Question

For a suitably chosen real constant a, let a function, $f:R-\{-a\}\to R$ be defined by $f\left(x\right)=\frac{(a-x)}{(a+x)}$ . Further suppose that for any real number $x\ne -aandf\left(x\right)\ne -a,\left(fof\right)\left(x\right)=x$. Then $f\left(\frac{-1}{2}\right)$ is equal to:

Answer 1

Finding the value of $f\left(\frac{-1}{2}\right)$:

Given that $f:R-\{-a\}\to R$ and $f\left(x\right)=\frac{(a-x)}{(a+x)}$

$$\begin{gathered} f\left(x\right)=\frac{(a-x)}{(a+x)} \\ \Rightarrow f\left(f\right(x\left)\right)=\frac{(a-f(x)}{(a+f(x)}=x \\ \Rightarrow \frac{(a-ax)}{(1+x)}=f\left(x\right)=\frac{(a-x)}{(a+x)} \\ \Rightarrow a\frac{(1-x)}{(1+x)}=\frac{(a-x)}{(a+x)} \\ \Rightarrow (a-ax)(a+x)=(a-x)(1+x) \\ \Rightarrow a^2+ax-a^{2}x-a{x}^2=a+ax-x-x^2 \\ \Rightarrow x^2(1-a)+x(1-a)(1+a)+a(a-1)=0 \\ \Rightarrow (1-a)(x^2+x(1+a)-a)=0 \\ \Rightarrow a=1 \end{gathered}$$

So, $f\left(x\right)=\frac{(1-x)}{(1+x)}$

$f\left(\frac{-1}{2}\right)=3$

Hence the value of $\mathit{f}\mathbf{\left(}\frac{\mathbf{-}\mathbf{1}}{\mathbf{2}}\mathbf{\right)}$ is $\mathbf{3}\mathbf{.}$