Question

Solve $\frac{|x-2|-1}{|x-2|-2}\le 0,xϵR$.

Answer 1

We have either $x-2\ge 0$ or $x-2<0$

Case I

When $x-2\ge 0$.

In this case, $x\ge 2$ and $|x-2|=x-2$.

So, the given inequation becomes

$\frac{x-2-1}{x-2-2}\ge 0\Rightarrow \frac{x-3}{x-4}\le 0$.

$\therefore (x-3\ge 0$ and $x-4<0\left)or\right(x-3\le 0andx-4>0)$

$\Rightarrow (x\ge 3andx-4<0)or(x<3andx>4)$

$\Rightarrow 3\le x<4$ [$\because$ x<3 and x>4 is not possible]

$\Rightarrow xϵ[3,4)$ [ this includes $x\ge 2$].

Case II

When x-2< 0.

In this case, $x<2$ and $|x-2|=-(x-2)=-x+2$.

So, the given inequation becomes

$\frac{-x+2-1}{-x+2-2}\le 0\Rightarrow \frac{-x+1}{-x}\le 0$

[multiplying num. and denom. by -1]

$\therefore (x-1\le 0andx>0)or(x-1\ge andx<0)$

$\Rightarrow (x\le 1andx>0)or(x\ge 1andx<0)$

$\Rightarrow 0<x\le$ [$\because x\ge 1andx<0$ is not possible]

$\Rightarrow xϵ(0,1]$ [this includes x<2].

Hence, from the above two case, we get

solution set $=(0,1]\cup [3,4)$.