 So, guys, last class we ended up on a Rolls theorem, right, I'll just repeat Rolls theorem once again. In fact, I had not taken enough questions on it, as I would like to do. So just a quick recap of Rolls theorem, a Rolls theorem, just a quick recap. So in Rolls theorem, we had discussed that if there is a function which is continuous, in closed interval a to b, okay, and differentiable, differentiable in open interval a to b, and the value of the function at a and b are equal, okay, then, then, there exists, then, there exists, at least one c, at least one, c, lying in the open interval a to b, where the derivative of the function will become zero, where the derivative of the function will become zero. So this is what we had discussed with respect to the Rolls theorem definition. And I had also given you an alternative version of the same. The alternative version of the same is, if at all you have alternately, if you have an equation which involves a function which is continuous and differentiable. So this function is a continuous and differentiable function, continuous and differentiable function, okay, thereby meeting the first two criteria. And this equation has, let's say, I am just assuming two of the roots has roots alpha and beta. Okay, then, then, the derivative of this function equated to zero, this equation will have, this equation will have, or this has at least one root, which belongs to the open interval alpha and beta. So between alpha and beta, there will be at least one root of f dash x equal to zero. Okay, so this is what was Rolls theorem. And we had, I think, then do two questions based on that, one or two questions. Okay, so we'll resume today with more questions and then we'll move on to Langrange's mean value theorem. And of course, I will give you the Kushi mean value theorem as well. And then we can move on to the concept of monotonicity. So please note this down so that we can take a few questions. Okay, let's take questions. I think we have taken this one. Let's take this one. Use those serums to find the condition for the polynomial equation f of x equal to zero to have repeated roots, hence or otherwise prove that f of x, hence prove that this equation cannot have repeated roots. I'll forget the initial part of the question. Okay, no need to work on the initial part of the question because we'll not be using any proof that in fact, if you want, you can answer that as well, not initial. So use those serums to find the condition for the polynomial equation to have repeated real roots, hence or otherwise prove that this equation cannot have repeated roots. The repeated roots, repeated root means the same root is occurring more than once. For example, if I have x minus one whole square equal to zero, this is a quadratic equation. So quadratic equation as per fundamental theorem of algebra must have the same number of roots as the degree of the polynomial. So it has two repeated roots one comma one. Okay, so one one is a repeated root of this polynomial equation. It's a Maclaurin series actually is an infinite series can shook it doesn't, it doesn't stop for a limited number of terms. Right now you see n is basically a fixed number of, you know, terms are there. So, yes, if it had, if it had gone to infinity, it would have been e to the power x. Got it. Sorry, I took your name. By the way, what was that that room? Was it your private cinema theater? Oh, you take classes from there or you just watch movies there. Nice, nice concept. Okay guys, what is the answer for the first part? If at all a polynomial has a repeated root. Let's say there's a polynomial f of x equal to zero, which has a repeated root. Let's say the repeated root is alpha. Okay. Now please understand if it has a repeated root of alpha that means you can say there are two roots alpha alpha. And the derivative of the function must have at least I mean let's say I have repeated root alpha alpha. Okay, there may be other roots also I'm not denying it alpha beta gamma dot dot dot I don't deny it. But if I differentiate it and put it to zero as per roles theorem, this must have have at least one root root between alpha and alpha. And of course, when you have the same number alpha and alpha having something in between is basically the same value. So basically it'll have at least one root which will be equal to alpha apart from that it can have more roots. I don't deny it. But the derivative of the function will also become zero at alpha. This is very, very important. So if alpha is an repeated root, let's say repeated at least once, sorry, at least twice, then the derivative of the function will definitely have one of the roots as the alpha itself. I'll give you a simple example to verify this a quick verification. So let's say if you have a polynomial function which looks like this x minus one whole square x minus three whole square. You see this polynomial equation has roots one one and three. Okay, now differentiate this and put it to zero. If you differentiate this, let us use our product rule and equate it to zero. You can clearly see that you can clearly see that it will have at least one of the roots as one. You can clearly see the one was a repeated root over here that also turned out to be the root of this guy which is actually f dash x equal to zero. Okay, so this is very important. Many times it has it is asked a lot of formative exams as well. So this proves the first part of the concept. Hence otherwise prove that this equation cannot have repeated roots. Let's do that. Let's say that this equation cannot have repeated roots. Just sit down. Just sit down. You don't have to write anything down for it. Should we do it? Okay. Okay, let's say let alpha be a repeated root of this equation one plus x by one factorial x square by two factorial dot dot dot till x to the power n by n factorial equal to zero. So what does it mean? It means alpha will satisfy this equation. That means one plus alpha by one factorial alpha square by two factorial dot dot dot dot dot. It'll alpha to the power n by n factorial will be zero. Let's call it as one. Okay. Now, as per this particular fact that if alpha is a repeated root of f of x equal to zero, it must also be the root of f dash x equal to zero. Correct. So can I say, can I say f dash x, let me call this as f of x. So f dash x equal to zero must have alpha as its source. Okay, so now if you differentiate it, you will realize that you'll end up getting zero. This will be one to x by two factorial will be back to x. And if you realize it starts giving you the same expression. But just the last term will now become x to the power n minus one by n minus one factorial. Okay. So this must also have one of the roots as one of the roots as alpha. Correct. That means one plus alpha plus alpha square by two factorial dot dot dot dot alpha to the power n minus one by n minus one factorial must be zero. Okay, let me call this as two. Okay. Now from one and two, you can realize that what does to say to says that the term which is just before it. Here it is giving you zero right till here it is giving you zero. Correct. So what does it mean from one and two can I say that alpha to the power n by n factorial must be zero. Because the whole thing is adding up to zero and until until the last term everything adds up to zero that means the last one will also be zero. Which means alpha is has to be zero. Right. But zero does not satisfy this equation. Correct. So let me call this equation as E. But zero does not satisfy this equation. But zero is not a root of zero is not a root of root of f of x equal to zero. Okay, because if I put x as zero, I get one equal to zero, which is not possible. So zero will not satisfy this. And hence, the whole fact that this can have a repeated root itself is null and void. Isn't it. So this whole thing cannot have repeated roots because if it does, it leads to that repeated root being zero and zero is actually not a root of f of x equal to zero. Make sense. Any questions. Any questions any concerns. Clear sedan everybody. But this is not e to the power x though it had it is a polynomial. Can we explain how you differentiate it of a differentiation the normal way I normally do differentiation one derivative is zero x derivative is one x square by two factorial is to x by two factorial that will be x. Then x cube by three factorial three x square by three factorial which is. See, if I differentiate x cube by three factorial, let me do this. What do you end up getting three x square by three factorial. Three factorial is three into two factorial so you may write this as three into two factorial so three and this gets cancelled so you get the second term. Okay, and so on and so forth. Clear. No. Any questions anybody. Okay, should you now move on to our next question based on the same topic. Any questions any concerns. Okay, cool. Let's move on. I think some of them are on languages mean value theorem. This we already did this we already did. Okay, let's do this one. This question says f of x is twice differentiable says that f of a zero f of these two f of C is minus one f of these zero f of ease also zero. A, B, C, D are such that is less than B is less than C is less than D less than the minimum number of zeros of gfx given by this expression in the interval a to e is which of the following is a twice differentiable function. And of course if it is twice differentiable it has to be continuous. It's giving you these values f of a as zero f of BS to f of CS minus one f of DS to f of ease zero again. ABCD is related like this find the minimum number of zeros of this polynomial which is given in terms of F, of course with respect to its derivatives. It's an IT question came in some some year, year is not mentioned here. Any idea. Any idea how to do it a little okay. And don't get intimidated by the structure of the question. Good, good. Try it out. If you even if you have a little idea or somewhat idea, just give it a try. I'm ready to wait for another minute. I don't know it's a polynomial, but I know it's a continuous function can take a polynomial doesn't make any difference. Either ways you should be able to get the answer. Okay. Very good. So Ruchita has given one response. Okay. Okay. Okay. Should we discuss it now. Anybody. Howard Archit. Rebov. Dave. I am. Okay. Let me run a poll. I would request you to put your response in the poll, even if you have not solved it. Take it. I guess. I would like to see your response. Just take a guess. So that we can discuss it out with respect to your. Popular. Option. Take a guess. Fast fast guys enough time was given. Everybody, whatever you can, you think is the appropriate answer for this question. You can put it on the poll. Cool. Okay. I'm closing it now. Five. Four. Three. Two. One. Okay. Now, most of you have gone with option number C. Sorry, option number B. Whereas C has also got substantial amount of votes. Comparable votes. Let's check it out. See the function is a continuous function for sure. Right. And it has one of the roots at a. So this, this route is clear. This is obviously given that there is a root at a. And there is a root at E also. So any are definitely, you know, two of the roots of the function. Okay. So what I'm trying to see, what is the minimum number of roots? This guy can have f of x equal to zero can have. So let's try to figure it out. So I'm trying to keep my things minimalistic. Okay. Minimalistic. That isn't my idea. I want minimum number of. Okay. Let's keep minimum number of roots over here. So let's say if my function is says that at eight was zero. And beat went to two. At seat went to minus one. There you go. So there is one root between B and C for sure. At least one root. There may be more as well. I don't know. Then. The function goes to. Two. Two means it has to again cross the X axis and go to two. Guys, just give me a second. My main wants to sweep the floor. All right. So at D it has gone to two. That means at least one more root is created between C and D. At least mind my word. At least one root. There could be more. But my aim is to find the least number of roots. Right. Now again at E it has come to come back to zero. Right. Now from this whole scenario, I can figure out that. There are at least five roots. I can see one. Sorry. Four roots. I can see one, two, three, four. So this will have at least have at least four roots. Have at least four roots. Do you all agree with this or not? Okay. Now. If f of X is twice differentiable, then F dash X must be continuous. So F dash X will also be. A continuous function. And can I say F dash X equal to zero will have at least. At least. Three roots. Because between every two root. Of F of X equal to zero, there has to be at least one root of F dash X equal to zero by roles. Is this convincing enough? Is this part. Understood by everybody. If F of X equal to zero have four roots. Let's say a. This guy. I'll call it as. I'll call it as. I'll call it as. I'll call it as. I'll call it as. I'll call it as. Let's say a. This guy. I'll call it as. Let's say alpha this guy beta and E. Then the derivative of the function equated to zero should have at least three roots. Okay. Because you can see there will be at least three X summums formed in this. Right. X summums are what those are the positions where the derivative becomes zero. And hence those are the positions where the root of F dash X equal to zero will lie. Correct. Now. You must be wondering why you said doing all these