# NCERT Solutions for Class 11 Maths

Share**NCERT Solutions for Class 11 Maths **consist of chapter-wise solutions for all 16 chapters of the textbook. NCERT Class 11 Maths Solutions will not only help you resolve your doubts but also make you understand even the most complex topics. Goprep's mathematics experts have solved the questions using simplest and relevant methods, which will make your exam preparation a lot easier.

Apart from NCERT Solutions, we have also provided the list of topics and formulas for each chapter of NCERT Class 11 Maths textbook. If you find our solutions helpful, then you should also share the link of solutions with your classmates so that they can also resolve their doubts and score good marks. Below, we have provided links to chapter-wise solutions for Maths textbook.

## NCERT Class 11 Maths Solutions - All Chapters

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Goprep’s CBSE Class 11th Maths Solutions give students an edge to deal with difficult questions. These solutions not only help students in their exam preparation, but also assist students in competitive exams such as JEE, VIT, and BITS. Our CBSE NCERT solutions are prepared in sync with the understanding level of the students, which enable students to practice regularly and develop mastery over a specific topic.

Class 11 Maths NCERT Solutions systematically include the detailed solutions for all **16 chapters** that are the part of Maths syllabus. Each topic within the 16 chapters is explained step-by-step to make learning simpler and quicker. Browse the chapter-wise solutions for Class 11th Maths by clicking the above-given links.

## NCERT Class 11 Maths Solutions (Chapter-wise description)

**Chapter 1: Sets**

**Introduction: **In the first chapter of NCERT Class 11 Maths textbook “Sets”, you will learn to study the geometrical shapes, sequences, probability, etc. using the concept of set. The idea behind this concept is that we often speak of collections of objects in daily life such as a pack of cards, a group of students etc. Georg Cantor, a German mathematician in his theory, termed such collection of objects as sets.

**Topics-**

- Empty set
- Finite and infinite set
- Equal set
- Subsets
- Power set
- Union and intersection of two sets
- Complement of a subset

**Important Formulas of Set**1. In the case of two sets A and B,

(A ∪ B)’ = A’ ∩ B’ and (A ∩ B)’ = A’ ∪ B’

2. If A and B are finite sets such that A ∩ B = ϕ, then

n (A ∪ B) = n (A) + n (B)

3. If A ∩ B ≠ ϕ, then

n (A ∪ B) = n (A) + n (B) – n (A ∩ B)

**Chapter 2: Relations and Functions**

**Introduction:**In chapter 2 “Relations and Functions”, you will learn to form pairs of objects from two sets and introduce relations between the objects in the pair. To master this chapter completely, you must have basic conceptual knowledge of chapter-1 sets. After a thorough understanding of relations, you will study special relations called functions.

**Topics-**

- Ordered pair
- Cartesian product
- Relations
- Image of an element
- Domain of a relation
- Range of a relation
- Function
- Domain and codomain
- Real function
- Algebra of functions

**Important Formulas of Relations and Functions**

1. **Cartesian product**

A × B of two sets A and B can be represented as-A × B = {(a, b): a ∈ A, b ∈ B}

In particular,

R × R = {(x, y): x, y ∈ R}and R × R × R = (x, y, z): x, y, z ∈ R}

2. **Some key concepts**

- If (a, b) = (x, y), then a = x and b = y.
- If n(A) = p and n(B) = q, then n(A × B) = pq.
- A × φ = φ
- In general, A × B ≠ B × A

3. **Algebra of functions**

For functions f : X → R and g : X → R, we have

- (f + g) (x) = f (x) + g(x), x ∈ X
- (f – g) (x) = f (x) – g(x), x ∈ X
- (f.g) (x) = f (x) .g (x), x ∈ X
- (kf) (x) = k ( f (x) ), x ∈ X, where k is a real number.
- ( f / g ) (x) = f (x)/g (x), x ∈ X, g(x) ≠ 0

**Chapter 3: Trigonometric Functions**

**Introduction: **Trigonometry is no longer an unfamiliar topic, as you have already studied its basic concepts in class 10. So far, you have learnt trigonometric ratios of an acute angle, the trigonometric ratio of the sides of a right-angled triangle, trigonometric identities and its applications. In NCERT Class 11 Maths textbook, you will be able to extend the definition of trigonometric ratios to any angle in terms of radian measure.

