Let f: R→ R be defined as f(x) = 100x + 1, where R is a set of real n…

The correct answer is Option (3) → f is both one-one and onto

Given: $f : \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = 100x + 1$

1. Injective (One-one): Suppose $f(x_1) = f(x_2)$

$\Rightarrow 100x_1 + 1 = 100x_2 + 1 \Rightarrow x_1 = x_2$

So, $f$ is injective.

2. Surjective (Onto): Let $y \in \mathbb{R}$

Then $y = 100x + 1 \Rightarrow x = \frac{y - 1}{100} \in \mathbb{R}$

So, for every $y \in \mathbb{R}$, there exists $x \in \mathbb{R}$ such that $f(x) = y$

Hence, $f$ is surjective.