 # RD Sharma Solutions for Class 11 Maths

## RD Sharma Solutions for Class 11 Maths – Chapter wise Free PDF Download

RD Sharma Solutions for Class 11 Maths are based on the latest syllabus of Central Board of Secondary Education according to the CCE guidelines. These solutions are designed for Class 11 CBSE students to help them face the exam with confidence. Students who are not able to answer the complex questions can refer to these solutions and get their doubts cleared. The experts at BYJU’S formulate the RD Sharma Solutions so that students will be able to answer the tricky questions that would be asked in the final exam. It will also provide a strong base of fundamental concepts which would help them in their higher levels of education.

As the subject involves lots of formulas and theorems, it is important for the students to have a strong grip on each chapter. The solutions will change the approach of every student towards the subject and help them to perform well in the Class 11 exam. Here are the complete RD Sharma Class 11 Solutions to all questions in the textbook given in a detailed way, so that students can understand the concepts from CBSE Textbook with ease.

Apart from the solutions, students can also access other study materials like books, notes, exemplar problems, question papers, sample papers, worksheets etc. at BYJU’S. To understand the exam pattern and marks weightage, students are highly recommended to solve the textbook questions on a regular basis. By doing this, students will be able to solve the questions of varying difficulty level.

Download the Class 11 Maths RD Sharma Solutions chapter wise and exercise wise from the links provided below.

### RD Sharma Solutions for Class 11 Maths – Exercise-wise Solutions

Students who wish to score well in the Mathematics exam are advised to refer to the solutions PDF available at BYJU’S. It not only helps in solving the chapter wise problems but also improves the conceptual knowledge among students. The solutions are present in both online text and offline PDF format which can be accessed by the students anywhere and at any time.

### RD Sharma Solutions for Class 11 Maths Chapter 1 – Sets

Chapter 1 of RD Sharma Class 11 Solutions explains about the concept of sets. All the important terms covered in the textbook are discussed in this chapter as per the latest CBSE syllabus. Students are advised to solve the exercise wise problems by referring to the solutions PDF for a better hold on the important concepts. The exercise wise problems are solved in a comprehensive manner so that students obtain a grip on topics which are important from the exam point of view.

Sets: Sets are represented as a collection of well-defined objects or elements and it does not change from person to person. A set is represented by a capital letter.
Roster Form: All the elements of a set are listed.
Set Builder Form: The general form is, A = { x : property }.
Operations on Sets: The basic operations on sets are:

• Union of sets
• Intersection of sets
• A complement of a set
• Cartesian product of sets.
• Set difference
• Provided below are the links to the exercise wise solutions of this chapter.

Exercise 1.1

Exercise 1.2

Exercise 1.3

Exercise 1.4

Exercise 1.5

Exercise 1.6

Exercise 1.7

Exercise 1.8

Also access the following resources for Class 11 Chapter 1 Sets at BYJU’S:

• NCERT Solutions for Class 11 Maths Chapter 1 – Sets
• NCERT Exemplar Solutions for Class 11 Maths Chapter 1 Sets
• ### RD Sharma Solutions for Class 11 Maths Chapter 2 – Relations

The cartesian product of sets is the main operation which is discussed under this chapter. The brief explanation of important concepts enables students to understand the concept of relations effortlessly. Students who have doubts in the important concepts can refer to the solutions designed by the experts at BYJU’S having vast experience in the education industry. The solutions are created chapter wise and exercise wise to fulfill the needs of the students.

Relations: A relation in mathematics defines the relationship between two different sets of information. If two sets are considered, the relation between them will be established if there is a connection between the elements of two or more non-empty sets.
Types of Relations: There are 8 main types of relations which include:
• Empty Relation
• Universal Relation
• Identity Relation
• Inverse Relation
• Reflexive Relation
• Symmetric Relation
• Transitive Relation
• Equivalence Relation

This chapter contains three exercises and the solution links are available below.

Exercise 2.1

Exercise 2.2

Exercise 2.3

Also access the following resources for Class 11 Chapter 2 Relations at BYJU’S:

• NCERT Solutions For Class 11 Maths Chapter 2 Relations and Functions
• NCERT Exemplar Solutions for Class 11 Maths Chapter 2 Relations and Functions
• ### RD Sharma Solutions for Class 11 Maths Chapter 3 – Functions

The concept of function as a corresponding between two sets, and function having a relation from one set to the other set is discussed in brief under this chapter. The tutor team at BYJU’S have prepared solutions to help students attain good marks in the annual exam. The textbook problems are solved in a simple and well structured manner based on the latest CBSE syllabus and guidelines. The solutions help students to understand their areas of weakness and work on them for a better academic performance.

Function: A function is defined as a relation between a set of inputs having one output each. In simple words, a function is a relationship between inputs where each input is related to exactly one output.
Types of Functions: There are several types of functions in Maths. Some important types are:
• Injective function or one to one function: When there is mapping for a range for each domain between two sets.
• Surjective functions or onto function: When there is more than one element mapped from domain to range.
• Polynomial function: The function which consists of polynomials.
• Inverse Functions: The function which can invert another function.

The exercise wise solution links of this chapter are mentioned here.

