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

Options

A. [-3, 0] `cup` [1, 3]
B. [-2, 0] `cup` [1,4]
C. [- 2 ,0 ] `cup` [ 2 ,4 ]
D. [ -2 ,0 ] `cup` [ l,2 ]

Answer

To find the domain of the function \( f(x) = \sin^{-1} (|(x-1)| - 2) \), we need to ensure that the argument of the sine inverse function is within the range \([-1, 1]\). This means we need to solve the inequality:

\[ -1 \leq |(x-1)| - 2 \leq 1 \]

### Step 1: Solve the left inequality
First, we solve the left part of the inequality:

\[ |(x-1)| - 2 \geq -1 \]

Adding 2 to both sides gives:

\[ |(x-1)| \geq 1 \]

This implies two cases for the absolute value:

1. \( x - 1 \geq 1 \) (which simplifies to \( x \geq 2 \))
2. \( x - 1 \leq -1 \) (which simplifies to \( x \leq 0 \))

### Step 2: Solve the right inequality
Now, we solve the right part of the inequality:

\[ |(x-1)| - 2 \leq 1 \]

Adding 2 to both sides gives:

\[ |(x-1)| \leq 3 \]

Again, this leads to two cases:

1. \( x - 1 \leq 3 \) (which simplifies to \( x \leq 4 \))
2. \( x - 1 \geq -3 \) (which simplifies to \( x \geq -2 \))

### Step 3: Combine the results
Now we combine the results from both inequalities:

From the left inequality, we have:
- \( x \geq 2 \) or \( x \leq 0 \)

From the right inequality, we have:
- \( -2 \leq x \leq 4 \)

### Step 4: Determine the valid intervals
Now we need to find the intersection of these conditions:

1. For \( x \geq 2 \):
- The valid range is \( [2, 4] \).

2. For \( x \leq 0 \):
- The valid range is \( [-2, 0] \).

### Final Step: Write the domain
Thus, the domain of the function \( f(x) \) is the union of the two intervals:

\[ \text{Domain of } f(x) = [-2, 0] \cup [2, 4] \]