 Hello, good evening. Good evening from Cinema Academy. Welcome to the YouTube live session on limits, continuity, and differentiability. So all those students who have joined in, please type in your names in the chat box so that I know who all are attending this session. So in this session, we are going to basically touch upon JEE main and advanced level problems on limits, continuity, differentiability, and methods of differentiation if time permits. So let's give a minute or more because it's just one minute more to four o'clock. But those who have joined in, I would request you to start working on the problems that you see. So here comes the first problem for you. Find the limit of X minus ln e to the power X minus plus e to the power minus X by two, X tending to infinity. And as you solve it, please type in your response on the chat box. Hey, everyone. Sai, Vaishnavi, Shweta, Sanjana, shares. Welcome all. So I'll wait for at least three people to respond before I start solving it. Okay, so Vaishnavi has given the response for the first question. Okay. So that's Purvik and Sai. Okay, great. So now since three people have responded, let me start giving the solution for this question. So first of all, what I'm going to do here is that from this expression, I'm going to take e to the power X common out. Okay, so first of all, I'm going to write it as ln. ln, let me take e to the power X common. So I'll have one plus e to the power minus two X by two. Okay. Which I can write it as limit X tending to infinity. X minus, I can separate it out by using the formula ln AB as ln A plus ln B. So it'll become X minus ln one plus e to the power minus two X by two. Okay. X, X will go for a toss. And we know that if X tends to infinity, okay, e to the power minus two X will tend to zero because it is as good as saying one by e square raised to the power of X. Okay. And one by e square has a magnitude which is lesser than one. Right. So such quantities as X tends to infinity will start going towards zero. So this is going to be zero. So what will happen? This term over here will get cancelled off and I'll be left with limit X tending to infinity minus of ln half. Okay. And that's going to be ln of two, which is nothing but log two to the base of e, which is option number three is correct. The first one to answer this was Vaishnavi. Vaishnavi, well done. Very good. And she answered this question well within one minute. That's good. That's a good speed. So guys, hope there's no problem, no question with respect to this. Let me move on to the next question now. Next is this question, question number two. Find the limit of e to the power X plus pi to the power X, whole h to the power one by X under curly brackets. And this curly brackets is basically representing the fractional part of X. I would request you all who are writing the response to these questions. Please first type the question number and then your response. Yeah. Anyone? Guys, a very simple question. We have many problems of a similar type. People who are joining in, please type in your names so that I know who are there in the session. Welcome, Sondarya, to the live session. These are all conceptual questions. I mean, Jay is trying to test off your understanding whether you know how to deal with the fractional part and how to take care of limits of these types. Basically, it's based on limit of this type, b to the power X, whole to the power one by X. And one of the limits here, one of the values here is greater than the other values. Let's say b, a is greater than b or b is greater than a. So how do you deal with such kind of limits? So for this case, I'll just quickly help you out so that you can actually solve this problem also. So here, whichever is greater, let's say, out of a and b, a is greater. So what we'll do is first we'll take a to the power X common out. So when you take a to the power X out, under the influence of one by X, we'll come out as a, right? And this will become this. Now remember, as X tends to infinity, of course, b by a is having a modulus less than one, right? And X is very, very large, so this term becomes zero. So it's as good as saying, you know, one plus something tending to zero, raise to a power which is almost very, very small. So one by infinity is also tending to zero, right? So here, this will actually become a, okay? In a similar way, I know that, forget about the fractional part first. I know that out of e and pi, pi is the greater of the two, correct? So what I will do is when I'm evaluating this, I will take pi out, right? And I'll have e by pi, raise to the power of X plus one, whole raise to the power of one by X. Again, this term over here will tend to zero, okay? So my answer will be pi, but my answer is going to be fractional part of pi. Since pi is approximately 3.14 something, your fractional part of this will be pi minus of three. So your option number two becomes the correct option over here. Is that clear, guys? Is it clear? I think it was just the lack of confidence that you could not solve it, as the concept was already conveyed to you during the classroom sessions. Great. So we'll now move on to the third problem for the day. If cos X by sine AX is a periodic function, is a periodic function, then find the value of this limit. Yes, guys. So these problems are just meant to be solved within a minute. It's like those touch and go problems. You know it or you don't know it. By the way, let me just complete the bracket over here. So I know most of you would be wondering about what is A over here. So see guys, cos of X, cos of X is a periodic function, right? It's period is two pi, okay? A sine of AX period will be two pi by, okay? So if cos X by sine X has to be periodic, right? That means the period of that will be LCM of two pi by two pi by. Okay, so this would be the period of cos X by sine AX, correct? Now if LCM has to exist, that means it has to be periodic, right? It clearly implies A must be a rational number, right? Because if A is irrational, what will happen? You will get one rational and one irrational and we cannot find an LCM of a rational and irrational quantity, right? So if A has to be irrational, then the function will not be periodic. That's what I'm trying to convey over here. So for it to be periodic, my A has to be a rational number, correct? So it has to be of the form P by Q. It has to be of the form P by Q form, okay? That means you are integers, Q is not