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(D) The function is defined as $f(x) = x^2$.$A$. $f$ is one-one and onto, if $A = B = R^{+}$.For $A = R^{+}$, $f(x) = x^2$ is strictly increasing, so

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(D) The function is defined as $f(x) = x^2$.$A$. $f$ is one-one and onto, if $A = B = R^{+}$.For $A = R^{+}$, $f(x) = x^2$ is strictly increasing, so it is one-one. Since $B = R^{+}$, for every $y \in R^{+}$, there exists $x = \sqrt{y} \in R^{+}$ such that $f(x) = y$, so it is onto. Thus, $A \rightarrow 4$.$B$. $f$ is one-one but not onto, if $A = R^{+}, B = R$.For $A = R^{+}$, $f(x) = x^2$ is one-one. Since $B = R$, the range is $R^{+}$, which is a proper subset of $B$, so it is not onto. Thus, $B \rightarrow 1$.$C$. $f$ is onto but not one-one, if $A = R, B = R^{+}$.For $A = R$, $f(1) = 1$ and $f(-1) = 1$, so it is not one-one. Since $B = R^{+}$, for every $y \in R^{+}$, there exists $x = \pm \sqrt{y} \in R$ such that $f(x) = y$, so it is onto. Thus, $C \rightarrow 3$.$D$. $f$ is neither one-one nor onto, if $A = B = R$.For $A = R$, $f(1) = f(-1) = 1$, so it is not one-one. Since $B = R$, the range is $R^{+}$, which is a proper subset of $B$, so it is not onto. Thus, $D \rightarrow 2$.Therefore, the correct matching is $A \rightarrow 4, B \rightarrow 1, C \rightarrow 3, D \rightarrow 2$.

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