Let f(x)=@x/x+1, x is not equal.to -1.Then write the value of @ stisfying f(f(x))=x for all x is not equal to -1.

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{ax}}{\mathrm{x}+1} ,\mathrm{x}\ne 1 \\ \mathrm{f}\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}=\frac{\mathrm{af}\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)+1} \\ =\frac{\mathrm{a}\frac{\mathrm{ax}}{\mathrm{x}+1}}{\frac{\mathrm{ax}}{\mathrm{x}+1}+1} \\ =\frac{{\mathrm{a}}^2\mathrm{x}}{\mathrm{ax}+\mathrm{x}+1} \\ \Rightarrow \\ \mathrm{f}\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}=\frac{{\mathrm{a}}^2\mathrm{x}}{\mathrm{ax}+\mathrm{x}+1} \\ \mathrm{We} \mathrm{have} \\ \mathrm{f}\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}=\mathrm{x} \\ \Rightarrow \frac{{\mathrm{a}}^2\mathrm{x}}{\mathrm{ax}+\mathrm{x}+1}=\mathrm{x} \\ \Rightarrow {\mathrm{a}}^2\mathrm{x}={\mathrm{ax}}^2+{\mathrm{x}}^2+\mathrm{x} \\ \Rightarrow {\mathrm{a}}^2\mathrm{x}={\mathrm{x}}^2\left(\mathrm{a}+1\right)+\mathrm{x} \\ \mathrm{Comparing} \mathrm{coefficients} \mathrm{of} \mathrm{x} \mathrm{and} {\mathrm{x}}^2 \mathrm{we} \mathrm{get}, \\ \mathrm{a}+1=0 \\ \Rightarrow \mathrm{a}=-1 \\ \mathrm{and} \\ {\mathrm{a}}^2=1 \\ \Rightarrow \mathrm{a}=\pm 1 \\ \mathrm{So} \mathrm{we} \mathrm{have} \mathrm{a}=-1 \mathrm{and} \mathrm{a}=\pm 1, \mathrm{taking} \mathrm{common} \mathrm{value} \mathrm{we} \mathrm{get}, \\ \mathbf{a}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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