Question

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

Answer 1

We have,
$f\left(x\right)=\frac{ax}{x+1}$
Now,
$f\left(f\right(x\left)\right)=xf\left(\frac{ax}{x+1}\right)=x\Rightarrow f\frac{a\left(\frac{ax}{x+1}\right)}{\frac{ax}{x+1}+1}=x\Rightarrow \frac{\frac{a^{2}x}{x+1}}{\frac{ax+x+1}{x+1}}=x\Rightarrow \frac{a^{2}x}{ax+x+1}=x\Rightarrow \frac{a^2}{ax+x+1}=1\Rightarrow a^2=ax+x+1\Rightarrow a^2-ax-(x+1)=0\Rightarrow a^2-a(x+1)+a-(x+1)=0\Rightarrow a\left[\right(a)-(x+1)+1[a-(x+1)]=0\Rightarrow [a-(x+1)\left]\right[a+1]=0\Rightarrow a+1=0$
[$\because a=x+1$ does not satisfies f(f(x)) = x]
$\Rightarrow a=-1$