Q61 / 180 · COMEDK 2025 (10 May Shift 1) · Mathematics
Identify the correct statement
- A$A \cup A^{\prime}=\emptyset$
- B$(A \cup B)^{\prime}=A^{\prime} \cup B^{\prime}$
- Correct answerC$A \subseteq B \Rightarrow B^{\prime} \subseteq A^{\prime}$
- D$A-B=A^{\prime} \cap B$
Solution
Analyze each option based on set theory principles.
Option (1): $A \cup A^{\prime} = U$, where $U$ is the universal set. Thus, $A \cup A^{\prime} = \emptyset$ is false.
Option (2): According to De Morgan's Law, $(A \cup B)^{\prime} = A^{\prime} \cap B^{\prime}$. Thus, $(A \cup B)^{\prime} = A^{\prime} \cup B^{\prime}$ is false.
Option (3): If $A \subseteq B$, then every element of $A$ is an element of $B$. Consequently, any element not in $B$ must also not be in $A$. This implies $B^{\prime} \subseteq A^{\prime}$. This statement is true.
Option (4): The set difference $A - B$ is defined as $A \cap B^{\prime}$. The statement $A - B = A^{\prime} \cap B$ is false.
Answer: $A \subseteq B \Rightarrow B^{\prime} \subseteq A^{\prime}$
Option (1): $A \cup A^{\prime} = U$, where $U$ is the universal set. Thus, $A \cup A^{\prime} = \emptyset$ is false.
Option (2): According to De Morgan's Law, $(A \cup B)^{\prime} = A^{\prime} \cap B^{\prime}$. Thus, $(A \cup B)^{\prime} = A^{\prime} \cup B^{\prime}$ is false.
Option (3): If $A \subseteq B$, then every element of $A$ is an element of $B$. Consequently, any element not in $B$ must also not be in $A$. This implies $B^{\prime} \subseteq A^{\prime}$. This statement is true.
Option (4): The set difference $A - B$ is defined as $A \cap B^{\prime}$. The statement $A - B = A^{\prime} \cap B$ is false.
Answer: $A \subseteq B \Rightarrow B^{\prime} \subseteq A^{\prime}$
Hence, the correct answer is option: 'C'.