Let `P = { x : x in R , |x| < 2}`
`Q = {x:x in R , |x -1 |ge 2}`
`P uu Q = R - S` then set S is : (where R is the set of real number)

Options

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

Explanation

To solve the problem, we need to find the sets \( P \) and \( Q \) and then determine the set \( S \) such that \( P \cup Q = \mathbb{R} - S \).

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

The set \( P \) is defined as:
\[ P = \{ x : x \in \mathbb{R}, |x| < 2 \} \]
This means that \( P \) includes all real numbers \( x \) such that the absolute value of \( x \) is less than 2.

To express this in interval notation:
\[ P = (-2, 2) \]

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

The set \( Q \) is defined as:
\[ Q = \{ x : x \in \mathbb{R}, |x - 1| \geq 2 \} \]
This can be rewritten as two inequalities:
1. \( x - 1 \geq 2 \) which simplifies to \( x \geq 3 \)
2. \( x - 1 \leq -2 \) which simplifies to \( x \leq -1 \)

Thus, the set \( Q \) can be expressed in interval notation as:
\[ Q = (-\infty, -1] \cup [3, \infty) \]

### Step 3: Find the Union of Sets \( P \) and \( Q \)

Now, we need to find the union of sets \( P \) and \( Q \):
\[ P \cup Q = (-2, 2) \cup \left( (-\infty, -1] \cup [3, \infty) \right) \]

To visualize this:
- The interval \( (-2, 2) \) covers all numbers between -2 and 2.
- The interval \( (-\infty, -1] \) covers all numbers less than or equal to -1.
- The interval \( [3, \infty) \) covers all numbers greater than or equal to 3.

Combining these intervals, we can write:
\[ P \cup Q = (-\infty, -1] \cup (-2, 2) \cup [3, \infty) \]

### Step 4: Identify the Set \( S \)

Given that \( P \cup Q = \mathbb{R} - S \), we need to find \( S \). The union \( P \cup Q \) covers all real numbers except for the interval \( (-1, 2) \).

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

### Conclusion

The final answer is:
\[ S = (-1, 2) \]