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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

$\frac{8 π \sqrt{3}}{5}$
$\frac{2 π \sqrt{3}}{9}$
$\frac{4 π^{2} \sqrt{3}}{9}$
$\frac{π^{2}}{6 \sqrt{3}}$

77.
The solution of the differential equation $(1 + y^{2}) + (x - e^{\tan^{- 1} (y)}) \frac{dy}{dx} = 0$ , is

${xe}^{\tan^{- 1} (y)} = \tan^{- 1} (y) + C$

${xe}^{2 \tan^{- 1} (y)} = \tan^{- 1} (y) + C$

$2 {xe}^{\tan^{- 1} (y)} = e^{2 \tan^{- 1} (y)} + C$

$x^{2} e^{\tan^{- 1} (y)} = 4 e^{2 \tan^{- 1} (y)} + C$

$2 {xe}^{\tan^{- 1} (y)} = e^{2 \tan^{- 1} (y)} + C$

$We have, (1 + y^{2}) + (x - e^{\tan^{- 1} (y)}) \frac{dy}{dx} = 0 \frac{dy}{dx} + \frac{x}{1 + y^{2}} = \frac{e^{\tan^{- 1} (y)}}{1 + y^{2}} This equation is form of linear differential equation \frac{dx}{dy} Px = Q where, P = \frac{1}{1 + y^{2}}, Q = \frac{e^{\tan^{- 1} (y)}}{1 + y^{2}} Now, IF = e^{\int Pdy} = e^{\int \frac{1}{1 + y^{2}} dy} = e^{\tan^{- 1} (y)} ∴ Solution of differential equation is x . IF = \int Q . IFdy + C \Rightarrow {xe}^{\tan^{- 1} (y)} = \int \frac{e^{\tan^{- 1} (y)}}{1 + y^{2}} dy + C \Rightarrow {xe}^{\tan^{- 1} (y)} = \frac{e^{2 \tan^{- 1} (y)}}{2} + C \Rightarrow {xe}^{\tan^{- 1} (y)} = \frac{e^{2 \tan^{- 1} (y)}}{2} + C \Rightarrow 2 {xe}^{\tan^{- 1} (y)} = e^{2 \tan^{- 1} (y)} + C$