Modulus inequality

$f(x) = \begin{cases}{\dfrac{(x - 2) - 2}{(x-1) - 1}} && {\forall\>x\>\geq\>2} \\ {\dfrac{-(x - 2) - 2}{(x - 1) - 1}} && {\forall\>1\>\leq\>x\><\>2} \\ {\dfrac{-(x - 2) - 2}{-(x - 1) - 1}} && {\forall\>x\><\>1}\end{cases}$

Case 1:-
$\begin{aligned} \dfrac{x - 4}{x - 2} & \leq 0\\ \\ \implies x & \in (2,4] \end{aligned}$

Case 2:-
$\begin{aligned} \dfrac{x}{x - 2} \geq 0\\ \\ \implies x & \in (-\infty , 0] \cup (2 , \infty)\\ \\ \text{Bute this equation ia true only} \forall 1 \leq x < 2\\ \\ \therefore x & \in \phi \end{aligned}$

Case 3:-
$\begin{aligned} \dfrac{x}{x} & \leq 0\\ \\ \implies 1 & \geq 0 , x \neq 0\\ \\ \implies x & \in \phi \end{aligned}$

Therefore the final solution set is $(2,4]$ .
Thus, $a + b = 2 + 4 = \color{#3D99F6}{\boxed{6}}$ .