Question

Let $A=\{x:x\in R,\left|x\right|<1\}$
$B=\{x:x\in R,|x-1|\ge 1\}$
and $A\cup B=R-D$, then set $D$ is

• A. $\{x:1<x\le 2\}$
• B. $\{x:1\le x<2\}$
• C. $\{x:1\le x\le 2\}$
• D. None of these

Answer 1

Answer: (B)

$\{x:1\le x<2\}$

Given $A=\{x:x\in R\left|x\right|<1\}$

If $x$ is positive, $\left|x\right|=x$

$\therefore x<1$

If $x$ is negative, $\left|x\right|=-x$

$\therefore -x<1$

$\therefore x>-1$

$\therefore A=\{-1<x<1\}$ (1)

Given $B=\{x:x\in R,|x-1|\ge 1\}$

If $|x-1|$ is positive, $|x-1|=x-1$

$\therefore x-1\ge 1$

$\therefore x\ge 2$

If $|x-1|$ is negative, $|x-1|=-(x-1)$

$\therefore -(x-1)\ge 1$

$\therefore (x-1)\le -1$

$\therefore x\le 0$

$\therefore B=\{x\le 0andx\ge 2\}$ (2)

Thus, $A\cup B=\{-1<x<1\}\cup \{x\le 0andx\ge 2\}$

The region which is not included in this domain has to be subtracted from remaining domain of real numbers.

$\therefore A\cup B=R-D$

Thus, set D is set of excluded region which is the region between 1 and 2 including 1.

$\therefore D:-1\le x<2$