The domain of definition of `f(x) = sqrt((1-|x|)/(2-|x|))` is

Options

A. `(-oo ,oo) -[-2,2]`
B. `(-oo,oo)-[-1,1]`
C. `[-1,1] uu (-oo ,-2) uu (2,oo)`
D. none of these

Answer

To find the domain of the function \( f(x) = \sqrt{\frac{1 - |x|}{2 - |x|}} \), we need to ensure that the expression inside the square root is non-negative and that the denominator is not zero.

### Step 1: Set the condition for the square root
The expression inside the square root must be greater than or equal to zero:
\[ \frac{1 - |x|}{2 - |x|} \geq 0 \]

### Step 2: Identify the critical points
To solve the inequality, we first find the points where the numerator and denominator are zero.

1. **Numerator**: Set \( 1 - |x| = 0 \)
\[ |x| = 1 \implies x = -1 \text{ or } x = 1 \]

2. **Denominator**: Set \( 2 - |x| = 0 \)
\[ |x| = 2 \implies x = -2 \text{ or } x = 2 \]

### Step 3: Determine the intervals
The critical points are \( -2, -1, 1, 2 \). We will test the intervals created by these points:
- \( (-\infty, -2) \)
- \( (-2, -1) \)
- \( (-1, 1) \)
- \( (1, 2) \)
- \( (2, \infty) \)

### Step 4: Test each interval
1. **Interval \( (-\infty, -2) \)**: Choose \( x = -3 \)
\[ \frac{1 - 3}{2 - 3} = \frac{-2}{-1} = 2 \quad (\text{positive}) \]

2. **Interval \( (-2, -1) \)**: Choose \( x = -1.5 \)
\[ \frac{1 - 1.5}{2 - 1.5} = \frac{-0.5}{0.5} = -1 \quad (\text{negative}) \]

3. **Interval \( (-1, 1) \)**: Choose \( x = 0 \)
\[ \frac{1 - 0}{2 - 0} = \frac{1}{2} \quad (\text{positive}) \]

4. **Interval \( (1, 2) \)**: Choose \( x = 1.5 \)
\[ \frac{1 - 1.5}{2 - 1.5} = \frac{-0.5}{0.5} = -1 \quad (\text{negative}) \]

5. **Interval \( (2, \infty) \)**: Choose \( x = 3 \)
\[ \frac{1 - 3}{2 - 3} = \frac{-2}{-1} = 2 \quad (\text{positive}) \]

### Step 5: Combine the results
From the tests, we find that the inequality is satisfied in the intervals:
- \( (-\infty, -2) \)
- \( (-1, 1) \)
- \( (2, \infty) \)

### Step 6: Exclude points where the denominator is zero
The points \( x = -2 \) and \( x = 2 \) must be excluded from the domain because they make the denominator zero.

### Final Domain
Thus, the domain of \( f(x) \) is:
\[ \boxed{(-\infty, -2) \cup [-1, 1] \cup (2, \infty)} \]