Sets and their representations

The word “and” means that x must satisfy both conditions simultaneously. The positive divisors of 60 include 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60. Among these, the numbers that also divide 15 are 1, 3, 5, and 15. Therefore A = {1, 3, 5, 15}. Option B lists many divisors of 60 that do not divide 15, while option D omits 1 and 15.

The relation A ⊆ B means that every element of A is already an element of B. The union A ∪ B contains all elements that belong to A or B. Since A contributes no element outside B, taking the union adds nothing to B. Therefore, A ∪ B = B. Notice that A ∩ B = A, but the question asks for the union, so option B is correct.

Solve the inequality: 2x + 1 < 10 gives 2x < 9 and hence x < 4.5. Since x belongs to N = {1, 2, 3, ...}, the possible natural-number values are 1, 2, 3, and 4. The value 5 fails because 2(5)+1 = 11, and 0 is not in the stated definition of N. Therefore A = {1, 2, 3, 4}, option A.

The set \(A\) has four elements, so it has \(2^4=16\) total subsets. It is easier to count the complement: subsets containing neither 1 nor 2 can use only the elements 3 and 4, giving \(2^2=4\) subsets. Hence the number containing 1 or 2 is \(16-4=12\). Here “or” is inclusive, so subsets containing both 1 and 2 are also counted. Therefore, option B is correct.

The condition requires 1 to be included and 2 to be excluded, so both of these elements have fixed choices. The remaining elements 3, 4, and 5 are unrestricted; each can either be included or left out independently. Therefore, the number of possible subsets is \(2\times2\times2=2^3=8\). For example, \{1\}, \{1,3\}, and \{1,4,5\} are valid, while any subset containing 2 is invalid. Thus option B is correct.

The governing concept is that a set contains distinct elements, so repetition does not create new members. The letters in LEVEL are L, E, V, E, L; after removing repeated occurrences, the membership list is {L, E, V}. Therefore option A is correct. Options B and D treat a set like an ordered list or multiset by repeating letters, while C incorrectly omits E.

The positive divisors of 18 are 1, 2, 3, 6, 9, and 18. Applying the condition “less than 10” removes 18, so A = {1, 2, 3, 6, 9}. This is exactly the same collection of elements as B. In a set, order does not matter, but every element must match. Therefore A = B, making option A correct. Option D incorrectly includes 18, which does not satisfy the given inequality.

For an absolute-value inequality, |x − 2| < 2 is equivalent to −2 < x − 2 < 2. Adding 2 throughout gives 0 < x < 4. Since x must be an integer, the only possible values are 1, 2, and 3. Thus A = {1, 2, 3}, which is exactly B. Therefore A = B is correct; neither set is a proper subset of the other, and their intersection is not empty.

The notation specifies that x must be an integer and must satisfy −2 ≤ x < 2. The symbol ≤ includes −2, while the symbol < excludes 2. Listing all integers in this half-closed interval gives −2, −1, 0, and 1. Therefore A = {−2, −1, 0, 1}, which is exactly option A. Option B omits −2, option C incorrectly includes 2, and option D omits valid integers and includes an invalid endpoint.

Consider the pair 2 and 3. There are two allowed cases: both are included, or both are excluded. In either case, the remaining elements 1 and 4 can each be independently included or excluded, giving 2^2 = 4 choices. Thus the total number is 4 + 4 = 8, or equivalently 2 × 2^2 = 8. Hence option C is correct.

A universal set is the set that contains every object or element being considered in a particular problem or discussion. It is usually denoted by U. Its elements depend on the stated context; therefore, it need not be empty, infinite, or restricted to numbers. For example, if the discussion concerns English vowels, U may be the set of all vowels. Hence option B is correct.

The set A consists of the part only in A and the common part. Therefore, n(A ∩ B) = n(A) − n(A − B) = 58 − 21 = 37. Then n(A ∪ B) = 58 + 49 − 37 = 70. The complement contains the elements of U outside this union, so n((A ∪ B)′) = 110 − 70 = 40. Option A is correct.

In a set, the order in which elements are written does not matter, and repeated elements are counted only once. Thus \(A\) and \(B\) contain exactly the same elements: 1, 2, and 3. Their union therefore contains these three distinct elements only: \(A\cup B=\{1,2,3\}\). No element such as 4 or 6 can be added because it appears in neither set.

Use the inclusion–exclusion formula first: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Hence 50 = 35 + 27 − n(A ∩ B), so n(A ∩ B) = 12. The set A is made up of the disjoint parts A − B and A ∩ B. Therefore n(A − B) = n(A) − n(A ∩ B) = 35 − 12 = 23. Option B is the intersection size, not the difference size; the other values do not satisfy the given relationships.

For every \(x\in A\), calculate \(y=2x\) and check whether the result belongs to \(B\). The values are \(2,4,8,16\) for \(x=1,2,4,8\), respectively. Therefore, the valid ordered pairs are \((1,2),(2,4),(4,8),(8,16)\). All four satisfy the condition and belong to \(A\times B\), so the answer is 4, option C.

Fix each value of \(x\) and count the allowed values of \(y\). For \(x=1\), all four values of \(y\) work. For \(x=2,3,4\), the numbers of choices are 3, 2, and 1, respectively. Hence the total is \(4+3+2+1=10\). Because ordered pairs are being counted, each distinct position of \(x\) and \(y\) is included, so option C is correct.

The condition says that the second coordinate is one more than the first, so \(y=x+1\). Substituting each element of \(A\) gives \(1,3,5\) for \(x=0,2,4\), respectively. Every resulting value belongs to \(B\), producing the pairs \((0,1),(2,3),(4,5)\). Thus there are exactly 3 pairs, and option C is correct.