Set, Relation and FunctionMedium
Question
Let $A = \left\{ x:\left| x^{2} - 10 \right|\ 6 \right\}$ and $B = \{ x:|x - 2| > 1\}$. Then
Options
A.$A \cup B = ( - \infty,1\rbrack \cup (2,\infty)$
B.$A - B = \lbrack 2,3)$
C.$A \cap B = \lbrack - 4, - 2\rbrack \cup \lbrack 3,4\rbrack$
D.$B - A = ( - \infty, - 4) \cup ( - 2,1) \cup (4,\infty)$
Solution
$\left| x^{2} - 10 \right| \leq 6$
$${- 6 \leq x^{2} - 10 \leq 6 }{4 \leq x^{2} \leq 16 }{A = \lbrack - 4, - 2\rbrack \cup \lbrack 2,4\rbrack }{|x - 2| > 1 }{B = ( - \infty,1) \cup (3,\infty) }{A \cup B = ( - \infty,1) \cup \lbrack 2,\infty) }{A \cap B = \lbrack - 4, - 2\rbrack \cup (3,4\rbrack }{A - B = \lbrack 2,3\rbrack }{B - A = ( - \infty, - 4) \cup ( - 2,1) \cup (4,\infty)}$$
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