NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 (Ex 1.1)
The importance of Mathematics scores in Class 12 cannot be underestimated. Students must prepare from the NCERT textbook to perform well. Having access to the NCERT Solutions of various chapters including the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 plays a key role in scoring high marks.
Students must focus on the NCERT textbooks. These textbooks contain several questions that are relevant for examinations and essential for practice. These questions can be confusing, making the subject challenging to understand. It is crucial to have a strong understanding of the concepts included in the subject. Having access to the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 help students in understanding questions given in the NCERT textbooks while having a ready solution for every practised question.
Also, Class 12 is a crucial year as it lays the foundation for the future academic careers of students. Students of both the science and commerce streams, need to obtain outstanding scores to pursue their desired professions. Therefore, it is critical to score well in Mathematics. Students must study from the NCERT textbooks multiple times to get used to the format of the questions that can appear. Students must aim to solve the textbook questions early on in their practice schedules. They can start from the first chapter and then move on to consecutive ones. The NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 are available on the Extramarks website to help with the preparation.
Aspirants of competitive and entrance examinations begin their preparation in advance to complete their board exams with high scores. This can help them get admission to their preferred colleges. Furthermore, they should check the NCERT solutions for various chapters, including the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1, which are readily available on the Extramarks website to retain what they have practiced, after revising from the textbook questions.
Studying for multiple hours every day, trying to remember all the important concepts, and remembering all the difficult calculus formulae takes a lot of hard work. Students work tirelessly for hours on end to revise. They should also study regularly from the NCERT textbook and solve the NCERT questions at the end of each chapter. The NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 and other chapters, are available via Extramarks and can be used for selfanalysis and improvement by the students.
Students must strive to give their best, and to achieve this, they can utilise multiple learning resources. They must solve every textbook question. They should also practice the NCERT Mathematics textbook multiple times to understand fundamental concepts. The NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1, available on the Extramarks’ website, provides detailed solutions for Exercise 1.1.
NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 (Ex 1.1)
NCERT is known for its role in developing the syllabus and putting together textbooks for students at the primary and secondary levels. These books focus on fundamental concepts to help students understand the basic studying resources. NCERT sets the curriculum for all schools that are affiliated with CBSE. The NCERT books are a significant part of the curriculum and are necessary to prepare for the board exams. Extramarks provides multiple readily available learning archives for students such as the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1. Other learning material that is accessible online via Extramarks is given below:
NCERT Solutions Class 12
NCERT Solutions Class 11
NCERT Solutions Class 10
NCERT Solutions Class 9
NCERT Solutions Class 8
NCERT Solutions Class 7
NCERT Solutions Class 6
NCERT Solutions Class 5
NCERT Solutions Class 4
NCERT Solutions Class 3
NCERT Solutions Class 2
NCERT Solutions Class 1
Class 12 is a significant academic year for high school students. If they do not revise fundamental concepts which are part of the Class 12 Mathematics syllabus, they can encounter problems while studying the subject in the future. The NCERT books serve as the best reading material for students to gain comprehensive knowledge of the subject. It also aids in revision for multiple examinations that one might want to succeed in, later on. Aspirants of such examinations must solve the problems in the NCERT textbooks. These textbook solutions for all classes, and specifically for the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1, are available on the Extramarks website which provides authentic material.
Extramarks makes available learning resources which provide reliable and detailed solutions. Students can easily access these solutions whenever they want, from anywhere. Extramarks offers interactive virtual lessons, doubt sessions, live classes and much more, to guide students in their preparation for various exams. This helps students learn quickly and efficiently.
Various solutions for NCERT textbooks are made accessible on the Extramarks website for students, ranging from material for primary and middle school to high school. Class 8 covers a variety of subjects like Hindi, English, Mathematics, Science, and Social Science. The NCERT Solutions Class 8, similar to the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1, helps students not only achieve good scores but will also help students compete in exams such as National Olympiads, NTSE etc. Apart from this, students can also take guidance from Extramarks to clarify the doubts in their problem areas. The resources available on the Extramarks’ website are created by subjectmatter experts that deliver authentic and meticulous solutions. Learning, practice, and testing is the way that guarantees result for students.
Class 9 also forms an essential academic year of secondary education. This year lays the groundwork for the years to follow. Students who can successfully manage to prepare for the Mathematics syllabus are bound to find the Class 10 syllabus much easier. Extramarks presents the NCERT Solutions Class 9, like the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1, to build a solid foundational and conceptual understanding.
Class 10 is an important year in a student’s life. Class 10 plays a key role in helping students decide on the subject stream that they want to pursue later during their high school. Students should focus on the NCERT books to score well in the board exams and get their desired stream. Extramarks provides the NCERT Solutions For Class 10, just like the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1, to prepare for the board exams and successfully revise topics that could appear in the exam. Extramarks also offers the option of selfassessment via their learning resources, which is an absolute necessity for senior classes, like Class 10.
Students in Class 11 can be admitted into a number of streams. Subjects like Mathematics, Physics, Chemistry, English, Commerce, Business Studies, Computer Science etc. comprise the multitude of options available to the students. Extramarks provides the NCERT Solutions Class 11 which helps students not only secure a good rank but also aids them to prepare for engineering and medical competitive exams. The Extramarks’ website also provides the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 for the students to practice and score well in the exams.
Class 12, as the last year of a student’s school life, is a vital year for students to showcase their academic performances. It is a crucial and critical year that lays the foundation for a student’s career. Students must study hard as securing high scores becomes a priority, and admission to colleges is often based on these scores. Students must study according to a revision schedule. They should practice daily to prepare for their final year exams. They should make a study plan and subsequently, solve the NCERT textbook problems. Particularly for Mathematics, they should prepare a revision schedule as the subject can be challenging to prepare at the last moment. Students could compare their answers for Mathematics chapters, specifically for Chapter 1 Exercise 1.1, with the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1, available via Extramarks. They can also access solutions to other chapters to selfassess their preparation and work on their doubts.
It is critical for students to study, practice, and revise their NCERT books. Extramarks offers the NCERT Solutions for Class 12, for every subject. Further, practising the NCERT solutions, such as the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1, prepares students for revising with the help of such authentic and detailed solutions. These solutions are prepared by experts on the Extramarks’ website and provide detailed and stepbystep solutions for each question in the NCERT textbook. The answers are written in easy language that students can understand without any confusion.
What Are The Topics Covered?
The NCERT syllabus for Class 12 Mathematics is divided into 5 units which are Relations and Functions, Algebra, Calculus, Vectors and ThreeDimensional Geometry, Linear Programming and Geometry.
Students must practice every unit to increase their selfconfidence to do well in board exams. Confidence plays an important role in the performance of students in exams. Practising with the NCERT textbook questions and having access to exercises, such as the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1, gives students a complete guide to practice and selfevaluation. The solutions given by experts are crosschecked for accuracy by multiple experts, making sure that the students’ knowledge base remains accurate and foolproof.
Furthermore, students should prepare thoroughly and not let the following mistakes reduce their chances of scoring well in examinations:
1) Studying for Long Durations
The most common mistake students commit while preparing for their exams, is studying at a massive stretch. Topics such as Relations and Functions require focus and concentration, so they should avoid studying for long hours. If students keep studying continuously, without breaks, they might encounter problems in retaining the topics they’re practising. This is why they’re advised to take breaks in between and study more efficiently. They should practice the NCERT exercises, take a break, and then compare their answers with the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 and other such chapters’ solutions, to retain the methods for solving multiple problemspecific exercises.
2) Lack of Writing Practice
The Class 12 Mathematics exam is considered to be quite long. If students encounter difficulties with a particular question, they may face a lack of time in being able to finish their exam. To address this issue, they need to practice their answers regularly. The NCERT textbook exercises aid students in inculcating the habit of writing practice. Students should start by solving the NCERT Mathematics books. Solutions for exercises, such as the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1, are available via the Extramarks experts, with whose guidance students can easily solve various types of questions.
3) Inability to Understand the Question Paper Properly and Writing Unplanned Answers
Students often make the mistake of not reading the entire question paper thoroughly and hence, take time in planning how to structure their answers. For a subject like Mathematics, mathematical logic and understanding of the variety of questions, play a major part in scoring high marks.