**Topics-**

- Degree and radian measure
- Relation between radian and real numbers
- Relation between radian and degree
- Trigonometric functions
- Domain and range of trigonometric functions
- Trigonometric Functions of Sum and Difference of Two Angles
- Trigonometric Equations

**Important Formulas of Trigonometric Functions**

- Radian measure = π/ 180 × Degree measure
- Degree measure = 180/ π × Radian measure
- cos²x + sin²x = 1
- 1 + tan² x = sec² x
- 1 + cot²x = cosec² x
- cos (2nπ + x) = cos x
- sin (2nπ + x) = sin x
- sin (– x) = – sin x
- cos (– x) = cos x
- cos (x + y) = cos x cos y – sin x sin y
- cos (x – y) = cos x cos y + sin x sin y
- cos (x + y) = cos x cos y – sin x sin y
- cos (x – y) = cos x cos y + sin x sin y
- cos ( π/ 2 − x ) = sin x
- sin ( π/ 2 − x ) = cos x
- sin (x + y) = sin x cos y + cos x sin y
- sin (x – y) = sin x cos y – cos x sin y

**Some more important formulas**

- sin (π/2 + x ) = cos x
- cos (π/2 + x ) = – sin x
- cos (π – x) = – cos x
- cos (π – x) = – cos x
- cos (π + x) = – cos x
- sin (π + x) = – sin x
- cos (2π – x) = cos x
- sin (2π – x) = – sin x

19. If angle (x), (y) and (x ± y) are not odd multiples of π/2, then

- tan (x + y) = tan x + tan y/ 1 - tan x tan y
- tan (x - y) = tan x - tan y/ 1 + tan x tan y

20. If angle (x), (y) and (x ± y) are multiples of π, then

- cot (x + y) = cot x cot y - 1/ cot y + cot x
- cot (x - y) = cot x cot y + 1/ cot y - cot x

21. cos 2x = cos² x – sin² x = 2cos²x – 1 = 1 – 2 sin² x = 1 - tan²x / 1 + tan²x

22. sin 2x = 2 sin x cos x = 2 tan x/ 1 + tan ²x

23. tan 2x = 2 tan x/ 1 - tan ²x

24. sin 3x = 3sinx – 4sin³ x

25. cos 3x = 4cos³ x – 3cos x

26. 2cos x cos y = cos ( x + y) + cos ( x – y)

27. – 2sin x sin y = cos (x + y) – cos (x – y)

28. 2sin x cos y = sin (x + y) + sin (x – y)

29. 2 cos x sin y = sin (x + y) – sin (x – y).

**Chapter 4: Principle of Mathematical Induction**

**Introduction: **In this chapter, you will learn deductive reasoning, which is a key aspect of scientific reasoning. The deduction is referred to as converting a general case to a particular case. On the other hand, induction can be defined as a generalization from particular cases or facts.

**Topics-**

- Deduction
- Induction
- Principle of induction

**Chapter 5: Complex Number and Quadratic Equations**

**Introduction: **In NCERT Class 10 Maths book, you have already studied linear equations in one and two variables and quadratic equations in one variable. This year, we will extend the real number system to a larger system. In mathematics, a complex number is said to be number in the form of a + ib, where (a) and (b) are real numbers.

**Topics-**

- Complex numbers
- Algebra of complex numbers
- The Modulus and the Conjugate of a Complex Number
- Argand Plane and Polar Representation
- Quadratic Equations

**Important Formulas of Complex Number and Quadratic Equations**1. Let z₁ = a + ib and z₂ = c + id. Then

- z₁ + z₂ = (a + c) + i (b + d)
- z₁ z₂ = (ac – bd) + i (ad + bc)

2. To find the solutions of the quadratic equation ax² + bx + c = 0, where a, b, c ∈ R, a ≠ 0, b^{2} – 4ac < 0, x is given by

x = -b ± √4ac - b²i/ 2a

**Chapter 6: Linear Inequalities**

**Introduction: **In earlier NCERT Maths textbooks, you built your concepts based on equations in one variable and two variables. Also, you deduced complex statement problems to simple equations. Now the question arises, is it always possible to convert a statement into an equation? Get to know whether it is possible or not in NCERT Maths class 11 textbook.

**Topics-**

- Inequalities
- Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation
- Graphical Solution of Linear Inequalities in Two Variables
- Solution of System of Linear Inequalities in Two Variables

**Chapter 7: Permutations and Combinations**

**Introduction: **Permutations and Combinations will introduce you to a few basic techniques, which will allow you to determine the different ways of arranging and selecting objects. Statement problems of this chapter are entirely based on the fundamental principle of counting. P & C is the easiest chapter featured in NCERT class 11 maths syllabus.

