Answer :

(i): $X={1,3,5} \quad Y={1,2,3}$

$X \cup Y={1,2,3,5}$

(ii): $A={a, e, i, o, u} \quad B={a, b, c}$

$\quad A \cup B={a, b, c, e, i, o, u}$

(iii): $ A={x: x}$ is a natural number and multiple of $3={3,6,9 \ldots }$

$ \qquad B = {x: x$ is a natural number less than $6 }={1,2,3,4,5,6}$

$A \cup B={1,2,4,5,3,6,9,12 \ldots}$

$\therefore \ \ A \cup B={x: x=1,2,4,5}$ or a multiple of $3$

(iv): $A={x: x$ is a natural number and $1<x \leq 6}={2,3,4,5,6}$

$B={x: x$ is a natural number and $6<x<10}={7,8,9}$

$A \cup B={2,3,4,5,6,7,8,9}$

$\therefore \ \ A \cup B={x: x \in N$ and $1<x<10}$

(v): $A={1,2,3}, B=\Phi$

$A \cup B={1,2,3}$

Answer:

Here, $A={a, b}$ and $B={a, b, c}$

Yes, $A \subset B$

$A \cup B={a, b, c}=B$

Answer :

If $A$ and $B$ are two sets such that $A \subset B$, then $A \cup B=B$.

Answer :

$A={1,2,3,4}, B={3,4,5,6}, C={5,6,7,8}$ and $D={7,8,9,10}$

(i): $A \cup B={1,2,3,4,5,6}$

(ii): $A \cup C={1,2,3,4,5,6,7,8}$

(iii): $B \cup C={3,4,5,6,7,8}$

(iv): $B \cup D={3,4,5,6,7,8,9,10}$

(v): $A \cup B \cup C={1,2,3,4,5,6,7,8}$

(vi): $A \cup B \cup D={1,2,3,4,5,6,7,8,9,10}$

(vii): $B \cup C \cup D ={3,4,5,6,7,8,9,10}$

Answer :

(i): $X={1,3,5}, Y={1,2,3}$

$\quad X \cap Y={1,3}$

(ii): $A={a, e, i, o, u}, B={a, b, c}$

$\quad \ A \cap B={a}$

(iii): $ A= {x : x \text { is a natural number and multiple of 3} } ={3,6,9 }$

$\quad \ B={x : x \text{ is a natural number less than} \ 6 } ={1,2,3,4,5}$

$\therefore \ \ A \cap B={3}$

(iv): $ \ A={x$ $: x$ is a natural number and $1<x < 6 }={2,3,4,5,6}$

$\quad \ \ B={x$ $ : x $ is a natural number and $6<x<10}={7,8,9}$

$\therefore \ A \cap B=\Phi$

(v): $ \ A={1,2,3}, B=\Phi$

$\quad \ \ A \cap B=\Phi$

Answer :

$ A={3,5,7,9,11}, B={7,9,11,13}, C={11,13,15} \text { and } D={15,17} $

(i): $ \ \mathrm{A} \cap \mathrm{B}={7,9,11}$

(ii): $ \ \mathrm{B} \cap \mathrm{C}={11,13}$

(iii): $ \ \mathrm{A} \cap \mathrm{C} \cap \mathrm{D}=(\mathrm{A} \cap \mathrm{C}) \cap \mathrm{D}={11} \cap{15,17}=\phi$

(iv): $ \ \mathrm{A} \cap \mathrm{C}={11}$

(v): $ \ \mathrm{B} \cap \mathrm{D}=\phi$

(vi): $ \ A \cap(B \cup C)=(A \cap B) \cup(A \cap C)$ $ ={7,9,11} \cup{11}={7,9,11} $

(vii): $ \ \mathrm{A} \cap \mathrm{D}=\phi$

(viii): $ \ A \cap(B \cup D)=(A \cap B) \cup(A \cap D) ={7,9,11} \cup \phi={7,9,11} $

(ix): $ \ (A \cap B) \cap(B \cup C)={7,9,11} \cap{7,9,11,13,15}={7,9,11} $

(x): $ \ (A \cup D) \cap(B \cup C)={3,5,7,9,11,15,17} \cap{7,9,11,13,15} ={7,9,11,15}$

Answer :

$ A={x: x \text { xis a natural number }}={1,2,3,4,5 \ldots \ldots} $

$ B={x: x \text { is an even natural number }}={2,4,6,8 \ldots \ldots} $

$ C={x: x \text { is an odd natural number }}={1,3,5,7,9 \ldots \ldots} $

$ D={x: x \text { xis a prime number }}={2,3,5,7 \ldots}$

(i): $\mathrm{A} \cap \mathrm{B}={\mathrm{x}: \mathrm{x}$ is a even natural number $}=\mathrm{B}$

(ii): $\mathrm{A} \cap \mathrm{C}={\mathrm{x}: \mathrm{x}$ is an odd natural number $}=\mathrm{C}$

(iii): $\mathrm{A} \cap \mathrm{D}={\mathrm{x}: \mathrm{x}$ is a prime number $}=\mathrm{D}$

(iv): $\mathrm{B} \cap \mathrm{C}=\phi$

(v): $\mathrm{B} \cap \mathrm{D}={2}$

(vi): $\mathrm{C} \cap \mathrm{D}={\mathrm{x}: x $ is odd prime number $}$

Answer :

(i): ${1,2,3,4}$

${x: x$ is a natural number and $4 \leq x \leq 6 }={4,5,6}$

Now, ${1,2,3,4} \cap{4,5,6}={4}$

Therefore, this pair of sets is not disjoint.

(ii): ${a, e, i, o, u} \cap{c, d, e, f}={e}$

Therefore, ${a, e, i, o, u}$ and ${c, d, e, f}$ are not disjoint.

(iii): ${x: x$ is an even integer $} \cap{x: x$ is an odd integer $}=$ $\Phi$

Therefore, this pair of sets is disjoint.

Answer :

(i): $A-B={3,6,9,15,18,21}$

(ii): $A-C={3,9,15,18,21}$

(iii): $A-D={3,6,9,12,18,21}$

(iv): $B-A={4,8,16,20}$

(v): $C-A={2,4,8,10,14,16}$

(vi): $D-A={5,10,20}$

(vii): $B-C={20}$

(viii): $B - D ={4,8,12,16}$

(ix): $C-B={2,6,10,14}$

(x): $D-B={5,10,15}$

(xi): $C- D ={2,4,6,8,12,14,16}$

(xii): $D-C={5,15,20}$

Answer :

(i): $X-Y={a, c}$

(ii): $Y-X={f, g}$

(iii): $X \cap Y=$ ${b, d}$

Answer :

$R$ : set of real numbers

Q: set of rational numbers

Therefore, $R-Q$ is a set of irrational numbers.

Answer :

(i): False

As $3 \in{2,3,4,5}, 3 \in{3,6}$

$\Rightarrow{2,3,4,5} \cap{3,6}={3}$

(ii): False

As $a \in{a, e, i, o, u}, a \in{a, b, c, d}$

$\Rightarrow{a, e, i, o, u} \cap{a, b, c, d}={a}$

(iii): True

As ${2,6,10,14} \cap{3,7,11,15}=\Phi$

(iv): True

As ${2,6,10} \cap{3,7,11}=\Phi$