NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1

Mathematics is a very vast field but it also plays an important role in other fields such as Physics, Chemistry, Economics, Accounts, Professional Engineering courses, etc. Mathematics is helpful in improving analytical skills and boosting a student’s reasoning ability. It also fosters the daily development of skills like data analysis, evidence gathering, and pattern recognition. It offers individuals the chance to better grasp or interpret information.

Regular practice is needed in Mathematics. A thorough understanding of theories and the application of formulas help students in mastering Mathematical concepts. Mathematics at the Class 12 level includes many topics and it is helpful in building concepts that will enable students to pursue various courses such as Mechanical Engineering and so on.

Understanding the basic theories and practising questions help students in scoring good marks in Class 12 Mathematics. Focusing on units that carry more topic weightage can help students in understanding the exam pattern. Students are advised to go through the Class 12 Mathematics syllabus to get an idea about the concepts which are covered, along with the weightage. To enhance their preparation, students need to solve sample question papers and sample papers that are based on the Class 12 NCERT Mathematics syllabus. It gives them an idea about the questions that can appear in final exams.

There are many books available in the market that promise to increase your knowledge in Class 12 Mathematics. It is advisable to stick to the NCERT textbook since it has an extensive range of exercise questions that will be sufficient for your board preparations.

The topic of Class 12 Chapter 5 is Continuity and Differentiability. It is one of the very important chapters in Class 12 from the exam point of view. There are various concepts such as Continuity, Algebra of Continuous Functions, Differentiability, etc that are explained in an elaborative manner in the NCERT Class 12 Mathematics textbook.

Extramarks provides very reliable NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 on its website and mobile application.

To get solutions for Exercise 5.1, students can click the link below to download it in PDF format. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 guides students in solving each question of the exercise without any confusion.

NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 (Ex 5.1)

Chapter 5 in Class 12 Mathematics is related to Continuity and Differentiability. To understand this topic, students are expected to have a thorough understanding of topics from Chapter 1 to Chapter 5. Chapter 5 is based on concepts of Differentiation and Functions of Class 11 Mathematics NCERT syllabus, so it is important to have an overview of exercises and solutions that were covered in the topic of Class 11 Differentiation and Functions.

Every chapter in Class 12 needs to be covered with proper planning and unique strategies. Chapter 5 has eight exercises and one miscellaneous exercise. Each one of them has questions based on theories and formulas given in Chapter 5 of the Class 12 NCERT Mathematics textbook. Textbook solutions are a good reference while solving these exercises but sometimes students have problems with some questions. In that case, students can use the Extramarks website for downloading the detailed and step-wise solutions to all questions mentioned in the exercises. To get detailed solutions and cover Class 12 Maths Chapter 5 Exercise 5.1, students are advised to download NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. The solutions provided by Extramarks are of optimum quality and are designed by teachers who are experts in the Mathematics field.

Solving exercise questions directly helps in exams. Students get an idea of how to break their answers into various steps. This leads them to score good marks in exams.

Some Mathematical questions are only mastered after a lot of practice. Practise sessions help students structure and organise their answers.

When students solve more problems they understand that they can use mathematical concepts to solve exercise problems and it gives them the confidence to attempt more questions during their exam preparation. To effectively prepare for the Exercise 5.1 questions, it becomes important to keep referring to the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1.

Students having issues with solving questions related to Continuous Functions can look for NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. Extramarks has very exact and detailed NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 to the NCERT textbook Chapter 5 Exercise 5.1 questions. This acts as a guide for them in solving problems related to Continuous Functions and other questions that are included in Exercise 5.1.

Questions that need students to find solutions for examining the Continuity of the Function can be solved using NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. Extramarks provides the best solutions for Class 12 students. Students can easily download NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 from the Extramarks website and application.

In Chapter 5 Exercise 5.1, students might have some difficulty solving problems. Downloading NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 from the Extramarks website will give students accurate solutions to problems they are facing and it helps them save time. NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 are easily accessible in PDF format from the Extramarks website. Students can download NCERT Solutions Class 12 of all subjects from Extramarks.

If any student is having a problem in solving questions of finding the point of Discontinuity in an Equation, they can take a look at NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. It will help them understand the pattern of the questions and enable them to give step-wise answers.

If students are looking for NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1, they can get themselves registered with Extramarks. Answers provided on the Extramarks website are curated by experts and are meant to help students in their exam preparation. Chapter 5 consists of a very important topic in the Class 12 Mathematics syllabus and a student needs to have various tools that give quick solutions to problems. NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 is one such tool that students can use to solve questions from Exercise 5.1 of Chapter 5. After practising using NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1, students will be well prepared to answer correctly in exams. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 can be easily accessed from the Extramarks website and mobile application.

NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 provides solutions to all the questions from Exercise 5.1 that can be referred to at any time for better preparation. Students can view them in offline mode and practise Exercise 5.1 anytime at their convenience.

The content presented in NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 provides students with better knowledge of topics covered in Chapter 5. It also gives them an idea for answering other exercise questions in the chapter.

NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 can be used for overcoming the challenges from Chapter 5 Exercise 5.1. Students will know how to answer in a stepwise, logical manner and use this technique while solving questions in final exams.

NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 help students in self-assessment leading to improvisation in their problem-solving ability. NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 are available in PDF format. Students can use the links provided below to download them.

Solving equations through the help of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 helps students keep track of their practice and also brushes up their understanding of theorems and formulas from Chapter 5. Before solving Chapter 5 of the Class 12 Mathematics syllabus, students need to have a good grasp of previous chapters, it is helpful in gaining momentum when solving exercises from Chapter 5. Extramarks has all the resources for Class 12 students that are designed to enable them to easily access solutions to all their academic problems within seconds. Students can also ask questions while accessing resources from the Extramarks website and mobile application.

Exercise 5.1 includes problems consisting of Algebra of Continuous Functions. These concepts can be understood easily with timely practice. Students while solving problems have to remember theorems related to the topic and apply them to the problem to get desired output. Students can also look for NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 if they face any difficulty going through the questions.

The chapter named Continuity and Differentiability is one of those chapters that has more topic weightage in exams. Having command over Chapter 5 of the Class 12 syllabus will fetch high scores in the Mathematics exam. Students require a thorough understanding of the concepts and with regular practice of exercise questions, they can develop a smart strategy for Chapter 5.

Analyzing one’s practice methods using NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 is one of the many ways to begin Mathematics preparations. Extramarks provides the best solutions for Chapter 5 questions. Students having issues in solving questions from Exercise 5.1 can download the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 using the download PDF button given on the Extramarks website. Students can also download solutions for other chapters of Class 12 to give a boost to their Mathematics study.

After Chapter 4 of Class 12 Mathematics, students have to learn one of the most important topics from the Class 12 syllabus which is Continuity and Differentiability. It is seen from past year’s experiences that some students might have problems solving questions from Chapter 5, which has eight exercises from 5.1 to 5.8 and there is also one miscellaneous exercise question. The miscellaneous exercise is made up of all the questions covered in the previous exercises and is very good to have a quick review of all the questions covered in Chapter 5. Students having any doubt when starting to solve Exercise 5.1 can make use of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 to get easy and step-wise solutions to questions from Exercise 5.1. All these solutions are available on the Extramarks website in PDF format and are accessible for download for all students from class 12.

Exercise 5.1 includes questions that are based on Continuity and Differentiability fundamentals and are very essential for solving questions that will come in the later exercises of Chapter 5. Students are advised to revise the concepts before starting to solve the Exercise 5.1 problems, it allows them to understand the questions and they will be able to solve them afterwards. Even after some practice, if students think it is daunting to solve those problems, they can always refer to NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1.

Initially, the problems in Exercise 5.1 require students to practise questions related to proving the continuity of a function. The question has a function and students are asked to prove the continuity of that function on given points on real numbers line. The questions related to Continuity are important for understanding the fundamental theories of Chapter 5. To practise these questions students are expected to remember definitions and theorems.

Students can look for NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 to get the better practice of Exercise 5.1 questions. Students can easily find the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 on the Extramarks website and mobile application. The solutions help students in breaking down complex problems into simpler and smaller steps that are easy to understand.

Class 12 students can also download the solutions to exercises of other chapters in the Class 12 Mathematics syllabus from Extramarks. The study material for other classes is also available on the Extramarks website.

To solve Exercise 5.1 questions, it is necessary to revise topics from Class 11 such as Limits and Functions. These topics are used in Chapter 5 to explain the concepts of Continuity and Differentiability. Having a revision of Class 11 chapters will also help students in attempting questions of Exercise 5.1 and other exercises. NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 are mainly to aid students in solving questions from Exercise 5.1. These solutions are helpful in giving structure to solutions written by students. Class 12 students can download NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 from the Extramarks website to get an idea of solving questions and what is expected of them to write in a particular question. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 are also available for download on the Extramarks mobile application.

Mathematics at the Class 12 level includes lessons that will clear Mathematical concepts in detail and it is important for students who would like to pursue courses like Engineering and Bachelor of Science. To enhance their Mathematical knowledge, regular practice is needed for Class 12 students. Practising NCERT exercises is the best way to start your Mathematics preparation.

The Class 12 Mathematics Chapter 5 comprises exercise questions that will help in getting a good grasp of topics such as Continuity and Differentiability. When starting to solve questions based on these topics students have to start by solving Exercise 5.1. Some questions might seem difficult to answer in Exercise 5.1. To overcome this, students can refer to NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. Looking at solutions given by expert teachers helps in improving answer writing skills. Extramarks is helping students in getting the best solutions to problems given in the NCERT textbooks. Class 12 students can download NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 from the Extramarks website. The solutions given by Extramarks are comprehensive and easy to understand. Students can also find other study material on Extramarks.

NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1

When preparing for the Mathematical concepts of Class 12, students need to develop smart planning strategies that help in time management and are proven effective. The chapters of Class 12 Mathematics are thoroughly explained and demand a better approach than Class 11 Mathematics. Looking at solutions provided by expert teachers helps students in getting clarity in core topics and expand their knowledge of the subject. Problem-solving is one of the most important strategies for dealing with Mathematical problems.

There are in total six chapters in the Class 12 Mathematics Part 1 textbook namely Relations and Functions, Inverse Trigonometric Functions, Matrices, Determinants, Continuity and Differentiability and Application of Derivatives.

