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Let $A, B$ and $C$ be sets such that $\phi \ne A \cap B \subseteq C$. Then which of the following statements is not true ?
If $\left( {A - C} \right) \subseteq B$ then $A \subseteq B$
If $\left( {A - B} \right) \subseteq C$ then $A \subseteq C$
$\left( {C \cup A} \right) \cap \left( {C \cup B} \right) = C$
$B \cap C \ne \phi $
Solution
For $A\, = \,C,\,A – C\, = \,\phi $
$ \Rightarrow \phi \, \subseteq \,B$
But $A\, \subseteq \,B$
$ \Rightarrow \,$ option $A$ is NOT true
Let $x\, \in \,\,(C\,x\, \in \,(C\, \cup \,A)\,\, \cap (C\, \cup \,B)\,)$
$ \Rightarrow \,x\,(C\, \cup \,A)$ and $x\, \in \,\,(C\, \cup \,B)$
$ \Rightarrow \,(x\, \in \,C$ or $x\, \in \,A)$ and $(x\, \in \,C$ or $x\, \in \,B)$
$ \Rightarrow \,(x\, \in \,C$ or $x\, \in \,\,(A\, \cap \,B)$
$ \Rightarrow \,(x\, \in \,C$ or $x\, \in \,C$ (as $A\, \cup \,B \subseteq \,C\,$ )
$ \Rightarrow \,x\, \in \,C$
$ \Rightarrow (C\, \cup \,A)\,\, \cap (C\, \cup \,B)\, \subseteq \,C\,\,\,(1)$
Now $x\, \in \,C\, \Rightarrow \,x\, \in \,(C\, \cup \,A)$ and $x\, \in \,\,(C\, \cup \,B\,)$
$ \Rightarrow x\, \in (C\, \cup \,A)\,\, \cap (C\, \cup \,B)$
$ \Rightarrow C\, \subseteq \,(C\, \cup \,A)\,\, \cap (C\, \cup \,B)\, \subseteq \,C\,\,\,(2)$
$ \Rightarrow $ from $(1)$ and $(2)$
$C\, = \,(C\, \cup \,A)\,\, \cap (C\, \cup \,B)$
$ \Rightarrow $ option $B$ is true
Let $x\, \in \,A$ and $x\, \notin \,B$
$\Rightarrow \,x\, \in \,(A – B)$
$ \Rightarrow \,x\, \in \,C$ (as $A – B \subseteq \,C$ )
Let $x\, \in \,A$ and $x\, \in \,B$
$ \Rightarrow \,x\, \in \,(A \cap B)$
$ \Rightarrow \,x\, \in \,C$ (as $A \cap B \subseteq \,C$ )
Hence $x\, \in \,A\,\, \Rightarrow \,x\, \in \,C$
$ \Rightarrow A\, \subseteq \,C$
$ \Rightarrow $ option $C$ is true
As $C \supseteq \,(A \cap B)$
$ \Rightarrow B \cap C \supseteq \,(A \cap B)$
As $A \cap B \ne \phi $
$ \Rightarrow B \cap C \ne \phi $
Hence the correct answer is option $(A)$
