Question

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

• A. $\sqrt{2}$
• B. $-\sqrt{2}$
• C. 1
• D. -1.

Answer 1

Answer: (D)

$y=f(x)=\frac{\alpha x}{x+1};x\ne -1$
$f[f(x)]=x$
$f\left[\frac{\alpha x}{x+1}\right]=\frac{\alpha \cdot \alpha x}{(x+1)\left(\frac{\alpha x}{n+1}+1\right)}=x$
$\Rightarrow \frac{{\alpha }^{2}x}{\alpha x+x+1}=\frac{x}{1}$
$\frac{{\alpha }^{2}x}{x(\alpha +1)+1}=\frac{x}{1}$
$\text{ comparing }{\alpha }^2=1\&\alpha +1=0$
$\therefore \alpha =-1$