First question for the class said A has three elements said B has four elements the number of injective mappings that can be formed from A to B number of injective mappings that we can have from A to B injective means one one mappings. I'm putting the poll on. It's a one minute question. Yes, yes, I have some section B questions towards the end we'll take that up. Excellent. Most of you have are getting this right because I can see the right option getting the most number of votes. Yes, so we'll be stopping in the next 20 seconds. Number of one one mappings. Okay. Five, four, three, two, one. Go. Okay. So most of you have said option number C, which is definitely the right answer. Let's check this out. So please understand here. If your M happens to be less than N that means number of elements in B is lesser than the number of elements in a there will be 011 mapping. If M is equal or greater than and the number of mappings are MP and we had already discussed this. We had already discussed this. So our question B is for that means m is for and m is three so answer is for P3 for P3 is four factorial by four minus three factorial, which happens to be 24 by one. So option number C is correct. I would say both for cool application of derivatives and section B part both are equally important. At least for you section B part has got, you know, quite a lot of weightage. So please don't ignore either of the two. All right. Next question. Number of equivalent solution containing one comma two. Find the number of equivalent solution containing one comma two from a set having one, two, three. I'm relaunching the poll. Oh, okay. Okay. Okay. Yeah. See, I would not. I would not recommend you doing one or only one of the sections will be evaluated. If you feel you're not good in vectors and 3d at all up till now, then maybe you can give it a try. But if you know your concepts, well, don't dilute your preparation by going for multiple chapters. Yeah, you stick with section B stick with section B. Yes. These numbers are not that big one, two, three and four. Okay. So maybe you can figure out the number of equivalence relations by actually writing them now. Oh, you're not able to see the screen. Now you're able to see the screen. Okay. Okay. Yes. Let's discuss this five, four, three, two, one, go. Okay. So most of you have gone with option number B. Let's check whether B is right or no.  It has to be symmetric and it has to be transitive. Okay. Now, if I just write a relation like this, let's say one of the relation R1, I write it. See, it has to be reflexive first of all. So these elements must be there one, one, two, three, three. And apart from that, they also mentioned that one comma two also has to be there. Okay. So if one comma two is there, okay, I have to have two comma one as well. That's the minimum I can introduce to make it an equivalent solution. Okay. And you can check it is reflexive also. It is symmetric also, and it will be transitive also. Right. So as you can see, one, two, two, one is there. So one, one is already there. And similarly, two, one, one, two is there. So two, two is already there. So that is one relation. So one is definitely your answer. Okay. Let's try to introduce few more elements. Okay. So one, two is there. So two, one has to be there. Okay. Now let's try to introduce one, three. If I introduce one, three, please understand because of that, I have to introduce three, one also. Correct. And because of that, I have to introduce, I have to introduce two, three also. Am I right? Yes or no. Because of this two, three, I need to introduce three, two as well. Does it make sense? Yes or no. Yes. Anything that I'm missing. Anything that I'm missing, which will stop it from being reflexive. Sorry, stop it from being equivalent solution. Okay. Now if you see one, two, three, four, five, six, seven, eight, nine. If you use all the nine elements and maximum number of elements or maximum number of relations that this can have is the Cartesian product of A with itself, which is actually nine. Okay. Because R is a subset of A cross A and A has got three elements. Okay. So maximum is nine. Now, the minimum was this, the maximum was this. There's nothing intermediate here. Right. So there is only two possible equivalent solutions that you can write from set A to itself. Is it fine? Now people who answered with three or four. Is there anything that you could figure out that we have overlooked? If yes, please bring it to our notice. We would like to know it. Anything, anything that you feel we have overlooked. No, right. So what was the reason for mentioning three and the four? Okay. So it was only two equivalent solution. Let's take the next one. If you only read one word, yes, that is also an equivalent solution. One, one, two, two, three, three. That is also an equivalent solution. But in our question, we also had to introduce one comma two also. So that was creating an issue. Yeah. All right. So let's take this one. Very easy question. Two to the power X plus two to the power mod X. Is this function one, one and onto. Is this function many one and onto one, one and into many one and into just five of you have responded so far. Last 30 seconds. Last 30 seconds. Five. Three. Two. One. All right. So again, not a very convincing response. There was a very close call between option C and D. Okay. But still option C has got most number of votes. So many of you think it's one, one and into. Okay. See into is very obvious. Why? Because this is going to be a positive function. Undoubtedly.  In fact, the minimum value it can take is two. Okay. The minimum value it will take is actually a two. See this function two to the power X. That's two to the power mod X. It will behave as two to the power X plus one. If your X is greater than equal to zero. Okay. So if you see the minimum value here is two. And if you have your X as negative also. Right. Then we can easily say that. The terms are positive. So their arithmetic mean will be greater than equal to their geometric. Correct. So two to the power X was two to the power minus X will be greater than equal to two root one, which is two. So minimum value is two. And after that will go up for sure. Okay. So it cannot exhaust all the numbers in the code of it. So it has to be into this option and this option is gone. Okay. So either it will be C or it will be D. Okay. So people who have said B, I think five of you have said B. And one of you have said A that is not going to work out. Okay. Next. Now, whether it is one one or whether it is a many one. Okay. For that, we can, you know, do some derivative tests. So now here, please be careful that this function is defined in a piecewise manner. Okay. How would you sketch the graph? This side is fine. This side I can easily sketch. How would you sketch this side? Two to the power X plus one by two to the power X. Both are rising functions for sure. Okay. We'll have that's why I'm trying to show through calculus. Yeah. So if you see this function is definitely a function which will be rising like this. Okay. And this function, if you see this function also the derivative. Okay. The derivative of this is two to the power minus X into, into L and two with a minus sign. Okay. So what will it become? It will become. No, just let's, let's try to analyze this. Is this increasing or is it decreasing? Is it a positive quantity or is it a negative quantity? Let's try to understand this. Two to the power X. Two. I lent you if you take common. Okay. You will get something like this. Now, when your x is negative, when your X is negative, please, please remember this guy, this guy will have a higher value as compared to the other person. So this fellow will be more than two to the power X. Are you getting my point for the simple reason is that now this will be 2 to the power negative of a negative quantity means you will have a positive power on it and this 2 to the power this x will be a negative number. So overall this quantity will be negative. What does it means? It has a decreasing slope. That means the function will have a decreasing slope. So something like this will happen. I mean not exactly of the symmetrical shape but it will be a falling shape. If it is a falling shape any horizontal line is going to cut the graph at more than two points. That means it has to be many one function can't be one one. Okay. So derivative on this side will always be negative f dash x will always be negative derivative on this side is always positive. In other words this function derivative is showing both the characteristic of positive negative in the domain of the function and because of that the function is going to be a many one. Should we see the graph of it also? Let's try to check the graph of it to be very very sure of what we are doing. Let me open my GeoGebra y is equal to 2 to the power x plus 2 to the power of ab6. Almost what we predicted. So this is the graph of this function. Of course starting from 2 going upwards so cannot be onto and as you can see if you draw a horizontal line it's going to cut the graph at two points for sure. Okay. So option number option number c is the right option number d is the right option. It's a many one. It's a many one into function. Okay. How many of you got this right? Six of you got this right majority of you were wrong actually. What is happening? Don't give me shocks one week before the exam. Normally I'm a very cool person. Okay. This is a repetitive question maybe maybe. If a set A contains five elements set B contains six elements the number of one one and onto mappings. In fact, they're asking you how many subjective mappings are happening here. So let me. Sorry. How many bijective mappings are happening here? So I'm just launching the poll once