equal to zero. Now, when N is very, very large, when N is very, very large, N factorial into pi A, okay? This will become an integral multiple of pi. This will be an integral multiple of pi, right? Because N factorial is a super large number and whatever is A, let's say it's of the form P by Q, Q will definitely appear on the numerator part of it. That means N factorial will definitely contain that Q. So this will ultimately become an integral multiple of pi, correct? So cos of N factorial pi A will either be plus or minus one, because we know that cos of any multiple of pi can either take a value of, I'll write it down, because cos of k pi can either be plus one or a minus one, isn't it? But the moment I raise it to the power of an even number, okay, right? And it doesn't matter how large is that even number. Then this will become one, isn't it? So ultimately, your answer is going to be one plus one, which is going to be two. So option number three is going to be correct. Guys, is it clear? Any question with respect to the solution? Is it fine? Just a simple analysis, nothing else. This is the analysis that Jay wants you to develop. It was not a question which was based on intense calculation. The moment it stuck to you that A is a rational number and the moment it stuck to you that n factorial pi A will be an integral multiple of pi and then things will become super easy, right? Great. So let us move on to the next question now. Question number four. So we have been given a function f of x as a piecewise function. So this is the definition of your f of x and there's another function g of x which is again a piecewise function and this is the definition of it. You have to find the limit of g of f of x as x tends to zero. Guys, just focus about zero point. What happens before zero? What happens after zero? Are the value equal? Just focus about that. All other things that you see around you may not be as important, okay? Vaishnavi says option four, okay? I acknowledge the response. What about others? Anyone else? Sai also thinks doesn't exist. Sondarya, Sanjana, Shweta, Vaishnavi, Purvek. Guys, back up, back up. Okay. Okay. So let me discuss this first. So since we are talking about finding the limit as x tends to zero, we'll take the case when your x is tending to zero plus. Now remember you are feeding x to f of x, correct? So we'll see that, which is the definition that the function will take as you're feeding zero plus? As you're feeding zero plus, basically you'll be talking about this, x by sine x, isn't it? I'll not write it over here. I'll just write f of g of x, okay? Now zero plus, this function is responsible for zero plus, isn't it? And we all know that x by sine x will be slightly higher than zero, right? So this will be almost one plus, yes or no? Because sine x is lesser than x in the neighborhood of zero, zero plus I would say. So x by sine x will be slightly higher than one. So when you're feeding this into your g of x function, that means your output is slightly greater than one, that means this function would be responsible for giving the output to you, correct? So this will be giving you one square minus two into one minus two, which if I'm not wrong is going to be minus of three, right? So I can say this result is going to be minus of three for us. Now let us talk about limit of this function as x tends to zero minus. At zero minus, this would be the one which is responsible. Now two minus zero, two minus a quantity which is slightly less than zero will actually be greater than two. So greater than two is basically going to be taken up by this part of g of x. So I should not write, I've written it in a ulta way. So this was g of f of x, not f of g of f x. Yeah, sorry about that, yeah. Now, so two minus of zero minus would be greater than two and that would be actually fed to x minus of five, correct? So that my answer is going to become two minus of five which is going to be minus of three. So left-hand limit and right-hand limit both are equal and the value is minus three. So the answer for this question will be option number two and not four contrary to what many people have stated. Okay, so unfortunately none of you got it right. So option number two is going to be the right answer. Guys, is this clear? So be very, very careful while dealing with such scenarios, okay? One plus and one minus can make a difference to the function where you are feeding it. Is that okay? Nevermind, you have a lot of chances today. So we'll move on to the next question. That's question number five for the day. If A n, B n are positive integers and A n plus root two B n is two plus root two to the power of n, then find the limit A n by B n as n tends to infinity. Just a hint actually think in lines of binomial expansion. Think as if when you binomally expand this, you get this expression. So can you somehow figure out what is your A n and what is your B n? Please remember they are positive integers. Anyone, any idea, any clue? Okay, guys, just now as I discussed with you, would you all agree with me if I say that if I just make this minus, this plus as a minus, can I say a minus sign will come over here? And you would have all realized this in binomial expansion that whenever you change the sign in between, you know, you can, you'll start realizing that alternate terms will start becoming negative. And those terms would be having a factor of root two in them. Okay, so this is just a hint for you so that you can actually find out what is your A n and B n. So once you know these two equations, you can figure out what is your A n and B n. First of all, you all agree with me on this, right? Is there anybody who thinks that how, how come I have written this expression? Great, so can you just figure out what is A n and B n? And see what is this ratio? Sure, sure, sure, take your time. Use this hint and I'm sure all of you know how to deal with when your n is becoming a very, very large value, okay? Three to the power n plus one by two, okay? But that's not there in the option also, Swetha. Let's wait for this to respond. I think many of you would be in the middle of your working. Okay, what we can do here is clearly, we can add these two expressions and you get your two A n as, or you can find your A n value as half of two plus root two to the power of n plus two minus root two to the power of n, correct? And when you subtract these two and divide by root two, you can get your B n as two