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## NCERT Solutions Class 11 Maths

Table of Contents

Class: | 11 |

Subject: | Mathematics (NCERT Solutions Class 11 Maths) |

Contents: | NCERT Solutions |

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### NCERT Solutions Class 11 Maths

NCERT Solutions for class 11 Maths all chapters are given below updated for new academic session 2022. Download NCERT Solutions 2022 for other subjects also. If you are having any suggestion for the improvement, your are welcome. The improvement of the website and its contents are based on your suggestion and feedback.

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### Important Topics In NCERT Solutions Class 11 Maths All Chapters

**NCERT Sols for Class 11**

NCERT Sols for Class 11 Maths, Physics, Chemistry and Biology are given below to study online or download. Notes and Study material for class 11 Physics, Chemistry, Biology, Business Studies and Physical Education are also given to download.

- Class 11 Maths
- Class 11 Physics
- Class 11 Computer Science
- Class 11 Economics
- Class 11 Business Studies
- Class 11 Financial Accounting
- Class 11 English Woven Words Essay
- Class 11 English
- Class 11 Biology PDF Download
- Class 11 Chemistry
- Class 11 Book
- Ask Your Questions

### How To Prepare Class 11 Maths All Chapters

#### Chapter 1: Sets

Set is the foundation of today’s Maths. In Chapter 1 Sets of Class 11, we will learn sets and their representation. It includes the definition as well as properties of various type of sets. A set which does not contain any element is called the empty set or the null set or the void set. Similarly, a set which is empty or consists of a definite number of elements is called finite. Otherwise, the set is called infinite. We should know about the operations of set like Union, Intersection, Adjoint and Disjoint Sets, Equal and equivalent set. Concepts of subsets, universal sets and Venn Diagram is widely used in Class 11 Maths. The Miscellaneous Exercise of Chapter 1 Class 11 Maths describes the applications of the set and its theorems.

#### Chapter 2: Relations And Functions

Relations and Functions of Class 11 Maths prepare a solid based for** Class 12 Maths Chapter 1 Relations and Functions**. In Class 11, we will learn about Cross Product of two or three sets to get ordered pair. This fact helps us to understand the concepts of Relations and Functions in Exercise 2.2 and Miscellaneous of Class 11 Maths. Do you know why students face difficulties in Class 12 Maths Chapter 1? The only reason is that they have not done the concepts of Chapter 2 of Class 11 Maths properly. Class 11 Maths Chapter 2 is the base of Class 12 Chapter 1 Relation and Functions.

#### Chapter 3: Trigonometric Functions

The first exercise of**NCERT Solutions Class 11 Maths** Chapter 3 Trigonometric Functions explains how to convert Radians to Degree or Degree to Radians. Students are more comfortable with degree measure so far. In higher classes, most often we use radians measure only. That is why now start practising in radian rather than degree. Exercise 3.2 explains how the signs of trigonometric functions change from one quadrant to others. All the formulae in Exercise 3.3, will be frequently used in most of the chapters of **NCERT Solutions Class 11 Maths**. If someone is not able to grasp these formulae, he will face the problem in Class 12 Maths and Physics.

#### Chapter 4: Principle Of Mathematical Induction

The Principle of Mathematical Inductions tells us if an expression is true for two consecutive natural numbers, then it will be true for all. To verify this, we take it is true for some positive integer k and try to prove that it is also true for k + 1. In this way, we try to prove for two consecutive positive integers. PMI is useful for not only in Class 12 Maths but in higher studies also. There are two sections of questions from 1 to 18 and from 19 to 24. In Exams, question papers consist of at least one question from each section.

#### Chapter 5: Complex Numbers And Quadratic Equations

In **NCERT Solutions Class 11 Maths** Chapter 4 Quadratic Equations, if the discriminant is negative, we were not able to find the solutions. We will deal here the same situation but getting the solutions. Now, we can find out the roots of equations even though the D is negative. It is become possible just because of Complex Numbers. In Class 11 Chapter 5 Complex numbers, we can find the real as well as imaginary roots of any quadratic equation. Square roots of a complex number also can be obtained in the supplementary section. Class 11 Maths Chapter 5 Miscellaneous Exercise is most important to know Complex numbers completely.

