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SubjectsMathematics
समुच्चयों पर संक्रियाएँ (संघ, प्रतिच्छेद और अंतर)
In Class 11 Mathematics, the Sets chapter introduces Operations on Sets (Union, Intersection, Difference). Students learn to combine sets using union, identify common elements through intersection, and find elements belonging to one set but not another using difference. They also apply these operations to subset relations, Venn diagrams, and problems involving the number of elements in sets.
Correct answer: A. A − B
Explanation: A − (A ∩ B) consists of elements that are in A but not in the intersection A ∩ B. An element of A is excluded from A ∩ B precisely when it is not in B. Therefore, the remaining elements are those in A but outside B, which is A − B. Equivalently, A − (A ∩ B) = A ∩ Bᶜ = A − B. Option A is correct.
Correct answer: B. B − A
Explanation: The union A ∪ B contains all elements from A and B. Removing A eliminates every element that belongs to A, including the common elements A ∩ B. What remains are precisely the elements that belong to B but not to A, which is B − A. Therefore, (A ∪ B) − A = B − A, so option B is correct.
Correct answer: A. {1, 3}
Explanation: The complement A′ is taken with respect to the universal set U. Removing the elements of A from U gives A′ = {1, 3, 5, 7, 9}. We then find the intersection of A′ with B = {1, 2, 3, 4}. The elements common to both sets are 1 and 3, so A′ ∩ B = {1, 3}. Therefore, option A is correct.
Correct answer: B. {1, 3, 5, 6}
Explanation: First calculate the union and intersection. A ∪ B = {1, 2, 3, 4, 5, 6}, while A ∩ B = {2, 4}. Subtracting the intersection from the union removes the common elements 2 and 4, leaving {1, 3, 5, 6}. This operation gives the elements belonging to exactly one of the two sets, also called their symmetric difference. Hence, option B is correct.
Correct answer: A. {1, 2, 3, 5, 7}
Explanation: The union A ∪ B consists of three mutually separate regions: elements in A but not B, elements in B but not A, and elements common to both sets. These are respectively A − B = {1, 5}, B − A = {7}, and A ∩ B = {2, 3}. Combining all distinct elements gives A ∪ B = {1, 2, 3, 5, 7}. Therefore, option A is correct.
Correct answer: A. {2}
Explanation: The prime numbers less than 15 are A = {2, 3, 5, 7, 11, 13}. The odd numbers less than 15 are B = {1, 3, 5, 7, 9, 11, 13}. A − B contains elements in A that are not in B. Every prime in A except 2 is odd and therefore belongs to B. Removing those common elements leaves only 2, so option A is correct.
Correct answer: A. A − B = {1, 3, 5}
Explanation: Every element of B is also an element of A, so B is a subset of A. Removing B’s elements 2, 4, and 6 from A leaves A − B = {1, 3, 5}. Also, B − A would be empty, A ∩ B would equal B = {2, 4, 6}, and A ∪ B would equal A, not B. Therefore, only statement A is correct.
Correct answer: A. \(B\subseteq A\)
Explanation: Let \(x\) be any element of \(B\). By the definition of union, \(x\in A\cup B\). Since \(A\cup B=A\), this means \(x\in A\). Thus every element of \(B\) belongs to \(A\), so \(B\subseteq A\). The reverse inclusion or equality is not necessary; for example, \(A=\{1,2\}\) and \(B=\{1\}\) satisfy the condition but are not equal.
Correct answer: A. \(\{2,3,4\}\)
Explanation: First evaluate the expression inside parentheses: \(B\cup C=\{2,3,4,6,7\}\). Next find the elements common to this set and \(A=\{1,2,3,4,5\}\). The common elements are 2, 3, and 4, so \(A\cap(B\cup C)=\{2,3,4\}\). Option D is only the union and includes 6 and 7, which are not in \(A\); option B contains elements excluded from the union.
Correct answer: A. \(\{2,4,6,8\}\)
Explanation: The parentheses require the intersection first. The elements common to \(A=\{1,2,4,8\}\) and \(B=\{2,4,6\}\) are 2 and 4, so \(A\cap B=\{2,4\}\). Taking the union with \(C=\{4,6,8\}\) gives \(\{2,4\}\cup\{4,6,8\}=\{2,4,6,8\}\). Repeated elements such as 4 are written only once in a set.
Correct answer: A. \(\{15\}\)
Explanation: Multiples of 3 from 1 through 20 are \(\{3,6,9,12,15,18\}\), while multiples of 5 are \(\{5,10,15,20\}\). The only number appearing in both lists is 15. Equivalently, the least common multiple of 3 and 5 is 15, and the next common multiple is 30, which exceeds 20. Therefore \(A\cap B=\{15\}\).
Correct answer: A. 8
Explanation: The multiples of 4 up to 25 are \(\{4,8,12,16,20,24\}\), so \(|A|=6\). The multiples of 6 are \(\{6,12,18,24\}\), so \(|B|=4\). Their common elements are \(\{12,24\}\), giving \(|A\cap B|=2\). By inclusion–exclusion, \(|A\cup B|=6+4-2=8\). The subtraction prevents 12 and 24 from being counted twice.
