01_Sets.pdf

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Definition
A set is a collectionof well defined objects.
Definition and Description of Sets

(a) Sets are generally denotedby capital letters
like A, B, X, Y etc.
Elements of the set are denotedby small letters
like a, b, x, y etc.
(a) If a is an element of the set A, then it’s written as a ∈ A.
(b) Elements are not repeated in the set.
Note
Definition
A set is a collectionof well defined objects.

Description
Sets are describedin two ways.
(i) Roster form (ii) SetBuilderform
Elements are listed using
commas and brackets
For example,
{a, e, i, o, u}
Elements are told by
telling the property or
rule they follow
For example,
{x : x is a vowel}

Match the following
(a) {1, 2, 4} (i) {x : x is a prime divisor of 6}
(b) {2, 3} (ii) {x : x ∈ N and x is a divisor of 4}
(c) {A, C, E, H, I, M, S, T} (iii) {x : x is a letter of the word MATHEMATICS}
(iv) {x : x ∈ N and x < 7}
(v) {x : x is a letter of the word ALCHEMIST}
Q
Question!

Solution:
A ➝ II; B ➝ I; C ➝ III

Ex
Write the set {1, 2, 4, 8, 16, 32, 64} in set-builder form.
Q
Question!

Solution:
Given set is A = {1, 2, 4, 8, 16, 32, 64}
By observation, we can say
A = {2n - 1 : n ≤ 7 and n ∈ N}
Alternatively, we can say
A = {2n : n ≤ 6 and n ∈ W}

Write the set in set-builder form.
Q
Question!

Write the following set in the roster form.
C = (x : x2 + 7x - 8 = 0, x ∈ R}
Q
Question!

Nullset orEmptyset or Voidset
A set which does not contain any element is called Empty set
or Null set or Void set. It’s denoted by 𝜙 or { }.
Eg. (i)
(ii)

The set A = {x : x ∈ R, x2 = 16 and 2x = 6] equal-
Q
Question!
𝜙
{3}
{4}
A
B
C
D
{14, 3, 4}

The set A = {x : x ∈ R, x2 = 16 and 2x = 6] equal-
Q
Question!
𝜙
{3}
{4}
B
C
D
{14, 3, 4}
A

SingletonSet
A set containingsingle element is called singleton set.
Remark : Curly brackets ie { . } are use to write singletons.
Eg. {z} , {0} , {𝜙} , { {1,2,3,4,5} } are singleton sets.

FiniteSet
A set which is either empty or has finite number of elements is
called a finite set.
Number of elements in a finite set A is called the order or
the cardinality of the set A, generally denoted by o(A) or (A).
Remark

If A = {1, 2, {3, 4} }, then n(A) = _____.
Q
Question!
2
4
None of these
A
B
C
D
3
Recall
Eg. {z} , {0} , {𝜙} , { {1,2,3,4,5} } are singleton sets.

If A = {1, 2, {3, 4} }, then n(A) = _____.
Q
Question!
2
4
None of these
A
C
D
3
B

Given A = {1, 2, {3, 4} }
Here, {3, 4} is an element of A
So, n(A) = 3
Solution:

InfiniteSet
A set having infinite number of elements is called infiniteset.
Eg. (i) {1, 2, 3, 4, ...},
(ii) {..., -2, -1, 0, 1, 2, ...} etc.

Equal Sets
Two sets are said to be equal if they have exactly the same
elements.
Eg. If A = {0, 1, 2} , B = {2, 1, 0} , C = {0, 1, 1, 2}, then A = B = C.
Equivalent Sets
Two finite sets are said to be equivalent if their cardinalities are
equal.
Eg. {1, 2, 3} is equivalent to {dog, cat, parrot}.

Definition
Set A is said to be a subset of set B if every element of A is
also an element of B. It is generally denotedas A ⊂ B.
Ex. If A = {1, 2, 3} and B = {1, 2, 3, 4}, then A ⊂ B.
Subsets

Note
(a) Every set is a subset of itself.
(b) ɸ is a subset of all sets.
(c) A ⊂ B and B ⊂ A ⇒ A = B .

If A = { p , q , {2, 3} } , then try to observe that
(i) { p , q } ⊂ A
(ii) { { 2, 3} , p } ⊂ A
Observations

Write the subsets of {1, 2}
Q
Question!

Solution:
Let A = {1, 2}
The subsets of A are: 𝜙 , {1}, {2}, {1,2}

Definition
The collectionof all the subsets of a set A is called the power
set of A, denoted by P(A).
Power set

Write down the power sets of the following sets.
(a) A = {1, 2}
(b) A = {1, 2, 3}
Q
Question!

(a) Given A = {1, 2}
Power set of A = {ɸ, {1}, {2}, {1, 2}}
Solution:
(b) Given A = {1, 2, 3}
P(A) = {ɸ, {1}, {2}, {3}, {1, 2} {2, 3}, {1, 3}, {1, 2, 3}}

If n(A) = p, then the number of subsets of A is 2p.
Clearly, same will be n(P(A)).
Result

Two finite sets have p and q number of elements Respectively.
The number of elements in the power set of the first set is 12 more
than the number of elements in the power set of the second set.
Find p and q.
Q
Question!

