Equal sets and Subsets

The union \(A\cup B\) contains every element of \(B\). If this union is equal to \(A\), then every element of \(B\) must also be an element of \(A\). This is precisely the definition of \(B\subseteq A\). The reverse inclusion, \(A\subseteq B\), is not necessary; for example, \(A=\{1,2\}\) and \(B=\{1\}\) satisfy the condition. The intersection need not be empty, and a complement relation cannot be inferred.

The equality \(A\cap B=B\) means that intersecting \(B\) with \(A\) does not remove any element from \(B\). Therefore, every element of \(B\) must already belong to \(A\), which gives \(B\subseteq A\). The reverse relation \(A\subseteq B\) does not necessarily hold, because \(A\) may contain additional elements. The complement statement is unrelated, and the intersection cannot be empty unless \(B\) is empty.

The intersection A ∩ B consists of all elements that occur in both A and B. The elements 2 and 3 are common to the two sets, while 1 occurs only in A and 4 occurs only in B. Therefore A ∩ B = {2, 3}. Hence option A is correct. The intersection is not the union, so elements occurring in only one set must not be included.

The positive multiples of 4 are 4, 8, 12, 16, 20, and so on. The condition “less than 20” excludes 20 and every larger multiple, while “positive” excludes 0 and negative numbers. Thus the members of A are exactly 4, 8, 12, and 16. Therefore option A represents a set equal to A. Option B includes 0, option C includes 20, and option D omits 4.

The prime factorization of 30 is 2 × 3 × 5, so the set of its prime divisors is A = {2, 3, 5}. A proper subset must contain only elements of A but must not be equal to A. The set {2, 5} satisfies both conditions: its elements belong to A and it omits 3. Therefore option A is the proper subset. Option B equals A, while C and D contain elements that are not in A.

Solving x² = 16 over the integers gives x = 4 or x = −4. Therefore A = {−4, 4}. The set B is also defined as {−4, 4}, so A and B contain exactly the same elements and are equal. Thus option A is true. Option B omits −4, option C incorrectly claims a proper inclusion even though the sets are equal, and option D is false because A has two elements.

A proper subset of B means that every element of A belongs to B, but A and B are not equal. For finite sets, this necessarily gives n(A) < n(B). Since n(A) = 6, B must contain at least one additional element and therefore must have more than six elements. Hence n(B) = 6 is impossible. Values 7, 8, and 10 are all possible because B can be formed by adding one, two, or four new elements to A, respectively.

For finite sets, if A ⊆ B, then every element of A is already in B. If B had even one additional element, its cardinality would be greater than that of A. Here both sets have cardinality 4, so B cannot contain any extra element. Consequently, A and B contain exactly the same elements and are equal. Option B is false because the inclusion is not proper, option C is false because B − A is empty, and option D is false because their intersection is A, not the empty set.

A set is a subset of A only when every one of its elements is also an element of A. The elements of A are 1, 2, 5, and 10. Options A, B, and C contain only elements from this list, so each is a subset of A. Option D contains 4, and 4 is not a divisor of 10 and is not an element of A. Therefore {2,4} is not a subset, making option D correct.

Equality of sets means that A and B contain exactly the same elements. Therefore, every element of A is necessarily an element of B, which is precisely the definition of A ⊆ B. The assertion is consequently true. The reason is also true because identical membership is what set equality means, and it directly explains why the subset relation follows. In fact, if A = B, then both A ⊆ B and B ⊆ A hold.

The assertion is false because equal cardinality only tells us that two sets have the same number of elements; it does not tell us that the elements themselves are identical. For example, A = {1,2} and B = {3,4} both have cardinality 2, but A ≠ B. The reason is also false for the same reason: equal size does not imply equal membership. To conclude A = B, we must show that every element of A is in B and every element of B is in A.

The complement A′ is defined relative to the universal set U. It contains every element of U that does not belong to A. Since U includes all students in the school and A includes the Class 11 students, A′ represents all school students who are not in Class 11. This could include students from several other classes, not necessarily only Class 12 students.

The equality A ∩ B = A says that intersecting B with A leaves all of A unchanged. This can happen only when every element of A is also contained in B. Therefore, A is a subset of B, written A ⊆ B. The other options do not follow from the given condition and may hold only in special cases.

The intersection contains elements common to both sets. Since \(A\subseteq B\), every element of A is also an element of B. Thus all elements of A are common to A and B, while B may contain additional elements that are not in A. Therefore, \(A\cap B=A\). The answer does not depend on the particular elements, only on the subset relationship.

Since A⊆B and A⊆C, every element of A belongs to all three sets. Conversely, an element in A∩B∩C must certainly belong to A. Thus the intersection cannot contain anything outside A and contains every element of A, so A∩B∩C=A. B and C may contain additional elements, while C−A contains elements specifically excluded from A.

Y − X contains elements that are in Y but not in X. The elements of Y are 3, 7, and 9. Since 3 and 7 also belong to X, they must be removed. The element 9 is not in X, so it remains. Consequently, Y − X = {9}. The set {3, 7} is Y ∩ X, while {1, 5} belongs to X − Y, not Y − X.

The relation \(A\subseteq B\) means that every element of \(A\) is already contained in \(B\). The union collects all elements from both sets, but adding the elements of \(A\) contributes nothing new because they are already in \(B\). Consequently, \(A\cup B=B\). In contrast, \(A\cap B=A\) and \(A\setminus B=\varnothing\) under this condition.

Because \(A\subseteq B\), every element of \(A\) is also an element of \(B\). The intersection contains elements common to both sets, so it contains every element of \(A\) and no element outside \(A\) is required. Therefore, \(A\cap B=A\). Set \(B\) may have additional elements, the union is generally larger, and \(B\setminus A\) contains elements of \(B\) excluded from \(A\).

Taking the union of a set with itself does not introduce any new elements. Every element of A is already in the first copy and the second copy, so A ∪ A = A. This is called the idempotent law. In contrast, A − A = ∅ and A ∩ ∅ = ∅, so those expressions are not generally equal to A.

The equality \(A\cap B=A\) means that intersecting A with B removes nothing from A. Consequently, every element of A must already be in B, so \(A\subseteq B\). The reverse relation \(B\subseteq A\) is not required; for example, A may be a proper subset of B. The complement and empty-union statements also do not follow. Therefore option A is the uniquely necessary conclusion.

The union \(A\cup B\) contains all elements of A and all elements of B. If the union is equal to A, then B cannot contain any element outside A; otherwise that element would appear in the union and make it larger than A. Hence every element of B belongs to A, which means \(B\subseteq A\). The other choices are not necessary and may contradict this containment.

The difference A \ B consists of elements that belong to A but do not belong to B. If this difference is empty, there is no element of A outside B. Therefore, every element of A is also an element of B, which is exactly the definition of A ⊆ B. The other statements do not necessarily follow; for example, A and B may overlap without being equal.

Two ordered pairs are equal if and only if their corresponding components are equal. Therefore, compare the first components and the second components separately: 2x - 1 = 5 and y + 3 = 10. The first equation gives 2x = 6, so x = 3. The second gives y = 7. Hence (x, y) = (3, 7), making option B correct. The order of components must be preserved while comparing ordered pairs.