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JEE Mathematics Solved Question Paper 2014

Multiple Choice Questions

71.

The value of $\lim_{x \to 0} \frac{\int_{0}^{x^{2}} \cos (t^{2}) d x}{xsin (x)}$

1
- 1
2
$\log_{e} (2)$

72.

The curve $y = {(\cos (x) + y)}^{1 / 2}$ satisfies the differential equation

$(2 y - 1) \frac{d^{2} y}{d x^{2}} + 2 {(\frac{d y}{d x})}^{2} + \cos (x) = 0$
$\frac{d^{2} y}{d x^{2}} + 2 {(\frac{d y}{d x})}^{2} + \cos (x) = 0$
$(2 y - 1) \frac{d^{2} y}{d x^{2}} - 2 {(\frac{d y}{d x})}^{2} + \cos (x) = 0$
$(2 y - 1) \frac{d^{2} y}{d x^{2}} - {(\frac{d y}{d x})}^{2} + \cos (x) = 0$

73.

The solution of the differential equation

$\frac{d y}{d x} + \frac{y}{{xlog}_{e} (x)} = \frac{1}{x}$

under the condition y = 1 when x = e is

$2 y = \log_{e} (x) + \frac{1}{\log_{e} (x)}$
$y = \log_{e} (x) + \frac{2}{\log_{e} (x)}$
${ylog}_{e} (x) = \log_{e} (x) + 1$
$y = \log_{e} (x) + e$

74.

Let f(x) = $\max \{x + |x|, x - [x]\}$ , where [x] denotes the greatest integer $\leq x$ . Then, the values of $\int_{- 3}^{3} f (x) d x$ is

0
51/2
21/2
1

75.
Suppose $M = \int_{0}^{π / 2} \frac{\cos (x)}{x + 2} d x, N = \int_{0}^{π / 4} \frac{\sin (x) \cos (x)}{{(x + 1)}^{2}} d x$ . Then, the values of (M - N) equals

$\frac{3}{π + 2}$

$\frac{2}{π - 4}$

$\frac{4}{π - 2}$

$\frac{2}{π + 4}$

$\frac{2}{π + 4}$

$Given, M = \int_{0}^{π / 2} \frac{\cos (x)}{x + 2} d x and N = \int_{0}^{π / 4} \frac{\sin (x) \cos (x)}{{(x + 1)}^{2}} d x \Rightarrow N = \int_{0}^{π / 4} \frac{1}{2} \frac{\sin (2 x)}{{(x + 1)}^{2}} d x Put 2 x = t \Rightarrow dx = \frac{dt}{2}$

$∴ N = \int_{0}^{π / 2} \frac{\sin (t)}{4 {(\frac{t}{2} + 1)}^{2}} d t = \int_{0}^{π / 2} \frac{\sin (t)}{{(t + 2)}^{2}} d t = \int_{0}^{π / 2} \frac{\sin (x)}{{(x + 2)}^{2}} d x ∴ M - N = \int_{0}^{π / 2} \cos (x) \times \frac{1}{(x + 2)} d x - \int_{0}^{π / 2} \frac{\sin (x)}{{(x + 2)}^{2}} d x = {[\frac{\sin (x)}{(x + 2)}]}_{0}^{π / 2} - \int_{0}^{π / 2} - \frac{\sin (x)}{{(x + 2)}^{2}} d x - \int_{0}^{π / 2} \frac{\sin (x)}{{(x + 2)}^{2}} d x = \frac{\sin (\frac{π}{2})}{\frac{π}{2} + 2} = \frac{1}{\frac{π + 4}{2}} = \frac{2}{π + 4}$