Solve the following

Solution:

As, $\frac{|x-2|-1}{|x-2|-2} \leq 0$

Case I: When $x \geq 2,|x-2|=x-2$,

$\frac{x-2-1}{x-2-2} \leq 0$

$\Rightarrow \frac{x-3}{x-4} \leq 0$

$\Rightarrow(x-3 \leq 0$ and $x-4>0)$ or $(x-3 \geq 0$ and $x-4<0)$

$\Rightarrow(x \leq 3$ and $x>4)$ or $(x \geq 3$ and $x<4)$

$\Rightarrow \phi$ or $(3 \leq x<4)$

$\Rightarrow 3 \leq x<4$

So, $x \in[3,4)$

Case II : When $x \leq 2,|x-2|=2-x$,

$\frac{2-x-1}{2-x-2} \leq 0$

$\Rightarrow \frac{1-x}{-x} \leq 0$

$\Rightarrow \frac{x-1}{x} \leq 0$

$\Rightarrow(x-1 \leq 0$ and $x>0)$ or $(x-1 \geq 0$ and $x<0)$

$\Rightarrow(x \leq 1$ and $x>0)$ or $(x \geq 1$ and $x<0)$

$\Rightarrow(0

$\Rightarrow 0

So, $x \in(0,1]$

$\therefore$ From both the cases, we get

$x \in(0,1] \cup[3,4)$