Question

If the function \( f : \mathbb{R} \to \mathbb{R} \) is defined as \( f(x) = 3x \) then f is

• A. one-one and onto
• B. many-one and onto
• C. one-one but not onto
• D. neither one-one nor onto

Hint:

Linear functions of the form \( f(x) = ax + b \) with \( a \neq 0 \) are always bijective (one-one and onto) when the domain and codomain are \( \mathbb{R} \). Recognizing this pattern can provide a quick answer.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to check if the function \( f(x) = 3x \) is injective (one-one) and surjective (onto) for the domain and codomain of all real numbers (\(\mathbb{R}\)).
One-one (Injective): A function is one-one if every distinct element in the domain maps to a distinct element in the codomain. Mathematically, \( f(x_1) = f(x_2) \implies x_1 = x_2 \).
Onto (Surjective): A function is onto if for every element \( y \) in the codomain, there exists at least one element \( x \) in the domain such that \( f(x) = y \).
Step 2: Detailed Explanation:
Checking for one-one:
Let \( x_1, x_2 \in \mathbb{R} \) such that \( f(x_1) = f(x_2) \).
\[ 3x_1 = 3x_2 \] Dividing both sides by 3, we get: \[ x_1 = x_2 \] Since \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), the function is one-one.
Checking for onto:
Let \( y \) be any arbitrary element in the codomain \( \mathbb{R} \). We need to find if there is an \( x \) in the domain \( \mathbb{R} \) such that \( f(x) = y \).
\[ f(x) = y \] \[ 3x = y \] \[ x = \frac{y}{3} \] Since \( y \) is a real number, \( \frac{y}{3} \) is also a real number. Thus, for any \( y \in \mathbb{R} \) (codomain), there exists a pre-image \( x = \frac{y}{3} \in \mathbb{R} \) (domain).
Therefore, the function is onto.
Step 3: Final Answer:
Since the function is both one-one and onto, the correct option is (i).