Q.Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be defined as $f(x) = 3x$. Choose the correct answer. (A) $f$ is one-one onto (B) $f$ is many-one onto (C) $f$ is one-one but not onto (D) $f$ is neither one-one nor onto.

Concept: Linear Function Properties — A function $f(x) = ax$ with $a \neq 0$ is always one-one and onto on $\mathbb{R}$.

Step 1 — One-one (injective):

If $f(x_1) = f(x_2)$, then $3x_1 = 3x_2 \implies x_1 = x_2$. So $f$ is one-one.

Step 2 — Onto (surjective): …

Why this approach works

The question tests two fundamental properties of functions: one-one (injective) and onto (surjective). For a function $f: \mathbb{R} \to \mathbb{R}$, being one-one means different inputs give different outputs; being onto means every real number is hit as an output.

The function $f(x) = 3x$ is a simple linear function with slope $3 \neq 0$. Linear functions with non-zero slope are always one-one on $\mathbb{R}$ because they are strictly increasing (or decreasing). They are also onto on $\mathbb{R}$ because you can solve $y = 3x$ for any real $y$ to get $x = y/3$, which is always a real number.

Let’s verify both properties step by step.

1. Checking one-one (injective)

   A function is one-one if $f(a) = f(b)$ implies $a = b$.

   Suppose $f(a) = f(b)$. Then $3a = 3b$. Dividing both sides by $3$ (which is allowed since $3 \neq 0$), we get $a = b$.

   So $f$ is one-one.

2. Checking onto (surjective)

   A function is onto if for every $y \in \mathbb{R}$, there exists some $x \in \mathbb{R}$ such that $f(x) = y$.

   Given any $y \in \mathbb{R}$, we need $3x = y$. Solving gives $x = \frac{y}{3}$, which is a real number for every real $y$.

   Hence, every real number has a preimage, so $f$ is onto. …

Method: Classifying a linear function on R (MCQ)

For $f(x)=mx$ (or $mx+c$) with $m\neq 0$ and codomain $\mathbf{R}$, the function is always a bijection — verify both parts, then choose the option.

Steps

Step 1: One-one — $f(x_1)=f(x_2)$ gives $mx_1=mx_2$; dividing by $m\neq 0$ yields $x_1=x_2$. …

Common Mistakes

Mistake 1: Thinking $f(x)=3x$ is not onto because it "gives only multiples of 3".

Why it's wrong: that would be true if the codomain were $\mathbf{Z}$, but here it is $\mathbf{R}$, and $\frac{y}{3}$ is a real number for every real $y$ …