NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions (Ex 2.3)
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A student’s educational career is greatly impacted by their performance in Class 12. For entrance into many educational institutions, the board exams for Class 12 serve as the benchmark. An essential requirement for receiving admission to various courses is the marks attained in the board exams. Therefore, it is important for students to perform well in the board exam. However, getting ready for the board exam is not a simple task. The preparation for the Class 12 board exams starts right from Class 11. The concepts that are required for the board exams in Class 12 are a continuation of the concepts in Class 11. Hence, students in Class 11 need to understand the concepts in Class 11 thoroughly, which will not only benefit them in the Class 11 annual exams but also in the board exams in Class 12. It calls for a significant amount of effort and careful planning. Particularly for a subject like Mathematics, students must practisee answering several questions. For this reason, teachers advise students to practise NCERT questions. Additionally, the CBSE advises using the NCERT textbook for its board exams. The NCERT exercises should hence be completed by students.
One of the most important and difficult subjects for students in Class 11 is Mathematics. In addition to solid conceptual knowledge, Mathematics also necessitates a lot of problemsolving practice. Practising lots of questions is the key to getting good grades in Mathematics in the board examinations. In fact, the majority of board exam questions are taken from the NCERT textbooks. As a result, the NCERT book is the most crucial book for learning problemsolving techniques for board exams. For students in Class 11 to advance in their preparation for the Class 11 annual examinations in Mathematics, the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 are a crucial resource. The CBSE suggests using the Mathematics textbook from the National Council of Educational Research and Training (NCERT) for Class 11. The NCERT textbook is also suggested in the syllabi of several other boards. Hence, Extramarks provides the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 which can be beneficial for many students.
In the NCERT textbook for Class 11, Relations and Functions are covered in Chapter 2. The concepts in this chapter are continued from the previous chapter on Sets. Exercise 2.3 covers the concepts of Functions. Extramarks provides various study materials to understand this topic and practise questions from the exercises. Along with other study materials offered by Extramarks, the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 cover all of the key ideas given in the CBSE syllabus for the Class 11 annual exams. Expert teachers have created the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 to help students be ready for questions found in exams. Students can benefit from the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 provided on Extramarks in a variety of ways. They are aided in resolving difficult problems by using the comprehensive, stepbystep solutions provided in the NCERT Solutions for Class 11 Maths Chapter 2 Exercise 2.3.It is required to split challenging mathematical problems in order to solve them. The NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 provide the answer to this problem. Using the NCERT textbook boosts their confidence and advances their understanding of the most challenging topics. Students can evaluate themselves and their progress by comparing their answers to the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3.
Before exams, students need access to a range of questions as well as clearly explained solutions. The majority of the question banks mostly use NCERT textbook problems as their content. When students have access to NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3, answering questions becomes easier. They pick up on the ideas more quickly as a result. Students who are able to process information quickly feel more prepared for any exam. This gives them more time to focus on other aspects of their schooling while retaining their mental focus. Learning the concepts offered in the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 will help students advance in their preparation. Despite the fact that the chapter’s material can occasionally be challenging and confusing, having access to the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 that is tailored to their needs can be a useful study resource. The NCERT questions can be used as a practise by students who desire to proceed steadily while finishing other papers or problems. Study guides like the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 by Extramarks concentrate on a specific chapter.
NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions (Ex 2.3) Exercise 2.3
Lack of practise is one of the most frequent causes of students’ inability to finish the question paper within the allotted time. The exam paper also contains a number of questions from the NCERT book. Students who have access to NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 learn how to formulate their answers so that they can finish the question paper before the exam’s specified deadline. Hence, Extramarks provides the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3. The NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 are provided in PDF format so that students can access them whenever they want. The NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 can be easily downloaded from the Extramarks website and mobile application. Access NCERT Solutions for Class 11 Maths Chapter 2 – Relations and Functions
By having access to the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3, students may evaluate their own understanding and discover their strong and weak areas.Furthermore, the NCERT Solutions for Class 11 Maths, Chapter 2, Exercise 2.3, provide reviewed and analysed learning while assisting students in the development of a variety of abilities, including logical and reasoning abilities. Additionally, the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 supports students in their successful and effective preparation for the annual exams. Hence, it is regarded as a significant stage in the preparation for the exam.
