Let f be real valued function from N to N satisfying. The relation f(m+n)=f(m)+f(n) for all `m,n in N`.
If domain of f is first 3m natural numbers and if the number of elements common in domain and range is m, then the value of f(1) is

Options

A. 2
B. 3
C. 6
D. Can't say

Answer

To solve the problem, we need to find the value of \( f(1) \) given the properties of the function \( f \).

### Step-by-step Solution:

1. **Understanding the Functional Equation**:
We are given that \( f(m+n) = f(m) + f(n) \) for all \( m, n \in \mathbb{N} \). This property suggests that \( f \) is a linear function.

2. **Assuming a Linear Form**:
Based on the functional equation, we can assume that \( f(n) = k \cdot n \) for some constant \( k \). This is a common form for functions satisfying the given property.

3. **Determining the Domain**:
The domain of \( f \) is the first \( 3m \) natural numbers, which means the function is defined for \( \{1, 2, 3, \ldots, 3m\} \).

4. **Finding the Range**:
If \( f(n) = k \cdot n \), then the range will also be a set of natural numbers. The number of elements common in the domain and the range is given to be \( m \).

5. **Analyzing the Common Elements**:
For \( f(n) \) to have \( m \) common elements with the domain \( \{1, 2, \ldots, 3m\} \), we need to find a suitable \( k \).

If we take \( k = 3 \), then:
\[ f(n) = 3n \]
The range will be \( \{3, 6, 9, \ldots, 9m\} \) (which corresponds to \( 3 \times \{1, 2, \ldots, 3m\} \)).

6. **Counting Common Elements**:
The common elements between the domain \( \{1, 2, \ldots, 3m\} \) and the range \( \{3, 6, 9, \ldots, 9m\} \) are \( 3, 6, 9, \ldots, 3m \). The number of these common elements is \( m \) (since \( 3, 6, \ldots, 3m \) gives \( m \) terms).

7. **Calculating \( f(1) \)**:
Now, substituting \( n = 1 \) into our assumed function:
\[ f(1) = 3 \cdot 1 = 3 \]

### Conclusion:
Thus, the value of \( f(1) \) is \( 3 \).