 one second sir one second good morning thank you so much thank you let us begin methods of differentiation all set for tomorrow's exam? kind of sir some five chapters I love just five chapters okay yeah five six I guess you don't have permutation and combinations no oh maths got over sir oh maths is over yeah oh I thought it's tomorrow how was my speech okay sir some chapters I did pretty well in some chapters like partial fractions sets around I didn't do well can you send me the question people okay sir I'll send fine okay so methods of differentiation we'll talk about it so we only know how to find the derivative of all the standard functions and I'm sure you know also the derivative of finding the finding the derivative of inverse trig functions are you aware of that sine inverse cos inverse inverse no sir no okay so I'll just give you the list first before I begin with that okay sir we'll start with derivatives of of standard functions this is your f of x this is your f dash x again f of x f dash x f of x f dash x okay so we'll start with the constant function we know its derivative is zero x to the power n it's n x to the power n minus one this is called the power rule of differentiation okay exponential function e to the power x is e to the power x e to the power x is e to the power x log 8 sin x is cos x cos x is minus sin x tan x derivative is secant square x secant derivative is secant tan x cos secant derivative is minus cos secant cot x cot x derivative is minus cos secant square x no this is the list probably we which will be not known to you sine inverse x derivative is one by under root one minus x square yes sir how is it negative one by under root one minus x square tan inverse is one by one plus x square okay yes sir sec inverse x is one by x under root x square minus one cos secant inverse x cos secant inverse x is minus one by x under root x square minus one and one more last thing is your let me write it here itself cot inverse x is one minus one by one plus x square okay now Venkat I would like you to derive few of them by the user first principles okay okay so just once again I'll load down these yeah yeah yeah please hi Dheeraj good morning Vidyota good morning by the way which exam is there tomorrow bio and computers bio and computer today you can speak Dheeraj there are very few students so you can speak you can unmute yourself how was the math super Dheeraj yeah yeah I can hear you math was okay okay two sums I didn't get okay this was proper yeah this was fine okay Vidyota it is okay sir okay okay so I just started with giving the formula for the derivatives of some standard functions so a special mention to the inverse trick function derivative so please make a note of it here is the complete list on the screen left hand side of the red line is your function the right hand side is your derivative so I would request you all to give me the derivative of the following function by first principle one is your cosec x okay other is your a to the power x and other is your tan inverse x let's see whether you are able to solve the derivatives or find the derivatives of these three functions by first principles again in the class I told you there are various manifestation of the first principle you can use the one which is most convenient and simpler to look yes sir can I switch to the next board yes sir find the derivative of find the derivative of the following by first by the first principles number one a to the power x okay so this is your function number two cosec x and number three for a change inverse trick function I will help you out with the third one but for the first two I'm sure you should be able to get it just remember recall the definition of the derivative by first principle which is dash x is defined as limit h tending to zero f of x plus h minus f of x by h you may want you can use f of x minus h minus f of x by minus h also doesn't matter so I'll give you two minutes for the first one let's work under the time constraints yes sir yes sir done just a doubt sir yeah limit if you do this operation a to the power h minus one by h limit h tending to zero it's l n a right yeah yeah yeah correct yes sir according to the first okay according to the first principles it'll be a to the power x plus h minus a to the power x by h if you take a to the power x common you will get a to the power h minus one by h correct yes sir try to recall that this limit over here is l n of a so your answer becomes a to the power x l n a done with the first one second one right out sir is there a formula for cosec x plus h no you have to convert it in terms of sine sine right yeah yes like this you have to do from here you can proceed oh yes sir any fixes uh one second sir trying stuff um please note this term is a harmless term why harmless because as you put h is zero it just becomes sine square x so you can pull it out as sine square x uh one second sir almost done yes sir done done this thing is negative like minus derivative of sine x using first principles absolutely absolutely if you can recognize that that's well and good as you rightly said this is the derivative of negative sine x which is actually negative cos x correct but yes uh nevertheless i will do it by you know expansion by using my transformation formula so this is the limit extending to zero uh two cos uh x plus x plus h by two into sine of x minus x minus h by two okay uh let me bring this to as a denominator of this h okay now this is a harmless term because when you put h as zero it just becomes cos of x just put it out so my uh cos of x by sine square x and this becomes limit extending to zero a sine minus h by two divided by h by two okay just put a negative sign here and this term will actually become a one the answer would be negative cos x by sine square x and if you simplify this further it will become negative cortex into cosec x or negative cosec x into cortex so what did you do in this step yeah sir could you explain like the second last step onwards this step this okay i use the transformation formula sine a minus sine b is two cos a plus b minus b that's what i did correct okay yeah for the last one i'm going to help you out because it involves inverse trig functions and you're not been introduced to it before so let me move on to the next page see uh inverse trig function so when you're finding the derivative of inverse trig function of course by use of first principles you would write this formula tan inverse of x plus h minus tan inverse of x by h limit h tending to zero but my problem is i don't know what is tan inverse x plus h minus tan inverse x for that matter i don't know what is tan inverse a minus tan inverse b correct yes or no okay so nothing to worry let's say this is your angle theta and this is your angle phi correct see at the end of the day tan inverses are some angles isn't it yeah right any inverse trig function is an angle at the end of the day so let's say that angle is theta and this angle is phi okay now let's begin with the formula of tan theta minus phi what is that tan theta minus tan phi by tan tan phi now if tan inverse is theta can i say a is tan theta yes sir and can i say b is tan phi yes sir can i put it over here and write it as a minus b by one plus a b that is equal to tan of theta minus phi okay and if you take the tan inverse on both the sides so theta minus phi is going to be tan inverse of a minus b by one plus a b correct now treat this itself as your theta and treat this itself as your phi do right there but i like it here this is as phi okay so according to this formula that we derived can i say i can write tan inverse of let me write the whole expression tan inverse of x plus h minus tan inverse of x by h as tan inverse of a minus b by one plus a b am i right whole divided by h now this is limit h tending to zero tan inverse of h by one plus xx plus h whole divided by h now this is a function which this is a function of h which tends to a zero as h tends to zero so try yeah try to recall this formula where i had told you that limit of tan inverse of x by x is extending to zero is one and this will be holding true even for a broader interpretation that means you needn't have just x over here you can have any function which tends to zero as x tends to whatever correct and here this is my function so what i'll do is i have to create the same function in the numerator and denominator both so within this i have to create down also one plus xx plus h but i can only do when i do this activity i cannot just multiply and divide i cannot just divide with one number okay i have to multiply also does this make sense diraj vidyota does it make sense okay now yeah this fits into our definition right this definition and this is one so the answer is one by one plus xx plus h and when you put h as