Question

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

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

Answer 1

Answer: (B)

$f(x)=\frac{a-x}{a+x}$
$x\in R-\{-a\}\to R$
$f(f(x))=\frac{a-f(x)}{a+f(x)}=\frac{a-\left(\frac{a-x}{a+x}\right)}{a+\left(\frac{a-x}{a+x}\right)}$
$f(f(x))=\frac{\left(a^2-a\right)+x(a+1)}{\left(a^2+a\right)+x(a-1)}=x$
$\Rightarrow \left(a^2-a\right)+x(a+1)=\left(a^2+a\right)x+x^2(a-1)$
$\Rightarrow a(a-1)+x\left(1-a^2\right)-x^2(a-1)=0$
$\Rightarrow a=1$
$f(x)=\frac{1-x}{1+x}$
$f\left(\frac{-1}{2}\right)=\frac{1+\frac{1}{2}}{1-\frac{1}{2}}=3$