Q105 / 180 · COMEDK 2024 (Afternoon Shift) · Mathematics
$\text { Let } \mathrm{A} \text { and } \mathrm{B} \text { be two sets then } A-(A \cap B) \text { is equal to }$
- A$(A \cap B)^{\prime}$
- B$\emptyset$
- C$B-A$
- Correct answerD$A-B$
Solution
By the definition of set difference, the set $A - (A \cap B)$ consists of all elements that are in $A$ but not in $A \cap B$.
Using the distributive law or the property of set difference, $A - (A \cap B) = A \cap (A \cap B)'$.
Applying De Morgan's Law, $(A \cap B)' = A' \cup B'$.
Thus, $A \cap (A' \cup B') = (A \cap A') \cup (A \cap B') = \emptyset \cup (A \cap B') = A \cap B'$.
The expression $A \cap B'$ is equivalent to $A - B$.
Answer: $A-B$
Using the distributive law or the property of set difference, $A - (A \cap B) = A \cap (A \cap B)'$.
Applying De Morgan's Law, $(A \cap B)' = A' \cup B'$.
Thus, $A \cap (A' \cup B') = (A \cap A') \cup (A \cap B') = \emptyset \cup (A \cap B') = A \cap B'$.
The expression $A \cap B'$ is equivalent to $A - B$.
Answer: $A-B$
Hence, the correct answer is option: 'D'.