Let $f(x)=\frac{\alpha x}{x+1}, x \neq-1$. Then, for what value of $\alpha$ is $f(f(x))=x ?$

Solution:
(d) $-1$
$f(f(x))=x$
$\Rightarrow f\left(\frac{\alpha x}{x+1}\right)=x$
$\Rightarrow \frac{\alpha\left(\frac{a x}{x+1}\right)}{\left(\frac{a x}{x+1}\right)+1}=x$
$\Rightarrow \frac{\alpha^{2} x}{\alpha x+x+1}=x$
$\Rightarrow \alpha^{2} x=\alpha x^{2}+x^{2}+x$
$\Rightarrow \alpha^{2} x-\alpha x^{2}-x^{2}-x=0$
$\Rightarrow \alpha^{2} x-\alpha x^{2}-\left(x^{2}+x\right)=0$
Solving for the $\alpha$ we get,
$\alpha=\frac{-\left(-x^{2}\right) \pm \sqrt{\left(-x^{2}\right)^{2}-4 \times x \times\left[-\left(x^{2}+x\right)\right]}}{2 x}$
$=\frac{x^{2} \pm \sqrt{x^{4}+4 x^{3}+4 x^{2}}}{2 x}$
$=x+1,-1$
Here, $-1$ is independent of $x$,
$\therefore$ for, $\alpha=-1, f(f(x))=x$