Explanation

(d)
$\because f(x) = ax + b$
$\therefore f^{'}(x) = a > 0$
$\Rightarrow f(x)$ is an increasing function.
$\therefore f( - 1) = 0\ and\ f(1) = 2$
Or $–a + b = 0$
$and\ a + b = 2$
then, $a = b = 1$
$\Rightarrow f(x) = x + 1$
Now, cot $\lbrack\cot^{- 1}{7 + \cot^{- 1}{8 + \cot^{- 1}{18\rbrack}}}$
$= \cot\left\{ \tan^{- 1}{\left( \frac{1}{7} \right) + \tan^{- 1}{\left( \frac{1}{8} \right) + \tan^{- 1}\left( \frac{1}{18} \right)}} \right\}$
$= \cot\left\{ \tan^{- 1}\left( \frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7} \bullet \frac{1}{8}} \right) + \tan^{- 1}\left( \frac{1}{18} \right) \right\}$
$= \cot\left\{ \tan^{- 1}{\left( \frac{15}{35} \right) + \tan^{- 1}\ \left( \frac{1}{18} \right)} \right\}$
$= \cot\left\{ \tan^{- 1}{\left( \frac{3}{11} \right) + \tan^{- 1}\left( \frac{1}{18} \right)} \right\}$
$= \cot\left\{ \tan^{- 1}\left( \frac{\frac{3}{11} + \frac{1}{18}}{1 - \frac{3}{1 \bullet 1} \bullet \frac{1}{18}} \right) \right\}$
$= \cot\left\{ \tan^{- 1}\left( \frac{65}{195} \right) \right\}$
$= \cot\left\{ \tan^{- 1}\left( \frac{1}{3} \right) \right\}$
$= \cot{(\cot^{- 1}{3) = 3 = 1 + 2 = f(2)}}$