 Hello, good afternoon, welcome to Centrum Academy, live class. So those who have joined in the session, I would request you to type in your names in the chat box so that I know who all are attending the session. So I would request people who have joined in the session to type in your names in the chat box so that I know who all are attending the session. All right, so let us begin with today's class. So today my agenda would be two things. In the first one, one and a half hours we'll be talking about how to find out the range of functions. So the first thing is, we'll be talking about how to find the range of real functions, real functions, real valued functions. They're all the same things. And secondly, we are going to talk about a new chapter today which is called the principle of mathematical induction. Principle of mathematical induction, PMI, right? Very easy chapter. So last class we had an offline class and we discussed about how to find out the domain of the function, right? So today we'll be carrying forward with the concept of finding the range of a function. Now what is the range of a function? What is the definition of range of a function? Basically it's the span of the function along the y-axis. So it is the span of the function along the y-axis. So the values of the function for the different inputs or different real inputs, or you can say the domain of the inputs is basically called the range. Okay, now there's a different range and co-domain, right? Please remember range is a subset of the co-domain. Okay, so range are those values which are actually taken up by the function when you put the inputs which are belonging to the domain of the function. So a very simple rule here is domain is nothing but the span of the function along the x-axis, whereas the range is the span of the function along the y-axis. Now before finding the range of the function, there are certain steps which we normally follow. Number one, we should always first find out the domain of the function because there is no existence of range without a domain. So step number one, you should always find out the domain of the function because domain actually restricts the range of the function. Okay, so first find out the domain. Now, if your domain, let's say I call it as DF, okay? If your DF function happens to be a discrete set, it happens to be a discrete set. Discrete set means let's say your domain has values like A, B, C, okay? Then what would be the range of the function? Range of the function would be nothing but it would be again a set which would be containing FA, FB and FC, okay? So all you have to do is you have to put the values of A, B and C into the function and just see what is the outcome that you get from it. That would be your nothing but the range of the function, okay? However, such cases are asked very less but you may get a case like this to deal with. Let's take an example on this type, okay? So I'll just take a pause over here and take an example where you can actually find such a question. Let's say I ask you, find the range of, find the range of F of X which is a function which is given by seven minus X, PX minus three. Now most of you would not have done permutation combination but there is something which we call as NPR, the number of ways in which you can arrange R objects chosen from N objects, okay? NPR formula is nothing but N factorial by N minus R factorial, okay? So now we have a function where your N role is now being played by seven minus X and R role is being played by X minus three, okay? Okay, so the definition of NPR is clear, Venkat. So NPR means the number of ways of, it's the number of ways of arranging R objects selected from N distinct objects, yeah. So can you try this out? What will be the range of this function? Just remember the first step, you have to first find out the domain of the function. Just a minute for you to try it out. So the first thing that you would do is you would first find out the domain. Now, when you know that you have to find the domain, you have to take care of two things. One is this quantity must be always greater than or equal to zero and this quantity must also be greater than or equal to zero. We can't have negative quantities, okay? So this means X is less than or equal to seven and X is greater than or equal to three. In other words, X lies between three and seven, okay? Now, this is not sufficient. We have to also take into account that X minus three must be less than or equal to seven minus X, all right? That is two X should be less than or equal to 10. That means X should be less than or equal to five, okay? Now, the combined or the intersection of these two intervals we have to take, right? If you take the intersection of these two intervals, let me call it as one, let me call this as two. So if you do one intersection two, you'll end up getting, I mean, I normally use a number line for it. Here it is too simple for you to use a number line, but still I would recommend always use a number line. So the first interval is from three to seven. Second interval is from five and below, right? So it clearly implies that your intersection is happening in this zone. That is your X should lie actually between three and five, okay? Now, again, here my X value has to be integers. So it implies X could either be three, four, or five. So these are the three values. So this is acting like your domain of the function. So domain of the function will contain three, four, and five. Okay? So my first step is accomplished. My first step is accomplished. I have found out the domain of the function and I realize that it is coming out to be a discrete set which has a certain number of elements. Okay? So now what I will do is the range of the function would be nothing but I'll start putting these values in place of X. So seven minus three, P, three minus three. Okay? This will be one answer. Other would be seven minus four, P, four minus three, and seven minus five, P, five minus three. Okay? Let's calculate these values. They are easy to calculate. This is going to be four, P, zero. Okay? This is going to be three, P, one. Okay? This is going to be two, P, two. By the way, four, P, zero is one. Three, P, three is going to be three. And two, P, two is going to be two factorial, which is two. So this will become the range of the function. Is that fine? How it works? But apart from JE, don't expect such questions to be asked in school exams. In school exams, normally we don't see questions which are based on those type of problems where the domain comes out to be a discrete set. Okay? What are the next scenario? Scenario number two. Scenario number two is where your domain comes out to be a continuous set, but a finite something like this. Continuous, but finite set like this. So this is a set of all numbers, all real numbers, let's say from A to B, but it's a finite set. I know there are infinitely many real numbers between A and B, but it is not open-ended. Just like, you know, all real numbers or it doesn't go all the way to infinity or minus infinity. Okay? In such cases, what do we do? In such cases, your range is defined as the minimum value of the function to the maximum value of this function in this interval. Okay? To a certain extent, you may have to use maximum and minimum also for this. Okay? Or there may be some cases where you can actually speculate the maximum and the minimum values, by knowing the fact that you know the restrictions. For example, in case of trigonometric functions, we know that in trigonometric functions, the sine x function, the cos x function, they're all limited between minus one to one. So in those scenarios, it's very easy to find out the range of such functions. Okay? By looking at the maximum and the minimum values. Let's take an example based on this. Let's say I ask you this question, find the range of c-coff, five by four cos square x. Find the range of c-coff, five by four cos square x. Can you guys try it and tell me the answer? Again, requesting people who have joined in the session to type in your names, I could just see two names so far. Any idea? Gee, it's very simple. You know that your cos x lies between minus one to one. So cos square x will lie between zero to one, correct? So pi times, sorry, pi by four times cos square x will lie between again zero to pi by four, right? Correct? So think as if seek has been fed values from zero to pi by four. So if you quickly draw the graph of seek, I'm just drawing a part of the graph from zero onwards, zero to pi by two, let's say this is your pi by two line, asymptotic to the graph. So pi by four would be somewhere over here, correct? So its value will be from here to here. And this value is one and this value is root two. So you can directly say that the range of the function would be from one to root two, done, correct? So here I know that you would say that the range of this function would be, sorry, the domain of this function would be all real values, right? But normally we see that the cos square function is restrictive function. This is acting like a input for seek function. So even if x is taking all real numbers, your cos square x is restricted between zero and one, right? So more or less we are talking of the same scenario as what we discussed in the theory right over here, okay? So you know there's a restriction in the input or indirectly there is a restriction in the input and hence