Chapter 5, which is Continuity and Differentiability, is considered very important from the exam perspective. To get command over Chapter 5 topics, students are advised to practise exercise questions given in the chapter. Exercise 5.1 includes questions like examining the continuity of the given function, proving that the function is continuous at given points, and finding all points of discontinuity.

After proper practice, these exercise questions can be easily mastered but if students face any difficulty while solving the exercise questions, they can always look for NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. The Extramarks website is the best place to get NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. Here students can download the solutions easily in PDF format.

Practising exercise problems often and well helps in understanding the theory concepts better and it gives students an upper hand in their final exams. Concepts become very clear after practising a lot of the exercises based on them and it becomes easier to learn new concepts and solve their questions.

Exercise 5.1 deals with questions that are important for solving questions from later exercises of Chapter 5.

Students who are having issues in getting solutions for Exercise 5.1 can download NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 from the Extramarks website and mobile application. Extramarks also provides other resources like past years’ papers, sample question papers, etc. for students. The resources available on Extramarks are designed by expert teachers and strictly follow the official NCERT syllabus and guidelines.

Exercise 5.1 consists of 34 questions that are asked on topics covered under topics – Continuity and Algebra of Continuous Functions. Even though there are various examples given in the textbook, the questions provided in the exercise can be a little tricky to solve. In that case, students are advised to refer to NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. It will help them in solving the questions after referring to solutions mentioned in the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 provided by Extramarks. You can download them in PDF format from the Extramarks website or mobile application.

To help students understand the questions and give comprehensive answers to questions asked in Exercise 5.1, Extramarks provides the best NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. The solutions are given below each question so that students can easily access them and practise them as needed.

The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 are designed to give students all the essential information on problem-solving skills and to enhance their capabilities to solve more questions. It boosts your confidence in a subject like Mathematics that sometimes becomes difficult to understand.

When you get better solutions to your existing practised question, you understand the ways of giving solutions to any question from anywhere on the topic. To solve Exercise 5.1 and speed up your preparation, it is advised that students look at NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. You can access the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 in PDF format and it is advised to keep this PDF in front of you while solving questions for better understanding.

After Chapter 4, with the topic Determinants, the Class 12 NCERT Mathematics Part 1 textbook has a Continuity and Differentiability chapter that is mainly based on the Differentiation of Functions topic of the Class 11 Mathematics syllabus.

Students have to study mainly Continuity, Differentiability and Relation between them. While going through the questions of Exercise 5.1, some students might face struggles or issues. Hence, to pull through this, they can use NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 which give step-by-step solutions to all the questions asked in Exercise 5.1. Extramarks helps you in getting NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. It is available on the Extramarks website and mobile app.

Exercise 5.1 is based on the following topics –

Introduction,

Continuity and

Algebra Of Continuous Functions.

Therefore, questions require a good understanding of these concepts. When these concepts are clear, students tend to solve the exercise questions easily. If there is any difficulty regarding problem-solving for Exercise 5.1 questions, students can get NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 from Extramarks. These solutions will give direction to your answer writing skills and help you in solving problems from miscellaneous exercises too.

Apart from NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1, students can also obtain NCERT solutions for other exercises of Chapter 5 and the rest of the chapters of Class 12 CBSE Mathematics syllabus.

Practising questions with their solutions helps students in getting a deeper understanding of the concepts. Chapter 5 has topics that are essential for understanding the topics included in Chapter 6 such as Applications of Derivatives in various disciplines such as Engineering, Science, Social Science, etc. So, it becomes very important for one to understand the topics covered in Chapter 5. Solving the examples and exercise questions helps students in grasping the concepts of Chapter 5.

Exercise 5.1 helps in understanding the Continuity And Algebra of Continuous Functions in depth. When you get a grip on how to solve Exercise 5.1, it will become a lot easier to start new topics in the chapter. Students having doubts about the questions of Exercise 5.1 can go through the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. The Extramarks website provides you with the best NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. The solutions are properly designed and made according to the Class 12 Mathematics level. Apart from NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1, students are also able to get NCERT Solutions for other exercises and other resources such as sample question papers, past years’ question papers, etc from the Extramarks website and mobile application.

Chapter 5 of Class 12 Mathematics is Continuity and Differentiability.

The important topics covered in this are as follows:

Continuity and Differentiability
Introduction
Algebra of continuous functions
Differentiability
Derivatives of composite functions
Derivatives of implicit functions
Derivatives of inverse trigonometric functions
Exponential and Logarithmic Functions
Logarithmic Differentiation
Derivatives of Functions in Parametric Forms
Second-Order Derivative
Mean Value Theorem
Summary

Exercise 5.1 helps in understanding Continuity and Algebra of Continuous Functions. When solving problems, students are required to include theorems and definitions of Continuity that will add value to their answers. These are the important points that are included in the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. These solutions are very reliable and can be followed by students along with the steps mentioned in the solutions to improve their own problem-solving skills. Extramarks provides NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 for all the thirty-four questions asked in Exercise 5.1. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 have been revised and are recommended by many expert teachers and senior students of Class 12.

The lessons offered by Extramarks help students from all over India in concept building and improving their knowledge of the subject. The concepts of Mathematics given on the Extramarks website and mobile application also help students in preparation for the JEE Mains and other competitive exams.

Class 12 students regularly need guidance and support in their Mathematics exam preparation.

To get good marks in the Chapter 5 portion that will appear in the final exam, students need to have a good grip on Exercise 5.1 questions. The questions require students to use their knowledge of Chapter 5 concepts and solve them in a stepwise manner. To help students in understanding the way of giving solutions to Exercise 5.1 questions, Extramarks is giving NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 in PDF format. The solutions are very authentic and will definitely help students wherever they feel themselves struggling while solving Exercise 5.1. Apart from NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1, students can also download NCERT solutions for other chapters of Class 12 Mathematics subject.

In order to score higher marks in Chapter 5, having a thorough practice of exercise questions is needed. Exercise 5.1 is the very first exercise of Chapter 5. Students can refer to examples given before Exercise 5.1 to help them in solving the questions on their own but if they come across any problems, they can even use the doubt-sessions provided by Extramarks. Moreover, they can also use NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 to overcome those problems. Extramarks is helping students in getting very precise NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 to assist them in solving Exercise 5.1.

Exercise 5.1 questions have been designed to help students in retaining concepts of Continuity and Algebra of Continuous Functions. These concepts are very useful for the Differentiability topic that lies exactly after Exercise 5.1. Students who would like to improve their practice of Exercise 5.1 can use NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 that is being provided by Extramarks. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 is available on the Extramarks website and mobile application.

Exercise 5.1 question four requires students to prove if a given function is continuous at a given value. To solve this question, students need to remember the concepts of Continuity and Algebra of Continuous Functions. If a student faces any difficulty in solving question 4, he/she can refer to NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 given here on the Extramarks website.

The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 will help in solving all the questions from Exercise 5.1 and boost their exam preparations for the Mathematics subject.

Question 2 in Exercise 5.1 is asking to examine the continuity of the given function. To approach questions like this, students have to remember the exact theorems that question needs.

Remembering theorems while solving such questions helps students to practise in an effective way. Students can refer to NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 given by expert Mathematics teachers at the Extramarks learning platform. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 are available on Extramarks.

NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1

PDF of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 is being provided by the Extramarks according to CBSE (NCERT) guidelines. These solutions are curated in a way to help Class 12 students in getting proper practice of Exercise 5.1 questions and they are prepared carefully keeping in mind the question’s requirements and concepts explained in Chapter 5.

If students come across any doubt while going through the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1, they are free to ask questions anytime using the doubt-solving platform.

Chapter 5 has topics that are important from the exam point of view, therefore it is necessary for the students to give more focus on the concepts and solve all the questions of the exercises given in Chapter 5.

The Introduction part includes information about the importance of learning Chapter 5 and then topics are explained. It consists of theorems and some examples to explain the concept of Continuity. After that, Exercise 5.1 is given for students to practise questions. And all the questions in Exercise 5.1 are really helpful for students to get good marks in the final exams.

The kind of questions asked in Exercise 5.1 serves as a basic building block for other important questions that will appear later in the chapter. If students get a good grip on questions of Exercise 5.1, they will find it easier to solve other exercises in Chapter 5.

Students can practise Exercise 5.1 on their own and try to implement the concepts of Continuity and Algebra of Continuous Functions. During the process, if they face any difficulty with the exercise questions they can make use of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. Students can access the best NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 from the Extramarks platform. These NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 are recommended not only due to their high-quality answers but also because they are easy to understand and master.

Class 12 Maths Chapter 5 Exercise 5.1 questions are also crucial while preparing for the examination. Therefore, it is necessary to use NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 while practising.

NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 has NCERT Class 12 solutions and are written in simple language. NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 is an important tool for Class 12 students.

Learning the fundamentals of Continuity and Differentiability and solving questions based on them will build major concepts of Calculus in Class 12 students. To effectively solve Exercise 5.1, it is advisable to make use of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. Practising using NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 has various advantages, it gives stepwise solutions to questions and concepts become more clear in the student’s mind.

To get the best NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1, students can make use of the guidance provided by Extramarks. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 available on the Extramarks platform are designed by expert teachers and they are in a simple format.

Learn this before Exercise 5.1:

Chapter 5 is basically a continuation of what students have already learned in the Differentiation of Functions in the Class 11 Mathematics NCERT textbook. So, it is essential for students to remember the concepts of Differentiation of Functions taught in Class 11.

To solve questions based on it, one thing which is to be kept in mind is learning the definitions as well in the Continuity and Algebra of Continuous Functions topic. Students are required to read the explanations given in the NCERT textbook. It is also important to carefully learn the theorems and definitions given under the Algebra of Continuous Functions because it will help students in solving the questions from Exercise 5.1.

To get exact solutions for Exercise 5.1 questions, students are advised to get help from NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. When going through these solutions students will understand how to frame solutions for the kind of problems asked in Exercise 5.1 and they will be able to practise them in a better way. Extramarks provides very authentic NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 for Class 12 students to help in their preparation for final exams. If students are searching for the best solutions for NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1, they can easily access it on both the Extramarks website and mobile application.

Students are expected to learn Continuity and Algebra of Continuous Functions by going through the explanation provided in the Class 12 Mathematics Part 1 textbook, then only they will be able to practise Exercise 5.1 questions.

Exercise 5.1 has questions that require students to examine the continuity of the function, proving that a function is continuous at a given value, finding all points of discontinuity in a function, etc. The questions can sometimes become tricky to solve, so students can make use of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. Students can download the best NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 from the Extramarks website and mobile application. Extramarks is the best place to get NCERT Class 12 Solutions based on the student’s needs. The study material provided on Extramarks is very reliable and easy to understand for school students.

