# Let A , B and C be sets such that $\phi \neq A\cap B\subseteq C.$  Then which of the following statements is not true ?  Option 1) $B\cap C\neq \phi$ Option 2) $If\: \: (A-B)\subseteq C, then\: \: A\subseteq C$ Option 3) $(C\cup A)\cap(C\cup B)=C$ Option 4) $If\: \: (A-C)\subseteq B, then\: \: A\subseteq B$

As $(A\cap B)\subseteq C$

=> $(A\cap B)\subseteq (B\cap C)$

as $(A\cap B)\neq \phi$ => $(B\cap C)\neq \phi$

So, option (1) is true

let $x\epsilon A \: \: and\: \: x\notin B$ => $x\epsilon (A- B)$=> $x\epsilon C$

let $x\epsilon A \: \: and\: \: x\epsilon B$ => $x\epsilon (A\cap B)$=> $x\epsilon C$

Hence $x\epsilon A \: \: and\: \:$  $x\epsilon C$  => $A\subseteq C$

So, option (2) is  true.

Let $x\in C , x\in(C\cup A)\cap(C\cup B))$

$=>x\epsilon (C\cup A)\: and\: x\epsilon (C\cup B))$

$\left (x\in C\;or\;x\in A \right )\;\;and\;\;\left ( x\in C\;or\;x\in B \right )$

$=>x\epsilon C\: or\: x\epsilon (A\cap B)$

$=>x\epsilon C\;\;\;As,\;A\cup B\subseteq C$

$(C\cup A)\cap(C\cup B)\subseteq C$................(1)

Now,

$=>x\epsilon C$

$=>x\epsilon (C\cup A)\cap(C\cup B)$

$=>C\subseteq (C\cup A)\cap(C\cup B)$..............(2)

From (1) and (2)

$(C\cup A)\cap(C\cup B)=C$

=> option (3) is  correct.

For A = C , $A-C=\phi$

$=>\phi\subseteq B$  but  $=>A\nsubseteq B$

So, option (4) is not true.

Option 1)

$B\cap C\neq \phi$

Option 2)

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

Option 3)

$(C\cup A)\cap(C\cup B)=C$

Option 4)

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

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