my output cannot go beyond a certain range, okay? So this is how you can actually find out the answer for such type of scenarios where your domain is a continuous but a finite set. So you can find the minimum and the maximum value of that function in that interval and that becomes your range. Let's have another one, let's have another example. Let me call this as one. Let's take another question. Find the range of, find the range of one by and a root of four plus three sine x. Just type done on your screen if you are done on the chat box so that I can start discussing it. All right, so here also the situation is pretty much similar. You know that your sine x will lie between minus one to one, correct? So let us try building on this expression. So three sine x will lie between minus three to three, correct? Now add a four, so four plus three sine x would lie between one to seven, correct? So under root of four plus three sine x will lie between one to root seven. So reciprocal of four plus three sine x under root will lie between one by root seven to one. One by root seven to one, which implies that this function range would be from one by root seven to one. Absolutely correct, absolutely correct. Venkat, let's take another one. Find the range of under root pi square by nine minus x square. Any idea? First of all, again step number one is find the domain of the function, correct? Now in order to find the domain of the function we have to take into account that the expression within the under root sine, this should always be greater than equal to zero, right? Which means x square minus pi square by nine should be less than equal to zero. Now this is factorizable as this, correct? Which means x lies between, I'm sure you all are aware of the wavey curve sine scheme. So minus pi by three, pi by three, the signs will be plus minus plus. So you want it to be negative, that means it should be between minus pi by three to pi by three, okay? Now, if you want to know what is the maximum and the minimum value that this function can take, so we first need to know what is the maximum and the minimum that pi square by nine minus x square can take in this interval minus pi by three to pi by three. So what is the value of, or what is the value will not be a right term? What would be the range of this expression when x lies between minus pi by three to pi by three? So let us first find that out, okay? Now it's very clear that this is actually a parabola which is opening downwards. It's an inverted parabola and that parabola will actually cut the x-axis between pi by three to minus pi by three, all right? And since you're in the interval minus pi by three to pi by three, as you can see the range would be from zero all the way till this point, correct? And this point is the point where your x is zero, so this value is going to be pi square by nine, correct? So the range of this expression, the range of, or you can write it in terms of symbolic representation. So pi square by nine minus x square lies between zero to pi square by nine, okay? So under root of pi square by nine minus x square will lie between zero to pi by three. So tan of this expression that is under root pi square by nine minus x square will lie between tan zero to tan pi by three. Remember tan is an increasing function, okay? So this expression would be between zero to root three. Correct, right? So this will become our final answer. So my final answer would become the range of this function would be from zero to root three. This would become your answer. Is that clear how it works? Right, now we'll be touching upon the next aspect. So I'm going back to my board three. In case of any doubt, please type it in your chat box so that I can address it. So now we'll be talking about the third scenario. The third scenario where your domain of the function is an open interval, right? Something like a to infinity or something like minus infinity to a, correct? Or something like all real numbers or something like all real numbers excluding certain values, okay? These are called continuous intervals but they are open, open in the sense that they're not restricted unlike your previous situation, okay? So in these situations, what do we do is we use the concept and I'll be talking about that concept in detail, but just as of one liner here, we say that the range of the function is equal to the domain of its inverse. So we take the approach where we find out the domain of the inverse of that function and the domain of the inverse of that function actually acts like the range of the function. Yeah, yeah, I'll give you Bharat, no worries, sorry, Venkat. Okay, so let me just complete this scenario. This is a very important thing because all your school level questions would be directed towards this type of question. Okay, what do I mean by inverse of a function? First of all, let me talk about that. See, I'll take a very simple example. Let's say I have a function which is 2x plus three, okay? I want to find out, sorry, 2x plus three, let's say divided by 3x minus one, okay? Let's say this is my function and I want to find the range of this function, okay? Now what I do is we first find, we try to find an expression for the inverse of this function. Now what is inverse of a function? In a simple and plain language, if function is a machine where if you put in x value and you get a y value, then inverse of a function would be such a machine where if you put this y value here, it will give you the x value back, okay? A few typical examples would be like log to the base ex and e to the power x. These are inverses of each other. These are inverses of each other. Why they call inverses of each other? And this is the way to symbolically represent an inverse. Please note, this doesn't mean one by f. It is just a symbolic representation. This is a symbolic representation, not to be treated as one by f, okay? Now what is the inverse of a function? Inverse means what the function does, inverse will do exactly opposite of that. For example, if I put my x as e square, my function will actually give me two as the answer, correct? If I put this output in this function, I will get e square back. Is that clear now? What is the meaning of inverse? Another example would be sine x and sine inverse x. Here if you put in pi by six, let's say it gives you half, correct? In this, if you put half, it'll give you pi by six back, right? Now because of this relation between f and f inverse, normally we say that if f acts from a and gives you b as the answer, f inverse will act on b and give you a as the answer, okay? So now if you want to find out the range of f, is it not better that we find the domain of f inverse because both are same things, correct? Therefore we take this approach while finding out the range of functions in this case. So let me apply this approach in this case. By the way, let me tell you, this inside story may not be conveyed in the school, in the classes in the school because they don't want you to get into the, no aspect of understanding what is inverse because inverse is the class 12th topic, right? So they would not actually take you to the inverse concept in order to make you explain this. They will just tell you the mechanism, what we normally follow to find out the range, right? So let us find out how the inverse of this function. So for the inverse of this function, what do we do first? So in general, how do we find out the inverse? In general, how do we find the inverse? Okay. So I'll just tell you the approach for it and then I'll start solving this problem for you. So if you have been given any function, we normally first write it as y is equal to f of x, okay? Then what do we do is, we make x the subject of the formula. So when you make x the subject of the formula, automatically you end up getting x as f inverse y, okay? Now treating this as my new function, let's say g of y, I start finding the domain of g of y and when you're finding the domain of g of y, you automatically end up getting the range of f of x, okay? Let's apply it to this question. So step number one, I will take y as 2x plus 3 by 3x minus 1, right? Step number one is done. Step number two is I make x the subject of the formula. So for making that, I have to cross multiply. Yeah, absolutely Venkat. Normally when we define a function's inverse, there is a criteria that we have to follow. The criteria is the function must be a bijective function. Bijective function means a one one and onto function. Again, it has a lot of theory involved. I have just shown you the tip of the iceberg. We cannot find inverse of every function and even if you are finding the inverse of every function, you have to ensure that the domain is such that the function remains a one one function in that particular domain. Now what is one one, what is onto? That's again a subject matter of class 12th. But I'll definitely talk about it in some time. In fact, I can talk about it right now. So we can park this problem for the time being and let's talk about it. When you say a function is one one, it means that the function doesn't take a turn back. That means it's always increasing or always decreasing. That means for one input, there has to be one, the one output should come only from one input. It cannot come from multiple inputs. A typical arrow diagram for a one one function would be like this. Let's say a, b, c, one, two, three, four. This