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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 4 questions from this page. Select your focus, then start.
The set B contains the multiples of 9 from 1 to 54: B = {9, 18, 27, 36, 45, 54}. The complement A′ contains all elements of U that are not divisible by 6. Among the elements of B, 18, 36, and 54 are divisible by 6, so they are excluded from A′. The remaining elements are 9, 27, and 45. Therefore, A′ ∩ B = {9, 27, 45}, and its cardinality is n(A′ ∩ B) = 3.
Solving x² = 4 over the real numbers gives x = −2 or x = 2. Thus A = {−2, 2}. The complement is always determined relative to the stated universal set. Since U = ℝ, remove these two elements from all real numbers: A' = ℝ \ {−2, 2}. Option B is A itself, while options C and D incorrectly ignore the two excluded real numbers.
By De Morgan’s law, A' ∩ B' = (A ∪ B)'. Taking the complement once more gives (A' ∩ B')' = ((A ∪ B)')' = A ∪ B, using the double-complement law. The question already gives n(A ∪ B) = 50, so n((A' ∩ B')') = 50. The values of n(U), n(A), and n(B) are not needed after this identity is applied.
By De Morgan’s law, A′ ∩ B′ = (A ∪ B)′. In U, the multiples of 10 are 10, 20, ..., 90, so there are 9; the multiples of 18 are 18, 36, 54, 72, 90, so there are 5. Their only common element is 90 because lcm(10,18) = 90. Thus n(A ∪ B) = 9 + 5 - 1 = 13, and n(A′ ∩ B′) = 90 - 13 = 77. Therefore, option A is correct.
QUIZ COMPLETE