Having access to the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 helps the student to structure their answers. Furthermore, it helps them in learning how not to hurry as they begin to write their answers. Reading the questions carefully, understanding them, and then composing answers, seems like the proper way to solve the questions methodically. Relations and Functions can be a challenging topic and the ability to properly understand the question primarily helps in arriving at the correct answer.
Solving the NCERT books and keeping them as a benchmark for proper solutions could aid students in framing answers that are detailed and credible. The NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1, and various other chapters, are made accessible for the students through the Extramarks’ website, where solutions are prepared by the experts.
4) Not Solving the Questions using Stepwise Formulae
Students must always ensure that they use proper stepwise formulas and techniques, to attain a good score in that particular question. Formulas form a major portion of mathematical concepts and are required for solving different types of questions. They should be practised multiple times by students to ensure that they do not fail to solve questions due to a lack of practising them. Students require meticulous practice to make informative and neat diagrams. Students need to practice all such formulae multiple times to perform their best and not commit mistakes during their final exam.
In a topic like Relations and Functions, stepwise marks are allotted by the examiners. Skipping the steps of formulae might result in examiners deducting marks for the same. Students must habituate themselves to practising from the NCERT books. The NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 and other exercises provide them with a proper understanding of writing a stepwise detailed answer to score full marks.
5) Time Mismanagement
Time management skills are required for students to complete their exam preparation in time. The Mathematics examination for Class 12 requires students to allot time to each section, in accordance with the marks distribution. Thorough practice of the NCERT textbook exercises aids students to solve the exam paper in the stipulated time. The NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 and other solutions available on the Extramarks’ website, help students to structure their answers and manage time while composing answers and thus, achieve a high score.
6) Last Minute Revision
Applying the wrong formula or the inability to remember the correct formula in a particular sequence often results in students being unable to frame the right answers. Losing their focus and concentration is typically a result of hurried preparation. If students prepare their subjects early, they will not have to revise multiple topics together at the last moment, and thus, be able to remember and compose answers effectively.
The topic of Relations and Functions requires students to have a good understanding of the concept to avoid the need to cram last minute. The NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 provided by Extramarks helps students to get better at such topical preparation as students get to practise with specific and detailed solutions for their topics and thus, score well.
7) Submitting the Paper without Revising
Students also commit mistakes while submitting their final answer papers. Revising answer sheets plays a key part in avoiding mistakes while trying to score well. Students must practice daily NCERT textbooks to improve their writing speed. Students must practice these exercises and compare their results with the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 and other solutions for selfassessment and improvement.
What is Relation in Mathematics?
Before deep diving into Relations in Mathematics of Class 12, a clear, crisp, and sound understanding is necessary to get good at Relations and Functions. Now, students should know what is a set. A set is a welldefined collection of objects whose elements cannot vary and are fixed. NCERT solutions of Class 11 of Sets and Functions and NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 help students know the concept better to perform well in exams.
Having access to NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 helps students in understanding the questions given in the NCERT textbooks while having a clear solution and approach to every question.
The word relation is drawn from the English language, which means the relationship between two individuals, bodies, and people. It means a connection between two things. It means there is a recognizable relation between the two. The connection is formed if there is a connection between elements between two or more empty sets.
To get a better understanding of these topics, questions from NCERT books should be practised in Class 11. The NCERT solutions for Class 11 and NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 offered by Extramarks, delivers genuine content. Students can avail this material to get top scores in examinations. The concepts of the chapter can be confusing and challenging at times, but having NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 solutions for students and their needs can be a good learning tool.
Knowledge of various Types of Sets further aids preparation skills:
1) Empty Sets
The set which contains no elements or null elements is called an empty set. It is also known as a Null set or Void Set.
2) Singleton Set
A set containing one element is called a singleton set.
3) Finite and Infinite Sets
A set with a finite number of elements is called a finite element. Those sets whose number of elements cannot be estimated, but it has some huge figure or number, which is very large to express in a set.
The number of States in India is a finite set as there is a finite number of states, whereas the number of animals in India is an infinite set as it has infinite elements in India. Make use of NCERT books to get better at it. Also, NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 gives students detailed explanations to get better at core understanding.
4) Equal Sets
If every element of set A is also an element of set B and if every element of set B is also an element of set A, then two sets A and B are said to be equal. It means A and B have the same elements. Go through the NCERT books of Class 11 again to get a feel of it. Also, NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 provided by Extramarks gives you detailed answers.
5) Subsets
A set A is said to be a subset of set B if the elements of set A belong to set B, or you can say each element of set A is present in set B. It is denoted with the symbol of A ⊂ B. To know more about the subset, connect to the Extramarks’ website to get genuine solutions. Extramarks provides NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 to get students ready for the ultimate test.
6) Power Sets
A set of all subsets is known as a power set. An empty set is a subset of all sets, and every set is a subset of itself. NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 guides students about the power sets.
7) Universal Sets
A set which contains all elements of other sets is universal.
Students can get a step by step answers from a trustworthy source such as Extramarks for NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1.
Types of Relations
Different types of relations are covered in this topic. Empty relation, universal relation, reflexive relation, symmetric relation, transitive relation, and equivalent relation.
NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 covers all the above relations.
The relations are explained in detail below:
1) Empty Relation
If there is no relation between the elements of A, then it’s called empty relation. Students can gain guidance from NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 about empty sets.
2) Universal Relation
If each element of A is related to every element of A, then the relation is said to be a universal relation. Students can improve their learnings from NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 about universal relations.
Sometimes empty relations and universal relations are called trivial relations.
3) Reflexive Relation
If every element of the set is related to itself, then the binary relation of A is said to be reflexive relation. Students can gain knowledge from NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 about reflexive relation.
4) Symmetric Relation
Symmetric relation is the relation that if the first element is related to the second, then the second element is also related to the first similarly. E.g. if A=B, then B=A. NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 can assist students in learning about symmetric relations.
5) Transitive Relation
If the first element is related to the second element, the second element is related to the third, then the third element is related to the first element. Students can gain guidance from NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 about transitive relations.
6) Equivalent Relation
The relation is binary, reflexive, and transitive. It has the following properties: they are reflexive A is related to A, symmetricIf A is related to B, B is related to A, and transitive relation if A is related to B, B is related to C, then C is related to A.
Extramarks provides NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 to clear any doubts you have about the above topic. The Extramarks’ website provides the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 which is an example of one such learning module. Students can understand and learn about the solutions and the steps from the beginning while having the scope of practice with the learning modules.
Access NCERT Solutions for Class 12 Maths Chapter 1–Relations and Functions
NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1
The NCERT solutions are easily available online and offline, but students need to be sure of the steps and the formulas used in the solutions to reach the answer. Some platforms might not be reliable to access the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 The chapter can be challenging and the formulas can be confusing. Having a credible source to access the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 can fasttrack the learning and preparation process. Students can access NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 on the Extramarks website. The solutions are curated by academic experts, making them a reliable source of knowledge. The solutions are explained stepwise and align with the answers mentioned in the NCERT textbooks. Students can access the Extramarks website if they want to access the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1
The NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 is an excellent exercise to help students practice the concepts that are explained thoroughly throughout the first chapter. It consists of various questions that help students practice understanding the concept of Relation and Function completely.
NCERT books give you detailed knowledge of a subject. Each chapter is explained in a way that helps them understand the essence of the subject to the core.
There are several benefits to solving the NCERT textbooks. NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1. available through Extramarks helps students, so their preparation remains wellrounded. Further, it helps students in several ways, which are explained below:
1) Indepth understanding of complex topics
Mathematics requires breaking down complex topics into small parts to be able to solve them with ease. NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 help in this regard.
The practice of the NCERT textbook boosts their confidence and takes their understanding of the most difficult topic to another level. Students must compare their answers with NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 for selfassessment and improvement.
2) Follows the CBSE Curriculum
NCERT textbooks follow the CBSE curriculum completely. The majority of the board exams’ questions are picked from the NCERT textbook. If students have practised the exercises rigorously, there are high chances they are going to excel in their board exams. Furthermore, they can check out the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 available via Extramarks. The curriculum and study material on the platform is brought together by the inhouse experts of the industry. The information provided to the students is factchecked and factbased.
3) Prepared by Extramarks’ Experts
NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 is prepared by experts, so there is no doubt regarding the genuineness of the solutions.
They are the best experts selected from all over the country with decades of experience. The NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 are explained in simple language and stepbystep process. Students would be able to understand and excel in these topics in no time.
