Key Concepts and Formulas

1. Absolute Value Inequalities:

   • $|x| < k \iff -k < x < k$ (for $k > 0$)
   • $|x| \ge k \iff x \le -k \text{ or } x \ge k$ (for $k > 0$)
   • $|x - c| < k \iff c - k < x < c + k$
   • $|x - c| \ge k \iff x - c \le -k \text{ or } x - c \ge k$

2. Set Operations:

   • Union ($A \cup B$): Elements in $A$ or $B$ or both.
   • Intersection ($A \cap B$): Elements in both $A$ and $B$.
   • Difference ($A - B$): Elements in $A$ but not in $B$. This is equivalent to $A \cap B^c$.
   • Complement ($B^c$): Elements in the universal set (here $\mathbb{R}$) but not in $B$.

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Step-by-Step Solution

Step 1: Define Set A in Interval Notation

• What we are doing: We are converting the definition of set A, which is given as an absolute value inequality, into a standard interval notation on the real number line.
• Why: To easily perform set operations, it is essential to represent sets as intervals.
• Working: The set A is given by $A = \{x \in \mathbb{R} : |x| < 2\}$. Using the absolute value inequality rule $|x| < k \iff -k < x < k$, with $k=2$, we get: $-2 < x < 2$
• Result: $A = (-2, 2)$.

Step 2: Define Set B in Interval Notation

• What we are doing: Similar to Step 1, we convert the absolute value inequality defining set B into interval notation.
• Why: This allows for straightforward comparison and combination with set A.
• Working: The set B is given by $B = \{x \in \mathbb{R} : |x – 2| \ge 3\}$. Using the absolute value inequality rule $|x - c| \ge k \iff x - c \le -k \text{ or } x - c \ge k$, with $c=2$ and $k=3$, we get: $x - 2 \le -3 \quad \text{or} \quad x - 2 \ge 3$ Solving the first inequality: $x - 2 \le -3 \implies x \le -3 + 2 \implies x \le -1$ Solving the second inequality: $x - 2 \ge 3 \implies x \ge 3 + 2 \implies x \ge 5$
• Result: $B = (-\infty, -1] \cup [5, \infty)$.

Step 3: Evaluate Option (A): A – B = [–1, 2)

• What we are doing: We are calculating the set difference $A - B$ and comparing it to the given interval.
• Why: To determine if option (A) is a correct statement.
• Working: The set difference $A - B$ contains elements that are in $A$ but not in $B$. This is equivalent to $A \cap B^c$. First, we find the complement of $B$, $B^c$. $B^c = \mathbb{R} \setminus B = \mathbb{R} \setminus ((-\infty, -1] \cup [5, \infty))$. The complement of $(-\infty, -1]$ is $(-1, \infty)$. The complement of $[5, \infty)$ is $(-\infty, 5)$. Therefore, $B^c = (-1, \infty) \cap (-\infty, 5) = (-1, 5)$. Now, we compute $A - B = A \cap B^c$: $A - B = (-2, 2) \cap (-1, 5)$ The intersection of these two intervals is the set of numbers that are greater than -1 and less than 2. $A - B = (-1, 2)$.
• Comparison: The calculated $A - B$ is $(-1, 2)$. Option (A) states $A - B = [-1, 2)$.
• Conclusion for Option (A): Since $(-1, 2) \ne [-1, 2)$, Option (A) is incorrect.

Step 4: Evaluate Option (B): A $\cup$ B = R – (2, 5)

• What we are doing: We are calculating the union $A \cup B$ and comparing it to the given set $\mathbb{R} - (2, 5)$.
• Why: To check the validity of option (B).
• Working: The union $A \cup B$ contains all elements that are in $A$ or in $B$ or in both. $A \cup B = (-2, 2) \cup ((-\infty, -1] \cup [5, \infty))$ Combining the intervals: $(-2, 2) \cup (-\infty, -1] = (-\infty, 2)$ (since $(-1, 2)$ is contained in $(-\infty, 2)$ and $(-\infty, -1]$ covers everything up to -1). So, $A \cup B = (-\infty, 2) \cup [5, \infty)$. Now, let's evaluate $\mathbb{R} - (2, 5)$: $\mathbb{R} - (2, 5) = (-\infty, 2] \cup [5, \infty)$.
• Comparison: The calculated $A \cup B$ is $(-\infty, 2) \cup [5, \infty)$. Option (B) states $A \cup B = (-\infty, 2] \cup [5, \infty)$.
• Conclusion for Option (B): Since $(-\infty, 2) \cup [5, \infty) \ne (-\infty, 2] \cup [5, \infty)$ (due to the endpoint at 2), Option (B) is incorrect.

