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JEE Mathematics Solved Question Paper 2008

Multiple Choice Questions

If $|\frac{z - 25}{z - 1}| = 5$ , find the value of $|z|$

Argument of the complex number $(\frac{- 1 - 3 i}{2 + i})$ is

45°
135°
225°
240°

The shortest distance between the straight lines through the points A₁ = (6, 2, 2) and A₂ = (- 4, 0, - 1) in the directions of (1, - 2, 2) and (3, - 2, - 2) is

The centre and radius of the sphere x² + y² + z² + 3x - 4z + 1 = 0 are

$(- \frac{3}{2}, 0, - 2); \frac{\sqrt{21}}{2}$
$(\frac{3}{2}, 0, 2); \sqrt{21}$
$(- \frac{3}{2}, 0, 2); \frac{\sqrt{21}}{2}$
$(- \frac{3}{2}, 0, 2); \frac{21}{2}$

Let A and B are two fixed points in a plane, then locus of another point Con the same plane such that CA + CB = constant, (> AB) is

circle
ellipse
parabola
hyperbola

The directrix of the parabola y² + 4x + 3 = 0 is

$x - \frac{4}{3} = 0$
$x + \frac{1}{4} = 0$
$x - \frac{3}{4} = 0$
$x - \frac{1}{4} = 0$

If g(x) is a polynomial satisfying g(x) g(y) = g(x) + g(y) + g(xy) - 2 for all real x and y and g(2) = 5, then $\lim_{x \to 3}$ g(x) is

8.
The length of the parabola y² = 12x cut off by the latusrectum is

$6 [\sqrt{2} + \log (1 + \sqrt{2})]$

$3 [\sqrt{2} + \log (1 + \sqrt{2})]$

$6 [\sqrt{2} - \log (1 + \sqrt{2})]$

$3 [\sqrt{2} - \log (1 + \sqrt{2})]$

$6 [\sqrt{2} + \log (1 + \sqrt{2})]$

Given equation of parabola is

y² = 12x ...(i)

and equation of latusrectum is

x = 3 ...(ii)

From Eqs. (i) and (ii), we get

y² = 36

$\Rightarrow y = \pm 6$

$∴$ Coordinates of end points of a latusrectum are (3, 6) and (3, - 6).

$∴ Required length = 2 \int_{0}^{3} \sqrt{1 + {(\frac{dy}{dx})}^{2}} d x = 2 \int_{0}^{3} \sqrt{1 + {(\frac{6}{y})}^{2}} = 2 \int_{0}^{3} \sqrt{\frac{12 x + 36}{12 x}} dx = 2 \int_{0}^{3} \frac{x + 3}{\sqrt{x^{2} + 3 x}} dx = 2 {[\sqrt{x^{2} + 3 x} + \frac{3}{2} \log |(x + \frac{3}{2})| + \sqrt{x^{2} + 3 x}]}_{0}^{3} = 2 [3 \sqrt{2} + \frac{3}{2} \log (\frac{9}{2} + 3 \sqrt{2}) - \frac{3}{2} \log (\frac{3}{2})] = 2 [3 \sqrt{2} + 3 \log (\sqrt{2} + 1)] = 6 [\sqrt{2} + \log (1 + \sqrt{2})]$