Miscellaneous

Misc 1

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 Important You are here

Misc 7 Important

Misc 8

Misc 9

Misc 10 Important

Question 1 Deleted for CBSE Board 2025 Exams

Question 2 Important Deleted for CBSE Board 2025 Exams

Question 3 Important Deleted for CBSE Board 2025 Exams

Question 4 Deleted for CBSE Board 2025 Exams

Question 5 Important Deleted for CBSE Board 2025 Exams

Question 6 Important Deleted for CBSE Board 2025 Exams

Chapter 1 Class 11 Sets

Serial order wise

Last updated at April 16, 2024 by Teachoo

Misc 6 - Introduction Show that for any sets A and B, A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B) Let U = {1, 2, 3, 4, 5} A = {1, 2} B = {2, 3, 4} A – B = A – (A ∩ B) = {1, 2} – {2} = {1} We use the result A – B = A ∩ B’ in this question Also, B’ = U – B = {1, 2, 3, 4, 5} – {2, 3, 4} = {1, 5} A – B = A ∩ B’ = {1, 2} ∩ {1, 5} = {1} Misc 6 Show that for any sets A and B, A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B) To prove : A = (A ∩ B) ∪ (A – B) Solving R.H.S (A ∩ B) ∪ (A – B) Using A – B = A – (A ∩ B) = A ∩ B’ = (A ∩ B) ∪ (A ∩ B’) = A ∩ (B ∪ B’) = A ∩ (U) = A = L.H.S Hence proved To prove : A ∪ (B – A) = (A ∪ B) Taking L.H.S A ∪ (B – A) Using B – A = B – (A ∩ B) = B ∩ A’ = A ∪ (B ∩ A’) Using distributive law :A ∪ (B ∩ C)= (A ∪ B) ∩ (A ∪ C) = (A ∪ B) ∩ (A ∪ A’) = (A ∪ B) ∩ (U) = (A ∪ B) = R.H.S Hence proved