Let $f(x)=\frac{αx}{x+1},x≠-1$. If $f(f(x))=x$, then value of $α$ is

$\sqrt{2}$ $-\sqrt{2}$ 1 -1

-1

$f(f(x))=\frac{αf(x)}{f(x)+1}=\frac{α^2x}{(α+1)x+1}$ Thus $f ( f (x)) = x⇔α^2x = (α + 1)x^2 + x$ $⇔(α^2-1)x = (α + 1)x^2$ $⇔(α+1)((α - 1)-x)=0$ This equation is true for all x, if $α = –1$