When in doubt about the authenticity of NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1, students can access the Extramarks website and clarify their doubts in no time. Apart from this, Extramarks also provides live classes to prepare you for exams in no time.
4) Preparation at Multiple Levels
NCERT books prepare you at multiple levels. The students who study NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 and other solutions well are capable of cracking the JEE exam as well.
NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 brings about a positive change in students so that they can crack these tough exams.
Class 12 board exams are designed in such a way as to test the problemsolving skills of the students to the core. It remains one of the most strenuous exams, and whoever can crack this one can clear JEE as well.
5) Important Questions at the End of Each Chapter in NCERT Textbooks
NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 to these important questions are extremely helpful for students.
These include fillintheblanks, true or false, oneword answers, or descriptive questions. Students need to be thorough with these questions since these questions could be a part of CBSE board exam question papers.
Practising NCERT exercises and checking the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 is an excellent way to improve the knowledge of the subject and prepare for the board exams.
6) CBSE Recommends NCERT Textbooks
NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 guides students in preparation for CBSE board exams.
According to CBSE, NCERT textbooks are more than enough to achieve high marks in board exams. To achieve stellar marks, students can go through the NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 provided by Extramarks’ experts to make sure they ace the exams.
NCERT textbooks are a great way to start preparing for the board exams. Students must solve the exercises at the end. NCERT solutions for every exercise and NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 assist students to structure their answers and explain in a detailed manner.
7) Efficient Utilization of Time
The time management skills of students get better with the practice of NCERT books.
Students must practice various NCERT textbooks to get better at it. The biggest reason they fail to solve the exam in the allotted time frame amounts to a lack of practice. The best solution remains to compare their answers with NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1, analyse their mistakes and improve.
Start practising various chapters from the beginning and view NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.1 and other solutions prepared by Extramarks’ experts. It would make certain that students can ace their board exams.
8) Increase in Self Confidence by guidance from NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1
The practice of NCERT textbooks of Class 12 Mathematics remains students’ best friend to boost their confidence while preparing for board exams.
The key to confidence is preparation, and to prepare well, students need to solve the NCERT books daily to give their best in board exams. Confidence leads to students composing better answers in their exams. By solving the NCERT Mathematics textbooks with detailed Class 12 Maths Chapter 1 Exercise 1.1 Solutions available on the Extramarks’ website.
NCERT Solutions for Class 12 Maths Chapter 1 Exercises
Having access to several questions and question banks before the examinations is not enough for students, they also need wellexplained solutions. NCERT textbook questions are the primary contributors to most of the question banks. Having NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 for students and their needs makes the process of solving questions easier. Having a fast mode of understanding and learning makes students more confident for any examination. This also gives students more time to focus on the other aspects of learning while keeping their minds cool. Understanding the concepts of the NCERT solutions can help students move ahead with the preparation. Certain concepts of the chapter can be confusing and challenging at times, but having NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 for students and their needs can be a good learning tool. Utilizing these learning resources can make the subject of Mathematics easy.
The syllabus of Class 12 and that of Class 11 are interlinked with each other. Students need to have their concepts clear from Class 11. It ensures they do not face any problems in the biggest test of their life. Daily practice of NCERT books guarantees concept clearance. NCERT Solutions for Class 11 Maths and NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 provide authentic solutions without having to look elsewhere.
Class 12 Syllabus consists of six units which are Relations and Functions, Algebra, Calculus, Vector and ThreeDimensional Geometry, Linear Programming, and Probability.
Class 11 Syllabus covers I. Sets and Functions, II. Algebra III. Coordinate Geometry IV. Calculus V. Mathematical Reasoning VI. Statistics and Probability.
Both have a strong connection, so the knowledge of Class 11 topics aids in doing great in Class 12. Going through the NCERT books of Class 11 helps students. NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 available through Extramarks provides detailed solutions to the students.
Now, the following points reiterate the connection or differences between the two classes:
 The basics of Relation and Functions are covered in Set and Functions:
A strong percentage of students tend to overlook this chapter in Class 11 and tend to suffer in Class 12. It is better to devote some quality time to understanding these basic concepts. Otherwise, students would be forced to devote more time to Class 12. Students should also practice Trigonometry because it will help in Differentiation and Integration.
NCERT books of Class 11 should be used in this case. NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 and NCERT solutions for Class 11 are available on the Extramarks website. It even provides India’s best teachers to make sure students remember Class 11 concepts and do not need to revise their concepts in Class 12.
 Algebra of Class 11 and 12 is quite different.
The Class 11 chapters in Algebra are Principle of Mathematical Induction, Complex Numbers and Quadratic Equations, Linear Inequalities, Permutation and Combinations, Binomial Theorem, and Sequence and Series whereas Class 12 chapters in Algebra are Matrices and Determinants.
The only similarity remains is that students might need to know the basic concepts of linear inequalities in Vectors. Practice NCERT textbooks and view NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1 and NCERT solutions for Class 12 Algebra to thoroughly understand the technicalities of Algebra concepts.
 Coordinate Geometry and Calculus are vital:
Chapters such as Straight lines and Introduction to 3D form the basis of Vectors and 3D.
Chapters such as Limit and Derivatives are vital for Calculus of Class 12 and students should know that a major portion of the Class 12 syllabus is from Calculus. So be diligen
NCERT Solutions for Class 12 Maths Chapter 1 Exercises
Chapter 1 Relations And Functions Other Exercises  
Exercise 1.2 
12 Questions And Solutions (5 Short Answer, 7 Long Answers)

Exercise 1.3 
14 Questions And Solutions (4 Short Answer, 10 Long Answers)

Exercise 1.4 
13 Questions And Solutions (6 Short Answer, 7 Long Answers)

Q.1 Determine whether each of the following relations are reflexive, symmetric and transitive:
(i ) Relation R in the set A = {1, 2, 3…13, 14}
defined as R = {(x, y): 3x − y = 0}
(ii) Relation R in the set N of natural numbers
defined as R = {(x, y): y = x + 5 and x < 4}
(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as
R = {(x, y): y is divisible by x}
(iv) Relation R in the set Z of all integers defined as
R = {(x, y): x − y is as integer}
(v) Relation R in the set A of human beings in a
town at a particular time given by
(a) R = {(x, y): x and y work at the same place}
(b) R = {(x, y): x and y live in the same locality}
(c) R = {(x, y): x is exactly 7 cm taller than y}
(d) R = {(x, y): x is wife of y}
(e) R = {(x, y): x is father of y}
Ans
i) A = {1, 2, 3 … 13, 14}
R = {(x, y): 3x − y = 0 or y=3x}
∴ R = {(1, 3), (2, 6), (3, 9), (4, 12)}
Given relation R is not reflexive because
(1, 1), (2, 2), (3,3)… (14, 14) ∉ R.
Also, R is not a symmetric relation as
(2, 6) ∈R, but (6,2) ∉ R.
Also, R is not transitive as (1, 3), (3, 9) ∈R,
but (1, 9) ∉ R.
Hence, R is neither reflexive, nor symmetric, nor transitive.
