Solve 1≤|x−2|≤3

As, $1 \leq|x-2| \leq 3$

$\Rightarrow|x-2|>1$ and $|x-2|<3$

$\Rightarrow((x-2) \leq-1$ or $(x-2) \geq 1)$ and $(-3 \leq(x-2) \leq 3)$

(As, $|x| \geq a \Rightarrow x \leq-a$ or $x \geq a ;$ and $|x| \leq a \Rightarrow-a \leq x \leq a$ )

$\Rightarrow(x \leq 1$ or $x \geq 3)$ and $(-3+2 \leq x \leq 3+2)$

$\Rightarrow(x \leq 1$ or $x \geq 3)$ and $(-1 \leq x \leq 5)$

$\Rightarrow x \in(-\infty, 1] \cup[3, \infty)$ and $x \in[-1,5]$

$\therefore x \in[-1,1] \cup[3,5]$