Question
If $$x$$ satisfies $$\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6,$$ then
A.
$$0 \leqslant x \leqslant 4$$
B.
$$x \leqslant - 2\,{\text{or}}\,x \geqslant 4$$
C.
$$x \leqslant 0\,{\text{or}}\,x \geqslant 4$$
D.
None of these
Answer :
$$x \leqslant 0\,{\text{or}}\,x \geqslant 4$$
Solution :
$$\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6,$$
Consider $$f\left( x \right) = \left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right|$$
$$f\left( x \right) = \left\{ {_{_{3x - 6,}^{x,}}^{_{4 - x,}^{6 - 3x,}}} \right.\,_{_{x \geqslant 3}^{2 \leqslant x < 3}}^{_{1 \leqslant x < 2}^{x < 1}}$$
NOTE THIS STEP:
Graph of $$f\left( x \right)$$ shows $$f\left( x \right)\, \geqslant 6$$ for $$x \leqslant 0$$ or $$x \geqslant 4$$
$$\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6,$$
Consider $$f\left( x \right) = \left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right|$$
$$f\left( x \right) = \left\{ {_{_{3x - 6,}^{x,}}^{_{4 - x,}^{6 - 3x,}}} \right.\,_{_{x \geqslant 3}^{2 \leqslant x < 3}}^{_{1 \leqslant x < 2}^{x < 1}}$$
NOTE THIS STEP:
Graph of $$f\left( x \right)$$ shows $$f\left( x \right)\, \geqslant 6$$ for $$x \leqslant 0$$ or $$x \geqslant 4$$