Question:
If \(A = \{x \in \mathbb{R} : |x| \lt 2\}\) and \(B = \{x \in \mathbb{R} : |x - 2| \ge 3\}\); then :
If \(A = \{x \in \mathbb{R} : |x| \lt 2\}\) and \(B = \{x \in \mathbb{R} : |x - 2| \ge 3\}\); then :
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Write A and B as intervals: \(A = (-2, 2)\) and \(B = (-\infty, -1] \cup [5, \infty)\), then test each option.
Updated On: Nov 5, 2026
- \(A - B = [-1, 2)\)
- \(A \cup B = \mathbb{R} - (2, 5)\)
- \(A \cap B = (-2, -1)\)
- \(B - A = \mathbb{R} - (-2, 5)\)
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The Correct Option is D
Solution and Explanation
\(|x| \lt 2 \Rightarrow A = (-2, 2)\).
\(|x - 2| \ge 3 \Rightarrow x - 2 \le -3\) or \(x - 2 \ge 3 \Rightarrow x \le -1\) or \(x \ge 5\), so \(B = (-\infty, -1] \cup [5, \infty)\).
\(A - B = (-1, 2)\), which is not \([-1, 2)\) because \(-1 \in B\). So (1) is false.
\(A \cup B = (-\infty, 2) \cup [5, \infty) = \mathbb{R} - [2, 5)\), not \(\mathbb{R} - (2, 5)\). So (2) is false.
\(A \cap B = (-2, -1]\), not \((-2, -1)\). So (3) is false.
\(B - A\): remove \((-2, 2)\) from B, leaving \((-\infty, -2] \cup [5, \infty) = \mathbb{R} - (-2, 5)\). So (4) is true.
Answer: (4) \(B - A = \mathbb{R} - (-2, 5)\)
\(|x - 2| \ge 3 \Rightarrow x - 2 \le -3\) or \(x - 2 \ge 3 \Rightarrow x \le -1\) or \(x \ge 5\), so \(B = (-\infty, -1] \cup [5, \infty)\).
\(A - B = (-1, 2)\), which is not \([-1, 2)\) because \(-1 \in B\). So (1) is false.
\(A \cup B = (-\infty, 2) \cup [5, \infty) = \mathbb{R} - [2, 5)\), not \(\mathbb{R} - (2, 5)\). So (2) is false.
\(A \cap B = (-2, -1]\), not \((-2, -1)\). So (3) is false.
\(B - A\): remove \((-2, 2)\) from B, leaving \((-\infty, -2] \cup [5, \infty) = \mathbb{R} - (-2, 5)\). So (4) is true.
Answer: (4) \(B - A = \mathbb{R} - (-2, 5)\)
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