The function `f(x)=e^(sinx+cosx)AA x in [0, 2pi]` attains local extrema at `x=alpha and x= beta,` then `alpha+beta` is equal to

Options

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

Answer

To solve the problem, we need to find the local extrema of the function \( f(x) = e^{\sin x + \cos x} \) on the interval \( [0, 2\pi] \). The local extrema of \( f(x) \) will occur at the points where the derivative \( f'(x) \) is equal to zero.

### Step-by-Step Solution:

1. **Define the Function**:
We start with the function:
\[ f(x) = e^{\sin x + \cos x} \]

2. **Find the Derivative**:
To find the local extrema, we first need to differentiate \( f(x) \):
\[ f'(x) = e^{\sin x + \cos x} \cdot (\cos x - \sin x) \]
Here, we used the chain rule where the derivative of \( \sin x + \cos x \) is \( \cos x - \sin x \).

3. **Set the Derivative to Zero**:
For local extrema, we set the derivative equal to zero:
\[ f'(x) = 0 \implies e^{\sin x + \cos x} \cdot (\cos x - \sin x) = 0 \]
Since \( e^{\sin x + \cos x} \) is never zero, we focus on:
\[ \cos x - \sin x = 0 \]
This simplifies to:
\[ \cos x = \sin x \]

4. **Solve for x**:
The equation \( \cos x = \sin x \) can be rewritten as:
\[ \tan x = 1 \]
The solutions for \( \tan x = 1 \) in the interval \( [0, 2\pi] \) are:
\[ x = \frac{\pi}{4} \quad \text{and} \quad x = \frac{5\pi}{4} \]

5. **Identify Local Extrema**:
Let \( \alpha = \frac{\pi}{4} \) and \( \beta = \frac{5\pi}{4} \).

6. **Calculate \( \alpha + \beta \)**:
Now, we find:
\[ \alpha + \beta = \frac{\pi}{4} + \frac{5\pi}{4} = \frac{6\pi}{4} = \frac{3\pi}{2} \]

### Final Answer:
Thus, the value of \( \alpha + \beta \) is:
\[ \boxed{\frac{3\pi}{2}} \]