Let f : R → R be defined by

Given: The function $f: R \rightarrow R$ be defined by $f(x)=\frac{1}{x}$.

To check f is one-one:

Let $f\left(x_{1}\right)=f\left(x_{2}\right)$

$\Rightarrow \frac{1}{x_{1}}=\frac{1}{x_{2}}$

$\Rightarrow x_{1}=x_{2}$

Hence, $f$ is one-one.

To check $f$ is onto:\

Since, $y=\frac{1}{x}$

$\Rightarrow x=\frac{1}{y}$

$\Rightarrow y \in R-\{0\} \neq R$

There is no pre-image of $y=0$.

Hence, $f$ is not onto.

Hence, the correct option is (a).