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JEE Mathematics Solved Question Paper 2016

Multiple Choice Questions

If 2^x + 2^y = 2^x then the value of $\frac{dy}{dx}$ at (1, 1) is equal to

Let y = $\tan^{- 1} (\sec (x) + \tan (x))$ . Then $\frac{dy}{dx}$ is equal to

$\frac{1}{4}$
$\frac{1}{2}$
$\frac{1}{\sec (x) + \tan (x)}$
$\frac{1}{\sec^{2} (x)}$

If s = $\sec^{- 1} (\frac{1}{2 x^{2} - 1})$ and t = $\sqrt{1 - x^{2}}$ , then $\frac{ds}{dt}$ at x = $\frac{1}{2}$ is

The minimum valuje of 2x³ - 9x² + 12x + 4 is

5.
The slope of the curve y = e^xcos(x), x $\in (- π, π)$ is maximum at

x = $\frac{π}{2}$

$x = - \frac{π}{2}$

$x = \frac{π}{4}$

x = 0

x = 0

Let m be the slope of the tangent to the curve

y = e^xcos(x)

$Then, m = \frac{dy}{dx} = e^{x} (\cos (x) - \sin (x)) \Rightarrow \frac{dm}{dx} = e^{x} (\cos (x) - \sin (x)) + e^{x} (- \sin (x) + \cos (x)) = - 2 e^{x} \sin (x) and \frac{d^{2} m}{{dx}^{2}} = - 2 e^{x} (\sin (x) + \cos (x)) ∴ \frac{dm}{dx} = 0 \Rightarrow \sin (x) = 0 \Rightarrow x = 0 Cleraly, \frac{d^{2} m}{{dx}^{2}} < 0 for x = 0$

If y = f(x) is continuous on [0, 6], differentiable on (0, 6), f(0) = - 2 and f(6) = 16, then at some point between x = 0 and x = 6, f'(x) must be equal to

- 18
- 3
3
14

The equation of tangent to the curve y = x³ - 6x + 5 at (2, 1) is

6x - y - 11 = 0
6x - y - 13 = 0
6x + y + 11 = 0
6x - y + 11 = 0

Let f(x) = 2x³ - 5x² - 4x + 3, $\frac{1}{2} \leq x \leq 3$ . The point at which the tangent to the curve is parallel to the X-axis, is

(1, - 4)
(2, - 9)
(2, - 4)
(2, - 1)

Two sides of triangle are 8 m and 56 m in length. The angle between them is increasing at the rate 0.8=08 rad/s. When the angle between sides of fixed length is $\frac{π}{3}$ , the rate at which the area of the triangle is increasing, is