Given: The set A=1,2,3,4,5,6. The symbol in means "is an element of" and notin means "is not an element of". (i) 5 is an element of set A. So, 5 in A. (ii) 8 is not an element of set A. So, 8 notin A. (iii) 0 is not an element of set A. So, 0 notin A. (iv) 4 is an element of set A. So, 4 in A. (v)…

Q3 (i): A=x: x is an integer and -3 ≤ x<7

In roster form, we list all the elements of a set, separated by commas, within braces. (i) A=x: x is an integer and -3 ≤ x<7 The integers satisfying the condition are -3, -2, -1, 0, 1, 2, 3, 4, 5, 6. So, A = -3, -2, -1, 0, 1, 2, 3, 4, 5, 6. (ii) B=x: x is a natural number less than 6 The natural…

Q3 (ii): B=x: x is a natural number less than 6 (iii) C=x: x is a two-digit natural number such that the sum of its digits is 8

In roster form, we list all the elements of a set, separated by commas, within braces. (i) A=x: x is an integer and -3 ≤ x<7 The integers satisfying the condition are -3, -2, -1, 0, 1, 2, 3, 4, 5, 6. So, A = -3, -2, -1, 0, 1, 2, 3, 4, 5, 6. (ii) B=x: x is a natural number less than 6 The natural…

Q3 (iv): D=x: x is a prime number which is divisor of 60 (v) E= The set of all letters in the word TRIGONOMETRY (vi) F=$ The set of all letters in the word…

In roster form, we list all the elements of a set, separated by commas, within braces. (i) A=x: x is an integer and -3 ≤ x<7 The integers satisfying the condition are -3, -2, -1, 0, 1, 2, 3, 4, 5, 6. So, A = -3, -2, -1, 0, 1, 2, 3, 4, 5, 6. (ii) B=x: x is a natural number less than 6 The natural…

Sets Class 11 Mathematics NCERT Solutions

Q1EXERCISE 1.1

Which of the following are sets? Justify your answer.

(i)

The collection of all the months of a year beginning with the letter J.

(ii)

The collection of ten most talented writers of India.

(iii)

A team of eleven best-cricket batsmen of the world.

(iv)

The collection of all boys in your class.

(v)

The collection of all natural numbers less than 100.

(vi)

A collection of novels written by the writer Munshi Prem Chand.

(vii)

The collection of all even integers.

(viii)

The collection of questions in this Chapter.

(ix)

A collection of most dangerous animals of the world.

Solution

A collection of objects is a set if it is well-defined, meaning we can definitively determine whether a given object belongs to the collection or not.

(i)

This is a set. The months of a year beginning with the letter J are January, June, and July. This is a well-defined collection.

(ii)

This is not a set. The criterion for determining a writer's talent can vary from person to person. Therefore, the collection is not well-defined.

(iii)

This is not a set. The term "best-cricket batsmen" is subjective and the criteria for selection can vary. Therefore, the collection is not well-defined.

(iv)

This is a set. The collection of all boys in a specific class is well-defined. One can definitively identify whether a student is a boy in that class.

(v)

This is a set. The natural numbers less than 100 are

1, 2, 3, \dots, 99

. This is a well-defined collection.

(vi)

This is a set. The collection of novels written by Munshi Prem Chand is well-defined. We can definitively determine if a novel was written by him.

(vii)

This is a set. The collection of all even integers is well-defined. An integer is either even or not.

(viii)

This is a set. The collection of questions in this chapter is well-defined and can be clearly identified.

(ix)

This is not a set. The criterion for what constitutes a "most dangerous" animal is subjective and can vary based on different factors (venom, strength, aggressiveness, etc.). Therefore, the collection is not well-defined.

Q2EXERCISE 1.1

Let $\mathrm{A}=\{1,2,3,4,5,6\}$ . Insert the appropriate symbol $\in$ or $\notin$ in the blank spaces:

(i)

$5 \dots \mathrm{A}$

(ii)

$8 \dots \mathrm{A}$

(iii)

$0 \dots \mathrm{A}$

(iv)

$4 \dots \mathrm{A}$

(v)

$2 \dots \mathrm{A}$

(vi)

$10 \dots \mathrm{A}$

Solution

Given: The set

\mathrm{A}=\{1,2,3,4,5,6\}

.

The symbol

\in

means "is an element of" and

\notin

means "is not an element of".

(i)

5 is an element of set A. So,

5 \in \mathrm{A}

.

(ii)

8 is not an element of set A. So,

8 \notin \mathrm{A}

.

(iii)

0 is not an element of set A. So,

0 \notin \mathrm{A}

.

(iv)

4 is an element of set A. So,

4 \in \mathrm{A}

.

(v)

2 is an element of set A. So,

2 \in \mathrm{A}

.

(vi)

10 is not an element of set A. So,

10 \notin \mathrm{A}

.

Q3EXERCISE 1.1

Write the following sets in roster form:

(i)

$\mathrm{A}=\{x: x$ is an integer and $-3 \leq x<7\}$

(ii)

$\mathrm{B}=\{x: x$ is a natural number less than 6} $(iii)$ \mathrm{C}={x: x $is a two-digit natural number such that the sum of its digits is 8\}$

(iv)

$\mathrm{D}=\{x: x$ is a prime number which is divisor of 60} $(v)$ \mathrm{E}= $The set of all letters in the word TRIGONOMETRY (vi)$ \mathrm{F}=$ The set of all letters in the word BETTER

Solution

In roster form, we list all the elements of a set, separated by commas, within braces { }.

(i)

\mathrm{A}=\{x: x

is an integer and

-3 \leq x<7\}

The integers satisfying the condition are -3, -2, -1, 0, 1, 2, 3, 4, 5, 6. So,

\mathrm{A} = \{-3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}

.

(ii)

\mathrm{B}=\{x: x

is a natural number less than 6}

The natural numbers less than 6 are 1, 2, 3, 4, 5. So,

\mathrm{B} = {1, 2, 3, 4, 5}$.

(iii)

\mathrm{C}=\{x: x

is a two-digit natural number such that the sum of its digits is 8}

The two-digit numbers whose digits sum to 8 are 17, 26, 35, 44, 53, 62, 71, 80. So,

\mathrm{C} = {17, 26, 35, 44, 53, 62, 71, 80}$.

(iv)

\mathrm{D}=\{x: x

is a prime number which is divisor of 60}

The divisors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. The prime numbers among these are 2, 3, 5. So,

\mathrm{D} = {2, 3, 5}$.

(v)

\mathrm{E}=

The set of all letters in the word TRIGONOMETRY We list each unique letter from the word. The letters are T, R, I, G, O, N, M, E, Y. So,

\mathrm{E} = \{\mathrm{T, R, I, G, O, N, M, E, Y}\}

.

(vi)

\mathrm{F}=

The set of all letters in the word BETTER We list each unique letter from the word. The letters are B, E, T, R. So,

\mathrm{F} = \{\mathrm{B, E, T, R}\}

.

Q4EXERCISE 1.1

Write the following sets in the set-builder form :

(i)

$(3,6,9,12)$

(ii)

$\{2,4,8,16,32\}$

(iii)

$\{5,25,125,625\}$

(iv)

$\{2,4,6, \dots\}$

(v)

$\{1,4,9, \dots, 100\}$

Solution

In set-builder form, we describe the elements of the set by a common property.

(i)

(3,6,9,12)

Note: The source uses parentheses, but this is likely a typo for braces. Assuming the set is

\{3, 6, 9, 12\}

. The elements are multiples of 3. Specifically,

3 \times 1, 3 \times 2, 3 \times 3, 3 \times 4

. Set-builder form:

\{x: x = 3n, n \in \mathbf{N} \text{ and } 1 \leq n \leq 4\}

.

(ii)

\{2,4,8,16,32\}

The elements are powers of 2. Specifically,

2^1, 2^2, 2^3, 2^4, 2^5

. Set-builder form:

\{x: x = 2^n, n \in \mathbf{N} \text{ and } 1 \leq n \leq 5\}

.

(iii)

\{5,25,125,625\}

The elements are powers of 5. Specifically,

5^1, 5^2, 5^3, 5^4

. Set-builder form:

\{x: x = 5^n, n \in \mathbf{N} \text{ and } 1 \leq n \leq 4\}

.

(iv)

\{2,4,6, \dots\}

The elements are all even natural numbers. Set-builder form:

\{x: x \text{ is an even natural number}\}

.

(v)

\{1,4,9, \dots, 100\}

The elements are squares of natural numbers from 1 to 10. Specifically,

1^2, 2^2, 3^2, \dots, 10^2

. Set-builder form:

\{x: x = n^2, n \in \mathbf{N} \text{ and } 1 \leq n \leq 10\}

.

Q5EXERCISE 1.1

List all the elements of the following sets :

(i)

$\mathrm{A}=\{x: x$ is an odd natural number} $(ii)$ \mathrm{B}={x: x $is an integer,$ -\frac{1}{2}<x<\frac{9}{2}} $(iii)$ \mathrm{C}={x: x $is an integer,$ x^{2} \leq 4} $(iv)$ \mathrm{D}={x: x $is a letter in the word "LOYAL"\}$

(v)

$\mathrm{E}=\{x: x$ is a month of a year not having 31 days} $(vi)$ \mathrm{F}={x: x $is a consonant in the English alphabet which precedes$ k}$

Solution

(i)

\mathrm{A}=\{x: x

is an odd natural number}

The elements are

1, 3, 5, 7, \dots

So,

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

(ii)

\mathrm{B}=\{x: x

is an integer,

-\frac{1}{2}<x<\frac{9}{2}\}

We have

-0.5 < x < 4.5

. The integers in this range are 0, 1, 2, 3, 4. So,

\mathrm{B} = \{0, 1, 2, 3, 4\}

.

(iii)

\mathrm{C}=\{x: x

is an integer,

x^{2} \leq 4\}

The integers whose square is less than or equal to 4 are -2, -1, 0, 1, 2. So,

\mathrm{C} = \{-2, -1, 0, 1, 2\}

.

(iv)

\mathrm{D}=\{x: x

is a letter in the word "LOYAL"}

The unique letters are L, O, Y, A. So,

\mathrm{D} = {\mathrm{L, O, Y, A}}$.

(v)

\mathrm{E}=\{x: x

is a month of a year not having 31 days}

Months with 31 days are January, March, May, July, August, October, December. The remaining months are February, April, June, September, November. So,

\mathrm{E} = {\text{February, April, June, September, November}}$.

(vi)

\mathrm{F}=\{x: x

is a consonant in the English alphabet which precedes

k\}

The letters preceding 'k' are a, b, c, d, e, f, g, h, i, j. The vowels are a, e, i. The consonants are b, c, d, f, g, h, j. So,

\mathrm{F} = \{\mathrm{b, c, d, f, g, h, j}\}

.

Q6EXERCISE 1.1

Match each of the set on the left in the roster form with the same set on the right described in set-builder form:

(i)

$\{1,2,3,6\}$

(ii)

$\{2,3\}$

(iii)

{M,A,T,H,E,I,C,S}

(iv)

$\{1,3,5,7,9\}$

(a)

\{x: x

is a prime number and a divisor of 6}

(b)

\{x: x

is an odd natural number less than 10}

(c)

\{x: x

is natural number and divisor of 6}

(d)

\{x: x

is a letter of the word MATHEMATICS}

Solution

Let's analyze each option in set-builder form: (a)

\{x: x

is a prime number and a divisor of 6}: The divisors of 6 are 1, 2, 3, 6. The prime numbers among these are 2, 3. The set is

\{2,3\}

. (b)

\{x: x

is an odd natural number less than 10}: The odd natural numbers less than 10 are 1, 3, 5, 7, 9. The set is

\{1,3,5,7,9\}

. (c)

\{x: x

is natural number and divisor of 6}: The natural number divisors of 6 are 1, 2, 3, 6. The set is

\{1,2,3,6\}

. (d)

\{x: x

is a letter of the word MATHEMATICS}: The unique letters in the word are M, A, T, H, E, I, C, S. The set is {M,A,T,H,E,I,C,S}.

Now we can match them:

(i)

\{1,2,3,6\}

matches with (c)

\{x: x

is natural number and divisor of 6}$.

(ii)

\{2,3\}

matches with (a)

\{x: x

is a prime number and a divisor of 6}$.

(iii)

{M,A,T,H,E,I,C,S} matches with (d)

\{x: x

is a letter of the word MATHEMATICS}$.

(iv)

\{1,3,5,7,9\}

matches with (b)

\{x: x

is an odd natural number less than 10}$.

