(A) Given $(A \cap C) \cup (B \cap C^{\prime}) = \phi$. Since the union of two sets is the empty set $\phi$, each set must be empty. Therefore, $A \ca

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(A) Given $(A \cap C) \cup (B \cap C^{\prime}) = \phi$. Since the union of two sets is the empty set $\phi$, each set must be empty. Therefore, $A \cap C = \phi$ and $B \cap C^{\prime} = \phi$. From $B \cap C^{\prime} = \phi$, we have $B \subseteq C$. Since $A \cap C = \phi$ and $B \subseteq C$, it follows that $A \cap B = \phi$.

Class 11 Mathematics Sets Word Problem - Set Theory

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Let $A, B, C$ be three non-void subsets of set $S$. Let $(A \cap C) \cup (B \cap C^{\prime}) = \phi$, where $C^{\prime}$ denotes the complement of set $C$ in $S$. Then:

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A
$A \cap B = \phi$
B
$A \cap B \neq \phi$
C
$A \cap C = A$
D
$A \cup C = A$

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Let $A, B, C$ be three non-void subsets of set $S$. Let $(A \cap C) \cup (B \cap C^{\prime}) = \phi$, where $C^{\prime}$ denotes the complement of set $C$ in $S$. Then:

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