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JEE Mathematics Solved Question Paper 2016

Multiple Choice Questions

21.

The value of x satisfying the equation $\tan^{- 1} (x) + \tan^{- 1} (\frac{2}{3}) + \tan^{- 1} (\frac{7}{4})$ is equal to

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

22.

If $\tan^{- 1} (x) + \tan^{- 1} (y) = \frac{2 π}{3}$ , then $\cot^{- 1} (x) + \cot^{- 1} (y)$ is equal to

$\frac{π}{2}$
$\frac{1}{2}$
$\frac{π}{3}$
$\frac{\sqrt{3}}{2}$

23.

$\int {(\sec (x))}^{m} (\tan^{3} (x) + \tan (x)) dx$ is equal to

$\sec^{m + 2} (x) + C$
$\tan^{m + 2} (x) + C$
$\frac{\sec^{m + 2} (x)}{m + 2} + C$
$\frac{\tan^{m + 2} (x)}{m + 2} + C$

24.

$\int \frac{1}{7} \sin (\frac{x}{7} + 10) dx$ is equal to

$\frac{1}{7} \cos (\frac{x}{7} + 10) + C$
$- \frac{1}{7} \cos (\frac{x}{7} + 10) dx$
$- \cos (\frac{x}{7} + 10) + C$
$- 7 \cos (\frac{x}{7} + 10) + C$

25.

$\int (\frac{x - a}{x} - \frac{x}{x + a}) dx$ is equal to

$\log |\frac{x + a}{x}| + C$
$alog |\frac{x + a}{x}| + C$
$alog |\frac{x}{x + a}| + C$
$\log |\frac{x + a}{x}| + C$

26.

$\int x^{4} e^{x^{5}} \cos (e^{x^{5}}) dx$ is equal to

$\frac{1}{3} \sin (e^{x^{5}}) + C$
$\frac{1}{4} \sin (e^{x^{5}}) + C$
$\frac{1}{5} \sin (e^{x^{5}}) + C$
$\sin (e^{x^{5}}) + C$

27.

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

$\log |\tan (x)| + C$
$\log |\sin (2 x)| + C$
$\log |\sec (x)| + C$
$\log |\cos (x)| + C$

28.

$\int \frac{2 x + \sin (2 x)}{1 + \cos (2 x)} dx$ is equal to

$x + \log (\tan (x)) + C$
$xlog |\tan (x)| + C$
$xtan (x) + C$
$x + \tan (x) + C$

29.

$\int \frac{1}{8 \sin^{2} (x) + 1} dx$ is equal to

$\sin^{- 1} (\tan (x)) + C$
$\frac{1}{3} \sin^{- 1} (\tan (x)) + C$
$\frac{1}{3} \tan^{- 1} (3 \tan (x)) + C$
$\tan^{- 1} (3 \tan (x)) + C$

30.
$\int_{0}^{\frac{π}{2}} \log (\frac{\cos (x)}{\sin (x)}) dx$ is equal to

$\frac{π}{2}$

$\frac{π}{4}$

$π$

0

$Let I = \int_{0}^{\frac{π}{2}} \log (\frac{\cos (x)}{\sin (x)}) d x = \int_{0}^{\frac{π}{2}} \log (\cot (x)) d x . . . (i) Then, I = \int_{0}^{\frac{π}{2}} \log (\cot (\frac{π}{2} - x)) d x [∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x] = \int_{0}^{\frac{π}{2}} \log (\tan (x)) d x . . . (ii) On adding Eqs . (i) and (ii), we get 2 I = \int_{0}^{\frac{π}{2}} (\log (\cot (x)) + \log (\tan (x))) d x 2 I = \int_{0}^{\frac{π}{2}} 0 d x = 0 \Rightarrow I = 0$