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JEE Mathematics Solved Question Paper 2013

Multiple Choice Questions

Each of a and b can take values 1 or 2 with equal probability. The probability that the equation ax² + bx + 1= 0 has real roots, is equal to

$\frac{1}{2}$
$\frac{1}{4}$
$\frac{1}{8}$
$\frac{1}{16}$

Cards are drawn one-by-one without replacement from a well shuffled pack of 52 cards. Then, the probability that a face card (jack, queen or king) will appear for the first time on the third turn is equal to

$\frac{300}{2197}$
$\frac{36}{85}$
$\frac{12}{85}$
$\frac{4}{51}$

There are two coins, one unbiased with probaility $\frac{1}{2}$ or getting heads and the other one is biased with probability $\frac{3}{4}$ of getting heads. A coin is selected at random and tossed. It shows heads up. Then, the probability that the unbiased coin was selected is

$\frac{2}{3}$
$\frac{3}{5}$
$\frac{1}{2}$
$\frac{2}{5}$

Lines x + y = 1 and 3y = x + 3 intersect the ellipse x² + 9y² = 9 at the points P,Q and R. The area of the $∆$ PQR is

$\frac{36}{5}$
$\frac{18}{5}$
$\frac{9}{5}$
$\frac{1}{5}$

5.
For the variable , the locus of the point of intersection of the lines 3tx - 2y + 6t = 0 and 3x + 2ty - 6 = 0 is

$the ellipse \frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$

$the ellipse \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$

$the hyperbola \frac{x^{2}}{4} - \frac{y^{2}}{9} = 1$

$the hyperbola \frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$

$the ellipse \frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$

Given equation of lines are

3tx - 2y + 6t = 0 ...(i)

and 3x + 2ty - 6 = 0 ...(ii)

On multiplying Eq. (i) by t and then adding in Eq. (ii), we get

(3t² + 3)x + 6t² - 6 = 0

$\Rightarrow x = \frac{2 (1 - t^{2})}{(1 + t^{2})} \Rightarrow x + {xt}^{2} = 2 - 2 t^{2} \Rightarrow (x + 2) t^{2} = (2 - x) \Rightarrow t^{2} = \frac{2 - x}{2 + x} . . . (iii)$

On multiplying Eq. (ii) by t and then subtract from Eq. (i), we get

$(- 2 - 2 t^{2}) y + 6 t + 6 t = 0 \Rightarrow 12 t = 2 (1 + t^{2}) y On squaring both sides, we get 144 t^{2} = 4 y^{2} {(1 + t^{2})}^{2} \Rightarrow 144 (\frac{2 - x}{2 + x}) = 4 y^{2} {(1 + \frac{2 - x}{2 + x})}^{2} [∵ from Eq . (iii)] \Rightarrow 36 (\frac{2 - x}{2 + x}) = y^{2} {(\frac{4}{2 + x})}^{2} \Rightarrow 36 \frac{2 - x}{2 + x} = \frac{16 y^{2}}{{(2 + x)}^{2}} \Rightarrow 36 (4 - x^{2}) = 16 y^{2} \Rightarrow 9 (4 - x^{2}) = 4 y^{2} \Rightarrow 36 - 9 x^{2} = 4 y^{2} \Rightarrow 9 x^{2} + 4 y^{2} = 36$

$∴ \frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$ , which represents an ellipse.

The locus of the mid-points of the chords of an ellipse x² + 4y² = 4 that are drawn from the positive end of the minor axis, is

a circle with centre $(\frac{1}{2}, 0)$ and radius 1
a parabola with focus $(\frac{1}{2}, 0)$ and directrix x = - 1
an ellipse with centre $(0, \frac{1}{2})$ , major axis $\frac{1}{2}$ and minor axis
a hyperbola with centre $(0, \frac{1}{2})$ , transverse axis 1 and conjugate axis $\frac{1}{2}$

A point P lies on the circle x² + y² = 169. If Q = (5, 12) and R = (-12, 5) then the $∠$ QPR is

$\frac{π}{6}$
$\frac{π}{4}$
$\frac{π}{3}$
$\frac{π}{2}$

A point moves, so that the sum of squares of its distance from the points (1, 2) and (- 2, 1) is always 6. Then, its locus is

the straight line $y - \frac{3}{2} = - 3 (x + \frac{1}{2})$
a circle with centre $(- \frac{1}{2}, \frac{3}{2})$ and radius $\frac{1}{\sqrt{2}}$
a parabola with focus (1, 2) and directrix passing through (- 2, 1)
an ellipse with foci (1, 2) and (- 2, 1)