ 2
⎪ 1 5
⎨
⎠ ⎟ ⋅ 1⎫⎪
⎬
6 6 6
⎩⎪
⎭⎪
⎧
5
1 1−
( 5/ 6)
⎫
1−
⎨ ⋅ ⎬
−
=
⎩6
1 ( 5/ 6)
⎭
= ⎛ 5
⎧
3 ⎜ ⎞ 2
1 1−
( 5/ 6)
⎫ ⎝ 6⎠ ⎟ 25
=
36
1−
⎨ ⋅ ⎬⎭
⎩ 6 1−
( 5/ 6)
88 CHEMISTRY TODAY | APRIL ‘17