\begin{array}{l}\text{(ii) R}=\text{}\left\{\left(x,y\right):y=x+\text{5 and}x<\text{4}\right\}\\ \text{R}=\text{}\left\{\left(\text{1},\text{6}\right),\text{}\left(\text{2},\text{7}\right),\text{}\left(\text{3},\text{8}\right)\right\}\\ \text{Since}\left(\text{1},\text{1}\right)\notin \text{R}.\\ \therefore \text{R is not reflexive}.\\ \left(\text{1},\text{6}\right)\in \text{R But}\left(\text{6},\text{1}\right)\notin \text{R}.\\ \therefore \text{R is not symmetric}.\\ \text{Now},\text{since there is no pair in R such that}\left(a,\text{}b\right)\text{and}\\ \left(b,c\right)\in \text{R},\text{then}\left(a,c\right)\text{cannot belong to R}.\\ \therefore \text{R is not transitive}.\\ \text{Hence},\text{R is neither reflexive},\text{nor symmetric},\text{nor transitive}.\\ \text{(iii) A = {1, 2, 3, 4, 5, 6}}\\ \text{R = {(x, y): y is divisible by x}}\\ \text{Since every number is divisible by itself}.\\ \therefore \text{(x, x)}\in \text{R}\\ \text{Hence R is reflexive}\text{.}\\ \text{Now,}\\ \text{(2, 4)}\in \text{R [as 4 is divisible by 2]}\\ \text{But,}\\ \text{(4, 2)}\notin \text{R}\text{. [as 2 is not divisible by 4]}\\ \therefore \text{R is not symmetric}\text{.}\\ \text{Let (x, y), (y, z)}\in \text{R}\text{. Then, y is divisible by x and z is divisible}\\ \text{by y}\text{.}\\ \therefore \text{z is divisible by x}\text{.}\\ \text{Hence (x, z)}\in \text{R}\\ \therefore \text{R is transitive}\text{.}\\ \text{Hence, R is reflexive and transitive but not symmetric}\text{.}\\ \text{(iv) R = {(x, y): x}\text{y is an integer}}\\ \text{For every x}\in \text{Z, (x, x)}\in \text{R as x}\text{x = 0 which is an integer}\text{.}\\ \therefore \text{R is reflexive}\text{.}\\ \text{Now, for every x, y}\in \text{Z if (x, y)}\in \text{R, then x}\text{y is an integer}\text{.}\\ \text{For every (y,x)}\in \text{R by definition}\\ \left(yx\right)=\left(xy\right)\text{\hspace{0.17em}}\text{is also an integer}\text{.}\\ \Rightarrow \left(yx\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{is also an integer}\text{.}\\ \therefore \text{(y, x)}\in \text{R}\\ \therefore \text{R is symmetric}\text{.}\\ \text{Now,}\\ \text{Let (x, y) and (y, z)}\in \text{R, where x, y, z}\in \text{Z}\text{.}\\ \Rightarrow \text{(x}\text{y) = integer}\\ \Rightarrow \text{(y}\text{z) = integer}\\ \text{On adding both, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(xz\right)=\text{integer}\text{.}\\ \therefore \left(x,z\right)\in R\\ \therefore \text{R is transitive}\text{.}\\ \text{Hence, R is reflexive, symmetric, and transitive}\text{.}\\ \\ \text{(v) (a) R = {(x, y): x and y work at the same place}}\\ \text{Every individual worker belongs to itself}\\ \Rightarrow \text{(x, x)}\in \text{R}\\ \text{\hspace{0.17em}}\therefore \text{R is reflexive}\text{.}\\ \text{\hspace{0.17em}}\text{If (x, y)}\in \text{R, then x and y work at the same place}\text{.}\\ \text{Similarly y and x work at the same place}\text{.}\\ \therefore \text{(y, x)}\in \text{R}\text{.}\\ \text{Hence R is symmetric}\text{.}\\ \text{Now, let (x, y), (y, z)}\in \text{R}\\ \Rightarrow \text{x and y work at the same place and y and z work at}\\ \text{the same place}\text{.}\\ \therefore \text{x and z also work at the same place}\text{.}\\ \therefore \text{(x, z)}\in \text{R}\\ \therefore \text{R is transitive}\text{.}\\ \text{Hence, R is reflexive, symmetric and transitive}\text{.}\\ \text{(b) R = {(x, y): x and y live in the same locality}}\\ \text{Every individual belongs to itself}\\ \text{Thus (x, x)}\in \text{R}\\ \therefore \text{R is reflexive}\text{.}\\ \text{Since (x, y)}\in \text{R it means x and y live in the same locality}\text{.}\\ \text{Which implies y and x also live in the same locality}\text{.}\\ \therefore \text{(y, x)}\in \text{R}\\ \text{Hence R is symmetric}\text{.}\\ \text{Now, let (x, y)}\in \text{R and (y, z)}\in \text{R}\text{.}\\ \Rightarrow \text{x and y live in the same locality and y and z live in the same}\\ \text{locality}\text{.}\\ \text{Which implies that x and z also live in the same locality}\text{.}\\ \therefore \text{(x, z)}\in \text{R}\\ \text{Hence R is transitive}\text{.}\\ \text{Hence, R is reflexive, symmetric and transitive}\text{.}\\ \text{(c) R = {(x, y): x is exactly 7 cm taller than y}}\\ \xe2\u02c6\mu \text{An individual height can be equal to itself but cannot}\\ \text{be more by 7 cm}\\ \text{Hence (x, x)}\notin \text{R}\\ \therefore \text{R is not reflexive}\text{.}\\ \text{Now, let (x, y)}\in \text{R}\text{.}\\ \Rightarrow \text{x is exactly 7 cm taller than y}\\ \text{Then y should be shorter than x}\\ \therefore \text{(y, x)}\notin \text{R}\\ \therefore \text{R is not symmetric}\text{.}\\ \text{Again,}\\ \text{Let (x, y), (y, z)}\in \text{R}\text{.}\\ \Rightarrow \text{x is exactly 7 cm taller than y and y is exactly 7 cm}\\ \text{taller than z}\text{.}\\ \\ \text{Which implies x is exactly 14 cm taller than z}\text{.}\\ \therefore \text{(x, z)}\notin \text{R}\\ so,\text{R is not transitive}\text{.}\\ \text{Hence, R is neither reflexive nor symmetric nor transitive}\text{.}\\ \text{(d) R = {(x, y): x is the wife of y}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Since x cannot be the wife of herself}\text{.}\\ \therefore \text{(x, x)}\notin \text{R}\\ \text{Then, R is not reflexive}\text{.}\\ \text{Let}\text{\hspace{0.17em}}\left(x,y\right)\in R\\ which\text{implies that x is wife of y}\text{.}\\ i.e.,\text{y is husband of x}\text{.}\\ \text{so, y can not be wife of x}\text{.}\\ \Rightarrow \left(y,x\right)\notin R\\ Then,\text{\hspace{0.17em}}\text{R is not symmetric}\text{.}\\ \text{Let}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{(x, y), (y, z)}\in \text{R}\\ \Rightarrow \text{x is the wife of y and y is the wife of z}\text{.}\\ \text{A husband can never become wife of anyone}.\\ so,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(x,z\right)\notin R\\ Thus,\text{\xe2\u20ac\u2039}\text{R is not transitive}\text{.}\\ \text{Hence, R is neither reflexive nor symmetric nor transitive}\text{.}\\ \text{(e) R = {(x, y): x is the father of y}}\\ \text{Since, x cannot be the father of himself}\text{.}\\ \text{So,}\left(x,x\right)\notin R\\ Then,\text{\hspace{0.17em}}\text{R is not reflexive}\text{.}\\ \text{Now, let (x, y)}\in \text{R}\text{.}\\ \\ \text{If x is the father of y}\text{.}\\ \text{Then y cannot be the father of x}\text{.}\\ \text{so,}\left(y,x\right)\notin R\\ Then,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{R is not symmetric}\text{.}\\ \text{Again, let (x, y)}\in \text{R and (y, z)}\in \text{R}\text{.}\\ \Rightarrow \text{x is father of y and y is father of z}\text{.}\\ \Rightarrow \text{x can not be father of z}\text{.}\\ \Rightarrow \text{x will be grandfather of z}\\ \therefore \left(x,z\right)\notin R\\ \therefore \text{\hspace{0.17em}}R\text{is not transitive}\text{.}\\ \text{Hence, R is neither reflexive nor symmetric not transitive}\text{.}\end{array}
