Let A, B and C be sets such that phi ne A cap | vedclass

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Question

(A) Given that $\phi 
e A \cap B \subseteq C$.
Check option (A): If $(A - C) \subseteq B$ then $A \subseteq B$.
Let $A = \{1, 2\},\ B = \{2, 3\},\

Vedclass · Accepted Answer

If $(A - C) \subseteq B$ then $A \subseteq B$

Vedclass · Answer

(A) Given that $\phi 
e A \cap B \subseteq C$.
Check option (A): If $(A - C) \subseteq B$ then $A \subseteq B$.
Let $A = \{1, 2\},\ B = \{2, 3\},\ C = \{2\}$.
Here $A \cap B = \{2\} 
e \phi$ and $A \cap B = \{2\} \subseteq C = \{2\}$.
Now,
$A - C = \{1, 2\} - \{2\} = \{1\}$.
Since $\{1\} 
ot\subseteq B = \{2, 3\}$, the condition $(A - C) \subseteq B$ is false for this example.
However, if we take $A = \{1, 2\},\ B = \{1, 3\},\ C = \{1, 2\}$, then
$A \cap B = \{1\} \subseteq C$.
Here $A - C = \phi \subseteq B$ is true, but
$A = \{1, 2\} 
ot\subseteq B = \{1, 3\}$.
Thus, statement (A) is not true.
Check option (B): If $(A - B) \subseteq C$ then $A \subseteq C$.
Let $x \in A$. If $x \in B$, then $x \in A \cap B \subseteq C$, so $x \in C$.
If $x 
otin B$, then $x \in A - B \subseteq C$, so $x \in C$.
Thus $A \subseteq C$. This statement is true.
Check option (C):
$(C \cup A) \cap (C \cup B) = C \cup (A \cap B)$.
Since $A \cap B \subseteq C$, we get $C \cup (A \cap B) = C$.
Thus this statement is true.
Check option (D): Since $A \cap B \subseteq C$ and $A \cap B 
e \phi$, there exists $x \in A \cap B$.
So $x \in B$ and $x \in C$, hence $x \in B \cap C$.
Thus $B \cap C 
e \phi$. This statement is true.
Therefore, the statement that is not true is (A).

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?

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