The domain of the function
`f(x)=sqrt(log_(sinx+cosx)(abs(cosx)+cosx)), 0 le x le pi` is

Options

A. `(0,pi)`
B. `(0,pi/2)`
C. `(0,pi/3)`
D. None of these

Answer

To find the domain of the function \( f(x) = \sqrt{\log_{\sin x + \cos x}(|\cos x| + \cos x)} \) for \( 0 \leq x \leq \pi \), we need to ensure that the expression inside the square root is non-negative, as the square root function is only defined for non-negative values.

### Step 1: Ensure the argument of the square root is non-negative
The expression inside the square root is \( \log_{\sin x + \cos x}(|\cos x| + \cos x) \). For the square root to be defined, we need:
\[ \log_{\sin x + \cos x}(|\cos x| + \cos x) \geq 0 \]

### Step 2: Analyze the logarithmic condition
The logarithm \( \log_a(b) \) is non-negative when \( b \geq 1 \) if \( a > 1 \) or \( 0 < b \leq 1 \) if \( 0 < a < 1 \). Thus, we need to analyze the base \( \sin x + \cos x \) and the argument \( |\cos x| + \cos x \).

### Step 3: Determine when the base is valid
1. **Base Condition**: \( \sin x + \cos x > 1 \)
- We know that \( \sin x + \cos x \) achieves its maximum value of \( \sqrt{2} \) at \( x = \frac{\pi}{4} \) and is at least \( 1 \) for \( 0 \leq x \leq \frac{\pi}{2} \).
- Thus, we need to check when \( \sin x + \cos x = 1 \):
\[ \sin x + \cos x = 1 \implies \sqrt{2}\sin\left(x + \frac{\pi}{4}\right) = 1 \implies x = \frac{\pi}{4} \]
- Therefore, \( \sin x + \cos x > 1 \) for \( 0 \leq x < \frac{\pi}{4} \).

### Step 4: Analyze the argument of the logarithm
2. **Argument Condition**: \( |\cos x| + \cos x \geq 1 \)
- For \( 0 \leq x < \frac{\pi}{2} \), \( |\cos x| = \cos x \), so:
\[ \cos x + \cos x = 2\cos x \geq 1 \implies \cos x \geq \frac{1}{2} \implies x \leq \frac{\pi}{3} \]
- For \( \frac{\pi}{2} < x \leq \pi \), \( |\cos x| = -\cos x \), so:
\[ -\cos x + \cos x = 0 \quad \text{(not applicable)} \]

### Step 5: Combine conditions
From the above analysis:
- For \( 0 \leq x < \frac{\pi}{2} \), we have \( x \in [0, \frac{\pi}{3}] \) to satisfy both conditions.
- For \( \frac{\pi}{2} < x \leq \pi \), the logarithm becomes undefined since \( |\cos x| + \cos x \) is not greater than or equal to 1.

### Conclusion
Thus, the domain of the function \( f(x) \) is:
\[ \boxed{[0, \frac{\pi}{3}]} \]