## CBSE CLASS 9 MATHS NCERT SOLUTIONS

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## CBSE CLASS 10 MATHS NCERT SOLUTIONS

Dear  students  i am very happy to announce that  on demands of many students from various parts of country  i am today here giving CBSE CLASS 10 MATHS NCERT SOLUTIONS . Soultions of 9,11,12 Class is already available on this blog  i hope it is helping lot of students from villages who have lack of study material resources  and common students who really needs helps in their study i wish  and pray to god for their sucess   i am happy to serve students by helping them , i hopes my works in maths will always helps them in many ways and they will not only get good marks but also they will understand the concepts of maths and  will help them in development of their maths skills . i had prepared this solutions in step wise manner ,so that students can easily understand it.

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## UNION AND INTERSECTION OF SETS

Consider the sets A = {7, 9, 5} and B = {4, 6, 8}
What will we obtain, if we write all the elements of the sets A and B together in another set?
We will obtain the set {4, 5, 6, 7, 8, 9}.
This set, which we have obtained by writing all the elements of the sets A and B together, is the union of the two given sets A and B.
The union of two sets is defined as follows:
The union of two sets A and B is the set that consists of all the elements of A, all the elements of B, and the common elements taken only once.
The symbol ‘∪’ is used for denoting the union. For example, if X = {2, 4, 6, 8, 10} and = {4, 8, 12}, then the union of X and Y is given by X ∪ Y = {2, 4, 6, 8, 10, 12}
There are some properties of union of two sets:
1. A ∪ B = B ∪ A  (Commutative Law)
2. A ∪ Φ = A         (Law of identity element Φ )
3. A ∪ A = A         (Idempotent Law)
4. (A ∪ B) ∪ C = A ∪ (B ∪ C)    (Associative Law)
5. U ∪ A = U    (Law of universal set, U)
Now, consider the sets A = {4, 5, 9, 14} and B = {2, 4, 8, 10, 12, 14}
Are there any elements, which are common to both the sets A and B?
We can observe that the elements 4 and 14 are common to both the sets A and B. The set, which consists of the common elements i.e., the set {4, 14}, is the intersection of the sets A and B.
The intersection of two sets is defined as follows:
The intersection of sets A and B is the set of all elements that are common to both A and B.
The symbol ‘∩’ is used for denoting the intersection. For example, if X = {A, E, I, O, U} and Y = {A, B, C, D, E}, then the intersection of the sets X and Y is given by X ∩ Y = {A, E}
The properties of the intersection of two sets are given as follows:
1. A ∩ B = B ∩ A  (Commutative Law)
2. Φ ∩ A = Φ        (Law of identity element Φ )
3. A ∩ A = A        (Idempotent Law)
4. (A ∩ B) ∩ C = A ∩ (B ∩ C)    (Associative law)
5. U ∩ A = A     (Law of U)
6. A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) (Distributive law)
Have you ever observe some properties involving both union and intersection.
Some properties are given below.
• Distributive law of union over intersection of sets:
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
• Distributive law of intersection over union of sets
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
These can be verified by taking any three sets.
Now, let us again consider the above discussed sets X = {A, E, I, O, U} and Y = {A, B, C, D, E}. We found that the elements A and E are common to both these sets. Such type of sets, which have one or more elements in common, are called overlapping sets.
Two sets are called overlapping (or joint) sets, if they have at least one element in common.
Now consider the sets {1, 2, 3} and {4, 5}. Is there any element common to these sets?
We can observe that there is no element common to these sets. These types of sets are called by a special name, which is disjoint sets.
Disjoint sets are defined as follows:
If two sets A and B are such that ∩ B = Φ i.e., they have no element in common, then A and Bare called disjoint sets.
Now let us study some formulae used in set theory.
For any two finite sets A and Bn (A ∪ B) = n (A) + n (B) − n (A ∩ B)
If A and B are disjoint sets, i.e., A ∩ B = Φ then n (A ∪ B) = n (A) + n (B).
We know that if A and B are two sets, then there can be the following relationships between A and B:
1. A and B can be overlapping.
2. A and B can be disjoint.
3. Either of the sets A or B will be contained in the other set.
The Venn diagrams representing the intersection and union of the sets A and B in the above cases can be shown as follows:
1. When the sets are overlapping, they can be shown by two intersecting closed figures.
When the sets A and B are overlapping, the Venn diagram representing A ∪ B can be shown as:
When the sets A and B are overlapping, the set A ∩ B is the shaded portion of the following the Venn diagram.
1. When the sets are disjoint, they can be shown by two separate figures drawn side by side.
When the sets A and B are disjoint, the Venn diagrams representing A ∪ B can be shown as:
When the sets A and B are disjoint, the Venn diagrams representing A ∩ B can be shown as:
1. When all the elements of one set are present in the second set, they can be represented by drawing one circle inside the other.
When set B is fully contained in set A, the Venn diagrams representing A ∪ B can be shown as:
When set B is fully contained in set A, the Venn diagrams representing A ∩ B can be shown as:

