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 $x\ne -a$ and $f\left(x\right)\ne -a,\left(fof\right)\left(x\right)=x$. Then $f(-\frac{1}{2})$ is equal to:

• A. $-3$
• B. $3$
• C. $\frac{1}{3}$
  
• D. $-\frac{1}{3}$
  

Answer 1

The correct option is B $3$

$f\left(x\right)=\frac{a-x}{a+x}$
$f\left(f\right(x\left)\right)=\frac{a-f\left(x\right)}{a+f\left(x\right)}=x$
$\frac{a-ax}{1+x}=f\left(x\right)=\frac{a-x}{a+x}$
$a\left(\frac{1-x}{1+x}\right)=\frac{a-x}{a+x}$
$\Rightarrow a=1$
So $f\left(x\right)=\frac{1-x}{1+x}$
$\Rightarrow$ $f\left(\frac{-1}{2}\right)=3$