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JEE Mathematics Solved Question Paper 2014

Multiple Choice Questions

61.

$If k > 1 and the determinant of the matrix A^{2}, where A = [\begin{matrix} k & kα & α \\ 0 & α & kα \\ 0 & 0 & k \end{matrix}], is k^{2}, then |α| = ?$

$\frac{1}{k^{2}}$
k
k²
$\frac{1}{k}$

62.

$If x > 0, y > 0, z > 0, xy + yz + zx < 1 and if \tan^{- 1} (x) + \tan^{- 1} (y) + \tan^{- 1} (z) = π, then x + y + z = ?$

0
xyz
3xyz
$\sqrt{xyz}$

63.

${sech}^{- 1} (\frac{1}{2}) - {cosec}^{- 1} (\frac{3}{4}) = ?$

$\log_{e} (3 (2 + \sqrt{3}))$
$\log_{e} (\frac{1 + \sqrt{3}}{3})$
$\log_{e} (\frac{2 + \sqrt{3}}{3})$
$\log_{e} (\frac{2 - \sqrt{3}}{3})$

64.

$If f (x) = \frac{x}{1 + x} and g (x) = f (f (x)), then g' (x) =$

$\frac{1}{{(2 x + 3)}^{2}}$
$\frac{1}{{(x + 1)}^{2}}$
$\frac{1}{x^{2}}$
$\frac{1}{{(2 x + 1)}^{2}}$

65.
$If \int \frac{dx}{\sqrt{\sin^{3} (x) \cos (x)}} = g (x) + c, then g (x) = ?$

$- \frac{2}{\sqrt{\cot (x)}}$

$- \frac{2}{\sqrt{\tan (x)}}$

$\frac{2}{\sqrt{\cot (x)}}$

$\frac{2}{\sqrt{\tan (x)}}$

$- \frac{2}{\sqrt{\tan (x)}}$

$Given, \int \frac{dx}{\sqrt{\sin^{3} (x) \cos (x)}} = g (x) + c Now, \int \frac{dx}{\sqrt{\sin^{4} (x) \cot (x)}} = \int \frac{dx}{\sin^{2} (x) \sqrt{\cot (x)}} = \int \frac{\csc^{2} (x)}{\sqrt{\cot (x)}} dx Put \cot (x) = t \Rightarrow - \csc^{2} (x) dx = dt ∴ \int - \frac{1}{\sqrt{t}} dt = - \frac{t^{\frac{1}{2}}}{\frac{1}{2}} + C = - 2 \sqrt{\cot (x)} + C = - \frac{2}{\sqrt{\tan (x)}} + C ∴ g (x) = - \frac{2}{\sqrt{\tan (x)}}$

66.

$If \int \frac{dx}{(1 + \sqrt{x}) \sqrt{x - x^{2}}} = \frac{A \sqrt{x}}{\sqrt{1 - x}} + \frac{B}{\sqrt{1 - x}} + C, where C is real constant, then A + B = ?$

67.

For any integer n $\geq$ 2, let I_n = $\int$ tanⁿ(x)dx. If $I_{n} = \frac{1}{a} \tan^{n - 1} (x) - {bI}_{n - 2} for n \geq 2,$ then the ordered par (a, b) equals to

$(n - 1, \frac{n - 1}{n - 2})$
$(n - 1, \frac{n - 2}{n - 1})$
(n, 1)
(n - 1, 1)

68.

$If \frac{(x^{2} - 1)}{{(x + 1)}^{2} \sqrt{x (x^{2} + x + 1)}} dx = {Atan}^{- 1} (\sqrt{\frac{x^{2} + x + 1}{x}}) C, in which C is a constant, then A = ?$

$\frac{1}{2}$
3
2
1

69.

$By the defination of the definite integral, the value of \lim_{n \to \infty} (\frac{1^{4}}{1^{5} + n^{5}} + \frac{2^{4}}{2^{5} + n^{5}} + \frac{3^{4}}{3^{5} + n^{6}} + . . . + \frac{n^{4}}{n^{5} + n^{5}}) is$