is a one one function, okay? This is a one one function. On the other hand, there is something called many one function. Many one function, if I show an arrow diagram, you'll come to know by the arrow diagram itself why it is called a many one function. Let's say a, b, c, and we have one, two, three. So it is something like this. One is going towards, a is going towards one, b is also going towards one, okay? So something like this is a many one function. Now if you want your function to be invertible, it must be a one one function. Why is that so? Because if let's say I talk about sine x function, right? You know this function is not a one one function if it is defined from all real numbers to real numbers, right? So as you rightly said, when you put 30 also we get half. When we put 150 also we get half. When we put 390 also we get half. You're absolutely correct. So what happens is in that situation, there would be a confusion when we are finding the inverse of it. So let's say 30 degree gives you half, 150 degree also gives you half, 390 degree also gives you half. So this is when you are writing the function mapping. But when you're writing the inverse mapping, please note that inverse mapping is ulta, it is opposite. So you're mapping b to a now, set b to a. So please note in this case, what will happen? Half will map to 30 also, 150 also and 390 also. And we know from the basic definition that this is not a function. This is not a function. So first question would arise in your mind. So sine x is not invertible. How am I talking about sine inverse x? Now there is a basically a catch over here. So what do we do is we actually don't take all real numbers as our domain. We only take from minus 90 to 90, okay? And not only that, we also take our co-domain to be from minus one to one. So now if you are saying the mapping of f in this case, you will see that only 30 degree will give you a half. So it's inverse would only map to 30 degree. Are you getting this point? Anyway, these are things which are not required. As of now, we are going to anyway spend a whole class on the type of functions in class 12. But just to give you an idea that why we normally talk about inverse of sine x function, even though you can have multiple answers. So we restrict the domain of the function in order to make it a one-one function, okay? Let's go back. Yeah. Now in this, I'll make x the subject of the formula. So step number one is accomplished. In step number two, I will make x the subject of the formula. So let me multiply it with y, okay? So three x y minus y is equal to two x plus three. So take the x on the one side. So three x y minus two x is y plus three. Take x common, so three y minus two is equal to y plus three. So x will become y plus three divided by three y minus two, okay? Now treat as if you have been given a function gy, which is y plus three by three y minus two. Now, I have a simple question for you. What should be the domain of gy? What should be the domain of gy? So you would say it can be any real number, but it should not be two by three. Because if it is two by three, what will happen to the denominator? It would become zero, right? So I don't want my denominator to become a zero. That means my y should not be two by three, correct? Now exactly the domain of gy will actually become the range of the function, right? So this function, this function, range would be all real numbers except two by three, right? Now many people don't believe me. They say, sir, oh, do you mean to say this cannot become equal to two by three? I say yes. So they try to test it and they try to equate it to two by three and try to see whether it really works. So if you see this function, two x plus three by three x minus one, try to equate it to two by three, see what will happen. Let's try to see. We are trying to claim that this function cannot become two by three, right? So let's try to equate it to two by three and see what happens. So when you cross multiply, it becomes six x plus nine is equal to six x minus two. In other words, it'll give you the shock of your life when you see nine is equal to minus two, which is not possible, correct? So it is just like a verification that you may actually perform and see that whether this is coming out to be the right answer. So this becomes your range of the function. Any question regarding this? This is the most important type that you will get in your school exams, okay? So we'll take few more questions based on them. So let's take this question. Find the range of x by one plus x square. Always remember the first step is finding the domain. So domain of this function is all real numbers, right? Because your denominator will always be, you know, defined for all real x, numerator is also defined and denominator can never become zero. All right, so let's look into this. So first y is equal to x by one plus x square, okay? Now I have to make x the subject of the formula. So if I multiply with this, it becomes x square y plus y minus x equal to zero. And you can see here that there is a quadratic being formed, okay? Now, if you use the Shridharacharya formula, that is minus b plus minus under root b square, b square minus four ac by two a, okay, you get this, correct? Now, irrespective of whether there is a plus or there is a minus, there are few things that we need to take into account. One is since my x is real, this term should also give me a real answer, right? If my x is real, this term should also give me a real answer. And that can only happen when my one minus four y square is greater than equal to zero, okay? That is number one. Number two is people say that the denominator here, even that is not permittable to be zero, okay? Now, let us look into the second scenario first. Let us look into this first. You are saying y should not be zero. Now, my counter question is why? Because I can always get y as zero when my x is zero, right? And zero is very much in my real numbers, very much in my domain of the function. So, basically, you have to ignore this condition. Now, why does it come? I'll tell you in some time. Let us look into the first situation. So, when you say one minus four y square is greater than equal to zero, let us try to solve for the interval of y which satisfies this. So, I can use factorization over here, correct? Okay, let's use the real number line for this. So, the zero of this is going to be half, the zero of this is going to be minus half, okay? Let's take a value higher than half. Let's take one. In one, you'll get everything negative. So, negative, positive, negative, okay? Now, you're wanting the interval where it is greater than equal to zero. Greater than equal to zero means wherever you have written a positive sign. So, it would become minus half to half, okay? And this very much includes zero, okay? Now, this becomes my answer, actually. So, my range of the function should be from minus half to half. Now comes the point, why did not I remove zero? Because if y is zero, my denominator will become zero and everything will become undefined. Now, here is a catch which very less people know and many people tend to ignore also. When y is zero, you can see your numerator will also have a chance of becoming zero, right? There, my making it an indeterminate form. Thereby, making it an indeterminate form, which actually we can take any values, correct? So, blindly removing y equal to zero from our range set is not justified, okay? Because we can clearly see that my y has full right of becoming zero when x becomes zero. Now, in this case, zero is included. That doesn't mean it'll work for every other case. You have to be watchful. For example, let's say you had some value, which actually was genuinely not taken by y, then you have to remove it, okay? So, these are indications that the solution actually gives you, but you don't have to take all the indication very seriously in the sum that you don't have to react to that. You have to use your own discretion that, okay, if it is saying y is not equal to zero, let me check here. When I checked here, I realized that y could be zero. So, there's no point saying y cannot be zero, correct? You know, many, 90% of the students will write this answer as minus half to half excluding zero. They will exclude zero from this, but which is not correct, okay? So, don't do that. Just do a quick check. That is my suggestion in this case. Is it fine? Any question here? Vidyota, Venkat, Advaita? Let's move on to few more examples. Let's talk about a very simple function. Find the range of under root of 16 minus x square. It's pretty simple. So, first of all, when you say, what is the domain of the function? For finding the domain, you'll say 16 minus x square should be greater than equal to zero, which implies x square minus 16 is less than equal to zero. So, basically, you are going to use the wavey curve sine scheme over here plus minus plus, okay? Less than equal to zero means negative interval. So, this function would work only when your input is between minus four to four, correct? Now, first of all, I'll talk about 16 minus x square. 