 As discussed in the previous class, we are going to start with functions, we are going to start with functions because I personally believe and I believe this with conviction that without knowing functions, it's very difficult to understand the subsequent part of our class 11, sorry class 12 topics because everything is based on functions. Now the topic name is actually relations and functions, but relations part I have already done to, you can say complete extent in class 11. Okay, so in relations you have types of relations. So if you check your notes of class 11, I'd already discussed about void relation, universal relation, reflexive relation, symmetric relation, transitive relation, equivalence relation, and you're done enough questions on that as well. Okay. So today I'm going to start with functions. So let me just give you an overview of the functions chapter. In functions, the prerequisite for this is your class 11th domain and range. So the prerequisite is your concepts of domain, domain and range. I hope everybody who is sitting over here is fine with the idea of domain and range. Now in class 12th, of course domain and range will be useful. More importantly range will be useful. Right, because later on we're going to talk about something called onto functions where you will be asked to find out the range of the function. Don't worry if you are a new joiner you are not with us in class 11th. We will talk about finding the range as and when we reach this topic. So these are the two prerequisites for you from class 11th. So this you must have done in class 11. Okay. So 11th concepts. Okay. So what are we going to study in 12th? So in 12th we are going to first talk about number one types of functions. Types of functions. This is very important because it will be tested in your board exam as well. If at all it happens. No, I think it will happen. Something will happen in your time. Yeah, types of functions. Then we'll be talking about composition of functions. Composition of functions. Followed by inverse of a function. Okay. Now composition of function is very, very important for your, your competitive level exam point of view, but not very important from your school point point of view. So from your school point of view type of functions. Inverse of a function. These are the two main concepts that will be tested. Wouldn't be supposed to start for Conics. No, I told you I'll be starting with functions. And after I complete functions, we'll be starting with the part of the clinic because clinic will not take clinic will take, you know, at least one month. So we are waiting for the right time to begin. Okay. So don't worry about it. Next concept that we're going to take in functions is periodic functions. Periodic functions. This part again is not for your school. Periodic function is not included under your school syllabus. Okay. Next, we're going to talk about the concept of even odd functions. Even odd functions. Though we know it, officially, we have not done it. So we are going to do it. So we are already aware of even function and odd function. We have been talking about it since our bridge course days. But officially, we have not done this concept. We are going to take this up officially in class 12. Okay. And finally, we are going to talk about functional equation. We are going to talk about functional, functional equations. Okay. Now functional equations also let me tell you, when you are solving questions in your DPPs, you would have definitely come across something like, you know, if f of x plus f of y is f of x, y, find so and so. Right. So what is this? This is actually a type of a functional equation. Okay. So we'll be dealing with problems based on functional equation as well. People who are prepared for RMOs, pre-RMOs or INMOs, they would have already done it during their preparation because this is a very integral part of olympiads. Okay. Never mind. Even if you have not attempted these exams before, you will be taught this concept from scratch. Don't worry much about it. Okay. So these are the concepts that we are going to cover under this. And yes, after this, we are going to definitely start with conics. So for the new joiners, let me tell you, we have done conics only till board level concept or for your class 11. We haven't done it from the main point of view. So we'll be taking that up as well. All right. So let us begin with types of functions, types of functions. Before that, before that, just a quick recap of functions are not to the extent that I will be teaching you how to find domain and range. What is a function first of all? Most of you already know it. So I would like to hear from you. What type of relations are basically called functions? Yes, it's a mathematical machine for sure. But formally speaking, how do you define a function? So what type of relations qualify to be called a function? Exactly. So functions, when you say functions, okay, so those relations, so those relations qualify to be called functions. If they satisfy the fact that every preimage has a unique image, has a unique image. Okay. So in this kind of a mapping, you would realize that every element of set A must be used. Okay. I'll just give you a simple example for the same. Let's say I have a relation from set A to set B. Okay. Where relation A has, let's say A1, A2, A3, A4 and the set B has B1, B2, B3, B4, B5. Okay. So let me ask you a simple question. If I make a mapping like this, let's say A1 goes to B1, A2 goes to B3. A3 goes to B2. Is this mapping a function or is this relation a function? The answer is no. Right? Because A4 is left out. That is why this word every is very important. Every preimage. Now preimage and image, I hope everybody's clear. So when there's a mapping like this, that means A1, B1 is a part of that relation. So A1 is called the preimage of B1 and B1 is called the image of A1. Okay. I hope the main preimage and image is clear to everybody. So this doesn't qualify to be a function. So this is not a function because A4 is left out. Okay. Now let us say A4 maps to B4. Okay. So is this a function? Is this a function? The answer is yes. Because every preimage, first of all, is undergoing a mapping and every preimage has a unique image. Right? So there is no element from set A which is mapping to multiple elements of set B. Okay. Now, even if let's say A4 goes and maps to, let's say I undo this, even let's say A4 goes and maps to B2. It is still a function because A4 has a unique image of B2. A3 also has a unique image of B2. Right? But something like this, let me again undo this. Something like this where let's say A4 goes to B4 and A4 also goes to B5. This is not a function because this uniqueness criteria, the unique image criteria is being not fulfilled by this relation. Okay. So I hope everybody is clear about the basic definition of what makes a relation a function. Every preimage, the word every has a unique image, the word unique. So these two words are very important. Every and unique. Okay. So in this case, this is not a function. So I'll just undo this part. Okay. So this is an example of a, so this relation can be called as a function. Okay. Now, a simple question I would like to ask you. Let us say there is a, oh, so sorry. Yeah. Let's say there is a function from set A to set B where A has got N number of elements, B has got R number of elements. Okay. Okay. First tell me. First tell me. Okay. Let's, let's not call it function as of now. Let's call it as relation only. Tell me how many relations are possible from set A to set B? Everybody should respond. How many relations are possible from set A to set B? How many relations, Gayathri? Okay. Havesh has given. Shitesh has given. Okay. Rithvik. Awesome. Everybody is giving different different answers. Okay. Bhumika. So as of now I can see just two of you giving the right answer. Just two of you. And almost I think 12, 15 people have responded. Okay. Nishant. Gaurav says there cannot be any relation at all. Okay. Okay. Anusha, Advik. Very good. So now let us understand this. Number of relations are nothing but what is the relation? Relation is a subset of the Cartesian product of A and B. Correct. Now recall your class 11 days when you were learning sets. So the number of subsets of a particular set. Okay. So let's say there is a set X. Okay. Just a quick recall. Okay. I'll just write a recall over it. So let's say there is a set X which has got X number of elements. Okay. So how many subsets are possible of this set X? The answer used to be two to the power X or sorry, two to the power small X not capital X or two to the power number of elements in the given set X. Isn't it? Now, from where does this result come? You're already aware of it. So to make a subset, you either need no element. Okay. So you can say X is zero or one element or one element. Or one element or two element or you could take all the elements. So as per your some of the binomial coefficients you have learned it is two to the power of X. Okay. So that's how this is actually comes. Anyways, so now relating this fact to the present question here, which asks you how many relations are possible from set A to set B? Now A cross B, the number of elements in this set. We all know is the product of the number of elements in A into number of elements in B. So as per our given scenario, it is N into R. Correct. So if I ask you how many relations are possible as per this formula relation is a subset of A cross B. So the number of relations possible will be two to the power N R. Okay. So this result must not be forgotten because you may get a direct question asking you for the number of relations even in class 12 that can happen. Okay. So normally this question is asked in class 11, but it may also come in class 12 as well. So number of relations is two to the power N R. So people who are saying different, different responses and our N R is basically true for the number of elements in A cross B, not the number of relations from A to B. So please be careful about it. Okay. Okay. Now extending this question, how many functions are possible from A to B? So let's say there's a function from A to B. Again, the same situation holds. There are N elements in A and there are R elements in B. How many functions are possible? How many functions are possible from A to B? Yes. What's your response? What's your response? Okay. Adit. Okay. Shritaj, Aditya, Anusha. Okay. Okay. Okay. What is that? R-P-O means, oh, R for the power. Okay. Okay. Okay. Good, good, good. Okay. Now if I'm a person, if I'm a layman and I have to derive this result, it's very easy to derive. So let us say this is my set A which has got N number of elements and this is my set B which has got R number of elements. So let's say I call this as A1, A2, A3 till let's say A, N. And this set B has B1, B2, B3 till B, R. Okay. Now in order to count how many functions are possible from A to B, let us look at how many choices of mapping does each of the element of set A have. So how many choices does A1 have? If you see A1 can map to any one of B1, B2, B3 till B, R. So this is what R choices. Okay. How many options A2 has? A2 again will have R choices. Now why? Because let's say if A1 goes and maps to B2, then you cannot stop A2 also from mapping to B2. A2 also can go and map to B2. Right? Please note this is not the unique criteria. The uniqueness criteria says one element of set A cannot have multiple images. Two elements can have the same image. That is fine. But one element of set A or one pre-image cannot have multiple images. Right? So which is not happening in this case. A1 has got only one image, B2. A2 has only got one image. Again B2. Are you getting my point? So many people say R minus 1 or many people give different responses. But the answer here is A2 also has R options. A3 will also have our options. So like that all these N elements will have RRRRRR each. So as per fundamental principle of multiplication. Why I am using fundamental principle of multiplication? Because my task will not be completed till all the elements of set A are getting mapped. Are you getting my point? So A1 has to get mapped. A2 has to get mapped. A3 has to get mapped. AN has to get mapped. Till all of them get mapped, your mapping or your functions will not be created. Because I cannot leave out any one of your A1, A2, A3. Are you getting my point? Okay. Now some people are saying NCR. Runoff. Why NCR? What if N was less than R? What will you do? Then you will say there is no function possible. Okay. Alright. You corrected yourself. So number of functions possible will be R into R into R. N number of times. So your answer will become R to the power N. So this is your final verdict. R to the power N. Not to be forgotten because you will not get time to derive all