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CBSE Worksheets for Class 11 Maths

CBSE Worksheets for Class 11 Maths: One of the best teaching strategies employed in most classrooms today is Worksheets. CBSE Class 11 Maths Worksheet for students has been used by teachers & students to develop logical, lingual, analytical, and problem-solving capabilities. So in order to help you with that, we at WorksheetsBuddy have come up with Kendriya Vidyalaya Class 11 Maths Worksheets for the students of Class 11. All our CBSE NCERT Class 11 Maths practice worksheets are designed for helping students to understand various topics, practice skills and improve their subject knowledge which in turn helps students to improve their academic performance. These chapter wise test papers for Class 11 Maths will be useful to test your conceptual understanding.

Board: Central Board of Secondary Education(www.cbse.nic.in) Subject: Class 11 Maths Number of Worksheets: 42

CBSE Class 11 Maths Worksheets PDF

All the CBSE Worksheets for Class 11 Maths provided in this page are provided for free which can be downloaded by students, teachers as well as by parents. We have covered all the Class 11 Maths important questions and answers in the worksheets which are included in CBSE NCERT Syllabus. Just click on the following link and download the CBSE Class 11 Maths Worksheet. CBSE Worksheets for Class 11 Math can also use like assignments for Class 11 Maths students.

Binomial Theorem

• CBSE Worksheets for Class 11 Mathematics Binomial Theorem Assignment 1
• CBSE Worksheets for Class 11 Mathematics Binomial Theorem Assignment 2

Complex Numbers and Quadratic Equation

• CBSE Worksheets for Class 11 Mathematics Complex Numbers and Quadratic Equation Assignment 1
• CBSE Worksheets for Class 11 Mathematics Complex Numbers and Quadratic Equation Assignment 2
• CBSE Worksheets for Class 11 Mathematics Complex Numbers and Quadratic Equation Assignment 3
• CBSE Worksheets for Class 11 Mathematics Complex Numbers and Quadratic Equation Assignment 4
• CBSE Worksheets for Class 11 Mathematics Complex Numbers and Quadratic Equation Assignment 5
• CBSE Worksheets for Class 11 Mathematics Complex Numbers and Quadratic Equation Assignment 6

Conic Sections

• CBSE Worksheets for Class 11 Mathematics Conic Sections Assignment 1
• CBSE Worksheets for Class 11 Mathematics Conic Sections Assignment 2

Introduction To 3Dimensional Geometry

• CBSE Worksheets for Class 11 Mathematics Introduction To 3Dimensional Geometry Assignment 1
• CBSE Worksheets for Class 11 Mathematics Introduction To 3Dimensional Geometry Assignment 2

Linear Inequalities

• CBSE Worksheets for Class 11 Mathematics Linear Inequalities Assignment 1
• CBSE Worksheets for Class 11 Mathematics Linear Inequalities Assignment 2

Permutations and Combinations

• CBSE Worksheets for Class 11 Mathematics Permutations and Combinations Assignment 1
• CBSE Worksheets for Class 11 Mathematics Permutations and Combinations Assignment 2

Principle of Mathematical Induction

• CBSE Worksheets for Class 11 Mathematics Principle of Mathematical Induction Assignment 1
• CBSE Worksheets for Class 11 Mathematics Principle of Mathematical Induction Assignment 2

Probability

• CBSE Worksheets for Class 11 Mathematics Probability Assignment 1
• CBSE Worksheets for Class 11 Mathematics Probability Assignment 2

Relations and Functions

• CBSE Worksheets for Class 11 Mathematics Relations and Functions Assignment

Sequences and Series

• CBSE Worksheets for Class 11 Mathematics Sequences and Series Assignment 1
• CBSE Worksheets for Class 11 Mathematics Sequences and Series Assignment 2
• CBSE Worksheets for Class 11 Mathematics Sequences and Series Assignment 3
• CBSE Worksheets for Class 11 Mathematics Sequences and Series Assignment 4
• CBSE Worksheets for Class 11 Mathematics Sequences and Series Assignment 5
• CBSE Worksheets for Class 11 Mathematics Sequences and Series Assignment 6
• CBSE Worksheets for Class 11 Mathematics Sequences and Series Assignment 7
• CBSE Worksheets for Class 11 Mathematics Sequences and Series Assignment 8
• CBSE Worksheets for Class 11 Mathematics Set Theory Assignment 1
• CBSE Worksheets for Class 11 Mathematics Set Theory Assignment 2
• CBSE Worksheets for Class 11 Mathematics Statistics Assignment 1
• CBSE Worksheets for Class 11 Mathematics Statistics Assignment 2

Straight Lines

• CBSE Worksheets for Class 11 Mathematics Straight Lines Assignment 1
• CBSE Worksheets for Class 11 Mathematics Straight Lines Assignment 2

Trigonometric Ratios

• CBSE Worksheets for Class 11 Mathematics Trigonometric Ratios Assignment 1
• CBSE Worksheets for Class 11 Mathematics Trigonometric Ratios Assignment 2
• CBSE Worksheets for Class 11 Mathematics Trigonometric Ratios Assignment 3
• CBSE Worksheets for Class 11 Mathematics Sample Paper 2014 Assignment 1
• CBSE Worksheets for Class 11 Mathematics Sample Paper 2014 Assignment 2
• CBSE Worksheets for Class 11 Mathematics Sample Paper 2014 Assignment 3
• CBSE Worksheets for Class 11 Mathematics Mathematical Reasoning Assignment

Advantages of CBSE Class 11 Maths Worksheets

• By practising NCERT CBSE Class 11 Maths Worksheet , students can improve their problem solving skills.
• Helps to develop the subject knowledge in a simple, fun and interactive way.
• No need for tuition or attend extra classes if students practise on worksheets daily.
• Working on CBSE worksheets are time-saving.
• Helps students to promote hands-on learning.
• One of the helpful resources used in classroom revision.
• CBSE Class 11 Maths Workbook Helps to improve subject-knowledge.
• CBSE Class 11 Math Worksheets encourages classroom activities.

Worksheets of CBSE Class 11 Maths are devised by experts of WorksheetsBuddy experts who have great experience and expertise in teaching Maths. So practising these worksheets will promote students problem-solving skills and subject knowledge in an interactive method. Students can also download CBSE Class 11 Maths Chapter wise question bank pdf and access it anytime, anywhere for free. Browse further to download free CBSE Class 11 Maths Worksheets PDF .

Now that you are provided all the necessary information regarding CBSE Class 11 Maths Worksheet and we hope this detailed article is helpful. So Students who are preparing for the exams must need to have great solving skills. And in order to have these skills, one must practice enough of Class 11 Math revision worksheets . And more importantly, students should need to follow through the worksheets after completing their syllabus.  Working on CBSE Class 11 Maths Worksheets will be a great help to secure good marks in the examination. So start working on Class 11 Math Worksheets to secure good score.

CBSE Worksheets For Class 11

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Important Questions for CBSE Class 11 Maths (2024-25)

• Class 11 Important Question

CBSE Class 11 Maths Chapter wise Important Questions - Free PDF Download

One of the most important subjects that students attempt in the science stream is Mathematics. For those taking exams in their eleventh grade, having the important questions for class 11 maths beforehand can help prepare better. We at Vedantu provide you with the opportunity to gain a lot of advantages when it comes to studying before the exam:

We provide the important questions for class 11 maths laid out chapter wise with solutions. Students can accordingly set their schedule

All the questions for the examination are available in a free PDF download any time of the year so you can choose when you want to include it in your study schedule

The questions are all prepared in the latest CBSE syllabus of the latest year of the CBSE board and are updated every annual academic year

Solving these questions will give students the advantage of preparing themselves better before their class 11 examinations

The NCERT important questions for class 11 Maths with answers are present for all chapters from Sets to Probability

There are a whole host of other advantages that come when you want to download the class 11 Maths NCERT important questions available as a free PDF download

Download CBSE Class 11 Maths Important Questions 2024-25 PDF

Also, check CBSE Class 11 Maths Important Questions for All chapters:

Chapter wise Important Questions for CBSE Class 11 Maths

Important Questions for Class 11 Maths CBSE Board

There are a lot of advantages that come with our different services for education. We provide private tutors for students that will help them understand the important questions for class 11 Maths that come chapter wise with free PDF solutions better. Here, they can solve problems at their own pace and also clear their doubts with the tutor online.

Our tutors are experienced and will be able to streamline themselves to the student's schedule. Students can also decide a schedule that works for them so that they can study at a pace that suits them. Our teachers will be able to guide you better when it comes to how you must pace your answers, tackling the right questions and practising well so you can perform better.

CBSE Class 11 Maths Weightage 2024-25

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At Vedantu, you will be able to also tap into various that we can provide to you such as CBSE Important question papers , revision notes , CBSE syllabus and so much more.

Students must enjoy their studies, and we want to help them understand their lessons better instead of just studying by rote.

We provide students with personal teachers for subjects such as Maths, Physics, Chemistry and Biology . They know exactly what sort of questions appear in exams due to their vast years of experience and this can be very helpful for students who are slow learners.

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Other CBSE Class 11 Important Questions Links

Important Related Links for CBSE Class 11

FAQs on Important Questions for CBSE Class 11 Maths (2024-25)

Q 1: How Vedantu’s answers assist us in scoring good marks in the exam?