16 minus x square is a function which opens, it's a parabola which opens downwards, okay? With these values, there's four and minus four, correct? So, in the interval minus four to four, as you can see, the function is between zero to 16. So, this lies between zero to 16, correct? So, when you take the under root, basically you're taking the under root on either side of it because they're all positive quantities. So, 16 minus x square would lie between zero to four, and that's how your job is done, right? So, the range of this function is from zero to four. This is a very favorite school question that they normally ask, okay? So, mark it with a star. Now, you may get a situation where you may be asked a question which is based on modulus function, okay? So, let's take another type of questions where we find out the range by redefining a modulus function. So, consider the question, find the range of mod x minus one plus mod x plus two plus three x minus four. Now, in such questions, it is preferred that you actually plot this. It's preferred that or is suggested that find the range by plotting the function. Find the range of f of x by plotting the graph of the function. In school, they would say redefine this function and hence find the range of the function. Okay, so for such kind of functions, we'll first redefine the function. So, step number one is, we'll spend time redefining the function, okay? Let's redefine f of x. Now, I've already discussed whenever modulus functions are involved, they actually take different definitions under different intervals, okay? So, let's say if you talk about mod of x minus one, okay? It'll change its definition about one, okay? And mod of x plus two will change its definition about minus two, okay? So, what I'll do is, I will plot minus two and one on my real number line. Many people ask, sir, why didn't you plot four by three? Please note that this doesn't change its definition. This remains, right? It doesn't change its sign, okay? In the neighborhood of four by three. So, only when we have whatever other functions which are subjected to modulus, only the zeros of those functions would be plotted, right? So, there's no need to plot four by three on the number line, right? Now, this divides the interval into three parts. Now, for each part, we'll have to redefine these modulus functions. So, first, let us take interval where x is less than equal to minus two. So, let's see what is the definition of the function. When x is less than equal to minus two, mod of x minus one will actually become negative x minus one, right? Mod of x plus two will become negative x plus two and there would be no change in three x minus four. It will remain three x minus four like that. You don't have to change the sign of anything in three x minus four, okay? Let me just space this out a bit. It's becoming too close to my, I think. So, x less than equal to minus two, right? Now, let us talk about interval which is between minus two to one. In minus two to one, mod of x minus one will again be a negative x minus one, right? Whereas, mod x plus two will remain x plus two. So, there will be no change in the function. And again, three x minus four as discussed would remain three x minus four. Now, when x is greater than equal to one, what will happen? Everything will become positive in this case. So, this will also become x minus one. This will become x plus two. This will become three x minus four, okay? So, what have I done? I have redefined the function in different intervals of x. Now, normally we can simplify it also and rephrase our function. So, this will become a negative two x plus three x, which is x, and we'll have one minus two minus four, okay? One minus two minus four, that will become a minus five. This will become x and x will get canceled. So, we'll have three x, okay? One, three, and this will become a minus one, okay? This will, on the other hand, become five x minus three. Now, let us plot this on the graph. So, minus two is somewhere over here, okay? One is somewhere over here. So, from x minus infinity to minus two, it will follow the graph of x minus five, right? So, x minus five graph would be a line having a positive slope. And when you reach minus five, sorry, when you reach minus two, its value would actually be minus of seven. So, it comes all the way like this. Let me just slightly drag it up, yeah. So, it comes like this and stops at minus seven, okay? Remember, this value is, this value is minus seven, okay? Post that it will follow three x minus one graph till one. So, I'll just continue from here, three x, yeah. And at one, it will stop at two. So, this value will be two. After that, it'll follow the graph of five x minus three, which will be a line with a further slope like this. So, it'll extend like this. And this will extend indefinitely. So, what is the range of this graph? Range of this graph would be from minus infinity to infinity. All real numbers are included in this, okay? So, this is how you find out sometimes the range of functions which are defined under modulus sign. Make sense? Okay, Vidyatha, is it clear? I have another question for you. Plot the function x minus two minus three minus x plus four. Okay? And state the range of f of x. Please try it out. Okay, let's approach this. Now, first of all, don't be scared of three minus x mod. It is the same as saying, mod of x minus three, right? There's no difference at all in there. So, now the two moduli x minus two mod and x minus three suggests that you need to plot two n three on the real number line. So, there are three zones again, one, two, and three zones. So, one is x greater than, sorry, x less than equal to two. Other is between two and three. And other is greater than equal to three. I think we had done this in a bridge course also to a certain extent. So, x less than equal to two, this guy is going to be negative of x minus two, okay? And same with this guy, this is going to be negative of negative which is plus x minus three plus four. Between two to three, the first one would be positive x minus two, while the second one will still be negative. So, plus x minus three plus four. Greater than equal to three, everything would be positive. Is that fine? So, if you redefine this, it will become negative x, x will get canceled. So, two plus four, six, six minus, three which is going to be three again. This is going to be two x minus one. Here again, I will have minus two, three, seven is going to be five. So, if you plot the graph for this, this is your two, this is your three. Less than two, it is following a simple graph like this. So, your f of x is always three in this. Between two to three, it is following the graph like this, correct? And after three, it is following the graph like this, okay? So, this is f of x equal to five line and in between this line is f of x is equal to two x minus one. Okay? So, you can say the range of this function would be from three to five. Check your calculations, Venkat. Did you find your mistake? Okay, so proceeding, now we'll take a different type of questions. The questions where actually discriminant of quadratic equation is involved. So, this is another type of question which normally comes in school as well as in J exams. Find the range of x square plus x plus two by x square plus x plus one. Okay, try this question out. Okay, so let us discuss this quickly. Now, first of all, domain. What is the domain in this kind of function? Domain is basically all real numbers because the denominator here can never be zero, okay? In fact, x square plus x plus one will always be a positive term, right? Why is that so? It's because you can actually complete the square and realize that it's always, right? In fact, it's always greater than equal to three by four. Correct? So, this is all real numbers. This is also all real numbers. So, the domain of the combined function will always be real numbers. So, the domain is all real numbers, okay? Now, let's take this expression as y, okay? One thing I can do here is that I can first write this as x plus one plus one divided by x square plus x plus one, which is nothing but x square plus x plus one by x square plus x plus one plus one over x square plus x plus one, okay? So, y becomes, by the way, you can cancel it off and say y is equal to one plus. So, this gives you y minus one is equal to one by x square plus x plus one. Now, let us write things in terms of x. So, let's cross-multiply. So, y minus one times x square plus x plus one minus one is equal to zero. So, if you collect your terms, it becomes y minus one x square plus y minus one x and here you'll have a y minus one minus one equal to zero. Okay? So, it gives you y minus one x square plus y minus one x plus y minus two is equal to zero, okay? Now, if I use the Shridhar Acharya formula or the quadratic equation formula to make x the subject of the formula, I get minus b, which is going to be one minus y plus minus and the root b square minus four ac, okay? By so far, so good. No problem so far. Now, if you want your x to be real, if you want this guy x to be real, the first thing that I need to take care is that the term which is under the square root should always be greater than equal to zero. So, this guy must be greater than equal to zero, okay? That means I need to solve this inequality here. So, I can always take y minus one common. So, it becomes y minus one minus four times y minus two greater than equal to zero, okay? Which will simplify to y minus one times minus three y plus seven greater than equal to zero, correct? Let's use our wave curve science scheme to know the interval where it is greater than equal to zero. So, I'll plot the zeros on the number line. So, zeros