Muft Shiksha™ एक 100% Free Education Portal है 🇮🇳, जिसका उद्देश्य Class 9–12 के हर विद्यार्थी तक High-Quality Education को पूरी तरह मुफ्त पहुँचाना है। 🇮🇳 हम मानते हैं कि अच्छी शिक्षा किसी student की आर्थिक स्थिति पर निर्भर नहीं होनी चाहिए। 🇮🇳 हर विद्यार्थी को वही Quality Study Material, MCQs, Quizzes, Exam Preparation, Concept-Based Learning और Bilingual Support मिलना चाहिए, जो आमतौर पर महंगी Coaching या Premium Platforms में मिलता है। Muft Shiksha™ 🇮🇳 इसी सोच के साथ बनाया गया है
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
Quiz this set
Up to 4 questions from this page. Select your focus, then start.
If U = {1, 2, ..., 22}, A = {x : x ∈ U, x is even}, and B = {x : x ∈ U, x is a perfect square}, what is (A ∩ B)'?
Correct answer: A
The perfect squares in U are 1, 4, 9, and 16. The even perfect squares, which belong to both A and B, are therefore A ∩ B = {4, 16}. The complement is taken within U, so remove 4 and 16 from {1, 2, ..., 22}. The remaining 20 elements are exactly the set listed in option A. Option B is the intersection itself, not its complement.
If U = {1, 2, ..., 81}, A = {x : x ∈ U, 3 divides x}, and B = {x : x ∈ U, 27 divides x}, how many elements are in B' ∩ A?
Correct answer: A
The set A contains all multiples of 3 from 1 to 81, so n(A) = 81 ÷ 3 = 27. The set B contains the multiples of 27: 27, 54, and 81, so n(B) = 81 ÷ 27 = 3. Since every multiple of 27 is also a multiple of 3, B is a subset of A. Therefore B' ∩ A consists of the elements of A not in B, and its cardinality is 27 − 3 = 24.
Let U = {x ∈ ℕ : x ≤ 120}, A = {x ∈ U : 8 divides x}, B = {x ∈ U : 15 divides x}, and C = {x ∈ U : 20 divides x}. What is n((A ∪ B ∪ C)′)?
Correct answer: A
Count the union by inclusion–exclusion. There are 120/8 = 15 multiples of 8, 120/15 = 8 multiples of 15, and 120/20 = 6 multiples of 20. The pairwise intersections contain 2 multiples of lcm(8,15)=120, 1 multiple of lcm(8,20)=40, and 2 multiples of lcm(15,20)=60. The triple intersection contains only 120. Hence n(A ∪ B ∪ C) = 15 + 8 + 6 - 1 - 3 - 2 + 1 = 24. Therefore, its complement has 120 - 24 = 96 elements.
Let U = ℝ, A = {x ∈ ℝ : (x - 2)(x + 5) ≤ 0}, and B = {x ∈ ℝ : |x - 1| < 3}. What is (A′ ∩ B)′?
Correct answer: A
The inequality (x - 2)(x + 5) ≤ 0 holds between the roots, including both roots, so A = [-5, 2]. Also, |x - 1| < 3 gives -3 < x - 1 < 3, hence -2 < x < 4 and B = (-2, 4). Therefore A′ = (-∞, -5) ∪ (2, ∞), and A′ ∩ B = (2, 4). Taking the complement in ℝ gives (A′ ∩ B)′ = (-∞, 2] ∪ [4, ∞), which is option A.
Google Analytics helps us understand site usage. Google may send limited cookie-free signals before your choice. The Live Visitors widget operates independently of this analytics choice; see the privacy policy for its provider and fallback details. Essential site features work without analytics cookies. You can change your choice later in Privacy choices. Privacy policy