All our solutions are framed with the aid of our subject matter specialists of maths and that they have given every part of the bankruptcy at the side of the exercise questions which might be referred to at the give up of the bankruptcy. Our Solutions for Class 11 Maths will provide an explanation for to you all the essential major points of the chapter. This will basically help you in building up a sturdy basis of the challenge and eventually you can also enhance your grades with it. The answers supplied through us will help in revision before the examination. These paintings as last-minute have a look at notes for your exam which could be visible just earlier than the examination to memorise all the key factors.

Q 2: What are the important concepts in chapter 11 conic sections?

The chapter covers topics like parts of a cone which teaches about circle, ellipse, parabola, hyperbola and degenerate conic sections . You will also learn about Relationship between the semi-major axis, semi-minor axis and the distance of the focus from the centre of the ellipse, Standard equations of an ellipse, Standard equations of a parabola and Special cases of an ellipse .

The chapter also discusses on Eccentricity, Latus rectum, Standard equation of Hyperbola and minor axis of the distance . This is one of the important chapters of class 11 maths which also explains about the equilateral triangle and equations of the hyperbola.

Q 3: What are the principal topics in relation and functions?

The chapter introduces you to the relations and features. You will particularly encounter the Cartesian Product of Sets and Relations . Here, you may, in particular, realize approximately the real-life model of cartesian products which are explained with the assist of Lindt chocolates are available in 5 shapes, three flavours and in six colours.

You can also be taught of functions which encompass Some capabilities and their graphs along with Identity feature, Constant function, Polynomial feature, Rational features, The Modulus function, Signum feature and Greatest integer characteristic. You will also study about the Algebra of real functions which incorporates Addition of two actual features, Subtraction of a real character from another, Multiplication with the aid of a scalar, Multiplication of two real capabilities and The quotient of two real capabilities.

Q 4: What will I learn in chapter four Mathematical Induction?

Mathematical Induction is a specific technique that is primarily used to prove a given statement or a theorem. The important section of this method is that the theorem should always be true for every natural number given. It is one of the useful methods as you do not have to solve an equation or a statement for every possible value it could take.

You will get familiar with the following pointers while studying Mathematical Induction:

Proving the given illustration or statement is the primary motive.

The proof should always stand true for all the values consisting of natural numbers.

The statement should be true for the initial value as well.

The statement should be true for all other values till nth iteration.

Every step involved in the proof must be justified and has to be true.

Q5. Name the important concepts discussed in Chapter 11 Conic Sections of NCERT Solutions for Class 11 Maths.

Chapter 11 "Conic Sections" deals with various conic sections like circles, ellipses, parabolas, and hyperbolas. The important topics of this chapter are as follows:

a) Sections of a cone

b) Circles

c) Ellipses

d) Special cases of an ellipse

e) Parabolas

f) Standard equations of a parabola

g) Hyperbolas

h) Degenerated Conic Sections

i) Latus rectum

j) Eccentricity 

k) Standard equation of an ellipse

Important questions on all these topics are available here.

These solutions are available on Vedantu's official website( vedantu.com ) and mobile app free of cost.

Q6. What are the main topics covered in important questions of “Relations and Functions” for Class 11 Maths?

"Relations and Functions" is Chapter 2 in the NCERT textbook for Class 11. This is an important chapter and includes the following topics:

Cartesian products of sets

Functions and their graphs

   i) Identity function

            ii) Polynomial function

            iii) Constant function

            iv) Rational function

            v) Modulus function

            vi) Signum function

            vii) Greatest integer function

Algebra of real functions

Students can find all the important questions from this chapter here.

Q7. Which book is best for Class 11 Maths?

Ans: The most important book for Class 11 Mathematics is the main NCERT textbook. Almost all questions in the CBSE exam are directly asked from the NCERT textbook. For this very reason, it is extremely important to understand the concepts mentioned in the NCERT textbook well and practice the NCERT example, and practice questions thoroughly.

Apart from this, students can also practice Vedantu's important questions, NCERT Solutions, sample papers, and previous years' question papers for thorough practice.

Q8 . How many chapters are there in Class 11 Maths?

Ans:   Class 11 CBSE Maths curriculum consists of a total of 16 chapters.  Vedantu has carefully enlisted the important questions from all these chapters. Vedantu's Important Questions for Class 11 Mathematics will allow the students to practice questions that are significant from the exam point of view. It will also teach them the right way to compose step-by-step answers to these questions so they do not lose important marks in exams due to incomplete answers.

Q9. Is Class 11 Maths tough?

Ans: Class 11 Mathematics is definitely not a child's play. It has some complicated topics as well. However, this does not mean that with proper guidance, planning, and effort you cannot ace this subject. You can even top in Class 11 Maths if you follow the following steps:

a) Understand and practice NCERT Solutions and example questions thoroughly.

b) Refer to Vedantu's Important questions and practice

c) Solve loads of sample papers and previous years' question papers.

d) Practice over and over again.

Important Questions for CBSE Class 11

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Chapter 3 Class 11 Trigonometric Functions

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NCERT Solutions of Chapter 3 Class 11 Trigonometry is available free at teachoo. You can check the detailed explanation of all questions of exercises, examples and miscellaneous by clicking on the Exercise link below.

We had learned Basics of Trigonometry in Class 10. In this chapter, we will learn

• What is a positive or a negative angle
• Measuring angles in Degree , Minutes and Seconds
• Radian measure of an angle
• Converting Degree to Radians , and vice-versa
• Sign of sin, cos, tan in all 4 quadrants
• Finding values of trigonometric functions when one value is given (Example: Finding value of sin, cot, cosec, tan, sec, when cos x = -3/5 is given)
• Finding Value of trigonometric functions, given angle
• Solving questions by formula like  (x + y) formula, 2x 3x formula, Cos x + cos y formula , 2 sin x sin y formula 
• Finding principal and general solutions of a trigonometric equation
• Sin and Cosine Formula with supplementary Questions

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NCERT Solutions for Class 11 Maths Chapter 1 – Sets

Ncert solutions for class 11 maths chapter 1 – sets pdf.

Free PDF of NCERT Solutions for Class 11 Maths Chapter 1 – Sets includes all the questions provided in NCERT Books prepared by Mathematics expert teachers as per CBSE NCERT guidelines from Mathongo.com. To download our free pdf of Chapter 1 – Sets Maths NCERT Solutions for Class 11 to help you to score more marks in your board exams and as well as competitive exams.

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Class 11 Mathematics Trigonometry Assignments

We have provided below free printable Class 11 Mathematics Trigonometry Assignments for Download in PDF. The Assignments have been designed based on the latest NCERT Book for Class 11 Mathematics Trigonometry . These Assignments for Grade 11 Mathematics Trigonometry cover all important topics which can come in your standard 11 tests and examinations. Free printable Assignments for CBSE Class 11 Mathematics Trigonometry , school and class assignments, and practice test papers have been designed by our highly experienced class 11 faculty. You can free download CBSE NCERT printable Assignments for Mathematics Trigonometry Class 11 with solutions and answers. All Assignments and test sheets have been prepared by expert teachers as per the latest Syllabus in Mathematics Trigonometry Class 11. Students can click on the links below and download all Pdf Assignments for Mathematics Trigonometry class 11 for free. All latest Kendriya Vidyalaya Class 11 Mathematics Trigonometry Assignments with Answers and test papers are given below.

Mathematics Trigonometry Class 11 Assignments Pdf Download

We have provided below the biggest collection of free CBSE NCERT KVS Assignments for Class 11 Mathematics Trigonometry . Students and teachers can download and save all free Mathematics Trigonometry assignments in Pdf for grade 11th. Our expert faculty have covered Class 11 important questions and answers for Mathematics Trigonometry as per the latest syllabus for the current academic year. All test papers and question banks for Class 11 Mathematics Trigonometry and CBSE Assignments for Mathematics Trigonometry Class 11 will be really helpful for standard 11th students to prepare for the class tests and school examinations. Class 11th students can easily free download in Pdf all printable practice worksheets given below.

Topicwise Assignments for Class 11 Mathematics Trigonometry Download in Pdf

Advantages of Class 11 Mathematics Trigonometry Assignments

• As we have the best and largest collection of Mathematics Trigonometry assignments for Grade 11, you will be able to easily get full list of solved important questions which can come in your examinations.
• Students will be able to go through all important and critical topics given in your CBSE Mathematics Trigonometry textbooks for Class 11 .
• All Mathematics Trigonometry assignments for Class 11 have been designed with answers. Students should solve them yourself and then compare with the solutions provided by us.
• Class 11 Students studying in per CBSE, NCERT and KVS schools will be able to free download all Mathematics Trigonometry chapter wise worksheets and assignments for free in Pdf
• Class 11 Mathematics Trigonometry question bank will help to improve subject understanding which will help to get better rank in exams

Frequently Asked Questions by Class 11 Mathematics Trigonometry students

At https://www.cbsencertsolutions.com, we have provided the biggest database of free assignments for Mathematics Trigonometry Class 11 which you can download in Pdf

We provide here Standard 11 Mathematics Trigonometry chapter-wise assignments which can be easily downloaded in Pdf format for free.

You can click on the links above and get assignments for Mathematics Trigonometry in Grade 11, all topic-wise question banks with solutions have been provided here. You can click on the links to download in Pdf.

We have provided here topic-wise Mathematics Trigonometry Grade 11 question banks, revision notes and questions for all difficult topics, and other study material.

