Educational

For any three sets A, B and C, (A∪B∪C) ∩ (A∩B’∩C’) ∩ C’ is equal to __________ Solution \[ (A \cup B \cup C) \cap (A \cap B’ \cap C’) \cap C’ \] \[ = (A \cap B’ \cap C’) \cap (A \cup B \cup C) \] Using distributive law, \[ = (A \cap B’ […]

If A and B are two sets, then (A ∩ B’)’ ∪ (B ∩ C) is equal to _________ Solution \[ (A \cap B’)’ \cup (B \cap C) \] \[ = (A’ \cup B) \cup (B \cap C) \] \[ = A’ \cup [B \cup (B \cap C)] \] \[ = A’ \cup B \]

For any three sets A, B and C, (A – B) ∩ (C – B) is equal to __________ Solution \[ (A – B) \cap (C – B) \] \[ = (A \cap B’) \cap (C \cap B’) \] \[ = A \cap C \cap B’ \] \[ = (A \cap C) – B \]

For any three sets A, B and C, (A – B) – (B – C) is equal to __________ Solution \[ (A – B) – (B – C) \] \[ = (A \cap B’) – (B \cap C’) \] \[ = (A \cap B’) \cap (B \cap C’)’ \] \[ = (A \cap B’) \cap

For any two sets A and B, [B’ ∪ (B’ – A)]’ is equal to……. Solution \[ [B’ \cup (B’ – A)]’ \] Using \[ X – Y = X \cap Y’ \] \[ = [B’ \cup (B’ \cap A’)]’ \] Using absorption law, \[ B’ \cup (B’ \cap A’) = B’ \] Therefore, \[

If A and B are two sets, then ((A’ ∪ B’) – A)’ is equal to………. Solution \[ ((A’ \cup B’) – A)’ \] Using \[ X – Y = X \cap Y’ \] we get \[ = ((A’ \cup B’) \cap A’)’ \] Since \[ A’ \cap (A’ \cup B’) = A’ \] therefore,

If A and B are two sets, then A ∩ (A ∪ B)’ is equal to ___________ Solution \[ A \cap (A \cup B)’ \] By De Morgan’s law, \[ = A \cap (A’ \cap B’) \] Using associative law, \[ = (A \cap A’) \cap B’ \] \[ = \Phi \cap B’ \] \[

If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 5}, B = {2, 4, 6, 7} and C = {2, 3, 4, 8}, Then (i) \((B \cup C)’\) = …. (ii) \((C – A)’\) = …. Solution \[ U = \{1,2,3,4,5,6,7,8,9,10\} \] \[ B = \{2,4,6,7\}

Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8} then the universal set of all the three sets A, B and C can be…. Solution \[ A = \{1,3,5\} \] \[ B = \{2,4,6\} \] \[ C = \{0,2,4,6,8\} \] The universal set

For all sets A and B, B – (A ∩ B) is equal to…… Solution \[ B – (A \cap B) \] \[ = B \cap (A \cap B)’ \] \[ = B \cap (A’ \cup B’) \] \[ = (B \cap A’) \cup (B \cap B’) \] \[ = B \cap A’ \] \[