Here, let the two sets be A and B,
such that n(A) = p and n(B) = q.
Number of elements in power set of A = 2p
Number of elements in power set of B = 2q
Now, according to question, 2p − 2q = 12,
that is 2q(2p - q - 1) = 22(22 -1)
∴ q = 2, p = 4
Solution:

To understand operations on sets, first we need to know the
term “universal set”.
Operations on Sets

UniversalSet
If we talk about any particular context in sets then the set
which is superset of all possible sets in the given context is
called universal set.
(Generally, it’s given in the question).
VennDiagram
Ex. A = {1, 2, 3}
B = {3, 4, 5}
U = {1, 2, 3, 4, 5, 6}
Operations on Sets

Followingare the operationson sets that we are now goingto
learn.
(a) Unionof two Sets
(b) Intersection of two Sets
(c) Difference of two Sets
(d) Symmetric Difference of two Sets
(e) Complement of a Set
Operations on Sets

Union of two Sets
A ∪ B = {x : x ∈ A or x ∈ B}
Ex. {1, 2, 3} ∪ {2, 3, 4} = {1, 2, 3, 4}
A B
Operations on Sets

Ex
1.5
Intersection of two Sets
A ∩ B = {x : x ∈ A and x ∈ B
simultaneously}
Ex. {1, 2, 3} ∩ {2, 3, 4} = {2, 3} A B
Operations on Sets

Note
(a) If C ⊂ B, then
(i) C ∪ B = B
(ii) C ∩ B = C

Note
(b) C ∩ (B ∪ A) = (C ∩ B) ∪ (C ∩ A); C ∪ (B ∩ A) = (C ∪ B) ∩ (C ∪ A)

Note
(a) If C ⊂ B, then
(i) C ∪ B = B
(ii) C ∩ B = C
(b) C ∩ (B ∪ A) = (C ∩ B) ∪ (C ∩ A); C ∪ (B ∩ A) = (C ∪ B) ∩ (C ∪ A)
(c) If C ∩ B = ɸ, then C and B are called disjointsets.

Ac or A’ = {x : x ∉ A and x ∈ U}
Eg. If U = {1, 2, 3, 4, 5, 6}, then complement of {1, 2, 3, 4} is {5, 6}.
A
Ac
Complement of a Set
Operations on Sets

Note
(a) (Ac)c = A
(b) A ∪ Ac = U

Note
(a) (Ac)c = A
(b) A ∪ Ac = U
(c) A ∩ B = 𝜙 ⇒ A ⊂ Bc

De-Morgan’s Laws
(i) (A ∪ B)c = Ac ∩ Bc
(ii) (A ∩ B)c = Ac ∪ Bc

Difference of two Sets
A – B (or) AB = {x : x ∈ A and x ∉ B}
Eg. {1, 2, 3} – {2, 3, 4} =
Eg. {1,2,3}{4,5} =
B
A
Operations on Sets

SymmetricDifference of two Sets
B
A
Operations on Sets

(i) n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
Result
Cardinality based Problems

(i) n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
(ii) n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(A ∩ C) - n(B ∩ C)
+ n(A ∩ B ∩ C)
Result
Cardinality based Problems

(a) n(A) = x1 + x2
(b) n(only A) = x1
(c) n(B) = x2 + x3
(d) n(only B) = x3
(e) n(none of A or B) = x4
Remarks
B
A
x1
x4
x2 x3

In a school, every teacher teaches either maths or physics. If 15
teach maths, 24 teach physics, while 6 teach both, then find
(a) the number of teachers.
(b) the number of teachers who teach physics only.
Q
Question!

(a) the number of teachers.
Q
Question!

Solution:
Let A be the set of teacher who teaches Maths
B be the set of teacher who teaches Physics
Then n(A) = 15, n(B) = 24
n(A ∩ B) = 6
∴ Total number of teachers
n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
= 15 + 24 – 6
= 33
A B

(b) the number of teachers who teach physics only.
Q
Question!

Solution:
Let x1 = num of teachers who teach physics only
Let x2 = num of teachers who teach maths only
Let x3 = num of teachers who teach both maths and physics
According to the question,
x1 + x3 = 24 and x2 + x3 = 15 and x3 = 6
Solving, we get x1 = 24 - 6 = 18
x1 x3 x2

In a class 60% passed their Physics examination and 58% passed
in Mathematics. What percentage of students passed both their
physics and Mathematics examination?
Q
A
B
C
D
Question!
18%
17%
16%
2%

In a class 60% passed their Physics examination and 58% passed
in Mathematics. What percentage of students passed both their
physics and Mathematics examination?
Q
B
C
D
Question!
18%
17%
16%
2%
A

Solution:
n(P) = 60,
n(M) = 58,
n(P ∪ M) = 100,
n(P ∩ M) = n(P) + n(M) - n(P ∪ M)
= 60 + 58 - 100
= 18

In a survey of 600 students in a school, 150 students were found to
be drinking Tea and 225 drinking Coffee, 100 were drinking both
Tea and Coffee. Find how many students were drinking neither
Tea nor Coffee.
Q
Question!

