Question

lf $x$ satisfies $|x-1|+|x-2|+|x-3|\ge 6$, then

• A. $0\le$ $x$ $\le 1$
• B. $x\le -2$ or $x\ge 4$
• C. $x\le 0$ or $x\ge 4$
• D. $R$

Answer 1

Answer: (C)

$x\le 0$ or $x\ge 4$

Let $f\left(x\right)=|x-1|+|x-2|+|x-3|$
For $x<1;|x-1|=-(x-1)$
$|x-2|=-(x-2)$
$|x-3|=-(x-3)$
$\therefore -(x-1)-(x-2)-(x-3)\ge 6$
$\Rightarrow -3x+6\ge 6\Rightarrow x\le 0.$
For $x\ge 1;f\left(x\right)=x-1-(x-2)-(x-3)$
$=x-1-x+2-x+3$
$=-x+4$
$-x+4\ge 6\Rightarrow -x\ge 2\Rightarrow x\le -2.$
No values of $x$ possible for this change.
Similarly, for $x\ge 2$
No values of $x$ possible for this change.
For $x\ge 3;|x-1|=x-1;|x-2|=x-2;|x-3|=x-3.$
$\therefore f\left(x\right)=3x-6\ge 6$
$3x\ge 12$
$x\ge 4.$
$\therefore x\le 0$ or $x\ge 4.$