Answer :

$A=\{x: x \in R $ and $x$ satisfies $.x^{2}-8 x+12=0\}$

$2$ and $6$ are the only solutions of $x^{2}-8 x+12=0$

$\therefore \ \ A=\{2,6\}$

$B=\{2,4,6\}, C=\{2,4,6,8 \ldots\}, D=\{6\}$

$\therefore \ \ D \subset A \subset B \subset C$

Hence, $A \subset B, A \subset C, B \subset C, D \subset A, D \subset B, D \subset C$

Answer :

(i) False

Let $A=\{1,2\}$ and $B=\{1,\{1,2\},\{3\}\}$

Now, $2 \in\{1,2\}$ and $\{1,2\} \in\{\{3\}, 1,\{1,2\}\}$

$A$ $\in$ $B$

However, $2 \notin\{\{3\}, 1,\{1,2\}\}$

(ii) False

Let $A=\{2\}, B=\{0,2\}$, and $C=\{1,\{0,2\}, 3\}$

As $A \subset B$

$B \in C$

However, $A \notin C$

(iii) True

Let $A \subset B$ and $B \subset C$.

Let $x \in A$

$\Rightarrow x \in B \quad[\because A \subset B]$

$\Rightarrow x \in C \quad[\because B \subset C]$

$\therefore \ \ A\subset C$

(iv) False

$ A=\{1,2\}, B=\{0,6,8\}, \text{ and } C=\{0,1,2,6,9\} $

Let Accordingly, $A \not \subset B$ and $B \not \subset C$.

However, $A \subset C$

(v) False

Let $A=\{3,5,7\}$ and $B=\{3,4,6\}$

Now, $ 5 \in A \text{ and } A \in B$

However, $5 \notin B$

(vi) True

Let $A \subset B$ and $x \notin B$.

To show: $x \notin A$

If possible, suppose $x \notin A$.

Then, $x \in B$, which is a contradiction as $x \notin B$

$\therefore \ \ x \notin A$

Answer :

Let, $A, B$ and $C$ be the sets such that $A \cup B=A \cup C$ and $A \cap B=A \cap C$

To show: $B=C$

Let $x \in B$

$ \begin{matrix} \Rightarrow x \in A \cup B & {[B \subset A \cup B]} \\ \\ \Rightarrow x \in A \cup C & {[A \cup B=A \cup C]} \\ \\ \Rightarrow x \in A \text{ or } x \in C & \end{matrix} $

Case:1

$\qquad x\in A ~ $ Also, $ ~ x \in B$

$ \begin{aligned} & \therefore \ \ x \in A \cap B \\ \\ & \Rightarrow x \in A \cap C \quad[\because A \cap B=A \cap C] \end{aligned} $

$\therefore \ \ x \in A$ and $x \in C$

$ \qquad x \in C$

$\therefore \ \ B \in C $

Similarly, we can show that $ C \in B $

$\therefore \ \ B=C$

Answer :

First, we have to show that $(i) \Leftrightarrow (ii).$

$(i) \Rightarrow (ii).$

Let $\mathrm{A} \subset \mathrm{B}$

To show: $\mathrm{A}-\mathrm{B}=\phi$

If possible, suppose $A-B \neq \phi$

This means that there exists $\mathrm{x} \in \mathrm{A}, \mathrm{x} \notin \mathrm{B}$, which is not possible as $\mathrm{A} \subset \mathrm{B}$.

$ \begin{aligned} & \therefore \ \ \mathrm{A}-\mathrm{B}=\phi \\ \\ & \therefore \ \ \mathrm{A} \subset \mathrm{ ~ B} \Rightarrow \mathrm{ ~ A}-\mathrm{B}=\phi \end{aligned} $

$(ii) \Rightarrow (i).$

Let $\mathrm{A}-\mathrm{B}=\phi$

To show: $\mathrm{A} \subset \mathrm{B}$

Let $x \in A$

Clearly, $x \in B$ because if $x \notin B$, then $A-B \neq \phi$

$\therefore \ \ \mathrm{A}-\mathrm{B}=\phi \Rightarrow \mathrm{A} \subset \mathrm{B}$

Hence, $(ii) \Leftrightarrow (i)$

$(i) \Rightarrow (iii).$

Let $\mathrm{A} \subset \mathrm{B}$

To show: $\mathrm{A} \cup \mathrm{B}=\mathrm{B}$

Clearly, $B \subset A \cup B$

Let $x \in A \cup B$

$\Rightarrow \mathrm{x} \in \mathrm{A}$ or $\mathrm{x} \in \mathrm{B}$

Case I:

$ x \in A $

$ \begin{aligned} & \Rightarrow \mathrm{x} \in \mathrm{ ~ B} \quad[\because \mathrm{ ~ A} \subset \mathrm{ ~ B}] \\ \\ & \therefore \ \ \mathrm{A} \cup \mathrm{ ~ B} \subset \mathrm{ ~ B} \end{aligned} $

Case II:

$\mathrm{x} \in \mathrm{B}$

Then $A \cup B \subset B$

So, $A \cup B=B$

$(iii) \Rightarrow (i).$

Conversely, let $\mathrm{A} \cup \mathrm{B}=\mathrm{B}$

To show: $\mathrm{A} \subset \mathrm{B}$

Let $xe \ A$

$ \begin{aligned} & \Rightarrow x \in A \cup B[\because A \subset A \cup B] \\ \\ & \Rightarrow x \in B \quad [\because A \cup B=B] \\ \\ & \therefore \ \ A \subset B \end{aligned} $

