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

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

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

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

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

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

$Let I = \int \frac{x}{{(x^{2} + 2 x + 2)}^{2}} dx I = \int \frac{x}{{[{(x + 1)}^{2} + 1]}^{2}} dx Put x + 1 = \tan (θ) \Rightarrow dx = \sec^{2} (θ) ∴ I = \int \frac{(\tan (θ) - 1) \sec^{2} (θ)}{{(\tan^{2} (θ) + 1)}^{2}} dθ I = \int \frac{(\tan (θ) - 1) \sec^{2} (θ)}{\sec^{4} (θ)} dθ = \int \frac{(\tan (θ) - 1)}{\sec^{2} (θ)} dθ = \int (\frac{\tan (θ)}{\sec^{2} (θ)} - \frac{1}{\sec^{2} (θ)}) dθ = \int (\sin (θ) \cos (θ) - \cos^{2} (θ)) dθ$

$= \frac{1}{2} \int (2 \sin (θ) \cos (θ) - 2 \cos^{2} (θ)) dθ = \frac{1}{2} (\sin (2 θ) - 1 - \cos (2 θ)) dθ = \frac{1}{2} [\frac{- \cos (2 θ)}{2} - θ - \frac{\sin (2 θ)}{2}] + C = - \frac{1}{4} [\cos (2 θ) + \sin (2 θ)] - \frac{1}{2} θ + C = - \frac{1}{4} [\frac{1 - {(x + 1)}^{2} + 2 (x + 1)}{1 + {(x + 1)}^{2}}] - \frac{1}{2} \tan^{- 1} (x + 1) + C = \frac{1}{4} (\frac{x^{2} - 2}{x^{2} + 2 x + 2}) - \frac{1}{2} \tan^{- 1} (x + 1) + C$

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