Question:
If $\mathrm{A}=\{\mathrm{x} \in \mathbf{R}:|\mathrm{x}-2|>1\}, \mathrm{B}=\left\{\mathrm{x} \in \mathbf{R}: \sqrt{\mathrm{x}^{2}-3}>1\right\}$, $\mathrm{C}=\{\mathrm{x} \in \mathbf{R}:|\mathrm{x}-4| \geq 2\}$ and $\mathbf{Z}$ is the set of all integers, then the number of subsets of the set $(A \cap B \cap C)^{C} \cap \mathbf{Z}$ is
Solution:
$\mathrm{A}=(-\infty, 1) \cup(3, \infty)$
$\mathrm{B}=(-\infty,-2) \cup(2, \infty)$
$\mathrm{C}=(-\infty, 2] \cup[6, \infty)$
So, $A \cap B \cap C=(-\infty,-2) \cup[6, \infty)$
$\mathrm{z} \cap(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})^{\prime}=\{-2,-1,0,-1,2,3,4,5\}$
Hence no. of its subsets $=2^{8}=256$