For a suitable chosen real constant $a$, let a 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 a 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

$\begin{aligned} & (f \circ f)(x)=\frac{a-\frac{a-x}{a+x}}{a+\frac{a-x}{a+x}}=x \ & \Rightarrow \frac{a^2+a x-a+x}{a^2+a x+a-x}=x \ & \Rightarrow(a+1) x+\left(a^2-aight)=\left(a^2+aight) x+(a-1) x^2 \ & \Rightarrow a+1=a^2+a, a^2-a=0, a-1=0 \ & \Rightarrow a=1 \ & \Rightarrow f(x)=\frac{1-x}{1+x} \ & \Rightarrow f\left(-\frac{1}{2}ight)=\frac{1+\frac{1}{2}}{1-\frac{1}{2}}=3\end{aligned}$

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

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