We have provided the best collection of question bank and practice tests for Class 11 for all subjects. You can download them all and use them offline without the internet.

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NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

September 22, 2019 by phani

Get Free NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1, Ex 3.2, Ex 3.3, Ex 3.4, and Miscellaneous Exercise PDF in Hindi and English Medium. Trigonometric Functions Class 11 Maths NCERT Solutions are extremely helpful while doing your homework. Trigonometric Functions All Exercises Class 11 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers.

Class 11 Maths Trigonometric Functions NCERT Solutions in English Medium and Hindi Medium

• Trigonometric Functions Class 11 Ex 3.1
• Trigonometric Functions Class 11 Ex 3.2
• Trigonometric Functions Class 11 Ex 3.3
• Trigonometric Functions Class 11 Ex 3.4
• Trigonometric Functions Class 11 Miscellaneous Exercise

त्रिकोणमितीय फलन प्रश्नावली 3.1 का हल हिंदी में

• त्रिकोणमितीय फलन प्रश्नावली 3.2 का हल हिंदी में
• त्रिकोणमितीय फलन प्रश्नावली 3.3 का हल हिंदी में
• त्रिकोणमितीय फलन प्रश्नावली 3.4 का हल हिंदी में
• त्रिकोणमितीय फलन विविध प्रश्नावली का हल हिंदी में
• Trigonometry Formulas
• Trigonometry Functions Class 11 Notes
• NCERT Exemplar Class 11 Maths Chapter 3 Trigonometric Functions
• Trig Cheat Sheet
• JEE Main Trigonometry Previous Year Questions

Free download NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1, Ex 3.2, Ex 3.3, Ex 3.4, and Miscellaneous Exercise PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.1

Ex 3.1 Class 11 Maths Question 1: Find the radian measures corresponding to the following degree measures: (i) 25° (ii) – 47° 30′ (iii) 240° (iv) 520° Ans:

(i) \(\frac{11}{16}\) We know that: π radian = 180° ∴ \(\frac{11}{16}\) radain = \(\frac{180}{\pi} \times \frac{11}{16}\) × degree

= \(\frac{45 \times 11}{\pi \times 4}\) degree

= \(\frac{45 \times 11 \times 7}{22 \times 4}\) degree

= \(\frac{315}{8}\) degree

= 39 \(\frac{3}{8}\) degree

= 39° + \(\frac{3 \times 60}{8}\) minutes [1° = 60′]

= 39° + 22′ + \(\frac{1}{2}\) minutes

= 39°22’30” [1′ = 60°].

More Resources for CBSE Class 11

NCERT Solutions

• NCERT Solutions Class 11 Maths
• NCERT Solutions Class 11 Physics
• NCERT Solutions Class 11 Chemistry
• NCERT Solutions Class 11 Biology
• NCERT Solutions Class 11 Hindi
• NCERT Solutions Class 11 English
• NCERT Solutions Class 11 Business Studies
• NCERT Solutions Class 11 Accountancy
• NCERT Solutions Class 11 Psychology
• NCERT Solutions Class 11 Entrepreneurship
• NCERT Solutions Class 11 Indian Economic Development
• NCERT Solutions Class 11 Computer Science

Ex 3.1 Class 11 Maths Question 3: A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second? Ans: Number of revolutions made by the wheel in 1 minute = 360 ∴ Number of revolutions made by the wheel in 1 second = \(\frac{360}{6}\) = 6 In one complete revolution, the wheel turns an angle of 2π radian. Hence, in 6 complete revolutions, it will turn an angle of 6 × 2π radian, i.e., 12π radian Thus, in one second, the wheel turns an angle of 12π radian. Ex 3.1 Class 11 Maths Question 4: Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm y an arc of length 22 cm (Use π = \(\frac{22}{7}\)). Ans:

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = \(\frac{l}{r}\) Therefore, for r = 100 cm, l = 22 cm, we have

Thus, the required angle is 12°36′.

Ex 3.1 Class 11 Maths Question 5: In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord. Ans:

Given, diameter = 40 cm ∴ radius (r) = \(\frac{40}{2}\) = 20 cm and length of chord, AB = 20 cm Thus, ∆OAB is an equilateral triangle. We know that, θ = \(\frac{\text { Arc } A B}{\text { radius }}\) ⇒ Arc AB = θ × r = \(\frac{\pi}{3}\) × 20 . = \(\frac{20}{3}\) π cm.

Ex 3.1 Class 11 Maths Question 6: If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii. Ans:

Let the radii of the two circles be r 1  and r 2 . Let an arc of length l subtend an angle of 60° at the centre of the circle of radius r 1 , while let an arc of length l subtend an angle of 75° at the centre of the circle of radius r 2 . Now, 6o° = \(\frac{\pi}{3}\) radian and 75° = \(\frac{5 \pi}{12}\) radian We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = \(\frac{l}{r}\) or l = rθ ∴ l = \(\frac{r_{1} \pi}{3}\) and

l = \(\frac{r_{2} 5 \pi}{12}\)

⇒ \(\frac{r_{1} \pi}{3}=\frac{r_{2} 5 \pi}{12}\)

⇒ r = \(\frac{r_{2} 5}{4}\)

\(\frac{r_{1}}{r_{2}}=\frac{5}{4}\) Thus, the ratio of the radii is 5 : 4.

Ex 3.1 Class 11 Maths Question 7: Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length (i) 10 cm (ii) 15 cm (iii) 21 cm. Ans:

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = \(\frac{l}{r}\). It is given that r = 75 cm

(i) Here, l = 10 cm θ = \(\frac{10}{75}\) radian = \(\frac{2}{15}\) radian

(ii) Here, l = 15 cm θ = \(\frac{15}{75}\) radian θ = \(\frac{1}{5}\) radian

(iii) Here, l = 21 cm θ = \(\frac{21}{75}\) radian = \(\frac{7}{75}\) radian.

NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.2

sin x = \(\frac{3}{5}\)

cosec x = \(\frac{1}{\sin x}=\frac{1}{\left(\frac{3}{5}\right)}=\frac{5}{3}\) sin 2  x + cos 2  x = 1 ⇒ cos 2  x = 1 – sin 2  x ⇒ cos 2  x = 1 – (\(\frac{3}{5}\)) 2

⇒ cos 2  x = 1 – \(\frac{9}{25}\)

⇒ cos 2  x = \(\frac{16}{25}\)

⇒ cos x = ± \(\frac{4}{5}\) Since x lies in the 2nd quadrant, the value of cos x will be negative

⇒ \(\frac{4}{3}=\frac{\sin x}{\frac{-3}{5}}\) ⇒ sin x = \(\left(\frac{4}{3}\right) \times\left(\frac{-3}{5}\right)=-\frac{4}{5}\) ⇒ cosec x = \(\frac{1}{\sin x}=-\frac{5}{4}\).

tan x = – \(\frac{5}{12}\)

cot x = \(\frac{1}{\tan x}=\frac{1}{\left(-\frac{5}{12}\right)}=-\frac{12}{5}\)

Ex 3.2 Class 11 Maths Question 6: Find the value of the trigonometric function sin 765°. Ans: It is known that the values of sin x repeat after an interval of 2π or 360°. ∴ sin 765° = sin (2 × 360° + 45°) = sin 45° = 1 Ex 3.2 Class 11 Maths Question 7: Find the value of the trigonometric function cosec (- 1410°) Ans: It is known that the values of cosec x repeat after an interval of 2π or 360°. ∴ cosec (- 1410°) = cosec (- 1410° + 4 x 360°) = cosec (- 1410° + 1440°) = cosec 30° = 2. Ex 3.2 Class 11 Maths Question 8: Find the value of the trigonometric function tan \(\frac{19 \pi}{3}\). Ans:

It is known that the values of tan x repeat after an interval of π or 180°. ∴ \(\tan \frac{19 \pi}{3}=\tan 6 \frac{1}{3} \pi\)

= \(\tan \left(6 \pi+\frac{\pi}{3}\right)=\tan \frac{\pi}{3}\)

= tan 60° = √3.

Ex 3.2 Class 11 Maths Question 9: Find the value of the trigonometric function sin \(\left(-\frac{11 \pi}{3}\right)\). Ans:

It is known that the values of cot x repeat after an interval of π or 180°.

∴ \(\sin \left(\frac{11 \pi}{3}\right)=\sin \left(-\frac{11 \pi}{3}+2 \times 2 \pi\right)\)

= \(\sin \left(\frac{\pi}{3}\right)=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\)

Ex 3.2 Class 11 Maths Question 10: Find the value of the trigonometric function cot \(\left(-\frac{15 \pi}{4}\right)\). Ans:

It is known that the values of cot x repeat after an interval of ir or 1800. ∴ \(\cot \left(-\frac{15 \pi}{4}\right)=\cot \left(-\frac{15 \pi}{4}+4 \pi\right)=\cot \frac{\pi}{4}\) = 1.

NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.3

Ex 3.3 Class 11 Maths Question 1:

Prove that: sin 2  \(\frac{\pi}{6}\) + cos 2  \(\frac{\pi}{3}\) – tan 2  \(\frac{\pi}{4}\) = – \(\frac{1}{2}\) Ans:

L.H.S.= sin 2  \(\frac{\pi}{6}\) + cos 2  \(\frac{\pi}{3}\) – tan 2  \(\frac{\pi}{4}\)

= \(\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}\) – (1) 2

= \(\frac{1}{4}+\frac{1}{4}-1=-\frac{1}{2}\)

= R.H.S. Hence proved.

