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JEE Mathematics Solved Question Paper 2003

Multiple Choice Questions

21.

$\int_{0}^{π} \frac{xdx}{a^{2} \cos^{2} (x) + b^{2} \sin^{2} (x)} dx$ is equal to

$\frac{π}{2 ab}$
$\frac{π}{ab}$
$\frac{π^{2}}{2 ab}$
$\frac{π^{2}}{ab}$

22.

The differential equation for which sin^-1(x) + sin^-1(y) = c is given by

$\sqrt{1 - x^{2}} dy + \sqrt{1 - y^{2}} dx = 0$
$\sqrt{1 - x^{2}} dx + \sqrt{1 - y^{2}} dy = 0$
$\sqrt{1 - x^{2}} dx - \sqrt{1 - y^{2}} dy = 0$
$\sqrt{1 - x^{2}} dy - \sqrt{1 - y^{2}} dx = 0$

23.

$\int e^{x} (\frac{1 + \sin (x)}{1 + \cos (x)}) dx$ is equal to

$e^{x} \sec^{2} (\frac{x}{2}) + c$
$e^{x} \tan (\frac{x}{2})$ + c
$e^{x} \sec (\frac{x}{2}) + c$
$e^{x} \tan (x) + c$

24.
$\int \sqrt{1 + \sin (\frac{x}{4})} dx$ is equal to

$8 (\sin (\frac{x}{8}) + \cos (\frac{x}{8})) + C$

$8 (\sin (\frac{x}{8}) - \cos (\frac{x}{8})) + C$

$8 (\cos (\frac{x}{8}) - \sin (\frac{x}{8})) + C$

$\frac{1}{8} (\sin (\frac{x}{8}) - \cos (\frac{x}{8})) + C$

$8 (\sin (\frac{x}{8}) - \cos (\frac{x}{8})) + C$

$We have, \int \sqrt{1 + \sin (\frac{x}{4})} dx = \int \sqrt{(1 + 2 \sin (\frac{x}{8}) \cos (\frac{x}{8}))} dx [∵ \sin (2 x) = 2 \sin (x) \cos (x)] = \int \sqrt{(\sin^{2} (\frac{x}{8}) + \cos^{2} (\frac{x}{8})) + 2 \sin (\frac{x}{8}) \cos (\frac{x}{8})} dx = \sqrt{{(\sin (\frac{x}{8}) - \cos (\frac{x}{8}))}^{2}} dx [∵ {(a + b)}^{2} = a^{2} + b^{2} + 2 ab] = \int (\sin (\frac{x}{8}) + \cos (\frac{x}{8})) dx = \frac{- \cos (\frac{x}{8})}{\frac{1}{8}} + \frac{\sin (\frac{x}{8})}{\frac{1}{8}} = 8 (- \cos (\frac{x}{8}) + \sin (\frac{x}{8})) + C = 8 (- \cos (\frac{x}{8}) + \sin (\frac{x}{8})) = 8 (\sin (\frac{x}{8}) - \cos (\frac{x}{8})) + C$