Q.2
$\begin{array}{l}\mathrm{Show}\mathrm{}\mathrm{that}\mathrm{}\mathrm{the}\mathrm{}\mathrm{relation}\mathrm{}\mathrm{R}\mathrm{}\mathrm{in}\mathrm{}\mathrm{the}\mathrm{}\mathrm{set}\mathrm{}\mathrm{R}\mathrm{}\mathrm{of}\mathrm{}\mathrm{real}\mathrm{}\mathrm{numbers},\mathrm{}\\ \mathrm{defined}\mathrm{}\mathrm{as}\mathrm{}\mathrm{R}=\mathrm{}\{\mathrm{}(\mathrm{a},\mathrm{b}):\mathrm{\hspace{0.17em}}\mathrm{a}\le {\mathrm{b}}^{2}\}\mathrm{\hspace{0.17em}}\mathrm{is}\mathrm{neithe}\mathrm{rreflexive}\mathrm{nor}\\ \mathrm{symmetric}\mathrm{nor}\mathrm{transitive}.\end{array}$Ans
\begin{array}{l}\text{We have R = {(a, b): a}\le {b}^{\text{2}}\text{} where a, b}\in \text{R}\\ \text{We can see that}\frac{\text{1}}{3}\le {\left(\frac{\text{1}}{3}\right)}^{\text{2}}\text{is not valid}\text{. So,}\left\{\frac{\text{1}}{3}\text{,}\frac{\text{1}}{3}\right\}\notin R\\ Then,\text{\hspace{0.17em}}\text{R is not reflexive}\text{.}\\ \text{Since,}\left(2,3\right)\in R\text{\hspace{0.17em}}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\text{\xe2\u20ac\u2039}{\text{2<3}}^{\text{2}}\\ But{\text{3}}^{\text{2}}\text{is not less than}\text{2}\text{.}\\ \end{array}
$\begin{array}{l}\mathrm{Thus},\mathrm{R}\mathrm{is}\mathrm{not}\mathrm{symmetric}\mathrm{.}\\ \mathrm{Again},\mathrm{let}(5,6),(6,2)\in \mathrm{R}\\ \mathrm{As}\mathrm{\hspace{0.17em}}5<{(6)}^{2}\mathrm{and}6{<2}^{\mathrm{2}}\\ \mathrm{But}5\mathrm{is}\mathrm{not}\mathrm{less}\mathrm{than}4\mathrm{.}\\ \mathrm{So},(5,2)\notin \mathrm{R}\\ \mathrm{Thus},\mathrm{R}\mathrm{is}\mathrm{not}\mathrm{transitive}\mathrm{.}\\ \mathrm{Therefore},\mathrm{R}\mathrm{is}\mathrm{neither}\mathrm{reflexive}\mathrm{nor}\mathrm{symmetric}\mathrm{nor}\mathrm{transitive}.\end{array}$
Q.3 Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Ans
Given A = {1, 2, 3, 4, 5, 6}. A relation R is defined
on A as: R = {(a, b): b = a + 1}
\begin{array}{l}\therefore \text{R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}}\\ \text{We can find (1,1)}\notin \text{R, where 1}\in \text{A}\text{.}\\ \therefore \text{R is not reflexive}\text{.}\\ \text{It can be observed that (1, 2)}\in \text{R, but (2, 1)}\notin \text{R}\text{.}\\ \therefore \text{R is not symmetric}\text{.}\\ \text{Now, (1, 2), (2, 3)}\in \text{R but}\left(1,3\right)\notin R.\\ \therefore \text{R is not transitive}\text{.}\\ \text{Hence, R is neither reflexive nor symmetric nor transitive}\text{.}\end{array}
Q.4
$\begin{array}{l}\mathbf{Show}\mathbf{}\mathrm{}\mathbf{that}\mathbf{}\mathrm{}\mathbf{the}\mathbf{}\mathrm{}\mathbf{relation}\mathbf{}\mathrm{}\mathbf{R}\mathbf{}\mathrm{}\mathbf{in}\mathrm{}\mathbf{R}\mathrm{\hspace{0.17em}}\mathbf{defined}\mathrm{}\mathbf{as}\mathrm{}\mathbf{R}=\mathrm{}\{(\mathbf{a},\mathbf{b}):\mathrm{\hspace{0.17em}}\mathrm{a}\le \mathrm{b}\}\mathrm{\hspace{0.17em}}\\ \mathrm{is}\mathrm{reflexive},\mathrm{symmetric}\mathrm{or}\mathrm{transitive}.\end{array}$Ans
$\begin{array}{l}\mathrm{We}\mathrm{have}\\ \mathrm{R}=\; \{(\mathrm{a},\mathrm{b});\mathrm{a}\le \mathrm{b}\}\\ \mathrm{We}\mathrm{can}\mathrm{see}\mathrm{that}(\frac{\mathrm{1}}{2}\mathrm{,}\frac{\mathrm{1}}{2}\mathrm{)}\in \mathrm{R}\mathrm{as}\frac{\mathrm{1}}{2}=\frac{\mathrm{1}}{2}\\ \therefore \mathrm{R}\mathrm{is}\mathrm{reflexive}\mathrm{.}\\ \mathrm{Since}(1,2)\in \mathrm{R}\mathrm{as}2\mathrm{is}\mathrm{greater}\mathrm{than}1\mathrm{.}\\ \mathrm{but}1\mathrm{\xe2\u20ac\u2039}\mathrm{is}\mathrm{not}\mathrm{greater}\mathrm{than}2,\mathrm{so}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}(2,1)\notin \mathrm{R}\\ \therefore \mathrm{R}\mathrm{is}\mathrm{not}\mathrm{symmetric}\mathrm{.}\\ \mathrm{Again},\mathrm{let}(\mathrm{a},\mathrm{b}),\; (\mathrm{b},\mathrm{c})\in \mathrm{R}\mathrm{.}\\ \mathrm{Then}\mathrm{according}\mathrm{to}\mathrm{condition},\\ \mathrm{a}\le \mathrm{b}\mathrm{and}\mathrm{b}\le \mathrm{c}\hspace{0.17em}\hspace{0.17em}\Rightarrow \mathrm{a}\le \mathrm{c}\\ \Rightarrow (\mathrm{a},\mathrm{c})\in \mathrm{R}\end{array}$
∴
\text{Hence,R is reflexive and transitive but not symmetric}\text{.}
Q.5
$\begin{array}{l}\mathrm{Check}\mathrm{whether}\mathrm{the}\mathrm{relation}\mathrm{R}\mathrm{in}\mathrm{R}\mathrm{defined}\mathrm{as}\mathrm{R}=\left\{\right(\mathrm{a},\mathrm{b}):\mathrm{a}\le {\mathrm{b}}^{3}\}\\ \mathrm{is}\mathrm{reflexive},\mathrm{symmetric}\mathrm{or}\mathrm{transitive}.\end{array}$Ans
$\begin{array}{l}\mathrm{We}\mathrm{have}\\ \mathrm{R}=\; \{(\mathrm{a},\mathrm{b});\mathrm{a}\le {\mathrm{b}}^{\mathrm{3}}\mathrm{\}}\\ \mathrm{We}\mathrm{can}\mathrm{see}\mathrm{that}\frac{\mathrm{1}}{2}\le {\left(\frac{\mathrm{1}}{2}\right)}^{3}\mathrm{\hspace{0.17em}}\end{array}$
Q.6 Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.
Ans
\begin{array}{l}Since,\\ A=\left\{1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}3\right\}\\ \text{A relation R on A is defined as R = {(1, 2), (2, 1)}}\text{.}\\ \text{Since (1, 1), (2, 2), (3, 3)}\notin \text{R}\text{.}\\ \therefore \text{R is not reflexive}\text{.}\\ \text{Now, as (1, 2)}\in \text{R and (2, 1)}\in \text{R,}\\ \therefore \text{R is symmetric}\text{.}\\ \text{Now, (1, 2) and (2, 1)}\in \text{R}\\ But\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(1,1\right)\notin R\\ \therefore R\text{is not transitive}\text{.}\\ \text{Hence, R is symmetric but neither reflexive nor transitive}\text{.}\end{array}
Q.7 Show that the relation R in the set A of all the books in a library of a college, given by
R = {(x, y): x and y have same number of pages} is an equivalence relation.
Ans
\begin{array}{l}\text{Set A is the set of all books in the library of a college}\text{.}\\ \text{R = {x, y): x and y have the same number of pages}}\\ \text{since (x, x)}\in \text{R as x and x has the same number of pages}\text{.}\end{array}
\begin{array}{l}\text{If (x, y)}\in \text{R where x and y have the same number of pages}\text{.}\\ \Rightarrow \text{y and x have the same number of pages}\text{.}\\ \Rightarrow \text{(y, x)}\in \text{R}\\ \therefore \text{R is symmetric}\text{.}\\ \text{Again, let (x, y)}\in \text{R and (y, z)}\in \text{R}\text{.}\\ \Rightarrow \text{x and y and have the same number of pages and y and z}\\ \text{have the same number of pages}\text{.}\\ \Rightarrow \text{x and z have the same number of pages}\text{.}\\ \therefore \text{(x, z)}\in \text{R}\\ \text{Therefore, R is transitive}\text{.}\\ \text{Since, R is reflexive, symmetric and transitive, so R is an}\\ \text{equivalence relation}\text{.}\end{array}
Q.8 Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b): a – b is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.
