`(|x|-1)/(|x|-2) <= 0` then `x` lies in the interval

Options

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

Explanation

To solve the inequality \(\frac{|x|-1}{|x|-2} \leq 0\), we will follow these steps:

### Step 1: Identify critical points
We start by determining where the expression is equal to zero or undefined. This occurs when the numerator is zero or the denominator is zero.

- **Numerator**: \(|x| - 1 = 0\) gives \(|x| = 1\), which means \(x = 1\) or \(x = -1\).
- **Denominator**: \(|x| - 2 = 0\) gives \(|x| = 2\), which means \(x = 2\) or \(x = -2\).

Thus, the critical points are \(x = -2, -1, 1, 2\).

### Step 2: Test intervals
Next, we will test the intervals formed by these critical points:
- \((- \infty, -2)\)
- \((-2, -1)\)
- \((-1, 1)\)
- \((1, 2)\)
- \((2, \infty)\)

We will check the sign of the expression \(\frac{|x|-1}{|x|-2}\) in each interval.

1. **Interval \((- \infty, -2)\)**: Choose \(x = -3\)
\[ \frac{|-3|-1}{|-3|-2} = \frac{3-1}{3-2} = \frac{2}{1} > 0 \]

2. **Interval \((-2, -1)\)**: Choose \(x = -1.5\)
\[ \frac{|-1.5|-1}{|-1.5|-2} = \frac{1.5-1}{1.5-2} = \frac{0.5}{-0.5} < 0 \]

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

4. **Interval \((1, 2)\)**: Choose \(x = 1.5\)
\[ \frac{|1.5|-1}{|1.5|-2} = \frac{1.5-1}{1.5-2} = \frac{0.5}{-0.5} < 0 \]

5. **Interval \((2, \infty)\)**: Choose \(x = 3\)
\[ \frac{|3|-1}{|3|-2} = \frac{3-1}{3-2} = \frac{2}{1} > 0 \]

### Step 3: Determine valid intervals
From our tests, we find:
- The expression is negative in the intervals \((-2, -1)\) and \((1, 2)\).
- The expression is zero at the points \(x = -1\) and \(x = 1\) (since the numerator is zero).

However, \(x = 2\) is not included since it makes the denominator zero.

### Step 4: Combine intervals
Thus, the solution to the inequality \(\frac{|x|-1}{|x|-2} \leq 0\) is:
\[ x \in [-1, -2) \cup [1, 2) \]

### Final Answer
The intervals where \(x\) lies are:
\[ [-2, -1] \cup [1, 2) \]