these things in your examination hall. Okay. So here is a verdict. If you have any function from A to B, then the number of functions. I will just generalize it. The number of functions will always be, number of functions will always be, number of elements, or let's say the number of elements in the co-domain raised to the power of, raised to the power of, number of elements in the domain. Okay. So very easy to remember. Co-domain to the power domain. Co-domain to the power domain. Number of elements in co-domain, here is R. Number of elements in the domain is N, R to the power N. I hope everybody knows what's a co-domain, what's a domain, what's a range. Okay. Everybody's clear with that. Anybody who has an element of doubt about what is domain, co-domain, and range, do let me know. I'll be most happy to explain you again. Everybody's fine. Okay. Good. Everybody's clear. Good. Alright. So with this quick recap, I'm ready to start with types of functions. So let us start with types of functions. Types of functions. So the first type that we are going to look at is called one one function. Okay. One one function. It is also given an alternate name called injective functions. Injective functions. Injective functions. Many times, inject word and ION word is combined to make it injections. Okay. Okay. So injective plus ION functions, there is a, you can say there is a combination of this word and they're called injections also. Okay. But mostly you will hear one one and injective functions in your, in your books also. Now what is an injective function? Any function where, where every image has a unique pre-image. Okay. So in those functions where every image has a unique pre-image. Now what is image? Image are basically members of the range. Getting my point. So whatever mapping is happening with the elements of set B, those elements will be comprising the range. Right. So whatever elements are getting mapped in B, it should have a unique pre-image. Now recall from our definition of functions. Multiple pre-images could go to the same image. Right. So functions basically basically can have a possibility that multiple elements of set A can map to a single element of set B. That is possible. Right. But one one functions are those functions where this thing, where this mapping is such that the element of B, which is getting mapped has a unique pre-image rather than having a multiple pre-image. So a typical example I can show you are a typical arrow diagram I can show you. So this is an example of a one one function mapping. So let's say there is a set A to B and there is a function which is one one. So this mapping will look like this. Okay. So let's say A1 maps to B2, A2 maps to B3 and A3 maps to B1. Okay. So as you can see that every pre, every image. So B1, B2, B3. Okay. They are all images of some pre-image. Right. So there is a unique pre-image for B1, which is only A3. There's a unique pre-image for B2, which is only A1. There's a unique pre-image for B3, which is only A2. Okay. So these kinds of functions are called one one functions or injective functions. Yes, you want to very good point brought him on to the same sir. You said every but before is left out. What you tell me is before any image is before an image. If yes, what is its pre-image is before qualified to be called an image. Then why are you asking? I hope your concern is addressed. Every image before is not an image. Image is basically elements of your range is before a part of your range. No. Right. So then why is the question arising? So whatever is a part of your range, those are called images. Correct. So they should have a pre-image because unless a pre-image is there, there cannot be an image. Correct. It's like there has to be an object. Then only its image will be cast. Right. Image cannot be cast without an object. So the pre-image is like the object. So if before is an image, then worry about this definition. Before is not an image. Okay. Is this fine? Any questions here? Now, basically examples of these functions are seen in our daily solving of problems. So I can give you some examples. Let's say I take a simple linear function or linear polynomial 2x plus 3. Okay. This is a one-one function. This is a one-one function because if you have any output coming from it, it will come only from one particular input. For example, if I compare this to 10, I can only have one x for that 10 and that is 7 by 2. There is no other x for which I will get a 10. That is what it says. Every image has a unique pre-image. Okay. So there cannot be two values of x for which I will get a 10. For that matter, any number you take from the image, you will find that that is coming from a unique pre-image. It cannot come from multiple pre-images. Is it yet? Let me give you another example which will shock you. x square. x square. Now you must be thinking x square. How can it be a one-one function? Because if I want a nine, it can come from three also and it can come from minus three also. But please note, what have I written for the domain of the function? I have written r plus. That means if I want a nine, it will come only from three. Minus three is not a part of your domain. Now, what message I want to give through this example is whenever you are trying to decide whether a function is one-one or not, please take its domain into consideration. Don't start judging the function by only its definition. Because when I ask this question, 99% of the people first response is, sir, it doesn't seem to be a one-one function. Why? Because they just judge it on the basis of its definition. No. You have to also take into account what is the permissible domain. Are you getting my point? So under this restriction, if you see x square function, x square will only have this kind of a graph. So if I want a nine, it can come only from three. That means nine as an image can have only one unique or a unique pre-image which is three, nothing else. So I can cite multiple examples of one-one function. However, the opposite of one-one or complementary of one-one is many one, which I will talk in some time. I'm not going to discuss about it in detail, but just wanted to give you a heads up that what is not one-one is actually called a many one. So had I included r over here instead of r plus, then x square would have become a many one function. Many one separately I will talk about it. So let us talk in detail about one-one because this is going to be very important part for us. Any question related to the definition of a one-one function? No, it doesn't include r plus. It doesn't include r plus. Sorry, it doesn't include zero. r plus doesn't include zero. Yes, everybody's fine with what is a one-one function. Any question, any concerns? Iman Shu, Anusha Pranav, Gaurav, Shrita, Jaditya, Pratik, Gomika, everybody's fine. Everybody's happy. Clear, clear, clear, clear. Okay, very good. Now, if it is clear, I have a question and please address my doubt, my question. How many one-one functions are possible? Okay, so I'm introducing a permutation combination idea within functions. Okay, so how many one-one functions are possible? Number of one-one functions. So let's say I have a function defined from set A to set B. Again, let's say A has n number of elements, B has got r number of elements. Okay, who will tell me what is the number of one-one functions possible from set A to set B? What is the number of one-one functions possible from set A to set B? Think carefully and then answer, do not be in a rush. Give your answer in a well-thought-upon manner because I'm going to cross-question you also. Aram, say take your time. Okay, so two people have already answered. Okay, Nishant, that's a very unique answer from you. Do you mean to say our factory? Nishant? Okay, Gaurav, Gomika. So Nishant, you wrote r minus one, r minus two, dot, dot, dot. So I think you'll go to one, right? So that becomes our factory. Isn't it, Nishant? Okay, okay. Aditya is saying it will change depending upon the situation. Okay, Aditya, very good. Thankfully, you took that into account because everybody is giving a very flat answer for it. Okay, so Aditya is actually correct. Now let's try to analyze the situation. Okay, I'll take one first situation where let's say number of elements in the co-domain is less than the number of elements in the domain. Okay, then how many one-one functions are possible? Let's try to take a situation, a very small situation. So let's take a case study where r is less than n. Okay, I will assume that r is three and let's say n is five. That means the number of elements in the co-domain is three and the number of elements in the domain is five. Let's call it as a1, a2, a3, a4 and a5. Okay, now let's start the mapping. Let's say I'm trying my best to generate a one-one function. Okay, I'll try my best to generate or create or produce a one-one function. So let's say a1 maps to b1. Okay, now a2 could have mapped to b1 but as I told you, I'm trying my best to create a one-one function out of it. So a2 will go to b2 or b3 for that matter. Now a3 could go to b1, b2, b3, anything but again as I told you, I am trying my best to generate a one-one function so I'll map it to b3. But what will I do with a4 and a5? I don't think so there's an option left for a4 other than mapping to one of b1, b2, b3. So if I go and map to any one of them, then gone. It doesn't remain a one-one function. So it actually becomes a many one function. And no matter how hard you try, you will not be able to generate one. So what is the verdict here? What is the conclusion that we can draw? That if r is less than n, you cannot get any one-one function. So number of one-one functions is zero. Correct Gaurav, Gaurav now you have corrected yourself. Anusha sir, I think it depends on what you consider as the range. Anusha, does that point get clarified over here or still any element of doubt? Anusha Venkat? Clear? In this case at least it is clear. Okay, at least your doubt is clear in this case. Okay, now let's talk about the next situation where your r is equal to n. Where your r is equal to n. So let's take a case study of this as well. So next case study where I take r is equal to n. Let's say both are three-three. Okay, so again, let me draw an arrow diagram. A has got a1, a2, a3. B has got b1, b2, b3. Again, I'll be trying my best to generate a one-one function. So a1 can map to let's say any one of them b1. So a1 has got actually three options. Okay, it can map to b1 or b2 or b3. So if you talk about options available for a1, three options are available. How many options are available for a2? Now, if you try to make a one-one function, a2 cannot map to b1 then. So a2 can map either to b2 or b3. So it has got two options now. So a2 will have two options and let's say it chooses b3 as its mapping partner. Then a3 will be left with no choice but to go to b2. So this guy will have only one option or one choice. Now, this was only for a very small case where I considered only three elements in A and B. If you scale this up, if you scale this up to let's say r elements in each or n elements in each, what do you think should be the answer? In this case, your answer was three into two into one, which is three factorial. And if you scale this up to r elements, your answer will become r factorial or yes, rpr, whatever you want to call it. Fancy names are calling the same thing. So r factorial or n factorial or rpr or npn, all of them mean the same thing. So let me include that as well. So you have npn also included. So any one of them basically all mean the same thing, r factorial, n factorial, rpr, npn because r is equal to n. Now, let us take the last case, which is the only case left off for us. What happens when r is more than, let me write it in yellow, when r is more than n. And again, let us consider a case study on that. So if r is more than n, for the purpose of making our life simpler, we'll take this five and we'll take this as three. So let me make an arrow diagram. So three elements in set A, five elements in set B. So now let's calculate how many one-one functions are possible. Again, let us start with the mappings possible for A1. A1 can have how many options? Five options, exactly. So this guy will have five options. A2, let's say A1 goes and maps to B3. Okay. A2 will have how many options now? Four. Exactly. Only four options. So let's say this goes and maps to B1. Okay. See, I'm trying to generate a one-one function. So I'm trying my best only to create one-one functions. So I'm trying to see how many possibilities exist for me to create one-one functions. What about A3? A3? A3? Three options. Exactly. Absolutely. So this can go map to anyone of B2, B4 and B5, the left-outs. Okay. So this generates the one-one function. So as for the given scenario, the