The Empty Set, Finite and Infinite Sets, Equal Sets

A composite number is a natural number greater than 1 that has more than two positive divisors. The natural numbers less than 11 are 1 through 10. Among them, 4, 6, 8, 9, and 10 are composite. The numbers 2, 3, 5, and 7 are prime, while 1 is neither prime nor composite. Therefore A = {4, 6, 8, 9, 10}.

An integer that leaves remainder 1 on division by 2 is odd. The set can be written as A = {2k + 1 : k ∈ Z}, which includes ..., -5, -3, -1, 1, 3, 5, ... . Since k can be any integer and there is no largest or smallest member, the set contains infinitely many elements. Therefore, option A is correct.

In Euclidean division by 4, the remainder must be one of 0, 1, 2, or 3; it must always be less than the divisor. Therefore, no integer can leave remainder 5 when divided by 4. The defining condition has no solution, so A has no elements and is the empty set. Hence, option A is correct.

The conditions x < 100 and x > 100 must hold simultaneously. No number, whether natural, integer, or real, can be both less than 100 and greater than 100 at the same time. Thus, there is no natural number satisfying the definition of A. Consequently, A contains no elements and is the empty set, so option A is correct.

When n = 1, A consists of all one-digit natural numbers. Under the standard school convention N = {1, 2, 3, ...}, these numbers are 1, 2, 3, 4, 5, 6, 7, 8, and 9. The number 0 is not a one-digit natural number in this convention. Therefore, A has 9 elements, making option A correct.

Solving x² = 49 gives x = 7 or x = -7, because both numbers have square 49. The additional condition x > 0 eliminates -7 and retains only 7. Therefore, the set of all integers satisfying both conditions is A = {7}. It is a singleton set, so option A is correct.

To determine A, solve x² = 49. The integer solutions are x = 7 and x = −7. The additional condition x < 0 excludes 7 and retains only −7, so A = {−7}. Since B is also defined as the singleton set {−7}, A and B contain exactly the same element. Therefore, A = B. The braces show that each is a set, not merely the number −7.

Complete the square: x² - 2x + 2 = (x - 1)² + 1. For every real x, (x - 1)² is at least 0, so (x - 1)² + 1 is at least 1 and can never equal 0. Therefore, the equation has no real solution, and the set of real solutions is empty. Hence option A is correct.

The positive factors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30. Among these factors, the prime numbers are 2, 3, and 5 because each has exactly two positive divisors: 1 and itself. The number 1 is not prime, and 30 is composite because it has more than two positive divisors. Hence the required set in roster form is A = {2, 3, 5}.

The only positive natural-number factor of 1 is 1 itself. However, the definition also requires x > 1, which excludes 1. No natural number can therefore satisfy both conditions simultaneously. Thus, the set contains no elements and is the empty set, A = ∅. Hence, option A is correct.

The powers of 2 are 2¹, 2², 2³, 2⁴, and so on, giving 2, 4, 8, 16, 32, ... . The exponent can be any positive natural number, and there is no greatest exponent. Consequently, new elements continue to appear without end. Therefore, A is infinite and option A is correct.

Every integral power of 2 is positive: for any integer n, 2ⁿ > 0, including negative exponents such as 2⁻¹ = 1/2. Therefore no power of 2 can satisfy x < 0. Since the set contains no element, it is the empty set, written as ∅. The negative numbers listed in option B are not powers of positive 2.

The positive multiples of 7 that do not exceed 50 are 7, 14, 21, 28, 35, 42, and 49. The next multiple, 56, is greater than 50, so it is excluded. Thus A has seven elements, and its cardinality is n(A) = 7. The explicit positive-number convention also prevents 0 from being counted.

For an integer x, the inequality x² ≤ 1 means −1 ≤ x ≤ 1. The integers in this interval are exactly −1, 0, and 1. Therefore A = {−1, 0, 1}, which is precisely the set B. Hence A and B have the same elements and are equal; omitting 0 or treating the inequality as strict would produce an incorrect result.

Rearrange the equation as x² - x - 2 = 0. Factoring gives (x - 2)(x + 1) = 0, so x = 2 or x = -1. Both values are real and therefore A = {-1, 2}. Since B contains exactly these same elements, A = B. The order in which elements are written does not affect a set.

Under the convention stated in the question, the natural numbers begin with 1. Every natural number is therefore at least 1, so no natural number satisfies x < 1. Consequently A contains no elements and is the empty set, ∅. The empty set has cardinality 0, and it is classified as finite because its number of elements is bounded.

For every real number x, x² is non-negative, so x² + 4 is at least 4 and can never equal zero. Equivalently, solving the equation gives x² = -4, which has no real solution. Although complex numbers ±2i solve the corresponding equation in the complex number system, the domain here is ℝ, so they are not elements of A. Hence A = ∅.

A set records membership, not the order or frequency in which an element is written. Thus A = {2, 3, 5}, because the repeated 3 and 2 do not create new elements. Set B also contains exactly 2, 3, and 5. Since A and B have precisely the same elements, they are equal, so option C is correct.

The equation x² = 9 has two integer solutions: x = 3 and x = -3, because both 3² and (-3)² equal 9. Therefore A = {-3, 3}. This is exactly the roster used to define B. Since two sets are equal when they contain the same elements, A = B. The negative root must be included as well as the positive root.

The prime factorisation of 12 is 12 = 2² × 3. Its only prime factors are therefore 2 and 3, so A = {2, 3}. This set contains exactly two elements. Because the elements are limited and can be completely listed, A is a finite set, not an infinite set and certainly not all of ℕ.