#### Chapter 6: Linear Inequalities

**NCERT Solutions Class 11 Maths** Chapter 6 Linear Equalities, we shall study of linear inequalities in one and two variables only. Representing on a number line, finding solutions sets as Real Numbers, Integers and Problems based on word problems are the common topics of the chapter. Exercise 6.3 is the main exercise for the exams point of view. Its weightage is maximum with respect to other exercises. This exercise is also helpful in **Class 12 Maths Chapter 12 LPP**. This chapter is easier to score good enough in the CBSE exams.

#### Chapter 7: Permutations And Combinations

**NCERT Solutions Class 11 Maths**1 Maths Chapter 7, we shall learn some basic counting techniques. It will enable us to answer so many questions without actually listing or arranging elements. Permutations will be useful to know the number of different ways quickly. As a first step, we shall know to the learning of Permutations tricks. Class 11 Maths Chapter 7 Miscellaneous Exercise questions need the application of both Permutation and Combination for answers. These topics are helpful for the calculation of Probability in **Class 11 Maths Chapter 16.**

#### Chapter 8: Binomial Theorem

**NCERT Solutions Class 11 Maths** Chapter 8 Binomial Theorem contains two exercises. Exercise 8.1 has simple sums on Binomial Expansions with two terms. The applications of general terms are in exercise 8.2. Just like the other chapters, miscellaneous of this chapter is also scoring for exams or tests. In 11th Maths,

we know the formula of expansion of 3 or 4 powers but using Binomial Theorem we can expand the expression with any number of powers. In Class 11 Maths, there are only natural numbers as indices. Binomial allows the negative powers also but not in Class 11 CBSE Syllabus 2022.

#### Chapter 9: Sequences And Series

**NCERT Solutions Class 11 Maths** Chapter 9 Sequence and Series is the next step of **Class 10 Maths Chapter 5** A P. We have done A P in Class 10, so about A.P. only a few sums are there in Class 11 Maths. Now we shall study about the arithmetic mean, geometric mean, the relationship between A.M. and G.M. and based questions. We also learn about special series in forms of the sum to n terms of consecutive natural numbers, sum to n terms of squares of natural numbers and sum to n terms of cubes of natural numbers. Class 11 Maths Chapter 9 needs more practice to know the concepts well.

#### Chapter 10: Straight Lines

**NCERT Solutions Class 11 Maths** Chapter 10 Straight Lines provides brief recall of 2 – D from earlier classes. In the supplementary part, we will learn about the shifting of origin, the slope of a line and angle between two lines. Exercise 10.1 includes simple questions, mostly based on the concepts of class 10 Maths. In the next exercises, we know about lines parallel to the axis, point-slope form, slope-intercept form, two-point form, intercept form and Normal form. Obtaining the distance of a point from a line is scoring for exams. This chapter is a little bit difficult as compared to others. So it needs more focus in practice to get good marks in tests or exams.

#### Chapter 11: Conic Sections

**NCERT Solutions Class 11 Maths** Chapter 11 Conic Sections, we shall study about curves like circles, ellipses, parabolas and hyperbolas. Just try to know who named these curves? Why is the study of these curves under the heading Conics? The 11 Maths Exercise 11.1 explains the terms and equations related to circle only whereas Exercise 11.2 for Parabola. Exercise 11.3 and 11.4 of Class 11 Maths explains about Ellipse and Hyperbola, respectively. We should know about the foci, vertices, length of the major & minor axis, eccentricity and the length of the latus rectum of the Ellipse as well as Hyperbola. In modern time, these curves are in fields such as planetary motion, design of telescopes, and so many other fields.

#### Chapter 12: Introduction To Three Dimensional Geometry

In Class 10 Maths we have learnt about the distance between the two points in 2 D plane. Here, in Class 11 Maths Chapter 12, we can find the same distance in 3 D plane. At the beginning of the chapter, the basics of quadrant octant are described. Whatever we have done in class 10 Maths Exercise 7.2, the same thing has to be repeated with the concept of 3 D geometry. Only one thing is different, that is Section Formula. In **NCERT Solutions Class 11 Maths**, we divide a line or line segment into a particular ratio internally, but in class 11 Maths internally and externally both are applicable.

