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JEE Mathematics Solved Question Paper 2016

Multiple Choice Questions

71.

$If \int x^{3} e^{5 x} dx = \frac{e^{5 x}}{5^{4}} [f (x)] + C, then f (x) = ?$

$\frac{x^{3}}{5} - \frac{3 x^{2}}{5^{2}} + \frac{6 x}{5^{3}} - \frac{6}{5^{4}}$
$5 x^{3} - 5^{2} x^{2} + 5^{3} x - 6$
$5^{3} x^{3} - 15 x^{2} + 30 x - 6$
$5^{3} x^{3} - 75 x^{2} + 30 x - 6$

72.

$\int \frac{x}{{(x^{2} + 2 x + 2)}^{2}} dx = ?$

$\frac{x^{2} + 2}{x^{2} + 2 x + 2} - \frac{1}{2} \tan^{- 1} (x - 1) + C$
$\frac{x^{2} - 2}{4 (x^{2} + 2 x + 2)} - \frac{1}{2} \tan^{- 1} (x + 1) + C$
$\frac{x^{2} + 2}{2 (x^{2} + 2 x + 2)} - \frac{1}{2} \tan^{- 1} (x + 1) + C$
$\frac{2 (x - 1)}{x^{2} + 2 x + 2} + \frac{1}{2} \tan^{- 1} (x + 1) + C$

73.

$If \int \log (a^{2} + x^{2}) dx = h (x) + C, then h (x) = ?$

$xlog (a^{2} + x^{2}) + 2 \tan^{- 1} (\frac{x}{a})$
$x^{2} \log (a^{2} + x^{2}) + x + {atan}^{- 1} (\frac{x}{a})$
$xlog (a^{2} + x^{2}) - 2 x + 2 {atan}^{- 1} (\frac{x}{a})$
$x^{2} \log (a^{2} + x^{2}) + 2 x - a^{2} \tan^{- 1} (\frac{x}{a})$

74.

$For x > 0, if \int (\log {(x)}^{5}) dx = ? x [A (\log {(x)}^{5}) + B {(\log (x))}^{4} + C {(\log (x))}^{3} + D {(\log (x))}^{2} + E (\log (x)) + F] + Constant, then A + B + C + D + E + F = ?$

- 44
- 42
- 40
- 36

75.

By the definition of the definite integral, the value of $\lim_{n \to \infty} (\frac{1}{\sqrt{n^{2} - 1}} + \frac{1}{\sqrt{n^{2} - 2^{2}}} + . . . \frac{1}{\sqrt{n^{2} - {(n - 1)}^{2}}})$ is equal to

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

76.
$\int_{\frac{π}{4}}^{\frac{π}{4}} (\frac{x + \frac{π}{4}}{2 - \cos (2 x)}) dx$ is equal to

$\frac{8 π \sqrt{3}}{5}$

$\frac{2 π \sqrt{3}}{9}$

$\frac{4 π^{2} \sqrt{3}}{9}$

$\frac{π^{2}}{6 \sqrt{3}}$

$\frac{π^{2}}{6 \sqrt{3}}$

$Let I = \int_{- \frac{π}{4}}^{\frac{π}{4}} (\frac{x + \frac{π}{4}}{2 - \cos (2 x)}) dx = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{x}{2 - \cos (2 x)} + \frac{π}{4} \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{dx}{2 - \cos (2 x)} = 0 + 2 \times \frac{π}{4} \int_{0}^{\frac{π}{4}} \frac{dx}{2 - \frac{1 - \tan^{2} (x)}{1 + \tan^{2} (x)}} 0 = \frac{π}{2} \int_{0}^{\frac{π}{4}} \frac{1 + \tan^{2} (x)}{2 + 2 \tan^{2} (x) - 1 + \tan^{2} (x)} dx = \frac{π}{2} \int_{0}^{\frac{π}{4}} \frac{\sec^{2} (x)}{1 + 3 \tan^{2} (x)} dx Put \tan (x) = t \Rightarrow \sec^{2} (x) dx = dt ∴ \frac{π}{2} \int_{0}^{1} \frac{dt}{1 + 3 t^{2}} = \frac{π}{2} \times \frac{1}{\sqrt{3}} {[\tan^{- 1} (\sqrt{3} t)]}_{0}^{1} = \frac{π}{2 \sqrt{3}} [\tan^{- 1} (\sqrt{3}) - \tan^{- 1} (0)] = \frac{π}{2 \sqrt{3}} (\frac{π}{3}) = \frac{π^{2}}{6 \sqrt{3}}$