Correct answer: C. \(A\cap B\subseteq A\)
Explanation: The intersection \(A\cap B\) is defined as the set of elements that belong to both \(A\) and \(B\). Since every element in the intersection necessarily belongs to \(A\), it follows that \(A\cap B\subseteq A\) for all sets \(A\) and \(B\). The other statements are not universally true: the two differences may differ, equality of union and intersection is exceptional, and \(A\subseteq A-B\) generally fails when the sets overlap.
Correct answer: B. 4
Explanation: The difference A − B contains elements that are in A but not in B. Since 3 and 4 are common to both sets, A − B = {1, 2}, so n(A − B) = 2. Similarly, B − A contains elements in B but not in A, giving B − A = {5, 6} and n(B − A) = 2. Hence the required sum is 2 + 2 = 4. The value 2 counts only one directional difference, while 6 and 8 result from overcounting or using the wrong set formula.
Correct answer: A. 20
Explanation: The union A ∪ B can be partitioned into three mutually disjoint parts: the elements only in A, represented by A − B; the elements only in B, represented by B − A; and the elements common to both, represented by A ∩ B. Therefore, n(A ∪ B) = n(A − B) + n(B − A) + n(A ∩ B). Substituting the given values gives 60 = 18 + 22 + n(A ∩ B), so n(A ∩ B) = 20. Thus option A is correct.
Correct answer: B. {2, 4, 6, 8, 10, 12, 14}
Explanation: A contains all natural numbers up to 15, while B contains the odd natural numbers up to 15. Subtracting B from A removes 1, 3, 5, 7, 9, 11, 13, and 15. The elements left in A are precisely the even natural numbers up to 15: {2, 4, 6, 8, 10, 12, 14}. Therefore option B is correct. Option A is B itself, option C incorrectly includes 16, which is greater than 15, and option D would mean every element of A were odd.
Correct answer: B. (1, 4]
Explanation: The intersection contains the real numbers that belong to both intervals. A includes every number from -1 through 4, including both endpoints. B contains numbers greater than 1 and less than 6; therefore, 1 is excluded, while 4 is included because A includes 4 and B also contains 4. Hence the common interval is (1, 4], which is option B.
Correct answer: A. {3, 4, 5}
Explanation: Factor the first equation: x² − 7x + 12 = (x − 3)(x − 4) = 0, so A = {3, 4}. Factor the second equation: x² − 9x + 20 = (x − 4)(x − 5) = 0, so B = {4, 5}. The union contains every distinct element appearing in either set, so A ∪ B = {3, 4, 5}. Therefore, option A is correct.
Correct answer: B. {1, 2, 8, 9, 25, 32}
Explanation: A − B contains elements present in A but absent from B, so A − B = {1, 9, 25}. Similarly, B − A = {2, 8, 32}. Taking their union gives {1, 9, 25} ∪ {2, 8, 32} = {1, 2, 8, 9, 25, 32}. This operation is also called the symmetric difference because common elements 4 and 16 are excluded. Thus option B is correct.
Correct answer: A. A − B = {4}
Explanation: Since x is an integer and −2 ≤ x < 5, A = {−2, −1, 0, 1, 2, 3, 4}. The inequality x² ≤ 9 is equivalent to −3 ≤ x ≤ 3, so B = {−3, −2, −1, 0, 1, 2, 3}. Set difference A − B retains members of A that are absent from B. Every member except 4 is removed, giving {4}; therefore option A is correct.
Correct answer: A. A ⊂ B
Explanation: A − B = ∅ means that no element of A lies outside B; therefore, every element of A belongs to B, so A ⊆ B. The additional condition B − A ≠ ∅ says that at least one element belongs to B but not to A. Hence A is a proper subset of B, written A ⊂ B. Equality and the disjointness condition are therefore impossible.
Correct answer: A. {a, c, e, f}
Explanation: The symmetric difference contains elements that occur in exactly one of the two sets. From A, the elements not in B are A − B = {a, c}. From B, the elements not in A are B − A = {e, f}. Their union is therefore A △ B = {a, c, e, f}. The common elements b and d are excluded because they belong to both sets.
Correct answer: A. [0, 5]
Explanation: A contains every real number from 0, including 0, up to but not including 3. B contains every real number greater than 1 through 5, including 5. The intervals overlap on (1, 3), so there is no gap between them. Their union therefore begins at 0 and ends at 5, with both endpoints included. Hence A ∪ B = [0, 5].
Correct answer: A. [−3, −1]
Explanation: The difference A − B consists of elements that are in A but not in B. Set A is the interval [−3, 2], while B contains all numbers strictly between −1 and 4. Thus, every point from just greater than −1 through 2 is removed from A. The point −1 remains because −1 belongs to A but is excluded from B. Therefore, A − B = [−3, −1].
Correct answer: A. 28
Explanation: The union is divided into three disjoint regions: elements only in A, counted by n(A − B) = 12; elements only in B, counted by n(B − A) = 9; and common elements, counted by n(A ∩ B) = 7. Therefore n(A ∪ B) = 12 + 9 + 7 = 28. The common elements must be included once in the union.
