If `2le||x-2|-1|le3`, then `x` belongs to interval

Options

A. `[-2, -1]uu[5, 6]`
B. `[-4, 6]`
C. `(-oo, -4]uu[6, oo]`
D. `[0, 1]uu[5,6]`

Explanation

To solve the inequality \( 2 \leq | |x - 2| - 1 | \leq 3 \), we will break it down into two parts:

1. **Finding the values for \( ||x - 2| - 1| \geq 2 \)**
2. **Finding the values for \( ||x - 2| - 1| \leq 3 \)**

### Step 1: Solve \( ||x - 2| - 1| \geq 2 \)

This can be split into two cases:

**Case 1:**
\[ |x - 2| - 1 \geq 2 \]
This simplifies to:
\[ |x - 2| \geq 3 \]
This further breaks down into two subcases:
- \( x - 2 \geq 3 \) which gives \( x \geq 5 \)
- \( x - 2 \leq -3 \) which gives \( x \leq -1 \)

**Case 2:**
\[ |x - 2| - 1 \leq -2 \]
This simplifies to:
\[ |x - 2| \leq -1 \]
Since the absolute value cannot be negative, this case does not yield any valid solutions.

### Step 2: Solve \( ||x - 2| - 1| \leq 3 \)

This can also be split into two cases:

**Case 1:**
\[ |x - 2| - 1 \leq 3 \]
This simplifies to:
\[ |x - 2| \leq 4 \]
This further breaks down into:
- \( x - 2 \leq 4 \) which gives \( x \leq 6 \)
- \( x - 2 \geq -4 \) which gives \( x \geq -2 \)

**Case 2:**
\[ |x - 2| - 1 \geq -3 \]
This simplifies to:
\[ |x - 2| \geq -2 \]
Since the absolute value is always non-negative, this case is always true.

### Step 3: Combine the results

From Step 1, we found:
- \( x \geq 5 \) or \( x \leq -1 \)

From Step 2, we found:
- \( -2 \leq x \leq 6 \)

Now we need to find the intersection of these results:

1. **For \( x \geq 5 \)** and \( -2 \leq x \leq 6 \):
- This gives \( x \in [5, 6] \).

2. **For \( x \leq -1 \)** and \( -2 \leq x \leq 6 \):
- This gives \( x \in [-2, -1] \).

### Final Solution

Thus, the solution for \( x \) belongs to the intervals:
\[ [-2, -1] \cup [5, 6] \]