will be seven by three and one, okay? If you take a number greater than seven by three, let's say if you take a number three, this everything will become a negative, positive, negative, okay? So, it becomes this interval that is our required answer because in this interval only it is greater than equal to zero. So, your y should lie between minus, sorry, one to seven by three. Now, a point to be noted over here is that we cannot ignore the fact that y minus one should not be zero. Now, again, this is subject to a test that we can actually perform. Because last time we had an example where my denominator, even it was zero, it was included in my range, correct? Now, is the same situation repeated over here or in this case, do I actually have to remove y one from my answer? Let's check. A very simple way to check it. If you're saying y cannot be one, just try to give a function a value of one, okay? Let's say I say this function is equal to one, okay? Let me assign this a value of one. If I do that, again, something very strange is going to happen with me. You'll realize two is going to be equal to one, which is not possible, okay? And therefore, it gives you an indication that boss, we cannot include one in our answer, right? So, y cannot be one. If y cannot be one, they cannot be a square bracket over here. So, you have to further redefine this and say y belongs to open interval one to seven by three. Is this clear or not? At any step, if there is an ambiguity, please let me know so that I can clarify it because domain and range are very integral part of your class 11 functions. And this is going to be the forerunner of all the concepts related to functions later on. Now, let's move on to some more problems. So, we'll have at least five, six more problems, which we'll be taking up on these concepts. Let's start with this question. Find the range of x square plus 14x plus nine by x square plus 2x plus three. If you're done, you can type in your response in the chat box so that I can comment whether you're right or wrong. All right, so let's discuss this. If you talk about the domain of the function, x square plus 2x plus three should not be zero, which actually cannot be zero because if you see, it's actually made up of a perfect square like this. Okay, in fact, this will always be greater than equal to two. Right? Now, so domain is all real numbers. That means x can take any real value. So now for finding the range, we equate it to y. That's the normal procedure that we follow for finding out the range. And we try to make x the subject of the formula. Because we need to know the inverse of the function and the domain of the inverse will actually act as your range. Okay, so if you collect your x square terms together, you'll get y minus one. Okay, if you collect your x terms together, you'll get two y minus 14. And your constants will give you three y minus nine equal to zero. Okay, now this is a quadratic in x. This is a quadratic in x. And your domain is all real numbers. So the discriminant must be greater than equal to zero. Absolutely correct, Venkat. You're absolutely correct. The answer is minus five to four. Absolutely, great. Yeah, so discriminant greater than equal to zero means b square minus four a c must be greater than equal to zero. Okay, now if you simplify this, let's simplify this. First of all, I can pull out a factor of four from here. So it's y minus seven the whole square minus four. Y minus, in fact, you can pull out a three over here. You can drop a factor of four from everywhere. So it gives you y minus seven the whole square minus three times y minus one y minus three greater than equal to zero. So if you simplify this, you get a minus two y square and minus 14, and here you'll get a minus 12. So that gives you minus two y, okay? And constants would be 49 minus nine, which is plus 40, yeah. Okay, drop the factor of minus two, but when you drop a factor of minus two, make sure you switch the inequality like this. Now, this is easily factorizable. I think its factors are y plus five and y minus four. Okay, on the wavey curve, let's assign the sign to the intervals. So minus five and this is four plus minus plus. Since we want a negative interval, that means I will state this interval. So y belongs to minus five to four, okay? And this says y cannot be one, y cannot be one, okay? Let's try out whether y actually cannot be one, right? If we do that, we're actually trying to assign whether it can take a value of one. So let's check. So x square plus 14x plus nine is equal to x square plus 2x plus three, x square, x square goes off. And I'll get 12x is equal to minus six. That means x could be equal to minus half. So of course I can have a value of one, okay? So unlike the previous argument that we had, many people when they write x in terms of x as the, by the use of Schilderach area formula, they'll get two times a in the denominator and they say a should not be zero, but that situation you must verify again with the question. Right? So here y can be one. So this condition is not valid. So this answer holds good. So this is my range for the function. Is that fine? Which exam do you have tomorrow? Which UT's do you have tomorrow? All right, so let's have this function. Find the range of this function. Oh, English. Yeah, try this out. See everything depends upon this. Now first of all, I have to find the domain of the function. Okay? For the domain of the function, this guy must always be greater than zero. Correct? Okay, if you see this closely, this will always be actually true. Why? Because you can actually complete a square over here like this. So this becomes three x minus two by three the whole square and this becomes 11 by three. Okay? So this expression will always be greater than equal to 11 by three. Right? So the domain here would be all real numbers. Correct? Correct? Now, log is a continuously increasing function. Right? What is the meaning of increasing function? Increasing function means if you have a continuously increasing function and if you feed it a value which is between A and B, then the output would be from f of A to f of B. Right? This is the meaning of the function being increasing function. So something which is always on the rise. So if it's something is always on the rise that means if you feed the value from A to B, its range will also be from f of A to f of B. Right? This is a common sense thing that we can all apply. Okay? So now, log will only operate on what is the range of this function? Correct? The range of this function is from 11 by three to infinity. See, it is always greater than 11 by three, right? So I can say that three x square minus four x plus five will always be greater than, in fact, I should always write it like this, greater than 11 by three and go all the way till infinity. Correct? You can always draw the graph of it and check it out that it's a quadratic. Okay? Which is opening upwards. This point is the vertex which occurs at two by three comma 11 by three. So 11 by three is the minimum value of this function and it goes all the way till plus infinity. Okay? So log of this will always be between log of 11 by three which implies the range of the function would be from ln of 11 by three to infinity. This is going to be your answer. Are you getting it? So you have to take into consideration what type of composite function. This is actually a composite function, okay? So you have to see what is the range of the function which is fed inside the outside function. What is the nature of the outside function? Is that fine? In a similar way, let's say if I give you this question, find the range of ln of four minus x square. I'm sure all of you would be able to answer this. So please type in your response in the chat box. ln zero, see there's nothing like ln zero. You can say minus infinity to ln two, that is fine. Okay? So let's check this out again. Let's talk about four minus x square first. First of all, here there's a restriction that if you want this function to be a valid function, this term must be greater than zero. That means four minus x square must be greater than zero. Which means x square minus four should be less than zero. That means x minus two, x plus two should be less than zero. Which means if I use the wavey curve sine scheme plus minus plus, your x should belong to minus two to two. Is that clear? Now four minus x square is a function whose graph is like this. Okay? Cutting the x-axis at minus two and two respectively. That means the range of this function is from zero till you reach four. Okay? So four minus x square will lie somewhere between zero to four. Four can be included. Okay? So under root will lie between zero to two. Okay? And again, as you know, log is a increasing function. So when you take ln zero, it actually becomes minus infinity to ln two. So you can say, so a slight change you have to make in your answer, when that range would be from minus infinity to ln two or log two to the base E. Is that fine? Any question? Let's take this one. Find the range of, find the range of four to the power x plus two to the power x plus one. Real numbers from one to infinity, can it be one? Is one included in your range? Venkat? Oh, then it's correct. All right, so let's talk about this. Domain, first of all, is all the numbers. No doubt about it because each of these functions is defined for all the numbers. Now for the range, let's talk