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

Options

A. `[2,-1) cup (1,2]`
B. `[-2,2]`
C. `(-1,1)`
D. `(-infty,-1) cup (2,infty)`

Answer

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

### Step 1: Identify the conditions for the square root to be defined
The expression inside the square root must be non-negative:
\[ \frac{2 - |x|}{|x| - 1} \geq 0 \]
This means both the numerator and the denominator must be either both positive or both negative.

### Step 2: Analyze the numerator
The numerator \( 2 - |x| \) is non-negative when:
\[ 2 - |x| \geq 0 \implies |x| \leq 2 \]
This gives us the interval:
\[ -2 \leq x \leq 2 \]

### Step 3: Analyze the denominator
The denominator \( |x| - 1 \) must be positive (since it cannot be zero):
\[ |x| - 1 > 0 \implies |x| > 1 \]
This gives us two intervals:
\[ x < -1 \quad \text{or} \quad x > 1 \]

### Step 4: Combine the conditions
Now we combine the conditions from Steps 2 and 3:
1. From the numerator: \( -2 \leq x \leq 2 \)
2. From the denominator: \( x < -1 \) or \( x > 1 \)

### Step 5: Find the intersection of the intervals
- For \( x < -1 \): The intersection with \( -2 \leq x \leq 2 \) gives:
\[ -2 \leq x < -1 \quad \text{(Interval 1)} \]

- For \( x > 1 \): The intersection with \( -2 \leq x \leq 2 \) gives:
\[ 1 < x \leq 2 \quad \text{(Interval 2)} \]

### Step 6: Write the final domain
Combining both intervals, we find the domain of \( f(x) \):
\[ \text{Domain} = [-2, -1) \cup (1, 2] \]

### Final Answer
The domain of \( f(x) = \sqrt{\frac{2 - |x|}{|x| - 1}} \) is:
\[ [-2, -1) \cup (1, 2] \]

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