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JEE Mathematics Solved Question Paper 2016

Multiple Choice Questions

51.

Suppose f(x) = x(x + 3)(x - 2), x [- 1, 4]. Then, a value of c in (- 1, 4) satisfying f'(c) = 10 is

2
3
$\frac{7}{2}$

52.

The area included between the parabola $y = \frac{x^{2}}{4 a} and the curve y = \frac{8 a^{3}}{(x^{2} + 4 a^{2})} is$

$a^{2} (2 π + \frac{2}{3})$
$a^{2} (2 π - \frac{8}{3})$
$a^{2} (π + \frac{4}{3})$
$a^{2} (2 π - \frac{4}{3})$

53.

If a, b, c are distinct positive real numbers, then the value of the determinant $|\begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix}| is$

$< 0$
$> 0$
0
$\geq 0$

54.

$The equations x - y + 2 z = 4 3 x + y + 4 z = 6 x + y + z = 1 have$

unique solution
infinitely many solutions
no solution
two solutions

55.

$If x = \sin (2 \tan^{- 1} (2)) and y = \sin (\frac{1}{2} \tan^{- 1} (\frac{4}{3})), then$

$x > y$
x = y
x = 0 = y
$x < y$

56.

$If \cosh (x) = \frac{5}{4}, then \cosh (3 x) = ?$

$\frac{61}{16}$
$\frac{63}{16}$
$\frac{65}{16}$
$\frac{61}{63}$

57.

$In a ∆ ABC, if < A = 90 °, then \cos^{- 1} (\frac{R}{r_{2} + r_{3}}) = ?$

$90 °$
$30 °$
$60 °$
$45 °$

58.

The value (s) of x for which the function

$f (x) \{= 1 - x, x < 1 = (1 - x) (2 - x), 1 \leq x \leq 2 3 - x, x > 2\} fails to be continuous is (are)$

1
2
3
all real numbers

59.

$If y = \log_{2} (\log_{2} (x)), then \frac{dy}{dx} = ?$

$\frac{\log_{e} (2)}{{xlog}_{e} (x)}$
$\frac{1}{\log_{e} {(2 x)}^{x}}$
$\frac{1}{({xlog}_{e} (x) \log_{e} (2))}$
$\frac{1}{x {(\log_{2} (x))}^{2}}$

60.
The angle of intersection between the curves y² + x² = a² $\sqrt{2}$ and x² - y² = a² is

$\frac{π}{3}$

$\frac{π}{4}$

$\frac{π}{6}$

$\frac{π}{12}$

$\frac{π}{4}$

$Given, y^{2} + x^{2} = a^{2} \sqrt{2} . . . (i) and x^{2} - y^{2} = a^{2} . . . (ii) On solving eqs (i) and (ii), we get x = a \sqrt{\frac{\sqrt{2} + 1}{2}}, y = a \sqrt{\frac{\sqrt{2} - 1}{2}} Now, y^{2} + x^{2} = a^{2} \sqrt{2} 2 y {(\frac{dy}{dx})}_{1} + 2 x = 0 \Rightarrow {(\frac{dy}{dx})}_{1} = - \frac{x}{y} ∴ {(\frac{dy}{dx})}_{1} = \frac{- a \sqrt{\frac{\sqrt{2} + 1}{2}}}{a \sqrt{\frac{\sqrt{2} - 1}{2}}} = - \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2} - 1}} and x^{2} - y^{2} = a^{2} On differentiating$

${(\frac{dy}{dx})}_{2} = \frac{x}{y} = \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2} - 1}} Then, \tan (θ) = \frac{{(\frac{dy}{dx})}_{1} - {(\frac{dy}{dx})}_{2}}{1 + {(\frac{dy}{dx})}_{1} {(\frac{dy}{dx})}_{2}} = \frac{- \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2} - 1}} - \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2} - 1}}}{1 + (- \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2} - 1}} \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2} - 1}})} = \frac{- 2 \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2} - 1}}}{1 - \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2} - 1}}} = \frac{- 2 \sqrt{(\sqrt{2} + 1) (\sqrt{2} - 1)}}{\sqrt{2} - 1 - \sqrt{2} - 1} = \frac{- 2 \sqrt{2 - 1}}{- 2} = 1 ∴ θ = \tan^{- 1} (1) = \frac{π}{4}$