total number of functions possible is five into four into three, according to fundamental principle of multiplication. And this we normally call as five P3. If you recall, when we were talking about permutation combinations, this is generally called as five P3. Isn't it? Other way around, Anusha. So the number of functions possible here, which are one-one in nature, will be RPN. Okay. RPN. Let me just remove this and write it. Yeah. So you could actually club the second and the third case. Okay. So both these cases actually can be clubbed and you can write it like this. So number of one-one functions possible is zero if R is less than N. And it is RPN if R is greater than or equal to N. This is to be noted and kept in mind because this is going to be a direct question. For you in board exam also or in JEE or other CET, Comet K, VIT exams also. This has been a very, you can say, regularly seen questions in competitive exams also. Is it fine? So let's do a quick exercise over here. Very small question I have for every one of you. So let us say there is a relation from set A to set B. There's a relation from set A to set B. There's a relation from set A to set B where A has got two elements, B has got four elements. Okay. First of all, how many relations are possible? That's number one question. Then if let's say this relation happens to be a function also, how many functions are possible? Okay. And how many one-one functions are possible? Okay. Let's answer these questions. Everybody, let me call this as one, two and three so that you can number your answer as well. Quickly. I want everybody to respond. If they give the range, if they give the range, then the range becomes your core domain. Okay. So core domain is something which is not a very mandatory field. Domain is more important. Right. It's very important to know what you can input in a machine. What do you expect from the machine that is not that important? So if somebody mentions you A and B, it is always considered as if he has mentioned domain and core domain. Even though B was a range, it will be called as a core domain. Okay. Aditya, does this answer your question? Okay. Yes. So number of relations as most of you have rightly mentioned is two to the power eight. Now two to the power eight is almost 256. Okay. 256 relations are possible from set A to set B. Okay. How many functions are possible? The answer is four to the power two, which is only 16. So what do you see here? You see that out of 256 relations possible, only 16 qualified to be called functions. Rest 240 are non-functional relations. Okay. See the ratio very less getting the point. So number of functions is drastically less. How many of them are 11 functions? As you can see here, R is more than N. So it is four C, sorry, four P two. My bad. Four P two. So four P two is for factorial by two factorial, which is nothing but 12. So out of 16 out of 16 functions, only 12 of them are 11. That means just four of them are many one. Okay. We'll talk about many one separately. Exactly like Jay. Yes. So out of so many people who write Jay only few of you qualify and out of few of you only some again selected few of you qualify Jay at once. So it's a, it's a rejection procedure. Okay. Is it fine? Any questions here? All right. Now the vital question comes in your mind that if somebody gives me a function, how do you identify it's a one one function or not? So let me name the other concept as identification of one one function. How does one one identify? Sorry. How does one identify a one one function? Okay. How does one identify a one one function? So if I give you a function. Okay. And of course I define the function. How will you be able to identify whether it's a one one function or not? Okay. VIT see VIT exam is no different from your CT exam or other regional interest exams. Okay. So do see don't focus on too many exams. There was some guy who said, sir, any specialty for VIT, any specialty for CT, any specialty for PSAT, any special tip for XYZ regional exam. See, only one tip is there. Prepare well for Jay main. I'm not asking you to go for Jay advance. Prepare well for Jay main. Other exams are going to be covered under that preparation. Yes. Casting may be different for these exams. Number of questions asked the time duration, the negative marks involved. They will be different. But the idea, the concept remains the same for all these exams. Of course, Jay advance requires a bit of more training. You're some of the seniors who have scored 99.5 percentile and above. They are undergoing Jay advance training. So that is different. But if you are planning for any competitive level exam in India, your target should only be CT. That's right. The target should only be Jay main. No need to separately venture out for any other preparation. If you prepare separately for some exam, then what will happen? You know, you will be useless for other deviations from that particular paper. For example, let's say I only prepared for CT where there was no negative mark. So I don't keep marking even if I don't know how to solve it. I just take a guesswork also. If you do the same in Jay main, a bit sad or exams which has got negative mark, you will end up, you know, making yourself rejected from the competition. So prepare for the most difficult part. You will all, you will automatically be well prepared for these exams. Is it fine? Akash don't lower your guard. Okay. So Jay main seems to be out of my this thing. Jay main is not a difficult exam. It's an exam qualified by 2.5 lakh people. Even if you get 100 marks out of 300, you qualify Jay main. Okay. So let's keep our ambitions at least to that part. Okay. Now coming to the question. How do you identify a one one function? So there are three ways to identify a one one function. One is graphically. So I'll talk about graphical method first. So let us say you are aware of the graph of this. So this graph is known to you known to you or you can draw it or you can sketch it. Okay. So let's say it is a very sketchable function. You can easily sketch it by using your bridge course understandings of sketching a graph. So let's say you could sketch this graph. Then if you sketch this graph, then the process is just one simple, you know, a test which we call as horizontal line test, horizontal line test. What does this test say? This test says that if this function is a one one, so if f of x is a one one function is a one one function, then a horizontal line. Okay. So it's drawn to test the function not to save the function. Are you getting my point? So when you're drawing a horizontal line, draw it in order to critique the function. Okay. In order to test it not to save it. Okay. Then a horizontal line must cut the graph of f of x exactly at one point. Okay. Now, why is this test working? First of all, let us try to understand it. If you have a graph like this, let us say. Okay. So let's say this graph is the graph of the function which you have been able to sketch. If I draw a horizontal line. Okay. I can see that this line is only cutting the function at one point. So what does it indicate? It indicates that if there is an output B, it comes from a unique input A. But on the other hand, if I had a graph of a function like this, let me draw it in the same color as what I have drawn for the previous one. Yeah. So let's say there's a graph like this. Okay. And here if I draw a horizontal line, what is happening in this case? In this case, the horizontal line is going to cut the graph at multiple places. Okay. It is cutting here. It is cutting here. It is cutting here. What does it indicate? It indicates that a particular output. Let's say B is coming from multiple inputs. A1, A2, A3. Thereby suggesting that a pre-made, sorry, image B doesn't have a unique pre-image. That means multiple inputs are giving you the same output. Correct. So in such cases, your function will no longer remain a one one. In fact, it is called many one. And in this case, it will be a one one. Okay. So please read this test carefully. The test says that any horizontal line, horizontal line means line parallel to the X axis. Okay. Now don't start drawing a line over here. Right. Don't like, sir. See if I draw it here, it is not cutting at all. No, you don't have to save the function. You have to, you know, test the function. Okay. So I can give you all the easy questions and make you all get 300 out of 300 in J main, but that is not testing you then. In the one. So in order to test you, I have to give you some questions, which are challenging, which I feel you may not be able to solve. Okay. So any horizontal line drawn to test the function, if it cuts the function at exactly one point, then it's a one one function. Else it is a many one function. Okay. But this test is not to be used in school because in school, they do not honor this test. Right. So in the school exam, please do not try to solve a question by sketching its graph and doing a horizontal line tests on it. You may not be awarded marks for it. This may be used as a supplementary test, not as a primary test. Are you getting my point? So horizontal line tests can be used in competitive exams. Okay. Can be used in a regional entrance exam, but do not use it for school exams as a primary test. Use it as a supplementary test. Is it fine? Now this method has a drawback for the, the drawback is you should know how to sketch it because many times they may not give you the graph of the function. You may have to sketch it and sometimes it becomes too challenging to sketch a graph of a function. Of course, if it is a, if it is based on our basic transformations, we'll be able to do it like horizontal shifting or vertical shifting or mirror imaging. Those are fine. But some graphs are beyond our scope to sketch it. Okay. So nevertheless, we have two more methods to take care of that. Meanwhile, any questions, any questions, any concerns with respect to what is a horizontal line test and why it is used in a horizontal line test and why it is used. Is it clear to everybody? Clear. Everyone clear. Aditya, Gayatri, Havish, Anusha, Dora, Himanshu, Karthik are clear. Everybody is fine. Happy. If you are happy, let's move on. So we'll move on to the second test, which is called the calculus test. Okay. Even though most of us are not very good in our calculus, but we will talk about it. Even though we'll be using it only when we learn our application of derivatives. Okay. So what is this test? Let us try to understand. Now, if I go back to the, to the graphical tests. Okay. Okay. So let me draw two cases for you. In fact, let me draw three cases for you. Sorry. Sorry. Let me draw three cases for you. Okay. So first is the case where a function is increasing like this. Okay. Do you think it is going to pass the horizontal line test? It's always increasing, increasing, increasing, increasing, increasing. Increasing means everybody understands, right? Increasing means if you put more input, output will also grow. That is called increasing function. Do you think it is going to pass HLT? Yes, sir. Yes, sir. Yes, sir. Everybody says yes, sir. I also agree it is going to pass HLT. Okay. Let's say another example of a function always falls. I don't know how many of you are following. So yeah, it's all, it's falling down. That means if you put more input, output is going to fall off. Do you think it's going to pass HLT? It will pass. Yes. There you go. What if I take something like this? I think this is what I had drawn in the previous slide also. Do you think this is going to pass HLT? And here the function is exhibiting both type of nature. It is increasing also, decreasing also, increasing also, decreasing also means in some interval of its domain, it is rising. In some interval of its domain, it is falling. Okay. So you will say this will pass HLT. HLT pass. But this fellow will be HLT fail. As if it was like metric fail. Okay. This will fail the test. Right. Now calculus tries to answer this in a very interesting way. It says that in the first two cases, if I pick any point on this curve and I sketch a tangent, the tangent at that point, let's say I take any point X, the slope of the tangent at any point X. The slope of the tangent at any point X on this function will be given my f dash X. Just now we completed, we concluded our differentiation chapter. And in this case, you realize that any point if you take the slope will always be positive. So slope will always be positive. So in such situation, you'll see that your derivative of the function will always be positive for all X belonging to the domain of the function. Okay. In situation number one, I'm just trying to fit in a bit of calculus. In situation number one, when your graph is always on the rise, the derivative of the function