zero this becomes one by one plus x squared just just look at it for a few minutes and let me know if you want explanation any of this particular solution yes sir can you screw it you want me to go off okay a little bit yes sir one second sir no what the process just a second here almost done so just the last step can you go down a bit yeah sure thank you sir when is your last exam everyone 28th oh chemistry sir last exams chemistry 10 days of school will go on oh yeah from 21st to first we have school okay and then first onwards is the shara vacation so okay yes it done okay done very good so now we'll move on to the methods of differentiation yes sir this is something which you have already known so i'll be slightly faster and we'll focus more on problem solving okay so the first method is basically when you have a function multiplied to a constant let's say c okay and f is the function then it is just c into the derivative of the function second when you have a sum and difference of two or more function and you're differentiating the result it is as good as finding the sum and difference of their respective derivatives this rule you must be aware of by the way let me write the name sum slash difference rule so this is f derivative of g plus g derivative of f so one at a time you have to differentiate the other functions will remain as such and this can be extended even to let's say you have three functions okay i'll write extension so even if you have three functions let me say f g and h it will be f g derivative of h plus f h derivative of g plus g h derivative of f okay this rule is called the product rule we all know that fourth rule is when you have a quotient of two function f by g it is g times derivative of f minus f times derivative of g by g square okay remember the order in which you are writing it do not mess up with the order else your answer will become negative of the previous answer this is the quotient rule okay now comes the chain rule chain rule is applied when you're finding the derivative of composition of functions so f of g so everybody knows their composition of function yes sir when one function is fed as an input to the other that is called composition of function okay so this rule says it will be derivative of f of g with respect to g into derivative of g with respect to x okay and this can extend let's say even if you have a composition of three functions let me write it as extension let's say f of g of h of x okay so first you'll write derivative of f of g of h of x with respect to g of h of x into derivative of g of h of x with respect to h of x into derivative of h of x with respect to x okay by the way there is some notation that we normally associate with this please be careful about the notation notation that we use is f of g of x let me not write x yeah f of g of x whole dash is given as f dash g into g dash there's a difference between these two but normally people think they are the same notation there's a difference between them what is the difference here you are differentiating with respect to what x yeah here you are differentiating with respect to g okay let me ask this question what do you understand by f dash x cube differentiating x cube with respect to x no no no no no we are differentiating f of x cube with respect to x cube this is the meaning of but if i write f of x cube whole dash what does this mean d by dx of x cube yes f of x cube d by dx of f of x cube getting my yes sir okay so don't get confused there are different notations one second sir yes sir so i think product tool question tool you are well aware of chain rule in case you're forgotten we can take few examples of chain rule let's say i ask you to differentiate find the derivative of under root of and x square with respect to x okay so here there are three functions which are being you know composed to form this function one is root x other is tan x and the third one is x square correct so here if you see f of g of h of x this itself is your given function root of tan of x square correct so first this says you have to differentiate under root of tan x square with respect to tan of x square it's like saying differentiating under root u with respect to u what is the result for that you say 1 by 2 under root u and u itself is this into now the derivative of tan of x square with respect to x square what is that he can square of that into now derivative of x square itself which is 2x okay so this is how the chain rule works is that fine yes now chain rule has one more advantage it actually helps us to differentiate a function which is present in some variable with respect to another variable let us say i ask you what the derivative of sine y with respect to x remember here y is a different variable x is a different variable how do you do this differentiate sine y with respect to y differentiate into y with respect to x excellent yes one second sir that's a question cos y into dy by dx absolutely correct as as Venkat rightly said this is nothing but first you have to differentiate sine y with respect to y into derivative of y with respect to it this itself is a chain rule right so chain rule is one of the most important methods which actually is useful in finding the derivatives of any function which is first of all in a composite form and second of all if let's say there are the function is in a different variable vis-a-vis the variable with respect to which you are differentiating so this will be cos y dy by dx is that fine yes it is very also useful in finding the derivative of derivative itself for example if i say dy by dx whole cube if i want to find its derivative with respect to x tell me what will happen yeah correct first you treat this as something cube something cube is 3 now once you're three second care of derivative of dy by dx with respect to x is d2 y by dx square this will be useful next year in class 12 getting my point yes sir as a rule what I say is in order to make your life simple imagine just try to recall in your bridge course I had done this so first treat this as gates you're crossing these gates to reach x okay first is the root gate then is the tan gate then is again the square gate etc okay so let's say you are standing near the root gate so you'll read it as root something root something is one by two root something then remove that gate once you have crossed it then tan something tan something is secant square something once you remove tan gate you are left with a function which you can dip easily differentiate so two x so that is the method that we normally suggest for solving chain rule now time for some questions so I'll be bombarding you with a lot of questions today yeah let us begin with this question hope you can read this cos y is equal to x cos a plus y prove that dy by dx is cos square a plus y by sin a and I'll just give you three minutes for this time starts now how would you rate your paper overall was it easy moderate difficult moderate actually it was easy sir but then uh some like not prepared enough like some chapters I forgot differentiation in the last minute so that's one thing I always tomorrow's preparation I'm studying I'm studying it's not getting over sir there's so much I know sir 11 chapters in bio it's nice but it's fun I guess sir is the y in the cos and like is the y in LHS and the y in RHS the same of course yeah why is same throughout again okay sir one second then I it's more complicated than idealized by the way this is a very potential question for you in your exams it may come sooner or later to you uh jay school school okay sir oh one second sir then yeah yeah oh yeah can I help you on no sir no sir please this should be ideally at two minutes two and a half minutes not more than that last steps now applying the formulas okay okay fine dy by dx low d high minus high d okay let me take this over let's say x in terms of y is given as cos y by cos a plus y okay uh let us find dx by dy for a change because I can see x is completely a function of y so let me just just differentiate with respect to y on both the sides okay so when you're differentiating this with respect to y you can use your quotient rule so it's cos of a plus y derivative of cos of y is minus sine y please remember you are differentiating cos y with respect to y not x and you'd be differentiating cos y with respect to x you would have written minus sine y dy by dx yes we're differentiating cos y with respect to y so it is just minus sine y dy by dy but dy by dy is one so we don't write it actually minus cos y derivative of this is going to be minus sine a plus y into derivative of a plus y derivative of a plus y will be zero plus one isn't it because a derivative is zero and y derivative with respect to y is one okay whole divided by cos square a plus y correct basically I use