plus root two to the power of n minus two minus root two to the power of n. And of course we'll have by two root two or one by two root two coming up over here, okay? Right, now when you take the ratio A n by B n, okay? You will get root two times two plus root two to the power of n plus two minus root two to the power of n divided by two plus root two to the power of n minus two minus root two to the power of n, okay? Now, the term which is higher, as I already told you in the previous problem as well, the term which is higher among these two, we should always take that common out because that is the one which will be tending to infinity. So when I take that out, that means if I take two plus root two to the power n out, I will get one plus two minus root two by two plus root two to the power of n, okay? And in the denominator also, I can take the same out, so two plus root two to the power of n and one minus two minus root two by two plus root two to the power of n, okay? So these two will get cancelled off and now, my dear students, please listen to this very, very carefully. Since this expression mod is less than one, right? What will happen to these two terms? What will happen to these two terms? They will become zero, right? Ultimately, what will be left? Ultimately, it will be left just with root two, so which makes option number one as the correct option over here. Is that fine? Is that fine? Shweta, Sanjana, Sai, Purvik, I don't think so. It was a difficult problem after you cracked this step because many of you would not be able to find, realize how would I get a n and b n terms. So the moment you are through with this step, I think it was just a simple task and of course, you all know how to deal with limit n tending to infinity cases, okay? So all clear about this? No question? Good. So we can now move on to the next question of the day. That's question number six. So limit n tending to infinity, root of n square plus n minus one, whole divided band. This root is only till n, okay? Not till minus of one. And this entire thing is raised to the power of two times under root n square plus n minus one. Again, this minus one is outside the root. So don't include it within the root. By the way, this is one by e square. There's a typo error over there. Yes, anyone? So guys, we all know first of all that this form is actually one to the power infinity form, right? Because if you evaluate this under root of n square plus n minus one by n, okay? When you divide both the numerator and denominator by n, you would realize you get n square plus n by n square minus one by n, right? This is actually under root of one plus one by n minus one by n. And as your n tends to infinity, okay? This term will become zero, this term will become zero, and you'll get one as your answer. So this part is becoming one. And of course, you know that as n tends to infinity, this part will be becoming infinity, right? So hope this is clear in everybody's mind that we are dealing with one to the power infinity kind of a problem, okay? The moment we know we are dealing one to the power infinity kind of problem, we know the formula for that. That is going to be e to the power limit n tending to infinity, this minus one. So we'll have under root of n square plus n minus one minus n by n. I'm directly jumping a step. Instead of doing minus one and taking the LCM and all, I'm just writing it like this. And multiply it with this expression, okay? Two under root of, okay? So just try to recall limit of this form, limit of f of x to the power of one by g of x. You can say one plus f of x to the power of g of x. The answer for this limit is e to the power limit x tending to a f of x by g of x, okay? So in light of that formula, I have written this expression. So here the first step that I'm going to do is I'm going to rationalize this part, okay? And how will I do that? n square plus n, this is minus of n plus one, okay? Divided by n. So what I'm going to do is I'm going to rationalize this by multiplying it with under root of n square plus n plus n plus one. So this part I can just make it as minus n plus one square, right? And if I expand it up in the numerator, remember this is all limit n tending to infinity and whole raise to the power of e. So if I expand this on the numerator, I'm going to get minus n minus one and to under root of n square plus n minus one by n, n square plus n plus n plus one, okay? Limit n tending to infinity, whole raise to the power of e, okay? Now the highest power occurring throughout is n square, right? So I will multiply and I'll divide both the numerator and denominator by n square, giving one n to this, one n to this, one n to this and one n to this. So that's going to become, correct me if I'm wrong, minus one, minus one by n, okay? Here it will become under root of one plus one by n, minus of one by n, okay? And here I will get rid of this n and within the under root I will again get one plus one by n and then again one plus one by n. So when n becomes very, very large terms like one by n will start becoming zero. So this will result into minus one and this will become minus one and this will become minus of, this will become two minus zero and in the denominator also I'll have two, correct? So this will be minus one and whole raised to the power of e. So e raised to the power of minus one, okay? So the limit of this as n tends to infinity raised over e. So option number two is going to be the right option over here. Again, none of you could answer this correctly. It's not four, it's not e to the power four, it's not e to the power minus two, it's e to the power minus one. Is that fine, guys? Okay, so now let's move on to the next question which is the seventh question. So we have a function in n which is given by limit x tending to zero one plus sine x by two, one plus sine x by two square, etc. Till one plus sine x by two to the power n whole raised to the power one by x then you have to find the limit of f of n as n tends to infinity. So I'll write down once again for you over here f of n is limit x tending to zero one plus sine x by two, one plus sine x by two square, till one plus sine x by two to the power n and this whole thing is raised to the power of one by x. And then you have to find out this expression, limit of f of n as n tends to infinity. So Vaishnavi has given the response Okay, Shweta and Sai have different opinions. Okay, so let's discuss this now. So again, if