To start solving Exercise 5.1, students need to have a proper understanding of concepts taught in Chapter 5 and also need to revise the topics covered under the Limits and Derivatives Chapter in Class 11 Mathematics NCERT textbook.

The topics covered in the Limits and Derivatives chapter are-

Intuitive Idea of Derivatives
Limits
Algebra of limits
Limits of polynomial and rational functions
Derivatives
Derivative of polynomials and trigonometric functions

There are two exercises and one miscellaneous exercise given in Chapter 13 of the Class 11 Mathematics textbook.

After having revised the Limits and Derivatives Class 11 topic, students can start practising Exercise 5.1. With all the previous concepts in mind, it will become easier to solve the questions mentioned in Exercise 5.1. To effectively practise these questions, students are advised to solve them using NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. The Extramarks website and mobile application have the precise NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 for the students to download.

Learning the basics of any topic and regular practice of the questions framed using those topics is an important aspect of Mathematics preparation in higher classes.

Chapter 5 of Class 12 Mathematics has some of the important concepts that form the basis for some subjects that are studied in some fields like Engineering, etc. Getting a thorough understanding of these concepts will help students in solving questions that are asked in exercises.

Exercise 5.1 is the very first exercise of the Continuity and Differentiability chapter and requires good practice. To practise these questions of Exercise 5.1, it is advisable to refer to NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 provided by Extramarks. To download the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1, students can make use of the Extramarks website or the Extramarks mobile application available on the Google Play Store.

The solutions provided by Extramarks are designed by expert Mathematics teachers from all over India and are easy to understand. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 is one of the important resources for Class 12 Mathematics final exam preparation.

Students can also access NCERT Solutions Class 12 to get solutions for other exercises of Chapter 5 and also of other chapters in the Class 12 Mathematics syllabus.

Students can practise the example questions based on Continuity and Algebra of Continuous Functions before starting with Exercise 5.1 for a better understanding of the concepts. These examples given in the NCERT textbook are easy to comprehend and will give hints for solving exercise questions.

Some of the benefits of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 are given below –

These solutions enhance the understanding of Continuity and Differentiability concepts.
The solutions guide students to give structure to their answers and enhance their answer writing skills.
The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 give an idea of solving the type of questions that will appear in the final exams.
The solutions help students in evaluating their ongoing Mathematics practice.
These solutions help students give step-wise answers to questions and include only necessary steps in their answers.

Extramarks is providing NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 for Class 12 students of English and Hindi Medium both. Students can also access solutions for other exercises of Chapter 5 and other chapters of Class 12 Mathematics as well. These solutions will help in their exam preparation effectively and they will be able to score good marks in the Class 12 Mathematics final exam. Referring to solutions helps in getting a deeper understanding of the concepts covered in the chapters.

Exercise 5.1 has questions that expect students to find the point of discontinuity of a function using left-hand limits and right-hand limits. Students will get an idea of solving such questions using NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 provided by Extramarks. The solutions are of the best quality and very reliable. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 can be downloaded easily and can be accessed anytime.

Continuity and Differentiability

Continuity and Differentiability is an important topic of Calculus and it needs the right focus while studying its concepts. It is covered under Chapter 5 of the Class 12 Mathematics syllabus. These topics can be well understood when practised along with exercise questions and example questions.

A function is said to be continuous if it can be drawn without lifting up the pencil. Continuity of a function is dependent on three conditions–

The function is expressed at x = a.
The limit of the function as the approaching of x takes place, a exists.
The limit of the function as the approaching of x takes place, a is equal to the function value f(a).

To explain continuity well enough, the concept of limits is used. Understanding of limits topic is crucial for explaining continuity. Continuity and Differentiability are considered complementary to each other. Students are required to first prove if a function given by y=f(x) is continuous at x=a, before proving its differentiability at a point x=a. Both the graphical and algebraic methods can be used to prove if a function is continuous and also differentiable or not.

The topics covered under Continuity and Differentiability are important for building a base for advanced Mathematics concepts of Calculus.

These concepts are explained with the help of various examples and exercises. Chapter 5 preparation strategy can be strengthened by making use of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. The solutions give students a better approach to the exercise questions and make them exam-ready.

NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 are available on the Extramarks website and mobile application and are necessary for building a foundation for some of the opening concepts in the chapter.

The concepts given in the Continuity and Differentiability chapter may seem difficult to understand at first. But when it is practised by solving sums and evaluating answers with NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1, students gradually can have a better understanding of the concepts.

Extramarks is helping Class 12 students to solve Exercise 5.1 with error-free and easy NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. Students can make use of the study material given by the Extramarks learning platform to enhance their course preparation strategy and boost their confidence for final examinations.

Continuity as a topic is a basic building block for students to proceed to learn Differentiability. The explanations given in the NCERT textbook are very easy to understand and are supported by definitions and theorems that are also thoroughly explained for students to easily understand. It is advisable to remember these theorems and definitions for an in-depth understanding of the topics and subtopics.

The Continuity and Differentiability chapter consists of eight exercises and one miscellaneous exercise that will test concepts taught under the topics and subtopics.

The complete topics in Chapter 5 are given below-

Introduction
Continuity
Algebra of continuous functions
Differentiability
Derivatives of composite functions
Derivatives of implicit functions
Derivatives of inverse trigonometric functions
Exponential and Logarithmic Functions
Logarithmic Differentiation
Derivatives of Functions in Parametric Forms
Second Order Derivative
Mean Value Theorem

Apart from reading textbooks and notes on the Continuity and Differentiability chapter, it is also recommended to practise examples and exercises given in the textbook. Solving questions regularly gives students an advantage in their exams and they are able to attempt questions in the final exam without any hesitation. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 are designed in a way to guide students in answering properly to questions asked and improve their Mathematical preparation. Extramarks gives very authentic NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 for students to have a stepwise analysis of their self-written answers and it is considered part of the important study material for Mathematics exam preparation.

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Continuity and Differentiability Exercise 5.1 contains thirty-four questions that are important from the exam point of view and practising each question in detail will increase your grasp over Chapter 5. The Extramarks website and mobile application are providing very reliable NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1. Students will be truly benefited from NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1 given by Extramarks since it is designed by subject experts and is checked multiple times to avoid any error in the solution.

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NCERT Solutions for Class 12 Maths Chapter 5 Other Exercises

Chapter 5 – Continuity and Differentiability Exercises
Exercise 5.2	2 Short Questions 8 Long Questions
Exercise 5.3	9 short Questions and 6 long Questions
Exercise 5.4	5 Short Questions and 5 Long Questions
Exercise 5.5	4 Short Questions and 14 Long Questions
Exercise 5.6	1 Short Question and 1 Long Question
Exercise 5.7	10 Short Questions and 7 Long Questions
Exercise 5.8	Questions with Solutions

Q.1

\begin{array}{l} Prove that the function f (x) = 5 x - 3 is continuous at x = 0, \\ at x = - 3 and at x = 5 . \end{array}

Ans

$\begin{array}{l} Given function is f (x) = 5 x - 3 \\ At x = 0, f (0) = 5 (0) - 3 = - 3 \\ \lim_{x \to 0} f (x) = \lim_{x \to 0} 5 x - 3 \\ = 5 (0) - 3 \\ = - 3 \\ ∴ \lim_{x \to 0} f (x) = f (0) \\ Therefore, f is continuous at x = 0 . \\ At x = - 3, f (- 3) = 5 (- 3) - 3 = - 18 \\ \lim_{x \to - 3} f (x) = \lim_{x \to 0} 5 x - 3 \\ = 5 (- 3) - 3 \\ = - 18 \\ ∴ \lim_{x \to - 3} f (x) = f (- 3) \\ Therefore, f is continuous at x = - 3 . \\ At x = 5, f (5) = 5 (5) - 3 = 22 \\ \lim_{x \to 5} f (x) = \lim_{x \to 0} 5 x - 3 \\ = 5 (5) - 3 \\ = 22 \\ ∴ \lim_{x \to 5} f (x) = f (5) \\ Therefore, f is continuous at x = 5 . \end{array}$

Q.2 Examine the continuity of the function
f(x) = 2x² – 1 at x = 3.

Ans

$\begin{array}{l} The given function is f (x) = 2 x^{2} - 1 \\ At x = 3, f (3) = 2 {(3)}^{2} - 1 = 17 \\ \lim_{x \to 3} f (x) = \lim_{x \to 3} 2 x^{2} - 1 \\ = 2 {(3)}^{2} - 1 \\ = 17 \\ ∴ \lim_{x \to 3} f (x) = f (3) \\ Therefore, f is continuous at x = 3 . \end{array}$

Q.3

\begin{array}{l} Examine the following functions for continuity . \\ (a) f (x) = x - 5 (b) f (x) = \frac{1}{x - 5}, x \neq 3 \\ (c) f (x) = \frac{x^{2} - 25}{x + 5}, x \neq - 5 (d) f (x) = |x - 5| \end{array}

Ans

$\begin{array}{l} (a) The given function is f (x) = x - 5 \\ It is evident that f is defined at every real number k, \\ then f (k) = k - 5 . \\ \lim_{x \to k} f (x) = \lim_{x \to k} (x - 5) = k - 5 \\ ∴ \lim_{x \to k} f (x) = f (k) \\ Hence, f is continuous at every real number and therefore, \\ it is a continuous function . \\ (b) The given function is f (x) = \frac{1}{x - 5}, x \neq 5 \\ For any real number k \neq 5 \\ then f (k) = k - 5 . \\ \lim_{x \to k} f (x) = \lim_{x \to k} (x - 5) = k - 5 \\ ∴ \lim_{x \to k} f (x) = f (k) \\ Hence, f is continuous at every real number and therefore, \\ it is a continuous function . \\ (c) The given function is f (x) = \frac{x^{2} - 25}{x + 5}, x \neq - 5 \\ For any real number k \neq - 5 \\ then f (k) = \frac{(k - 5) (k + 5)}{(k + 5)} = k - 5 \\ \lim_{x \to - k} f (x) = \lim_{x \to k} \frac{(k - 5) (k + 5)}{(k + 5)}) = k - 5 \\ ∴ \lim_{x \to k} f (x) = f (k) \\ Hence, f is continuous at every real number and therefore, \\ it is a continuous function . \\ (d) The given function is f (x) = | x - 5 | = {\begin{cases} - x + 5, if x < 5 \\ x - 5, if x \geq 5 \end{cases} \\ This function f is defined at all points of the real line . \\ Let c be a point on a real line . Then, k < 5 or k = 5 or k > 5 \\ Case I : c < 5 \\ Then, f (c) = 5 - c \\ \lim_{x \to c} f (x) = \lim_{x \to c} (5 - x) \\ = 5 - c \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all real numbers less than 5 . \\ Case II : c = 5 \\ Then, f (c) = f (5) \\ = 5 - 5 = 0 \\ \lim_{x \to 5^{-}} f (x) = \lim_{x \to 5} (5 - x) \\ = 5 - 5 = 0 \\ \lim_{x \to 5^{+}} f (x) = \lim_{x \to 5} (x - 5) \\ = 5 - 5 = 0 \\ ∴ \lim_{x \to 5^{-}} f (x) = \lim_{x \to 5^{+}} f (x) = f (c) \\ Therefore, f is continuous at x = 5 \\ Case III : c > 5 \\ Then, f (c) = c - 5 \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x - 5) = c - 5 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all real numbers greater than 5 . \\ Hence, f is continuous at every real number and therefore, \\ it is a continuous function . \end{array}$

Q.4 Prove that the function f x = x n is continuous at x = n, where n is a positive integer.

