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If $f(x)=\frac{\alpha x}{x+1}, x \neq-1$. Then, for what value of $\alpha$ is $f(f(x))=x$
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Verified Answer
The correct answer is:
$-1$
$\begin{aligned}
& f(f(x))=\frac{\alpha f(x)}{f(x)+1}=\frac{\alpha\left(\frac{\alpha x}{x+1}\right)}{\left(\frac{\alpha x}{x+1}+1\right)}=\frac{\alpha^2 \cdot x}{\alpha x+x+1} \\
& \quad x=\frac{\alpha^2 \cdot x}{(\alpha+1) x+1} \text { or } x\left((\alpha+1) x+1-\alpha^2\right)=0
\end{aligned}$
or $(\alpha+1) x^2+\left(1-\alpha^2\right) x=0$. This should hold for all $x$.
$\Rightarrow \alpha+1=0,1-\alpha^2=0, \quad \therefore \alpha=-1 \text {. }$
& f(f(x))=\frac{\alpha f(x)}{f(x)+1}=\frac{\alpha\left(\frac{\alpha x}{x+1}\right)}{\left(\frac{\alpha x}{x+1}+1\right)}=\frac{\alpha^2 \cdot x}{\alpha x+x+1} \\
& \quad x=\frac{\alpha^2 \cdot x}{(\alpha+1) x+1} \text { or } x\left((\alpha+1) x+1-\alpha^2\right)=0
\end{aligned}$
or $(\alpha+1) x^2+\left(1-\alpha^2\right) x=0$. This should hold for all $x$.
$\Rightarrow \alpha+1=0,1-\alpha^2=0, \quad \therefore \alpha=-1 \text {. }$
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