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JEE Mathematics Solved Question Paper 2013

Multiple Choice Questions

61.

Two numbers are chosen at random from{1, 2, 3, 4, 5, 6, 7, 8} at a time. The probability that smaller of the two numbers is less than 4 is

$\frac{7}{14}$
$\frac{8}{14}$
$\frac{9}{14}$
$\frac{10}{14}$

62.

Two fair dice are rolled. The probability of the sum of digits on their faces to be greater than or equal to 10 is

$\frac{1}{5}$
$\frac{1}{4}$
$\frac{1}{8}$
$\frac{1}{6}$

63.

A bag contains 2n + 1 corns. It is known that n of these coins have a head on both sides, whereas the remaining n + 1 coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $\frac{31}{42}$ , then n is equal to

64.

The random variable takes the values 1, 2, 3, 1 ..., m. If P(X = n) = $\frac{1}{m}$ to each n, then the variance of X is

$\frac{(m + 1) (2 m + 1)}{6}$
$\frac{m^{2} - 1}{12}$
$\frac{m + 1}{2}$
$\frac{m^{2} + 1}{12}$

65.

$If X is a poisson variate P (X = 1) = 2 P (X = 2), then P (X = 3) = ?$

$\frac{e^{- 1}}{6}$
$\frac{e^{- 2}}{2}$
$\frac{e^{- 1}}{2}$
$\frac{e^{- 1}}{3}$

66.

The direction ratios of the two lines AB and AC are 1, - 1, - 1 and 2, - 1, 1. The direction ratios of the normal to the plane ABC are

2, 3, -1
2, 2, 1
3, 2, - 1
- 1, 2, 3

67.

A plane passing through(- 1, 2, 3) and whose normal makes equal angles with the coordinate axes is

x + y + z + 4 = 0
x - y + z + 4 = 0
x + y + z - 4 = 0
x + y + z = 0

68.

A variable plane passes through a fixed point (1, 2, 3). Then, the foot of the perpendicular from the origin to the plane lies on

a circle
a sphere
an ellipse
a parabola

69.

$If \cos^{- 1} (\frac{y}{b}) = 2 \log (\frac{x}{2}), where x > 0, then x^{2} \frac{d^{2} y}{{dx}^{2}} + x \frac{dy}{dx} = ?$

4y
- 4y
0
- 8y

70.
If the curves x² + py²= l and qx² + y² = l are orthogonal to each other, then

p - q = 2

$\frac{1}{p} - \frac{1}{q} = 2$

$\frac{1}{p} + \frac{1}{q} = - 2$

$\frac{1}{p} + \frac{1}{q} = 2$

$\frac{1}{p} + \frac{1}{q} = 2$

$Given curves are x^{2} + {py}^{2} = 1 . . . (i) and {qx}^{2} + y^{2} = 1 . . . (ii) On differentiating Eq (i), w r t, x we get 2 x + 2 yp \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = m_{1} = - \frac{x}{py} On differentiating Eq (ii), w r t, x we get 2 qx + 2 y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = m_{2} = \frac{- qx}{y} Since, both the curves are orthogonal to each other, Then, m_{1} m_{2} = - 1 \Rightarrow \frac{- x}{py} . \frac{- qx}{y} = - 1 \Rightarrow {qx}^{2} = - {py}^{2} (iii) \Rightarrow q (1 - {pqy}^{2}) = - {py}^{2} [from eq (i)] \Rightarrow q - {pqy}^{2} = - {py}^{2} \Rightarrow q = (pq - p) = y^{2} ∴ y^{2} = \frac{q}{pq - p} On putting x^{2} and y^{2} in eq (ii) we get - \frac{pq}{pq - p} + \frac{q}{pq - p} = 1 \Rightarrow - pq + q = pq - p \Rightarrow p + q = 2 pq \Rightarrow \frac{1}{p} + \frac{1}{q} = 2$