it, call it as y. So two to the power two x plus two to the power x plus one. So you can actually make a, okay? Equation like this and if you talk, if you call two to the power x as, let's say z for the time being, it actually sounds like, looks like a quadratic in z, okay? And I can use my Schroeder-Acharya formula to solve it, which is minus b plus minus under root b square minus four ac by two a, okay? Which actually becomes minus one plus minus under root four y minus three by two. Now please note at the end of the day, your z is two to the power x, okay? So a couple of things to be noted over here. Two to the power x is an exponential function, right? And an exponential function is always positive. If it is always positive, there is no chance that I can have a minus over here because if I do, everything will become a negative on the numerator and thereby giving me a negative answer. Therefore, my two to the power x has only one option that is to be plus under root four y minus three by two, right? Okay, next thing is four y minus three should also be greater than equal to zero, else I will have imaginary values coming up over here. Is this fine? So this gives me y greater than equal to three by four, okay? And there's one more important thing that has to be kept in mind is that since you are saying two to the power x is this and two to the power x is greater than equal to zero, it implies this should also be greater than zero, correct? So minus one plus under root four y minus three by two must be greater than zero. That means minus one plus under root four y minus three must be greater than zero. No, no, no, it cannot be equal to zero because for it to be equal to zero, x has to reach minus infinity. Remember, two to the power x graph is like this. Two to the power x graph is a graph like this. So you have to reach minus infinity in order to become zero, which you can never attain, okay? So this means four y minus three under root should be greater than one. That means four y minus three should be greater than one, which means four y should be greater than four. That means y should be greater than one. So one condition I get from here, another condition I get from here. Now we have to take the overlapping conditions of these two, okay? The overlapping condition here would be, again, I normally use a number line for it. Is the screen visible? Okay, so this becomes your answer. Let's take a couple of more questions and then we'll stop this. We take a break of five to seven minutes and then I'll start principle of mathematical induction with you. Normally I see here that the strength is not too much today. So I was slightly skeptical of starting a new topic. So let's do some more questions. Find the range of, find the range of function f of x given as log to the base of root five, root two, sine x minus cos x plus three. These are normal brackets. Don't take them as fraction part brackets. These are normal brackets. Yes, any idea how it works? Again, so focus on this term first. Focus on this term first. So we know that sine x minus cos x, okay? All of you would recall that I had taken, I had talked about the maximum and the minimum values of these type of expressions, right? So the max value for this, let me call it as f of x. So the max value of f of x is under root of a square plus b square. And the min value of f of x is negative of under root a square plus b square, correct? I hope you remember this, please recall it. So max value of this is going to be root two and min value is going to be negative root two, right? Now let us try to build up this particular expression over here. So let us multiply everywhere with root two. So root two times sine x minus cos x would lie between minus two to two, correct? Let's add a three everywhere. So minus two plus three is less than equal to root two sine x minus cos x plus three less than equal to five. So actually lies between minus five to five. Of course you can't go to minus five, there's a restriction, okay? So you can actually go from zero onwards, isn't it? I'm sorry, it's plus one, not minus five. Yeah. All right, okay? Now if I take a log to the base root five on everywhere, okay, this is what I'll get. This is always zero, okay? And I'm copying this ditto. And this is going to be, what is log five to the base root five? That's going to be two, right? Which clearly implies that the range of this function would be from zero to two. The range of this function would be from zero to two. Is that fine? So let's take another one. Change of the function three sine x minus x plus greatest integer of x. Three sine x minus x plus greatest integer of x. Let me make your life a little bit simple because this may be slightly tricky for you. So let's remove this three over here. So let's remove the sine x over here and let me just add a three over here. Okay, try this first. Now I'm sure it must be quite simple. Any idea? This is your gif of x function. This is your greatest integer function. So guys, it's very simple. If you see this function closely, actually what I have given you is this function, right? This is nothing but fractional part of x, isn't it? Now all of us know that the fractional part of x lies between zero to one. So minus fractional part of x would lie again between minus one to zero. Can you see the screen, guys? So this will lie between minus one to zero. So add a three on all the sides. So three minus, this will lie between three minus one to three plus zero. That means it will lie between two to three. So the range of this function is going to be, the range of this function is going to be from two to three, including three, excluding two. Now let's take a break for five minutes. Okay, let's break out for five to seven minutes break. We'll meet at 6.30 p.m., okay? All right, welcome back. Hope you can see my screen now. So I'm now going to introduce a simple topic to you which is called the principle of mathematical induction. Okay, very easy, simple topic. Even though in JEE, the type of problems asked on this concept is very, very less. And mostly this is a topic which is pertinent from school syllabus point of view. But nevertheless, it's very scoring. So let's try to understand this topic. Now, what is this topic all about? This topic is all about validating mathematical statements, mathematical statements through the process of induction, okay? Or what we call as the inductive reasoning. Now, there are two types of reasoning that we normally follow. When we call as the deductive reasoning, another is called the inductive reasoning, okay? Deductive reasoning is basically something which we'll study in mathematical reasoning chapter, okay? Where you try to deduce something through logic, through the use of logics like conjunction, disjunction, if, then, if and only if, not, all those types. That is called the deductive way of validating a statement. Inductive is what we are going to study in this present chapter, principle of mathematical induction, okay? Now, first of all, it's very important to understand what the meaning of the word validating. Validating means proving something is true. Proving that a statement is true. Second thing is what is the meaning of statement? Now, many people, they confuse statement with sentence, okay? So, let us try to understand the difference between sentence and a statement. Difference between sentence and a statement. If you would like to take a shot, what is the difference between a sentence and a statement? Okay, let me tell you the difference here. Sentence are of various types. It can be an assertion. It can be an assertion, also called proposition, okay? Something like, I say three is a prime number. This is actually an assertion, okay? I'm asserting something, okay? Or I say five is greater than two. So, I'm trying to assert something, okay? Moon is bigger than Earth. Here also, I'm trying to assert something, and in this case, my assertion is actually not correct. So, these are called assertions or propositions. Second type of sentences that we normally come across is called imperative sentence. Imperative sentence is where you give a request or a command to somebody, right? For example, or something like, please get me a glass of water, okay? This is also an imperative sentence because you are requesting something to be done. Next type of sentence is interrogative sentence. In this kind of sentence, you are questioning something, okay? Like, how was your exam? How was your UT? Or, what are you going to study today? So, this is an interrogative sentence. Four type of sentence is exclamatory sentence. Somewhere where you express some kind of excitement, like, wow, what an easy paper, okay? What an easy paper, okay? Oh, what a day. Oh, or you can say, wish I could get 25 on 25 in math's UT, okay? So, these are like exclamatory sentences. Now, out of these four types of sentences, in whichever sentence you can actually say true or a false, but not both, that would be called as a statement. So, those type of sentences where you can actually reply to that sentence by saying true or a false, but not both would be called a statement. For example, let's talk about exclamatory sentence. If I say, oh, wow, what an easy paper, you cannot say true or a false to it, because it's somebody's emotions, right? Wish I could get 25 on 25 in math's UT, you cannot say true. You're not granting a wish actually, right? Like a genie