Option : (B)
|x-2|/(x-2) ≥ 0
Case I : x > 2
\(\frac{x-2}{x-2}\) ≥ 0
1 ≥ 0
It is true that 1 is always greater than 0 so case I is also true x > 2
Case II : x < 2
\(\frac{-(x-2)}{x-2}\) ≥ 0
-1 ≥ 0 It is false that -1 is not greater than 0 so case II is also false.
So, the final solution is x > 2
i.e., x ∈ (2, ∞)