**Topics-**

- Fundamental principle of counting
- Permutations
- Combinations
- Theorems based on permutations and combinations

**Important Formulas of Permutations and Combinations**

^{n}P_{r}= n!/ (n-r)!, 0 ≤ r ≤ n (For permutation)^{n}C_{r }= n!/ r! (n-r)!, 0 ≤ r ≤ n (For combination)

**Chapter 8: Binomial theorem**

**Introduction: **In earlier NCERT textbooks, you learnt the method to find the squares and cubes of binomials like a + b and a - b. Using the method, you could easily solve the numerical values of numbers like (98)², (999)³ etc. However, it was not possible to find the values of numbers with higher powers like (98)⁵, (101)⁶ etc. This academic session, you will be able to calculate such values using a theorem based on repeated multiplication.

**Topics-**

- Binomial Theorem for Positive Integral Indices
- Pascal’s Triangle
- General and Middle Terms

**Chapter 9: Sequences and Series**

**Introduction: **A sequence can be termed as a collection in an ordered manner where it has an identified 1st member, 2nd member, 3rd member, and so on. For example, the population of human beings at different times, the amount of money deposited in a bank, etc. In chapter 9, you will revisit topics from earlier NCERT textbooks such as arithmetic progression (A.P) and geometric progression (G.P), etc. Besides, you will study new topics which are mentioned below.

**Topics-**

- Sequences
- Series
- Arithmetic progression
- Geometric progression
- Arithmetic mean
- Geometric mean
- Relationship between A.M. and G.M.

**Important Formulas of Sequences and Series**

1. To find the nth term of the A.P

a_{n }= a + (n-1)d

2. To find the sum of first n terms of an A.P

S_{n }= n/2. [2a + (n-1)d] = n/2. (a+l)

3. The arithmetic mean of two numbers a and b = (a + b)/ 2

4. To find the sum of first n terms of a G.P

S_{n }= a (r^{n }- 1)/ (r-1) or a (1 - r^{n})/ (1 - r), if r ≠ 1

5. The geometric means of any two positive numbers a and b = √ab, where a, G, b is G.P.

Chapter 10: Straight Lines

Chapter 10: Straight Lines

**Introduction: **You are already familiar with two-dimensional coordinate geometry which was covered in NCERT class 10 maths syllabus. The topics that were covered in the previous class were coordinate axes, coordinate plane, the distance between two points, section formulae, etc. apart from these basic concepts, you also studied a few important formulae. In NCERT class 11 maths textbook, you will study the properties of straight line. Most importantly, you will learn to represent the line algebraically.

**Topics-**

- Slope of a line
- Various Forms of the Equation of a Line
- General Equation of a Line
- Distance of a Point From a Line

**Important Formulas of Straight Lines**

1. Slope of a line (m) passing through the points (x_{1 }, y_{1}) and (x_{2}, y_{2}) = (y_{2}- y_{1})/ (x_{2}- x_{1 }) = (y_{1}- y_{2})/ (x_{1 }- x_{2}), x_{1 }≠ x_{2}

2. If a line forms an angle with the positive direction of the x-axis, then the slope of the line is given by- m = tan α, α ≠ 90°

3. Equation of the line passing through the points (x_{1}, y_{1}) and (x_{2}, y_{2}) can be expressed as

y - y_{1}= (y_{2}- y_{1})( x - x_{1})/ ( x_{2}- x_{1})

4. A line with coordinate (x, y) with slope (m) and y-intercept (c) lies on the line only if y = mx + c

5. A line with slope (m) and x-intercept (d) lies on the line only if y = m(x - d)

6. An equation of a line having intercept a and b on x- and y-axis can be expressed as (x/a) + (y/b) = 1

**Chapter 11: Conic Sections**

**Introduction: **In chapter 11 “Conic Sections” of NCERT class 11 maths Solutions, you will study equations of curved shapes such as circles, ellipses, hyperbola and parabola. Apollonius, a great mathematician, called such curved shapes as conic sections. These conic sections can be obtained by the intersection of a plane with a double-napped right circular cone. These curves find its applications in various fields such as the design of telescopes and antennas, planetary motion, reflectors in flashlights, etc.

