The domain of definition of `f(x)=sin^(-1)(|x-1|-2)` is

Options

A. `[-2,0]uu[2,4]`
B. `(-2,0)uu(2,4)`
C. `[-2,0]uu[1,3]`
D. `(-2,0)uu(1,3)`

Explanation

To find the domain of the function \( f(x) = \sin^{-1}(|x-1| - 2) \), we need to ensure that the expression inside the inverse sine function lies within its valid range, which is from -1 to 1.

### Step-by-Step Solution:

1. **Set up the inequality for the inverse sine function:**
\[ -1 \leq |x-1| - 2 \leq 1 \]

2. **Solve the left side of the inequality:**
\[ |x-1| - 2 \geq -1 \]
Adding 2 to both sides:
\[ |x-1| \geq 1 \]

3. **Solve the right side of the inequality:**
\[ |x-1| - 2 \leq 1 \]
Adding 2 to both sides:
\[ |x-1| \leq 3 \]

4. **Break down the absolute value inequalities:**
- For \( |x-1| \geq 1 \):
- This gives us two cases:
1. \( x - 1 \geq 1 \) which simplifies to \( x \geq 2 \)
2. \( x - 1 \leq -1 \) which simplifies to \( x \leq 0 \)

- For \( |x-1| \leq 3 \):
- This also gives us two cases:
1. \( x - 1 \leq 3 \) which simplifies to \( x \leq 4 \)
2. \( x - 1 \geq -3 \) which simplifies to \( x \geq -2 \)

5. **Combine the results:**
- From \( |x-1| \geq 1 \), we have:
\[ x \leq 0 \quad \text{or} \quad x \geq 2 \]
- From \( |x-1| \leq 3 \), we have:
\[ -2 \leq x \leq 4 \]

6. **Find the intersection of the intervals:**
- For \( x \leq 0 \) and \( -2 \leq x \leq 4 \), the intersection is:
\[ -2 \leq x \leq 0 \]
- For \( x \geq 2 \) and \( -2 \leq x \leq 4 \), the intersection is:
\[ 2 \leq x \leq 4 \]

7. **Combine the two intervals:**
- The domain of \( f(x) \) is:
\[ x \in [-2, 0] \cup [2, 4] \]

### Final Answer:
The domain of the function \( f(x) = \sin^{-1}(|x-1| - 2) \) is:
\[ [-2, 0] \cup [2, 4] \]