again. Okay. This is what I'm scared for. Jumping into the conclusion too early. Both the habits should be avoided. One jumping too early to a conclusion and lurking with a problem for very long. Both are not good for examination on. So in your mind you have. Okay. I have to spend one minute maybe 30 seconds 45 seconds. You can just give yourself that amount of time. This answer is going to be just touch and go. Nothing I have to do. Okay. Should we stop the poll now? Five, four, three, two, one, go. Okay. Most of you have got it right. Option C is the right option. Okay. It's zero amount, zero bijective functions could be formed. Please remember this that in order to have a bijective function on discrete sets. So A and B are discrete sets. Discrete set means they have some fixed number of elements. Unlike continuous sets, the number of elements in A should be equal to the number of elements in B. Then only you can form a bijective functions. Okay. For bijections, this is the criteria. And the answer to this is same as the number of one one functions that you can form. That is factorial of NA or factorial of NB. But in our present situation, since these elements are not mapping, the answer will be zero. They cannot be a bijective function. Option C is correct. Why? Because if you give, if you do a one one mapping, one element of set B will be left. So it will not become onto at the same time. And if you want to do onto mapping, then what will happen? It will no longer remain a function because a particular element has to map to two elements in order to satisfy that it's going to be onto, but that will not remain a function because one to many mapping will happen. Okay. So in either of the case, there will be zero bijective functions formed. Next. f of x is sine square x and the composite function g of f of x is more sine x. What is g of x? Again, a very easy question. I'm relaunching the poll. All of you. Okay. So does it even require discussion? Should I end the poll? Okay. The answer is, the answer is under root of x. Absolutely. So under root of sine square, we all know it's more of sine x. Okay. So it has to be under root of x. Okay. No point discussing this. So we'll move on quickly. So some questions you'll find that easy also. Next one. How was the physics paper for the NAFLA's 12 IC people physics paper? Was it easy than what you were getting from the school or was it difficult? It was easier. Okay. Good to know that. Okay. Okay. No worries. Don't check the answer key during between the exam. Okay. Check it after your papers are over. Never ever try to see how much you're scoring in the, in the mid of your exams that has a lot of psychological impact to some of you, it will not make a difference. But to many of you, it would make a difference. So if you realize that in the previous exam, you're losing so much of marks. Normally we recommend this for J advance exams because on the same day two exams happen. So previous paper, you should not never try to find out the answers from anybody.  Then you sit and analyze. Maybe you'll be sad for a few hours, but that is fine. It will not going to impact your next day paper. Don't discuss it with your friends also. Many of you have the habit no question paper, you know, take and say, Hey, what, what do you got? What is this? Don't do this. Just put the question paper in your bag. Just get to home, eat, take a good nap and prepare for the next thing. Yeah, it depends on people to people can check. Yeah. So let it let the you can say uncertainty be there till the examination is over till all the exams are over.  Yes. So should we discuss this a five, four, three, two, one go. Okay. I can see most of you have got option A. Okay, let's check it out. See, f of f of x, that clearly means you're putting the same function in in place of x like this. Okay. And this is giving you an x. If you simplify this, you get something like this. All right. So let's collect your x terms together. So you'll have a square plus BC x, and you'll have AB plus BD. And here you will have AC plus C DC, AC plus DC x, and you have BC plus D square. Am I right? This is equal to anything I've left here, please do let me know in order to get an x. Please understand here, AB plus BD, this should be zero, and BC plus D square should be zero. Okay. So now, from here, I can say either B is zero, or A plus D is zero. Okay. B zero or A plus D is zero. Okay. Now, if this is zero, and apart from it, A square plus BC should be equal to AC plus DC, so that they cancel each other's out and leave a, leave a x behind. Am I right? Oh, sorry. This is BC plus D square. Correct. One second, one second. A small error from my side. Yeah, small error from my side. This should be zero. This should be zero. Okay. And this should be same as this, and this should be zero. Yeah. Sorry about that confusion. So AC plus DC should be zero. And A square plus BC should be BC plus D square. Okay. Okay. Then only what will happen? This term and this term will get cancelled off, and these two anyways be zero zero each, so they will not appear giving you an x. So from here, you end up getting BA plus D as zero. From here, you end up getting CA plus D is equal to zero. And from here, you end up getting A plus D, A minus D equal to zero. Okay. Now, something which is true in all the three is actually A plus D being equal to zero. Correct. Which is nothing but option number A, which says D is minus A. And B, C and D are not going to fulfill our conditions. So option number A is definitely going to be right. Yes, you can do that as well, Siddhant. Okay. Any questions, any concerns here? Any questions, any concerns here? Especially if this part is clear to everybody, why I chose these two to be zero zero, this to be zero and this to be zero. And why I chose these two to be the same to in order to get an x. Okay. Any questions, any concerns? All right. So we'll move on to the next question. Again, an easy question. You have already done this in our functions inverse topic. So I'll put the poll on. There's a function defined from the interval two comma three open to interval zero comma one open as x minus gif of x. F inverse is which of the following? Touch and go. 30 seconds is also too much for this question. Officially inverse is not there in the syllabus, right? Finding inverse of the function is not there, right? Okay. Okay. This one one on to a kind of a question would be there. Should I stop the poll because it is a super easy question. I can see half the class only has voted 20 more seconds I can give. Okay. Five, four, three, two, one, go. Okay. Option number D is the most chosen response that is actually the correct answer. See in this interval your gif of x can only give you a two. So your function is actually x minus two. Okay. So the question is asking you for the inverse of this function. So make x the subject of the formula. So this is actually your f inverse y. So f inverse x is x plus two, which is option number D. Okay. Easy. Should we go to the next one? What happens to the domain of f inverse? Okay. Let's go back. The domain of f inverse will be same as a range of this domain of f inverse is a range of the given functions. See this is going to be this functions domain is going to be zero to one only. Okay. So in the interval zero to one, this answer will be from two to three. Relation R. Oh, sorry. I didn't put the poll on relation R on the set Z defined by all ordered pairs a comma B says that a minus B is divisible by five divides the set Z into how many disjoint equivalence classes into how many disjoint equivalence classes. It's a very easy question. Don't get scared by the warnings of it. Done. Oh, only three people responded. Are you see in this question? I'll just explain this. Yeah, it's a touch and go question actually. If you see this equivalence classes, I hope you are all of you are aware of the equivalence classes concept. So to count the number of equivalence classes, you can only have equivalence classes. Let's say the number zero is there as a part of how many equivalence classes will be there for zero. So you can have, okay, like this.  So this will be an equivalence class. Okay, let's say zero. Let's say equivalence class of one, that will also be separate. If I'm not mistaken. Sorry, this is minus four, minus four, minus nine, and so on. And this will be okay. So this is another equivalence class. So you can say this is the equivalence class of zero. This is the equivalence class of one. Similarly, there would be an equivalence class of two, three, four, and they will all be disjoint from each other. Okay, I'm not writing them down, but similarly, there will be classes for these three also. Equivalence class for five will match with equivalence class for zero. Correct. Are you getting my point? Equivalence class for six will match with equivalence class for one. Okay, so if you write the equivalence class for all the integers, there will be only five disjoint equivalence classes. There will be only five. These five will be disjoint. After this, there will be repetitions or after this, the equivalence classes will have some elements in common. In fact, there will be many elements in common, because they will have, they will become the same sets. Okay, so the answer is only five disjoint equivalence classes can happen. Okay, so if you start writing the equivalence class for all the integers, there will be only five distinct or disjoint equivalence classes coming. After five, you will realize, sorry, fifth one will match with zero equivalence class. Six equivalence class will match with the first equivalence class and so on and so forth. If the set was defined for, no, but if the set is defined for all real numbers, then I mean, for which element would you write? So they would be, in that case, they would be infinitely many. Yes, I can check that you are right. They would be infinitely many. Is it fine? Any questions, any concerns here? Next. I'm relaunching the poll. So function f is from a rational number to rational number defined as 2x, q is x plus 2, find the value of fog inverse 20. Simple question, touch and go. Very good. Most of you have got it right. Should we stop the poll in the next 15 seconds? Five, four, three, two, one. All right. So D is the most chosen response. It's very obvious. See, you can use this property G inverse, OF inverse. Okay, also because the inverse follows the law of reversal. Okay. Yeah. So now see, this is G inverse of f inverse of 20. f inverse here would be x by 2 and G inverse here is x minus 2. So f inverse 20 is going to be 20 by 2, which is 10. And G inverse of 10 is going to be x minus 2, which is going to be option number. D is correct. Is it fine? Any questions? 