Final Answer:

(i)

\rightarrow

(c)

(ii)

\rightarrow

(a)

(iii)

\rightarrow

(d)

(iv)

\rightarrow

(b)

Q1EXERCISE 1.2

Which of the following are examples of the null set

(i)

Set of odd natural numbers divisible by 2

(ii)

Set of even prime numbers

(iii)

$\{x: x$ is a natural numbers, $x<5$ and $x>7\}$

(iv)

$\{y: y$ is a point common to any two parallel lines}

Solution

The null set (or empty set) is a set that contains no elements.

(i)

This is a null set. An odd natural number cannot be divisible by 2. There are no such numbers.

(ii)

This is not a null set. The number 2 is an even prime number. The set is

\{2\}

, which is a singleton set, not a null set.

(iii)

This is a null set. A number cannot simultaneously be less than 5 and greater than 7. There are no natural numbers that satisfy this condition.

(iv)

This is a null set. Parallel lines, by definition, never intersect. Therefore, there is no point common to any two parallel lines.

Q2EXERCISE 1.2

Which of the following sets are finite or infinite

(i)

The set of months of a year

(ii)

$\{1,2,3, \dots\}$

(iii)

$\{1,2,3, \dots 99, 100\}$

(iv)

The set of positive integers greater than 100

(v)

The set of prime numbers less than 99

Solution

A set is finite if it is empty or consists of a definite number of elements. Otherwise, it is infinite.

(i)

Finite. There are exactly 12 months in a year.

(ii)

Infinite. This is the set of all natural numbers, which continue indefinitely.

(iii)

Finite. This set contains exactly 100 elements, from 1 to 100.

(iv)

Infinite. The set of positive integers greater than 100 is

\{101, 102, 103, \dots\}

, which goes on forever.

(v)

Finite. There is a definite number of prime numbers less than 99. We can count them (there are 25 such primes).

Q3EXERCISE 1.2

State whether each of the following set is finite or infinite:

(i)

The set of lines which are parallel to the $x$ -axis

(ii)

The set of letters in the English alphabet

(iii)

The set of numbers which are multiple of 5

(iv)

The set of animals living on the earth

(v)

The set of circles passing through the origin $(0,0)$

Solution

(i)

Infinite. There are infinitely many lines parallel to the x-axis. For any real number

c

, the line

y=c

is parallel to the x-axis.

(ii)

Finite. The English alphabet has exactly 26 letters.

(iii)

Infinite. The multiples of 5 are

\{5, 10, 15, 20, \dots\}

, and this list continues indefinitely.

(iv)

Finite. Although the number of animals living on Earth is very large, it is a definite, countable number. Therefore, the set is finite.

(v)

Infinite. Infinitely many circles can be drawn that pass through the origin. For any point

(a,b)

not at the origin, there is a circle passing through

(0,0)

and

(a,b)

with center on the perpendicular bisector of the segment connecting them. There are infinite choices for such points.

Q4EXERCISE 1.2

In the following, state whether $\mathrm{A}=\mathrm{B}$ or not:

(i)

$\mathrm{A}=\{a, b, c, d\} \quad \mathrm{B}=\{d, c, b, a\}$

(ii)

$\mathrm{A}=\{4, 8, 12, 16\} \quad \mathrm{B}=\{8, 4, 16, 18\}$

(iii)

$\mathrm{A}=\{2, 4, 6, 8, 10\} \quad \mathrm{B}=\{x: x$ is positive even integer and $x \leq 10\}$

(iv)

$\mathrm{A}=\{x: x$ is a multiple of 10}, $\quad \mathrm{B}=\{10, 15, 20, 25, 30, \dots\}$

Solution

Two sets A and B are equal if they have exactly the same elements.

(i)

\mathrm{A}=\{a, b, c, d\}, \mathrm{B}=\{d, c, b, a\}

A = B. The order of elements in a set does not matter. Both sets contain the same four elements.

(ii)

\mathrm{A}=\{4, 8, 12, 16\}, \mathrm{B}=\{8, 4, 16, 18\}

A $\neq$ B. The element 12 is in A but not in B. Also, the element 18 is in B but not in A.

(iii)

\mathrm{A}=\{2, 4, 6, 8, 10\}, \mathrm{B}=\{x: x

is positive even integer and

x \leq 10\}

First, let's write set B in roster form. The positive even integers less than or equal to 10 are 2, 4, 6, 8, 10. So,

\mathrm{B} = \{2, 4, 6, 8, 10\}

. A = B. Both sets contain the same elements.

(iv)

\mathrm{A}=\{x: x

is a multiple of 10},

\mathrm{B}=\{10, 15, 20, 25, 30, \dots\}

Set A in roster form is

\mathrm{A} = \{10, 20, 30, 40, \dots\}

. A $\neq$ B. Set B contains numbers like 15, 25, etc., which are not multiples of 10 and are therefore not in set A. Also, all elements of A (except 10, 20, 30...) are not in B.

Q5EXERCISE 1.2

Are the following pair of sets equal ? Give reasons.

(i)

$\mathrm{A}=\{2,3\}, \quad \mathrm{B}=\{x: x$ is solution of $x^{2}+5 x+6=0\}$

(ii)

$\mathrm{A}=\{x: x$ is a letter in the word FOLLOW} $\mathrm{B}=\{y: y$ is a letter in the word WOLF}

Solution

(i)

\mathrm{A}=\{2,3\}, \quad \mathrm{B}=\{x: x

is solution of

x^{2}+5 x+6=0\}

First, we find the solution set for B.

x^2 + 5x + 6 = 0

(x+2)(x+3) = 0

x = -2 \text{ or } x = -3

So, the set B in roster form is

\mathrm{B} = \{-2, -3\}

. A $\neq$ B. The elements of A are 2 and 3, while the elements of B are -2 and -3. They are not the same.

(ii)

\mathrm{A}=\{x: x

is a letter in the word FOLLOW},

\mathrm{B}=\{y: y

is a letter in the word WOLF}

First, we write both sets in roster form. Remember that elements in a set are not repeated. For set A, the unique letters in FOLLOW are F, O, L, W. So,

\mathrm{A} = {\mathrm{F, O, L, W}}

. For set B, the letters in WOLF are W, O, L, F. So,

\mathrm{B} = {\mathrm{W, O, L, F}}$. A = B. Both sets contain exactly the same elements. The order of listing does not matter.

Q6EXERCISE 1.2

From the sets given below, select equal sets : $A=\{2,4,8,12\}, & B=\{1,2,3,4\}, & C=\{4,8,12,14\}, & D=\{3,1,4,2\} \nE=\{-1,1\}, & F=\{0, a\}, & G=\{1,-1\}, & H=\{0,1\} \end{array}$$$

Solution

We need to compare each set with every other set to find pairs that have exactly the same elements.

$\mathrm{A}=\{2,4,8,12\}$
$\mathrm{B}=\{1,2,3,4\}$
$\mathrm{C}=\{4,8,12,14\}$
$\mathrm{D}=\{3,1,4,2\}$
$\mathrm{E}=\{-1,1\}$
$\mathrm{F}=\{0, a\}$
$\mathrm{G}=\{1,-1\}$
$\mathrm{H}=\{0,1\}$

Comparing the sets:

Is A equal to any other set? No.
Is B equal to any other set? Let's check D. B has elements 1, 2, 3, 4. D has elements 3, 1, 4, 2. Since the order does not matter, B = D.
Is C equal to any other set? No.
Is E equal to any other set? Let's check G. E has elements -1, 1. G has elements 1, -1. Since the order does not matter, E = G.
Is F equal to any other set? No (unless $a=0$ or $a=1$ , but we assume 'a' is a distinct literal element).
Is H equal to any other set? No.

Final Answer: The pairs of equal sets are B = D and E = G.

Q1EXERCISE 1.3

Make correct statements by filling in the symbols $\subset$ or $\not\subset$ in the blank spaces :

(i)

$\{2,3,4\} \dots \{1,2,3,4,5\}$

(ii)

$\{a, b, c\} \dots \{b, c, d\}$

(iii)

$\{x: x$ is a student of Class XI of your school} $\dots \{x: x$ student of your school}

(iv)

$\{x: x$ is a circle in the plane} $\dots \{x: x$ is a circle in the same plane with radius 1 unit}

(v)

$\{x: x$ is a triangle in a plane} $\dots \{x: x$ is a rectangle in the plane}

(vi)

$\{x: x$ is an equilateral triangle in a plane} $\dots \{x: x$ is a triangle in the same plane}

(vii)

$\{x: x$ is an even natural number} $\dots \{x: x$ is an integer}

Solution

The symbol

\subset

means "is a subset of" and

\not\subset

means "is not a subset of". A set P is a subset of Q if every element of P is also an element of Q.

(i)

\{2,3,4\} \subset \{1,2,3,4,5\}

Reason: Every element of

\{2,3,4\}

(namely 2, 3, and 4) is also an element of

\{1,2,3,4,5\}

.

(ii)

\{a, b, c\} \not\subset \{b, c, d\}

Reason: The element 'a' is in the first set but not in the second set.

(iii)

\{x: x

is a student of Class XI of your school}

\subset \{x: x

student of your school}$ Reason: Every student of Class XI is also a student of the school.

(iv)

\{x: x

is a circle in the plane}

\not\subset \{x: x

is a circle in the same plane with radius 1 unit} Reason: A circle with a radius of 2 units is in the first set but not in the second set. The second set is actually a subset of the first.

(v)

\{x: x

is a triangle in a plane}

\not\subset \{x: x

is a rectangle in the plane} Reason: Triangles and rectangles are different geometric shapes. No element of the first set is an element of the second set.

(vi)

\{x: x

is an equilateral triangle in a plane}

\subset \{x: x

is a triangle in the same plane} Reason: Every equilateral triangle is a type of triangle.

(vii)

\{x: x

is an even natural number}

\subset \{x: x

is an integer}$ Reason: Every even natural number (like 2, 4, 6) is also an integer.

Q2EXERCISE 1.3

Examine whether the following statements are true or false:

(i)

$\{a, b\} \not\subset \{b, c, a\}$

(ii)

$\{a, e\} \subset \{x: x$ is a vowel in the English alphabet}

(iii)

$\{1,2,3\} \subset \{1,3,5\}$

(iv)

$\{a\} \subset \{a, b, c\}$

(v)

$\{a\} \in \{a, b, c\}$

(vi)

$\{x: x$ is an even natural number less than 6} $\subset \{x: x$ is a natural number which divides 36}

Solution

(i)

\{a, b\} \not\subset \{b, c, a\}

False. Every element of

\{a, b\}

(namely 'a' and 'b') is present in

\{b, c, a\}

. Therefore,

\{a, b\}

is a subset of

\{b, c, a\}

.

(ii)

\{a, e\} \subset \{x: x

is a vowel in the English alphabet}

The set of vowels is

{a, e, i, o, u}

. **True.** Every element of

{a, e}$ is a vowel.

(iii)

\{1,2,3\} \subset \{1,3,5\}

False. The element 2 is in the first set but not in the second set.

(iv)

\{a\} \subset \{a, b, c\}

True. Every element of the set

\{a\}

(which is just 'a') is in the set

\{a, b, c\}

.

(v)

\{a\} \in \{a, b, c\}

False. The symbol

\in

means "is an element of". The elements of

\{a, b, c\}

are

a, b, c

. The set

\{a\}

is not an element of this set. The correct statement would be

a \in \{a, b, c\}

.

(vi)

\{x: x

is an even natural number less than 6}

\subset \{x: x

is a natural number which divides 36}

Let's write both sets in roster form. First set: The even natural numbers less than 6 are 2, 4. So, the set is

{2, 4}

. Second set: The natural numbers that divide 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36. So, the set is

{1, 2, 3, 4, 6, 9, 12, 18, 36}

. **True.** Every element of

{2, 4}$ is present in the set of divisors of 36.

Q3EXERCISE 1.3

Let $\mathrm{A}=\{1,2,\{3,4\}, 5\}$ . Which of the following statements are incorrect and why?

(i)

$\{3,4\} \subset \mathrm{A}$

(ii)

$\{3,4\} \in \mathrm{A}$

(iii)

$\{\{3,4\}\} \subset \mathrm{A}$

(iv)

$1 \in \mathrm{A}$

(v)

$1 \subset \mathrm{A}$

(vi)

$\{1,2,5\} \subset \mathrm{A}$

(vii)

$\{1,2,5\} \in \mathrm{A}$

(viii)

$\{1,2,3\} \subset \mathrm{A}$

(ix)

$\phi \in \mathrm{A}$

(x)

$\phi \subset \mathrm{A}$ (xi) $\{\phi\} \subset \mathrm{A}$

Solution

Given: The set

\mathrm{A}=\{1,2,\{3,4\}, 5\}

. The elements of A are 1, 2, the set

\{3,4\}

, and 5.

