The domain of `f(x) = sqrt(pi/2 - sin^(-1) (x+|x|)/3)` is

Options

A. `[0, 2/3]`
B. `[0,1]`
C. `[0,3/2]`
D. `(-infty, 3/2]`

Answer

To find the domain of the function \( f(x) = \sqrt{\frac{\pi}{2} - \frac{\sin^{-1}(x + |x|)}{3}} \), we need to ensure that the expression inside the square root is non-negative. This means we need to solve the inequality:

\[ \frac{\pi}{2} - \frac{\sin^{-1}(x + |x|)}{3} \geq 0 \]

### Step 1: Rearranging the Inequality
First, we rearrange the inequality to isolate the inverse sine function:

\[ \frac{\sin^{-1}(x + |x|)}{3} \leq \frac{\pi}{2} \]

Multiplying both sides by 3 gives:

\[ \sin^{-1}(x + |x|) \leq \frac{3\pi}{2} \]

### Step 2: Understanding the Range of \( \sin^{-1} \)
The range of the function \( \sin^{-1}(y) \) is from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) for \(y\) in the interval \([-1, 1]\). Therefore, we need to ensure that:

\[ x + |x| \in [-1, 1] \]

### Step 3: Analyzing \( x + |x| \)
The expression \( x + |x| \) behaves differently based on whether \(x\) is positive or negative:

1. **Case 1: \(x \geq 0\)**
- Here, \( |x| = x \), so:
\[ x + |x| = x + x = 2x \]
Thus, we need:
\[ 2x \in [-1, 1] \implies -1 \leq 2x \leq 1 \implies -\frac{1}{2} \leq x \leq \frac{1}{2} \]

2. **Case 2: \(x < 0\)**
- Here, \( |x| = -x \), so:
\[ x + |x| = x - x = 0 \]
Thus, we need:
\[ 0 \in [-1, 1] \quad \text{(which is always true)} \]

### Step 4: Combining the Results
From Case 1, we found that \( x \) must be in the interval \([-0.5, 0.5]\) when \(x\) is non-negative. From Case 2, we found that any negative \(x\) is valid.

Thus, combining both cases, the domain of \( f(x) \) is:

\[ (-\infty, 0) \cup \left[-\frac{1}{2}, \frac{1}{2}\right] \]

### Final Domain
The final domain of the function \( f(x) \) is:

\[ (-\infty, 0) \cup \left[-\frac{1}{2}, \frac{1}{2}\right] \]