Question:
Mark the correct alternative in each of the following:
If $|x+2| \leq 9$, then
(a) $x \in(-7,11)$
(b) $x \in[-11,7]$
(c) $x \in(-\infty,-7) \cup(11, \infty)$
(d) $x \in(-\infty,-7) \cup[11, \infty)$
Solution:
$|x+2| \leq 9$
$\Rightarrow-9 \leq x+2 \leq 9$
$\Rightarrow-9-2 \leq x+2-2 \leq 9-2$
$\Rightarrow-11 \leq x \leq 7$
$\Rightarrow x \in[-11,7]$
Hence, the correct option is (b).