`A={x in R : abs(x+1) lt 2} , B={x in R: abs(x-1) ge 2}` then

Options

A. `AuuB=R-[1,3]`
B. `AnnB=(-1,1)`
C. `AnnB=(-3,1]`
D. `B-A=R-(-3,1]`

Answer

To solve the problem, we need to analyze the sets A and B defined by the given conditions.

### Step 1: Define Set A
Set A is defined as:
\[ A = \{ x \in \mathbb{R} : |x + 1| < 2 \} \]

To solve this, we can break down the absolute value inequality:
\[ -2 < x + 1 < 2 \]

Now, we can subtract 1 from all parts of the inequality:
\[ -2 - 1 < x < 2 - 1 \]
\[ -3 < x < 1 \]

Thus, we can express Set A in interval notation:
\[ A = (-3, 1) \]

### Step 2: Define Set B
Set B is defined as:
\[ B = \{ x \in \mathbb{R} : |x - 1| \geq 2 \} \]

We can break this down into two cases:
1. \( x - 1 \geq 2 \)
2. \( x - 1 \leq -2 \)

For the first case:
\[ x - 1 \geq 2 \implies x \geq 3 \]

For the second case:
\[ x - 1 \leq -2 \implies x \leq -1 \]

Thus, we can express Set B in interval notation:
\[ B = (-\infty, -1] \cup [3, \infty) \]

### Step 3: Find A ∪ B (Union of A and B)
Now, we need to find the union of sets A and B:
\[ A \cup B = (-3, 1) \cup (-\infty, -1] \cup [3, \infty) \]

To combine these intervals, we note:
- The interval \( (-\infty, -1] \) extends to the left of \( -1 \).
- The interval \( (-3, 1) \) covers from \( -3 \) to \( 1 \).
- The interval \( [3, \infty) \) starts from \( 3 \).

Thus, the union can be represented as:
\[ A \cup B = (-\infty, -1] \cup (-3, 1) \cup [3, \infty) \]

### Step 4: Find A ∩ B (Intersection of A and B)
Next, we find the intersection of sets A and B:
\[ A \cap B = (-3, 1) \cap \left((- \infty, -1] \cup [3, \infty)\right) \]

The intersection with \( (-\infty, -1] \) gives:
\[ (-3, 1) \cap (-\infty, -1] = (-3, -1] \]

The intersection with \( [3, \infty) \) gives:
\[ (-3, 1) \cap [3, \infty) = \emptyset \]

Thus, the intersection is:
\[ A \cap B = (-3, -1] \]

### Step 5: Find B - A (Difference of B and A)
Finally, we find \( B - A \):
\[ B - A = \left((- \infty, -1] \cup [3, \infty)\right) - (-3, 1) \]

For the interval \( (-\infty, -1] \):
\[ (-\infty, -1] - (-3, 1) = (-\infty, -3] \cup (-3, -1] \]

For the interval \( [3, \infty) \):
\[ [3, \infty) - (-3, 1) = [3, \infty) \]

Thus, we can express \( B - A \) as:
\[ B - A = (-\infty, -3] \cup [3, \infty) \]

### Summary of Results
- \( A = (-3, 1) \)
- \( B = (-\infty, -1] \cup [3, \infty) \)
- \( A \cup B = (-\infty, -1] \cup (-3, 1) \cup [3, \infty) \)
- \( A \cap B = (-3, -1] \)
- \( B - A = (-\infty, -3] \cup [3, \infty) \)