Ans

$\begin{array}{l} The given function is f (x) = x^{n} \\ It is evident that f is defined at all positive integers, n, and \\ its value at n is n^{n} . \\ Then, \lim_{x \to n} f (n) = \lim_{x \to n} x^{n} = n^{n} \\ ∴ \lim_{x \to n} f (x) = f (n) \\ Therefore, f is continuous at n, where n is a positive integer . \end{array}$

Q.5

\begin{array}{l} Is the function f defined by \\ f (x) = \{\begin{cases} x, if x \leq 1 \\ 5, if x > 1 \end{cases} \\ continuous at x = 0 ? Atx = 1 ? Atx = 2 ? \end{array}

Ans

$\begin{array}{l} The given function f is \\ f (x) = {\begin{cases} x, if x \leq 1 \\ 5, if x >1 \end{cases} \\ At x =0 \\ It is evident that f is defined at 0 and its value at 0 is 0 . \\ Then, \lim_{x \to 0} f (x) = \lim_{x \to 0} x = 0 \\ ∴ \lim_{x \to 0} f (x) = f (0) \\ Therefore, f is continuous at x = 0 \\ At x = 1, \\ f is defined at 1 and its value at 1 is 1 . \\ The left hand limit of f at x = 1 is, \\ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} x = 1 \\ The right hand limit of f at x = 1 is, \\ \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} (5) = 5 \\ ∴ \lim_{x \to 1^{-}} f (x) \neq \lim_{x \to 1^{+}} f (x) \\ Therefore, f is not continuous at x = 1 . \\ At x = 2, \\ f is defined at 2 and its value at 2 is 5 . \\ Then, \lim_{x \to 2} f (x) = \lim_{x \to 2} (5) = 5 \\ ∴ \lim_{x \to 2} f (x) = f (2) \\ Therefore, f is continuous at x = 2 . \end{array}$

Q.6 Find all the points of discontinuity of f, where f is defined by

$f(x)={\begin{array}{l} 2 x + 3, if x \leq 2 \\ 2 x - 3, if x > 2 \end{array}$

Ans

$\begin{array}{l} The given function f is f (x) = {\begin{cases} 2x+3, if x \leq 2 \\ 2x - 3, if x>2 \end{cases} \\ It is evident that the given function f is defined at all \\ the points of the real line . \\ Let c be a point on the real line . Then, three cases arise . \\ (i) c < 2 \\ (i i) c > 2 \\ (i i i) c = 2 \\ C a s e 1 : c < 2 \\ T h e n, f (c) = 2 c + 3 \\ \lim_{x \to c} f (x) = \lim_{x \to c} (2x+3) = 2c+3 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < 2 . \\ C a s e 2 : c > 2 \\ T h e n, f (c) = 2 c - 3 \\ \lim_{x \to c} f (x) = \lim_{x \to c} (2x - 3) = 2c - 3 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x > 2 . \\ Case (iii) c = 2 \\ Then, the left hand limit of f at x = 2 is, \\ \lim_{x \to 2^{-}} f (x) = \lim_{x \to 2} (2x+3) = 7 \\ \lim_{x \to 2^{+}} f (x) = \lim_{x \to 2} (2x - 3) = 1 \\ ∴ \lim_{x \to 2^{-}} f (x) \neq \lim_{x \to 2^{+}} f (x) \\ Therefore, f is not continuous at x = 2 \\ Hence, x = 2 is the only point of discontinuity of f . \end{array}$

Q.7 Find all the points of discontinuity of f, where f is defined by

$f(x)={\begin{array}{l} | x | + 3, if x \leq - 3 \\ - 2 x, if - 3 < x < 3 \\ 6 x + 2, if x \geq 3 \end{array}$

Ans

$\begin{array}{l} The given function f is f (x) = {\begin{cases} | x | + 3 = - x +3, if x \leq - 3 \\ - 2 x, if - 3 < x < 3 \\ 6 x + 3, if x \geq 3 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case 1 : If c < - 3 \\ Then, f (c) = - c + 3 \\ \lim_{x \to c} f (x) = \lim_{x \to c} (- x +3) = - c +3 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < - 3 . \\ Case 2 : c = - 3 \\ Then, f (- 3) = - (- 3) + 3 = 6 \\ \lim_{x \to 3^{-}} f (x) = \lim_{x \to 3^{-}} (- x +3) = - (- 3) + 3 = 6 \\ \lim_{x \to 3^{+}} f (x) = \lim_{x \to 3^{+}} (- 2 x) = - 2 (- 3) = 6 \\ ∴ \lim_{x \to 3^{-}} f (x) = \lim_{x \to 3^{+}} f (x) = f (- 3) \\ Therefore, f is continuous at x = - 3. \\ Case 3 : If - 3 < c < 3, then \\ f (c) = - 2 c and \\ \lim_{x \to c} f (x) = \lim_{x \to c} (- 2 x) = - 2 c \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous in (- 3,3) . \\ Case 4: If c = 3, then \\ L . H . L . = \lim_{x \to 3^{-}} f (x) \\ = \lim_{x \to 3^{-}} (- 2 x) \\ = - 2 (- 3) = 6 \\ R . H . L . = \lim_{x \to 3^{+}} f (x) \\ = \lim_{x \to 3^{+}} (6 x +2) \\ = 6 (3) + 2 = 20 \\ ∴ \lim_{x \to 3^{-}} f (x) = \lim_{x \to 3^{+}} f (x) \\ Therefore, f is not continuous at x = 3 \\ Case 5: If c > 3, then \\ f (c) = 6 c + 2 and \lim_{x \to c} f (x) = \lim_{x \to c} (6 x + 2) = 6 c + 2 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x > 3 \\ Hence, x = 3 is the only point of discontinuity of f . \end{array}$

Q.8 Find all the points of discontinuity of f, where f is defined by

$f(x)={\begin{array}{l} \frac{| x |}{x}, if x \neq 0 \\ 0, if x = 0 \end{array}$

Ans

$\begin{array}{l} The given function f is \\ f (x) = {\begin{cases} \frac{| x |}{x}, if x \neq 0 \\ 0, if x =0 \end{cases} \\ = {\begin{cases} \frac{| x |}{x} = \frac{x}{x} = 1, if x > 0 \\ 0, if x = 0 \\ \frac{| x |}{x} = \frac{- x}{x} = - 1, if x < 0 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case I : If c <0, then f (c) = - 1 \\ \lim_{x \to c} f (x) = \lim_{x \to c} (- 1) = - 1 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x < 0 . \\ Case II : If c = 0, then L . H . L . at x = 0 is, \\ L . H . L . = \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} (- 1) = - 1 \\ L . H . L . at x = 0 is \\ R . H . L . = \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} (1) = 1 \\ L . H . L . \neq R . H . L . \\ Therefore, f is not continuous at x = 0 \\ Case III : If c > 0, then f (c) = 1 \\ \lim_{x \to c} f (x) = \lim_{x \to c} (1) = 1 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x > 0 . \\ Hence, x = 0 is the only point of discontinuity of f . \end{array}$

Q.9 Find all the points of discontinuity of f, where f is defined by

$f(x)={\begin{array}{l} \frac{x}{| x |}, if x < 0 \\ - 1, if x \geq 0 \end{array}$

Ans

$\begin{array}{l} The given function f is \\ f (x) = {\begin{cases} \frac{x}{| x |} = \frac{x}{- x} = - 1, if x < 0 \\ - 1, if x \geq 0 \end{cases} \\ \Rightarrow f (x) = - 1 for all x \in R \\ Let c be any real number . Then, \lim_{x \to c} f (x) = \lim_{x \to c} (- 1) = - 1 \\ And f (c) = - 1 for all x \in R \\ Therefore, the given function is a continuous function . \\ Hence, the given function has no point of discontinuity . \end{array}$

Q.10 Find all the points of discontinuity of f, where f is defined by

$f(x)={\begin{array}{l} x + 1, if x \geq 1 \\ x^{2} + 1, if x < 1 \end{array}$

Ans

$\begin{array}{l} The given function f is f (x) = {\begin{cases} x +1, if x \geq 1 \\ x^{2} + 1, if x <1 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case I : If c <1, then f (c) = c^{2} + 1 and \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x^{2} + 1) \\ = c^{2} + 1 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < 1 . \\ Case II : If c = 1, then f (c) = f (1) = 1 + 1 = 2 \\ The left hand limit of f at x = 1 is, \\ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} (x^{2} + 1) = 1^{2} + 1 = 2 \\ The right hand limit of f at x = 1 is, \\ \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} (x + 1) = 1 + 1 = 2 \\ ∴ \lim_{x \to 1} f (x) = f (1) \\ Therefore, f is continuous at x = 1 . \\ Case III : If c > 1, then f (c) = c + 1 \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x + 1) \\ = c + 1 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x > 1 \\ Hence, the given function f has no point of discontinuity . \end{array}$

Q.11 Find all the points of discontinuity of f, where f is defined by

$f(x)={\begin{array}{l} x^{3} - 3, if x \leq 2 \\ x^{2} + 1, if x > 2 \end{array}$

