Question

A function \( f: \mathbb{R} \to \mathbb{R} \) is defined by \( f(x) = 3x \) for all \( x \in \mathbb{R} \). Then function \( f \) will be:

• A. Not one-one
• B. Not onto
• C. Onto
• D. Many-one

Hint:

For a function \( f: A \to B \), to check if it's onto, pick an arbitrary element \( y \in B \) and try to solve the equation \( f(x) = y \) for \( x \). If you can always find a solution for \( x \) that is in the domain A, then the function is onto.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to determine if the function \( f(x) = 3x \) is one-to-one (injective) and/or onto (surjective).
- One-to-one (injective): A function is one-to-one if every distinct element in the domain maps to a distinct element in the codomain. Formally, \( f(x_1) = f(x_2) \implies x_1 = x_2 \).
- Onto (surjective): A function is onto if every element in the codomain has a corresponding element in the domain. Formally, for every \( y \in \mathbb{R} \) (codomain), there exists an \( x \in \mathbb{R} \) (domain) such that \( f(x) = y \).
Step 2: Detailed Explanation or Calculation:
Checking for One-to-one:
Let \( f(x_1) = f(x_2) \) for some \( x_1, x_2 \in \mathbb{R} \).
Then, \( 3x_1 = 3x_2 \).
Dividing both sides by 3 gives \( x_1 = x_2 \).
Since \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), the function is one-to-one. This eliminates options (A) and (D).
Checking for Onto:
Let \( y \) be an arbitrary element in the codomain \( \mathbb{R} \). We need to see if we can find an \( x \) in the domain \( \mathbb{R} \) such that \( f(x) = y \).
\[ 3x = y \] Solving for x, we get:
\[ x = \frac{y}{3} \] For any real number y, \( y/3 \) is also a real number. Thus, for any y in the codomain, there exists an \( x = y/3 \) in the domain that maps to it.
Therefore, the function is onto. This eliminates option (B).
Step 3: Final Answer:
The function is both one-to-one and onto. From the given options, (C) "Onto" is a correct description of the function.