If 1, 2, 3 and 4 are the roots of the equation x⁴ + ax³ + bx² + cx + d = 0, then a + 2b + c is equal to

• - 25

• 0

• 10

• 24

If $\mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma }$ are the roots of x³ - 2x² + 3x - 4 = 0, then the value of ${\mathrm{\alpha }}^2{\mathrm{\beta }}^2 + {\mathrm{\beta }}^2{\mathrm{\gamma }}^2 + {\mathrm{\gamma }}^2{\mathrm{\alpha }}^2$ is

• - 7

• - 5

• - 3

• 0

$\mathrm{The} \mathrm{locus} \mathrm{of} \mathrm{the} \mathrm{point} \mathrm{z} = \mathrm{x} + \mathrm{iy} \mathrm{satisfying} \left|\begin{array}{c}\frac{\mathrm{z} - 2\mathrm{i}}{\mathrm{z} + 2\mathrm{i}}\end{array}\right| = 1 \mathrm{is}$

• x-axis

• y-axis

• y = 0

• x = 2

$\mathrm{A} \mathrm{value} \mathrm{of} \mathrm{n} \mathrm{such} \mathrm{that} {\left(\frac{\sqrt{3}}{2} +\hspace{0.17em}\frac{\mathrm{i}}{2}\right)}^{\mathrm{n}} = 1 \mathrm{is}$

• 12

• 3

• 2

• 1

If a = $\frac{1 - \mathrm{i}\sqrt{3}}{2}$, then the correct matching of List-I from List-II is

         List-I                       List-II

(i)      $\mathrm{a}\overline{\mathrm{a}}$                            $- \frac{\mathrm{\pi }}{3}$

(ii)    $\arg \left(\frac{1}{\overline{\mathrm{a}}}\right)$                       $- \mathrm{i}\sqrt{3}$

(iii)    $\mathrm{a} - \overline{\mathrm{a}}$                        $\raisebox{1ex}{$2\mathrm{i}$}\!\left/ \!\raisebox{-1ex}{$\sqrt{3}$}\right.$ 

(iv)     $\mathrm{Im}\left(\frac{4}{3\mathrm{a}}\right)$                    1

                                        $\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$

                                        $\frac{2}{\sqrt{3}}$

correct match is 

The points in the set $\left\{\mathrm{z} \in \mathrm{C} : \arg \left(\frac{\mathrm{z} - 2}{\mathrm{z} - 6\mathrm{i}}\right) = \frac{\mathrm{\pi }}{2}\right\}$ (where C denotes the set of all complex numbers) lie on the curve which is a

• circle

• pair of lines

• parabola

• hyperbola

If w is a complex cube root of unity, then $\sin \left\{\left({\mathrm{w}}^{10} + {\mathrm{w}}^{23}\right)\mathrm{\pi } - \frac{\mathrm{\pi }}{4}\right\}$ is equal to

• $\frac{1}{\sqrt{2}}$

• $\frac{1}{2}$

• 1

• $\frac{\sqrt{3}}{2}$

If m₁, m₂, m₃ and m₄ respectively denote the moduli of the complex numbers 1 + 4i, 3 + i, 1 - i and 2 - 3i, then the correct one, among the following is

• m₁ < m₂ < m₃ < m₄

• m₄ < m₃ < m₂ < m₁

• m₃ < m₂ < m₄ < m₁

• m₃ < m₁ < m₂ < m₄

$$\begin{gathered} \mathrm{If} \mathrm{\alpha } + \mathrm{\beta } = - 2 \mathrm{and} {\mathrm{\alpha }}^3 + {\mathrm{\beta }}^3 = - 56, \mathrm{then} \\ \mathrm{the} \mathrm{quadratic} \mathrm{equation} \mathrm{whose} \mathrm{roots} \mathrm{are} \mathrm{\alpha } \mathrm{and} \mathrm{\beta } \end{gathered}$$

• x² + 2x - 16 = 0

• x² + 2x + 15 = 0

• x² + 2x - 12 = 0

• x² + 2x - 8 = 0

310.

The cubic equation whose roots are thrice to each of the roots of x³ + 2x² - 4x + 1 = 0 is

• ${\mathrm{x}}^3 + 6{\mathrm{x}}^2 - 36\mathrm{x} + 27 = 0$

• ${\mathrm{x}}^3 + 6{\mathrm{x}}^2 + 36\mathrm{x} + 27 = 0$

• ${\mathrm{x}}^3 - 6{\mathrm{x}}^2 - 36\mathrm{x} + 27 = 0$

• ${\mathrm{x}}^3 - 6{\mathrm{x}}^2 + 36\mathrm{x} + 27 = 0$