1
Sets
The set theory was developed by German Mathematician Georg Cantor
(1845-1918). He first encountered sets while working on ‘problems on
trigonometric series’. This concept is used in every branch of Mathematics i.e.,
Geometry, Algebra, etc.
1 Sets and Their Types
In our daily life, while performing our regular work, we often come across a
variety of things that occur in groups.
e.g., Team of cricket players, group of tall boys, group of teachers, etc.
The words used above like team, group, etc., convey the idea of certain
collections.
3 Venn Diagrams and Operations
Well-defined Collection of Objects
on Sets
If any given collection of objects is in such a way that it is possible to tell
without any doubt whether a given object belongs to this collection or not,
then such a collection of objects is called a well-defined collection of objects.
e.g., ‘The rivers of India’ is a well-defined collection. Since, we can say that the
river Nile does not belong to this collection. On the other hand, the river
Ganga does belong to this collection.
Difference between Not Well-defined Collection and Well-defined Collection
Not Well-defined Collection
Well-defined Collection
A group of intelligent students.
A group of students scoring more than 95% marks of
your school.
A group of most talented writers A group of odd natural numbers less than 25.
of India.
Group of pretty girls.
3 Sets and Their Types
3 Subsets of Set
Group of girls of class XI of your school.
Here, a group of intelligent students, a group of most talented writers of India,
group of pretty girls are not well-defined collections, because we can not
decide whether a given particular object belongs to the given collection or not.
1
3 Applications of Set Theory
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2
Definition of Set
A well-defined collection of objects, is called a set.
Sets are usually denoted by the capital letters A, B, C, X, Y
and Z etc.
The elements of a set are represented by small letters a, b, c,
x, y and z etc.
If a is an element of a set A, then we say that a belongs to A.
The word belongs to denoted by the Greek symbol
∈(epsilon).
Thus, in notation form, a belongs to set A is written as
a ∈ A and b does not belongs to set A is written as b ∉ A.
e.g.,
(i) 6 ∈ N , N being the set of natural numbers and 0 ∉ N .
(ii) 36 ∈ P , P being the set of perfect square numbers, so
5 ∉ P.
* Some examples of sets used particularly in Mathematics are
N → The set of natural numbers. {1, 2, 3, …}
Z → The set of all integers. {..., −3, − 2, − 1, 0, 1, 2, 3, ... }
Q → The set of all rational numbers.
3 2
1 1 2 3 

..., − , − , − , , , , ...
4
3
2
2 3 4 

R → The set of real numbers.
(rational and irrational numbers)
Z + → The set of positive integers. {1, 2, 3, … }
Q + → The set of positive rational numbers.
1 2 3 4 
 , , , , ...
2 3 4 5 
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Mathematics Class 11th
(iii) Here, we can definitely decide whether a given novel
belongs to this collection or not.
So, this collection is well-defined. Hence, it is a set.
(iv) The term most dangerous is not a well-defined term. An
animal may be most dangerous for one person and may
not be for the other.
So, it is not well-defined. Hence, it is not a set.
Representation of Sets
Sets are generally represented by following two ways
1. Roster or Tabular Form or Listing Method
2. Set-builder Form or Rule Method
1. Roster or Tabular Form or Listing Method
In this form, all the elements of a set are listed, the
elements are being separated by commas and are
enclosed within curly braces { }.
e.g.,
(i) The set of all natural numbers less than 10 is
represented in roster form as {1, 2, 3, 4, 5, 6, 7,
8, 9}.
(ii) The set of prime natural numbers is {2, 3, 5, 7, K}.
Here, three dots tell us that the list of prime
natural numbers continue indefinitely.
Note (i) In roster form, order in which the elements are listed is not
important.
Hence, the set of natural numbers less than 10 can also be
written as {2, 4, 1, 3, 5, 6, 8, 7, 9} instead of {1, 2, 3, 4, 5,
6, 7, 8, 9}.
Here, the order of listing elements is not important.
(ii) In roster form, element is not repeated i.e., all the elements
are taken as distinct.
Hence, the set of letters forming the word
‘MISCELLANEOUS’ is {M, I, S, C, E, L, A, N, O, U}.
R + → The set of positive real numbers.
(positive rational and irrational numbers)
Note Objects, elements and members of a set are synonymous terms.
2. Set-builder Form or Rule Method
In this form, all the elements of set possess a single
common property p ( x ) , which is not possessed by any
other element outside the set.
In such a case, the set is described by { x : p ( x ) holds}.
e.g.,
In the set {a , e , i , o, u}, all the elements possess a
common property, i.e., each of them is a vowel of the
English alphabet and no other letter possesses this
property.
If we denote the set of vowels by V, then we write
V = {x : x is a vowel in the English alphabet}
The above description of the set V is read as ‘‘the
set of all x such that x is a vowel of the English
alphabet.’’
Example 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 collection of novels written by the writer Munshi Prem
Chand.
(iv) A collection of most dangerous animals of the world.
Solution
(i) We are sure that members of this collection are January, June
and July.
So, this collection is well-defined. Hence, it is a set.
(ii) A writer may be most talented for one person and may not be
for other. Therefore, we cannot definitely decide which writer
will be there in the collection.
So, this collection is not well-defined. Hence, it is not a set.
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3
Sets
Some examples are given below
(i) Set of all natural numbers less than 10,
A = { x : x ∈ N , x < 10}
(ii) Set of all real numbers,
R = {x : x ∈ R }
Example 2. Describe the set
(i) “The set of all vowels in the word EQUATION” in
Roster form.
(ii) “The set of reciprocals of natural numbers” in
set-builder form.
Solution
(i) The word “EQUATION” has following vowels i.e., A, E,
I, O and U.
Hence, the required set can be described in roster form as
{A, E, I, O, U}.
(ii) Given set can be described in set-builder form as { x : x
is a reciprocal of a natural number}
1


or
x : x = , n ∈ N 
n


* Working Rule to Write the Set in the
Set-builder Form
To convert the given set in set-builder form, we use the following
steps
Step I Describe the elements of the set by using a symbol x or any
other symbol y , z etc.
Step II Write the symbol colon ‘:’.
Step III After the sign of colon, write the characteristic property
possessed by the elements of the set.
Example 3. Describe the following set in Roster
form.
{x : x is positive integer and a divisor of 9}
Step IV Enclose the whole description within braces i.e., { }.
Worked out Problem
Write the following set A = {14, 21, 28, 35, 42, ..., 98}
in set-builder form.
Solution Here, x is a positive integer and a divisor of 9.
So, x can take values 1, 3, 9.
∴ {x : x is a positive integer and a divisor of 9} = {1, 3, 9}
Step I Describe the elements of the set by using a symbol.
Let x represent the elements of given set.
Example 4. Write the set of all natural numbers x such
that 4x + 9 < 50 in roster form.
Step II Write the symbol colon.
Write the symbol ‘:’ after x.
/
Step III Find the characteristic property possessed by the
Firstly, simplify the inequality and then listed all the natural
numbers under given condition.
Solution We have, 4x + 9 < 50
elements of the set.
Given numbers are all natural numbers greater than 7 which are
multiples of 7 and less than 100.
⇒
Step IV Enclose the whole description within braces.
Thus, A = {x : x is a set of natural numbers greater than 7 which
are multiples of 7 and less than 100}.
Which is the required set- builder form of given set.
⇒
∴
x < 10.25
Since, x is a natural number, so x can take values 1, 2, 3,
4, 5, 6, 7, 8, 9, 10.
∴ Required set = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Representation of a Statement in Both Form
Statement
Roster form
Set- builder form
The set of all
natural numbers
between 10
and 14.
{11, 12, 13}
{x : x ∈ N , 10 < x < 14}
The set of all
schools in Delhi
beginning with
letter D.
{DAV School, DPS School, {x : x is name of school
Decent School, ...}
in Delhi beginning with
letter D}
4x + 9 − 9 < 50 − 9
[subtracting 9 from both sides]
41
4x < 41 ⇒ x <
4
Example 5. Describe the following set in set-builder
form.
B = {53, 59, 61, 67, 71, 73, 79, 83, 89, 97}
The set of all
{F, r, i, e, n, d}
distinct letters
used in the word
‘Friend’.
/
Firstly, find the characteristic property possessed by the
elements of the set, then write in set-builder form.
Solution The given set is
{53, 59, 61, 67, 71, 73, 79, 83, 89, 97}.
We observe that these numbers are all prime numbers
between 50 and 100.
∴ Given set, B = { x : x is a prime number and
50 < x < 100}
{x : x is the distinct
letters used in the word
‘Friend’}
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Types of Sets
1. Finite Set
A set, which is empty or consists of a definite
number of elements, is called a finite set.
e.g.,
(i) The set {1, 2, 3, 4} is a finite set, because it
contains a definite number of elements
i.e., only 4 elements.
(ii) The set of solutions of x 2 = 25 is a finite
set, because it contains a definite number
of elements i.e., 5 and − 5.
(iii) An empty set, which does not contain any
element is also a finite set.
The number of distinct elements in a finite set A
is called cardinal number of set and it is denoted
by n ( A).
e.g., If A = { − 3, − 1, 8, 10, 13}, then n ( A) = 5.
2. Infinite Set
A set which consists of infinite number of
elements is called an infinite set.
When we represent an infinite set in the roster
form, it is not possible to write all the elements
within braces { } because the number of elements
of such a set is not finite, so we write a few
elements which clearly indicate the structure of
the set following by three dots.
e.g., Set of squares of natural numbers is an
infinite set, because such natural numbers are
infinite and it can be represented as
{4, 9, 16, 25 ...}.
Note All infinite sets cannot be described in the roster
form. For example, the set of real numbers cannot be
described in this form, because the elements of this
set do not follow any particular pattern.
3. Empty Set
A set, which does not contain any element, is
called an empty set or null set or void set. It is
denoted by φ or { }.
e.g., A = { x : x is a natural number less than 1}
We know that, there is no natural number less
than one. Therefore, set A contains no element
and hence it is an empty set.
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Mathematics Class 11th
4. Singleton Set
A set, consisting of a single element, is called a singleton set.
e.g., (i) The sets {0}, {5}, {− 7} are singleton sets.
(ii) A = { x : x + 8 = 0, x ∈ Z } is a singleton set, because this
set contains only one integer, namely − 8.
5. Equivalent Sets
Two sets A and B are equivalent, if their cardinal numbers are
same i.e., n ( A) = n (B) .
e.g., Let A = { a , b, c , d } and B = {1, 2, 3, 4}, then n( A) = 4 and
n( B ) = 4.
Therefore, A and B are equivalent sets.
6. Equal Sets
Two sets A and B are said to be equal, if they have exactly the same
elements and we write A = B . Otherwise, two sets are said to be
unequal and we write A ≠ B .
e.g., Let A = { a , b, c , d } and B = {c , d , b, a }, then A = B , because
each element of A is in B and vice-versa.
Note A set does not change, if one or more elements of the set are repeated.
e.g., The sets A = {1, 4, 5} and B = {1, 1, 4, 5, 5} are equal because elements
of A is in B and vice-versa. That’s why, we generally do not repeat any
element in describing a set.
Example 6. Identify which of the following set is an empty
set, singleton set, infinite set or equal sets.
(i) A = {x : x is a girl being living on the Jupiter}
(ii) B = {x : x is a letter in the word ‘‘MARS’’}
(iii) C = { y : y is a letter in the word ‘‘ARMS’’}
(iv) D = {x : 3x − 2 = 0, x ∈ Q }
(v) E = {x : x ∈ N and x is an odd number}
Solution
(i) A = { x : x is a girl being living on the Jupiter}
We know that, there is no human being or any girl being living on
the Jupiter. Hence, A is an empty set.
(ii) B = { x : x is a letter in the word ‘‘MARS’’}
⇒ B = {M, A, R, S}
(iii) C = { y : y is a letter in the word ‘‘ARMS’’}
⇒ C = {A, R, M, S}
Here, we observe that the elements of sets B and C are exactly
same, hence these sets are equal.
(iv) D = { x : 3x − 2 = 0, x ∈ Q }
2
 2


