Question

Solve $|x-2|\ge 5$

• A. $(-\infty ,-3]\cup [7,\infty )$
• B. $(-\infty ,-7]\cup [7,\infty )$
• C. $(-\infty ,\infty )$
• D. $[7,\infty )$

Answer 1

The correct option is A $(-\infty ,-3]\cup [7,\infty )$

$|x-2|\ge 5$
When $x\ge 2,$ When $x<2$
$x–2\ge 5-x+2\ge 5$
$\Rightarrow x\ge 7\Rightarrow x\le -3$
$\therefore x\in (-\infty ,-3]\cup [7,\infty )$