Question

Let $U$ be universal set. If $A$ and $B$ are subsets of $U$, then match the following and choose the correct answer. {ll} List - I & List - II
(a) $A \cup A'$ & (i) $A$
(b) $(A \cup B)'$ & (ii) $A' \cup B'$
(c) $(A \cap B)'$ & (iii) $A' \cap B'$
(d) $(A')'$ & (iv) $U$ Codes:

• A. a - ii, b - i, c - iv, d - iii
• B. a - ii, b - iii, c - iv, d - i
• C. a - iii, b - iv, c - ii, d - i
• D. a - iv, b - iii, c - ii, d - i

Hint:

A useful memory trick for De Morgan's Laws is: “Break the bar, change the operation.” When the complement sign is distributed, union changes to intersection and intersection changes to union.

Answer 1

The Correct Option is D

Concept: This problem is based on the fundamental laws of Set Theory. The most important laws required are:
• Complement Law: \[ A\cup A'=U \]
• Double Complement Law: \[ (A')'=A \]
• First De Morgan's Law: \[ (A\cup B)'=A'\cap B' \]
• Second De Morgan's Law: \[ (A\cap B)'=A'\cup B' \] Let us match each item carefully.

Step 1: Match (a) $A \cup A'$. A set together with its complement contains every element of the universal set. Hence, \[ A\cup A'=U \] This corresponds to item (iv). Therefore, \[ a \rightarrow iv \]

Step 2: Match (b) $(A \cup B)'$. Applying the first De Morgan's Law: \[ (A\cup B)'=A'\cap B' \] This corresponds to item (iii). Therefore, \[ b \rightarrow iii \]

Step 3: Match (c) $(A \cap B)'$. Applying the second De Morgan's Law: \[ (A\cap B)'=A'\cup B' \] This corresponds to item (ii). Therefore, \[ c \rightarrow ii \]

Step 4: Match (d) $(A')'$. Taking complement twice returns the original set. \[ (A')'=A \] This corresponds to item (i). Therefore, \[ d \rightarrow i \]

Step 5: Final matching. \[ a \rightarrow iv \] \[ b \rightarrow iii \] \[ c \rightarrow ii \] \[ d \rightarrow i \] Thus, \[ \boxed{\text{a - iv,\; b - iii,\; c - ii,\; d - i}} \] which matches option \[ \boxed{(D)} \]