NCERT Exemplar for Class 11 Maths Chapter 1 - Sets (Book Solutions)

Solved Examples

Short Answer Type

1. Write the following sets in roster form.

(i) $A=\{x\mid x$ is a positive integer less than 10 and $2^x-1$ is an odd number $\}$

Given: Sets in set-builder form.

To find: Sets in roster form.

Ans: $2^x-1$ is always an odd positive integer for $x$ as a positive integer. Therefore, the set in roster form is,

$A=\{1,2,3,4,5,6,7,8,9\}$

(ii) $C=\{x:x^2+7x-8=0,x\in R\}.$

Given: Sets in set-builder form.

To find: Sets in roster form.

Ans: Here, by using factorization method

$x^2+7x-8=0$

$\Rightarrow x^2+8x-x-8=0$

$\Rightarrow x(x+8)-1(x+8)=0$

$\Rightarrow (x+8)(x-1)=0$

$\Rightarrow x=-8,1$

$\therefore C=\{-8,1\}$

2. State which of the following statements are true and which are false. Justify your answer.

(i) $37\notin \{x\mid x$ has exactly two positive factors $\}$

Given: Statements in set-builder form.

To find: True or false. 

Ans: There are exactly two factors 1 and 37 . But 37 belongs to a given set Therefore, the statement is false.

(ii) $28\in \{y\mid$ the sum of all positive factor of $y$ is $2y\}$

Given: Statements in set-builder form.

To find: True or false.

Ans: The positive factors of 28 are 1, 2, 4, 7, 14, 28 . The sum of positive factors $1+2+4+7+14+28=56$ or $2(28)$ Therefore, the statement is true.

(iii) $7,747\in \{\mathrm{t}\mid \mathrm{t}$ is a multiple of 37$\}$

Given: Statements in set-builder form.

To find: True or false.

Ans: The statement is false. Since, 7,747 is not multiple of $37.$

3. If $X$ and $Y$ are subsets of the universal set $U$, then show that

(i) $\mathrm{Y}\subset \mathrm{XUY}$

Given: $\mathrm{X}$ and $\mathrm{Y}$ are subsets of the universal set $\mathrm{U}$.

To find: Show that  $\mathrm{Y}\subset \mathrm{X}\cup \mathrm{Y}$

Ans: Here,

$X\cup Y=\{x\mid x\in X\text{ or }x\in Y\}$

$\Rightarrow x\in Y$

$\Rightarrow x\in X\cup Y$

$\text{ Hence, }Y\subset X\cup Y$ .

(ii) $\mathrm{X}\cap \mathrm{Y}\subset \mathrm{X}$

Given: $\mathrm{X}$ and $\mathrm{Y}$ are subsets of the universal set U.

To find: Show that  $\mathrm{X}\cap \mathrm{Y}\subset \mathrm{X}$

Ans: Here,

$X\cap Y=\{x\mid x\in X\text{ and }x\in Y\}$

$\Rightarrow \mathrm{x}\in \mathrm{X}\cap \mathrm{Y}$

$\Rightarrow x\in X$

Hence, $X\cap Y\subset X$.

(iii) $X\subset Y\Rightarrow X\cap Y=X$

Given: $X$ and $Y$ are subsets of the universal set U.

To find: Show that  $X\subset Y\Rightarrow X\cap Y=X$

Ans: Here,

$\text{ As, }x\in X\cap Y$

$\Rightarrow X\in X$

$\text{ Also, }X\cap Y\subset X$

$\text{ since, }X\subset Y$

$\Rightarrow X\in X\text{ and }x\in Y$

$\Rightarrow X\in X\cap Y$

$\therefore X\subset X\cap Y$

$\text{ Hence, }X\subset Y\Rightarrow X\cap Y=X.$

4. Given that $N=\{1,2,3,\dots \dots 100\}$, then

(i) Write the subset $A$ of $N$, whose elements are odd numbers.

Given: $\mathrm{N}=\{1,2,3,\dots \dots 100\}$.

To find: Subsets of $\mathrm{N}$, for given conditions.

Ans: $\mathrm{A}=\{1,3,5,7\dots .99\}$.

(ii) Write the subset $B$ of $N$, whose elements are represented by $x+2$, where $x\in N$. 

Given: $\mathrm{N}=\{1,2,3,\dots \dots 100\}$. To find: Subsets of $\mathrm{N}$, for given conditions.

Ans: Here,

$y=x+2,$

$\text{ for }x=1,y=3$

$x=2,y=4\text{ and so on. }$

$\therefore B=\{3,4,5,6\dots \dots 100\}$.

