The domain of the function
`f(x)=sqrt(abs(sin^(-1)(sinx))-cos^(-1)(cosx))` in `[0,2pi]` is

Options

A. `[0,pi/2] cup [(3pi)/2,2pi]`
B. `[pi,2pi]`
C. `[0,pi]-{pi/2}`
D. `[0,2pi]-{pi/2,(3pi)/2}`

Answer

To find the domain of the function \( f(x) = \sqrt{|\sin^{-1}(\sin x) - \cos^{-1}(\cos x)|} \) in the interval \([0, 2\pi]\), we will analyze the components of the function step by step.

### Step 1: Understanding the components
1. **Sine and Cosine Inverses**:
- The function \( \sin^{-1}(\sin x) \) will return values based on the range of \( x \). Specifically, it will yield:
- \( x \) for \( 0 \leq x \leq \frac{\pi}{2} \)
- \( \pi - x \) for \( \frac{\pi}{2} < x \leq \pi \)
- \( x - 2\pi \) for \( \pi < x \leq \frac{3\pi}{2} \)
- \( 2\pi - x \) for \( \frac{3\pi}{2} < x \leq 2\pi \)

- The function \( \cos^{-1}(\cos x) \) will yield:
- \( x \) for \( 0 \leq x \leq \pi \)
- \( 2\pi - x \) for \( \pi < x \leq 2\pi \)

### Step 2: Analyzing the intervals
Now we will analyze the function \( f(x) \) in the four intervals of \([0, 2\pi]\):

#### Interval 1: \( [0, \frac{\pi}{2}] \)
- Here, \( \sin^{-1}(\sin x) = x \) and \( \cos^{-1}(\cos x) = x \).
- Thus, \( f(x) = \sqrt{|x - x|} = \sqrt{0} = 0 \).
- This is valid, so this interval is in the domain.

#### Interval 2: \( [\frac{\pi}{2}, \pi] \)
- Here, \( \sin^{-1}(\sin x) = \pi - x \) and \( \cos^{-1}(\cos x) = x \).
- Thus, \( f(x) = \sqrt{|(\pi - x) - x|} = \sqrt{|\pi - 2x|} \).
- For \( f(x) \) to be defined, \( \pi - 2x \geq 0 \) must hold, leading to \( x \leq \frac{\pi}{2} \).
- However, since we are in the interval \( [\frac{\pi}{2}, \pi] \), the only valid point is \( x = \frac{\pi}{2} \).
- Thus, this interval contributes only the endpoint \( \frac{\pi}{2} \) to the domain.

#### Interval 3: \( [\pi, \frac{3\pi}{2}] \)
- Here, \( \sin^{-1}(\sin x) = x - 2\pi \) and \( \cos^{-1}(\cos x) = x \).
- Thus, \( f(x) = \sqrt{|(x - 2\pi) - x|} = \sqrt{| -2\pi |} \), which is not valid since it results in a negative value under the square root.
- Therefore, this interval is not in the domain.

#### Interval 4: \( [\frac{3\pi}{2}, 2\pi] \)
- Here, \( \sin^{-1}(\sin x) = 2\pi - x \) and \( \cos^{-1}(\cos x) = 2\pi - x \).
- Thus, \( f(x) = \sqrt{|(2\pi - x) - (2\pi - x)|} = \sqrt{0} = 0 \).
- This is valid, so this interval is in the domain.

### Conclusion
The valid intervals for the domain of \( f(x) \) in \([0, 2\pi]\) are:
- \( [0, \frac{\pi}{2}] \)
- \( [\frac{3\pi}{2}, 2\pi] \)

Thus, the domain of the function \( f(x) \) is:
\[ \text{Domain of } f(x) = [0, \frac{\pi}{2}] \cup [\frac{3\pi}{2}, 2\pi] \]