Let A ∪ B = {x|(x - a)(x - b) > 0, where a < b}. What are A and B equal to?
A = {x|x < a} and B = {x|x > b}
Understanding the Set Union Defined by an Inequality
The problem asks us to identify the individual sets A and B given their union, \(A \cup B\), is defined by the inequality \((x - a)(x - b) > 0\), where it is specified that \(a < b\). To solve this, we first need to understand the set described by the given inequality.
Solving the Inequality \((x - a)(x - b) > 0\)
The inequality \((x - a)(x - b) > 0\) means that the product of the two factors, \((x - a)\) and \((x - b)\), must be positive. This can happen in two possible ways:
- Both factors are positive.
- Both factors are negative.
Case 1: Both factors are positive
- \(x - a > 0 \implies x > a\)
- \(x - b > 0 \implies x > b\)
Since we are given that \(a < b\), for both conditions \(x > a\) and \(x > b\) to be true simultaneously, x must be greater than the larger value, which is \(b\). So, this case gives us the condition \(x > b\).
Case 2: Both factors are negative
- \(x - a < 0 \implies x < a\)
- \(x - b < 0 \implies x < b\)
Since we are given that \(a < b\), for both conditions \(x < a\) and \(x < b\) to be true simultaneously, x must be less than the smaller value, which is \(a\). So, this case gives us the condition \(x < a\).
Combining the conditions from both cases, the solution to the inequality \((x - a)(x - b) > 0\) is \(x < a\) or \(x > b\).
Therefore, the given set \(A \cup B\) is equal to \(\{x \mid x < a \text{ or } x > b\}\).
Analyzing the Options for Sets A and B
We are given four possible pairs of sets A and B. We need to find the pair whose union matches the set \(\{x \mid x < a \text{ or } x > b\}\).
| Option | Set A | Set B | Union \(A \cup B\) | Matches \( \{x \mid x < a \text{ or } x > b\} \)? |
|---|---|---|---|---|
| 1 | \(\{x \mid x > a\}\) | \(\{x \mid x > b\}\) | \(\{x \mid x > a\} \cup \{x \mid x > b\} = \{x \mid x > a\}\) (since \(a < b\)) | No |
| 2 | \(\{x \mid x < a\}\) | \(\{x \mid x > b\}\) | \(\{x \mid x < a\} \cup \{x \mid x > b\}\) | Yes |
| 3 | \(\{x \mid x < a\}\) | \(\{x \mid x < b\}\) | \(\{x \mid x < a\} \cup \{x \mid x < b\} = \{x \mid x < b\}\) (since \(a < b\)) | No |
| 4 | \(\{x \mid x > a\}\) | \(\{x \mid x < b\}\) | \(\{x \mid x > a\} \cup \{x \mid x < b\} = \mathbb{R}\) (the set of all real numbers, since \(a < b\)) | No |
As shown in the table, only the second option, where \(A = \{x \mid x < a\}\) and \(B = \{x \mid x > b\}\), results in the union \(A \cup B = \{x \mid x < a \text{ or } x > b\}\).
Therefore, the correct definitions for sets A and B are \(A = \{x \mid x < a\}\) and \(B = \{x \mid x > b\}\).
Revision Table: Key Concepts in Set Theory and Inequalities
| Concept | Explanation | Example |
|---|---|---|
| Set Union (\(\cup\)) | The union of two sets A and B is the set containing all elements that are in A, OR in B, or in both. Denoted as \(A \cup B\). | If A = {1, 2} and B = {2, 3}, then \(A \cup B\) = {1, 2, 3}. |
| Inequality: \((x-a)(x-b) > 0\) | This inequality holds when both factors \((x-a)\) and \((x-b)\) have the same sign (both positive or both negative). For \(a < b\), this corresponds to \(x < a\) or \(x > b\). | For \((x-1)(x-2) > 0\), the solution is \(x < 1\) or \(x > 2\). |
| Set Builder Notation | A way to describe a set by stating the properties its members must satisfy. Written as \(\{x \mid \text{property of } x\}\). | \(\{x \mid x > 5\}\) is the set of all real numbers greater than 5. |
Additional Information on Solving Quadratic Inequalities
The inequality \((x - a)(x - b) > 0\) is a simple form of a quadratic inequality. Quadratic inequalities involving a quadratic expression \(px^2 + qx + r\) can often be solved by factoring the quadratic expression into the form \(p(x - \alpha)(x - \beta)\) or by finding the roots of the corresponding quadratic equation \(px^2 + qx + r = 0\).
The roots (\(\alpha\) and \(\beta\)) divide the number line into intervals. The sign of the quadratic expression \(p(x - \alpha)(x - \beta)\) is constant within each interval.
For an inequality like \((x - a)(x - b) > 0\) with roots \(a\) and \(b\) and \(a < b\), the number line is divided into three intervals: \((-\infty, a)\), \((a, b)\), and \((b, \infty)\).
- In \((-\infty, a)\), both \((x-a)\) and \((x-b)\) are negative, so \((x-a)(x-b)\) is positive.
- In \((a, b)\), \((x-a)\) is positive and \((x-b)\) is negative, so \((x-a)(x-b)\) is negative.
- In \((b, \infty)\), both \((x-a)\) and \((x-b)\) are positive, so \((x-a)(x-b)\) is positive.
Therefore, \((x - a)(x - b) > 0\) holds for \(x \in (-\infty, a) \cup (b, \infty)\), which is the set \(\{x \mid x < a \text{ or } x > b\}\).
Important Questions from Union and Intersection of Sets
Consider the following statements
1. A = (A ∪ B) ∪ (A - B),
2. A ∪ (B - A) = (A ∪ B)
3. B = (A ∪ B) - (A - B)
Which of the statements given above are correct?
If a set A contains 3 elements and another set B contains 6 elements, then what is the minimum number of elements that (A∪B) can have?
If A = {x : 0 ≤ x ≤ 2} and B = {y; y is a prime number}, then what is A ∩ B equal to?
If A = {x ϵ Z : x 3– 1 = 0} and B = {x ϵ Z: x 2+ x + 1 = 0}, where Z is set of complex numbers, then what is A ∩ B equal to?
In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper I, 26 read newspaper T, 9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find the number of students who read exactly one newspaper.
