Consider a region R = {(x, y) $\in$ R² : x^{2 $\le \mathrm{y} \le 2\mathrm{x}$}}. If a line y = $\mathrm{\alpha }$ divides the area of region R into two equal parts, then which of the following is true ?

• $3{\mathrm{\alpha }}^2 - 8\mathrm{\alpha } + 8 = 0$

• ${\mathrm{\alpha }}^3 - 6{\mathrm{\alpha }}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} - 16 = 0$

• ${\mathrm{\alpha }}^3 - 6{\mathrm{\alpha }}^2 + 16 = 0$

• ${\text{3α}}^2 - 8{\mathrm{\alpha }}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} + 8 = 0$

$\mathrm{The} \mathrm{area} \left(\mathrm{in} \mathrm{sq} \mathrm{unit}\right) \mathrm{of} \mathrm{the} \mathrm{region} \left\{\left(\mathrm{x}, \mathrm{y}\right); \frac{1}{2} \le \mathrm{y} \le \sin \left(\mathrm{x}\right), 0 \le \mathrm{x} \le \mathrm{\pi }\right\}\mathrm{is} \mathrm{equal} \mathrm{to}$

• $3 - 2\mathrm{\pi }$

• $\sqrt{3} - \frac{\mathrm{\pi }}{6}$

• 3 - $\frac{\mathrm{\pi }}{3}$

• $\sqrt{3} - \frac{\mathrm{\pi }}{3}$

The area (in sq. units) of the region A = {x, y) : (x –1) [x] $\le \mathrm{y} \le 2\sqrt{\mathrm{x}}, 0 \le \mathrm{x} \le 2$,where [t] denotes the greatest integer function, is :

• $\frac{4}{3}\sqrt{2} - \frac{1}{2}$

• $\frac{8}{3}\sqrt{2} - 1$

• $\frac{4}{3}\sqrt{2} + 1$

• $\frac{8}{3}\sqrt{2} - \frac{1}{2}$

$\mathrm{The} \mathrm{area} (\mathrm{in} \mathrm{sq}. \mathrm{units}) \mathrm{of} \mathrm{the} \mathrm{region} \mathrm{A} = \left\{\left(\mathrm{x}, \mathrm{y}\right) :\hspace{0.17em}\left|\mathrm{x}\right| + \left|\mathrm{y}\right| \le 1, 2{\mathrm{y}}^2 \ge \left|\mathrm{x}\right|\right\} :\hspace{0.17em}$

• $\frac{5}{6}$

• $\frac{7}{6}$

• $\frac{1}{3}$

• $\frac{1}{6}$

The area (in sq. units) of the region enclosed by the curves y = x² – 1 and y = 1 – x² is equal to :

• $\frac{8}{3}$

• $\frac{7}{2}$

• $\frac{4}{3}$

• $\frac{16}{3}$