The function y = $\left|2\sin \left(\mathrm{x}\right)\right|$ is continuous for any x but it is not differentiable at

• only x = 0

• only x = $\mathrm{\pi }$

• only x = $\frac{\mathrm{\pi }}{2}$

• x = $\mathrm{k\pi } \left(\mathrm{k} \mathrm{is} \mathrm{integer}\right)$

If $\mathrm{y} = {\mathrm{e}}^{\log \left(1 + \mathrm{x} +\hspace{0.17em}{\mathrm{x}}^2 + {\mathrm{x}}^3 + \right)}, \mathrm{where} \left|\mathrm{x}\right| < 1$, then $\frac{\mathrm{dy}}{\mathrm{dx}}$ is equal to

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

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

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

• None of these

233.

Let f (x + y) = f(x) + f(y) for all x and y. If the function f(x) is continuous at x = 0, then f(x) is continuous

• only at x = 0

• $\mathrm{at} \mathrm{x} \in \mathrm{R} - \left\{0\right\}$

• for all x

• None of these

Let $$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right) = \left\{{\mathrm{x}}^2\sin \left(\frac{1}{\mathrm{x}}\right), \mathrm{x} \ne 0 \\ 0, \mathrm{x} = 0\right\} \end{gathered}$$. Then, f(x) is continuous but not differentiable at x = 0, if

• $\mathrm{n} \in \left(0, 1\right)$

• $\mathrm{n} \in [1, \infty )$

• $\mathrm{n} \in \left(- \infty , 0\right)$

• n = 0

The function f(x) = $$\begin{gathered} \left\{\raisebox{1ex}{${\mathrm{x}}^2$}\!\left/ \!\raisebox{-1ex}{$\mathrm{a}$}\right., 0 \le \mathrm{x} < 1 \\ \mathrm{a}, 1 \le \mathrm{x} < \sqrt{2} \\ \frac{2{\mathrm{b}}^2 - 4\mathrm{b}}{{\mathrm{x}}^2}, \sqrt{2} \le \mathrm{x} < \infty \right\} \end{gathered}$$ is continuous for $0 \le \mathrm{x} < \infty$,then the most suitable values of a and b are

• a = 1, b = - 1

• a = - 1, b = 1 + $\sqrt{2}$

• a = - 1, b = 1

• None of the above

If f(x) = $$\begin{gathered} \left\{1, \mathrm{x} < 0 \\ 1 + \sin \left(\mathrm{x}\right), 0 \le \mathrm{x} < \frac{\mathrm{\pi }}{2}\right\} \end{gathered}$$,then at x = 0 the derivative f'(x) is

• 1

• 0

• infinite

• not defined

If f(x) = $$\begin{gathered} \left\{\frac{1 - \cos \left(4\mathrm{x}\right)}{{\mathrm{x}}^2}, \mathrm{when} \mathrm{x} < 0 \\ \mathrm{a}, \mathrm{when} \mathrm{x} = 0 \\ \frac{\sqrt{\mathrm{x}}}{\sqrt{\left(16 + \sqrt{\mathrm{x}}\right)}{\displaystyle }{\displaystyle -}{\displaystyle }{\displaystyle 4}}, \mathrm{when} \mathrm{x} > 0\right\} \end{gathered}$$ is continuous at x = 0, then the value of a will be

• 8

• - 8

• 4

• None of these

If f is a real-valued differentiable function satisfying $\left|\mathrm{f}\left(\mathrm{x}\right) - \mathrm{f}\left(\mathrm{y}\right)\right| \le$ (x - y)² , x, y $\in$ R and f(0) = 0, then f(1) is equal to

• 2

• 1

• - 1

• 0

Let f(x) = $$\begin{gathered} \left\{- 2\sin \left(\mathrm{x}\right), - \mathrm{\pi } \le \mathrm{x} \le - \frac{\mathrm{\pi }}{2} \\ \mathrm{asin}\left(\mathrm{x}\right) + \mathrm{b}, - \frac{\mathrm{\pi }}{2} \le \mathrm{x} \le \frac{\mathrm{\pi }}{2} \\ \cos \left(\mathrm{x}\right), \frac{\mathrm{\pi }}{2} \le \mathrm{x} \le \mathrm{\pi }\right\} \end{gathered}$$. If f(x) is continuous on $\left[- \mathrm{\pi }, \mathrm{\pi }\right]$, then

• a = 1, b = 1

• a = - 1, b = - 1

• a = - 1, b = 1

• a = 1, b = - 1

If f(x) = $$\begin{gathered} \left\{\frac{\mathrm{x}}{1 + \exp \left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.}\right) }, \mathrm{x} \ne 0 \\ 0 , \mathrm{x} = 0\right\} \end{gathered}$$, then f(x) at x = 0 is

• continuous

• not continuous

• differentiable

• not differentiable