(i)

\{3,4\} \subset \mathrm{A}

Incorrect. For this to be true, the elements 3 and 4 must be elements of A. However, 3 and 4 are not elements of A. The element is the set

\{3,4\}

itself.

(ii)

\{3,4\} \in \mathrm{A}

Correct. The set

\{3,4\}

is one of the four elements of A.

(iii)

\{\{3,4\}\} \subset \mathrm{A}

Correct. This is a set containing a single element, which is

\{3,4\}

. For this to be a subset of A, its element

\{3,4\}

must be an element of A. This is true.

(iv)

1 \in \mathrm{A}

Correct. 1 is an element of A.

(v)

1 \subset \mathrm{A}

Incorrect. The symbol

\subset

is used for sets. 1 is an element, not a set. The correct statement would be

\{1\} \subset \mathrm{A}

.

(vi)

\{1,2,5\} \subset \mathrm{A}

Correct. Each element of the set

\{1,2,5\}

(namely 1, 2, and 5) is also an element of A.

(vii)

\{1,2,5\} \in \mathrm{A}

Incorrect. The set

\{1,2,5\}

is not an element of A. The elements of A are 1, 2,

\{3,4\}

, and 5.

(viii)

\{1,2,3\} \subset \mathrm{A}

Incorrect. For this to be true, the element 3 must be in A. But 3 is not an element of A.

(ix)

\phi \in \mathrm{A}

Incorrect. The empty set

\phi

is not one of the elements of A.

(x)

\phi \subset \mathrm{A}

Correct. The empty set is a subset of every set by definition.

(xi)

\{\phi\} \subset \mathrm{A}

Incorrect. For this set to be a subset of A, its element,

\phi

, must be an element of A. But as stated in (ix),

\phi

is not an element of A.

Q4EXERCISE 1.3

Write down all the subsets of the following sets

(i)

$\{a\}$

(ii)

$\{a, b\}$

(iii)

$\{1,2,3\}$

(iv)

$\phi$

Solution

A set with

n

elements has

2^n

subsets.

(i)

\{a\}

This set has 1 element. It will have

2^1 = 2

subsets. The subsets are:

\phi, \{a\}

.

(ii)

\{a, b\}

This set has 2 elements. It will have

2^2 = 4

subsets. The subsets are:

\phi, \{a\}, \{b\}, \{a, b\}

.

(iii)

\{1,2,3\}

This set has 3 elements. It will have

2^3 = 8

subsets. The subsets are:

Subsets with 0 elements: $\phi$
Subsets with 1 element: $\{1\}, \{2\}, \{3\}$
Subsets with 2 elements: $\{1,2\}, \{1,3\}, \{2,3\}$
Subsets with 3 elements: $\{1,2,3\}$ Total subsets: $\phi, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}$ .

(iv)

\phi

This set has 0 elements. It will have

2^0 = 1

subset. The only subset of the empty set is the empty set itself:

\phi

.

Q5EXERCISE 1.3

Write the following as intervals :

(i)

$\{x: x \in \mathrm{R},-4<x \leq 6\}$

(ii)

$\{x: x \in \mathrm{R},-12<x<-10\}$

(iii)

$\{x: x \in \mathrm{R}, 0 \leq x<7\}$

(iv)

$\{x: x \in \mathrm{R}, 3 \leq x \leq 4\}$

Solution

We use parentheses () for endpoints that are not included (strict inequality

<

or

>

), and square brackets [] for endpoints that are included (inequality

\leq

or

\geq

).

(i)

\{x: x \in \mathrm{R},-4<x \leq 6\}

This interval is open at -4 and closed at 6. Interval form:

(-4, 6]

.

(ii)

\{x: x \in \mathrm{R},-12<x<-10\}

This interval is open at both ends. Interval form:

(-12, -10)

.

(iii)

\{x: x \in \mathrm{R}, 0 \leq x<7\}

This interval is closed at 0 and open at 7. Interval form:

[0, 7)

.

(iv)

\{x: x \in \mathrm{R}, 3 \leq x \leq 4\}

This interval is closed at both ends. Interval form:

[3, 4]

.

Q6EXERCISE 1.3

Write the following intervals in set-builder form :

(i)

$(-3,0)$

(ii)

$[6,12]$

(iii)

$(6,12]$

(iv)

$[-23,5)$

Solution

We convert the interval notation back to inequalities for real numbers

x

.

(i)

(-3,0)

This represents all real numbers

x

such that

x

is greater than -3 and less than 0. Set-builder form:

\{x: x \in \mathrm{R}, -3 < x < 0\}

.

(ii)

[6,12]

This represents all real numbers

x

such that

x

is greater than or equal to 6 and less than or equal to 12. Set-builder form:

\{x: x \in \mathrm{R}, 6 \leq x \leq 12\}

.

(iii)

(6,12]

This represents all real numbers

x

such that

x

is greater than 6 and less than or equal to 12. Set-builder form:

\{x: x \in \mathrm{R}, 6 < x \leq 12\}

.

(iv)

[-23,5)

This represents all real numbers

x

such that

x

is greater than or equal to -23 and less than 5. Set-builder form:

\{x: x \in \mathrm{R}, -23 \leq x < 5\}

.

Q7EXERCISE 1.3

What universal set(s) would you propose for each of the following :

(i)

The set of right triangles.

(ii)

The set of isosceles triangles.

Solution

A universal set is a basic set that contains all the elements and subsets relevant to a particular context.

(i)

For the set of right triangles, a suitable universal set would be the set of all triangles in a plane. Another possible universal set could be the set of all polygons in a plane or the set of all two-dimensional geometric shapes.

(ii)

For the set of isosceles triangles, a suitable universal set would also be the set of all triangles in a plane. As in the previous case, the set of all polygons in a plane or the set of all two-dimensional geometric shapes would also be appropriate.

Q8EXERCISE 1.3

Given the sets $\mathrm{A}=\{1,3,5\}, \mathrm{B}=\{2,4,6\}$ and $\mathrm{C}=\{0,2,4,6,8\}$ , which of the following may be considered as universal set ( s ) for all the three sets $\mathrm{A}, \mathrm{B}$ and C

(i)

$\{0,1,2,3,4,5,6\}$

(ii)

$\phi$

(iii)

$\{0,1,2,3,4,5,6,7,8,9,10\}$

(iv)

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

Solution

A universal set for A, B, and C must contain all the elements of A, B, and C. The elements to be included are from A: {1, 3, 5}, from B: {2, 4, 6}, and from C: {0, 2, 4, 6, 8}. Combining all unique elements, the universal set must contain at least

\{0, 1, 2, 3, 4, 5, 6, 8\}

.

Let's check each option:

(i)

\{0,1,2,3,4,5,6\}

This set does not contain the element 8, which is in set C. So, it cannot be a universal set.

(ii)

\phi

The empty set cannot be a universal set for non-empty sets A, B, and C. So, it cannot be a universal set.

(iii)

\{0,1,2,3,4,5,6,7,8,9,10\}

This set contains all the elements of A (

\{1,3,5\}

), B (

\{2,4,6\}

), and C (

\{0,2,4,6,8\}

). So, it can be considered a universal set.

(iv)

\{1,2,3,4,5,6,7,8\}

This set does not contain the element 0, which is in set C. So, it cannot be a universal set.

Final Answer: Only the set in option (iii) can be considered a universal set for A, B, and C.

Q1EXERCISE 1.4

Find the union of each of the following pairs of sets :

(i)

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

(ii)

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

(iii)

$\mathrm{A}=\{x: x$ is a natural number and multiple of 3} $\mathrm{B}=\{x: x$ is a natural number less than 6} $(iv)$ \mathrm{A}={x: x $is a natural number and$ 1<x \leq 6}\mathrm{B}={x: x $is a natural number and$ 6<x<10} $(v)$ \mathrm{A}={1,2,3}, \mathrm{B}=\phi$

Solution

The union of two sets, denoted by

\cup

, is the set of all elements that are in either set (or in both), with common elements taken only once.

(i)

\mathrm{X}=\{1,3,5\}, \mathrm{Y}=\{1,2,3\}

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

(ii)

\mathrm{A}=\{a, e, i, o, u\}, \mathrm{B}=\{a, b, c\}

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

(iii)

\mathrm{A}=\{x: x

is a natural number and multiple of 3}

= \{3, 6, 9, 12, \dots\}

\mathrm{B}=\{x: x

is a natural number less than 6}

= \{1, 2, 3, 4, 5\}

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

or in set-builder form:

\{x: x \in \{1,2,4,5\} \text{ or } x \text{ is a multiple of } 3\}

.

(iv)

\mathrm{A}=\{x: x

is a natural number and

1<x \leq 6\} = \{2, 3, 4, 5, 6\}

\mathrm{B}=\{x: x

is a natural number and

6<x<10\} = \{7, 8, 9\}

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

. This can also be written as

\{x: x \in \mathbf{N} \text{ and } 1 < x < 10\}

.

(v)

\mathrm{A}=\{1,2,3\}, \mathrm{B}=\phi

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

. The union of any set with the empty set is the set itself.

Q2EXERCISE 1.4

Let $\mathrm{A}=\{a, b\}, \mathrm{B}=\{a, b, c\}$ . Is $\mathrm{A} \subset \mathrm{B}$ ? What is $\mathrm{A} \cup \mathrm{B}$ ?

Solution

Given:

\mathrm{A}=\{a, b\}

and

\mathrm{B}=\{a, b, c\}

.

Is $\mathrm{A} \subset \mathrm{B}$ ? Yes,

\mathrm{A} \subset \mathrm{B}

. This is because every element of set A (namely 'a' and 'b') is also an element of set B.

What is $\mathrm{A} \cup \mathrm{B}$ ? The union

\mathrm{A} \cup \mathrm{B}

contains all elements from both sets without repetition.

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

. We can observe that

\mathrm{A} \cup \mathrm{B} = \mathrm{B}

.

Q3EXERCISE 1.4

If A and B are two sets such that $\mathrm{A} \subset \mathrm{B}$ , then what is $\mathrm{A} \cup \mathrm{B}$ ?

Solution

Given: A and B are two sets such that

\mathrm{A} \subset \mathrm{B}

. This means that every element of A is also an element of B.

The union

\mathrm{A} \cup \mathrm{B}

is the set of all elements that are in A or in B or in both. Since all elements of A are already in B, the collection of all elements from both sets is simply the collection of all elements in B.

Therefore, if

\mathrm{A} \subset \mathrm{B}

, then

\mathrm{A} \cup \mathrm{B} = \mathrm{B}

.

Q4EXERCISE 1.4

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

(i)

$\mathrm{A} \cup \mathrm{B}$

(ii)

$\mathrm{A} \cup \mathrm{C}$

(iii)

$\mathrm{B} \cup \mathrm{C}$

(iv)

$\mathrm{B} \cup \mathrm{D}$

(v)

$\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}$

(vi)

$\mathrm{A} \cup \mathrm{B} \cup \mathrm{D}$

(vii)

$\mathrm{B} \cup \mathrm{C} \cup \mathrm{D}$

Solution

Given:

\mathrm{A}=\{1,2,3,4\}

\mathrm{B}=\{3,4,5,6\}

\mathrm{C}=\{5,6,7,8\}

\mathrm{D}=\{7,8,9,10\}

(i)

\mathrm{A} \cup \mathrm{B} = \{1,2,3,4\} \cup \{3,4,5,6\} = \{1,2,3,4,5,6\}

(ii)

\mathrm{A} \cup \mathrm{C} = \{1,2,3,4\} \cup \{5,6,7,8\} = \{1,2,3,4,5,6,7,8\}

(iii)

\mathrm{B} \cup \mathrm{C} = \{3,4,5,6\} \cup \{5,6,7,8\} = \{3,4,5,6,7,8\}

(iv)

\mathrm{B} \cup \mathrm{D} = \{3,4,5,6\} \cup \{7,8,9,10\} = \{3,4,5,6,7,8,9,10\}

(v)

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

(vi)

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

(vii)

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

Q5EXERCISE 1.4

Find the intersection of each pair of sets of question 1 above.

Solution

The intersection of two sets, denoted by

\cap

, is the set of all elements that are common to both sets.

