A relation $f: N→ N$ be defined by $f(x) = x^2,x∈ N$ (Set of Natural numbers), Then $f(x)$ is

injective only

The correct answer is Option (2) → injective only

Given: Function $f : \mathbb{N} \to \mathbb{N}$ defined by $f(x) = x^2$, where $x \in \mathbb{N}$.

Injectivity (One-to-One):

Suppose $f(x_1) = f(x_2)$. Then:

$x_1^2 = x_2^2 \Rightarrow x_1 = x_2$ (since $x_1, x_2 \in \mathbb{N}$ and squares of distinct natural numbers are distinct).

Therefore, $f$ is injective.

Surjectivity (Onto):

To be surjective, for every $y \in \mathbb{N}$, there must exist $x \in \mathbb{N}$ such that $f(x) = y$. But $f(x) = x^2$ only produces perfect squares like 1, 4, 9, 16, etc., not all natural numbers.

For example, there is no $x \in \mathbb{N}$ such that $f(x) = 2$ or $f(x) = 3$.

Therefore, $f$ is not surjective.