 All right, so welcome to another session on application of derivatives So today we are going to take up a two important application of derivative one being the mean value theorems and The other is the monotonicity that is the increasing decreasing nature of functions Now both these concepts are pretty important I Understand you have done mean value theorems in school our old theorem languages mean value theorem But let me tell you its interpretation in J is very very tricky Some of the most challenging questions have come from these topics Okay So let's let's discuss today mean value theorems. Let me start with the mean value theorems mean value theorems in Mean value theorem, I would start with the rose theorem Rose theorem was basically first given by Michelle rule. Okay, so there was a guy by the name of Michelle rule You don't have to know exactly about this game Michelle Michelle rule Okay, he lived in the 17th century so he gave a very, you know interesting theorem which was actually Trivial in the present frame of calculus, but yes, it was a very important breakthrough then So what he said was if let's say f is a real valued function Let's say f is a real valued function I'll write the name of the topic Misch Michelle source rose theorem rose theorem So he said that if f is a real valued function, which is continuous in the closed interval a to b Okay, it is Differentiable in the open interval a to b it also satisfies this characteristic that f of a is equal to f of b F of a is equal to f of b Okay, then there exists Then then there exists at least one see Belonging to the open interval a comma b such that The derivative of the function at that point becomes a zero Okay, such that the derivative of the function at that point becomes zero now. How do we prove this theorem? How do we prove this theorem? The proof of this theorem basically can be seen in two ways. There are two cases under it If your function is a constant function Then this particular theorem is obviously correct Correct. So if I take a constant function, let me take a case number one If my function is a constant function something like this Okay Not a very straight line. Let me draw Okay, so this is your a this is your b and f of a and f of b values are equal to each other So these values are equal f of a and f of b values both are equal then obviously you would realize that At least one point will result into so many points. So every point you can say Blying between a to b the slope of the tangent at that point would be zero take any point in this particular zone The slope would be zero Correct. So his theorem is definitely true for such a case But what about a case where the function is not a constant function? So for the purpose of simplicity, I'll draw a function like this Okay, I'm just drawing a function like this Which is showing increasing and decreasing characteristic both So let's say this is your a This is your b Okay, and f of a and f of b values are equal f of a value and f of b values are equal Okay, now remember his Criteria was it should be continuous. So as you can see there is no break in the function Okay, it should be differentiable That means you can find the derivative at every point in the open interval a to b that means the function is smooth Right the function is smooth Okay Now if this function is showing a increasing and decreasing characteristic Now it can show several increments and decrements. I'm just taking the simplest of all case Okay, that's why the criteria says at least one point. So I've taken a criteria of exactly one point right now just for explanation So if you realize that let's say there's a point C Okay, this point C you would realize is At the maximum point of the function Now I could have taken an example where this curve would have gone down. So that could have been a minimum point Okay, or there could have been several maximum and minimum points in that function right Now, can I say since this point is at its local maxima The value of the function at this point can I say the value of the function at C Would be greater than equal to what it has before it correct Yes, I know and at the same time will be greater than what it has after it. Is this fine. Everybody's fine with this Yes, correct Venkat so that it is either constant or if it increases then it has to decrease back F of a equal to f of b is a condition which is kept to fulfill this requirement. You are absolutely right. Okay Now, let us say I want to find out the derivative of the function at C What does first principle say First principle says the derivative of the function at C can be found out by finding its left hand derivative and right hand derivative and No equating both of them. Okay, so let us find out the left hand derivative first Let us find out the left hand derivative first left hand derivative that is f dash C minus This would be what limit H tending to zero plus F of C minus H minus f of C by minus H Correct now since All of you, please pay attention to this condition Since f of C minus H is lesser than f of C Can I say the quantity on the numerator is A negative quantity Yes or no Yes, sir, correct. What about the quantity in the denominator? That's also negative quantity that means overall this quantity is a positive quantity Can I write it greater than equal to zero? Okay, so this is your left-hand derivative If you look at the right-hand derivative in a similar way It will be f of C plus H minus f of C by H correct Now f of C plus H is also lesser than f of C Right, so the quantity on the top. Let me just put this quantity slightly above Yeah, so the quantity here on the top This quantity is a negative quantity and this quantity is a positive quantity So overall this quantity is a negative quantity less than equal to zero Okay Now, you know very well that this function is Differentiable at every point in the open interval a to b that means it must be also differentiable at a point C Now if it is differentiable at a point C and here you have a situation where the left-hand derivative is greater than equal to zero and Right-hand derivative is less than equal to zero and they both have to be equal also and they both have to be equal also because The value of the function derivative at C should be same as Should be same as f dash C minus And should be same as f dash C plus So the only way they can be same is When it is equal to zero Because the quantity cannot be less than and greater than at the same time So equality sign will be used in both these conditions Getting the point So this proof-souls theorem okay Actually, it's a very trivial theorem which says that which says that if you I mean, I'm just giving a very physical representation of it So let's say this is our displacement versus time graph Let's say this is a displacement versus time graph Okay, that means you Accelerated I mean not accelerated you had a velocity Okay Positive velocity and then you started having a negative velocity so as to reach to the same point back So let's say a car is moving forward and Then it has to come back again to the same position. So it'll stop somewhere Correct. It'll stop somewhere and then start coming backwards Right. So overall his you can say His average speed is zero because his next net displacement is zero, right? So there would definitely be a point in the journey Where the average speed will be equal to the instantaneous speed So the car must stop somewhere So as you can see, this is the point where the car has stopped So this is the point where DS by DT will become a zero then it can only come back to the previous position Are you getting my point? So in order to come back to its initial position, it must stop somewhere And then only come back to the previous position. Is that clear? Any questions here? Now I have given you an alternative explanation of Roll's theorem and I was doing Theory of equations with you where I was primarily dealing with polynomial equations correct so remember the another way to look at Roll's theorem and One of the important application for us is One of the important application The local maxima minima occurs with that thing is derivative zero If there is a polynomial function or for that matter if there is any function, which is known to be continuous and Differentiable, okay, and if such a equation has roots Has roots alpha and beta let's say I just taking two roots for the sake of For the sake of you know easy concept so that you can understand it So what does this theorem say if this function f of x is known to be continuous? Normally Exponential functions polynomial functions Logarithmic functions etc. They are known to be continuous. So you can apply this concept there So if you know your function is continuous everywhere And you know your function is differentiable also