NCERT Solutions for Chapter 2 Class 11 Exercise 2.3 Maths – What it covers
Class 11 Maths NCERT Solutions Chapter 2 Exercise 2.3 cover the topic named – Functions. As students might be aware, Functions are a special type of relation. It is important that students understand the concepts of functions, as they are widely used in various other concepts in Mathematics. Students need to understand this topic, as functions are a fundamental part of numerous other topics in other chapters of the book. They should understand the various types of functions given in the chapter, like Identity Function, Polynomial Function, Constant Function, Modulus Function, etc. It is also vital to learn the addition of two real functions, subtraction of a real function from another, the multiplication of two real functions, multiplication of a function by a scalar, and the quotient of two real functions. It is crucial that students solve the Class 11 Maths Chapter 2 Exercise 2.3.
Other Chapters in Class 11 Maths
The class 11 NCERT textbook contains numerous important chapters. There are 16 chapters in total in the NCERT. All these chapters are equally important for the Class 11 annual exams. Students should also check the latest CBSE syllabus for the chapters and topics included in it. Being thorough with all the chapters will not only help students in Class 11 annual exams but also help them comprehend the concepts in the Class 12. Hence, understanding the concepts thoroughly in Class 11 can also help students score better marks in the board exams in Class 12.
NCERT Solutions for Class 11 Maths Chapter 2 PDF and others
Apart from the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3, Extramarks also provides NCERT Solutions for other exercises in Chapter 2. Also, NCERT Solutions are available for all the other chapters as well.
Exam Preparation Tips For CBSE Class 11 Maths
Students preparing for the Class 11 annual exams need to work hard. In addition to working hard, students preparing for the board examinations need to think critically. Students must build a strong approach if they want to achieve extraordinary achievements from their preparation. For each topic, they should construct a study schedule and try to adhere to it at all times. They need to have everything planned out in advance. The course syllabus should be reviewed first, and each topic should be thoroughly studied in accordance with it. Students should practise the textbook questions while taking notes on important information. They should also attempt to solve all the exercises’ questions on their own. They can refer to Extramarks’ NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 for assistance. The formulas used in the questions should also be noted down. Additionally, they can keep a record of the steps they took to answer the questions. The topics should then be revised, with a focus placed more heavily on those areas where they fall short. Finally, they should take mock tests in order to be thorough with the material. Additionally, they can solve sample question papers and past years’ papers that are available on Extramarks.
Extramarks – Your Digital StudyMate
The NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 provided by Extramarks are only one of the indepth, genuine NCERT solutions that are available on the internet. Even though numerous platforms provide the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3, some students and teachers have concerns about their reliability. Students must use reliable study resources because annual exams are crucial. The NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 and other exercises are described in simple language. Students can learn the proper format for structuring their answers. The NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 offer the most effective and simple solutions to mathematical numerical problems. In addition to the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3, Extramarks also provides various study tools for Class 11 Mathematics. The study materials make it easier to fully comprehend the ideas required to do Exercise 2.3 Class 11th Maths. The chapter revision process is aided by these study materials. They are authored by highly experienced educators with years of expertise in a clear, understandable format. Extramarks provides practise worksheets, solved exemplar problems, revision notes, and more. Because of this, students may discover all they need for their studies in one place and save time by not having to search elsewhere.
Apart from the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3, Extramarks offers NCERT solutions for all classes and subjects.
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Q.1 Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
(ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}
(iii) {(1, 3), (1, 5), (2, 5)}.
Ans.
(i) Given relation is a function, because there is no two different images of same elements.
Domain = {2, 5, 8, 11, 14, 17} and Range = {1}
(ii) Given relation is a function, because each element has unique image.
Domain = {2, 4, 6, 8, 10, 12, 14} and
Range = {1, 2, 3, 4, 5, 6, 7}
(iii) The given relation is not a function, because here one element has two different images,
i.e., 2 has two images 3 and 5.
Q.2
$\begin{array}{l}\text{Find the domain and range of the following real functions:}\\ \begin{array}{l}\left(\text{i}\right)\text{\hspace{0.33em}f}\left(\text{x}\right)\text{= \u2013}\left\text{x}\right\\ \left(\text{ii}\right)\text{\hspace{0.33em}f}\left(\text{x}\right)\text{=}\sqrt{\text{9}{\text{x}}^{\text{2}}}\end{array}\end{array}$
Ans.