things because I've been asked for the minimum number of zeros of G of X. Now all of you please look at G of X. Looking at G of X would remind you of something. Do you see some kind of a product rule hidden there? Yes. You see it is actually F of X into F dash X derivatives. Am I right? Correct me if I'm wrong. Correct. Okay. Now answer the simple question of mine. If F of X equal to zero has at least four roots. Has at least four roots. Nah. Now if people started saying ah who and all has at least four roots. And F dash X equal to zero has at least three roots. Then can I say. This guy equal to zero has at least seven roots. So the derivative of this guy, the derivative of this guy. So that means this guy equal to zero has at least six roots. Correct. Which clearly means G of X has at least six zeros. Has at least six zeros. So the answer is six, which is option number C. Is it fine? Understood. Okay. Even those will be counted as different roots. Even if it has a repeated roots, they will be counted as same number of roots. Okay. And I want it to be minimalistic. Right. I wanted to be minimalistic. So I will not assume that they are C. Right now what I've assumed is that it is basically going directly from A to B. Okay. But it could have gone up or come down cross 50 times and then go on to that position. Right. But that is not my aim. My aim is to get minimum number of zeros for G of X. In fact, I want minimum number of roots for F of X times F dash X equal to zero. So I'm keeping things the least possible. That's why the question says minimum number. But how do you know F dash X is continuous and differentiable? It is given. No. It is twice different. F of X is twice differentiable. So you can't differentiate F of X till it is continuous and differentiable. Right. So if you want to find double derivative of something, the single derivative must be continuous and differentiable. That's why this information twice differentiable is given to us. Is it fine? So basically I have used the concept of intermediate value theorem as along with roles theorem. So intermediate value theorem says that if F of alpha and F of beta have opposite signs and you know the function is continuous, then there must be at least one root between alpha and beta. That let me to believe that F of X equal to zero has at least four roots. And then after that roles theorem took over and you can see the conclusion. Is it fine? Any questions? Now we'll move on to LMVT. Languages mean value theorem. You want to copy this down? You want to write it out? Please do so so that I can go on to the next slide. Okay. Done everybody. Let's move on. Next concept that we're going to talk is Languages mean value theorem. So you must have all heard of this guy's Languages. Okay. French guy. And he was one of the, you can say very well known mathematician of his time. Of course, not as big as Euler was. But yes, he tried to fill in the gap left by Euler. So he also did a lot of work in the field of calculus in the field of number theory. You would have, you would hear his name, you know, many a times in your future studies of mathematics. By the way, just a trivia for you. Euler's name appears 96 times in the mathematical works that he has done so far, which is no mathematician name has, I mean, when I say name means he has given some kind of a theorem and people have given the name of that theorem as, you know, in his name. Like there is an Euler's identity. There is an Euler's line. All those names are given after Euler. So Euler name appears in the, you can say the entire repository of mathematical theory, 96 times. Languages name appears only 70 times. Okay. And many mathematicians are not even close to 10 also. Anyways, so languages mean value theorem. What this guy did, he basically was smart enough. And he basically generalized those theorem. Okay, he made it a little bit more generic. Okay. So what did he do? He said that he basically copied few things from a rose theorem as it is. He said that if a function is continuous in the closed interval. A to B. And this function is differentiable in the open interval A to B. And he said there is no need of F of A equal to F of B. Okay. Then, then there exists at least one C in open interval A to B. Such that F dash C is equal to F of B minus F of A by B minus A. Now he was really smart. He wanted it. He just took the rose theorem graph and he just tilted it a bit. Okay. So let me show you pictorially what he actually meant. Geometrically what he actually meant. And then I'll also prove it non-geometrically as well. So if you just recall what rose theorem basically said it said that if the function was continuous in a close interval A to B and differentiable in the open interval A to B between X equal to A till X equal to B. And he gave an extra condition that the value of the function at A and B must be equal. Okay. F of A and F of B are equal. There exists at least one C. There exists at least one C where the derivative of the function at C will vanish. Right. Okay. That means if you are traveling right in a continuous path and you stop somewhere. Right. Or you come back to the original position that means you should have stopped somewhere. Right. That is what basically rose theorem says. Right. Okay. What did Langrange do? This guy he said that let the function be continuous and differentiable that means smooth function from X equal to A till X equal to B. Okay. X equal to A till X equal to B. And that there must be at least one point in the entire path or entire graph where the derivative of the function, let me, let me use a, let me use a green color, where the derivative of the function. Let me make this line. Yeah, where the derivative of the function will be nothing but the slope of the chord connecting these two points. So let's say this is point A, this is point B. So this derivative will be giving you the slope of this tangent and this will match with the slope of this particular chord. So this particular fact is based on a physical phenomena that if you are, you know, traveling along a continuous path, you realize that the average velocity of your entire journey will match with at least instantaneous velocity during one of the time periods of your path. So let's say at least once in your journey, your instantaneous velocity would actually be same as the average velocity of the entire journey. Correct. So if you treat like a concept of, you know, you can say motion of a body. If you see if you move from a point to be point the slope of that line the slope of this chord AB. Okay. It basically gives you the average velocity. Right. Change in displacement by changing time. Isn't it. So basically this is your change in the displacement. Let's say I consider this to be S versus T graph. Okay. So this is a change in the displacement. This is a change in the time. So this is actually your average velocity. So average velocity value will match with instantaneous velocity at some instant during the path. Okay. So M AB will match with F dash or you can say DS by DT calculated at DS by DT calculated at X equal to C. This is what actually language is said. Okay. Now, this is very obvious because there would be a point here where you would realize that the tangent drawn is parallel to this chord. So if you see in his formula, he has said exactly the same thing. A point is actually a comma F of a B point is B point is B comma F of B. So this is nothing but the slope of the slope of the chord AB, right. Why two minus Y one by X two minus X one like you already have done in your straight lines how to find the slope and F dash X is the derivative at C. Okay. So as you can see, it is a very special case of you can say it is a generic case of rose theorem where if you have F of a equal to F of B, that means the graph would have come back to the same level of you can say the displacement, then you would actually see that F dash G would become a zero. So if F of a is equal to F of B, then LMVT will convert to rose theorem LMVT will transform to rose theorem. Okay. So this is a more generic you can say a version of rose theorem, where F of a and F of B conditions need not be met. So if that is not met, you can say F dash G will be equal to F of B minus F of a by B minus a. Okay. Now this is a you can say geometrical interpretation of the theorem that language it actually gave us. Let's try to prove it. Let's try to prove it. Mathematic I mean non geometrically. Is it clear. Okay. So if you see what this guy did was he took this graph. Here also if you see this line has a slope of zero right. So what he did he took the same graph he just rotated it a bit. That's it. Nothing else he did he was very smart and he basically got a slightly more generalized version of rose theorem. Okay. Now let us try to prove this let us try to prove this math non geometrically. Now, let us say my f of x is a function which is continuous. Okay, in a comma B and let's say it is differentiable. It is differentiable in open interval a comma B. Okay. Now I'm going to devise a function I'm going to make a new function. I'm going to make a function phi of x like this. I'm going to make a function phi of x minus f of B minus f of a by B minus a times x. So why only see the slope is equal if it is a straight line then it should be parallel throat. That's what that's what it says no sit dance there exists at least one C read the theorem properly at least one C. If the path was straight away from A to B then they would be infinitely many C's but what he said that there must be at least one C. It may be one two three five hundred two million whatever but at least one C should be there that is what he said. Okay. Now how do you prove it so basically I have assumed another function let phi of x be a function like this. So phi of x function is made up of you can say f of x and some lambda you can say this is a lambda times x. Okay. Now subtracting a lambda x to a function which is already continuous doesn't change its continuity. We have already learned in our property of continuity that some of two continuous function lambda x is a linear function or you can say it's a polynomial. So f of x plus lamb negative lambda x is still continuous. It is still differentiable because both these functions are differentiable in themselves. So can I say phi of x is continuous in the interval a to b. Okay. So about phi of x can I say this phi of x is continuous in the interval a to b agreed. Anybody has any doubt related to that. Okay. Can I say if phi of x will also be differentiable in the open interval a to b. Okay. Now you realize that the function has been chosen in such a way that Phi a and Phi b will be equal. Check it out. Check it out. The function has been chosen in such a way that Phi a and Phi b will be equal. Check it out. So if I put a Phi a, what do you see? It will be f of a minus fb minus f a by b minus a times a. Correct. On simplification, this will give you, correct me if I'm wrong, bf of a minus af of a minus af of b plus af of a. I think this and this will get cancelled. So it'll give you bf of a minus af of b by b minus a. Okay. This is your Phi a. Check Phi b also. Phi b will be fb minus f of b minus f of a by b minus a times b. So on taking the LCM, you'll end up getting bf of b minus af of b minus bf of b minus bf of b plus bf of a. So now if you cancel this off, you will see that you will be left with, you will be left with bf of a minus af of b by b minus a. So aren't they the same? Thereby meeting the third condition of Rose theorem as well. So by assuming such a function, what did Langrej basically ensure that he's meeting all the three conditions of Rose theorem because those two was already established. Okay. Much before language was born also. So what he did was he basically took that as a basis and he formulated a function Phi x by choosing the given function f and of course doing some extra manipulations add on which was the brainchild of language. And then he basically as ensured that these three conditions are met. If these three conditions are met, my dear, then you would all agree with me. Where should I write? Where should I raise this part because this part is already understood by you. If a is equal to b, a cannot be equal to b. A and b are separate points. Okay. And b are different points. The point function, the point function as I told you it is a debatable function whether it's continuous. Okay. So let's not apply it to a very, very special case of a equal to b because the conditions may not be met. Okay, anyways, I should I raise this part anybody wants to copy this part, because I don't want to go to the next part of the page for the simple thing. So I'm just erasing this part. This is just to ensure that Phi a and Phi b are equal and you all understand that. Okay. Now, if that is true, if all the conditions of rose serum are met. So it implies that there exists at least one C at least one C lying in the open interval a to b where the derivative of Phi will become a zero. So basically when you differentiate this and put it to zero f dash C, you'll end up getting f dash. By the way, I'm just writing down the substituted part. And this will be nothing but f of b minus f of a by b minus a x will go off because the derivative of x is going to be one. And since this is zero, you can say that you can say that f dash C minus this part which I'm writing. It will be zero, thereby proving the Langranges mean value theorem. Okay. This is what nine inches mean