Ans
\begin{array}{l}Since,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{A = {1, 2, 3, 4, 5}}\\ R=\left\{\left(a,b\right):\leftab\right\text{\hspace{0.17em}}is\text{\hspace{0.17em}}even\right\}\\ For\text{any element a}\in \text{A, we have}\leftaa\right=0,\text{\hspace{0.17em}}which\text{is even number}\text{.}\\ \therefore \text{R is reflexive}\text{.}\\ \text{Let (a, b)}\in \text{R}\text{.}\end{array}
\begin{array}{l}\Rightarrow \text{\hspace{0.17em}}\left\left(ba\right)\right\text{\hspace{0.17em}}is\text{\hspace{0.17em}}even.\\ \Rightarrow \leftba\right\text{\hspace{0.17em}}is\text{\hspace{0.17em}}even.\\ \Rightarrow \left(b,a\right)\in R\\ So,\text{\hspace{0.17em}}\text{R is symmetric}\text{.}\\ \text{Again, let}\left(a,b\right)\in R\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\left(b,c\right)\in R.\\ \Rightarrow \leftba\right\text{\hspace{0.17em}}is\text{even number and}\leftcb\right\text{\hspace{0.17em}}is\text{also even number}\text{.}\\ \Rightarrow \left(ab\right)\text{\hspace{0.17em}}is\text{\hspace{0.17em}}even\text{and}\left(bc\right)\text{is even}\text{.}\\ \Rightarrow \left(ac\right)=\left(ab\right)+\left(bc\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}is\text{even}\text{.}\\ \Rightarrow \leftac\right\text{\hspace{0.17em}}is\text{even}\text{.}\\ \Rightarrow \left(a,c\right)\text{\hspace{0.17em}}is\text{even}\text{.}\\ \Rightarrow \left(a,c\right)\in R\\ \therefore R\text{is transitive}\text{.}\\ \text{Hence, R is an equivalence relation}\text{.}\\ \text{Now, all elements of the set {1, 3, 5} are related to each}\\ \text{other as the modulus of difference between any two elements}\\ \text{of this set is even}\text{.}\\ \text{Similarly, all elements of the set {2, 4} are related to each}\\ \text{other as the difference between the two elements is even}\text{.}\\ \text{Also, no element of the subset {1, 3, 5} can be related to any}\\ \text{element of {2, 4} as all elements of {1, 3, 5} are odd and all}\\ \text{elements of {2, 4} are even}\text{.}\text{\hspace{0.17em}}\text{The modulus of difference between}\\ \text{odd and even is again odd}.\end{array}
Q.9
$\begin{array}{l}\mathrm{Which}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{values}\mathrm{of}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{make}\mathrm{the}\mathrm{following}\\ \mathrm{pair}\mathrm{of}\mathrm{matrice}\mathrm{sequal}\\ \left[\begin{array}{l}3\mathrm{x}+75\\ \mathrm{y}+12\u20133\mathrm{x}\end{array}\right],\mathrm{\hspace{0.17em}}\left[\begin{array}{l}0\mathrm{y}\u20132\\ 84\end{array}\right]\\ \left(\mathrm{A}\right)\mathrm{x}=\frac{1}{3},\hspace{0.17em}\hspace{0.17em}\mathrm{y}=7\left(\mathrm{B}\right)\mathrm{Not}\mathrm{possiblet}\mathrm{of}\mathrm{ind}\\ \left(\mathrm{C}\right)\mathrm{y}=7,\mathrm{x}=\frac{2}{3}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left(\mathrm{D}\right)\mathrm{x}=\frac{1}{3},\mathrm{\hspace{0.17em}}\mathrm{y}=\frac{2}{3}\end{array}$Ans
\begin{array}{l}Here,\text{\hspace{0.17em}}\text{\hspace{0.17em}}A=\left\{x\in Z,0\le x\le 12\right\}=\left\{0,1,2,3,4,5,6,7,8,9,10,11,12\right\}\\ \left(i\right)\text{\hspace{0.17em}}R=\left\{\left(a,b\right):\leftab\right\text{is a multiple of 4}\right\}\\ For\text{any element a}\in \text{A,}\\ \text{we have}\left(a,a\right)\in R\text{\hspace{0.17em}}\text{\hspace{0.17em}}as\text{\hspace{0.17em}}\leftaa\right=0\text{\hspace{0.17em}}which\text{\hspace{0.17em}}\text{\hspace{0.17em}}is\text{multiple of 4}\text{.}\\ \therefore \text{\hspace{0.17em}}\text{R is reflexive}\text{.}\\ \text{Let}\left(a,b\right)\in R\text{\hspace{0.17em}}as\text{\xe2\u20ac\u2039}\text{}\leftab\right=multiple\text{\hspace{0.17em}}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\text{\hspace{0.17em}}4\\ and\text{\hspace{0.17em}}\leftba\right=\left\left(ab\right)\right\\ \text{}\text{}\text{\hspace{0.17em}}=\leftab\right=\text{multiple of 4}\\ \text{So,}\text{\hspace{0.17em}}\left(b,a\right)\in R\\ Therefore,\text{\hspace{0.17em}}R\text{is symmetric}\text{.}\\ Let\text{\xe2\u20ac\u2039}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\xe2\u20ac\u2039}\left(a,b\right),\text{\hspace{0.17em}}\left(b,c\right)\in R\\ \Rightarrow \leftab\right=multiple\text{\hspace{0.17em}}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\text{\hspace{0.17em}}4\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\text{\hspace{0.17em}}\leftcb\right=multiple\text{\hspace{0.17em}}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\text{\hspace{0.17em}}4\\ \Rightarrow \left(a,b\right)\text{is multiple of 4}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(b,a\right)\text{is multiple of 4}\end{array}
\begin{array}{l}\Rightarrow \left\left(ac\right)\right\text{is also multiple of 4}\\ \Rightarrow \left(a,c\right)\in R\\ \therefore R\text{is transitive}\text{.}\\ \text{Hence, R is equivalence relation}\text{.}\\ \text{Let x be an element of A such that (x,1)}\in \text{R}\\ \text{Then}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x}\text{1 is a multiple of 4}\\ \Rightarrow \text{\hspace{0.17em}}\text{x}\text{1}=0,4,8,12\\ \Rightarrow \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=1,5,9\text{}\text{}\left[13\text{is not a part of set}\text{\hspace{0.17em}}\text{A}\right]\\ \text{Hence the set of all elements of A which are related to 1}\\ \text{is {1,5,9}}.\\ \left(ii\right)\text{R = {(a, b): a = b}}\\ \text{For any element a}\in \text{A, we have (a, a)}\in \text{R, since a = a}\text{.}\\ \therefore \text{R is reflexive}\text{.}\\ \text{Now, let (a, b)}\in \text{R}\text{.}\\ \Rightarrow \text{a}=\text{b}\\ \Rightarrow \text{b}=\text{a}\\ \left(b,a\right)\in \text{R}\\ \therefore \text{R is symmetric}\text{.}\\ \text{Let (a, b)}\in \text{R and (b, c)}\in \text{R}\\ \Rightarrow \text{a = b and b = c}\\ \Rightarrow \text{a = c}\\ \Rightarrow \text{(a, c)}\in \text{R}\\ \therefore \text{R is transitive}\text{.}\\ \text{Hence, R is an equivalence relation}\text{.}\end{array} \begin{array}{l}\text{The elements in R that are related to 1 will be elements}\\ \text{from set A which are equal to 1}\text{.}\\ \text{Hence, the set of elements related to 1 is {1}}\text{.}\end{array}
Q.10 Show that the relation R in the set A of points in a plane given by R = {(P, Q): distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all point related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.
