Question

Solve $|x-1|+|x-2|+|x-3|\ge 6$

• A. $(-\infty ,0]\cup [6,\infty )$
• B. $(-\infty ,0]\cup [4,\infty )$
• C. $[4,\infty )$
• D. $(-\infty ,0]$

Answer 1

The correct option is B $(-\infty ,0]\cup [4,\infty )$

$|x-1|+|x-2|+|x-3|\ge 6$
If $x<1$,
$-x+1-x+2-x+3\ge 6\Rightarrow x\le 0\Rightarrow x\in (-\infty ,0]$

If $1\le x<2$,
$x-1-x+2-x+3\ge 6$
$\Rightarrow x\le -2$ which is absurd.

If $2\le x<3$,
$x-1+x-2-x+3\ge 6$
$\Rightarrow x\ge 6$ which is absurd.

If $x\ge 3$,
$x-1+x-2+x-3\ge 6$
$\Rightarrow x\ge 4\Rightarrow x\in [4,\infty )$

Hence, $x\in (-\infty ,0]\cup [4,\infty )$