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Example 22 - Chapter 1 Class 11 Sets
Last updated at Dec. 13, 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
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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
- Proof - Using properties of sets
- Proof - where properties of sets cant be applied,using element
Transcript
Example 22 Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}. Find A′, B′ , A′ ∩ B′, A ∪ B and hence show that (A ∪ B)′ = A′ ∩ B′. A′ = U – A = {1, 2, 3, 4, 5, 6} – {2, 3} = {1, 4, 5, 6} B′ = U – B = {1, 2, 3, 4, 5, 6} – {3, 4, 5} = {1, 2, 6} Now, A′ ∩ B′ = {1, 4, 5, 6} ∩ {1, 2, 6} = {1, 6} ∩ Intersection – Common of two sets ∪ Union - Combination of two sets Also, A ∪ B = {2, 3} ∪ {3, 4, 5} = { 2, 3, 4, 5 } Now, we need to prove (A ∪ B)′ = A′ ∩ B′ (A ∪ B)′ = U – (A ∪ B ) = {1, 2, 3, 4, 5, 6} – {2, 3, 4, 5} = {1, 6} Now, A′ ∩ B′ = {1, 6} & (A ∪ B)′ = {1, 6} Thus, (A ∪ B)′ = A′ ∩ B′ Hence proved
