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JEE Mathematics Solved Question Paper 2011

Multiple Choice Questions

61.
${(\tan^{- 1} (x))}^{2} + {(\cot^{- 1} (x))}^{2} = \frac{5 π^{2}}{8} \Rightarrow x =$

- 1

1

0

$π \sqrt{\frac{5}{8}}$

- 1

$Given equation is {(\tan^{- 1} (x))}^{2} + {(\cot^{- 1} (x))}^{2} = \frac{5 π^{2}}{8} {(\tan^{- 1} (x))}^{2} + {(\cot^{- 1} (x))}^{2} - 2 (\tan^{- 1} (x)) (\cot^{- 1} (x)) = \frac{5 x^{2}}{8} [∵ (\tan^{- 1} (x)) + (\cot^{- 1} (x)) = \frac{π}{2}] \Rightarrow \frac{π^{2}}{4} - 2 \tan^{- 1} (x) (\frac{π}{2} - \tan^{- 1} (x)) = \frac{5 π^{2}}{8} \Rightarrow - 2 . \frac{{πtan}^{- 1} (x)}{2} + 2 {(\tan^{- 1} (x))}^{2} = \frac{5 π^{2}}{8} - \frac{π^{2}}{4} \Rightarrow {(\tan^{- 1} (x))}^{2} - \frac{{πtan}^{- 1} (x)}{2} = \frac{3 π^{2}}{16} ∴ \tan^{- 1} (x) = \frac{\frac{π}{2} \pm \sqrt{\frac{π^{2}}{4} + \frac{3 π^{2}}{4}}}{2} \Rightarrow \tan^{- 1} (x) = \frac{\frac{π}{2} \pm π}{2} = \frac{3 π}{4}, - \frac{π}{4} \Rightarrow x = \tan (\frac{3 π}{4}), \tan (- \frac{π}{4}) \Rightarrow x = - 1, - 1$

62.

$For 0 < x \leq π, \sinh^{- 1} (\cot (x)) = ?$

$\log (\cot (\frac{x}{2}))$
$\log (\tan (\frac{x}{2}))$
$\log (1 + \cot (x))$
$\log (1 + \tan (x))$

63.

$\int (\sqrt{\frac{a + x}{a - x}} + \sqrt{\frac{a - x}{a + x}}) dx = ?$

$2 \sin^{- 1} (\frac{x}{a}) + c$
$2 {asin}^{- 1} (\frac{x}{a}) + C$
$2 \cos^{- 1} (\frac{x}{a}) + C$
$2 a \cos^{- 1} (\frac{x}{a}) + C$

64.

$If \int \frac{\sin^{8} (x) - \cos^{8} (x)}{1 - 2 \sin^{2} (x) \cos^{2} (x)} dx = Asin (2 x) + B, then A = ?$

$- \frac{1}{2}$
- 1
$\frac{1}{2}$
1

65.

$\int \frac{1 + \cos (4 x)}{\cot (x) - \tan (x)} dx$

$- \frac{1}{4} \cos (4 x) + C$
$\frac{1}{8} \cos (4 x) + C$
$\frac{1}{4} \sin (4 x) + C$
$- \frac{1}{8} \cos (4 x) + C$

66.

The area (in square units) of the region bounded by the curves x = y² and x = 3 - 2y² is

$\frac{3}{2}$
2
3
4

67.

$If I_{n} = \int_{0}^{\frac{π}{4}} \tan^{n} (dθ) for n = 1, 2, 3 . . ., then I_{n + 1} + I_{n + 1} = ?$

0
1
$\frac{1}{n}$
$\frac{1}{n + 1}$

68.

$Let f (0) = 1, f (0.5) = \frac{5}{4}, f (1) = 2, f (1.5) = \frac{13}{4}, and f (2) = 5 . Using Simson' s rule, \int_{0}^{2} f (x) dx = ?$

$\frac{14}{3}$
$\frac{7}{6}$
$\frac{14}{9}$
$\frac{7}{9}$

69.

$The solution of the differential equation \frac{dy}{dx} = \frac{y}{x} + \frac{ϕ (\frac{y}{x})}{ϕ' (\frac{y}{x})} is$

$xϕ (\frac{y}{x}) = k$
$ϕ (\frac{y}{x}) = kx$
$yϕ (\frac{y}{x}) = k$
$ϕ (\frac{y}{x}) = ky$

70.

If y = y(x) is the solution of the differential equation $(\frac{2 + \sin (x)}{y + 1}) \frac{dy}{dx} + \cos (x) = 0$ then y( $\frac{π}{2}$ )is equal to