Let `f:[a, b] rarrR` be such that f is differentiable in (a, b), f is continuous at x = a & x = b and moreover `f(a)=0=f(b)`. Then

Options

A. there exists at least one point c in (a, b) such that `f'(c)=f(c)`
B. `f'(c)=f(x)` does not hold at any point in (a, b)
C. at every point of `(a, b), f'(x) gt f(x)`
D. at every point of `(a, b), f'(x) lt f(x)`

Answer

To solve the problem, we will use the properties of differentiable and continuous functions along with Rolle's Theorem.

### Step-by-Step Solution:

1. **Define the function**: Let \( f: [a, b] \to \mathbb{R} \) be a function that is differentiable on the open interval \( (a, b) \) and continuous on the closed interval \( [a, b] \). We know that \( f(a) = 0 \) and \( f(b) = 0 \).

2. **Construct a new function**: Define a new function \( h(x) = e^{-x} f(x) \). This function is continuous on \( [a, b] \) and differentiable on \( (a, b) \) because \( f(x) \) is differentiable and \( e^{-x} \) is continuous and differentiable everywhere.

3. **Apply Rolle's Theorem**: Since \( h(a) = e^{-a} f(a) = e^{-a} \cdot 0 = 0 \) and \( h(b) = e^{-b} f(b) = e^{-b} \cdot 0 = 0 \), we can apply Rolle's Theorem. According to Rolle's Theorem, since \( h \) is continuous on \( [a, b] \) and differentiable on \( (a, b) \), there exists at least one point \( c \in (a, b) \) such that \( h'(c) = 0 \).

4. **Differentiate \( h(x) \)**: To find \( h'(x) \), we use the product rule:
\[ h'(x) = \frac{d}{dx}(e^{-x}) \cdot f(x) + e^{-x} \cdot \frac{d}{dx}(f(x)) = -e^{-x} f(x) + e^{-x} f'(x) = e^{-x}(f'(x) - f(x)). \]

5. **Set the derivative to zero**: Since \( h'(c) = 0 \), we have:
\[ e^{-c}(f'(c) - f(c)) = 0. \]
Since \( e^{-c} \) is never zero, we can conclude that:
\[ f'(c) - f(c) = 0 \quad \Rightarrow \quad f'(c) = f(c). \]

6. **Conclusion**: Therefore, there exists at least one point \( c \in (a, b) \) such that \( f'(c) = f(c) \).