Now, if three sets AB and C are given, then how will we represent the union and the intersection of these three sets?
Let’s see.
The union of the three sets AB and C, i.e., A ∪ B ∪ C, is represented by the shaded portion of the following Venn diagram.
The intersection of the three sets AB and C, i.e., A ∩ B ∩ C is represented by the shaded portion of the following Venn diagram.

Example 1:
Three sets AB, and C are defined as A = {3, 6, 8, 2, 11, 13, 12}, B = {7, 9, 3, 2, 10, 14, 15} and C= {1, 2, 3, 6, 8, 10, 11}. Find ∩ (∪ C). Also, prove the associative law of intersection and union using these sets.
Solution:
We have to find A ∩ (B ∪ C). Hence, let us first find the union of B and C, and then its intersection with A.
B = {7, 9, 3, 2, 10, 14, 15}
C = {1, 2, 3, 6, 8, 10, 11}
B ∪ C = {1, 2, 3, 6, 7, 8, 9, 10, 11, 14, 15}
Let us represent the set B ∪ C as D.
D = {1, 2, 3, 6, 7, 8, 9, 10, 11, 14, 15}
Now, we have to find A ∩ D.
A = {3, 6, 8, 2, 11, 13, 12}
∴ A ∩ D = {3, 6, 8, 2, 11}
Hence, we have
A ∩ (B ∪ C) = {3, 6, 8, 2, 11}
Now, let us prove the associative laws:
We have,
A = {3, 6, 8, 2, 11, 13, 12}
B = {7, 9, 3, 2, 10, 14, 15}
C = {1, 2, 3, 6, 8, 10, 11}
A ∪ B = {2, 3, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
A ∩ B = {2, 3}
B ∪ C = {1, 2, 3, 6, 7, 8, 9, 10, 11, 14, 15}
B ∩ C = {2, 3, 10}
Now, (A ∩ B) ∩ = {2, 3} ∩ {1, 2, 3, 6, 8, 10, 11} = {2, 3}
and A ∩ (B ∩ C) =  {3, 6, 8, 2, 11, 13, 12} ∩ {2, 3, 10} = {2, 3}
Thus, (A  B) ∩ C = A ∩ (B  C)
Also, (A ∪ B) ∪ C = {2, 3, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} ∪ {1, 2, 3, 6, 8, 10, 11}
={1, 2, 3, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
and A ∪ (B ∪ C) = {3, 6, 8, 2, 11, 13, 12} ∪ {1, 2, 3, 6, 7, 8, 9, 10, 11, 14, 15}
= {1, 2, 3, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
Thus, (A  B) ∪ C A ∪ (B  C)
Example 2:
Three sets A, and C are defined as A = {1, 2, 3, 4, 5, 6, 7, 8}, B = {4, 8, 12, 16, 20, 24} and C = {1, 4, 12, 15}. Their universal set is given as U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 18, 20, 22, 24}.
Verify the following results:
1. U  A = A and U ∪ B = U
2. n(∪ B) = n(A) + n(B) − n(∩ B)
3. n(∪ ∪ C) = n(A) + (B) + n(C) − n(∩ B) − n(∩ C) − n(∩ A) + n(∩ ∩ C)
Solution:
1.
A = {1, 2, 3, 4, 5, 6, 7, 8}, B = {4, 8, 12, 16, 20, 24} and
U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 18, 20, 22, 24}
∴ ∩ A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 18, 20, 22, 24} ∩ {1, 2, 3, 4, 5, 6, 7, 8}
= {1, 2, 3, 4, 5, 6, 7, 8}
= A
Similarly,
∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 18, 20, 22, 24} ∪ {4, 8, 12, 16, 20, 24}
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 18, 20, 22, 24}
= U
2.
A = {1, 2, 3, 4, 5, 6, 7, 8} and B = {4, 8, 12, 16, 20, 24}
Therefore, n(A) = 8 and n(B) = 6
We have:
∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 12, 16, 20, 24}
∩ B = {4, 8}
∴ n(A ∪ B) = 12 and n(AB) = 2
Now, n(A) + n(B) − n(AB) = 8 + 6 − 2 = 12 = n(∪ B)
Thus, n(A ∪ B) = n(A) + n(B) − n(∩ B)
3.
We have
A = {1, 2, 3, 4, 5, 6, 7, 8}, B = {4, 8, 12, 16, 20, 24} and C = {1, 4, 12, 15}
Therefore, n(A) = 8, n(B) = 6 and n(C) = 4
Also, ∩ B = {4, 8}, ∩ C = {4, 12}, ∩ = {1, 4}, ∪ ∪ = {1, 2, 3, 4, 5, 6, 7, 8, 12, 15 16, 20, 24} and ∩ ∩ C = {4}
Therefore, n(∪ ∪ C) = 13, n(∩ B) = 2, n(∩ C) = 2, n (∩ A) = 2 and  n (∩ ∩ C) = 1
Now,
n(A) + n(B) + n(C) − n(∩ B) − n(∩ C) − n (∩ A) + n (∩ ∩ C
= 8 + 6 + 4 − 2 − 2 − 2 + 1
= 13
n(∪ ∪ C)
Thus, n(∪ ∪ C) = n(A) + n(B) + n(C) − n(∩ B) − n(∩ C) − n (∩ A) + n (∩ ∩ C
Example 3:
Classify the following sets as disjoint or overlapping sets:
1. A = {x : 17 < x≤ 30} and B = {Prime numbers lying less than 20}
2. X = {x : x is a letter in the word COMPUTER} and Y = {y : y is the letter in the word BAG}
Solution:
1. We have A = {18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30} and
B = {2, 3, 5, 7, 11, 13, 17, 19}
A ∩ B = {19}
Therefore, the sets A and B have one element in common. Hence, they are overlapping sets.
1. We have X = {C, O, M, P, U, T, E, R} and Y = {B, A, G}
The sets X and Y have no element in common. Hence, they are disjoint sets.