$\begin{array}{l}\begin{array}{l}\left(\text{i}\right)\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}f}\left(\text{x}\right)=\left\text{x}\right,\forall \text{x}\in \text{R}\\ \mathrm{Since},\text{}\left\text{x}\right=\{\begin{array}{l}\text{x},\text{\hspace{0.33em}x}\ge 0\\ \text{x},\text{\hspace{0.33em}x}<0\end{array}\\ \mathrm{So},\text{\hspace{0.33em}\hspace{0.33em}f}\left(\text{x}\right)=\left\text{x}\right=\{\begin{array}{l}\text{x},\text{\hspace{0.33em}x}\ge 0\\ \text{\hspace{0.33em}x},\text{\hspace{0.33em}x}<0\end{array}\\ \mathrm{Since},\text{x}\in \text{R},\text{so domain of f}\left(\text{x}\right)\text{is R.}\\ \mathrm{Here},\text{}\left\text{x}\right\text{is negative number for all real number, R.}\end{array}\\ \text{Therefore, Range of f}\left(\text{x}\right)\text{is (}\infty ,\text{0].}\\ \begin{array}{l}\left(\mathrm{ii}\right)\mathrm{We}\text{have,}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}f}\left(\text{x}\right)=\sqrt{9{\text{x}}^{2}}\\ \text{f}\left(\text{x}\right)\text{is defined for all x satisfying}\end{array}\\ \begin{array}{l}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}9}{\text{x}}^{\text{2}}\ge 0\\ \Rightarrow {\text{x}}^{2}9\le 0\\ \Rightarrow \left(\text{x}3\right)\left(\text{x}+3\right)\le 0\end{array}\\ \begin{array}{l}\Rightarrow \text{\hspace{0.33em}}3\le \text{x}\le 3\\ \Rightarrow \text{\hspace{0.33em}x}\in \left[3,3\right]\\ \mathrm{Hence},\text{Domain of f}\left(\text{x}\right)=\left[3,3\right]\end{array}\\ \mathrm{Let}\text{y}=\text{f}\left(\text{x}\right).\mathrm{Then},\\ \begin{array}{l}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}y}=\sqrt{9{\text{x}}^{2}}\\ \Rightarrow {\text{\hspace{0.33em}y}}^{2}=9{\text{x}}^{2}\\ \Rightarrow {\text{\hspace{0.33em}x}}^{2}=9{\text{y}}^{2}\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}x}=\sqrt{9{\text{y}}^{2}}\end{array}\\ \mathrm{It}\text{is clear that x will take real values, if}\\ \begin{array}{l}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}9{\text{y}}^{2}\ge 0\\ \Rightarrow {\text{\hspace{0.33em}y}}^{2}9\le 0\\ \Rightarrow \text{\hspace{0.33em}}\left(\text{y}3\right)\left(\text{y}+3\right)\le 0\end{array}\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}3\le \text{y}\le 3\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}y}\in \left[3,3\right]\\ \mathrm{Since},\text{y}=\sqrt{9{\text{x}}^{2}}\ge 0\text{for all x}\in \left[3,3\right]\\ \therefore \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}y}\in \left[0,3\right]\text{for all x}\in \left[3,3\right]\\ \mathrm{Hence},\text{range of f}\left(\text{x}\right)=\left[0,3\right]\end{array}$
Q.3 A function f is defined by f(x) = 2x –5. Write down the values of
(i) f (0),
(ii) f (7),
(iii) f (–3).
Ans.
f(x) = 2x –5
(i) Substituting x = 0 in f(x), we get
f(0) = 2(0) – 5
= – 5
(ii) Substituting x = 7 in f(x), we get
f(7) = 2(7) – 5
= 9
(iii) Substituting x = – 3 in f(x), we get
f(– 3) = 2(– 3) – 5
= – 11
Q.4
$\begin{array}{l}\text{The function \u2018t\u2019 which maps temperature in degree Celsius\hspace{0.33em}}\mathrm{into}\\ \text{temperature in degree Fahrenheit is defined by t}\left(\text{C}\right)=\frac{\text{9C}}{\text{5}}\text{+32.}\\ \text{Find}\\ \text{(i) \hspace{0.33em}t(0)}\\ \text{(ii)\hspace{0.33em}t(28)}\\ \text{(iii)\hspace{0.33em}t(\u201310)}\\ \text{(iv)\hspace{0.33em}The value of C, when t(C) = 212.}\end{array}$
Ans.