Q 3x − 2 = 0 ⇒ x = ∈ Q
⇒D =  

3
3


Hence, D is a singleton set.
(v) E = { x : x ∈ N and x is an odd number}
Clearly, it is an infinite set because there are infinite natural
numbers which are odd.
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5
Sets
Example 7. Which of the following pairs of sets are
equal? Justify your answer.
Here, we see that both sets have exactly the same elements.
∴
A=B
(i) A = {x : x is a letter of the word ‘‘LOYAL’’},
B = {x : x is a letter of the word ‘‘ALLOY’’}
(ii) Given, A = { x : x ∈ Z and x 2 < 8}
= { − 2, − 1, 0, 1, 2}
(ii) A = {x : x ∈ Z and x 2 < 8},
B = {x : x ∈ R and x 2 − 4x + 3 = 0}
/
[Qx 2 < 8 ⇒ − 2 2 < x < 2 2 and as x ∈ Z
∴ x ∈ { − 2, − 1, 0, 1, 2}]
Firstly, describe the given sets in the roster form and check
whether they have exactly same elements.
Solution
(i) Given,
and
and
B = { x : x ∈ R and x − 4x + 3 = 0} = {1, 3}
2
[Q x 2 − 4x + 3 = 0 ⇒( x − 1)( x − 3) = 0 ⇒ x = 1, 3]
A = { x : x is a letter in the word ‘‘LOYAL’’}
= {L, O, Y, A} = {A, L, O, Y}
B = { x : x is a letter of the word ‘‘ALLOY’’}
= {A, L, O, Y}
Here, we see that set A has 5 distinct elements and set B has
2 distinct elements. So, they do not have same elements.
∴
A≠B
EXAM
Very Short Answer Type Questions
[1 Mark each]
A = {1, 3, 5, 7, 9, 11, 13, 15}, then insert the
appropriate symbol ∈or ∉in each of the following
blank spaces.
1. If
(i) 1... A
(iii) 9 ... A
6. Use listing method to express the set
A = {x : x = n3 , n ∈ N and x < 80}.
Sol. A = {1, 8, 27, 64}
(ii) 6 ... A
(iv) 14 ... A
7. List the elements of the set
Sol. Given set is A = {1, 3, 5, 7, 9, 11, 13, 15}
(i) 1 ∈ A
(ii) 6 ∉ A
(iii) 9 ∈ A
(iv) 14 ∉ A
1
9

A =  x : x is an integer, − < x < 
2
2

Sol. A = {0, 1, 2, 3, 4}

n2 − 1

D = x : x = 2
, n ∈ N and n < 4  in roster form.
n +1


 3 4
[1]
Sol. D = 0, , 
 5 5
2. Write {x : x is an integer and − 3 ≤ x < 7} in roster
form.
[1]
3. Write set A = {3, 6, 9, 12, 15} in set-builder form.
9. Describe
Sol. A = {x : x is a natural number multiple of 3 and x < 18}. [1]
(i) “The
set
of
vowels
in
the
word
MATHEMATICS” in roster form.
(ii) “The set of all odd natural numbers’’ in set
builder form.
4. Write set A={1, 4, 9, ..., 100} in set-builder form.
Sol. A = { x : x = n 2 , n ∈ N and n < 11}
Sol. (i) The word “MATHEMATICS” has following vowels
i.e., A, E, I.
Hence, the required set can be described in roster form
as {A, E, I}.
[1/2]
(ii) An odd natural number can be written in the form
( 2n − 1).
Hence, given set can be described in set-builder form as
[1/2]
{ x : x = 2n − 1, n ∈ N }
[1]
3 4 5 6 7
, ,
,
,  in
 2 5 10 17 26 37 50 
5. Write set D =  1 , 2 ,
set-builder form.


n
Sol. D =  x : x = 2
, n ∈ N and n ≤ 7
n +1


[1]
8. Express the set
[1]
Sol. { − 3, − 2, − 1, 0, 1, 2, 3, 4, 5, 6}
[1]
[1]
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Short Answer Type Questions
10.
12. Which of the following sets are finite and which
Which of the following sets are empty?
are infinite?
(i) The set of the months of a year.
(ii) {1, 2, 3, ...}
(iii) The set of prime numbers less than 99.
A set which does not contain any element is called the
empty set.
Sol. (i) It is a finite set, as there are 12 members in the set which
are months of the year.
[1]
(ii) It is an infinite set, since there are infinite natural numbers.
Sol. (i) A = {x : x ∈ N and x ≤ 1}
Since, x ≤ 1i.e., x < 1and x = 1, which is a natural number.
∴
A = {1}
[1]
So, this is not an empty set.
(ii) B = {x : 3x + 1 = 0, x ∈ N }
Since, 3x + 1 = 0
−1
⇒
x=
∉N
3
Thus, set B does not contain any element.
Hence, set B is an empty set.
[1 12 ]
(iii) C = {x : 2 < x < 3, x ∈ N }
Since, there is no natural number between 2 and 3.
Hence, set C is an empty set.
[1 12 ]
11.
[1 12 ]
(iii) It is a finite set, because the set is {2, 3, 5, 7, ..., 97}. [1 12 ]
13. From the following sets, select equal sets.
A = {2, 4, 6, 8},
C = {−2, 4, 6, 8},
/
 1
1 
⇒ ( 4x − 40) 
+
 =0
(13 − x ) ( x − 7)
⇒
Two sets are said to be equal, if they have exactly the same
elements.
[1]
Therefore, A and E are equal sets.
[1]
Similarly, B = {1, 2, 3, 4, 5} and D = { 2, 3, 5, 4, 1}
Since, each element of set B presents in set D and vice-versa.
Therefore, B and D are also equal sets.
14.
[1]
[1]
Are the following pair of sets equal? Give reason.
(i) A = {2, 3} and B = {x : x is a solution of
x 2 + 5 x + 6 = 0}
(ii) A = {x : x is a letter in the word ‘‘FOLLOW’’}
and B = {y : y is a letter in the word ‘‘WOLF’’}
[1]
/ Firstly, convert the given sets in roster form and then
check whether they have exactly the same elements.
Sol. (i) Here, A = { 2, 3}
and B = {x : x is a solution of x 2 + 5x + 6 = 0}
First, we find the solution of x 2 + 5x + 6 = 0.
Now,
x 2 + 3x + 2x + 6 = 0
⇒
x ( x + 3) + 2( x + 3) = 0
⇒
( x + 2)( x + 3) = 0 ⇒ x = − 2, − 3
[1]
∴
B = { −2, − 3}
Since, the elements of A and B are not same, therefore
[1]
A ≠ B.
(ii) Here, A = {x : x is a letter of the word ‘‘FOLLOW’’}
= {F, O, L, W}
and B = { y : y is a letter of the word ‘‘WOLF’’}
[1]
= {W, O, L, F}
Since, every element of A is in B and every element of B
is in A i.e., both have exactly same elements.
[1]
∴
A=B
[1]
[1]


6
( 4x − 40) 
 =0
(13 − x ) ( x − 7)
⇒
4x − 40 = 0
∴
x = 10
Hence, T is not an empty set.
B = {1, 2, 3, 4, 5},
D = {2, 3, 5, 4, 1}, E = {8, 6, 2, 4}
Sol. A = { 2, 4, 6, 8} and E = {8, 6, 2, 4}
Since, each element of set A presents in set E and vice-versa.
 x +5
4 x − 40 
Let T =  x :
−5=
.
13 − x 
 x −7
Is T an empty set? Justify your answer.
Sol. The given set T is not an empty set.
Justification
4x − 40
x +5
−5=
Q
13 − x
x−7
x + 5 5 4x − 40
− =
⇒
x − 7 1 13 − x
x + 5 − 5x + 35 4x − 40
=
⇒
13 − x
x−7
40 − 4x 4x − 40
=
⇒
x−7
13 − x
− ( 4x − 40) 4x − 40
=0
−
⇒
x−7
13 − x
Mathematics Class 11th
[4 Marks each]
(i) A = {x : x ∈ N and x ≤ 1}
(ii) B = {x : 3 x + 1 = 0, x ∈ N}
(iii) C = {x :2 < x < 3, x ∈ N}
/
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[1]
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TRY YOURSELF
1 Mark Questions
1.
2.
3.
Write set A = {x : x is an integer and x 2 < 20} in the roster form.
4.
Write the following sets in the roster form.
(i) B = { x : x 2 = x, x ∈ R}
(ii) C = {x : x is a positive factor of a prime number p}
5.
Write the following sets in the set-builder form.
(i) { 5, 25, 125, 625}
(ii) {2, 4, 6, 8, 10}
Write set A = {x : x is a planet} in the roster form.
Which of the following is/are the examples of an empty set?
(i) Set of all even prime numbers.
(ii) {x : x is a point common to any two parallel lines}
4 Marks Questions
6.
From the sets given below, pair the equivalent sets.
A = {1, 2, 3}, B = { t, p, q, r, s}, C = {α, β, γ } and D = {a, e, i, o, u}
7.
From the sets given below, select empty set, singleton set, infinite set and equal sets.
(i) A = {x : x < 1and x > 3}
(ii) B = { x : x 3 − 1 = 0, x ∈ R}
(iii) C = {x : x ∈ N and x is a prime number}
(v) E = {x : x is positive even integer and x ≤ 10}
(iv) D = {2, 4, 6, 8, 10}
8.
From the sets given below, select equal sets and equivalent sets.
A = { 0, a},
B = {1, 2, 3, 4}, C = { 4, 8, 12},
D = { 3, 1, 2, 4}, E = {1, 0},
F = { 8, 4, 12},
G = {1, 5, 7, 11} and H = {a, b}
9.
State which of the following statements are true and which are false? Justify your answer.
(i)
(ii)
(iii)
(iv)
35 ∈{x : x has exactly four positive factors}
128 ∈{y : the sum of all the positive factors of y is 2y}
3 ∉{ x : x 4 − 5x 3 + 2 x 2 − 112 x + 6 = 0}
496 ∉{y : the sum of all the positive factors of y is 2y}
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Subsets of Set
Subset
Let A and B be two sets. If every element of A is an
element of B, then A is called a subset of B.
If A is a subset of B, then we write A ⊂ B , which is
read as ‘‘A is a subset of B ” or A is contained in B.
In other words, A ⊆ B , if whenever a ∈ A, then
a ∈ B . It is often convenient to use the symbol
‘‘⇒’’ which means implies. Using this symbol, we
can write the definition of subset as follows
A ⊂ B , if x ∈ A
⇒
x ∈B
The above statement is read as
A is subset of B, if x is an element of A, then it implies
that x is also an element of B.
If A is not a subset of B, then we write A ⊆
/ B.
If A ⊂ B , then B is called a superset of A written as
B ⊃ A.
e.g., Consider the sets A and B, where set A denotes
the set of all students in your class, B denotes the
set of all students in your school. We observe that,
every element of A is also an element of B.
Therefore, we can say that A is subset of B i.e.,
A ⊂ B.
If it happens for both sets A and B, i.e., every
element of A is in B and every element of B is in A,
then in this case, A and B are same sets. Thus we
have,
A ⊆ B and B ⊆ A ⇔ A = B, where ‘‘⇔’’ is a
symbol for two ways implications and is usually
read as if and only if (iff ).
Note (i) {1} ⊂ {1, 2, 3}
(ii) 2 ⊂ {1, 2, 3}, which is not possible.
Proper Subset
If A ⊆ B and A ≠ B , then A is called a proper
subset of B, written as A ⊂ B and B is called proper
superset of A.
e.g., Let A = {x : x is an even natural number}
and
B = { x : x is a natural number}
Then, A = {2, 4, 6, 8,...}
and
B = {1, 2, 3, 4, 5,...}
∴
A⊂B
❖
Some Important Results
1. Every set is a subset of itself.
Proof Let A be any set. Then, each element of A is clearly in A. Hence,
A ⊆ A.
2. The empty set φ is a subset of every set.
Proof Let Abe any set and φ be the empty set. To show that φ ⊆ A, i.e., to
show that every element of φ is an element of Aalso. But we know that
empty set φ contains no element. So, every element of φ is in A. Hence,
φ ⊆ A.
3. A set itself and an empty set are always subsets of every set and set itself
is called improper subset of the set.
4. The total number of subsets and proper subset of a finite set containing n
elements is 2n and 2n − 1, respectively.
5. If A ⊆ B andB ⊆ C, then A ⊆ C.
6. A = B, if and only if A ⊆ B andB ⊆ A.
Example 1. Let A = {1, 2, {3, 4}, 5}. Which of the following
statements are incorrect and why?
(i) {3, 4} ⊂ A
(iv) 1 ∈ A
(ii) {3, 4} ∈ A
(iii) {{3, 4}} ⊂ A
(v) 1 ⊂ A
(vi) {1, 2, 5} ⊂ A
(vii) {1, 2, 5} ∈ A (viii) φ ⊂ A
(ix) φ ∈ A
(x) {φ} ⊂ A
Solution We have, A = {1, 2, {3, 4}, 5}
(i) Since, {3, 4} is a member of set A.
{3, 4} ∈ A
∴
Hence, {3, 4} ⊂ A is incorrect.
(ii) Since, {3, 4} is a member of set A.
Hence, {3, 4} ∈ A is correct.
(iii) Since, {3, 4} is a member of set A.
So,
{{3, 4}} is a subset of A.
Hence, {{3, 4}} ⊂ A is correct.
(iv) Since, 1 is a member of A.
Hence, 1 ∈ A is correct.
(v) Since, 1 is a member of set A.
Hence, 1 ⊂ A is incorrect.
(vi) Since, 1, 2, 5 are members of set A.
So,
{1, 2, 5} is a subset of set A.
Hence, {1, 2, 5} ⊂ A is correct.
(vii) Since, 1, 2 and 5 are members of set A.
So,
{1, 2, 5} is a subset of A.
Hence, {1, 2, 5} ∈ A is incorrect.
(viii) Since, φ is subset of every set. Hence, φ ⊂ A is correct.
(ix) Since, φ is not a member of set A.
Hence, φ ∈ A is incorrect.
(x) Since, φ is not a member of set A.
Hence, { φ} ⊂ A is incorrect.
8
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9
Sets
Subsets of the Set of Real
Numbers
5. Irrational Numbers
A number which cannot be written in the form p/q, where
p and q both are integers and q ≠ 0, is called an irrational
number i.e., a number which is not rational is called an
irrational number.
The set of irrational numbers is denoted by T.
T = { x : x ∈ R and x ∉Q }
Rational numbers and irrational numbers taken together,
are known as real numbers. Thus, every real number is
either a rational or an irrational number. The set of real
numbers is denoted by R.
There are many important subsets of set of real numbers which
are given below
e.g., 0.535335333..., 2, 3 are irrational numbers.
Above subsets can be represented diagramatically as given
below
1. Natural Numbers
Non-negative
rational
number
The numbers being used in counting as 1, 2, 3, 4,...,
called natural numbers.
The set of natural numbers is denoted by N.
N = {1, 2, 3, 4, 5, 6, ...}
Real
number
Negative rational
number
Whole
number
Irrational
number
Natural
number
2. Whole Numbers
The natural numbers along with number 0 (zero) form
the set of whole numbers i.e., 0, 1, 2, 3, ... , are whole
numbers. The set of whole numbers is denoted by W.
W = {0, 1, 2, 3, ...}
Set of natural numbers is the proper subset of the set of
whole numbers.
⇒
N ⊂W
Intervals as Subsets of R
Let a, b belongs to R and a < b. Then, the set of real
numbers { x : a < x < b} is called an open interval and is
denoted by ( a , b ).
All the real numbers between a and b belongs to the open
interval (a, b) but a and b do not belong to this set
(interval).
The interval which contains the first and last members of
the set, is called closed interval and it is denoted by [a, b].
Here,
[ a , b ] = { x : a ≤ x ≤ b}
3. Integers
The natural numbers, their negatives and zero make the
set of integers and it is denoted by Z.
Z = {..., − 5, − 4, − 3, − 2, − 1, 0, 1, 2, 3, 4, ...}
Set of whole numbers is the proper subset of integers.
⇒
W ⊂Z
Semi-open or Semi-closed Interval
Some intervals are closed at one end and open at the
other, such intervals are called semi-closed or semi-open
intervals.
[ a , b) = { x : a ≤ x < b} is an open interval from a to b
which includes a but excludes b.
( a , b ] = { x : a < x ≤ b} is an open interval from a to b
which excludes a but includes b.
e.g.,
(i) (2, 8) is a subset of ( −1, 11).
(ii) [4, 6) is a subset of [4, 6].
(iii) The set [ 0, ∞) defines the set of non-negative real
numbers.
(iv) The set ( − ∞, 0) defines the set of negative real
numbers.
(v) ( −∞, ∞) is the set of real numbers.
4. Rational Numbers
p
A number of the form , where p and q both are integers
q
and q ≠ 0, is called a rational number (division by 0 is not
permissible).
The set of rational numbers is generally denoted by Q.
p