Example 4:
Let A = {1, 2, 4, 6, 8}, B = {1, 3, 6, 9, 12} and C = {1, 3, 5, 7, 9, 11}.
Verify the following.
(a) Distributive law of union over intersection of sets:
(b) Distributive law of intersection over union of sets
Solution:
(a)
B ∩ C = {1, 3, 6, 9, 12} ∩ {1, 3, 5, 7, 9, 11} = {1, 3, 9}
A ∪ (B ∩ C) = {1, 2, 4, 6, 8} ∪ {1, 3, 9} = {1, 2, 3, 4, 6, 8, 9}

A ∪ B = {1, 2, 4, 6, 8} ∪ {1, 3, 6, 9, 12} = {1, 2, 3, 4, 6, 8, 9, 12}
A ∪ C = {1, 2, 4, 6, 8}∪{1, 3, 5, 7, 9, 11} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 11}
(A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4, 6, 8, 9, 12} ∩ {1, 2, 3, 4, 5, 6, 7, 8, 9, 11} = {1, 2, 3, 4, 6, 8, 9}
Thus, A ∪ (B ∩ C) = (A ∪B) ∩ (A ∪C)

Hence, distributive law of union over intersection of sets is verified.

(b)
B ∪ C = {1, 3, 6, 9, 12} ∪ {1, 3, 5, 7, 9, 11} = {1, 3, 5, 6, 7, 9, 11, 12}
A ∩ (B ∪ C) = {1, 2, 4, 6, 8} ∩ {1, 3, 5, 6, 7, 9, 11, 12} = {1, 6}
A ∩ B = {1, 2, 4, 6, 8} ∩ {1, 3, 6, 9, 12} = {1, 6}
A ∩ C = {1, 2, 4, 6, 8} ∩ {1, 3, 5, 7, 9, 11} = {1}
∴ (A ∩ B) ∪(A ∩ C) = {1, 6} ∪{1} = {1, 6}
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Hence, distributive law of intersection over union of sets is verified.

