Solution:
(i) Neither one-one nor onto
$\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ given by $\mathrm{f}(\mathrm{x})=|\mathrm{x}|=\left\{\begin{array}{l}\mathrm{x}, \text { if } \mathrm{x} \geq 0 \\ -\mathrm{x}, \text { if } \mathrm{x} \leq 0\end{array}\right.$
For One-one
$\begin{array}{l}
f(x)=f(y) \\
|x|=|y| \\
x=y \text { or } x=-y
\end{array}$
Therefore, it is not one-one.
For Onto
We know that $\mathrm{f}(\mathrm{x})=|\mathrm{x}|$ is always non-negative. So, there won’t be any element in domain $\mathrm{R}$ for which $\mathrm{f}(\mathrm{x})$ is negative.
Therefore, it is not onto.
Hence, $\mathrm{f}(\mathrm{x})=|\mathrm{x}|$ is neither one-one nor onto.
(ii) Onto but not one-one
$\mathrm{f}(\mathrm{x})=|\mathrm{x}|$ from the set of Real numbers to the Set of Whole numbers.
For one-one
$\begin{array}{l}
f(x)=f(y) \\
|x|=|y|
\end{array}$
$\mathrm{x}=\mathrm{y}$ and $-\mathrm{x}=\mathrm{y}(\mathrm{x}$ is a real number, so can be positive or negative)
Therefore, it is not one-one.
For onto
Every element is set of Real numbers will have a value in set of Whole numbers, as a result, it is onto.