5. Give that $E=\{2,4,6,8,10\}$. If $n$ represents any number of $E$, then, write the following sets containing all numbers represented by

Given: $\mathrm{E}=\{2,4,6,8,10\}$ and $\mathrm{n}$ represent its elements.

To find: Sets containing all numbers represented by,

(i) $\mathrm{n}+1$

Ans: Here,

$A=\{x\mid x=n+1,n\in E\}$

$\text{ for }n=2,x=3$

$n=4,x=5$

$\therefore A=\{3,5,7,9,11\}$

(ii) ${\mathrm{n}}^2$

Given: $\mathrm{E}=\{2,4,6,8,10\}$ and $\mathrm{n}$ represent its elements.

To find: Sets containing all numbers represented by, ${\mathrm{n}}^2$

 Ans: Here,

$B=\{x\mid x=n^2,n\in E\}$

$\text{ for }n=2,x=4,$

$n=4,x=16\text{ and so on. }$

$B=\{4,16,36,64,100\}$.

6. Let $X=\{1,2,3,4,5,6\}$. If $n$ represents any member of $X$, express the following as sets. 

Given: $X=\{1,2,3,4,5,6\}$ and $n$ represent its elements.

To find: Express sets as,

(i) $\mathrm{n}\in \mathrm{X}$ but $2\mathrm{n}\notin X$

Ans:  Here,

$A=\{x\mid x\in X\text{ and }2x\notin X\}$

$4\in A\text{ and }2\times 4=8\notin X,$

$5\in A\text{ and }2\times 5=10\notin X,$

$6\in A\text{ and }2\times 6=12\notin X.$

$\therefore A=\{4,5,6\}$.

(ii) $\mathrm{n}+5=8$

Given: $X=\{1,2,3,4,5,6\}$ and $n$ represent its elements.

To find: Express sets as,

$\mathrm{n}+5=8$

Ans: Here,

$B=\{x\mid x\in X\text{ and }x+5=8\}$

$\Rightarrow x=3$

$\therefore B=\{3\}$.

(iii) n is greater than $4.$

Given: $X=\{1,2,3,4,5,6\}$ and n represent its elements.

To find: Express sets as  $\mathrm{n}$ are greater than 4 .

Ans: Here,

$C=\{x\mid x\in X,x>4\}$

$\text{ Cleary, }C=\{5,6\}$.

7. Draw the Venn diagrams to illustrate the following relationship among sets $\mathbf{E},\mathbf{M}$ and ${\mathbf{U}}_1$, where $\mathbf{E}$ is the set of students studying English in a school, $M$ is the set of students studying Mathematics in the same school, $U$ is the set of all students in that school.

(i) All the students who study Mathematics study English, but some students who study English do not study Mathematics.

Given: $\mathrm{E},\mathrm{M}$ and $\mathrm{U}$ are sets.

To find: Relation between $\mathrm{E},\mathrm{M}$ and $\mathrm{U}$ using Venn diagram.

Ans: Here, $\mathrm{M}\subset \mathrm{E}\subset \mathrm{U}$.

Therefore, Venn diagram is 

(ii) There is no student who studies both Mathematics and English.

Given: $\mathrm{E},\mathrm{M}$ and $\mathrm{U}$ are sets.

To find: Relation between $\mathrm{E},\mathrm{M}$ and $\mathrm{U}$ using Venn diagram.

Ans: Here, $\mathrm{E}\cap \mathrm{M}=\varphi$.

Therefore, Venn diagram is

(iii) Some of the students study Mathematics but do not study English, some study English but do not study Mathematics, and some study both.

Given: E, M and U are sets.

To find: Relation between $\mathrm{E},\mathrm{M}$ and $\mathrm{U}$ using Venn diagram. 

Ans: As there are some students who study English, Mathematics or both. Therefore, Venn diagram is  

(iv) Not all students study Mathematics, but every student studying English studies Mathematics.

Given: $\mathrm{E},\mathrm{M}$ and $\mathrm{U}$ are sets.

To find: Relation between E, M and U using Venn diagram.

Ans: Here, $\mathrm{E}\subset \mathrm{M}\subset \mathrm{U}$.

Therefore, Venn diagram is 

8. For all sets $A,B$ and $C$. Is( $A\cap B)\cup C=A\cap B\cup C?$ Justify your statement.

Given: $\mathrm{A},\mathrm{B}$ and $\mathrm{C}$ are sets.

To find: Justify $(A\cap B)\cup C=A\cap (B\cup C)$.

