Q47 / 125 · BITSAT 2013 · Mathematics
If x satisfies $|3 x-2|+|3 x-4|+|3 x-6| \geq 12$, then
- A$0 \leq \mathrm{x} \geq \frac{8}{3}$
- B$x \geq \frac{8}{3}$ or $\frac{-4}{3}$
- Correct answerC$x \leq 0$ or $x \geq \frac{8}{3}$
- D$x \geq 2$ only
Solution
Dividing $\mathrm{R}$ at $\frac{2}{3}, \frac{4}{3}$ and $2,$ analyse 4 cases. When $\mathrm{x} \leq \frac{2}{3},$ the inequality becomes $2-3 x+4-3 x+6-3 x \geq 12$
implying $-9 x \geq 0 \Rightarrow x \leq 0$
when $\mathrm{x} \geq 2$ the ineqality becomes $3 x-2+3 x-4+3 x-6 \geq 12$
Implying $9 \mathrm{x} \geq 24 \Rightarrow \mathrm{x} \geq 8 / 3$
The inequality in invalid in the other two sections.
$\therefore$ either $x \leq 0$ or $x \geq 8 / 3$
Hence, the correct answer is option: 'C'.