Ans

$\begin{array}{l} The given function f is f (x) = {\begin{cases} x^{3} - 3, if x \leq 2 \\ x^{2} + 1, if x >2 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case I : If c <1, then f (c) = c^{2} + 1 and \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x^{2} + 1) \\ = c^{2} + 1 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < 1 . \\ Case II : If c = 1, then f (c) = f (1) = 1 + 1 = 2 \\ The left hand limit of f at x = 1 is, \\ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} (x^{2} + 1) = 1^{2} + 1 = 2 \\ The right hand limit of f at x = 1 is, \\ \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} (x + 1) = 1 + 1 = 2 \\ ∴ \lim_{x \to 1} f (x) = f (1) \\ Therefore, f is continuous at x = 1 . \\ Case III : If c > 1, then f (c) = c + 1 \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x + 1) \\ = c + 1 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x > 1 \\ Hence, the given function f has no point of discontinuity . \\ The given function f is f (x) = {\begin{cases} x +1, if x \geq 1 \\ x^{2} + 1, if x <1 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case I : If c <1, then f (c) = c^{2} + 1 and \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x^{2} + 1) \\ = c^{2} + 1 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < 1 . \\ Case II : If c = 1, then f (c) = f (1) = 1 + 1 = 2 \\ The left hand limit of f at x = 1 is, \\ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} (x^{2} + 1) = 1^{2} + 1 = 2 \\ The right hand limit of f at x = 1 is, \\ \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} (x + 1) = 1 + 1 = 2 \\ ∴ \lim_{x \to 1} f (x) = f (1) \\ Therefore, f is continuous at x = 1 . \\ Case III : If c > 1, then f (c) = c + 1 \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x + 1) \\ = c + 1 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x > 1 \\ Hence, the given function f has no point of discontinuity . \end{array}$

Q.12 Find all the points of discontinuity of f, where f is defined by

$f(x)={\begin{array}{l} x^{10} - 1, if x \leq 1 \\ x^{2}, if x > 1 \end{array}$

Ans

$\begin{array}{l} The given function f is f (x) = {\begin{cases} x^{10} - 1, if x \geq 1 \\ x^{2}, if x <1 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case I : If c <1, then f (c) = c^{10} - 1 and \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x^{10} - 1) \\ = c^{10} - 1 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < 1 . \\ Case II : If c = 1, \\ The left hand limit of f at x = 1 is, \\ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} (x^{10} - 1) = 1^{10} - 1 = 0 \\ The right hand limit of f at x = 1 is, \\ \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} (x^{2}) = 1^{2} = 1 \\ ∴ \lim_{x \to 1^{-}} f (x) \neq \lim_{x \to 1^{+}} f (x) \\ It is observed that the left and right hand limit of f at x = 1 do \\ not coincide . Therefore, f is not continuous at x = 1 \\ Case III : If c > 1, then f (c) = c^{2} \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x^{2}) \\ = c^{2} \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x > 1 \\ Thus, from the above observation, it can be concluded that \\ x = 1 is the only point of discontinuity of f . \end{array}$

Q.13

\begin{array}{l} Is the function defined by \\ f (x)= \{\begin{matrix} x + 5 & if x \leq 1 \\ x - 5 & if x >1 \end{matrix} \\ a continuous function ? \end{array}

Ans

$\begin{array}{l} The given function f is f (x) = {\begin{cases} x + 5, if x \leq 1 \\ x - 5, if x >1 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case I : If c <1, then f (c) = c + 5 and \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x + 5) \\ = c + 5 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < 1 . \\ Case II : If c = 1, then f (1) = 1 + 5 = 6 \\ The left hand limit of f at x = 1 is, \\ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} (x + 5) = 1 + 5 = 6 \\ The right hand limit of f at x = 1 is, \\ \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} (x - 5) = 1 - 5 = - 4 \\ ∴ \lim_{x \to 1^{-}} f (x) \neq \lim_{x \to 1^{+}} f (x) \\ Therefore, f is not continuous at x = 1 \\ Case III : If c > 1, then f (c) = c - 5 and \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x - 5) \\ = c - 5 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x > 1 \\ Thus, from the above observation, it can be concluded that \\ x = 1 is the only point of discontinuity of f . \end{array}$

Q.14 Discuss the continuity of the function f, where f is defined by

$f(x)={\begin{array}{l} 3, if 0 \leq x \leq 1 \\ 4, if 1 < x < 3 \\ 5, if 3 \leq x \leq 10 \end{array}$

Ans

$\begin{array}{l} The given function f is f (x) = {\begin{cases} 3, & i f 0 \leq x \leq 1 \\ 4, & i f 1 < x < 3 \\ 5, & i f 3 \leq x \leq 10 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ C a s e I : I f 0 \leq c < 1, t h e n f (c) = 3 a n d \\ lim_{x \to c} f (x) = lim_{x \to c} (3) \\ = 3 \\ ∴ lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous in the interval [0,1) . \\ C a s e I I : I f c = 1, t h e n f (1) = 3 \\ The left hand limit of f at x = 1 is, \\ lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{-}} (3) = 3 \\ The right hand limit of f at x = 1 is, \\ lim_{x \to 1^{+}} f (x) = lim_{x \to 1^{+}} (4) = 4 \\ ∴ lim_{x \to 1^{-}} f (x) \neq lim_{x \to 1^{+}} f (x) \\ Therefore, f is not continuous at x = 1 \\ C a s e I I I : I f 1 < c < 3, t h e n f (c) = 4 a n d \\ lim_{x \to c} f (x) = lim_{x \to c} (4) \\ = 4 \\ ∴ lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points of the interval (1, 3) . \\ C a s e I V : I f c = 3, t h e n f (c) = 5 \\ T h e l e f t h a n d l i m i t o f f a t x = 3 i s, \\ lim_{x \to 3^{-}} f (x) = lim_{x \to 3^{-}} (4) = 4 \\ The right hand limit of f at x = 3 is, \\ lim_{x \to 3^{+}} f (x) = lim_{x \to 3^{+}} (5) = 5 \\ ∴ lim_{x \to 3^{-}} f (x) \neq lim_{x \to 3^{+}} f (x) \\ Therefore, f is not continuous at x = 3 . \\ C a s e I V : I f 3 < c \leq 10, t h e n f (c) = 5 a n d \\ lim_{x \to c} f (c) = lim_{x \to c} (5) = 5 \\ lim_{x \to c} f (c) = f (c) \\ Therefore, f i s c o n t i n u o u s a t a l l p o i n t s o f t h e i n t e r v a l (3, 10] . \\ H e n c e, f i s n o t c o n t i n u o u s a t x = 1 a n d x = 3 . \end{array}$

Q.15 Discuss the continuity of the function f, where f is defined by

$f(x)={\begin{array}{l} 2 x, if x < 0 \\ 0, if 0 \leq x < 1 \\ 4 x, if x > 1 \end{array}$

Ans

$\begin{array}{l} The given function f is f (x) = {\begin{cases} 2 x, if x < 0 \\ 0, if 0 \leq x <1 \\ 4 x, if x > 1 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case I : If c >0, then f (c) = 2 c \\ \lim_{x \to c} f (x) = \lim_{x \to c} (2 x) \\ = 2 c \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < 0 . \\ Case II : If c = 0, then f (c) = f (0) = 0 \\ The left hand limit of f at x = 0 is, \\ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} (2 x) = 2 \times 0 = 0 \\ The right hand limit of f at x = 0 is, \\ \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} (0) = 0 \\ ∴ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) = f (c) \\ Therefore, f is continuous at x = 0 . \\ Case III : If 0< c < 1, then f (c) = 0 . \\ \lim_{x \to c} f (x) = \lim_{x \to c} (0) = 0 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points of the interval (0, 1) . \\ Case IV : If c = 1, then f (c) = f (1) = 0 . \\ The left hand limit of f at x = 1 is, \\ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} (0) = 0 \\ The right hand limit of f at x = 1 is, \\ \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} (4 x) = 4 \times 1 = 4 \\ \Rightarrow \lim_{x \to 1^{-}} f (x) \neq \lim_{x \to 1^{+}} f (x) \\ Therefore, f is not continuous at x = 1 . \\ Case V : If c < 1, then f (c) = 4 c and \\ \lim_{x \to c} f (x) = \lim_{x \to c} (4 x) = 4 c \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < 1 \\ Hence, f is not continuous only at x = 1 . \end{array}$

Q.16 Discuss the continuity of the function f, where f is defined by

$f(x)={\begin{array}{l} - 2, if x < - x \\ 2 x, if 0 \leq x < 1 \\ 2, if x > 1 \end{array}$

Ans

$\begin{array}{l} The given function f is f (x) = {\begin{cases} - 2, if x \leq - 1 \\ 2 x, if - 1 \leq x <1 \\ 2, if x > 1 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case I : If c < - 1, then f (c) = - 2 and \\ \lim_{x \to c} f (x) = \lim_{x \to c} (- 2) \\ = - 2 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < - 1 . \\ Case II : If c = - 1, then f (c) = f (- 1) = - 2 \\ The left hand limit of f at x = - 1 is, \\ \lim_{x \to - 1^{-}} f (x) = \lim_{x \to - 1^{-}} (- 2) = - 2 \\ The right hand limit of f at x = - 1 is, \\ \lim_{x \to - 1^{+}} f (x) = \lim_{x \to - 1^{+}} (2 x) = 2 \times (- 1) = - 2 \\ ∴ \lim_{x \to - 1} f (x) = f (- 1) \\ Therefore, f is continuous at x = - 1. \\ Case III : If - 1 < c < 1, then f (c) = 2 c . \\ \lim_{x \to c} f (x) = \lim_{x \to c} (2 x) = 2 c \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points of the interval (-1, 1) . \\ Case IV : If c = 1, then f (c) = f (1) = 2 \times 1 = 2 . \\ The left hand limit of f at x = 1 is, \\ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} (2 x) = 2 \times 1 = 2 \\ The right hand limit of f at x = 1 is, \\ \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} (2) = 2 \\ \Rightarrow \lim_{x \to 1} f (x) = f (c) \\ Therefore, f is not continuous at x = 1 . \\ Case V : If c > 1, then f (c) = 2 and \\ \lim_{x \to c} f (x) = \lim_{x \to c} (2) = 2 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x > 1 \\ Thus, from the above observations, it can be concluded that \\ f is continuous at all points of the real line . \end{array}$