2x plus sine x, x from r to r, is which type of function? 1, 1 on 2, 1, 1 but not on 2. On 2 but not 1, 1. Neither 1, 1 nor on 2. Yeah. Yes. Should we stop this in the next 30 seconds? Okay. Five, four, three, two, one. All right. So this answer is very obvious. It is actually a 1, 1 and on 2 function. It is a 1, 1 function because if you differentiate it, you'll end up getting 2 plus cos x, which is always positive. So it's always an increasing function. Okay. And if you see, because of this 2x, without, because of this 2x, it will be able to achieve all the real values. Okay. So it has to be on 2 as well. If I'm not mistaken, I have shown you the graph of it sometimes. I mean, it goes like this actually, slide and like this it'll go. I mean, it appears to be flat, but it is not actually. So I'll just show it how it looks like in the graph. Y is equal to 2x plus sine x. Yeah. This is how it goes. Next one. All right. So I'm asking you a range question because it may be required for you in your exam. What is the range of this function? 4 is on. All right. I can give you 30 more seconds. Only eight of you have responded. Supposedly a class 11th question, finding the range. So it's a range of a rational function. Okay. Five, four, three, two, one. Go. Only 12 of you have responded and out of which 10 of you say option C, let's check. So this function is clearly one plus one by x square plus x plus one. Correct. Now here you don't have to, you know, do anything very special. This is actually this. Okay. Now this is a constant. This is not going to change. Only thing that is going to change is this guy. Okay. So what is the maximum it can go? The maximum it can go is decided by what is the minimum this can go. The minimum this can go is three by four. So overall the max value of this is going to be four by three. That means overall the max value of the function. Okay. That is, that is going to become seven by three and that is achievable. So this is inclusive. Okay. Now the minimum value of this will happen when your denominator becomes very, very large. So this tends to one, oh, sorry. This tends to zero. Okay. So when this guy becomes very, very large, infinitely big, this entire term will be tending to zero. So what will happen? The minimum value of this entire expression will be tending to zero plus one, which is tending to one. So it'll be open interval one to close interval seven by three. That will be the range of the function. So option number C is definitely right. Okay. So maybe helpful in your onto type of questions where they may ask you find the, or suggest a codomain of the function says that so-and-so function is onto. So in this case, if you have to suggest the codomain, you could suggest the codomain to be between one to seven by three inclusive of seven by three exclusive of one. Is it fine? Any questions? Any questions? Any concerns? Should we take the next one? Okay. Some inverse trick. Just four of you have responded so far. Last 30 seconds. Five, four, three, two, one. All right. Most of you have gone with option number C. Okay. So let's use this formula that tan of tan inverse P plus tan inverse Q is P plus Q by one minus PQ. Okay.  So tan of this expression is equal to tan of five by four, which is going to be one. But mind you, this may come with an extraneous roots. This may give you more roots than required. But we can always figure it out from the original equation, which of them are extraneous and which of them are desired. So as per this formula, this will become P plus Q by one minus PQ. Oh, sorry. One minus PQ. This was I think plus one. I wrote it plus two. And this is simple.  This is fine. So this is as good as a two X times X plus two. And this is going to give you X square, X square is going to get cancelled off. So you'll have a four X and four plus one. It's going to give you a five. Okay. In short, you get two X square plus four X as four X plus five. That means X is equal to plus minus and root five by two. Now don't waste your time checking which of the roots are extraneous because there is no separate root five by two. There's no separate minus root five by two. They have actually given them in the option C. None of the options are actually related to root five by two. So don't waste your time checking which is the extraneous, which is not because as for the question, option C will be the right answer. Okay. So even if let's say the question setter has not bothered to check which is it may happen, especially in school exams, it may happen that the question setter may not bother to check whether there's an extraneous root which has kept into the system because of the first step that you have done.  Don't be like, oh, I have learned it in my JEE classes that we should check and all. Okay. Keep it for JEE exams. In school exams, if it is not required, don't waste your time. Time is marks. Simple question. Sign inverse of this is a multiple of pipe. Seems to be a silly question actually. Anyways, let's complete this. Come on. People are getting this wrong. Unbelievable. No issues are up enough. Okay. I've noted down your response. If you're unable to see the poll anybody, please leave your response in the chat box. Okay. Five, four, three, two, one, go. Okay. 89% of you got it right. And that 89 is like just 1% less. So it's very obvious that sign inverse happens to lie between sign inverse of any distinct input. Of course, that should be a valid input between minus one to one. It lies between minus pi by two to pi by two. The only integral multiple of pi which lies between this is zero. Okay. So this has to be a zero. That means x square minus seven x plus 12 has to be zero. So this is factorizable as x minus three x minus four. So x could either be three and four, but three is not there in your answer. So only four is the legitimate answers. Option B is right. Tan of two tan inverse one by five minus pi by four. Tan of two tan inverse one by five minus pi by four. Easy question, but I have figured out many people get muddled up in this. They do a lot of mistakes in these type of simple questions also. So hence let's try this out. These are normal brackets, not GIFs, normal brackets. Let it be inside. What is the problem? Having a pi by four, tan operating on something, some angle minus pi by four, that's absolutely fine. So how is your typical day going like? So what are you doing? I mean, I'm assuming you're not going to school. You have some study holidays given to you, right? Yeah. So Archit, how is your typical day going like? Oh, you had school till today. Oh, okay. Okay, guys, please make full use of your time. Don't waste even a single minute. Eat, sleep, study, repeat. Try to make notes and take a test on one top. That's a good strategy, actually. Hey, hi, Raghav. How was your physics paper? Great. Good to know that. All right. So should we discuss this? Okay, five, four, three, two, one, go. Okay. Okay, Pranam. So 12 of you have responded only to this question. Out of which nine of you say option number C. All right. See, first of all, two tan inverse one by five, let's try to write it as tan inverse something and that something will be two into one by five minus one by five square. So that's going to give you tan inverse two by five divided by 24 by 25. If I'm not mistaken, that should give you tan inverse of five by 12. So now you have something like tan of, now this angle, let's say I call it as five. You have something like tan of five minus five by four. How to solve half? Yeah, you have to, if you have a half inside, yeah, that also I'll tell you. I'll come to that question. Okay. Yeah. So coming back to this question, so you have this expression to evaluate as per the question. Am I right? So as per our form, latest tan five by four, sorry, tan five minus tan five by four, which is one by one plus tan five. Okay. And tan five is going to be from here, five by five by 12. So put it over here, five by 12 minus one minus one upon one plus five by 12. So that's going to give you a negative seven by 17 option number C. Okay. Now, Prakul has a question. He is asking, how do you get tan of half tan inverse x? Okay. How do you find this? So this is going to see, let's say I take some value. I'll just take some value over here. Let's say you have been given something like half. Okay. So let's say tan inverse half is theta. Okay. That means tan theta is equal to half. And here this question that is asking you tan theta by two. Okay. So here using this expression, you can put two tan theta by two by one minus tan square theta by two. Okay. I'm just taking some dummy examples. Now here solve for solve for tan theta by two. Now, please, please make note over here that you need to be careful that this lies in which interval. So this is definitely be an, this will definitely be an angle in the first quadrant. Yes or no. So your answer has to be a positive one. So accept the one which is positive. So from this quadratic, accept the answer, which is going to be a positive answer. This is quadratic in tan theta by two. So while you're using your Sridharacharya formula or quadratic equation formula, accept the one which is positive. Is it fine? Now, does it answer that question of yours? Let's move on to the next one. Again, easy question based on inverse trick. Find the value of cost of tan inverse of tan of 15 pi by four. Many a times I feel that I should give you certain questions which are of slightly higher than board exam. But then I also get scared that many of you may not be very sound with the basics. So that purpose of revision may get lost if I give you slightly higher. So that's why some easy questions also have picked up so that everybody is clear with the basics. See, there is no such question where I'm getting 100% votes. That means somewhere somebody is doing a mistake. Sorry, Aditya. I could not read that message. You were talking about the previous question. Yeah, not necessarily. But again, you have to take a call. You have to take a call. Which of them would be depending upon the situation?  