**Topics-**

- Sections of a Cone
- Parabola
- Ellipse
- Hyperbola

**Important Formulas of Conic Sections**

**Circle**

A circle with centre (h,k) and the radius r can be represented as-(x – h)^{2 }+ (y – k)^{2}= r^{2}

**Parabola**

A parabola with focus (a,0), where a>0 and directrix x= -a can be expressed in the form as-y^{2}= 4ax

**Ellipse**

- An ellipse with foci on the x-axis can be written in the form of-x
^{2}/a^{2}+ y^{2}/b^{2}= 1 - To find the length of the latus rectum of the ellipse x
^{2}/a^{2}+ y^{2}/b^{2}= 1 ,the expression is given below-Length of latus rectum = 2b^{2}/a

**Hyperbola**

- A hyperbola with foci on the x-axis can be represented in the form of the equation as follows:x
^{2}/a^{2}- y^{2}/b^{2}= 1 - To find the length of the latus rectum of the hyperbola x
^{2}/a^{2}- y^{2}/b^{2}= 1, the expression is given below-Length of latus rectum = 2b^{2}/a

**Chapter 12: Introduction to Three Dimensional Geometry**

**Introduction: **In NCERT class 10 maths syllabus, you have already covered the basic concepts of two-dimensional geometry, including coordinate axes and coordinates of the point with respect to the axes. In this chapter, you will be able to extend your knowledge of coordinate geometry by studying the coordinates of the point with respect to the three coordinate planes.

**Topics-**

- Coordinate Axes and Coordinate Planes in Three Dimensional Space
- Coordinates of a Point in Space
- Distance between Two Points
- Section Formula

**Important Formulas of Three Dimensional Geometry**

1. The distance between two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) can be expressed as- √(x_{2}- x_{1})^{2 }+ (y_{2}- y_{1})^{2}+ (z_{2}- z_{1})^{2}

2. The coordinates of the point R when divides the line segment joining two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) both internally and externally in the ratio m:n, then, they are given by-

- [(mx
_{2}+ nx_{1})/ (m + n), (my_{2}+ ny_{1})/ (m + n), (mz_{2}+ nz_{1})/ (m + n)] and - [(mx
_{2}- nx_{1})/ (m - n), (my_{2}- ny_{1})/ (m - n), (mz_{2}- nz_{1})/ (m - n)]

3. The coordinates of the mid-point of the line segment joining two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) are given by-

[(x_{1}+ x_{2})/2, (y_{1}+ y_{2})/2, (z_{1}+ z_{2})/2]

4.To find the coordinates of the centroid of a triangle with vertices (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) and (x_{3}, y_{3}, z_{3}), use the following expressions.

[(x_{1}+ x_{2}+ x_{3})/3, (y_{1}+ y_{2}+ y_{3})/3, (z_{1}+ z_{2}+ z_{3})/3]

Chapter 13: Limits and Derivatives

Chapter 13: Limits and Derivatives

**Introduction: “**Limits and derivatives” is one of the most important chapters of NCERT class 11 maths from the exam point of view. This chapter will introduce you to calculus (integration), which involves the study of change in the value of a function with respect to change in domain. As you delve deeper into the chapter, you will study the algebra of limits and derivatives. Finally, you will learn to obtain derivatives of certain functions.

**Topics-**

- Intuitive Idea Of Derivatives
- Limits
- Limits Of Trigonometric Functions
- Derivatives
- Algebra Of Derivatives Of Functions
- Derivative Of Polynomials And Trigonometric Functions

**Important Formulas of Limits and Derivatives**

1. Different operations for functions *f *and *g* are given below:

- lim (x→a) [f (x) ± g (x)] = lim (x→a) f (x) ± lim (x→a) g (x)
- lim (x→a) [f (x). g (x)] = lim (x→a) f (x). lim (x→a) g (x)
- lim (x→a) [f (x)/ g (x)] = lim (x→a) f (x) / lim (x→a) g (x)

2. Standard limits

- lim (x→a) [(x n - a n )/ (x - a) = n a (n-1)
- lim (x →0) [Sin x/ x] = 1
- lim (x →0) [1 - cos x/ x] = 0

3. Derivative of a function *f *at *a *can be expressed as-

F’ (a) = lim (h →0) [ {f (a + h) - f (a)}/ h]

4. Derivative of a function *f *at *x *can be expressed as-

F’ (x) = lim (h →0) [ {f (x + h) - f (x)}/ h]

5. For functions *u *and *v, *use the following equations:

- ( u ± v)’ = u’ ± v’
- (uv)’ = u’v + uv’
- (u/v)’ = u’v - uv’ / v2

Chapter 14: Mathematical Reasoning

Chapter 14: Mathematical Reasoning

**Introduction: **As the title of the chapter suggests, you will learn a few basic ideas of mathematical reasoning. Human beings are considered superior to other species due to their ability to reason. How well can you use the power of reasoning? How to develop this power? You will get answers to all such questions in the context of mathematics.