everywhere Let me write it down Okay, and this Equation has roots alpha and beta What is the meaning of root alpha and beta? It means it satisfies the third condition also that is f alpha is equal to f beta Now it's a different thing that it is going to be equal to zero also Okay, but a role is not bothered about whether it is zero or whether they are both fixed It doesn't care about it having a value of zero just cares about them being equal okay, then Then f dash x equal to zero Has at least one root Has at least one root Which belongs to the interval alpha to beta right as you can see This is completely in line with the initial theorem that he mentioned so when you say fc is zero and See lies between a and b. It is like saying this is akin to saying or this is similar to saying F dash x equal to zero has a root x equal to c isn't it This is like saying that f dash x equal to zero has a root which is C. Isn't it? So that is what this particular, you know, alternative way of understanding those theorem says and this is very important Lot of questions have been framed on this Okay, so before I move on to Langranges mean value theorem. Do you have any question on this? So in the second definition, why should fx f of x be zero f dash x So f f of x So you've given if f of x has roots alpha beta Yeah, if f of x equal to zero has root alpha beta means this characteristic will be satisfied now then only What is what is the characteristic of rose theorem rose theorem says f a should be equal to fb? correct, so root is a you can say it's a In Hindi we say it's a bahana For making both of them same. It's an excuse for making both of them same, right? So alpha and beta being roots of f of x equal to zero establish the fact that f alpha and f beta are same values and both are equal to zero Okay, so this condition is met getting this point and really so we have to meet So if your function involved is continuous function If your function involved is differentiable function, which you can make out from the look and feel of the function itself Right and f of alpha and f of beta both value both are equal They may be zero also because root implies they are zero then this condition will hold to Then this condition will hold to So you can apply it all theorem even if there's missing point discontinuities except if c is the point where there's discontinuities Yes, this is something which I would like to discuss here. There's something called reverse of rose theorem Okay, now rose theorem says that if this is satisfied then this will happen vice versa. He never said That means if you realize that a function has a derivative of zero at a point C It doesn't mean it has to be continuous. It may have holes It doesn't mean it is differentiable. It may have corners Are you getting a point? So reverse may not be true Reverse of rose theorem may not be true. So please you don't get me wrong If these three criteria meant then this will happen Reverse is not to be considered to be true For example, as he said, I could have a situation like this whole Yeah, correct. I can still have a point C where that derivative becomes zero Right here F dash C is equal to zero and it has a hole here Correct, or I could have a point like this hole Okay, and then there is a there is a dip to zero Okay, so there's a dip to zero and there's a discontinuity here. Sorry. There's a non-differentiability here. This can also happen Right. So reverse may not be true Reverse may not be true clear Any questions? Okay. Now I would take the problems based on this little later on But I would like to take up before that Langrange is mean value theorem Okay, so that we cover all the theorems and then we start solving questions on them Else what will happen? I have to keep Coming back to the theory part again. The next theorem that we are going to talk is Langrange is mean value theorem NCRT just calls it mean value theorem Even though I call it as LMVT But in NCRT, they would just use the word mean value theorem. If they just use the word mean value theorem Means they are talking about Langrange is mean value theorem Okay Now what what does this theorem say this theorem is basically has better roles theorem So you can say it's a more generic version of what role Michel roll had actually stated So in this theorem, we say that again if F is the real valued function which satisfies the following characteristics Number one it is continuous in the open and sorry in the closed interval a to b It is differentiable Did you realize why I said it is differentiable in an open interval a to b? The ends it might be just the end of the function No, then no, that's fine But why does it open interval because if you see the proof of this once again? You always needed a point to the left of something and right of something right? How would you get C minus H and C plus H if you're trying to apply it at the corner points? So C cannot be a corner point That's what it's trying to say C can C has to be one of the points Which is between a and b excluding a and excluding b because you need a point left and right to it Then only left-hand derivative and right-hand derivative will both exist Right so the whole You can see the core principle on which the proof of rose theorem is base is that you need left-hand derivative and you need right-hand derivative So if you take it at a and b if you if you say it's differentiable in closed interval a to b How would you justify left-hand derivative at a and how do we find out right-hand derivative at b that we cannot justify? That's why this restriction of open interval differentiability has been imposed I in the school did they give you the proof for this? And no Raja jee nagar R and R No, I don't know people are not speaking. What happened? I don't know Tomorrow you tea So like it's cancelled from the portion so we didn't study. Okay. Oh They didn't do it when the syllabus reduce syllabus was announced Yeah, yeah, they would have finished it I also did the same mistake in in Napoli's are here shankin shankin yeah shawmick Anurag Okay Cool. Yeah, so what does this theorem say now in Wrangling's mean value theorem Only these two criteria are required. He basically You know remove the criteria of f of a equal to f of b So he did not require f of a equal to f of b criteria So what he said is that if a function satisfies these two characteristics then there exists then there exists at least At least one value C in the open interval a to b such that Such that F dash C is equal to f of b minus f of a by b minus a Got the point. Okay Now what he actually did I'll explain it to you graphically first and then I'll give you a mathematical proof for it He actually bettered roles theorem See in roles theorem. I'll just do a comparison of these old role theorem and mission and Languages mean value theorem. Let's say I take this graph for roles theorem. Okay, so what did roles theorem say is that? F dash C F dash C is actually zero for some point C Lying between a and b right. This is what he said. So at this point the derivative is zero Now what did mission role observe is that? He's basically calling it zero because the secant line that you draw over here So this point is actually a comma f a this point is actually b comma f b So this secant line a b will have a slope zero Right. So this zero is actually the slope of the secant line a b So what he did is said, okay, if this is the case that you want to use Then why do we have to have f of a equal to f of b? Let me just tilt it so let's say he Let's say made the graph like this Okay, so what he did he just rotated the whole thing. Let's say this is your a point now This is your b point now. So now he is no more adhering to f of a equal to f of b criteria But what he's still adhering is the fact that there would exist a point C There would still exist a point C where the slope would be same as Where the slope would be same as not very accurate Where the slope would be same as the slope of the secant so this and this will have the same slope Okay, correct. So he just rotated it So if you see the slope of the secant is actually f of b minus f of a By b minus this is the slope of this a b line And this is what will become the derivative at a point C sling somewhere between a and b line somewhere between a and b okay Again, this particular fact comes from the basic physical you can say comparison that Instantaneous velocity of a car at some instant at some instant Must be equal to the average velocity of the car Must be equal to sorry average velocity of the car. Okay, so let's say if you're traveling from Bangalore to Mysore and let's say you Your average speed was 30 kilometers per hour Then there must be a time in between the journey when the speedometer shows 30 kilometers per hour That's what this particular