A relation $f: N→ N$ be defined by $f(x) = x^2,x∈ N$ (Set of Natural numbers), Then $f(x)$ is
Answer & explanation
Correct answer: option 2
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.