If $A = \{x \in \mathbb{R} : |x-2|>1\}$, $B = \{x \in \mathbb{R} : \sqrt{x^2-3}>1\}$, $C = \{x \in \mathbb{R} : |x-4| \geq 2\}$ and $\mathbb{Z}$ is the set of all integers, then the number of subsets of the set $(A \cap B \cap C)^c \cap \mathbb{Z}$ is _________.
Show Hint
For modulus inequalities like $|x-a|>r$, always split into $x-a>r$ or $x-a<-r$. Use a number line to visualize the intersection of multiple sets quickly.
Step 1: Understanding the Concept:
This question requires finding the intersection of three sets defined by inequalities, taking the complement relative to the universal set of real numbers, and then finding the integer elements within that complement to calculate the total number of subsets. Step 2: Key Formula or Approach:
1. Solve each inequality to find sets $A$, $B$, and $C$.
2. Intersection $A \cap B \cap C$ is the common region of all three.
3. $(A \cap B \cap C)^c$ is the region not in the intersection.
4. Number of subsets of a set with $n$ elements is $2^n$. Step 3: Detailed Explanation:
Set $A$: $|x-2|>1 \implies x-2>1$ or $x-2<-1 \implies x>3$ or $x<1$.
Set $B$: $\sqrt{x^2-3}>1 \implies x^2-3>1 \implies x^2>4 \implies x>2$ or $x<-2$.
Set $C$: $|x-4| \geq 2 \implies x-4 \geq 2$ or $x-4 \leq -2 \implies x \geq 6$ or $x \leq 2$.
Now, find $A \cap B \cap C$:
For $x>0$: $x \in (3, \infty) \cap (2, \infty) \cap [6, \infty) = [6, \infty)$.
For $x<0$: $x \in (-\infty, 1) \cap (-\infty, -2) \cap (-\infty, 2] = (-\infty, -2)$.
So, $A \cap B \cap C = (-\infty, -2) \cup [6, \infty)$.
The complement is $(A \cap B \cap C)^c = [-2, 6)$.
The integer set is $S = (A \cap B \cap C)^c \cap \mathbb{Z} = \{-2, -1, 0, 1, 2, 3, 4, 5\}$.
The number of elements $n(S) = 8$.
The number of subsets $= 2^8 = 256$. Step 4: Final Answer:
The number of subsets is 256.
