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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

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

78.
The solution of the differential equation $(2 x - 4 y + 3) \frac{dy}{dx} + (x - 2 y + 1) = 0$ is

$[\log (2 x - 4 y) + 3] = x - 2 y + C$

$[\log [2 (2 x - 4 y) + 3]] = 2 (x - 2 y) + C$

$\log [2 (x - 2 y) + 5] = 2 (x + y) + C$

$\log [4 (x - 2 y) + 5] = 4 (x + 2 y) + C$

$\log [4 (x - 2 y) + 5] = 4 (x + 2 y) + C$

$We have, (2 x - 4 y + 3) \frac{dy}{dx} + (x - 2 y + 1) = 0 \Rightarrow \frac{dy}{dx} = \frac{(x - 2 y + 1)}{2 (x - 2 y) + 3} Put x - 2 y = v \Rightarrow 1 - 2 \frac{dy}{dx} = \frac{dv}{dx} ∴ \frac{1}{2} (1 - \frac{dv}{dx}) = - (\frac{v + 1}{2 v + 3}) \Rightarrow \frac{2 v + 3}{4 v + 5} dv = dx \Rightarrow \frac{1}{2} \int (\frac{4 v + 5}{4 v + 5} + \frac{1}{4 v + 5}) dv = \int dx \Rightarrow \frac{1}{2} v + \frac{1}{2 \times 4} \log (4 v + 5) = x + C \Rightarrow 4 v + \log (4 v + 5) = 8 x + C \Rightarrow 4 (x - 2 y) + \log (4 (x - 2 y) + 5) = 8 x + C \Rightarrow \log [4 (x - 2 y) + 5] = 4 (x + 2 y) + C$