Exercise 3.1

Exercise 3.2

Exercise 3.3

Exercise 3.4

Also access the following resources for Class 11 Chapter 3 Functions at BYJU’S:

• NCERT Solutions For Class 11 Maths Chapter 2 Relations and Functions
• NCERT Exemplar Solutions for Class 11 Maths Chapter 2 Relations and Functions
• ### RD Sharma Solutions for Class 11 Maths Chapter 4 – Measurement of Angles

Students in the previous classes would have learnt that trigonometry is a branch of Mathematics which deals with angles and sides of a triangle. Relation between degrees, radians and real numbers are the main concepts which are covered under each exercise. The expert faculty at BYJU’S have given the exercise wise solutions to help students gain more knowledge in the subject. By regular practice, students will be able to come up with flying colours in the annual exams. There is only one exercise in this chapter which can be downloaded by the students in PDF format.

Angle: An angle can be defined as the rotation from the initial point to an endpoint of a ray.
Angle measurement: Angle measurement is the amount of rotation from the initial to an endpoint of a ray. The angle is said to be a positive angle if the rotation is clockwise and negative if the rotation is anticlockwise.
Systems of measurement of angles: There are three systems for measuring angles:
1. Sexagesimal system
2. Centesimal system
3. Circular system

### RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions

Chapter 5 of RD Sharma Solutions explains about the definitions of trigonometric ratios to any angle and explains them as trigonometric functions. Shortcut tips and practical examples are explained in the solutions module in a simple language to help students score well in the exam. The solutions are designed in a lucid and accurate form to encourage students in solving complex problems in the most efficient ways.

Trigonometric Functions: Trigonometric functions are also known as a Circular Functions can be simply defined as the functions of an angle of a triangle.
Trigonometric Identities: There are various identities in trigonometry which are used to solve many trigonometric problems.
• Reciprocal Trigonometric Identities
• Pythagorean Trigonometric Identities
• Ratio Trigonometric Identities
• Opposite Angles Trigonometric Identities
• Complementary Angles Trigonometric Identities
• Supplementary Angles Trigonometric Identities
• Sum and Difference of Angles Trigonometric Identities
• Double Angle Trigonometric Identities
• Half Angle Trigonometric Identities
• Product Sum Trigonometric Identities
• Products Trigonometric Identities
Periodic Function: A function returning to the same value at regular intervals.

There are 3 exercises in this chapter and the solution links are provided here.

Exercise 5.1

Exercise 5.2

Exercise 5.3

Also access the following resources for Class 11 Chapter 5 Trigonometric Functions at BYJU’S:

• NCERT Solutions For Class 11 Maths Chapter 3 Trigonometric Functions
• NCERT Exemplar Solutions for Class 11 Maths Chapter 3 Trigonometric Functions
• ### RD Sharma Solutions for Class 11 Maths Chapter 6 – Graphs of Trigonometric Functions

In Chapter 5, students would have understood that trigonometric functions are periodic functions. But here we shall discuss the graphs on the intervals of lengths equal to their periods. We at BYJU’S have formulated the solutions to enhance the performance of students in the Class 11 annual exam. The graphs of sine, cosine, tangent, cosecant, cotangent and secant are the main concepts which are covered under this chapter.

This Chapter has 3 exercises and the solution links are as follows.

Exercise 6.1

Exercise 6.2

Exercise 6.3

Also access the following resources for Class 11 Chapter 6 Graphs of Trigonometric Functions at BYJU’S:

• NCERT Solutions For Class 11 Maths Chapter 3 Trigonometric Functions
• NCERT Exemplar Solutions for Class 11 Maths Chapter 3 Trigonometric Functions
• ### RD Sharma Solutions for Class 11 Maths Chapter 7 – Trigonometric Ratios of Compound Angles

The formulae which expresses the values of trigonometric functions at the sum or difference of two real numbers are derived in this chapter. The experts at BYJU’S have formulated these solutions in a lucid manner as per the latest CBSE syllabus. The shortcut tips explained in a simple language helps students to grasp the important concepts effortlessly. The maximum and minimum value of trigonometric expressions and to express the given expressions in designed form are the other topics which are highlighted in this chapter.

Sum and Difference Identities:
• sin(x+y) = sin(x). cos(y)+cos(x).sin(y)
• sin(x–y) = sin(x). cos(y)–cos(x).sin(y)
• cos(x+y) = cosx.cosy–sinx.siny
• cos(x–y) = cosx.cosy+sinx.siny
• tan(x+y) = [tan(x)+tan(y)]/[1-tan(x)tan(y)]

• tan(x-y) = [tan(x)-tan(y)]/[1+tan(x)tan(y)]

Chapter 7 has 2 exercises and the exercise wise links of the solutions are given here.

Exercise 7.1

Exercise 7.2

### RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae

The solutions for the RD Sharma textbook are designed in accordance with the latest CBSE syllabus. In Chapter 8, two sets of transformation formulae are explained in a simple language to make the concepts clear among the students. These two sets of formulae are of fundamental importance and the students should be thorough with them in order to solve the exercises efficiently. Practising the exercise wise solutions on a daily basis is an essential task among students to learn and score well in the annual exam.

Product into Sum or Difference formulae: To solve the trigonometry functions use below given Product to sum formula.
Product to Sum formulas are:

• cos(a)cos(b)= ½ [cos (a + b) + cos (a – b)]
• sin(a) sin(b)= ½ [cos (a – b) – cos (a + b)]
• sin (a) cos (b) = ½ [sin (a + b) + sin (a – b)]
• cos(a) sin (b) = ½ [sin(a + b) – sin (a – b)]
• Sum to Product formulas are:

• sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
• sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
• cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
• cos A – cos B = -2 sin (A+B)/2 sin (A-B)/2
• The exercise wise solution links of Chapter 8 is provided below.