L.H.S = \(2 \sin ^{2} \frac{3 \pi}{4}+2 \cos ^{2} \frac{\pi}{4}+2 \sec ^{2} \frac{\pi}{3}\)

= \(2\left\{\sin \left(\pi-\frac{\pi}{4}\right)\right\}^{2}+2\left(\frac{1}{\sqrt{2}}\right)^{2}+2(2)^{2}\)

= \(2\left\{\sin \frac{\pi}{4}\right\}^{2}+2 \times \frac{1}{2}+8\)

= 2 \(\left(\frac{1}{\sqrt{2}}\right)^{2}\) + 1 + 8

= 1 + 1 + 8 = 10 = R.H.S. Hence proved.

Ex 3.3 Class 11 Maths Question 5: Find the value of: (i) sin 75°, (ii) tan 15° Ans:

(i) sin 75° sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30° [∵ sin (x + y) = sin x cos y + cos x sin y] = \(\left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)\)

= \(\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}=\frac{\sqrt{3}+1}{2 \sqrt{2}}\)

(ii) tan 15° = tan (45° – 30°)

L.H.S = \(\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}\)

= \(\frac{[-\cos x][\cos x]}{(\sin x)(-\sin x)}=\frac{-\cos ^{2} x}{-\sin ^{2} x}\)

= R.H.S Hence proved.

It is known that cos A – cos B = \(-2 \sin \left(\frac{A+B}{2}\right) \cdot \sin \left(\frac{A-B}{2}\right)\)

∴ L.H.S.= \(=\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)\)

= \(– 2 \sin \left\{\frac{\left(\frac{3 \pi}{4}+x\right)+\left(\frac{3 \pi}{4}-x\right)}{2}\right\} \cdot \sin \left\{\frac{\left(\frac{3 \pi}{4}+x\right)-\left(\frac{3 \pi}{4}-x\right)}{2}\right\}\)

= – 2 sin (\(\frac{3 \pi}{4}\)) sin x

= – 2 sin (- \(\frac{\pi}{4}\)) sin x

= – √2 sin x = R.H.S. Hence proved.

Ex 3.3 Class 11 Maths Question 12: Prove that: sin 2  6x – sin 2  4x = sin 2x sin 10 x Ans:

It is known that sin A + sin B = 2 \(\sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)

sin A – sin B = 2 \(\cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\) L.H.S.= sin 2  6x – sin 2  4x = (sin 6x + sin 4x) (sin 6x – sin 4x) = (2 sin 5x cos x) (2 cos 5x sin x) = (2 sin 5x cos 5x) (2 sin x cos x) = sin 10x sin 2x = R.H.S. Hence proved.

Ex 3.3 Class 11 Maths Question 13: Prove that: cos 2  2x cos 2  6x = sin 4x sin 8x Ans:

It is known that cos A + cos B = 2 \(\cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)

cos A – cos = 2 \(\sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\)

∴ L.H.S = cos 2  2x – cos 2  6x = (cos 2x + cos 6x) (cos 2x – 6x) = \(\left[2 \cos \left(\frac{2 x+6 x}{2}\right) \cos \left(\frac{2 x-6 x}{2}\right)\right]\left[-2 \sin \left(\frac{2 x+6 x}{2}\right) \sin \frac{(2 x-6 x)}{2}\right]\) ∴ L.H.S.= cos 2  2x – cos 2  6x = (cos 2x + cos 6x) (cos 2x – 6x) = [2 cos 4x cos (-2x)] [- 2 sin 4x sin (- 2x)] = [2 cos 4x cos 2x] [- 2 sin 4x (- sin 2x)] = (2 sin 4x cos 4x) (2 sin 2x cos 2x) = sin 8x sin 4x = R.H.S. Hence proved.

It is known that sin A + sin = 2 \(\sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)

cos A + cos = 2 \(\cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)

∴ L.H.S = \(\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}\)

= \(\frac{2 \sin \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)}{2 \cos \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)}\)

= \(\frac{2 \sin 4 x \cos x}{2 \cos 4 x \cos x}=\frac{\sin 4 x}{\cos 4 x}\)

= tan 4x = R.H.S. Hence proved.

It is known that sin A – sin B = 2 \(\cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\)

cos 2  A – sin 2  A = cos 2A

∴ L.H.S. = \(=\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}\)

= \(\frac{2 \cos \left(\frac{x+3 x}{2}\right) \sin \left(\frac{x-3 x}{2}\right)}{-\cos 2 x}\)

= \(\frac{2 \cos 2 x \sin (-x)}{-\cos 2 x}\)

= – 2 × (- sin x) = 2 sin x

Hence proved.

Class 11 Maths NCERT Solutions Chapter 3 Exercise 3.4

cos 4x = cos 2x cos 4x – cos 2x = 0 – 2 sin \(\left(\frac{4 x+2 x}{2}\right)\) sin \(\left(\frac{4 x-2 x}{2}\right)\) = 0

[∵ cos A – cos B = 2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\(\)]

sin 3x sin x = 0 sin 3x = 0or sin x = 0 3x = nπ or x = nπ, where n ∈ Z x = \(\frac{n \pi}{3}\) or x = nπ, where n ∈ Z.

Ex 3.4 Class 11 Maths Question 6: Find the general solution of the equation cos 3x + cosx – cos 2x = 0 Ans:

cos 3x + cos x – cos 2x = 0 2 cos \(\left(\frac{3 x+x}{2}\right)\) cos \(\left(\frac{3 x-x}{2}\right)\) – cos 2x = 0

[∵ cos A + cos B = 2 \(\cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)]

2 cos 2x cos x – cos 2x = 0 cos 2x (2 cos x – 1) = 0 cos 2x = 0 or 2 cos x – 1 = 0 cos 2x = 0 or cos x = \(\frac{1{2}\) ∴ 2x = (2n + 1) \(\frac{\pi}{2}\) or cos x = cos \(\frac{\pi}{3}\), where n ∈ Z x = (2n + 1) \(\frac{\pi}{4}\) or x = 2nπ ± \(\frac{\pi}{3}\) where n ∈ Z.

Ex 3.4 Class 11 Maths Question 7: Find the general solution of the equation sin 2x + cos x = 0 Ans:

sin 2x + cos x = 0 ⇒ 2sin x cos x + cos x = 0 ⇒ cos x (2 sin x + 1) = 0 ⇒ cos x = 0 or 2 sin x + 1 = 0 Now, cos x = 0 ⇒ x = (2n + 1) \(\frac{\pi}{2}\) , where n ∈ Z. or 2 sin x + 1 = 0 ⇒ sin x = – \(\frac{1}{2}\)

= – sin \(\frac{\pi}{6}\)

= sin (π + \(\frac{\pi}{6}\))

= sin \(\frac{7 \pi}{6}\)

x = nπ + (- 1) n  \(\frac{7 \pi}{6}\) where n ∈ Z Therefore, the general solution is (2n + 1) \(\frac{\pi}{2}\) or nπ + (- 1) n  \(\frac{7 \pi}{6}\) where n ∈ Z.

Ex 3.4 Class 11 Maths Question 8: Find the general solution of the equation sec 2  2x = 1 – tan 2x. Ans:

sec 2  2x = 1 – tan 2x 1 + tan 2  2x = 1 – tan 2x tan 2  x + tan 2x = 0 => tan 2x (tan 2x + 1) = 0 => tan 2x = 0 or tan 2x + 1 = 0 Now, tan 2x = 0 => tan 2x = tan 0 2x = nπ + 0, where n ∈ Z x = \(\frac{n \pi}{2}\), where n ∈ Z or tan 2x + 1 = 0 = tan 2x = – 1 = – tan \(\frac{\pi}{4}\)

= tan (π – \(\frac{\pi}{4}\))

= tan \(\frac{3 \pi}{4}\)

2x = nπ + \(\frac{3 \pi}{4}\) where n ∈ Z

x = \(\frac{n \pi}{2}+\frac{3 \pi}{8}\), where n ∈ Z

Therefore, the general solution is \(\frac{n \pi}{2}\) or \(\frac{n \pi}{2}+\frac{3 \pi}{8}\) where n ∈ Z.

Ex 3.4 Class 11 Maths Question 9: Find the general solution of the equation sin x + sin 3x + sin 5x = 0 Ans:

sin x + sin 3x + sin 5x = 0 ⇒ (sin x + sin 5x) + sin 3x = 0

\(\left[2 \sin \left(\frac{x+5 x}{2}\right) \cos \left(\frac{x-5 x}{2}\right)\right]\) + sin 3x = 0

[∵ sin A + sin B = 2 sin \(\sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)]

2 sin 3x cos (2x) + sin 3x = 0 2 sin 3x cos 2x + sin 3x = 0 sin 3x (2 cos 2x +1) = 0 sin 3x = 0 or 2 cos 2x + 1 = 0 Now sin 3x = 0 ⇒ 3x = nπ, where n ∈ Z i.e., x = \(\frac{n \pi}{3}\) where n ∈ Z or 2 cos 2x + 1 = 0 cos 2x = \(-\frac{1}{2}\)

= – cos \(\frac{\pi}{3}\)

= cos (π – \(\frac{\pi}{3}\))

cos 2x = cos \(\frac{2 \pi}{3}\)

⇒ 2x = 2nπ ± \(\frac{2\pi}{3}\), where n ∈ Z

⇒ x = nπ ± \(\frac{\pi}{3}\), where n ∈ Z

Therefore, the general solution is \(\frac{n \pi}{3}\) or nπ ± \(\frac{\pi}{3}\), where n ∈ Z.

