IIT-JEE 2011 - Career Point

Sol. y = m(x – 4) ± 4
y = mx ± 9m
2 − 4
– 4m ± 4
2
m
2
1+ m
1+ = ± 9m
2 − 4
16m 2 + 16 + 16m 2 m 32 m
2
2
1+ m = 9m 2 – 4
m 32m 1+ m = – 23m 2 – 20
1024m 2 + 1024 m 4 = 529m 4 + 400 + 920 m 2
495 m 4 + 104 m 2 – 400 = 0
(5m 2 – 4) (99m 2 + 100) = 0
∴ m 2 4
2
= ∴ m = ± 5 5
So tangent with positive slope
y =
2 4 x ±
5 5
2x – 5 y ± 4 = 0
43. Equation of the circle with AB as its diameter is
(A) x 2 + y 2 – 12 x + 24 = 0
(B) x 2 + y 2 + 12 x + 24 = 0
(C) x 2 + y 2 + 24 x – 12 = 0
(D) x 2 + y 2 – 24x – 12 = 0
Ans.[A]
Sol. x 2 + y 2 – 8x = 0
4x 2 – 9y 2 = 36
⎛ ⎞
x 2 + ⎜
4x
2 − 36
⎟ – 8x = 0
⎝ 9 ⎠
13x 2 – 72x – 36 = 0
(x – 6) (13x + 6) = 0
−6
x = 6, 13
x = 6, y = ± 12
∴ Equation of required circle
(x – 6) 2 + (y – 12 ) (y + 12 ) = 0
x 2 + y 2 – 12x + 24 = 0
Paragraph for Question No. 44 to 46
Let p be an odd prime number and T p be the
following set of 2 × 2 matrices :
⎪⎧
⎡a
b⎤
⎪⎫
= ⎨A
= ⎢ ⎥ : a, b,c ∈ { 0,1,2,......, p −1} ⎬
⎪⎩ ⎣c
a⎦
⎪⎭
T p
44. The number of A in T p such that A is either
symmetric or skew-symmetric or both, and det
(A) divisible by p is -
(A) (p – 1) 2 (B) 2(p – 1)
(C) (p –1) 2 + 1 (D) 2p –1
Ans.[D]
Sol.
|A | = a 2 – bc
if A is symmetric b = c
then | A | = (a + b) (a – b)
So a & b can attain 2(p – 1) solution
It A is skew symmetric then a = b = c = 0
So total no. of solution = 2p – 2 + 1 = 2p – 1
45. The number of A in T p such that the trace of A is
not divisible by p but det (A) is divisible by p is
[Note : The trace of a matrix is the sum of its
diagonal entries.]
(A) (p – 1) (p 2 – p + 1) (B) p 3 – (p – 1) 2
(C) (p –1) 2 (D) (p –1) (p 2 – 2)
Ans.[C]
46. The number of A in T p such that det (A) is not
divisible by p is -
(A) 2p 2
(B) p 3 – 5p
(C) p 3 –3p (D) p 3 – p 2
Ans. [D]
SECTION – IV
Integer type
This section contains TEN paragraphs. The answer to
each question is a single-digit integer, ranging from 0 to
9. The correct digit below the question number in the
ORS is to be bubbled.
47. Let f be a real-valued differentiable function on R
(the set of all real numbers) such that f(1) = 1. If
the y-intercept of the tangent at any point P(x, y)
on the curve y = f(x) is equal to the cube of the
abscissa of P, then the value of f(–3) is equal to –
Ans.[9]
dy
Sol. Y – y = ( X – x) dx
dy
y – x = x
3
dx
dy
x – y = – x
3
dx
T y – = – x
2
T x
1
2
I.F. = x
1
y =
x ∫ − xdx + C
y − x 2
= + C
x 2
1
3
1 = − + C ∴ C =
2
2
∴ y =
x( −x
2
2 +
3)
XtraEdge for IIT-JEE 74
MAY 2010