Let f be a function such that `f(x+f(y)) = f(x) + y, AA x, y in R`, then find f(0). If it is given that there exists a positive real `delta` such that f(h) = h for `0 lt h lt delta`, then find f'(x)

Options

A. 0, 1
B. `-1, 0`
C. 2, 1
D. `-2, 0`

Answer

To solve the problem, we start with the functional equation given:

1. **Functional Equation**:
\[ f(x + f(y)) = f(x) + y \quad \forall x, y \in \mathbb{R} \]

### Step 1: Finding \( f(0) \)

We can find \( f(0) \) by substituting specific values into the functional equation. Let's set \( x = 0 \):

\[ f(0 + f(y)) = f(0) + y \]
This simplifies to:
\[ f(f(y)) = f(0) + y \quad \text{(1)} \]

### Step 2: Finding \( f(f(0)) \)

Next, we can substitute \( y = 0 \) into the original equation:

\[ f(x + f(0)) = f(x) + 0 \]
This simplifies to:
\[ f(x + f(0)) = f(x) \quad \text{(2)} \]

### Step 3: Analyzing Equation (2)

From equation (2), we see that \( f \) is periodic with period \( f(0) \). This means that for any \( x \):
\[ f(x + f(0)) = f(x) \]
This implies that \( f(0) \) must be equal to zero for the function to be consistent, otherwise, it would contradict the injectivity of \( f \).

### Step 4: Conclusion for \( f(0) \)

Thus, we conclude:
\[ f(0) = 0 \]

### Step 5: Finding \( f'(x) \)

We are given that there exists a positive real \( \delta \) such that \( f(h) = h \) for \( 0 < h < \delta \). This means that \( f \) behaves like the identity function in this interval.

### Step 6: Differentiating \( f(x) \)

Since \( f(h) = h \) for \( 0 < h < \delta \), we can differentiate both sides with respect to \( h \):
\[ f'(h) = 1 \quad \text{for } 0 < h < \delta \]

### Step 7: Conclusion for \( f'(x) \)

Thus, we conclude:
\[ f'(x) = 1 \quad \text{for } 0 < x < \delta \]

### Final Answers

- \( f(0) = 0 \)
- \( f'(x) = 1 \) for \( 0 < x < \delta \)