And one of the answer was lesser than one and other was more than one. Then you would always choose the one which is more than one. So it depends upon what is the nature of the angle that you're dealing with. Okay, 19 of you have already responded. Should I stop the poll now in the next 15 seconds? Okay, 5, 4, 3, 2, 1. Okay, 16 of you say option C. See, the simple way to evaluate this question is if at all this angle is such that you can find tan 15 pi by 4, then go for it. So 15 pi by 4 is what? 15 pi by 4 is 4 pi minus pi by 4. 4 pi minus 5 by 4 means you are in the fourth quadrant. So tan of this such an angle would be, first of all, sine will be negative and magnitude will be tan pi by 4, which is negative 1. So you are finding cos of tan inverse of negative 1. Tan inverse negative 1 is cos minus pi by 4. Okay, this is minus pi by 4. But cos is unaffected by negativity. So it is as good as cos pi by 4, which is option number C. And I could see 5 of you voting for A. That means there is some issue in understanding these concepts. Okay, first of all, please get this clear. Tan inverse tan theta is not theta. Tan inverse tan theta is not theta. If your theta is beyond the principal value branch, principal value branch for tan inverse is negative pi by 2 to pi by 2 open. 15 pi by 4 is much beyond it. See, tan of tan inverse x, that is x, agreed. Whatever is your x. Tan of tan inverse x is x always. But tan inverse of tan theta is not always theta. Right? When it is not theta, when theta is beyond the principal value branch. So in such cases, I would always recommend you to either follow the graph of tan inverse tan theta if at all your angle is not a very familiar one. But here, 15 pi by 4 is a very familiar angle. It's an allied angle. So you can easily figure out the value of tan 15 pi by 4 and then find tan inverse and then find cos of that angle. Okay. But those 5 of you who are making mistakes, please be careful. Okay. Let's not lose marks to these simple questions. Yeah. Sin inverse x is tan inverse y. What is the value of 1 by x square minus 1 by y square? Study properties of inverse stick for boards. Even properties of inverse stick has been deleted from your syllabus. That is what you're trying to say, Charan. So what property are you using here? You're just evaluating it. See, omission deletion is a very, very, I would say dicey thing. Okay. Even many times these resource people, they are also confused. If it is not been asked good, but I would not take a chance with the basic ones at least. At least I should know the basic formula. If not J level, I should know the school level, NC, RD prescribed this thing. Omission and addition, et cetera, they become very subjective at times and it'll do a long-term damage. Exactly, Harian. That's the issue. When you omit it from one topic and you require it in the next, then that becomes very, very tricky. Of course, there will be questions which may require it in some other topic, but if it does come, there might be a quote contention on this. People will contend this. Okay. This was not in the syllabus. How was it asked? Et cetera, et cetera. All right. So most of you have voted now. I think one of the options have, but should we discuss it? Any questions? Okay. Five, four, three, two, one, go. All right. So this is a simple concept. Most of you have got option number A as their choice. So let's say this, each of these angle is equal to theta. If this each angle is equal to theta at the end of the day, it's a manual, right? So X is sine theta, Y is tan theta. What are they asking you? One by X square minus one by Y square. So six square theta minus quad square theta. What's the answer one? Option number A. Easy question. This is a null matrix actually looks like zero, but it's a null. Okay. Should we close the poll now? Five, four, three, two, one, go. Okay. Again, 12 of you responded. Most of you are saying option number A. Okay. So let's expand this. So this will be A square. A into I is A. Two I into A is minus two A. So it's going to be plus A minus two A. And minus two I, this is equal to null set. So A square minus A minus two I is equal to a null set. Okay. Let's do a pre facto multiplication with A inverse. If at all A inverse exists. Okay. They should provide the fact that A inverse here exists. So A is an invertible matrix or non-singular square matrix. So that will give you something like this. Okay. So from here, you can say two A inverse is A minus I. So A inverse is going to be A minus I. Easy question. We will move on. I hope I have not done this question before. Anybody recalls having done this question before? I hope not. Similar one we have done. Okay. One of the DPPs. Okay. Seven of you have responded so far. I can give one more minute. Okay. Five, four, three, two, one, go. All right. So end of poll. 13 of you have voted and most of you have gone with option number, sorry, 14 of you have voted and most of you have gone with option number D. Okay. Let's discuss. So first of all, this property will be valid.  And now you're claiming it to be two AB. That will only happen when AB and BA are equal to each other. Okay. That means the matrix operation is commutative. So AB is BA. So let's compare a few elements. Let's see what happens. So here if you multiply, you get A minus B. Okay. And here if you multiply, you get A plus two. That means B is negative two. That means option number C and option number B are ruled out. Okay. Next, don't waste your time writing all the elements. See which elements are going to be helpful. Now let me multiply this row with this. So you get two A plus B. So this with this will give you B minus two. So A is negative one. Two A plus B is equal to, so A is negative one. Oh yeah. A is negative one. And B is negative two. Only option number D is right. A is wrong. So D ups and most of you have got it right. Very well done. Any questions, any concerns? So a couple of things in matrix based question, one important advice here is that don't sit and multiply everything. Just pick which multiplication is going to help you get the answer. In fact, even if you get one equation, try to see which option match it so that you eliminate your options. Next, A is this, B is this. M is A, B. Find the value of, find the value of M inverse. They may, they could have, they could have given you more where you can see us. But the problem is we'll also have to check the number of linearly independent equations. Many a times we get the same equation in multiple shapes. So that is not considered to be two unique equations. So the question while setting the question has to watch out for the fact that the student is getting unique or linearly independent equations to work with. Oh, I'm sorry. I think I did not put the poll on. I'm so sorry. Yeah. Okay. Should we stop now? Next 15, 20 seconds. Five, four, three, two, one, go. Okay. Most of you have gone with option number B, B for Bangalore. Let's check. So what is A, B first of all?  This into this will be zero, zero, two, zero minus two, and two, zero, zero. Am I right? This is what you will be getting for A, B. I hope I have not done any simplification error. Just check. So inverse of such a matrix, as I already told you, you just have to divide by the determinant which in this case is going to be six, two minus minus four is one, six, and you just flip the position of the leading diagonal and change the sign over here. That's it. So your answer is going to be one third minus one third, one third, one sixth, option number B is correct. So for a two by two adjoint is obtained by flipping the leading diagonal element positions and just changing the sign here. Okay. Next. If the matrix is singular, find the value of lambda. Again, a very simple question just to remind everybody what is a singular matrix? That's it. I know this is not a challenging question. You can easily do it. People making mistakes here also. Okay. 20 more seconds. Should we wrap this up? Five, four, three, two, one, go. Okay. You are the culprit. All right. So end of point. So again, a mixed response. I mean, not a mixed response, I would say, but he has six of you have given non B as your answer. Let's check. So singular matrix is a matrix whose determinant is equal to zero. Singular matrix means this does not exist. A inverse does not exist. Okay. So determinant of this will be 40 minus 40, which is zero minus three, 20 minus 24, which is minus four lambda plus two, 10 minus 12, which is minus two. So lambda plus six, lambda plus six is going