Explanation

(d)
$\because\ \ \ \ \ \ f\left( f(x) \right) = f\left( \frac{\alpha x}{x + 1} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{\alpha\left( \frac{\alpha x}{x + 1} \right)}{\left( \frac{\alpha x}{x + 1} \right) + 1} = \frac{\alpha^{2}x}{ax + x + 1}$
$\Rightarrow \ \ \ \ \ \ \frac{\alpha^{2}x}{ax + x + 1} = x$ [given]
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \alpha^{2} = \alpha x + x + 1$
$\Rightarrow \ \ \ \ \ \ \ \alpha^{2} - 1 = (\alpha + 1)x$
$\Rightarrow \ \ \ \ \ \ \ (\alpha + 1)(\alpha - 1 - x) = 0$
$\Rightarrow \ \ \ \ \ \ \ \ \alpha + 1 = 0\ \ \ \Rightarrow \ \ \ \ \ \alpha = - 1\ \ \ \ \ \ \ \ \ \ \ \lbrack\therefore\alpha - 1 - x \neq 0\rbrack$