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

$f\left(\frac{\alpha x}{x+1}\right)=\frac{\alpha \left(\frac{\alpha x}{x+1}\right)}{\frac{\alpha x}{x+1}+1}$

$=\frac{{\alpha }^{2}x}{\alpha x+x+1}$

As, it is given that $f\left(x\right)=x$, so,

$\frac{{\alpha }^{2}x}{\alpha x+x+1}=x$

${\alpha }^{2}x=x\left(\alpha x+x+1\right)$

${\alpha }^2=\alpha x+x+1$

${\alpha }^2-\alpha x-\left(x+1\right)=0$

${\alpha }^2-\alpha x-\alpha +\alpha -\left(x+1\right)=0$

$\alpha \left[\alpha -\left(x+1\right)\right]+1\left[\alpha -\left(x+1\right)\right]=0$

$\left(\alpha +1\right)\left[\alpha -\left(x+1\right)\right]=0$

$\alpha +1=0,\alpha -\left(x+1\right)=0$

$\alpha =-1,\alpha =x+1$

Therefore, the value of $\alpha$ is $-1$.