the formula g f dash minus f g dash by g square if you simplify this it becomes minus cos a plus y sine y plus sine a plus y cos y by cos square a plus y just rearrange the term in a proper way so that we can identify that oh it is the compound angle identity it is the formula of sine a minus b so if you look at this expression this is nothing but your a b a b so sine a cos b minus cos a sine b and what is that sine a minus sine a minus b right so sine a minus b okay this is your dx by dy so y y gets cancelled so it is sine a by cos square a plus y so dy by dx is the recipe procal guys remember this dy by dx is the recipe procal of dx by dy so that is going to be cos square a plus y by sine a simple a couple of things to be noted over here so I'll I'll write it down in a box structure here important something very important dy by dx and dx by dy are recipe procals of one another okay but d2 y by dx square is not the recipe procal of d2 x by dy square please don't have this misconception the double derivative do not follow the same relation as a single derivative follow okay so what is the actual relationship between d2 y by dx square and d2 x by dy square that we'll discuss next year when we are doing higher order derivatives okay meanwhile is this problem clear any doubt any question with respect to this so there was an element of trigonometry also involved in this yes here right let's move on next question here comes another problem for you if f of x is log of log x to the base x please note that this inside log has a base of e but outside log has a base of x so this is your e find f dash x at x equal to e by the way many books will also write it as find f dash of e okay that means the same thing as finding this so after differentiating put x as e tell me the answer again let's have just three minutes for it not more than that so was it derivative log a to the power b like log a to the base b very good question very good question see derivative of log x to the base e this is only one by x okay this is this formula is not valid if your base is not e but since you have asked what will happen if my base is something like a okay then first we have to convert this log x to the base a as log x to the base e divided by log a to the base e by use of change of base formula this is called the change of base formula once you have done this one by log a to the base e is just like a constant so you can just pull it out and it's derivative of log x to the base e which is actually one by x so you answer one by x log a to the base oh yes sir does this help you uh Venkat yes sir proceed with this problem yes sir one second now starting the problem now one second sir getting it yeah yeah so one no one is not the right answer okay so let me just attend this so first of all when you're finding the derivative of log x if you first change the base of it we'll write it as log e of log x to the base e by log x to the base okay so better to write this in terms of ln because too much of writing work is involved so ln of ln x by ln x okay so think as if your problem was this now if you want to differentiate it first the overall structure is there is a quotient rule involved one function let's say u is divided by another function let's say v so we all know the formula of quotient rule okay uv dash is v u dash minus u v dash by v square okay let us try to fill in these spaces over here so v will be nothing but ln x u dash u dash means derivative of ln of ln x so for this we'll use chain rule so read this as ln something first ln something is one by something into remove this ln from your mind then you see ln x ln x derivative is one by x okay put proper brackets once it is done just close the bracket so these two are done now u u is going to be ln of ln x into derivative of v is one by x whole divided by v square v square is ln x the whole square okay once this is done you just have to put your value of x as e okay so you have to calculate this at x equal to e so we normally put dash line and put x equal to e it suggests that you have to substitute e in place of x so ln e is one we all know that one by ln e is also one one by x is one by e minus ln of ln e ln of ln ln of one will be zero so this will be zero so we don't have to bother about this term anymore divided by ln e is again one one square okay so if you have confusion how this becomes zero ln e is one so ln of one is actually zero so the answer is this goes for zero this is one by e by one that is the answer is one by e is that fine any questions please ask me so it was a use of a quotient rule along with the use of chain rule in between so remember all these rules normally they work at in tandem to solve a given problem it's not like a question is completely solved by a product rule or a quotient rule or a chain rule all these rules work together depending upon how is your function structured and what are the functions involved in that structure can you move on to the next one or you want to copy this done sir others others i'm sure they are open with their bio book or computer book dams are also important i was actually under the impression that you have math exam tomorrow that's what else i would have you know just kept a session a little later on anyways so next is y is one plus x by one factorial x square by two factorial x cube by three factorial and so on till x to the power n by n factorial so that dy by dx is this so we're just saying you look up in the evening class 11 first semester is always a very important exam you know why because it gives sometimes it gives shocker to many of you okay you people become very shocked after this exam because three hours paper and new topics are there this is a life changing moment for me it was at least a life changing moment so first exam of my class 11th was the big shock of my life from 96 i went to 69 so after that i realized no boss this is not you know the class 10th anymore so i had to change my gear and and no i gave everything then second semester was good i ended up in 90s and so what did you change at gear two six gear first i went to first year and then slowly i started okay it is not the same as class 10 it is a continuous study game you can't study one day or two day and they don't do well in it yeah that's true i'm just saying maths test already changed my life good to know that village i took back this is a good learning for you press on the pedal and go full throttle so this thing is inclusively feeling true like dy by dx minus y a yeah i know that i know that but you have to prove it normally yeah it's simple it's not it's not difficult yes i'm done awesome yes sir okay let me discuss this quickly so dy by dx will be nothing but derivative of one will be zero derivative of x by one factorial will be one this will be two x by two factorial which is x this is three x square by three factorial by the way let me tell you three x square by three factorial is as good as saying by two factorial factorial similarly the derivative of x to the power x to the power four by four factorial is four x cube by four factorial which is x cube by three factorial so you'll get x cube by three factorial can i say this trend will go on till x to the power n minus one by n minus so had you added one more term it would have actually become your y so if you add x to the power n by n factorial in both the places so basically i'm adding x to the power n by n factorial in both the places then this term would have actually become a y right so dy by dx plus x to the power n by n factorial is actually y which implies which implies dy by dx minus y plus x to the power n by n factorial is equal to zero okay no question with respect to this very simple you just had to realize this part okay can i move on now yes sir simple question just to just test your name rule find dy by dx when y is sine of x square plus one just a 10 second problem yes sir what's the answer cos of x square plus one into two x cos of x square plus one into derivative of x square plus one is two x absolutely so two x cos of x square plus one simple oh yeah looks good that let's have this one it's why is under root of log of sine of x square by three minus one find dy by dx sir by the way the limits question right i went outside limits question i went outside right 729 cube something like that x bar 729 minus i still didn't get that sir and i'll try it again one last time and then i'll ask you for solution oh okay fine sir i think i got it you got it ridhuta uh one second yeah last part done uh yes you're done let's discuss something to the power half something to the power half is one by two the whole thing yeah the whole thing okay into forget the root log something is one by something into forget log also sine something is cos something forget sine also and this is going to be two