**Topics-**

- Statements
- New Statements from Old
- Compound Statement
- Special Words/Phrases
- Implications
- Validating Statements

**Chapter 15: Statistics**

**Introduction: **Statistics is another familiar topic and a mainstay in NCERT maths books from classes 6 to 11. Earlier, you have understood the basic concepts of statistics such as the representation of data in graphical and tabular form, finding a representative value for the given data, etc.

In class 11, you will revisit three measures of central tendency, including mean (average), median and mode. Further, you will study a measure of dispersion and their methods of calculation for ungrouped and grouped data.

**Topics-**

- Measures of Dispersion
- Range
- Mean Deviation
- Variance and Standard Deviation
- Analysis of Frequency Distributions

**Important Formulas of Statistics**

Mean deviation for ungrouped data (MD) can be found out using-

- 𝛔
^{2 }= h^{2}/ N^{2}. [ N ∑f_{i}y_{i}^{2 }- (∑ f_{i}y_{i})^{2}] - 𝛔 = h/ N . √[ N ∑f
_{i}y_{i}^{2}- (∑ f_{i}y_{i})^{2}] , where yi = xi - A/ h

**Chapter 16: Probability**

**Introduction: **Alongside statistics, the probability is another common topic found in NCERT maths textbooks. Previously, we studied that the probability is a measure of uncertainty of various phenomena. In class 9, you learnt to find the probability based on the observations and collected data.

In NCERT class 11 maths, you will revisit a few basic terms of probability from earlier classes such as random experiment, sample space, events, etc. Besides, you will also learn new topics which are given below.

**Topics**

- Random Experiments
- Event
- Axiomatic Approach to Probability

**Important Formulas of Probability**

1. For a finite sample space with equally likely outcomes, probability of an even will be given by-

P (A) = n (A)/ n (S), where n (A) = number of elements in the set A,n (S) = number of elements in the set S

2. If there are two events A and B, then

P(A or B) = P(A) + P(B) – P(A and B)equivalently, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

3. For two mutually exclusive events A and B,

P(A or B) = P(A) + P(B)

4. If A is any event, then

5. P(not A) = 1 – P(A)

## Benefits of Class 11 Maths NCERT Solutions

- Written by expert teachers of Goprep in sync with latest CBSE curriculum
- Help students to solve all the problems given in the Class 11 Maths NCERT Book
- Systematically arranged solutions to foster quick learning and save time
- Difficult problems are solved categorically to help students learn easily
- Available for free and can be accessed without any hassle
- Act as self-study material and provide quick revision facility
- Enable students to have an idea about different types of questions and how to tackle them

## FAQs |**NCERT Class 11 Maths Book**

### Q. Where can I get the best Solutions for NCERT 11th Class Maths?

A. You can get the best quality, reliable and effective NCERT Solutions for Class 11 from Goprep. By practising the Maths subject regularly with this solution, one can achieve good marks. Our CBSE NCERT Solutions include the solved questions which you can practise daily to improve your question-solving skills.

### Q. What should be my strategy to score 85% marks in Maths of 11th Class?

A. If you are looking to score above 85% marks in the maths exam, then you got to have a perfect strategy in place. Preparing with the help of Class 11 Maths NCERT Solutions is definitely a strong strategy that can enable you to come with good marks in the exam. These solutions foster a better understanding of difficult topics and can help you improve to solve the questions quickly.

Try completing the syllabus a few months before the exam so that you get time to practice extra questions from the following Class 11 Maths study material. You may choose any one reference book to prepare for your Maths exam.

**Q. How much time do I require to cover the entire NCERT Class 11 Maths Solutions?**

A. The amount of time required to cover the entire NCERT Class 11th Maths Solutions depends upon your study routine. If you are taking reference of these NCERT Solutions on a daily basis to solve the questions of Maths Textbook, then you can cover the solutions well before the exam.

### Q. How do I improve maths question-solving skills with the help of NCERT Solutions for 11th Class Maths?

A. In order to improve your question-solving skills, the best way is to practice difficult questions on a day to day basis. This thing you can do with the help CBSE NCERT Solutions for 11th Class Maths from Goprep. Regular practice is the key to understanding the logic behind every solution and improving your skills to solve such questions.

### Final Words

Maths in 11th Class is completely different from the Maths of Class 10th. With the introduction of new and difficult concepts, it gets quite challenging for students to master the subject and score good marks in the exam. However, when you decide to study with NCERT Solutions for Class 11th Maths, you get to practice difficult questions and solve them through a different and easy approach. Moreover, these solutions enable you to improve the skills which ultimately enhances the knowledge and approach the exam with more confidence.