Exercise 8.1

Exercise 8.2

### RD Sharma Solutions for Class 11 Maths Chapter 9 – Trigonometric Ratios of Multiple and Sub Multiple Angles

The formula expressing the values of trigonometric functions at multiples and submultiples of ‘x’ is the main concept discussed in this chapter. The solutions of chapter wise problems are completely exam oriented to help students score well in the annual exam. Practising the exercise wise problems on a regular basis will ensure students to excel in their final exams. The values of trigonometric functions at some important points are the other concepts discussed in this chapter.

Values of trigonometric functions at ‘2x’ in terms of values at ‘x’:

• sin(2x) = 2sin(x).cos(x) = [2tan x/(1+tan2 x)]
• cos(2x) = cos2(x)–sin2(x) = [(1-tan2 x)/(1+tan2 x)]
• cos(2x) = 2cos2(x)−1 = 1–2sin2(x)
• tan(2x) = [2tan(x)]/ [1−tan2(x)]
• sec (2x) = sec2 x/(2-sec2 x)
• csc (2x) = (sec x. csc x)/2
• Values of trigonometric functions at ‘3x’ in terms of values at ‘x’:
• Sin 3x = 3sin x – 4sin3x
• Cos 3x = 4cos3x-3cos x
• Tan 3x = [3tanx-tan3x]/[1-3tan2x]

The exercise solution links for the topics covered in this chapter are given here.

Exercise 9.1

Exercise 9.2

Exercise 9.3

### RD Sharma Solutions for Class 11 Maths Chapter 10 – Sine and Cosine Formulae and their Applications

Practising all the exercise wise questions on a regular basis helps students to achieve a good academic score in Mathematics. Chapter 10 of RD Sharma Solutions helps students to learn about some trigonometric relations with elements of a triangle. Students having doubts about the important concepts can cross check their answers referring to the RD Sharma Solutions prepared by the experts at BYJU’S. The laws of sines, cosines, projection formulae, Napier’s analogy and area of triangle are the important concepts covered in this chapter.

Law of sines or sine rule: The law of sines is generally used to find the unknown angle or side of a triangle. This law can be used if certain combinations of measurement of a triangle are given.
1. ASA Criteria: Given two angles and included side, to find the unknown side.
2. AAS Criteria: Given two angles and non-included side, to find the unknown side.
Law of cosines formula: As per the cosines law formula, to find the length of sides of triangle say △ABC, we can write as;
• a2= b2 + c2 – 2bc cos α
• b2 = a2 + c2 – 2ac cos β
• c2 = b2 + a22 – 2ba cos γ

This chapter consists of two exercises and the solution links for the exercise wise problems are available below.

Exercise 10.1

Exercise 10.2

### RD Sharma Solutions for Class 11 Maths Chapter 11 – Trigonometric Equations

Students get an overview of the trigonometric equations in this chapter. Trigonometric equations are nothing but equations having trigonometric functions of unknown angles. Students who are not able to understand the complex topics during class hours can refer to the solutions provided in PDF format. The solutions are curated with the main objective of helping students understand the fundamental concepts which are important from the exam perspective. This chapter has only one exercise which can be referred by the students to score well in the Class 11 exams.

Trigonometric Equations: The equations that involve the trigonometric functions of a variable are called trigonometric equations.

 Equations Solutions sin x = 0 x = nπ cos x = 0 x = (nπ + π/2) tan x = 0 x = nπ sin x = 1 x = (2nπ + π/2) = (4n+1)π/2 cos x = 1 x = 2nπ sin x = sin θ x = nπ + (-1)nθ, where θ ∈ [-π/2, π/2] cos x = cos θ x = 2nπ ± θ, where θ ∈ (0, π] tan x = tan θ x = nπ + θ, where θ ∈ (-π/2 , π/2] sin2 x = sin2 θ x = nπ ± θ cos2 x = cos2 θ x = nπ ± θ tan2 x = tan2 θ x = nπ ± θ

### RD Sharma Solutions for Class 11 Maths Chapter 12 – Mathematical Induction

This chapter mainly throws light on the problems based on principles of mathematical induction. The solutions are designed by the expert tutor at BYJU’S keeping in mind the understanding abilities of students. Each and every problem is solved in an explanatory manner to make learning fun for the students. Mathematical statements is another important concept which is explained in this chapter.

Mathematical Induction: Mathematical Induction is a technique of proving a statement, theorem or formula which is thought to be true, for each and every natural number n.
Principles of Mathematical Induction: The first step of the principle is a factual statement and the second step is a conditional one. According to this if the given statement is true for some positive integer k only then it can be concluded that the statement P(n) is valid for n = k + 1.

The PDF links of the exercise wise problems are provided below for easy access.

Exercise 12.1

Exercise 12.2

Also access the following resources for Class 11 Chapter 12 Mathematical Induction at BYJU’S:

• NCERT Solutions For Class 11 Maths Chapter 4 Principle of Mathematical Induction
• NCERT Exemplar Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction
• ### RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers

The combination of both real numbers and imaginary numbers are called complex numbers. Suitable examples are provided for each concept with detailed explanations to make the understanding of concepts easier among students. Integral powers of IOTA, imaginary quantities, equality, addition, subtraction, multiplication, division and modulus of complex numbers are various other concepts which are covered in this chapter.

