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Q.
Let $f(x)=\frac{\alpha x}{x+1}, x \neq-1$. Then for what value of $\alpha$ is $f(f(x))=x$ ?
Relations and Functions - Part 2
Solution:
$y = f ( x )=\frac{\alpha x }{ x +1} ; x \neq-1 $
$f [ f ( x )]= x $
$f \left[\frac{\alpha x }{ x +1}\right]=\frac{\alpha \cdot \alpha x }{( x +1)\left(\frac{\alpha x }{ n +1}+1\right)}= x$
$\Rightarrow \frac{\alpha^2 x }{\alpha x + x +1}=\frac{ x }{1} $
$\frac{\alpha^2 x }{ x (\alpha+1)+1}=\frac{ x }{1} $
$\text { comparing } \alpha^2=1 \& \alpha+1=0 $
$\therefore \alpha=-1$