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

• - 2

• - 1

• 0

• 1

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

• $\frac{1}{4}$

• $\frac{1}{2}$

• $\frac{1}{\sec \left(\mathrm{x}\right) + \tan \left(\mathrm{x}\right)}$

• $\frac{1}{{\sec }^2\left(\mathrm{x}\right)}$

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

• 1

• 2

• - 2

• 4

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

• 4

• 5

• 6

• 8

The slope of the curve y = e^xcos(x), x $\in \left(- \mathrm{\pi }, \mathrm{\pi }\right)$ is maximum at

• x = $\frac{\mathrm{\pi }}{2}$

• $\mathrm{x} = - \frac{\mathrm{\pi }}{2}$

• $\mathrm{x} = \frac{\mathrm{\pi }}{4}$

• 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, $\frac{1}{2} \le \mathrm{x} \le 3$. 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 $\frac{\mathrm{\pi }}{3}$, the rate at which the area of the triangle is increasing, is

• 0. 4 m²/s

• 0.8 m²/s

• 0 . 6 m²/s

• 0.04 m²/s

If y = 8x³ - 60x² + 144x + 27 is a decreasing function in the interval

• (- 5, 6)

• $\left(- \infty , 2\right)$

• (5, 6)

• (2, 3)