If A = {x ∈ R : |x - 2| 1}, B = {x ∈ R : √(x² - 3) 1}, C = {x ∈ R : |x - 4| ≥ 2} and Z is the se

Q: Wait, why do we flip the inequality sign for the negative case in modulus?

Actually, we do not flip the inequality sign itself; we use the definition of absolute value. For $|x - 2| > 1$, the distance from $x$ to $2$ must be greater than $1$. This means $x$ is either to the right of $3$ ($x - 2 > 1$) or to the left of $1$ ($x - 2 < -1$). It is a geometric interpretation, not a sign flip!

Q: Why is the complement of $(-\infty, -2) \cup [6, \infty)$ equal to $[-2, 6)$?

This is the most common trap! The complement includes everything NOT in the original set. Since $-2$ is NOT in $(-\infty, -2)$, it MUST be in the complement. Since $6$ IS in $[6, \infty)$, it is NOT in the complement. Thus, the interval is closed at $-2$ and open at $6$.

Q: Why is the number of subsets $2^n$?

This is a fundamental combinatorial result. For every element in a set of size $n$, you have two choices when forming a subset: either include it or exclude it. Since there are $n$ elements, you have $2 \times 2 \times \dots \times 2$ ($n$ times) choices, which equals $2^n$.

Question

If $A = \{x \in \mathbb{R} : |x - 2| > 1\}$, $B = \{x \in \mathbb{R} : \sqrt{x^2 - 3} > 1\}$, $C = \{x \in \mathbb{R} : |x - 4| \ge 2\}$ 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 .

Sigma Percentile · Accepted Answer

### Analyzing the Modulus of Set A We begin with Set $A = \{x \in \mathbb{R} : |x - 2| > 1\}$. The modulus function represents the distance between $x$ and $2$ on the number line. When we state $|x - 2| > 1$, we are identifying all points whose distance from $2$ is greater than $1$. This inequality splits into two distinct paths: $$x - 2 > 1 \Rightarrow x > 3$$ $$x - 2 < -1 \Rightarrow x < 1$$ Thus, our first territory is defined as $A = (-\infty, 1) \cup (3, \infty)$. Visually, this represents two open rays pointing away from each other. ### The Square Root Trap of Set B Next, we examine Set $B = \{x \in \mathbb{R} : \sqrt{x^2 - 3} > 1\}$. Since the square root function is defined to be non-negative and the right side is $1$, we can safely square both sides without changing the inequality: $$x^2 - 3 > 1$$ $$x^2 > 4$$ Taking the square root of both sides yields $|x| > 2$. This results in the intervals $x > 2$ or $x < -2$. Therefore, we have $B = (-\infty, -2) \cup (2, \infty)$. ### The Modulus with Equality in Set C Now, we consider Set $C = \{x \in \mathbb{R} : |x - 4| \ge 2\}$. This follows the same logic as Set $A$, but with the inclusion of the boundary points: $$x - 4 \ge 2 \Rightarrow x \ge 6$$ $$x - 4 \le -2 \Rightarrow x \le 2$$ Thus, $C = (-\infty, 2] \cup [6, \infty)$. Note the use of closed brackets, indicating that the boundary points are included in the set. ### Finding the Intersection We must now find the common ground for all three sets: $A \cap B \cap C$. We look for regions where all conditions are satisfied simultaneously. On the left side of the number line, we compare $(-\infty, 1)$, $(-\infty, -2)$, and $(-\infty, 2]$. The most restrictive condition is $x < -2$. On the right side, we compare $(3, \infty)$, $(2, \infty)$, and $[6, \infty)$. The most restrictive condition is $x \ge 6$. Therefore, the intersection is $(-\infty, -2) \cup [6, \infty)$. ### The Complement and Final Calculation The problem asks for the complement of the intersection, denoted as $(A \cap B \cap C)^c$. This represents the "gap" between our intersection intervals. Since $-2$ was excluded from the intersection, it is included in the complement. Since $6$ was included in the intersection, it is excluded from the complement. Our resulting set is $[-2, 6)$. We now identify the integers $\mathbb{Z}$ within this interval: $$\{-2, -1, 0, 1, 2, 3, 4, 5\}$$ Counting these values, we find exactly $n = 8$ elements. The number of subsets of a set with $n$ elements is given by $2^n$. Calculating for $n = 8$: $$2^8 = 256$$ The final answer is **256**.

Animated Solution for Mathematics - Sets and Relations: If $A = {x \in R : ∣ x - 2∣ > 1}$ , $B = {x \in R : x^{2} - 3 > 1}$ , $C = {x \in R : ∣ x - 4∣ \geq 2}$ 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 .

Enter Numerical Value:

Visualized Solution

The Goal: Subsets of a Complex Set

Set $A$ : Modulus Inequality

Solving for Set $A$

Set $B$ : Square Root Inequality

Solving for Set $B$

Set $C$ : Modulus Inequality with Equality

Solving for Set $C$

Finding $A \cap B \cap C$

The Intersection Set

The Complement $(A \cap B \cap C)^{c}$

Calculating the Complement

Intersection with Integers $Z$

Number of Subsets

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Modulus of Set A

The Square Root Trap of Set B

The Modulus with Equality in Set C

Finding the Intersection

The Complement and Final Calculation

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Enter Numerical Value:

Visualized Solution

The Goal: Subsets of a Complex Set

Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A: Modulus Inequality

Solving for Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B: Square Root Inequality

Solving for Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C: Modulus Inequality with Equality

Solving for Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A∩B∩C