(i)

\mathrm{X}=\{1,3,5\}, \mathrm{Y}=\{1,2,3\}

\mathrm{X} \cap \mathrm{Y} = \{1, 3\}

(ii)

\mathrm{A}=\{a, e, i, o, u\}, \mathrm{B}=\{a, b, c\}

\mathrm{A} \cap \mathrm{B} = \{a\}

(iii)

\mathrm{A}=\{x: x

is a natural number and multiple of 3}

= \{3, 6, 9, \dots\}

\mathrm{B}=\{x: x

is a natural number less than 6}

= \{1, 2, 3, 4, 5\}

\mathrm{A} \cap \mathrm{B} = \{3\}

(iv)

\mathrm{A}=\{x: x

is a natural number and

1<x \leq 6\} = \{2, 3, 4, 5, 6\}

\mathrm{B}=\{x: x

is a natural number and

6<x<10\} = \{7, 8, 9\}

\mathrm{A} \cap \mathrm{B} = \phi

. There are no common elements.

(v)

\mathrm{A}=\{1,2,3\}, \mathrm{B}=\phi

\mathrm{A} \cap \mathrm{B} = \phi

. The intersection of any set with the empty set is the empty set.

Q6EXERCISE 1.4

If $\mathrm{A}=\{3,5,7,9,11\}, \mathrm{B}=\{7,9,11,13\}, \mathrm{C}=\{11,13,15\}$ and $\mathrm{D}=\{15,17\}$ ; find

(i)

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

(ii)

$\mathrm{B} \cap \mathrm{C}$

(iii)

$\mathrm{A} \cap \mathrm{C} \cap \mathrm{D}$

(iv)

$\mathrm{A} \cap \mathrm{C}$

(v)

$\mathrm{B} \cap \mathrm{D}$

(vi)

$\mathrm{A} \cap(\mathrm{B} \cup \mathrm{C})$

(vii)

$\mathrm{A} \cap \mathrm{D}$

(viii)

$\mathrm{A} \cap(\mathrm{B} \cup \mathrm{D})$

(ix)

$(\mathrm{A} \cap \mathrm{B}) \cap(\mathrm{B} \cup \mathrm{C})$

(x)

$(\mathrm{A} \cup \mathrm{D}) \cap(\mathrm{B} \cup \mathrm{C})$

Solution

Given:

\mathrm{A}=\{3,5,7,9,11\}

\mathrm{B}=\{7,9,11,13\}

\mathrm{C}=\{11,13,15\}

\mathrm{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)

\mathrm{A} \cap(\mathrm{B} \cup \mathrm{C})

First,

\mathrm{B} \cup \mathrm{C} = \{7,9,11,13\} \cup \{11,13,15\} = \{7,9,11,13,15\}

. Then,

\mathrm{A} \cap (\mathrm{B} \cup \mathrm{C}) = \{3,5,7,9,11\} \cap \{7,9,11,13,15\} = \{7,9,11\}

.

(vii)

\mathrm{A} \cap \mathrm{D} = \phi

(viii)

\mathrm{A} \cap(\mathrm{B} \cup \mathrm{D})

First,

\mathrm{B} \cup \mathrm{D} = \{7,9,11,13\} \cup \{15,17\} = \{7,9,11,13,15,17\}

. Then,

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

.

(ix)

(\mathrm{A} \cap \mathrm{B}) \cap(\mathrm{B} \cup \mathrm{C})

From (i),

\mathrm{A} \cap \mathrm{B} = \{7,9,11\}

. From (vi),

\mathrm{B} \cup \mathrm{C} = \{7,9,11,13,15\}

. Then,

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

.

(x)

(\mathrm{A} \cup \mathrm{D}) \cap(\mathrm{B} \cup \mathrm{C})

First,

\mathrm{A} \cup \mathrm{D} = \{3,5,7,9,11\} \cup \{15,17\} = \{3,5,7,9,11,15,17\}

. From (vi),

\mathrm{B} \cup \mathrm{C} = \{7,9,11,13,15\}

. Then,

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

.

Q7EXERCISE 1.4

If $\mathrm{A}=\{x: x$ is a natural number }, $\mathrm{B}=\{x: x$ is an even natural number } $\mathrm{C}=\{x: x$ is an odd natural number } $and$ \mathrm{D}={x: x $is a prime number \}$ , find

(i)

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

(ii)

$\mathrm{A} \cap \mathrm{C}$

(iii)

$\mathrm{A} \cap \mathrm{D}$

(iv)

$\mathrm{B} \cap \mathrm{C}$

(v)

$\mathrm{B} \cap \mathrm{D}$

(vi)

$\mathrm{C} \cap \mathrm{D}$

Solution

Given:

\mathrm{A} = \text{Set of natural numbers} = \{1,2,3,4, \dots\}

\mathrm{B} = \text{Set of even natural numbers} = \{2,4,6,8, \dots\}

\mathrm{C} = \text{Set of odd natural numbers} = \{1,3,5,7, \dots\}

\mathrm{D} = \text{Set of prime numbers} = \{2,3,5,7,11, \dots\}

(i)

\mathrm{A} \cap \mathrm{B}

: The common elements between all natural numbers and even natural numbers are the even natural numbers themselves.

\mathrm{A} \cap \mathrm{B} = \{x: x \text{ is an even natural number}\} = \mathrm{B}

.

(ii)

\mathrm{A} \cap \mathrm{C}

: The common elements between all natural numbers and odd natural numbers are the odd natural numbers themselves.

\mathrm{A} \cap \mathrm{C} = \{x: x \text{ is an odd natural number}\} = \mathrm{C}

.

(iii)

\mathrm{A} \cap \mathrm{D}

: The common elements between all natural numbers and prime numbers are the prime numbers themselves.

\mathrm{A} \cap \mathrm{D} = \{x: x \text{ is a prime number}\} = \mathrm{D}

.

(iv)

\mathrm{B} \cap \mathrm{C}

: The common elements between even natural numbers and odd natural numbers. There are no numbers that are both even and odd.

\mathrm{B} \cap \mathrm{C} = \phi

.

(v)

\mathrm{B} \cap \mathrm{D}

: The common elements between even natural numbers and prime numbers. The only number that is both even and prime is 2.

\mathrm{B} \cap \mathrm{D} = \{2\}

.

(vi)

\mathrm{C} \cap \mathrm{D}

: The common elements between odd natural numbers and prime numbers. This is the set of all prime numbers except 2.

\mathrm{C} \cap \mathrm{D} = \{x: x \text{ is an odd prime number}\} = \{3,5,7,11,13, \dots\}

.

Q8EXERCISE 1.4

Which of the following pairs of sets are disjoint

(i)

$\{1,2,3,4\}$ and $\{x: x$ is a natural number and $4 \leq x \leq 6\}$

(ii)

$\{a, e, i, o, u\}$ and $\{c, d, e, f\}$

(iii)

$\{x: x$ is an even integer} $and$ {x: x$ is an odd integer}

Solution

Two sets are disjoint if their intersection is the empty set (

\phi

), meaning they have no elements in common.

(i)

Let

\mathrm{A} = \{1,2,3,4\}

and

\mathrm{B} = \{x: x

is a natural number and

4 \leq x \leq 6\} = \{4,5,6\}

.

\mathrm{A} \cap \mathrm{B} = \{4\}

. Since the intersection is not empty, these sets are not disjoint.

(ii)

Let

\mathrm{A} = \{a, e, i, o, u\}

and

\mathrm{B} = \{c, d, e, f\}

.

\mathrm{A} \cap \mathrm{B} = \{e\}

. Since the intersection is not empty, these sets are not disjoint.

(iii)

Let

\mathrm{A} = \{x: x

is an even integer} and

\mathrm{B} = \{x: x

is an odd integer}

. An integer cannot be both even and odd. Therefore, there are no common elements.

\mathrm{A} \cap \mathrm{B} = \phi$. These sets are disjoint.

Q9EXERCISE 1.4

If $\mathrm{A}=\{3,6,9,12,15,18,21\}, \mathrm{B}=\{4,8,12,16,20\}, \mathrm{C}=\{2,4,6,8,10,12,14,16\}, \mathrm{D}=\{5,10,15,20\}$ ; find

(i)

$\mathrm{A}-\mathrm{B}$

(ii)

$\mathrm{A}-\mathrm{C}$

(iii)

$\mathrm{A}-\mathrm{D}$

(iv)

$\mathrm{B}-\mathrm{A}$

(v)

$\mathrm{C}-\mathrm{A}$

(vi)

$\mathrm{D}-\mathrm{A}$

(vii)

$\mathrm{B}-\mathrm{C}$

(viii)

$\mathrm{B}-\mathrm{D}$

(ix)

$\mathrm{C}-\mathrm{B}$

(x)

$\mathrm{D}-\mathrm{B}$ (xi) $\mathrm{C}-\mathrm{D}$ (xii) $\mathrm{D}-\mathrm{C}$

Solution

The difference

\mathrm{P}-\mathrm{Q}

is the set of elements which belong to P but not to Q.

(i)

\mathrm{A}-\mathrm{B} = \{3,6,9,15,18,21\}

(Remove 12 from A)

(ii)

\mathrm{A}-\mathrm{C} = \{3,9,15,18,21\}

(Remove 6 and 12 from A)

(iii)

\mathrm{A}-\mathrm{D} = \{3,6,9,12,18,21\}

(Remove 15 from A)

(iv)

\mathrm{B}-\mathrm{A} = \{4,8,16,20\}

(Remove 12 from B)

(v)

\mathrm{C}-\mathrm{A} = \{2,4,8,10,14,16\}

(Remove 6 and 12 from C)

(vi)

\mathrm{D}-\mathrm{A} = \{5,10,20\}

(Remove 15 from D)

(vii)

\mathrm{B}-\mathrm{C} = \{20\}

(Remove 4, 8, 12, 16 from B)

(viii)

\mathrm{B}-\mathrm{D} = \{4,8,12,16\}

(Remove 20 from B)

(ix)

\mathrm{C}-\mathrm{B} = \{2,6,10,14\}

(Remove 4, 8, 12, 16 from C)

(x)

\mathrm{D}-\mathrm{B} = \{5,10,15\}

(Remove 20 from D)

(xi)

\mathrm{C}-\mathrm{D} = \{2,4,6,8,12,14,16\}

(Remove 10 from C)

(xii)

\mathrm{D}-\mathrm{C} = \{5,15,20\}

(Remove 10 from D)

Q10EXERCISE 1.4

If $\mathrm{X}=\{a, b, c, d\}$ and $\mathrm{Y}=\{f, b, d, g\}$ , find

(i)

$\mathrm{X}-\mathrm{Y}$

(ii)

$\mathrm{Y}-\mathrm{X}$

(iii)

$\mathrm{X} \cap \mathrm{Y}$

Solution

Given:

\mathrm{X}=\{a, b, c, d\}

and

\mathrm{Y}=\{f, b, d, g\}

.

(i)

\mathrm{X}-\mathrm{Y}

: Elements in X but not in Y. The common elements are b, d. Removing them from X gives

\{a, c\}

.

\mathrm{X}-\mathrm{Y} = \{a, c\}

.

(ii)

\mathrm{Y}-\mathrm{X}

: Elements in Y but not in X. The common elements are b, d. Removing them from Y gives

\{f, g\}

.

\mathrm{Y}-\mathrm{X} = \{f, g\}

.

(iii)

\mathrm{X} \cap \mathrm{Y}

: Elements common to both X and Y.

\mathrm{X} \cap \mathrm{Y} = \{b, d\}

.

Q11EXERCISE 1.4

If $\mathbf{R}$ is the set of real numbers and $\mathbf{Q}$ is the set of rational numbers, then what is $\mathbf{R}-\mathbf{Q}$ ?

Solution

Given:

\mathbf{R}

is the set of all real numbers.

\mathbf{Q}

is the set of all rational numbers.

To Find:

\mathbf{R}-\mathbf{Q}

The set

\mathbf{R}-\mathbf{Q}

represents the set of numbers that are in

\mathbf{R}

but not in

\mathbf{Q}

. By definition, a real number is either rational or irrational. The set of all real numbers is the union of the set of rational numbers and the set of irrational numbers.

The numbers that are real but not rational are, by definition, the irrational numbers.

Therefore,

\mathbf{R}-\mathbf{Q}

is the set of all irrational numbers, which is often denoted by

\mathbf{T}

or

\mathbf{I}

.

Final Answer:

\mathbf{R}-\mathbf{Q}

is the set of irrational numbers.

Q12EXERCISE 1.4

State whether each of the following statement is true or false. Justify your answer.

(i)

$\{2,3,4,5\}$ and $\{3,6\}$ are disjoint sets.

(ii)

$\{a, e, i, o, u\}$ and $\{a, b, c, d\}$ are disjoint sets.

(iii)

$\{2,6,10,14\}$ and $\{3,7,11,15\}$ are disjoint sets.

(iv)

$\{2,6,10\}$ and $\{3,7,11\}$ are disjoint sets.

Solution

Two sets are disjoint if their intersection is the empty set.