Ans:

Let,

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

$B=\{2,3\}$,

$C=\{3,4\}.$

Taking L.H.S,

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

$(A\cap B)\cup C=(\{2\})\cup \{3,4\}$

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

$(A\cap B)\cup C=\{2,3,4\}$

Now,

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

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

$A\cap (B\cup C)=\{2\}$

$\therefore (A\cap B)\cup C\ne A\cap (B\cup C).$

9. Use the properties of sets to prove that for all the sets $A$ and $B$, $A-(A\cap B)=A-B$. 

Given: $A$ and $B$ are sets.

To find: Prove $A-(A\cap B)=A-B$.

 Ans:

Here

$A-(A\cap B)=A\cap (A\cap B{)}^{\prime }\{As,(A-B)=(A\cap B^{\prime })\}$

$\Rightarrow A-(A\cap B)=A\cap (A^{\prime }\cup B^{\prime })\{As,(A\cap B{)}^{\prime }=(A^{\prime }\cup B^{\prime })\}$

$\Rightarrow A-(A\cap B)=(A\cap A^{\prime })\cup (A\cap B^{\prime })$

$\Rightarrow A-(A\cap B)=\varphi \cup (A\cap B^{\prime })$

$\Rightarrow A-(A\cap B)=A\cap B^{\prime }$

$\Rightarrow A-(A\cap B)=A-B$.

Long Answer Type

10. For all sets $A,B$ and $C$ . Is $(A-B)\cap (C-B)=(A\cap C)-B$ ? Justify your answer.

Given: $\mathrm{A},\mathrm{B}$ and $\mathrm{C}$ are sets.

To find: Justify $(A-B)\cap (C-B)=(A\cap C)-B$. 

Ans:

Let $x\in (A-B)\cap (C-B)$

$\Rightarrow \mathrm{x}\in (\mathrm{A}-\mathrm{B})$ and $\mathrm{x}\in (\mathrm{C}-\mathrm{B})$

$\Rightarrow (\mathrm{x}\in \mathrm{A}$ and $\mathrm{x}\notin \mathrm{B})$ and $(\mathrm{x}\in \mathrm{C}$ and $\mathrm{x}\notin \mathrm{B})$

$\Rightarrow (x\in A$ and $x\in C)$ and $x\notin B\{$ using distributive law $\}$

$\Rightarrow x\in (A\cap C)-B$

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

Now,

Let $y\in (A\cap C)-B$

$\Rightarrow (\mathrm{y}\in \mathrm{A}$ and $\mathrm{y}\in \mathrm{C})$ and $\mathrm{y}\notin \mathrm{B}$

$\Rightarrow (\mathrm{y}\in \mathrm{A}$ and $\mathrm{y}\notin \mathrm{B})$ and $(\mathrm{y}\in \mathrm{C}$ and $\mathrm{y}\notin \mathrm{B})\{$ using distributive law $\}$

$\Rightarrow y\in (A-B)$ and $y\in (C-B)$

$\Rightarrow y\in (A-B)\cap (C-B)$

$\Rightarrow (A\cap C)-B\subset (A-B)\cap (C-B)$-(ii)

from (i) and (ii), we get

$(A-B)\cap (C-B)=(A\cap C)-B.$

11. Let $A,B$ and $C$. Then show that $A\cup (B\cap C)=(A\cup B)\cap (A\cup C)$. 

Given: $\mathrm{A},\mathrm{B}$ and $\mathrm{C}$ are sets.

To find: Show that, $A\cup (B\cap C)=(A\cup B)\cap (A\cup C)$.

Ans: 

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

$\Rightarrow x\in A\text{ or }x\in B\cap C$

$\Rightarrow x\in A\text{ or }(x\in B\text{ and }x\in C)$

$\Rightarrow (x\in A\text{ or }x\in B)\text{ and }(x\in A\text{ or }x\in C)$

$\Rightarrow (x\in A\cup B)\text{ and }(x\in A\cup C)$

$\Rightarrow x\in (A\cup B)\cap (A\cup C)$

$\therefore A\cup (B\cap C)\subset (A\cup B)\cap (A\cup C)$---------(i)

Now,

Let $y\in (A\cup B)\cap (A\cup C)$

$y\in (A\cup B)\text{ and }(A\cup C)$

$\Rightarrow (y\in A\text{ or }y\in B)\text{ and }(y\in A\text{ or }y\in C)$

$\Rightarrow y\in A\text{ or }(y\in B\text{ and }y\in C)$

$\Rightarrow y\in A\text{ or }(y\in B\cap C)$

$\Rightarrow x\in A\cup (B\cap C)$

Therefore, $(A\cup B)\cap (A\cup C)\subset A\cup (B\cap C)$ ----------(ii)

From (i) and (ii), we get

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

12. Let $P$ be the set of prime numbers and let $S=\{t\mid 2^t-1$ is a prime $\}$. Prove that $S\subset P$.

Given:

P set of prime numbers,

$S=\{t\mid 2^t-1$ is a prime $\}.$

To find: Prove that $S\subset P$.

