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1.Set Theory
hard
If $\mathrm{A}=\{\mathrm{x} \in {R}:|\mathrm{x}-2|>1\}, \mathrm{B}=\left\{\mathrm{x} \in {R}: \sqrt{\mathrm{x}^{2}-3}>1\right\}$, $\mathrm{C}=\{\mathrm{x} \in {R}:|\mathrm{x}-4| \geq 2\}$ 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
$256$
B
$64$
C
$8$
D
$16$
(JEE MAIN-2021)
Solution
$\mathrm{A}=(-\infty, 1) \cup(3, \infty)$
$\mathrm{B}=(-\infty,-2) \cup(2, \infty)$
$\mathrm{C}=(-\infty, 2] \cup[6, \infty)$
So, $\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}=(-\infty,-2) \cup[6, \infty)$
$\mathrm{z} \cap(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})^{\prime}=\{-2,-1,0,-1,2,3,4,5\}$
Hence no. of its subsets $=2^{8}=256$.
Standard 11
Mathematics