`f(x) = e^(sinx+cosx)` is an increasing function in

Options

A. `(0, pi/4)`
B. `(0, pi/2)`
C. `(-pi/4, pi/4)`
D. none of these

Answer

To determine where the function \( f(x) = e^{\sin x + \cos x} \) is increasing, we need to find the derivative of the function and analyze where it is greater than or equal to zero.

### Step 1: Differentiate the function
The first step is to differentiate \( f(x) \).

\[ f'(x) = \frac{d}{dx}(e^{\sin x + \cos x}) \]

Using the chain rule, we have:

\[ f'(x) = e^{\sin x + \cos x} \cdot \frac{d}{dx}(\sin x + \cos x) \]

Now, we differentiate \( \sin x + \cos x \):

\[ \frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x \]

Thus, the derivative becomes:

\[ f'(x) = e^{\sin x + \cos x} (\cos x - \sin x) \]

### Step 2: Analyze the sign of the derivative
Since \( e^{\sin x + \cos x} > 0 \) for all \( x \), the sign of \( f'(x) \) depends only on \( \cos x - \sin x \).

We need to find where:

\[ \cos x - \sin x \geq 0 \]

This simplifies to:

\[ \cos x \geq \sin x \]

### Step 3: Solve the inequality
To solve \( \cos x \geq \sin x \), we can rearrange it:

\[ \frac{\cos x}{\sin x} \geq 1 \]

This can be rewritten as:

\[ \tan x \leq 1 \]

The solutions to \( \tan x = 1 \) occur at:

\[ x = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]

### Step 4: Determine intervals
The function \( \tan x \) is less than or equal to 1 in the intervals:

1. From \( -\frac{\pi}{4} \) to \( \frac{\pi}{4} \)
2. From \( \frac{3\pi}{4} \) to \( \frac{5\pi}{4} \)
3. And so on, for every period of \( \pi \).

### Conclusion
Thus, the function \( f(x) = e^{\sin x + \cos x} \) is increasing in the intervals:

- \( \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \)
- \( \left[\frac{3\pi}{4}, \frac{5\pi}{4}\right] \)
- and similar intervals for other periods.