If the Rolle's theorem for `f(x)=e^(x)(sin x-cosx)` is verified on `[(pi)/4,(5pi)/4]` then the value of `C` is

Options

A. `(pi)/3`
B. `(pi)/2`
C. `(3pi)/4`
D. `pi`

Explanation

To solve the problem, we need to verify Rolle's Theorem for the function \( f(x) = e^x (\sin x - \cos x) \) on the interval \(\left[\frac{\pi}{4}, \frac{5\pi}{4}\right]\) and find the value of \(C\).

### Step-by-Step Solution:

**Step 1: Check the conditions of Rolle's Theorem**
- **Continuity**: The function \( f(x) \) is a product of \( e^x \) (which is continuous everywhere) and \( \sin x - \cos x \) (which is also continuous everywhere). Therefore, \( f(x) \) is continuous on the closed interval \(\left[\frac{\pi}{4}, \frac{5\pi}{4}\right]\).
- **Differentiability**: Since \( e^x \) and \( \sin x - \cos x \) are both differentiable everywhere, \( f(x) \) is differentiable on the open interval \(\left(\frac{\pi}{4}, \frac{5\pi}{4}\right)\).

**Step 2: Evaluate \( f\left(\frac{\pi}{4}\right) \) and \( f\left(\frac{5\pi}{4}\right) \)**
- Calculate \( f\left(\frac{\pi}{4}\right) \):
\[ f\left(\frac{\pi}{4}\right) = e^{\frac{\pi}{4}} \left( \sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) \right) = e^{\frac{\pi}{4}} \left( \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right) = 0 \]
- Calculate \( f\left(\frac{5\pi}{4}\right) \):
\[ f\left(\frac{5\pi}{4}\right) = e^{\frac{5\pi}{4}} \left( \sin\left(\frac{5\pi}{4}\right) - \cos\left(\frac{5\pi}{4}\right) \right) = e^{\frac{5\pi}{4}} \left( -\frac{1}{\sqrt{2}} - \left(-\frac{1}{\sqrt{2}}\right) \right) = 0 \]

Since \( f\left(\frac{\pi}{4}\right) = f\left(\frac{5\pi}{4}\right) = 0 \), the first condition of Rolle's Theorem is satisfied.

**Step 3: Find the derivative \( f'(x) \)**
Using the product rule:
\[ f'(x) = \frac{d}{dx}(e^x) \cdot (\sin x - \cos x) + e^x \cdot \frac{d}{dx}(\sin x - \cos x) \]
Calculating the derivatives:
\[ f'(x) = e^x (\sin x - \cos x) + e^x (\cos x + \sin x) = e^x (2\sin x) \]

**Step 4: Set the derivative equal to zero**
According to Rolle's Theorem, there exists at least one \( C \) in \(\left(\frac{\pi}{4}, \frac{5\pi}{4}\right)\) such that \( f'(C) = 0 \):
\[ e^C (2 \sin C) = 0 \]
Since \( e^C \neq 0 \) for any real \( C \), we have:
\[ 2 \sin C = 0 \implies \sin C = 0 \]

**Step 5: Solve for \( C \)**
The solutions to \( \sin C = 0 \) are given by:
\[ C = n\pi \quad \text{for integers } n \]
We need \( C \) to be in the interval \(\left(\frac{\pi}{4}, \frac{5\pi}{4}\right)\). The possible values of \( n \) are:
- For \( n = 0 \): \( C = 0 \) (not in the interval)
- For \( n = 1 \): \( C = \pi \) (in the interval)
- For \( n = 2 \): \( C = 2\pi \) (not in the interval)

Thus, the only valid solution is:
\[ C = \pi \]

### Final Answer:
The value of \( C \) is \( \pi \).