Question

Let $f\left(x\right)=\frac{\mathrm{\alpha }x}{x+1},x\ne -1$. Then write the value of α satisfying f(f(x)) = x for all x ≠ −1.

Answer 1

Given:
$f\left(x\right)=\frac{\mathrm{\alpha }x}{x+1},x\ne -1$

$$\begin{gathered} \mathrm{Since}f\left(f\right(x\left)\right)=x, \\ \frac{\alpha \left({\displaystyle \frac{\alpha x}{x+1}}\right)}{{\displaystyle \frac{\alpha x}{x+1}}+1}=x \\ \Rightarrow \frac{{\alpha }^{2}x}{\alpha x+x+1}=x \\ \Rightarrow {\alpha }^{2}x-\alpha x^2-(x^2+x)=0 \\ \mathrm{Solving}\mathrm{the}\mathrm{quadratic}\mathrm{equation}\mathrm{in}\alpha : \\ \alpha =\frac{x^2\pm \sqrt{x^4+4x(x^2+x)}}{2x} \\ \Rightarrow \alpha =x+1\mathrm{or}-1 \\ \mathrm{Since},\alpha \ne x+1, \\ \alpha =-1. \end{gathered}$$