For a suitably chosen real constant a, let a

function, $f:R - \left\{ { - a} \right\} \to R$ be defined by

$f(x) = {{a - x} \over {a + x}}$. Further suppose that for any real number $x \ne - a$ and $f(x) \ne - a$,

(fof)(x) = x. Then $f\left( { - {1 \over 2}} \right)$ is equal to :

Given, $f(x) = {{a - x} \over {a + x}}$ <br><br>and (fof)(x) = x <br><br>$\Rightarrow$ f(f(x)) = ${{a - f\left( x \right)} \over {a + f\left( x \right)}}$ = x<br><br>$\Rightarrow$ $${{a - \left( {{{a - x} \over {a + x}}} \right)} \over {a + \left( {{{a - x} \over {a + x}}} \right)}}$$ = x <br><br>$\Rightarrow$ ${{{a^2} + ax - a + x} \over {{a^2} + ax + a + x}}$ = x <br><br>$\Rightarrow$ (a² – a) + x(a + 1) = (a² + a)x + x²(a – 1) <br><br>$\Rightarrow$ a(a – 1) + x(1 – a²) – x²(a – 1) = 0 <br><br>$\Rightarrow$ a = 1 <br><br>$\therefore$ f(x) = ${{1 - x} \over {1 + x}}$ <br><br>So, $f\left( { - {1 \over 2}} \right)$ = ${{1 + {1 \over 2}} \over {1 - {1 \over 2}}}$ = 3