Example 5:
If A and B are two disjoint sets such that n (A) = 17 and n (AB) = 25, then what is the cardinal number of the set B?
Solution:
Since A and B are disjoint sets,
n (A ∪ B) = n (A) + (B)
⇒ 25 = 17 + (B)
⇒ (B) = 25 − 17 = 8
Therefore, the cardinal number of set B is 8.

Example 6:
In a group of 60 people, 38 people prefer to take coffee and 29 people prefer to take tea. If a person can choose at least one between tea and coffee, then how many people prefer to take both tea and coffee?
Solution:
Let the sets A and B denote
A= {people who prefer to take coffee}
B = {people who prefer to take tea}
Clearly, A ∪ B = {all people}
A ∩ B = {people who prefer both tea and coffee}
Now, we have
n (A ∪ B) = 60, n (A) = 38, (B) = 29
We know that,
n (A ∪ B) = n (A) + (B) − n (A ∩ B)
⇒ 60 = 38 + 29 − n (A ∩ B)
⇒ 60 = 67 − n (A ∩ B)
⇒ 60 − 67 = −n (A ∩ B)
⇒ −7 = −n (A ∩ B)
⇒ n (A ∩ B) = 7
Therefore, 7 people prefer to take both tea and coffee.

Example 7:
In a survey, 150 people liked winter, 200 liked summer and 50 liked both summer and winter. Find the number of people who liked
1. Winter but not summer
2. Winter or summer
Solution:
1) Let A denote the set of people who like winter and B denote the set of people who like summer.
n (A) = 150 n (B) = 200 (A ∩ B) = 50
1) From the Venn diagram, we have
A = (A − B) ∪ (A ∩ B)
n (A) = (A − B) + n (A ∩ B)
(A − B) = n (A) − n (A ∩ B)
= 150 − 50
= 100
Thus, the number of people who like winter but not summer is 100.
2) The number of people who like winter or summer is simply the union of the two sets A and B.
n (A ∪ B) = n (A) + n (B) − n (A ∩ B)
= 150 + 200 − 50
= 300
Thus, the number of people who like winter or summer is 300.
Example 8:
The Department of Non-Conventional Energy Resources conducted a survey of 1110 families to know the number of families who use different types of fuels as the means for cooking food. In the survey, it was reported that 600 families use LPG, 400 families use coal and 300 families use wood. 20 families use all the three types of fuels, 100 families use both LPG and coal, and 80 families use both LPG and wood. Also, each family uses at least one of the three fuels. How many families use both coal and wood as a fuel for cooking food?
Solution:
Let ABC denote the sets of families who use LPG, coal, and wood respectively as a fuel for cooking food. Accordingly, we have
n (A ∪ B ∪ C) = 1110, n (A) = 600, n (B) = 400, (C) = 300, n (A ∩ B ∩ C) = 20,
n (A ∩ B) = 100, n (A ∩ C) = 80
We know that
n (A ∪ B ∪ C) = n (A) + n (B) + n (C) − n (A ∩ B) − n (A ∩ C) − n (B ∩ C) + n (A ∩ B ∩ C)
⇒ 1110 = 600 + 400 + 300 − 100 − 80 − n (∩ C) + 20
⇒ 1110 = 1140 − n (B ∩ C)
⇒ n (B ∩ C) = 30

Thus, 30 families use both coal and wood as a fuel for cooking food.

