If f is a real- valued differentiable function satisfying `|f(x) - f(y)| le (x-y)^(2) ,x ,y , in R and f(0) =0` then f(1) equals

Options

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

Answer

To solve the problem, we need to analyze the given condition and derive the value of \( f(1) \).

### Step-by-Step Solution:

1. **Given Condition**: We start with the condition provided in the problem:
\[ |f(x) - f(y)| \leq (x - y)^2 \quad \text{for all } x, y \in \mathbb{R} \]
and \( f(0) = 0 \).

2. **Dividing by \( |x - y| \)**: We can manipulate the inequality by dividing both sides by \( |x - y| \) (assuming \( x \neq y \)):
\[ \frac{|f(x) - f(y)|}{|x - y|} \leq |x - y| \]

3. **Taking the Limit**: As \( y \) approaches \( x \), we can analyze the left-hand side:
\[ \lim_{y \to x} \frac{|f(x) - f(y)|}{|x - y|} = |f'(x)| \]
The right-hand side approaches \( 0 \) since \( |x - y| \to 0 \).

4. **Conclusion from the Limit**: Therefore, we have:
\[ |f'(x)| \leq 0 \]
Since the absolute value of a function is non-negative, this implies:
\[ f'(x) = 0 \quad \text{for all } x \in \mathbb{R} \]

5. **Determining \( f(x) \)**: If the derivative \( f'(x) = 0 \) for all \( x \), then \( f(x) \) must be a constant function.

6. **Using the Initial Condition**: We know from the problem that \( f(0) = 0 \). Since \( f(x) \) is constant and equals \( f(0) \), we conclude:
\[ f(x) = 0 \quad \text{for all } x \in \mathbb{R} \]

7. **Finding \( f(1) \)**: Finally, we can find \( f(1) \):
\[ f(1) = 0 \]

### Final Answer:
Thus, the value of \( f(1) \) is:
\[ \boxed{0} \]