#### Chapter 13: Limits And Derivatives

The concept of Chapter 13 Limits and Derivatives of Class 11 Maths is new for the students. Here, they learn how to find the value of a function using limits. Derivatives using First Principle is one of the good topics for exams. We should try to do almost all the derivatives using the First Principle. The concepts of** LHL and RHL** is useful for testing a function whether its Limits exist or not. After doing Chapter 13 of Class 11 Maths, everyone should do the **Class 12 Maths Exercise 5.2**, 5.3 and 5.4 to understand better the concept of Derivatives.

#### Chapter 14: Mathematical Reasoning

In mathematical language, there are two kinds of reasoning – inductive and deductive. Chapter 14 of Class 11 Maths follows deductive reasoning. Here, we will learn about a SENTENCE which is called a mathematically acceptable statement if it is either true or false but not both. While dealing with statements, we usually denote them by small letters p, q, r like letters. This Chapter also deals with the negation of a statement; Compound statements with connecting word and decision making final statements.

#### Chapter 15: Statistics

Exercise 15.1 of Class 11 Maths contains the questions of Deviations about the mean, mode and median. We will deal here with all the measures of central tendencies like Mean Deviation and Range. The calculation will include both grouped and ungrouped data series. Class 11 Maths Chapter 15 will discuss the questions for desecrate and continuous frequency distribution. Coefficient of variation using Variance and Standard Deviation based questions frequently come in the exams. Shortcut method to find variance and the standard deviation is equally useful for exams.

#### Chapter 16: Probability

Probability is always termed as a scoring topic in all the classes. It is scoring also in CBSE Board exams. Class 11 Maths Chapter 16 is the base for the** Class 12 Maths Chapter 13 Probability**. If you want to score well in class 12 probability, must do by learning all the concepts in grade 11. Exercise 16.1 is based on Sample Space and simple ideas of Probability. Next exercises include events which are Exclusive, Exhaustive or concepts of sets. Miscellaneous Exercise, containing good questions for practice, is vital for the class test or final exams.

The solutions by us provided are well explained keeping in mind the minutest of details and covering all the aspects of the book. The answers are created in such a manner that even a slow grasper student can also easily grab the concepts and intelligent student can mange out his time well.

NCERT is the standard book followed by 90% of the CBSE (Central Board of Secondary Education) and many state board schools across India. All the solutions provided resonate well with the latest **NCERT books** and cover the entire 16 chapters.

As we all know preparation for classes such as XI and XII is incomplete if you have not done NCERT Books. The curriculum designed by NCERT (National Council of Education Research and Training) plays an important role in preparation as all other help books and books from other authors are designed taking the pattern and syllabus from NCERT text books.

Mathematics is just not a cup of tea for everyone but we make sure that it becomes one. Our aim is just not to make the students pass the School/Board examination with flying colors but also to make them competent enough to crack top Entrance exams as well such as IIT/JEE, AIEEE and many more.

We at **NCERT Solutions Class 11 Maths** offer hassle free and free download to all the aspiring candidates. The founders believe that every child has a right to education which should be affordable as well. There are many websites floating over the internet promising to provide complete NCERT textbook solutions but are either paid or have broken links or redirects them.

All this leave the student demotivated. So, the founders decided to make the complete process clear and easy to use. There is no login or signup required. Any student with an internet connection can download our eBooks for free and get access to the most well explained solutions.

##### Which book is best for class 11 maths?

**NCERT Solutions Class 11 Maths**Textbook is best and enough for CBSE Class 11 Exam preparation. RD Sharma class 11 Textbook is best reference book for Class 11 students.##### How many chapters are there in class 11 maths?

There are 16 chapters starting from Sets to Probability.

**educationleaenacademy.com**has included all the chapters Answers in this page.##### Is class 11 Maths tough?

No. If you have very good foundation in lower classes, Maths will be very easy. CBSE has designed curriculum based on NCERT guidelines. These NCERT Textbooks are very good in presenting concepts and application based on these concepts.

**educationleaenacademy.com**recommends you to study introduction part of each and every chapter get the firm grip on the concepts.##### Write down all the subsets of the following set: {1, 2, 3}.

The subsets of {1, 2, 3} are

Φ,

{1},

{2},

{3},

{1, 2},

{2, 3},

{1, 3}

{1, 2, 3}##### In a group of students 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?

Let U be the set of all students in the group.

Let E be the set of all students who know English.

Let H be the set of all students who know Hindi.

