Show that the function $f: R_* \rightarrow R_*$ defined by $f(x)=\frac{1}{x}$ is one-one and onto, where $R_*$ is the set of all non-zero real numbers. Is the result true, if the domain $R_*$ is replaced by $N$ with co-domain being same as $R _*$ ?

If $f(x) = \frac{x}{{x – 1}} = \frac{1}{y}$, then $f(y) = $

Let $f: R \rightarrow R$ be a function defined by $f(x)=\left\{\begin{array}{l}\frac{\sin \left(x^2\right)}{x} \text { if } x \neq 0 \\ 0 \text { if } x=0\end{array}\right\}$ Then, at $x=0, f$ is

If $f(x) = \log \frac{{1 + x}}{{1 – x}}$, then $f(x)$ is