Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2,5), (3, 6)} be a function from A to B. Based on the given information, f is best defined as:

Options

A. Surjective function
B. Injective function
C. Bijective function
D. function

Answer

To determine the nature of the function \( f \) from set \( A \) to set \( B \), we need to analyze the properties of the function based on the definitions of injective, surjective, and bijective functions.

### Step 1: Identify the Sets and Function
- We have the sets:
- \( A = \{1, 2, 3\} \)
- \( B = \{4, 5, 6, 7\} \)
- The function \( f \) is defined as:
- \( f = \{(1, 4), (2, 5), (3, 6)\} \)

### Step 2: Check if the Function is Injective
- A function is **injective** (or one-to-one) if different elements in the domain map to different elements in the codomain.
- Here, we see:
- \( f(1) = 4 \)
- \( f(2) = 5 \)
- \( f(3) = 6 \)
- Since all outputs (4, 5, 6) are distinct, the function is injective.

### Step 3: Check if the Function is Surjective
- A function is **surjective** (or onto) if every element in the codomain has at least one element from the domain mapping to it.
- The codomain \( B \) has elements \( \{4, 5, 6, 7\} \).
- The range of \( f \) is \( \{4, 5, 6\} \).
- The element \( 7 \) in the codomain does not have a pre-image in the domain \( A \). Therefore, the function is not surjective.

### Step 4: Check if the Function is Bijective
- A function is **bijective** if it is both injective and surjective.
- Since the function is not surjective, it cannot be bijective.

### Conclusion
- The function \( f \) is injective but not surjective, hence it is best defined as an **injective function**.

The correct answer is **Option 2: Injective**.

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