Define f on R into itself by

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right) = \left\{\mathrm{xsin}\left(\frac{1}{\mathrm{x}}\right), \mathrm{when} \mathrm{x} \ne 0 \\ 0, \mathrm{when} \mathrm{x} = 0\right\} \end{gathered}$$, then :

• f is continuous at 0 but not differentiable at 0

• f is both continuous and differentiable at 0

• f is differentiable but not continuous at 0

• None of these

If x = $\mathrm{Acos}\left(4\mathrm{t}\right) + \mathrm{Bsin}\left(4\mathrm{t}\right)$, then $\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}$ is equal to

• - 16x

• 16x

• x

• - x

If f(x) = $$\begin{gathered} \left\{\frac{1 - \cos \left(\mathrm{x}\right)}{\mathrm{x}}, \mathrm{x} \ne 0 \\ \mathrm{k} , \mathrm{x} = 0\right\} \end{gathered}$$ is continuous at x = 0, then the value of k is

• 0

• $\frac{1}{2}$

• $\frac{1}{4}$

• $\frac{1}{2}$

The derivative of ${\cos }^{-1}\left(\frac{1 - {\mathrm{x}}^2}{1 + {\mathrm{x}}^2}\right) \mathrm{w}. \mathrm{r}. \mathrm{t}. {\cot }^{-1}\left(\frac{1 - 3{\mathrm{x}}^2}{3\mathrm{x} - {\mathrm{x}}^3}\right)$ is

• $\frac{3}{2}$

• 1

• $\frac{1}{2}$

• $\frac{2}{3}$

If y = 1 - x + $\frac{{\mathrm{x}}^2}{2!} - \frac{{\mathrm{x}}^3}{3!} + \frac{{\mathrm{x}}^4}{4!} ..., \mathrm{then} \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}$ is equal to

• - x

• x

• y

• - y

The function f(x) = $\left|\mathrm{x}\right| + \frac{\left|\mathrm{x}\right|}{\mathrm{x}}$ is

• discontinuous at origin because $\left|\mathrm{x}\right|$ is discontinuous there

• continuous at origin

• discontinuous at origin because both $\left|\mathrm{x}\right|$ and $\raisebox{1ex}{$\left|\mathrm{x}\right|$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.$ are discontinuous there

• discontinuous at the origin because $\raisebox{1ex}{$\left|\mathrm{x}\right|$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.$ is discontinuous there

Let f(x) = $$\begin{gathered} \left\{\frac{\sin \left(\mathrm{\pi x}\right)}{5\mathrm{x}}, \mathrm{x} \ne 0 \\ \mathrm{k}, \mathrm{x} = 0\right\} \end{gathered}$$. If f(x) continuous at x = 0, the value of k is

• $\frac{5}{\mathrm{\pi }}$

• $\frac{\mathrm{\pi }}{5}$

• zero

• 1

If f(x) = $$\begin{gathered} \left\{\mathrm{xsin}\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.\right), \mathrm{if} \mathrm{x} \ne 0 \\ 0, \mathrm{if} \mathrm{x} = 0\text{'}\right\} \end{gathered}$$, then at x = 0 the function f is

• continuous but not differentiable

• differentiable but not continuous

• continuous and differentiable

• not continuous

889.

If f(x) = $$\begin{gathered} \left\{\frac{\sin \left(5\mathrm{x}\right)}{{\mathrm{x}}^2 + 2\mathrm{x}}, \mathrm{x} \ne 0 \\ \mathrm{k} + \frac{1}{2}, \mathrm{x} = 0\right\} \end{gathered}$$ is continuous at x = 0, then the value of k is

• 1

• - 2

• 2

• 1/2

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

• $\frac{{\mathrm{x}}^2}{\sqrt{1 - {\mathrm{x}}^4}}$

• $\frac{{\mathrm{x}}^2}{\sqrt{1 + {\mathrm{x}}^4}}$

• $\frac{\mathrm{x}}{\sqrt{1 + {\mathrm{x}}^4}}$

• $\frac{\mathrm{x}}{\sqrt{1 - {\mathrm{x}}^4}}$