Q =  : p , q ∈ Z and q ≠ 0
q

All the whole numbers are also rational numbers, since
they can be represented as the ratio.
3 6 18
e.g.,
3 = , , , etc.
1 2 6
The set of integers is the proper subset of the set of
rational numbers i.e., Z ⊂ Q and N ⊂ Z ⊂ Q .
9
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10
On real line, we can draw the interval, which is shown by the dark portion on
the number line
(a, b)
a
[a, b]
b
a
b
a
[a, b)
a
b
(a, b]
b
Length of an interval The number (b − a ) is called the length of
any of the intervals ( a , b), [ a , b ], [ a , b) or ( a , b ].
Example 2. (a) Write the following as intervals
(i) {x : x ∈ R, − 5 < x ≤ 6} (ii) {x : x ∈ R, − 11 < x < − 9}
(b) Write the following as intervals and also represent on the number
line.
(i){x : x ∈ R, 2 ≤ x < 8}
(ii) {x : x ∈ R, 5 ≤ x ≤ 6}
/
If an inequality is of the form ≤ or ≥, then we use the symbol of closed
interval, otherwise we use the symbol of open interval.
Solution
(a) (i) { x : x ∈ R, − 5 < x ≤ 6} is the set that does not contain −5 but
contains 6. So, it can be written as an interval whose first end point is
open and last end point is closed i.e., ( −5, 6].
(ii) { x : x ∈ R, − 11 < x < − 9} is the set that neither contains−11 nor −9,
so it can be represented as open interval i.e., ( −11, − 9).
(b) (i) { x : x ∈ R, 2 ≤ x < 8} is the set that contains 2 but not contain 8.So, it
can be represented as an interval whose first end point is closed and
the other end point is open i.e., [2, 8).
On the real line [2, 8) may be graphed as shown in figure given
below
0 2
8
The dark portion on the number line is the required set.
(ii) { x : x ∈ R, 5 ≤ x ≤ 6} is the set which contains 5 and 6 both. So, it is
equivalent to a closed interval i.e., [5, 6].
On the real line [5, 6] may be graphed as shown in the figure given
below
0
5
6
The dark portion on the number line is the required set.
Universal Set
If there are some sets under consideration, then there happens to be a
set which is a superset of each one of the given sets. Such a set is
known as the universal set and it is denoted by U.
e.g.,
(i) Let A = {2, 4, 6}, B = {1, 3, 5} andC = {0, 7}
Then,U = {0, 1, 2, 3, 4, 5, 6, 7} is an universal set.
(ii) For the set of all integers, the universal set can be the set of
rational numbers or the set of real numbers.
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Mathematics Class 11th
Example 3. What universal sets would you
propose for each of the following?
(i) The set of right triangles.
(ii) The set of isosceles triangles.
Solution
(i) The universal set for the set of right triangles
is set of triangles.
(ii) The universal set for the set of isosceles
triangle is set of equilateral triangles or set of
triangles.
Power Set
The collection of all subsets of a set A, is
called the power set of A and it is denoted by
P(A). In P(A), every element is a set.
e.g., Let A = {3, 4, 5}
Then, P ( A) = {φ, {3}, {4}, {5}, {3, 4}, {3, 5},
{4, 5}, {3, 4, 5}}
Note (i) If set A has n elements, then the total number
of elements in its power set is 2n .
(ii) If A is an empty set φ, then P ( A ) has just one
element i.e., P( A ) = {φ}.
Properties of Power Sets
(i)
(ii)
(iii)
(iv)
Each element of a power set is a set.
If A ⊆ B , then P ( A) ⊆ P ( B ).
P ( A) ∩ P (B ) = P ( A ∩ B )
P ( A ∪ B ) ≠ P ( A) ∪ P (B )
* Method to Write the Power Set of a
Given Set
Let a set having n elements is given, then for
writing its power set, we use the following steps
Step I Write all the possible subsets having single
element of given set.
Step II Write all the possible subsets having two
elements at a time of given set.
Step III Write all the possible subsets having three
elements at a time of given set. Repeat this
process for writing all possible subsets
having n elements at a time, as the given set
has n elements.
Step IV Form a set, with the help of the subsets
obtained from steps I, II and III and
element φ.
This set will give the power set of given set.
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11
Sets
Worked out Problem
If A ={ 1, 2 , 3 } then find the power set of A.
Step I Write all the possible subsets having single
Example 4. Let A = {1, 2, 3, 4}, B = {1, 2, 3} and C = {2, 4}.
Find all sets X satisfying each pair of conditions
(i) X ⊆ B and X ⊆/ C (ii) X ⊆ B, X ≠ B and X ⊆/ C
Sol. Given, A = {1, 2, 3, 4}, B = {1, 2, 3} and C = { 2, 4}
element of given set.
Now, P ( A ) = { φ, {1}, { 2}, {3}, { 4}, {1, 2}, {1, 3}, {1, 4},
{ 2, 3}, { 2, 4}, {3, 4}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, { 2, 3, 4}, {1, 2, 3, 4}}
...(i)
...(ii)
P ( B ) = { φ, {1}, { 2}, {3}, {1, 2}, {1, 3}, { 2, 3}, {1, 2, 3}}
and
...(iii)
P (C ) = { φ, { 2}, { 4}, { 2, 4}}
(i) Now,
X ⊆ B and X ⊆/ C
⇒
X ∈ P ( B )and X ∉ P (C )
From Eqs. (ii) and (iii), we get
X = {1}, {3}, {1, 2}, {1, 3}, { 2, 3}, {1, 2, 3}
(ii) Now, X ⊆ B, X ≠ B and X ⊆/ C
⇒
X ∈ P ( B ), X ≠ B, X ∉ P (C )
From Eqs. (ii) and (iii), we get
X = {1}, {3}, {1, 2}, {1, 3}, { 2, 3}
Here, X does not contain B = {1, 2, 3} as X ≠ B.
All possible subsets of a given set having single
element are {1}, {2}, { 3}.
Step II Write all the possible subsets of a given set
having two elements at a time.
All possible subsets of a given set having two
elements at a time are {1, 2}, {2, 3}, {3, 1}.
Step III Write all the possible subsets of a given set
having three elements at a time.
All possible subsets of a given set having three
elements at a time is {1, 2, 3}.
Step IV Form a set with the help of the subsets
obtained from steps I, II and III and element φ.
Hence, the required power set is
{{1}, {2}, {3}, {1, 2}, {2, 3}, {3, 1}, {1, 2, 3}, φ}.
EXAM
Very Short Answer Type Questions
1.
Consider the following sets φ, A = {1, 2}
and B = {1, 4, 8}
Insert the following symbols ⊂ or ⊄
between each of the following pair of sets.
(i) φ ... B
(ii) A ... B
Sol. (i) Since, null set is proper subset of every set.
[1/2]
∴
φ⊂B
(ii) Given, A = {1, 2} and B = {1, 4, 8}. Since,
element 2 ∉B.
[1/2]
∴
A⊄B
2.
Prove that A ⊆ φ implies A = φ.
Sol. Given,
...(i)
A⊆φ
But φ ⊆ A [empty set is a subset of each set] …(ii)
[1/2]
From Eqs. (i) and (ii), we get
A=φ
[1/2]
3. Write the following subset of R as interval.
Also find the length of interval and
represent on number line.
{x : x ∈ R, − 12 ≤ x ≤ − 10}
[1 Mark each]
/
If inequalities are of the form ≥ or ≤, then use the symbol of closed
interval and then find the length of the interval, which is equal to the
difference of its extreme values.
Sol. { x : x ∈ R, − 12 ≤ x ≤ − 10} = [ −12, − 10]
and length of interval = − 10 − ( −12) = 2
On the real line set [−12, − 10] may be graphed as shown in figure
given below
–12
–10
0
The dark portion on the number line is the required set.
[1]
A = {a, b, {c, d}, e}. Which of the following
statements is/are true?
4. Let
(i) {c, d} ∈ A
(ii) {{c, d }} ⊂ A
Sol. Given, A = {a, b, {c, d}, e}
(i) Since a, b, {c, d} and e are elements of A.
{c, d} ∈ A
∴
[1/2]
Hence, it is a true statement.
(ii) As {c , d } ∈ A and {{c , d }} represents a set, which is a subset of A.
{{c, d}} ⊂ A
∴
[1/2]
Hence, it is a true statement.
11
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12
5. Write down the subsets of the following sets.
(i) {1, 2, 3}
statement is true or false. If it is true, then prove it.
If it is false, then give an example.
(ii) {φ}
(i) If x ∈ A and A ∈ B, then x ∈ B.
(ii) If A ⊂ B and B ∈ C, then A ∈ C.
Sol. (i) False,
Let A = { 2}, B = {{2}, 3}
Clearly, 2 ∈ A and A ∈ B, but 2 ∉B.
So, x ∈ A and A ∈ B need not imply that x ∈ B. [1/2]
(ii) False,
Let A = { 2}, B = { 2, 3} and C = {{2, 3}, 4}
Clearly, A ⊂ B and B ∈ C , but A ∉C .
Thus, A ⊂ B and B ∈ C need not imply that A ∈ C .
6. Let A = {1, 3, 5} and B = {x : x is an odd natural
number < 6}.
(ii) Is A = B?
Sol. Given, A = {1, 3, 5}
and B = {x : x is an odd natural number < 6}
⇒
B = {1, 3, 5}
(i) Yes, here A ⊆ B ,because all elements of set A are
present in set B.
[1/2]
(ii) Yes, here, B ⊆ A, because all elements of set B are
present in set A.
Hence, A = B, because both sets contain equal and same
elements.
[1/2]
[1/2]
9.
Sol. (i) Given, B = {0, 1, 3}
∴ P ( B ) = {φ, {0}, {1}, {3}, {0, 1}, {0, 3}, {1, 3}, {0, 1, 3}}
[1/2]
Short Answer Type Questions
[1/2]
[4 Marks each]
10. If A = {3, {4, 5}, 6}, then find which of the following
11. Let A and B be two sets. Then, prove that
A = B ⇔ A ⊆ B and B ⊆ A.
Sol. Given A = B
statements are true.
(i)
(ii)
(iii)
(iv)
(ii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Sol. We know that, universal set for sets A, B and C is superset
of A, B and C i.e., universal set contains all elements of A, B
and C.
(i) φ cannot be considered as universal set.
[1/2]
(ii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is the universal set for the
given sets A, B and C as all the elements of sets A, B and
C are in this set.
[1/2]
(ii) C = {1, {2}}
(ii) Given, C = {1, { 2}}
P (C ) = {φ, {1}, {{2}}, {1, {2}}}