value theorem says. Hence. Is this fine? Any questions? Any concerns? And please Rajaji Nagar, have you done languages mean value theorem in school? Yes, no, maybe. Not yet. Okay. Now there is an alternate version of this, which is normally seen in higher grades. I mean, when you go to your undergrad. Let's say you are pursuing your highest studies in mathematics. There's an alternate version of languages mean value theorem, which says that if the function f of x. Okay. Again, I'm just assuming all the conditions are met. If the function f of x is continuous in close interval a to b. Okay. And normally they will write B as a plus H. Okay. Why? Because I should have an option to reduce my B and make it very close to a. Okay. Normally we use H when, if you remember, we use it in our first principles, LSD, RSD concept in our continuity and differentiability. So the same H. Let's say your B is now a plus H. Okay. Then if the function is continuous in this interval and is differentiable. Okay. Then if the function is continuous in this interval and is differentiable is differentiable in this interval. Okay. Then, then, then there exists. Then there exists at least, at least one theta lying in the interval. Zero to one such that such that f dash a plus theta H is equal to f of a plus H minus f of a by a plus H minus a. In other words, in other words, f of a plus H is nothing but f of a plus H times f dash a plus theta H. Okay. Theta lying between zero to one open interval. Okay. This is another version of language is mean value theorem. And this particular concept becomes actually the starting point of Taylor's expansion or Taylor's theorem, you can say it, which anyways is a subject matter of your highest studies of mathematics. So if you see, you know, if you start, you know, applying the same concept to, you know, a continuous basis on this function as well, you will end up getting a chain, you'll end up getting a series which is called the Taylor series. Okay. And this Taylor series is what was used by McLaurin to actually end up getting his McLaurin series as well. So there is a linkage. The link is you start with those theorem. Then also was generalized to the languages mean value theorem in languages mean value theorem uses alternative definition and you keep applying the same concept to the other functions. So there is something called languages form of remainder, et cetera, which is anyways not required for you in your J preparation. Maybe some of you who are writing AP calculus BC, if you see the last chapter sequences and series, you'll find information about languages form of remainder, which I'm not going to discuss it because it is irrelevant for us. It is not going to be asked in Indian competitive exams. Okay. AP people, yes, they will be tested on it. So this is alternate version. This is alternate version of languages mean value theorem. Okay. Just note it down. Now, so all of you will basically realize that, you know, these concepts that you have learned. Okay. They can further be generalized. So I will take one more step of generalization and I will introduce you now to who she's mean value theorem. So I will introduce you to now the cushy's mean value theorem. So first noted down any questions, any concerns here to let me know. So we'll now move on to, we'll now move on to cushy's mean value theorem. How did we get the alternative definition of this theorem? I just replaced B with A plus H. Okay. And now see, nobody asked me this question by the way. Thank you, guy three for asking it. If your theta is between zero to one, that means a plus theta H will be somewhere between a to a plus H, isn't it? If your theta is between zero to one, when it becomes exactly one, it will become a B when it becomes exactly zero, it is a. So if it is between zero to one, it will ensure that this guy is like your C. So does it not, does it not become F dash C is equal to F of B minus F of A by B minus A? Right. Sorry, I think you understood that. So I did not put emphasis on this. Okay. No issues guys. Now is it clear to you? Okay. All right. Now let's move on to the further generalization of this, which is the cushy's mean value theorem. Cushy's mean value theorem. This guy also is a French. Okay. Full name was Augustine Louis Cushy. He almost lived in the same era as these guys, you know, Langrange's and on. Now what did he say is that he said that you can further generalize Cushy Langrange's mean value theorem. So he said that if, let me write it like this, if let's say there are two functions f of x and g of x and both are continuous in the close interval a to b. Both are continuous in the close interval. Okay. a to b are differentiable. Let me write this and this are differentiable in open interval a to b. That means g dash and f dash x exist. Okay. One more restriction he gave that the derivative of the function g should not be zero for any value belonging to the open interval a to b. Then there exists at least one c. There exists at least one c in the open interval a to b such that f of b minus f of a f of b minus f of a by g of b minus g of a is equal to is equal to f dash c by g dash c. Now, how would he actually how would he have actually proved this theorem? Let us look into the proof of this. See, he did nothing different. He basically again relied on roles theorem only. So, what he did this guy was very smart. He created a function phi of x by incorporating f of x and g of x. See, how did he do that? See, basically made a function phi of x by using this function f of x minus f of b minus f of a by g of b minus g of a times g of x. Okay. Now, you would agree with me when I say that if f of x is continuous g of x is continuous then f of x plus some lambda times g of x will also be continuous. Correct. So, I can say with surety that this guy is continuous. This guy is continuous in this interval. Correct. I can also say that this guy will be differentiable in this interval because it is made up of two functions which are differentiable themselves. Correct. And not only that, you would realize that phi of a and phi of b will be equal. Phi of a and phi of b will be equal. Try it out. Everybody first do this step on your notebook. I will not. I'm not going to do it. It's a simple activity. Put a in place of x, put b in place of x and check whether are you getting the same result. Please verify this guys. Please verify this. I'm giving you one minute and just give me yes. On your chat box. Yes means yes. They are coming out to be the same. Quickly. One minute of air. Kinshuk says yes. Good. Yes. Yes. Yes. Everybody is saying yes. Very good. So, this basically means that rose theorem conditions are met. Rose theorem. Conditions are satisfied. Okay. So, if rose theorem conditions are satisfied, even rose theorem should be satisfied. So, this is a situation. It should be satisfied. It doesn't. It doesn't. It doesn't. It doesn't. It doesn't. It doesn't. So, as per also on the derivative of the function, okay. So, there exists a sea lying in the open interval. A comma B, where the derivative of this function must become a zero. So, if you differentiate it and put to zero by putting, let's let me first differentiate at both sides. Okay. This is what we'll get in on differentiating. Correct. This guy will become a 0, 5-C will become a 0 as per Rose's theorem and this will be F-C and this will be F of B minus F of A by G of B minus G of A and this side will become G-C, okay? So you can see from here that whatever kushi had given that thing comes out to be true and that thing is F-C by G-C should be equal to F of B minus F of A by G of B minus G of A. So this is what is kushi mean value theorem, kushi mean value theorem hence proved, okay? Now from this, we'll understand this from two aspects, how geometrically is this working, okay? And then we will come to a very important result that comes out from here which is actually a Lopital's rule. So how does Lopital rule come from this kushi mean value theorem? Let's try to understand that. By the way, even though I have proved Lopital's rule from this, don't start using this in your school exams, CVSC people, not ISC people, ISC people have full freedom to use it. Lopital rule is what they have to use actually, but many a times now your teachers will jokingly say, if you can prove it, you can use it, okay? But let me tell you, even if you prove it, they will not allow you to use it. There was one guy, yeah, but one of the exam is MCQ, writes engine, very well pointed out. Let's say subjective exam and all you are trying it, don't use it actually ever. Keep it for competitive exam, but don't try to prove it because they will not appreciate it, okay? So there was one instance that the teacher told in HSR, I think NPSHSR that you prove it and you can use it. And this guy went and proved it like this the same way. And then he started solving questions based on this, zero he got, really good to know that. All right, now coming to this fact, see, first of all, do you realize that there's a close connection between CMVT and LMVT. So if your GFX happens to be your X, this is one very important point to be noted down. If GFX happens to be X, then basically CMVT converts to LMVT, do you agree with me? So if you just go to this formula once again, and keep your GFX as an X, okay, if your GFX becomes X, what happens? Left side will become F dash C by one, because GFX derivative at any point will be one, whether C or not C. And this guy will become F of B minus F of A and GB will become B, GA will become A. So what is this? This is LMVT, okay? So what did Kushi do? He was smarter than Langrange, what he did, he basically generalize it even further and most of them basically use rose serum only. So rose serum is that you can see it is the father property. And these both are child property and CMVT is the elder child, means it is more powerful than LMVT. So Kushi was like, okay, Langrange is basically acting smart, let me be more smarter than him. So he further generalize it, okay? And of course, you know how LMVT becomes your rose serum, when your F of A and F of B are equal, it becomes rose serum, okay? Now second thing is, how do you geometrically prove it? See, this concept is not new to us. This concept is not new to us. See, if let's say I have a curve equation which is in parametric form, let's say this is just a second. Yeah. I'm sorry. All right. So let's say I have a curve which is parametrically defined as let's say GTFT, okay? So now what I've done, I have taken a curve which is continuous, differentiable, okay? And this curve is parametrically defined like this. That means X is G of T and Y is F of T, okay? Now what does basically Langrange's mean volume theorem say that? It says that if let's say you're talking about this point, this point is G, let's say parameter here is A. So this point has a parameter A. So GAFA and let's say this point has a parameter B. So GBFB. So as per Langrange's theorem, if I connect a chord between these two points, okay, there will exist a point, there will exist a point here where the tangent drawn will have the same slope as this chord, isn't it? So what he did? He did nothing. He just used Langrange's mean value theorem and he basically converted it to a parametric form. So you can say that how do you find derivative? How do you find dy by dx in a parametric form? You would all recall when we did differentiation chapter, it is nothing but derivative of Y which is F of T in this case with respect to T divided by derivative of G of T with respect to T, okay? Is it fine? And let's say if I put a point T equal to C parameter, so this has a parameter T equal to C, this must be equal to the slope of this line. Let me call this line AB line, okay? This must be equal to slope of AB line. And what is slope of AB line? What is slope of AB line? Slope of AB line, if you see you can use your Y2 minus Y1 by X2 minus X1 formula. So it will become F of B minus F of A by X2 minus X1 which is G of B minus G of A, okay? So both of them should be equal and hence you can see that this is how your kushi mean value theorem becomes life. It comes to life. Where C is some point between A and B. Is it clear? Any questions here? Any questions? Any concerns? By the way, I'm giving you a very, very deep insight. Maybe you will not find these insights in your regular textbooks that you are following for IITJE or JEAdvance. This is something which I'm telling you from my higher studies experience, okay? In school, of course, this will not be taught to you. But this is how you should be able to appreciate how CMVT has evolved from LMVT and how both of them are actually the brainchild of Rose, Michelle Ruhl. Now coming to how Lopital Ruhl has been derived from it. So how Lopital Ruhl basically comes from it, okay? So Lopital Ruhl, now Lopital Ruhl basically again is a very, you know, you can say a special version of Pushy Mean Value theorem. Okay, so this is nothing but let's say F dash C by G dash C. Okay, now C. What I'm going to do is I'm going to convert this scenario. This scenario, please all of you please focus. Okay, I'm going to convert this scenario in this way. I'm going to take make my B as an X. Okay, fine. I'm going to take B as X or X plus H, let me call it. Okay, A is X. A is X or let me write it like this, better to write it like this. So let's say I convert the scenario like this, F of X minus F of A by G of X minus G of A, where X is very close to A, okay, where X is very close to A. So