Let $R$ denote the set of all real numbers and $R^{+}$denote the set of all positive real numbers. For the subsets $A$ and $B$ of $R$ define $f: A \rightarrow B$ by $f(x)=x^2$ for $x \in A$. Observe the two lists given below

Given, $f(x)=-x^2, f: A \rightarrow B$ where, $A$ and $B \in R$ $R \rightarrow$ Real No's $x \in A$ Here, the domain of the function is positive real number only ie, Domain $=A \in R^{+}$ Example $\{1,2,3, \ldots\}$ and $B=\{1,4,9, \ldots\}$ Both set have $f: A \rightarrow B$ unique image. But, $A=\{-1,1,2, \ldots\}$ and $B=\{1,1,4, \ldots\}$ In $f: A \rightarrow B$ have not unique image and for Range $x^2=y, x= \pm \sqrt{y}$ ie, (Range $\in R^{+}$) In onto function the (Range $=$ Co-domain $=B$ ) ie, $\left(B \in R^{+}\right)$ So, (A) $f$ is one-one and onto, if $A=B=R^{+}$ (B $f$ is one-one but not onto, if $A=R^{+}, B=R$ (C) $f$ is onto but not one-one, if $A=R, B=R^{+}$ (D) $f$ is neither one-one nor onto, if $A=B=R$ Hence, the answer is $(\mathrm{A}) \rightarrow 4,(\mathrm{~B}) \rightarrow 1,(\mathrm{C}) \rightarrow 3,(\mathrm{D}) \rightarrow 2$