Question

Let $f\left(x\right)=\frac{x}{1+x}$ and $g\left(x\right)=\frac{rx}{1-x}$. Let $S$ be the set of all real numbers $r$ such that $f\left(g\right(x\left)\right)=g\left(f\right(x\left)\right)$ for infinitely many real $x$. The number of elements in set $S$ is

• A. $1$
• B. $2$
• C. $3$
• D. $5$

Answer 1

The correct option is B $2$

$f\left(x\right)=\frac{x}{1+x}$ and $g\left(x\right)=\frac{rx}{1-x}$
$f\left(g\left(x\right)\right)=f\left(\frac{rx}{1-x}\right)=\frac{rx}{1-x+rx}$
$g\left(f\left(x\right)\right)=g\left(\frac{x}{1+x}\right)=rx$
Given $f\left(g\left(x\right)\right)=g\left(f\left(x\right)\right)$
$\Rightarrow \frac{rx}{1-x+rx}=rx$
$\Rightarrow r(r-1)x^2=0$
Hence, two solutions are possible.