Hence, $(iii) \Leftrightarrow (i)$

Now, we have to show that $(i) \Leftrightarrow (iv).$

Let $\mathrm{A} \subset \mathrm{B}$

Clearly $\mathrm{A} \cap \mathrm{B} \subset \mathrm{A}$

Let $x \in A$

We have to show that $x \in A \cap B$

As $A \subset B, x \in B$

$ \begin{aligned} & \therefore \ \ \mathrm{x} \in \mathrm{ ~ A} \cap \mathrm{ ~ B} \\ \\ & \therefore \ \ \mathrm{ ~ A} \subset \mathrm{ ~ A} \cap \mathrm{ ~ B} \end{aligned} $

Hence, $\mathrm{A}=\mathrm{A} \cap \mathrm{B}$

Conversely, suppose $A \cap B=A$

Let $x \in A$

$ \begin{aligned} & \Rightarrow x \in A \cap B \\ \\ & \Rightarrow x \in A \text { and } x \in B \\ \\ & \Rightarrow x \in B \\ \\ & \therefore \ \ A \subset B \end{aligned} $

Hence, $(i) \Leftrightarrow (iv).$

Answer :

Let $A \subset B$

To show: $(C - B) \subset (C - A)$

Let $x \in (C-B)$

$\Rightarrow x \in C$ and $x \notin B $

$\Rightarrow x \in C$ and $x \notin A[A$ $\subset B]$

$\Rightarrow x \in C-A$

$\therefore \ \ (C-B )\subset (C-A)$

Answer :

To show:

$ \begin{aligned} (A \cap B) \cup(A-B) & =(A \cap B) \cup\left(A \cap B^{\prime}\right) \\ \\ & =A \cap\left(B \cup B^{\prime}\right) \quad\text { (By distributive law) } \\ \\ & =A \cap U=A \end{aligned} $

Hence, $ \ A=(A \cap B) \cup(A-B)$

Also,

$ \begin{aligned} A \cup(B-A)& =A \cup\left(B \cap A^{\prime}\right) \\ \\ & =(A \cup B) \cap\left(A \cup A^{\prime}\right) \text { (By distributive law) } \\ \\ & =(A \cup B) \cap U \\ \\ & =A \cup B \end{aligned} $

Hence, $ \ A \cup(B-A)=A \cup B$.

Answer :

(i) From the distributive property:

$ \begin{aligned} & A \cup(B \cap C)=(A \cup B) \cap(A \cup C) \\ \\ & A \cup(A \cap B)=(A \cup A) \cap(A \cup B)=A \cap(A \cup B)=A \end{aligned} $

Hence,

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

(ii) From the distributive property:

$ \begin{aligned} & A \cap(B \cup C)=(A \cap B) \cup(A \cap C) \\ \\ & A \cap(A \cup B)=(A \cap A) \cup(A \cap B)=A \cup(A \cap B) \end{aligned} $

From part $(i),$

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

Hence,

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

Answer :

$ \begin{aligned} & \text { Let } \mathrm{A}=\{0,1\} \\ \\ & \mathrm{B}=\{\mathrm{0}, 2,3\} \\ \\ & \mathrm{C}=\{0,4,5\} \end{aligned} $

So,

$A \cap B=\{0\} \ \ $ and $ \ \ \mathrm{A} \cap \mathrm{C}=\{\mathrm{0}\}$

Here, $\mathrm{A} \cap \mathrm{B}=\mathrm{A} \cap \mathrm{C}=\{\mathrm{0}\}$

However, $\mathrm{B} \neq \mathrm{C}$ as $2 \in \mathrm{B}$ and $2 \notin \mathrm{C}$

Answer :

Let $\mathrm{A}$ and $\mathrm{B}$ be two sets such that

$\mathrm{A} \cap \mathrm{X}=\mathrm{B} \cap \mathrm{X}=\phi$ and $\mathrm{A} \cup \mathrm{X}=\mathrm{B} \cup \mathrm{X}$ for some set $\mathrm{X}$.

To show: $\mathrm{A}=\mathrm{B}$

It can be seen that

$ \begin{aligned} A & =A \cap(A \cup X) \\ \\ & =A \cap(B \cup X)(A \cup X=B \cup X) \\ \\ & =(A \cap B) \cup(A \cap X) \quad \text { (Distributive law )} \\ \\ & =(A \cap B) \cup \phi \quad(\because A \cap X=\phi) \\ \\ & =A \cap B \qquad . . .(1) \end{aligned} $

Now,

$ \begin{aligned} \mathrm{B} & =\mathrm{B} \cap(\mathrm{B} \cup \mathrm{X}) \\ \\ & =\mathrm{B} \cap(\mathrm{A} \cup \mathrm{X})\qquad(\because \mathrm{A} \cup \mathrm{X}=\mathrm{B} \cup \mathrm{X}) \\ \\ & =(\mathrm{B} \cap \mathrm{A}) \cup(\mathrm{B} \cap \mathrm{X}) \qquad \text { (Distributive law) } \\ \\ & =(\mathrm{B} \cap \mathrm{A}) \cup \phi\qquad (\because \mathrm{B} \cap \mathrm{X}=\phi) \\ \\ & =\mathrm{B} \cap \mathrm{A} \\ \\ & =\mathrm{A} \cap \mathrm{B} \qquad . . .(2) \end{aligned} $

Hence, from $(1)$ and $(2),$ we get, $ \ \mathrm{A}=\mathrm{B} $

Answer :

Let $A=\{0,1\}, B=\{1,2\}$, and $C=\{2, 0\}$

Accordingly,

$A \cap B=\{1\}$

$B \cap C=\{2\}$

$A \cap C=\{0\}$

$\therefore \ \ \mathrm{A} \cap \mathrm{B}, \mathrm{B} \cap \mathrm{C}$, and $\mathrm{A} \cap \mathrm{C}$ are non-empty.

However, $ \ \mathrm{A} \cap \mathrm{B} \cap \mathrm{C}=\phi$