theorem says Okay Now, let me give you a mathematical proof of course. This is also mathematical proof, but I'll give you now a rigorous proof So what did Michel Lohl say he said that okay? Let's say I have a function which satisfies these characteristics So I have a function which is continuous. I have a function which is differentiable. Okay Now you formulated another function phi x This function was formulated by using your f of x and a linear function Okay, of course, there is a unknown lambda over here, which I'll replace in some time So try to listen to me here very carefully. So he said that let there be a function like this Okay, let there be a function like this Since f of x is a continuous function and of course x is a continuous function because it's a polynomial This itself will be a continuous function. So phi of x would be a continuous function. Okay, and Since both these functions are continuous polynomial is always continuous and this guy is continuous in a to b Can I say phi x will also be continuous in a comma b? Yes Since f of x is differentiable in open interval a comma b and polynomial function is always differentiable So he said that phi of x will also be differentiable in open interval a comma b Okay Now he chose lambda in such a way this guy was very smart He chose lambda in such a way that f sorry phi b and phi a values are equal So let's see how what value he would have chosen So if phi n phi b are equal that means you're trying to say f b plus lambda b is F a plus lambda a In other words f b minus f a is Lambda a minus b Right Which means he basically chose lambda to be f b minus f a by a minus b Okay Right, so he actually chose his function in this way. Let me just write it over it So his function phi of x was f of x plus Lambda x Okay, now what he said Since phi of x is a function which is continuous in close interval a to b Differentiable in open interval a to b and f of b and f of a are equal by the virtue of the fact that he chose This as your lambda value. Okay, that means that means Phi of x must satisfy Rose theorem must satisfy Rose theorem So if phi of t satisfies rose theorem, then phi dash c must be zero for some point to see For some point c lying between a and b Yes, sir, so what is phi dash? Phi dash x is what f dash x plus this constant That is f b minus f a By a minus b and if you put a c over here, that means if you replace this with a c Let me write a small C and this also as C Okay, then this should be zero as per him as For this point this this left-hand side should be zero if let's say zero that means f dash c will become negative of f of b Minus f of a by a minus b. So just use the negative to switch the position of the denominator terms So observe this negative sign here and just switch the position of Yes, and this is how this is how your Languages mean value theorem is proved Are you getting my point here? social For a fire effects will be a tilted graph of f of x. Yeah, because of this lambda x There would be a kind of you know Tilting happening. I'm not saying this lambda x is brought about a tilting. I just said he just you know Looked at it from a more broader perspective Okay, okay Adding lambda x to take a transformation to the graph to of course it will of course it will bring a transformation to the graph You are adding an increasing part to the function. So the function will rise But that depends on lambda also it may fall also So what he's trying to say is that if even if you don't maintain the same level of a and b right then This particular This particular relation will hold true If you're f of a and f of b are equal then Languages mean value theorem becomes over Getting my point So like we can do this addition of lambda x to any function to make it parallel to the x-axis Make it parallel to the x-axis Because that's what we're doing no sir effectively because Five of a is equal to five of b and we're adding lambda x so f of x was Essentially like tilted and then now we added lambda x and became parallel to the x-axis Huh Yes You can apply that to anything you did not make it parallel to the x-axis. You made f a and f b same You made the second line parallel to the x-axis. You didn't make the function parallel to the x-axis. Oh, yes Make the second line parallel. Yes Okay, so as We discussed here Remember one important thing Remember That if the function is such that f of a and f of b are equal Okay, then LMBT Would become RT That means languages mean value theorem will become the roles theorem But if f of a is not equal to f of b then you can use you can call it as Languages mean value theorem Okay, any questions here Now there's another representation of Languages mean value theorem which you might find in some books Sorry, I'm writing from I'm writing in a u-shaped actually I Started from here. I went down And then I'm going up now So while reading the notes, please be careful. Okay, don't read it like this read it in a u-shaped So another way of representing The languages mean value theorem which I call as the second form of LMBT is the same thing I'm just writing in a very fancy way Okay, so let's say your B is a plus H right, of course B is ahead of a right. So let's say B is a plus H Then basically what it says the C term which happens to be between a and B you can treat it as a plus theta H theta being some fraction Theta being some fraction between zero and one Theta being some fraction between zero and one Okay, so there would be a point a plus Theta H Where the derivative will become f of a plus H minus f of a by H Okay, so this is an alternate form of Languages mean value theorem many books will also quote it like this F of a plus H is F of a plus H f dash a plus theta H Okay Now you would realize that this has got a bit of resemblance to Taylor series and Later on in your undergraduate you will understand that your Taylor series comes from Languages mean value theorem nothing else So your Maclaurin series comes from Taylor series Taylor series comes from The Languages mean value theorem so all of them are linked Okay, there is something called Languages form of remainder and etc. Which is obviously not required for J But people who had written AP calculus would have studied it AP calculus BC Okay, so this is a precursor. I would say let me write it. This is a precursor To Taylor series a Taylor theorem first of all and then Taylor series. I don't want to get into Taylor series because not required Okay, let me take some questions. I've got some questions coming up Aditya Manjunath, can you explain the second form again? Okay, so Aditya very simple Instead of writing C. I have written C as a plus theta H Why because B is a plus H. So B is like a plus one H Correct. So if I if I take theta anywhere from zero to one, let's say half Then we A plus half H. So you're somewhere between A and B Just a fancy way of writing C Right, so you are basically trying to Linearly grade it from A to B. You are linearly increasing So this is like a slope Okay Yeah Sigh has a question says series for sin x cos x all are from the yes Yes, yes, they all come from Maclaurin series and Maclaurin series comes from Taylor series Taylor series comes from Taylor's theorem Taylor's theorem comes from mean value Okay, sir, so that is the linkage actually that is the linkage. Okay All right now this is something which I don't Teach To a J main aspirant, but if you're a J advance aspirant, you would require a Third type of mean value theorem, which I call as the Coochies mean value theorem Coochies mean value theorem cmvt Okay, cmvt You actually study a lot of you know value there is something called extreme value theorem Extreme value theorem says that if a function is continuous in a close interval a to b then it must have at least one minima and one maxima Okay, there's something called intermediate value theorem. I think I've already done intermediate value theorem and we were doing in the Theory of equations chapter. Okay Now people who are interested in writing J advance obviously most of you would be For those people this theorem will be required, which is the Coochies mean value theorem. The pronunciation is Coochie He was a French mathematician Some big name Louis Coochie is something is that I have forgotten his name So what does Coochie mean value theorem say it says that if if f of x and g of x are Continuous in the close interval a to b. Okay Secondly, both these functions Are differentiable in open interval a to b Okay, and one more small condition the derivative of g must not become zero For any point x lying in the open interval a to b then then there exists at Least one C in the open interval a to b such that Such that such that f of b minus f of a By g of b minus g of a is equal to f dash c by g dash c Okay, many of you would think sir. They