at every point on it or every point in the domain will be positive. Agreed or not. Agreed or not. Okay. There's one exception. I will talk about it. Okay. Later on. In the second situation, also you would see that if I take any point, let's say X and I draw a tangent at that point, the slope of the tangent here would be negative actually. Any point you take, let's say I take here also. Okay. And I draw a tangent, its slope will be negative. Right. I hope everybody knows the derivative of a function at the point gives you the slope of the tangent drawn to the curve at that point. Right. So in this case, excuse me, in this case, you realize that the derivative of the function will always be negative for all X for all X. Okay. Now these two cases are actually called monotonic functions. So these two are actually named as monotonic functions in the language of calculus. Monotonic functions, the word has come from the word monotonous, which means exhibiting the same type of nature always. So it is always rising, rising, rising. This guy is always falling, falling, falling. So there is a monotonicity in their behavior. Isn't it? Just like most of our lives are. We are, we are leading a very monotonous life. Morning, get up, attend the school, evening, attend Centrum. Then in the night we do some homework and we sleep off. It's a very monotonous lifestyle for us, isn't it? Okay. Yeah, of course. I understand the two years of your life. You are going to have this kind of, you can say this is a disciplined life. I would say. So in these two functions, your functions are exhibiting monotonic nature. So monotonic functions are always known to be one one. That means either it's derivative will always be positive or it will always be negative. It will not be like the third guy whose life is slightly different. That means some type of monotonic nature. It will not be like the third guy whose life is slightly different. That means sometimes it is having positive slope and sometimes. Oh, sorry. And sometimes it is having a negative slope. Okay. So here it is having both both positive and negative in the domain of the function. So whenever a function is exhibiting non monotonic nature, that means it is having positive as well as negative slope in the domain of the function. Then those kind of functions are not going to be one one in nature. Such functions normally exhibit a turning point in them. Turning point means they will have a point where this function will take a turn back. Okay. So this is basically called the turning point. You will come to know later on that this is called the local maxima and this is called the local minima. I would not go into the details of it because this is again a subject matter of maxima minima chapter coming up for you in application of derivatives. Some of you who may have started it in school would be aware of it. Okay. Now, while I was discussing the first two cases, I told you there are some exceptions. There could be equal to zero also sometimes in these two cases. Okay. Now be careful. Equal to zero will also come here because here the slope will become zero. Here also the slope will become zero. But these two equal to zero cases are different from the third one. Let me take an example to explain you. Let us consider a function x cube. x cube graph looks like this. Okay. Everybody knows x cube graph. This x cube graph is known to be a one-one function because if I draw any horizontal line, okay, it is going to cut it exactly at one point. Okay. So any horizontal line is going to cut exactly at one point. Right. Now, if I try to solve this from calculus test, there is a point zero at which if I draw a tangent, let me draw it in green color, then this tangent will have a slope of zero. Okay. Then you may start, you know, contesting this rule that I gave you. You will say, sir, you said it is always greater than zero. But here in this example, I could see even if it is equal to zero, it is still one-one function for us. So that is why to account for such cases, I slightly tweak my rule and include zero, but with a dotted line. Why with a dotted line? Because it depends upon the function. It depends on the function. Let me write on the function. It depends on the function. Why it depends on the function? Because if I give you a function like this. Okay. Let's say if I give you a function like this. Okay. Let's say like this. Now, here you would say, okay, let's say, let's say for the time being, let's say for the time being, you're trying to, you know, try to inquire the nature of this function. So here equal to zero is included. Because at this point, the slope is becoming zero. In this case, it is exhibiting a turning point. Whereas in this case, it is exhibiting a point of inflection. Now turning point point of inflection, et cetera. I don't want to go into details of it. I just want to understand it from a layman's point of view that equal to zero is accepted if the function is basically taking a momentarily turn and going back in the same direction. Let me explain you in a very simple word. Let us say this is you. Okay. You are walking. Right. You are walking. Suddenly you stopped. Why would you have stopped for two reasons? Either you want to take a turn back. Correct. Or you are resting yourself. And again, you want to move ahead. Okay. Yes. You want to take a rest. You want to have an ice cream and probably want to move ahead. Correct. So let's say your journey, your destination is not reached and you stopped somewhere. It may be because of two reasons. You may want to rest yourself or you want to go back. Well, you missed out something or you have forgotten your wallet or telephone at home. Sorry, my mobile phone at home and you want to go back. Right. So the same situation is happening with these two cases. Here this guy is stopping to take a rest and is again moving in the same direction. So here the slope would be zero, but it would not be called as a turning point. It would be called as a point of inflection. So such cases in such cases you can claim these functions to be one one. But in the second case, which is happening here, the third one or you can look at this diagram also whatever, let's say this diagram in this case, the guy is going to take a turn back. Right. So this is a turning point. This is not an inflection point. So whenever a turning point comes, any horizontal line will cut it at more than more than one point. Okay. So the turning point is not allowed, but inflection point is allowed. That is why I have put this equal to zero with a dotted line over here because I want you to use your discretion depending upon the function. Is it fine? So summarizing this, what do I conclude? Let us summarize this. So now somebody is asking, sir, if I cannot draw the function, how do I know whether the function is momentarily stopping or it is basically taking a turn back? Now, for that you use your method of intervals. If you see here, Aditya, just to answer your question, Aditya, if you draw your wavy curve for f dash x. Okay. And let's say you have a doubt on the point C, whether the point C is a point of inflection or a turning point, you would see that the sign scheme on the either side of the C will be either both positive or both negative. Okay. In such cases, C will be a point of inflection. Means it is taking a momentarily stop there. It is not going to take a turn. It is just stopping for rest. Okay. But if you realize that on making the wavy curve for f dash on the either side of C, there is a change in the sign. So whether from plus to minus or minus to plus, it means the curve is taking a turn back. So this is a turning point. Okay. Are you getting my point? Now there are higher derivative tests also, but I don't want to go into that because many of you would say, sir, what if there are transcendental functions and I'm not able to make a wavy curve for it? What do I do in that case? Then there are some tests which is beyond your scope right now, but we'll definitely take them up in a matter of, let's say four to five months when we start our application of derivatives, not that late also, three months early. Okay. So double derivative test, triple derivative test, all those tests are there. Okay. Don't worry too much about it. But let me tell you, Aditya, that the functions that will be given to you, they will be very simple for you to either analyze it or just doing the graph of f dash x equal to zero. Or there's one more test coming up. That is even more better than this. But let me just summarize this. What is calculus test? What are we actually doing in a calculus test? In a calculus test, the summary is you would be given a function. Okay. This function would be given to you. Okay. You find the derivative of this function and you realize that this function is always having a positive slope or always having a negative slope. Now this or is not inclusive or this is exclusive or exclusive means don't know, it is always positive or always negative in your domain of the function or you can say for all x belonging to the domain of the function. If this is happening, then you can conclude that the function is a one-one function. Okay. Let me give an example for it. Let's say I talk about x cube function in R2R. So let's say somebody gave me a function x cube and he said he told me that, okay, do a calculus test on this to figure out whether this function is a one-one or not. So what I did first step, I just differentiated the function. Okay. Now here I realize that this is always greater than zero, but it could be equal to if your x is zero, right? Now here I have to use my discretion whether that point was a point of inflection, zero was a point of inflection or zero was a turning point for me. Okay. So here it is very clear that it is always going to be greater than zero equal to zero can come in one occasion, which is zero point, isn't it? So for that I will make a wavy curve. We know that x square will show positive sign here also, positive sign here also. I have special attention to Aditya who asked this question. So Aditya here it is a situation where you realize that your derivative of the function if you draw the wavy curve for it, it is showing the same sign on either side of that number zero. See here in this case is zero. Then here you realize that this function, this point is actually a point of inflection. So inflection points are most welcome. So we will still call this function as a one-one function. Is this fine? Let me give you another example. Let's say I talk about x square function. And I say figure out whether this is one-one function or not using calculus test. Then what will you do? You will differentiate this function. You will get a 2x. Now 2x can be positive also and negative also in real numbers. So if x is positive, your 2x will be positive. If your x is negative, your 2x will be negative. So basically it is exhibiting both the nature. So sometimes it is positive, sometimes it is negative. That is clearly the case because when you look at the graph, here it is positive, positive, positive, positive. Here it is negative, negative, negative. So in this case it is a many-one function because it is not showing monotonic nature. Is that fine? Now again this test is to be used only when you have done application of derivatives to a good extent. Don't start attempting questions unless until you are very sure about it or unless until you have tried and tested your hand at finding the, you know, in your application of derivatives concept. But ask your school teacher, she would be the right person to tell whether you can use calculus test for finding whether a function is one-one or not. If she says yes, you can do it. Then go ahead, use it. So now we are moving on to the last test which is the most commonly used test. And that is called the uniqueness test. So the third test which we are going to talk about, that is called the uniqueness. As the word itself says, there should be a uniqueness of the pre-image for every image. So what is this test? Let us try to understand this. This is to be used in school exam. Use for school exams. Now this test is very simple. It first starts with an assumption. It says that let there be two inputs x1 and x2 coming from the domain of the function. Coming from the domain of the function, in this case the domain is your A. You all know this is your domain. So let's say there are two inputs. So what you do is you make an assumption in the beginning of the process that let's say there are two inputs x1 and x2 which come from the domain of the function such that these two inputs give you the same output. Are you getting my point? So what are you doing? You are basically having this assumption in your mind that let's say out of several inputs that I