The Empty Set, Finite and Infinite Sets, Equal Sets

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}.

We solve x² − 1 < 0 by adding 1 to both sides, obtaining x² < 1. For real numbers, this is equivalent to −1 < x < 1. Therefore, A is the open interval (−1, 1). Every non-empty real interval contains infinitely many real numbers, so A is an infinite set. The endpoints −1 and 1 are excluded because the inequality is strict.

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.

Complete the square: x² − 4x + 5 = (x − 2)² + 1. For every integer, and indeed every real number, (x − 2)² is at least zero, so the expression is at least 1. It can never equal zero. Therefore, the equation has no integer solution and the set A contains no elements; hence A = ∅.

Rewrite the inequality as x² − 2x < 0, or x(x − 2) < 0. Over the real numbers, this holds when 0 < x < 2. The natural numbers in this open interval depend on the usual school convention ℕ = {1, 2, 3, …}; only x = 1 belongs to it. Therefore, A = {1}. Neither 0 nor 2 satisfies the strict inequality.

Move all terms to one side: x² − x ≤ 0, which factors as x(x − 1) ≤ 0. The real solution interval is 0 ≤ x ≤ 1. Among integers, the only values in this closed interval are 0 and 1, so A = {0, 1}. Therefore, the number of elements is n(A) = 2. Both endpoints are included because the inequality is non-strict.

The phrase “strictly between 10 and 20” means 10 < x² < 20. The only perfect square in this range is 16, because 3² = 9 is too small and 5² = 25 is too large. Solving x² = 16 gives x = 4 or x = −4. Hence A = {−4, 4}, not the four-element set shown in option A. Therefore, the original options contain no correct answer; option A must be corrected to A = {−4, 4}.

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.

The natural numbers satisfying x > 50 are 51, 52, 53, 54, and so on. After every such number, a larger natural number can be found, so the list never ends. Therefore the set has infinitely many elements. The lower bound 50 only tells us where the elements begin; it does not impose an upper limit or make the set finite.

A three-digit natural number must be at least 100; the three-digit numbers begin with 100 and continue through 999. The additional condition x < 100 contradicts the requirement that x have three digits. Hence there is no natural number satisfying both conditions, so A contains no elements and is the empty set, written as ∅.

The inequality x² < 30 implies −√30 < x < √30, so the possible integers range from −5 to 5. Among these, the integers divisible by 3 are −3, 0, and 3. Therefore A = {−3, 0, 3}, which has three distinct elements. Zero is included because 0 = 3 × 0, so it is divisible by 3. Hence option A is correct; the other counts omit or add valid elements.

The word “Mathematics” contains the letters m, a, t, h, e, m, a, t, i, c, s. In a set, repeated occurrences are written only once, so the distinct-letter set is {m, a, t, h, e, i, c, s}. Set order is irrelevant, and therefore this set is exactly B. Hence option A is correct.

The outer set A has two elements: the empty set ∅ and the singleton set {∅}. These are different objects. The symbol ∅ denotes a set with no elements, whereas {∅} denotes a set whose one element is the empty set. Since the two outer elements are distinct, n(A) = 2.