Given, the sets A = {1, 3, 5}, B = {2, 4, 6} and
C = {0, 2, 4, 6, 8}. Which of the following may be
considered as universal set(s) for all three sets A,
B and C?
(i) φ
7. Write down the power set of the following sets.
(i) B = {0, 1, 3}
Mathematics Class 11th
8. In each of the following, determine whether the
Sol. (i) The subsets of {1, 2, 3} are
[1/2]
φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}.
(ii) Clearly, { φ} is the power set of empty set φ. Now, its
[1/2]
subsets are φ and {φ}.
(i) Is A ⊆ B?
ne
{4, 5} ⊂ A
{4, 5} ∈ A
φ⊂ A
{3, 6} ⊂ A
To prove A ⊆ B and B ⊆ A
Proof By definition of equal sets, every element of A is in B
[1]
and every element of B is in A.
∴
A ⊆ B and B ⊆ A
Thus,
A = B ⇒ A ⊆ B and B ⊆ A
[1]
Again, let A ⊆ B and B ⊆ A
By the definition of a subset, if A ⊆ B then it follows that
every element of A is in B and if B ⊆ A then it follows that
every element of B is in A.
[1]
∴
A=B
Hence,
[1]
A = B ⇔ A ⊆ B and B ⊆ A
Hence proved.
Sol. Given, A = {3, {4, 5}, 6}
Here, 3, {4, 5}, 6 all are elements of A.
(i) { 4, 5} ⊂ A, which is not true.
Since, element of any set is not a subset of any set and
[1]
here { 4, 5} is an element of A.
(ii) { 4, 5} ∈ A, is a true statement.
[1]
(iii) It is always true that φ ⊂ A.
[1]
(iv) {3, 6} makes a set, so it is a subset of A i.e., {3, 6} ⊂ A.
[1]
12
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13
Sets
(iii) False, since a ∈{a, b, c} and not {a}.
[1]
(iv) True, since all elements of both sets {1, 3, 5} and
{2, 4, 6} are present in {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
Hence, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is the universal
set for given two sets.
[1]
12. Examine
whether the following statements are
true or false.
(i) {a, b} ⊄ {b, c, a}
(ii) {a, e} ⊂ {x : x is a vowel in the English
alphabet}
(iii) {a} ∈{a, b, c}
(iv) {0,1,2,3,4,5,6,7,8,9,10} is the universal set for
the sets {1, 3, 5} and {2, 4, 6}.
13. Show that n {P [P (P (φ))]} = 4.
Sol. We have P ( φ) = { φ}
Sol. (i) Since, the elements of the set {a, b} are also present in the
set {b, c, a}.
[1]
So, { a, b } ⊂ {b, c , a }
⇒ {a, b} ⊄ {b, c, a} is false.
(ii) Vowels in the English alphabets are a, e, i, o, u.
∴{a, e} ⊂ {x : x is a vowel in English alphabet}is true. [1]
∴
P ( P ( φ)) = { φ, { φ}}
[1]
⇒ P [ P ( P φ))] = { φ, { φ}, {{ φ}}, { φ, { φ}}}
[1]
Hence, number of elements in P [ P ( P ( φ))] is 4.
[1]
i.e.,
n { P [ P ( P ( φ))]} = 4
[1]
Hence proved.
TRY YOURSELF
1 Mark Questions
Directions (Q.Nos. 1 to 3) Determine whether the statement is true or false.
1.
2.
3.
If A = {3, 6, 7}, B = {2, 3, 7, 8, 10}, then A ⊂ B.
4.
Let A = {4, 5, {6}, 7}, then which of the following statements are incorrect and why?
If A = {x : x 2 + 4x − 21 = 0, x ∈ N} and B = { −7, 3}, then A ⊆ B.
If A = {1, 7, 9}, B = {2, 4, 6} and C = { 0, 2, 4}, then U = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9} is the universal set for all three
sets.
(i) {6} ∈ A
5.
6.
7.
(ii) {6} ⊂ A
Find the power set of a set A = {0, 1, 2}.
If set A = {1, 3, 5}, then find the number of elements in P{P( A )}.
If A = { x : x = n2 , n = 1, 2, 3}, then find the number of proper subsets.
4 Marks Questions
8.
Write the following intervals in the set-builder form.
(i) (−6, 0)
(ii) [3, 21)
(iii) [2, 21]
(iv) ( − 23, 5]
9.
Write the following as intervals.
(i) { x: x ∈ R, − 3 < x ≤ 7}
(ii) { x: x ∈ R, − 11< x < −7}
(iii) { x: x ∈ R, 0 ≤ x < 11}
(iv) { x: x ∈ R, 2 ≤ x ≤ 9}
13
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3 Venn Diagrams and Operations on Sets
Venn Diagrams
(ii) Since, every student studying English
Mathematics.
Hence,
E ⊂ M ⊂U
Through, Venn diagram we represent it as
Venn diagrams are named after the English logician,
John Venn (1834-1883). Venn diagrams represent most
of the relationship between sets. These diagrams consist
of rectangles and closed curves usually circles.
In Venn diagrams, the universal set is represented by a
rectangular region and its subset are represented by
circle or a closed geometrical figure inside the universal
set. Also, an element of a set is represented by a point
within the circle of set.
e.g., If U = {1, 2, 3, 4,..., 10} and A = {1, 2, 3},
then its Venn diagram is as shown in the figure
U
M
E
Operations on Sets
There are some operations which when performed on
two sets give rise to another set. Here, we will define
certain operations on set and examine their properties.
U
10 A
4
1. Union of Sets
9
1
5
2
Let A and B be any two sets. The union of A and B is the
set of all those elements which belong to either in A or in
B or in both. It is denoted by A ∪ B and read as A union
B. The symbol ‘∪’ is used to denote the union.
∴
A ∪ B = { x : x ∈ A or x ∈ B }
e.g., Let A = {2 , 3} and B = {3, 4, 5},
then A ∪ B = {2, 3, 4, 5}
The union of sets A and B is represented by the following
Venn diagram.
8
3
7
6
Example 1. Draw the Venn diagrams to illustrate the
following relationship among sets E, M and U,
where E is the set of students studying English in a
school, M is the set of students studying
Mathematics in the same school and U is the set of
all students in that school.
(i) All the students who study Mathematics also study
English, but some students who study English do not
study Mathematics.
(ii) Not all students study Mathematics, but every
student studying English studies Mathematics.
/
studies
A
U
B
The shaded portion represents A ∪ B .
Firstly, make a relation between sets under given condition.
Then, it is easy to draw a Venn diagram.
Some Properties of Union of Sets
(i) A ∪ B = B ∪ A
[commutative law]
(ii) ( A ∪ B ) ∪ C = A ∪ ( B ∪ C )
[associative law]
Solution
Given, E = Set of students studying English
M = Set of students studying Mathematics
U = Set of all students
(i) Since, all of the students who study Mathematics also
study English, but some students who study English do
not study Mathematics.
∴
M ⊂ E ⊂U
Through, Venn diagram we represent it as
A
U
B
C
(iii) A ∪ φ = A
[law of identity element, φ is the identity of ∪]
(iv) A ∪ A = A
[idempotent law]
(v) U ∪ A = U
[law of U]
U
E
M
14
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15
Sets
(iii) φ ∩ A = φ, U ∩ A = A
[law of φ and U]
(iv) A ∩ A = A
[idempotent law]
(v) If A, B andC are any three sets, then
(a) A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C )
[distributive law i.e., ∩ distributes over ∪]
This can be shown with the Venn diagram as given below
For LHS
Example 2. Find the union of each of the following
pairs of sets.
(i) A = {a, e, i, o, u}, B = {a, c, d }
(ii) A = {1, 3, 5}, B = {2, 4, 6}
(iii) A = {x : x is a natural number and 1 < x ≤ 5}
and B = {x : x is a natural number and 5 < x ≤ 10}
/
Firstly, convert the given set in roster form, if it is not given
in that. Then union of two sets, is the set which consists of all
those elements which are either in A or in B.
A
Solution
(i)
A = { a, e, i, o, u }, B = { a, c, d }
⇒ A ∪ B = { a, c , d , e , i, o, u }
(ii)
A = {1, 3, 5}, B = { 2, 4, 6}
⇒ A ∪ B = {1, 2, 3, 4, 5, 6}
(iii)
A = { x : x is a natural number and 1 < x ≤ 5}
⇒
A = { 2, 3, 4, 5}
B = { x : x is a natural number and5 < x ≤ 10}
B = { 6, 7, 8, 9, 10}
⇒
∴ A ∪ B = { 2, 3, 4, 5} ∪ {6, 7, 8, 9, 10}
= { 2, 3, 4, 5, 6, 7, 8, 9, 10}
B
U
A
C
B
U
C
(ii) Shaded portion
represents A ∩ (B ∪ C )
(i) Shaded portion
represents (B ∪ C )
For RHS
A
B
U
B
A
C
U
C
2. Intersection of Sets
(iii) Shaded portion
represents (A ∩ B )
Let A and B be any two sets. The intersection of A and B
is the set of all those elements which belong to both A and
B. It is denoted by A ∩ B and read as A intersection B.
The symbol ‘∩’ is used to denote the intersection.
∴
A ∩ B = { x : x ∈ A and x ∈ B }
A
The intersection of sets A and B is represented by the
following Venn diagram.
A
B
(iv) Shaded portion
represents (A ∩ C )
B
U
C
(v) Shaded portion represents
(A ∩ B ) ∪ (A ∩ C )
U
(b) A ∪ ( B ∩ C ) = ( A ∪ B ) ∩ ( A ∪ C )
[∪ distributes over ∩]
Example 3.
The shaded portion represents A ∩ B .
e.g., Let A = {2, 3, 4, 5} and B = {1, 3, 6, 4}
Then, A ∩ B = {3, 4}
(i) If A = {1, 3, 5, 7, 9} and B = {2, 3, 6, 8, 9}, then find
A ∩ B.
(ii) If A = {e, f , g } and B = φ, then find A ∩ B.
(iii) If A = {x : x = 3n, n ∈ Z } and
B = {x : x = 4n, n ∈ Z }, then find A ∩ B.
Solution
(i) Given,
A = {1, 3, 5, 7, 9}
and
B = { 2, 3, 6, 8, 9}
⇒
A ∩ B = {3, 9}
[Q3 and 9 are only elements which are common]
(ii) Given, A = {e , f , g } and B = φ
⇒ A ∩ B = φ [since, there is no common element]
Some Properties of Intersection of Sets
(i) A ∩ B = B ∩ A
[commutative law]
(ii) ( A ∩ B ) ∩ C = A ∩ ( B ∩ C )
[associative law]
A
U
B
C
15
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16
(iii) Let x ∈ A ∩ B
⇒ x ∈ A and x ∈ B
⇒ x is a multiple of 3 and x is a multiple of 4.
⇒ x is a multiple of 3 and 4 both.
⇒ x is a multiple of 12.
⇒ x = 12n, n ∈ Z
Hence, A ∩ B = { x : x = 12n, n ∈ Z }
A − B = {1, 2, 3, 4, 5} − { 2, 4, 6}
A − B = {1, 3, 5}
[only those elements of A which do not belong to B]
