Question

Let $ A=\left[ x:x\in R,|x|<1 \right];B=\left[ x:x\in R,|x-1|\ge 1 \right] $ and $ A\cup B=R-D $ then the set D is
A. $ \left[ x:1 < x\le 2 \right] $
B. $ \left[ x:1\le x < 2 \right] $
C. $ \left[ x:1\le x\le 2 \right] $
D. None of these

Answer 1

Hint: We will first find the domain of ‘x’ In set ‘A’ and Set ‘B’ which are the elements of sets, from where we will find out the $ A\cup B $ , which will help us to find the set D by comparing the result we got from $ A\cup B $ and $ R-D $ .

Complete step by step answer:
Moving ahead with the question in step wise manner,
We need to find out the value of set D, so let us first find out the set A and Set B elements, so first in set A we are given with the info $ A=\left[ x:x\in R,|x|< 1 \right] $ to get the domain of ‘x’ we need to simplify $ |x|< 1 $ by using the mode property. So we have $ |x|< 1 $ by simplifying it using the mode inequality we will get;
 $ -1 < x < 1 $ .
Similarly simplifying the set B we have $ B=\left[ x:x\in R,|x-1|\ge 1 \right] $ , so to find the elements of set B, we need to simplify $ |x-1|\ge 1 $ , which on simplifying using property of mode we will get;
 $ \begin{align}
  & |x-1|\ge 1 \\
 & x\ge 2\cup x\le 0 \\
\end{align} $
So we got set A and set B which are equal to $ -1 < x < 1 $ and $ x\ge 2\cup x\le 0 $ .
So now we need to find out the $ A\cup B $ . So let us draw it on the number line and find out the union of both sets, so let us plot the number line and mark the points $ -1,0,1 $ and 2 as shown in figure.

As we can see, if we take the union of two sets A and B then all numbers get covered on the number line except the numbers from 1 to 2, including 1 and excluding 2. We can also say that $ A\cup B=R-[1,2) $ so by comparing it with $ R-D $ we can say that set D is $ [1,2) $ which can also be written as $ 1\le x < 2 $ .
So set D is $ 1\le x < 2 $ .

So, the correct answer is “Option B”.

Note: As in the set B the number 2 is included, so in the set $A\cup B$ the number will be included, same is with 1; as in set A number 1 is not included in set so it will be not present in the number line, hence we can say that in the number line $A\cup B$ all number will be there except the number 2 which is included and the number 1 will be not be counted in $A\cup B$ because it is not in set A.