NCERT Solutions for Class 11 Maths Chapter 3 Miscellaneous Exercise

L.H.S. = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = sin 3x sin x + sin 2  x + cos 3x cos x – cos 2  x = cos 3x cos x + sin 3x sin x – (cos 2  x – sin 2  x) = cos (3x – x) – cos 2x [∵ cos(A – B) = cos A cos B + sin A sin B] = cos 2x – cos 2x = 0 =R.H.S. Hence proved.

Miscellaneous Exercise Class 11 Maths Question 3:

Prove that: (cos x + cos y) 2  + (sin x – sin y) 2  = 4 cos 2  \(\left(\frac{x+y}{2}\right)\)

L.H.S.= (cos x + cos y) 2  + (sin x – sin y) 2 = cos 2  x + cos 2  y + 2 cos x cos y + sin 2  x + sin 2  y – 2 sin x sin y = (cos 2  x + sin 2  x) + (cos 2  y + sin 2  y) + 2 (cos x cos y – sin x sin y) = 1 + 1 + 2 cos (x + y) [∵ cos (A + B) = (cos A cos B – sin A sin B)] = 2 + 2 cos (x + y) = 2 [1 + cos (x + y)] = 2[1 + \(2 \cos ^{2}\left(\frac{x+y}{2}\right)\) – 1] [∵ cos 2A = 2 cos 2  A – 1] = 4 c0s 2  \(\left(\frac{x+y}{2}\right)\) = R.H.S. Hence proved.

Miscellaneous Exercise Class 11 Maths Question 4: Prove that: (cos x – cos y) 2  + (sin x – sin y) 2  = 4 sin 2  \(\frac{x-y}{2}\) Ans:

L.H.S.= (cos x – cos y) 2  + (sin x – sin y) 2 = cos 2  x + cos 2  y – 2 cos x cos y + sin 2  x + sin 2  y – 2 sin x sin y = (cos 2  x + sin 2  x) + (cos 2  y + sin 2  y) – 2 [cos x cos y + sin x sin y] = 1 + 1 – 2 [cos (x – y)] = 2 [1 – {1 – 2 sin 2  \(\left(\frac{x-y}{2}\right)\)}] [∵ cos 2A = 1 – 2 sin 2  A] = 4 sin 2  \(\left(\frac{x-y}{2}\right)\) = R.H.S. Hence proved.

Miscellaneous Exercise Class 11 Maths Question 5: Prove that: sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x Ans:

It is known that sin A + sin B = 2 \(\sin \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)\) ∴ L.H.S. = (sin x + sin 3x) + (sin 5x + sin 7x) = (sin x + sin 5x) + (sin 3x + sin 7x) = \(2 \sin \left(\frac{x+5 x}{2}\right)\) . \(\cos \left(\frac{x-5 x}{2}\right)+2 \sin \left(\frac{3 x+7 x}{2}\right) \cos \left(\frac{3 x-7 x}{2}\right)\) = 2 sin 3x cos (- 2x) + 2 sin 5x cos (- 2x) = 2 sin 3x cos 2x + 2 sin 5x cos 2x = 2 cos 2x [sin 3x + sin 5x] = 2 cos 2x [latex]2 \sin \left(\frac{3 x+5 x}{2}\right) \cdot \cos \left(\frac{3 x-5 x}{2}\right)[/latex] = 2 cos 2x [2 sin 4x . cos (- x)] = 4 cos 2x sin 4x cos x = R.H.S. Hence proved.

Miscellaneous Exercise Class 11 Maths Question 6: Prove that: \(\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}\) = tan 6x Ans: It is known that

प्रश्न 3. एक-पहिया एक मिनट में 360° परिक्रमण करता है तो एक सेकंड में कितने रेडियन माप का कोण बनाएगा? हल: परिक्रमण में पहिया द्वारा बना कोण = 27 रेडियन 360 परिक्रमण में पहिया द्वारा बना कोण = 360 x 2π रेडियन 1 मिनट अर्थात् 60 सेकण्ड में 360 x 2π रेडियन का कोण बनता है। 1 सेकण्ट में चहिया द्वारा बना कोण = \(\frac { 360\times 2\pi }{ 60 }\) = 12π रेडियन।

Exercise 3.1

Q.1: Calculate the radian measurement of the given degree measurement:

(ii). 240 ∘

(iii). − 47 ∘ 30 ‘

(iv). 520 ∘

Q.2: Calculate the degree measurement of the given degree measurement: [Use π = \(\\ \frac { 22 }{ 7 } \)]

(i) \(\\ \frac { 11 }{ 16 } \)

(iii) \(\frac { 5\pi }{ 3 } \)

(iv) \(\frac { 7\pi }{ 6 } \)

Q.3: In a minute, wheel makes 360 revolutions. Through how many radians does it turn in 1 second?

Q.4: Calculate the degree measurement of the angle subtended at the centre of a circle of radius 100 m by an arc of length 22 m.

Q.5: In a circle of diameter 40 m, the length of the chord 20 m. Find the length of minor arc of chord.

Q.6: In two circles, arcs which has same length subtended at an angle of 60 ∘ and 75 ∘ at the center. Calculate the ratio of their radii.

Q.7: Calculate the angle in radian through which a pendulum swings if the length is 75 cm and the tip describes an arc of length

(iii) 21 cm

Exercise 3.2

Q.1: Calculate the values of five trigonometric func. if cos y = \(– \frac { 1 }{ 2 } \) and y lies in 3 rd quadrant.

(iii) cosec y

Q.2: Calculate the other five trigonometric function if we are given the values for sin y = \(\\ \frac { 3 }{ 5 } \), where y lies in second quadrant.

Q.3: Find the values of other five trigonometric functions if c o t y =\(\\ \frac { 3 }{ 4 } \) , where y lies in the third quadrant.

Q.4: Find the values of other five trigonometric if s e c y =\(\\ \frac { 13 }{ 5 } \) , where y lies in the fourth quadrant.

Q.5: Find the values of other five trigonometric function if tan y = \(– \frac { 5 }{ 12 } \) and y lies in second quadrant.

Q.6: Calculate the value of trigonometric function sin 765°.

Q.7: Calculate the value of trigonometric function cosec [-1410°]

Q.8: Calculate the value of the trigonometric function tan \(\frac { 19\pi }{ 3 } \) .

Q.9: Calculate the value of the trigonometric function sin \(-\frac { 11\pi }{ 3 } \) .

Q.10: Calculate the value of the trigonometric function cot \(-\frac { 15\pi }{ 4 } \)

Exercise 3.3

Q.1: Prove:

sin²\(\frac { \pi }{ 6 } \) + cos² \(\frac { \pi }{ 3 } \) – tan² \(\frac { \pi }{ 4 } \) = \(– \frac { 1 }{ 2 } \)

Q.2: Prove:

2 sin² \(\frac { \pi }{ 6 } \) + c o s e c ² \(\frac { 7\pi }{ 6 } \) 6 cos ² \(\frac { \pi }{ 3 } \) =\(\\ \frac { 3 }{ 2 } \)

Q.3: Prove:

cot ² \(\frac { \pi }{ 6 } \) + c o s e c \(\frac { 5\pi }{ 6 } \) + 3 tan ² latex s=2]\frac { \pi }{ 6 } [/latex] = 6

Q.4: Prove:

2 sin ² \(\frac { 3\pi }{ 4 } \) + 2 cos ² \(\frac { \pi }{ 4 } \) + 2 sec ² \(\frac { \pi }{ 3 } \) = 10

Q.5: Calculate the value of:

(i). sin 75 ∘

(ii). tan 15 ∘

cos ( \(\frac { \pi }{ 4 } \) – x ) cos ( \(\frac { \pi }{ 4 } \) – y ) – sin ( \(\frac { \pi }{ 4 } \) – x ) sin ( \(\frac { \pi }{ 4 } \) – y ) = sin ( x + y )

Q.7: Prove:

\(\frac { tan(\frac { \pi }{ 4 } +x) }{ tan(\frac { \pi }{ 4 } -x) } ={ \left( \frac { 1+tanx }{ 1-tanx } \right) }^{ 2 }\)

Q.8: Prove:

\(\frac { cos(\pi +x)cos(-x) }{ sin(\pi -x)cos\left( \frac { \pi }{ 2 } +x \right) } ={ cot }^{ 2 }x\)

Q.9: Prove:

\(cos(\frac { 3\pi }{ 2 } +x)cos(2\pi +x)[cot(\frac { 3\pi }{ 2 } -x)+cot(2\pi +x)]=1\)

Q.10: Prove:

sin ( n + 1 ) x sin ( n + 2 ) x + cos ( n + 1 ) x cos ( n + 2 ) x = cos x

Q.11 Prove:

\(cos(\frac { 3\pi }{ 4 } +x)-cos(\frac { 3\pi }{ 4 } -x)\)= − √2 sin x

Q.12: Prove:

sin² 6 x – sin ² 4 x = sin2 x sin 10 x

Q.13: Prove:

cos ² 2 x – cos ² 6 x = sin 4 x sin 8 x

Q.14:Prove:

sin 2 x + 2 sin 4 x + sin 6 x = 4 cos ² x sin 4 x

Q.15: Prove:

cot 4 x ( sin 5 x + sin 3 x ) = cot x ( sin 5 x – sin 3 x )