have, there are two inputs which actually end up giving me the same output. So output is your f of x1 or f of x2 since they are same, I am equating it. So now using this assumption, if you simplify this equation a bit more and you end up getting x1 equal to x2 as the only possibility. This word only is very, very important. Then this function will be a one-one function. Now why? Let me explain this. See what are you doing here? You are first of all assuming that there are two separate inputs x1 and x2 which are giving you the same output. Later on, based on this assumption, you figure out that such a scenario will only exist if your inputs were equal actually. That means you had uniqueness of the inputs. Are you getting my point? So unless and until inputs are equal, output cannot be the same. That is what you are stumbling upon. That is what you are concluding with. So here you start with the assumption that two different inputs x1 and x2 that's why I have taken different names for them have given me same output and you later on realize that no, if that is happening, inputs also have to be the same. If that is happening, then your function is a one-one. But why do I write only over here? Because common sense says that if ever you are asked to solve an equation like this, then x1 equal to x2 will always be the solution. Does it mean that every function in the world will become one-one? No, right? But if that is the only solution, that means there is no other possibility between x1 and x2. x1 equal to x2 is the only possibility. Then that function is a one-one. Let me illustrate this with a simple example. Let's say I take example one where I have a function x2 again. This is a very easy function to give as an example. So I am taking that as an example. So here I am taking x2 as a function defined from r to r. So I will say let x1 and x2 come from the domain of the function such that f of x1 is equal to f of x2. So f of x1 means x1 squared, f of x2 means x2 squared. That means you are saying x1 squared is equal to x2 squared. Now for God's sake, for Jesus' sake, do not cancel out this. These are the mistakes which a 10th grader makes when he is coming to 11th. Not an 11th grader coming to 12th. Because if you do that, you are going to miss out on the solutions. Please do not do such mistakes of canceling out the powers on both sides. So what are we going to do instead? We are going to take both of them to one side. We are going to factorize them. And here we are seeing that we are left with two situations. One is x1 minus x2 is 0. And the other situation is x1 plus x2 is equal to 0. So this is not the only possibility. We have one more possibility and that possibility is x1 is equal to negative x2. And which is obvious also. So if you in an x square graph, the only way you can get the same output, let's say you want a 9 to come, then you can have the two inputs as 3, 3 or the two inputs as 3 and minus 3. That means negatives of each other. That is what this is trying to tell you. So these equations never lie. They will give you all possibilities. So either your two inputs are same. That means 3, 3 only has been put two times. Then only you will get a 9. Or you put 3 and a minus 3 then only you will get a 9. So what is happening in this case? This is not the only possibility unless until you are able to refute the second guy. You will not be able to refute the second guy because there may be two real numbers which are negative of each other. So here the conclusion will be this function is a many one function. It cannot be a one-one function because this is not the only possibility. However, that will always come out as one of your answers. Because it is common sense. No matter whatever function you take on this world, if you say f of x1 equal to f of x2, x1 equal to x2 will always be a solution. What you have to check is that the only solution. If that is not the only solution, then it is many one. Got the point? Now, for the fun I will take another example exactly of the same nature. This time my function will be defined from r plus to r. The very same function x squared. Now see how does it make a difference to your result? So again we will say let x1 and x2 belong to r plus. Such that f of x1 is equal to f of x2. I think all the steps would be the same. So I will be just hurrying up in writing it. Let me put a dash. Now here again two possibilities come. x1 minus x2 is 0 or x1 plus x2 is 0. So that means x1 equal to x2 comes out and the other one comes out to be x1 is equal to negative x2. But now here try to understand this is not possible. Why it is not possible? Why it is not possible? The answer is hidden in this step. Exactly it has a positive domain. You cannot have one positive number negative of the other. I cannot have a positive number equal to negative of another positive number. So what is happening in this case? In this case the other's condition or the other possibility is ruled out. Or that is impossible. The other possibility is impossible. So that means it leaves this with the only possibility. In such case your answer would be it's a one-one function. See the very same function I just changed its domain and I made it from menu one to one-one. That is exactly what used to happen when you learnt your inverse trig functions. All your trigonometric functions are known to be menu one. But do you remember we curtailed it in the principal value branch and we made them one-one. So this is what I mean when I say only possibility. I'll take a pause here and I would request you to just give me a clear remark whether you have understood this. Everybody. Then I'll move forward. Any element of doubt, any iota of doubt you can please take me over here. Okay. So before I move on I would like to take a small break because it is a two and a half hour session. We are only at 5.20. So we'll take a small break. Okay. Five minute break. Not more than that. So it's 5.18 p.m. now. 5.23 we'll resume again. Meanwhile you can use this break time to grab a quick snack. I'll have some water. I use the washroom whatever is your situation. Okay. We'll meet in five minutes time. Okay. So the first question that I would like to you to solve and this is a very funny and interesting question because it comes in school exam also. The question is you already know the answer to it but I would like you to solve it by your third method which is your uniqueness test method. Is the function x cube defined from R to R a one one function? I know you'll say yes because you have already seen the graph of it before but let's say assuming that you're not aware of the graph of this function and you're trying to apply uniqueness test to figure out whether it's one one or not. Let's try to do this. Why I'm giving you this question is because this comes very very frequently in school exams as well. Okay. And in school exams as you know they would not entertain your graph method. Okay. So everybody please do this and let me know. Just say done. Once you're done, let me know. On the chat box. Very good. Very good. Anusha. Let me take the attendance also. So dear all, how are you finding a two and a half hour session every day? Is it better than a three and a half hour on one day or was the previous mode better? Honest opinion. This is better. Okay. Don't worry. We are anyway going to go back to the previous mode. I was just trying to know your opinion. Okay. So fifty-fifty response. Some of you are saying this phone is better. You want two and a half hours of classes. Only maths four times a week. So how many days are there in a week? So four into three 12 days should be there in a week. You know where you to have it. I understand Akash. I can understand Havish. Midnight classes. This definitely increases the number of contact towers and hence we can cover more things. We can solve more questions and then we can understand the topics in more detail. But once the school reopens because some of the classes are also happening. Your seniors have not passed out yet. So we are trying to accommodate them in the morning time. Three times a day. Okay. We'll keep that in our consideration and we'll see what we can do. Okay. Yes. So most of you are saying done, done, done. We'll wait for one more minute. Okay. Himanshu done. Pranav done. Pradyan. Reshmika. Atharv. Adityanayar. Done, done, done, done. Okay. Where are you? Okay. Okay. Let's discuss this guys. Again, if you want to use uniqueness test, let's say there are two such inputs belonging to the domain of the function. Such that the output from the two is same. Okay. Now it's very unfortunate that in one of these schools, I will not name it. The teacher actually canceled out this powers. Okay. Please, please, even if your teacher in school does it avoid doing it. That is not a right way to look at things. Right. So you cannot just cut off the powers like that. Okay. If that was so easy, that means math would become half as challenging as it is right now. Okay. So what are the best way to solve it? Of course, we have to take the X2 cube on the other side. I'm sorry, X2 cube on the other side. We need to factorize it. Okay. Or like a good boy, I'm writing or so that you don't lose out any marks. So this implies either X1 minus X2 is equal to zero. That means X1 is equal to X2. Or you realize that this guy could be zero. Okay. So now two possibilities are arising. How do I show that the second possibility either will not exist or will point out to the first possibility only. So guys, we have to somehow convince the examiner that this is the only possibility. Other possibilities are actually not probable or not possible. Right. How do I do that? All of you please pay attention. Most of you would have tried different different ways to do it. But the best way to do it is write this term as X1 plus half X2. The whole square plus three by four X2 square. Okay. Now see if this is possible, then this is true. Correct. Because this expression is same as this. Again convince yourself. If you perform the, if you open the square, you'll have X1 square X1 X2 plus one fourth X2 square. And that one fourth X2 square will combine with this to give you an X2 square. Okay. Now see this term is greater than equal to zero. This term is also greater than equal to zero. So there's some will also be greater than equal to zero. If you want equal to zero, then only way is this should be zero and this should be zero. There's no other way to get a zero out of the sum of these two. If both are zero means X2 is zero. And if X2 is zero, X1 square will be zero, which means X1 is zero. Correct. So unless until X1 and X2 both are zero each, that means ultimately you are saying X1 equal to X2 again is the only output, is the only possibility coming. Right. So both of these scenarios are reinforcing the fact or are supporting the cause that you can only have X1 equal to X2 as the possibility. This is the only possibility. Okay. So even though I got another one or so as to say prima facie it looked out to be a different possibility. But we figured out slowly that even this guy points that X1 equal to X2. Of course both should be zero, but still it is X1 equal to X2 condition is being reinforced. Isn't it? So this becomes the only possibility and hence it is a one one function. So what I want to highlight over here, don't easily give up. Right. Okay. I'm getting one more possibility. One one. It has to be one many one. No, don't easily give up. Try to see whether the second possibility can be ruled out some way or even that can be converted to X1 equal to X2 as the only case. Are you getting my point? See what I did in the second part shares. I'll explain once again. Second possibility says job X1 square plus X1 X2 plus X2 square zero hoga tabby your output will be the same from both the inputs. Right. So what I did is say, okay, if this is the case, that means this is the case. And if this is the case, that means this is the case. I hope the intermediate steps are clear to you what I did. So ultimately what did you get to X1 and X2 are equal. Zero means what they're equal at the end of the day equal. So X1 equal to X2 is the only possibility. Some of you would have tried your this thing also complex number approach. Many of you know that this could be factorized as X1 minus Omega X2 and X1 minus Omega square X2. And if this is zero, you're trying to say X1 by X2 is either Omega or Omega square. But X1 and X2 both are real numbers. How can ratio of two real numbers be basically a complex number? Yes, if they become 00 each, it is a case of indeterminacy. In that case, your answer would be possible. Okay. So there's another way to justify it, but this is a better way. I would say this is always a better way. I would say this, this way is a better way. Okay, better method. This