Let `A={(x:x inR,|x|lt1}`,
`B={x:x inR,|x-1|ge1}`
and `AuuB=R-D`, then set D is

Options

A. `{x:1ltxle2}`
B. `{x:1lexlt2}`
C. `{x:1lexle2}`
D. None of these

Answer

To solve the problem, we need to analyze the sets \( A \) and \( B \) and find the set \( D \) such that \( A \cup B = \mathbb{R} - D \).

### Step 1: Define the Set \( A \)

The set \( A \) is defined as:
\[ A = \{ x \in \mathbb{R} : |x| < 1 \} \]
This means that \( A \) includes all real numbers \( x \) such that \( -1 < x < 1 \).

### Step 2: Define the Set \( B \)

The set \( B \) is defined as:
\[ B = \{ x \in \mathbb{R} : |x - 1| \geq 1 \} \]
To understand this set, we can break it down into two cases based on the definition of absolute value:

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

Thus, we can express \( B \) as:
\[ B = (-\infty, 0] \cup [2, \infty) \]

### Step 3: Find the Union of Sets \( A \) and \( B \)

Now, we need to find the union \( A \cup B \):
\[ A \cup B = (-1, 1) \cup \left( (-\infty, 0] \cup [2, \infty) \right) \]

To visualize this, we can see how the intervals combine:
- The interval \( (-1, 1) \) overlaps with \( (-\infty, 0] \), so we can combine them:
\[ A \cup B = (-\infty, 1) \cup [2, \infty) \]

### Step 4: Determine the Set \( D \)

According to the problem, we have:
\[ A \cup B = \mathbb{R} - D \]
This means that \( D \) consists of the elements in \( \mathbb{R} \) that are not in \( A \cup B \).

The complement of \( A \cup B \) in \( \mathbb{R} \) is:
\[ D = \mathbb{R} - \left( (-\infty, 1) \cup [2, \infty) \right) \]

This simplifies to:
\[ D = [1, 2) \]

### Final Answer

Thus, the set \( D \) is:
\[ D = [1, 2) \]