would do in Aladdin, correct? Okay? So, this type of sentence cannot be called as a statement. Okay? So, this definitely is not a statement. This is not a statement. Now, interrogative sentence. Integrative sentence also you cannot answer with a true or a false. For example, how was your UT? You cannot say true, correct? What are you going to study today, false, right? So, even this is not a statement because you cannot answer this with a true or a false. Okay? Imperative sentence. For example, if I say switch on the lights, you can say, I will or I won't, correct? You will not say true, I will not say false, right? Get me a glass of water, true. That sounds so weird, isn't it? So, even in this, you cannot answer this with a true or a false. So, it's not a statement. But when it comes to assertion, when you say three is a prime number, you can say, yes, it's true. When I say five is greater than two, you can say, yes, it's true. If I say moon is bigger than earth, you can say, no, it's false. So, this is the type of sentence which we normally call as statements. It's very important to understand this difference. So, those sentences which are actually assertions or which are actually propositions are actually statements, okay? Remember, a statement should always be unambiguous. There should not be any ambiguity. And it should be always be able to answer a statement with a true or a false, but not both, correct? So, because statements are assertions, we normally write it as p, pn, or some variable we put inside, pk, pm, et cetera, okay? Where p stands for proposition. What is nkm? Let me give you an example. Let's say, all of us know that one plus two plus three, all the way till n is equal to nn plus one by two. Now, normally, this is an identity. This is an identity. And basically, it's a statement because you're trying to say that the sum of all natural numbers from one to n is nn plus one by two. So, if you have to represent this as a statement, I would write it as pn, okay? Don't write equal to over here. We say pn is a statement or the proposition which states that one plus two plus three all the way till n is n into n plus one by two, correct? Now, let us say I have to validate this statement by the process of induction, okay? Now, try to understand this process of induction. So, what are the steps involved when you're trying to validate a process, a statement through induction? The first thing that we normally do is something which we call as the verification step, okay? Verification step is what? Verification step is where you try to choose a special value of n and see whether the left hand and the right hand side are equal or not. For example, if I take n as three, I would like to check whether p three is two or not. That is, I would like to see whether one plus two plus three actually is three into three plus one by two or not. Let's check. So, is six equal to three into four by two? In this case, yes. Six is equal to six means this is verified. Now, verification is just the initial step. You cannot base your judgment on just the verification step. It may work for three, it may not work for four. It may work for four, it may not work for five, correct? It may work for, let's say, less than 5,000, it may not work for 5,001, right? So, verification step is a necessary step, but it is not a sufficient step because if the verification step itself fails, you can directly invalidate the statement. But if the verification step passes, you cannot still claim that the statement is true, okay? So, we normally move on to other steps after we have done the verification step, which we call as the induction step. In fact, we call it as the assumption step and the induction step. We call it as the assumption step and the induction step, okay? Now, here, what do we actually do? Let us try to understand through the same example. So, what I will do is I will say, let my statement be true for some value K, okay? K is some, again, a natural number. Here, please note that N is a natural number, okay? K is some natural number, okay? Let PK be true for some K belonging to natural number, okay? Normally, this K is some value. It's one value which is lesser than N, N being the last number of the series, okay? Now, by using this assumption, this is called your assumption step, we try to prove that or we try to prove that the statement is true even for K plus one. So, even when your N is replaced with K plus one, we get a correct answer. We get a true situation for this. Now, what I want to say over here is that we assume that one plus two plus three, all the way till K is K, K plus one by two. So, basically, what I have done in place of N, I have replaced K, okay? And I'm assuming that this is true. I'm assuming that this is true, okay? Correct? Now, I have to prove that one plus two plus three all the way till I reach K plus one. This should be nothing but K plus one, K plus two by two. Now, what am I doing? I'm actually replacing my N over here. This N has been replaced with K plus one. So, everything here will be replaced with K plus one. This will also be replaced with K plus one. This will also be replaced with K plus one. So, it actually becomes K plus one, K plus two by two. That is what I have written over here. Now, this is to be proven. This is to be shown, to be proven. By using what assumption? By using this assumption. So, using one, I have to prove two, think like that. So, using one, we need to prove, yeah, sorry. Is it visible now? Now, it should be visible. Sorry, the internet connection got interrupted. Yeah. So, I was saying that, so what did I achieve by using the fact that if P K is true, by using the fact that if I assume that P K is true implies P K plus one is true. So, what did I achieve by doing this? See, it's like saying that if P four is true, then automatically P five will become true, correct? And if P five is true, then P six will be true, correct? Then P seven will also be true in the same thing, because now your K can keep on changing the values. From four, it has become five, five, it has become six, six has become seven, and it'll go all the way till the last value. So, P one will become true. Not only that, if this is true, then it means this should also have been true, correct? That means this also should have been true, correct? So, it all goes down to the very first statement, which actually we can prove by using verification step. So, it's something like a chain reaction, which is happening, right? We all burst crackers during Diwali, right? Nowadays, we have stopped because we have become environment conscious. But normally, when we use, you know, Chitai bombs, right? Where there are a lot of bombs connected to like a, you know, like this, correct? You must have all used these kinds of bombs in Diwali, correct? And there is a, you know, place to light over here, right? And everything starts bursting. So, basically, if you show that, if this guy bursts, it'll make this guy also burst, correct? So, if this guy burst, then this guy will also burst, this guy will also burst like that. It will continue till the end of that bomb, correct? And bursting of this will happen because of this. So, it'll go back also till you reach the first cracker. And who will light the first cracker? That first cracker is basically what you are doing by the verification step, okay? So, the same way induction also works. Induction means proving without actually proving it, just by inducing the proof, right? So, coming back to the question, so if I'm able to prove that by assuming the fact that it is true for K, that means PK is true. And by using the fact that PK is true, if I can prove that PK plus one is true, that means I'm setting a type of chain reaction. Okay? And if the K varies, right? It can prove till the last value of the series. Are you getting it? Now, let us complete this process. Let me erase unwanted stuff over here. Yeah, so let me just clear this off. Yeah. So now, if I want to prove this, first of all, I'll start with the left-hand side. Left-hand side contains one plus two plus three, all the way till K plus one, right? But before we write the last term, I will write K, okay? Now, this entire term is actually known to us as KK plus one by two. So from one, I can say this term could be written as KK plus one by two, okay? And this extra term has been now added. This is an extra term, which I'm adding it to here. Now, I have to show that this sum should actually be this. Let us try to show it. So if you take the LCM of two, it will become KK plus two, like this. Take a K plus one common, it becomes K plus two by two, which becomes equal to your right-hand side, okay? Okay, so this implies that, this implies that statement PK plus one is true. This implies that statement PK plus one is true, okay? Now, whenever you're trying to solve a mathematical induction question, there is a way to conclude it, okay? The way to conclude it is PK is true implies, PK plus one is true, and therefore PN is true for all N belonging to the natural number. This is the concluding statement that you have to write by principle of mathematical induction, by principle of mathematical induction. Now, guys, let me just go back and just tell you something about the verification step over here. Now, this verification step we proved for N equal to three, but ideally we actually do it for N equal to one, because as I told you the