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Miscellaneous
Misc 2 (i)
Misc 2 (ii) Important
Misc 2 (iii) Important
Misc 2 (iv)
Misc 2 (v)
Misc 2 (vi) Important
Misc 3
Misc 4 Important
Misc 5
Misc 6 Deleted for CBSE Board 2023 Exams
Misc 7 Important Deleted for CBSE Board 2023 Exams
Misc 8 Important
Misc 9 Important You are here
Misc 10
Misc 11
Misc 12 Important
Misc 13 Important
Misc 14
Misc 15 Important
Misc 16 Important
Last updated at March 22, 2023 by Teachoo
Misc 9 Using properties of sets show that A ∪ (A ∩ B) = A In order to prove A ∪ (A ∩ B) = A, we should prove A ∪ (A ∩ B) is a subset of A i.e. A ∪ (A ∩ B) ⊂ A & A is a subset of A ∪ (A ∩ B) i.e. A ⊂ A ∪ (A ∩ B) As set is a subset of itself, A ⊂ A Also, A is a subset of A ∩ B , i.e. A ⊂ A ∩ B as all elements of set A are in A ∩ B ∪ Union - Combination of two sets ∩ Intersection – Common of two sets Now, A ∪ (A ∩ B) Using distributive law :A ∪ (B ∩ C)= (A ∪ B) ∩ (A ∪ C) = (A) ∩ (A U B) = A = R.H.S Thus, A ∪ (A ∩ B) = A Hence proved Misc 9 Using properties of sets show that A ∪ (A ∩ B) = A Solving L.H.S A ∪ (A ∩ B) Using distributive law :A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) = = (A) ∩ (A U B) = A = R.H.S Thus, A ∪ (A ∩ B) = A Hence proved ∪ Union - Combination of two sets ∩ Intersection – Common of two sets As 2 × (3 + 4) = 2 × 3 + 2 × 4 Misc 9 Using properties of sets show that (ii) A ∩ (A ∪ B) = A. Taking L.H.S A ∩ (A ∪ B) Using distributive law :A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) = = (A) U (A ∩ B) = A = R.H.S Thus A ∩ (A ∪ B) = A. Hence proved ∪ Union - Combination of two sets ∩ Intersection – Common of two sets (From part (i) , s A ∪ (A ∩ B)= A)