$Iff\left(x\right)=\frac{\alpha x}{x+1},x\ne -1,thenthevalueof\alpha forwhichf\left(f\left(x\right)\right)=xis$

$\begin{array}{l}f\left(x\right)=\frac{\alpha x}{x+1}\\ \Rightarrow f\left(f\left(x\right)\right)=x\Rightarrow f\left(\frac{\alpha x}{x+1}\right)=x\\ \Rightarrow \frac{\alpha \left(\frac{\alpha x}{x+1}\right)}{\left(\frac{\alpha x}{x+1}\right)+1}=x\\ \Rightarrow \frac{{\alpha }^{2}x}{\alpha x+x+1}=x\end{array}$

$\begin{array}{l}\Rightarrow {\alpha }^{2}x-\alpha x^2-x^2-x=0\\ \Rightarrow \left({\alpha }^2-1\right)x-x^2\left(\alpha +1\right)=0\\ \Rightarrow \left(\alpha +1\right)\left(\alpha -1\right)x-x^2\left(\alpha +1\right)=0\\ \Rightarrow \left(\alpha +1\right)\left[\left(\alpha -1\right)x-x^2\right]=0\\ \therefore \alpha +1=0\Rightarrow \alpha =-1\end{array}$