Question

The set of values of $x$ satisfying $1\le |x-1|\le 3$ is

• A. $[-2,0]\cup [2,\infty )$
• B. $(-\infty ,0]\cup [2,\infty )$
• C. $[-2,0]\cup [2,4]$
• D. $R-(0,2)$

Answer 1

The correct option is C $[-2,0]\cup [2,4]$

Condition $1:|x-1|\ge 1$
$\Rightarrow x-1\notin (-1,1)\Rightarrow x-1\in (-\infty ,-1]\cup [1,\infty )\Rightarrow x\in (-\infty ,0]\cup [2,\infty )\cdots \left(1\right)$

Condition $2:|x-1|\le 3$
$\Rightarrow -3\le x-1\le 3\Rightarrow -2\le x\le 4\cdots \left(2\right)$

Required solution is the intersection of conditions $\left(1\right)$ and $\left(2\right)$
Hence, $x\in [-2,0]\cup [2,4]$