will not raise eyebrows from your school teacher also. This probably my school teacher will say, you are where Omega, Omega square and all is coming from this gate. And said, okay, so avoid this, but this is for your understanding in an alternate way. Is it fine? Okay. So before I move on to questions, bit more of questions. Uh, I would just take the case of the complimentary of one one, which is many one functions. Okay. And then we can take problems. So plain and simple. If anything is not one one, it is many one. It's the complimentary breakup. Right. Are you getting my point? If, if an integer is not even, it is odd. Simple as that. Okay. If a function is not one one, it is a many one. It cannot be both 50 50. It cannot be like it is 50 50 one one 50 50 many one. So it is nothing but it is complimentary of one one. So functions which are not one one are many one. Okay. Formal definition here would be at least one image. Has a non unique or you can say multiple pre images. Okay. So if you have at least one image, which is having more than one pre image, then that function will be called a many one function. So there is an existence of just one image which has got more than one pre image. It will be called as a many one. Okay. Even one example you are able to find out that a particular output is coming from more than one input, which are not equal. Then it is a many one for me. Is it clear or not? Arrow diagram wise, I just give you a simple arrow diagram to just have a pictorial view in your mind. So this is a typical pictorial diagram of a function which is a many one. So let's say something like this. Okay. Okay. A constant function is a very good example for it. So let's say I talk about a constant function. So let's say a function which only gives you three. Okay. So this is an example of a many one function because no matter whatever input you have, whatever input you have, let's say any real number you take x1, x2, da, da, da, da, xn, all of them will point to the same fellow. Okay. So this is a typical example. A few more examples I can give as I have been always been telling x square. Okay. But I remember the domain will be all are. Okay. All your trig functions. Okay. Trig functions, sine x function, cos x function, they're all many one functions. Right. So is there any separate test for a one one function? The answer is no. The test for many one function is just the tests which are for one one. See, is there any separate test for somebody who is corona negative? No, right. Same test is there. If the test comes out to be positive, you are having that virus. If the test comes out to be negative, you are not having that virus separate test. Correct. So in the same way, if a function is not one one, that means if it fails, it should fail. Horizontal line test. Second, its derivative should be both positive and negative in the domain of the function. In the domain of the function. Okay. And if you do your uniqueness test, if you do do your uniqueness test, then this may imply x1 may not be equal to x2. Okay. Or x1 x2 may be distinct. Let me write it like this. x1 and x2 may be distinct. Okay. So whatever doesn't qualify to be one one becomes a many one. So there's no separate test required for it. Okay. So that one one test is only what we use. If it fails it, that test we say it's many one. Is it fine? Any questions here? Similarly, if I talk about number of, let me go to the next slide. I hope you're done with the copying part of it. Such functions normally they have up turning point in the graph of it. So whenever there's a many one function, they will always have a turning point in the graph. Done with this. Can I move on to number of many one functions? Okay. So number of many one functions. Again, a easy concept. Since we already know many one is complimentary of one one, we can always find out the number of many one functions by deducting the number of one one functions. So if your r happens to be less than n, please note that when this was a situation, there used to be no one one functions possible, right? I hope you recall that when your number of elements in the core domain is lesser than the number of elements in the domain, there cannot be any one one function, which means the total number of functions itself will be many one. That means all the functions will be many one in nature. So total number of functions is r to the power n and subtract from it number of one one functions, which is zero. So that leaves you with the answer as r to the power n only. That means if r is less than n, all functions that you can make will be many one in nature. Okay. If r is more than equal to n, then by using the same logic, total number of functions minus r p n. This will be your number of many one functions. Is this clear to everybody? Any questions, any concerns? Do let me know. Okay. Now time to take some questions. Let us take some questions. No concerns. Let's take some questions. So let's begin with some problem solving. I think we have discussed a lot of theories so far. Okay. We'll start with a very conceptual question. Which of the following statements are incorrect? Which of the following statements are incorrect? Now read the three statements carefully, very, very carefully. The first one says if f and g are one one, then f plus g is also one one. Okay. Could you repeat the previous slide I got disconnected maybe after this question Anusha can I can I take you to the next previous slide after this question. Okay. Second statement says if f and g are one one, then f into g is also one one. Yes, in their common domain. Exactly. So kind of remember f plus g will exist in the overlap of the domain of f and g that you have already done in class 11. And the third statement says if f is an odd function, then it is necessarily a one one function by the way what's an odd function. Many of you would already be knowing about it, but those who are not aware about it odd functions are those functions which satisfy this functional equation f of minus x is negative of f of x. Exactly. So I'll put the poll on. Please think and then respond. Don't be in a hurry to answer. I'll give you around two and a half minutes for it time starts now. Today drizzled a bit over here. What about your area? Did it drizzle a bit? Slight rain. Yeah, big relief. It was becoming hotter day by day. But rain heavily. Oh, in my area just drizzled. Okay. It will rain very soon. Yes. Fingers crossed. Hsr. Was it training there? Anybody from Hsr layout? Drizzle. Okay. Just 30 seconds. I can give you very few responses. Only 11 of you have responded. Come on, guys, everybody. Rohit, Rashmika, Parvati, Omkar, Vaidya. Rohit, Gurman, Charan, Akash, Akash, Bhoomika, number of O's are increasing every day in your name. By the time we end Bhoom. Okay. Let's begin the countdown. I think most of you have responded. Five, four, three, two, one. Okay. So almost 91% of you have responded to it. Most of you say option number B, B for Bangalore. However, there has been a close competition with option number D. 10 of you say option number D. Okay. Let's check it out. Guys, all these questions you can actually refute by examples. Very simple example. Let's say for the first case, I take my f of x as x and g of x as a minus x. Okay. I will take minus x plus two also, not an issue. But the moment I add them, I get a zero. It happens to be a constant function. So this happens to be one one. Correct. Because if you draw the graph of y equal to x, you will never find it taking a turn anywhere. That means any horizontal line that you draw is going to cut the graph exactly at one point. Same is true for this guy as well, one one. But the moment you add it, it's going to give you a constant function, which is unfortunately a many one. So whichever has... See the question is which of the following are incorrect. Okay. So one is definitely incorrect. So whichever has one, there still stand a chance to be your right option. B doesn't have, so B will be ruled out. So people who said B, bye-bye. Most of you said B. B for bye-bye means your answer is wrong. Okay. Next. If f of x and g of x are one one, f of x into g of x will be one one. Now same example. Same example. If I multiply f of x into g of x, I'll get negative x square. Negative x square graph, if you see, it is an inverted parabola like this. That means if I draw a horizontal line, it is going to cut it at more than one point. So definitely not one one. So option two is also an incorrect statement. So whichever doesn't have option to say bye-bye to that option, which is option C. Okay. So that leads to a close competition between A and D option. Okay. Next is if f of x is an odd function, it is necessarily one one. Now we all know from our childhood that this is not correct because sin x function is clearly an odd function. Correct. And this is not a one one function because if I draw a horizontal line, it is going to cut at so many points. Okay. So third statement is also incorrect. So whichever answer says one two and three, that will be your right option. Option number D hence becomes right. So out of 31 of you who said the 10 of you who said D. Absolutely correct. Isn't sin x domain minus five by two to why sin x domain only minus five by two? It is kept as minus five by two to five by two to make it invertible. Here there is no question of making it invertible. Why are you unnecessarily restricting it to minus five by two to five by two? We can use f of x equal to zero. Why not? In this case your function f of x equal to zero. f of x equal to zero is a many one function. That is what I'm trying to say. What is the question saying? It is necessarily a one one function. So you are justifying what I'm trying to say. The question is if the function is odd. Now zero is both a one one and sorry zero is both even and odd. We'll come to see in the chapter when we are doing even odd function. Zero is both one one and sorry both even and odd. Okay. Zero is the only function which is both even and odd simultaneously. You can use to disprove it. So is it fine? Any questions here? Anybody? Let's move on to the next one. Okay. Now this is a single option correct question. Which of the following? Oh yes Anusha. I'm so sorry. Anusha I'll have to go to the previous slide. Just give me one second guys and girls. Anusha wants to copy something. Nothing very serious here Anusha. We just talked about number of one one, sorry many one functions possible. So these two you can take either a snapshot. Okay. What happened? What happened to your computer? Shut down. What all you missed? I said I was about to say missed. What did you miss? Entire problem. Okay. Wait for the recordings to come to see it. Unfortunately and get a good computer man. Yeah. Don't worry. There is nothing very hard and fast. You can solve it when you see the notes also you can solve it. Okay. So let me put the poll on for this. Let's have around three and a half four minutes. Max to max to solve this question. Not a difficult one, but still I'm giving you enough time to crack this. Which of the following is our one one function. Single option correct. Three minutes almost over. I can give you one more minute because I can see very few people have responded. Just eight of you have responded. So last minute. Good boys and girls. Please. Fuck up. Okay. The countdown begins now. Five. Four. Three. Two. One. I'm calling off the polar. 