x by three simple one next okay here is a constant done sir done sir yes sir first log something is one by something into now once log is gone one plus this you have to differentiate it simultaneously so x derivative is one for derivative of under root of x square plus y square you have to again use a chain rule so which is one by two under root x a square plus x square into just multiply this to this term okay not with one okay into derivative of a square plus x square which is zero plus two x okay let us simplify this so it becomes one by x under root a square plus x square and this will become one and this two and this two will get cancelled so x by under root of a square plus x square just take an lcm over here just take an lcm over here it'll become x plus under root a square plus x square by a under root a square plus x square this will get cancelled off listen this gets cancelled off leaving you with the answer of 1 by under root a square plus x square. By the way, later on, you will learn in class 12 that when we integrate 1 by a square plus x square, the answer that we write is log of mod x plus under root a square plus x square. Okay, so this is a well-known result actually, which you'll learn in class 12th also. Yes, sir. Any questions so far? No, sir. Sir, will we do a bit of different integration also like for learning? I think I've only shared the videos with you. The videos, yeah, but like like this. Yeah, yeah, see, once your semester exams are over in class 11, the month of holidays will do a lot of things. The summer holidays are this one. Can we just do a bit of. The number of chapters are huge. We yet have to do complex numbers. We have to do conic sections. We have to use binomial, sequence series, statistics, mathematical reasoning, permutation, permutation combination, of course. A lot of chapters are there. Okay, sir. Here are the nice permutation combinations. Sir, this is just a lot of product rule, right? Yeah, but you have to be smart here. You don't have to use actually the product rule. First simplify this. That's my trick. That's my hint. Sir, once some of them get cancelled, it becomes 0. But you don't have to put pi by 4 initially, that is a mistake. You don't have to put the value and then differentiate. You have to differentiate and then put the value. Yeah, even after differentiating it, sometimes it's called like 8x. Okay, let it cancel. No worries. Oh, sir. Yes. Like we have a formula and trigger, right? Cos x is sin 2 NA by 2 N sin A. Absolutely. That was what I was looking for in this. Okay. Just now it came into my head. I've been doing something, you know, just okay. See, if you would have practiced signometry very well, you know that if your angles are in GP with a ratio of 2, see, these are in GP, geometric progression with a common ratio of 2, then your life becomes very simple. You can just initiate a chain reaction over here. See how the chain reaction works. Multiply and divide by 2 sin x. Okay. So when you multiply and divide by 2 sin x, what does 2 sin x cos x become? Sin 2 x. Sin 2 x. So may I raise that with your permission? Yes, sir. Which one? This term only. Yes, sir. It becomes sin 2 x. Now, just provide it with the two here. So and provide it with the two here. What does this term become? Sin 4 x. Isn't it? 2 sin theta cos theta is always sin 2 theta. So sin 2 x and cos 2 x will become a sin 4 x. So you don't have to mug up the formula also. This is sin 4 x. Again, put a 2 here. Again, put a 2 here. What does this term become now? Sin 8 x. Absolutely. Sin 8 x. No, let me write sin 8 x. Again, put a 2 here, 2 here. Again, put a 2 here, 2 here. What does it become? Sin 16 x. Again, put a 2 here, 2 here. What does this become? Sin 32 x. So ultimately, your answer is sin 32 x by 32 sin x. Now, ultimately the question is asking, find the derivative of this function at x equal to pi by 4. Can we not do that? Let's do that. Tell me the answer once we're done. Oh yes sir. What is sin pi by 4? 1 by root 2. 1 by 32, you can just keep it outside. Don't involve it. Oh yes sir. Square. So sin x cos 32 x into 32 minus sin 32 x into cos x by sin square x. This is 1 by root 2. Okay. Cos 8 pi. Cos 8 pi is 1. Yeah. Okay. But sin 8 pi is 0 by sin square is 1 by 2. So, your answer is 1 by, this will be 1 by 16 into 32 by root 2, which is actually root 2. So, no need to use product rule for 1, 2, 3, 4, 5, 5 times. That would be too lengthy. So, you just use a simplification of this. So, it's a more of a question on trigonometry rather than anything else. Can we move on now? One second sir. Sure. Yes sir. Done sir. Yes, I'm sure you'll be able to do it. Yes, I'm sure you'll be able to do it. If y is under root of 1 minus x by 1 plus x, prove that 1 minus x square dy by dx plus y is equal to 0. This is called differential equation based questions, which will be asked a lot in your school exams also. This is actually a differential equation, which we learn in classrooms. This is called a differential equation. It's a differential equation. It is 0. Equal to 0, you have to prove it. Oh, yeah. One second. No, no, no, I'm not happy. No, I'm not happy. No, I'm not happy. It's a lengthy question. One second, sir. Got it. Almost, almost. Last, last step, sir. You're sure. See, first of all, you do this thing. Multiply and divide with 1 minus x. 1 minus x square and this becomes 1 minus x square. This becomes y is equal to 1 minus x by under root of 1 minus x square. We'll do one thing. I will do take under root of 1 minus x square on the other side and this is 1 minus x. Yes, sir. So far, so good. No problem in this? No, sir. Now, what I'll do is we'll differentiate both sides with respect to x, treating this as two functions. So we'll use product rule over here. So under root of 1 minus x square into dy by dx plus y into derivative of this will be 1 by 2 under root 1 minus x square into minus 2x and this side will have a negative 1. Is that fine? Yes, sir. Okay. So this two will get cancelled. So we'll have under root of 1 minus x square dy by dx minus x by y under root 1 minus x square is equal to negative 1. Yes, sir. Yes or no? Okay. So now we'll do one thing. We'll multiply it through under root 1 minus x square. So this becomes 1 minus x square. This becomes minus x y and this becomes minus under root 1 minus x square. Yeah. Correct? Yes or no? Yes, sir. What do you want to prove? This plus y is equal to 0. Correct? Oh. So if you take this term, it'll be x y, okay? Add a y to both the sides. So y into 1 plus x minus under root 1 minus x square. Let's try to evaluate this. y under root of 1 minus x by 1 plus x into 1 plus x. Correct? Yeah. If you introduce this within the root sign, it will become under root of 1 minus x 1 plus x into 1 plus x the whole square. Basically, I'm introducing this term within the under root sign. These two terms will get cancelled. Yes, sir. We'll be left with under root 1 minus x square minus under root 1 minus x square which is obviously 0. That means 1 minus x square d y by d x plus y will be equal to 0. That's proved. Just a second, sir. The type of questions will come across because see the methods are very simple. Derivative differentiation is a very conventional process. Yeah. You just have to follow. So how can they make the question a little bit more challenging only by making such kind of questions? Okay. Yes, sir. Just a second, sir. So can you screw it up a bit? My target was to create that 1 minus x square d y by d x somehow. Yeah. So that we can move on. Yes, sir. Just a second. Done, sir. Okay. By the way, this is my blurred image. This is 1 plus x to the power 1 by 4. I'll write it again. Minus 1, sir. Answer? Yeah. It was too fast. Let's check. This is 1 by 4 inside. Yeah, so you club this up. What does it become? Yes, sir. 1 minus x square half. And you club this up. It becomes 1 minus x and d y by d x is minus 1. Done. Yeah. Next. So 1 plus x, 1 plus x square, 1 plus x 4, 1 plus x to the power 2 to the power n find d y by d x at x equal to 0. One second, sir. 2 to the power n plus 2 to the power 2 n. Let's check. First of all, in this question, we can set in a chain reaction. Chain reaction by multiplying and dividing with 1 minus x. Okay. So these two will club up to give you 1 minus x square. So let me write it down over here. Can I make the changes here itself or do you want me to write for every step? Here it is. No worries. I can make the change here. 1 minus x square. Okay. 1 minus x is there in the denominator. Again, these two terms will club up to become 1 minus x to the power 4. So now I'm making the change here itself. 