(i)

\{2,3,4,5\}

and

\{3,6\}

False. The intersection of these two sets is

\{3\}

, which is not empty.

(ii)

\{a, e, i, o, u\}

and

\{a, b, c, d\}

False. The intersection of these two sets is

\{a\}

, which is not empty.

(iii)

\{2,6,10,14\}

and

\{3,7,11,15\}

True. The first set contains only even numbers. The second set contains only odd numbers. There are no common elements. Their intersection is

\phi

.

(iv)

\{2,6,10\}

and

\{3,7,11\}

True. There are no common elements between the two sets. Their intersection is

\phi

.

Q1EXERCISE 1.5

Let $U=\{1,2,3,4,5,6,7,8,9\}, A=\{1,2,3,4\}, B=\{2,4,6,8\}$ and $\mathrm{C}=\{3,4,5,6\}$ . Find (i) $\mathrm{A}^{\prime}$ (ii) $\mathrm{B}^{\prime}$ (iii) $(\mathrm{A} \cup \mathrm{C})^{\prime}$ (iv) $(\mathrm{A} \cup \mathrm{B})^{\prime}$ (v) $(\mathrm{A}^{\prime})^{\prime}$ (vi) $(\mathrm{B}-\mathrm{C})^{\prime}$

Solution

Given:

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

A=\{1,2,3,4\}

B=\{2,4,6,8\}

C=\{3,4,5,6\}

The complement of a set P, denoted

\mathrm{P}^{\prime}

, is the set of all elements in the universal set U that are not in P. So,

\mathrm{P}^{\prime} = U - P

.

(i)

\mathrm{A}^{\prime} = U - A = \{1,2,3,4,5,6,7,8,9\} - \{1,2,3,4\} = \{5,6,7,8,9\}

(ii)

\mathrm{B}^{\prime} = U - B = \{1,2,3,4,5,6,7,8,9\} - \{2,4,6,8\} = \{1,3,5,7,9\}

(iii)

(\mathrm{A} \cup \mathrm{C})^{\prime}

First, find

\mathrm{A} \cup \mathrm{C} = \{1,2,3,4\} \cup \{3,4,5,6\} = \{1,2,3,4,5,6\}

. Then,

(\mathrm{A} \cup \mathrm{C})^{\prime} = U - (\mathrm{A} \cup \mathrm{C}) = \{1,2,3,4,5,6,7,8,9\} - \{1,2,3,4,5,6\} = \{7,8,9\}

.

(iv)

(\mathrm{A} \cup \mathrm{B})^{\prime}

First, find

\mathrm{A} \cup \mathrm{B} = \{1,2,3,4\} \cup \{2,4,6,8\} = \{1,2,3,4,6,8\}

. Then,

(\mathrm{A} \cup \mathrm{B})^{\prime} = U - (\mathrm{A} \cup \mathrm{B}) = \{1,2,3,4,5,6,7,8,9\} - \{1,2,3,4,6,8\} = \{5,7,9\}

.

(v)

(\mathrm{A}^{\prime})^{\prime}

From (i),

\mathrm{A}^{\prime} = \{5,6,7,8,9\}

. The complement of

\mathrm{A}^{\prime}

is

(\mathrm{A}^{\prime})^{\prime} = U - \mathrm{A}^{\prime} = \{1,2,3,4,5,6,7,8,9\} - \{5,6,7,8,9\} = \{1,2,3,4\} = A

. (Note: The law of double complementation states

(\mathrm{P}^{\prime})^{\prime} = P

for any set P).

(vi)

(\mathrm{B}-\mathrm{C})^{\prime}

First, find

\mathrm{B}-\mathrm{C} = \{2,4,6,8\} - \{3,4,5,6\} = \{2,8\}

. Then,

(\mathrm{B}-\mathrm{C})^{\prime} = U - (\mathrm{B}-\mathrm{C}) = \{1,2,3,4,5,6,7,8,9\} - \{2,8\} = \{1,3,4,5,6,7,9\}

.

Q2EXERCISE 1.5

If $\mathrm{U}=\{a, b, c, d, e, f, g, h\}$ , find the complements of the following sets :

(i)

$\mathrm{A}=\{a, b, c\}$

(ii)

$\mathrm{B}=\{d, e, f, g\}$

(iii)

$\mathrm{C}=\{a, c, e, g\}$

(iv)

$\mathrm{D}=\{f, g, h, a\}$

Solution

Given: Universal set

\mathrm{U}=\{a, b, c, d, e, f, g, h\}

.

(i)

\mathrm{A}=\{a, b, c\}

\mathrm{A}^{\prime} = U - A = \{d, e, f, g, h\}

(ii)

\mathrm{B}=\{d, e, f, g\}

\mathrm{B}^{\prime} = U - B = \{a, b, c, h\}

(iii)

\mathrm{C}=\{a, c, e, g\}

\mathrm{C}^{\prime} = U - C = \{b, d, f, h\}

(iv)

\mathrm{D}=\{f, g, h, a\}

\mathrm{D}^{\prime} = U - D = \{b, c, d, e\}

Q3EXERCISE 1.5

Taking the set of natural numbers as the universal set, write down the complements of the following sets:

(i)

$\{x: x$ is an even natural number}

(ii)

$\{x: x$ is an odd natural number}

(iii)

$\{x: x$ is a positive multiple of 3}

(iv)

$\{x: x$ is a prime number}

(v)

$\{x: x$ is a natural number divisible by 3 and 5}

(vi)

$\{x: x$ is a perfect square}

(vii)

$\{x: x$ is a perfect cube}

(viii)

$\{x: x+5=8\}$

(ix)

$\{x: 2 x+5=9\}$

(x)

$\{x: x \geq 7\}$ (xi) $\{x: x \in \mathrm{N}$ and $2 x+1>10\}$

Solution

Given: Universal set

U = \mathbf{N} = \{1, 2, 3, 4, \dots\}

. Let A be the given set. We need to find

\mathrm{A}^{\prime} = U - A

.

(i)

Let

\mathrm{A} = \{x: x \text{ is an even natural number}\}

.

\mathrm{A}^{\prime} = \{x: x \text{ is an odd natural number}\}

.

(ii)

Let

\mathrm{A} = \{x: x \text{ is an odd natural number}\}

.

\mathrm{A}^{\prime} = \{x: x \text{ is an even natural number}\}

.

(iii)

Let

\mathrm{A} = \{x: x \text{ is a positive multiple of 3}\}

.

\mathrm{A}^{\prime} = \{x: x \in \mathbf{N} \text{ and } x \text{ is not a multiple of 3}\}

.

(iv)

Let

\mathrm{A} = \{x: x \text{ is a prime number}\}

.

\mathrm{A}^{\prime} = \{x: x \text{ is a positive composite number or } x=1\}

.

(v)

Let

\mathrm{A} = \{x: x \text{ is a natural number divisible by 3 and 5}\}

. This means

x

is a multiple of 15.

\mathrm{A}^{\prime} = \{x: x \in \mathbf{N} \text{ and } x \text{ is not a multiple of 15}\}

.

(vi)

Let

\mathrm{A} = \{x: x \text{ is a perfect square}\}

.

\mathrm{A}^{\prime} = \{x: x \in \mathbf{N} \text{ and } x \text{ is not a perfect square}\}

.

(vii)

Let

\mathrm{A} = \{x: x \text{ is a perfect cube}\}

.

\mathrm{A}^{\prime} = \{x: x \in \mathbf{N} \text{ and } x \text{ is not a perfect cube}\}

.

(viii)

Let

\mathrm{A} = \{x: x+5=8\}

. Solving the equation,

x=3

. So

\mathrm{A} = \{3\}

.

\mathrm{A}^{\prime} = \{x: x \in \mathbf{N} \text{ and } x \neq 3\}

.

(ix)

Let

\mathrm{A} = \{x: 2x+5=9\}

. Solving,

2x=4

, so

x=2

. So

\mathrm{A} = \{2\}

.

\mathrm{A}^{\prime} = \{x: x \in \mathbf{N} \text{ and } x \neq 2\}

.

(x)

Let

\mathrm{A} = \{x: x \geq 7\} = \{7, 8, 9, \dots\}

.

\mathrm{A}^{\prime} = \{x: x \in \mathbf{N} \text{ and } x < 7\} = \{1, 2, 3, 4, 5, 6\}

.

(xi) Let

\mathrm{A} = \{x: x \in \mathrm{N} \text{ and } 2x+1>10\}

. Solving the inequality,

2x > 9

, so

x > 4.5

. Since

x

is a natural number,

\mathrm{A} = \{5, 6, 7, \dots\}

.

\mathrm{A}^{\prime} = \{x: x \in \mathbf{N} \text{ and } x \leq 4.5\} = \{1, 2, 3, 4\}

.

Q4EXERCISE 1.5

If $\mathrm{U}=\{1,2,3,4,5,6,7,8,9\}, \mathrm{A}=\{2,4,6,8\}$ and $\mathrm{B}=\{2,3,5,7\}$ . Verify that

(i)

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

(ii)

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

Solution

Given:

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

A=\{2,4,6,8\}

B=\{2,3,5,7\}

First, let's find the complements of A and B.

\mathrm{A}^{\prime} = U - A = \{1,3,5,7,9\}

\mathrm{B}^{\prime} = U - B = \{1,4,6,8,9\}

(i) Verify $(\mathrm{A} \cup \mathrm{B})^{\prime}=\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}$ (De Morgan's First Law)

LHS:

(\mathrm{A} \cup \mathrm{B})^{\prime}

First,

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

. Then,

(\mathrm{A} \cup \mathrm{B})^{\prime} = U - (\mathrm{A} \cup \mathrm{B}) = \{1,9\}

.

RHS:

\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}

\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime} = \{1,3,5,7,9\} \cap \{1,4,6,8,9\} = \{1,9\}

.

Since LHS = RHS, the law is verified.

(ii) Verify $(\mathrm{A} \cap \mathrm{B})^{\prime}=\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}$ (De Morgan's Second Law)

LHS:

(\mathrm{A} \cap \mathrm{B})^{\prime}

First,

\mathrm{A} \cap \mathrm{B} = \{2,4,6,8\} \cap \{2,3,5,7\} = \{2\}

. Then,

(\mathrm{A} \cap \mathrm{B})^{\prime} = U - (\mathrm{A} \cap \mathrm{B}) = \{1,3,4,5,6,7,8,9\}

.

RHS:

\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}

\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime} = \{1,3,5,7,9\} \cup \{1,4,6,8,9\} = \{1,3,4,5,6,7,8,9\}

.

Since LHS = RHS, the law is verified.

Q5EXERCISE 1.5

Draw appropriate Venn diagram for each of the following :

(i)

$(\mathrm{A} \cup \mathrm{B})^{\prime}$ ,

(ii)

$\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}$ ,

(iii)

$(\mathrm{A} \cap \mathrm{B})^{\prime}$ ,

(iv)

$\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}$

Solution

Since diagrams cannot be drawn, a description of the shaded region in a Venn diagram is provided for each case. The universal set U is represented by a rectangle, and sets A and B are represented by two overlapping circles inside the rectangle.

(i)

$(\mathrm{A} \cup \mathrm{B})^{\prime}$ This represents the complement of the union of A and B. The union,

\mathrm{A} \cup \mathrm{B}

, is the entire area covered by both circle A and circle B. The complement is everything outside of this combined area. Description of shaded region: The region inside the rectangle but outside of both circles A and B.

(ii)

$\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}$ This represents the intersection of the complement of A and the complement of B.

\mathrm{A}^{\prime}

is everything outside circle A.

\mathrm{B}^{\prime}

is everything outside circle B. The intersection is the region that is outside A AND outside B. Description of shaded region: The region inside the rectangle but outside of both circles A and B. (This is the same region as in (i), illustrating De Morgan's law).

(iii)

$(\mathrm{A} \cap \mathrm{B})^{\prime}$ This represents the complement of the intersection of A and B. The intersection,

\mathrm{A} \cap \mathrm{B}

, is the overlapping (lens-shaped) region of the two circles. The complement is everything outside of this specific region. Description of shaded region: The entire region inside the rectangle except for the overlapping part of circles A and B.

(iv)

$\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}$ This represents the union of the complement of A and the complement of B.

\mathrm{A}^{\prime}

is everything outside circle A.

\mathrm{B}^{\prime}

is everything outside circle B. The union includes any region that is outside A OR outside B (or both). Description of shaded region: The entire region inside the rectangle except for the overlapping part of circles A and B. (This is the same region as in (iii), illustrating the other De Morgan's law).

Q6EXERCISE 1.5

Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from $60^{\circ}$ , what is $\mathrm{A}^{\prime}$ ?

