Rolle's theorem is not applicable to the function f(x) = $\left|\mathrm{x}\right|$ for $- 2 \le \mathrm{x} \le 2$ because

• f is continuous for $- 2 \le \mathrm{x} \le 2$

• f is not derivable for x = 0

• f(- 2) = f(2)

• f is not a constant function

The function f(x) which satisfies

$\mathrm{f}\left(\mathrm{x}\right) = \mathrm{f}\left(- \mathrm{x}\right) = \frac{\mathrm{f}\text{'}\left(\mathrm{x}\right)}{\mathrm{x}}$ is given by

• f(x) = $\frac{1}{2}{\mathrm{e}}^{{\mathrm{x}}^2}$

• f(x) = $\frac{1}{2}{\mathrm{e}}^{- {\mathrm{x}}^2}$

• f(x) = ${\mathrm{x}}^2{\mathrm{e}}^{{\mathrm{x}}^2/2}$

• f(x) = ${\mathrm{e}}^{{\mathrm{x}}^2/2}$

A function f(x) is defined as follows for real x

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right) = \left\{1 - {\mathrm{x}}^2, \mathrm{for} \mathrm{x} < 1 \\ 0, \mathrm{for} \mathrm{x} = 1 \\ 1 - {\mathrm{x}}^2, \mathrm{for} \mathrm{x} > 1\right\} \end{gathered}$$

Then,

• f (x) is not continuous at x = 1

• f (x) is continuous but not differentiable at x = 1

• f(x) is both continuous and differentiable at x = 1

• None of the above

Select the correct statement from (a), (b), (c), (d). The function f(x) = xe^{1 - x}

• strictly increases in the interval $\left(\frac{1}{2}, 2\right)$

• increases in the interval (0, $\infty$)

• decreases in the interval (0, 2)

• strictly decreases in the interval (1, $\infty$)

The function f(x) = ${\mathrm{e}}^{\mathrm{ax}} + {\mathrm{e}}^{- \mathrm{ax}}$, a > 0 is monotonically increasing for

• - 1 < x < 1

• x < - 1

• x > - 1

• x > 0

If y = a^x . b^{2x - 1}, then $\frac{d^2\mathrm{y}}{d{\mathrm{x}}^2}$ is

• ${\mathrm{y}}^2\log \left({\mathrm{ab}}^2\right)$

• $\mathrm{ylog}\left({\mathrm{ab}}^2\right)$

• y²

• $\mathrm{y}{\left(\log \left({\mathrm{ab}}^2\right)\right)}^2$

Let f(x) = e^x, g(x) = ${\sin }^{-1}\left(\mathrm{x}\right)$ and h(x) = f[g(x)], then $\frac{\mathrm{h}\text{'}\left(\mathrm{x}\right)}{\mathrm{h}\left(\mathrm{x}\right)}$ is equal to

• ${\mathrm{e}}^{{\sin }^{-1}\left(\mathrm{x}\right)}$

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

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

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

698.

If $\mathrm{f}\left(\mathrm{x}\right) = \left|\log \left|\mathrm{x}\right|\right|$, then

• f(x) is continuous and differentiable for all x in its domain

• f(x) is continuous for all x in its domain but not differentiable at x = ± 1

• f(x) is neither continuous nor differentiable at x = ± 1

• None of the above

The derivative of ${\tan }^{-1}\left(\frac{\sqrt{1 + {\mathrm{x}}^2} - 1}{\mathrm{x}}\right)$ with respect to ${\tan }^{-1}\left(\frac{2\mathrm{x}\sqrt{1 - {\mathrm{x}}^2}}{1 - 2{\mathrm{x}}^2}\right)$ at x = 0 is

• $\frac{1}{8}$

• $\frac{1}{4}$

• $\frac{1}{2}$

• 1

Let f(x + y) = f(x) f(y) and f(x) = 1 + sin(2x) g(x) where g(x) is continuous. Then, f'(x) equals

• f(x) g(0)

• 2f(x) g(0)

• 2g(0)

• None of the above