and
B − A = { 2, 4, 6} − {1, 2, 3, 4, 5} = {6}
[only those elements of B which do not belong to A]
4. Symmetric Difference of Two Sets
Let A and B be any two sets. The symmetric difference of A
and B is the set ( A − B) ∪ ( B − A ). It is denoted by A∆B
and read as A symmetric difference B. The symbol ‘∆’ is
used to denote the symmetric difference.
∴
A∆B = ( A − B ) ∪ ( B − A)
= { x : x ∈ A or x ∈ B but x ∉ A ∩ B }
The symmetric difference of sets A and B is represented
by the following Venn diagram
Let A and B be any two sets. The difference of sets A and
B in this order is the set of all those elements of A which
do not belong to B. It is denoted by A − B and read as A
minus B.The symbol ‘−’ is used to denote the difference
of sets.
∴
A − B = { x : x ∈ A and x ∉ B }
Similarly,
B − A = { x : x ∈ B and x ∉ A}
The difference of two sets A and B can be represented by
the following Venn diagram
U
B
Mathematics Class 11th
∴
⇒
3. Difference of Sets
A
ne
A
B
A–B
B–A
U
U
A
B
A–B
Hence, the shaded portion represents the symmetric
difference of sets A and B.
e.g., Let A = {1, 2, 3, 4} and B = {3, 4, 5, 6}
Now, A − B = {1, 2}
B − A = {5, 6}
∴
A∆B = ( A − B ) ∪ ( B − A) = {1, 2, 5, 6}
B–A
The shaded portion represents the difference of two sets
A and B.
e.g., Let
A = {1, 2, 3, 4, 5}
and
B = {3, 5, 7, 9}
Then,
A − B = {1, 2, 4}
and
B − A = { 7, 9}
Some Properties of Intersection of Sets
(i) A − B = A ∩ B ′
(ii) B − A = B ∩ A ′
(iii) A − B ⊆ A
(iv) B − A ⊆ B
* Method of Finding Symmetric Difference of Two
Sets
If two sets A and B (say) are given, then we find their symmetric
difference with the help of following steps
Step I Write the given sets in tabular form (if not given in
tabular form) and assume them A and B (say).
Step II Find the difference of sets A and B i.e., A − B.
Example 4.
Step III Find the difference of sets B and A i.e., B − A.
(i) If X = {a, b, c, d, e, f}
and Y = { f, b, d, g, h, k}, then find X – Y and Y – X.
(ii) If A = {1, 2, 3, 4, 5} and B = {2, 4, 6}, then find
A − B and B − A.
Solution
(i) Given, X = { a, b, c , d , e , f } and Y = { f , b, d , g , h, k }
∴ X − Y = { a, b, c , d , e , f } − { f , b, d , g , h, k } = { a, c , e }
[only those elements of X which do not belong to Y ]
and Y − X = { f , b, d , g , h, k } − { a, b, c , d , e , f }
= { g , h, k }
[only those elements of Y which do not belong to X ]
(ii) Given,
A = {1, 2, 3, 4, 5}
and
B = { 2, 4, 6}
Step IV Find the union of sets obtained from steps II and III,
which will give the required symmetric difference.
Worked out Problem
Find the symmetric difference
A = { 1, 3, 5 , 6 , 7 } and B = { 3, 7 , 8, 9 }.
of
sets
Step I Write the given sets in tabular form
Given sets are A = {1, 3, 5 , 6 , 7 } and B = {3, 7 , 8 , 9} which are
in tabular form.
Step II Find the difference of sets A and B i.e., A − B.
We know that, difference of sets A and B is the set of
those elements of A, which are not present in B.
So,
A − B = {1, 3, 5 , 6 , 7 } − {3, 7 , 8 , 9} = {1, 5 , 6 }
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Sets
/
Step III Find the difference of sets B and A i.e., B − A.
We know that, difference of sets B and A is the set of
those elements of B, which are not present in A.
So,
B − A = {3, 7 , 8 , 9} − {1, 3, 5 , 6 , 7 } = {8 , 9}
Solution
(i) Given,
A = {1, 2, 3, 4, 5, 6}
and
B = { 4, 5, 6}
∴
A ∩ B = {1, 2, 3, 4, 5, 6} ∩ { 4, 5, 6}
= { 4, 5, 6} ≠ φ
Hence, this pair of sets is not disjoint.
(ii) Here, A = {b1 , b2 , ... } and B = { g 1 , g 2 , ... },
where b1 , b2 , ... , are the boys and g 1 , g 2 ,…, are the girls
of school.
Clearly,
A∩B = φ
Hence, this pair of set is disjoint set.
Step IV Find the union of sets obtained from steps II and III,
which will give the required symmetric difference.
∴Required symmetric difference,
A ∆ B = ( A − B ) ∪ (B − A ) = {1, 5 , 6 } ∪ {8 , 9} = {1, 5 , 6 , 8 , 9}
Example 5. If A = {1, 3, 5, 7, 9} and
B = {2, 3, 5, 7, 11},
then find A∆B.
/
Firstly, convert all the sets in roster form, if it is not given in
that. Then use the condition for disjoint sets i.e., A ∩ B = φ.
Firstly, determine the sets A − B and B − A and then find the
union of these sets.
Solution We know that, A∆B is the symmetric difference of
sets A and B.
∴
A∆B = ( A − B ) ∪ ( B − A )
Given,
A = {1, 3, 5, 7, 9},
and
B = { 2, 3, 5, 7, 11}
Then,
A − B = {1, 9}
and
B − A = { 2, 11}
∴
A∆B = ( A − B ) ∪ ( B − A )
= {1, 9} ∪ { 2, 11} = {1, 2, 9, 11}
Complement of a Set
Let U be the universal set and A be any subset of U,
then complement of A with respect to U is the set of all
those elements of U which are not in A.
It is denoted by A or A ′ and read as
‘A complement’.
Thus,
A = { x : x ∈U and x ∉ A}
or
A′ = U − A
The complement of set A is represented by the following
Venn diagram
Disjoint Sets
Two sets A and B are said to be disjoint sets, if they have
no common element i.e., A ∩ B = φ.
A
The disjoint of two sets A and B can be represented by
the following Venn diagram
A
B
U
A′
Hence, the shaded portion represents the complement
of set A.
e.g., Let
U = {1, 2, 3, 4, 5, 6}
and
A = {2, 4}
Then,
A′ = U − A
= {1, 2, 3, 4, 5, 6} − {2, 4}
= {1, 3, 5, 6}
U
Hence, separation of sets represents the two disjoint sets.
e.g., Let A = {2, 4, 6} and B = {1, 3, 5}
Then, A ∩ B = φ. Hence, A and B are disjoint sets.
Note If A is a subset of the universal set U, then its complement A′ is
also a subset of U.
Note The sets A − B, A ∩ B and B − A are mutually disjoint sets i.e.,
Some Properties of Complement of Sets
(i) ( A ′ ) ′ = A
[law of double complementation]
(ii) (a) A ∪ A ′ = U
(b) A ∩ A ′ = φ
[complement law]
(iii) (a) φ′ = U
(b) U ′ = φ [laws of empty set and universal set]
the intersection of any of these two sets is an empty set.
Example 6. Which of the following pairs of sets are
disjoint?
(i) A = {1, 2, 3, 4, 5, 6}
and B = {x : x is a natural number and 4 ≤ x ≤ 6}
(ii) A = {x : x is the boys of your school}
B = {x : x is the girls of your school}
17
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18
(iv) (a) ( A ∪ B ) ′ = A ′∩ B ′
(b) ( A ∩ B ) ′ = A ′ ∪ B ′
U
A
B′ =U − B
= {1, 2, 3, 4, 5, 6, 7, 8, 9} − { 2, 3, 5, 7}
= {1, 4, 6, 8, 9}
∴ A ′∩ B ′ = {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9}
...(ii)
= {1, 9}
From Eqs. (i) and (ii), we get
Hence verified.
( A ∪ B )′ = A ′ ∩ B ′
[De-Morgan’s law]
B
U
B
(ii) Here, A ∩ B = { 2, 4, 6, 8} ∩ { 2, 3, 5, 7} = { 2}
∴ ( A ∩ B )′ = U − ( A ∩ B )
= {1, 2, 3, 4, 5, 6, 7, 8, 9} − { 2}
…(i)
⇒ ( A ∩ B )′ = {1, 3, 4, 5, 6, 7 ,8, 9}
Now, A ′∪ B ′ = {1, 3, 5, 7, 9} ∪ {1, 4, 6, 8, 9}
…(ii)
= {1, 3, 4, 5, 6, 7, 8, 9}
From Eqs. (i) and (ii), we get
Hence verified.
( A ∩ B )′ = A ′∪ B ′
(b)
i.e., the complement of the union of two sets is the
intersection of their complements [figure (a)] and the
complement of the intersection of two sets is the union of
their complements [figure (b)].
All the operations between two sets can be represented in
a single Venn diagram, as given below
(Shaded Part) A ∪ B
(Portion shaded
by vertical lines
in circle A)
A–B
A
B
U
(Portion shaded by
B − A vertical lines in
A ∩ B circle B)
(Portion shaded by horizontal
line between A and B)
Example 8. If
A = {2, 4, 6, 8, 10}, B = {1, 2, 3, 4, 5, 6, 7},
C = {2, 6, 7, 10} and
U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then verify that
(i) (B ∪ C )′ = B ′∩ C ′
(ii) A ∪ (B ∪ C ) = ( A ∪ B) ∪ C
(iii) A ∩ (B ∩ C ) = ( A ∩ B) ∩ C
(iv) A ∪ (B ∩ C ) = ( A ∪ B) ∩ ( A ∪ C )
(v) A ∩ (B ∪ C ) = ( A ∩ B) ∪ ( A ∩ C )
Solution Given, A = {2, 4, 6, 8, 10}, B = {1, 2, 3, 4, 5, 6, 7},
(A ∪ B)′ (Without shaded
part)
A ∆ B (Portion shaded by vertical Line
in both circles)
Example 7. Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9},
A = {2, 4, 6, 8} and B = {2, 3, 5, 7}.
Verify that
C = { 2, 6, 7, 10} and U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(i) B ∪ C = {1, 2, 3, 4, 5, 6, 7} ∪ { 2, 6, 7, 10}
= {1, 2, 3, 4, 5, 6, 7, 10}
∴ ( B ∪ C )′ = U − ( B ∪ C )
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} − {1, 2, 3, 4, 5, 6, 7, 10}
...(i)
= {8, 9}
Similarly,
B ′ = U − B = {8, 9, 10}
and
C ′ = U − C = {1, 3, 4, 5, 8, 9}
... (ii)
∴
B ′∩C ′ = {8, 9}
From Eqs. (i) and (ii), we get
Hence verified.
( B ∪ C )′ = B ′∩ C ′
(i) ( A ∪ B)′ = A′∩ B ′
(ii) ( A ∩ B)′ = A′∪ B ′
/
Use the formulae,
(A ∪ B)′ = U − (A ∪ B), A′ = U − A and B ′ = U − B and then
simplify it.
Solution
(i) We have,
Mathematics Class 11th
and
A
(a)
ne
U = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A = { 2, 4, 6, 8}
B = { 2, 3, 5, 7}
∴
A ∪ B = { 2, 4, 6, 8} ∪ { 2, 3 ,5, 7}
⇒
A ∪ B = { 2, 3, 4, 5, 6, 7, 8}
Now,( A ∪ B )′ = U − ( A ∪ B )
= {1, 2, 3, 4, 5, 6, 7, 8, 9} − { 2, 3, 4, 5, 6, 7, 8}
...(i)
= {1, 9}
Now, A ′ = U − A
= {1, 2, 3, 4, 5, 6, 7, 8, 9} − { 2, 4, 6, 8}
= {1, 3, 5, 7, 9}
(ii) Here,
A = { 2, 4, 6, 8, 10}
Now,
B ∪ C = {1, 2, 3, 4, 5, 6, 7, 10}
∴ A ∪ ( B ∪ C ) = {1, 2, 3, 4, 5, 6, 7, 8, 10}
Now,
A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 10}
...(iii)
