Match List-I with List-II

Let $f: A → B$ be a function given by $f(x) = x^2$

Choose the correct answer from the options given below:

(A)-(IV), (B)-(III), (C)-(I), (D)-(II)

The correct answer is Option (2) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

Given $f(x)=x^{2}$.

(A) $A=\mathbb{R},\;B=\mathbb{R}$: $f$ is not one-one since $f(1)=f(-1)$ and not onto since negative values are not in the range.

So (A) → (IV).

(B) $A=\mathbb{R},\;B=[0,\infty)$: not one-one but onto because every non-negative real has a square root.

So (B) → (III).

(C) $A=B=[0,\infty)$: one-one and onto since $x^{2}$ is strictly increasing on $[0,\infty)$ and covers all $[0,\infty)$.

So (C) → (I).

(D) $A=[0,\infty),\;B=\mathbb{R}$: one-one but not onto since negative values are not attained.

So (D) → (II).

Final answer: (A)–(IV), (B)–(III), (C)–(I), (D)–(II)