Let `f:R->R` be a function such that `f(x+y)=f(x)+f(y),AA x, y in R.`

Options

A. f(x) is differentiable only in a finite interval containing zero
B. f(x) is continuous for all `x in R`
C. f'(x) is constant for all `x in R`
D. f(x) is differentiable except at finitely many points

Answer

To solve the functional equation \( f(x+y) = f(x) + f(y) \) for all \( x, y \in \mathbb{R} \), we will follow these steps:

### Step 1: Differentiate the functional equation
We start with the given functional equation:
\[ f(x+y) = f(x) + f(y) \]
We will partially differentiate this equation with respect to \( y \).

Using the chain rule, we differentiate both sides:
\[ \frac{\partial}{\partial y} f(x+y) = \frac{\partial}{\partial y} (f(x) + f(y)) \]
This gives us:
\[ f'(x+y) = 0 + f'(y) \quad \text{(since \( f(x) \) is constant with respect to \( y \))} \]
Thus, we have:
\[ f'(x+y) = f'(y) \]

### Step 2: Substitute a value for \( y \)
Now, we will set \( y = 0 \) in the differentiated equation:
\[ f'(x+0) = f'(0) \]
This simplifies to:
\[ f'(x) = f'(0) \]
Let \( k = f'(0) \), which is a constant. Therefore, we can write:
\[ f'(x) = k \]

### Step 3: Integrate to find \( f(x) \)
Now we integrate \( f'(x) \) to find \( f(x) \):
\[ f(x) = \int k \, dx = kx + C \]
where \( C \) is the constant of integration.

### Step 4: Determine the constant \( C \)
To find the value of \( C \), we will use the original functional equation. We can substitute \( x = 0 \) and \( y = 0 \):
\[ f(0+0) = f(0) + f(0) \]
This simplifies to:
\[ f(0) = 2f(0) \]
This implies:
\[ f(0) = 0 \]
Thus, substituting \( x = 0 \) into our expression for \( f(x) \):
\[ f(0) = k(0) + C = C \]
Since \( f(0) = 0 \), we have \( C = 0 \).

### Final Result
Now we have:
\[ f(x) = kx \]
for some constant \( k \).

### Conclusion
The function \( f(x) \) is of the form \( f(x) = kx \), where \( k \) is a constant.