have basically applied Langranges mean value theorem on f and g separately and they have divided the result. Of course, I agree with you on that But let me tell you few interesting things here These these concepts are basically used when your functions are represented in a parametric form Okay, so this makes sense when your function is represented in a parametric form So I would like you to give a like you to understand a geometrical significance of this Let's say I define a function y as f of t Okay, and I define a function. Sorry. I define a function parametrically as f of t and g of t. Sorry This is a parametric definition of a function. So if you look at Two parameters a and b, let's say there's a parameter a at this point. That means this point is G a f a Okay, that means your parameter t here is a okay, and let's say this point be the parameter is b That means this point is G b f b Okay, so what did mr. Kushi say he said that? There would definitely be a point with a parameter C Where the slope of the tangent will become equal to the Slope of the secant line Now what is the derivative of this function at C? You have all done differentiation, right? So what is the dy by dx at t equal to C? You'll say sir simple. It is f dash C by g dash C. Isn't it? This is what you have learned Okay, so this slope would become the slope of the secant line, which is fb minus f a by gb minus g Okay, so these two will become equal in other words This will become equal to f dash C By g dash C This is a sophisticated way to understand it. Otherwise many people they do a calm Chilau way You understand the meaning of calm Chilau You got You got away, huh? This is are you you have just divided to LMVD and you've got this Yes, you can treat it like that if you want. Okay, a couple of things we need to understand here If your g of x becomes x Then your CMVT will convert to LMVT Got it You can clearly see that happening over here if g of x was x gb minus ga would have become b minus a simply Correct g dash C would have become one because derivative of x would have become one and then it would have shown you LMVT Actually f dash C is equal to fb minus f a by gb minus g. Okay In other words if this point would have become a comma f a and this point would have become b comma f b CMVT would have converted to LMVT correct if your function had satisfied That f of a is equal to f of b somehow Okay, then whole of this numerator will vanish Correct and your CMVT would have become Getting up on third thing your Lopital's rule Officially came from CMVT Lopital rule is actually derived officially derived from CMVT Right, so if you have let's say be very close to a Right, and you want to find out the limit as x tends to a Then since b and a are very close to each other, that's why fb and fa are 0 and gb and ga are 0 Then your limit can also be evaluated by differentiating the function and putting that value a remember When a is very close to b Okay, right, then extending to a will be basically Implying that C is also tending to a isn't it? right, so this whole thing will convert to the Lopital's rule so if somebody asks you to derive Lopital rule it comes from kuchi's mean value theorem Are you getting my point? Now what is the mathematical proof for this? This is all about you know, okay? I explained you the concept. I gave you a graphical way of looking at it Second form of this theorem, which is the second from the theta and also For this one. Yeah for this one. We don't write theta. I know the previous one only I meant sir We'll come back to it. Okay. Sure. Sure now How do we how do we prove? Kuchi's mean value theorem again the proof is coming from rose theorem itself No, sir. I'm itself. Now. See what happens. How everybody please pay attention I'll do it in the next page. In fact, I Hope you remember these conditions right because I'll be asking you proof for CMVT See if you know your function f is continuous if you know your function g is also continuous, correct? Can I say? This will be also be a continuous function. Let me let me first name this function. Let me name it as phi of x So let's say I Formulated a function phi of x which is like this since in the Kuchis mean value theorem we have this conditions mentioned that f and g are continuous. It implies Phi of x is also continuous Okay Yes or no, it means phi of x will also be differentiable because your f and g are differentiable okay, and I'm sure you would have understood why I left a lambda there because I wanted to choose such a lambda For which Phi a is equal to Phi b, okay? So if I choose Phi a is equal to Phi b That means you're trying to say f a plus lambda g a is Fb plus lambda gb In other words your lambda will come out to be correct me if I'm wrong F a minus Fb By gb minus g a correct Okay now This function will be reshaped so I'll reshape this function as f of x plus Fb minus F a by Or by mistake. I switched the position of the tops. Okay, let me switch the position of the denominator doesn't make it ga minus gb Okay So now this function must satisfy Roast Theorem It has to satisfy because Jabardasthi we had made it satisfy So this condition Jabardasthi we made it you know forcefully we made it satisfy that condition so that I get a lambda value So it has to satisfy it can't skip So if this satisfies Roast Theorem that means there would be a point C Where the derivative of Phi x will vanish Correct. No So F dash C must sorry Phi dash C must become a zero so Phi dash C Will become actually F dash C correct me if I'm wrong and this term will be a constant. So I'm just I'm just copying it and this will become G dash C. Okay, and this guy is a zero as per this Okay, so from here I can say Your F dash C by G dash C Just check it out. It will become Fb minus F a by gb minus G This is what your Kuchy mean value theorem says Sir, could you please show the graph one minute? I think I made a mistake Which graph? Let's do a graph Teumograph, okay, so This Teumograph Yes, sir. Thank you What some new name you have given to the graph Teumograph Unfortunately, none of the J books talk about CMVT, but while asking questions, they will ask Yeah, basically it's an extension that's why many books they don't mention it so that they want students to apply it But trust me, they will teach you all this thing in your first year of engineering. First year of engineering, we had all these concepts So I remember my teacher having derived all these things in first year Answer the second form Second form one thing. I think it was in one board number two Yeah, this was a second. Just a minute, sir Sure Well, just a minute the last Okay Done, sir. Okay Let's take questions Okay, let's take this question Read this question carefully Okay, I hope all of you have read this question So the question says there's a function defined from the domain 0 to 4 to real numbers is differentiable Then for some value a and b lying between 0 to 4 Actually, they should have given a comma here Books they don't treat comma very seriously F of 4 square minus f of 0 square may be equal to which of the following now again There can be multiple debates regarding this question But you have to give an answer which is possible from the theorems you have studied Because whenever I give such question people say, sir, how do you know that? Others are not true. There could be some a and b in this interval which will satisfy this also this also this also this also But is there any valid support point for that? If yes, then you provide me and accept your answer. Are you getting my point? See many people when I give such question, they say, sir, let's say I give I say option number C is correct Let's say let's say I don't know whether C is correct or not, but it's a how do you know a is not correct? So my question to them is how do you know a is correct? Right, so I can only mark what I know is correct about others. I cannot comment. Yes, it is only one option aditya Okay, this this requires my intervention. I don't think so many of you would get it, but still I'll put the poll Let's see what response I get Then I'll discuss it. These type of questions are very tricky. Let me tell you I I've seen very good students also not getting these kind of questions. Correct. It is a very very Lallu Punju topic in school, but it is equally tough and challenging when it comes to school level. Sorry when it comes to competitive level Okay, good Let's wrap this up in another 30 seconds so that we can discuss it as soon as possible. Okay, five four three one Most of you have responded with option number b Okay Now this question was basically Framed by using two rules in mind. One was intermediate value theorem And other was The lang ranges mean value theorem Right, so what's what is intermediate value theorem? See by intermediate