basically I've done what I've kept my A and B in such a way that B and A, they're very close to each other. That means the gap between X and A is infinitely small, is infinitely small. Okay, delta is a infinitely small quantity. Let me write equal here. Okay, so what is going to happen, if you see this problem basically converts to a zero by zero form, if you substitute, isn't it? Okay, and if you try to equate it to this, it will be nothing but F dash, now see here, some value between A and X, which is between A and X and X itself is very close to A. So A is very, very close to X, very close to X. That means there is a very, very small gap between them, right? So you can say that it is as good as A and A plus delta, delta being an infinitismal quantity. Okay, delta is an infinitismal quantity. Okay, so if you want to find a quantity C between A and A plus delta, that quantity has to be actually be very close to A and as good as an A, okay? So this entire expression, this entire expression, F dash C will actually get converted to F of A by G of A and that's what rose theorem, sorry, that's what L'Hopital rule also teaches us. It teaches us that you have to differentiate this and put your A in place of X, isn't it? That's what we normally do when we are applying L'Hopital and we differentiate this and we put A in place of X. So if you take the A and B here to be very, very close to each other, then the C will be have to be your A itself because B is a quantity which is very, very close to it. So you can say L'Hopital rule is basically derived out of, L'Hopital rule is basically derived out of Kushi mean value theorem, okay? So this concept, this concept evolves from there. Any question? Any concerns? However, there's no combative exam which will ask you for these kind of proofs. It is just for your extra knowledge. It is just for your extra knowledge. Nobody is going to ask you about these. Copied? No, it's not about taking limit on both the sides. When automatically your X becomes very close to A, this limit becomes an obvious term over there. So even if you don't write it, the idea is the same, right? I'm not taking limit on both the sides. The idea is when X becomes very close to A or your B becomes very close to A, right? And you may also add an extra condition that F of A is equal to 0 because you can call this entire function to be, you can call this entire function to be let's say P of X and this function to be Q of X and you can say you're actually trying to apply, you're actually trying to apply L'Opital rule to something like this where where where P of A and P of B are both 0 or limiting case of both of them are 0, correct? So it becomes a more generic version and this will be nothing but F dash A will be nothing but P dash X or P dash A by Q dash A. That is what I'm trying to say. Clear? Any questions? Any concerns? Now, time to take some questions based on this. Let us see what type of questions will be coming on. LMVT and Kushi's mean value theorem. Let's get started with some problem solving. Can I go to the next page? Next slide. If you have copied this. Okay. Let's take questions. We can probably start with this question. This seems to be a simpler one. IITJ 2003. This question says that in the interval 0 to 1 closed, language's mean value theorem is not applicable to which of the following functions. One more thing I would like to add over here. Very important. When you when you say rules theorem, that is, let's say if a function is continuous in closed interval, differentiable in open interval and f of A equal to f of B, then there exists at least one C in the open interval A to B, where the derivative becomes zero. Why so are some may not be true? Are you getting my point? So reverse of rules theorem may not be true. That means that means sorry about that sound. That means if a function is having the derivative at a point in an interval as zero, right? It doesn't mean the function is continuous in the open interval, closed interval and differentiable in the open interval and f of B is equal to f of A. Why so are size not true? So rules theorem basically is in a one directional. If it is continuous, if it is differentiable, if f of A equal to f of B, then this condition will be true. The converse is not true. Right. So don't don't apply converse to it. If there is derivative at a point C is zero, then there is some interval, right? Where the function is continuous. No need not be. For example, I may have a function like this. Okay. And there is a whole over there. Okay. So let's say this is my A and B. And I could have a point here where the derivative is becoming zero. But the function is not continuous in A and B. Okay. So vice versa need not be true. So don't apply reverse or the converse of those theorems. That may not be true. All right. Let's do this question. I would request you to give me a response on the chat box. To which of the following is LMVT not applicable. Is not applicable. Okay. Okay. Okay. Okay. Should we discuss it? Let's look at the first function. Now this function is basically made up of polynomials. What are polynomials? Is it continuous? First of all, now the only critical point where I will be interested in checking this function will be half. Right. So in zero to one, only one critical point is there, which is half. So what is the left hand limit for this? You'll say it's a zero. Right hand limit that also zero. Correct. So left hand limit at X equal to half is equal to right hand limit at X equal to half is equal to the value. At half and all of them are zero. So this is continuous. No issue. We continue D. So I'm talking about a option. So it is continuous at X equal to half. Is it differentiable? So let's check F dash half minus is minus one. And F dash half plus F dash half plus. If I'm not mistaken, it is two times half minus X into minus one. And at half it will be zero. No, non-differentiable. Non-differentiable at X equal to half. So the initial condition of languages mean value theorem is not met. The theorem says it must be differentiable. In this case, it must be differentiable in an open interval zero to one. So which is not in this case. So this guy is our answer. Okay. Is there any more? Let's try to figure it out. F of X. So second one, B part. In this case, I don't have to do anything because I know this function very well. This function is like ripples. Okay. Ripples. This gap which was at zero has been filled up by one. So there is no problem with continuity. It is continuous. It is smooth also. Correct. So it is differentiable also. So this guy has no issue at all. So this cannot be my answer. Okay. Now what about C part? C also I will normally use a graph. X mod X graph is like this. Half parabola up, half parabola down like this. Okay. Continuous zero to one. Differentiable open interval zero to one. So this also cannot be my answer. Okay. Now what about D? D is mod of X. Now remember you have to analyze only between zero to one. Only between zero to one, this graph will actually not come into picture. This part of the graph will not come into picture. I don't see any continuity and differentiability issue with this, this graph. Okay. So this graph is perfectly continuous and perfectly differentiable. So D also cannot be my answer. Only one option, option number A is correct. Now most of you who said D basically did a very small, you can say, you can say conceptual error. You started checking the differentiability at zero. Whereas languages mean value theorem says it must be differentiable in open interval zero to one. So open interval, you can't check it at zero. Got the point. So some of you said A and D because of that, but the answer is only A. Okay. I knew people would make mistakes here. Anyways, can I go on to the next question? It's just a second. I may un-share my screen for something. Can you see my screen? Chhattah, I don't know. I will be un-sharing it for something. Actually, the problem is when you remove the charger from the MacBook, the screen automatically gets, yeah, the screen actually gets unshared automatically. I don't know why. All right. Let's take another question. Question is, let f of x and g of x be differentiable for x line between zero to two and f of zero is two, g of zero is one, f of two is eight. Let there exists a real number C in the closed interval zero to two says that f dash C is equal to three G dash C. Then the value of G two must be. It looks like Cushi mean value theorem. Okay. Just solve it. Very good. I mean, okay. Okay. Guy three guys easy. This is just under 30 seconds. Okay. Okay. Okay. All right. Let's discuss it simple. So it's already given that f dash C by G dash C is equal to three and we know by Cushi mean value theorem. This is equal to f of B minus f of A by G of B minus G of A. And here A and B are is respectively two and B is actually you're sorry. A is zero and B is actually two. So it's f of two minus f of zero by G of two minus G of zero f of two is what f of two f of two is eight. Okay. F of zero is two by G of two G of two. Okay. Simple. So solve this simple equation. So that gives you G two minus one as six by three, which is two. So G two is going to be three. So option number B is correct. Okay. Simple question. Let's take another one. Can I go to the next question now? FB a function from two comma seven close to zero comma infinity close at zero continuous differentiable function then the value of this is where C is some value between two and seven open. Okay. I can see some pattern here. It's like X minus Y X square plus Y square plus X Y. Okay. Let's let me not disclose anything. Please solve this and give me a response on the chat box. Let's take a clue from this expression because this expression tells you a lot of things. Anybody. Okay. Cool. Anybody else. Okay. Let's. Okay. Okay. Very good. See, if you look at this expression, I mean, a lot of things, I get a lot of hints from this expression. This seems to be like, if you call this guy as, let's say. Yeah. So if you see this is like a, or you can say P minus Q and this is P square Q square plus PQ. Right. So overall this gives me a feeling that this expression is nothing but F of seven whole cube minus F of two whole cube by three, isn't it? Now, how do I reach this expression for that? What I will do, I will take a clue from the fact that there's some cubing happening. So if I consider this to be a function fed with seven and this to be a function fed with two, I can assume that there is another function G, which is actually F of X whole cube. Okay. Since F is a continuous function and differentiable function, even it's Q will be continuous and differentiable. Correct. Why? Because Q is nothing but the same function multiplied to itself three times, isn't it? So product of two continuous function is continuous product of two differentiable function is differentiable. So if you apply it two times, it will be basically coming to the fact that F of X whole cube or G of X, whatever you want to call it, this is also basically a continuous and differentiable function. Now, if this is the case, can I say if I apply LMVT in the interval two to five, sorry, two to seven. So I can say there exists a C belonging to the open interval two to seven such that G dash C is equal to G of seven minus G of two by seven minus two. So what is G dash C? For that, we have to differentiate this and put X as C. So derivative of this will be three F square X in pace of X, you put a C, F dash X. So I've already differentiated it and put a C already. I hope you are able to understand it. I don't want to do it again. And this is nothing but F cube seven and this is F cube two and this is going to be a five. Now I can get a fair bit of an idea that how they would have actually got this expression. So F cube seven minus F cube two is already sitting there and they got a three somehow. There is evidence that they have actually stopped the position of five and three. So they bought the three there and send the five to the other side. So you can do the same. So this is going to be this expression. So whichever option says five F square C times F dash C, that is correct. So I think option number C. I think most of you got it right. Kinshukh, your answer is right or wrong? Or you didn't answer it? Okay. So Kittu was correct. Rohan correct. Bebha, Gayatri, Akshet also was in progress. So we have already progressed. Okay. Let's move on. Which of the following options is correct? Multiple options may be correct. Yaprakul also got it. Which of the following is correct? Maybe multiple options may be correct. Choose all your options. Now, many times when we look at this, okay, we get an idea that somewhere, LMVT is being used, right? So, yeah, because it looks like that F of X minus F of Y by X minus Y4. Okay, Kinshukh. Okay, Archit. Okay. No problem. We do it. No issues. So two people have responded so far. Okay, Rohan. Okay, Archit. Akshet. Bebha, Anand. Rohit. Okay. One more minute I'll give you and then we can discuss it. Okay. Let's discuss it out. So we already know our LMVT which says that if I have a function F of X, which is continuous in the close interval A to B and differentiable in the open interval A to B, then there exists a point C lying between open interval A to B, where this condition will be met. So at least one point like that would be there. Now, all of you please pay attention. For the first case, I will consider that function to be tan inverse X. Okay. And I will pick up