Let $A = \{x : |x^{2} - 10| \le 6\}$ and $B = \{x : |x - 2| > 1\}$. Then:

  • A
    $A \cup B = (-\infty, 1) \cup [2, \infty)$
  • B
    $A - B = [2, 3]$
  • C
    $A \cap B = [-4, -2] \cup (3, 4]$
  • D
    $B - A = (-\infty, -4) \cup (-2, 1) \cup (4, \infty)$

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