Complex Numbers: Combination of both the real number and imaginary number is a complex number.
Properties of Complex Numbers: The properties of complex numbers are listed below:
• The addition of two conjugate complex numbers will result in a real number
• The multiplication of two conjugate complex number will also result in a real number
• If x and y are the real numbers and x+yi =0, then x =0 and y =0
• If p, q, r, and s are the real numbers and p+qi = r+si, then p = r, and q=s
• The complex number obeys the commutative law of addition and multiplication.
z1+z2=z2+z1
z1. z2=z2. z1
• The complex number obeys the associative law of addition and multiplication.
(z1+z2) +z3=z1+ (z2+z3)
(z1.z2).z3= z1.(z2.z3)
• The complex number obeys the distributive law
z1.(z2+z3) = z1.z2+ z1.z3
• If the sum of two complex number is real, and also the product of two complex number is also real, then these complex numbers are conjugate to each other.
• For any two complex numbers, say zand z2, then |z1+z2| ≤ |z1|+|z2|
• The result of the multiplication of two complex numbers and its conjugate value should result in a complex number and it should be a positive value.

The solution links of all four exercises in this chapter are given below.

Exercise 13.1

Exercise 13.2

Exercise 13.3

Exercise 13.4

Also access the following resources for Class 11 Chapter 13 Complex Numbers at BYJU’S:

• NCERT Solutions For Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations
• NCERT Exemplar Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations
• ### RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations

In the previous classes, students would have learnt about quadratic equations with real coefficients and real roots only. But in this chapter, the important concepts like quadratic equations with real coefficients and complex roots are discussed in brief. Students can solve the exercise wise problems on a daily basis referring to the solutions designed by the experts at BYJU’S to score well in the exams.

Quadratic Equation: Quadratics or quadratic equations can be defined as a polynomial equation of a second degree, which implies that it comprises of a minimum of one term that is squared. The general form of the quadratic equation is:
ax² + bx + c = 0
where x is an unknown variable and a, b, c are numerical coefficients.
Quadratic Formula: The formula for a quadratic equation is used to find the roots of the equation. Since quadratics have a degree equal to two, therefore there will be two solutions for the equation. Suppose, ax² + bx + c = 0 is the quadratic equation, then the formula to find the roots of this equation will be:
x = [-b±√(b2-4ac)]/2
The sign of plus/minus indicates there will be two solutions for x.

There are 2 exercises in this chapter and the links of the solutions PDF is provided here.

Exercise 14.1

Exercise 14.2

Also access the following resources for Class 11 Chapter 14 Quadratic Equations at BYJU’S:

• NCERT Solutions For Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations
• NCERT Exemplar Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations
• ### RD Sharma Solutions for Class 11 Maths Chapter 15 – Linear Inequations

Linear Inequations in one and two variables is the main concept which is covered in this chapter. The knowledge of this concept helps students to solve problems in various streams like Engineering, Mathematics, Science and Linear Programming etc. The comprehensive explanations of solutions help students in solving complex problems in the most efficient ways possible. Graphical representation of problems helps students to understand the concepts without any difficulty.

Linear Inequations: A linear inequation in one variable is an equation which has a maximum of one variable of order 1. It is of the form ax + b = 0, where x is the variable.
Steps to solve Linear Equations in One Variable:
For solving an equation having only one variable, the following steps are followed
• Step 1: Using LCM, clear the fractions if any.
• Step 2: Simplify both sides of the equation.
• Step 3: Isolate the variable.
Linear equations in two variables: An equation is said to be linear equation in two variables if it is written in the form of ax + by + c=0, where a, b & c are real numbers and the coefficients of x and y, i.e a and b respectively, are not equal to zero.

All six exercise problems of this chapter are solved by subject matter experts at BYJU’S and the solution links are provided below.

Exercise 15.1

Exercise 15.2

Exercise 15.3

Exercise 15.4

Exercise 15.5

Exercise 15.6

Also access the following resources for Class 11 Chapter 15 Linear Inequations at BYJU’S:

• NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities
• NCERT Exemplar Solutions for Class 11 Maths Chapter 6 Linear Inequalities
• ### RD Sharma Solutions for Class 11 Maths Chapter 16 – Permutations

The term and notation of a factorial is introduced in this chapter. The solution module involves various shortcut tricks and illustrations to explain the exercise problems in an understandable manner. The faculty at BYJU’S have simplified the complex problems into simpler steps, which can be easily solved by the students. The factorial, fundamental principles of counting, permutations, permutations under certain conditions and permutations of objects not all distinct are the other concepts which are explained in this chapter.

Permutation: Permutation relates to the act of arranging all the members of a set into some sequence or order.
A permutation is the choice of r things from a set of n things without replacement and where the order matters.
nPr = (n!) / (n-r)!

Students can hereby access the solutions of 5 exercises from the links given here.

Exercise 16.1

Exercise 16.2

Exercise 16.3

Exercise 16.4

Exercise 16.5

Also access the following resources for Class 11 Chapter 16 Permutations at BYJU’S:

• NCERT Solutions for Class 11 Chapter 7- Permutations and Combinations
• NCERT Exemplar Solutions for Class 11 Maths Chapter 7 Permutations and Combinations
• ### RD Sharma Solutions for Class 11 Maths Chapter 17 – Combinations

Every different selection made by considering some or all the number of objects, irrespective of their arrangements is termed as combinations. Combinations, Properties of C, practical problems of combinations and mixed problems on permutations and combinations are the main concepts explained in this chapter. The RD Sharma Solutions for Class 11 Maths can be used as a reference material by the students while solving the textbook problems to score well in the annual exams.

Combination: The combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter.
A combination is the choice of r things from a set of n things without replacement and where order does not matter.
nCr = (n/r) = nPr/ r! = (n!) / r! (n-r)!

