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

• $\frac{7}{2}$

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

• $\mathrm{IR} - \left\{1, 5\right\}$

• $\left\{1, 3, 5\right\}$

• $\left\{1, 5\right\}$

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

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

• $\frac{1}{5}$

• $\frac{1}{3}$

• $\frac{1}{7}$

• $\frac{1}{4}$

$\mathrm{The} \mathrm{domain} \mathrm{of} \mathrm{the} \mathrm{function} \mathrm{f}\left(\mathrm{x}\right) = {\sin }^{-1}\left(\frac{\left|\mathrm{x}\right| + 5}{{\mathrm{x}}^2 + 1}\right) \mathrm{is} \left( - \infty , - \mathrm{a}\right) \cup \left(\mathrm{a}, \infty \right), \mathrm{Then} \mathrm{a} =?$

• $\frac{1 + \sqrt{17}}{2}$

• $\frac{\sqrt{17} - 1}{2}$

• $\frac{\sqrt{17}}{2}$

• $\frac{\sqrt{17}}{2} + 1$

105.

Let f(x) be a quadratic polynomial such that f( – 1) + f(2) = 0. If one of the roots of f(x) = 0 is 3, then its other root lies in :

• (1, 3)

• ( - 1, 0)

• ( - 3, - 1)

• (0, 1)

$$\begin{gathered} \mathrm{If} \mathrm{the} \mathrm{tangent} \mathrm{tothe} \mathrm{curve}, \mathrm{y} = \mathrm{f}\left(\mathrm{x}\right) = {\mathrm{xlog}}_{\mathrm{e}}\left(\mathrm{x}\right), (\mathrm{x} > 0) \mathrm{at} \mathrm{a} \mathrm{point} (\mathrm{c}, \mathrm{f}(\mathrm{c}\left)\right) \mathrm{is} \mathrm{parallel} \\ \mathrm{to} \mathrm{the} \mathrm{line}-\mathrm{segment} \mathrm{joining} \mathrm{the} \mathrm{points} (1, 0) \mathrm{and} (\mathrm{e}, \mathrm{e}), \mathrm{then} \mathrm{c} \mathrm{is} \mathrm{equal} \mathrm{to} : \end{gathered}$$

• $\frac{\mathrm{e} - 1}{\mathrm{e}}$

• $\frac{1}{\mathrm{e} - 1}$

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

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