Consider regions for \(x\) around 1 and 2.

1) \(x\le 1:\quad |x-1|+|x-2|=(1-x)+(2-x)=3-2x \ge 4 \Rightarrow x\le -\tfrac12.\)

2) \(1\le x\le 2:\quad |x-1|+|x-2|=(x-1)+(2-x)=1\) (not \(\ge4\)). No solutions.

3) \(x\ge 2:\quad |x-1|+|x-2|=(x-1)+(x-2)=2x-3 \ge 4 \Rightarrow x\ge \tfrac{7}{2}.\)

Final Answer: \(x \in (-\infty,\,-\tfrac12] \,\cup\, [\tfrac{7}{2},\,\infty)\).

Correct Option: 1