Ans:

$P=\{2,3,5,7,11,13,17,19\dots .\}$

$S=\{2,3,5,7\dots \dots .\}.$

Clearly, every element of $\mathrm{S}$ is also the element in $\mathrm{P}$.

Therefore, $\mathrm{S}\subset \mathrm{P}$.

Hence proved.

13. From 50 students taking examinations in Mathematics, Physics and Chemistry, each of the students has passed in at least one of the subjects, 37 passed Mathematics, 24 Physics and 43 Chemistry. At most 19 passed Mathematics and Physics, at most 29 Mathematics and Chemistry and at most 20 Physics and Chemistry. What is the largest possible number that could have passed all three examinations?

Ans: Let M be set of students passed in Mathematics,

$P$ be set of students passed in Physics,

C be a set of students passed in Chemistry.

Substitute values in the relation,

$n(M\cup P\cup C)=n(M)+n(P)+n(C)-n(M\cap P)-n(M\cap C)-n(P\cap C)+n(M\cap P\cap C)$

$37+24+43-19-29-20+n(M\cap P\cap C)\le 50$

$n(M\cap P\cap C)\le 50-36$

$\therefore \mathrm{n}(\mathrm{M}\cap \mathrm{P}\cap \mathrm{C})\le 14$

Objective Type Questions

Choose the correct answer from the given four options in each of the Examples 14 to 16 (M.C.Q.)

14. Each set $X_r$ contains 5 elements and each set $Y_r$ contains 2 elements and $\bigcup _{r=1}^{20}{X}_r=S=\bigcup _{r=1}^n{Y}_r.$ If each element of $S$ belongs to exactly 10 of the $X_r^{\prime }s$ and exactly 4 of $Y_r^{\prime }s$. Then, $n$ is

(A) 10

(B) 20

(C) 100

(D) 50

Ans: The Correct Answer is option B.

Given:

$X_r$ contains 5 elements,

${\mathrm{Y}}_{\mathrm{r}}$ contains 2 elements and,

$\bigcup _{r=1}^{20}{X}_r=S=\bigcup _{r=1}^n{Y}_r.$

To find: $\mathrm{n}=$ ? 

If elements are not repeated, then number of elements in

${\mathrm{X}}_1\cup {\mathrm{X}}_2\cup {\mathrm{X}}_3\dots \dots \cup {\mathrm{X}}_{20}$ is $20\times 5$. But each element is used 10 times, so

$\mathrm{S}={\displaystyle \frac{20\times 5}{10}}$

S=10

If each element in ${\mathrm{Y}}_1,{\mathrm{Y}}_2,{\mathrm{Y}}_3\dots \dots {\mathrm{Y}}_{\mathrm{n}}$ are not repeated then total number of element in $2\mathrm{n}$ but each element is repeated 4 times. So,

$\Rightarrow \mathrm{S}={\displaystyle \frac{2\mathrm{n}}{4}}$

$\Rightarrow 10={\displaystyle \frac{2\mathrm{n}}{4}}$

$\Rightarrow \mathrm{n}=20$

15. Two finite sets have $m$ and $n$ respectively. The total number of subsets of the first set is 56 more than the total number of subsets of the second set. The value of $\mathrm{m}$ and $\mathrm{n}$ respectively are:

(A) 7,6

(B) 5,1

(C) 6,3

(D) 8,7 .

Ans: The Correct Answer is option C.

Given: Two finite sets with $m$ and $n$ elements.

To find: The values of $m$ and $n$ are

$\mathrm{m}$=?

$\mathrm{n}$=?

Since, the number of subsets of a containing $m$ elements is 56 more than the subsets of the set containing $\mathrm{n}$ elements. 