∴ H ∪ E = U

Accordingly, n(H) = 100 and n(E) = 50

n( H U E ) = n(H) + n(E) – n(H ∩ E)

= 100 + 50 – 25

= 125

Hence, there are 125 students in the group.##### If A = {-1, 1}, find A × A × A.

A = {-1,1},

Therefore,

A×A

={-1,1}×{-1,1}={(-1,-1),(-1,1),(1,-1),(1,1)}

and

A×A×A

={(-1,-1),(-1,1),(1,-1),(1,1)} × {-1,1}

={(-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1),(1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1)}##### Write the relation R = {(x, x^3): x is a prime number less than 10} in roster form.

R = {(x, x^3): x is a prime number less than 10}

The prime numbers less than 10 are 2, 3, 5, and 7.

∴ The roster form R = {(2, 8), (3, 27), (5, 125), (7, 343)}##### मान ज्ञात कीजिए: sin〖765°〗

sin〖765°〗

=sin(2×360° + 45°)

〖=sin 45°〗

[∵ पहले चतुर्थांश में sin धनात्मक होता है]

=1/√2##### Prove the following by using the principle of mathematical induction for all n ∈ N: 〖10〗^(2n-1)+1 is divisible by 11.

Let the given statement be P(n), therefore,

P(n):〖10〗^(2n-1)+1 is divisible by 11.

For n = 1, we have

〖10〗^(2-1)+1=11,which is divisible by 11.

So, P(1) is true.

Let P(k) be true for some positive integer k, such that

P(k):〖10〗^(2k-1)+1 is divisible by 11.

Let 〖10〗^(2n-1)+1 =11m … (1)

Where, m is any natural number.

Now, to prove that P(k + 1) is true. i.e.

P(k+1):〖10〗^(2k+1)+1 is divisible by 11.

Consider 〖10〗^(2k+1)+1

=〖10〗^(2k-1+2)+1

=〖10〗^2.〖10〗^(2k-1)+1

=〖10〗^2.(11m-1)+1

[From the equation (1),〖10〗^(2n-1)=11m-1]

=100.(11m-1)+1

=1100m – 100 + 1

=1100m – 99

=11[100m – 9],which is divisible by 11.

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the Principle of Mathematical Induction, statement P(n) is true for all natural numbers.##### Solve the following equation: 2x^2 + x + 1 = 0.

The given quadratic equation is 2x^2 + x + 1 = 0.

On comparing the given equation with 〖ax〗^2+bx+c=0,

we obtain a = 2, b = 1, and c = 1

Therefore, the discriminant of the given equation is given by

D = b^2-4ac

= 1^2-4×2×1

=-7

Therefore, the required solutions are

x = (-b±√D)/2a=(-1±√(-7))/(2×2)

= (-1±√7.√(-1))/4

= (-1±√7 i)/4 [∵ √(-1)=i]##### Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.

Let x be the smaller of the two consecutive odd positive integers. Then, the other integer is x + 2.

Since both the integers are smaller than 10, therefore

x + 2 < 10 ⇒ x < 10 – 2 ⇒ x < 8 … (i) Also, the sum of the two integers is more than 11. ∴ x + (x + 2) > 11

⇒ 2x + 2 > 11

⇒ 2x > 11 – 2

⇒ 2x > 9

⇒x>9/2

⇒x>4.5 …(ii)

From (i) and (ii), we obtain that the value of x can be 4,5,6 or 7. .

Since x is an odd number, x can take the values, 5 and 7.

Hence, the required possible pairs of numbers are (5, 7) and (7, 9).##### Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?

Each signal requires the use of 2 flags.

There will be as many flags as there are ways of filling in 2 vacant places _ _ in succession by the given 5 flags of different colours.

The upper vacant place can be filled in 5 different ways by any one of the 5 flags following which, the lower vacant place can be filled in 4 different ways by any one of the remaining 4 different flags.

Thus, by multiplication principle, the number of different signals that can be generated is 5 × 4 = 20##### Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, 115 … 995.

Here, first term, a = 105 and common difference, d = 5

Now,a_n=a+(n-1)d

⇒ 995 = 105+(n-1)×5

⇒ 890=(n-1)×5

⇒ 178=(n-1)

⇒ n=179

S_n = n/2 [2a+(n-1)d]

⇒ S_179 = 179/2 [2×105+(179-1)×5]

⇒ S_179 = (179)[550]

= 98450

Hence, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.

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**NCERT Solutions Class 11 Maths**

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