∴ ( A ∪ B ) ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 10} ∪ { 2, 6, 7, 10}
...(iv)
⇒ ( A ∪ B ) ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 10}
From Eqs. (iii) and (iv), we get
Hence verified.
A ∪ (B ∪ C ) = ( A ∪ B ) ∪ C
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19
Sets
(iii) Here, B ∩ C = {1, 2, 3, 4, 5, 6, 7} ∩ { 2, 6, 7, 10}
∴ ( A ∪ B ) ∩ ( A ∪ C ) = {1, 2, 3, 4, 5, 6, 7, 8, 10}
∩ { 2, , 4, 6, 7, 8, 10}
...(viii)
= { 2, 4, 6, 7, 8, 10}
From Eqs. (vii) and (viii), we get
A ∪ (B ∩ C ) = ( A ∪ B ) ∩ ( A ∪ C )
Hence verified.
(v) Now, B ∪ C = {1, 2, 3, 4, 5, 6, 7} ∪ { 2, 6, 7, 10}
= { 2, 6, 7}
A ∩ ( B ∩ C ) = { 2, 4, 6, 8, 10} ∩ { 2, 6, 7}
...(v)
= { 2, 6}
Now, A ∩ B = { 2, 4, 6, 8, 10} ∩ {1, 2, 3, 4, 5, 6, 7}
= { 2, 4, 6}
∴ ( A ∩ B ) ∩ C = { 2, 4, 6} ∩ { 2, 6, 7, 10}
...(vi)
= { 2, 6}
From Eqs. (v) and (vi), we get
A ∩ (B ∩ C ) = ( A ∩ B ) ∩ C
Hence verified.
(iv) Here, B ∩ C = {1, 2, 3, 4, 5, 6, 7} ∩ { 2, 6, 7, 10}
= {1, 2, 3, 4, 5, 6, 7, 10}
A ∩ ( B ∪ C ) = { 2, 4, 6, 8, 10} ∩ {1, 2, 3, 4, 5, 6, 7, 10}
...(ix)
⇒ A ∩ ( B ∪ C ) = { 2, 4, 6, 10}
Now,
( A ∩ B ) = { 2, 4, 6, 8, 10} ∩ {1, 2, 3, 4, 5, 6, 7}
= { 2, 4, 6}
( A ∩ C ) = { 2, 4, 6, 8, 10} ∩ { 2, 6, 7, 10}
= { 2, 6, 10}
∴ ( A ∩ B ) ∪ ( A ∩ C ) = { 2, 4, 6} ∪ { 2, 6, 10}
...(x)
= { 2, 4, 6, 10}
From Eqs. (ix) and (x), we get
A ∩ (B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C )
Hence verified.
= { 2, 6, 7}
A ∪ ( B ∩ C ) = { 2, 4, 6, 8, 10} ∪ { 2, 6, 7}
... (vii)
= { 2, 4, 6, 7, 8, 10}
Now, ( A ∪ B ) = { 2, 4, 6, 8, 10} ∪ {1, 2, 3, 4, 5, 6, 7}
and
= {1, 2, 3, 4, 5, 6, 7, 8, 10}
and ( A ∪ C ) = { 2, 4, 6, 8, 10} ∪ { 2, 6, 7, 10}
= { 2, 4, 6, 7, 8, 10}
EXAM
Very Short Answer Type Questions
1. If A = {3, 5, 7, 9, 11},
[1 Mark each]
Sol. (i) We have, A = { 2, 4, 6, 8} and B = {1, 3, 5}
B = {7, 9, 11, 13} and
Now, A ∩ B = { 2, 4, 6, 8} ∩ {1, 3, 5}
⇒
A∩B = φ
Therefore, A and B are disjoint sets.
[1/2]
Hence, given statement is true.
(ii) We have, A = { a, e , i, o, u } and B = { a, b, c , d }
Now, A ∩ B = { a }
i.e.,
A∩B ≠ φ
Therefore, A and B are not disjoint sets.
Hence, given statement is false.
[1/2]
C = {11, 13, 15}, then find
(i) A ∩ B
(ii) B ∩ C
Sol. We have, A = {3, 5, 7, 9, 11}
B = { 7, 9, 11, 13} and C = {11, 13, 15}
(i) A ∩ B = {3, 5, 7, 9, 11} ∩ { 7, 9, 11, 13}
= { 7, 9, 11}
(ii) B ∩ C = { 7, 9, 11, 13} ∩ {11, 13, 15}
= {11, 13}
[1/2]
[1/2]
2. Find A − B and B − A when A = {1, 2, 3, 4, 5, 6} and
4. Prove that( A ∩ B′ )′ ∪ (B ∩ C) = A ′∪ B.
B = {2, 4, 6, 8, 10}.
Sol. Now, A − B = {1, 2, 3, 4, 5, 6} − { 2, 4, 6, 8, 10} = {1, 3, 5}
Sol.
[1/2]
and B − A = { 2, 4, 6, 8, 10} − {1, 2, 3, 4, 5, 6}
= {8, 10}
LHS = ( A ∩ B ′ )′ ∪ ( B ∩ C )
= { A ′∪ ( B ′ ) ′ } ∪ ( B ∩ C ) [by De-Morgan’s law]
[Q( B ′ ) ′ = B]
= ( A ′∪ B ) ∪ ( B ∩ C )
= (( A ′∪ B ) ∪ B ) ∩ (( A ′ ∪ B ) ∪ C )
= ( A ′∪ ( B ∪ B )) ∩ ( A ′ ∪ B ∪ C )
= ( A ′ ∪ B) ∩ ( A ′ ∪ B ∪ C )
[1]
= ( A ′∪ B ) = RHS
Hence proved.
[1/2]
3. State whether each of the following statement is
true or false.
(i) A = {2, 4, 6, 8} and B = {1, 3, 5} are disjoint sets.
(ii) A = { a, e, i, o, u} and B = { a, b, c, d} are disjoint
sets.
19
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20
Set of points inside the square, T = Set of
points inside the triangle and C = Set of points
inside the circle. If the triangle and circle
intersect each other and are contained in a
square. Then, prove that S ∪ T ∪ C = S, by Venn
diagram.
T
U = {a, b, c, d, e, f , g, h}, then
complement of the following sets.
(i) A = { a, b, c}
(iii) C = { a, c, e, g}
Sol. We have,
S
[4 Marks each]
find
Then, ( A − B ) = {1, 3, 6, 11, 12} − {1, 6}
= {3, 11, 12}
and ( B − A ) = {1, 6} − {1, 3, 6, 11, 12} = φ
∴
A∆B = ( A − B ) ∪ ( B − A )
= {3, 11, 12} ∪ φ = {3, 11, 12}
the
(ii) B = { d, e, f , g}
(iv) D = { f , g, h}
U = { a, b, c , d , e , f , g , h }
(i) Complement of A is A′.
∴ A ′ = U − A = { a, b, c , d , e , f , g , h } − { a, b, c }
= { d , e, f , g , h }
8. If
[1]
[1]
U = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11},
A = {2, 4, 7}, B = {3, 5, 7, 9, 11} and
C = {7, 8, 9, 10, 11}, then compute
[1]
(i) ( A ∩ U) ∩ ( B ∪ C)
(iii) B − C
(ii) Complement of B is B′.
∴ B ′ = U − B = { a, b, c , d , e , f , g , h } − { d , e , f , g }
[1]
= { a, b, c , h }
(ii) C − B
(iv) ( B − C)′
Sol. Given,
U = { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11},
(iii) Complement of C is C ′.
∴ C ′ = U − C = { a, b, c , d , e , f , g , h } − { a, c , e , g }
[1]
= {b, d , f , h }
A = { 2, 4, 7}, B = {3, 5, 7, 9, 11} and C = { 7, 8, 9, 10, 11}
(i) ( A ∩ U ) = { 2, 4, 7} ∩ { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
= { 2, 4, 7}
B ∪ C = {3, 5, 7, 9, 11} ∪ { 7, 8, 9, 10, 11}
= {3, 5, 7, 8, 9, 10, 11}
∴( A ∩ U ) ∩ ( B ∪ C ) = { 2, 4, 7} ∩ {3, 5, 7, 8, 9, 10, 11}
[1]
= { 7}
(ii) C − B = { 7, 8, 9, 10, 11} − {3, 5, 7, 9, 11}
[1]
= {8, 10}
(iii) B − C = {3, 5, 7, 9, 11} − { 7, 8, 9, 10, 11}
[1]
= {3, 5}
(iv) ( B − C )′ = U − ( B − C )
= { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} − {3, 5}
[1]
= { 2, 4, 6, 7, 8, 9, 10, 11}
(iv) Complement of D is D′.
∴ D ′ = U − D = { a, b, c , d , e , f , g , h } − { f , g , h }
[1]
= { a, b, c , d , e }
7. Find A∆B, if
(i) A = {1, 3, 4} and B = {2, 5, 9, 11}.
(ii) A = {1, 3, 6, 11, 12} and B = {1, 6}.
Sol. (i) We have,
A = {1, 3, 4}
and
B = { 2, 5, 9, 11}
Then, ( A − B ) = {1, 3, 4} − { 2, 5, 9, 11}
= {1, 3, 4}
and ( B − A ) = { 2, 5, 9, 11} − {1, 3, 4}
= { 2, 5, 9, 11}
∴
A∆B = ( A − B ) ∪ ( B − A )
= {1, 3, 4} ∪ { 2, 5, 9, 11}
= {1, 2, 3, 4, 5, 9, 11}
(ii) We have,
A = {1, 3, 6, 11, 12}
and
B = {1, 6}
C
It is clear from the Venn diagram that, S ∪ T ∪ C = S [1]
Hence proved.
Sol. Given, S = Set of points inside the square
T = Set of points inside the triangle
C = Set of points inside the circle
6. If
Mathematics Class 11th
According to the given condition, the Venn diagram is given
below
5. Let S =
Short Answer Type Questions
ne
9. Verify( A ∩ B)′ =
A ′ ∪ B′, where A = {3, 4, 5, 6} and
B = {3, 6, 7, 8} are subsets of
U = {1, 2, 3, 4, 5, 6, 7, 8}.
Sol. We have, A = {3, 4, 5, 6}, B = {3, 6, 7, 8}
[1]
[1]
and
U = {1, 2, 3, 4, 5, 6, 7, 8}
Now, A ∩ B = {3, 4, 5, 6} ∩ {3, 6, 7, 8}
= {3, 6}
20
[1]
ll
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21
Sets
∴ ( A ∩ B )′ = U − ( A ∩ B ) = {1, 2, 3, 4, 5, 6, 7, 8} − {3, 6}
... (i) [1]
⇒ ( A ∩ B )′ = {1, 2, 4, 5, 7, 8}
Also,
A′ = U − A
= {1, 2, 3, 4, 5, 6, 7, 8} − {3, 4, 5, 6}
[1]
= {1, 2, 7, 8}
and
B′ =U − B
= {1, 2, 3, 4, 5, 6, 7, 8} − {3, 6, 7, 8}
[1]
= {1, 2, 4, 5}
∴
A ′∪ B ′ = {1, 7, 8} ∪ {1, 2, 4, 5}
... (ii)
= {1, 2, 4, 5, 7, 8}
From Eqs. (i) and (ii), we get
[1]
( A ∩ B )′ = A ′ ∪ B ′
Hence verified.
12. Prove that A − (B ∪ C) = ( A − B) ∩ ( A − C).
Sol. Let x be any element of A − ( B ∪ C ).
Then, x ∈ A − ( B ∪ C )
⇒
x ∈ A and x ∉ B ∪ C
[1/2]
⇒
x ∈ A and (x ∉ B and x ∉C )
(x ∈ A and x ∉ B) and (x ∈ A and x ∉C )
⇒
⇒
x ∈ ( A − B ) and x ∈ A − C
⇒
x ∈( A − B ) ∩ ( A − C )
∴
A − ( B ∪ C ) ⊆ ( A − B ) ∩ ( A − C ) ...(i) [1]
Now, let x be any element of ( A − B ) ∩ ( A − C ).
Then,
x ∈( A − B ) ∩ ( A − C )
⇒
x ∈( A − B )
and
[1/2]
x ∈( A − C )
10. Let A and B be any two sets. Using properties of
(x ∈ A and x ∉ B)
⇒
and
(x ∈ A and x ∉C )
⇒
x ∈ A (and x ∉ B and x ∉C )
⇒
x ∈ A and x ∉ B ∪ C
⇒
x ∈ ( A − ( B ∪ C ))
∴
( A − B ) ∩ ( A − C ) ⊆ A − ( B ∪ C ) ...(ii) [1]
From Eqs. (i) and (ii), we get
[1]
A − (B ∪ C ) = ( A − B ) ∩ ( A − C )
Hence proved.
sets prove that
(i) ( A − B) ∪ B = A ∪ B (ii) ( A − B) ∪ A = A
(iii) ( A − B) ∩ B = φ
(iv) ( A − B) ∩ A = A ∩ B′
Sol. (i) ( A − B ) ∪ B = ( A ∩ B ′ ) ∪ B
[Q A − B = A ∩ B ′ ]
= ( A ∪ B ) ∩ ( B ′∪ B )
[since, ∪ is distributes over ∩]
= ( A ∪ B ) ∩U
[Q B ′ ∪ B = U ]
[1]
= A∪B
(ii) ( A − B ) ∪ A = A
[Q A − B ⊆ A ] [1]
(iii) ( A − B ) ∩ B = ( A ∩ B ′ ) ∩ B
[Q A − B = A ∩ B ′ ]
= A ∩ ( B ′∩ B ) [since, ∩ is associative]
= A∩ φ= φ
[Q B ′∩ B = φ] [1]
(iv) ( A − B ) ∩ A = A − B = A ∩ B ′
[1]
Hence proved.
13. Suppose
A1 , A2 , ..., A30 are thirty sets each
having 5 elements and B1 , B2 , ..., Bn are n sets
each having 3 elements.
Let
11. For
30
n
i=1
j =1
U A i = U Bj = 5 and each element of S
belongs to exactly 10 of A i ’s and exactly 9 of Bj ’s.
Then, find the value of n.
any two sets A and B, prove by using
properties of sets that
( A ∪ B) − ( A ∩ B) = ( A − B) ∪ (B − A)
Sol. LHS = ( A ∪ B ) − ( A ∩ B )
Sol. If elements are not repeated, then number of elements
A1 ∪ A 2 ∪ A3 ∪ K ∪ A30 is 30 × 5 but each element is
used 10 times, so
30 × 5
... (i) [1]
n (S ) =
= 15
10
Similarly, if elements in B1 , B 2 , ..., Bn are not repeated,
then total number of elements is 3n, but each element is
repeated 9 times, so
3n
[1]
n (S ) =
9
3n
[from Eq. (i)]
⇒
15 =
9
15 × 9
[2]
n=
= 45
∴
3
= ( A ∪ B ) ∩ ( A ∩ B )′
[Q X − Y = X ∩ Y ′ ]
= X ∩ ( A ′∪ B ′ ), where X = A ∪ B
[Q ( A ∩ B )′ = A ′∪ B ′ ] [1]
[∩ distributes over ∪]
= ( X ∩ A′ ) ∪ ( X ∩ B′ )
[1]
= (B ∩ A ′ ) ∪ ( A ∩ B ′ )
Q X ∩ A ′ = ( A ∪ B ) ∩ A ′ = ( A ∩ A ′ ) ∪ ( B ∩ A ′ )