## CONCEPTS OF SETS

In our daily lives, we talk about different collections such as collection of maths books in a cupboard, collection of toys in a shop, collection of shirts in a shop, students in a school, collection of all natural numbers, etc.
These collections are said to be sets.
 A set is a well-defined collection of objects.
Sets are usually represented by capital letters ABCDXYZ, etc. The objects inside a set are called  elements or members of a set. They are denoted by small letters a, bcdxyz, etc.
Now, let us consider the set of natural numbers. We know that 4 is a natural number. However, −1 is not a natural number. We denote it as 4 ∈ N and −1 ∉ N.
If is an element of a set A then we say that “a belongs to A” and mathematically we write it as “aA”; if is not an element of A then we say that “does not belong to A” and represent it as“bA”.
There are three different ways of representing a set:
1. Description method
2. Roster method or listing method or tabular form
3. Set-builder form or rule method
Let us study about them one by one.
Description method: In this method, a description about the set is made and it is enclosed in curly brackets { }.
For example: The set of composite numbers less than 30 is written as
{Composite numbers less than 30}
Roster method or listing method or tabular form : In the roster form, all the elements of a set are listed in such a manner that different elements are separated by commas and enclosed within the curly brackets { }. The roster form enables us to see all the members of a set at a glance.
For example: A set of all integers greater than 5 and less than 9 will be represented in roster form as {6, 7, 8}. However, it must be noted that in roster form, the order in which the elements are listed is immaterial. Hence, the set {6, 7, 8} can also be written as {7, 6, 8}.
Set-builder form or rule method: In set-builder representation of a set,all the elements of the set have a single common property that is exclusive to the elements of the set i.e., no other element outside the set has that property.
We have learnt how to write a set of all integers greater than 5 and less than 9 in roster form. Now, let us understand how we write the same set in set-builder form. Let us denote this set by L.
L = {x : x is an integer greater than 5 and less than 9}
Hence, in set-builder form, we describe an element of a set by a symbol (though we may use any other small letter), which is followed by a colon (:). After the colon, we describe the characteristic property possessed by all the elements of that set.
Note:
1. The order of listing the elements in a set can be changed.
2. If one or more elements in a set are repeated, then the set remains the same.
3. Each element of the set is listed once and only once.
Now, consider the following three sets.
A = {xx ∈ Z, −18 < x ≤ 5}
B = {xx ∈ W}
C = {xx ∈ N, −7 < x < −1}
Did you observe anything about the number of elements of these sets?
Observe that if we count the elements of set A, then we find that the number of elements is limited in this set. However, the number of elements in set B is not limited and we cannot count the number of elements of this set. Also, observe that set C does not contain any element as there does not exist any natural number lying between −7 and −1.
Therefore, on this basis i.e., on the basis of number of elements, the sets are classified into following categories:
(a) Finite set
(b) Infinite set
(c) Empty set
(d) Singleton set
Let us now study about them one by one.
(a) Finite set − A set that contains limited (countable) number of different elements is called a finite set.
(b) Infinite set − A set that contains unlimited (uncountable) number of different elements is called an infinite set.
(c) Empty set − A set that contains no element is called an empty set. It is also called null (or void) set. An empty set is denoted by Φ or {}. Also, since an empty set has no element, it is regarded as a finite set.
(d) Singleton set − A set having exactly one element is known as singleton set.
Therefore, we can now classify the above discussed sets as follows:
A = {−17, −16, −15, …, 0, 1, 2, 3, 4, 5} → Finite set
B = {0, 1, 2, 3, 4, 5 …} → Infinite set
C = Φ or {} → Empty set
D = {4} → Singleton set
Now, again consider set A. We have seen that it is a finite set.
Can you find the number of elements in this set?
We have A = {−17, −16, −15, −14, −13, −12, −11, −10, −9, −8, −7, −6, −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5}
We see that the number of elements in set A is 23. This number 23 is known as the cardinal number of set A.
Cardinal number of a set is defined as:
 The number of distinct elements in a finite set A is called its cardinal number. It is denoted byn (A).
Note: The cardinal number of an infinite set is not defined.
Now, can you find what the cardinal number of an empty set is?
As the empty set has no elements, therefore, its cardinal number is 0 i.e., n (Φ) = 0
We have learnt different ways of representing a set such as description method, roster method, and set-builder method. However, there is one more way of representing a given set and that is through Venn diagrams.