value theorem basically this theorem says that there would be an intermediate point a Lying in the interval zero to four Where the value would be the Average of the values of the corner points Okay, so intermediate value theorem says Now, how did I figure this out that intermediate value is going to be used if you see this term? It is actually f of Four square and f of zero square Which is actually factorizable as f of four plus f of zero Into f of four minus f of zero So from here, I got an idea that I could use intermediate value theorem So intermediate value theorem says that if there is a function Okay, which is continuous. How do I know it's the continuous because they have already said it is differentiable So continuous it has to be continuous, right? So if there's a continuous function then There would be some point between the interval zero to four Where the value of the function would be average of the value that it has at the at the corner points Okay, this theorem works for any continuous and differentiable function. Is that fine? Yes Now people may say sir, there may be a point which is equal to f zero plus f four But does it work for all continuous function? Will that work for a constant? No, right So if you try to justify it, sir, why only divided by two it could be just f zero plus f four also But does it work for all continuous function? No, right, but this can work for any continuous function no matter what continuous function you take Now by Langrange's mean value theorem by lm vt by lm vt we can say f of four minus f of zero by four minus zero Would be equal to f dash b where b is some point lying in the interval zero to four Now many times people ask me this question sir Can lm vt and intermediate value theorem occur at the same point? It may But it is not necessary Are you getting my point? lm vt Points b and intermediate value theorem point a can be the same point also, but it is not necessary Are you getting my point? So it is it would be a very very high coincidence that lm vt occurs at the same point where intermediate value theorem is occurring Okay, so from here I can say f of four minus f of zero is four times f dash b and from here I can say f of four plus f of zero Is two times f of a Okay, so when you multiply these two, let's say I multiply one and two I get f of four square f of zero square Is equal to eight f of a f dash b So with a lot of surety I can only comment on a being right Weird, yes, of course. That's why these these problems become very challenging in when they come in the exam Is that fine? So option number a was correct. I think a got the second lowest vote 23 percent Let's look into more questions Let us go into more weirdness of the concept Let me find an easier one. Actually, I picked up a difficult one. Okay. Let me take let me start with RT first Then we'll go on to LMVT and CMVT. Let me put the poll on Think carefully and then answer So reading this question, everybody think will think that oh, it's a quadratic equation question But actually it is a rule zero question. Okay, one person has responded Uh, another 45 seconds for the same Last 10 seconds Very few a few have responded. Uh, I've got just 10 votes five four three two one So I'll stop the poll now Almost equal number of votes cast for a b and c big confusion, huh? Okay, let's discuss See the moment I read this word at least one root in the a and b Now what all images come in my mind? First of all, I start treating this as F dash x it must be derivative of some function It must be derivative of a continuous and differentiable function whose roots are zero and one right Look, please try to recall the second that you know, you can say interpretation of rows zero. What did I say? I said that if there was a function Where this function is continuous and differentiable Okay, and it has got roots of let's say alpha and beta then The derivative of the function equated to zero will have at least one root between alpha and beta At least one root between alpha and beta So now what is given to us is this fact Correct. So there is some f dash x equal to zero which has at least one root in zero to one So my alpha is zero beta is one. So now what I do is start going backwards Okay, that means my function would have been My function would have been the integral of this Which is actually x cube Plus two ax square plus bx plus some c. Yes or no Yes or no Roots doesn't mean you have to make it zero. Basically This is just trying to justify that from here. I just want to justify that f of alpha and f of beta are same Need not to be equal to zero. Please do not impose extra restrictions Okay, so this function should satisfy f zero is equal to f one Correct Do you all agree with me on this or not? So f zero is c f one will be one plus two a plus b plus c Correct Which means which means Two a plus b plus one should be zero That is option number b should be correct. This is how you are supposed to know rose theorem for competitive exams I understand in school the type of questions are pretty different in school There are two types of questions that are normally asked One is prove whether rose theorem can be applied to a function That means you have to check on its continuity and differentiability in the interval specified And secondly, they'll ask you verify rose theorem for a function in an interval Correct So these are the two types of questions that will come but in competitive exam They have a different way of testing you does it make sense because I understand this is tricky This is tricky Sir, sir, basically you said if uh f of x has two roots alpha and beta Let's not say roots also. Let's say it has two values alpha and beta will satisfy this Oh, okay being root is a extra restriction, which I don't want to unnecessarily put here It's like yes, sir But it can also be incorporated under the same thing Special case yes, those are special cases of rose theorem Oh, okay, sir Well, you can think that there is a continuous and differentiable function and there exist two values alpha beta such that such that this condition is met Such that this condition is met Think only that equal to zero is an extra restriction that you may or may not want to impose It's your call that depends upon the question also Make sense Okay, let's take more questions See, I would like to take several questions on this concept because It's tricky in the first instance Okay, let's take this question alpha beta are any two roots of this equation e to the power x cos x equal to one Then the equation e to the power x sin x minus one equal to zero has Now so many times you have seen rose theorem that at least one Clause is always there in your mind. So most of you will mark c in this case. I'm very sure Let me put the poll on see c may be correct. I don't think like that Okay, sir has pointed out that many of you will mark c so I should not mark c My job is to scare you What sir you are scaring us there Good Okay, let's close this up in another 45 seconds five four three one Go Half the class has not voted. Please vote. Please vote. Please vote Don't worry about right wrong Let's say time is up and you have to mark something and you have to submit the omar sheet Oh, sorry, you know it's not cbt paper Yeah It's fine. You have to take a guess you are in desperate need of a one mark or sorry a four mark. Okay All right, so let me stop so much motivation also did not motivate 11 of you See this. Thank god. I'm not on kbc seat Mix this phones. Okay, let's check See I have to somehow link this guy to this guy as its derivative. Okay. See what I'll do Can I write e to the power x cos x? equal to one as cos x is equal to e to the power minus x right so can I say e to the power minus x minus cos x equal to zero this function has Roots alpha and beta Which means Let's say I call this as f of x Which means f of alpha and f of beta are equal. Yeah, it's a different thing that they're equal to zero also So that's fine. I mean, I'm happy to have that information also But what I needed was not this guy. What was what I needed was I needed this day Okay Secondly, you know that it is made up of two continuous functions. So it has to be continuous Correct, we have all learned in our continuity differentiability chapter like that exponential functions and trick functions are continuous in the Entire real number line. So there's some indifference will also be continuous Same goes with the differentiability also, right? It's continuous and differentiable everywhere. So the sum and difference will also be continuous So now by rose theorem, can I say f dash x equal to zero has at least has at least One root between alpha and beta Correct now see what will happen when you differentiate it. This becomes minus e to the power minus x This will become plus sine x equal to