This chapter contains 3 exercises and the PDF solution links are as follows.

Exercise 17.1

Exercise 17.2

Exercise 17.3

Also access the following resources for Class 11 Chapter 17 Combinations at BYJU’S:

• NCERT Solutions for Class 11 Chapter 7- Permutations and Combinations
• NCERT Exemplar Solutions for Class 11 Maths Chapter 7 Permutations and Combinations
• ### RD Sharma Solutions for Class 11 Maths Chapter 18 – Binomial Theorem

An algebraic expression possessing two terms is known as binomial expressions. Binomial theorem, some important conclusions, general and middle terms in a binomial expansion are the main concepts discussed in this chapter. These solutions provide the students with guidance in solving complex problems without any difficulty. Students under the CBSE board can download the solutions PDF and refer to them while solving the textbook problems.

Binomial Theorem: The Binomial Theorem is the method of expanding an expression which has been raised to any finite power.
Let n ∈ N,x,y,∈ R then
(x + y)n = nΣr=0 nCr xn – r · yr where,
nCr = (n!) / r! (n-r)!

2 exercises are present in this chapter and the PDF links are given below.

Exercise 18.1

Exercise 18.2

Also access the following resources for Class 11 Chapter 18 Binomial Theorem at BYJU’S:

• NCERT Solutions for Class 11 Maths Chapter 8- Binomial Theorem
• NCERT Exemplar Solutions for Class 11 Maths Chapter 8 Binomial Theorem
• ### RD Sharma Solutions for Class 11 Maths Chapter 19 – Arithmetic Progressions

A sequence in which the difference between the term and previous term is always a constant is called an arithmetic progression. Students often have difficulty in understanding the concepts during class hours. For this purpose, the teachers at BYJU’S have designed the exercise wise problems as per the grasping abilities of students. By solving these problems on a regular basis, students will improve their problem solving skills which are important from the exam perspective.

Arithmetic Progression: A mathematical sequence in which the difference between two consecutive terms is always a constant and it is abbreviated as AP.
nth term of an AP: The formula for finding the n-th term of an AP is:
an = a + (n − 1) × d
Where,
a = First term
d = Common difference
n = number of terms
an = nth term
Sum of n terms of an AP: The formula for the arithmetic progression sum is explained below:
Consider an AP consisting “n” terms.
S = n/2[2a + (n − 1) × d]

There are 7 exercises in this chapter and the links to the exercise wise solutions are provided here.

Exercise 19.1

Exercise 19.2

Exercise 19.3

Exercise 19.4

Exercise 19.5

Exercise 19.6

Exercise 19.7

Also access the following resources for Class 11 Chapter 19 Arithmetic Progression at BYJU’S:

• NCERT Solutions for Class 11 Maths Chapter 9- Sequences and Series
• NCERT Exemplar Solutions for Class 11 Maths Chapter 9 Sequence and Series
• ### RD Sharma Solutions for Class 11 Maths Chapter 20 – Geometric Progressions

A sequence in which the ratio of the term and the preceding term is alway a constant is called a geometric progression. The subject experts have designed the exercise wise solutions based on the latest syllabus of the CBSE board. The solutions contain explanations in a step wise manner to improve the conceptual knowledge in the minds of students. RD Sharma Solutions curated by the BYJU’S tutors are of immense help to the students who aim to score good marks in the Class 11 exams.

Geometric Progression: A geometric progression or a geometric sequence is the one, in which each term is varied by another by a common ratio.
It is represented by:
a, ar, ar2, ar3, ar4, and so on.
Where a is the first term and r is the common ratio.
General Term or nth term of GP: Let a be the first term and r be the common ratio for a G.P.
Then the second term, a2 = a × r = ar
Third term, a3 = a2 × r = ar × r = ar2
Similarly, nth term, an = arn-1
Therefore, the formula to find the nth term of GP is:
an = arn-1
Sum of n terms of GP: The formula to find sum of n terms of GP is:
Sn = a[(rn-1)/(r-1)] if r ≠ 1
Where
a is the first term
r is the common ratio
n is the number of terms

The exercise wise solution links for the problems present under this chapter are available below.

Exercise 20.1

Exercise 20.2

Exercise 20.3

Exercise 20.4

Exercise 20.5

Exercise 20.6

Also access the following resources for Class 11 Chapter 20 Geometric Progression at BYJU’S:

• NCERT Solutions for Class 11 Maths Chapter 9- Sequences and Series
• NCERT Exemplar Solutions for Class 11 Maths Chapter 9 Sequence and Series
• ### RD Sharma Solutions for Class 11 Maths Chapter 21 – Some Special Series

The sum to ‘n’ terms of some other special series is the main concept explained in this chapter. RD Sharma Solutions are designed for the students under the CBSE board as per the latest syllabus and guidelines. Method of difference and summation of some special series are the other topics covered here. Students are advised to thoroughly solve the textbook problems before the final exams and intensify their abilities to solve complex problems.

Sum of first n natural numbers: The sum of first n natural numbers is
Sum = n (n + 1)/2
Sum of squares of first n natural numbers: The formula for the addition of squares of natural numbers is given below:
Σn2 = [n(n+1)(2n+1)]/6

Chapter 21 has two exercises and the links for the exercise wise solutions designed by the subject experts at BYJU’S are provided here.

