If the coordinate axes are rotated through an angle $\frac{\mathrm{\pi }}{6}$ about the origin, then the transformed equation of $\sqrt{3}{\mathrm{x}}^2 - 4\mathrm{xy} + \sqrt{3}{\mathrm{y}}^2 = 0 \mathrm{is}$

• $\sqrt{3}{\mathrm{y}}^2 + \mathrm{xy} = 0$

• ${\mathrm{x}}^2 - {\mathrm{y}}^2 = 0$

• $\sqrt{3}{\mathrm{y}}^2 - \mathrm{xy} = 0$

$$\begin{gathered} \mathrm{The} \mathrm{harmonic} \mathrm{conjugate} \mathrm{of} (2, 3, 4) \mathrm{with} \\ \mathrm{respect} \mathrm{to} \mathrm{the} \mathrm{points} (3, - 2, 2), (6, - 17, - 4) \mathrm{is} \end{gathered}$$

• $\left(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\right)$

• $\left(\frac{18}{5}, - 5, \frac{4}{5}\right)$

• $\left(\frac{- 18}{5}, \frac{5}{4}, \frac{4}{5}\right)$

• $\left(\frac{18}{5}, - 5, \frac{- 4}{5}\right)$

The harmonic mean of two numbers is $- \frac{8}{5}$ and their geometric mean is 2. The quadratic equation whose roots are twice those numbers is

• ${\mathrm{x}}^2 + 5\mathrm{x} + 4 = 0$

• ${\mathrm{x}}^2 + 10\mathrm{x} + 16 = 0$

• ${\mathrm{x}}^2 - 10\mathrm{x} + 16 = 0$

• ${\mathrm{x}}^2 - 5\mathrm{x} + 4 = 0$

$\mathrm{If} \mathrm{z} \mathrm{is} \mathrm{a} \mathrm{complex} \mathrm{number} \mathrm{with} \left|\mathrm{z}\right| \ge 5. \mathrm{Then} \mathrm{the} \mathrm{least} \mathrm{value} \mathrm{of} \left|\mathrm{z} + \frac{2}{\mathrm{z}}\right| \mathrm{is}$

• $\frac{24}{5}$

• $\frac{26}{5}$

• $\frac{23}{5}$

• $\frac{29}{5}$

$\mathrm{If} \mathrm{\alpha } \mathrm{is} \mathrm{a} \mathrm{non}-\mathrm{real} \mathrm{root} \mathrm{of} {\mathrm{x}}^7 = 1, \mathrm{then} \mathrm{\alpha }(1 + \mathrm{\alpha }) (1 + {\mathrm{\alpha }}^2 + {\mathrm{\alpha }}^4) =$

• 1

• 2

• - 1

• - 2

$$\begin{gathered} \mathrm{If} \mathrm{\omega } \mathrm{is} \mathrm{a} \mathrm{complex} \mathrm{root} \mathrm{of} \mathrm{unity}, \mathrm{then} \mathrm{for} \mathrm{any} \\ \mathrm{n} > 1, \sum _{\mathrm{r} = 1}^{\mathrm{n} - 1}\mathrm{r}\left(\mathrm{r} + 1 - \mathrm{\omega }\right)\left(\mathrm{r} +\hspace{0.17em}1 - {\mathrm{\omega }}^2\right) = \end{gathered}$$

• $\frac{{\mathrm{n}}^2{\left(\mathrm{n} + 1\right)}^2}{4}$

• $\frac{\mathrm{n}\left(\mathrm{n} +\hspace{0.17em}1\right)\left(2\mathrm{n} +\hspace{0.17em}1\right)}{6}$

• $\frac{\mathrm{n}\left(\mathrm{n} - 1\right)}{4}\left({\mathrm{n}}^2 + 3\mathrm{n} +\hspace{0.17em}4\right)$

• $\frac{\mathrm{n}\left(\left(\mathrm{n} +\hspace{0.17em}1\right)\left(2\mathrm{n} + 1\right)\right)}{4}$

$$\begin{gathered} \mathrm{If} \mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma } \mathrm{are} \mathrm{the} \mathrm{roots} \mathrm{of} {\mathrm{x}}^3 + {\mathrm{px}}^2 + \mathrm{qx} +\hspace{0.17em}\mathrm{r} = 0 \\ \mathrm{then} \mathrm{the} \mathrm{value} \mathrm{of} \left(1 + {\mathrm{\alpha }}^2\right)\left(1 + {\mathrm{\beta }}^2\right)\left(1 + {\mathrm{\gamma }}^2\right) \mathrm{is} \end{gathered}$$

• ${\left(\mathrm{r} - \mathrm{p}\right)}^2 + {\left(\mathrm{r} - \mathrm{q}\right)}^2$

• ${\left(1 + \mathrm{p}\right)}^2 + {\left(1 + \mathrm{q}\right)}^2$

• ${\left(\mathrm{r}\hspace{0.17em}+ \mathrm{p}\right)}^2 + {\left(\mathrm{q} + 1\right)}^2$

• ${\left(\mathrm{r} - \mathrm{p}\right)}^2 + {\left(\mathrm{q} - 1\right)}^2$

$$\begin{gathered} \mathrm{Let} \mathrm{\alpha } \mathrm{and} \mathrm{\beta } \mathrm{be} \mathrm{the} \mathrm{roots} \mathrm{of} \mathrm{the} \mathrm{equation}, 5{\mathrm{x}}^2 + 6\mathrm{x} – 2 = 0. \\ \mathrm{If} \mathrm{Sn} = {\mathrm{\alpha }}^{\mathrm{n}}+ {\mathrm{\beta }}^{\mathrm{n}}, \mathrm{n} = 1, 2, 3,..., \mathrm{then} : \end{gathered}$$

• $6{\mathrm{S}}_6 +\hspace{0.17em}5{\mathrm{S}}_5 = 2{\mathrm{S}}_4$

• $5{\mathrm{S}}_6 +\hspace{0.17em}6{\mathrm{S}}_5 = 2{\mathrm{S}}_4$

• $5{\mathrm{S}}_6 +\hspace{0.17em}6{\mathrm{S}}_5 +\hspace{0.17em}2{\mathrm{S}}_4 = 0$

• $6{\mathrm{S}}_6 + 5{\mathrm{S}}_5 +\hspace{0.17em}2{\mathrm{S}}_4 = 0$

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

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

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

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

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

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