zero So this must have at least one root between alpha and beta If you see this very very carefully You have basically written e to the power minus x is equal to sine x that means you have written e to the power x sine x is equal to one That means this equation must have at least one root between alpha and beta And that's what the question center wanted Okay minus one you bring this minus one to the other side. What a big issue So yes, the answer is option c So you're working your way backwards actually Right, so you have to work your way backwards So c was the correct option Unfortunately, it lost to b Okay, next question. Oh, sorry sheshan. Uh, do you want me to go back sheshan probably after this question After this question. Okay. Thanks Yeah No response zero person has voted so far Good good good Okay, let's wrap this up in another 30 seconds Only six of you have voted so far five four three one Please vote 50 percent of the class has voted only Please vote. Please vote. Please vote fast fast fast press something I mean press on the poll button something. Yeah, okay, so Option b is what most of you are Saying option b. Okay, let's check 50 percent of you who have voted say option b Now let us discuss this it is known that This is a polynomial equation. I don't think so anybody would deny the fact that it's a polynomial equation, right? It has two distinct roots in zero comma one. So let's say Let's say the two distinct roots be Let the roots be alpha and beta Okay, let's say alpha is smaller of beta Okay, assume that alpha is smaller of beta And both these roots lie between zero to one that means This kind of relation is definitely existing between alpha and beta. Okay So now f dash Sorry f of x equal to zero has roots alpha and beta And it is known to be a polynomial and hence continuous and differentiable and also f of alpha and f of beta equal so I can apply Rolls theorem on it Right, so can I say the derivative of this function? The derivative of this function equated to zero must have at least One root between alpha and beta But if you differentiate it But if you differentiate it, you would realize you get three x square minus three and when you equate it to zero When you equate it to zero you get two roots, which is plus one and a minus one Okay, let's say You're looking at minus one and minus two and minus three and minus three and minus three Minus one doesn't fall in this interval because minus one should lie between alpha and beta. So it should lie between zero and one also, right? correct So this cannot be a Answer it can't be even one because one also doesn't lie between alpha and beta because alpha beta themselves are between zero and one So how can one lie between alpha and beta? So neither one Nor minus one can lie between alpha and beta that means That means There cannot be any value of k for which this equation has got two distinct roots between zero and one So your answer would be null set D is the right option Are you getting my point? Let's see how many of you gave d as the right option. Oh 15% only set only four of you got d Most of you went for b Sir, could you please repeat? see First of all, you have admitted that f of x is a continuous indifferentiable function because it's a polynomial Yes, correct And since alpha beta are roots between since alpha beta are roots f of alpha is equal to f of beta, correct That means rt should be applicable on rt can be applied to f of x That means the derivative of the function must be zero for some point between alpha and beta So when I put it zero and I realize the point that I'm getting is not between alpha and beta neither minus one is between alpha and beta Because alpha beta it still is between itself between zero and one So how can one lie between alpha and beta? Yes, sir Minus one lie between alpha and beta of course minus one can definitely not lie because alpha beta both are positive Okay So none of these point is satisfying rose theorem, but rose theorem must be satisfied. It's the theorem which has to be true Which means that the initial fact itself that it has a root between zero and one that is false That means you cannot have any real value of k for this equation has got two distinct roots between zero and one So k can be k has to be an ulcer Oh, okay, sir So you assumed x cube minus 3x to be the integrative Integral like the f of x and then you did the thing for the derivative No, this is the original function. Then I worked with the derivative. Yes, sir. Not like the previous question that we had taken Oh, yeah. Yes, sir Got it. Yes, sir. Okay next question This should be easy This should be a easy question very easy question They should have written the wrong bracket here Okay, last 30 seconds Please vote Okay, five four three one, please vote Let me stop the poll. Most of you have gone with option b 75 percent of you have voted for b. Okay, let's check. See this is a clear-cut case of kuchi mean value theorem, isn't it? So f dash c by g dash c If you know that your f of x is a continuous indifferentiable function Then this should be equal to f b minus f a by g b minus g a Okay, this ratio is given to us as a two So two is equal to f one minus f zero by g one minus g zero plain and simple Okay, now in this case, there is a small assumption that we need to make That the derivative of g doesn't vanish I'll tell you an alternate way to think of this So f one is given to us as six f zero is given to us as two Uh, g one. I don't know g zero is given to us as zero. So that gives you G one value as four by two that is actually two so option number b is correct Alternately what we can do is Alternately what we can do is we could have formulated a function phi x Which is made up of f of x minus two g of x Okay now now If you see phi one Phi one will be how much Phi one will be f one minus two g one Since g one is not known. Let it be as as it is Phi zero will be how much Phi zero will be how much f zero Let me write six for this Yeah f zero is given to us as two G zero is given to us as zero. Okay now this function Phi x Is continuous and differentiable And not only that there exists a point c such that phi dash c is equal to zero That means f dash c minus two g dash c is equal to zero. That's how this condition is actually satisfied, isn't it? So what are you doing? You're doing basically You're taking the first condition. You're taking the second condition and you're taking the The consequent to be true So the third condition must be true That implies f phi one should be equal to phi zero then only rose theorem will get completed See what i'm trying to say here Phi is a continuous f of x is a continuous function g of x is a continuous function. So phi will be continuous function f of x is a differentiable function g of x is a differentiable function. So phi will also be differentiable function And because this is true Because of the given condition in the equation This is true. It means phi dash c is equal to zero For a point c between One and zero isn't it? That means Rose theorem to be completely work Phi of one and phi of zero should give you the same values That means these two should be same That means two should be equal to six minus two g one That means two g one is equal to four So g one is equal to two. That's another way of looking at it Say you're coming from very first you said that if you can't apply the converse of rose theorem if we know that rose theorem is true No See converse of rose theorem will be supported here because you know your function is continuous and differentiable Oh, so that's only for the continuous and differentiable parts. Yes. Yes. Yes. Yes. Okay So what if in the first case first method of solving you assume that c was the point which we are talking about, right? C is any point in general between between a and b Oh, right. There will exist some point. Which point we never know. We don't know that point, but there will be some point Oh, right. Right. Okay, sir Oh, excuse me, sir Yes, Sushami Sir, why did you change the brackets to round brackets for that zero and one in the question? I mean it's been given that it is differentiable in the closed Yeah, sorry. Sorry. I did not see that Then this will work. Oh Now there's no point. It's fine. Even if square bracket is there. That's fine Sometimes what happened they they tend to do some mistakes. So yes, you are absolutely correct It's given that it's differentiable. So you can apply there Next question Okay, let's take this as a proof that question Actually here it should be round brackets prove that For lambda greater than one this equation Has at least one solution in one to lambda So if you can prove this by thinking of a function Whose roots are one and lambda And such that the derivative of that function tries to generate such kind of an