two points X comma Y and I'll just take X comma Y as to some inputs. Okay, of course the name, the variable X here matches with the variable of the function. If you want to can call this as Z for the time being. Okay. So what I've done, I have basically taken two real numbers A, X and Y. X and Y are two real numbers as stated in the problem which are not equal to each other. Where this function has to be continuous because you know tan inverse X is a continuous function. It is differentiable function. No, no corners, no cuts, no breakage anywhere in the function. So can I say F of B or let's say I switched the position of these two ways doesn't make a difference. Okay. So F of B minus F of A by B minus A. Okay. Should be equal to the derivative of this function at some point C. So C is a value in the open interval Y comma X. So that will be one by one plus Z square and replace your Z with a C. Okay. Now it is very obvious that this quantity will be greater than equal to one. That means this whole quantity will be less than equal to one. Do you agree with me? If I take any C. Which is between Y X, Y comma X open interval one by one plus C square will be less than equal to one. Correct. Which means tan inverse X minus tan inverse Y by X minus Y will be less than equal to one for all X comma Y belonging to R where X is not equal to Y. So option number A is correct for sure. A is paka paka correct. Let us apply the same concept to this, this fellow as well. Now all of you please pay attention for the B part. For the B part, I will assume my function to be, let's say my function is sin inverse, sorry for writing X here. I should write a Z over it. Yeah. So let my function be sin inverse Z. Okay. We all know that this function is continuous in close interval minus one to one differentiable in open interval minus one to one. Correct. So let us analyze it in the interval Y comma X, Y and X are not equal. So sin as per LMVT sin inverse X minus sin inverse Y by X minus Y should be equal to derivative of, now there is a C value which is in the open interval Y to X. So derivative of this guy at C, which is actually this. Now all of you please pay attention. The one minus C square under root will be definitely a quantity less than one. Isn't it? This quantity will be definitely be less than one because okay less than equal to one I can say. Yes or no? For all this quantity should be greater than equal to one. Yes or no? So I end up getting sin inverse X minus sin inverse Y by X minus Y greater than equal to one. Now my option says greater than one. Should I mark it or should I not mark it? Should I mark it or should I not mark it? What do you think? Mark it as it is included in the interval. See it is trying to say that there exists a C for which it will be greater than equal to one. Of course X is not equal to Y. If X is equal to Y the expression will not be existing first of all whatever. But my point is not that C equal to zero. Yes that's what I'm trying to say. C equal to zero actually can lie in that interval whichever you have taken. So greater than one will be actually a further restriction. Right? So they should have actually written greater than equal to one. If it is equal to one then basically you're trying to say sin inverse X minus sin inverse Y. So basically finding the derivative at zero. So meanwhile I personally feel that I should not include it. Yeah, I should not include this fact. Okay. So they should have written greater than equal to. So I would not include it. I will not go for it. Okay. Now if you talk about the second instance. If you talk about second instance. Sorry. C in C case you'll have a similar result. In C case you'll end up getting cos inverse X cos inverse Y by X minus Y is equal to negative of under root one minus C square. That again brings you to the fact that cos inverse X cos inverse Y by Y minus X is equal to one by under root one minus C square. And again, this quantity. This quantity is greater than equal to one. So this is definitely wrong. C option is definitely wrong. Because it should be greater than equal to one. Correct. What about the option? The option. The function is caught inverse actually so caught inverse X minus caught inverse Y caught inverse Y by X minus Y is equal to negative of one by one per C square. Okay. So because of that it will become caught inverse X minus caught inverse Y. By Y minus X is equal to one by one per C square. And this quantity has to be less than equal to one. So this D option is again a wrong option. So only option A is the right option. What happens with the answer won't sign change when you put the negative on the other side. No, I switched the position of X and Y. No. So that's why it became Y minus X here. That's what the question also says. Okay. So only option A is correct. But if the examiner makes option B correct, it will become a controversial question. So I mean I would go with A only if at all. All right. So let's move on to another question. Okay. Let's take this question. Let f of X be sign X. A is equal to alpha B is equal to alpha plus H. If there exists a real number T lying between zero to one. Where five dash alpha plus T H is equal to zero. And this is equal to lambda times this. Then what is lambda? Now, see this question has actually tried to indirectly use languages mean value theorem in the next second format, which I gave you. So if you're aware of this, you'll be able to crack this question immediately. Many people are not aware of this, that format and that gives an opportunity for the question center to frame a question in the second format, which many of the students are not aware of. Done. Okay. Anybody else? Anybody else? Okay guys, I would like you to pay attention to a small concept, which I would like to tell you. I think I've already mentioned about it casually also. See, I said this that in the one of our discussions earlier on, in the second form of in the second form of languages mean value theorem, there exists a T lying between zero to one says that the derivative of this will be nothing but f of a plus H minus f of a by a plus H minus a, right? There's nothing but the languages mean value theorem. Now, if you look at it more closely, you realize that f of a plus H is actually f of a plus H times f dash a plus a plus T H. Okay. And you can basically, you know, keep writing this to more, I mean, you can just extend this feature to even further. Right. How you can extend this feature even further. So you can write this something like f of a plus H is equal to f of a plus H times f dash a plus half or h square by two s square by two factorial f double dash a plus T H. Let me write clearly T H. Okay. I can keep on doing this. Okay. So basically formulation of the Taylor series. Taylor series actually evolves from here. Now, many times student asked me, sir, from here to here, how do you get? Right. Even though many times I normally tell to remember this result to the students not to go into the derivation part. But let me attempt to do the derivation part of it. Okay. For you to make sense. So let's do a proof for this. Let's let me try to do a proof for this. All of you please pay attention. Let me consider the function f to be a function which is continuous in close interval a to b and differentiable in open interval a to b. Okay. And now I'm trying to formulate a new function phi of x. All of you please pay attention. So I'm trying to formulate a new function phi of x in this way. f of b minus f of x minus b minus x f dash x minus b minus x the whole square times a big number. I like that number down. And that number is f of b minus f of a minus b minus a f dash a upon b minus a the whole square. Okay. So I've created a very, very, you can say unusual type of function, which is this big. So phi of x is phi f of b minus f of x minus b minus x f dash x minus b minus x whole squared times this particular number. And that number is f of b minus f of a minus b minus a f dash a by b minus a whole square. Now again, this is the reason why I don't give the proofs for this because it becomes very, very complicated for me to convince the child or the student. How did I take this function? What benefit it basically serves? Actually, this process comes from reverse engineering. Okay. So I reverse engineered and I got this expression. Now, let me tell you why did I choose this function phi? What benefit it serves? First of all, do you all agree that phi of x will be continuous in close interval a to b? Agreed? Why? Because it is made up of function and its derivative and some polynomial function. So it has to be continuous function. Okay. Because all the functions involved are continuous. Second thing is you would realize that this will be differentiable in open interval a to b. Now for this to be differentiable f dash x must be differentiable. So here I have to basically keep into my account that this function is twice differentiable. Okay. Because if this is not twice differentiable, finding this kind of a thing will not be possible. So we can't do this if it is not twice differentiable. So it has to be twice differentiable. Okay. So let's say function is continuous and it is twice differentiable. And you would also realize that I have chosen the function in such a way that f phi of a and phi of b should be equal. I would request you all to verify this. At your end everybody please do this verification and tell me yes on the chat box for this giant function. If you put first this guy is anyways a zero. I can tell you for sure because the way I formulated it will be zero for sure. Test whether phi of a is also coming out to be zero or not. Please test it out. Please verify it has to be actually because these two will get cancelled off and this will be exactly negative of this game. But convince yourself before I move on convince yourself. Convince sedan saying yes it is coming out to be equal and both are zero each. It's just a one second affair. You know, this is not a big expression to solve. Can everybody's convinced say yes. Yes. Okay. Great. So that means all those theorem condition are basically satisfied. Those theorem conditions are satisfied. Okay. If those theorem conditions are satisfied, I can say phi dash X will be zero for some X lying in the open interval a to b or you can say phi dash C would be zero for some C lying between. Okay. Not for all for at least for at least at least one C lying between a to b. Okay. Now, so if I differentiate this and put C, so I am going to differentiate this guy and put X equal to C. Okay. So when I do that, the left hand side will be phi dash C, which is zero anyways as per the row's theorem. This guy will give you minus F dash C. Correct me if I'm wrong. Okay. This guy will give me. This guy will give me. What is this term? This is going to be the derivative of F dash will be F double dash C. Correct. Yes or no? And this will be, this will be, I think, negative, negative F dash C. Am I right? Anything I'm missing out, do let me know. This part will become minus. This anyways is a constant. So minus two times B minus C. Of course, this will become a plus and this whole thing F of B minus F of a B minus a F dash a by B minus a the whole square. I hope I have not missed out any expression. If I have, do let me know. Now here, I'm going to do some simplification. But before that, please confirm whether everything that I have written is correct. I don't want to rewrite the whole thing again. In the second term, won't it be F dash C whole square? Second term, this term, this time I'm differentiating this with this, this fellow. Okay. So I'm applying product rule. So first B minus C as it is derivative of this will be F double dash. Then F dash X as it is derivative of this will be minus one. Okay. All right. So in the second bracket, you will see, first of all, some terms will get cancelled off. Okay. One term that will be answered off is F dash C and plus F dash C. So that will get cancelled off and you'll end up getting something like negative. Negative B minus C F double dash C. Okay. What will happen over here? You'll end up getting and I'm no simplification will happen. So it'll be two to B minus C. F of B minus F of A minus B minus A F dash A by B minus A the whole square. Okay. Let's do one simple activity. Drop of B minus C also from both the places B minus C from here, B minus C from here. So can I say, can I say F double dash C by two is equal to F of B minus F of A minus B minus A F dash A by B minus A whole square B minus A whole square. Also should I keep it over here? Okay. Now we are almost in the last step. I know it's a giant expression to look at. But if you see, you are finally at the last step. The last step is F of B is equal to F of A plus B minus A F dash A plus plus half of B minus A whole square into F double dash C. Okay. Now check this out. If you take your B as A plus H and if you take your C as A plus T H where T is some number between zero to one and write it down over here. Doesn't this expression become F of A plus H is equal to F of A plus H time F dash A plus half H square. Half is what? Half is basically one by two factorial. You can write it down. F double dash C and C is A plus T H. Right. This is what I was actually talking about over here. This is what I was trying to talk about. This is what I proved right now. Do you see this? Okay. Now this expression leads to Taylor series. Okay. And Taylor series will lead to Maclaurin series. The one which you learned in the