Key Concepts and Formulas

• Solving absolute value inequalities:
  • $|x| < a \iff -a < x < a$
  • $|x| \ge a \iff x \le -a \text{ or } x \ge a$
• Set operations:
  • $A - B = \{x \mid x \in A \text{ and } x \notin B\}$ (Set difference)
  • $A \cap B = \{x \mid x \in A \text{ and } x \in B\}$ (Intersection)
  • $A \cup B = \{x \mid x \in A \text{ or } x \in B\}$ (Union)

Step-by-Step Solution

Step 1: Determine the set A

• What & Why: We need to find all real numbers x that satisfy the inequality $|x + 1| < 2$. We use the rule for $|x| < a$ to solve this.
• Math: $|x + 1| < 2$ $-2 < x + 1 < 2$ $-2 - 1 < x < 2 - 1$ $-3 < x < 1$
• Reasoning: We subtracted 1 from all parts of the inequality to isolate x. This gives us the interval for set A.
• Result: $A = \{x \in \mathbb{R} \mid -3 < x < 1\} = (-3, 1)$

Step 2: Determine the set B

• What & Why: We need to find all real numbers x that satisfy the inequality $|x - 1| \ge 2$. We use the rule for $|x| \ge a$ to solve this, which will result in two separate inequalities.
• Math: $|x - 1| \ge 2$ This gives us two cases: Case 1: $x - 1 \ge 2$ $x \ge 3$ Case 2: $x - 1 \le -2$ $x \le -1$
• Reasoning: We solved each case separately by adding 1 to both sides of the inequalities.
• Result: $B = \{x \in \mathbb{R} \mid x \le -1 \text{ or } x \ge 3\} = (-\infty, -1] \cup [3, \infty)$

Step 3: Calculate A - B and check option (A)

• What & Why: We need to find the set difference A - B, which contains elements in A but not in B. Then, we check if it matches option (A).
• Math: $A = (-3, 1)$ $B = (-\infty, -1] \cup [3, \infty)$ $A - B = \{x \mid x \in A \text{ and } x \notin B\}$ We are looking for x such that $-3 < x < 1$ and ($x > -1$ and $x < 3$). Combining these gives $-1 < x < 1$.
• Reasoning: A - B consists of the part of A where x is not less than or equal to -1, and not greater than or equal to 3. This is equivalent to x being strictly greater than -1 and strictly less than 3. Combining this with the condition that x is between -3 and 1, we get -1 < x < 1.
• Result: $A - B = (-1, 1)$
• Option Check: Option (A) states $A - B = (-1, 1)$. This is TRUE.

Step 4: Calculate B - A and check option (B)

• What & Why: We need to find the set difference B - A, which contains elements in B but not in A. Then, we check if it matches option (B).
• Math: $B = (-\infty, -1] \cup [3, \infty)$ $A = (-3, 1)$ $B - A = \{x \mid x \in B \text{ and } x \notin A\}$ We are looking for x such that ($x \le -1$ or $x \ge 3$) and (x <= -3 or x >= 1 is false). This means that x <= -1 and x <= -3 or x >=3 . The numbers in A are strictly between -3 and 1. The overlap between $(-\infty, -1]$ and $(-3, 1)$ is $(-3, -1]$. Thus, $(-\infty, -1] - (-3, 1) = (-\infty, -3]$. The interval $[3, \infty)$ has no overlap with $(-3, 1)$. Thus, $[3, \infty) - (-3, 1) = [3, \infty)$. Therefore, $B - A = (-\infty, -3] \cup [3, \infty)$.
• Reasoning: B - A consists of the part of B where x is not strictly between -3 and 1. This means either x is less than or equal to -3, or x is greater than or equal to 1.
• Result: $B - A = (-\infty, -3] \cup [3, \infty)$
• Option Check: Option (B) states $B - A = \mathbb{R} - (-3, 1) = (-\infty, -3] \cup [1, \infty)$. Since $(-\infty, -3] \cup [3, \infty) \ne (-\infty, -3] \cup [1, \infty)$, this is NOT TRUE.

Step 5: Calculate A ∩ B and check option (C)

• What & Why: We need to find the intersection of A and B, which contains elements in both A and B. Then, we check if it matches option (C).
• Math: $A = (-3, 1)$ $B = (-\infty, -1] \cup [3, \infty)$ $A \cap B = \{x \mid x \in A \text{ and } x \in B\}$ We are looking for x such that $-3 < x < 1$ and ($x \le -1$ or $x \ge 3$). The intersection of $(-3, 1)$ and $(-\infty, -1]$ is $(-3, -1]$. The intersection of $(-3, 1)$ and $[3, \infty)$ is empty. Therefore, $A \cap B = (-3, -1]$.
• Reasoning: A ∩ B consists of the elements common to both A and B.
• Result: $A \cap B = (-3, -1]$
• Option Check: Option (C) states $A \cap B = (-3, -1]$. This is TRUE.

Step 6: Calculate A ∪ B and check option (D)

• What & Why: We need to find the union of A and B, which contains elements in either A or B or both. Then, we check if it matches option (D).
• Math: $A = (-3, 1)$ $B = (-\infty, -1] \cup [3, \infty)$ $A \cup B = \{x \mid x \in A \text{ or } x \in B\}$ We are looking for x such that $-3 < x < 1$ or ($x \le -1$ or $x \ge 3$). Combining these intervals, we get $(-\infty, 1) \cup [3, \infty)$.
• Reasoning: A ∪ B consists of all elements in A or B.
• Result: $A \cup B = (-\infty, 1) \cup [3, \infty)$
• Option Check: Option (D) states $A \cup B = \mathbb{R} - [1, 3) = (-\infty, 1) \cup [3, \infty)$. This is TRUE.

Common Mistakes & Tips

• Pay close attention to whether the inequalities are strict ($<, >$) or non-strict ($\le, \ge$). This determines whether the endpoints are included in the intervals.
• Visualizing the intervals on a number line can be extremely helpful for performing set operations and avoiding errors.
• Remember the definitions of set operations: union, intersection, and difference.

Summary

We found that A = (-3, 1) and B = $(-\infty, -1] \cup [3, \infty)$. After performing the set operations, we determined that option (B) is the statement that is NOT true.

Final Answer

The final answer is \boxed{B}, which corresponds to option (B).