you see guys very, very clearly it is actually one to the power infinity form, right? Undoubtedly. So I can write this as e to the power limit. Okay, x tending to zero. If you start multiplying, you'd realize you get a one term. Okay, and you will have the sum of these all powers of sum of all these sine terms. Okay, sine x by two to the power n. Okay, and you'll have product two at a time. Okay, so product two at a time. Okay, and product three at a time dot dot dot. I'm not going to write all of them. And finally you'll have a minus one coming up because in the formula itself we have a minus one coming up. Okay, whole multiplied with one by x. So one by x will come out. So this one and this one will get cancelled, right? So you'll have e to the power limit extending to zero. Sine x by two, sine x by two to the power two, etc. Till sine x by two to the power n whole divided by x. Plus, now when you talk about terms which have two sine terms or three sine terms multiplied or four sine terms multiplied when you divide them by x you would realize that these terms will start becoming zero. Correct? Okay, and here you'll start getting the answer as e to the power half. So when you do sine x by two by x you get a half. When you do sine x by two squared divided by x you get one by two squared and so on till one by two to the power n. Correct? So this is your function of n. Right? Now what is the question asking you? Question asking you what is the limit of this expression as x tends to infinity. That means e to the power half e to the power half squared up till infinity. Which is going to be e to the power half divided by one minus half. That's going to be e to the power one. So that's e which is option number two is going to be correct in this case. So the first one to answer this correctly was Shweta. Well done, very good. So we'll move on to the next question. That's the eighth question over here. This should be a simple question. I think many of you should be able to answer this. Remember the limit is applied to n not to x. This is the nth root of e. So you can also write this as one minus x plus x e to the power one by n. Always to the power. Guys you should not be confused about the form of this because we all know the form of this is going to be one to the power infinity form. Because as n is very very large this will tend to one because it's almost e to the power zero. So it will be one minus x plus x which is one and this is infinity. So it is one to the power infinity form. So tell me the right answer now. So option number one. One more response I need. So we'll have e to the power this and we'll have one minus x plus x e to the power one by n minus one multiplied with n. One can get cancelled. Take an x out. If you take an x out it will become e to the power one by n minus one into n. Okay. We can write this as e to the power one by n minus one divided by one by n. Okay. Now this is very similar to that expression where we have learned that limit e to the power x minus one by x extending to zero that's going to be one. Okay. So this term is going to become one for us. So it's e to the power x. Okay. So option number one is definitely going to be the right answer. Simple question just to boost your confidence when you have few could not answer the previous one. Sorry. The previous to previous one. Next. So over here which is defined from R to R. It's a differentiable function not a different functions. So again excuse for the type over here. It's a differentiable function at x equal to zero satisfying f of zero is zero and f dash zero is one. Then find the value of this has to be zero over here. Yeah. Limit extending to zero one by x summation from n equal to one to infinity minus one to the power n f of x by n. This is the ninth question so you can just use that number for your response. Okay. So I'm getting response from Gaurav, Sai, Purvik. So let's start solving this. So first of all if I write it in a proper way it's limit extending to zero and it's a summation one to infinity minus one to the power n f of x by n divided by x. Okay. Now since f of zero is zero we can clearly say that it's zero by zero form. Right. So this is of zero by zero form and hence I can apply okay LH rule. So if I apply LH rule I will get limit extending to zero summation from one to infinity this will be minus one to the power n f dash x by n into one by n. Okay. And of course denominator will vanish off because the derivative of x is going to be one. Right. Now as you put zero it becomes summation from n equal to one to infinity minus one to the power f dash zero by n and f dash zero is given to us as one. Correct. So it's summation of minus one to the power n by n. Correct. Now if you look at this term carefully the terms are like this. Minus one by one plus one by two minus one by three up to infinity. Correct. Now most of us have already studied the Maclaurin Taylor expansion for ln one plus x which is x minus x square by two plus x cube by three minus x four by four and so on till infinity. Right. If you realize here when you put x as one when you put x as your one ln two becomes one minus one by two plus one by three etc. This is exactly negative of what is required. So this expression is nothing but negative of ln two. So option number two is going to be correct and I think the person to answer this correctly first person to answer this correctly was Saim Mir. Okay. Well done Saim. Very good. Again simple, simple things. So basically Lopital rule got tested your understanding of the Maclaurin Taylor expansion got tested over here. So it's not a difficult problem. You just have to apply your mind. J main is a very simple exam 220, 230 is very easy to score with ordinary effort. Next question. So a limit of extending to three pi by four one plus cube root of tan x by one minus there is a small error. This has to be two over it. So I will rewrite the question once again so that you are not confused. Extending to three pi by four one plus cube root of tan x by one minus two cos square x. There was a typo this two was missing so I just added it. Again a very very simple question. This is question number ten. Okay. How about others? Okay. So ten people are saying fourth option. Alright. So let's discuss this. So now as x is two three pi by four this is definitely going to be a zero by zero form. So this is going to be a zero by zero form problem. So I can apply Lopital. So when I apply