If $\mathrm{y} = {\sin }^{-1}\left(2\mathrm{x}\sqrt{1 - {\mathrm{x}}^2}\right), - \frac{1}{\sqrt{2}} \le \mathrm{x} \le \frac{1}{\sqrt{2}}$, then $\frac{\mathrm{dy}}{\mathrm{dx}}$ is equal to

• $\frac{\mathrm{x}}{\sqrt{1 - {\mathrm{x}}^2}}$

• $\frac{1}{\sqrt{1 - {\mathrm{x}}^2}}$

• $\frac{2}{\sqrt{1 - {\mathrm{x}}^2}}$

• $\frac{2\mathrm{x}}{\sqrt{1 - {\mathrm{x}}^2}}$

The function $$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right) = \left\{2{\mathrm{x}}^2 - 1, \mathrm{if} 1 \le \mathrm{x} \le 4 \\ 151 - 30\mathrm{x}, \mathrm{if} 4 < \mathrm{x} \le 5\right\} \end{gathered}$$ is not suitable to apply Rolle's theorem, since

• f(x) is not continuous on [1, 5]

• f(1) $\ne$ f(5)

• f(x) is continuous only at x = 4

• f(x) is not differentiable at x = 4

The number of points at which the function $\mathrm{f}\left(\mathrm{x}\right) = \frac{1}{{\log }_{\mathrm{e}}\left|\mathrm{x}\right|}$ is discontinuous, is

• 1

• 2

• 3

• infinitely many

The functions f, g and h satisfy the relations f'(x) = g (x + 1) and g'(x) = h(x - 1). Then, f"(2x) is equal to

• h(2x)

• 4h(2x)

• h(2x - 1)

• h(2x + 1)

If f(x) = ${\sin }^{-1}\left(\frac{1 - \cos \left(2\mathrm{x}\right)}{2\sin \left(\mathrm{x}\right)}\right)$, then $\left|\mathrm{f}\text{'}\left(\mathrm{x}\right)\right|$ is equal to

• $\left|\sin \left(\mathrm{x}\right)\right|$

• x

• 0

• 1

If $\mathrm{y} \frac{\mathrm{x}}{\mathrm{x} + 1} + \frac{\mathrm{x} + 1}{\mathrm{x}}$, then $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}$ at x = 1 is equal to

• $\frac{7}{4}$

• $\frac{7}{8}$

• $\frac{1}{4}$

• $\frac{- 7}{8}$

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

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