Ans
\begin{array}{l}\text{Let O denote the origin in the given plane}\text{. Then}\\ \text{R = {(P, Q): OP=OQ}}\\ \text{We can observe that for any point P in set A we have}\\ \text{OP=OP}\\ \Rightarrow \text{(P,P)}\in \text{R}\end{array}
\begin{array}{l}\text{So R is reflexive}\\ \text{Now,}\\ \text{Let (P, Q)}\in \text{R}\text{.}\\ \Rightarrow \text{OP=OQ}\\ \Rightarrow \text{OQ=OP}\\ \Rightarrow \left(Q,\text{\hspace{0.17em}}\text{\hspace{0.17em}}P\right)\in \text{R}\\ \therefore R\text{is symmetric}\text{.}\\ \text{Again, let}\left(P,Q\right),\left(Q,S\right)\in \text{R}\\ \Rightarrow \text{OP=OQ and OQ=OS}\\ \Rightarrow \text{OP=OS}\\ \Rightarrow \left(P,S\right)\in R\\ \therefore R\text{is transitive}\text{.}\\ \text{Thus, R is an equivalence relation}\text{.}\end{array} \begin{array}{l}Again,\text{Let P be a fixed point in set A and Q be a point in set A}\\ \text{such that (P,Q)}\in \text{R}\text{. Then}\\ \Rightarrow \text{OP=OQ}\\ \Rightarrow \text{Q moves in the plane in such a way that its distance from}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{the origin (0,0) is always}\text{\hspace{0.17em}}\text{equal and is equal to OP}\\ \Rightarrow \text{Locus of Q is a circle with centre at the origin and radius OP}\text{.}\end{array}
Q.11 Show that the relation R defined in the set A of all triangles as R = {(T_{1}, T_{2}): T_{1 }is similar to T_{2}}, is equivalence relation. Consider three right angle triangles T_{1 }with sides 3, 4, 5, T_{2 }with sides 5, 12, 13 and T_{3 }with sides 6, 8, 10. Which triangles among T_{1}, T_{2 }and T_{3 }are related?
Ans
\begin{array}{l}Since,\text{every triangle is similar to itself}\text{.}\\ {\text{R = {(T}}_{\text{1}}{\text{, T}}_{\text{2}}{\text{): T}}_{\text{1}}{\text{is similar to T}}_{\text{2}}\text{}}\\ \text{(T,T)}\in \text{R for all T}\in \text{A}\\ \therefore \text{R is Reflexive}.\\ {\text{Let (T}}_{\text{1}}{\text{, T}}_{\text{2}}\text{)}\in \text{R,}\\ \Rightarrow {\text{T}}_{\text{1}}{\text{is similar to T}}_{\text{2}}\text{.}\\ \Rightarrow {\text{T}}_{\text{2}}{\text{is similar to T}}_{\text{1}}\text{.}\\ \Rightarrow {\text{(T}}_{\text{2}}{\text{, T}}_{\text{1}}\text{)}\in \text{R}\\ \therefore \text{R is symmetric}\text{.}\\ \text{Now,}\\ {\text{(T}}_{\text{1}}{\text{,T}}_{\text{2}}{\text{,T}}_{\text{3}}\text{)}\in {\text{A such that (T}}_{\text{1}}{\text{,T}}_{\text{2}}\text{)}\in {\text{R and (T}}_{\text{2}}{\text{,T}}_{\text{3}}\text{)}\in \text{R}\\ \Rightarrow {\text{T}}_{\text{1}}{\text{is similar to T}}_{\text{2}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{T}}_{\text{2}}{\text{is similar to T}}_{\text{3}}\text{.}\\ \Rightarrow \text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{T}}_{\text{1}}{\text{is similar to T}}_{\text{3}}\text{.}\\ \Rightarrow \text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{(T}}_{\text{1}}{\text{, T}}_{\text{3}}\text{)}\in \text{R}\\ \therefore \text{R is transitive}\text{.}\\ \text{Thus, R is an equivalence relation}\text{.}\\ {\text{In Triangles T}}_{\text{1}}\text{\hspace{0.17em}}{\text{and T}}_{\text{3}}\text{we observe that the corresponding angles}\\ \text{are equal and the}\text{\hspace{0.17em}}\text{corresponding sides are proportional}\\ \text{i}\text{.e}\text{.}\text{\hspace{0.17em}}\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\frac{1}{2}\\ Hence,\text{\hspace{0.17em}}{\text{T}}_{\text{1}}\text{is related to}\text{\hspace{0.17em}}{\text{T}}_{\text{3}}\text{.}\end{array}
Q.12 Give an example of a relation. Which is
(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive.
Ans
\begin{array}{l}\text{(i) Let A = {1,2,3}}\text{.}\\ \text{Define a relation R on A as R = {(1, 2), (2, 1)}}\text{.}\\ \text{Relation R is not reflexive as (1,1), (2, 2), (3, 3)}\notin \text{R}\text{.}\\ \text{Now, as (1, 2)}\in \text{R and also (2, 1)}\in \text{R, R is symmetric}\text{.}\\ \Rightarrow \text{(1, 2), (2, 1)}\in \text{R, but (1, 1)}\notin \text{R}\\ \therefore \text{R is not transitive}\text{.}\\ \text{Hence, relation R is symmetric but neither reflexive or transitive}\text{.}\\ \text{(ii) Consider a relation R in R defined as:}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{R = {(x, y): x < y}}\\ \text{Since x cannot be less than x}.\\ \therefore \text{R is not reflexive}\text{.}\\ Now,\text{\hspace{0.17em}}\text{}\text{(1, 2)}\in \text{R}\text{}\text{}\text{(as1 < 2)}\\ \text{But, 2 is not less than 1}\text{.}\\ \therefore \text{(2, 1)}\notin \text{R}\\ \therefore \text{R is not symmetric}\text{.}\\ \\ \text{Again, let (a, b), (b, c)}\in \text{R}\text{.}\\ \Rightarrow \text{a < b and b < c}\\ \Rightarrow \text{a < c}\\ \Rightarrow \left(a,c\right)\in R\\ \therefore \text{\hspace{0.17em}}\text{}R\text{is transitive}\text{.}\\ \text{Hence, relation R is transitive but neither reflexive nor symmetric}\text{.}\\ \left(iii\right)\text{\hspace{0.17em}}\text{Let A = {2, 4, 6}}\text{.}\\ \text{Define a relation R on A as:}\\ \text{A = {(2,2), (4,4), (6,6), (2,4), (4,2), (4,6), (6,4)}}\\ \text{Relation R is reflexive as (2,2), (4,4), (6,6)}}\in \text{R}\text{.}\\ \text{Relation R is symmetric as (2,4), (4,2), (4,6), (6,4)}\in \text{R}\text{.}\\ \text{Relation R is not transitive since (2,4), (4,6)}\in \text{R, but (2,6)}\notin \text{R}\text{.}\\ \text{Hence, relation R is reflexive and symmetric but not transitive}\text{.}\\ \left(iv\right)\text{A = {1, 2, 3, 4, 5, 6}}\\ \text{R = {(x, y): y is divisible by x}}\\ \text{Since every number is divisible by itself}\\ \therefore \text{(x, x)}\in \text{R}\\ \text{Hence R is reflexive}\text{.}\\ \text{Now,}\\ \text{(2, 4)}\in \text{R}\text{}\text{[as 4 is divisible by 2]}\\ \text{But,}\\ \text{(4, 2)}\notin \text{R}\text{.}\text{}\text{[as 2 is not divisible by 4]}\\ \therefore \text{R is not symmetric}\text{.}\\ \text{Let (x, y), (y, z)}\in \text{R}\text{.}\\ \\ \text{Then, y is divisible by x and z is divisible by y}\text{.}\\ \therefore \text{z is divisible by x}\text{.}\\ \text{Hence (x, z)}\in \text{R}\text{.}\\ \therefore \text{R is transitive}\text{.}\\ \text{Hence, R is reflexive and transitive but not symmetric}\text{.}\\ \text{(v) Let A = {1, 2}}\text{.}\\ \text{Define a relation R on A as:}\\ \text{R = {(1,2), (2,1), (1,1)}}\\ \text{Relation R is not reflexive as (2,2)}\notin \text{R}\text{.}\\ \text{Relation R is symmetric as (1,2)}\in \text{R and (2,1)}\in \text{R}\text{.}\\ \text{It is seen that (1,2), (2,1)}\in \text{R}\text{. Also, (1,1)}\in \text{R}\text{.}\\ \therefore \text{The relation R is transitive}\text{.}\\ \text{Hence, relation R is symmetric and transitive but not reflexive}\text{.}\end{array}
Q.13 Show that the relation R defined in the set A of all polygons as R = {(P_{1}, P_{2}): P_{1 }and P_{2 }have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?
