Question

If x satisfies $|3x-2|+|3x-4|+|3x-6|\ge 12$, then

• A. $0\le x\ge \frac{8}{3}$
• B. $x\ge \frac{8}{3}$ or $\frac{-4}{3}$
• C. $x\le 0$ or $x\ge \frac{8}{3}$
• D. $x\ge 2$ only

Answer 1

Answer: (C)

Dividing $R$ at $\frac{2}{3},\frac{4}{3}$ and $2$, analyses $4$ cases.
When $x\le \frac{2}{3}$, the inequality becomes
$2-3x+4-3x+6-3x\ge 12$.
implying $-9x\ge 0\Rightarrow x\le 0$.
when $x\ge 2$ the ineqality becomes
$3x-2+3x-4+3x-6\ge 12$
Implying $9x\ge 24\Rightarrow x\ge 8/3$
The inequality in invalid in the other two sections.
$\therefore$ either $x\le 0$ or $x\ge 8/3$