1 minus x to the power 4. These two terms will club up to become 1 minus x to the power 8. And can I say if this trend continues, if this trend continues, what will happen till you reach the last term? 1 minus x to the power 2 n. 1 minus x to the power 2 to the power n square, right? Yeah. n square will become this into 2, correct? So this will become n plus 1 if I'm not wrong. Whole divided by 1 minus x. So I think that with this given expression, this unique expression will actually disturb. For differentiation of this, we can use our quotient rule. Let's use our quotient rule for this. So dy by dx will be nothing but minus x times. Derivative of this will be 0 minus 2 to the power n plus 1 into x into 2 to the power n plus 1 minus 1. 1 minus x to the power 2 to the power n plus 1 into derivative of 1 minus x is minus 1. Whole divided by 1 minus x to the whole square. Now you have been asked to put x as 0. Actually we want the derivative of this function at x equal to 0. Let's see what happens when you put x as 0. It becomes 1 minus 0. This becomes 0 minus, this entire thing will collapse. This will become 1 minus 0 into minus 1 divided by 1 minus 0 the whole square. So ultimately giving you the answer as 1 by 1 square which is nothing but 1. No n will be there in your answer. I found it at n equal to 1. That's my mistake. Next question, is that nice? Yes sir. If x under root 1 plus y under root 1 plus x is 0, prove that dy by dx is minus 1 by x plus 1 the whole square. Looks somewhat like the product rule. Any idea anyone Venkat? Kind of sir but it's not helping me. This is actually an NCRT question by the way. Then wait sir, I'll try. NCRT question not able to solve. No sir. Okay. So first thing we'll do is, we'll write this as negative y under root 1 plus x. Okay. We'll square both the sides. Okay. Let's expand it. Let's take x square, y square on one side. So I can say x plus y is negative xy. So x plus y is negative xy. You can find the derivative. You can make y the subject of the formula. So y 1 plus x is minus x. So y is minus x by 1 plus x. So dy by dx is going to be use the quotient rule 1 plus x into minus 1 minus of minus x, which is plus x into 1 by 1 plus x the whole square. So if you open this up, it becomes minus 1 by 1 plus x the whole square. Did you want to copy that or? Yes sir, just a second. Okay. I'll just paste the question here and then I'll move back. Done sir. Try this out. It is given that cos x by 2 cos x by 4 cos x by and so on till infinity is sin x by x. Then find the sum of 1 by 2 square, 6 square x by 2, 1 by 2 to the power 4, 6 square x by 4. If you want subsequent terms, I can provide it. 1 by 2 to the power 6 secant square x by 6 and so on, x by 8. This goes all the way till infinity. You must be wondering how does derivative feature in our air because you don't see a derivative term anywhere in this question. But that's the trick actually. But why is it sin x by x given? Just to make your life easy. If they wanted, they would have not even given that term. Oh. They want you to use that expression actually, that's it. This looks like product rule again, like Ultaf product rule. But I don't know, it's not speaking back. Any idea how to proceed? I'll give you a hint. Take log on both the sides and differentiate. Log? Yeah. Log to the basic. Oh. Take log on both the sides and differentiate. Try that out. Okay, one second sir. One second sir. Did you do that, Venkat? I'm doing sir. No sir, not getting. Okay, let's see. If you take log, it becomes ln cos x by 2. This side will become ln sin x minus ln x. Yeah. If you differentiate this, what's the derivative of ln of cos x by 2? Which rule will you use? Chain rule. So it's 1 by cos x by 2 into minus sin x by 2 into half. Yeah. This term will be what? 1 by cos x by 4 minus x by 4 into 1 by 4. Keep on doing this. Here it will become 1 by sin x into cos x minus 1 by x. Yeah. So this term is minus half tan x by 2. This term is minus 1 by 4 tan x by 4. Next term would have been minus 1 by 8 tan x by 8 and so on. This term would have become cot x minus 1 by x. Okay. Differentiate both sides again with respect to x. So differentiate again with respect to x. This becomes minus half secant square x by 2 into half again so 2 square will come. This will become minus 1 by 4 secant square x by 4 into 1 by 4 again which is 4 square. Now 4 square is actually 2 to the power 4 also. Okay. So do you start realizing that we have started getting these terms? Okay. And this side will be minus cos secant square x plus 1 by x square. Just take the negative sign. Just cancel out the negative sign from both the sides. So half secant square x by 2. Sorry. Half square secant square x by 2 plus 1 by 2 to the power 4 secant square x by 4. And so on. Will be nothing but cos secant square x minus half x square. So this becomes your answer. Is that fine? Sir, could you scroll down a bit? This is nice. Sir, just for additional information, can you do differentiation of x power x and all today? Yeah. Yeah. Why not? Yes, sir. Yeah. Just a second. Done, sir. Y dash means dy by dx. Okay. This y dash, it means dy by dx. Y dash by y is 1 by x times that expression. Ax square plus bx plus c. Degree of denominator is always 1 more than degree of numerator. Yeah. Is that of any user? Partial fractions. Yeah. And not exactly partial fractions. Just thinking. Okay. Try it out. Yes, sir. Sorry. You don't learn when you look at the solution. You only learn when you struggle with the question. Yes, sir. Any idea? I'll give you a hint. Yeah. Hint, sir. Start combining these two terms. See what will you get? Oh. Oh. x by x minus c. If you take this common, you'll get c plus x minus c, which is x by x minus c. Correct. So let me just remove this unnecessary things. Now try to combine these two terms. See what do you get? You can take x by x minus c common also. You'll get b by x minus b plus one. So yeah. Oh. Which is x by x minus b. So it becomes x square by x minus b x minus c. You'll get a by x minus a plus one. Can I say it will result into x cube by x minus a x minus b x minus c. Yes, sir. Now. Take log on both the sides. So ln y is equal to ln x cube. X cube minus ln that. ln x cube is like three ln x. Right. Yeah. So one by y dy by dx is equal to three by x minus one by x minus a minus one by x minus b minus one by x minus c. Okay. Ooh. Now let's see what is asked in the expression. They've asked you to get this. Okay. So what I'll do here. I'll write this as one by x. Yeah. You need to get this term right one by x a by one a minus x b by b minus x and c by c minus x. First of all, three by x and I'll make it plus one by a minus x plus one by b minus x plus one by c minus x. So this is nothing but y dash by y. Yeah. Now how do I get to that desired expression? Contribute one by x to each of the others. Okay. How will that? Yes. One by x plus one by x plus one by x plus one by a minus c. I'm not sure if that works. It might work. Yeah. Just try it out after that. Just a second sir. Just a second sir. Yeah, you get. No. Yes sir, you got it. B minus x plus c by c minus c minus x. One by x. Yes sir, you got it. You got it? So we have to distribute one by x to each one of them. Yes sir. So this will give you a minus x plus x by x a minus x. Again b minus x plus x by x b minus x. And again c minus x plus x by x c minus x. Correct. So x x x x goes off. And then one by x common. Take one by x is common. So it will become a by a minus x b by b minus x and c by c minus. That is the desired expression. Super question. Yes sir. One more question. Easy? Okay. That was a cute problem. This is f of x is mod of ln of mod x. Find f dash x, which of the following four options? Sir, how will mod function affect the process of finding derivative? It'll affect it a lot. If you just try to find the graph of this and see how is this function defined? Yes sir. One second sir. Sure. So figuring it out, two, three mods and all, taking some time to figure stuff out. One second. Just redefine, try to redefine the function without having mod. Try to write down the function. That's just lnx. Yeah, just lnx. What type of manifestation will you see in that? One is when the graph like y is equal to mod f of x and y is equal to f of mod x. Both the things are combined into this. See, it's very graphical point of view. See, how will I do it graphically? First of all, if you draw the graph of lnx, lnx graph is like this. Now let's go step by step. This is the graph of y equal to lnx. Y is equal to lnx. Correct. So if you take mod on x, what will happen? If you take mod on x, the graph just gets mirror image about y axis. Y axis, yes sir. Correct. If you take mod on everything, what will happen? The part of the graph will come up. It will come up like this. Correct. So you're retaining the word? No, I'm not retaining the downward, down part a little bit. Okay. Now see, this itself will tell you a lot of things. This point, first of all, is 1, right? X is 1 and this is minus 1. Yeah. Yes sir. You see this blue part, yellow part, this yellow part is still a part of lnx graph, isn't it? Yes sir. Would you agree with me? Yes sir. Correct. This is a part which is a graph of negative lnx. Negative lnx. Oh yeah. Yes sir. This part is the graph of y is equal to negative ln negative x because it's the mirror image of this white part. Yes sir. When you're taking a mirror image, you just replace the sign of x with the minus x. What about this? This is the part of the graph y is equal to ln. Negative lnx minus x. Correct. So the definition is well, you know, understood over here. The definition is when you are below minus 1, when your x is less than equal to minus 1, you are actually following the graph of ln of minus x. Yes or no? Yes or no? Yes sir. When you are from x to 0, not including 0, you are following the graph of minus ln minus x. Correct. When you are from 0 to 1, excuse me, 0 will not be included, 0 to 1, you are following the graph of minus lnx. Yeah. Oh beauty. And when you are above 1. Normal lnx only. Right, absolutely. No doubt about this redefinition? No sir. Now let us differentiate this entire thing. All I have to do is find the derivative of each one of them. For example here, what will be the answer? I am sure it is 1 by x only. Try chain rule. 1 by minus x into minus 1. 