Let f(x) = 15 - $\left|\mathrm{x} - 10\right|$ ; x $\in$ R. Then the set of all values of x, at which the function, g(x) = f(f(x) is not differentiable, is:

• {5, 10, 15}

• {10}

• {10, 15}

• {5, 0, 15, 20}

If $$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right) = \left[\frac{\sin \left(\mathrm{p} + 1\right)\mathrm{x} + \sin \left(\mathrm{x}\right)}{\mathrm{x}}, \mathrm{x} < 0 \\ \mathrm{q}, \mathrm{x} = 0 \\ \frac{\sqrt{{\mathrm{x}}^2 + \mathrm{x}} - \sqrt{\mathrm{x}}}{{\mathrm{x}}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}, \mathrm{x} > 0\right] \end{gathered}$$ is continuous at x = 0, then the ordered pair (p, q) is equal to :

• $\left(- \frac{1}{2}, \frac{3}{2}\right)$

• $\left(- \frac{3}{2}, - \frac{1}{2}\right)$

• $\left(\frac{5}{2}, \frac{1}{2}\right)$

• $\left(- \frac{3}{2}, \frac{1}{2}\right)$

Let f R $\to$ R : be differentiable at c $\in$ R  and f(c) = 0. If g(x) = $\left|\mathrm{f}\left(\mathrm{x}\right)\right|$, then at x = c , g is

• not differentiable

• differentiable if f'(c) = 0

• not differentiable if f'(c) = 0

• differentiable if f'(c) $\ne$ 0

If f(x) is continuous on $\left[- \mathrm{\pi }, \mathrm{\pi }\right]$, where

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right) = \left\{- 2\sin \left(\mathrm{x}\right), \mathrm{for} - \mathrm{\pi } \le \mathrm{x} \le - \frac{\mathrm{\pi }}{2} \\ \mathrm{\alpha sin}\left(\mathrm{x}\right) + \mathrm{\beta }, \mathrm{for} - \frac{\mathrm{\pi }}{2} < \mathrm{x} < \frac{\mathrm{\pi }}{2} \\ \cos \left(\mathrm{x}\right), \mathrm{for} \frac{\mathrm{\pi }}{2} \le \mathrm{x} \le \mathrm{\pi }\right\} \end{gathered}$$

then $\mathrm{\alpha } \mathrm{and} \mathrm{\beta }$ are

• - 1, - 1

• 1, - 1

• 1, 1

• - 1, 1

If f(x) = $$\begin{gathered} \left\{{\log }_{\left(1 - 3\mathrm{x}\right)}\left(1 +\hspace{0.17em}3\mathrm{x}\right), \mathrm{for} \mathrm{x} \ne 0 \\ \mathrm{k}, \mathrm{for} \mathrm{x} = 0\right\} \end{gathered}$$ continuous at x = 0, then k is equal to

• - 2

• 2

• 1

• - 1

If $\mathrm{x} = \log \left(1 + {\mathrm{t}}^2\right) \mathrm{and} \mathrm{y} = \mathrm{t} - {\tan }^{-1}\left(\mathrm{t}\right)$. Then, $\frac{\mathrm{dy}}{\mathrm{dx}}$ is equal to

• e^x - 1

• t² - 1

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

• e^x - y

$\frac{\mathrm{d}}{\mathrm{dx}}\left[\sec \left\{{\cos }^{-1}\left(\frac{\mathrm{x}}{8}\right)\right\}\right]$ is equal to

• $\frac{1}{8}$

• $- \frac{1}{8}$

• $\frac{8}{{\mathrm{x}}^2}$

• - $\frac{8}{{\mathrm{x}}^2}$

If f(x) = $\sqrt{1 + {\cos }^2\left({\mathrm{x}}^2\right)}$, then $\mathrm{f}\text{'}\left(\frac{\sqrt{\mathrm{\pi }}}{2}\right)$ is

• $\frac{\sqrt{\mathrm{\pi }}}{6}$

• $- \sqrt{\frac{\mathrm{\pi }}{6}}$

• $\frac{1}{\sqrt{6}}$

• $\frac{\mathrm{\pi }}{\sqrt{6}}$

189.

If y = ${\mathrm{asin}}^3\left(\mathrm{\theta }\right)$ and x = ${\mathrm{acos}}^3\left(\mathrm{\theta }\right)$, then at $\mathrm{\theta } = \frac{\mathrm{\pi }}{3}, \frac{\mathrm{dy}}{\mathrm{dx}}$ is equal to

• $\frac{\sqrt{\mathrm{\pi }}}{6}$

• $- \sqrt{3}$

• $\frac{- 1}{\sqrt{3}}$

• $\sqrt{3}$

${∆}^2{\mathrm{y}}_0$ is equal to

• 2y₂ - 2y₁ - y₀

• y₂ - 2y₁ - y₀

• 2y₂ - 2y₁ + y₀

• y₂ - 2y₁ + y₀