Venn diagrams are closed figures such as square, rectangle, circle, etc. inside which some points are marked. The closed figure represents a set and the points marked inside it represent the elements of the set.
For example, consider the set of all letters in the word AMERICA. This set consists of the letters A, M, E, R, I, and C.
This set can be represented by a Venn diagram as follows:
Sometimes, in Venn diagrams, points are not marked, only the elements are written inside the closed figure. For example, the set of letters in the word AMERICA can also be shown as follows:
Now, consider the set of all natural numbers. How will we represent this set by a Venn diagram?
In such cases, when the number of elements in a set is large, the description of the set is written in the closed figure.
Therefore, the set of all natural numbers can be shown by a Venn diagram as follows:
Let us now look at some examples to understand the above discussed concepts better.
Example 1:
Which of the following collection are sets?
1. Collection of rivers in India
2. Collection of good dancers in a locality
3. Collection of integers which are less than 21
4. Collection of best runners
5. Collection of all states of America
6. Collection of all vowels
Solution:
1. The collection of rivers is a set because every river of India will be included in it.
2. The collection of good dancers in a locality is not a set because some dancers of the locality may be good from the point of view of one person, but the same may not be good from the point of view of another person.
3. The collection of integers which are less than 21 is a set as the range of integers in the collection is defined.
4. The collection of best runners is not a set because some runners may be good from the point of view of one person, but they may not be good from the point of view of another person.
5. The collection of states of America is a set because all the states of America will be included in it.
6. The collection of all vowels is a set because all the five vowels will be included in it.
Example 2:
Write the roster form for the set A = {x : x is a letter in the word AEROPLANE which has vowels just before and after it}.
Solution:
In the word AEROPLANE, the vowels are A, E, and O.
Now, the third letter (i.e., R) has a vowel (i.e., E) just before it and a vowel (i.e., O) just after it. Hence, it satisfies the given condition.
Now, look at letter N, which has vowel (i.e., A) just before it and a vowel (i.e., E) just after it. Hence, this letter also satisfies the given condition.
Thus, the set can be written in roster form as
A = {R, N}
Example 3:
State whether each of the following sets is finite or infinite:
1. Set of multiples of 7
2. Set of lines passing through the point (1,1) as well as the origin
Solution:
1. The multiples of 7 are 7, 14, 21, 28, 35 …
Hence, the number of elements in set A = {7, 14, 21, 28, 35…} is not definite. Hence, it is an infinite set.
1. The two given points are (1,1) and (0,0) and we know that there is one and only one line passing through two fixed points. Hence, there will be only one line that passes through the given points.
Thus, the set contains only one element. Hence, it is a finite set.
2.
Example 4:
Write the following sets in set builder form.
1. {1, 8, 27, 64, 125}
2. {0, 1, 2, 3, 4, 5, 6 ,7}
3. {w, x, y, z}
4. {3, 6, 9, 12, 15, 18, 21}
5. {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
Solution:
1. {x : x is the cube of first five natural numbers}
2. {x : x ∈ Wx < 8}
3. {x : x is a letter amongst the last four letters of the English alphabet}
4. {x : x is a multiple of 3, x ≤ 21}
5. {x : x is a day of a week}
Example 5:
Write the following sets in roster and descriptive forms:
1. {x : x is a letter in the word MATHEMATICS}
2. {y : y≤ 23 and it is odd}
Solution:
1. Roster form: {M, A, T, H, E, I, C, S}
Descriptive form: {letters of the word MATHEMATICS}
1. Roster form: {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23}
Descriptive form: {Odd numbers less than and equal to 23}
Example 6:
Classify the following sets into finite set, infinite set, and empty sets. Also, find the cardinal number in case of finite sets.
(a) A = {xx is a letter in the word ENGINEER}
(b) B = {xx is a multiple of 9}
(c) C = {xx is a factor of 48}
(d) D = {xx is a vowel in the word RHYTHM}
(e) E = {xx is a vowel in the word SKY}
(f) F = {xx∈ Z}
(g) G = {all stars in universe}
(h) H = {x > 2, where x is an even prime number}
Solution:
(a) A = {E, N, G, I, R}
It is a finite set and n (A) = 5
(b) B = {9, 18, 27, 36, 45, …}
It is an infinite set.
(c) C = {2, 3, 4, 6, 8, 12, 16, 24}
It is a finite set and n (C) = 8
(d) D = Φ
It is an empty set and n (D) = 0
(e) E = Φ
It is an empty set and n (E) = 0
(f) F = {…, − 2, − 1, 0, 1, 2, …}
It is an infinite set.
(g) G = {all stars in universe}
The number of elements in set G is not defined and hence it is an infinite set.
(h) H = {x > 2, where x is an even prime number}
We can see that no value of x will satisfy the given property as 2 is the only even prime number and no even number greater than 2 will be a prime number. Hence, A will be an empty set as it has no elements.

