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`.
The range of f contains all the even numbers, the value of f(1) is

Options

A. 1
B. 2
C. 1 or 2
D. 4

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 is a well-known property of linear functions.

2. **Assuming a Form for \( f \)**:
Given the nature of the equation, we can assume that \( f \) is a linear function of the form:
\[ f(x) = kx \]
where \( k \) is a constant.

3. **Analyzing the Range of \( f \)**:
We know that the range of \( f \) contains all even numbers. This means that \( kx \) must yield even numbers for all natural numbers \( x \).

4. **Determining the Value of \( k \)**:
For \( f(x) = kx \) to produce only even numbers, \( k \) must be an even integer. The simplest choice for \( k \) that satisfies this condition is \( k = 2 \). Thus, we can write:
\[ f(x) = 2x \]

5. **Verifying the Functional Equation**:
We need to check if our assumption \( f(x) = 2x \) satisfies the original functional equation:
\[ f(m+n) = 2(m+n) = 2m + 2n = f(m) + f(n) \]
Since both sides are equal, our assumption is valid.

6. **Finding \( f(1) \)**:
Now, we can find the value of \( f(1) \):
\[ f(1) = 2 \cdot 1 = 2 \]

### Conclusion:
Thus, the value of \( f(1) \) is \( 2 \).