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JEE Mathematics Solved Question Paper 2005

Multiple Choice Questions

61.

$\int \cos^{\frac{- 3}{7}} (x) \sin^{\frac{- 11}{7}} (x) dx$ is equal to:

$\log |\sin^{\frac{4}{7}} (x)| + c$
$\frac{4}{7} \tan^{\frac{4}{7}} (x) + c$
$\frac{- 7}{4} \tan^{\frac{- 4}{7}} (x) + c$
$\log |\cos^{\frac{3}{7}} (x)| + c$

62.

$\int \frac{(\sin (θ) + \cos (θ))}{\sqrt{\sin (2 θ)}} dθ$ is equal to :

$\log |\cos (θ) - \sin (θ) + \sqrt{\sin (2 θ)}|$
$\log |\sin (θ) - \cos (θ) + \sqrt{\sin (2 θ)}|$
$\sin^{- 1} (\sin (θ) - \cos (θ))$
$\sin^{- 1} (\sin (θ) + \cos (θ))$

63.

$\int_{\frac{π}{6}}^{\frac{π}{3}} \frac{dx}{1 + \sqrt{\tan (x)}}$ is equal to :

$\frac{π}{12}$
$\frac{π}{2}$
$\frac{3 π}{2}$
$2 π$

64.

$\int_{- π}^{π} \frac{\sin^{4} (x)}{\sin^{4} (x) + \cos^{4} (x)}$ dx is equal to :

$π$
$\frac{π}{2}$
$\frac{3 π}{2}$
$2 π$

65.

The value of $\frac{2^{\sin (x)}}{2^{\sin (x)} + 2^{\cos (x)}} dx$ is :

2
$π$
$\frac{π}{4}$
$2 π$

66.

If f is continuous function, then :

$\int_{- 2}^{2} f (x) dx = \int_{0}^{2} [f (x) - f (- x)] dx$
$\int_{- 3}^{5} 2 f (x) d x = \int_{- 6}^{10} f (x - 1) dx$
$\int_{- 3}^{5} f (x) dx = \int_{- 4}^{4} f (x - 1) dx$
$\int_{- 3}^{5} f (x) dx = \int_{- 2}^{6} f (x - 1) dx$

67.

The area of the region bounded by y² = 4ax and x² = 4ay, a > 0 in sq unit, is :

16 $\frac{a^{2}}{3}$
$14 \frac{a^{2}}{3}$
$13 \frac{a^{2}}{3}$
16a²

68.
An integrating factor of the differential equation $x \frac{dy}{dx} + ylog (x) = {xe}^{x} x^{\frac{1}{2} \log (x)}$ , (x > 0) is :

x^log(x)

${(\sqrt{x})}^{\log (x)}$

${(\sqrt{e})}^{{(\log (x))}^{2}}$

$e^{x^{2}}$

${(\sqrt{e})}^{{(\log (x))}^{2}}$

Given differential equation is

$x \frac{dy}{dx} + ylog (x) = {xe}^{x} = {xe}^{x} x^{- \frac{1}{2} \log (x)} \frac{dy}{dx} + y \frac{1}{x} \log (x) = e^{x} x^{- \frac{1}{2} \log (x)} Here, P = \frac{1}{x} \log (x) and Q = e^{x} x^{- \frac{1}{2} \log (x)} ∴ IF = e^{\int \frac{1}{x} \log (x) dx} = e^{\frac{{(\log (x))}^{2}}{2}} = {(\sqrt{e})}^{{(\log (x))}^{2}}$