The imaginary part of

${\left(3 +\hspace{0.17em}2\sqrt{ - 54}\right)}^{\frac{1}{2}} - {\left(3 - 2\sqrt{ - 54}\right)}^{\frac{1}{2}} \mathrm{can} \mathrm{be}$

• $\sqrt{ - 6}$

• $\sqrt{6}$

• $- 2\sqrt{6}$

• 6

$$\begin{gathered} \mathrm{Let} \mathrm{\alpha }, \mathrm{\beta } \mathrm{are} \mathrm{roots} \mathrm{of} {\mathrm{x}}^2 + \mathrm{px} + 2 = 0 \mathrm{and} \frac{1}{\mathrm{\alpha }}, \frac{1}{\mathrm{\beta }} \mathrm{are} \mathrm{the} \mathrm{roots} \mathrm{of} 2{\mathrm{x}}^2 - 2\mathrm{qx} + 1 = 0. \\ \mathrm{Then} \mathrm{find} \mathrm{the} \mathrm{value} \mathrm{of} \left(\mathrm{\alpha } + \frac{1}{\mathrm{\beta }}\right)\left(\mathrm{\beta } + \frac{1}{\mathrm{\alpha }}\right)\left(\mathrm{\alpha } - \frac{1}{\mathrm{\alpha }}\right)\left(\mathrm{\beta } - \frac{1}{\mathrm{\beta }}\right) \end{gathered}$$

• $\frac{9}{4}\left(9 - {\mathrm{p}}^2\right)$

• $\frac{9}{4}\left(9 + {\mathrm{p}}^2\right)$

• $\frac{4}{9}\left(9 - {\mathrm{q}}^2\right)$

• $\frac{9}{4}\left(9 - {\mathrm{q}}^2\right)$

If z₁, z₂ are complex numbers such that Re(z₁) = |z₁ = 1| and Re(z₂) = |z₂ = 1| and arg (z₁ – z₂) = $\frac{\mathrm{\pi }}{6}$, then Im (z₁ + z₂) is equal to :

• $2\sqrt{3}$

• $\frac{\sqrt{3}}{2}$

• $\frac{1}{\sqrt{3}}$

• $\frac{2}{\sqrt{3}}$

If a and b are real numbers suchthat (2 + $\mathrm{\alpha }$)⁴ = a + b$\mathrm{\alpha }$, where $\mathrm{\alpha } = \frac{ - 1 + \mathrm{i}\sqrt{3}}{2}$ then a + b is equal to :

• 33

• 24

• 9

• 57

$$\begin{gathered} \mathrm{Let} \mathrm{\lambda } \ne 0 \mathrm{be} \mathrm{in} \mathrm{R}. \mathrm{If} \mathrm{\alpha } \mathrm{and} \mathrm{\beta } \mathrm{are} \mathrm{the} \mathrm{roots} \mathrm{of} \mathrm{the} \mathrm{equation}, {\mathrm{x}}^2 – \mathrm{x} + 2\mathrm{\lambda } = 0 \mathrm{and} \\ \mathrm{\alpha } \mathrm{and} \mathrm{\gamma } \mathrm{are} \mathrm{the} \mathrm{roots} \mathrm{of} \mathrm{the} \mathrm{equation}, 3{\mathrm{x}}^2 – 10\mathrm{x} + 27\mathrm{\lambda } = 0, \mathrm{then} \frac{\mathrm{\beta \gamma }}{\mathrm{\lambda }} \mathrm{is} \mathrm{equal} \mathrm{to} \end{gathered}$$

• 27

• 36

• 9

• 18

286.

$$\begin{gathered} \mathrm{If} \mathrm{\alpha } \mathrm{is} \mathrm{the} \mathrm{positive} \mathrm{root} \mathrm{of} \mathrm{the} \mathrm{equation}, \mathrm{p}\left(\mathrm{x}\right) = {\mathrm{x}}^2 – \mathrm{x} – 2 = 0, \mathrm{then} \\ \underset{\mathrm{x}\to \mathrm{\alpha }}{\lim }\frac{\sqrt{1 - \cos \left(\mathrm{p}\left(\mathrm{x}\right)\right)}}{\mathrm{x} +\hspace{0.17em}\mathrm{\alpha } - 4} = ? \end{gathered}$$

• $\frac{3}{2}$

• $\frac{3}{\sqrt{2}}$

• $\frac{1}{\sqrt{2}}$

• $\frac{1}{2}$

The product of the roots of the equation 9x² – 18|x| + 5 = 0 is

• $\frac{25}{81}$

• $\frac{5}{27}$

• $\frac{25}{9}$

• $\frac{5}{9}$

If the four complex numbers $\mathrm{z}, \overline{\mathrm{z}}, \overline{\mathrm{z}} - 2\mathrm{Re}\left(\overline{\mathrm{z}}\right) \mathrm{and} \mathrm{z} - 2\mathrm{Re}\left(\mathrm{z}\right)$ represent the vertices of a square of side 4 units in the Argand plane, then |z| is equal to :

• 4

• 2

• $4\sqrt{2}$

• $2\sqrt{2}$

$\mathrm{If} \mathrm{\alpha } \mathrm{and} \mathrm{\beta } \mathrm{are} \mathrm{the} \mathrm{roots} \mathrm{of} \mathrm{the} \mathrm{equation}, 7{\mathrm{x}}^2 – 3\mathrm{x} – 2 = 0, \mathrm{then} \mathrm{the} \mathrm{value} \mathrm{of} \frac{\mathrm{\alpha }}{1 - {\mathrm{\alpha }}^2} + \frac{\mathrm{\beta }}{1 - {\mathrm{\beta }}^2} \mathrm{is} \mathrm{equal} \mathrm{to}$

• $\frac{1}{24}$

• $\frac{27}{36}$

• $\frac{27}{16}$

• $\frac{3}{8}$

$\mathrm{The} \mathrm{value} \mathrm{of} {\left(\frac{ - 1 + \mathrm{i}\sqrt{3}}{1 - \mathrm{i}}\right)}^{30} \mathrm{is} :$

• 6⁵

• 2¹⁵i

• - 2¹⁵

•  - 2¹⁵i