Lopital it becomes one third of tan x to the power of minus two third into secant square x and down in the denominator I will get two sin two x. Correct. Okay. Now when you put the value of x as three pi by four it becomes this minus one to the power of minus two third. Okay. So secant square three pi by four by two sin three pi by two. So this is going to be one third. This is going to become one. This is going to become two by two and this is going to become a minus one. Sin three pi by four is going to be sin three pi by two is going to be minus one. So your answer will be minus one third which is option number four is correct. Number four is the correct answer. Again the first one to answer this correctly was Vaishnavi. Never mind Gaurav. Keep trying. Good to see you on this group. Next. Question number 11. So here is the function g of x given as one minus sorry x minus one to the power of n a log of cos log of this entire thing cos of x minus one to the power of m x lying between zero to two m and n are integers. This is a comma over n. m and n are integers m not equal to zero n is greater than zero and if the limit of g of x as x tends to one plus right is minus one then which of the following are the values for m and n. Not a difficult problem. We have already done many problems of the similar nature. So be accurate. Don't be in a hurry. Just a reminder, tomorrow you have a test at the center. Hope the schedule and the location when you have been shared with you. It is in HSR. Alright, so mostly Janta is going with the third option. So let's discuss this. So what is given to us is limit of x tending to one plus of x minus one to the power i log cos to the power m x minus one. This is minus of one. Right. So first of all, I would like to just do a simple activity over here. Let me replace x with one plus h and say h tending to zero. Okay. So this entire limit converts to limit h tending to zero. h to the power n by I can write it as m log of cos h. Okay. This answer is given to me as minus of one. Right. Since it is zero by zero I can apply Lopital over here. So when I apply Lopital I get n h to the power n minus one. And in the denominator I will get m times sin h by cos h. So I can write this as n h to the power n minus one into cos of h by m sin h. Okay. Now since I see a sin h and h is tending to zero what I will do is I will borrow a h from here and I will do this. Correct. Now this answer should be minus of one. Now guys remember few things. This is going to be one undoubtedly. Okay. This is going to be one undoubtedly. Correct. So what I am getting is I am getting n or limit h tending to zero n h to the power n minus one. Okay. By the way I will get a minus over here as well. So this minus will make it plus. So this will become plus. Now if this h exists on the numerator it will actually make your answer zero. And you don't want that to happen. So I can safely say that this h to the power n minus two is zero. And if it should not have been there that means n minus two should have been zero. That means clearly your n value is equal to two. Okay. Right. So only option number three says that but let's verify even further. If n is two you get two by m as your one so m also has to be two. So of course it restablishes the fact that option number three has to be correct in this case. So let us see who gives the right answer to this. Shweta is the first one again to give the right answer to this. Well done. Very good. Okay. Never mind. Keep trying. So let's move on to the next one. Again a limit problem. Question number 12. Okay. So I'll start getting the answer. Ramcharan. Okay. So mostly Janta is going with three, three, three, three. Okay. So time to discuss. I think this is an easy question many of you got this right. Again if this is finite, the best way to solve these kind of problems is Lopital. So when you know you put zero on the numerator everything becomes zero. So it's a perfect case for a Lopital to happen. So applying LH rule it becomes limit extending to zero. A e to the power ax minus e to the power x 1 by 2x. Again since the numerator is becoming zero you can again apply Lopital but for it to be applicable, Lopital to be applicable when you put x as zero in the numerator you should get the answer as zero. So when you put a zero over here you get A minus 1 minus 1 equal to zero. So that means A has to be 2. Okay. Now putting it back, so limit extending to zero two e to the power 2x minus e to the power x minus 1 by 2x. Again applying Lopital it gives you limit extending to zero four e to the power 2x minus e to the power x by 2 which becomes 4 minus 1 by 2 which is 3 by 2. So B has to be 3 by 2 and A has to be 2. Absolutely. Option number 3 becomes the right option. First one to answer this correctly was Ramcharan. Well done. Well done Ram. So let's move on to the next one. This was a very very easy question. So limit of x cubed by under root of A plus x times Bx minus sinx is equal to 1 extending to zero we have to find the value of A plus B. Guys tomorrow again I am reminding this those who have not received the message there is a test at ourHSR center from 9 to 4 p.m. 10 to 12 we will have the first test then there would be a one hour of lunch break from 12 to 1 and then from 1 to 4 you will have another test and both of these tests are of J main level I would request you all please do come and write it nobody should be missing out on these tests. If anybody is not available tomorrow because of some urgent reason or emergency please let me know now only on the group in fact you can respond who all are coming tomorrow for the test just write I am coming great great Purvik all must come see at home again the same you know I mean sense of relaxation comes when you are taking it at home okay so you should always take it at a place which is familiar to you but if you come early morning Sunday it will not take that much of a time I think anyways I will leave it up to you yes anyone side things option number 2 yes guys anyone else see again if you see this term as X sends to 0 this will just become root A so when you are applying Lopita rule do not unnecessarily complicate your function because even differentiation of such cases may take a lot of time so basically it's as good as finding the limit of this function now this is 0 by 0 form undoubted D right if I take this out I can definitely say 0 by 0 so since it is 0 by 0 form okay we can apply Lopita so when you