Misc 6 - Chapter 1 Class 11 Sets
Last updated at April 16, 2024 by Teachoo
Chapter 1 Class 11 Sets
Concept wise
- Depiction and Definition
- Depicition of sets - Roster form
- Depicition of sets - Set builder form
- Intervals
- Null Set
- Finite/Infinite
- Equal sets
- Subset
- Power Set
- Universal Set
- Venn Diagram and Union of Set
- Intersection of Sets
- Difference of sets
- Complement of set
- Number of elements in set - 2 sets (Direct)
- Number of elements in set - 2 sets - (Using properties)
- Number of elements in set - 3 sets
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Proof - Using properties of sets
- Proof - where properties of sets cant be applied,using element
Transcript
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
