1. Xtra Edge February 2012 - Career Point

7. f(A) = A 2 – 5A + 7 I2
⎡ 3 1⎤
⎡ 3 1⎤
⎡ 3 1⎤
⎡1
0⎤
= ⎢ ⎥ .
⎣−1
2
⎢ ⎥ – 5
⎦ ⎣−1
2
⎢ ⎥ + 7.
⎦ ⎣−1
2
⎢ ⎥
⎦ ⎣0
1⎦
⎡ 8 5⎤
⎡15
5 ⎤ ⎡7
0⎤
= ⎢ ⎥ –
⎣−
5 3
⎢ ⎥ +
⎦ ⎣−
5 10
⎢ ⎥
⎦ ⎣0
7⎦
⎡0
0⎤
= ⎢ ⎥ = Null Matrix.
⎣0
0⎦
8. Area of ∆ = 0
1
⇒
2
k
− k + 1
− 4 − k
2 − 2k
2k
6 − 2k
1
1
1
= 0
k . [2k – 6 + 2k] – (2 – 2k) [–k + 1 + 4 + k]
+ 1 [(–k + 1) (6 – 2k) –2k (–4 – k)] = 0
k [4k – 6] – (2 – 2k) (5) + [2k 2 – 8k + 6 + 8k + 2k 2 ] = 0
4k 2 – 6k – 10 + 10k + 4k 2 + 6 = 0
8k 2 + 4k – 4 = 0
2k 2 + k – 1 = 0
(2k – 1) (k + 1) = 0 ⇒ k = 1/2, – 1
9. y = f (e x )
y' = f '(e x ). e x
x
x
y " = f ' ( e ). e + e . f " ( e ). e
10. f (g(x)) = g ( x)
= x 1
x
2 −
g(f (x)) = f (x) 2 – 1 = ( x ) 1 = x – 1
x
2 −
Section B
11. A : 3 cards have the same number
⎛ 4 ⎞
n(A) = 13 C(4, 3) = 13 ⎜ ⎟
⎜ ⎟
= 13 (4) = 52
⎝ 3 1⎠
52
n(S) = C(52, 3) = = 22100
3 49
Required probability
n(
A)
52 13 1
= P(A) = = = =
n(
S)
22100 5525 425
OR
Let E : Candidate Reaches late
A1 = Candidate travels by bus
A2 : Candidate travels by scooter
A3 : Candidate travels by other modes of
transport
x
3 1 3
P(A1) = , P(A2) = , P(A3) =
10 10 5
1 1
P(E/A1) = , P(E/A2) = , P(E/A3) = 0
4
3
∴ By Baye's Theorem
P(A1/E) =
) P(
E / A ) +
) P(
E / A1)
) P(
E / A ) +
) P(
E / A
XtraEdge for IIT-JEE 83 FEBRUARY 2012
=
P(
A
1
3 1
×
10 4
3 1
+ + 0
40 30
→
1
9
=
13
P(
A
P(
A
1
2
2
P(
A
12. Here, A = 3 iˆ
+ 2 ˆj
+ 9kˆ
; B = iˆ
+ λˆj
+ 3kˆ
+ → →
⇒ a b = ( 3iˆ
+ 2 ˆj
+ 9kˆ
) + ( iˆ
+ ˆj
+ kˆ
)
= 4 iˆ
+ ( 2 + λ)
ˆj
+ 12 kˆ
− → →
a b = ( 3iˆ
+ 2 ˆj
+ 9kˆ
) − ( iˆ
+ λˆj
+ 3kˆ
)
= 2 iˆ
+ ( 2 − λ)
ˆj
+ 6kˆ
⎡ → →⎤
⎡→ →⎤
Since ⎢a
+ b ⎥ ⊥ ⎢a
− b ⎥ we have
⎣ ⎦ ⎣ ⎦
→
⎡ → →⎤
⎡→ →⎤
⎢a
+ b ⎥ ⋅ ⎢a
− b ⎥ = 0
⎣ ⎦ ⎣ ⎦
⇒ [ 4i
ˆ + ( 2 + λ)
ˆj
+ 12kˆ
] [ 2iˆ
+ ( 2 − λ)
ˆj
+ 6kˆ
] = 0
⇒ 4 × + (2 + λ) × (2 – λ) + 12 × 6 = 0
⇒ 8 + 4 – λ 2 + 72 = 0
⇒ λ 2 = 84 ⇒ λ = ± 2 21
13. Equation of plane passing through the inter sections
of planes
x + 2y + 32 – 4 = 0 and 2x + 4 – z + 5 = 0
(x + 2y + 32 – 4) + λ (2x y – z + 5) = 0 …(i)
x + 2y + 32 – 4 + 2λx + λy –λz + 5λ = 0
(1 + 2λ) x + (2 + λ) y + (3 –λ) z – 4 + 5λ = 0
Since the plane (i) is perpendicular to
5x + 3y + 6z + 8 = 0
∴ (1 + 2λ) ⋅ 5+ (2 + λ) ⋅ 3 + ( 3 – λ) ⋅ 6 = 0
5 + 10λ + 6 + 3λ + 18 – 6 λ = 0
7λ + 29 = 0 ⇒
−29
λ =
7
∴ required equation of plane is
−29
(x + 2y + 3z – 4) (2x + y – z + 5) = 0
7
7x + 14 y + 21z – 28 – 54x – 29y + 29z – 145 = 0
– 47 x – 15y + 50z –173 = 0
47x + 15y – 50z + 173 = 0
3
3
)