apply Lopita we get limit limit 3 X square by B minus cos X okay again denominator is numerator is 0 but denominator also has to be 0 right so that means B has to be B minus cos 0 has to be 1 which means B sorry I am sorry it has to be 0 which means B has to be cos of 0 which is 1 so I can write this as limit X ending to 0 3 X square by 1 minus cos X okay and this answer is 1 so one thing is for sure that your B value is 1 now we all know that we know that limit of 1 minus cos X by X square as X ends to 0 this is equal to half correct so this part is half okay so it's actually reciprocal of half so it's going to be 1 by root A 3 divided by half and this is equal to 1 which implies 6 by root A is equal to 1 which means root A is equal to 6 which means A is equal to 36 so if A is 36 and B is 1 then A plus B has to be 37 which means option number 2 is going to be the right answer again the moral of the story is many of you would have wasted time applying product rule over here not required once you know that this limit is evaluatable that means you can always put the value of X over here just evaluate it and take it out don't apply Lopital rule on the entire problem so make your life as easy as possible while applying Lopital so moving on to the next question now that's question number 14 please let me tell you this is determinant this is determinant determinant of this expression you have taken ln X extending to infinity this is minus 5 then we have to find the value of alpha, beta and gamma ok so Purvik and Ram go with option number 3 ok so let me now discuss this since 3 people have already responded on this first of all let us evaluate this determinant itself let me expand it with respect to you know any one of the let's say expanded with respect to the first column so I get alpha by X and I have 1 by X square and plus 1 I will have beta minus gamma by X ok so ultimately my limit problem is limit extending to infinity X ln of alpha by X plus beta minus gamma by X right now when X tends to infinity ok I can see that this tends to infinity so this must tend to 0 actually then only my limit will exist isn't it because for limit to exist you should have any one of the 7 indeterminate forms right it should be either 0 by 0 infinity by infinity infinity minus infinity 0 to the power 0 infinity to the power 0 or tending to 1 to the power infinity or tending to 0 into infinity so these are the 7 indeterminate forms under which your limit falls correct so for that to happen you can see as X tends to infinity this will tend to 0 this will tend to 0 so it will be left with beta now beta definitely has to be 1 because if beta is not 1 then you will not get a 0 over here correct so clearly option number 2 is ruled out because beta has to be 1 no doubt about that right so now your limit can be written as limit extending to infinity X times ln this will become 1 plus alpha by X cube minus gamma by X here I can use my expansions if I want we all know the expansion of ln 1 plus X expansion is X minus X square by 2 plus X cube by 3 etc so this will become X times the same expression alpha by X cube minus gamma by X okay and you may have other expressions as well like minus half whole you can say minus X by 2 dot dot dot okay but as you can see here you realize that since X tends to infinity all these terms after this will start becoming 0 because you will have more X in the denominator as compared to X in the numerator only the first term here will be alpha by X square minus gamma and as X tends to infinity your answer will be minus of gamma gamma would be your final answer which is minus 5 so gamma has to be 5 so if gamma is 5 alpha is nowhere in the picture so alpha can be any real value so option number 4 is going to be correct in this case option number 4 is going to be correct in this case and none of you have answered this question correctly surprisingly I think it's the dominos effect one person says 3 and all of you follow the same answer okay so option number D is correct answer in this case guys you got your mistake let's now take up questions from continuity and differentiability so this is your 15th question so g of X is the composite function f of f of X and f of X is the piecewise function we have to find the number of points of discontinuity of this function so goharav says option 2 okay need two more responses okay purbik also goes with option 2 only okay first of all we should all be familiar with the fact that when we are talking about the continuity or differentiability of composite functions we also check at the point where your interior function is continuous or not right so the f of f of f of f of X will be discontinuous even at those points where f of X is discontinuous so first of all let us see where is f of X discontinuous okay so f of X to check the discontinuity points of f of X we will have to only look at a critical point which is your 2 okay so left hand limit of this function as X tends to 2 is 3 and right hand limit of the function as X tends to 2 is 1 since they are not equal that means the function is discontinuous so f of X is discontinuous at X equal to 2 okay now since f of X is been supplied as an input to f of f itself I have to see when is these two are equal to 2 also because at 2 this function outer function will also become discontinuous right so the outer f of X will also become discontinuous when the input is 2 right so I have to check when is this 2 when is 1 plus X and 3 minus X both are equal to 2 so even such point should be avoided that means X cannot be 1 and here X also same answer X cannot be 1 okay so for these points your f of f of X will be discontinuous okay so all together we have 2 points 1 and 2 where f of f of X is going to be discontinuous so option is option number 3 so there are 2 points we cannot feed 1 we cannot feed 2 because if we feed 1 what will happen the output of this will be 2 and again we cannot feed 2 to the outer function f of X so I think the first person to get this right again is Shweta guys these are simple concepts don't mess this up we have already discussed this ample number of times that for checking the discontinuity or non-differentiability of composite functions we have to take care of the continuity of both the functions involved the inner and the outer functions okay great so we will move on to the next question now that's question number 16 if this function is continuous at X equal to 0 then the value of A