bridge course with me. It all comes from this fact. Okay. Now how is this question? How is this fact useful in solving this question that I have given you right now to solve? If I go back to the question, the question says something very similar. If you try to rewrite this expression, taking sine alpha plus H as the subject of the formula, you'll end up getting something like this. Sine alpha plus H is equal to sine alpha plus H cos alpha plus lambda times lambda times H square sine alpha plus T H. Correct. Yes or no? Right. Now try to compare this guy with F of A plus H. Okay. This is F of A. This is H F dash A. And this should actually be H square by two factorial F double dash C. Okay. Just do a direct comparison. Compare these two. Correct. So if, if you do the comparison, this definitely brings the fact that F of X is actually sine of X, which is there in the given problem as well. No surprise here. It's already mentioned here. So this is F of alpha, F of A is your alpha by the way. You can write alpha here, alpha here, alpha here. Okay. So this is this. This is this. This is definitely this. So ideally F double dash C, that means if you double differentiate sine, you'll get cos and then minus sine. So you ideally should have got, you ideally should have got lambda H square. And this term, this term should have been negative H square by two factorial. Isn't it? Am I right? Isn't it? So C is what? C is actually this term, right? C is, where is C? Where is C? C they have not written, but C is actually this term. Alpha plus T H. So double derivative, let me pull the screen to this side. The double derivative of sine X function, the double derivative of sine X function. Okay. At X equal to A plus or alpha plus T H, alpha plus T H. What will it be? Sine is cos, cos is minus sine. So it'll be minus sine alpha plus T H. So basically this guy, this guy here should have been negative sine alpha plus T H. So when you compare, you get lambda H square is negative H square by two. Thereby your lambda will become negative one by two factorial, which is actually a negative half. So option number B is going to be correct. Okay. Is it fine? So now you don't have to go to this, you know, this complicated function to get to this result. You can actually use this, you know, Taylor series from now onwards. So this is, you can say J advance concept, but still I will feel, I don't, I don't think so. This is too difficult for anybody to understand. So this is how the process goes on. Okay. So many people asked me, sir, had I generalized it even further, could I have written it like this? F of A H F dash A. S square by two factorial F double dash A. Then H cube by three factorial H triple dash A plus T H. Okay. If at all it is required, you could have gone to this extent also. Okay. So this can move on and on forever. Okay. So with this, we will close this concept of LMVT mean value theorems and we can take a break now. As of now, the time on my watch is six 15. Let's take a break. 616 to 631. So let's meet at 631 p.m. On the other side of the break, I will discuss with you monotonicity. Enjoy your break. See you after the break. Monotonicity. Okay. Monotonicity. Monotonicity is basically the study of increasing, decreasing nature of function. Okay. So basically calculus is helping us to figure out whether the function is increasing at a point or increasing at an interval or decreasing at a point or decreasing in an interval by the use of derivatives. Okay. So NPS, have you done this concept in school or this has not been taken up in school? How are you placed with respect to monotonicity in school? Done. Not done. NPS. Have you done this concept in school? Nobody's responding. Kinshuk is saying yes. Shraddha is saying no. Are you same school only now? Yes. Okay. Fine. So you must be knowing a bit about monotonicity. So when you talk about monotonicity, we will talk about the increasing, decreasing nature in an interval and in a, at a point. So first we'll talk about monotonicity at a point, at a point, and then slowly we'll extend this to and an interval. In an interval. Okay. Now, since you've done this in school, tell me what do you understand by a function strictly increasing at a point? So there are two concepts that we're going to talk about. The concept of increasing and decreasing. So first I'm going to begin with at a point. So let me just put the heading here. So let me just put the heading here. Okay. Okay. Okay. Okay. Okay. So let me just put the heading here. Study of monotonicity of a function at a point. So monotonicity at a point. So when is the function said to be strictly increasing at a point? So when is, when is a function strictly increasing? Okay. Some, some books will call it as monotonically increasing at a point X equal to C. Let's say point X equal to C. Okay. Kinshok has given me a response. What if the function is not differentiable at C? You're saying the derivative of the function should be positive. Right. What if it is not differentiable? Right. It may also happen. The function is increasing at a point and there is no derivative. Right. Right. It doesn't exist. Then what will you do? Slow open derivative. Same thing here. What is the more, you can say granular level definition of a function increasing strictly increasing at a point X equal to C. That is what I'm asking. How can a function be increasing at a point? How do you define the increment? Or how do you define the increasing nature of a function at a point? A function is said to be increasing at a point X equal to C. If you find that the value of the function before C, let me write C minus H first. Yeah. The value of the function before C should be less than the value of the function at C. And this in turn should be less than the value of the function at C plus H. That means after C where H is a very, very small. You can say positive quantity. Okay. So if any function is exhibiting this criteria. Okay. Of course, I'm assuming that the function C is not the end point. It is not the extreme point of the interval or extreme point of its domain. So a function is said to be increasing strictly at a point X equal to C. If the value of the function at C minus H is lesser than what it has at C and the value of the function at C is in turn lesser than what it has at C plus H. Then the function is said to be strictly increasing at X equal to C. Is it clear? This definition doesn't involve any derivatives. This is the primary definition when a function is strictly increasing at X equal to C. Are you getting my point? Now if you drop the word strictly, if you drop the word strictly, that means if you say a function is increasing at X equal to C what does it mean? It means that the value of the function at C minus H may be less than equal to the value of the function at C may be equal to less than equal to the value of the function at C plus H. Okay. Such cases are also called non-decreasing at X equal to C. Non-decreasing at X equal to C. Now there is no, there's no role of calculus over there. There's no role of derivative there. This is the primary definition. This is the primary definition of a function being strictly increasing at a point X equal to C or just increasing at point X equal to C. Now try to understand the difference between these two concepts. When you say strictly increasing, that means the value of the function must keep on increasing in the vicinity of C. But when you just say increasing, it may be stagnant also. Okay. So it must not decrease actually. Right. It may be stagnant, it may be increasing, but it should not be decreasing. Are you getting my point? Okay. So that is what we call as strictly increasing and increasing. Understand the difference between these two. Yes. Yes. Okay. Thank you. In this definition, what would a function that is a jump for discontinuity? The Fc function actually be less than F of C. See, if there is a jump discontinuity, if the function is like this, let's say I have a function like this. Okay. Now let's say this is your point X equal to C. Correct. Is the value of the function at C, let's say this is your F of C. Is it higher than what it has at C minus H? Let's say this point is X equal to C minus H. Is it higher than F of C minus H? Yes. And this is your F of C plus H. This point is your C plus H. Is the value of the function at C plus H higher than F of C? Yes. Then in such cases, we will say this function is strictly increasing or monotonically increasing at X equal to C. Okay. But if jump is downward, I didn't get you. You want to say something like this? The jump goes downward. The jump is below. Like this? Yes. Now in this case, you cannot say it is increasing at C. This is not a case. Yes. So it cannot satisfy this condition. So I cannot say this function is neither increasing nor decreasing at C. Is it decreasing? That also we cannot say, because this is actually a maxima position. If you see it's actually at C, there is a local maxima happening. But that's a subsequent part of our discussion. We will not go into the maxima minima part. Okay. Now, when it is called just increasing or non-decreasing, if let's say the function is like this. Okay. Here you realize the value of the function at C is same as the value of the function at C minus H is same as the value of the function at C plus H or maybe the function maybe like this. Oh, sorry. Maybe it is like this. Yeah. So here also, if you see C minus H, it is same as F of C. And this is more than F of C. So in this case, all of them are equal. In this case, all of them are equal. F of C minus H, F of C and F of C plus H, they all are equal. So this is a case of, these are cases of non-decreasing. Okay. Non-decreasing. That means it is not increasing, but it is not decreasing nevertheless. Okay. This is a case where F of C minus H is equal to F of C and this in turn is lesser than F of C plus H. This will also be covered in the non-decreasing case. Okay. So all these cases as strictly increasing, these are all cases of non-decreasing case. Is it clear? Any questions? So I think Shraddha, your doubt is clarified. Ananyaar, clear. So here this stagnant for some time and then it is shooting up. Is it fine? Now, this definition which I have given you doesn't involve the use of calculus anywhere because the condition of strictly decreasing and increasing, sorry, strictly increasing and increasing basically depends upon these primary definition. Of course, we start using calculus whenever we find that the function is differentiable. That means it is continuous, doesn't have any common. There we start introducing our calculus concept. That is why the condition which I think Kinshuk told in the beginning of the question, that is actually not a necessary condition because sometimes the derivative may not exist also. But still the function can be increasing or strictly increasing just like the way I gave this example. So at this position you can say the function is strictly increasing. Okay. Is it clear? Any questions? Now, if I talk about strictly decreasing or just decreasing in the same way, let me talk about that. Let me check whether I have space over there. Yeah, I have space over there. So a function is said to be strictly decreasing, strictly decreasing at x equal to c. If it follows the reverse trend, that is c minus h is more than f of c and this is more than c plus h. Okay. Okay. Now, many people ask me this question, sir. Somebody's, yeah, Mike is on. Somebody's watching it in two devices. Okay. Anyways, somebody asked me this the last time, sir, what if the c point is at the end of the interval or at the end of the domain of the function? There's nothing c plus h, let's say. Then what do you do? If there is nothing c plus h, then you just have to follow the first two conditions. That means the value of the function at c minus h must be less than the value of the function at c. And if let's say c is the initial point of the question, then it must follow the these two definition. So basically when you are talking about the extreme positions of the domain of the function, where there is nothing after or before that particular point, then you can follow this particular primary definition. You can say in parts or partly. So if let's say c was, c was the rightmost point of the interval. Then f of c minus h must be less than f of c for it to be strictly increasing at c. If c was the leftmost part of the interval, then f of c must be less than f of c plus h for it to be strictly increasing at c. Okay. I hope you can make those minor changes depending upon whether c is actually the extreme ends of your, you can say domain of the given function. Okay. Anyways, just for your notes, I will erase this unnecessary markings. Okay. Now coming to strictly decreasing, I already gave you the definition. Now, if somebody says a function is just decreasing, decreasing at a given point x equal to c, he actually means it is non-increasing. Okay. So decreasing case may also include those cases where the function is even stagnant. Okay. But should not be increasing. So in such cases, the definition is more or less the same f of c minus h should be greater than equal to f of c. And this should be greater than equal to f of c plus h. Okay. So I am giving you these four conditions that you should not always think in terms of derivatives because derivatives will exist only for few cases where it is differentiable, where the function is differentiable. But that doesn't mean if the function is not differentiable