Complement of a Set and Its Properties

The complement of a set is taken with respect to the stated universal set. Here, \(A'=\{2,4,6,8,10\}\), because these are the elements of \(U\) that are not in \(A\). Taking the complement again returns all the original elements of \(A\). Therefore, the double-complement law gives \((A')'=A=\{1,3,5,7,9\}\), so option A is correct. Option B represents only the first complement.

De Morgan’s law states that the complement of a union is the intersection of the complements. Thus, \((A\cup B)'=A'\cap B'\). An element is outside \(A\cup B\) only when it is outside both \(A\) and \(B\), which explains the intersection symbol. Therefore, option A is correct. Option B incorrectly uses a union, and options C and D do not complement both original sets correctly.

The correct De Morgan identity is \((A\cup B)'=A'\cap B'\). The complement of a union contains precisely those elements that belong to neither \(A\) nor \(B\), so they must lie in both complements. Statement A is incorrect because \((A\cap B)'=A'\cup B'\). Also, \((A')'=A\), not the empty set, and \(A\cup A'=U\), not the empty set. Hence option B is the only always-true statement.

Because the universal set is all real numbers, the complement contains every real number outside the interval \([-2,4)\). The endpoint \(-2\) belongs to \(A\) because the left bracket is closed, so it is excluded from the complement. The endpoint \(4\) does not belong to \(A\) because the right parenthesis is open, so \(4\) belongs to the complement. Therefore, \(A'=(-\infty,-2)\cup[4,\infty)\), option A.

A set and its complement are disjoint and together contain every element of the universal set. Hence, \(n(A)+n(A')=n(U)\). Substituting the given values gives \(n(A)+47=120\), so \(n(A)=120-47=73\). Therefore, option A is correct. The value 47 is the cardinality of the complement, while 120 is the cardinality of the entire universal set, not of \(A\).

The condition \(A\cap B=\varnothing\) means that no element is common to both \(A\) and \(B\). Therefore, every element of \(B\) lies outside \(A\). By the definition of complement, all such elements belong to \(A'\), so \(B\subseteq A'\) must hold. The other statements need not always be true: \(B\) may be smaller than \(A'\), and the union need not equal \(U\).

Within the universal set \(U=\{1,2,\ldots,24\}\), set \(A\) contains exactly the even numbers. Its complement therefore contains every element of \(U\) that is not even, namely the odd numbers \(\{1,3,5,\ldots,23\}\). Set \(B\) is defined as precisely this collection of odd numbers. Consequently, \(A'=B\). Option B is false because the two sets are equal, not a proper subset.

By De Morgan’s law, \(A'\cap B'=(A\cup B)'\). Thus, we count numbers from 1 to 40 that are not divisible by 2 or 5. There are 20 multiples of 2 and 8 multiples of 5, while 4 numbers are multiples of both, namely multiples of 10. Therefore, \(n(A\cup B)=20+8-4=24\), and \(n(A'\cap B')=40-24=16\). Hence, option A is correct.

First solve the defining inequality: \(x+2>4\) gives \(x>2\). Since \(x\) must be an integer in \(U=\{-3,-2,-1,0,1,2,3,4,5,6,7\}\), we obtain \(A=\{3,4,5,6,7\}\). The complement consists of the remaining elements of \(U\), namely \(A'=\{-3,-2,-1,0,1,2\}\). Thus option A is correct. The endpoint 2 is included in the complement because the inequality is strict.

Use the given identity \(A-B=A\cap B'\). Taking complements on both sides gives \((A-B)'=(A\cap B')'\). De Morgan’s law changes the complement of an intersection into the union of the complements: \((A\cap B')'=A'\cup(B')'\). Since the complement of \(B'\) is \(B\), the result is \(A'\cup B\). Therefore, option A is correct.

A contains all real numbers less than 1, while B contains all real numbers greater than 4. Therefore, A union B excludes the endpoints 1 and 4 as well as every number between them. The complement in R consequently contains 1, 4, and all real numbers between them, so (A union B)' = [1, 4]. Thus option A is correct.

A complement is defined relative to a specified universal set U. The elements outside A may change when U changes, even if A itself remains the same. Hence A' is not independent of U. The other statements are standard complement identities: A union A' equals U, A intersection A' is empty, and A' equals U minus A. Therefore option A is false.

The elements of U divisible by 4 are 4, 8, 12, and 16, so A = {4, 8, 12, 16}. The complement A' consists of every element of U that is not in A. Removing those four multiples of 4 from U leaves {1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15}. Thus option A is correct; option B is A itself.

For a subset A of U, the relative difference U minus A is exactly the complement A'. Taking the complement once more gives (A')'. By the double-complement law, the complement of the complement of A is A itself. Therefore (U minus A)' = (A')' = A, making option A correct.

An element belongs to A intersection B exactly when it is divisible by both 3 and 4. Such numbers are multiples of lcm(3, 4) = 12. From 1 through 60, the multiples are 12, 24, 36, 48, and 60, so the intersection has 5 elements. Since U has 60 elements, its complement has 60 - 5 = 55 elements. Option A is correct.

Using De Morgan’s law, A' ∪ B' = (A ∩ B)'. The common elements of A = {1, 2, 5} and B = {2, 4, 6} are A ∩ B = {2}. Taking the complement of {2} in the universal set U = {1, 2, 3, 4, 5, 6, 7, 8} removes 2 and retains every other element, giving {1, 3, 4, 5, 6, 7, 8}. Therefore, option A is correct.

The interval A = (-1, 6] contains numbers strictly greater than -1 and includes 6. Therefore -1 is excluded from A and must be included in its complement, while 6 is included in A and must be excluded from the complement. All real numbers less than -1 and greater than 6 belong to A', giving (-infinity, -1] union (6, infinity).

A contains every number in U that is not prime. Therefore its complement contains exactly the prime numbers in U. The primes from 1 through 25 are 2, 3, 5, 7, 11, 13, 17, 19, and 23. Note that 1 is neither prime nor composite, so it remains in A rather than A'. Hence option A is correct.

A union B = U means every element of U is in A, in B, or in both. If an element is in A', it is not in A. To still belong to U = A union B, it must therefore be in B. Hence every element of A' belongs to B, so A' is a subset of B. Equality is not guaranteed because B may contain elements that are also in A.

For every subset A of a universal set U, the elements of A and the elements outside A together cover all of U. Therefore A union A' = U, which is the complement law of excluded alternatives. In this question, A' contains 6 through 12, and combining it with A gives every element from 1 through 12. Thus option A is correct.