expression Then you are through and that is not difficult to guess. Let me tell you Guessing that is not difficult. This is a prove that question Okay, so let me take this up because it might One minute, sir One minute, sir, please. Yeah, yeah, okay Yes. Yes, Aditya See, let us say there is a solution c lying between One to lambda which satisfies this equation Okay, so let's see lying between one to lambda satisfy x log x Plus x equal to lambda that means C l and c plus c is equal to lambda Correct. In other words, can I write this like this? l and c Plus c minus lambda equal to zero divided by c. So you have Now, why did I divide by c is because? Zero had to bring because I want to compare this with some f dash c Equal to zero. So I wanted to compare it to this. So that's why I bought a zero Why I divided by c because in order to know your f of x you need to integrate this Correct. So integration will become easier for me if I divide by c else I have to apply integration by parts Okay, so if you're claiming this to be f dash c Oh, sorry plus That means your f dash x is actually ln x Plus x minus lambda by x Correct. You can write it as ln x plus 1 minus lambda by x. This is your f dash x So if your f dash x is this If your f dash x is this means your f of x is Just integrate x ln x Uh minus x if I'm not mistaken and this will be minus lambda ln x Plus some constant of integration Now from a given scenario If I'm able to show that If I'm able to show that This is a continuous function This is a continuous function which anyways it will be because it is made up of all continuous function And it's a differentiable function And f of 1 and f of lambda are equal So let us choose k in such a way that f of 1 and f of lambda are equal So what is f of 1 here? What is f of 1 here? Can I say this will become zero this will become minus this will become zero and this will become k Okay What will be f of lambda? Now f of lambda will be lambda ln lambda minus lambda minus lambda plus k Sorry Minus lambda plus k plus k Minus lambda ln lambda plus k But yes, sir. But that will get cancelled off, right? Yeah. Yes, sir Yes, I know. Yeah. Okay So k here is getting cancelled. So you can take any k you want actually No problem with that Right But we just have to check That the k that we have One second When you take lambda, what do you get minus lambda ln lambda minus lambda? And this will get cancelled plus a k you get right But then we're getting lambda and one is the same huh You're getting if you substitute one in f of x and lambda and f of x we are somehow proving that one equal to lambda So but that need not be that is not true actually. Yeah, or it's like telling one and lambda at the same points Then we can't have any question of the interval of one and lambda, right? Okay. Let's let's do this thing. Let us say Let us say That oh my mistake. Uh, there's a small mistake. There was x over here also This derivative is this. Sorry. This integration is this Then I directly integrated this. I forgot this one. I'm sorry. That's why a mistake is arising You got you got the mistake The mistake is When you're trying to get f of x You are integrating this term. So ln x integration is x ln x minus x Correct Who will integrate one that we forgot now it's proper. So now this x and this x will go off So now you'll get okay. Let me put a k Now you get x minus lambda ln x Plus k equal to zero Now you want this equation to have a root f one root x equal to one So if you put x equal to one Anyways, this term is going to become a zero. So we don't need a k actually So we don't need a k. So k will be zero again So you can directly assume your function to be x minus lambda ln x Now you can see This will have roots One And lambda check it out Isn't it f one will also give you zero F lambda will also give you zero correct That means f of one and f of lambda are equal satisfying your rose theorem condition Correct and not only that it is continuous and differentiable So if you go backwards now That means if you differentiate it Then the derivative of it will have at least one root between open interval one to lambda Which is what the question is asking us to prove Right So the end result Is given to us and we are trying to figure out whether all the conditions of rose theorem are being met by some function whose derivative is the given expression So I have been able to successfully find a function Of course, I did a lot of reverse engineering on this Which was not only continuous Which was not only differentiable But which also satisfied f of one is equal to f of lambda And hence the derivative of this function, which was actually this step Okay Will be satisfied by a value c between one and alpha So this will be satisfied by a value c Between one and alpha Okay, so these were the type of questions that you know, we also practice when the j was Uh not objective type So I wrote a subjective j So we had to practice these kind of questions now people will ask that Hasn't it become irrelevant in the present scenario because we get objective questions No, they can direct they can make a objective question out of it Right, they can give you this has at least one root In one of the options it would be at least one root between one and lambda So sir, basically in this question the point was We somehow proved we found a function in which f of one equal to f of lambda Yeah, we found a legal function Yes, we found a legal function which was differentiable and f of one equal to f of lambda So its derivative equal to zero will have at least one root between open interval one to lambda Oh, okay, sir Right, so you you're working your way backwards to arrive at a function on which rt is applicable Yes, sir At the point Yeah, yes, sir One more Let's take Some katana problems After this chapter many people say sir, I never expected this chapter to be So difficult in j because it's so easy in school So this is slightly different. I mean it is not like what you would be expecting it to be It is not fb minus f a by gb minus ga Is equal to f dash c by g dash c that probably will work Uh, uh, probably will work in cmbt Here instead of a b you have a c here. Okay, I'll put the poll on Okay, uh Can we discuss it now? Let's close the poll in another 10 seconds It's fine. I understand if you're not able to get it Please please port something so that we have some data points in front of us If you had to take a smart guess What would it be? What would it be five? four three two One go Most of you have gone for c Okay, I think it was a desperate Move that we had to take Something had to be voted. Okay Might not be correct. So let's check See here, uh, many of us How they solve this question I'll tell you Okay, they'll solve it like this By kuchi's mean value theorem By kuchi's mean value theorem Is f dash c by g dash t, right? Okay for c belonging to the interval a to b open. Okay So by cmbt This is true. Okay Now I'm telling you the mistakes which people do Okay And they say sir, uh, can we say fc is half of F a plus fb By ibt Right So this b you can replace it with 2fc Sorry, sorry my mistake This fc here you can replace it with half of f of a f of b Minus f of a And here gb Similarly for gc also you can replace it with Half of sorry Half of g of b minus g of a sorry g of b plus g of a Okay So when you do that automatically your numerator will become f of b minus f of a and denominator will become g of b minus g of a Correct So this term here that you have in your question Will miraculously get converted to this term Okay, and hence you can write your answer as f dash c by g dash c g dash c because of lmbt Right and your answer is b And that is correct also right, but What is the what what could have gone wrong over here? Luckily this answer is correct But Right Yes, your cmbt and lmbt did not occur at the same point So you're trying to say c is a point which is also satisfying cmbt And c is also a point which is satisfying ibt Which need not be correct But luckily the answer is correct actually Okay, so this is what people do and they get the answer also for this question I'm not saying they will be as lucky In other questions. Okay So this is not a right approach Not a right approach Because you're assuming that at x equal to c Because you're assuming that at x equal to c the function is satisfying Is satisfying lmbt as well as sorry satisfying cmbt and ibt both Okay, see there may be equal i'm not denying that But the probability for that to happen is very rare Yes, sir Now what is the right way to solve this? So this is not a right way. What is the right way to solve it? Okay, so what I'll do is now this is something which you know comes with a little bit of experience also Let me assume a function Let me assume a function In such a way which involves f and g Right