Ans
\begin{array}{l}Since,\\ {\text{R = {(P}}_{\text{1}}{\text{, P}}_{\text{2}}{\text{): P}}_{\text{1}}{\text{and P}}_{\text{2}}\text{have same the number of sides}}\\ {\text{Since (P}}_{\text{1}}{\text{, P}}_{\text{1}}\text{)}\in \text{R polygon with same number of sides belongs}\\ \text{to itself}\text{.}\\ \therefore \text{R is Reflexive}\text{.}\\ {\text{Let (P}}_{\text{1}}{\text{, P}}_{\text{2}}\text{)}\in \text{R}\text{.}\\ \Rightarrow {\text{P}}_{\text{1}}{\text{and P}}_{\text{2}}\text{have the same number of sides}\text{.}\\ \Rightarrow {\text{P}}_{\text{2}}{\text{and P}}_{\text{1}}\text{have the same number of sides}\text{.}\\ \Rightarrow {\text{(P}}_{\text{2}}{\text{, P}}_{\text{1}}\text{)}\in \text{R}\\ \therefore \text{R is symmetric}\text{.}\\ \text{Again, let}\left({P}_{1},{P}_{2}\right),\left({P}_{2},{P}_{3}\right)\in \text{R}\\ \Rightarrow {\text{P}}_{\text{1}}{\text{and P}}_{\text{2}}\text{have the same number of sides}{\text{. Also, P}}_{\text{2}}{\text{and P}}_{\text{3}}\text{have}\\ \text{the same number of sides}\text{.}\\ \Rightarrow {\text{P}}_{\text{1}}{\text{and P}}_{\text{3}}\text{have the same number of sides}\text{.}\\ \Rightarrow {\text{(P}}_{\text{1}}{\text{, P}}_{\text{3}}\text{)}\in \text{R}\\ \therefore \text{R is transitive}\text{.}\\ \text{Therefore, R is an equivalence relation}\text{.}\\ \text{Hence, the set of all elements in A related to triangle T is the}\\ \text{set of all triangles}\text{.}\end{array}
Q.14 Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L_{1}, L_{2}): L_{1 }is parallel to L_{2}}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.
Ans
\begin{array}{l}Since,\\ {\text{R = {(L}}_{\text{1}}{\text{, L}}_{\text{2}}{\text{): L}}_{\text{1}}{\text{is parallel to L}}_{\text{2}}\text{}}\\ {\text{(L}}_{\text{1}}{\text{,L}}_{\text{1}}\text{)}\in \text{R as every line is parallel to itself}\\ \therefore \text{R is reflexive}\text{.}\\ \text{Now,}\\ {\text{Let (L}}_{\text{1}}{\text{, L}}_{\text{2}}\text{)}\in \text{R}\text{.}\\ \Rightarrow {\text{L}}_{\text{1}}{\text{is parallel to L}}_{\text{2}}\text{.}\\ \Rightarrow {\text{L}}_{\text{2}}{\text{is parallel to L}}_{\text{1}}\text{.}\\ \Rightarrow {\text{(L}}_{\text{2}}{\text{, L}}_{\text{1}}\text{)}\in \text{R}\\ \therefore \text{R is symmetric}\text{.}\\ {\text{Let (L}}_{\text{1}}{\text{, L}}_{\text{2}}{\text{), (L}}_{\text{2}}{\text{, L}}_{\text{3}}\text{)}\in \text{R}\text{.}\\ \Rightarrow {\text{L}}_{\text{1}}{\text{is parallel to L}}_{\text{2}}{\text{. And L}}_{\text{2}}{\text{is parallel to L}}_{\text{3}}\text{.}\\ \Rightarrow {\text{L}}_{\text{1}}{\text{is parallel to L}}_{\text{3}}\text{.}\\ \therefore \text{R is transitive}\text{.}\\ \text{Hence, R is an equivalence relation}\text{.}\\ \text{The set of all lines related to the line y = 2x + 4 must be parallel}\\ \text{line only and we know that in parallel lines\u2019 equations only the}\\ \text{constant value changes the coefficient of x and y remains same}\\ \therefore \text{set of parallel lines is y=2x+}\text{\hspace{0.17em}}\text{c, where c can be any constant}\text{.}\end{array}
Q.15
$\begin{array}{l}\mathrm{Let}\mathrm{R}\mathrm{be}\mathrm{the}\mathrm{relation}\mathrm{in}\mathrm{the}\mathrm{set}\{1,2,3,4\}\mathrm{given}\mathrm{by}\\ \mathrm{R}=\left\{\right(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2\left)\right\}.\mathrm{}\end{array}$$\begin{array}{l}\left(\mathrm{A}\right)\mathrm{R}\mathrm{is}\mathrm{reflexive}\mathrm{and}\mathrm{symmetric}\mathrm{but}\mathrm{not}\mathrm{transitive}.\\ \left(\mathrm{B}\right)\mathrm{R}\mathrm{is}\mathrm{reflexive}\mathrm{and}\mathrm{transitive}\mathrm{but}\mathrm{not}\mathrm{symmetric}.\\ \left(\mathrm{C}\right)\mathrm{R}\mathrm{is}\mathrm{symmetric}\mathrm{and}\mathrm{transitive}\mathrm{but}\mathrm{not}\mathrm{reflexive}.\\ \left(\mathrm{D}\right)\mathrm{R}\mathrm{is}\mathrm{an}\mathrm{equivalence}\mathrm{relation}.\end{array}$
Ans
$\begin{array}{l}\left(\mathrm{B}\right)\mathrm{is}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{Since}\{(1,1),(2,2),(3,3),(4,4)\}\in \mathrm{R}\\ \therefore \mathrm{R}\mathrm{is}\mathrm{reflexive}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{Now}(1,2)\in \mathrm{R}\mathrm{but}(2,1)\notin \mathrm{R}\\ \therefore \mathrm{R}\mathrm{is}\mathrm{not}\mathrm{symmetric}\\ \mathrm{And}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(1,2)\mathrm{and}(2,2)\in \mathrm{R}\Rightarrow (1,2)\in \mathrm{R}\\ \therefore \mathrm{R}\mathrm{is}\mathrm{transitive}\\ \mathrm{Hence}\mathrm{the}\mathrm{given}\mathrm{Relation}\mathrm{R}\mathrm{is}\mathrm{reflexive}\mathrm{and}\mathrm{transitive}\\ \mathrm{but}\mathrm{not}\mathrm{symmetric}\end{array}$
Q.16
$\begin{array}{l}\mathrm{Let}\mathrm{R}\mathrm{be}\mathrm{the}\mathrm{relationin}\mathrm{theset}\hspace{0.17em}\hspace{0.17em}\mathrm{N}\hspace{0.17em}\hspace{0.17em}\mathrm{given}\mathrm{by}\\ \mathrm{R}=\{(\mathrm{a},\mathrm{b}):\mathrm{a}=\mathrm{b}2,\hspace{0.17em}\hspace{0.17em}\mathrm{b}>6\}.\\ \mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer}.\\ \left(\mathrm{A}\right)(2,4)\in \mathrm{R}\hspace{0.17em}\hspace{0.17em}\left(\mathrm{B}\right)(3,8)\in \mathrm{R}\hspace{0.17em}\hspace{0.17em}\left(\mathrm{C}\right)(6,8)\in \mathrm{R}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left(\mathrm{D}\right)(8,7)\in \mathrm{R}\end{array}$Ans
$\begin{array}{l}\mathrm{R}=\; \{(\mathrm{a},\mathrm{b}):\mathrm{a}=\mathrm{b}2,\mathrm{b}>\; 6\}\\ \mathrm{Now},\mathrm{since}\mathrm{b}>\; 6,\; (2,\; 4)\in \mathrm{R}\\ \therefore \mathrm{Option}\left(\mathrm{A}\right)\mathrm{cannot}\mathrm{be}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ \mathrm{Also}\mathrm{by}\mathrm{putting}8\mathrm{we}\mathrm{will}\mathrm{not}\mathrm{get}3\mathrm{as}\mathrm{difference}\end{array}$ $\begin{array}{l}\therefore \mathrm{Option}\left(\mathrm{B}\right)\mathrm{is}\mathrm{also}\mathrm{not}\mathrm{the}\mathrm{correct}\mathrm{option}\\ \mathrm{And},\mathrm{as}8\ne \mathrm{7}2\\ \therefore (8,\; 7)\notin \mathrm{R}\\ \mathrm{Now},\mathrm{consider}(6,\; 8)\mathrm{.}\\ \mathrm{We}\mathrm{have}8\; >\; 6\mathrm{and}\mathrm{also},\; 6\; =\; 82.\\ \therefore (6,\; 8)\in \mathrm{R}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{option}\hspace{0.17em}\hspace{0.17em}\left(\mathrm{C}\right)\mathrm{.}\end{array}$
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