$\Rightarrow 2^{\mathrm{m}}-2^{\mathrm{n}}=56$

$\Rightarrow 2^{\mathrm{n}}(2^{\mathrm{m}-\mathrm{n}}-1)=2^3\cdot 7$

$\Rightarrow 2^{\mathrm{n}}=2^3\text{ and }{2}^{\mathrm{m}-\mathrm{n}}-1=7$

$\Rightarrow \mathrm{n}=3\text{ and }{2}^{\mathrm{m}-\mathrm{n}}=8$

$\Rightarrow \mathrm{n}=3\text{ and }{2}^{\mathrm{m}-\mathrm{n}}=2^3$

$\Rightarrow \mathrm{m}-\mathrm{n}=3\text{ and }\mathrm{n}=3$

$\Rightarrow \mathrm{m}=6\text{ and }\mathrm{n}=3$

16. The set $(A\cup B\cup C)\cap {(A\cap B^{\prime }\cap C^{\prime })}^{\prime }\cap C^{\prime }$ is equal to

(A) $B\cap C^{\prime }$

(B) $A\cap C$

(C) $B\cup C^{\prime }$

(D) $A\cap C$

 Ans: The correct option is A.

Given: A set.

To find: $(A\cup B\cup C)\cap {(A\cap B^{\prime }\cap C^{\prime })}^{\prime }\cap C^{\prime }=?$

The set operations are as follows,

$=(A\cup (B\cup C))\cap (A^{\prime }\cup (B\cup C))\cap C^{\prime }$

$=(A\cap A^{\prime })\cup (B\cup C)\cap C^{\prime }$

$=\varphi \cup (B\cup C)\cap C^{\prime }$

$=B\cap C^{\prime }\cup \varphi$

$=B\cap C^{\prime }$

Fill in the blanks 17 to $18.$

17. If $\mathbf{A}$ and $\mathbf{B}$ are finite sets, then $\mathbf{n}(\mathbf{A})+\mathbf{n}(\mathbf{B})$ is equal to

Ans: Given: $A$ and $B$ are finite sets.

To find: $n(A)+n(B)=?$

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$\therefore n(A)+n(B)=n(A\cup B)+n(A\cap B)$.

18. If $\mathbf{A}$ is a finite set containing $\mathrm{n}$ element, then the number of subsets of $\mathbf{A}$ is ____________ .

Ans: Given: Set $\mathrm{A}$ with $\mathrm{n}$ elements.

To find: Number of subsets of $\mathrm{A}$.

Let,

$\mathrm{A}=\{1\}$,

Subsets $=\{\varphi \},\{1\}$

number of subsets $=2^1$.

$\mathrm{A}=\{0,1\}$

Subsets $=\{\varphi \},\{0\},\{1\},\{0,1\}$

number of subsets $=2^2$.

$\therefore$ for $\mathrm{n}$ elements, number of subsets $=2^{\mathrm{n}}$.

State True or False 19 and $20.$

19. Let $R$ and $S$ be the sets defined as follows,

$\mathrm{R}=\{\mathrm{x}\in \mathrm{Z}\mid \mathrm{x}$ is divisible by 2$\}$,

$\mathrm{S}=\{\mathrm{y}\in \mathrm{Z}\mid \mathrm{y}$ is divisible by 3$\}.$

then, $R\cap S=\varphi$.

Ans: Given:

$\mathrm{R}=\{\mathrm{x}\in \mathrm{Z}\mid \mathrm{x}$ is divisible by 2$\}$,

$\mathrm{S}=\{\mathrm{y}\in \mathrm{Z}\mid \mathrm{y}$ is divisible by 3$\}.$

To find: True or false.

$\mathrm{R}\cap \mathrm{S}\ne \varphi$.

Since, $\{6,12,18\dots .\}$ are divisible by both 2 and $3.\mathrm{R}\cap \mathrm{S}$ can never be an empty set. Therefore, false. 

20. $Q\cap R=Q$, where $Q$ is the set of rational numbers and $R$ is the set of real numbers.

Ans: Given:

Q set of rational numbers,

R set of real numbers.

To find: True or false.

Since, every element of set $Q$ is also the element of set $R$. Every rational number is also a real number.

$\therefore Q\subset R$

$\Rightarrow \mathrm{Q}\cap \mathrm{R}=\mathrm{Q}$ .

Therefore, true.

EXERCISE 1.3

1. Write the following sets in roster form:

(i) $A=\{x:x\in R,2x+11=15\}.$

Ans: Given: Sets are given in set-builder form

$A=\{x:x\in R,2x+11=15\}$

To find: The following sets in roster form.

$A=\{x:x\in R,2x+11=15\}$

$2x+11=15$

$\Rightarrow 2x=4$

$x=2$

Therefore, the set in roster form $\mathrm{A}=\{2\}$.

(ii) $B=\{x\mid x^2=x,x\in R\}.$

Ans: Given: Set is given in set-builder form

To find: The following sets in roster form.