Q.16: Prove:

\(\frac { cos9x-cos5x }{ sin17x-sin3x } =-\frac { sin2x }{ cos10x } \)

Q.17: Prove:

\(\frac { sin5x+sin3x }{ cos5x+cos3x } =tan4x\)

Q.18: Prove:

\(\frac { sinx-siny }{ cosx+cosy } =tan\frac { x-y }{ 2 } \)

Q.19: Prove:

\(\frac { sinx+sin3x }{ cosx+cos3x } =tan2x\)

Q.20: Prove:

\(\frac { sinx-sin3x }{ { sin }^{ 2 }x-{ cos }^{ 2 }x } =2sinx\)

Q.21: Prove:

\(\frac { cos4x+cos3x+cos2x }{ sin4x+sin3x+sin2x } =cot3x\)

Q.22: Prove:

cot x cot 2 x – cot 2 x cot 3 x – cot 3 x cot x = 1

Q.23: Prove:

\(tan4x=\frac { 4tanx(1-{ tan }^{ 2 }x) }{ 1-6{ tan }^{ 2 }x+{ tan }^{ 4 }x } \)

Q.24: Prove:

cos 4 x = 1 – 8 sin² x cos² x

Q.25: Prove:

cos 6 x = 32 cos 6 x – 48 cos 4 x + 18 cos 2 x − 1

Exercise 3.4

Q.1: Find general solutions and the principle solutions of the given equation: tan x = √3

Q.2: Find general solutions and the principle solutions of the given equation: sec x = 2

Q.3: Find general solutions and the principle solutions of the given equation: cot = − √3

Q.4: Find general solutions and the principle solutions of the given equation: cosec x = -2

Q.5: Find the general solution of the given equation: cos 4x = cos 2x

Q.6: Find the general solution of the given equation: cos 3x + cos x – cos 2x = 0

Q.7: Find the general solution of the given equation: sin 2x + cos x = 0

Q.8: Find the general solution of the given equation: sec² 2 x = 1 – tan 2 x

Q.9: Find the general solution of the given equation: sin x + sin 3x + sin 5x = 0

Miscellaneous Exercise

Q.1: Prove that:

\(2cos\frac { \pi }{ 13 } cos\frac { 9\pi }{ 13 } +cos\frac { 3\pi }{ 13 } +cos\frac { 5\pi }{ 13 } =0\)

Q.2: Prove that:

( sin 3 x + sin x ) sin x + ( cos 3 x – cos x ) cos x = 0

Q-3: Prove that:

( cos x + cos y )² + ( sin x – sin y ) ² = 4 cos ²\(\\ \frac { x+y }{ 2 } \)

Q-4: Prove that:

( cos x – cos y ) ² + ( sin x – sin y ) ² = 4 sin ² \(\\ \frac { x-y }{ 2 } \)

Q-5: Prove that:

sin x + sin 3 x + sin 5 x + sin 7 x = 4 cos x cos 2 x cos 4 x

Q-6: Prove that:

\(\frac { (sin7x+sin5x)+(sin9x+sin3x) }{ (cos7x+cos5x)+(cos9x+cos3x) } =tan6x\)

Q-7: Show that: sin 3 y + sin 2 y – sin y = 4 sin y cos\(\\ \frac { y }{ 2 } \) cos\(\\ \frac { 3y }{ 2 } \)

Q-8: The value of tan y =\(– \frac { 4 }{ 2 } \) where y in in 2 nd quadrant then find out the values of sin\(\\ \frac { y }{ 2 } \) , cos\(\\ \frac { y }{ 2 } \) a n d tan\(\\ \frac { y }{ 2 } \) .

Q-9: The value of cos y = \(– \frac { 1 }{ 3 } \) where y in in 3 rd quadrant then find out the values of sin \(\\ \frac { y }{ 2 } \) , cos \(\\ \frac { y }{ 2 } \) a n d tan \(\\ \frac { y }{ 2 } \) .

Q-10: The value of sin y = \(\\ \frac { 1 }{ 4 } \) where y in in 2 nd quadrant then find out the values of sin \(\\ \frac { y }{ 2 } \) , cos \(\\ \frac { y }{ 2 } \) a n d tan \(\\ \frac { y }{ 2 } \) .

NCERT Solutions for Class 11 Maths All Chapters

• Chapter 1 Sets
• Chapter 2 Relations and Functions
• Chapter 3 Trigonometric Functions
• Chapter 4 Principle of Mathematical Induction
• Chapter 5 Complex Numbers and Quadratic Equations
• Chapter 6 Linear Inequalities
• Chapter 7 Permutation and Combinations
• Chapter 8 Binomial Theorem
• Chapter 9 Sequences and Series
• 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

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• NCERT Solutions
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• Chapter 2: Relations And Functions

NCERT Solutions For Class 11 Maths Chapter 2 Relations and Functions

Ncert solutions class 11 maths relations and functions – free pdf download.

NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions are solved in detail in the PDF given below. All the solutions to the problems in the exercises are created in such a way that it enables the students to prepare for the exam and ace it. The NCERT Solutions are prepared by the most experienced teachers in the education space, making the explanation of each solution simple, understandable, and according to the latest CBSE Syllabus 2023-24. The solution helps Class 11 students to master the concept of Relations and Functions.

The solutions provide a good understanding of the fundamental concepts before they solve the equations. Through regular practice, students will know the difference between relations and functions, which are included under the syllabus, and become well-versed in its concepts. Numerous examples are present in the textbook before the exercise questions to help them understand the methodologies to be followed while solving the problems. Referring to the NCERT Class 11 Solutions PDF, students can get a glimpse of the important concepts before facing their final exams.

NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions

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Access answers of Maths NCERT Class 11 Chapter 2 – Relations and Functions

Exercise 2.1 Page No: 33

As the ordered pairs are equal, the corresponding elements should also be equal.

x/3 + 1 = 5/3 and y – 2/3 = 1/3

Solving, we get

x = 2 and 3y = 3

x = 2 and y = 1

2. If set A has 3 elements and set B = {3, 4, 5}, then find the number of elements in (A × B).

Given, set A has 3 elements, and the elements of set B are {3, 4, and 5}.

So, the number of elements in set B = 3

Then, the number of elements in (A × B) = (Number of elements in A) × (Number of elements in B)

= 3 × 3 = 9

Therefore, the number of elements in (A × B) will be 9.

3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.

Given, G = {7, 8} and H = {5, 4, 2}

We know that,

The Cartesian product of two non-empty sets P and Q is given as

P × Q = {( p ,  q ):  p ∈ P,  q  ∈ Q}

G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}

H × G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}

4. State whether each of the following statements is true or false. If the statement is false, rewrite the given statement correctly.

(i) If P = { m ,  n } and Q = { n ,  m }, then P × Q = {( m ,  n ), ( n ,  m )}

(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs ( x ,  y ) such that  x  ∈ A and  y  ∈ B.

(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ Φ) = Φ

(i) The statement is false. The correct statement is

If P = { m ,  n } and Q = { n ,  m }, then

P × Q = {( m ,  m ), ( m ,  n ), ( n,   m ), ( n ,  n )}

5. If A = {–1, 1}, find A × A × A.

The A × A × A for a non-empty set A is given by

A × A × A = {( a ,  b ,  c ):  a ,  b ,  c  ∈ A}

Here, it is given A = {–1, 1}

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)}

6. If A × B = {( a ,  x ), ( a ,  y ), ( b ,  x ), ( b ,  y )}. Find A and B.

A × B = {( a ,  x ), ( a,   y ), ( b ,  x ), ( b ,  y )}

We know that the Cartesian product of two non-empty sets, P and Q is given by:

P × Q = {( p ,  q ):  p  ∈ P,  q  ∈ Q}

Hence, A is the set of all first elements, and B is the set of all second elements.

Therefore, A = { a ,  b } and B = { x ,  y }

7. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that

(i) A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) A × C is a subset of B × D

A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)

Now, B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ

L.H.S. = A × (B ∩ C) = A × Φ = Φ

A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}

A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

R.H.S. = (A × B) ∩ (A × C) = Φ

Therefore, L.H.S. = R.H.S.

Hence verified

(ii) To verify: A × C is a subset of B × D

B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}

Now, it’s clearly seen that all the elements of set A × C are the elements of set B × D.

Thus, A × C is a subset of B × D.

8. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.

A = {1, 2} and B = {3, 4}

A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}

Number of elements in A × B is  n (A × B) = 4

If C is a set with  n (C) =  m , then  n [P(C)] = 2 m .

Thus, the set A × B has 2 4  = 16 subsets.

And these subsets are as given below:

Φ, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2, 4)}, {(1, 4), (2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 3), (2, 3), (2, 4)}, {(1, 4), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2, 4)}

9. Let A and B be two sets such that  n (A) = 3 and  n  (B) = 2. If ( x , 1), ( y , 2), ( z , 1) are in A × B, find A and B, where  x ,  y  and  z  are distinct elements.

n (A) = 3 and  n (B) = 2; and ( x , 1), ( y , 2), ( z , 1) are in A × B.

A = Set of first elements of the ordered pair elements of A × B

B = Set of second elements of the ordered pair elements of A × B

So, clearly, x ,  y , and  z  are the elements of A; and

1 and 2 are the elements of B.

