Question

Let$A=[x:x\in R,|x|<1]$; $B=[x:x\in R,|x-1|\ge 1]$ and $A\cup B=R-D$, then the 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

The correct option is B

$[x:1\le x<2]$

Explanation for the correct option:

The sets $A=[x:x\in R,|x|<1]$ and $B=[x:x\in R,|x-1|\ge 1]$ .

To find:

The set $D$.

Consider the set

$$\begin{gathered} A=[x:x\in R,|x|<1] \\ =[x:x\in R,-1<x<1] \\ =(-1,1) \end{gathered}$$

Also the set,

$B=[x:x\in R,|x-1|\ge 1]$

Now

$$\begin{gathered} \therefore \left|x-1\right|\ge 1 \\ \Rightarrow x\ge 2orx\le 0 \end{gathered}$$

Therefore, the range of $B=R-(0,2)=(-\infty ,0]\cup [2,\infty )$

Now consider,

$$\begin{gathered} \Rightarrow A\cup B=\left(-1,1\right)\cup (-\infty ,0]\cup [2,\infty ) \\ =(-\infty ,1)\cup [2,\infty ) \\ =R-[x:1\le x<2]....\left(i\right)\left[\because R=\left(-\infty ,\infty \right)\right] \end{gathered}$$

Comparing (i) with the given expression $A\cup B=R-D$ we get

$\Rightarrow D=[x:1\le x<2]$

Therefore, the value of $D$ is $[x:1\le x<2]$.

Hence, the correct answer is option (B).