The region represented by $\left\{\mathrm{z} = \mathrm{x} + \mathrm{iy} \in \mathrm{C} : \left|\mathrm{z}\right| - \mathrm{Re}\left(\mathrm{z}\right) \le 1\right\}$ is also given by the inequality

• ${\mathrm{y}}^2 \ge 2\left(\mathrm{x} + 1\right)$

• ${\mathrm{y}}^2 \le \mathrm{x} + \frac{1}{2}$

• ${\mathrm{y}}^2 \le 2\mathrm{x} +\hspace{0.17em}1$

• ${\mathrm{y}}^2 \ge \mathrm{x} + 1$

If $\mathrm{\alpha } \mathrm{and} \mathrm{\beta }$ be two roots of the equation x² – 64x + 256 = 0. Then the value of ${\left(\frac{{\mathrm{\alpha }}^3}{{\mathrm{\beta }}^5}\right)}^{\frac{1}{8}} + {\left(\frac{{\mathrm{\beta }}^3}{{\mathrm{\alpha }}^5}\right)}^{\frac{1}{8}} = ?$

• 2

• 4

• 4

• 3

293.

$\mathrm{If} \mathrm{\alpha } \mathrm{and} \mathrm{\beta } \mathrm{are} \mathrm{the} \mathrm{roots} \mathrm{of} \mathrm{the} \mathrm{equation} 2\mathrm{x}(2\mathrm{x} + 1) = 1, \mathrm{then} \mathrm{\beta } \mathrm{is} \mathrm{equal} \mathrm{to} :$

• $2\mathrm{\alpha }\left(\mathrm{\alpha } - 1\right)$

• $2\mathrm{\alpha }\left(\mathrm{\alpha } + 1\right)$

• $2{\mathrm{\alpha }}^2$

• $- 2\mathrm{\alpha }\left(\mathrm{\alpha } + 1\right)$

If the complex numbers z₁, z₂, z₃ and z₄ denote the vertices of a square taken in order. If z₁ = 3 + 4i and z₃ = 5 + 6i, then the other two vertices z₂ and z₄ are respectively

• 5 + 4i, 5 + 6i

• 5 + 4i, 3 + 6i

• 5 + 6i, 3 + 5i

• 3 + 6i, 5 + 3i

$\mathrm{If} \frac{\left(1 + \mathrm{i}\right)\mathrm{x} - \mathrm{i}}{2 + \mathrm{i}} + \frac{\left(1 + 2\mathrm{i}\right)\mathrm{y} + \mathrm{i}}{2 - \mathrm{i}} = 1, \mathrm{then} \left(\mathrm{x}, \mathrm{y}\right) =?$

• $\left(\frac{7}{3}, \frac{- 7}{15}\right)$

• $\left(\frac{7}{3}, \frac{7}{15}\right)$

• $\left(\frac{7}{5}, \frac{- 7}{15}\right)$

• $\left(\frac{7}{5}, \frac{7}{15}\right)$