Let `f(x)` be a differentiable function and `f(alpha)=f(beta)=0(alpha< beta),` then the interval `(alpha,beta)`

Options

A. `f(x)+f'(x)=0` has at least one root
B. `f(x)-f'(x)=0` has at least one real root
C. `f(x)xxf'(x)=0` has at lease one real root
D. none of these

Answer

To solve the problem step by step, we can use Rolle's Theorem, which is applicable here since we have a differentiable function that meets the conditions of the theorem.

### Step-by-Step Solution:

1. **Identify the Given Information**:
We have a differentiable function \( f(x) \) such that \( f(\alpha) = f(\beta) = 0 \) where \( \alpha < \beta \).

2. **Apply Rolle's Theorem**:
Since \( f(x) \) is continuous on the closed interval \([ \alpha, \beta ]\) and differentiable on the open interval \( ( \alpha, \beta ) \), we can apply Rolle's Theorem.

According to Rolle's Theorem, if \( f(a) = f(b) \) for some \( a < b \), then there exists at least one \( c \) in the interval \( (a, b) \) such that:
\[ f'(c) = 0 \]

3. **Conclusion**:
Therefore, there exists at least one point \( c \) in the interval \( (\alpha, \beta) \) such that \( f'(c) = 0 \).

4. **Final Answer**:
The interval \( (\alpha, \beta) \) contains at least one point \( c \) where the derivative \( f'(c) = 0 \).