• $\mathrm{x} \in \left(- \frac{3}{2}, \frac{1}{4}\right)$

• $\mathrm{x} \in \left(- \frac{3}{4}, \frac{1}{4}\right)$

• $\mathrm{x} \in \left(- \frac{3}{4}, \frac{1}{2}\right)$

• $\mathrm{x} < \frac{1}{4}$

12.

$$\begin{gathered} \mathrm{If} \mathrm{R}\to \mathrm{R} \mathrm{and} \mathrm{is} \mathrm{defined} \mathrm{by} \mathrm{f}\left(\mathrm{x}\right) = \frac{1}{2 - \cos \left(3\mathrm{x}\right)} \\ \mathrm{fpr} \mathrm{each} \mathrm{x} \in \mathrm{R}, \mathrm{then} \mathrm{the} \mathrm{range} \mathrm{of} \mathrm{f} \mathrm{is} \end{gathered}$$

• $\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right., - 1\right)$

• $\left[\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right., 1\right]$

• (1, 2)

• $\left[1, 2\right]$

$\sqrt{2+ \sqrt{5} - \sqrt{6 - 3\sqrt{5} + \sqrt{14 - 6\sqrt{5}}}} = ?$

• 1

• 2

• 3

• 4

If $\frac{3\mathrm{x}}{\left(\mathrm{x} - \mathrm{a}\right)\left(\mathrm{x} - \mathrm{b}\right)} = \frac{2}{\mathrm{x} - \mathrm{a}} + \frac{1}{\mathrm{x} - \mathrm{b}}$, then a : b is equal to

• 1 : 2

• - 2 : 1

• 1 : 3

• 3 : 1

$$\begin{gathered} \mathrm{f} : [-6, 6] \mathrm{R} \mathrm{is} \mathrm{defined} \mathrm{by} \mathrm{f}\left(\mathrm{x}\right) = {\mathrm{x}}^2 - 3 \\ \mathrm{for} \mathrm{x} \in \mathrm{R}, \mathrm{then} \\ \left(\mathrm{fofof}\right) (-1) + \left(\mathrm{fofof}\right) \left(0\right) + \left(\mathrm{fofof}\right)\left(1\right) \mathrm{is} \mathrm{equal} \mathrm{to} \end{gathered}$$

• $\mathrm{f}\left(4\sqrt{2}\right)$

• $\mathrm{f}\left(3\sqrt{2}\right)$

• $\mathrm{f}\left(2\sqrt{2}\right)$

• $\mathrm{f}\left(\sqrt{2}\right)$

$\mathrm{If} \frac{{\mathrm{x}}^2 + \mathrm{x} + 1}{{\mathrm{x}}^2 + 2\mathrm{x} + 1} = \mathrm{A} +\hspace{0.17em}\frac{\mathrm{B}}{\mathrm{x} +\hspace{0.17em}1} + \frac{\mathrm{C}}{{\left(\mathrm{x} + 1\right)}^2}, \mathrm{then} \mathrm{A} - \mathrm{B} \mathrm{is} \mathrm{equal} \mathrm{to}$

• 4C

• 4C + 1

• 3C

• 2C

$$\begin{gathered} \mathrm{If} \mathrm{f} : \left[2, 3\right]\to \mathrm{R} \mathrm{is} \mathrm{defined} \mathrm{by} \mathrm{f} \left(\mathrm{x}\right) = {\mathrm{x}}^3 + 3\mathrm{x} - 2, \mathrm{then} \mathrm{the} \mathrm{range} \mathrm{f}\left(\mathrm{x}\right) \mathrm{is} \\ \mathrm{contained} \mathrm{in} \mathrm{the} \mathrm{interval} \end{gathered}$$

• [1, 12]

• [12, 34]

• [35, 50]

• [- 12, 12]

The longest distance of the point (a, 0) from the curve 2x² + y² = 2x is

• 1 + a

• $\left|1 - \mathrm{a}\right|$

• $\sqrt{1 - 2\mathrm{a} +2{\mathrm{a}}^2}$

• $\sqrt{1 - 2\mathrm{a} + 3{\mathrm{a}}^2}$

If f : R $\to$ R is defined by f(x) = $\left[\frac{\mathrm{x}}{5}\right]$ for x ∈ R, where [y] denotes the greatest integer not exceeding y, then $\left\{\mathrm{f}\left(\mathrm{x}\right) : \left|\mathrm{x}\right| < 71\right\}$ is equal to

• $\left\{- 14, - 13, . . . 0, . . . 13, 14\right\}$

• $\left\{- 14, - 13, . . . 0, . . . 14, 15\right\}$

• $\left\{- 15, - 14, . . . 0, . . . 14, 15\right\}$

• $\left\{- 15, - 14, . . . 0, . . . 13, 14\right\}$

$$\begin{gathered} \frac{{\mathrm{x}}^2 + \mathrm{x} + 1}{\left(\mathrm{x} - 1\right)\left(\mathrm{x} - 2\right)\left(\mathrm{x} - 3\right)} = \frac{\mathrm{A}}{\mathrm{x} - 1} + \frac{\mathrm{B}}{\mathrm{x} - 2} + \frac{\mathrm{C}}{\mathrm{x} - 3} \\ \Rightarrow \mathrm{A} + \mathrm{C} = \end{gathered}$$

• 4

• 5

• 6

• 8