1 by minus x into minus 1, which is 1 by x only. This will be minus of minus of 1 by minus x into minus 1, which is minus x. Correct. This also will be minus 1 by x. I am sorry. And this will be 1 by x. Now let us try to see what is happening in this. If you see clearly, whenever your x is less than minus 1 or greater than 1, which is like saying mod x is greater than 1, your answer for the derivative is 1 by x. And when you are between these two zones, these two zones means you are between minus 1 to 1. Of course excluding 0. Of course excluding 0. Your answer is minus 1 by x. So I can say mod x less than 1. Your answer is minus 1 by x. Let us see what does the option say. By the way, we cannot include 1 and minus 1. Why? Because there is a corner getting formed. In the corners, the function is not differentiable. So at these points, I have to remove it. So basically it becomes less than a greater than. So now look at the options. The option says 1 by x when x is not equal to 0. This is not right. It says 1 by x when x is greater than 1. So let me check. 1 by x when mod x is greater than 1 and minus 1 by x when mod x is less than 1. This is the right option. This cannot be there and this cannot be there. So option number 2 is the right option. Is that clear? Are you guys there with me? Yes, sir. Any questions? How we solved this question? This question was actually not difficult. Had you known how to redefine the function? And for redefining the function, it's good to take help of the graph. Especially when multiple mods are involved. Log of tan x to the base of sin x is y. Find dy by dx at pi by 4. Sir, what is the standard of these questions? C, D, J, whatever. They're slightly less than C, D, C. J will not ask you finding the derivative. Right now you're learning this chapter. That's why I'm giving these questions. J will always incorporate a multi-conceptual question. Two, three concepts simultaneously will be tested. Oh, yes, sir. Just like the previous one. The previous one was the J main question. Oh, then nice. Yeah. This is not. This is the super simple. One second, sir. Wait, sir, getting something. This is second, sir. Almost done. You're in there. I know, sir, it became too lengthy unnecessarily. Okay, did you get this first? Ln of tan x by ln of sin x. Apply quotient rule for this. ln of sin x. Into derivative of ln tan x is 1 by tan x into secant square x. Yeah. ln of tan x. Derivative of this is 1 by sin x into cos x. By ln sin x the whole square. Okay. Now put x as 5 by 4. This becomes ln 1 by root 2. Okay, this becomes 2 by 1. 4. This becomes ln 1 which is 0. So everything will become 0 after this. By ln 1 by root 2 the whole square. This and this will go off. So it becomes 2 divided by ln. You can write this as negative half of 2. So that is minus 4 by ln 2. 3 is correct. I know my approach. I messed up somewhere like simplification. If I am equal to this then find the value of 1 plus x squared by dx plus xy plus 1. One second, sir. Sure. Sir, just curious. Yeah. Find f of x is equal to ln of ln of mod x, right? Yeah. Was it possible to solve that without the graph? Yeah. Why not? How, sir? Do you want me to do it after this or right now? Yes, sir. After this. Should I help you with this? Just a hill, sir. Can you give? You differentiate both sides with respect to x. Yes, sir. Done. This product will use chain rule over here. Yeah. That I did. So, into derivative of this will be 1 by 2 into 2x. Okay. Plus under root of x squared plus 1 into dy by dx. Derivative of the right side will be 1 by into. Derivative of this term will be 1 by 2 under root x squared plus 1 into 2x minus 1. 2 to 1. This also 2 to 1. So, xy by, this will be x under root x squared plus 1 by under root x squared plus 1. I'm sure you would have realized that this term and this term can be cancelled with a minus 1 here. Okay. Multiply throughout, multiply throughout with x squared plus 1. So, this will become xy. This will become x squared plus 1 dy by dx. This will become minus 1. So, x squared plus 1 dy by dx plus xy plus 1 is actually 0. I think this is what we need. Yeah. What we need for option 1 is correct. Zero. One second, sir. Yes, sir. Done. Done. Okay. So, now I'll solve that question without a thought. See, first of all, log of anything. For two cases when that, but then it's E, right? Yes. We have to first take into account what happens when x is greater than zero, what happens when x is less than zero. Yes, sir. Greater than zero. This function becomes mod log x and this function becomes mod log minus. Correct. Yes, sir. Yes, sir. Now, ln x is known to be negative when x is less than one, right? Yeah. Okay. So, greater than zero, but less than one. I can say like this, greater than zero, but less than one. This function will behave as minus ln x. Yeah. Correct. And greater than one, the function will behave as ln x. ln x it will. Yes, sir. Oh, yes, sir. Now, less than zero, but less than zero, let's say this is the graph. If you see log of minus x graph, this is the graph. Hmm. Less than zero, but before one. Hmm. Less than zero before one. X is negative, right? This is already negative term. So, negative into negative is a positive term. Yes, sir. Correct. So, can I say between zero and one, less than zero, but greater than one. Correct. This function will behave as negative of log of negative x. Because if you take a quantity which is between minus one and zero, let's say I take a minus half. Yes, sir. Log of minus of minus half is 0.5. Oh, yeah. And log 0.5 is a negative quantity. So, it will give you negative 0.5. Yes, sir. Correct? Oh, yeah. Isn't it like saying negative, negative log negative x? Oh, yes, sir. And if x is less than minus one, this will behave as positive log of negative x. Oh, yeah. And now you can differentiate it and get your answer. That's basically instead of taking graph, graph you're taking values. Oh, okay. Got it, sir. If y is equal to sine inverse by under root of one minus x square, then one plus x square dy by dx is equal to? Yeah. Yes. Let's hope that we can solve this. Wait, sir. I hope I'm getting this. Okay. One plus xy. Options there. Two. Yes. Yes, option two is absolutely correct. Absolutely. Yes. We can do this. Just to check whether whatever you have done is the right way. Okay. Yes, sir. So we can apply. Transpost. Oh, yeah. Okay. We can apply product rule over here. The derivative of this will be one by two under root one minus x square into minus two x. This is one by under root one minus x square. Okay. So this will get cancelled off multiplied throughout with one minus x square. This will become a minus xy and this will become a one. That means one minus x square dy by dx is equal to? Okay. Take it. Yes. Next question. Oh, f of x is x plus one by two x plus one by two x plus one by two x plus one by one plus infinity? Find the value of f of f of 15 to f dash of f. How do you differentiate this? That's the trick. If you know how to differentiate everything. Yeah. After this problem and take a break and then we'll start with kvBy. Okay. Yes, and a derivative x in after db by also if you're free. Oh, Mr. Okay, sir. sir. Nothing is working in this question. Okay. See here, you have to observe a pattern here. You can see that after 1 plus 2x, this part is getting repeated, right? This part, which is actually also this part, is getting repeated. Yes, sir. Now, see, can I write f of x minus x as 1 by this time? Yeah. The same term here, can I say this is actually f of x minus x itself? Oh, yeah. f of x minus x is