= φ ∪ (B ∩ A ′ ) = B ∩ A ′


 [1]
Similarly,


X ∩ B ′ = ( A ∪ B ) ∩ B ′ = ( A ∩ B ′) ∪ (B ∩ B ′ ) 
= ( A ∩ B ′ ) ∪ φ = ( A ∩ B ′)


= ( A ∩ B ′ ) ∪ (B ∩ A ′ ) = ( A − B ) ∪ (B − A )
[Q A − B = A ∩ B ′ and B − A = B ∩ A ′] [1]
Hence proved.
21
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TRY YOURSELF
1 Mark Questions
1.
2.
3.
4.
Let A = {2, 4, 6, 8} and B = {6, 8, 10, 12}. Find A ∪ B.
5.
6.
If A and B are two sets, then verify A ∩ ( A ∪ B ) = A.
Let A = {a, e, i, o, u} and B = {a, i, u}. Show that A ∪ B = A.
If A = {1, 3, 5, 7, 9 } and B = {2, 3, 5, 7, 11}, then find A − B.
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4} and B = {2, 4, 6, 8}. Find ( A ∪ B )′.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 5}, B = {2, 4, 6, 7} and C = {2, 3, 4, 8}, then find (B ∪ C )′.
4 Marks Questions
7.
If A = {2, 4, 6, 8, 10}, B = {1, 2, 3, 4, 5, 6, 7} and C = {2, 6, 7, 10}, then verify that
(i) A − (B ∪ C ) = ( A − B ) ∩ ( A − C )
(ii) A − (B ∩ C ) = ( A − B ) ∪ ( A − C )
8.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 3, 4} and B = { 5, 6}, then verify that
A − B = A ∩ B′ = B′ − A ′
9.
If U = {1, 2, 3, 4, 5, 6, 7, 8}, A = {1, 2, 3, 4}, B = { 3, 4, 6} and C = { 5, 6, 7, 8}, then verify that
(i) A ∪ (B ∩ C ) = ( A ∪ B ) ∩ ( A ∪ C )
(ii) ( A ∩ B )′ = A ′∪B ′
10.
If A = {1, 3, 5, 7, 9, 11, 13, 15, 17}, B = {2, 4, ..., 18} and N, the set of natural numbers is the universal set, then
prove that A ′ ∪ {( A ∪ B ) ∩ B ′ } = N.
11.
(i) If L = {1, 2, 3, 4}, M = { 3, 4, 5, 6} and N = {1, 3, 5}, then verify that L − (M ∪ N) = (L − M ) ∩ (L − N)
(ii) If A and B are subsets of the universal set U, then show that A ⊆ A ∪ B .
12.
13.
If X and Y are two sets and X′ denotes the complement of X, then verify that X ∩ ( X ∪ Y )′ = φ.
14.
Determine, whether each of the following statement is true or false. Justify your answer.
(i) For all sets A and B, ( A − B ) ∪ ( A ∩ B ) = A.
(ii) For all sets A, B and C, if A ⊂ C and B ⊂ C, then A ∪ B ⊂ C.
A, B and C are subsets of universal set U. If A = {2, 4, 6, 8, 12, 20}, B = { 3, 6, 9, 12, 15}, C = { 5, 10, 15} and U is
the set of all whole numbers, then draw a Venn diagram showing the relation of U, A, B and C.
22
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4 Applications of Set Theory
In this topic, we discuss some practical problems
related to our daily life, which are based on union and
intersection of two sets.
For solving these types of problem, we have following
formulae.
If A and B are two finite sets, then two cases arise
Case I If A and B are disjoint sets, then there is no
common element in A and B i.e., A ∩ B = φ.
Then,
n( A ∪ B ) = n( A) + n( B )
A
B
Using the formula,
n ( X ∪ Y ) = n ( X ) + n(Y ) − n ( X ∩ Y ), we get
60 = 40 + n(Y ) − 10
⇒
60 = 30 + n(Y )
⇒ n (Y ) = 60 − 30
∴ n(Y ) = 30
Hence, total number of elements in Y have 30.
❖
Important Results
1. n(X ∪Y ) = n(X ∩Y ′) + n(X ′∩Y ) + n(X ∩ C)
2. n (A∆B) = n [(A − B) ∪ (B − A)] = n (A − B) + n(B − A)
U
[since, (A − B) and (B − A) are disjoint sets]
= n (A) − n (A ∩ B) + n (B) − n (A ∩ B)
= n (A) + n (B) − 2n(A ∩ B)
3. n (A′∪ B′ ) = n [(A ∩ B)′] = n (U ) − n (A ∩ B)
4. n (A′∩ B ′) = n [(A ∪ B)′] = n (U ) − n (A ∪ B)
Case II If A and B are not disjoint sets, then there are
common elements in A and B. Then,
n ( A ∪ B ) = n ( A) + n (B ) − n ( A ∩ B )
A
A–B
B
5. If A, B and C are finite sets, then
n (A ∪ B ∪ C) = n [(A ∪ B ) ∪ C ]
= n (A) + n (B) + n(C) − n(A ∩ B)
− n (B ∩ C) − n (A ∩ C) + n (A ∩ B ∩ C)
U
6. Let M = Set of students who have Mathematics.
P = Set of students who have Physics.
C = Set of students who have Chemistry.
On applying the different operations on these three sets, we get
following important results.
(i) M ′ = Set of students who have no Mathematics.
(ii) P′ = Set of students who have no Physics.
(iii) C′ = Set of students who have no Chemistry.
(iv) M ∪ P = Set of students who have atleast one subject
Mathematics or Physics.
(v) P ∪ C = Set of students who have atleast one subject
Physics or Chemistry.
(vi) C ∪ M = Set of students who have atleast one subject
Chemistry or Mathematics.
(vii) M ∩ P = Set of students who have both subjects
Mathematics and Physics.
(viii) P ∩ C = Set of students who have both subjects Physics
and Chemistry.
(ix) C ∩ M = Set of students who have both subjects Chemistry
and Mathematics.
(x) (M ∩ P ∩ C) = Set of students who have all the subjects.
(xi) (M ∪ P ∪ C) = Set of all students who have three subjects.
A∩ B B – A
It is called addition theorem for two sets.
We know that, sets A − B , A ∩ B and B − A are
disjoint and their union is A ∪ B . Then,
(i) n ( A ∪ B ) = n ( A − B ) + n ( B − A) + n( A ∩ B )
(ii) n ( A) = n ( A − B ) + n ( A ∩ B )
(iii) n ( B ) = n ( B − A) + n ( A ∩ B )
Example 1. If X and Y are two sets such that X has
40 elements, X ∪ Y has 60 elements and X ∩ Y
has 10 elements, then how many elements does Y
have?
/
Use the formula, n (X ∪Y ) = n (X ) + n (Y ) − n (X ∩Y )
and simplify it.
Solution Given,
n( X ) = 40,
n( X ∪ Y ) = 60,
n( X ∩ Y ) = 10,
23
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Solution of Word Problems
ne
Mathematics Class 11th
Example 2. In a class of 60 students, 25 students
play Cricket, 20 students play Tennis and
10 students play both the games. Then, find the
number of students who play neither games.
Word problems related to sets can be solved by following two
methods
1. By Using Operations
Solution Let C and T denote the students who play Cricket,
and Tennis, respectively and U denotes the total number of
students in a class.
Then,
n(C ) = 25, n(T ) = 20 and n(U ) = 60
We know that,
n(C ∪ T ) = n(C ) + n(T ) − n(C ∩ T )
On putting the values, we get
n (C ∪ T ) = 25 + 20 − 10 = 45 − 10 = 35
∴Number of students who play neither games,
[by De-Morgan’s Law]
n(C ′∩T ′ ) = n(C ∪ T ) ′
= n(U ) − n(C ∪ T ) = 60 − 35 = 25
Hence, 25 students play neither games.
In this method, we use the operations related to set
according to the given information.This method can be
understand with the help of an worked out problem
which is given below.
Worked out Problem
Each student in a class of 40 students study atleast
one of the subjects English, Mathematics and
Economics. 16 students study English, 22
Economics and 26 Mathematics, 5 study English
and Economics, 14 Mathematics and Economics
and 2 English, Economics and Mathematics. Find
the number of students who study English and
Mathematics.
2. By Using Venn Diagram
Sometimes a word problem contains three sets and many
information related to these sets but not exact
information related to such set separately. Then, that
problem become difficult to solve by previous method.
Such problems can be solved by considering a variable on
each part of the venn diagram containing three sets and
then make some equations, which are given below
(i) Union of three sets
a + b + c + d + e + f + g = Constant
Step I Consider the set which represent students in given
subjects.
Let A, B and C denote the set of students who study English,
Economics and Mathematics, respectively.
Step II Find the relation between number of students of each
subject given in question.
Given, total number of students, n ( A ∪ B ∪ C ) = 40
Number of students who study English, n ( A ) = 16
Number of students who study Economics, n (B ) = 22
Number of students who study Mathematics, n (C ) = 26
Number of students who study English and Economics,
n (A ∩ B) = 5
Number of students who study Mathematics and
Economics, n (B ∩ C ) = 14
and number of students who study all subjects,
n (A ∩ B ∩ C ) = 2
B
A
a
d
b
e
U
c
f
g
C
Step III Use the suitable formula, for three sets. Then,
substitute all the given values and simplify it.
For Set A a + b + d + e = Constant
For Set B b + c + e + f = Constant
For Set C d + e + f + g = Constant
Intersection of two sets at a time
(a) b + e = Constant
(b) d + e = Constant
(c) e + f = Constant
Now, solve these equations one by one by using given
values and get separate values of each set.
(ii)
(iii)
(iv)
(v)
We know that,
n( A ∪ B ∪ C ) = n( A ) + n(B ) + n(C )
− n( A ∩ B ) − n(B ∩ C )− n( A ∩ C ) + n( A ∩ B ∩ C )
On putting the values from step II, we get
40 = 16 + 22 + 26 − 5 − 14 − n(C ∩ A ) + 2
⇒
40 = 66 − 19 − n (C ∩ A )
⇒ n (C ∩ A ) = 47 − 40
=7
Hence, number of students who study English and
Mathematics are 7.
24
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25
Sets
⇒
b +e =9
[Qe = 3]
⇒
b +3=9
⇒
b =6
∴ Number of students who had taken Physics and
Chemistry,
n( P ∩ C ) = 4
⇒
e+ f =4
[Qe = 3]
⇒
3+ f =4
⇒
f =1
Number of students who had taken Maths and Chemistry,
n( M ∩ C ) = 5
⇒
d +e =5
[Qe = 3]
⇒
d +3=5
⇒
d=2
Number of students who had taken Maths,
n( M ) = 15
⇒
a + b + d + e = 15
⇒
a + 6 + 2 + 3 = 15 [Qb = 6, d = 2, e = 3]
⇒
a = 15 − 11 ⇒ a = 4
Number of students who had taken Physics.
n( P ) = 12
⇒
b + c + e + f = 12
⇒
6 + c + 3 + 1 = 12 [Qb = 6, e = 3, f = 1]
⇒
c = 12 − 10 ⇒ c = 2
Number of students who had taken Chemistry,
n(C ) = 11
⇒
d + e + f + g = 11
⇒
2 + 3 + 1 + g = 11 [Q d = 2, e = 3, f = 1]
⇒
g = 11 − 6 ⇒ g = 5
Now, the number of students in various cases are as follows
(i) Only Chemistry, g = 5
(ii) Only Maths, a = 4
(iii) Only Physics, c = 2
(iv) Physics and Chemistry but not Maths, f = 1
(v) Maths and Physics but not Chemistry, b = 6
(vi) Only one of the subject, a + c + g = 4 + 2 + 5 = 11
(vii) Atleast one of the three subjects
a + b + c + d + e + f + g = 4 + 6 + 2 + 2 + 3 + 1 + 5 = 23
(viii) None of the subjects,
25 − ( a + b + c + d + e + f + g ) = 25 − 23 = 2
Example 3. In a survey of 25 students, it was found
that 15 had taken Maths, 12 had taken Physics,
11 had taken Chemistry, 5 had taken Maths and
Chemistry, 9 had taken Maths and Physics, 4 had
taken Physics and Chemistry and 3 had taken all
the three subjects. Find the number of students that
had taken
(i) only Chemistry.
(ii) only Maths.
(iii) only Physics.
(iv) Physics and Chemistry but not Maths.
(v) Maths and Physics but not Chemistry.
(vi) only one of the subject.
(vii) atleast one of the three subjects.
(viii) None of the subjects.
Solution Let M, P and C denote the set of students who had
taken Mathematics, Physics and Chemistry, respectively.
Let in Venn diagram, a, b, c, d, e, f and g denote the number
of elements in respective regions.
M
P
a
d
b
e
U
c
f
g
C
From the Venn diagram, we get
n( M ) = a + b + d + e
n( P ) = b + c + e + f
n(C ) = d + e + f + g
n( M ∩ P ) = b + e
n( P ∩ C ) = e + f
n( M ∩ C ) = d + e
and
n( M ∩ P ∩ C ) = e
It is given that, number of students who had taken all three
subjects,
n( M ∩ P ∩ C ) = 3
⇒
e =3
∴ Number of students who had taken Maths and Physics,
n( M ∩ P ) = 9
25
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EXAM
Very Short Answer Type Questions
