Question

Let \( A = \{1,2,3\} \) and \( B = \{2,4,6\} \). A function \( f : A \to B \) is defined by \( f(x)=2x \). The function is:

• A. One-one only
• B. Onto only
• C. Both one-one and onto
• D. Neither one-one nor onto

Hint:

To verify one-one: Check if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).
To verify onto: Check if the range of the function equals its codomain. If \( |A| = |B| \) and the function is one-one, it must also be onto.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question defines a set A (domain), a set B (codomain), and a function \( f(x) = 2x \) mapping from A to B. We need to determine if the function is one-one, onto, both, or neither.

Step 2: Key Formula or Approach:
1. One-one (Injective): A function \( f: A \to B \) is one-one if every distinct element in A maps to a distinct element in B. That is, if \( f(x_1) = f(x_2) \Rightarrow x_1 = x_2 \).
2. Onto (Surjective): A function \( f: A \to B \) is onto if every element in B (codomain) has at least one corresponding element in A (domain). That is, the range of the function must be equal to its codomain.

Step 3: Detailed Explanation:
Given:
Domain \( A = \{1,2,3\} \)
Codomain \( B = \{2,4,6\} \)
Function \( f(x) = 2x \)
Let's find the image of each element in the domain A:
- For \( x=1 \): \( f(1) = 2 \times 1 = 2 \)
- For \( x=2 \): \( f(2) = 2 \times 2 = 4 \)
- For \( x=3 \): \( f(3) = 2 \times 3 = 6 \)
The range of the function is \( \text{Range}(f) = \{2,4,6\} \).
Check for One-one:
Each distinct element in A (1, 2, 3) maps to a distinct element in B (2, 4, 6).
\( f(1) = 2 \)
\( f(2) = 4 \)
\( f(3) = 6 \)
Since no two distinct elements in A map to the same element in B, the function is one-one.
Check for Onto:
The codomain is \( B = \{2,4,6\} \).
The range of the function is \( \text{Range}(f) = \{2,4,6\} \).
Since the range is equal to the codomain, every element in B has a pre-image in A. Therefore, the function is onto.
Since the function is both one-one and onto, option (C) is the correct answer.

Step 4: Final Answer:
The function is Both one-one and onto.