actually 1 by 2x plus f of x minus x, which means f of x minus x is 1 by f of x plus x. Yes, sir. Cross multiply. So f square x minus x square is actually 1. Square x. Yeah. Square x is 1 plus x square. Okay, now let's differentiate both sides with respect to x. What are the derivatives of f square x with respect to x? 2 f of x into df of x by dx. Which is actually x only, right? Yes, sir. That is equal to 2x. 2, 2 gets cancelled, f of x has 50, then this is the answer. Oh, once, let me just process that. Process. f of x. Oh, wow. Yes, sir. Awesome. Awesome. Let's take a break for some time. Can you see the screen now? Yes, sir. So let's say we have a quadratic with us x square minus m minus 3x plus m equal to 0. Okay. m is some real number. m is some real number. Okay. My question to you is first, find the values of m, if you write it here, find the values of m says that this quadratic equation has got real and distinct root. Okay. I'm sure you can do this. Yes, sir. Just a second, sir. I'm coming. Yeah, yeah, sure. From minus infinity to 1 and 9 to infinity. Open interval, close interval. All open. All open. Very good. So you would have got this. You just did b square minus 4ac greater than 0. Yes. So your answer here would be m should belong to minus infinity to 1 union 9 to infinity. Okay. Yes, sir. Next question. Good. I'll give you a series of questions. Okay. So almost 13, 14 questions I'll give you in the same thing. Okay. Oh, yes, sir. Tell me the value of values of m or interval of m such that the roots are equal. Equal to 0 then. Equal, equal. Yeah. Okay. 1 and 9. 1 and 9. Very good. Chalo, you passed this also. Next is your roots are not equal. Oh, sorry. Not real. Minus 1 to, sorry, 1 to 9 open. 1 to 9 open. Very good. As you move on, the complication will increase. Okay. Yes, sir. Next is the roots are opposite in sign. Does that the roots are opposite in sign? Same as real and distinct, but hopefully not. One second, sir. Opposite in sign. Think carefully. Yes, sir. How do we analyze for this? Think that's the, that's the thinking part, right? Oh. Without actually finding the roots, how do you find this condition that the roots are opposite in sign? Depends on where they got the x-axis, like left side of y, y-axis or right side of y-axis. One will be left, one will be right, then only it will be opposite, right? Yes, that's what, that's what should happen. So if that is happening, okay, that means you have a vertex. One second there. One second. I think I figured it out. There's the fourth quadrant. So just tell this, if this is right, the vertex should be in the fourth quadrant. Not necessarily. Oh. There's much easier method to do it. You don't have to figure out where the vertex lies. Just by using the coefficients, you can figure it out. Is it? How, sir? If I say both the roots are opposite in sign, can I say a product of the roots will be negative? Yes, sir. That means m, which is nothing but product of the root, c by a. Yeah. Yeah. Product of the root is c by a, right? So c by a here is m, right? So m should be negative. Oh. So at the same time m must belong to minus infinity to one union nine to infinity because it must be real also. Yeah. So it doesn't have m. Correct. So the answer would be m should belong to less than, less than zero means it is well under this zone. So it is minus infinity to zero. Yes, sir. Getting the point. Yes, sir. Wasn't it easy? It was. Same question. Same question. Now the fifth question is they should be equal in magnitude but opposite in sign. Equal in magnitude but opposite in sign. Okay. Think carefully. Think carefully. Oh, okay. That is m is equal to three. Is that wrong? Wrong. Yeah. The agenda is to make it to a perfect square. That's the thing. But that, no, it's substituted in C. So sum of root alpha beta alpha alpha is zero sum. So sum of root should be zero, right? Yeah. So minus b by a. So m minus three by one is equal to zero. At the same time, it must be real also. So it should lie in this interval and three doesn't lie in this interval. Yeah. So the answer is no such m exists. Oh, that way. Okay. For any condition to be fulfilled, root must exist first of all, right? Yes, sir. Before you study in some school, you should be alive first of all, correct? So first you have to make sure that this condition is met and then whatever condition is coming, you have to take the overlap with it. Yes, sir. This was actually a googly question. Okay. Next one. Tell me the condition says that both roots are positive. Both roots are positive. Same sign. So can you just go up a bit? Oh, this is the upward part. I cannot go up. You want to go to the previous page? Yeah. Yes, sir. Opposite sign was discovered. Both are positive. Okay, sir. Nice. Just a second, sir. Yeah. Nine to infinity. Nine to infinity. Nine to infinity is absolutely correct. Yeah. Yes, sir. Extruding nine, right? Yeah, both open. Now both roots are negative. Minus infinity to one. Oh, that will work. Oh, wait. Yes, sir. No such M exists because both are negative. That means some should be negative and product is positive. Correct, correct, correct. Nothing else. One is beta. Sir, no such M exists. No such M exists, okay. What did you do for making it both negative? Both negative. Thing like that. And both are positive and nine to infinity. So that interval is gone. And now we have the thing only from minus infinity to one. See, can I say both negative is possible only when some is negative and the product is positive. Yes, sir. Correct. I did that along with the fact that discriminant must be less than equal to zero because I have a negative and sorry. Greater than equal to zero. I'm sorry. Because it can have equal also and negative also. Like zero, zero. No, no, minus two, minus two like that. Oh, yeah. So alpha plus beta is minus B by which is this, correct? Yeah. Correct. This should be less than zero. Alpha beta is M. M should be greater than zero. And this, this we have already solved. It's minus infinity to one union nine to infinity. Yes, sir. Correct. Yes, sir. M should be less than three and should be greater than zero also. Correct. And it belongs to minus infinity to one union nine to infinity also. Okay. So can I say overall, if you collect all these scenarios into one, it says M should lie between zero and one less than three is also satisfied. Greater than zero is also satisfied. And this condition is also satisfied. You can make a number line to find the overlap. See, if you make a number line, one is there. Three is here. Zero is here. Nine is here. So first thing says less than three. So this zone. This is greater than zero. This zone. This says minus infinity to one minus infinity to one is this zone. Nine to infinity is this zone. I've made a very grave mistake. This is okay. I'm sorry. Yeah. Even in classes that I always do that. Less than three is this zone. Yeah. Then zero is this zone. Less than one and greater than nine are these ones. So basically this is the overlapping part. So between zero and one, this is the answer. Oh, okay. Next question is at least one root is positive. At least one root is positive. Yeah. Okay. So is this all the real things minus the reason in which both are negative. Awesome. So what is the answer? That one. Infinity to one union nine to infinity minus zero to one. What is the answer? Oh, one second. Simplified further. Yes. Minus infinity to one. What is that? Minus infinity to zero. Union is a separate. Remove that. Minus infinity to zero union nine to infinity. Brilliant. Absolutely correct. Minus infinity to zero. Union nine to infinity. Awesome. Next one. Ninth one. If they ask one is positive, we don't take the sign of zero. Right. Like at least one root is positive. Like the thing will may be zero at nine. Right. If you consider open interval. See one root is positive means M should not lie between zero to one. Correct. And at the same time M should lie between minus infinity to one union. Union nine to infinity because your root can be equal also. Yes, sir. Correct. So let me just write it down. One nine. So excluding zero can stay back. Zero to one has to be removed. Yeah. So this will be removed here. Zero to one will be zero can stay back. Okay. So your answer will be minus infinity to zero. Zero included and nine to infinity nine included. That's what you told. Right. Yes, sir. The thing is if I substitute nine right, it becomes zero. The, the entire quadratic equation will become 90 minus 90 zero. So then how can we. After that you're substituting M value. Right. Yeah. Yeah. Not X value. Yes. I got it. Nine. One root is smaller than two and the other root is greater than two. One root is smaller than two and the other root is greater than two. One root is smaller than two. One root is smaller than two. Other root is greater than two. Just