by B is just a small correction over here this is limit over here sorry that's fine question is correct Vaishnavi has given the first response yes guys back up back up I just got one response so far Ramcharan, Purvik Sai, Saunarya Gaurav so if it is continuous at X equal to 0 first of all limit must exist right so if limit must exist it must be 0 by 0 form right so when you put X as 0 you get 128 8 to the power 1 by 8 minus 2 and of course in the denominator you will get 32 to the power 1 by 5 minus 2 okay so both have to be 0 0 of course the numerator has is already 0 32 raise to the power 1 5th is 2 2 minus 2 is equal to 0 so numerator has to be 0 means 128 A is equal to 2 to the power 8 which is again 128 so A has to be 1 I'm sorry sorry sorry this is 256 not 128 so A has to be 2 so if A has to be 2 your function becomes 256 plus 2X to the power of 1 by 8 minus 2 divided by 32 plus BX to the power of 1 by 5 minus 2 so let's find the limit of this and equate it to f of 0 because it has been mentioned that the function is continuous at X equal to 0 so this is equal to limit of this function as X tends to 0 we can apply Lopital over here because it's 0 by 0 form right now so it's 18th 26 plus 2X to the power of minus 7 by 8 times 2 divided by 1 5th 32 plus BX to the power of minus 4 5th into B now when you put X as 0 in this case it will become 2 by 8th B by 5 this will become 32 to the power of this is 2 to the power minus 4 this will become 2 to the power of minus 7 and this will be f of 0 right now this is 2 to the power of minus 3 so I can find 5 by 64 into 2 32 so 5 by 32 B will be your f naught right so from here I can say your B is going to be 5 by 32 f naught so A by B would be what A by B will be 2 divided by 5 by 32 f naught which is just going to be 64 f naught by 5 so option number 3 is correct I thought it was a pretty simple question so only one person could answer this Vaishnavi well done Vaishnavi very good so without wasting much time let's move on to the next question which is your question number 17 for the day so if f of X is 1 minus cos of 1 minus cos X by 2 divided by 2 to the power m X to the power n when X is not equal to 0 and 1 when X is 0 and this function is continuous at X equal to 0 then find the value of 1 plus n yes anyone guys okay so on the other side option 3 I need two more responses okay Sai also goes with the same option alright so Purvik also says the same thing great so this function has to be continuous at 0 of course we know that the limit of the function as X tends to 0 should be equal to 1 so as X tends to 0 1 minus cos of 1 minus cos X by 2 by 2 to the power m X to the power n should be equal to the value of the function at 0 which is 1 right now the moment I see 1 minus cos theta you know this formula comes to my mind limit 1 minus cos X by X square X tending to 0 this is half right this formula runs in my mind so what I will do is I will multiply and divide the expression by this let me just write it little bit far so this is equal to 1 now the moment I do this I can make my life simple because this term will become half remember 1 minus cos X by 2 as X tends to 0 is becoming 0 hence I could use this particular formula don't start using it everywhere it has to meet the fact that whichever function is over here the same function should be here and the function must be tending to 0 as X tends to whatever okay so this simplifies to very very simple expression 1 minus cos X by 2 the whole square okay and we will have 2 to the power m X to the power n and we will have a 2 factor which I can take it on the other side if I want okay so I can make this limit as 2 correct now 1 minus cos X by 2 I can write it as 2 sin square X by 4 so it will become 4 sin 4 X by 4 okay whole divided by 2 to the power m X to the power n and this is equal to 2 now if you want your limit to exist there should be as many X's in the denominator as the power on the sin isn't it so n should be clearly 4 no doubt about it alright now when n is 4 let me just cancel this factor of 2 so you will have sorry this is 2 to the power m how can I cancel it now yeah so n is clearly 4 now this I can write it as 4 by 2 to the power m this is going to be 1 by 4 to the power of 4 okay this is going to be your 2 so we will have in the denominator 2 to the power this is 2 to the power of 6 so it's m plus 6 correct and this is going to be 2 correct which clearly implies 2 to the power minus of m plus 6 is going to be 2 to the power of 1 so minus of m plus 6 is going to be 1 which means m is going to be minus of 7 so if m is minus of 7 and n is 4 then m plus n will be minus 7 plus 4 which is minus 3 so minus 3 will become your answer which is option number 3 is correct in this case is that fine so the first one to answer this correctly was Sondarya very good moving on to the next question now looks to be a pretty big question so you have a piecewise function f of x defined as this and g of x defined as this where signum function denotes you all know the signum function and you have a function h of x which is given as f of x plus g of x and it is discontinuous at exactly one point then which of the following is not possible yes of course sorry Purvik that's correct ok Purvik thinks option 4 Purvik just a question what is the molecular weight of acetic acid just to Purvik and you know why I am asking this question to you yes guys so anybody else apart from Purvik who wants to give answer to this question so before we add these two functions it's better to redefine it without mod right for example if you look at the first function f of x you know that mod x changes its form at 0 ok so what I will do is first of all I will redefine this function let's say I take it 0 and from 0 to 1 ok so this will become minus 3 minus x in this interval and this becomes minus 3 plus x in this interval and again mod of 2 minus x changes its definition about 2 so from 1 to 2 and from 2 to infinity I will have a different definition so from 1 to 2 it will be a plus 2 minus x and from 2 to infinity it will be a plus x minus 2 and let me just now define g of x in a similar way so this minus doesn't have any meaning because it's under the mod so it's as good as 2 minus mod of x and again we have to redefine it at 0 so minus infinity to 0 let me write g of x over here so minus infinity to 0 this will behave as 2 minus x and 2 plus x from sorry 2 plus x here and 2 minus x over here from 0 to ok because any x in