Question

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

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

Answer 1

The correct option is D $-1$

We have
$f\left\{f\right(x\left)\right\}=x$ for all $x\ne -1$
$\Rightarrow f\left(\frac{\alpha x}{x+1}\right)=x$ for all $x\ne -1$
$\Rightarrow \frac{\alpha \left(\frac{\alpha x}{x+1}\right)}{\left(\frac{\alpha x}{x+1}\right)+1}=x$ for all $x\ne -1$
$\Rightarrow \frac{{\alpha }^{2}x}{\alpha x+x+1}=x$ for all $x\ne -1$
$\Rightarrow (\alpha +1)x^2+(1-{\alpha }^2)x=0$
$\Rightarrow \alpha +1=0;1-{\alpha }^2=0$
$\therefore \alpha =-1$