$\begin{array}{l}\begin{array}{l}\mathrm{The}\text{given function is:}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}t}\left(\text{C}\right)=\frac{9\text{C}}{5}+32\\ \left(\text{i}\right)\text{\hspace{0.33em}Putting C}=0\text{in t}\left(\text{C}\right),\text{we get}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}t}\left(0\right)=\frac{9\left(0\right)}{5}+32\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}=32\\ \left(\mathrm{ii}\right)\text{\hspace{0.33em}Putting C}=28\text{in t}\left(\text{C}\right),\text{we get}\end{array}\\ \begin{array}{l}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}t}\left(28\right)=\frac{9\left(28\right)}{5}+32\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}=82.4\\ \left(\mathrm{iii}\right)\text{\hspace{0.33em}Putting C}=10\text{in t}\left(\text{C}\right),\text{we get}\end{array}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}t}\left(10\right)=\frac{9\left(10\right)}{5}+32\\ \begin{array}{l}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}=14\\ \left(\mathrm{iv}\right)\text{\hspace{0.33em}t}\left(\text{C}\right)=212\\ \text{\hspace{0.33em}}\frac{9\text{C}}{5}+32=212\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\frac{9\text{C}}{5}=21232\end{array}\\ \begin{array}{l}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}=180\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}C}=180\times \frac{5}{9}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}=100\\ \mathrm{Thus},\text{the value of t is 100 when t}\left(\text{C}\right)=212.\end{array}\end{array}$
Q.5 Find the range of each of the following functions.
(i) f (x) = 2 – 3x, x ∈ R, x > 0.
(ii) f (x) = x^{2} + 2, x is a real number.
(iii) f (x) = x, x is a real number.
Ans.
(i) Since, x > 0
So, – 3x < 0
2 – 3x < 2
⇒ f(x) < 2
Therefore, Range of f = (–∞, 2)
$\begin{array}{l}\left(\text{ii}\right)\text{f}\left(\text{x}\right)={\text{x}}^{2}+2,\text{\hspace{0.33em}x is a real number.}\\ \text{Let \hspace{0.33em}y}={\text{x}}^{2}+2\\ \Rightarrow {\text{x}}^{2}=\text{y}2\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}x}=\sqrt{\text{y}2}\\ \mathrm{Since},\text{x is a real number.}\\ \text{So, y}\text{2}\ge 0\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}y}\ge 2\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}y}\in \left[2,0\right)\\ \mathrm{Therefore},\text{the range of function}\mathrm{is}\text{}\left[2,0\right).\\ \left(\mathrm{iii}\right)\text{f}\left(\text{x}\right)=\text{x},\text{x is a real number.}\\ \text{x is real number}\Rightarrow \text{f}\left(\text{x}\right)\text{is also a real number.}\\ \text{So, range of function}=\text{R}.\end{array}$
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FAQs (Frequently Asked Questions)
1. Where can Class 11 students find the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3?
Students of Class 11 can find the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 on the Extramarks website and mobile app.
2. What advantages can one expect from using the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3?
The NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 are useful for students getting ready for board exams since they are written in simple terms that all students can understand. In order to help them succeed in their exams, the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 have been written in a stepwise manner. The NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 have been developed in accordance with the most recent syllabus. So, students can learn to write answers in the exams correctly by following the detailed answers given by Extramarks.
3. How many questions have been solved in the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 provided by Extramarks?
All five questions in Exercise 2.3 have been solved in the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3 provided by Extramarks.
4. Are the concepts in chapter 2 clarified by using the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3?
For students in Class 11, NCERT solutions are crucial for annual exam preparation. In order to prepare for the exam, students can use learning modules like the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.3. Since students need to know the procedures that were taken to arrive at the answer, answers alone are insufficient. Students can improve their ability to respond to any similar question that appears in the exam by becoming more adept at following clear steps and procedures as given in the solutions.