Domain of the function `f(x) = sin^(-1) [1 + cos x]- sqrt(16-x^(2))` ([.) denotes the greatest integer function) is

Options

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

Answer

To find the domain of the function \( f(x) = \sin^{-1}(1 + \cos x) - \sqrt{16 - x^2} \), we need to analyze both components of the function separately.

### Step 1: Analyze the square root component
The first component we analyze is \( -\sqrt{16 - x^2} \). For the square root to be defined, the expression inside the square root must be non-negative:

\[ 16 - x^2 \geq 0 \]

Rearranging gives:

\[ x^2 \leq 16 \]

Taking the square root of both sides, we find:

\[ -4 \leq x \leq 4 \]

Thus, the domain from this part is:

\[ x \in [-4, 4] \]

### Step 2: Analyze the sine inverse component
Next, we analyze the component \( \sin^{-1}(1 + \cos x) \). The sine inverse function is defined for inputs in the range \([-1, 1]\). Therefore, we need:

\[ -1 \leq 1 + \cos x \leq 1 \]

Breaking this down into two inequalities:

1. **First Inequality**:
\[ -1 \leq 1 + \cos x \implies -2 \leq \cos x \]
Since the cosine function has a range of \([-1, 1]\), this inequality is always satisfied for all \( x \).

2. **Second Inequality**:
\[ 1 + \cos x \leq 1 \implies \cos x \leq 0 \]
This means that \( x \) must be in the intervals where the cosine function is non-positive. The cosine function is non-positive in the intervals:

\[ x \in \left[\frac{\pi}{2} + 2n\pi, \frac{3\pi}{2} + 2n\pi\right], \quad n \in \mathbb{Z} \]

### Step 3: Combine the conditions
Now we need to find the intersection of the two conditions:

1. From the square root condition, we have \( x \in [-4, 4] \).
2. From the sine inverse condition, we have \( x \) values where \( \cos x \leq 0 \).

The critical points in the interval \( [-4, 4] \) where \( \cos x = 0 \) are \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \).

Calculating these values:
- \( \frac{\pi}{2} \approx 1.57 \)
- \( \frac{3\pi}{2} \approx 4.71 \)

Thus, within the interval \( [-4, 4] \), the relevant interval for \( x \) where \( \cos x \leq 0 \) is:

\[ x \in \left[\frac{\pi}{2}, 4\right] \]

### Step 4: Exclude points where the function is undefined
Next, we check for points where the function is undefined. The term \( \sin^{-1}(1 + \cos x) \) becomes undefined when \( 1 + \cos x > 1 \), which occurs when \( \cos x = 1 \). This happens at \( x = 2n\pi \).

In the interval \( [-4, 4] \), the only point to exclude is \( x = 0 \) (since \( 2n\pi = 0 \) when \( n = 0 \)).

### Final Domain
Thus, the domain of the function \( f(x) \) is:

\[ x \in \left[\frac{\pi}{2}, 4\right] \setminus \{0\} \]

### Summary
The domain of the function is:

\[ \left[\frac{\pi}{2}, 4\right] \text{ excluding } 0 \]