Let f be twice differentiable function such that f"(x) = - f(x) and f'(x) = g(x), h(x) = {f(x)}² + {g(x)}². If h(5) = 11, then h(10) is equal to :

• 22

• 11

• 0

• 20

A differentiable function f(x) is defined for all x > 0 and satisfies f(x³) = 4x⁴ for all x > 0. The value of f'(8) is :

• $\frac{16}{3}$

• $\frac{32}{3}$

• $\frac{16\sqrt{2}}{3}$

• $\frac{32\sqrt{2}}{3}$

If the derivative of the function f(x) is every where continuous and is given by

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right) = \left\{{\mathrm{bx}}^2 + \mathrm{ax} + 4; \mathrm{x} \ge - 1 \\ {\mathrm{ax}}^2 + \mathrm{b}; \mathrm{x} < - 1\right\} \end{gathered}$$, then :

• a = 2, b = - 3

• a = 3, b = 2

• a = - 2, b = - 3

• a = - 3, b = - 2

If f(x + y) = f(x)f(y) for all real x and y, f(6) = 3 and f'(0) = 10, then f'(6) is :

• 30

• 13

• 10

• 0

Derivative of ${\sec }^{-1}\left(\frac{1}{1 - 2{\mathrm{x}}^2}\right)$ w . r. t ${\sin }^{-1}\left(3\mathrm{x} - 4{\mathrm{x}}^3\right)$ is :

• $\frac{1}{4}$

• $\frac{3}{2}$

• 1

• $\frac{2}{3}$

Let f : $\mathrm{R} \to \mathrm{R}$ be a diiferentiable function Satisfyingf'(3) + f'(2) = 0. Then $\underset{\mathrm{x}\to 0}{\lim }{\left(\frac{1 + \mathrm{f}\left(3 + \mathrm{x}\right) - \mathrm{f}\left(3\right)}{1 + \mathrm{f}\left(2 - \mathrm{x}\right) - \mathrm{f}\left(2\right)}\right)}^{\frac{1}{\mathrm{x}}}$ is equal to :

• 1

• e

• e^{- 1}

• e²

Let f : [- 1, 3] $\to$ R be defined as f(x) = $$\begin{gathered} \left\{\left|\mathrm{x}\right| + \left[\mathrm{x}\right], - 1 \le \mathrm{x} < 1 \\ \mathrm{x} + \left|\mathrm{x}\right|, 1 \le \mathrm{x} < 2 \\ \mathrm{x} +\hspace{0.17em}\left[\mathrm{x}\right], 2 \le \mathrm{x} \le 3\right\} \end{gathered}$$ where [t] denotes the greatest integer less than or equal to t. Then, f is discontinuous at :

• only three points

• only one point

• only two points

• four or more points

178.

If the function f(x) = $$\begin{gathered} \left\{\mathrm{a}|\mathrm{\pi } - \mathrm{x}| + 1, \mathrm{x} \le 5 \\ \mathrm{b}\left|\mathrm{x} - \mathrm{\pi }\right| + 3, \mathrm{x} > 5\right\} \end{gathered}$$ is continuous at x = 5, then the value of a - b is

• $\frac{- 2}{\mathrm{\pi } + 5}$

• $\frac{2}{\mathrm{\pi } + 5}$

• $\frac{2}{\mathrm{\pi } - 5}$

• $\frac{2}{5 - \mathrm{\pi }}$

If $\mathrm{f}\left(\mathrm{x}\right) = \left[\mathrm{x}\right] - \left[\frac{\mathrm{x}}{4}\right], \mathrm{x} \in \mathrm{R}$ , where [x] denotes the greatest integer function, then:

• Both $\underset{\mathrm{x}\to 4^{-}}{\lim }\mathrm{f}\left(\mathrm{x}\right)$ and $\underset{\mathrm{x}\to 4^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right)$ exist but noot equal

• f is continuous at x = 4

• $\underset{\mathrm{x}\to 4^{-}}{\lim }\mathrm{f}\left(\mathrm{x}\right)$ exist but $\underset{\mathrm{x}\to 4^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right)$ does not exist

• $\underset{\mathrm{x}\to 4^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right)$ exist but $\underset{\mathrm{x}\to 4^{-}}{\lim }\mathrm{f}\left(\mathrm{x}\right)$ does not exist

If the function f defined on $\left(\frac{\mathrm{\pi }}{6}, \frac{\mathrm{\pi }}{3}\right)$ by f(x) = $$\begin{gathered} \left\{\frac{\sqrt{2}\cos \left(\mathrm{x}\right) - 1}{\cot \left(\mathrm{x}\right) - 1}, \mathrm{x} \ne \frac{\mathrm{\pi }}{4} \\ \mathrm{k}, \mathrm{x} = \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 4}}\right\} \end{gathered}$$ is continuous, then k is equal to :

• 2

• $\frac{1}{2}$

• $\frac{1}{\sqrt{2}}$

• 1