For a suitable chosen real constant a, let the function $f: R-\{-a\} ightarrow R$ be defined by $f(x)=\frac{a-x}{a+x}$. Further, suppose that for any real number $x
eq-a$ and $f(x)
eq-a,(f o f)(x)=x$. Then, $f\left(-\frac{1}{2} ight)$ is equal to

Question

For a suitable chosen real constant a, let the function $f: R-\{-a\} ightarrow R$ be defined by $f(x)=\frac{a-x}{a+x}$. Further, suppose that for any real number $x 
eq-a$ and $f(x) 
eq-a,(f o f)(x)=x$. Then, $f\left(-\frac{1}{2}ight)$ is equal to

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { fof }(x)=f(f(x))=f\left(\frac{a-x}{a+x}ight)=x \ & \Rightarrow \frac{a-\frac{a-x}{a+x}}{a+\frac{a-x}{a+x}}=x \ & \Rightarrow \frac{a^2+a x-a+x}{a^2+a x+a-x}=x \ & \Rightarrow a^2+a x-a+x=a^2 x+a x^2+a x-x^2 \ & \Rightarrow(a-1) x^2+\left(a^2-1ight) x-a(a-1)=0 \ & \Rightarrow(a-1)(x+a)(x-1)=0 \ & \Rightarrow a=1[\operatorname{as} x 
eq-a] \ & \Rightarrow f(x)=\frac{1-x}{1+x} \ & \Rightarrow f\left(-\frac{1}{2}ight)=\frac{1+\frac{1}{2}}{1-\frac{1}{2}}=3\end{aligned}$

For a suitable chosen real constant a, let the function $f: R-\{-a\} \rightarrow R$ be defined by…

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