at a given point c, there cannot be a case of strictly increasing or increasing or strictly decreasing or decreasing. It can be depending upon whether these conditions are satisfied. Okay. So most of you would be thinking, sir, we are doing application of derivatives. So derivative has to come in. Yes, it will come in, but let's not have a parochial view about increasing, decreasing concepts. Okay. Yes, if a function is basically satisfying the condition that they all are equal, then you can basically say any of the two cases. Okay. So you can say it's a case of both increasing and decreasing, or you can say it's a case of neither increasing nor decreasing like many of the books actually mentioned. If a function exhibits the criteria that f of c minus h is equal to f of c, is equal to f of c plus h means it's a constant function. You can call it in, you can make it fall in any one of the category. You can say it's neither increasing nor decreasing. Okay. Or you can say it's neither non-increasing case nor non-decreasing case. It makes the same thing. Okay. Is it fine? Any question with respect to the primary definition of increasing, decreasing, strictly increasing, strictly decreasing? Fine. Now when we try to put calculus into this, we have to first assume that my function is differentiable at c. Differentiable at x equal to c. Then only you can use calculus at c. Or you can use derivative concept and differentiate the function at c. So in such cases, if your function is strictly increasing, you realize that the derivative of the function at c must be positive. Now here is a very small catch, which I want to introduce here. Many a times equal to zero also means the function may be strictly increasing at that point, but it has to use, it has to be used with discretion. So use with discretion. Discretion means analyze the function completely before using it. Now let me give you an example for this. XQ function. What is the nature of the function? What is the nature of XQ function? Let me write give you option. This function, option number A is strictly increasing at x equal to zero. Is just increasing at x equal to zero. Of course other way around, decreasing, strictly decreasing at x equal to zero. And decreasing at x equal to zero. Which option do you think is the right option in this case? I am getting two types of options. Of course C and D many people are not saying. That's obvious. Yes, everybody please participate. I want everybody to participate. Very good. The answer is actually A. The answer is actually A. Okay. This function XQ, if you see the graph of the function, let me draw the graph of the function. Okay. Let me draw the graph of the function. Now if you try to zoom in this part, if you try to zoom in this part, the curve is basically like this. Okay. Of course there is a point of inflection there. The curve is changing the concavity from down to up. Okay. Point of inflection I will talk about in our maximum enema concept. So here you realize at x equal to zero, the derivative of the function actually becomes zero, isn't it? The derivative of the function at zero is equal to zero. So here it doesn't satisfy greater than zero. But you will be surprised to know that this function is still strictly increasing at x equal to zero because it basically satisfies our primary definition. And what are the primary definition? That the value of the function at zero minus H is lesser than what it has at zero and this in turn is lesser than what it has at zero plus H. Despite the derivative being equal to zero. Okay. So if f zero minus is less than f zero is lesser than f zero plus, it is a case of strictly decreasing. This is a case of strictly increasing function. Okay. So that is why I'm very basically careful about you being including zero in your interval also. So you have to, you know, watch out for the nature of the graph also. Okay. If a function is an injective function, it is compulsorily monotonic throughout. I think we had a discussion about it when we were doing functions. Yes or no? So Pakul had a question. He says if a function is a one-one function or an injection, is it compulsorily monotonic throughout its domain? What do you think? So if a function is monotonic, then it is compulsorily one-one. Is the converse necessarily true? Is the converse necessarily true? Tell me, is this possible? He has a very good question actually. He's saying that the function is one-one. Does it have to be monotonic? Compulsorily. Okay. Consider a situation like this. Here if you see, I have taken a situation in this case where if you draw any horizontal line, it is going to cut the graph only at one point. Correct? Anywhere you draw. Okay. Even if you draw through this, it is not going to cut here but it is going to cut at the other point. Okay. So is it monotonic? You'll say, sorry, is it one-one? You'll say yes. Because it is passing the horizontal line test. But is it monotonic? Is it monotonic? Means is it in the entire interval, is the function always, let me just put a sorry dot here. Yeah. Sorry. In the entire interval of this particular domain of the function, is this function always increasing at every point? The answer is no. Okay. And this question that you have asked for cool. It has been asked couple of times in many comparative exams also. Okay. So if it is monotonic, it is definitely one-one. But if it is one-one, is it monotonic? No. Now you got your answer to the question. For cool. Okay. Now, now coming to this, many of you would be again, judging this from your derivative point of view. Now the derivative is zero. You will be thinking that, this may, this is not strictly increasing, but no, it is strictly increasing here. Right. So what I want you to understand is, if the point is a maxima or a minima, and because of that, the derivative is becoming zero. Then of course that point is not your point where the function is increasing. But if the point is an inflection point, right, in those cases, you may have that point as your point where the function is increasing or, sorry, strictly increasing or strictly decreasing. Are you getting my point? So if there was a situation like this, let's say there was some dummy graph, which looked like this. Okay. Now, will you include this point? Will you include this point C into your strictly increasing point? You say this is your strictly increasing point just because your derivative is becoming zero here. You say no, sir, because it doesn't meet the criteria of this. So this criteria is not met. This criteria is not met in such cases. This criteria is violated. So this criteria is not met. Okay. But the same fact that here, the derivative is becoming zero. This is accepted because this criteria is met. This criteria is true. Let me write true. Okay. So you have to use your discretion. I can't give you a ready made formula or a ready made one stop solution for all the problems. Right. So the moral of the story is you should never take your eyes off the graph of the function. So graph becomes a very vital element when it comes to problem solving, because you may not be able to make it, but at least have an idea about what will happen to the left hand, right hand limit of that particular function. Is it increasing? Is it going to increase or not? Are you getting my point? Is it clear? Okay. So this is something which I want you to note down. So if a function is differentiable at X equal to C, then the function to be strictly increasing at C, the derivative of the function should be greater than zero. And in some cases where there is an inflection point, especially you will have equal to zero also included in your inequality. So it will be an impure inequality. That is what I want to say. Is it clear? So now many people ask me, sir, you know, if I, if I can't make the graph, then how to do it? Then you basically can use method of interval also. Right. What is method of interval? Method of interval is your wave of science scheme. So if you realize that about a point, the left-hand derivative and the right-hand derivative are positive, right? The left-hand derivative, right-hand derivative is positive, but exactly at C, the derivative is zero. Then still that point is a point where the function is strictly increasing. Does it make sense? See, if let's say you made a wavy curve for F dash X. Okay. Wavy curve, there was a point C. Okay. So this was the point where it became zero. And here the sign is plus, here the sign is plus. That means at C, the function is, at C, the function is Si, strictly increasing, correct? Because, because the derivative of the function is basically like this. So it is basically positive here and here also it is positive. So this is positive slope and here also it is positive slope. Right. But if it is changing sign there, then you should understand that it is a point of extrema. It is a point of local maxima or a local minima. Okay. Then that point should not be included. Then that point should not be included in your, that point should not be called as a point where it is strictly increasing or strictly decreasing depending upon the situation. Are you getting my point? Are you getting this analysis because this is important. Just knowing the formula F dash X greater than zero at that point is a point where it is strictly increasing. No. You may miss out. Infection cases. Okay. Non-monotonic at X equal to zero. No, it's monotonic. It's strictly increasing. Why? What happened? X mod X is strictly increasing. I minus two X. And if you put excess slightly negative, it is still positive. No. Cool. Think again before you say, before you ask any question. If you differentiate it, left-hand derivative is minus two X. Right-hand derivative is plus two X. If you put X in the left-hand derivative, you're putting an X, which is negative already. So negative two of negative quantity is a positive quantity. Right. And for two X, you're putting excess slightly positive quantity. So that will again be a positive. So this positive on both sides of zero. So it's a strictly increasing position. What are you writing? I'm not able to understand. What is the derivative of X mod X to the left of zero? So Dhanth has a question. No. It is minus two X. I'm not able to answer the question. Sir, can you repeat? What do you say about the point of inflection being C? Many times what will happen to Dan? At point of inflection, the derivative becomes zero. And derivative also becomes zero at maximum minimum positions. Maybe you have done it in school, local maximum, local minimum. Have you done that part? Correct. So how do you differentiate? Which is a point which I have to keep? Or which is the point where the function is strictly increasing and which is a point where the function is not strictly increasing. You need to understand whether that point was a point of inflection or a point of maximum or a minimum. If it is a point of inflection, the sign change of the derivatives or the sign of the left-hand derivative and the right-hand derivative in the neighborhood of that point will both be positive. Then it will be a strictly increasing point. If it is both negative, it will be strictly decreasing point, which I will write it in some time. Don't worry. But if the sign is different, that means to the left of C, the sign is positive. To the right of C, the sign is negative. Then it is actually a point of maximum, which I will discuss with you later on. So there should not be any sign change in case you are able to make the wavy curve for that particular function derivative in the neighborhood of C. Make sense? Okay. Let's carry on with our discussion. So this is something which I wanted to discuss. And please note down if the function is just increasing, if a function is just increasing at X equal to C, then of course the derivative of the function at C must be greater than or equal to zero. But here again, let me tell you, you can't include those zero cases where there is a maxima or a minima coming up. Many of the books I have seen, they make these mistakes. Including some of the trainers, including some of the teachers, some of the, I was shocked to see this in one of my seminars I attended, conducted by one of the very old faculties. They're just putting arbitrary saying include this, put sound round bracket, put square bracket. I will tell you what they actually mean to say. See, don't take your eye off the basic primary definition of increasing, decreasing or strictly increasing or strictly decreasing. When they say greater than equal to zero, that equal to zero is basically a case of inflection or your case where it has got a stagnation. This equal to zero cannot be a case where there is a maxima or a minima. Are you getting my point? Let me give you an example. Let's say the function was like this. It went here and it basically stagnated here. And basically, let's say it may have moved up. Now here, if you take this point as C, X equal to C, is this a point of increasing? Yes, it is a point where the function is increasing, agreed. Even if you had something like this, that means there was an inflection point. Is this a point of increasing? You will say yes. But if I have a case like this, is this a point of increasing? No. Is this a point of increasing? No. Why? Because the primary definition is not getting satisfied here. For it to be a point of increasing, the value of the function, at least to the left, it should be lesser than equal to and to the right of it should also be