As  n (A) = 3 and  n (B) = 2, it is clear that set A = { x ,  y ,  z } and set B = {1, 2}

10. The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0, 1). Find the set A and the remaining elements of A × A.

If  n (A) =  p  and  n (B) =  q,  then  n (A × B) =  pq .

Also,  n (A × A) =  n (A) ×  n (A)

n (A × A) = 9

So, n (A) ×  n (A) = 9

Thus, n (A) = 3

Also, given that the ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A × A.

And, we know in A × A = {( a, a ):  a ∈ A}

Thus, –1, 0, and 1 have to be the elements of A.

As  n (A) = 3, clearly A = {–1, 0, 1}

Hence, the remaining elements of set A × A are as follows:

(–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), and (1, 1)

Exercise 2.2 Page No: 35

1. Let A = {1, 2, 3, … , 14}. Define a relation R from A to A by R = {( x ,  y ): 3 x  –  y  = 0, where  x ,  y  ∈ A}. Write down its domain, codomain and range.

The relation R from A to A is given as:

R = {( x ,  y ): 3 x  –  y  = 0, where  x ,  y  ∈ A}

= {( x ,  y ): 3 x  =  y , where  x ,  y  ∈ A}

R = {(1, 3), (2, 6), (3, 9), (4, 12)}

The domain of R is the set of all first elements of the ordered pairs in the relation.

Hence, Domain of R = {1, 2, 3, 4}

The whole set A is the codomain of the relation R.

Hence, Codomain of R = A = {1, 2, 3, …, 14}

The range of R is the set of all second elements of the ordered pairs in the relation.

Hence, Range of R = {3, 6, 9, 12}

2. Define a relation R on the set N of natural numbers by R = {( x ,  y ):  y  =  x  + 5,  x  is a natural number less than 4;  x ,  y  ∈ N}. Depict this relationship using roster form. Write down the domain and the range.

The relation R is given by:

R = {( x ,  y ):  y  =  x  + 5,  x  is a natural number less than 4,  x ,  y  ∈  N }

The natural numbers less than 4 are 1, 2, and 3.

R = {(1, 6), (2, 7), (3, 8)}

Hence, Domain of R = {1, 2, 3}

Hence, Range of R = {6, 7, 8}

3. A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {( x ,  y ): the difference between  x  and  y  is odd;  x  ∈ A,  y  ∈ B}. Write R in roster form.

A = {1, 2, 3, 5} and B = {4, 6, 9}

The relation from A to B is given as

R = {( x ,  y ): the difference between  x  and  y  is odd;  x  ∈ A,  y  ∈ B}

R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}

4. The figure shows a relationship between the sets P and Q. Write this relation

(i) in set-builder form (ii) in roster form

What is its domain and range?

From the given figure, it’s seen that

P = {5, 6, 7}, Q = {3, 4, 5}

The relation between P and Q:

Set-builder form

(i) R = {( x, y ):  y = x  – 2;  x  ∈ P} or R = {( x, y ):  y = x  – 2 for  x  = 5, 6, 7}

Roster form

(ii) R = {(5, 3), (6, 4), (7, 5)}

Domain of R = {5, 6, 7}

Range of R = {3, 4, 5}

5. Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by

{( a ,  b ):  a ,  b  ∈ A,  b  is exactly divisible by  a }.

(i) Write R in roster form

(ii) Find the domain of R

(iii) Find the range of R

A = {1, 2, 3, 4, 6} and relation R = {( a ,  b ):  a ,  b  ∈ A,  b  is exactly divisible by  a }

(i) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}

(ii) Domain of R = {1, 2, 3, 4, 6}

(iii) Range of R = {1, 2, 3, 4, 6}

6. Determine the domain and range of the relation R defined by R = {( x ,  x  + 5):  x  ∈ {0, 1, 2, 3, 4, 5}}.

Relation R = {( x ,  x  + 5):  x  ∈ {0, 1, 2, 3, 4, 5}}

R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}

Domain of R = {0, 1, 2, 3, 4, 5} and,

Range of R = {5, 6, 7, 8, 9, 10}

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

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

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

R = {(2, 8), (3, 27), (5, 125), (7, 343)}

8. Let A = { x ,  y , z} and B = {1, 2}. Find the number of relations from A to B.

Given, A = { x ,  y , z} and B = {1, 2}

A × B = {( x , 1), ( x , 2), ( y , 1), ( y , 2), ( z , 1), ( z , 2)}

As  n (A × B) = 6, the number of subsets of A × B will be 2 6 .

Thus, the number of relations from A to B is 2 6 .

9. Let R be the relation on Z defined by R = {( a ,  b ):  a ,  b  ∈ Z,  a  –  b  is an integer}. Find the domain and range of R.

Relation R = {( a ,  b ):  a ,  b  ∈ Z,  a  –  b  is an integer}

We know that the difference between any two integers is always an integer.

Domain of R = Z and Range of R = Z

Exercise 2.3 Page No: 44

1. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.

(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

(iii) {(1, 3), (1, 5), (2, 5)}

As 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their unique images, this relation can be called a function.

Here, domain = {2, 5, 8, 11, 14, 17} and range = {1}

As 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation can be called a function.

Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}

It’s seen that the same first element, i.e., 1, corresponds to two different images, i.e., 3 and 5; this relation cannot be called a function.

2. Find the domain and range of the following real function:

(i)  f ( x ) = –| x | (ii) f(x) = √(9 – x 2 ) 

f ( x ) = –| x |,  x  ∈ R

As  f ( x ) is defined for  x  ∈ R, the domain of  f  is R.

It is also seen that the range of  f ( x ) = –| x | is all real numbers except positive real numbers.

Therefore, the range of  f  is given by (–∞, 0].

(ii) f(x) = √(9 – x 2 )

As √(9 – x 2 ) is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, for 9 – x 2 ≥ 0.

So, the domain of  f ( x ) is { x : –3 ≤  x  ≤ 3} or [–3, 3].

For any value of  x  in the range [–3, 3], the value of  f ( x ) will lie between 0 and 3.

Therefore, the range of  f ( x ) is { x : 0 ≤  x  ≤ 3} or [0, 3].

3. A function  f  is defined by  f ( x ) = 2 x  – 5. Write down the values of

(i)  f (0), (ii)  f (7), (iii)  f (–3)

Function, f ( x ) = 2 x – 5

(i)  f (0) = 2 × 0 – 5 = 0 – 5 = –5

(ii)  f (7) = 2 × 7 – 5 = 14 – 5 = 9

(iii)  f (–3) = 2 × (–3) – 5 = – 6 – 5 = –11

Find (i)  t  (0) (ii)  t  (28) (iii)  t  (–10) (iv) The value of C, when  t (C) = 212

5. Find the range of each of the following functions:

(i)  f ( x ) = 2 – 3 x ,  x  ∈ R,  x > 0

(ii)  f ( x ) =  x 2  + 2,  x is a real number

(iii)  f ( x ) =  x ,  x is a real number

f(x) = 2 – 3 x ,  x  ∈ R,  x > 0

2 – 3x < 2

Therefore, the value of 2 – 3x is less than 2.

Hence, Range = (–∞, 2)

(ii) Given,

f ( x ) =  x 2  + 2,  x is a real number

Therefore, the value of x 2 + 2 is always greater or equal to 2, for x is a real number.

Hence, Range = [2, ∞)

(iii) Given,

f ( x ) =  x, x  is a real number

Clearly, the range of  f  is the set of all real numbers.

Range of  f  = R

Miscellaneous Exercise Page No: 46

Show that  f  is a function and  g  is not a function.

The given relation  f  is defined as:

It is seen that for 0 ≤  x  < 3,

f ( x ) =  x 2 and for 3 <  x  ≤ 10,

f ( x ) = 3 x

Also, at  x  = 3

f ( x ) = 3 2  = 9 or  f ( x ) = 3 × 3 = 9

Hence, for 0 ≤  x  ≤ 10, the images of  f ( x ) are unique.

Therefore, the given relation is a function.

In the given relation,  g  is defined as

It is seen that, for  x  = 2

g ( x ) = 2 2  = 4 and  g ( x ) = 3 × 2 = 6

Thus, element 2 of the domain of the relation  g corresponds to two different images, i.e., 4 and 6.

Therefore, this relation is not a function.

2. If  f ( x ) =  x 2 , find

f ( x ) =  x 2

3. Find the domain of the function 

Given function,

It’s clearly seen that the function f  is defined for all real numbers except at  x  = 6 and  x = 2, as the denominator becomes zero otherwise.

Therefore, the domain of  f  is R – {2, 6}.

4. Find the domain and the range of the real function  f  defined by f (x) = √(x – 1).

Given real function,

f (x) = √(x – 1)

Clearly, √(x – 1) is defined for ( x – 1) ≥ 0

So, the function f (x) = √(x – 1) is defined for  x ≥ 1

Thus, the domain of  f  is the set of all real numbers greater than or equal to 1.

Domain of  f = [1, ∞)

As  x  ≥ 1 ⇒ ( x  – 1) ≥ 0 ⇒ √(x – 1) ≥ 0

Thus, the range of  f  is the set of all real numbers greater than or equal to 0.

Range of  f = [0, ∞)

5. Find the domain and the range of the real function  f  defined by  f  ( x ) = | x  – 1|.

Given a real function,

f  ( x ) = | x  – 1|

Clearly, the function | x  – 1| is defined for all real numbers.