Q.17

\begin{array}{l} Find the relationship between a and b so that the function \\ f defined by \\ f (x) = \{\begin{cases} ax + 1, if x \leq 3 \\ bx +, if x > 3 \end{cases} \\ tinuous at x = 3 . \end{array}

Ans

$\begin{array}{l} The given function f is \\ f (x) = {\begin{cases} ax + 1, if x \leq 3 \\ bx + 3, if x > 3 \end{cases} \\ If f is continuous at x = 3, then \\ \lim_{x \to 3^{-}} f (x) = \lim_{x \to 3^{+}} f (x) = f (3) . .. (i) \\ \lim_{x \to 3^{-}} f (x) = \lim_{x \to 3^{-}} (ax + 1) = 3 a + 1 \\ \lim_{x \to 3^{+}} f (x) = \lim_{x \to 3^{+}} (bx + 3) = 3 b + 3 \\ f (3) = 3 a + 1 \\ Therefore, from (1), we obtain \\ 3 a + 1 = 3 b + 3 = 3 a + 1 \\ \Rightarrow 3 a + 1 = 3 b + 3 \\ \Rightarrow 3 a = 3 b + 3 - 1 \\ \Rightarrow a = b + \frac{2}{3} \\ Therefore, the required relationship is given by \\ a = b + \frac{2}{3} . \end{array}$

Q.18

\begin{array}{l} For what value of λ is the function f defined by \\ f (x) = \{\begin{cases} λ (x^{2} - 2 x), if x \leq 0 \\ 4 x +, if x > 0 \end{cases} \\ tinuous at x = 0 ? What about continuity at x = 1 ? \end{array}

Ans

$\begin{array}{l} The given function f is \\ f (x) = {\begin{cases} λ (x^{2} - 2 x), if x \leq 0 \\ 4 x +1, if x >0 \end{cases} \\ If f is continuous at x = 0, then \\ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) = f (0) \\ \Rightarrow \lim_{x \to 0^{-}} λ (x^{2} - 2 x) = \lim_{x \to 0^{+}} 4 x +1 = λ (0^{2} - 2 \times 0) \\ \Rightarrow λ (0^{2} - 2 \times 0) = 4 (0) + 1 = 0 \\ \Rightarrow 0 = 1 = 0, which is not possible . \\ Therefore, there is no value of λ for which f is continuous at x = 0 \\ At x = 1, \\ f (1) = 4 x + 1 = 4 \times 1 + 1 = 5 \\ \lim_{x \to 1} (4 x + 1) = 4 \times 1 + 1 = 5 \\ ∴ \lim_{x \to 1} f (x) = f (1) \\ Therefore, for any values of λ, f is continuous at x = 1 . \end{array}$

Q.19 Show that the function defined by g(x) = x – [x] is discontinuous at all integral
points. Here [x] denotes the greatest integer less than or equal to x.

Ans

$\begin{array}{l} The given function is g (x) = x - [x] \\ Since, g is defined at all integral points . \\ Let n be an integer . \\ Then, \\ g (n) = n - [n] = n - n = 0 \\ The left hand limit of f at x = n is, \\ \lim_{x \to n^{-}} g (x) = \lim_{x \to n^{-}} (x - [x]) \\ = \lim_{x \to n^{-}} x - \lim_{x \to n^{-}} [x] \\ = n - (n - 1) \\ = 1 \\ The right hand limit of f at x = n is, \\ \lim_{x \to n^{+}} g (x) = \lim_{x \to n^{+}} (x - [x]) \\ = \lim_{x \to n^{+}} x - \lim_{x \to n^{+}} [x] \\ = n - n \\ = 0 \\ ∴ \lim_{x \to n^{-}} g (x) \neq \lim_{x \to n^{+}} g (x) \\ Therefore, f is not continuous at x = n . \\ Hence, g is discontinuous at all integral points . \end{array}$

Q.20

\begin{array}{l} Is the function defined by f (x) = x^{2} - sinx + 5 continuous \\ at x = π ? \end{array}

Ans

$\begin{array}{l} The given function is f (x) = x^{2} - 6 sinx + 5 \\ ∵ f is defined at x = π \\ At x = π, \\ f (π) = π^{2} - 6 sinπ + 5 \\ = π^{2} - 6 (0) + 5 \\ = π^{2} + 5 \\ \lim_{x \to π} f (x) = \lim_{x \to π} (x^{2} - 6 sinx + 5) \\ Put x \to π + h \\ If x \to π, then h \to 0 \\ ∴ \lim_{x \to π} f (x) = \lim_{x \to π} (x^{2} - 6 sinx + 5) \\ = \lim_{h \to 0} {{(h + π)}^{2} - 6 \sin (h + π) + 5} \\ = \lim_{h \to 0} {(h + π)}^{2} - 6 \lim_{h \to 0} \sin (h + π) + \lim_{h \to 0} 5 \\ = \lim_{h \to 0} {(h + π)}^{2} - 6 \lim_{h \to 0} (sinhcosπ + sinπcosh) + 5 \\ = {(0 + π)}^{2} - 6 \lim_{h \to 0} (\sinh \times - 1 + 0 \times \cosh) + 5 \\ = π^{2} + 6 \sin 0 + 5 \\ = π^{2} + 5 \\ ∴ \lim_{x \to π} f (x) = f (π) \\ Therefore, the given function f is continuous at x = π . \end{array}$

Q.21 Discuss the continuity of the following functions:
(a) f(x) = sin x + cos x
(b) f(x) = sin x − cos x
(c) f(x) = sin x . cos x

Ans

$\begin{array}{l} If two functions (f and g) are continuous then their sum, \\ difference and product (i . e ., f + g, f - g and f . g) are also \\ continuous . \\ Let us prove that g (x) = sinx and h (x) = cosx are continuous \\ functions . \\ It is clear that g (x) = \sin x is defined for every real number . \\ Let c be a real number . Put x = c + h \\ If x \to c, then h \to 0 \\ g (c) = sinc and \\ \lim_{x \to c} g (x) = \lim_{x \to c} sinx = \lim_{h \to 0} \sin (c + h) \\ = \lim_{h \to 0} (sinccosh + sinhcosc) \\ = sinccos 0 + \sin 0 cosc \\ = sinc \times 1 + 0 \times cosc \\ = sinc \\ ∴ \lim_{x \to c} g (x) = g (c) \\ Therefore, g is a continuous function . \\ Let h (x) = \cos x \\ It is also evident that h (x) = \cos x is defined for every \\ real number . \\ Let c be a real number . Put x = c + h \\ If x \to c, then h \to 0 \\ h (c) = cosc \\ \lim_{x \to c} h (x) = \lim_{x \to c} cosx = \lim_{h \to 0} \cos (c + h) \\ = \lim_{h \to 0} (cosccosh - sinhsinc) \\ = cosccos 0 + \sin 0 sinc \\ = cosc \times 1 + 0 \times sinc \\ = cosc \\ ∴ \lim_{x \to c} h (x) = h (c) \\ Therefore, h is a continuous function . \\ Therefore, it can be concluded that \\ (a) f (x) = g (x) + h (x) = \sin x + \cos x is a continuous function \\ (b) f (x) = g (x) - h (x) = \sin x - \cos x is a continuous function \\ (c) f (x) = g (x) \times h (x) = \sin x \times \cos x is a continuous function \end{array}$

Q.22 Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Ans

$\begin{array}{l} Firstly we have to prove that g (x) = sinx and h (x) = \cos x \\ are continuous functions . \\ It is clear that g (x) = \sin x is defined for every real number . \\ Let c be a real number . Put x = c + h \\ If x \to c, then h \to 0 \\ g (c) = sinc and \\ \lim_{x \to c} g (x) = \lim_{x \to c} sinx = \lim_{h \to 0} \sin (c + h) \\ = \lim_{h \to 0} (sinccosh + sinhcosc) \\ = sinccos 0 + \sin 0 cosc \\ = sinc \times 1 + 0 \times cosc \\ = sinc \\ ∴ \lim_{x \to c} g (x) = g (c) \\ Therefore, g is a continuous function . \\ Let h (x) = \cos x \\ It is also evident that h (x) = \cos x is defined for every \\ real number . \\ Let c be a real number . Put x = c + h \\ If x \to c, then h \to 0 \\ h (c) = cosc \\ \lim_{x \to c} h (x) = \lim_{x \to c} cosx = \lim_{h \to 0} \cos (c + h) \\ = \lim_{h \to 0} (cosccosh - sinhsinc) \\ = cosccos 0 + \sin 0 sinc \\ = cosc \times 1 + 0 \times sinc \\ = cosc \\ ∴ \lim_{x \to c} h (x) = h (c) \\ Therefore, h is a continuous function . \\ It can be concluded that, \\ cosecx = \frac{1}{sinx}, sinx \neq 0 is continuous . \\ \Rightarrow cosecx, nx \neq 0 (n \in Z) is continuous . \\ Therefore, cosecant is continuous except at x = nπ, n \in Z . \\ secx = \frac{1}{cosx}, cosx \neq 0 is continuous . \\ \Rightarrow secx, x \neq (2 n + 1) \frac{π}{2} (n \in Z) is continuous . \\ Therefore, secant is continuous except at x = (2 n + 1) \frac{π}{2} (n \in Z) . \\ cotx = \frac{cosx}{sinx}, sinx \neq 0 is continuous . \\ \Rightarrow cotx, x \neq nπ (n \in Z) is continuous . \\ Therefore, cotangent is continuous except at x = nπ, n \in Z . \end{array}$

Q.23

\begin{array}{l} Find the points of discontinuity of f, where \\ f (x) = \{\begin{cases} \frac{sinx}{x}, if x < 0 \\ x +, if x \geq 0 \end{cases} \end{array}

Ans

$\begin{array}{l} The given function f is f (x) = {\begin{cases} \frac{sinx}{x}, if x < 0 \\ x + 1, if x \geq 1 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case I : If c <0, then f (c) = \frac{sinc}{c} and \\ \lim_{x \to c} f (x) = \lim_{x \to c} \frac{sinx}{x} \\ = \frac{sinc}{c} \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < 0 . \\ Case II : If c > 0, then f (c) = c + 1 \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x + 1) \\ = c + 1 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x > 0 . \\ Case III : If c = 0, then \\ f (c) = f (0) \\ = 0 + 1 = 1 \\ The left hand limit of f at x = 0 is, \\ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} \frac{sinx}{x} = 1 \\ The right hand limit of f at x = 0 is, \\ \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} (x + 1) = 1 \\ ∴ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) = f (0) \\ Therefore, f is not continuous at x = 0 \\ From the above observations, it can be concluded that f is \\ continuous at all points of the real line . \\ Thus, f has no point of discontinuity . \end{array}$