greater than equal to that value. Sorry, that value should be lesser than... See, this definition must be satisfied. This must be satisfied. Okay. So while this is getting satisfied over here, F of C minus H is at a lesser value than F of C, but F of C plus H, this criteria is not satisfied. So this guy will also satisfy F dash C equal to zero. Will you include that point there? No. So only point where there is a stagnancy happening or where there is an inflection happening, they can be included, but these guys cannot be included. So that doesn't mean you'll start including maxima, minima. Values, okay. So don't blindly follow the textbook. Some of the textbooks, I don't want to name them. They are non-books. I don't want the publishers to slap me a notice of 50 lakh rupees trying to defame their books. But that is not how it should be. And we had basically a good discussion on it with the teacher group as well, all over the country. They were almost our 800 teachers and we had a discussion about this. Okay. Anyways. So is the idea clear when to include zero? So this guy has to be used with discretion. Okay. This guy, use it with discretion. Use it with discretion. That means do some other, you can say analysis before you jump to that conclusion. Is it clear? Any questions, any concerns? So so far I was just talking about at a point. So I gave you a definition about monotonicity at a point primary definition. And if the function was differentiable at that point, then I also use the derivative. But in the derivative also there are a lot of ifs and buts. Okay. Be careful about it. Can I have a similar analysis for strictly decreasing and decreasing? Anyways, I've spoken enough about increasing strictly increasing and increasing. So similar concept will be valid for strictly decreasing and decreasing as well. So strictly decreasing again. The fact is the same. You have to have your F dash C should be less than. And again, I'll just put a dotted, dotted equal to sign because it has to be used with discretion. Okay. Then same decreasing. See, things look very easy when you do it in school. Okay. When you do it in school. I mean, you are, you are given some straight away formulas to be used, right? But things are not as easy or simple looking when you're basically dealing with a serious exam like J main, J advance, all those exams. There you have to be careful. Is it fine? Any questions? So should we not talk about in an interval? Let's talk about monotonicity in an interval. So when you say a function is strictly increasing in an interval, strictly increasing, let's say in an interval. I. Okay. I'll take an interval. I, when you say a function is strictly increasing. Where is my word increasing? Sometimes I eat up my word. Okay. So when you say a function is strictly increasing in an interval. No, see. A function is said to be strictly increasing in an interval. If you take any two X one and X two belonging to that interval. Such that such that. If X two is more than X one. Then f of X two should be more than f of X one. For any. X one X two belonging to that interval. Right. Any X one X two pair belonging to that interval, pick up any two numbers in that interval. If you realize that X two is more than X one, then the output from f of X two should also be more than the output from X one. That is f of X two should be more than f of X one. Then only you can say the function is strictly increasing in that entire interval. And this is the primary definition, basic definition. Okay. So what are the cases where the function can be said to be strictly increasing? So there are three cases that comes up. What is the case of like this? I'll call this as first case. Okay. So let's say I talk about this interval, whatever is this interval from here to here. Okay. Here if you see the function is going to always satisfy this condition, that means if you take X one here and let's say X two here, then the value of the function at X one or value of the function at X two will always be more than the value of the function at X one. Okay. So this is a case where the function is strictly increasing in that interval. Next is a case. Next is a case which is like this. This is the second case. Okay. Again, if you take any two X one and X two value such that X two is more than X one, then f of X two will be more than f of X one. And the other is a constant case like this, which is a stagnant case. Not stagnant case. It's a uniformly increasing case. You can say that. Okay. So in all these cases, you will realize that the function is strictly increasing cases. Okay. Strictly increasing in I. Okay. I'm not made I everywhere, but I hope you can figure out how this is your I. This is also your I. Is it fine? Any questions? Any concerns? This is the primary definition. This is the primary definition. Now, having said that this is impractical. Right? Why does it practical? How will you figure out for every pair? Right? Is it possible that you pick up every pair in that interval? Let's say I give an interval. Two to five. Two to five. There are so many infinite pairs you can form. Is it practical to do this? No. This is good for your basic primary understanding. Right? So how does how do you practically solve this question? Now, let us say if my function is differentiable in that interval. So if your function is differentiable in I. Okay. Then basically you can say that if you take any X in that particular interval, then the derivative of the function at that point should be greater than. Now, what should I say? Greater than equal to or greater than? You'll say greater than but equal to you have to use with discretion. Use with your discretion. Are you getting my point? Is it clear? So if it is a point of inflection, that means that is also a point where the function is strictly increasing. So that is allowed to be in that interval. Yes or no? Because this primary definition is satisfied even for the infection point. Yes or no? So greater than equal to but equal to has to be used with discretion. It should be a mindful inclusion or exclusion, whatever you're doing for that zero. Is this fine? Okay. Now a small thing I would like you to answer here. For all these three cases, the derivative is. The derivative is positive. Okay. For all X. Isn't it? Okay. This also derivative is positive. But you can see clearly that there is a difference in their concavity. This is concave downwards. This is concave downwards. This is concave upwards. Okay. And this is neither concave up nor concave down. So you can say it's a straight line. Okay. It's a straight increase. Okay. How do you distinguish? How does calculus help you to distinguish between these three cases? How can you figure out that the function is strictly increasing but with concave downwards or just strictly increasing but with concave upwards? And it is a linear increase. How do you figure it out? Okay. Yes. So we do that by the use of double derivative provided the function is twice differentiable. Okay. So guys, please do not take the differentiability criteria for granted. Right. I've seen many places that Stainless starts using derivative. What if it is not derivative there? Differentiable there. Can't I say the function is strictly increasing in that interval? It can be. The example which I showed you this example. In this interval it is strictly whatever interval is this in this interval the function is strictly increasing. But it is not differentiable here. Right. So if your function is differentiable you can say you can use the derivative to figure out the increasing nature of the function or decreasing nature of the function in that interval. But if the function is twice differentiable you can also talk about its concavity. So in such cases you would realize that the function double derivative at any point X will actually be negative. Why? Why? Can anybody tell me? Why? Let me write it slightly below. Because the slope is decreasing absolutely. As you can see at every point the slope is positive. Agreed. Positive, positive, positive, positive, positive. But the slope function the gradient function itself is a decreasing function. Yes or no? F dash X which we call as the gradient function that function is a decreasing function. Yes or no? So if I increase my X from X1 to X2 the value falls down. Isn't it? F dash, F double dash X1 is more than F double dash X2. So value falls down. So in that case your double derivative will become negative for all X belonging to that interval. If the curve is concave upwards so it is strictly increasing following the pattern it is concave upwards then the double derivative here will be positive. And in this case you will realize that the double derivative will always be a zero. That means there is no change that is F dash X that is your gradient function is a constant function. There is no change in the slope at any point. Okay. So please understand these you know finer nuances. So these three are all cases of strictly increasing but strictly increasing and concave downward how it is different strictly increasing and concave upwards how it is different strictly increasing and linear growth or a linear increase how it is different. Is it fine any questions any concerns so I don't think so we need to you know go any deeper into this so let us now move on to strictly decreasing in an interval. Maybe I will use this space. So a function is said to be strictly decreasing decreasing in an interval if for any pair X1 and X2 belonging to that interval such that X2 is more than X1 the value of the function by the way I did not write for one second just give me I will just take increasing also sorry about that I did not talk about increasing if X1 if X2 is greater than X1 then F of X2 will be greater than equal to F of X1 this is the primary definition okay in this case it was pure inequality F of X2 should be greater than F of X1 but in this case I'm so sorry so sorry strictly what has to be dropped sorry thanks again yeah so in this case F of X2 should be greater than equal to F of X1 okay is it fine so how do you use derivative over here again derivative will give you the same result so if the function is differentiable is differentiable in that interval then F of F dash X should be greater than equal to 0 again equal to 0 has to be used with discretion okay for all X belonging to that interval by the way the question asks is in actually reverse order the question is asked as what is the interval where it is increasing the question asks is actually in the reverse order they don't give you the interval and they ask you to verify it of course there are questions like that but the question mainly asked 90% of the question is actually in a reverse order they will say in which interval is the function strictly increasing okay or in which interval is the function increasing okay like that so interval will be sought for not the other way round of course other cases are also there but very less okay so let us now talk about strictly decreasing and decreasing cases also and let's see how calculus helps us there however I find calculus is a toothless you can say you can say tool in these cases at many times okay calculus fails let's talk about strictly decreasing in an interval so strictly decreasing in an interval if you take any set of points x1 and x2 belonging to that interval such that x2 is more than x1 then f of x2 will actually be less than f of x1 that means the value of the function will go down if you increase the input so if you put more input value will fall down okay of course you know many of such functions like cos x function from 0 to 2 as you increase the angle the value will fall down in fact 0 to pi you can go actually okay anyways so if the function is differentiable then basically what do we say if your f of x is differentiable differentiable in i then basically you can say that the derivative of the function should be less than again equal to 0 must be used by discretion please use it with discretion don't blindly say greater than equal to 0 for all x belonging to the interval similarly if the function is increase decreasing decreasing means non-increasing in the interval i then you can say that for any pair x1 and x2 belonging to the interval if let's say such that f of sorry such that x2 is more than x1 will imply that f of x2 will be less than equal to f of x1 I should actually not write then because then means if has to be there okay anyways similarly if the function is differentiable differentiable in that interval then basically again f dash f dash x must be less than equal to 0 for all x belonging to that interval okay see theory wise there is nothing great in it okay just like our other applications it is not theory heavy like your mean value theorems where mean value theorems were really heavy with respect to theory but these are very light actually you can understand it from the nature of their slopes okay so let us look into problems let us look into problems based on that that will be that will be giving us more clarification with respect to this topic so so let's take this one find the interval in which this function is increasing okay just increasing please solve this and give me a response on the chat box very simple question based on your basic understanding one thing I forgot to tell you a primary requirement for doing increasing decreasing into finding increasing decreasing intervals is basically knowing your wavey curve method of difference sorry method of intervals