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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

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

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

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

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

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

Given curve is,

$y = {(\cos (x) + y)}^{1 / 2}$

On differentiating both sides w.r.t. x, we get

$\frac{d y}{d x} = \frac{1}{2 {(\cos (x) + y)}^{1 / 2}} \times (- \sin (x) + \frac{d y}{d x}) (- \sin (x) + \frac{d y}{d x}) \Rightarrow 2 y \frac{d y}{d x} = - \sin (x) + \frac{d y}{d x} \Rightarrow (2 y - 1) \frac{d y}{d x} = - \sin (x) + \frac{d y}{d x}$

Again, differentiating both sides w.r.t. x, we get

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

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