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JEE Mathematics Solved Question Paper 2012

Multiple Choice Questions

41.

The integrating factor of the differential equation $x \frac{dy}{dx} - y = 2 x^{2}$ is

$\frac{1}{x}$
x
e^-x
e^-y

42.

By Simposon's rule taking n = 4, the value of the integral $\int_{0}^{1} \frac{dx}{1 + x^{2}}$ is equal to

0.785
0.788
0.781
None of these

43.
Thevalue of integral $\int_{0}^{\frac{π}{2}} (\sqrt{\tan (x)} + \sqrt{\cot (x)}) dx$ is

$\sqrt{2} π$

$π$

$\frac{π}{2}$

0

$\sqrt{2} π$

$Ley I = \int_{0}^{\frac{π}{2}} (\sqrt{\tan (x)} + \sqrt{\cot (x)}) dx [∵ If \int_{0}^{2 a} f (x) = \int_{0}^{2 a} f (2 a - x), then \int_{0}^{2 a} f (x) dx = 2 \int_{0}^{a} f (x) dx] ∴ I = 2 \int_{0}^{\frac{π}{4}} (\sqrt{\tan (x)} + \sqrt{\cot (x)}) dx = 2 \int_{0}^{\frac{π}{4}} (\frac{\sqrt{\sin (x)}}{\sqrt{\cos (x)}} + \frac{\sqrt{\cos (x)}}{\sqrt{\sin (x)}}) dx = 2 \int_{0}^{\frac{π}{4}} \frac{(\sin (x) + \cos (x))}{\sqrt{\sin (x) \cos (x)}} dx = 2 \sqrt{2} \int_{0}^{\frac{π}{4}} \frac{\sin (x) + \cos (x)}{\sqrt{2 \sin (x) \cos (x)}} dx$

$(on multiplying numerator and denominator by \sqrt{2}) = 2 \sqrt{2} \int_{0}^{\frac{π}{4}} \frac{\sin (x) + \cos (x)}{\sqrt{1 - {(\sin (x) - \cos (x))}^{2}}} dx [∵ {(\sin (x) - \cos (x))}^{2} = \sin^{2} (x) + \cos^{2} (x) - 2 \sin (x) \cos (x) ∴ 2 \sin (x) \cos (x) = 1 - {(\sin (x) - \cos (x))}^{2}] Put \sin (x) - \cos (x) = t \Rightarrow \cos (x) + \sin (x) = \frac{d t}{d x} \Rightarrow dx = \frac{dt}{\cos (x) + \sin (x)} ∴ I = 2 \sqrt{2} \int_{- 1}^{0} \frac{dt}{\sqrt{1 - t^{2}}} = 2 \sqrt{2} {[\sin^{- 1} (t)]}_{- 1}^{0} = 2 \sqrt{2} [\sin^{- 1} (0) - \sin^{- 1} (- 1)] = 2 \sqrt{2} [0 + \sin^{- 1} (1)] = 2 \sqrt{2} (\frac{π}{2}) = \sqrt{2} π$

44.

The value of $\int_{0}^{1} xdx$ by Trapezoidal rule taking x = 4 is

0.34375
0.5
0.38387
0.353367

45.

The area of the figure bounded by the curves $y = |x - 1| and y = 3 - |x|$ is

1 sq units
2 sq units
3 sq units
4 sq units

46.

If $\frac{d}{dx} f (x) = 4 x^{3} - \frac{3}{x^{4}}$ such that f(2) = 0. Then, f(x) is

$x^{3} + \frac{1}{x^{4}} - \frac{129}{8}$
$x^{4} + \frac{1}{x^{3}} + \frac{129}{8}$
$x^{3} + \frac{1}{x^{4}} + \frac{129}{8}$
$x^{4} + \frac{1}{x^{3}} - \frac{129}{8}$

47.

The order and degree of the differential equation ${(\frac{d^{3} y}{{dx}^{3}})}^{2} - 3 \frac{d^{2} y}{{dx}^{2}} + 2 {(\frac{dy}{dx})}^{4} = y^{4}$ are