example of that firecracker, you have to make the first cracker, you know, you have to light up the first cracker if you want to make the second cracker burst, isn't it? So, taking a verification step at three is not a good idea. You should take it at the very initial value of N, right? So in this case, if your N is natural number, you can start it as early as one. So you have to show that one is equal to one into one plus one by two, right? Which is actually true, right? Therefore P one is true, and this should be used as your verification step. Avoid using any other arbitrary number that you feel like. Is the idea clear from this example? Let me take another example to show this by principle of mathematical induction that one square plus two square plus three square all the way till N square is N, N plus one to N plus one by six for all N belonging to natural number. Now follow the steps very closely. It's very important that first you name the statement. So let PN be one square plus two square, three square all the way till N square be NN plus one to N plus one by six. Now many people get confused, they start naming it as PN square. See guys, it's just a function that you are writing in terms of N. It has nothing to do with the last term. See if you write it in terms of K, that means one square plus two square plus three square all the way till K square is equal to KK plus one by six, you would have called that statement as PK. If you write it in terms of R, you will write it as PR. If you write it in terms of M, you would write it as PM. So don't get confused from the previous example that since there is an N square over here, I should write this as N square. No, not required. Now, what is the very first step? First step is the verification step. Even though in the exam, it may not be required that you write verification step, but since you are learning it for the very first time, I'm going to do it by writing all the steps which I'm doing. So first step is the verification step. Verification step means I have to see whether P1 is true or not. So what does P1 state? P1 states, never write P1 equal to. This is something which another type of mistake which people do. They say P1 is equal to. No, P1 is not equal to. P1 states, okay? So P1 states that one square is equal to one into one plus one into two into one plus one by six. Now I want to see whether this actually is a true statement or not. So one square means one. This becomes one into two into three by six, which means one is equal to six by six, which means one is equal to one, correct? Is this true? Yes, which is clearly true, which is true. Therefore, P1 is true. So at least by doing this, you have ensured that you will get one marks. Yeah, if it comes false, then you have to say, you have to stop at the verification step itself and you have to say the statement is not true. Clear, Venkat? Okay. Next is the assumption step. Normally, we do the assumption and induction step together. Assumption step is where you have to write this statement. Let Pk be true, okay? For some n value equal to k, and k is also a natural number, okay? Normally, this k is some random value that we normally take between one to n, okay? That is, you are trying to say that one square plus two square plus three square, all the way till k square is equal to k, k plus one, two k plus one by six. Let me call this as one, okay? Now we have to do the induction step. In induction step, we have to prove that Pk plus one is true using statement number one. Is that clear? So induction step, it's clear. What are we doing in the induction step? In the induction step, we are trying to prove that Pk plus one is true using our assumption step, okay? Now, first of all, I have to write what does Pk plus one say? Pk plus one says one square plus two square, three square, all the way till I reach. Yeah, you have to replace your n with k plus one. So it'll become k plus one, k plus one plus one into two k plus one plus one by six. In other words, you have to prove that this is equal to, this is equal to k plus one, k plus two, two k plus three by six, okay? This is, I have to prove. I have not yet proven this. Now, I will start from step number one. Sorry, I'll start with left-hand side of this expression. Let's start with the left-hand side. So left-hand side says this term, correct? So I'll write it down one square plus two square plus three square, all the way till I reach k plus one square. Now, before I reach k plus one square, I would obviously cross k square also, correct? Now, till k square, till this part, I know the answer. The answer is this, because you have assumed this to be true, correct? Remember, so till k square, you know the answer. The answer is k, k plus one, two k plus one by six. Now you have to add an extra term, which is k plus one the whole square, and you have to simplify it to bring it to this form. We need to simplify it to get to this form, okay? So let's try to do that. So let us simplify it. Let's take an LCM of six. So I'll have k, k plus one, two k plus one, plus six times k plus one square. Let's take k plus one common. I will get k, two k plus one plus six k plus one by six. Okay, let's further simplify this. So it becomes k plus one. Two k square plus seven k plus six. And this is factorizable as two k square plus three k plus four k plus six. So k plus one, you can take a k common here, two k plus three. Take a two common here, again, two k plus three by six. Which is nothing but k plus one into k plus two into two k plus three by six. Isn't this what we wanted to achieve? Isn't this what we wanted to achieve? See this term. See this term over here. And see this term over here, aren't they equal? That means this is equal to my RHS, okay? Now again, don't stop abruptly. You have to make the concluding statements. The concluding statement is hence, this is the concluding statement. Please note that in your exam, one marks will be allotted for this concluding statement. One mark will be allotted for your verification step and at least two marks will be allotted for your induction step. So concluding statement is hence, pk is true implies pk plus one is true. And therefore, pn is true for all n belonging to natural number by principle of mathematical induction. By, you can write in the short form like by PMI. By PMI. This is something which you need to write at the end. Is the entire process clear? Verification induction, using induction, sorry, verification assumption and using assumption, you have to prove induction step. Is all the three steps clear? Is it clear? If yes, I would like you to solve a question then. Prove by principle of mathematical induction that one cube plus two cube plus three cube all the way till n cube is n square n plus one square by four for all n belonging to natural number. Yeah, it is the square of the first problem that we did. nn plus one by two the whole square. Yeah, correct. Done. Okay, so let's call this as pn for, okay, the timing. So first is the verification step. Verification step means I have to first show what is my p1, is it true or not? So p1 states one cube is one square, one plus one square by four, which means one is equal to one into four by four, which is actually one is equal to one, which is true. Therefore, p1 is true. Therefore, p1 is true. Guys, can you see the screen? Now, next is the assumption step, where you say, let, let, okay, for some n equal to k and k belonging to natural number. So now I have to do the induction step, where I have to prove that pk plus one is true, right? That is I need to prove one cube plus two cube plus three cube all the way till k plus one cube is equal to k plus one square, k plus one plus one square by four, okay? Let's start with LHS. LHS, you can see that we are actually going to k plus one cube via k cube, right? So this term is known from the assumption step. So till one, till k cube, the sum is known to us and that is k square k plus one whole square by four. I'm just adding this extra term, correct? So let us simplify this. So let's take an LCM of four, okay? Let's take k plus one whole square common. So that will be k square plus four k plus one, correct? So this could be written as k plus one the whole square and this would become k plus two the whole square by four, which is actually nothing but k plus one square into k plus one plus one square by four, which is actually your RHS, right? Hence, pk is true implies pk plus one is true. pk is true implies pk plus one is true. Therefore, pn is true for all n belonging to natural number by principle of mathematical induction, okay? So guys, we'll stop over here and we'll resume with this chapter next class where we'll talk about different types of problems. So under principle of mathematical induction, there are different types of problems that we normally get. One is called the identity type of problem which we did today, okay? Which we call as the identity problem, where you have to prove a well-known formula. Next type is the divisibility problem. This is the most difficult of all, okay? So we'll take a lot of divisibility problem. Finally, there is an inequality problem, inequality problem and if time permits, we'll take some recursive problem, okay? So this is what we'll be doing next class when we meet, okay? And all the best for your English beauty tomorrow, okay? Over and out from my side, thank you. Have a good night, bye-bye.