Question

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

• A.
• B.
• C. 1
• D. -1

Answer 1

The correct option is D

-1

$f\left(f\right(x\left)\right)=\frac{\alpha f\left(x\right)}{f\left(x\right)+1}=\frac{\alpha \left(\frac{az}{z+1}\right)}{(\frac{az}{z+1}+1)}=\frac{{\alpha }^2.x}{ax+x+1}$

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

Or $(\alpha +1)x^2+(1-{\alpha }^2)x=0$.This should hold for all $x$

$\Rightarrow \alpha +1=0,1-{\alpha }^2=0,\therefore \alpha =-1$