Solution

Given: U = The set of all triangles in a plane. A = The set of all triangles with at least one angle different from

60^{\circ}

.

To Find:

\mathrm{A}^{\prime}

\mathrm{A}^{\prime}

is the complement of A, which means

\mathrm{A}^{\prime} = U - A

. This is the set of all triangles in the plane that are NOT in set A.

Set A contains all triangles that are not equilateral. If a triangle has at least one angle different from

60^{\circ}

, it cannot be equilateral (since an equilateral triangle has all three angles equal to

60^{\circ}

). Conversely, if a triangle is not equilateral, it must have at least one angle different from

60^{\circ}

.

So, set A is the set of all non-equilateral triangles.

The complement,

\mathrm{A}^{\prime}

, will be the set of all triangles that do NOT have at least one angle different from

60^{\circ}

. This means it is the set of triangles where the statement "at least one angle is different from

60^{\circ}

" is false. The negation of "at least one" is "none". So, the complement is the set of triangles where no angle is different from

60^{\circ}

.

If no angle is different from

60^{\circ}

, it means all angles must be equal to

60^{\circ}

. A triangle with all angles equal to

60^{\circ}

is an equilateral triangle.

Final Answer:

\mathrm{A}^{\prime}

is the set of all equilateral triangles.

Q7EXERCISE 1.5

Fill in the blanks to make each of the following a true statement :

(i)

$\mathrm{A} \cup \mathrm{A}^{\prime}=\dots$

(ii)

$\phi^{\prime} \cap \mathrm{A}=\dots$

(iii)

$\mathrm{A} \cap \mathrm{A}^{\prime}=\dots$ .

(iv)

$\mathrm{U}^{\prime} \cap \mathrm{A}=\dots$

Solution

(i)

\mathrm{A} \cup \mathrm{A}^{\prime} = \mathbf{U}

Reason: The union of a set and its complement includes all elements in the set and all elements not in the set, which together make up the universal set.

(ii)

\phi^{\prime} \cap \mathrm{A} = \mathbf{A}

Reason: The complement of the empty set is the universal set,

\phi^{\prime} = U

. The statement becomes

U \cap A

. The intersection of the universal set and any of its subsets A is the set A itself.

(iii)

\mathrm{A} \cap \mathrm{A}^{\prime} = \mathbf{\phi}

Reason: The intersection of a set and its complement contains elements that are both in A and not in A, which is impossible. Thus, the intersection is the empty set.

(iv)

\mathrm{U}^{\prime} \cap \mathrm{A} = \mathbf{\phi}

Reason: The complement of the universal set is the empty set,

U^{\prime} = \phi

. The statement becomes

\phi \cap A

. The intersection of the empty set with any set A is the empty set.

Q1Miscellaneous Exercise on Chapter 1

Decide, among the following sets, which sets are subsets of one and another: $\mathrm{A}=\{x: x \in \mathbf{R}$ and $x$ satisfy $x^{2}-8 x+12=0\}$ , $\mathrm{B}=\{2,4,6\}, \quad \mathrm{C}=\{2,4,6,8, \dots\}, \mathrm{D}=\{6\}$ .

Solution

Step 1: Write all sets in roster form.

Set A: We need to solve the quadratic equation

x^2 - 8x + 12 = 0

.

x^2 - 6x - 2x + 12 = 0

x(x - 6) - 2(x - 6) = 0

(x - 2)(x - 6) = 0

The solutions are

x=2

and

x=6

. So,

\mathrm{A} = \{2, 6\}

.

Set B:

\mathrm{B} = \{2, 4, 6\}

.

Set C:

\mathrm{C} = \{2, 4, 6, 8, \dots\}

. This is the set of all positive even integers.

Set D:

\mathrm{D} = \{6\}

.

Step 2: Compare the sets to find subset relationships.

Compare A and B: A = $\{2, 6\}$ , B = $\{2, 4, 6\}$ . Every element of A is in B. Therefore, A $\subset$ B.
Compare A and C: A = $\{2, 6\}$ , C = $\{2, 4, 6, 8, \dots\}$ . Every element of A is in C. Therefore, A $\subset$ C.
Compare A and D: A = $\{2, 6\}$ , D = $\{6\}$ . Not all elements of A are in D. So A is not a subset of D.
Compare B and C: B = $\{2, 4, 6\}$ , C = $\{2, 4, 6, 8, \dots\}$ . Every element of B is in C. Therefore, B $\subset$ C.
Compare B and D: B = $\{2, 4, 6\}$ , D = $\{6\}$ . Not all elements of B are in D. So B is not a subset of D.
Compare D and A: D = $\{6\}$ , A = $\{2, 6\}$ . Every element of D is in A. Therefore, D $\subset$ A.
Compare D and B: D = $\{6\}$ , B = $\{2, 4, 6\}$ . Every element of D is in B. Therefore, D $\subset$ B.
Compare D and C: D = $\{6\}$ , C = $\{2, 4, 6, 8, \dots\}$ . Every element of D is in C. Therefore, D $\subset$ C.

Summary of subset relationships:

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

Q2Miscellaneous Exercise on Chapter 1

In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.

(i)

If $x \in \mathrm{A}$ and $\mathrm{A} \in \mathrm{B}$ , then $x \in \mathrm{B}$

(ii)

If $\mathrm{A} \subset \mathrm{B}$ and $\mathrm{B} \in \mathrm{C}$ , then $\mathrm{A} \in \mathrm{C}$

(iii)

If $\mathrm{A} \subset \mathrm{B}$ and $\mathrm{B} \subset \mathrm{C}$ , then $\mathrm{A} \subset \mathrm{C}$

(iv)

If $\mathrm{A} \not\subset \mathrm{B}$ and $\mathrm{B} \not\subset \mathrm{C}$ , then $\mathrm{A} \not\subset \mathrm{C}$

(v)

If $x \in \mathrm{A}$ and $\mathrm{A} \not\subset \mathrm{B}$ , then $x \in \mathrm{B}$

(vi)

If $\mathrm{A} \subset \mathrm{B}$ and $x \notin \mathrm{B}$ , then $x \notin \mathrm{A}$

Solution

(i)

False. Counterexample: Let

x = 1

,

\mathrm{A} = \{1\}

, and

\mathrm{B} = \{\{1\}, 2\}

. Here,

x \in \mathrm{A}

(since

1 \in \{1\}

) and

\mathrm{A} \in \mathrm{B}

(since

\{1\}

is an element of B). However,

x \notin \mathrm{B}

because the elements of B are

\{1\}

and 2, not 1.

(ii)

False. Counterexample: Let

\mathrm{A} = \{1\}

,

\mathrm{B} = \{1, 2\}

, and

\mathrm{C} = \{\{1, 2\}, 3\}

. Here,

\mathrm{A} \subset \mathrm{B}

(since

1 \in \{1,2\}

) and

\mathrm{B} \in \mathrm{C}

(since

\{1,2\}

is an element of C). However,

\mathrm{A} \notin \mathrm{C}

because the elements of C are

\{1,2\}

and 3, not

\{1\}

.

(iii)

True. Proof: Let

x

be an arbitrary element of A, i.e.,

x \in \mathrm{A}

. Since

\mathrm{A} \subset \mathrm{B}

, it implies that if

x \in \mathrm{A}

, then

x \in \mathrm{B}

. Since

\mathrm{B} \subset \mathrm{C}

, it implies that if

x \in \mathrm{B}

, then

x \in \mathrm{C}

. Combining these, if

x \in \mathrm{A}

, then

x \in \mathrm{B}

, and consequently

x \in \mathrm{C}

. Thus, every element of A is also an element of C. Therefore,

\mathrm{A} \subset \mathrm{C}

.

(iv)

False. Counterexample: Let

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

,

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

, and

\mathrm{C} = \{1, 2, 4\}

. Here,

\mathrm{A} \not\subset \mathrm{B}

(because

1 \in \mathrm{A}

but

1 \notin \mathrm{B}

). And

\mathrm{B} \not\subset \mathrm{C}

(because

3 \in \mathrm{B}

but

3 \notin \mathrm{C}

). However,

\mathrm{A} \subset \mathrm{C}

(because both 1 and 2 are in C).

(v)

False. Counterexample: Let

x = 1

,

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

, and

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

. Here,

x \in \mathrm{A}

(since

1 \in \{1,2\}

) and

\mathrm{A} \not\subset \mathrm{B}

(since

1 \notin \mathrm{B}

). However,

x \notin \mathrm{B}

(since

1 \notin \{2,3\}

).

(vi)

True. Proof (by contradiction): We are given

\mathrm{A} \subset \mathrm{B}

and

x \notin \mathrm{B}

. We want to prove that

x \notin \mathrm{A}

. Let's assume the opposite, that

x \in \mathrm{A}

. Since

\mathrm{A} \subset \mathrm{B}

, every element of A must also be an element of B. So, if

x \in \mathrm{A}

, it must follow that

x \in \mathrm{B}

. But this contradicts our given information that

x \notin \mathrm{B}

. Therefore, our assumption that

x \in \mathrm{A}

must be false. Hence,

x \notin \mathrm{A}

.

Q3Miscellaneous Exercise on Chapter 1

Let $\mathrm{A}, \mathrm{B}$ , and C be the sets such that $\mathrm{A} \cup \mathrm{B}=\mathrm{A} \cup \mathrm{C}$ and $\mathrm{A} \cap \mathrm{B}=\mathrm{A} \cap \mathrm{C}$ . Show that $\mathrm{B}=\mathrm{C}$ .

Solution

To Prove: B = C

Proof: To show that B = C, we need to show that B

\subset

C and C

\subset

B.

Part 1: Show B $\subset$ C Let

x

be an arbitrary element of B. So,

x \in \mathrm{B}

. This implies

x \in \mathrm{A} \cup \mathrm{B}

. Since we are given

\mathrm{A} \cup \mathrm{B} = \mathrm{A} \cup \mathrm{C}

, it follows that

x \in \mathrm{A} \cup \mathrm{C}

. This means

x \in \mathrm{A}

or

x \in \mathrm{C}

.

Case 1: If

x \in \mathrm{A}

. Since we also know

x \in \mathrm{B}

, it must be that

x \in \mathrm{A} \cap \mathrm{B}

. We are given

\mathrm{A} \cap \mathrm{B} = \mathrm{A} \cap \mathrm{C}

, so

x \in \mathrm{A} \cap \mathrm{C}

. This implies

x \in \mathrm{C}

.

Case 2: If

x \in \mathrm{C}

. This directly shows that

x

is in C.

In both cases, if

x \in \mathrm{B}

, then

x \in \mathrm{C}

. Therefore, B $\subset$ C.

Part 2: Show C $\subset$ B Let

y

be an arbitrary element of C. So,

y \in \mathrm{C}

. This implies

y \in \mathrm{A} \cup \mathrm{C}

. Since we are given

\mathrm{A} \cup \mathrm{B} = \mathrm{A} \cup \mathrm{C}

, it follows that

y \in \mathrm{A} \cup \mathrm{B}

. This means

y \in \mathrm{A}

or

y \in \mathrm{B}

.

Case 1: If

y \in \mathrm{A}

. Since we also know

y \in \mathrm{C}

, it must be that

y \in \mathrm{A} \cap \mathrm{C}

. We are given

\mathrm{A} \cap \mathrm{B} = \mathrm{A} \cap \mathrm{C}

, so

y \in \mathrm{A} \cap \mathrm{B}

. This implies

y \in \mathrm{B}

.

Case 2: If

y \in \mathrm{B}

. This directly shows that

y

is in B.

In both cases, if

y \in \mathrm{C}

, then

y \in \mathrm{B}

. Therefore, C $\subset$ B.

Since B

\subset

C and C

\subset

B, we can conclude that B = C.

Hence Proved.

Q4Miscellaneous Exercise on Chapter 1

Show that the following four conditions are equivalent :

(i)

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

(ii)

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

(iii)

$\mathrm{A} \cup \mathrm{B}=\mathrm{B}$

(iv)

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

Solution

To show that the four conditions are equivalent, we will prove the following chain of implications: (i)

\Rightarrow

(ii)

\Rightarrow

(iii)

\Rightarrow

(iv)

\Rightarrow

(i).

(i) $\Rightarrow$ (ii): Assume A $\subset$ B. Show A - B = $\phi$ . Let's assume A - B is not empty. This means there exists an element

x

such that

x \in \mathrm{A} - \mathrm{B}

. By definition of set difference, this means

x \in \mathrm{A}

and

x \notin \mathrm{B}

. But this contradicts our assumption that A

\subset

B (which states that every element of A must be in B). Therefore, our assumption that A - B is not empty must be false. Hence, A - B =

\phi

.

