Match List-I with List-II Where R is set of real numbers List-I List-…

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

List-I	List-II
(A) $f: R→R$ s.t $f(x) = x^4$ is	(II) many-one, into
(B) $f: R→[0, ∞)$ s.t $f(x) = x^4$ is	(IV) many-one, onto
(C) $f: [0, ∞)→R$ s.t $f(x) = x^4$ is	(I) one-one, Into
(D) $f: [0, ∞)→[0, ∞)$ s.t $f(x) = x^4$ is	(III) one-one, onto

Explanation:

(A) $f:\mathbb{R} \to \mathbb{R}$, $f(x)=x^{4}$

For $x$ and $-x$, $f(x)$ is the same, so it is many-one. The range is $[0,\infty)$, which is a subset of $\mathbb{R}$, so not onto. Matches (II).

(B) $f:\mathbb{R} \to [0,\infty)$, $f(x)=x^{4}$

Still many-one because $x$ and $-x$ map to the same value, but now the range exactly matches the codomain $[0,\infty)$, so it is onto. Matches (IV).

(C) $f:[0,\infty) \to \mathbb{R}$, $f(x)=x^{4}$

For $x\ge0$, $f$ is strictly increasing, so one-one. The range is $[0,\infty)$, which is a subset of $\mathbb{R}$, so not onto. Matches (I).

(D) $f:[0,\infty) \to [0,\infty)$, $f(x)=x^{4}$

For $x\ge0$, $f$ is strictly increasing, so one-one. The range is $[0,\infty)$, which matches the codomain, so onto. Matches (III).