For a suitable chosen real constant a, let a function \(\mathrm{f}: \mathbb{R}-\{-\mathrm{a}\} \rightarrow \mathbb{R}\) be defined by \(\mathrm{f}(x)=\frac{\mathrm{a}-x}{\mathrm{a}+x}\). Further suppose that for any real number \(x \neq-\mathrm{a}\) and \(\mathrm{f}(x) \neq-\mathrm{a}\), (fof) \((x)=x\). Then \(\mathrm{f}\left(-\frac{1}{5}\right)\) is equal to

Given: \(\mathrm{f}(x)=\frac{\mathrm{a}-x}{\mathrm{a}+x}\)
\(\begin{aligned}
& \because \quad \mathrm{f}(\mathrm{f}(x))=x \\
& \quad \Rightarrow \frac{\mathrm{a}-\mathrm{f}(x)}{\mathrm{a}+\mathrm{f}(x)}=x \\
& \quad \Rightarrow \frac{\mathrm{a}-\left(\frac{\mathrm{a}-x}{\mathrm{a}+x}\right)}{\mathrm{a}+\left(\frac{\mathrm{a}-x}{\mathrm{a}+x}\right)}=x \\
& \quad \Rightarrow \frac{\mathrm{a}^2+\mathrm{a} x-\mathrm{a}+x}{\mathrm{a}^2+\mathrm{a} x+\mathrm{a}-x}=x \\
& \quad \Rightarrow\left(\mathrm{a}^2-\mathrm{a}\right)+(\mathrm{a}+1) x=\left(\mathrm{a}^2+\mathrm{a}\right) x+(\mathrm{a}-1) x^2 \\
& \quad \Rightarrow(\mathrm{a}-1) x^2+\left(\mathrm{a}^2-1\right) x-\mathrm{a}^2+\mathrm{a}=0 \\
& \quad \Rightarrow(\mathrm{a}-1)\left[x^2+(\mathrm{a}+1) x-\mathrm{a}\right]=0
\end{aligned}\)
This is possible when \(\mathrm{a}=1\)
\(\begin{aligned} & \therefore \quad f(x)=\frac{1-x}{1+x} \\ & \begin{aligned} f\left(\frac{-1}{5}\right) & =\frac{1-\left(\frac{-1}{5}\right)}{1+\left(\frac{-1}{5}\right)} \\ & =\frac{1+\frac{1}{5}}{1-\frac{1}{5}} \\ & =\frac{\frac{6}{5}}{\frac{4}{5}}=\frac{6}{4}=1.5\end{aligned}\end{aligned}\)