26 of you have responded out of which most of you have said option number C. Okay. Of course you would have gone to be as well. Let's try to check these functions. In detail one by one. Okay. See again, I will use a mixed approach. If I feel I can do it with graph. I'll do it by graph. If I feel I can do it by calculus test. I'll do it by calculus test. Okay. Let's try to look into the very first function a. So a says you have a function from R to R. Defined as e to the past signum function. So this is a signum function. We all know the characteristic of a signum function. So signum function gives you a one. So let me just recall everybody. What's the signum function? It gives you a one when X is positive zero when X is zero and minus one when X is negative. Right. We all know the characteristic of a signum function. Right. So in light of that, in light of this definition, I can redefine my function as correct me if I'm wrong. E plus e to the power X square when X is positive. Oh, sorry. X is positive. One plus. Now, since it is zero, I can put zero here also. So one plus one, which is a two. Okay. So two and X is two. Sorry, X is zero. Okay. And e to the power minus one plus e to the power X square when X is negative. Correct. Now, if I sketch this graph, I think things will clear out very easily. First, let me talk about e to the power X square graph. Anybody knows e to the power X graph or I can guess how would e to the power X square graph look like. First of all, e to the power X square graph. Since there is a even power on X, it has to be symmetrical about y axis. Why does symmetrical about y axis? Because if you change your X with a minus X, nothing will happen to the function. Correct. And normally when you write about X as a minus X, the graph gets reflected about the y axis. So something which is already symmetrical about y axis, only that function will not get affected. Right. So this graph, I predict that the graph will actually look like this. Okay. Any doubt regarding that? Okay. So let me show you on the GOG graph. So it's very easy to guess this. Okay. So as I told you, there are some graphs which are very easy to sketch. So this is one of them. So I'll just show you e to the power X square graph. Cool. Yes. Y is equal to e to the power X square. Yeah. There you go. Okay. So we're almost on the right track. Okay. Now let's go back. Let us try to sketch the graph of this function. So I'll again make my X and Y axis. Don't worry. So the first graph is plotted when X is more than zero and it is e plus e to the power X square. That means the same graph will get a kick up. That means this point, instead of being zero comma one, it will be zero comma one plus e. Okay. So the same graph is getting a kick up by e. Okay. So here this point was zero comma one. Now it has gone up by e. Okay. Now put a whole over here. Why? Because zero is not included here. It is only starting from right side of zero from zero plus. That is what we call it in a language of limits. Isn't it? Exactly at zero. It is at two. So two will be somewhere over here. Not that. So one plus e is 3.7 ish. So almost half you can say. Yeah. Somewhere there. Roughly. You don't have to know exact position. Okay. And when X is less than zero, it is e to the power X square graph. That means look at the left arm of this graph. Just kick it up by one by one by his point. Forish. Correct. So the graph will start from here. Okay. And we'll go up like this. Please note that this point will be zero comma. One by e plus one. And don't forget to put a whole because. Yeah. So there will be a whole over it. Now look at this graph and try to tell me is this graph or one one function or a many one function. The answer is loud and clear. If I draw a horizontal line like this, it is going to cut it at two points. Right. So it is clearly a many one. It is clearly a many one. Okay. Another way to look at it is most people will say, sir, this is anyways a constant because it is either going to take e or one or one by right. So if you differentiate it, you are going to get the derivative of one plus something which is zero plus derivative of this guy is e to the power x square into two x. Okay. That all depends on this x. If x is positive, everything is positive. If x is negative, everything is negative. So in your domain, it is going to show both the nature of positive and negative. So that is anyways, non monotonic. And if it is non monotonic, it is many one. So this is not my answer. Okay. So people who said a how many of you said a by the way, only two of you said a. Okay. So those two of you, sorry, my dears. Is not the correct answer. Now talking about B, B is an even a simpler case to deal with. Okay. If you see, I can guess two values for which I get the same output and that is one and a minus one. Correct. So if you see here, let's say I take, I take one. I'll get e to the part two. If I take minus one also, I get e to the part two. What does it mean? It means that they're two separate inputs for which the output is e square. So this is clearly a many one. Okay. What I want to exhibit from this, you know, approach is that if you are able to figure out an example or if it, if an example strikes you that, okay, I can put a one and a minus one. I can get the same output. Then you can easily say it's a many one function. So one example is sufficient to disprove it. Calculus method for a you're talking about Aditya or the a option you're saying is very obvious. If you differentiate this guy, okay, it is, it's going to give you e to the power x square into two x. Correct. Right. So when x is positive, this will be positive effects is positive and this will be negative effects is negative and both belongs to your domain of the function. So it is showing both the characteristic, both the characteristic. It should not show if you want it to be one one, it should show only one characteristic either it is always positive or it is always negative. It should not show both the nature. Got it Aditya? Aditya, do you understand Hindi? Svalpa, Svalpa. Little bit you understand? Little bit, okay. Panikkar, you are from Kerala, I believe, right? Originally? Yeah. Okay. So your option number B is also gone for a toss. Let's see who all gave B. B, B, B, B, B, B, six of you said B. Sorry, my ideas. Done. Oh, Akash is also from Kerala. Okay. Coming to option number C. Let's check. See, again, read the function in light of the domain. Only from three to four you want to analyze this function. Are you getting my point? Never, never overlook the domain of the function. Again, I'm telling you, it is not important. It is absolutely mandatory for you to look at the domain. Okay. So when you talk about C, let me just copy this function. Okay. So we need to talk about three C option. Your function was defined from three to four. So let's redefine this function from three to four. So your domain is from three to four. And output is from our core domain is from four to six. So in three to four, if you redefine the function, this guy will become X minus one. This guy will become X minus two. Or this guy will become X minus three. But the last guy will become four minus six. Right. Remember how to redefine a function. We had already done this in our class 11. Okay. So mod X minus one is going to remain X minus one in that interval. What X minus two is going to remain X minus two. What X minus three is going to remain X minus three. But what X minus four is going to become four minus six. Isn't it on simplification? If I'm not trying to define X minus one, the function, if I'm not mistaken, this is going to cancel with this. It's going to give you two X minus two, which means it's a linear function. So if I sketch this graph between three to four, it's going to be a linear line like this. Okay. Probably this end will be at four. This end will be at six. Okay. So this is your graph. And if I draw any horizontal line, it only cuts the graph at exactly one point. So this is one one for sure. So option number C is correct. That means your search ends over here, but we'll still solve the D part just to know what type of a function it was. Why four minus six? Hariharan, you're asking this question too late in your life. See, if I take any value between three and four, let's say three point two. Okay. What is mod of three point two minus four give you? You'll say, sir, it gives you point eight. Correct. So if you get point eight from three point two, you have to do three minus. So you have to do four minus three point two. That's why four minus six. Is it fine? Here. So here, I didn't just revisit your modulus part of class 11, how to redefine the function. It'll be very clear to you what, or you can watch my bridge course notes, our lectures on your learners. Okay. Let's go back to your D part under root of Ellen, cause of sign X. Okay. Let's talk about D. Since nothing is mentioned about the domain, et cetera, we'll take the most exhaustive domain possible. What was it? By the way, I forgot. Ellen cost sign X. Ellen cost sign X. Okay. Now everybody please understand it. See, since there is an outside root, and you want this to be a real valued function, then this guy must be greater than equal to zero agreed, because there is an outside square root over it. Yes or no? Yes or no? That means you're saying cost of sign X is greater than equal to one. I hope you remember your logarithmic in equations that we did under the chapter theory of equation last year. Correct. But we know that cost sign X cost of anything cannot exceed one, right? So this implies cost of sign X can only be equal to one. Because this is out of question. Agreed. So far fine with everybody. So far fine with everybody. Okay. All right. Now this means sign X has to be only zero. But here many people will say, so why zero? Why not two pi? Why not minus two pi? Why not four pi? Why not minus four pi? For the simple reason it cannot be two pi, because two pi is roughly 6.28. Sign cannot become 6.28, right? Sign has a limit. Minus one to one. That limit cannot be crossed. Right? It has a limitation. Sorry, I spoke too much Hindi. It has a limitation to be between minus one to one. It cannot surpass that. It cannot surpass these values also. So it only can become a zero. And if sign X is zero, your X can only be multiples of pi. Correct? So it can be zero pi, two pi, three pi, four pi, minus pi, minus two pi, minus three pi, and so on and so forth. Now, what do I understand from this entire set of process which I did? I realize that this function will only exist for such values of inputs which are multiples of pi. And if you put any multiple of pi, sign X is going to become a zero. Cos X is going to become a one. Ellen is going to become a zero. That means your output will always be a under, will always be a zero. What kind of a function is this? What kind of a function is this? It's a constant function. A constant function is a many one function. We all know that. So the verdict, the final judgment is many one function. So this guy is also not correct. Just for fun, we solved it. We could have stopped at C also, but I wanted you to have a complete picture of the scenario. Is this fine? Any questions, any concerns? Any questions, any concerns? Okay. So we'll now move on to another typification of functions. So what we did was one type of typification. What was that type? One one, many one. This is another typification which is called on to into. You must have heard of this on to or read of this on to many places while solving the problems. So on to functions, what is on to functions? First of all, I'll tell you the name alternate name for it. It is also called surjective functions. Okay. This name surjective. I mean, many people ask me, sir, what does it mean? Okay. It is actually it's given by a French guy called Nicholas Warbaki. So it was given by a French guy by a French mathematician called Nicholas Warbaki. Okay. So you use this word surjective while talking about these kind of functions. What are these functions we'll discuss, but there is no, I mean, I at least I do not know of any meaning of the word surject. Okay. If you're able to find a meaning to it, do let me know. I just know that this guy French guy, he was a French mathematician. He came up with this word surjective and that's why it is called surjective. You may also call it a surjection. Again, a mixture of the word surject and I own surjection. Okay. What is the, what is the onto function or what is the surjective function in plain simple words, functions where your core domain is equal to the range. That is to say every element of the core domain must have at least one pre-image, must have at least one pre-image. That means there cannot be any element of the core domain which is left unmapped, plain and simple. So all the elements of your core domain must enter into the mapping or must have up at least one pre-image that gives that answer. Are you getting my point? So normally we know that your range is a subset of the core domain, right? So if that subset is an improper subset, it becomes an onto function. Okay. If it is a proper subset, it will become an into function. I'll talk about it in something. Okay. So is it clear what's an onto function where every element of the core domain, I'm not saying every image. I'm not saying every image. I'm saying every element of the core domain must have at least one pre-image. Okay. A typical arrow diagram, if I have to draw for the same, just a typical arrow diagram so that there is an image left in your mind. So let's say A1, A2, A3, A4 and let's say this is B1, B2, B3. So basically something like this is a onto function. You can see that none of the elements of set V, which happens to be your core domain, is left out. By the way, if I ask you a complete name for this function, you will say many one onto function. Yes. So we'll have a permutation combination of these two typifications. It is not like it is either only one one or only many one or only onto or only into. There can be a combination also. So this is an example of a many one onto. Okay. So from one one and many one, you can pick one of them and from