Question
Let $A = \{x : |x^2 - 10| \le 6\}$ and $B = \{x : |x - 2| > 1\}$. Then
The correct answer is
$A \cup B = (-\infty, 1] \cup (2, \infty)$
Let's analyze the given sets \(A\) and \(B\) and determine the correct option.
Set A: The set is defined as \(A = \{x : |x^2 - 10| \le 6\}\).
- This implies \(-6 \le x^2 - 10 \le 6\).
- Solving the inequalities:
- For \(x^2 - 10 \ge -6\), we get \(x^2 \ge 4\). Thus, \(x \le -2\) or \(x \ge 2\).
- For \(x^2 - 10 \le 6\), we get \(x^2 \le 16\). Thus, \(-4 \le x \le 4\).
- Combining these, we have \(A = [-4, -2] \cup [2, 4]\).
Set B: The set is defined as \(B = \{x : |x - 2| > 1\}\).
- This implies \(x - 2 > 1\) or \(x - 2 < -1\).
- Solving the inequalities:
- \(x - 2 > 1\) implies \(x > 3\).
- \(x - 2 < -1\) implies \(x < 1\).
- Thus, \(B = (-\infty, 1) \cup (3, \infty)\).
We need to determine \(A \cup B\).
From \(A = [-4, -2] \cup [2, 4]\) and \(B = (-\infty, 1) \cup (3, \infty)\), we can find the union:
- \([-4, -2]\) is within \((-\infty, 1)\).
- \([2, 4]\) overlaps with \((3, \infty)\).
- The union gives us \((-\infty, 1] \cup (2, \infty)\).
Thus, the correct answer is \(A \cup B = (-\infty, 1] \cup (2, \infty)\).
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