Question

If $A=\{x\in R:|x-2|>1\},B=\{x\in R:\sqrt{x^2-3}>1\},$ $C=\{x\in R:|x-4|\ge 2\}$ and $Z$ is the set of all integers, then the number of subsets of the set $(A\cap B\cap C{)}^c\cap Z$ is

• A. 256.0
• B. 256.00
• C. 256

Answer 1

Answer: (A), (B), (C)

$A=\{x\in R:|x-2|>1\}$
$|x-2|>1$
$\Rightarrow x-2>1$ or $x-2<-1$
$\Rightarrow x>3$ or $x<1$
$\therefore A=(-\infty ,1)\cup (3,\infty )$

$B=\{x\in R:\sqrt{x^2-3}>1\}$
$x^2-3\ge 0$ and $x^2-3>1$
$\Rightarrow x^2-4>0$
$\Rightarrow x>2$ or $x<-2$
$\therefore B=(-\infty ,-2)\cup (2,\infty )$

$C=\{x\in R:|x-4|\ge 2\}$
$x-4\ge 2$ or $x-4\le -2$
$x\ge 6$ or $x\le 2$
$\therefore C=(-\infty ,2]\cup [6,\infty )$
So, $(A\cap B\cap C)=(-\infty ,-2)\cup [6,\infty )$
Now, $(A\cap B\cap C{)}^c\cap Z=\{-2,-1,0,1,2,3,4,5\}$
$\Rightarrow n\left(\right(A\cap B\cap C{)}^c\cap Z)=8$
$\therefore$ Number of subset $=2^8=256$