Consider the function $f: R→R$ defined by $f(x)=\frac{x}{x^2+1}$ then

$f$ is neither one-one nor onto

The correct answer is Option (4) → $f$ is neither one-one nor onto

Given $f(x)=\frac{x}{x^{2}+1}$ for $x\in\mathbb{R}$.

$f'(x)=\frac{(x^{2}+1)-2x^{2}}{(x^{2}+1)^{2}}=\frac{1-x^{2}}{(x^{2}+1)^{2}}$.

$f'(x)=0\Rightarrow x=\pm1$.

At $x=-1$, $f(-1)=-\frac{1}{2}$; at $x=1$, $f(1)=\frac{1}{2}$.

Hence range of $f(x)$ is $\left[-\frac{1}{2},\frac{1}{2}\right]$.

Since domain $\mathbb{R}$ and range is bounded, $f$ is not onto $\mathbb{R}$.

Also, $f(x)$ is not one-one because $f(-x)=-f(x)$ and takes same absolute values for $x$ and $-x$ (e.g., $f(2)=\frac{2}{5}, f(1/2)=\frac{1/2}{1.25}=\frac{2}{5}$).

$f$ is neither one-one nor onto.