The Empty Set, Finite and Infinite Sets, Equal Sets

The two-digit natural numbers are 10, 11, 12, and so on up to 99. Both the lower bound and upper bound are fixed, so only finitely many natural numbers satisfy the condition. In fact, the number of elements is 99 − 10 + 1 = 90. Therefore A is a finite set, not an empty set, an infinite set, or a singleton. Hence option B is correct.

For option B, solving x² = 1 over the integers gives x = −1 or x = 1. Thus A = {−1, 1}, which has two elements, while B = {1}, which has one element. Both sets are finite, but they are not equal because −1 belongs to A and does not belong to B. Option A describes equal sets, option C has an infinite first set, and option D has two equal empty sets.

The integers from −5 to 5 that are congruent to 1 modulo 3 are −5, −2, 1, and 4. For example, each differs from 1 by a multiple of 3: −6, −3, 0, and 3 respectively. Hence A = {−5, −2, 1, 4} = B. Set equality depends on having exactly the same elements, not on their order or on whether the elements are positive or negative. Therefore option A is correct.

Factor the quadratic expression as x² − 7x + 12 = (x − 3)(x − 4). Therefore the equation is satisfied when x = 3 or x = 4. Both values are integers, so A = {3, 4}. Since B is also {3, 4}, the two sets contain exactly the same elements and are equal. The numbers 7 and 12 are coefficients and are not the solutions themselves. Thus option A is correct.

The condition 2x + 1 = 2x can be simplified by subtracting 2x from both sides, giving 1 = 0. This statement is impossible and is independent of the value of x. Therefore no natural number satisfies the defining condition of A. A set containing no elements is called the empty set, denoted by ∅. Hence option B is correct; it is not a singleton, a two-element set, or an infinite set.

Because x is an integer, the values in the interval are −4, −3, −2, −1, 0, 1, and 2. The lower boundary −4 is included because the symbol is ≤, whereas 3 is excluded because the symbol is <. Counting the listed integers gives 7 elements. Thus A is a finite set with cardinality 7, and option B is correct.

The natural numbers not exceeding 50 that are divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, and 50. There are only ten such numbers because the condition includes the upper bound x ≤ 50. Since the list ends at 50, it cannot continue indefinitely. Therefore A is a finite set, so option B is correct.

The congruence x ≡ 2 (mod 5) means that x has the form x = 5k + 2, where k is any integer. Taking k = 0, 1, 2 gives 2, 7, 12, while k = −1, −2 gives −3, −8. Since there are infinitely many possible integer values of k, there are infinitely many elements in A. Thus option C is correct.

The integers between −6 and 6 that are congruent to 2 modulo 5 can be found from x = 5k + 2. For k = −1, x = −3; for k = 0, x = 2; while k = 1 gives 7, which is outside the interval. Therefore the only eligible elements are −3 and 2, so A = {−3, 2}. Hence option A is correct.

The symbol ∅ denotes the empty set, which contains no elements at all. In contrast, {∅} is a set whose only element is the empty set itself. Therefore, {∅} has exactly one element and is a singleton set, whereas ∅ has zero elements. They are different sets, and neither statement about both being equal or infinite is correct.

An absolute value is zero only when its inside expression is exactly zero. Thus |x − 1| = 0 implies x − 1 = 0, giving x = 1. Since the variable is restricted to real numbers, the solution set contains only 1. Consequently, A = {1}, which is a singleton set. The pair {−1, 1} would arise from an equation such as |x| = 1, not from |x − 1| = 0.

The two-digit natural numbers begin at 10 and end at 99. The even members are 10, 12, 14, and so on, up to 98. This is a bounded list with a fixed first and last value, so it has only finitely many elements. In fact, there are 45 such numbers. Option D lists only one-digit even numbers and therefore does not represent A.

Factor 84 into primes: 84 = 2² × 3 × 7. Its positive divisors include 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84, but the question asks only for prime divisors. The prime divisors are therefore 2, 3, and 7, so A = {2, 3, 7}. The number 1 is not prime, and 4 is composite, so neither belongs in A.

Complete the square: x² + 2x + 2 = (x + 1)² + 1. For every real x, (x + 1)² is at least zero, so (x + 1)² + 1 is always at least one and can never equal zero. Equivalently, the discriminant is 2² − 4(1)(2) = −4, which is negative. Hence there is no real solution and A is the empty set.

Subtracting x from both sides of x + 5 = x gives 5 = 0, which is a contradiction. Thus no natural number, or any number at all, satisfies the condition. A set containing no elements is the empty set, so A = ∅. It is not {5}, because x = 5 gives 10 = 5.

For natural numbers, test the values around the boundary: 1³ = 1, 2³ = 8, and 3³ = 27, so 1, 2, and 3 satisfy x³ ≤ 27. The next natural number does not, because 4³ = 64 > 27. Therefore B = {1, 2, 3}, which is exactly the same collection of elements as A. Hence A = B; both sets are finite and equal.

Use the standard absolute-value inequality: |x + 2| < 1 is equivalent to −1 < x + 2 < 1. Subtracting 2 throughout gives −3 < x < −1. Among integers, the only number strictly between −3 and −1 is −2. The endpoints −3 and −1 are excluded because the inequality is strict. Thus A contains exactly one element and A = {−2}.

The natural numbers greater than 100 are 101, 102, 103, and so on without end, so the first set in option B is infinite. The negative integers less than zero are −1, −2, −3, and so on, which also continue indefinitely; therefore the second set is infinite. The other options contain bounded sets, divisors of 100, or sets with only zero or one element.

The positive natural-number divisors of 36 are obtained from factor pairs: 1×36, 2×18, 3×12, 4×9, and 6×6. Therefore the complete divisor set is {1, 2, 3, 4, 6, 9, 12, 18, 36}, exactly the set B. Repetition is not used in a set, so 6 appears only once. Hence A = B.

The positive divisors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. Among these, the odd numbers are 1, 3, and 9; every other divisor is even. Thus the set satisfying both conditions is A = {1, 3, 9}. Notice that 1 is odd and is also a divisor of every nonzero integer.