Match List I with List - II
List - I (Set Laws) List - II (Form) A. Idempotent law III. \(A \cup A = A\) B. Commutative law I. \(A \cup B = B \cup A\) C. Associative law II. \(A \cap (B \cap C) = (A \cap B) \cap C\) D. De-Morgan law IV. \((A \cap B)' = A' \cup B'\)
Match List I with List - II
| List - I (Set Laws) | List - II (Form) |
|---|---|
| A. Idempotent law | III. \(A \cup A = A\) |
| B. Commutative law | I. \(A \cup B = B \cup A\) |
| C. Associative law | II. \(A \cap (B \cap C) = (A \cap B) \cap C\) |
| D. De-Morgan law | IV. \((A \cap B)' = A' \cup B'\) |
Show Hint
- A-III, B-I, C-II, D-IV
- A-III, B-II, C-I, D-IV
- A-I, B-IV, C-III, D-II
- A-I, B-III, C-IV, D-II
The Correct Option is A
Solution and Explanation
Step 1: Match Idempotent and Commutative laws.
- Idempotent Law: States that an operation applied to a set and itself results in the same set. $A \cup A = A$. (A-III) - Commutative Law: States that the order of the sets does not change the result. $A \cup B = B \cup A$. (B-I)
Step 2: Match Associative and De-Morgan laws.
- Associative Law: States that when performing the same operation on three sets, the grouping does not matter. $A \cap (B \cap C) = (A \cap B) \cap C$. (C-II) - De-Morgan's Law: Describes how the complement of an intersection or union is distributed. $(A \cap B)' = A' \cup B'$. (D-IV)
Step 3: Conclusion.
The mapping is A-III, B-I, C-II, D-IV, which corresponds to Option (1).
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