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JEE Mathematics Solved Question Paper 2012

Multiple Choice Questions

21.

The value of $\int \frac{\sin (2 x)}{\sin^{4} (x) + \cos^{4} (x)} dx$ is

$\tan^{- 1} (\cot^{2} (x)) + C$
$\tan^{- 1} (\cos (2 x)) + C$
$\tan^{- 1} (\sin (2 x)) + C$
$\tan^{- 1} (\tan^{2} (x)) + C$

22.

The value of $\int \sqrt{1 + \sec (x)} dx$ is

$\sin^{- 1} (\sqrt{2} \sin (x)) + C$
$2 \sin^{- 1} (\sqrt{2} \sin (x / 2)) + C$
$2 \sin^{- 1} (\sqrt{2} \sin (x)) + C$
$2 \sin^{- 1} (\sqrt{2} x / 2) + C$

23.

The value of $\int \frac{(x^{2} + 1)}{x^{4} + x^{2} + 1} dx$ is

$\frac{1}{\sqrt{3}} \tan^{- 1} \{\frac{x - 1 / x}{\sqrt{3}}\} + C$
$\frac{1}{2 \sqrt{3}} \log \{\frac{(x - 1 / x) - \sqrt{3}}{(x - 1 / x) + \sqrt{3}}\} + C$
$\tan^{- 1} \{\frac{x + 1 / x}{\sqrt{3}}\} + C$
$\tan^{- 1} \{\frac{x - 1 / x}{\sqrt{3}}\} + C$

24.

The value of $\int_{0}^{1} x^{2} {(1 - x^{2})}^{\frac{3}{2}} dx$ is

$\frac{1}{32}$
$\frac{π}{8}$
$\frac{π}{16}$
$\frac{π}{32}$

25.

The value of $\int_{0}^{\infty} \frac{x}{(1 + x) (x^{2} + 1)} dx$ is

$2 π$
$\frac{π}{4}$
$\frac{π}{16}$
$\frac{π}{32}$

26.
The area of the region bounded by the curves, y² = 8x and y = x is

$\frac{64}{3}$

$\frac{32}{3}$

$\frac{16}{3}$

$\frac{8}{3}$

$\frac{32}{3}$

Given curves,

y² = 8x ...(i)

and y = x ...(ii)

On solving Eqs. (i) and (ii), we get

x² - 8x = 0

x(x - 8) = 0

$∴ Required area = \int_{0}^{8} (\sqrt{8 x} - x) dx$

$= {[2 \sqrt{2} . \frac{2}{3} . x^{\frac{3}{2}} - \frac{x^{2}}{2}]}_{0}^{8} = \frac{4 \sqrt{2}}{3} . {(8)}^{\frac{3}{2}} - \frac{{(8)}^{2}}{2} = \frac{4 \sqrt{2}}{3} . {(2)}^{3} . 2 \sqrt{2} - \frac{64}{2} = \frac{16}{3} \times 8 - \frac{64}{2} = 64 (\frac{2}{3} - \frac{1}{2}) = 64 \times \frac{(4 - 3)}{6} = 64 \times \frac{1}{6} = \frac{32}{3}$