$$\begin{gathered} \mathrm{Given} \mathrm{that} \mathrm{a}, \mathrm{b} \in \left\{0, 1, 2, . . . , 9\right\} \mathrm{with} \\ \mathrm{a} +\hspace{0.17em}\mathrm{b} \ne 0 \mathrm{and} \mathrm{that} {\left(\mathrm{a} + \frac{\mathrm{b}}{10}\right)}^{\mathrm{x}} = {\left(\frac{\mathrm{a}}{\mathrm{b}} + \frac{\mathrm{b}}{100}\right)}^{\mathrm{y}} = 1000. \mathrm{Then}, \\ \frac{1}{\mathrm{x}} - \frac{1}{\mathrm{y}} \mathrm{is} \mathrm{equal} \mathrm{to} \end{gathered}$$

• 1

• $\frac{1}{2}$

• $\frac{1}{3}$

• $\frac{1}{4}$

12.

$\mathrm{If} \mathrm{x} = \frac{1}{2}\left(\sqrt{7} + \frac{1}{\sqrt{7}}\right), \mathrm{then} \frac{\sqrt{{\mathrm{x}}^2 - 1}}{\mathrm{x} - \sqrt{{\mathrm{x}}^2 - 1}} \mathrm{is} \mathrm{equal} \mathrm{to}$

• 1

• 2

• 3

• 4

$\mathrm{For} \mathrm{any} \mathrm{integer} \mathrm{n} \ge 1, \mathrm{the} \mathrm{sum} \sum _{\mathrm{k} = 1}^{\mathrm{n}}\mathrm{k}\left(\mathrm{k} + 2\right) \mathrm{is} \mathrm{equal} \mathrm{to}$

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

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

• $\frac{\mathrm{n}\left(\mathrm{n} + 1\right)\left(2\mathrm{n} + 7\right)}{6}$

• $\frac{\mathrm{n}\left(\mathrm{n} +\hspace{0.17em}1\right)\left(2\mathrm{n} + 9\right)}{6}$

9 balls are to be placed in 9 boxes and 5 of the balls cannot fit into 3 small boxes. The number of ways of arranging one ball in each of the boxes is

• 18720

• 18270

• 17280

• 12780

$\mathrm{If} {}^{\mathrm{n}}\mathrm{P}_{\mathrm{r}} = 30240 \mathrm{and} {}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}} = 252, \mathrm{then} \mathrm{the} \mathrm{ordered} \mathrm{pair} \left(\mathrm{n}, \mathrm{r}\right) \mathrm{is} \mathrm{equal} \mathrm{to}$

• (12, 6)

• (10, 5)

• (9, 4)

• (16, 7)

$\mathrm{If} {\left(1 +\hspace{0.17em}\mathrm{x} + {\mathrm{x}}^2 + {\mathrm{x}}^3\right)}^5 = \sum _{\mathrm{k} = 0}^{15}{\mathrm{a}}_{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}, \mathrm{then} \underset{\mathrm{k} = 0}{\overset{7}{\sum {\mathrm{a}}_2}}\mathrm{k} = 0$

• 128

• 256

• 512

• 1024

$$\begin{gathered} \mathrm{If} \mathrm{\alpha } = \frac{5}{2 ! 3} + \frac{5 . 7}{3 ! 3^2} + \frac{5 . 7. 9}{4! 3^3} +\hspace{0.17em} . . . , \mathrm{then} \\ {\mathrm{\alpha }}^2 + 4\mathrm{\alpha } \mathrm{is} \mathrm{equal} \mathrm{to} \end{gathered}$$

• 21

• 23

• 25

• 27

$\mathrm{If} \frac{{\mathrm{x}}^2 + \mathrm{x} + 1}{{\mathrm{x}}^2 + 2\mathrm{x} + 1} = \mathrm{A} +\hspace{0.17em}\frac{\mathrm{B}}{\mathrm{x} +\hspace{0.17em}1} + \frac{\mathrm{C}}{{\left(\mathrm{x} + 1\right)}^2}, \mathrm{then} \mathrm{A} - \mathrm{B} \mathrm{is} \mathrm{equal} \mathrm{to}$

• 4C

• 4C + 1

• 3C

• 2C

$\sum _{\mathrm{k} = 1}^{\infty }\frac{1}{\mathrm{k}!}\left(\sum _{\mathrm{n} = 1}^{\mathrm{k}}{2}^{\mathrm{n} - 1}\right) \mathrm{is} \mathrm{equal} \mathrm{to}$

• e

• e² + e

• e²

• e² - e

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

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

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

• $2{\log }_{\mathrm{e}}\left(4\right)$

• ${\log }_{\mathrm{e}}\left(4\right)$