to be, sorry, lambda plus two is going to be six, lambda is four. Option number B is correct. If ABCR cube roots of unity, then this determinant is equal to which of the following ABCR cube roots of unity minus one is down e to the power three a then down minus one. Got it. And here, of course, it is a three by three. This is a part of this terminal. Okay. Okay. So this is basically the scenario. These are the ABC. Anyways, I'll wait for a few more seconds. e to the power omega will be a complex sum of see what is omega omega is minus half plus i root three by two, right? So it is e to the power minus half into e to the power i root three by two. Okay. So it's a complex number whose modulus is this and argument is root three by two. So it is going to be cos root three by two plus i sine root three by two. Why you want to differentiate it as something? Yeah, it's a complex constant. It's a constant. Okay. Should we discuss it now? Five, four, three, two, one, go. Just nine of you voted out of which eight of you say option number eight. Okay. This is the result. Okay. Let's let's discuss it. See, first of all, we will split this determinant as two separate determinants e to the power a e to the power b e to the power c e to the power two a e to the power two b e to the power two c e to the power three a e to the power three b e to the power three c and you have negative of correct me if I'm wrong. It will be e to the power a e to the power b e to the power c e to the power two a e to the power two b e to the power two c one one one. Okay. Just take take e to the power a common from row number one here. Take e to the power b common from no number B. Okay, by the way, it will become something like this. If I do need not write it. Okay, so this will be 111. This will lose one of the powers. So let me write it as A, B, C, 2A, 2B, 2C. Anyways, this is going to be 1 plus omega plus omega square kind of a thing, right? So this whole thing is going to be a zero. That means this whole thing is going to be a one. So you can safely ignore this and just do a simple activity, switch the position of these two and switch the position of one in this. Oh, that means you can easily bring this 111 over here. Okay, e to the power A and this is e to the power 2A. That means twice you are switching the columns here and the determinant is not getting changed because of that. So overall, everything will become a zero. Is it fine? Any questions? Any questions? Any questions, any concerns? Can you explain the separation once again? Oh, yes, why not? So if you recall, there was a property that if a particular row or a column is written as a sum of two elements, what you can do, you can take this, this and this together and then finally this, this and the last one. That's what I did. See, it's something like this. I'll just explain again. ABC, PQR and let's say if it is X plus T, Y plus U, Z plus W, you can break it like this. ABC, PQR, XYZ and again ABC, PQR and TW. This is the property which I utilized. Okay. And then in the last determinant, I stopped the position of the two columns twice. So even a number of times if you're doing the stopping, you'll get back the same determinant and that is what happened here. Should we go to the next question? If the system of linear equations has a non-zero solution, then ABC satisfy which of the following? ABC are non-zero numbers here. Hello? Am I audible? Can you all see the poll? I mean, I just ended it. Oh, one second. Now can you see the poll? Because nobody answered. That's why I was wondering. Okay. All right. So let's discuss it. Meanwhile, please give your response on the poll as well. So we already know that if a non-trivial solution has to exist, this determinant should be zero. So you can take the last or you can take the first row and just subtract it from the other rows. So that will give you 1, 2A, 0, 3B minus 2A, A minus B. And again, okay, before that, why don't we do this operation? Why don't we create a zero here? Okay. Let's let me do one more operation before I go to this operation. So first I will do this operation. I will try to do C2 as C2 minus 2C3. So 1, 1, 1, this will be 0A. Okay. And this will be B, B. And this will be 2C. C. Yeah. Now let us do this operation. So take the first row and subtract it. So 0, B, A minus B, again as 0, 2C, C minus A. Okay. Now let us expand this with respect to the first column. So it will give you B times C minus A minus A minus B times 2C. Okay. That is equal to 0 as per the question. So BC minus AB minus 2AC plus 2BC is equal to 0. Is it fine? Anything that I'm missing out? Do highlight. Oh yeah, this is B minus. So I think this will become minus and this will become plus. Thanks a lot. So this is going to become 2AC is equal to BC plus AB divided by ABC divided by ABC. So when you do that, you get 2 by B is equal to 1 by A plus 1 by C, which is definitely indicating that ABC are in, ABC are in HP, which is option number, option number D. Is it fine? Any questions? Let's move on to the next one. The value of this determinant is which of the following. Again, this doesn't require you to actually solve the question. Just think carefully what will help you to solve the question and then proceed. Yes. Should we discuss it? Should we end the poll now in the next 15 seconds? See, in this kind of a question, many a times none of these is not an option. But even if you want to be very sure, most of you have gone with C by the way, if you want to be very sure, you can take two different scenarios and then see which of ABC is basically giving you the options. For example here, I would first try to see whether 0 is actually my answer or not by taking ABC, all of them as 1 each. So when I do that, I end up getting something like this. And when I expand it, I'll end up getting minus 1, 1 minus 1, which is 0, again a minus 1, minus 1, minus 1, which is minus 2, and then a 1, 1 minus 1, which is a 2. So answer is a 4. So 0 is out of the question. ABC is also out of the question, but yes, this is still in the game. But there is something like none of these also sitting over it. So most of the time, none of these is not the right option unless until you have a bad day running there. And so in such cases, even if you had to find it out by putting one more set of substitution, let's say you can try for A as 1, maybe B as a minus 1 and C as a 1, maybe. Let's just check whether it is going to again satisfy 4A square B square C square. Try it out. So this is minus 1. This is again a minus 1. This is 1. This is minus 1. This is minus 1. This is minus 1. This is 1. This is minus 1 and this is minus 1. Let's try this out. So this is minus 1 into 0. This is going to cancel each other out. Plus 1 and 1 plus 1, which is 2. And again, 1, 1 minus 1, which is 2, again a 4. So both the times you realize that C is our right option. But again, it could be a matter of coincidence that again, C is matching with this is it. Again, it's a matter of coincidence. Probability is still there. Is it like that? Yes. I mean, even I have got to hear that the board is not going to give you none. Did you get any none of these kind of a situation in your previous exams? No, okay. Those options will be very, I mean, I will not rule it out. But even in J main, they avoid even none of these as the option. Okay, anyways, even if you have to solve this properly, what I will do is I will take out, let's say a common form column number one. Okay, so if I take a common from column number one, it loses out, I'll just write it down once again. First of all, so we have minus a, b, c, you still have a, b, a, c minus b square, b, c, b, c minus c square. So again, take a common from column row number one. So that will be a square. Okay, take b common from column number two, and b common from column number row number one. Okay, so where there's nothing, there's a one coming up. Okay, again, take c common from this and c common from this. So that'll give you one, one, minus one, one. Okay, so b square, c square will come. And if you evaluate this, this has to come out to be, this will be zero anyhow, minus one, this will be minus, minus two, and plus one, one, one, minus one, which is two, yeah, four a square, b square, c square. So in case there is a none of these option, you, instead of trying out two different modes, you can also evaluate it. This is not a difficult problem for you to evaluate. Yeah, because I want to make situations difficult now. Okay. Let's take this question. Why is e to the power x plus e to the power x plus e to the power x and so on till infinity, then divide by dx is which of the following. Simple basic differentiation function, maybe you would have done it in the very first or second class of differentiation. Very good. Six of you have responded. Same option. Okay, should we solve the poll now in the next 15 seconds? Five, four, three, one, 21 people responded to this out of which something wrong with the poll, yeah, out of which 17 of you say option number c. So this is very simple. Here there is a kind of an infinite series. So you just write y as e to the power. Now if you see from here on the series is repeated, from here on the series is repeated, isn't it? So this is back to y. Okay, so differentiate it once again with respect to x, differentiate it with respect to x. So this will become something like this. And this is back to y again. Is it fine? So it will be dy by dx 1 minus y is equal to y. So dy by dx is equal to y by 1 minus y. Option number c is correct. If you're aware that you are missing out on the sign, then the chances of you getting it wrong is rare. Okay, after this question, we'll take a break. Minus y by 1 minus y. One second, give me a second. Pranav is saying he used log on both the sides and he got a