Given, $B=\{x\mid x^2=x,x\in R\}$

$\Rightarrow x^2=x$

$\Rightarrow x^2-x=0$

$\Rightarrow x(x-1)=0,$

$\Rightarrow x=0,1$ .

Therefore, the set in roster form $\mathrm{B}=\{0,1\}$.

(iii) $C=\{x\mid x$ is a positive factor of a prime number $p\}$

Ans: Given: Set is given in set-builder form

To find: The following sets in roster form.

Given, $C=\{x\mid x$ is a positive factor of a prime number $p\}.$

The positive factor of any prime number $p$ is 1 and the number itself that is $p$.

Therefore, the set in roster form $C=\{1,p\}$.

2. Write the following sets in the roaster form:

(i) $D=\{t\mid t^3=t,t\in R\}$

Ans: Given: Set is in set builder form,

$D=\{t\mid t^3=t,t\in R\}$,

To find: The following sets in roster form.

Given, $D=\{t\mid t^3=t,t\in R\},$

$\Rightarrow t^3=t$

$\Rightarrow t^3-t=0$

$\Rightarrow t(t^2-1)=0$

$\Rightarrow t=-1,0,1$ .

Therefore, the set in roster form $D=\{-1,0,1\}$.

(ii) $E=\{w\mid {\displaystyle \frac{w-2}{w+3}}=3,w\in R\}$  

Ans: Given,

$E=\{w\mid {\displaystyle \frac{w-2}{w+3}}=3,w\in R\},$

$\Rightarrow {\displaystyle \frac{w-2}{w+3}}=3$

$\Rightarrow w-2=3w+9$

$\Rightarrow 2w=-11$

$\Rightarrow w=-{\displaystyle \frac{11}{2}}$ .

Therefore, the set in roster form $\mathrm{E}=\{-{\displaystyle \frac{11}{2}}\}$.

(iii) $F=\{x\mid x^4-5x^2+6=0,x\in R\}$

Given set is in set builder form,

$F=\{x\mid x^4-5x^2+6=0,x\in R\}$ .

To find: The following sets in roster form.

Ans: Given,

$F=\{x\mid x^4-5x^2+6=0,x\in R\}$

$\Rightarrow x^4-5x^2+6=0$

$\Rightarrow x^4-3x^2-2x^2+6=0$

$\Rightarrow x^2(x^2-3)-2(x^2-3)=0$

$\Rightarrow (x^2-3)(x^2-2)=0$

$\Rightarrow x=\pm \sqrt{3},\pm \sqrt{2}$

Therefore, the set in roster form $\mathrm{F}=\{-\sqrt{3},-\sqrt{2},\sqrt{2},\sqrt{3}\}$.

3. If $Y=\{x\mid x\text{ is a positive factor of the number }{2}^{p-1}(2^p-1),\text{where }{2}^p-1\text{ is a prime number }\}$. Write $\mathrm{Y}$ in the roaster form.

Given: A set

$\mathrm{Y}=\{\mathrm{x}\mid \mathrm{x}$ is a positive factor of the number $2^{\mathrm{p}-1}(2^{\mathrm{p}}-1)$, where $(2^{\mathrm{p}}-1)$ is a prime number $\}$ in set builder form.

To find: The given set in roster form.

Ans: The factor of $2^{p-1}$ are $1,2,2^2,2^3\dots \dots {.2}^{p-1}.$ since, $(2^p-1)$ is the prime number with factors 1 and $(2^p-1)$.

Therefore, the set in roster form $Y=\{1,2,2^2,2^3\dots \dots .{.2}^{p-1},2^p-1\}$ 

4. State which of the following statements are true and which are false. Justify your answer.

(i) $35\in \{x\mid x$ has exactly four positive factors $\}$.

Ans: Given,

$35\in \{x\mid x$ has exactly four positive factors $\}$

The factors of 35 are 1, 5,7 and 35 .

Hence, the statement is true.

(ii) $128\in \{y\mid$ the sum of all the positive factors of $y$ is $2y\}$.

Ans: Given,

$128\in \{y\mid$ the sum of all the positive factors of $y$ is $2y\}$. The factors of 128 are $1,2,4,8,16,32,64,128$.

The sum of the factors is $1+2+4+8+16+32+64+128=255$ which is not equals to $2y=2\times 128=256$.

Therefore, the statement is false.

(iii) $3\notin \{x\mid x^4-5x^3+2x^2-112x+6=0\}.$

Ans: Given,

$3\notin \{x\mid x^4-5x^3+2x^2-112x+6=0\}.$

Substitute $x=3$, in $x^4-5x^3+2x^2-112x+6=0$. At $x=3$,

$(3{)}^4-5(3{)}^3+2(3{)}^2-112(3)+6=0$

$\Rightarrow -346=0$

Therefore, the statement is true.