onto and into you can pick one of them and you can combine it to create another type of function. So it could be like one one onto one one into many one onto many one into so four combinations can result from it. But it cannot be one one many one simultaneously. Okay. Don't say that it is a one one many one. No, that cannot happen because they're complimentary. But one one can pair up with either onto into many one can pair up with onto it. Similarly, onto into cannot go together. They're complimentary. What is into I will discuss with you in some time. Now, what is the method to find? What is the method to find? What is the test to find or identify? I should use the word identify and not find test to identify and onto function. Okay. The test is simple. Find the range. So let's say there's a function given to you. So find the range of F. How will you find out? You already know the trick. You have done it in class 11. Sir, we don't remember now, sir. Then please go and revise. Simple. Open your book, solve few questions from the MCQs. Okay. After how to find out the range because that is not going to be taught. Of course, we are going to do examples and in that process, we are going to recall it. And if your range happens to be so this is step number one and if your range happens to be equal to set B, then this implies it's a onto function. Okay. Else it will be into function. Okay. We'll talk about into in some time. So is this test clear? Is this test clear to you? Any questions? So what's an onto function in plain and simple word where co domain is equal to the range? Correct. What are the tests to figure out whether a function is onto or not? Find out the range by whatever process you know. See whether it is equal to your set to be given in the question. So B will be given to you in the question. Okay. So this will be given. This will be given to you in the question. Right. Check whether they're equal. If they're equal, it's onto. If they're not, it's into simple as that. Okay. So since I talked about onto, I'll also talk about into into is just a complimentary or onto. So what's an into function? Play them simple words. It's complimentary to onto. Okay. That means at least at least one element of set B must not have any pre-image. Okay. So if there is an element of set B, which is left out, let me give you a diagrammatic representation of it so that the picture is well set in your mind. So something like this is an into function. So let's say this goes to B1. This goes to B2. This goes to B4. So what is happening? B3 is left out is left unmarked. Correct. So at least one element of B doesn't have any pre-image that means it is, it is not an image for that matter. That means in this case, your range happens to be a proper subset, a proper subset of your code domain. Okay. So in such cases, your function will be a into function. So this is an example of an into function or you can say non-subjective function. Okay. Is it clear? Now one exercise that we did for the previous cases was how many number of onto into function. So let us talk about that as well. Meanwhile, if you want to copy something, please do that. Done. So everybody is now clear with what's onto domain, sorry, code domain and range. They are same. Okay. And into range is a proper subset of the code. Okay. Now coming to the last, sorry, coming to the next part, how many onto functions are there? So number of onto functions, number of onto functions. Now here, let me first ask this question to you. Very simple question. How many onto functions are possible if let's say the number of elements in set A is N number of elements in set B is R. How many onto functions are possible if N happens to be lesser than R? Think carefully and answer. How many onto functions are possible if N happens to be lesser than R? Please give me a response on the chat box. Excellent. Excellent. Excellent. Okay. Okay. Most of you have given the right answer. The answer here is zero. Okay. Now see how see when N happens to be less than R. Okay. So let's say I take a case study. I take a case study where this is three and this is four. Okay. So let us say, let us say there are three elements A1, A2, A3 and A and there are four elements B1, B2, B3 and B4 in B. Okay. Now my attempt would be to make an onto function. That means I would not like to miss out on any one of B1, B2, B3. Okay. So let's start the mapping. And again, I'm starting with the, I can say an attempt to create an onto function. Okay. So let's say A1 goes and maps to B1. Correct. Now if I want to create onto function, I would like as much as B1, B2, B3, B4 to get mapped. Right? So I would not like A2 to map to B1 again. So I'll choose a different, you know, element to map. But however hard I try, I would realize that B4 is left out. Now you'll say, sir, map A3 to B4, but this is not a function, right? I cannot map A3 to B4 because it will violate the definition of a function in function. Any preimage cannot have multiple images, isn't it? So whatever I do, one of them here will be left out. Correct. That means there will be no onto function possible. This will actually become into. So the answer here is zero. At this situation, you can think as if, I think as if there are, there are. So this situation is analogous to saying that you have to distribute. You have to distribute N different balls or N different objects to our people or our persons such that such that every person gets at least one object. Okay. So finding the number of onto function is like meeting this criteria where you're trying to distribute N objects which are distinct among our people says that every person gets an object, at least one object he should get. So what is happening when N is less than R? So number of objects are less, number of people are more. Then how can you ensure that every person gets at least one object? You cannot ensure. There will be one person at least who will go empty handed. Okay. That is exactly happening in this case also. Okay. So here you cannot have any onto functions meet. Now tell me what will happen if N is equal to R? What is your result? Some people are already with the answer R factorial or N factorial absolutely correct. Okay. Very obvious if you take N as R. So let's say I take N is equal to R scenario. Let's say I take both of them as three three just for just for figuring it out. Let's figure it out. So let's say I have three three elements in each one of them. A1, A2, A3, B1, B2, B3. Now if you want to create an onto function, you must try to use single single mapping as far as possible. So this maps to this. Let's say this maps to this. Let's say and this maps to this. In this case, you will realize that you will automatically end up having one one mapping. Because if you do many one mapping, there is a chance that an element of set B will be left out. Isn't it? So let us say I decide to not map A3 to B3. I decided to map to B2. Right. So what will happen? This guy will be crying. It is left. Nobody's attending to it. Okay. So this will no longer remain up onto function. So when R is equal to N, it tantamounts to you having a one one function. So how many one one functions were possible? When we talked about N and R being equal, it was clearly N factorial or R factorial. So this becomes your total number of onto functions. Okay. Now guys and girls. The third case where N happens to be more than R. This is tricky actually. This is tricky. Some people are saying R to the power N. But this is tricky actually. Okay. And that's why I decided I will discuss it in the next class with you because it really requires around 20 minutes, at least 15 minutes of discussion, which we actually don't have. So this is something which is slightly tricky. Okay. Why it is tricky? And those were giving the answers that you feel. Please try it out with some small example. Okay. It will not work. No, no, it is not zero. It is going to be a formula which comes from principle of inclusion and exclusion. We'll talk about it in our Friday's class. Don't worry. So Friday's class will start with this. But before we end today's session, a lot of things changes. Aditya will discuss in our next class. Okay. Don't worry about it. But I would like to summarize today's class with one question on identification or one question on onto function, because I don't want to end the class without it. Else you'll not be able to solve your DPPs. So just one question I will take up on this. Okay. So that later part we'll discuss it in our next class. Don't worry about it. Don't worry about it. Let's focus on this question for the next three minutes. There's a function defined from R to a as X square by X square plus one. If this function is a surjection, find your set A or what is the most exhaustive set A possible? Let's solve this. I'll give you around two minutes. Yeah. YPR, your teacher will start integration. I don't know why. Very good point of guys. Let me tell you integration is a chapter which is not one week, not two week, not three week. It takes around one and a half months to finish. It's huge. The chapter is huge because you have almost 20 to 25 types of integration. Yeah. You have a lot of types of integration and for every type you need to at least practice five to six problems. Okay. So many people are saying, okay, okay, okay, I would love everybody to respond. This is our last question for the day. So please participate. Okay. So what is the set A? Set A is the range, isn't it? So basically you have to find out the range of this guy. Okay. So let's figure out what is the range. Even if you have not done it earlier, we'll discuss it out. So what do we do in the range? We first equate it to Y. Okay. And then we make X the subject of the formula. So here if you want to make X the subject of the formula, let's cross multiply. Okay. So this boils down to X square minus one Y. Or you can write it like this. X square Y. X square Y minus one plus Y equal to zero. Okay. So this gives you X square as Y by one minus Y. Yes or no? Okay. So X being a real number, X square will always be greater than equal to zero. That means Y by one minus Y should always be greater than equal to zero. This thing can easily be done by our baby curve or what we call as method of intervals. Okay. So let's plot down the zeros of this. So zeros of this is zero and one. This side will be negative idea. Please be careful. Don't be like blindly putting positive sign to the right most interval. You can always do a sample check. For example, if I pick up a two and I put over here, I'll always get a negative answer two by one minus two. So this is negative. So this will become positive because all these factors are subjected to odd powers. Okay. And again, this will be negative. So there will be a switching of the signs. Again, people who were not with us in the last year, when we did this baby curve, I would request you to watch the lecture on theory of equations. Okay. If you cannot find it on learners, you can always think me, I will send you the entire video series for that. Okay. Now here you want it to be greater than equal to zero. So please note that between zero to one, but one cannot be included. Zero can be included. So this will become your range and hence this will become your set A. So those who have answered with zero to one, zero inclusive, one exclusive, thumbs up. Absolutely correct. Okay. So we have just started with this concept. We are going to take many more examples in the coming class in the Friday's class. We will, first of all, we'll do how many onto functions are there. And then we'll also talk about how many into functions and then we'll take a mix of all the types. Okay. As I told you, there can be permutation combinations. So we'll talk about mix of both the types. Meanwhile, today I'll also send you the DPPs. Try to submit it. Okay. Let's, let's try to submit it next Wednesday. Okay. I'll give you one week time to submit it. And next class, we are also going to talk about more concepts like a composition of functions and all. Now you may hear of some word called bijective in your DPP. So I'll just talk about it for one minute. I hope I'm not taking much of your time. You will come across a word called bijective many a times in DPP bijective functions are basically those functions which are both one one and onto. Okay. So a combination of injective plus surjective becomes bijective. Okay. So one one and onto functions. They are called to be also they're called to be bijective functions bijections also is correct. Okay. They're called to be bijective functions. We will discuss about it a little later on in our next class, which is on Friday. Okay. So this will be continued. Thanks so much for joining in for today's class. Thank you so much. Thank you. Bye bye. Take care.