this interval is going to be positive so signum of that will be 1 so it will be minus b plus 1 I am writing it over here minus b plus 1 for x lying between 2 to infinity now when you add these two functions we have to be very careful while adding them so h of x is the addition of these two functions so let me take the smallest interval possible so minus infinity to 0 my function is going to behave as sum of these two which is going to be minus of 1 and from 0 to 1 it is going to behave again as minus of 1 so I can directly club and save from minus infinity to 1 my function is going to behave as minus 1 ok from 1 to 2 1 to 2 my function is going to behave as a plus 2 minus x and plus 2 minus x which is a plus 4 minus 2x a plus 4 minus 2x from 1 to 2 ok and from 2 to infinity and from 2 to infinity it will be the sum of these two so if I am not wrong it is going to be a minus b a minus b minus 1 plus x ok now if this function has to be discontinuous at exactly one point it could be discontinuous at 1 or 2 so any one of them it can be discontinuous right ok so let's say it is discontinuous at 1 if it has to be discontinuous at x equal to 1 left hand limit which is minus 1 should not match with a plus 4 minus 2 that means a should not be minus 3 it should be continuous at 2 so continuous at 2 means it should be continuous at x equal to 2 so this implies left hand limit at 2 which is a plus 4 minus 4 should match with a minus b minus 1 minus 2 so a cancels this will become 0 so b should be equal to oh this is plus so b should be equal to 1 so your first possibility is your a should be not minus 3 and b should be 1 or the other way round so if it is continuous at 1 a should be equal to minus 3 and if it is discontinuous at 2 b should not be 1 so these are the 2 set of possibilities that we can have so either a is minus 3 b is 1 or a is minus a is not minus 3 b is 1 or a is minus 3 b is not 1 so a is minus 3 b is 1 this is possible so this cannot be my right answer a is not minus 3 b is 1 this is also possible and if b is 1 a should not be minus 3 this is also possible but this simultaneously it is not possible that a should be minus 3 and b is 1 so this is not possible and hence it becomes my right answer because my question says which is not possible okay so option number 4 is the right option in this case and the first one to answer this was again the trick was to redefine the function alright and once you redefine the function things were quite easy after that great so we will move on to the next question now which is 19th question so here we have a question f of x be a continuous function defined from r to r satisfying f of 0 is 1 f of 2x minus f of x is x then the value of f of 3 okay so I have started getting so Ramchalan, Purvik all say option 3 okay guys I think you have guessed the function in this case it is actually easy also I think you have guessed the function you think the function is going to be x plus 1 right so f of 2x will actually become 2x plus 1 and this will also satisfy the inequality okay of course if I have to do this problem I will always guess the function so your answer is 4 definitely correct okay but let's say if I want to do this without guessing the question now without guessing the function how will I do that so let me begin with this criteria that f of 2x minus x is x f of 2x minus f of x is x okay if I replace your x with x by 2 then can I say this is equal to x by 2 again if I replace x with x by 2 I can say this is x by 4 correct if I keep on doing this if I keep on doing this till my let's say this term it will become x to the power 2 to the power okay now from here on if I start adding the function in this part if I start adding the function you will get f of x in fact alternate terms will start getting cancelled off you will get f of x minus f of x to the power 2n as x by 2 x by 4 all the way till x by 2 to the power n correct now if I take my n tending to infinity if I take my limit n tending to infinity what will happen will become f of x minus f of 0 because as n tends to infinity this will almost become 0 right and right hand side will become a geometric progression like this so f of x minus f of 0 is going to be x that means f of x is going to be x plus 1 the moment I get this I can always find the value of f of 3 which is 3 plus 1 which is going to be 4 3 is correct but again this is the longer route should not be followed in competitive exams but just to know how to get to the function you can use this ok so without wasting much time we will move on to the next question now question number 20 the number of points at which this function g of x is not differentiable given that your f of x is 1 by 1 plus 1 by x so I am getting a mixed response option 2, option 3, option 2 ok so let's discuss this not a difficult question actually so if you want your f of if you want your g of x to be differentiable first of all we have to see which is the point where the function f of x is not differentiable ok now as you can see clearly it is not differentiable when x is 0 correct and if you little bit simplify f of x it actually becomes x by x plus 1 so it is also not differentiable when x is minus of 1 so at this point f of x is dc and hence non-differentiable correct what about g of x now now g of x will definitely include those points where f of x is discontinuous so 2 is your answer so 2 points will definitely be there so now let us redefine this first so we will have 1 by 1 plus ok 2 by f of x so 2 by f of x ok now if you simplify this of course 0 will be a point which is already included over here if you simplify this it becomes x by 3x plus 2 so this will also be discontinuous at minus of 2 by 3 ok so this is the third point so one point is this another point is this where your g of x will become non-differentiable so total there will be 3 points so option number 3 is going to be correct and the first person to answer this correctly was Sondarya Sondarya gave the right answer to this so I just want to answer this correctly guys again you have not learnt from your past mistakes so be very very careful while dealing with such cases next moving on to the 21st question now sorry guys there was just a power cut can you see the screen now you can just hear my voice can't see the screen once again just once again and I am trying it out is it visible now ok just a second I will send you a new link I am stopping this broadcast now I will send you a new link ok