Question

The function : $R\to [-\frac{1}{2},\frac{1}{2}]$ defined as $f\left(x\right)=\frac{x}{1+x^2}$ is

• A. invertible
• B. injective but not surjective
• C. surjective but not injective
• D. neither injective nor surjective

Answer 1

The correct option is C surjective but not injective

We have, $f\left(x\right)=\frac{x}{1+x^2}$
$\therefore f\left(\frac{1}{x}\right)=\frac{\frac{1}{x}}{1+\frac{1}{x^2}}=\frac{x}{1+x^2}=f\left(x\right)\therefore f\left(\frac{1}{2}\right)=f\left(2\right)orf\left(\frac{1}{3}\right)=f\left(3\right)andsoon$
So, f(x) is many – one function.
Again, let $y=f\left(x\right)\Rightarrow y=\frac{x}{1+x^2}$
$\Rightarrow y+x^{2}y=x\Rightarrow y{x}^2-x+y=0AsxϵR\therefore (-1{)}^2-4(y\left)\right(y)\ge 0\Rightarrow 1-4y^2\ge 0\Rightarrow yϵ[-\frac{1}{2},\frac{1}{2}]$
$\therefore$ Range = Codomain $=[-\frac{1}{2},\frac{1}{2}]$