If A = {1,2,3}, B = {4,5,6,7} and f = {(1, 4), (2,5), (3,6)} is a function from A to B.
Assertion (A): f(x) is a one-one function.
Reason (R): f(x) is an onto function.

Options

A. Both A and R are true and R is the correct explanation of A
B. Both A and R are true but R is NOT the correct explanation of A
C. A is true but R is false
D. A is false and R is True

Answer

To solve the problem, we need to analyze the assertions and reasons regarding the function \( f \) defined from set \( A \) to set \( B \).

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

### Step 2: Check if \( f \) is a one-one function
A function is one-one (injective) if each element in the domain maps to a unique element in the codomain. In our case:
- \( 1 \) maps to \( 4 \)
- \( 2 \) maps to \( 5 \)
- \( 3 \) maps to \( 6 \)

Since all three elements in set \( A \) map to different elements in set \( B \), we conclude that \( f \) is a one-one function.

### Step 3: Check if \( f \) is an onto function
A function is onto (surjective) if every element in the codomain is mapped by at least one element from the domain. The codomain of \( f \) is set \( B = \{4, 5, 6, 7\} \).

From the function \( f \):
- The range of \( f \) is \( \{4, 5, 6\} \) (the outputs of the function).
- The codomain is \( \{4, 5, 6, 7\} \).

Since \( 7 \) in the codomain does not have a pre-image in set \( A \), \( f \) is not onto.

### Conclusion
- Assertion (A): \( f \) is a one-one function. **True**
- Reason (R): \( f \) is an onto function. **False**

Thus, the assertion is true, but the reason is false.

### Final Answer
The correct conclusion is that Assertion (A) is true and Reason (R) is false.

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