Key Concepts and Formulas

• Absolute Value Inequalities:
  • For $|u| > k$ where $k > 0$, the solution is $u > k$ or $u < -k$.
  • For $|u| \ge k$ where $k > 0$, the solution is $u \ge k$ or $u \le -k$.
• Square Root Inequalities: For an inequality of the form $\sqrt{f(x)} > c$:
  1. Domain: The expression under the square root must be non-negative: $f(x) \ge 0$.
  2. Squaring: If $c \ge 0$, square both sides: $f(x) > c^2$.
  3. Intersection: The solution is the intersection of the domain and the result from squaring.
• Set Operations:
  • Intersection ($\cap$): Elements common to all sets.
  • Complement ($^c$): Elements in the universal set (here $\mathbb{R}$) not in the set.
• Number of Subsets: A set with $n$ distinct elements has $2^n$ subsets.

Step-by-Step Solution

Our objective is to find the number of subsets of the set $(A \cap B \cap C)^c \cap Z$. We will first determine the sets A, B, and C, then compute their intersection, find the complement of the intersection, and finally extract the integers and count the subsets.

Step 1: Determine Set A

Set A is given by $A = \{x \in \mathbb{R} : |x - 2| > 1\}$.

• Reasoning: We use the property of absolute value inequalities: $|u| > k \iff u > k$ or $u < -k$. Here, $u = x - 2$ and $k = 1$.
• Solving the inequalities:
  • $x - 2 > 1 \implies x > 3$
  • $x - 2 < -1 \implies x < 1$
• Result: Set A is the union of these two intervals: $A = (-\infty, 1) \cup (3, \infty)$.

Step 2: Determine Set B

Set B is given by $B = \{x \in \mathbb{R} : \sqrt{x^2 - 3} > 1\}$.

• Reasoning (Domain): For the square root to be defined, $x^2 - 3 \ge 0$, which means $x^2 \ge 3$. This implies $x \le -\sqrt{3}$ or $x \ge \sqrt{3}$. So, the domain is $(-\infty, -\sqrt{3}] \cup [\sqrt{3}, \infty)$.
• Reasoning (Squaring): Since both sides of $\sqrt{x^2 - 3} > 1$ are non-negative, we can square both sides: $x^2 - 3 > 1^2 \implies x^2 - 3 > 1 \implies x^2 > 4$. This implies $x < -2$ or $x > 2$. So, this condition gives $(-\infty, -2) \cup (2, \infty)$.
• Reasoning (Intersection): The solution for B must satisfy both the domain and the squared inequality. We find the intersection of $(-\infty, -\sqrt{3}] \cup [\sqrt{3}, \infty)$ and $(-\infty, -2) \cup (2, \infty)$.
  • $(-\infty, -\sqrt{3}] \cap (-\infty, -2) = (-\infty, -2)$ (since $-2 < -\sqrt{3}$)
  • $(-\infty, -\sqrt{3}] \cap (2, \infty) = \emptyset$
  • $[\sqrt{3}, \infty) \cap (-\infty, -2) = \emptyset$
  • $[\sqrt{3}, \infty) \cap (2, \infty) = (2, \infty)$ (since $\sqrt{3} < 2$)
• Result: Set B is $B = (-\infty, -2) \cup (2, \infty)$.

Step 3: Determine Set C

Set C is given by $C = \{x \in \mathbb{R} : |x - 4| \ge 2\}$.

• Reasoning: We use the property of absolute value inequalities: $|u| \ge k \iff u \ge k$ or $u \le -k$. Here, $u = x - 4$ and $k = 2$.
• Solving the inequalities:
  • $x - 4 \ge 2 \implies x \ge 6$
  • $x - 4 \le -2 \implies x \le 2$
• Result: Set C is the union of these two intervals: $C = (-\infty, 2] \cup [6, \infty)$.

Step 4: Find the Intersection $A \cap B \cap C$

We need to find the elements common to A, B, and C.

• $A = (-\infty, 1) \cup (3, \infty)$
• $B = (-\infty, -2) \cup (2, \infty)$
• $C = (-\infty, 2] \cup [6, \infty)$
• Reasoning: We can find this by intersecting the intervals.
  • First, $A \cap B$:
    • $(-\infty, 1) \cap ((-\infty, -2) \cup (2, \infty)) = (-\infty, -2)$
    • $(3, \infty) \cap ((-\infty, -2) \cup (2, \infty)) = (3, \infty)$ So, $A \cap B = (-\infty, -2) \cup (3, \infty)$.
  • Now, intersect $(A \cap B)$ with C:
    • $((-\infty, -2) \cup (3, \infty)) \cap ((-\infty, 2] \cup [6, \infty))$
    • $(-\infty, -2) \cap (-\infty, 2] = (-\infty, -2)$
    • $(-\infty, -2) \cap [6, \infty) = \emptyset$
    • $(3, \infty) \cap (-\infty, 2] = \emptyset$
    • $(3, \infty) \cap [6, \infty) = [6, \infty)$
• Result: $A \cap B \cap C = (-\infty, -2) \cup [6, \infty)$.

Step 5: Find the Complement $(A \cap B \cap C)^c$

Let $S = A \cap B \cap C$. We need to find $S^c$, which is the set of all real numbers not in $S$.

• Reasoning: The complement of $(-\infty, a)$ is $[a, \infty)$, and the complement of $[b, \infty)$ is $(-\infty, b)$. For a union of disjoint intervals, the complement fills the "gaps".
• The complement of $(-\infty, -2)$ is $[-2, \infty)$.
• The complement of $[6, \infty)$ is $(-\infty, 6)$.
• The complement of $S = (-\infty, -2) \cup [6, \infty)$ is the set of numbers that are not in $(-\infty, -2)$ AND not in $[6, \infty)$. This means numbers $\ge -2$ AND $< 6$.
• Result: $(A \cap B \cap C)^c = [-2, 6)$.

Step 6: Find the Set of Integers $(A \cap B \cap C)^c \cap Z$

We need to find the integers within the interval $[-2, 6)$.

• Reasoning: $Z$ represents the set of all integers. We list the integers $x$ such that $-2 \le x < 6$.
• Result: The integers are $\{-2, -1, 0, 1, 2, 3, 4, 5\}$.

Step 7: Determine the Number of Subsets

The set $(A \cap B \cap C)^c \cap Z$ is $\{-2, -1, 0, 1, 2, 3, 4, 5\}$.

• Reasoning: This set contains $n=8$ distinct integers. The number of subsets of a set with $n$ elements is $2^n$.
• Result: The number of subsets is $2^8 = 256$.

Common Mistakes & Tips

• Domain for Square Roots: Always establish the domain of square root functions first to avoid extraneous solutions.
• Interval Endpoints: Pay close attention to whether interval endpoints are included (closed brackets []) or excluded (open parentheses ()) when performing operations, especially complements.
• Number Line Visualization: Drawing a number line can be very helpful for visualizing intersections and complements of intervals.

Summary

We systematically determined the sets A, B, and C by solving the given inequalities. We then found the intersection $A \cap B \cap C$, which resulted in the union of two intervals. Taking the complement of this intersection yielded a single interval. Finally, we identified all integers within this interval and calculated the number of subsets of this finite set of integers. The set of integers found was $\{-2, -1, 0, 1, 2, 3, 4, 5\}$, which has 8 elements. Therefore, the number of subsets is $2^8$.

The final answer is $\boxed{256}$.