[1 Mark each]
X and Y are two sets such that X ∪ Y has
18 elements. X has 8 elements and Y has
15 elements, then how many elements does X ∪ Y
have?
Sol. Given, n ( X ∪ Y ) = 18, n ( X ) = 8 and n (Y ) = 15
1. If
2. If
X and Y are two sets such that
n(X) = 17 , n(Y) = 23 and n(X ∪ Y ) = 38, then find
n(X ∩ Y ).
/
We will use the addition theorem,
n (A ∪ B) = n (A) + n (B) − n (A ∩ B) , to calculate anyone of
the four values, if three of them are given.
Sol. Given,
According to the addition theorem, we have
n ( X ∪ Y ) = n ( X ) + n (Y ) − n ( X ∩ Y )
n ( X ) = 17
n (Y ) = 23
and
n ( X ∪ Y ) = 38
According to the addition theorem, we have
n ( X ∪ Y ) = n ( X ) + n (Y ) − n ( X ∩ Y )
⇒
n( X ∩ Y ) = n( X ) + n(Y ) − n( X ∪ Y )
On putting the values, we get
n ( X ∩ Y ) = 17 + 23 − 38
= 40 − 38
=2
Short Answer Type Questions
⇒
⇒
18 = 8 + 15 − n ( X ∩ Y )
n ( X ∩ Y ) = 23 − 18 = 5
[1]
n( A ) = 4 , n (B) = 5 , n (U) = 7 and n ( A ∩ B) = 2,
then find the value of n (A ∪ B) ′.
Sol. Given,
n ( A ) = 4, n ( B ) = 5, n (U ) = 7
3. If
[1]
and
Q
∴
n ( A ∩ B) = 2
n ( A ∪ B ) = n ( A ) + n (B ) − n ( A ∩ B )
n ( A ∪ B) = 4 +5 − 2 = 7
Now,
n ( A ∪ B )′ = n (U ) − n ( A ∪ B ) = 7 − 7 = 0 [1]
[4 Marks each]
4. In a town of 840 persons, 450 persons read Hindi,
5.
300 read English and 200 read both newspapers.
Then, find the number of persons who read
neither of the newspapers.
Sol. Let H and E denote Hindi and English newspapers.
Given, total number of persons, n (U ) = 840
Total number of persons who read Hindi newspaper,
n ( H ) = 450
Total number of persons who read English newspaper,
n ( E ) = 300
and n ( H ∩ E ) = 200
We know that, n ( H ∪ E ) = n ( H ) + n ( E ) − n ( H ∩ E )
∴
n( H ∪ E ) = 450 + 300 − 200
[2]
= 750 − 200 = 550
∴ The number of persons who read neither of the
newspaper,
[by De-Morgan’s law]
n (H ′ ∩ E ′) = n (H ∪ E )′
= n(U ) − n( H ∪ E ) = 840 − 550 = 290
Hence, 290 persons read neither of the newspapers.
[2]
In a survey of 400 movie viewers, 150 were listed
as liking ‘Veer Zaara’, 100 were listed as liking
‘Aitraaz’ and 75 were listed as both liking ‘Aitraaz’
as well as ‘Veer Zaara’. Find how many people
were liking neither ‘Aitraaz’ nor ‘Veer Zaara’.
Sol. Let U denotes the set of movie viewers, V denotes the set of
people liking ‘Veer Zaara’ and A denotes the set of people
liking ‘Aitraaz’.
Then, we have, n(U ) = 400, n(V ) = 150, n( A ) = 100
and
[1]
n(V ∩ A ) = 75
Now, the number of people those were liking neither
‘Aitraaz’ nor ‘Veer Zaara’,
[by De-Morgan’s law]
n(V ′∩ A ′ ) = n(V ∪ A )′
= n(U ) − n(V ∪ A )
= n(U ) − n(V ) − n( A ) + n(V ∩ A )
[Qn ( A ∪ B ) = n ( A ) + n ( B ) − n ( A ∩ B )] [1]
= 400 − 150 − 100 + 75 = 225
Hence, 225 people were liking neither ‘Aitraaz’ nor ‘Veer
[2]
Zaara’.
26
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27
Sets
Now, n ( M ∪ P ) = n ( M ) + n ( P ) − n ( M ∩ P )
[1]
= 60 + 90 − 30 = 120
(iii) Most of the students study Physics in comparison of
Mathematics.
[1]
6. There are 60 students in a Mathematics class and
90 students in Physics class. Find the number of
students which are either in Physics class or
Mathematics class in the following cases
(i) Two classes meet at the same hour.
(ii) Two classes meet at different hours and
30 students are enrolled in both the courses.
(iii) What value is shown here?
7.
Sol. Let M be the set of students in Mathematics class and P be
the set of students in Physics class.
Given that, n( M ) = 60 and n( P ) = 90
(i) Two classes meet at same hour i.e.,
M ∩P = φ
[1]
⇒
n( M ∩ P ) = 0
Now, n ( M ∪ P ) = n ( M ) + n ( P ) − n ( M ∩ P )
= 60 + 90 − 0
= 150
Sol. Let T and C be the set of persons taking tea and coffee,
respectively.
T
8
90
[1]
(ii) In this case, two classes meet at different hours and
30 students are enrolled in both the courses.
M
30 30
∴
U
P
60
n ( M ∩ P ) = 30
Long Answer Type Questions
...(i)
n ( A ) = a + b + c + d = 21
…(ii) [1]
n ( B ) = b + c + f + g = 26
…(iii)
n (C ) = c + d + e + f = 29
…(iv)
n ( A ∩ B ) = b + c = 14
…(v)
n (C ∩ A ) = c + d = 12
…(vi)
n ( B ∩ C ) = c + f = 14
and n ( A ∩ B ∩ C ) = c = 8
…(vii) [1]
We have to find the number of people like product C only
i.e., value of e.
On putting the value of c from Eq. (vii) in Eq. (iv), we get
b + 8 = 14
[1]
⇒
b =6
Now, putting the value of c from Eq. (vii) in Eq. (v), we
get
8 + d = 12
[1]
⇒
d=4
Then,
A, 26 people like product B and 29 people like
product C. If 14 people like products A and B, 12
people like products C and A, 14 people like
products B and C, 8 people like all the three
products. Find how many like product C only.
Sol. Let A, B andC denote the set of people who liked products
A, B and C, respectively.
Let in Venn diagram a, b, c , d , e , f and g denote the
number of elements in respective regions.
C
a
b
d
e
10
[6 Marks each]
8. In a survey, it is found that 21 people like product
A
8
U
n (T ∪ C ) = 26
n (T ∩ C ′ ) = 8
and
[1]
n (T ) = 16
Using the identity,
n (T ∩ C ′ ) = n (T ) − n (T ∩ C )
⇒
8 = 16 − n (T ∩ C )
[1]
⇒
n(T ∩ C ) = 16 − 8 = 8
Using the identity,
n (T ∪ C ) = n (T ∩ C ′ ) + n (C ∩ T ′ ) + n (T ∩ C )
⇒
26 = 8 + n (C ∩ T ′ ) + 8
⇒ n (C ∩ T ′ ) = 26 − 16 = 10
Number of persons who take coffee but not tea,
[1]
n (C ∩ T ′ ) = 10
Here, we see that most people like coffee in comparison of
tea.
[1]
M
60
C
We have,
U
P
In a group of 26 persons, 8 take tea but not coffee
and 16 take tea. How many persons take coffee
but not tea? What value is shown here?
U
e
f
g
B
27
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28
Now, putting the value of c from Eq. (vii) in Eq. (vi), we
get
[1]
8 + f = 14 ⇒ f = 6
On putting the values of c, d and f in Eq. (iii), we get
8 + 4 + e + 6 = 29
∴
e = 29 − 18 = 11
Hence, number of people who like product C only is 11.
9. In
a survey of 60 people, it was found that
25 people read newspaper H, 26 read
newspaper T, 26 read newspaper Z, 9 read both H
and Z, 11 read both H and T, 8 read both T and Z
and 3 read all three newspapers. Find
(i) the number of people who read atleast one of
the newspaper.
(ii) the number of people who read exactly one
newspaper.
10. In
a survey of 200 students of a school, it was
found that 120 study Mathematics, 90 study
Physics and 70 study Chemistry, 40 study
Mathematics and Physics, 30 study Physics and
Chemistry, 50 study Chemistry and Mathematics
and 20 none of these subjects. Find the number of
students who study all the three subjects.
Sol. Given, set of people who read types of newspapers H, T and
Z.
Let in Venn diagram a, b, c , d , e , f and g denote the
number of elements in respective regions.
Z
a
b
d
e
Mathematics Class 11th
On putting the values of c, d and f in Eq. (iv), we get
3 + 6 + e + 5 = 26
[1]
⇒
e = 26 − 14 = 12
On putting the values of b, c and f in Eq. (iii), we get
8 + 3 + 5 + g = 26
⇒
g = 26 − 16 = 10
On putting the values of b, c and d in Eq. (ii), we get
a + 8 + 3 + 6 = 25
[1]
⇒
a = 25 − 17 = 8
(i) Number of people who read atleast one of the three
newspapers,
A(U ) = a + b + c + d + e + f + g
= 8 + 8 + 3 + 6 + 12 + 5 + 10 = 52
(ii) Number of people who read exactly one newspaper
[1]
= a + e + g = 8 + 12 + 10 = 30
[1]
H
ne
U
e
Sol. Let M, P and C denote the students studying Mathematics,
Physics and Chemistry, respectively.
Then, we have,
n (U ) = 200, n ( M ) = 120, n ( P ) = 90, n (C ) = 70,
[1]
n ( M ∩ P ) = 40, n ( P ∩ C ) = 30, n ( M ∩ C ) = 50
and
n ( M ∪ P ∪ C )′ = 20
Now,
n ( M ∪ P ∪ C )′ = n (U ) − n ( M ∪ P ∪ C )
[1]
∴
20 = 200 − n ( M ∪ P ∪ C )
[1]
⇒ n ( M ∪ P ∪ C ) = 200 − 20 = 180
We know that, n ( M ∪ P ∪ C ) = n ( M ) + n ( P ) + n (C )
− n ( M ∩ P ) − n ( P ∩ C ) − n (C ∩ M )
+ n (C ∩ M ∩ P )
∴
180 = 120 + 90 + 70 − 40 − 30 − 50
+ n (C ∩ M ∩ P )
[1]
⇒
180 = 280 − 120 + n (C ∩ M ∩ P )
[1]
⇒
180 + 120 − 280 = n ( P ∩ C ∩ M )
∴
n ( P ∩ C ∩ M ) = 300 − 280 = 20
[1]
Hence, 20 students study all the three subjects.
f
g
T
Here, n (U ) = a + b + c + d + e + f + g = 60
...(i) [1]
...(ii)
n ( H ) = a + b + c + d = 25
...(iii)
n (T ) = b + c + f + g = 26
...(iv)
n ( Z ) = c + d + e + f = 26
...(v)
n (H ∩ Z ) = c + d = 9
...(vi)
n ( H ∩ T ) = b + c = 11
...(vii)
n (T ∩ Z ) = c + f = 8
and
...(viii) [1]
n (H ∩T ∩ Z ) = c = 3
On putting the value of c in Eq. (vii), we get
3+ f =8⇒ f =5
On putting the value of c in Eq. (vi), we get
[1]
3 + b = 11 ⇒ b = 8
On putting the value of c in Eq. (v), we get
3+ d =9⇒d =6
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TRY YOURSELF
1 Mark Questions
1.
2.
If A and B are two sets, such that n( A ) = 28, n(B ) = 32 and n( A ∩ B ) = 10, then find the value of n( A ∪ B ).
If A and B are two sets, such that A ∪ B has 18 elements, A has 8 elements and B has 15 elements, then how
many elements does A ∩ B have?
4 Marks Questions
3.
In a school, there are 20 teachers who teach Maths or Physics. Out of these, 12 teach Maths and 4 teach
Physics and Maths. How many teach Physics?
4.
In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How
many people speak atleast one of these two languages?
5.
In a survey of 400 students in a school, 100 were listed as taking apple juice, 150 as taking orange juice and
75 were listed as taking both apple as well as orange juice. Find how many sutdents were taking neither apple
juice nor orange juice.
6.
In a group of 65 people, 40 like nutrition Indian food, 10 like both Indian nutrition food and fast food. How many
like only fast food? What is your opinion about this set of people?
6 Marks Questions
7.
In a town of 10000 families, it was found that 40% families buy newspaper A, 20% families buy newspaper B,
10% families buy newspaper C, 5% families buy A and B, 3% families buy B and C and 4% families buy A and
C. If 2% families buy all the three newspapers, then find
(i) the number of families which buy newspaper A only.
(ii) the number of families which buy newspaper of A, B and C only.
8.
In a group of 50 students, the number of students studying French, English and Sanskrit were found to be as
follows
French = 17, English = 13, Sanskrit = 15, French and English = 9, English and Sanskrit = 4, French and
Sanskrit = 5 and English, French and Sanskrit = 3.
Then, find the number of students who study
(i) French only
(ii) English only
(iii) French and English both
Which language is study by most of the students?
Answers for
TRY YOURSELF
Topic 1 Sets and Their Types
1. A = { − 4, − 3, − 2, − 1, 0, 1, 2, 3, 4}
2. A = {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune}
3. (ii)
4. (i) { 0, 1}
(ii) {1, p}
5. (i) { x : x = 5 , n ∈ N and 1 ≤ n ≤ 4}
n
(ii) { x : x = 2n, n ∈ N and 1 ≤ n ≤ 5}
6. A, C ; B, D
7. (i) Empty set (ii) Singleton set
(iii) Infinite set
8. Equal sets-B = D, C = F and Equivalent sets-A, E, H; B, D, G; C , F
(iv) and (v) Equal sets
9. (i) T
(iv) F
(ii) F
(iii) T
29