Exercise 21.1

Exercise 21.2

Also access the following resources for Class 11 Chapter 21 Some Special Series at BYJU’S:

• NCERT Solutions for Class 11 Maths Chapter 9- Sequences and Series
• NCERT Exemplar Solutions for Class 11 Maths Chapter 9 Sequence and Series
• ### RD Sharma Solutions for Class 11 Maths Chapter 22 – Brief Review of Cartesian System of Rectangular Coordinates

RD Sharma Solutions prepared by the experts at BYJU’S contains a variety of shortcut tips with the aim of helping students ace the exam without fear. This chapter deals with topics like cartesian coordinate system, distance between two points, area of a triangle, section formula, locus and shifting of origin. Students are advised to refer to the solutions PDF in order to gain a strong knowledge of fundamental concepts. It also provides students with an overview of the concepts which are important for the exams.

Cartesian Coordinate System: A Cartesian coordinate system or Coordinate system is used to locate the position of any point and that point can be plotted as an ordered pair (x, y) known as Coordinates.
Distance formula for two points: The distance formula for two points in a plane is written as:
$$PQ=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$$

This chapter has three exercises and the solution links below provides answers for all the questions present in the textbook.

Exercise 22.1

Exercise 22.2

Exercise 22.3

### RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines

A straight line is a curve where every point on the line segment joining any two points lies on it. This chapter discusses the concept of straight lines with numerous examples to provide better understanding in the minds of students. The faculty at BYJU’S have solved each and every problem accurately as per the textbook prescribed by the CBSE board. The solutions are detailed and considered to be the best reference guide for the students preparing for their Class 11 exams.

Slope Formula: If P(x1,y1) and Q(x2,y2) are the two points on a straight line, then the slope formula is given by:
Slope, m = Change in y-coordinates/Change in x-coordinates
m = (y2 – y1)/(x2 – x1)

Slope of a line equation: The equation for the slope of a line and the points also called point slope form of equation of a straight line is given by:
y − y1 = m(x − x1)
Whereas the slope-intercept form the equation of the line is given by:
y = mx + b
Where b is the y-intercept.

Students can download the PDF of 19 exercise solutions of this chapter from the links given below.

Exercise 23.1

Exercise 23.2

Exercise 23.3

Exercise 23.4

Exercise 23.5

Exercise 23.6

Exercise 23.7

Exercise 23.8

Exercise 23.9

Exercise 23.10

Exercise 23.11

Exercise 23.12

Exercise 23.13

Exercise 23.14

Exercise 23.15

Exercise 23.16

Exercise 23.17

Exercise 23.18

Exercise 23.19

Also access the following resources for Class 11 Chapter 23 The Straight Lines at BYJU’S:

• NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines
• NCERT Exemplar Solutions for Class 11 Maths Chapter 10 Straight Lines
• ### RD Sharma Solutions for Class 11 Maths Chapter 24 – The Circle

In this chapter, students will learn about the circles and the equation of any circle whose centre and radius are given. The experts at BYJU’S create the exercise wise solutions after conducting a vast research on each concept by keeping in mind the understanding of students. Shortcut techniques are used in order to help students in solving complex problems without any difficulty. RD Sharma Class 11 Solutions are considered to be one of the best study materials for the students in the current education market.

Standard Form of Circle Equation:
(x – a)2 + (y – b)2 = r2
Where,
• (a, b) are the coordinates of centre
General Form of Circle Equation:
x2 + y2 + Ax + By + C = 0

Chapter 24 has 3 exercises and the links of the PDF solutions are provided below.

Exercise 24.1

Exercise 24.2

Exercise 24.3

Also access the following resources for Class 11 Chapter 24 The Circle at BYJU’S:

• NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections
• NCERT Exemplar Solutions for Class 11 Maths Chapter 11 Conic Sections
• ### RD Sharma Solutions for Class 11 Maths Chapter 25 – Parabola

Chapter 25 explains students about the parabola and to find the general equation of a conic section when its directrix, focus and eccentricity are provided. Students who are not able to understand the important concepts during class hours are recommended to solve the problems using the solutions PDF as a reference guide. The solutions are available in both online and offline mode based on the comfort of students. The PDF can be downloaded by the students and used whenever they feel to study irrespective of time and place. This chapter has one exercise covering all the concepts which are important from the exam perspective.

Parabola: Section of a right circular cone by a plane parallel to a generator of the cone is a parabola.
Standard Equation of a Parabola: In general, if the directrix is parallel to the y-axis in the standard equation of a parabola is given as:
y2 = 4ax
If the parabola is sideways i.e., the directrix is parallel to x-axis, the standard equation of a parabola becomes,
x2 = 4ay

Also access the following resources for Class 11 Chapter 25 Parabola at BYJU’S:

• NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections
• NCERT Exemplar Solutions for Class 11 Maths Chapter 11 Conic Sections
• ### RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse

This chapter mainly deals with the concept of ellipse and to determine the equation of ellipse in standard and other forms. In order to boost the academic level of students in the annual exams, we at BYJU’S have designed exercise wise solutions for the problems available in the textbook. Students can download the PDF and start practising the exercises in order to obtain a grip on the fundamental concepts. The solutions contain explanations in a simple language by keeping in mind the understanding abilities of students irrespective of their intelligence quotient. Chapter 26 has only one exercise as per the latest CBSE syllabus.

Ellipse: The ellipse is one of the conic sections, that is produced, when a plane cuts the cone at an angle with the base.
Ellipse Equation: When the centre of the ellipse is at the origin (0,0) and the foci are on the x-axis and y-axis, then we can easily derive the ellipse equation.
The equation of the ellipse is given by;
x2/a2 + y2/b2 = 1
Standard Equation of Ellipse: The equation of the ellipse with center at origin and major axis along the y-axis is:
x2/b2 + y2/a2 = 1
where –b ≤ y ≤ b.

