The Empty Set, Finite and Infinite Sets, Equal Sets

Factor the quadratic expression: x² + 3x + 2 = (x + 1)(x + 2). Therefore, the equation is satisfied when x = -1 or x = -2. Since both values are integers, A = {-1, -2}. Set order is irrelevant, so {-1, -2} = {-2, -1} = B. Hence A = B. The other options either use incorrect roots, claim no solutions, or incorrectly describe a two-element set as infinite.

Complete the square: x² - 2x + 5 = (x - 1)² + 4. For every real number x, (x - 1)² is at least zero, so the expression is at least 4 and can never equal zero. Therefore, the equation has no real solution, and the set of its real solutions is empty: A = ∅.

A natural number divisible by both 2 and 3 is divisible by their least common multiple, 6. Conversely, every multiple of 6 is divisible by both 2 and 3. Thus A and B contain exactly the same numbers: 6, 12, 18, 24, and so on. Since there are endlessly many multiples of 6, both sets are infinite and A = B.

A product is zero when at least one factor is zero. Thus (x − 2)(x + 3) = 0 gives x − 2 = 0 or x + 3 = 0, so x = 2 or x = -3. Both values belong to the integers, and hence A = {2, -3}. Since the order of elements does not affect a set, {2, -3} = {-3, 2} = B. Therefore, A = B. The set has only two elements, so it is finite, not infinite.

For natural numbers, x² ≤ 30 implies x ≤ √30, and √30 is approximately 5.47. Therefore, the possible natural numbers are 1, 2, 3, 4, and 5. The odd numbers among them are 1, 3, and 5, so A = {1, 3, 5}. This set contains three elements. The correct answer is option B; the count must be made after applying both the square bound and the odd-number condition.

The expression |x - 4| + |x - 9| represents the sum of the distances from x to 4 and 9 on the number line. The distance between 4 and 9 is 5, so for every x in the closed interval [4, 9], the sum is exactly 5. Since an interval contains infinitely many real numbers, A is an infinite set.

The powers of 2 not exceeding 64 are obtained from 2⁰ through 2⁶: 2⁰ = 1, 2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32, and 2⁶ = 64. Thus A = {1, 2, 4, 8, 16, 32, 64}. The value 1 must be included because 1 = 2⁰, so option B is correct. The set is finite because the condition x ≤ 64 stops the powers.

For option A, the natural-number divisors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. Among them, the prime divisors are only 2 and 3. Therefore A = {2, 3}, which is exactly B. Two sets are equal when they contain the same elements, regardless of how their defining descriptions look. The other options produce different sets.

Since 0.25 = 1/4, the equation x² = 0.25 becomes x² = (1/2)². A square equation has both the positive and negative square roots, so x = 1/2 or x = -1/2. Both values are rational and must be included. Hence A = {-0.5, 0.5}, a finite set containing two elements.

We solve the absolute-value inequality first: |2x − 1| < 4 implies −4 < 2x − 1 < 4. Adding 1 and dividing by 2 gives −3/2 < x < 5/2. Since x must be an integer, the possible values are −1, 0, 1, and 2. Therefore A = {−1, 0, 1, 2}, which is exactly the set B. Hence the correct conclusion is A = B.

Among the positive natural numbers less than 12, the number 1 is neither prime nor composite. A prime number has exactly two distinct positive divisors, while a composite number has more than two; 1 has only one positive divisor, itself. Every other positive natural number below 12 is either prime or composite. Therefore A = {1}.

Using the standard school convention N = {1, 2, 3, ...}, every natural number is at least 1. Consequently, there is no natural number satisfying x < 1. A set that contains no element is called the empty set and is written as ∅. Therefore the given set is empty, not a singleton or an infinite set.

For every real number x, x² is non-negative, so x² + 1 is at least 1. It can therefore never equal zero. Equivalently, the equation x² + 1 = 0 would require x² = -1, which has no real solution. Since no real number satisfies the condition, the set contains no elements and is empty.

In set notation, repetition of an element does not create a new element, and the order in which elements are written is irrelevant. Thus A = {2, 3, 4}, because the repeated 3 is counted only once. Set B also contains exactly 2, 3, and 4. Therefore A and B are equal, while A does not have four distinct elements.

For x² < 10, we need −√10 < x < √10. Since √10 is approximately 3.16 and x must be an integer, the permitted values are −3, −2, −1, 0, 1, 2, and 3. There are seven distinct integers in this set. The set is finite because its elements are bounded between −√10 and √10. Therefore, its cardinality is 7, so option C is correct.

For every natural number n, the set contains the square n². Its beginning is 1, 4, 9, 16, 25, and so on. As n can be chosen arbitrarily large, the squares continue without an ending largest element. Also, different natural numbers give different squares, so infinitely many distinct elements occur. Therefore the set is infinite.

The notation {0} represents a set whose only element is the number 0, so its cardinality is 1 and it is a singleton set. The symbol ∅ represents a set with no elements at all, so its cardinality is 0 and it is the empty set. Zero is an element in {0}; it is not the same as having no elements. Hence option C is correct.

Two sets are equal only when they contain exactly the same elements; the order of elements does not matter. Set A already contains 2, so its other element, a + 1, must correspond to 3 in set B. Therefore, a + 1 = 3, and subtracting 1 from both sides gives a = 2. Hence option B is correct.

Interpreting “between 10 and 50” as strictly between the two endpoints, the multiples of 5 are 15, 20, 25, 30, 35, 40, and 45. The numbers 10 and 50 are excluded because they are the endpoints. This gives seven distinct elements, so the set is finite and option A is correct.

The ellipsis on the left indicates that the sequence continues with further negative integers, such as -6, -7, -8, and so on. There is no final first element at the negative end, so the list does not terminate and its elements cannot all be counted. Therefore, the set has infinitely many elements and is an infinite set. Option C is correct.