Match List-I with List-II

Where ℝ is set of real numbers

List-I	List-II
(A) f: ℝ → ℝ s.t f(x) = x⁴ is	(I) one-one, Into
(B) f: ℝ → [0, ∞) s.t f(x) = x⁴ is	(II) many-one, into
(C) f: [0, ∞) → ℝ s.t f(x) = x⁴ is	(III) one-one, onto
(D) f: [0, ∞) → [0, ∞) s.t f(x) = x⁴ is	(IV) many-one, onto

Choose the correct answer from the options given below:

For $f(x) = x^4$, recall:

One-one means every different input gives a different output.

Many-one means two or more different inputs give the same output.

Onto means Range = Codomain.

Into means Range ⊊ Codomain (some values in codomain are never reached).

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

Domain is all of $\mathbb{R}$, so:

$f(-1) = (-1)^4 = 1 \quad \text{and} \quad f(1) = (1)^4 = 1$

Two different inputs gave the same output → Many-one

When $x \in \mathbb{R}$, $x^4 \geq 0$ always, so:

$\text{Range} = [0, \infty), \quad \text{Codomain} = \mathbb{R}$

$[0, \infty) \neq \mathbb{R}$ → Into

(A) = Many-one, Into = (II)

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

Domain is still all of $\mathbb{R}$, so $f(-1) = f(1) = 1$ → Many-one

$\text{Range} = [0, \infty), \quad \text{Codomain} = [0, \infty)$

$\text{Range} = \text{Codomain}$ → Onto

(B) = Many-one, Onto = (IV)

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

Domain is $[0, \infty)$ — only non-negative numbers.

On $[0, \infty)$, $x^4$ is strictly increasing, so no two different non-negative inputs can give the same output → One-one

$\text{Range} = [0, \infty), \quad \text{Codomain} = \mathbb{R}$

$[0, \infty) \neq \mathbb{R}$ → Into

(C) = One-one, Into = (I)

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

Domain is $[0, \infty)$ → $x^4$ is strictly increasing here → One-one

$\text{Range} = [0, \infty), \quad \text{Codomain} = [0, \infty)$

$\text{Range} = \text{Codomain}$ → Onto

(D) = One-one, Onto = (III)

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Final matching:

(A) → (II) Many-one, Into

(B) → (IV) Many-one, Onto

(C) → (I) One-one, Into

(D) → (III) One-one, Onto