The Empty Set, Finite and Infinite Sets, Equal Sets

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.

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.

The natural-number multiples of 5 are 5, 10, 15, 20, 25, and so on. For every multiple 5n, where n is a natural number, the next multiple 5(n+1) also belongs to the set. Since this process never ends and no upper bound is imposed, A has infinitely many elements. Therefore A is an infinite set.

The empty set ∅ has no elements, and the notation {} also represents a set with no elements. However, {∅} is different: it contains one element, namely the empty set itself. Therefore, {∅} is a singleton set and is not empty. Also, no integer has square −1, so option D represents the empty set.

Because x must be an integer strictly greater than −2 and strictly less than 2, neither endpoint −2 nor 2 is included. The only integers satisfying the inequality are −1, 0, and 1. Therefore A = {−1, 0, 1}, which is exactly the same collection of elements as B. Sets are equal when they contain precisely the same elements, regardless of the order in which those elements are written.

The condition requires x to be an integer strictly greater than 0 and strictly less than 1. There is no integer between 0 and 1: 0 is excluded because the inequality is strict, and 1 is also excluded for the same reason. Hence no element satisfies the defining condition, so A has no elements and is the empty set, written as ∅.

Using the stated convention ℕ = {1, 2, 3, ...}, test the natural numbers against x² < 10. We have 1² = 1, 2² = 4, and 3² = 9, all less than 10, while 4² = 16, which is not less than 10. Therefore A = {1, 2, 3}. Since B contains exactly the same elements, A and B are equal, so option A is correct.

Two sets are equal when they contain exactly the same elements; order and repeated listing do not matter. In option A, the repeated 2 and 3 are counted only once, so {1, 2, 2, 3, 3} is simply {1, 2, 3}. The other pairs have different elements or different numbers of elements, so they are not equal.

A natural-number factor of 18 is a natural number that divides 18 exactly. The positive factors are 1, 2, 3, 6, 9, and 18, because each divides 18 without a remainder. These are precisely the elements listed in B. A number is always a factor of itself, so 18 belongs in the set. Hence A = B.

Assuming ℕ = {1, 2, 3, ...}, the condition x ≤ 100 gives the elements 1 through 100. Thus A = {1, 2, ..., 100}, which has exactly 100 elements. A set with a fixed, countable number of elements is finite. Therefore option B is correct; the set is not empty or infinite and is certainly not all integers.

The natural numbers greater than 100 are 101, 102, 103, and so on. There is no final natural number in this sequence; whenever one such number is chosen, the next integer is also greater than 100. Therefore the set has infinitely many elements. It is not empty, it is not finite, and 100 itself is excluded by the strict inequality.

Factor the quadratic equation: x² − 5x + 6 = (x − 2)(x − 3) = 0. Therefore, x = 2 or x = 3. Since both values are integers, both belong to A, so A = {2, 3}. This set has exactly two distinct elements, and every set with a fixed, countable number of elements is finite. Hence option A is correct. The value 6 is the constant term, not a solution of the equation.

Rewrite the equation as x² − 2x + 1 = (x − 1)² = 0. Hence x = 1 is the only real solution. Although the root is repeated algebraically, a set does not record repetition; it contains the element 1 only once. Therefore A = {1}, which is a singleton and a finite set. Option B is correct.

The condition |x| ≤ 2 means that x lies between −2 and 2, including both endpoints. Because x must be an integer, the possible values are −2, −1, 0, 1, and 2. Thus A = {−2, −1, 0, 1, 2}. The other choices omit valid integers or include too many numbers, so option A is correct.

A prime number has exactly two positive divisors, while an even number is divisible by 2. The only number that is both prime and even is 2; every other even number has at least 2 and another divisor, so it is composite. Therefore A = {2}. This set has one element and is a singleton finite set, making option B correct.

Two sets are equal when they contain exactly the same elements, regardless of the order or notation used to write them. Therefore, whenever A = B, their cardinalities must also be equal; they have the same number of elements. However, equal sets may be finite, infinite, or empty, and they need not be written in the same form. Thus option A is necessary.

Subtracting x from both sides of x = x + 1 gives 0 = 1. This is a contradiction and cannot be true for any real number x. Therefore no real number satisfies the defining condition of A. A set containing no elements is called the empty set, written as ∅. Hence A = ∅ and option C is correct.