(ii) $\Rightarrow$ (iii): Assume A - B = $\phi$ . Show A $\cup$ B = B. Let

x

be an element of A

\cup

B. This means

x \in \mathrm{A}

or

x \in \mathrm{B}

. We need to show that

x

must be in B. If

x \in \mathrm{B}

, we are done. If

x \in \mathrm{A}

, we must show it is also in B. Since A - B =

\phi

, there are no elements that are in A but not in B. This means every element of A must also be in B. So if

x \in \mathrm{A}

, then

x \in \mathrm{B}

. Thus, any element of A

\cup

B is also an element of B. So, A

\cup

B

\subset

B. Also, we know that for any two sets, B

\subset

A

\cup

B. Since A

\cup

B

\subset

B and B

\subset

A

\cup

B, we have A

\cup

B = B.

(iii) $\Rightarrow$ (iv): Assume A $\cup$ B = B. Show A $\cap$ B = A. We know that A

\cap

B

\subset

A is always true. Now, let's show A

\subset

A

\cap

B. Let

x \in \mathrm{A}

. This implies

x \in \mathrm{A} \cup \mathrm{B}

. Since A

\cup

B = B, it follows that

x \in \mathrm{B}

. So, we have

x \in \mathrm{A}

and

x \in \mathrm{B}

. This means

x \in \mathrm{A} \cap \mathrm{B}

. Thus, A

\subset

A

\cap

B. Since A

\cap

B

\subset

A and A

\subset

A

\cap

B, we have A

\cap

B = A.

(iv) $\Rightarrow$ (i): Assume A $\cap$ B = A. Show A $\subset$ B. Let

x

be an arbitrary element of A. So,

x \in \mathrm{A}

. Since A = A

\cap

B, we can say that

x \in \mathrm{A} \cap \mathrm{B}

. By definition of intersection, this means

x \in \mathrm{A}

and

x \in \mathrm{B}

. In particular,

x \in \mathrm{B}

. Since every element of A is also an element of B, we have A

\subset

B.

Since we have shown (i)

\Rightarrow

(ii)

\Rightarrow

(iii)

\Rightarrow

(iv)

\Rightarrow

(i), all four conditions are equivalent.

Hence Proved.

Q5Miscellaneous Exercise on Chapter 1

Show that if $\mathrm{A} \subset \mathrm{B}$ , then $\mathrm{C}-\mathrm{B} \subset \mathrm{C}-\mathrm{A}$ .

Solution

Given: A

\subset

B To Prove: C - B

\subset

C - A

Proof: Let

x

be an arbitrary element of the set C - B. So,

x \in \mathrm{C} - \mathrm{B}

. By the definition of set difference, this means: (1)

x \in \mathrm{C}

(2)

x \notin \mathrm{B}

We are given that A

\subset

B. This means that if an element is in A, it must also be in B. The contrapositive of this statement is also true: if an element is not in B, then it cannot be in A.

From (2), we know

x \notin \mathrm{B}

. Using the contrapositive of A

\subset

B, since

x \notin \mathrm{B}

, we can conclude that

x \notin \mathrm{A}

.

Now we have two pieces of information about

x

:

from (1), $x \in \mathrm{C}$
from our deduction, $x \notin \mathrm{A}$

By the definition of set difference, if

x \in \mathrm{C}

and

x \notin \mathrm{A}

, then

x \in \mathrm{C} - \mathrm{A}

.

We started by taking an arbitrary element

x

from C - B and have shown that it must also be in C - A. Therefore, every element of C - B is an element of C - A.

This proves that C - B $\subset$ C - A.

Hence Proved.

Q6Miscellaneous Exercise on Chapter 1

Show that for any sets A and B, $A=(A \cap B) \cup (A-B)$ and $A \cup (B-A)=(A \cup B)$

Solution

Part 1: Show A = (A $\cap$ B) $\cup$ (A - B)

We can prove this using properties of sets. Start with the Right Hand Side (RHS):

(A \cap B) \cup (A - B)

By definition of set difference, A - B = A

\cap

B'.

RHS = (A \cap B) \cup (A \cap B')

Using the distributive law (A

\cap

distributes over

\cup

is not what we have, but A is common):

RHS = A \cap (B \cup B')

By the complement law, B

\cup

B' = U (the universal set).

RHS = A \cap U

By the identity law, the intersection of any set with the universal set is the set itself.

RHS = A

Since RHS = A = LHS, the identity is proved.

Part 2: Show A $\cup$ (B - A) = (A $\cup$ B)

Start with the Left Hand Side (LHS):

A \cup (B - A)

By definition of set difference, B - A = B

\cap

A'.

LHS = A \cup (B \cap A')

Using the distributive law (

\cup

distributes over

\cap

):

LHS = (A \cup B) \cap (A \cup A')

By the complement law, A

\cup

A' = U.

LHS = (A \cup B) \cap U

By the identity law, the intersection of any set with the universal set is the set itself.

LHS = A \cup B

Since LHS = A

\cup

B = RHS, the identity is proved.

Hence Proved.

Q7Miscellaneous Exercise on Chapter 1

Using properties of sets, show that

(i)

$\mathrm{A} \cup(\mathrm{A} \cap \mathrm{B})=\mathrm{A}$

(ii)

$\mathrm{A} \cap(\mathrm{A} \cup \mathrm{B})=\mathrm{A}$ .

Solution

These are known as the absorption laws.

(i) To Show: A $\cup$ (A $\cap$ B) = A

Proof: We know that for any set X, X

\subset

A. So, (A

\cap

B)

\subset

A. When we take the union of a set with one of its subsets, the result is the larger set itself. Since (A

\cap

B) is a subset of A, their union must be A. Therefore, A

\cup

(A

\cap

B) = A.

Alternatively, using distributive laws:

LHS = A \cup (A \cap B)

We can write A as A

\cap

U.

LHS = (A \cap U) \cup (A \cap B)

Using the distributive law:

LHS = A \cap (U \cup B)

Since B is a subset of U, U

\cup

B = U.

LHS = A \cap U

LHS = A = RHS

(ii) To Show: A $\cap$ (A $\cup$ B) = A

Proof: We know that for any set X, A

\subset

A

\cup

X. So, A

\subset

(A

\cup

B). When we take the intersection of a set with one of its supersets, the result is the smaller set itself. Since A is a subset of (A

\cup

B), their intersection must be A. Therefore, A

\cap

(A

\cup

B) = A.

Alternatively, using distributive laws:

LHS = A \cap (A \cup B)

Using the distributive law:

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

By the idempotent law, A

\cap

A = A.

LHS = A \cup (A \cap B)

From part (i) of this question, we already proved that A

\cup

(A

\cap

B) = A.

LHS = A = RHS

Hence Proved.

Q8Miscellaneous Exercise on Chapter 1

Show that $\mathrm{A} \cap \mathrm{B}=\mathrm{A} \cap \mathrm{C}$ need not imply $\mathrm{B}=\mathrm{C}$ .

Solution

To Show:

\mathrm{A} \cap \mathrm{B}=\mathrm{A} \cap \mathrm{C}

does not necessarily mean

\mathrm{B}=\mathrm{C}

.

To show this, we need to provide a counterexample where the first condition is true, but the conclusion is false.

Counterexample: Let's choose three sets A, B, and C. Let

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

. Let

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

. Let

\mathrm{C} = \{2, 4\}

.

Now, let's check the condition

\mathrm{A} \cap \mathrm{B}=\mathrm{A} \cap \mathrm{C}

.

\mathrm{A} \cap \mathrm{B} = \{1, 2\} \cap \{2, 3\} = \{2\}

.

\mathrm{A} \cap \mathrm{C} = \{1, 2\} \cap \{2, 4\} = \{2\}

. So, the condition

\mathrm{A} \cap \mathrm{B}=\mathrm{A} \cap \mathrm{C}

is satisfied, as both are equal to

\{2\}

.

Now, let's check the conclusion B = C.

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

\mathrm{C} = \{2, 4\}

Clearly,

\mathrm{B} \neq \mathrm{C}

because

3 \in \mathrm{B}

but

3 \notin \mathrm{C}

, and

4 \in \mathrm{C}

but

4 \notin \mathrm{B}

.

Since we have found an example where

\mathrm{A} \cap \mathrm{B}=\mathrm{A} \cap \mathrm{C}

is true but

\mathrm{B}=\mathrm{C}

is false, we have shown that the implication is not always true.

Q9Miscellaneous Exercise on Chapter 1

Let A and B be sets. If $\mathrm{A} \cap \mathrm{X}=\mathrm{B} \cap \mathrm{X}=\phi$ and $\mathrm{A} \cup \mathrm{X}=\mathrm{B} \cup \mathrm{X}$ for some set X , show that $\mathrm{A}=\mathrm{B}$ . (Hints $\mathrm{A}=\mathrm{A} \cap(\mathrm{A} \cup \mathrm{X}), \mathrm{B}=\mathrm{B} \cap(\mathrm{B} \cup \mathrm{X})$ and use Distributive law)

Solution

Given:

$\mathrm{A} \cap \mathrm{X}=\phi$
$\mathrm{B} \cap \mathrm{X}=\phi$
$\mathrm{A} \cup \mathrm{X}=\mathrm{B} \cup \mathrm{X}$

To Prove: A = B

Proof: We start with set A. We know from the absorption law that

\mathrm{A} = \mathrm{A} \cap (\mathrm{A} \cup \mathrm{X})

.

Using the given condition (3), we can replace

(\mathrm{A} \cup \mathrm{X})

with

(\mathrm{B} \cup \mathrm{X})

.

A = A \cap (B \cup X)

Now, we use the distributive law to expand the expression:

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

Using the given condition (1), we know that

\mathrm{A} \cap \mathrm{X} = \phi

.

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

The union of any set with the empty set is the set itself.

A = A \cap B \quad ...(i)

This result implies that A is a subset of B (A

\subset

B).

Now, we repeat the process starting with set B. We know that

\mathrm{B} = \mathrm{B} \cap (\mathrm{B} \cup \mathrm{X})

.

Using the given condition (3), we can replace

(\mathrm{B} \cup \mathrm{X})

with

(\mathrm{A} \cup \mathrm{X})

.

B = B \cap (A \cup X)

Using the distributive law:

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

Using the given condition (2), we know that

\mathrm{B} \cap \mathrm{X} = \phi

.

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

Since intersection is commutative (

B \cap A = A \cap B

):

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

B = A \cap B \quad ...(ii)

This result implies that B is a subset of A (B

\subset

A).

From (i), we have

A = A \cap B

, which means A

\subset

B. From (ii), we have

B = A \cap B

, which means B

\subset

A.

Since A

\subset

B and B

\subset

A, it must be that A = B.

Hence Proved.

Q10Miscellaneous Exercise on Chapter 1

Find sets $\mathrm{A}, \mathrm{B}$ and C such that $\mathrm{A} \cap \mathrm{B}, \mathrm{B} \cap \mathrm{C}$ and $\mathrm{A} \cap \mathrm{C}$ are non-empty sets and $\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}=\phi$ .

Solution

Goal: Find three sets A, B, and C that satisfy two conditions:

Each pair of sets has a non-empty intersection (they overlap pairwise).
The intersection of all three sets is empty (there is no region common to all three).

We can think of this visually using a Venn diagram. We need three circles that overlap with each other, but not all at the same point.

Let's construct the sets by assigning elements to these overlapping regions.

Let's put an element common to only A and B. Let this be 1. Let's put an element common to only B and C. Let this be 2. Let's put an element common to only A and C. Let this be 3.

Now, let's define the sets based on these elements.

Let

\mathrm{A} = \{1, 3\}

. Let

\mathrm{B} = \{1, 2\}

. Let

\mathrm{C} = \{2, 3\}

.

Let's check the conditions:

Pairwise intersections:
- $\mathrm{A} \cap \mathrm{B} = \{1, 3\} \cap \{1, 2\} = \{1\}$ . This is non-empty.
- $\mathrm{B} \cap \mathrm{C} = \{1, 2\} \cap \{2, 3\} = \{2\}$ . This is non-empty.
- $\mathrm{A} \cap \mathrm{C} = \{1, 3\} \cap \{2, 3\} = \{3\}$ . This is non-empty. The first condition is satisfied.
Intersection of all three:
- $\mathrm{A} \cap \mathrm{B} \cap \mathrm{C} = (\mathrm{A} \cap \mathrm{B}) \cap \mathrm{C} = \{1\} \cap \{2, 3\} = \phi$ . The second condition is satisfied.

Final Answer: One possible example of such sets is:

\mathrm{A} = \{1, 3\}

\mathrm{B} = \{1, 2\}

\mathrm{C} = \{2, 3\}

SetsClass 11 Mathematics NCERT Solutions

Which of the following are sets? Justify your answer.