Solution:
Given,
Total number of students = 600
Number of students who were drinking Tea = n(T) = 150
Number of students who were drinking Coffee = n(C) = 225
Number of students who were drinking both Tea and Coffee = n(T ∩ C) = 100
n(T ∪ C) = n(T) + n(C) - n(T ∩ C)
= 150 + 225 - 100
= 375 - 100
= 275
Hence, the number of students who are drinking neither Tea nor Coffee
= 600 - 275 = 325

In a school, if 15 teachers teach maths and 25 teachers teach
physics, and if total number of teachers is 30, then try to
observe that minimum value of n(M ∩ P) is 10.
Observation

(a) If n(A) + n(B) ≥ n(U) , then minimum value of n(A ∩ B) is
given by n(A) + n(B) − n(U)
(b) If n(A) + n(B) < n(U), then minimum n(A ∩ B) = 0
Remarks

If n(A) = 5 and n(B) = 10 , then find maximum and minimum
possible values of n(A ∩ B) if
(a) n(U) = 8
(b) n(U) = 20
Q
Question!

Solution:
(a) Given, n(A) = 5 and n(B) = 10
Here, maximum possible value of n(A ∩ B) = 5
Now, as n(U) = 8 and n(A) + n(B) = 15
Thus, minimum value of n(A ∩ B) = 7

Solution:
(b) Given, n(A) = 5 and n(B) = 10
Here, maximum possible value of n(A ∩ B) = 5
Now, as n(U) = 20 and n(A) + n(B) = 15
Thus, minimum value of n(A ∩ B) = 0

There are 25 students. Every student reads 10 newspapers and
each newspaper is read by (i.e., shared by) 5 students. Find the
number of newspapers.
Q
A
B
C
D
Question!

There are 25 students. Every student reads 10 newspapers and
each newspaper is read by (i.e., shared by) 5 students. Find the
number of newspapers.
Q
B
C
D
Question!
A

Solution:
Since every student reads 10 newspaper
Thus, total number of newspaper read = 10 × 25 = 250
But each news paper is shared by 5 students
Thus actual number of newspapers =

Suppose A1, A2, ….., A30 are 30 sets, each with 5 elements and B1,
B2, …., Bn are n sets, each with 3 elements.
and each element of S belongs to exactly 10 of Ai‘s and 9 of Bj’s.
Find ‘n’.
Q
Question!

(a) n(A) = x1 + x4 + x6 + x7
(b) n(only A) = x1
(c) n(only A and B) = x1 + x4 + x2
(d) n(only one of A, B or C) = x1 + x2 + x3
(e) n(exactly two of A, B and C) = x4 + x5 + x6
(f) n(A, B or C) = x1 + x2 + x3 + x4 + x5 + x6 + x7
B
A
x1
x4
x2
x5
x6
x7
x3
Remarks
Cardinality based
Problems
C

Of the number of three athletic teams in a school, 21 are in the
basketball team, 26 in hockey team and 29 in the football team, 14
play hockey and basketball, 15 play hockey and football, 12 play
football and basketball and 8 play all the games. The total number
of members is
Q
Question!
A B C D
42 43 45 None of these
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(A ∩ C) - n(B ∩ C)
+ n(A ∩ B ∩ C)
Recall:

Of the number of three athletic teams in a school, 21 are in the
basketball team, 26 in hockey team and 29 in the football team, 14
play hockey and basketball, 15 play hockey and football, 12 play
football and basketball and 8 play all the games. The total number
of members is
Q
Question!
A C D
42 43 45 None of these
B

Solution:
Let B, H, F be the sets of the three teams respectively
So n(B) = 21, n(H) = 26, n(F) = 29,
n(H ∩ B) = 14, n(H ∩ F) = 15, n(F ∩ B) = 12,
n(B ∩ H ∩ F) = 8 and
n(B ∪ H ∪ F) = n(B) + n(H) + n(F) - n(H ∩ B) - n(H ∩ F)
- n(F ∩ B) + n(B ∩ H ∩ F)
= 21 + 26 + 29 - 14 - 15 - 12 + 8 = 43

In a class, every student plays either cricket or hockey or
badminton. 10 play cricket, 15 play hockey, 20 play badminton,
5 play cricket and hockey, 2 play hockey and badminton,
5 play badminton and cricket and 2 play all the games.
How many play exactly one of the games?
Q
Question!

Solution:
n(C) = 10 = x1 + x4 + x6 + x7
n(H) = 15 = x2 + x4 + x7 + x5
n(B) = 20 = x3 + x5 + x6 + x7
n(C ∩ H) = 5 = x4 + x7
n(H ∩ B) = 2 = x5 + x7
n(B ∩ C) = 5 = x6 + x7
n(C ∩ H ∩ B) = 2 = x7
∴ x4 = 3 ; x5 = 0 ; x6 = 3
x1 = 2 ; x2 = 10 ; x3 = 15
H
C
x1
x4
x2
x5
x6
x7
x3 B

Solution:
Number of students playing
exactly one game
= x1 + x2 + x3 = 27
H
C
x1
x4
x2
x5
x6
x7
x3 B

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