one moment. I'll be back from the washroom. Yes, sir. Yeah. I'm back. Yes. Any idea how to attack this? So two happens to be in between your roots. Correct. Yeah. So if you make a graph, things will be more evident there. Let's say this is your Y axis, X axis. So it's somewhat like this. Obviously it's an upward opening parabola, right? Because coefficient of X square is one. Okay. So two is somewhere between this. Hmm. Okay. See a couple of things that come out from there. First of all, your root must be real and distinct. Yes, sir. Second thing is that your f of two must be negative. Because if you find the value of the function at two, f of two is negative. This is clearly a negative quantity. Correct. Hmm. Let's solve them and take their intersection. Yes, sir. We have already done. M should belong to minus infinity to one union nine to infinity. Okay. F two will you what F two would be four four minus M minus three into two plus M. This should be negative. Hmm. So if you resolve 10 minus M should be less than zero. That means M should be greater than 10. Hmm. Correct. So the union, sorry, the intersection of these two, intersection, not union, intersection. What does it mean? Just draw the number and see sir. Nine to ten. Hmm. Yeah. Sorry, ten to infinity. Getting the point. Yes, sir. D greater than zero. Yes, sir. Okay. Next condition. Just a second. Just a second, sir. Yes, sir. Next is both roots are greater than two. Same graph will just shift a bit towards the right. Does that mean f of two does not exist? No, it doesn't mean that it can exist. Both roots are greater. Yeah. Got it. Got it. Got it. Got it. Got it. Got it. Got it. Got it. Got it. Got it. Got it. Got it. Got it. Got it. Got it. f of 2. So 10 minus m greater than 0. So m less than 10, 1, 9, 10. So minus infinity to 1 into infinity, minus than 10. Sir minus infinity to 1 close, union 9 to 10, 9 close. Answer is from 9 to 10 only, 9 close. Is it? What was your answer? Can you repeat once again? m belongs to minus infinity to 1, 1 closed. No, no, no, no. Answer is just m should belong to 9 to 10. 9 close, 10 open. That I got, but that also is coming, some wrong terms also coming. Yeah, correct. f of 2 greater than 0, right? Yes, sir. So 10 minus m greater than 0. Okay, I'll tell you now the drawback in that approach. See, of course, I'm sure you would have done this. d should be greater than equal to 0. Number one. Number two, you would have done f of 2 greater than 0. Yeah, draw it up and we'll get it. Now, this is necessary but not sufficient, Venkat, because even if 2 was here, these two conditions would have been satisfied. All right. So there's a part one which nails the entire thing. That is, your, this part which is actually minus b by 2a, this must also be greater than 2. Oh, okay, gotcha. Only when these three are simultaneously satisfied, your job will be done. Oh, yeah. Awesome. Now, can you correct the answer? Minus b by 2a. Minus b by 2a greater than 2. This will give you minus infinity to 1 union 9 to infinity. Hmm. f of 2 greater than 0, I think f of 2 we have found out earlier, right? That was 10 minus. 10 minus. So minus m should be greater than 0. That means m should be less than 10. And minus b is like m minus 3 by 2a is 2. That should be greater than 2. That means m should be greater than 7. Hmm. Correct. Now, let us take the overlap of these conditions. So you have 1, you have 7, you have 9 lined up, you have 10, okay? Yes, sir. 2 says 1, 2 minus infinity, 9 to infinity. The other says less than 10. Less than 10 is this and the other says greater than 7. Greater than 7 is again this. Okay, so I can see an overlap of three lines occurring over here. Yes, sir. Your answer will be m should belong to 9 included to 10. 10 not included. Yes, sir. Is that fine? Yes, sir. Next condition. 11th question. Both are smaller than 2. I think this should be easy. Yeah, same process and put that third condition of this. You can put the third condition. Minus b by 2a less than 2. Less than 2, yeah. Tell me the answer now. One second, sir. Less than 7, okay. 1, 7, 9, 10, infinity, 10 less than m and m less than 7. Sir, minus infinity to 7. Minus infinity to 7. No, no, no, no. Minus infinity to 1. Minus infinity to 1 is correct. Yes, sir. Yes, sir. Minus infinity to 1. Yes, sir. Next question. 12th question. Yes, sir. Exactly one root lies in the interval. Exactly one root lies in the interval 1, 2. I think you need to take two cases for this, I guess. Yes, sir. From the graph only it will be evident. What is one of the roots? So, yeah, that is one. Condition will not change. Condition will not change. So, let's say alpha lies between 1 and 2. The condition is not going to change. Okay, sir. So, now that minus infinity to 1, union 9 to infinity, f of 1 greater than 0. So, what if you get f of 1 as a constant? f of 1 as a constant. Okay. 0 to 3. Okay. Okay, then there will be a point, I guess. I don't know. Let's see. And f of 2 minus b by 2 is 11, 2. So, minus infinity to 1, 9 to infinity, f of 1 is equal to 3, m greater than 10. Sir, no m exists. No m exists. Okay, let's try this. First of all, if exactly one root lies between this, can I say the root must be distinct? Yes, sir. Number one condition d must be greater than 0, which we have already solved. Correct? Yes, sir. Can I say that f1 will be positive, but f2 would be negative? Yeah. So, I can say f of 1 into f of 2 would be negative. Oh, yeah. Now, this condition is helpful in both the cases. Even if let's say 1 was here and 2 was here and this we are targeting this root, f1 would be negative, f2 would be positive. So, this condition is all-income passing. You are saying, do we have to make cases? Hmm. This helps us to avoid making of those cases. Oh, yes, sir. Correct. Wow. Now, tell me f of 1. What is f of 1? I think you figured out f of 1. 3. f of 1 was 3. Okay. Yes, sir. How 2 was 10 minus m? m. Correct. So, 10 minus m less than 0. Yeah. So, m must be greater than 10. Hmm. Correct? Yes, sir. What do we get from here? Then 10 to infinity answer. Answer is 10 to infinity. Oh, yes, sir. I took separate cases. This is an f of 1 into f of 2, sir. Learning. Okay. Next question. One second, sir. Lesson 0. This is important. Full package. Now, both the roots lie in the interval 1 comma 2. Yes, sir. Then one second. 10 less than m. Sir, minus infinity to 1. No. No. And 9 to 10. Wait, sir. I figured one thing out. Minus b by 2 a less than 1. No, greater than 1. Minus b by 2 a less than 2. Oh, yes, sir. I forgot two conditions. m greater than 5. 5 to 7. Absolutely correct. 5 to 7. How do you get that? There's another two conditions also. Ah, correct. So, both the roots lie in 1 and 2. See, first of all, d should be greater than equal to 0. Yes, sir. Now, people think that if both the roots lie between 1 and 2. So, 1 and 2 is here. Both the roots lie between 1 and 2. They think that f of 1 into f of 2 should be positive, right? Yeah. But this is also possible if 1 and 2 were between the roots also. Correct. Oh, yeah. So, we have to take the third condition that minus b by 2 a must also be greater than 1 and less than 2. Yeah, between 1 and 2. Uh, ULTA, sir. Inequality. Yes, sir. Yeah, yes, sir. Okay. So, this, all the conditions combined will help you solve this. So, last one, let's take it. So, the inequality is ULTA. Oh, I'm sorry. Yes, sir. And your charge magically became 98 percent. I keep charging in between. Oh. One root is greater than 2 and other is smaller than 1. One root is greater than 2 and other is smaller than 1. So, what is the unique thing of this then? That is 2 and the thing should be less than the roots. See, one root is greater than 2 and the other is smaller than 1. That means 1 and 2 lie between the roots. Ah, 1 and 2 lie between them. There's a unique thing of this then. See, first of all, root should be distinct, so d should be greater than 0. Secondly, f of 1, f of 2 must be positive. Yes, sir. Correct. But this is not sufficient. One root is greater than 1 and other is smaller than 1. Smaller than 1, yeah. Yeah. Because if you do this, this condition is valid even for f1 and 2 lying over here also, right? Yes, sir. Correct. So, this is not sufficient. So, I'll remove this. Instead of that, I have to say separately that f of 1 is less than 0 and f of 2 is also less than 0. f of 1 is 3, though. f of 1 is 3? Yes, sir. So, basically, such a situation is not possible? Yeah. Okay, because this itself is a null set? Yes, sir. A null set intersection with all the set will make it a null set? Null set, yeah. No such m exists. Oh, no such m exists. Okay. Yes, sir. Yes, sir. Okay, Venkat, we'll stop here. So, you have some 10 minutes? I have another class to go, so I have to go and have my lunch also. Oh, yes, sir, yes, sir. Okay. Is there any doubt that you have? Expertics.