different result. Okay, if you'd use log Pranav, what will happen? It will become ln y is equal to x plus y. Okay, so when you differentiate it, this is what you'll get. Bring it to one side. You'll have this. Yeah, so this is back to dy by dx as y by 1 minus y. Check your working once again. All right, next. The number of points in which this function is discontinuous. The number of points in which this function is discontinuous. Yeah, touching your question. Any catch here? You should tell me. Done. Should we stop the poll? Five, four, three, two, one, go. It's obvious. Answer D is correct. So whenever x becomes an integer, this will start giving you zero. Okay, so none of these is the answer. They'll be infinitely many points. We'll take a break right now. Right now, the time is 6.4. We'll meet exactly at 6.15 is fine after a 10, 11 minute break. Okay. See you on the other side of the break. So let's take up this question as our next question. f of x is given to be a piecewise function. g of x is e to the power x. Find g o f derivative at zero. Find g o f derivative at zero. Let me on the poll. Okay, 30 more seconds and then we'll discuss it out. Okay, five, four, three, two, one. So I could see two major responses coming up here, which is option A and D. So some people say it's option A. Some people say option D. None of these. Let us check it out. First of all, what is g o f? So when x is greater than equal to zero, g o f is e to the power sin x. And when x is less than equal to zero, in fact, we could add zero, both of them give the same value. And so it could be included at both the positions, not to worry. So g o f, let us say I call this as a function k of x. Okay, so k zero is first of all a one. Now, let us try to find out the derivative of this function k to the left of zero. So left and derivative. So k dash zero minus. So limit h tending to zero plus k of zero minus h minus k of zero by minus h. So k of minus h. k of minus h, I will be using the second definition here. So it will be e to the power one minus cos of minus h minus one by minus h. Okay, let's try to evaluate this limit as h tends to zero plus. So in order to evaluate this limit, I'm going to multiply and divide. I'm going to multiply and divide this with one minus cos h. The reason is this function of h tends to zero as h tends to zero plus.  But one minus cos h by h will become a zero. Okay, so left hand derivative is a zero. Let's look into the right hand derivative. So it's going to be k zero plus h minus k zero by h. So it will be e to the power of sin h minus one by h. Okay, limit h tending to limit h tending to zero plus. Yes. So for this again, you have to do one thing. You have to multiply and divide with sin h. So I'll multiply and divide with the sin h. So this term is a one and this term is a one again. So this will be one. So left hand derivative, right hand derivative are not matching with each other. Anyhow, you could have also performed that thing by differentiating this literally. If you differentiate this literally, it will be e to the power sin x into derivative of sin, which is cos at x equal to zero. This gives you a one. Okay. And the derivative of this will be e to the power one minus cos x into sin x derivative at x equal to zero would have given you a zero. Left hand derivative, right hand derivative do not match. So option number d is correct. None of these. It's not differentiable. Is it fine? Any questions? Yeah, that's what I also checked and verified my result. Yeah, we can do that. We can do this as well. Next question. Yeah, poll is on f of x be a twice differentiable function and f double dash zero is five. Then what is the value of this limit? Especially helpful for the ISC students who have indeterminate forms. Simple question. Done. In the next 20 seconds, I'll stop the poll. Super easy question. Okay, five, four, three, two, one. Okay. So 13 of you responded out of which four of you say option a seven of you say option b. Now see, in the present shape, it is a zero by zero form because if you put x as zero, it becomes three f zero minus four f zero plus f zero, which makes zero and denominator is obviously a zero. So you can apply your L'Hopital's rule on this. So it'll become three f dash x minus 12 f dash three x plus nine f dash nine x. Okay, but it's still a zero by zero if you check because numerator will become 12 f dash zero minus 12 f dash zero. So zero and this is anyways a zero. So you have to apply L'H rule once again. So that gives us three f dash zero three f double dash x. This will give you minus 36 f double dash three x. This will give you 81 f double dash nine x upon two. Now you have to put your x as a zero. So that will give you, if I'm not mistaken, 84 minus 36, which is 48 f double dash zero by two, which is nothing but 24 f double dash zero f double as zero is given to us as a five. So this answer is going to be 120 option number B. Is it fine? Any questions? Next. A function has a second order derivative as six x minus one. If its graph passes through two comma one and at that point, the tangent to the curve is y is equal to three x minus five. Then which of the following is your function? This is highly scammable question. Cheren, yes, you can use integration for this question. Right, Kinshukh. So Cheren is asking whether he can use integration. I said yes, he can. But as I feel reading from the question that there is actually no need for it. Options themselves basically speak out. Okay. Should we conclude in the next 20 seconds? Okay, five, four, two, one, go. Great. I think 100 percent vote has gone to option number B, which is very obvious. But let me solve this question as if I'm not looking at the options, but that is not a good idea to approach a question. So when you differentiate it, you end up getting six x minus one whole square by two. And there will be a constant added to it. Okay. Now use the fact that the derivative, the slope of the tangent at two comma one, that means f dash two is going to be three because the slope of this line is three. That means if I put a two in place of x, I'll get three plus c is equal to three. That means c is actually a zero. Okay. In short, your f dash x is three x minus one whole square. So what does this mean? It means f of x is, again, integrate this. It'll get three x minus one whole cube by three plus another constant. Let's say called c dash. Okay. And use the fact that the curve passes through the curve passes through two comma one. So when x is two, this is one, which clearly implies that c dash is zero. In short, you just drop this, your curve will become x minus one whole cube. Option number B. Okay. But since here it is very obvious, you don't have to worry too much, but this is the actual way of solving it. Now, many of you would say that integration is coming into picture. Hence it might not come in the board exam. Okay. I can buy that argument. It might not come. I think this is not rate of change is not there one second. I'll pick up another question. On the interval zero to one closed, this function takes its maximum value at which of the point. So what's the answer, Bakul? Check again. You're doing a lot of mistakes. Bakul, everybody make everybody you're doing. We'll check Kendrick. Let everybody work this out. I can give 30 more seconds. Okay. We'll check. We'll check. All right. So let's stop here. Five, four, three, two, one, go. Okay. Six votes each to option B and C. Okay. We'll check it out. So let's say this is our function x to the power 25, one minus x to the power 75. Let's differentiate it. So this will be 25 x to the power 24, one minus x to the power 75. And you will have minus 75 x to the power 25, one minus x to the power of 74. Take 25 x to the power 24 common. In fact, you can take one minus x to the power 74 also common. So that will give you a one minus x from here. And from here, you'll end up getting minus three x. Am I right? So that gives you 24. This is one minus four x. Now this guy, this guy one fourth, zero, zero, one fourth, and one. Actually I have to analyze this only in the zero to one interval. Let's try to see what is going to happen. If I'm not mistaken, if I go higher than one, I'm going to get a negative number. This is negative positive. Okay. Oh, sorry, negative. And one is subjected to even power. So there'll be no change in the sign. Again, a negative. Okay. And this will be positive. And this will be positive. Okay. So there is a change of slope from positive to negative. Okay. Thereby creating a maxima at one by four. So the point where it is going to have a maxima is one by four. Okay. No, what about one by two? Oh, after differentiating. Yeah, I mean is it? Roles zero. Okay. How do you plan to use Roles zero here? Ginshukh, how do you plan to use the Roles zero? Selections terms after one are basically lesser than one. And terms before one will actually be greater than zero. And term, which means and one is a zero and zero also gives you a zero. So you can say like the term between has to be a value of the greater than both. So one by four is not in between. Half is between zero and one. Half is between zero to one. So I'm not saying exactly in between. So exactly in between. So exactly in between, how do you know that one by two, one by three cannot be. Why one by four? So after getting the derivative. After getting the derivative, just one by four is good enough. You don't have to check that out. That is what you're trying to say, right? But what is it? Yes, sir. So what is the use of the Roles zero? Okay, sir. Let's take this one. This line touches this curve at the point a comma b for which values of n then only two people have responded so far. Okay, let's discuss this now five, four, three, two, one, go. Okay. So first of all, the slope of most of you have got with option d by the way, the slope of this tangent, this line is minus b by a. Okay. Now let's find the derivative of this curve. Let us differentiate both sides. First of all, dy by dx equal to zero. Okay. Now at a comma b, if you put a comma b, it becomes n by a plus n by b, dy by dx equal to zero. So this becomes independent of n. So at a by b, the slope will be minus b by a and this is independent of x. Sorry, independent of does not depend on n. Whatever is your n, that is anyway is going to get cancelled off. Okay. So for any n, this is going to work. Is it fine? Any questions? Any questions, any concerns? All right. Okay, we are going to take some assertion reasoning questions. Okay.  