Solve for x, the inequalities in

According to the question,

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

Let $y=|x-2|$, then

$\Rightarrow \frac{y-1}{y-2} \leq 0$

Now, if $y<1$, then

$\mathrm{y}-1<0$ and $\mathrm{y}-2<0$

and, $\frac{y-1}{y-2}>0$ ,which is not required

if $y>2$, then

$y-1>0$ and $y-2>0$

and, $\frac{y-1}{y-2}>0$ , which is not required

if $1 \leq \mathrm{y}<2$, then

$\mathrm{y}-1 \geq 0$ and $\mathrm{y}-2<0$

and,

$\frac{y-1}{y-2}<0$ , which is the required answer,

Hence,

1 ≤ y < 2

⇒ 1 ≤ |x – 2| < 2

Here, there are two cases

⇒ 1 ≤ x – 2 < 2

⇒ 3 ≤ x < 4

And

⇒ 1 ≤ -(x – 2) < 2

⇒ 1 ≤ – x + 2 < 2

Multiplying each term by -1,

⇒ -2 ≤ x – 2 < -1

Adding 2 to each term,

⇒ 0 ≤ x < 1

∴ Hence, solution is [0, 1) ∪ [3, 4)