# For a suitably chosen real constant a, let a

Question:

For a suitably chosen real constant a, let a function, $f: \mathrm{R}-\{-\mathrm{a}\} \rightarrow \mathrm{R}$ be defined by

$f(x)=\frac{a-x}{a+x} .$ Further suppose that for any real

number $x \neq-a$ and $f(x) \neq-a,(f \circ f)(x)=x$. Then

$f\left(-\frac{1}{2}\right)$ is equal to :

1. $\frac{1}{3}$

2. 3

3. $-3$

4. $-\frac{1}{3}$

Correct Option: , 2

Solution:

$f(x)=\frac{a-x}{a+x}$

$\mathrm{x} \in \mathrm{R}-\{-\mathrm{a}\} \rightarrow \mathrm{R}$

$f(f(\mathrm{x}))=\frac{\mathrm{a}-f(\mathrm{x})}{\mathrm{a}+f(\mathrm{x})}=\frac{\mathrm{a}-\left(\frac{\mathrm{a}-\mathrm{x}}{\mathrm{a}+\mathrm{x}}\right)}{\mathrm{a}+\left(\frac{\mathrm{a}-\mathrm{x}}{\mathrm{a}+\mathrm{x}}\right)}$

$f(f(x))=\frac{\left(a^{2}-a\right)+x(a+1)}{\left(a^{2}+a\right)+x(a-1)}=x$

$\Rightarrow\left(a^{2}-a\right)+x(a+1)=\left(a^{2}+a\right) x+x^{2}(a-1)$

$\Rightarrow a(a-1)+x\left(1-a^{2}\right)-x^{2}(a-1)=0$

$\Rightarrow a=1$

$f(x)=\frac{1-x}{1+x}$

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