Q.24

\begin{array}{l} Determine if f defined by \\ f (x) = \{\begin{cases} x^{2} \sin \frac{}{x}, if x \neq 0 \\ , if x = 0 \end{cases} \\ is a continuous function ? \end{array}

Ans

$\begin{array}{l} The given function f is f (x) = {\begin{cases} x^{2} \sin \frac{1}{x}, if x \neq 0 \\ 0, if x = 1 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case I : If c \neq 0, then f (c) = c^{2} \sin \frac{1}{c} and \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x^{2} \sin \frac{1}{x}) \\ = c^{2} \sin \frac{1}{c} \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x \neq 0 . \\ Case II : If c = 0, then f (0) = 0 \\ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} (x^{2} \sin \frac{1}{x}) \\ Since, - 1 \leq \sin \frac{1}{x} \leq 1, x \neq 0 \\ \Rightarrow - x^{2} \leq x^{2} \sin \frac{1}{x} \leq x^{2} \\ \Rightarrow \lim_{x \to 0} - x^{2} \leq \lim_{x \to 0} x^{2} \sin \frac{1}{x} \leq \lim_{x \to 0} x^{2} \\ \Rightarrow 0 \leq \lim_{x \to 0} x^{2} \sin \frac{1}{x} \leq 0 \\ \Rightarrow \lim_{x \to 0} x^{2} \sin \frac{1}{x} = 0 \\ ∴ \lim_{x \to 0^{-}} f (x) = 0 \\ Similarly, \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} (x^{2} \sin \frac{1}{x}) \\ = \lim_{x \to 0} x^{2} \sin \frac{1}{x} = 0 \\ ∴ \lim_{x \to 0^{-}} f (x) = f (0) = \lim_{x \to 0^{+}} f (x) \\ Therefore, f is continuous at x = 0 \\ From the above observations, it can be concluded that f is \\ continuous at every point of the real line . \\ Thus, f is a continuous function . \end{array}$

Q.25

\begin{array}{l} Examine the continuity of f, where f is defined by \\ f (x) = \{\begin{cases} sinx - cosx, if x \neq 0 \\ -, if x = 0 \end{cases} \end{array}

Ans

$\begin{array}{l} The given function f is f (x) = {\begin{cases} sinx - cosx, if x \neq 0 \\ - 1, if x = 1 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ CaseI : If c \neq 0, then f (c) = sinc - cosc \\ \lim_{x \to c} f (x) = \lim_{x \to c} (sinx - cosx) = sinc - cosc \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x \neq 0. \\ CaseII : \\ If c =0, then f (0) = - 1 \\ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} (sinx - cosx) \\ = \sin 0 - \cos 0 \\ = 0 - 1 = - 1 \\ \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} (sinx - cosx) \\ = \sin 0 - \cos 0 \\ = 0 - 1 = - 1 \\ ∴ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) = f (0) \\ Therefore, f is continuous at x = 0 \\ From the above observations, it can be concluded that f is \\ continuous at every point of the real line . \\ Thus, f is a continuous function . \end{array}$

Q.26 Find the values of k so that the function f is continuous at the indicated point

$f(x)={\begin{array}{l} \frac{k cosx}{π - 2 x}, if x \neq \frac{π}{2} \\ 3, if x = 0 \end{array} atx = \frac{π}{2}$

Ans

$\begin{array}{l} The given function f is f (x) = {\begin{cases} \frac{kcosx}{π - 2 x}, if x \neq \frac{π}{2} \\ 3, if x = \frac{π}{2} \end{cases} \\ The given function f is continuous at x = \frac{π}{2}, if f is defined at \\ x = \frac{π}{2} and if the value of the f at x = \frac{π}{2} equals the limit of f at x = \frac{π}{2} . \\ Since, f is defined at x = \frac{π}{2} and f (\frac{π}{2}) = 3 \\ \lim_{x \to \frac{π}{2}} f (x) = \lim_{x \to \frac{π}{2}} \frac{kcosx}{π - 2 x} \\ Let x = \frac{π}{2} + h, then x \to \frac{π}{2} \Rightarrow h \to 0 \\ ∴ \lim_{x \to \frac{π}{2}} f (x) = \lim_{x \to \frac{π}{2}} \frac{kcosx}{π - 2 x} \\ = \lim_{h \to 0} \frac{kcos (\frac{π}{2} + h)}{π - 2 (\frac{π}{2} + h)} \\ = \lim_{h \to 0} \frac{- ksinh}{- 2 h} \\ = k \lim_{h \to 0} \frac{\sinh}{2 h} \\ = k \times \frac{1}{2} \\ = \frac{k}{2} \\ ∴ \lim_{x \to \frac{π}{2}} f (x) = f (\frac{π}{2}) \\ \Rightarrow \frac{k}{2} = 3 \\ \Rightarrow k = 6 \\ Therefore, the required value of k is 6 . \end{array}$

Q.27

\begin{array}{l} Find the values of k so that the function f is continuous at the indicated point . \\ f (x) = \{\begin{cases} x^{2}, if x \leq \\ , if x > 2 \end{cases} at x =2 \end{array}

Ans

$\begin{array}{l} The given function f is f (x) = {\begin{cases} {kx}^{2}, if x \leq 2 \\ 3, if x > 2 \end{cases} \\ The given function f is continuous at x = 2, if f is defined at \\ x =2 and if the value of f at x =2 equals the limit of f at x =2 . \\ It is clear that f is defined at x = 2 and f (2) = k {(2)}^{2} = 4 k \\ \lim_{x \to 2^{-}} f (x) = \lim_{x \to 2^{+}} f (x) = f (2) \\ \Rightarrow \lim_{x \to 2^{-}} ({kx}^{2}) = \lim_{x \to 2^{+}} (3) = 4 k \\ \Rightarrow k \times 2^{2} = 3 = 4 k \\ \Rightarrow 4 k = 3 \\ \Rightarrow k = \frac{3}{4} \\ Therefore, the required value of k is \frac{3}{4} . \end{array}$

Q.28 Find the values of k so that the function f is continuous at the indicated point

$f(x)={\begin{array}{l} k x + 1, if x \leq π \\ cosx, if x > π \end{array} atx = π$

Ans

$\begin{array}{l} The given function f is f (x) = {\begin{cases} kx + 1, if x \leq π \\ cosx, if x > π \end{cases} \\ The given function f is continuous at x = π, then \\ L . H . L . = R . H . L . = f (π) \\ It is clear that f is defined at x = π and \\ f (π) = k (π) + 1 = πk + 1 \\ \lim_{x \to π^{-}} f (x) = \lim_{x \to π^{+}} f (x) = f (π) \\ \Rightarrow \lim_{x \to π^{-}} (kx + 1) = \lim_{x \to π^{+}} (cosx) = 4 π + 1 \\ \Rightarrow kπ + 1 = cosπ = 4 π + 1 \\ \Rightarrow kπ + 1 = - 1 = 4 π + 1 \\ \Rightarrow k = - \frac{2}{π} \\ Therefore, the required value of k is - \frac{2}{π} . \end{array}$

Q.29 Find the values of k so that the function f is continuous at the indicated point

$f(x)={\begin{array}{l} k x + 1, i f x \leq 5 \\ 3 x - 5, i f x > 5 \end{array} a t x = 5$

Ans

$\begin{array}{l} The given function f is f (x) = {\begin{cases} kx + 1, if x \leq 5 \\ 3 x - 5, if x > 5 \end{cases} \\ The given function f is continuous at x = 5, then \\ L . H . L . = R . H . L . = f (π) \\ It is clear that f is defined at x = 5 and \\ f (5) = 5 k + 1 \\ \lim_{x \to 5^{-}} f (x) = \lim_{x \to 5^{+}} f (x) = f (5) \\ \Rightarrow \lim_{x \to 5^{-}} (kx + 1) = \lim_{x \to 5^{+}} (3 x - 5) = 5 k + 1 \\ \Rightarrow 5 k + 1 = 15 - 5 = 5 k + 1 \\ \Rightarrow 5 k + 1 = 10 \\ \Rightarrow k = \frac{9}{5} \\ Therefore, the required value of k is \frac{9}{5} . \end{array}$

Q.30

\begin{array}{l} Find the values of a and b such that the function defined by \\ f (x) = \{\begin{cases} , if x \leq \\ ax + b, if < x < \\ , if x \geq \end{cases} \\ is a continuous function . \end{array}

Ans

$\begin{array}{l} The given function f is f (x) = {\begin{cases} 5, if x \leq 2 \\ a x + b, if 2 < x < 10 \\ 21, i f x \geq 10 \end{cases} \\ It is evident that the given function f is defined at all points \\ of the real line .If f is a continuous function, then f is continuous \\ at all real numbers . \\ S o, f is continuous at x = 2 and x = 10 . \\ Since f is continuous at x = 2, we obtain \\ \lim_{x \to 2^{-}} f (x) = \lim_{x \to 2^{+}} f (x) = f (2) \\ \Rightarrow \lim_{x \to 2^{-}} (5) = \lim_{x \to 2^{+}} (a x + b) = 5 \\ \Rightarrow 5 = 2 a + b = 5 \\ \Rightarrow 2 a + b = 5 … (i) \\ Since f is continuous at x = 10, we obtain \\ \lim_{x \to 10^{-}} f (x) = \lim_{x \to 10^{+}} f (x) = f (10) \\ \Rightarrow \lim_{x \to 10^{-}} (a x + b) = \lim_{x \to 2^{+}} (21) = 21 \\ \Rightarrow 10 a + b = 21 = 21 \\ \Rightarrow 10 a + b = 21 … (i i) \\ On subtracting equation (i) from equation (ii), we obtain \\ 8a = 16 \Rightarrow a = \frac{16}{8} = 2 \\ B y putting a = 2 in equation (ii), we get \\ 10 (2) + b = 21 \\ \Rightarrow b = 21 - 20 = 1 \\ Therefore, the values of a and b for which f is a continuous \\ function are 2 and 1 respectively . \end{array}$

Q.31 Show that the function defined by f(x) = cos(x²) is a continuous function.