The collection of all the months of a year beginning with the letter J.

The collection of ten most talented writers of India.

A team of eleven best-cricket batsmen of the world.

The collection of all boys in your class.

The collection of all natural numbers less than 100.

A collection of novels written by the writer Munshi Prem Chand.

The collection of all even integers.

The collection of questions in this Chapter.

A collection of most dangerous animals of the world.

Solution

Let A={1,2,3,4,5,6}\mathrm{A}=\{1,2,3,4,5,6\}A={1,2,3,4,5,6}. Insert the appropriate symbol ∈\in∈ or ∉\notin∈/ in the blank spaces:

5…A5 \dots \mathrm{A}5…A

8…A8 \dots \mathrm{A}8…A

0…A0 \dots \mathrm{A}0…A

4…A4 \dots \mathrm{A}4…A

2…A2 \dots \mathrm{A}2…A

10…A10 \dots \mathrm{A}10…A

Solution

Write the following sets in roster form:

A={x:x\mathrm{A}=\{x: xA={x:x is an integer and −3≤x<7}-3 \leq x<7\}−3≤x<7}

B={x:x\mathrm{B}=\{x: xB={x:x is a natural number less than 6}(iii)(iii)(iii)\mathrm{C}={x: xisatwo−digitnaturalnumbersuchthatthesumofitsdigitsis8} is a two-digit natural number such that the sum of its digits is 8\}isatwo−digitnaturalnumbersuchthatthesumofitsdigitsis8}

D={x:x\mathrm{D}=\{x: xD={x:x is a prime number which is divisor of 60}(v)(v)(v)\mathrm{E}=ThesetofalllettersinthewordTRIGONOMETRY(vi)The set of all letters in the word TRIGONOMETRY (vi)ThesetofalllettersinthewordTRIGONOMETRY(vi)\mathrm{F}=$ The set of all letters in the word BETTER

Solution

Write the following sets in the set-builder form :

(3,6,9,12)(3,6,9,12)(3,6,9,12)

{2,4,8,16,32}\{2,4,8,16,32\}{2,4,8,16,32}

{5,25,125,625}\{5,25,125,625\}{5,25,125,625}

{2,4,6,… }\{2,4,6, \dots\}{2,4,6,…}

{1,4,9,…,100}\{1,4,9, \dots, 100\}{1,4,9,…,100}

Solution

List all the elements of the following sets :

E={x:x\mathrm{E}=\{x: xE={x:x is a month of a year not having 31 days}(vi)(vi)(vi)\mathrm{F}={x: xisaconsonantintheEnglishalphabetwhichprecedesis a consonant in the English alphabet which precedesisaconsonantintheEnglishalphabetwhichprecedesk}$

Solution

Match each of the set on the left in the roster form with the same set on the right described in set-builder form:

{1,2,3,6}\{1,2,3,6\}{1,2,3,6}

{2,3}\{2,3\}{2,3}

{M,A,T,H,E,I,C,S}

{1,3,5,7,9}\{1,3,5,7,9\}{1,3,5,7,9}

Solution

Which of the following are examples of the null set

Set of odd natural numbers divisible by 2

Set of even prime numbers

{x:x\{x: x{x:x is a natural numbers, x<5x<5x<5 and x>7}x>7\}x>7}

{y:y\{y: y{y:y is a point common to any two parallel lines}

Solution

Which of the following sets are finite or infinite

The set of months of a year

{1,2,3,… }\{1,2,3, \dots\}{1,2,3,…}

{1,2,3,…99,100}\{1,2,3, \dots 99, 100\}{1,2,3,…99,100}

The set of positive integers greater than 100

The set of prime numbers less than 99

Solution

State whether each of the following set is finite or infinite:

The set of lines which are parallel to the xxx-axis

The set of letters in the English alphabet

The set of numbers which are multiple of 5

The set of animals living on the earth

The set of circles passing through the origin (0,0)(0,0)(0,0)

Solution

In the following, state whether A=B\mathrm{A}=\mathrm{B}A=B or not:

A={a,b,c,d}B={d,c,b,a}\mathrm{A}=\{a, b, c, d\} \quad \mathrm{B}=\{d, c, b, a\}A={a,b,c,d}B={d,c,b,a}

A={4,8,12,16}B={8,4,16,18}\mathrm{A}=\{4, 8, 12, 16\} \quad \mathrm{B}=\{8, 4, 16, 18\}A={4,8,12,16}B={8,4,16,18}

A={2,4,6,8,10}B={x:x\mathrm{A}=\{2, 4, 6, 8, 10\} \quad \mathrm{B}=\{x: xA={2,4,6,8,10}B={x:x is positive even integer and x≤10}x \leq 10\}x≤10}

A={x:x\mathrm{A}=\{x: xA={x:x is a multiple of 10}, B={10,15,20,25,30,… }\quad \mathrm{B}=\{10, 15, 20, 25, 30, \dots\}B={10,15,20,25,30,…}

Solution

Are the following pair of sets equal ? Give reasons.

A={2,3},B={x:x\mathrm{A}=\{2,3\}, \quad \mathrm{B}=\{x: xA={2,3},B={x:x is solution of x2+5x+6=0}x^{2}+5 x+6=0\}x2+5x+6=0}

A={x:x\mathrm{A}=\{x: xA={x:x is a letter in the word FOLLOW} B={y:y\mathrm{B}=\{y: yB={y:y is a letter in the word WOLF}

Solution

From the sets given below, select equal sets : A=\{2,4,8,12\}, & B=\{1,2,3,4\}, & C=\{4,8,12,14\}, & D=\{3,1,4,2\} \nE=\{-1,1\}, & F=\{0, a\}, & G=\{1,-1\}, & H=\{0,1\} \end{array}$$

Solution

Make correct statements by filling in the symbols ⊂\subset⊂ or ⊄\not\subset⊂ in the blank spaces :

{2,3,4}…{1,2,3,4,5}\{2,3,4\} \dots \{1,2,3,4,5\}{2,3,4}…{1,2,3,4,5}

{a,b,c}…{b,c,d}\{a, b, c\} \dots \{b, c, d\}{a,b,c}…{b,c,d}

{x:x\{x: x{x:x is a student of Class XI of your school} …{x:x\dots \{x: x…{x:x student of your school}

{x:x\{x: x{x:x is a circle in the plane} …{x:x\dots \{x: x…{x:x is a circle in the same plane with radius 1 unit}

{x:x\{x: x{x:x is a triangle in a plane} …{x:x\dots \{x: x…{x:x is a rectangle in the plane}

{x:x\{x: x{x:x is an equilateral triangle in a plane} …{x:x\dots \{x: x…{x:x is a triangle in the same plane}

Let $\mathrm{A}=\{1,2,3,4,5,6\}$ . Insert the appropriate symbol $\in$ or $\notin$ in the blank spaces:

$5 \dots \mathrm{A}$

$8 \dots \mathrm{A}$

$0 \dots \mathrm{A}$

$4 \dots \mathrm{A}$

$2 \dots \mathrm{A}$

$10 \dots \mathrm{A}$

$\mathrm{A}=\{x: x$ is an integer and $-3 \leq x<7\}$

$\mathrm{B}=\{x: x$ is a natural number less than 6} $(iii)$ \mathrm{C}={x: x $is a two-digit natural number such that the sum of its digits is 8\}$

$\mathrm{D}=\{x: x$ is a prime number which is divisor of 60} $(v)$ \mathrm{E}= $The set of all letters in the word TRIGONOMETRY (vi)$ \mathrm{F}=$ The set of all letters in the word BETTER

$(3,6,9,12)$

$\{2,4,8,16,32\}$

$\{5,25,125,625\}$

$\{2,4,6, \dots\}$

$\{1,4,9, \dots, 100\}$

$\mathrm{E}=\{x: x$ is a month of a year not having 31 days} $(vi)$ \mathrm{F}={x: x $is a consonant in the English alphabet which precedes$ k}$

$\{1,2,3,6\}$

$\{2,3\}$

$\{1,3,5,7,9\}$

$\{x: x$ is a natural numbers, $x<5$ and $x>7\}$

$\{y: y$ is a point common to any two parallel lines}

$\{1,2,3, \dots\}$

$\{1,2,3, \dots 99, 100\}$

The set of lines which are parallel to the $x$ -axis

The set of circles passing through the origin $(0,0)$

In the following, state whether $\mathrm{A}=\mathrm{B}$ or not:

$\mathrm{A}=\{a, b, c, d\} \quad \mathrm{B}=\{d, c, b, a\}$

$\mathrm{A}=\{4, 8, 12, 16\} \quad \mathrm{B}=\{8, 4, 16, 18\}$

$\mathrm{A}=\{2, 4, 6, 8, 10\} \quad \mathrm{B}=\{x: x$ is positive even integer and $x \leq 10\}$

$\mathrm{A}=\{x: x$ is a multiple of 10}, $\quad \mathrm{B}=\{10, 15, 20, 25, 30, \dots\}$

$\mathrm{A}=\{2,3\}, \quad \mathrm{B}=\{x: x$ is solution of $x^{2}+5 x+6=0\}$

$\mathrm{A}=\{x: x$ is a letter in the word FOLLOW} $\mathrm{B}=\{y: y$ is a letter in the word WOLF}

From the sets given below, select equal sets : $A=\{2,4,8,12\}, & B=\{1,2,3,4\}, & C=\{4,8,12,14\}, & D=\{3,1,4,2\} \nE=\{-1,1\}, & F=\{0, a\}, & G=\{1,-1\}, & H=\{0,1\} \end{array}$$$

Make correct statements by filling in the symbols $\subset$ or $\not\subset$ in the blank spaces :

$\{2,3,4\} \dots \{1,2,3,4,5\}$

$\{a, b, c\} \dots \{b, c, d\}$

$\{x: x$ is a student of Class XI of your school} $\dots \{x: x$ student of your school}

$\{x: x$ is a circle in the plane} $\dots \{x: x$ is a circle in the same plane with radius 1 unit}

$\{x: x$ is a triangle in a plane} $\dots \{x: x$ is a rectangle in the plane}

$\{x: x$ is an equilateral triangle in a plane} $\dots \{x: x$ is a triangle in the same plane}

$\{x: x$ is an even natural number} $\dots \{x: x$ is an integer}

$\{a, b\} \not\subset \{b, c, a\}$

$\{a, e\} \subset \{x: x$ is a vowel in the English alphabet}

$\{1,2,3\} \subset \{1,3,5\}$

$\{a\} \subset \{a, b, c\}$

$\{a\} \in \{a, b, c\}$

$\{x: x$ is an even natural number less than 6} $\subset \{x: x$ is a natural number which divides 36}

Let $\mathrm{A}=\{1,2,\{3,4\}, 5\}$ . Which of the following statements are incorrect and why?

$\{3,4\} \subset \mathrm{A}$

$\{3,4\} \in \mathrm{A}$

$\{\{3,4\}\} \subset \mathrm{A}$

$1 \in \mathrm{A}$

$1 \subset \mathrm{A}$

$\{1,2,5\} \subset \mathrm{A}$

$\{1,2,5\} \in \mathrm{A}$

$\{1,2,3\} \subset \mathrm{A}$

$\phi \in \mathrm{A}$

$\phi \subset \mathrm{A}$ (xi) $\{\phi\} \subset \mathrm{A}$

$\{a\}$

$\{a, b\}$

$\{1,2,3\}$

$\phi$

$\{x: x \in \mathrm{R},-4<x \leq 6\}$

$\{x: x \in \mathrm{R},-12<x<-10\}$

$\{x: x \in \mathrm{R}, 0 \leq x<7\}$

$\{x: x \in \mathrm{R}, 3 \leq x \leq 4\}$

$(-3,0)$

$[6,12]$

$(6,12]$

$[-23,5)$

Given the sets $\mathrm{A}=\{1,3,5\}, \mathrm{B}=\{2,4,6\}$ and $\mathrm{C}=\{0,2,4,6,8\}$ , which of the following may be considered as universal set ( s ) for all the three sets $\mathrm{A}, \mathrm{B}$ and C

$\{0,1,2,3,4,5,6\}$

$\phi$

$\{0,1,2,3,4,5,6,7,8,9,10\}$

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

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

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

Let $\mathrm{A}=\{a, b\}, \mathrm{B}=\{a, b, c\}$ . Is $\mathrm{A} \subset \mathrm{B}$ ? What is $\mathrm{A} \cup \mathrm{B}$ ?

If A and B are two sets such that $\mathrm{A} \subset \mathrm{B}$ , then what is $\mathrm{A} \cup \mathrm{B}$ ?

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