For the circle C with the equation x² + y² - 16 - 12y + 64 = 0 match the List I with the List II given below.

The correct match is

$\mathrm{If} \mathrm{f}\left(\mathrm{x}\right) = {\left(\mathrm{p} - {\mathrm{x}}^{\mathrm{n}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{n}$}\right.}, \mathrm{p} > 0 \mathrm{and} \mathrm{n} \mathrm{is} \mathrm{a} \mathrm{positive} \mathrm{integer}, \mathrm{then} \mathrm{f}\left[\mathrm{f}\left(\mathrm{x}\right)\right] = ?$

• x

• xⁿ

• p^1/n

• p - xⁿ

$\mathrm{The} \mathrm{value} \mathrm{of} \left\{\mathrm{x} \in \mathrm{R} ⋮ \left[\log \left(\log {\left(1.6\right)}^{1 - {\mathrm{x}}^2} - {\left(0.625\right)}^{6\left(1 + \mathrm{x}\right)}\right)\right] \in \mathrm{R}\right\} \mathrm{is}$

• $\left(- \infty , - 1\right) \cup \left(7, 0\right)$

• (- 1, 5)

• (1, 7)

• (- 1, 7)

$\mathrm{If} { }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r} - 1} = 330, {}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}} = 462, \mathrm{and}{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r} + 1} = 462, \mathrm{then} \mathrm{r} = ?$

• 3

• 4

• 5

• 6

10 men and 6 women are to be seated in a row so that no two women sit together. The number of ways they can be seated, is

• 11! 10!

• $\frac{11!}{6! 5!}$

• $\frac{10! 9!}{5!}$

• $\frac{11! 10!}{5!}$

If t_n denotes the number of triangles formed with n points in a plane, no three of which are collinear and if t_{n + 1} - t_n= 36_, then n is equal to

• 7

• 8

• 9

• 10

$$\begin{gathered} \mathrm{The} \mathrm{term} \mathrm{independent} \mathrm{of} \mathrm{x}\left(\mathrm{x} > 0, \mathrm{x} \ne 1\right) \mathrm{in} \mathrm{the} \mathrm{expansion} \mathrm{of} \\ {\left[\frac{\left(\mathrm{x} + 1\right)}{{\mathrm{x}}^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}} - {\mathrm{x}}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}} + 1} - \frac{\left(\mathrm{x} - 1\right)}{\left(\mathrm{x} - \sqrt{\mathrm{x}}\right)}\right]}^{10} \mathrm{is} \end{gathered}$$

• 105

• 210

• 315

• 420

If x is small, so that x² and higher powers can be neglected, then the approximate value for $\frac{{\left(1 - 2\mathrm{x}\right)}^{ - 1}{\left(1 - 3\mathrm{x}\right)}^{ - 2}}{{\left(1 - 4\mathrm{x}\right)}^{ - 3}} \mathrm{is}$

• 1 - 2x

• 1 - 3x

• 1 - 4x

• 1 - 5x

$\mathrm{If} \frac{1}{{\mathrm{x}}^4 + {\mathrm{x}}^2 + 1} = \frac{\mathrm{Ax} +\hspace{0.17em}\mathrm{B}}{{\mathrm{x}}^2 + \mathrm{x} +\hspace{0.17em}1 } + \frac{\mathrm{Cx} + \mathrm{D}}{{\mathrm{x}}^2 - \mathrm{x} +\hspace{0.17em}1}, \mathrm{then} \mathrm{C} + \mathrm{D} = ?$

• - 1

• 1

• 2

• 0

$\frac{1}{2 . 3} + \frac{1}{4 . 5} + \frac{1}{6 . 7} + \frac{1}{8 . 9} + ... \infty = ?$

• $\log \left(\frac{2}{\mathrm{e}}\right)$

• $\log \left(\frac{\mathrm{e}}{2}\right)$

• $\log \left(2\mathrm{e}\right)$

• e - 1