Prove that the function $f: R \rightarrow R$, given by $f(x)=2 x,$ is one-one and onto.
$f$ is one-one, as $f\left(x_{1}\right)$ $=f\left(x_{2}\right) \Rightarrow 2 x_{1}$ $=2 x_{2} \Rightarrow x_{1}=x_{2} .$ Also, given any real number $y$ in $R$ there exists $\frac{y}{2}$ in $R$ such that $f\left(\frac{y}{2}\right)$ $=2 \cdot\left(\frac{y}{2}\right)=y .$ Hence, $f$ is onto.
Let $X$ be a non-empty set and let $P(X)$ denote the collection of all subsets of $X$. Define $f: X \times P(X) \rightarrow R$ by $f(x, A)=\left\{\begin{array}{ll}1, & \text { if } x \in A \\ 0, & \text { if } x \notin A^*\end{array}\right.$ Then, $f(x, A \cup B)$ equals
Let $\mathrm{f}: N \rightarrow N$ be a function such that $\mathrm{f}(\mathrm{m}+\mathrm{n})=\mathrm{f}(\mathrm{m})+\mathrm{f}(\mathrm{n})$ for every $\mathrm{m}, \mathrm{n} \in N$. If $\mathrm{f}(6)=18$ then $\mathrm{f}(2) \cdot \mathrm{f}(3)$ is equal to :
The function $f(x) = \;|px - q|\; + r|x|,\;x \in ( - \infty ,\;\infty )$, where $p > 0,\;q > 0,\;r > 0$ assumes its minimum value only at one point, if
If the domain of the function $f(x)=\sin ^{-1}\left(\frac{x-1}{2 x+3}\right)$ is $R-(\alpha, \beta)$ then $12 \alpha \beta$ is equal to :
If $\mathrm{R}=\left\{(\mathrm{x}, \mathrm{y}): \mathrm{x}, \mathrm{y} \in \mathrm{Z}, \mathrm{x}^{2}+3 \mathrm{y}^{2} \leq 8\right\}$ is a relation on the set of integers $\mathrm{Z},$ then the domain of $\mathrm{R}^{-1}$ is