Question

Let $f\left(x\right)=\frac{ax}{x+1}$, $x\ne -1$. Then value of the constant a such that $f(f(x))=x$ is

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

Answer 1

Answer: (D)

$f\left(x\right)=\frac{ax}{x+1}$
$\Rightarrow f\left(f\left(x\right)\right)=\frac{af\left(x\right)}{f\left(x\right)+1}$
$=\frac{a\left(\frac{ax}{x+1}\right)}{\frac{ax}{x+1}+1}$
$=\frac{a^{2}x}{\left(a+1\right)x+1}$
$=x$ if $a=-1$.