A=1,2,3andB=3,8; then (A∪B)×(A∩B) is
(3,1),(2,3),(3,3),(3,8)
(1,3),(2,3),(3,3),(8,3)
(1,2),(2,3),(3,3),(8,8)
(8,3),(8,2),(8,1),(8,8)
Explanation for correct answer:
Applying required operation on sets:
A=1,2,3andB=3,8
From the given question:
A∪B=1,2,3,8
and A∩B=3
(A∪B)×(A∩B)=(1,3),(2,3),(3,3),(8,3)
Hence, we can conclude that Option(B) is correct.
Prove that : (i) (A∪B)×C=(A×C)∪(B×C)
(ii) (A∩B)×C=(A×C)∩(B×C).
If A = {5,6,7,8,9}, B = {x:3 < x < 8 and x ∈ W} and C= {x:x ≤ 5 and x ∈ N}. Find:
(i)A∪B and (A∪B) ∪ C
(ii) B∪C and A∪(B∪C)
(iii) A∩B and (A∪B)∩C
(iv) B∩C and A∩(B∩C)
Is (A∪B)∪ C = A ∪(B∪C) ?
Is (A∩B)∩C = A ∩(B∩C) ?
Let A = {1,2,4,5}, B = {2,3,5,6} C = {4,5,6,7}. Verify the following identities :
(i)A∪(B∩C)=(A∪B)∩(A∪C)
(ii) A∩(B∪C)=(A∩B)∪(A∩C)
(iii) A∩(B−C)=(A∩B)−(A∩C)
(iv) A−(B∪C)=(A−B)∩(A−C)
(v) A−(B∩C)=(A−B)∪(A−C)
(vi) A∩(BΔC)=(A∩B)Δ(A∩C)