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SubjectsMathematics
समुच्चय का पूरक और उसके गुणधर्म
In Class 11 Mathematics, under the chapter Sets, Complement of a Set and Its Properties explains the elements of a universal set that are not in a given set. Students find complements using notation such as A′, determine complements from given sets, and understand basic relationships between a set, its complement, and the universal set.
TOPIC PRACTICE
Up to 13 questions from this page. Select your focus, then start.
The complement A' with respect to U contains every element of U except 1, 4, 7, and 10. The second set is {2, 4, 6, 8, 10, 12}. Removing the elements 4 and 10, which are not in A', leaves 2, 6, 8, and 12. Hence A' ∩ {2, 4, 6, 8, 10, 12} = {2, 6, 8, 12}, so option A is correct.
The elements of A are the positive multiples of 8 not exceeding 100. Their number is floor(100/8) = 12, since the multiples run from 8 through 96. The universal set U contains 100 elements. The complement A' therefore contains all elements of U that are not divisible by 8, so n(A') = 100 − 12 = 88. Thus, option A is correct.
Within the universal set U, the numbers that are not even are precisely the odd numbers, so A is the set of odd numbers from 1 to 50. The complement A' consists of all elements of U that are not in A. These are exactly the even numbers: {2, 4, 6, …, 50}. Therefore, A' is the set of even numbers in U, making option A correct.
The universal set U contains the natural numbers from 1 through 32, so n(U) = 32. Since k belongs to ℕ₀, the relevant powers are 2⁰, 2¹, 2², 2³, 2⁴ and 2⁵, giving A = {1, 2, 4, 8, 16, 32}. Thus n(A) = 6. The complement A′ contains all elements of U that are not in A, so n(A′) = n(U) − n(A) = 32 − 6 = 26.
A and A′ are complementary subsets of U, so A = U \ A′. Remove 2, 3, 5, 7, 11, 13, 17, and 19 from the numbers 1 through 20. The remaining elements are 1, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, and 20. Notice that 1 remains because it is neither prime nor listed in A′.
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 must also be an element of B, which is precisely the definition of A ⊆ B. The condition does not imply that B ⊆ A or that the two sets are complements.
The odd numbers from 1 through 22 are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, and 21, making 11 odd numbers. The odd primes in this range are 3, 5, 7, 11, 13, 17, and 19. Removing these seven primes leaves 1, 9, 15, and 21 in A′, so the required number is 4. Note that 1 is not prime.
A set and its complement together contain every element of the universal set, so A′∪A=U. Taking the complement of both sides with respect to U gives (A′∪A)′=U′. The complement of the universal set is the empty set because no element of U lies outside U. Thus, the expression equals ∅, and option A is correct.
A set and its complement are disjoint, so A′ ∩ A = ∅. The complement is taken relative to the universal set U. The complement of the empty set is the entire universal set, because every element of U is outside ∅. Therefore (A′ ∩ A)′ = (∅)′ = U, making option A correct.
List the numbers from 1 to 50 whose last digit is 2 or 7: 2, 7, 12, 17, 22, 27, 32, 37, 42, and 47. Thus n(A) = 10. The universal set U has 50 elements, and the complement-count formula is n(A′) = n(U) − n(A). Therefore n(A′) = 50 − 10 = 40. Option A is correct. Option B counts A itself, while options C and D result from incorrect counting of the qualifying last digits.
The universal set contains every integer from −9 to 9. Set A consists of integers whose absolute value is greater than 4, namely −9 through −5 and 5 through 9. Its complement therefore contains the integers in U that do not satisfy |x| > 4. The opposite condition is |x| ≤ 4, which gives −4, −3, −2, −1, 0, 1, 2, 3, and 4. Hence option A is correct; option B describes A itself.
The numbers from 1 to 49 that are divisible by 7 are 7, 14, 21, 28, 35, 42, and 49, giving seven numbers in total. Among them, only 49 is a perfect square because 49 = 7², so 49 belongs to A and is excluded from A′. The other six numbers are not perfect squares and therefore belong to the complement A′. Hence, the correct answer is 6, option D.
The double-complement law states that (A′)′ = A for every subset A of a universal set U. Here, A consists of all real numbers less than or equal to -1 together with all real numbers greater than or equal to 6. Its complement in ℝ is (-1, 6), because -1 and 6 already belong to A. Taking the complement again restores the original set, so the answer is option A.
QUIZ COMPLETE