Step 5: Evaluate Option (C): A $\cap$ B = (–2, –1)

• What we are doing: We are calculating the intersection $A \cap B$ and comparing it to the given interval.
• Why: To verify if option (C) is true.
• Working: The intersection $A \cap B$ contains elements that are common to both $A$ and $B$. $A \cap B = (-2, 2) \cap ((-\infty, -1] \cup [5, \infty))$ We find the intersection of $A$ with each part of $B$:
  1. $(-2, 2) \cap (-\infty, -1]$: The numbers must be greater than -2 and less than 2, AND less than or equal to -1. This gives the interval $(-2, -1]$.
  2. $(-2, 2) \cap [5, \infty)$: The numbers must be greater than -2 and less than 2, AND greater than or equal to 5. There is no overlap, so this intersection is the empty set $\emptyset$. Combining these, $A \cap B = (-2, -1] \cup \emptyset = (-2, -1]$.
• Comparison: The calculated $A \cap B$ is $(-2, -1]$. Option (C) states $A \cap B = (-2, -1)$.
• Conclusion for Option (C): Since $(-2, -1] \ne (-2, -1)$, Option (C) is incorrect.

Step 6: Evaluate Option (D): B – A = R – (–2, 5)

• What we are doing: We are calculating the set difference $B - A$ and comparing it to the given set $\mathbb{R} - (-2, 5)$.
• Why: To check the correctness of option (D).
• Working: The set difference $B - A$ contains elements that are in $B$ but not in $A$. This is equivalent to $B \cap A^c$. First, we find the complement of $A$, $A^c$. $A^c = \mathbb{R} \setminus A = \mathbb{R} \setminus (-2, 2) = (-\infty, -2] \cup [2, \infty)$. Now, we compute $B - A = B \cap A^c$: $B - A = ((-\infty, -1] \cup [5, \infty)) \cap ((-\infty, -2] \cup [2, \infty))$ We find the intersection of each part of $B$ with each part of $A^c$:
  1. $(-\infty, -1] \cap (-\infty, -2]$: The numbers must be less than or equal to -1 AND less than or equal to -2. This gives $(-\infty, -2]$.
  2. $(-\infty, -1] \cap [2, \infty)$: There is no overlap, so this is $\emptyset$.
  3. $[5, \infty) \cap (-\infty, -2]$: There is no overlap, so this is $\emptyset$.
  4. $[5, \infty) \cap [2, \infty)$: The numbers must be greater than or equal to 5 AND greater than or equal to 2. This gives $[5, \infty)$. Combining these results: $B - A = (-\infty, -2] \cup \emptyset \cup \emptyset \cup [5, \infty) = (-\infty, -2] \cup [5, \infty)$. Now, let's evaluate $\mathbb{R} - (-2, 5)$: $\mathbb{R} - (-2, 5) = (-\infty, -2] \cup [5, \infty)$.
• Comparison: The calculated $B - A$ is $(-\infty, -2] \cup [5, \infty)$. Option (D) states $B - A = \mathbb{R} - (-2, 5)$, which is also $(-\infty, -2] \cup [5, \infty)$.
• Conclusion for Option (D): Since both expressions are equal, Option (D) is correct.

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Common Mistakes & Tips

• Strict vs. Non-strict Inequalities: Pay very close attention to whether the inequality is strict ($<$, $>$) or non-strict ($\le$, $\ge$). This determines whether the endpoints are included in the interval (closed bracket ] or [) or excluded (open parenthesis ) or ().
• Visualizing on a Number Line: Drawing a number line and marking the intervals for sets A and B can greatly help in understanding intersections, unions, and differences.
• Complement of Intervals: Remember that the complement of an open interval $(a, b)$ is $(-\infty, a] \cup [b, \infty)$, and the complement of a union of intervals like $(-\infty, a] \cup [b, \infty)$ is $(a, b)$.

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Summary

We began by converting the absolute value inequalities defining sets A and B into interval notations: $A = (-2, 2)$ and $B = (-\infty, -1] \cup [5, \infty)$. We then systematically evaluated each of the given options by performing the specified set operations. After calculating $A - B$, $A \cup B$, $A \cap B$, and $B - A$, we compared our results with the statements in options (A), (B), (C), and (D). Our calculations showed that $B - A = (-\infty, -2] \cup [5, \infty)$, which is precisely equal to $\mathbb{R} - (-2, 5)$. Therefore, option (D) is the correct statement.

The final answer is $\boxed{\text{(D)}}$.