Also access the following resources for Class 11 Chapter 26 Ellipse at BYJU’S:

• NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections
• NCERT Exemplar Solutions for Class 11 Maths Chapter 11 Conic Sections
• ### RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola

In the earlier chapters, we have discussed that a hyperbola is a particular case of conic. But here in this chapter, students learn about hyperbola in detail by determining the equation in standard form. The solutions created by the experts help students improve their logical approach to solving difficult problems. The detailed explanations under each problem help students in preparing well for their annual exams. Students who are not able to understand the concepts covered in this chapter can download the solutions PDF and start practising offline for better results. This chapter has only one exercise covering all the topics based on the latest syllabus of CBSE board.

Hyperbola: A hyperbola is the locus of all those points in a plane such that the difference in their distances from two fixed points in the plane is a constant.
Standard Equation of Hyperbola: The equation with centre at origin and conjugate axis along the y-axis is:
x2/a2 – y2/b2 = 1
Hyperbola Eccentricity: The ratio of distances from the center of hyperbole from either focus to either of the vertices of the hyperbola is defined as eccentricity.
Eccentricity, e = c/a
Since c ≥ a, the eccentricity is always greater than 1 in the case of a hyperbola.

Also access the following resources for Class 11 Chapter 27 Hyperbola at BYJU’S:

• NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections
• NCERT Exemplar Solutions for Class 11 Maths Chapter 11 Conic Sections
• ### RD Sharma Solutions for Class 11 Maths Chapter 28 – Introduction to 3D Coordinate Geometry

The coordinates of a point in space, signs of coordinates of a point, distance formula and section formula are the important topics covered in this chapter. The subject matter experts having vast knowledge about these concepts create the exercise wise solutions by keeping in mind the understanding levels of students. Each and every problem is solved in a comprehensive way in order to help students analyse the type of problems that would appear in the exams.

Distance Formula: The formula to find the distance between two points PQ is given by:
PQ = √[(x2 – x1)2 + (y2 – y1)2] Where D is the distance between the points.
Signs of coordinates of a point:

Students can access the exercise wise solutions designed by the BYJU’S tutors from the links given below.

Exercise 28.1

Exercise 28.2

Exercise 28.3

Also access the following resources for Class 11 Chapter 28 Introduction to 3D Coordinate Geometry at BYJU’S:

• NCERT Solutions for Class 11 Maths Chapter 12- Introduction to Three Dimensional Geometry
• NCERT Exemplar Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry
• ### RD Sharma Solutions for Class 11 Maths Chapter 29 – Limits

The concept of limits and its derivations are explained in brief in this chapter. Students who find it difficult to grasp the topics can make use of the solutions designed by the highly knowledgeable experts at BYJU’S. The evaluation of algebraic limits, trigonometric limits, exponential and logarithmic limits are the other concepts which are covered in this chapter. Numerous solved examples are present before each exercise wise problems in order to help students understand the method of solving problems efficiently.

Limits: In Mathematics, a limit is defined as a value that a function approaches the output for the given input values.

There are 11 exercises in this chapter and the links of the solutions PDF designed by the experts at BYJU’S are as follows.

Exercise 29.1

Exercise 29.2

Exercise 29.3

Exercise 29.4

Exercise 29.5

Exercise 29.6

Exercise 29.7

Exercise 29.8

Exercise 29.9

Exercise 29.10

Exercise 29.11

Also access the following resources for Class 11 Chapter 29 Limits at BYJU’S:

• NCERT Solutions for Class 11 Maths Chapter 13- Limits and Derivatives
• NCERT Exemplar Solutions for Class 11 Maths Chapter 13 Limits and Derivatives
• ### RD Sharma Solutions for Class 11 Maths Chapter 30 – Derivatives

Chapter 30 of RD Sharma Class 11 Solutions provides students with the knowledge of derivatives at a point. The derivative at a point, derivative of a function, differentiation from first principles and fundamental rules for differentiation are the concepts of importance under this chapter. Students who aspire to score well in the annual exams are advised to solve the chapter wise problems using the solutions PDF devised by the experts at BYJU’S. It mainly helps students to understand the type of questions that would appear in the exams and work on them for better results.

Derivatives: Derivatives are defined as the varying rate of change of a function with respect to an independent variable.
Derivatives Formulas:
The formulas of derivatives for some of the functions such as linear, exponential and logarithmic are listed below:

• d/dx (k) = 0, where k is any constant
• d/dx(x) = 1
• d/dx(xn) = nxn-1
• d/dx (kx) = k, where k is any constant
• d/dx (√x) = 1/2√x
• d/dx (1/x) = -1/x2
• d/dx (log x) = 1/x, x > 0
• d/dx (ex) = ex
• d/dx (ax) = ax log a
• The PDF links of exercise wise solutions are provided below for further reference.

Exercise 30.1

Exercise 30.2

Exercise 30.3

Exercise 30.4

Exercise 30.5

Also access the following resources for Class 11 Chapter 30 Derivatives at BYJU’S:

• NCERT Solutions for Class 11 Maths Chapter 13- Limits and Derivatives
• NCERT Exemplar Solutions for Class 11 Maths Chapter 13 Limits and Derivatives
• ### RD Sharma Solutions for Class 11 Maths Chapter 31 – Mathematical Reasoning

This chapter helps students to learn about the concepts like statements, negation of a statement, compound statements, basic connectives, quantifiers, implications and validity of statements. The solutions are designed to facilitate easy learning and help students to understand the concepts covered in this chapter. The PDF of solutions can be used as a reference tool to quickly review all the topics which are important from the exam point of view.