(iv) $496\notin \{y\mid$ the sum of all the positive factors of y is $2y\}$.

Ans: Given, $496\notin \{y\mid$ the sum of all the positive factors of y is $2y\}$.

The factors of 496 are 1, 2, 4, 8, 16, 31, 62, 124, 248, 496 .

The sum of the factors is $1+2+4+8+16+31+62+124+248+496=992$ which is equal to $2y=2\times 496=992$.

Therefore, the statement is false.

5. Given, $\mathrm{L}=\{1,2,3,4\},\mathrm{M}=\{3,4,5,6\}$ and $\mathrm{N}=\{1,3,5\}$. Verify that $L-(M\cup N)=(L-M)\cap (L-N)$

Given: Three sets are given as,

$\mathrm{L}=\{1,2,3,4\},\mathrm{M}=\{3,4,5,6\}$ and $\mathrm{N}=\{1,3,5\}.$

To find: obtain $\mathrm{L}-(\mathrm{M}\cup \mathrm{N})$ and $(\mathrm{L}-\mathrm{M})\cap (\mathrm{L}-\mathrm{N})$ then verify them.

Ans: Calculate: L - $(M\cup N)$

$\Rightarrow \mathrm{L}-(\mathrm{M}\cup \mathrm{N})=\{1,2,3,4\}-\{3,4,5,6\}\cup \{1,3,5\}$

$\Rightarrow \mathrm{L}-(\mathrm{M}\cup \mathrm{N})=\{1,2,3,4\}-\{1,3,4,5,6\}$

$\Rightarrow \mathrm{L}-(\mathrm{M}\cup \mathrm{N})=\{2\}$ .

Calculate: $(\mathrm{L}-\mathrm{M})\cap (\mathrm{L}-\mathrm{N})$

$(\mathrm{L}-\mathrm{M})\cap (\mathrm{L}-\mathrm{N})=(\{1,2,3,4\}-\{3,4,5,6\})\cap (\{1,2,3,4\}-\{1,3,5\})$

$\Rightarrow (\mathrm{L}-\mathrm{M})\cap (\mathrm{L}-\mathrm{N})=\{1,2\}\cap \{2,4\}=\{2\}$.

Now, $L-(M\cup N)$ and $(L-M)\cap (L-N)$ Both sets are equal.

Verified $\mathrm{L}-(\mathrm{M}\cup \mathrm{N})=(\mathrm{L}-\mathrm{M})\cap (\mathrm{L}-\mathrm{N})$.

6. If $A$ and $B$ are subsets of the universal set $U$, then show that

(i) $A\subset A\cup B$ 

Given: The universal set $U,A$ and $B$ are subsets of the universal set $U$ which can be expressed as $\mathrm{A}\subset \mathrm{U}$ and $\mathrm{B}\subset \mathrm{U}.$

To find:

Show that: $A\subset A\cup B$

Ans: To show that $A\subset A\cup B$, .

Let us consider that $x\in A$..

If $\mathrm{x}\in \mathrm{A}$ then $\mathrm{x}$ is the element in set $\mathrm{A}$..

Now, $x\in (A\cup B)$ as $x\in A$ or $x\in B$.

This implies that $A\subset A\cup B$.

(ii) $A\subset B\iff A\cup B=B$

Given: The universal set $\mathrm{U},\mathrm{A}$ and $\mathrm{B}$ are subsets of the universal set $\mathrm{U}$ which can be expressed as $\mathrm{A}\subset \mathrm{U}$ and $\mathrm{B}\subset \mathrm{U}..$

To find:

Show that: $A\subset B\iff A\cup B=B$

Ans: It is given that $A\subset B$. So consider that $x\in (A\cup B)$.

From this condition either $\mathrm{x}\in \mathrm{A}$ or $\mathrm{x}\in \mathrm{B}$.

As $A\subset B$ this directly implies that $x\in B$.

From the above conclusions,

$A\cup B\subset B--------(1)$ and $B\subset A\cup B---------(2)$.

From -(1) and -(2) $A\cup B=B$.

If $A\cup B=B$ then consider $y\in A$, which implies that $y\in (A\cup B)$.

$y\in (A\cup B)$

$\Rightarrow y\in B$

$\Rightarrow A\subset B$

$\text{ Hence, }A\subset B\iff A\cup B=B$ .

