The Empty Set, Finite and Infinite Sets, Equal Sets

Subtracting x from both sides of the condition x = x + 1 gives 0 = 1. This is a contradiction and cannot be true for any real number x. Consequently, there is no real number belonging to A. A set containing no elements is called the empty set, denoted by ∅, so A = ∅.

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 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}.

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}.

Rewrite the equation as x² − 2x + 1 = 0. This expression is a perfect square: (x − 1)² = 0. Hence the only real solution is x = 1. Although the root is repeated algebraically, a set does not record multiplicity; it contains the value only once. Therefore the solution set is the singleton set A = {1}, not a two-element set and not the empty set.

The two-digit natural numbers begin at 10 and end at 99, inclusive. To count consecutive integers in an inclusive interval, use last number − first number + 1. Thus the number of elements is 99 − 10 + 1 = 90. Since the list has a definite first and last term, it is finite. Therefore A is a finite set with n(A) = 90.

Solve the condition x² = 4x by moving all terms to one side: x² − 4x = 0. Factoring gives x(x − 4) = 0. By the zero-product property, x = 0 or x = 4. Both values are integers and satisfy the original equation. Hence the complete solution set is A = {0, 4}, so A is equal to the given set {0, 4}.

The positive natural-number cubes are obtained as 1³ = 1, 2³ = 8, 3³ = 27, and 4³ = 64. The condition requires the cube to be less than 30, so 1, 8, and 27 are included, while 64 is excluded. Under the usual school convention that ℕ begins with 1, zero is not included. Therefore A = {1, 8, 27}.

The condition requires an integer to be positive and negative at the same time. No integer has both properties. Positive integers are greater than zero, while negative integers are less than zero; these two classes do not overlap. Zero does not provide an exception because it is neither positive nor negative. Hence no integer satisfies the defining condition, and A is the empty set ∅.

The natural numbers not exceeding 20 that are divisible by 5 are 5, 10, 15, and 20. Thus A = {5, 10, 15, 20}, which is exactly the same set as B. Since there are four distinct elements, n(A) = 4. The condition x ≤ 20 includes 20, and no other positive multiple of 5 lies within the stated bound.

Starting with x² + 2 = 2, subtract 2 from both sides to obtain x² = 0. The only real number whose square is zero is x = 0. Therefore, the solution set is A = {0}. This is a singleton set containing exactly one element, so it is finite. It is not empty, because zero satisfies the original equation exactly.

For every real number x, the absolute value |x − 3| is non-negative, so it can be zero or positive but never −1. Consequently, the equation |x − 3| = −1 has no real solution. Since A consists of real numbers satisfying this impossible condition, it contains no elements. Therefore, A is the empty set, written as ∅.

To determine A, solve x² = 4. Factoring gives (x − 2)(x + 2) = 0, so x = 2 or x = −2. Both numbers are rational, hence both belong to A. Therefore A = {-2, 2}, which is exactly the set B. A set does not repeat elements, and the order of elements is irrelevant. Thus the correct statement is A = B.

Solving x² = 3 over the real numbers gives x = √3 or x = −√3. Although √3 is irrational, it is still a real number, so both solutions satisfy the stated domain x ∈ ℝ. Consequently, A = {√3, −√3}, which is exactly B. The fact that the roots are irrational does not exclude them from a real-number set.