Domain of  f  = R

Also, for  x  ∈ R, | x  – 1| assumes all real numbers.

Therefore, the range of  f  is the set of all non-negative real numbers.

Substituting values and determining the images, we have

The range of  f  is the set of all second elements. It can be observed that all these elements are greater than or equal to 0 but less than 1.

We know that, for x ∈ R,

x 2 + 1 ≥ x 2

1 ≥ x 2 / (x 2 + 1)

Therefore, the range of  f  = [0, 1)

7. Let  f ,  g : R → R be defined, respectively by  f ( x ) =  x  + 1,  g ( x ) = 2 x  – 3. Find  f  +  g ,  f  –  g  and f/g .

Given the functions f ,  g : R → R is defined as

f ( x ) =  x  + 1,  g ( x ) = 2 x  – 3

( f  +  g ) ( x ) =  f ( x ) +  g ( x ) = ( x  + 1) + (2 x  – 3) = 3 x  – 2

Thus, ( f + g ) ( x ) = 3 x  – 2

( f – g ) ( x ) =  f ( x ) –  g ( x ) = ( x  + 1) – (2 x  – 3) =  x  + 1 – 2 x  + 3 = –  x  + 4

Thus, ( f – g ) ( x ) = – x  + 4

f/g (x) = f (x) /g (x), g(x) ≠ 0, x ∈ R

f/g (x) = x  + 1/ 2 x  – 3, 2 x  – 3 ≠ 0

Thus, f/g (x) = x  + 1/ 2 x  – 3, x ≠ 3/2

8. Let  f  = {(1, 1), (2, 3), (0, –1), (–1, –3)} be a function from Z to Z defined by  f ( x ) =  ax  +  b , for some integers  a ,  b . Determine  a ,  b .

Given, f  = {(1, 1), (2, 3), (0, –1), (–1, –3)}

And the function defined as,  f ( x ) =  ax  +  b

For (1, 1) ∈  f

We have,  f (1) = 1

So, a  × 1 +  b  = 1

a  +  b  = 1 …. (i)

And for (0, –1) ∈  f

We have f (0) = –1

a  × 0 +  b  = –1

On substituting  b  = –1 in (i), we get

a  + (–1) = 1 ⇒  a  = 1 + 1 = 2.

Therefore, the values of  a  and  b  are 2 and –1, respectively.

9. Let R be a relation from N to N defined by R = {( a ,  b ):  a ,  b  ∈ N and  a  =  b 2 }. Are the following true?

(i) ( a ,  a ) ∈ R, for all  a  ∈ N (ii) ( a ,  b ) ∈ R, implies ( b ,  a ) ∈ R

(iii) ( a ,  b ) ∈ R, ( b ,  c ) ∈ R implies ( a ,  c ) ∈ R

Justify your answer in each case.

Given relation R = {( a ,  b ):  a ,  b  ∈ N and  a  =  b 2 }

(i) It can be seen that 2 ∈ N; however, 2 ≠ 2 2  = 4.

Thus, the statement “( a ,  a ) ∈ R, for all  a  ∈ N” is not true.

(ii) Its clearly seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 3 2 .

Now, 3 ≠ 9 2  = 81; therefore, (3, 9) ∉ N

Thus, the statement “( a ,  b ) ∈ R, implies ( b ,  a ) ∈ R” is not true.

(iii) It’s clearly seen that (16, 4) ∈ R, (4, 2) ∈ R because 16, 4, 2 ∈ N and 16 = 4 2  and 4 = 2 2 .

Now, 16 ≠ 2 2  = 4; therefore, (16, 2) ∉ N

Thus, the statement “( a ,  b ) ∈ R, ( b ,  c ) ∈ R implies ( a ,  c ) ∈ R” is not true.

10. Let A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and  f  = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}. Are the following true?

(i)  f  is a relation from A to B (ii)  f is a function from A to B

A = {1, 2, 3, 4} and B = {1, 5, 9, 11, 15, 16}

A × B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}

Also, given that,

f  = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

(i) A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B.

It’s clearly seen that  f  is a subset of A × B.

Therefore, f  is a relation from A to B.

(ii) As the same first element, i.e., 2 corresponds to two different images (9 and 11), relation f  is not a function.

11. Let  f  be the subset of Z × Z defined by  f  = {( ab ,  a  +  b ):  a ,  b  ∈ Z}. Is  f  a function from Z to Z: justify your answer.

Given relation, f  is defined as

f  = {( ab ,  a  +  b ):  a ,  b  ∈ Z}

We know that a relation  f  from a set A to a set B is said to be a function if every element of set A has unique images in set B.

As 2, 6, –2, –6 ∈ Z, (2 × 6, 2 + 6), (–2 × –6, –2 + (–6)) ∈  f

i.e., (12, 8), (12, –8) ∈  f

It’s clearly seen that the same first element, 12, corresponds to two different images (8 and –8).

Therefore, the relation  f  is not a function.

12. Let A = {9, 10, 11, 12, 13} and let  f : A → N be defined by  f ( n ) = the highest prime factor of  n . Find the range of  f .

A = {9, 10, 11, 12, 13}

Now, f : A →  N  is defined as

f ( n ) = The highest prime factor of  n

Prime factor of 9 = 3

Prime factors of 10 = 2, 5

Prime factor of 11 = 11

Prime factors of 12 = 2, 3

Prime factor of 13 = 13

Thus, it can be expressed as

f (9) = The highest prime factor of 9 = 3

f (10) = The highest prime factor of 10 = 5

f (11) = The highest prime factor of 11 = 11

f (12) = The highest prime factor of 12 = 3

f (13) = The highest prime factor of 13 = 13

The range of  f  is the set of all  f ( n ), where  n  ∈ A.

Range of  f  = {3, 5, 11, 13}

NCERT Solutions for Class 11 Maths Chapter 2 – Relations and Functions

The following ideas from Chapter 2 Relations and Functions for Class 11, are given elaborately.

2.1 Introduction

This section introduces the concepts covered in the chapter Relations and Functions.

The combination of the register number of the student and their corresponding height is a relationship, which can be written as a set of ordered-pair numbers. Ordered-pair numbers are expressed as (x, y). The set of all elements of x is called the domain of the relation, and the set of all elements of y is called the range of the relation.

2.2 Cartesian Product of Sets

This section defines the Cartesian product and ordered pairs by giving a real-life model, its representation, and some worked examples.

Lindt chocolates come in five shapes, three flavours and six colours.

C :={circle, triangle, rectangle, rhombus, square}

N :={orange, vanilla, peach}

S :={red, blue, pink, white, yellow, purple}

C:={circle, triangle, rectangle, rhombus, square}, N:={orange, vanilla, peach}, S:={red, blue, pink, white, yellow, purple}

be the five shapes, three flavours and six colours, respectively. Then the set of all chocolates to be manufactured in the triple Cartesian product C×N×S and consists of 5⋅3⋅6=90 elements. As a manager, to sell this set of chocolates would have to make room for 90 heaps.

2.3 Relations

This section explains the mapping of set A to set B with a few solved problems. Definitions of domain and codomain are also introduced.

The idea of mapping a particular phone number to the respective person to whom the number belongs. That’s a relation – from phone number to person.

2.4 Functions

This section covers functions, the visualisation of functions, and how a relation is said to be a function, with a few examples. Meaning of image and preimage.

The height of a person can be determined by the length of his femur bone. Hence, it is an example of a function.

2.4.1 Some functions and their graphs

This section talks about different types of functions and their graphical representations. Some of the types of functions are listed below.

• Identity function
• Constant function
• Polynomial function
• Rational functions
• The Modulus function
• Signum function
• Greatest integer function

2.4.2 Algebra of real functions

This section includes the algebraic operations on functions.

• Addition of two real functions
• Subtraction of a real function from another
• Multiplication by a scalar
• Multiplication of two real functions
• The quotient of two real functions

Exercise 2.1 Solutions 10 Questions

Exercise 2.2 Solutions 9 Questions

Exercise 2.3 Solutions 5 Questions

Miscellaneous Exercise on Chapter 2 Solutions 12 Questions

Key Features of NCERT Solutions for Class 11 Maths Chapter 2 – Relations and Functions

• The two elements grouped in a particular order are called an ordered pair.
• Cartesian product A × B of two sets A and B is given by A × B = {(a, b): a ∈ A, b ∈ B}
• Relation R from a set A to a set B is a subset of the cartesian product A × B obtained by explaining the relationship between the first element x and the second element y of the ordered pairs in A × B.
• The image of an element x under a relation R is given by y, where (x, y) ∈ R.
• The domain of R is the set of all first elements of the ordered pairs in a relation R.
• The range of a relation R is the set of all second elements of the ordered pairs in a relation R.
• Function A from a set A to a set B is a specific type of relation for which every element x of set A has one and only one image y in set B. We write f: A→B, where f(x) = y.
• The range of the function is the set of images.
• A real function has a set of real numbers or one of its subsets both as to its domain and as its range.

Frequently Asked Questions on NCERT Solutions for Class 11 Maths Chapter 2

How to find which relation is a function in chapter 2 of ncert solutions for class 11 maths, explain the basic steps for the cartesian product of sets in ncert solutions for class 11 maths chapter 2 relations and functions., what is the meaning of relations in chapter 2 of ncert solutions for class 11 maths, leave a comment cancel reply.

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