JEE Main 2019 — Mathematics Algebra

For $A=C,A-C=\phi$A=C,A−C=ϕ

$$\Rightarrow \phi \subseteq B$$⇒ϕ⊆B

But $A⊈B$A⊈B

Hence, If $(A-C)\subseteq B,$(A−C)⊆B, then $A\subseteq B$A⊆B is not true

Let $x\in C, \Rightarrow x\in (C\cup A)\cap (C\cup B)$x∈C,⇒x∈(C∪A)∩(C∪B)

$\Rightarrow x\in (C\cup A)$⇒x∈(C∪A) and $x\in (C\cup B)$x∈(C∪B)

$\Rightarrow (x\in C$⇒(x∈C or $x\in A)$x∈A) and $(x\in C$(x∈C or $x\in B)$x∈B)

$\Rightarrow x\in C$⇒x∈C or $x\in (A\cap B)$x∈(A∩B)

$\Rightarrow x\in C$⇒x∈C or $x\in C (as A\cup B\subseteq C)$x∈C(asA∪B⊆C)

$$\Rightarrow x\in C$$⇒x∈C

$\Rightarrow (C\cup A)\cap (C\cup B)\subseteq C$⇒(C∪A)∩(C∪B)⊆C .... (i)

Now $x\in C\Rightarrow x\in (C\cup A)$x∈C⇒x∈(C∪A) and $x\in (C\cup B)$x∈(C∪B)

$$\Rightarrow x\in (C\cup A)\cap (C\cup B)$$⇒x∈(C∪A)∩(C∪B)

$\Rightarrow C\subseteq (C\cup A)\cap (C\cup B)$⇒C⊆(C∪A)∩(C∪B) ..... (ii)

$\Rightarrow$⇒ From (i) and (ii)

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

Hence, $(C\cup A)\cap (C\cup B)=C$(C∪A)∩(C∪B)=C is true

Let $x\in A$x∈A and $x\notin B$x∈/B

$$\Rightarrow x\in (A-B)$$⇒x∈(A−B)

$\Rightarrow x\in C$⇒x∈C $(as A-B\subseteq C)$(asA−B⊆C)

Let $x\in A$x∈A and $x\in B$x∈B

$$\Rightarrow x\in (A\cap B)$$⇒x∈(A∩B)

$\Rightarrow x\in C$⇒x∈C $(as A\cap B\subseteq C)$(asA∩B⊆C)

Hence $x\in A \Rightarrow x\in C$x∈A⇒x∈C

$$\Rightarrow A\subseteq C$$⇒A⊆C

Hence, If $(A-B)\subseteq C,$(A−B)⊆C, then $A\subseteq C$A⊆C is true

as $C\supseteq (A\cap B)$C⊇(A∩B)

$$\Rightarrow B\cap C\supseteq (A\cap B)$$⇒B∩C⊇(A∩B)

as $A\cap B\neq \phi$A∩B=ϕ

$$\Rightarrow B\cap C\neq \phi$$⇒B∩C=ϕ

Hence, $B\cap C\neq \phi$B∩C=ϕ is true

Alternate solution

Given

$$\phi \neq A\cap B\subseteq C$$ϕ=A∩B⊆C

Let answer that

$$C={1,2,3,4,5,6}$$C=1,2,3,4,5,6

$$A={1,2,3}$$A=1,2,3

$$B={3,4,5}$$B=3,4,5

$$A\cap B={3}$$A∩B=3

Option (a)

$$B\cap C={3,4,5}\neq \phi$$B∩C=3,4,5=ϕ

Option (b) $C\cup A={1,2,3,4,5,6}$C∪A=1,2,3,4,5,6

$$C\cup B={1,2,3,4,5,6}$$C∪B=1,2,3,4,5,6

$$(C\cup A)\cap (C\cup B)={1,2,3,4,5,6}=C$$(C∪A)∩(C∪B)=1,2,3,4,5,6=C

Option (c) $A-B={1,2,3}\subseteq CA\subseteq C$A−B=1,2,3⊆CA⊆C True

Option (d) $A-C=\phi \subseteq B A\subseteq C$A−C=ϕ⊆BA⊆C Not correct

$\therefore$∴ Option (d) is correct