Let `A = {1,2,3}, B {5,7,6} and f : A to B ` be defined as `f = {(1,7), (2,6),(3,5)}.` Then f is :

Options

A. one-one
B. onto
C. many-one
D. bijective

Answer

To determine the nature of the function \( f \) defined from set \( A \) to set \( B \), we need to analyze the given mapping:

1. **Identify the sets and the function**:
- Set \( A = \{1, 2, 3\} \)
- Set \( B = \{5, 6, 7\} \)
- The function \( f \) is given as \( f = \{(1, 7), (2, 6), (3, 5)\} \).

2. **Check if \( f \) is a function**:
- A relation \( f \) from set \( A \) to set \( B \) is a function if every element in \( A \) is associated with exactly one element in \( B \).
- In our case:
- \( 1 \) maps to \( 7 \)
- \( 2 \) maps to \( 6 \)
- \( 3 \) maps to \( 5 \)
- Each element in \( A \) has a unique corresponding element in \( B \), confirming that \( f \) is indeed a function.

3. **Determine if \( f \) is one-to-one (injective)**:
- A function is one-to-one if different elements in \( A \) map to different elements in \( B \).
- Here:
- \( 1 \) maps to \( 7 \)
- \( 2 \) maps to \( 6 \)
- \( 3 \) maps to \( 5 \)
- Since all the outputs \( 7, 6, \) and \( 5 \) are distinct, \( f \) is one-to-one.

4. **Determine if \( f \) is onto (surjective)**:
- A function is onto if every element in \( B \) is the image of at least one element in \( A \).
- The outputs of \( f \) are \( 7, 6, \) and \( 5 \), which cover all elements in \( B \).
- Therefore, \( f \) is onto.

5. **Conclusion**:
- Since \( f \) is both one-to-one and onto, we conclude that \( f \) is a bijective function.

**Final Answer**: The function \( f \) is a bijection (one-to-one and onto).