JEE Main 2021 — Basic of Mathematics Question with Solution

For First set $A=\{x\in R:|x-2|>1\}$

$\Rightarrow$ Value which $A$ contains lies from $x\in \left(-\infty ,1\right)\cup \left(3,\infty \right)$

For second set $B=\left\{x\in R:\sqrt{x^2-3}>1\right\}$

$\Rightarrow$ Value which $B$ contains lies from $x\in \left(-\infty ,-2\right)\cup \left(2,\infty \right)$

For third set $C=\{x\in R:|x-4|⩾2\}$

$\Rightarrow$ Value which $C$ contains lies from $x\in (-\infty ,2]\cup [6,\infty )$

Now, $\left(-\infty ,-2\right)\cup [6,\infty )\in \left(A\cap B\cap C\right)$

Therefore, ${\left(A\cap B\cap C\right)}^c\cap Z=\left\{-2,-1,0,1,2,3,4,5\right\}$

Number of Subsets of above set $=2^8=256$