Ans

$\begin{array}{l} The given function is f (x) = \cos (x^{2}) \\ This function f is defined for every real number and f can be \\ written as the composition of two functions as, \\ f = gοh, where g (x) = \cos x and h (x) = x^{2} \\ [∵ (gοh) (x) = g (h (x)) = g (x^{2}) = {cosx}^{2}] \\ First we have to prove that g (x) = \cos x and h (x) = x^{2} are \\ continuous functions . \\ It is evident that g is defined for every real number . \\ Let c be a real number . \\ Then, g (c) = \cos c \\ Put x = c + h \\ If x \to c, then h \to 0 \\ \lim_{x \to c} g (x) = \lim_{x \to c} cosx \\ = \lim_{h \to 0} \cos (c + h) \\ = \lim_{h \to 0} (cosccosh - sincsinh) \\ = \lim_{h \to 0} cosccosh - \lim_{h \to 0} sincsinh \\ = cosccos 0 - sincsin 0 \\ = cosc \times 1 - sinc \times 0 \\ = cosc \\ ∴ \lim_{x \to c} g (x) = cosc \\ Therefore, g (x) = \cos x is continuous function . \\ h (x) = x^{2} \\ Clearly, h is defined for every real number . \\ Let k be a real number, then h (k) = k^{2} \\ \lim_{x \to c} h (x) = \lim_{x \to c} x^{2} = k^{2} \\ \lim_{x \to c} h (x) = h (k) \\ Therefore, h is a continuous function . \\ It is known that for real valued functions g and h, such that \\ (goh) is defined at c, if g is continuous at c and if f is continuous \\ at g (c), then (f o g) is continuous at c . \\ Therefore, f (x) = (gοh) (x) = \cos (x^{2}) is a continuous function . \end{array}$

Q.32 Show that the function defined by f(x) = |cos x| is a continuous function.

Ans

$\begin{array}{l} The given function is f (x) = | cosx | \\ This function f is defined for every real number and f can be \\ written as the composition of two functions as, \\ f = gοh, where g (x) = | x | and h (x) = cosx \\ [∵ (gοh) (x) = g (h (x)) = g (cosx) = | cosx |] \\ First we have to prove that g (x) = | x | and h (x) = cosx are \\ continuous functions . \\ g (x) = | x | = {\begin{cases} x, if x < 0 \\ - x, if x \geq 0 \end{cases} \\ Clearly, g is defined for all real numbers . \\ Let c be a real number . \\ CaseI : \\ If c <0, then g (c) = - c \\ and \lim_{x \to c} g (x) = \lim_{x \to c} (- x) = - c \\ ∴ \lim_{x \to c} g (x) = g (c) \\ Therefore, g is continuous at all points x, such that x < 0 . \\ CaseII : \\ If c >0, then g (c) = c \\ and \lim_{x \to c} g (x) = \lim_{x \to c} (x) = c \\ ∴ \lim_{x \to c} g (x) = g (c) \\ Therefore, g is continuous at all points x, such that x > 0 . \\ CaseIII : \\ If c = 0, then g (c) = g (0) = 0 \\ \lim_{x \to 0^{-}} g (x) = \lim_{x \to 0^{-}} (- x) = 0 \\ \lim_{x \to 0^{+}} g (x) = \lim_{x \to 0^{+}} (x) = 0 \\ ∴ \lim_{x \to 0^{-}} g (x) = \lim_{x \to 0^{+}} g (x) = g (0) \\ Therefore, g is continuous at x = 0 \\ From the above three observations, it can be concluded \\ that g is continuous at all points . \\ Since, h (x) = \cos x is defined for every real number . \\ Let c be a real number . Put x = c + h \\ If x \to c, then h \to 0 \\ h (c) = \cos c \\ \lim_{x \to c} h (x) = \lim_{x \to c} cosx \\ = \lim_{h \to 0} \cos (c + h) \\ = \lim_{h \to 0} (cosccosh - sincsinh) \\ = \lim_{h \to 0} cosccosh - \lim_{h \to 0} sincsinh \\ = cosccos 0 - sincsin 0 \\ = cosc \times 1 - sinc \times 0 \\ = cosc \\ ∴ \lim_{x \to c} h (x) = cosc \\ Therefore, h (x) = \cos x is a continuous function . \\ It is known that for real valued functions g and h, such that \\ (gοh) is defined at c, if g is continuous at c and if f is continuous \\ at g (c), then (f o g) is continuous at c . \\ Therefore, f (x) = (gοh) (x) \\ = g (h (x)) \\ = g (cosx) = | cosx | is continuous function . \end{array}$

Q.33 Examine that sin|x| is a continuous function.

Ans

$\begin{array}{l} The given function is f (x) = \sin | x | \\ This function f is defined for every real number and f can be \\ written as the composition of two functions as, \\ f = gοh, where g (x) = | x | and h (x) = sinx \\ [∵ (gοh) (x) = g (h (x)) = g (| x |) = \sin | x | = f (x)] \\ First we have to prove that g (x) = | x | and h (x) = sinx are \\ continuous functions . \\ g (x) = | x | = {\begin{cases} x, if x < 0 \\ - x, if x \geq 0 \end{cases} \\ Clearly, g is defined for all real numbers . \\ Let c be a real number . \\ CaseI : \\ If c <0, then g (c) = - c \\ and \lim_{x \to c} g (x) = \lim_{x \to c} (- x) = - c \\ ∴ \lim_{x \to c} g (x) = g (c) \\ Therefore, g is continuous at all points x, such that x < 0 . \\ CaseII : \\ If c >0, then g (c) = c \\ and \lim_{x \to c} g (x) = \lim_{x \to c} (x) = c \\ ∴ \lim_{x \to c} g (x) = g (c) \\ Therefore, g is continuous at all points x, such that x > 0 . \\ CaseIII : \\ If c = 0, then g (c) = g (0) = 0 \\ \lim_{x \to 0^{-}} g (x) = \lim_{x \to 0^{-}} (- x) = 0 \\ \lim_{x \to 0^{+}} g (x) = \lim_{x \to 0^{+}} (x) = 0 \\ ∴ \lim_{x \to 0^{-}} g (x) = \lim_{x \to 0^{+}} g (x) = g (0) \\ Therefore, g is continuous at x = 0 \\ From the above three observations, it can be concluded \\ that g is continuous at all points . \\ Since, h (x) = sinx is defined for every real number . \\ Let c be a real number . Put x = c + h \\ If x \to c, then h \to 0 \\ h (c) = sinc \\ \lim_{x \to c} h (x) = \lim_{x \to c} sinx \\ = \lim_{h \to 0} \sin (c + h) \\ = \lim_{h \to 0} (sinccosh - coscsinh) \\ = \lim_{h \to 0} sinccosh - \lim_{h \to 0} coscsinh \\ = sinccos 0 - coscsin 0 \\ = sinc \times 1 - cosc \times 0 \\ = sinc \\ ∴ \lim_{x \to c} h (x) = sinc \\ Therefore, h (x) is a continuous function . \\ It is known that for real valued functions g and h, such that \\ (gοh) is defined at c, if g is continuous at c and if f is continuous \\ at g (c), then (f o g) is continuous at c . \\ Therefore, f (x) = (gοh) (x) \\ = g (h (x)) \\ = g (sinx) = | sinx | is continuous function . \end{array}$

Q.34 Find all the points of discontinuity of f defined by f(x) = |x| – |x+1|.

Ans

$\begin{array}{l} The given function is f (x) = | x | - | x + 1 | \\ The two functions, g and h, are defined as \\ g (x) = | x | and h (x) = | x + 1 | \\ Then, f = g - h \\ The continuity of g and h is checked first . \\ g (x) = | x | = {\begin{cases} x, if x < 0 \\ - x, if x \geq 0 \end{cases} \\ Clearly, g is defined for all real numbers . \\ Let c be a real number . \\ CaseI : \\ If c <0, then g (c) = - c \\ and \lim_{x \to c} g (x) = \lim_{x \to c} (- x) = - c \\ ∴ \lim_{x \to c} g (x) = g (c) \\ Therefore, g is continuous at all points x, such that x < 0 . \\ CaseII : \\ If c >0, then g (c) = c \\ and \lim_{x \to c} g (x) = \lim_{x \to c} (x) = c \\ ∴ \lim_{x \to c} g (x) = g (c) \\ Therefore, g is continuous at all points x, such that x > 0 . \\ CaseIII : \\ If c = 0, then g (c) = g (0) = 0 \\ \lim_{x \to 0^{-}} g (x) = \lim_{x \to 0^{-}} (- x) = 0 \\ \lim_{x \to 0^{+}} g (x) = \lim_{x \to 0^{+}} (x) = 0 \\ ∴ \lim_{x \to 0^{-}} g (x) = \lim_{x \to 0^{+}} g (x) = g (0) \\ Therefore, g is continuous at x = 0 \\ From the above three observations, it can be concluded \\ that g is continuous at all points . \\ h (x) = | x + 1 | can be written as \\ h (x) = {\begin{cases} - (x + 1), if x < - 1 \\ x + 1, if x \geq - 1 \end{cases} \\ Since, h is defined for every real number . \\ Let c be a real number . \\ CaseI : \\ If c < - 1, then h (c) = - (c + 1) \\ and \lim_{x \to c} h (x) = \lim_{x \to c} [- (x + 1)] \\ = - (c + 1) \\ ∴ \lim_{x \to c} h (x) = h (c) \\ Therefore, h is continuous at all points x, such that x < - 1. \\ CaseII : \\ Ifc > - 1, thenh (c) = c + 1 and \\ \lim_{x \to c} h (x) = \lim_{x \to c} [(x + 1)] \\ = (c + 1) \\ ∴ \lim_{x \to c} h (x) = h (c) \\ Therefore, h is continuous at all points x, such that x > - 1 \\ CaseIII : \\ Ifc = - 1, thenh (c) = h (- 1) = - 1 + 1 = 0 \\ \lim_{x \to - 1^{-}} h (x) = \lim_{x \to - 1^{-}} [- (x + 1)] = - (- 1 + 1) = 0 \\ \lim_{x \to - 1^{+}} h (x) = \lim_{x \to - 1^{+}} [(x + 1)] = (- 1 + 1) = 0 \\ ∴ \lim_{x \to - 1^{-}} h (x) = \lim_{x \to - 1^{+}} h (x) = h (- 1) \\ Therefore, h is continuous at x = - 1 \\ From the above three observations, it can be concluded \\ that h is continuous at all points of the real line . \\ g and h are continuous functions . Therefore, f = g - h \\ is also a continuous function . \\ Therefore, f has no point of discontinuity . \end{array}$

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NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1

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NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1

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