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

Answer

To solve the problem, we need to find the value of \( f(1) \) given that \( f \) is a differentiable function satisfying the condition:

\[ |f(x) - f(y)| \leq (x - y)^2 \]

for all \( x, y \in \mathbb{R} \) and that \( f(0) = 0 \).

### Step-by-Step Solution:

1. **Understanding the Condition**: The condition \( |f(x) - f(y)| \leq (x - y)^2 \) implies that the difference in function values is bounded by the square of the difference in their inputs. This suggests that \( f(x) \) is a very "flat" function.

2. **Setting Up the Limit**: We can rewrite the condition by substituting \( x = \theta + h \) and \( y = \theta \):
\[ |f(\theta + h) - f(\theta)| \leq h^2 \]

3. **Taking the Limit**: As \( h \) approaches 0, we can analyze the left side:
\[ \lim_{h \to 0} \frac{|f(\theta + h) - f(\theta)|}{|h|} \leq \lim_{h \to 0} \frac{h^2}{|h|} = \lim_{h \to 0} |h| = 0 \]
This means that the derivative \( f'(\theta) \) must be 0 for all \( \theta \).

4. **Conclusion About the Function**: Since \( f'(\theta) = 0 \) for all \( \theta \), it follows that \( f \) is a constant function.

5. **Using the Given Condition**: We know \( f(0) = 0 \). Since \( f \) is constant and equal to 0 at \( x = 0 \), it must be that:
\[ f(x) = 0 \quad \text{for all } x \]

6. **Finding \( f(1) \)**: Therefore, we can conclude:
\[ f(1) = 0 \]

### Final Answer:
\[ f(1) = 0 \]