And now I'll put the questions also. Let's start with this question. If the length of the three sides of a trapezium other than the base are equal to 10 centimeter, then the area of the trapezium, when it is maximum is 75 root three. And this happens. The reason is it happens at x equal to five. Which of the following options are correct? I'm relaunching the poll. They have actually not the length of the three sides of the trapezium. Yeah. That's what I was wondering. What is the extent they've given the length of the three sides of the base are equal to x. Okay. So let's consider this to be our x. This to be our x is referred to this diagram while solving the question. Done. Poll was on, right? Because I can see zero, zero people voted so far. Oh, now somebody has voted. Simple. You just have to follow this figure. Write down the expression for the area. Maximize it. That's it. Basic, basic, basic question of application of derivatives. Maximum minimum. Okay. So it's time. Half. This is also going to be 10. Sum of parallel sites into height. Should we close the poll now? Anybody who's on the verge of finishing it? Five, four. Oh, you want something? Okay. Enough time has been given. Let's discuss this out. Most of you have given option number A as your response. Okay. Let's check. So this area is as good as x plus 10 under root of 100 minus x square. So dA by dx should be zero. So when you differentiate this, you get x plus 10. Derivative of this will be minus x by root 100 minus x square plus under root of 100 minus x square is equal to zero. In short, minus x square minus 10x or let's put it to the positive side here on the other side is equal to 100 minus x square. So 2x square plus 10x minus 100 equal to zero. Just drop the value of 2 from both the sides. So this is as good as x plus 10 into x minus 5. Okay. So x has to be 5. And at this position, it will be maximum only. So area of the trapezium is maximum and x equal to 5. At least statement number two or the reason is actually correct. And what is the area? Area is going to be 15 under root of 75. Okay. Which is going to be, which is going to be, yes. So 25 will come out as 5, 75 root 3. So both statement, assertion and the reason are correct. So here I will go for option number eight. Is it fine? Any questions? Next. Okay. We'll take this question. Hole is on. So there's a function which is following this particular nature. f of x is mod of sine x when mod x lies between zero to pi by two. And this is okay. And half an x is equal to zero. So it has a local maximum at x equal to zero. And the reason is f dash zero is zero and f double dash zero is less than zero. That's your question. Right. Well, it's actually touching, sorry, kinship. After a few more questions, the CVSC people can leave with the ISC people. I'll do some vector and 3D questions. But anyways, you're all welcome to stay back and solve that. But just if you want to use that time for something else, you can always do that. So just two or three questions I will take on the topic which is common to both CVSC and ISC students. And then CVSC people may leave if they want. Okay, should we discuss it now? We need more time. Guys, just draw a graph. A lot of things will get clear because this is not a difficult graph to draw. See, when it says mod x is greater than zero and less than equal to pi by 2, it just means it is from minus pi by 2 to pi by 2 excluding zero. In this interval, you have to draw a mod of sine x graph. Before that, I would like to close the poll. If anybody wants to vote, please do so. Five, four, three, two, one. Okay. So most of you have said option C but A also has got three votes. Yeah. So mod sine x graph in this interval is going to appear like this. See, this part which was supposed to be down will go up. So it'll come like this.  This is how the graph will look between minus pi by 2 to pi by 2. And at zero it is at half. So it will be somewhere over here at zero. This will be half. Does it have a local maxima at x equal to zero? Yes or no? Just give me a response on the chat box. Does it have a local maxima at x equal to zero or not? It does. Yes, it has. Okay. So assertion is true. But is it because the derivative of the function is zero at that point and double derivative is negative? It is not even differentiable at zero. So this is completely bakwas, completely false. So assertion is correct but the reason is incorrect. Option number C is right. Is it fine? Any questions, any concerns? Yeah. Let's do this one. Should we discuss it now? So f dash x is given to you. All right. So 5, 4, 3, 2, 1, go. All right. So most of you have said option number C. Again, assertion is correct but the reason is incorrect. Okay. See again, if you do the wavy curve for this, this is definitely positive and crossing two, two is subjected to an even power. So there will be no change in the sign. And then one is subjected to an odd power. So there will be a negative here. Okay. So there's a minima at one, but there is an inflection point at two. Okay. So this is an inflection point. So it has neither maxima nor minima at two. This is a correct statement. And the reason is because f of x changes is signed from negative to positive. No, this is not a right statement. So option number C is correct. Assertion is correct, but the reason is incorrect. Is it fine? So two is a point of inflection and there is no change in the sign of the derivative in the neighborhood of two. All right. Now, CBC people, if you want, you can relax. You can study something else. We will do some vector questions now with the ISC CHAPs because they have ISC. Thank you. Thank you. Those who want to leave, thank you so much. Okay. If L, M, N are the direction cosines of a vector, alpha, beta, gamma are the angles with the vector make with X, Y and Z axes, then the unit vector in the direction of that vector is which of the following option is correct. Especially Prakul and Radha would request you to give me a response on the chat box if possible. I know if some CBC people might want to attempt it. Okay. Okay, Prakul. I got your response. This is easy. So which option is correct?  So option number B is definitely correct. Any questions here? Anybody? Any ISC student? Any question? Do let me know. Next. If three points A, B, C have position vectors as given to you on your screen and if they're collinear, what is X minus Y value? What is X minus Y value? If they're collinear. Okay, Prakul. What are they asking? X minus Y they're asking, right? Or are they asking X comma minus Y? I'm asking you, what are they asking? Because X minus Y, X minus Y should give them a fixed value. Okay, let's assume that they're asking X comma Y. Let's interpret the question in that one. Let's check. See, first of all, if they're collinear, the difference of, let's say the coordinates or the difference of these position vectors, that is nothing but 2Y, 4 minus XJ and 4K. This must be proportional to the difference of, let's say I take these two. So this will be Y minus 3I and minus 6J. And minus 12K. So 2, you can write it like this. Y minus 3 or 2 divided by Y minus 3 is 4 minus X divided by minus 6. And this is equal to 4 divided by minus 12. Okay, they should be proportional. So from here, if you check out, check this out. So X minus 4 by 6 is minus 1 by 3. So X minus 4 is minus 2. So X is equal to 2. And if you check these two ways, so Y, 2 by Y minus 3 is minus 1 by 3. So minus 6 is Y minus 3. So Y is negative 3. So X is 2 and Y is negative 3, which is option number A. Right, Hariharan. So Hariharan also attempted this question. Let's move on to the next one. Yes, A, B, C are linearly dependent vectors. See, linearly dependent just means they are coplanar. Okay. And modulus of C is root 3. Then alpha and beta are which of the following values? Sorry, the pole was running. I'll relaunch the pole. Was that you, Prakul, who responded on the chat box or somebody else? Anybody else who wants to answer? Okay, I did. Fine. So if they're linearly dependent, as I already told you, they are coplanar. Okay. So the first thing that I would like to state here is that this determinant is going to be a 0. Second thing they have given is that the modulus of this vector is root 3. That means this is equal to 3. Of course, all of them satisfy the second equation. So the first equation is the deal breaker for us. So let's do the simplification here. You can just take the first column and subtract it from the other two before you start simplifying it. 0 minus 1, alpha minus 1, 0, 0, beta minus 1. Okay. So let us expand it. So with respect to the first row, it'll only give you 1 minus beta equal to 0. So beta only can be 1. Okay. So all these possibilities are straight away ruled out. And alpha could be, of course, alpha square will be 1. So alpha could be plus minus 1. Option number D is correct. Is it fine? Any questions? Okay. A unit vector perpendicular to the plane formed by these three points. I mean, don't treat this as a plane question. Treat this as a vector