Let f: R → R be defined as

$f: \mathbf{R} \rightarrow \mathbf{R}$ is defined as $f(x)=x^{4}$.

Let $x, y \in \mathbf{R}$ such that $f(x)=f(y)$.

$\Rightarrow x^{4}=y^{4}$

$\Rightarrow x=\pm y$

$\therefore f\left(x_{1}\right)=f\left(x_{2}\right)$ does not imply that $x_{1}=x_{2}$.

For instance,

$f(1)=f(-1)=1$

∴ f is not one-one.

Consider an element 2 in co-domain R. It is clear that there does not exist any x in domain R such that f(x) = 2.

∴ f is not onto.

Hence, function f is neither one-one nor onto.

The correct answer is D.