The function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \dfrac{x}{x^2+1}\ \forall x \in \mathbb{R}$ is

Given $f(x) = \dfrac{x}{x^2+1}$

To check for one-one, let $f(x_1) = f(x_2)$

$\dfrac{x_1}{x_1^2+1} = \dfrac{x_2}{x_2^2+1}$

$x_1 x_2^2 + x_1 = x_2 x_1^2 + x_2$

$x_1 x_2(x_2 - x_1) - (x_2 - x_1) = 0$

$(x_2 - x_1)(x_1 x_2 - 1) = 0$

This gives $x_1 = x_2$ or $x_1 x_2 = 1$.

For example, $f(2) = \dfrac{2}{5}$ and $f\left(\dfrac{1}{2}\right) = \dfrac{2}{5}$. Since $f(2) = f\left(\dfrac{1}{2}\right)$ for $2 \neq \dfrac{1}{2}$, the function is not one-one.

To check for onto, let $y = \dfrac{x}{x^2+1}$

$y x^2 - x + y = 0$

For $x$ to be real, the discriminant must be non-negative:

$1 - 4y^2 \ge 0$

$y^2 \le \dfrac{1}{4}$

$-\dfrac{1}{2} \le y \le \dfrac{1}{2}$

The range of $f(x)$ is $\left[-\dfrac{1}{2}, \dfrac{1}{2}\right]$, which is a proper subset of the codomain $\mathbb{R}$. Thus, the function is not onto.

Therefore, $f(x)$ is neither one-one nor onto.

Answer: Neither one-one nor onto