$\mathrm{If} \mathrm{A} = \left\{\mathrm{x} \in \mathrm{Rl}\frac{\mathrm{\pi }}{4} \le \mathrm{x} \le \frac{\mathrm{\pi }}{3}\right\} \mathrm{and} \mathrm{f}\left(\mathrm{x}\right) = \sin \left(\mathrm{x}\right) - \mathrm{x}, \mathrm{then} \mathrm{f}\left(\mathrm{A}\right) =?$

• $\left[\frac{\sqrt{3}}{2} - \frac{\mathrm{\pi }}{3}, \frac{1}{\sqrt{2}} - \frac{\mathrm{\pi }}{4}\right]$

• $\left[\frac{- 1}{\sqrt{2}} - \frac{\mathrm{\pi }}{4}, \frac{\sqrt{3}}{2} - \frac{\mathrm{\pi }}{3}\right]$

• $\left(- \frac{\mathrm{\pi }}{3}, - \frac{\mathrm{\pi }}{4}\right)$

• $\left[\frac{\mathrm{\pi }}{4}, \frac{\mathrm{\pi }}{3}\right]$

$\mathrm{In} \mathrm{a} ∆\mathrm{ABC}, \frac{\mathrm{a}}{\tan \left(\mathrm{A}\right)} \hspace{0.17em}+ \frac{\mathrm{b}}{\tan \left(\mathrm{B}\right)} + \frac{\mathrm{c}}{\tan \left(\mathrm{C}\right)} = ?$

• 2r

• r +2R

• 2r +R

• 2(r + R)

If f is defined in [1, 3] by f(x) = x³ + bx² + ax,such that f(1) - f(3) = 0 and f'(c) = 0, where c = 2 + $\frac{1}{\sqrt{3}}$, then (a, b) is equal to

• ( - 6, 11)

• $\left(2 - \frac{1}{\sqrt{3}}, 2 + \frac{1}{\sqrt{3}}\right)$

• (11, - 6)

• (6, 11)

The domain of the function f(x) = $\sqrt{{\log }_{0.5}\left(\mathrm{x}!\right)} \mathrm{is}$

• $\left\{0, 1, 2, 3, ...\right\}$

• $\left\{0, 1, 2, 3, ...\right\}$

• $\left(0, \infty \right)$

• $\left\{0, 1\right\}$

185.

$\mathrm{If} \mathrm{f}\left(\mathrm{x}\right) = \left|\mathrm{x} - 1\right| + \left|\mathrm{x} - 2\right| + \left|\mathrm{x} - 3\right|, 2 < \mathrm{x} < 3, \mathrm{then} \mathrm{f} \mathrm{is}$

• an onto function but not one-one

• one-one function but not onto

• a bijection

• neither one-one nor onto

If x = a is a root of multiplicity two of a polynomial equation f(x) = 0, then

• f'(a) = f''(a) = 0

• f''(a) = f(a) = 0

• $\mathrm{f}\text{'}\left(\mathrm{a}\right) \ne 0 \ne \mathrm{f}\text{'}\text{'}\left(\mathrm{a}\right)$

• $\mathrm{f}\left(\mathrm{a}\right) = \mathrm{f}\text{'}\left(\mathrm{a}\right) = 0, \mathrm{f}\text{'}\text{'}\left(\mathrm{a}\right) \ne 0$

Suppose f(x) = x(x + 3)(x - 2), x $\in$ [- 1, 4]. Then, a value of c in (- 1, 4) satisfying f'(c) = 10 is

• 2

• $\frac{5}{2}$

• 3

• $\frac{7}{2}$

Let A = {- 4, - 2, - 1, 0, 3, 5} and f : A $\to$ IR be defined by

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right) = \left\{3\mathrm{x} - 1 \mathrm{for} \mathrm{x} > 3 \\ {\mathrm{x}}^2 + 1 \mathrm{for} - 3 \le \mathrm{x} \le 3 \\ 2\mathrm{x} - 3 \mathrm{for} \mathrm{x} < - 3\right\} \\ \mathrm{Then} \mathrm{the} \mathrm{range} \mathrm{of} \mathrm{f} \mathrm{is} \end{gathered}$$

• $\left\{- 11, 5, 2, 1, 10, 14\right\}$

• $\left\{- 11, - 7, 2, 1, 8, 14\right\}$

• $\left\{- 11, 5, 2, 1, 8, 14\right\}$

• $\left\{- 11, - 7, - 5, 1, 10, 14\right\}$

$$\begin{gathered} \mathrm{If} \mathrm{f} : \mathrm{IR} \to \mathrm{IR} \mathrm{is} \mathrm{defined} \mathrm{by} \\ \mathrm{f}\left(\mathrm{x}\right) = \left\{\mathrm{x} - 1, \mathrm{for} \mathrm{x} \le 1 \\ 2 - {\mathrm{x}}^2, \mathrm{for} 1 < \mathrm{x} \le 3 \\ \mathrm{x} - 10, \mathrm{for} 3 < \mathrm{x} < 5 \\ 2\mathrm{x}, \mathrm{for} \mathrm{x} \ge 5\right\} \\ \mathrm{then} \mathrm{the} \mathrm{set} \mathrm{of} \mathrm{points} \mathrm{of} \mathrm{discontinuity} \mathrm{of} \mathrm{f} \mathrm{is} \end{gathered}$$

• $\mathrm{IR} - \left\{1, 5\right\}$

• $\left\{1, 3, 5\right\}$

• $\left\{1, 5\right\}$

• $\mathrm{IR} - \left\{1, 3, 5\right\}$

For the function f(x) = (x - 1)(x - 2) defined on [0, ½] the value of c satisfying Lagrange's mean value theorem is

• $\frac{1}{5}$

• $\frac{1}{3}$

• $\frac{1}{7}$

• $\frac{1}{4}$