Negotiation of a Statement: In Mathematics, the negation of a statement is the opposite of the given mathematical statement. If “P” is a statement, then the negation of statement P is represented by ~P.
Compound Statement: The statements in reasoning can be compound i.e., they can be composed of two or more than two statements together.

To understand the various methods of solving problems, students can download the solutions PDF from the exercise wise links provided below.

Exercise 31.1

Exercise 31.2

Exercise 31.3

Exercise 31.4

Exercise 31.5

Exercise 31.6

Also access the following resources for Class 11 Chapter 31 Mathematical Reasoning at BYJU’S:

• NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning
• NCERT Exemplar Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning
• ### RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics

Students would have learnt about the methods of representing the given data graphically and in tabular form. In this chapter, students understand the methods of determining a representative value of the given data. Measures of dispersion, range, mean deviation, limitations of mean deviation, variance and standard deviation and analysis of frequency distribution are the other important topics discussed here. RD Sharma Solutions help students to understand the shortcut tips and tricks which can be used in solving complex problems without any difficulty.

Mean Deviation: The mean deviation is defined as a statistical measure that is used to calculate the average deviation from the mean value of the given data set.
The formula to calculate the mean deviation for the given data set is given below.
Mean Deviation = [Σ |X – µ|]/N
Here,
Σ represents the addition of values
X represents each value in the data set
µ represents the mean of the data set
N represents the number of data values
| | represents the absolute value, which ignores the “-” symbol
Variance: The variance is a measure of how far a set of data are dispersed out from their mean or average value.
Standard Deviation: The spread of statistical data is measured by the standard deviation.

This chapter contains 7 exercises and the PDF links of the solutions are as follows.

Exercise 32.1

Exercise 32.2

Exercise 32.3

Exercise 32.4

Exercise 32.5

Exercise 32.6

Exercise 32.7

Also access the following resources for Class 11 Chapter 32 Statistics at BYJU’S:

• NCERT Solutions for Class 11 Maths Chapter 15 Statistics
• NCERT Exemplar Solutions for Class 11 Maths Chapter 15 Statistics
• ### RD Sharma Solutions for Class 11 Maths Chapter 33 – Probability

In the previous classes, students would have learnt about the statistical and classical approach. But in this chapter, the serious deficiencies and limitations are discussed as per the CBSE syllabus. The experts at BYJU’S have curated the chapter wise solutions in an interactive manner to help students understand the concepts. The topics like sample spaces, random experiments, events, algebra and types of events are also covered according to the prescribed board.

Probability: It is a measure of the likelihood of an event to occur.
Formula for Probability: The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.
Probability of event to happen P(E) = Number of favourable outcomes/Total Number of outcomes

The exercise wise solution links of Chapter 33 are provided below.

Exercise 33.1

Exercise 33.2

Exercise 33.3

Exercise 33.4

Also access the following resources for Class 11 Chapter 33 Probability at BYJU’S:

• NCERT Solutions for Class 11 Maths Chapter 16 Probability
• NCERT Exemplar Solutions for Class 11 Maths Chapter 16 Probability
• ### Features of RD Sharma Solutions for Class 11 Maths

1. Detailed illustrations with more than solutions.

2. Brief summary consisting of formulae and concepts.

3. An algorithmic approach to solve each problem

4. Complete answers to all the questions in RD Sharma class 11 maths textbook.

RD Sharma Solutions for Class 11 for all maths chapter are given in the table mentioned below. There are numerous important topics in class 11, which are all discussed here in-depth. Students can get the complete solutions for all RD Sharma questions to prepare for their examination.

## Frequently Asked Questions on RD Sharma Solutions for Class 11 Maths

### How many Chapters are there in RD Sharma Maths Class 11 Maths?

There are 33 Chapters are there in Class 11 RD Sharma Textbook. Viz, Sets, Relations, Functions, Measurement Of Angles, Trigonometric Functions, Graphs Of Trigonometric Functions, Trigonometric Ratios Of Compound Angles, Transformations formulae, Trigonometric Ratios Of Multiple And Sub Multiple Angles, Sine And Cosine Formulae And Their Applications, Trigonometric Equation, Mathematical Induction, Complex Numbers, Quadratic Equations, Linear Inequations, Permutations, Combinations, Binomial Theorem, Arithmetic Progression, Geometric Progression, Some Special Series, Brief Review Of Cartesian System Of Rectangular Coordinates, The Straight Lines, The Circle, Parabola, Ellipse, Hyperbola, Introduction To 3D Coordinate Geometry, Limits, Derivatives, Mathematical Reasoning, Statistics and Probability.

### Why are BYJU’S RD Sharma Solutions favoured by Class 11 students?

The fidelity and correctness of the solutions provided by BYJU’S are the main reason to opt for it. The solutions provided on the website will be in a student-friendly language which makes learning easier.

### How to achieve good marks in Class 11 Mathematics?

By referring to BYJU’S RD Sharma Solutions for Class 11 Maths students can achieve high marks in finals. Practise is an essential task to learn and score well in Mathematics. Students are required to go through RD Sharma Solutions thoroughly before the final exams to score well and intensify their problem-solving abilities.

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