(iii) $(A\cap B)\subset A$

Given: The universal set U, A and B are subsets of the universal set U which can be expressed as $\mathrm{A}\subset \mathrm{U}$ and $\mathrm{B}\subset \mathrm{U}..$

To find:

Show that: $(A\cap B)\subset A$

Ans: Let us consider $x\in (A\cap B)$,

$\Rightarrow \mathrm{X}\in \mathrm{A}$ and $\mathrm{x}\in \mathrm{B}$

$\Rightarrow \mathrm{X}\in \mathrm{A}$

Hence, $(A\cap B)\subset A$.

7. Given that $\mathbf{N}=\{1,2,3\dots \dots \dots 100\}$. Then write

(i) The subset of $\mathrm{N}$ whose elements are even numbers.

Given: The set in Roster form $\mathrm{N}=\{1,2,3\dots \dots \dots .100\}$.

To find:

The subset of $\mathrm{N}$ whose elements are even numbers

Ans: The subset of $\mathrm{N}$ whose elements are even numbers is $\{2,4,6\dots \dots \dots 100\}$. This set is the subset of $\mathrm{N}=\{1,2,3\dots \dots \dots 100\}$.

(ii) The subset of $\mathrm{N}$ whose elements are perfect square numbers

Given: The set in Roster form $\mathrm{N}=\{1,2,3\dots \dots \dots .100\}$.

To find:

The subset of $\mathrm{N}$ whose elements are perfect square numbers.

Ans: The set of perfect square numbers is $\{1,4,9,16,25,36,49,64,81,100\}$ and this set is subset of $\mathrm{N}=\{1,2,3\dots \dots \dots ..100\}$.

8. If $X=\{1,2,3\}$, if $n$ represents any member of $X$, write the following sets containing all numbers represented by 

(i). $4n$

Given: A set $X=\{1,2,3\}$ and $\mathrm{n}$ represents any member of $X$.

To find: The set which contains the numbers represented by the given relation.

Ans: Substituting each and every member $X=\{1,2,3\}$ in the relation $4\mathrm{n}$. We get

$\{4n\mid n\in X\}=\{4\times 1,4\times 2,4\times 3\}=\{4,8,12\}.$

(ii) $\mathbf{n}+\mathbf{6}$

Given: A set $X=\{1,2,3\}$ and $n$ represents any member of $X$.

To find: The set which contains the numbers represented by the given relation.

 Ans: Substituting each and every member of $X=\{1,2,3\}$ in the relation $n+6$. We get,

$\{\mathrm{n}+6\mid \mathrm{n}\in \mathrm{X}\}=\{1+6,2+6,3+6\}=\{7,8,9\}.$

(iii) ${\displaystyle \frac{\mathrm{n}}{2}}$

Given: A set $X=\{1,2,3\}$ and $\mathrm{n}$ represents any member of $X$.

To find: The set which contains the numbers represented by the given relation.

Ans: Substituting each and every member of $X=\{1,2,3\}$ in the relation ${\displaystyle \frac{n}{2}}$. We get,

$\{{\displaystyle \frac{n}{2}}\mid n\in X\}=\{{\displaystyle \frac{1}{2}},{\displaystyle \frac{2}{2}},{\displaystyle \frac{3}{2}}\}=\{{\displaystyle \frac{1}{2}},1,{\displaystyle \frac{3}{2}}\}.$

(iv) $\mathrm{n}-1$

Given: A set $X=\{1,2,3\}$ and $n$ represents any member of $X$.

To find: The set which contains the numbers represented by the given relation.

Ans: Substituting each and every member $X=\{1,2,3\}$ of in the relation $n-1$. We get,

$\{\mathrm{n}-1\mid \mathrm{n}\in \mathrm{X}\}=\{1-1,2-1,3-1\}=\{0,1,2\}$.

9. If $\mathbf{Y}=\{1,2,3,\dots ..10\}$, and a represents any element of $\mathbf{Y}$, write the following sets, containing all the elements satisfying the given conditions.

(i) $a\in Y$ but $a^2\notin Y$

Given: A set $\mathrm{Y}=\{1,2,3,\dots ..10\}$ and a represents any element of $\mathrm{Y}$.

To find: The set which contains the numbers represented by the given relation.

Ans: Here $1^2,2^2,3^2\in \mathrm{Y}$. Therefore, the set $\{\mathrm{a}\mid \mathrm{a}\in \mathrm{Y}$ but ${\mathrm{a}}^2\notin \mathrm{Y}\}=\{4,5,6,7,8,10\}$.

(ii) $a+1=6,a\in Y$