Which involves let me write in a very crude manner which involves Which involves f of x and g of x Okay Right So I'm using some kind of a combination of f of x and g of x it may be addition It may be linear combination. It can be multiplied also. I'm not denying that it cannot be multiplied Okay um This function is such that When you put f dash c equal to zero Okay, it generates a term like fc minus f a Okay It generates a term. I'm not saying this is the only term it'll generate some term like this Okay, and h a and h b are equal So think of a function which involves f of x g of x Right so is that when you when you put its derivative as zero This term is generated because of that other terms may be generated. I'm not denying it and g h of a is equal to h of b because If you make a function involving f and g it has to be a continuous function because f and g will be continuous So it has to be a continuous function It has to be a differentiable function Okay And it is meeting the third condition of roll's theorem And its derivative should generate such a term Are you getting my point? Are you getting my point? So what could be this function now getting this is not an easy task It may come, you know with experience a bit, but it is not very easy for everybody to do it Okay Yeah, thank you tell me So how exactly do you ensure that When h dash c equal to zero you're getting such a point like f of c minus f of a and all No, I'm not getting a point h dash c expression Yeah expression will lead to this will lead to this kind of a scenario See for example, I see fc and gc and ultimately there should be x there, right? So it is generating something like like this f of x minus f a By g of b minus g x so such kind of a thing should be generated in its derivative Now, of course when you're putting it to zero you bring something to the left Yes, so let me just do some bit of reverse engineering Can I can I do something like this? f a Okay And let me write Let me write let me write G dash x over here Okay And this term I will write it like this Plus g x Minus g b into f dash x Okay, let me just take a Random guess. I'm just working my way backward. Okay. I'm working my way backward So if such a derivative is there that means your f of x should have been f of x minus f of a times times g of No g of g of x Minus g of b Will that work? Will that work? Let me check So derivative will will produce this Oh, my my my mistake. It should be gb minus gb minus gx Sorry Okay, this will be gb minus gx now see Forget about this reverse engineering thing. Let's say if you choose this as your function Right, what will be h dash x if you apply product rule It will be f of x minus f of a Into derivative here, which will be minus g dash x plus plus f dash x into g of b minus g of x Correct And at c it should become a zero correct that means This will generate fc minus f a minus g dash c plus f dash c Into gb minus gc Okay, and as you can see the term which I desired That is fc minus f a By gb minus gc I've been able to get as f dash c by g dash c So this term appeared right And if this term appeared the right side will give you this And this will become your answer there But to get to this function. It is not easy Right, you have to think backwards And sometimes you need to be aware of the answers also Sometimes you need to be aware of your options also what options you have Right, this is the right way to solve it, but doesn't come very easily to us Yes So if you want to copy something then don't copy from here just Copy from here. Let me erase the unnecessary things Okay, copy from this part Let So how did you come up with the skeletal structure of the function? Hey, I didn't understand that part Which one? uh starting from f of x minus f of a into g of b minus g of x. What was the reasoning behind that sir? Come again. Come again. Sorry f of Uh, h of x let know the one below that. Yes, sir that one. Yeah, this one. No above that This one. Yeah, this is the main point How did you get this itself is a very big question? I actually worked my way backwards. I wanted this expression to come up somehow So what I did was what I did was So there is there is some function There is some function When you put its derivative as zero This term should make its appearance that you should be able to get it on the other side Uh, there is some function There is some function Yes, sir. This derivative at c if you put to zero Then this term appears Isn't that the same as telling uh, you're indirectly telling that intermediate value theorem and kuchi is being satisfied in the same place, right? No, this this doesn't involve any intermediate value theorem anyway Intermediate value theorem says that there should be a point between the uh end point Yeah, where the the the value of the function is the average of the value of the function value at those end points That is the intermediate value theorem Okay, sir Okay Okay Okay So this is something which is difficult. That's why I took this as a last problem so that you can understand the complexity of the situation Oh, yes. Yes, sir Fine Yeah, yes, sir. Do you want more problems? Okay, we'll take more problems But I think we have spent around two hours doing lmbt So, okay, sir worth it Okay, we'll do this one. Actually, this is not a lmbt question. Actually, this would be oh one second. Sorry. This is not a required question That is the increasing decreasing nature function. Okay, let's do this question poll is on Yeah, any response anybody I'll say two minutes So why are ffx and gfx given in the first place? So that you can apply lmbt or rose theorem whatever you want to apply Okay Oh one minute, sir Sure Okay, let's let's uh conclude this in another 15 seconds all right five four three two one go okay Most of you have gone with option c Let's discuss see first of all Let's have an eye on these expressions this result is made up of four to the power something and five to the power something and four and five are also in the a and b values right That means it has to do with something like You know some there is some function Okay, there is some function. I don't know what is that function? Okay, and you have done something like This which has resulted into this kind of an expression Okay Because let's say if even if I do this this will generate a one There the derivative of this function Is having a c they should have a c in the denominator Sir too much thing you are asking you know So first of all There should be a c coming in the denominator of it. That's why this c will go up and this term Should give rise to this term something like this you need to think of Okay So how would you think of such term? Of course you have to make a function out of these two functions given to you correct right So if you have to use these two given functions because log is there Something related to log is there Okay, so how can you make a function like this think about it 25 is coming on the four That means there should be something like x and there's a four on the 25 like four squared five five square four Yeah, there's something called x square ln four correct and there is 16 coming on the power of five so 16 coming on the power of five So so so so so so you should have something of the nature minus 16 ln x Okay, so I've involved x square. I've involved ln x and I've got constants along with them So constants will not change its continuity and differentiability So now if you see this term carefully, let's say I call this as my phi of x Okay, so what is Phi of five Phi of five will be 25 ln four minus 16 ln five correct Correct. No any questions in that No, sir Okay And what is phi of four? What is phi of four? That would be zero I guess Yeah correct So thankfully I can just write ln four to the power 25 by five to the power 16 minus zero as derivative of this which is two C ln four minus 16 by c so you can write it like this right That means all my requirements are being met by this function now. There may be some Hitting trial also it may not come in first go. Let me tell you the way i'm solving it It may not happen the same with you in first go Okay So your answer would be the one which shows two c square ln four minus 16 So option number b is the right option Option number b is the right option So how did you get the c in the down step? I differentiated it and I took a c as the lcm If you differentiate it, what do you get? You get two x ln four minus 16 by x no Yeah, and you have to put x value as c also, right? Oh, right Two c ln four minus 16 by c Oh, yeah, yeah, correct So you took a c lcm you got two c square ln four minus 16 Okay, so option number b is correct option number b is correct I think very few of you voted for b not 26 percent of you voted for me okay, so We'll stop this topic now. We'll take a break