C.

ad = bc

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right) = \frac{\mathrm{ax} + \mathrm{d}}{\mathrm{cx} + \mathrm{d}} ... \left(\mathrm{i}\right) \\ \mathrm{Now} \mathrm{take} \mathrm{option} \left(\mathrm{c}\right) \\ \mathrm{i}.\mathrm{e}. \frac{\mathrm{a}}{\mathrm{b}} = \frac{\mathrm{c}}{\mathrm{d}} = \mathrm{k} \\ \mathrm{From} \mathrm{equation} \left(\mathrm{i}\right) \\ \mathrm{f}\left(\mathrm{x}\right) = \frac{\mathrm{bkx} + \mathrm{b}}{\mathrm{dkx} + \mathrm{d}} \\ = \frac{\mathrm{b}\left(\mathrm{kx} + 1\right)}{\mathrm{d}\left(\mathrm{kx} + 1\right)} = \frac{\mathrm{b}}{\mathrm{d}} = \mathrm{constant} \end{gathered}$$

D.

$\left\{- 15, - 14, . . . 0, . . . 13, 14\right\}$

$$\begin{gathered} \mathrm{Given}, \mathrm{f} :\hspace{0.17em}\mathrm{R} \to \mathrm{R} \mathrm{and} \mathrm{f}\left(\mathrm{x}\right) = \left[\frac{\mathrm{x}}{5}\right] \\ \mathrm{Also}, \\ \left\{\mathrm{f}\left(\mathrm{x}\right) : \left|\mathrm{x}\right| < 71\right\} = \left\{\mathrm{f}\left(\mathrm{x}\right) : - 71 < \mathrm{x} < 71\right\} \\ = \left\{\left[\frac{\mathrm{x}}{5}\right] : - 71 < \mathrm{x} < 71\right\} \\ = \left\{\left[\frac{- 70 - 0.1}{5}\right], . . . ,\left[\frac{- 70}{5}\right], . . . , \left[\frac{0}{5}\right], . . . ,\left[\frac{65}{5}\right], . . . , \left[\frac{70}{5}\right], . . . ,\left[\frac{70 + 0.999}{\mathrm{h}}\right]\right\} \\ = \left\{- 15, - 14, . . . , 0, . . . , 13, 14\right\} \end{gathered}$$

A.

22.5

$$\begin{gathered} \sum _{\mathrm{k} = 1}^5\frac{1^3 + 2^3 + .... + {\mathrm{k}}^3}{1 + 3 + 5 + ... + \left(2\mathrm{k} - 1\right)} \\ = \sum _{\mathrm{k} = 1}^5\frac{{\left({\displaystyle \frac{\mathrm{k}\left(\mathrm{k} + 1\right)}{2}}\right)}^2}{{\mathrm{k}}^2} \\ \left[\because \sum {\mathrm{n}}^3 = {\left(\frac{\mathrm{n}\left(\mathrm{n} + 1\right)}{2}\right)}^2 \mathrm{and} 1 + 3 + 5 + ... + \left(2\mathrm{k} - 1\right) = {\mathrm{k}}^2\right] \\ = \sum _{\mathrm{k} = 1}^5\frac{{\left(\mathrm{k} + 1\right)}^2}{4} = \frac{2^2 + 3^2 + 4^2 + 5^2 + 6^2}{4} \\ = \frac{4 + 9 + 16 + 25 + 36}{4} \\ = \frac{90}{4} = 22.5 \end{gathered}$$

C.

3

$$\begin{gathered} \mathrm{Since}, \mathrm{d}\left(\mathrm{n}\right)\mathrm{represents} \mathrm{number} \mathrm{of} \mathrm{the} \mathrm{divisors} \mathrm{of} \mathrm{n}. \\ \therefore \mathrm{d}\left({\mathrm{p}}^7\right) = 8 \\ \mathrm{d}\left(8\right) = \mathrm{d}\left(2^3\right) = 4 \\ \mathrm{d}\left(4\right) = \mathrm{d}\left(2^2\right) = 3 \\ \mathrm{d}\left(\mathrm{d}\left(\mathrm{d}\left({\mathrm{p}}^7\right) \right)\right) = 3 \end{gathered}$$