Let $X = \{x^2 : x \in \mathbb{N}\}$ and the relation $f: \mathbb{N} \rightarrow X$ is defined by $f(x) = x^2, x \in \mathbb{N}$. Then, this function is:

bijective

The correct answer is Option (4) → bijective

Injective (one-one):

If $f(x_1) = f(x_2) \Rightarrow x_1^2 = x_2^2 \Rightarrow x_1 = x_2$ (since $x \in \mathbb{N}$, no negative values). Hence, the function is injective

Surjective (onto):

The codomain is $X = \{x^2 : x \in \mathbb{N}\}$, i.e., all perfect squares.

Every element in $X$ has a pre-image in $\mathbb{N}$.

Hence, the function is also surjective

Therefore, the function is actually bijective