n (H) = 0

## Complement of a set

Let us consider a set as
X = {2, 3, 6, 8}
Let us consider its universal set ξ as
ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
Can we find the set of elements in set ξ, which are not in X?
We can find this by taking the difference of X from ξ as
ξ − X = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} − {2, 3, 6, 8} = {0, 1, 4, 5, 7, 9}
This set (consisting of all the elements ξ, which do not belong to X) is known as the complement of set Xand we denote it by or.
 Let X be any set and ξ be its universal set. The complement of set X is the set consisting of all the elements of ξ, which do not belong to X. It is denoted by X′ or Xc (read as complement of set X). Thus, X′ = {x|x∈ξ and x∉X} or X′ = ξ−X
For the above sets X and ξ, we may observe that (X) = 4, (ξ) = 10, and n () = 6.
Can we find any relation among them?
We observe that:
 n() = n (ξ) − n (X)
This relation holds true for a set, its complement, and a universal set.
The other properties of a set and its complement are as follows.
(a)
(b)
(c)
(d)
(e)
(f) If X ⊆ Y then
Apart from these properties, there are two more properties for two sets A and B. They are:
(a)
(b)
These are also known as De Morgan’s laws.
Let us prove the first one.
Let A = {1, 2, 3}, B = {2, 3, 4}, and ξ = {1, 2, 3, 4, 5, 6}
Now, A ∩ B = {2, 3}
Therefore, (A ∩ B)′ = {1, 4, 5, 6}
Also, A′ = {4, 5, 6} and B′ = {1, 5, 6}
∴ A ′ ∪ B′ = {1, 4, 5, 6}
Clearly, we have
Similarly, we can prove the second one.
Now, how will we represent the complement of a set A with the help of a Venn diagram?
We know that if A is a set and ξis a universal set for the set A, then the complement of the set A is Ac = ξ− A.
If we represent the sets ξand A by a Venn diagram, then we can easily represent Ac on it.
For this, we represent the set A by using a circle and ξ by using a rectangle (or a square which is bigger and encloses the circle). Now, the portion outside the set A, but inside the set ξ, represents the set Ac. This can be shown as follows:
Let us look at some examples in order to understand these concepts better.
Example 1:
If A and B are two sets and ξ is their universal set such that , and
(B) = 6, then how many elements are there in the complement of set B?
Solution:
We know that,
We also know that,
Therefore, the complement of set B contains 2 elements.
Example 2:
If A = {x, 1, 2, 3, y}, B = {2, 4, 5, y}, and ξ= {xyz, 1, 2, 3, 4, 5, 6}, then show that
Solution:
Now, Ac = ξ − A = {x, y, z, 1, 2, 3, 4, 5, 6} − {x, 1, 2, 3, y}= {z, 4, 5, 6}
Bξ − B = {x, yz, 1, 2, 3, 4, 5, 6} − {2, 4, 5, y} = {x, y, 1, 3, 6}
A ∪ B = {x, y, 1, 2, 3, 4, 5}
∴(A ∪ B)c = {z, 6}
Now, Ac ∩ Bc = {z, 6}
Clearly, (A ∪ B)c = Ac ∩ Bc
Example 3:
Find the following sets from the adjoining Venn-diagram.
(i) (A ∩ B)c
(ii) Ac
(iii) (A  C)c
(iv) ξ
Solution:
From the given Venn-diagram, we find that
(i) (A ∩ B){1, 2, 3, 4, 6, 7, 9, 10, 11, 12, 13, 14, 15, 19}
(ii) A= {3, 4, 6, 7, 9, 11, 12, 13, 14, 15}
(iii) (A ∪ C)= {3, 4, 7, 9, 11, 13}
(iv)  ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 19}
Example 4:
Taking the set of first ten natural numbers as the universal set, find the set
(B− A)′ ∩ B′, where A = {1, 2, 4, 9} and B = {2, 5, 7, 9, 8, 10, 1, 3}
Solution:
B − A = {5, 7, 8, 10, 3}
U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(B − A) ′ = {1, 2, 4, 6, 9}
B′ = {4, 6}
∴ (B− A)′ ∩ B′ = {4, 6}

## CBSE NCERT CLASS 11 MATHS SOLUTIONS

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chapter 1
sets

chapter 2

relations and functions

chapter 3
trigonometric functions

chapter 4
_principle_of_mathematical_induction

chapter 5

chapter 6

linear_inequalities

chapter 7
permutations_and_combinations.

chapter 8

binomial_theorem

chapter 9

chapter 10

_straight_lines.

chapter 11

conic_sections

chapter 12

introduction_to_three_dimensional_geometry

chapter 13

limits and derivatives

chapter 14

mathematical reasoning

chapter 15

statistics

chapter  16

probability