The Empty Set, Finite and Infinite Sets, Equal Sets

Since x belongs to ℕ, take the positive natural numbers under the convention used in the options: 1, 2, 3, and so on. The inequality x² ≤ 49 gives x ≤ 7 for natural numbers. Thus the possible values are 1 through 7, so A = {1, 2, 3, 4, 5, 6, 7}. Negative integers are excluded because the domain is ℕ.

From x² = 16, we obtain x = 4 or x = −4. However, the definition of A also requires x to be an odd integer. Both 4 and −4 are even, so neither value satisfies all the stated conditions simultaneously. Therefore, there is no element in A, and A is the empty set, written as ∅. The answer is option A.

The months with exactly 30 days are April, June, September, and November. There are four such months, so the set contains a fixed, limited number of distinct elements. It is therefore finite. Because these four months actually exist, the set is also non-empty. Hence the correct classification is finite and non-empty.

Using the usual school convention ℕ = {1, 2, 3, ...}, the natural numbers less than 10 are exactly 1 through 9. Thus A = {1, 2, 3, 4, 5, 6, 7, 8, 9}, which is precisely the set B. Since this list contains nine elements, both A and B are finite. The fact that 10 is not included is correct because the condition says “less than 10,” not “less than or equal to 10.” Therefore, option A is correct.

Factor the quadratic expression: x² + 2x + 1 = (x + 1)². Therefore, the equation becomes (x + 1)² = 0, which has the single real solution x = −1. A set records an element only once, even when that element is a repeated root of an equation. Hence A contains exactly one element and A = {−1}. Option A is the only correct answer.

The set begins with -2 and continues as -2, -1, 0, 1, 2, 3, and so on. Although it has a lower bound, there is no greatest integer in the set because integers continue indefinitely to the right. Therefore the set has infinitely many elements. A starting boundary alone does not make a set finite.

The natural-number squares less than 50 are 1², 2², 3², 4², 5², 6², and 7², giving {1, 4, 9, 16, 25, 36, 49}. The next square, 8² = 64, is already greater than 50. Thus A and B contain exactly the same elements, so A = B. The set is finite, and 50 itself is not a square.

For every real number x, the square x² is always non-negative: it is positive when x is nonzero and equal to zero when x = 0. Therefore every real number satisfies x² ≥ 0. Hence A contains all real numbers and A = ℝ. Since the real-number set has infinitely many elements, A is infinite. Option A is therefore correct.

A natural number divisible by 2 is even by definition. An odd number is a natural number that is not divisible by 2. Therefore no natural number can satisfy both conditions simultaneously. Since there are no elements that belong to A, the set has zero elements and is the empty set, written as ∅. Thus option A is correct.

Solving x² = 25 gives x = 5 or x = −5, so A = {5, −5}. This is exactly the same collection of elements as B. In set notation, repeated listing of an element does not create a new element; therefore C = {−5, 5, 5} is simply {−5, 5}. Consequently, all three sets have the same elements and A = B = C.

The set contains every real number strictly between 1 and 2. It includes numbers such as 3/2, 4/3, 7/5, and infinitely many other decimals and fractions. Between any two distinct real numbers there are infinitely many real numbers. Because the endpoints are excluded, the set is the open interval (1, 2), but excluding endpoints does not make it finite or empty. Thus A is infinite.

The equation x² = 2 has the real solutions x = √2 and x = −√2. However, √2 is irrational, and its negative is also irrational; neither number belongs to ℚ, the set of rational numbers. Since the defining domain is ℚ, not ℝ, there are no admissible solutions. Therefore A contains no elements and A = ∅.

The inequality x² ≤ 4 is equivalent to |x| ≤ 2, or −2 ≤ x ≤ 2. The integers in this closed interval are −2, −1, 0, 1, and 2. Hence A = {−2, −1, 0, 1, 2}, which is exactly the set B. The endpoints are included because the inequality is ≤, and all intermediate integers must also be counted.

The prime factorisation of 18 is 18 = 2 × 3 × 3 = 2 × 3². The only prime numbers that divide 18 are 2 and 3. Numbers such as 6, 9, and 18 are factors, but they are not prime, so they do not belong to A. Therefore A = {2, 3}, which is exactly B, and A = B is correct.

Subtracting x from both sides of x + 2 = x gives 2 = 0. This is a contradiction, because 2 is not equal to 0. Consequently, no number—natural or otherwise—can satisfy the equation. The solution set therefore contains no elements, so it is the empty set, denoted by ∅. Hence option A is correct.

The condition −5 < x < 5 allows the integers −4, −3, −2, −1, 0, 1, 2, 3, and 4. From these, the even integers are −4, −2, 0, 2, and 4. The endpoints −5 and 5 are excluded because the inequalities are strict. Zero is even because it is divisible by 2, so it must be included. Therefore, in roster form, A = {−4, −2, 0, 2, 4}.

No natural number can be greater than every natural number. If x is any natural number, then x + 1 is also a natural number and is greater than x. Thus x fails the required condition. This argument works for every possible natural number, so the set has no elements and is the empty set: A = ∅.

An element of A would have to be less than every natural number, including itself. However, no number is less than itself because the relation x < x is always false. Therefore, regardless of whether a textbook begins N with 0 or 1, no natural number satisfies the condition, so A is the empty set.

We solve the defining equation x³ = x by bringing all terms to one side: x³ − x = 0. Factoring gives x(x² − 1) = 0, and then x(x − 1)(x + 1) = 0. Thus x can be 0, 1, or −1. Each of these is an integer and satisfies the original equation. Since a set lists each element only once, A = {−1, 0, 1}.

The positive divisors of 36 can be found in pairs: 1 and 36, 2 and 18, 3 and 12, 4 and 9, and 6 and 6. Removing the repeated middle entry gives the nine distinct divisors 1, 2, 3, 4, 6, 9, 12, 18, and 36. These are exactly the elements listed in B. Therefore A contains nine elements, so n(A) = 9, and because both sets have the same elements, A = B.