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  4. Let f (x)=(ax/x+1), x ≠-1. Then value of the constant a such that f(f(x)) = x is

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Q. 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

Relations and Functions - Part 2

Solution:

$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 $a=-1$.