Let f be any continuously differentiable function on [a, b] and twice differentiable on (a, b) such that f(a) = f"(a) = 0 and f(b) = 0. Then,

• f''(a) = 0

• f'(x) = 0 for some $\mathrm{x} \in \left(\mathrm{a}, \mathrm{b}\right)$

• $\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right) \ne 0 \mathrm{for} \mathrm{some} \mathrm{x} \in \left(\mathrm{a}, \mathrm{b}\right)$

• $\mathrm{f}\text{'}\text{'}\text{'}\left(\mathrm{x}\right) = 0 \mathrm{for} \mathrm{some} \mathrm{x} \in \left(\mathrm{a}, \mathrm{b}\right)$

Let f : $\mathrm{R} \to \mathrm{R}$  be such that f(2x - 1) = f(x) for all $\mathrm{x} \in \mathrm{R}$. If f is continuous at x = 1 and f(1) = 1, then

• f(2) = 1

• f(2) = 2

• f is continuous only at x = 1

• f is continuous at all points

Let f(x) be a differentiable function in [2, 7]. If f(2) = 3 and f'(x) $\le$ 5 for all x in (2, 7), then the maximum possible value of f(x) at x = 7 is

• 7

• 15

• 28

• 14

The function $\mathrm{f}\left(\mathrm{x}\right) = \frac{\tan \left\{\mathrm{\pi }\left[\mathrm{x} - {\displaystyle \frac{\mathrm{\pi }}{2}}\right]\right\}}{2 + {\left[\mathrm{x}\right]}^2}$, where $\left[\mathrm{x}\right]$ denotes the greatest integer $\le$ x, is

• continuous for all values of x

• discontinuous at x = $\frac{\mathrm{\pi }}{2}$

• not differentiable for some values of x

• discontinuous at x = - 2

The function f (x) = $\mathrm{f}\left(\mathrm{x}\right) = \mathrm{asin}\left|\mathrm{x}\right| + {\mathrm{be}}^{\left|\mathrm{x}\right|}$ is differentiable at x = 0 when

• 3a + b = 0

• 3a - b = 0

• a + b = 0

• a - b = 0

Let R be the set of all real numbers and f : [- 1, 1] $\to$ R be defined by

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

• f satisfies tile conditions of Rolle's theorem on [-1, 1]

• f satisfies the conditions of Lagrange's mean value theorem on [-1, 1]

• f satisfies the conditions of Rolle's theorem on [0, 1]

• f satisfies the conditions of Lagrange's mean value theorem on [0, 1]

Suppose that f(x) is a differentiable function such that f'(x) is continuous, f'(0) = 1 and f''(0) does not exist. Let g(x) = xf'(x), Then,

• g'(0) does not exist

• g'(0) = 0

• g'(0) = 1

• g'(0) = 2

Applying Lagrange's Mean Value Theorem for a suitable function f(x) in [0, h], we have f(h) = f(0) + hf'($\mathrm{\theta h}$), $0 < \mathrm{\theta } < 1$. Then, for f(x) = cos(x), the value of $\underset{\mathrm{h}\to 0^{+}}{\lim }\mathrm{\theta }$ is

• 1

• 0

• 1/2

• 1/3

$$\begin{gathered} \mathrm{Let} \mathrm{f}\left(\mathrm{x}\right) = \left\{{\int }_0^{\mathrm{x}}\left|1 - \mathrm{t}\right|d\mathrm{t}, \mathrm{x} > 0 \\ \mathrm{x} - \frac{1}{2}, \mathrm{x} \le 1\right\} \end{gathered}$$. Then

• f(x) is continuous at x = 1

• f(x) is not continuous at x = 1

• f(x) is differentiable at x = 1

• f(x) is not differentiable at x = 1

670.

If f(x) = $$\begin{gathered} \left\{{\mathrm{x}}^3 - 3\mathrm{x} + 2, \mathrm{x} < 2, \\ {\mathrm{x}}^3 - 6{\mathrm{x}}^2 + 9\mathrm{x} + 2, \mathrm{x} \ge 2\right\} \end{gathered}$$

then

• $\underset{\mathrm{x}\to 2}{\lim }\mathrm{f}\left(\mathrm{x}\right)$ does not exist

• f is not continuous at x = 2

• f is continuous but not differentiable at x = 2

• f is continuous and differentiable at x = 2