Question

If $|(2|x-1|-2)|\ge 1$, then $x\in$

• A. $(-\infty ,-\frac{1}{2}]\cup [\frac{5}{2},\infty )$
• B. $(-\infty ,-\frac{1}{2}]\cup [\frac{1}{2},\frac{3}{2}]\cup [\frac{5}{2},\infty )$
• C. $[\frac{1}{2},\frac{3}{2}]\cup [\frac{5}{2},\infty )$
• D. $(-\infty ,-\frac{1}{2}]\cup [\frac{1}{2},\frac{3}{2}]$

Answer 1

The correct option is B $(-\infty ,-\frac{1}{2}]\cup [\frac{1}{2},\frac{3}{2}]\cup [\frac{5}{2},\infty )$

$|(2|x-1|-2)|\ge 1\Rightarrow |(|x-1|-1)|\ge \frac{1}{2}$
Let $|x-1|-1=t\Rightarrow |t|\ge \frac{1}{2}$
$\Rightarrow t\le -\frac{1}{2}\text{or}t\ge \frac{1}{2}$
$\Rightarrow |x-1|-1\le -\frac{1}{2}\text{or}|x-1|-1\ge \frac{1}{2}$
$\Rightarrow |x-1|\le \frac{1}{2}\text{or}|x-1|\ge \frac{3}{2}$
$\Rightarrow (x-1)\in [-\frac{1}{2},\frac{1}{2}]\text{or}(x-1)\in (-\infty ,-\frac{3}{2}]\cup [\frac{3}{2},\infty )$
$\Rightarrow x\in [\frac{1}{2},\frac{3}{2}]\text{or}x\in (-\infty ,-\frac{1}{2}]\cup [\frac{5}{2},\infty )$

$\therefore x\in (-\infty ,-\frac{1}{2}]\cup [\frac{1}{2},\frac{3}{2}]\cup [\frac{5}{2},\infty )$