question. Okay. I'll discuss this once you have attempted it and once others have attempted it. I have relaunched the pole. I think the old pole was still going on. Okay. No, unless until I say what is the equation of a plane, it doesn't require the plane concept. It just requires a vector understanding. That's it. Okay. So you want a vector, you want a unit vector perpendicular to a plane which contains three points. Okay. So let's say I call this point as a B and C. So if you want a vector perpendicular to the plane containing this, it basically lies along the cross product. So this is a vector proportional to the cross product of any two vectors you take on it. Let's say I take AB and AC. Okay. So you just find out just AB cross AC only you find out and take the modulus of it and divide it. Plus minus that could be your answer. Okay. So let's check which of the options match it. But before that, I will end the pole because only two of you have answered it. Okay. So AB. Let's find out AB. AB is, I'll write it in a shortcut notation. AB is B minus A. So it's going to be minus 1, 2, 1. AC. Normally, you can save your time by not writing IJK because you can write this as a column vector as a column matrix. Yeah. So C minus A, that will give you a 2, 3 minus 1. So cross product will be IJK. Take the components. So let's find out the cross product. So that's going to give you a minus 5, 1 minus 1. Okay. And K will give you minus 3, minus 4, which is minus 7. So it comes out to be minus 5I plus J minus 7K. And you have to divide it by the magnitude of this vector. So to get a unit vector, you have to divide by the magnitude. So magnitude will be, if I'm not mistaken, 49 plus 150, 75, 75 is 5 root 3. So it's minus 5I plus J minus 7K by 5 root 3. Any vector which is proportional to this, I don't think so. First one is proportional to this. Second one, however, is proportional. Yeah. Third one is not. So out of the options which are present, B could be a unit vector perpendicular to the plane. Is it trying any questions? Any questions? Say nothing to guess here. If it was a difficult question, I would have understood. Nothing to guess. Okay. I'll take some question. Just give me a second from the previous year papers also. I'll be unsharing my screen for some time. Okay. Can you see my screen? You can all see my screen. Okay. So since we are doing 3D questions, let us take this question, question number 9. First part, if a line makes alpha, theta 1, theta 2, theta 3 with the x, y, z axis, find the value of cos 2 theta 1, cos 2 theta 2 plus cos 2 theta 3. This actually is a CBC paper which came. This is how, I mean, I'll just show you the paper how it actually appears. This is how it used to come earlier. They used to be two parts, part 1. This is for CBC. Okay. And there were two markers, three markers and five markers, etc. So this is how a typical question paper used to look like. Anyways, we don't have to worry too much about the entire paper because only limited topics are there. Yes. Give me a response on the chat box, especially ISC people. Okay. So this part is quite easy. We can easily do it.   So this is anyways a 1. So it is 2 into 1 minus 3. Answer is a negative 1. The next part of the question is quite trivial. If the equation of a line is this, then the numbers to which the direction ratios of the line parallel to AB is proportional, that is 1 minus 2, 4 only. Some of the direction cosines are actually 1. We will study that, Aditya. We will study that in our 3D. 3D, we have not started yet. So some of the direction cosines of a line add up to give you a 1. Seriously, people, don't worry too much about it. We will do it. We will do it once. Some of the squares, yes, some of the squares of the direction cosines add up to give you a 1. I think apart from this, we don't have much here. Okay. Can we try this question? Give me a second. Yeah. Can we try solving this question? Question number 37. This part, not this part. It was a 2 marker. Oh, I think even this question involves a plane. Find the distance from the plane passing through this. Okay. Just give me a second. We will not do this one also. Okay. Yeah. Let's do this one. Show that these two lines intersect and find out their point of intersection. In fact, let's take this question on the screen only. Find out the point of intersection of these two lines. Give me a response on the chat box. Okay. Okay. Should we discuss it now? All right. So let's discuss it. So assume that the point where they both intersect for this line, that point has a parameter of let's say R. So X is 3R minus 1. Y is 5R minus 3 and Z is 7R minus 5. And for this line, it is let's say T. So X minus 2 by 1. Y minus 4 by 3. Z minus 6 by 5. This is equal to T. Right. In short, X is T plus 2. Y is 3T plus 4. And Z is 5T plus 6. Now just compare these two. So 3R minus 1 is T plus 2. 5R minus 3 is 3T plus 4. And the third equation is like a test or like a check whether we have got the right values of T and R. All right. So let's do one simple activity here. Let's multiply this with a 3 and subtract it from this equation. So that's going to give us a 4R is equal to 2. So R is half. Okay. So if R is half, this is going to be half is equal to T plus 2. So T is going to be negative 3 by 2. Does it satisfy the second equation? So 7 by 2 minus 5. Is it equal to minus 15 by 2 plus 6? Yes. It is correct. So R is equal to half. If you put it in this equation, this gives you 3 by 2 minus 1, which is half. Again, 5 by 2 minus 3, which is minus half. And here, if I put, I'll get minus 3 by 2. So point of intersection is half minus half minus 3 by 2. Is that fine? Any questions? Exactly. This question itself was objectified in one of the exams. Okay. That is how your question is going to come. Okay. Anyway, so all the best to all of you. Thank you. I will stop here. Do well. We are always there for any kind of a doubt that you have. Please use the personal chats or group as well to ask your doubts, so that if I am not able to reach out to you, some peer can help you out. Okay. And just keep your calm during the exam. Don't rush. You have enough time. You have enough skills to solve those questions. So don't doubt yourself. Don't get nervous. Okay. Thank you. Bye-bye. Take care. Good night. Stay safe. Thank you. Bye-bye.


live from the Computer History Museum in the heart of Silicon Valley. It's theCUBE, covering OpenStack Silicon Valley 2016. Brought to you by Morantis. Now, here are your hosts, John Furrier and Lisa Martin. Welcome back, we are live in Silicon Valley for day one of two days of live coverage of OpenStack SV, SV Silicon Valley. This is Silicon Angles theCUBE, our flagship program. We go out to the events and extract the signal from the noise. I'm John Furrier, my co-host Lisa Martin. Our next guest is Shannon Williams, co-founder of Rancher Labs, hot startup. I've been on theCUBE before, you guys. We've talked to the other co-founder and hot space.  You saw the middle of the fairway docker containers. Now, certainly they're playing through huge. Kubernetes on the doorstep. How does this all fit into what you guys are doing? You know, it's a really good insight because I think 18 months ago when we were getting our company started, we were trying to decide if the market was going to move really quickly or if this might be a sort of a longer burning thing, which as a startup is really important because you've got to figure out how quickly to grow, how much to invest in a team, how quickly to hire people. And there were a lot of really good indicators. We talked to a lot of large companies that were just sort of really impressed at the potential for containers to shake things up. And my co-founders and I, when we started building Rancher, we wanted to build an infrastructure platform that could basically consume, you know, cloud resources from anywhere, virtual machines, physical machines, Amazon, Google, wherever, and implement container management on top of those. And getting started early as the more space was developing allowed us to get our feet dirty with tools like Kubernetes and Mesos when they were just in their earliest stage, you know, contribute, learn about these platforms. And it's really helped us a lot because we started an alpha program back in the fall of 2014 and ran that for about nine months working with a lot of larger, you know, people we'd known from previous lives who were interested in using containers. And very quickly what we found was they were pushing us to move faster. You know, they were saying, we need this stuff now, we've got teams that are already using Docker. We want to put it in production. And so it just spiraled. So I got to ask you, one of the things Lisa and I were talking before, you came on, we're talking about, you know, the Docker madness or someone said Docker sanity. I think Docker madness, we have Cube madness, which is our March madness promotion. And Cube folks know that, but Docker is watching sanity to develop our community, but they don't have a revenue model. And Kubernetes is coming in, OpenStack is shifting, VMware has changed, just saw a rumor come across the grid that Pat Gelsinger might step down. So it's all this, VMware is in turmoil with virtual machines. So are you guys in the middle of all this action? Are you guys in the center of the tornado? Because if you think about like Docker and the revenue model potential of where this is going, it seems that you guys actually might be in the sweet spot, having a nice management platform for containers. You guys see yourself that way? We're gonna open source companies. So it's always, you know, you're always kind of, I think being in the middle of the tornado is always a good place to be as a small startup because you want to get in and talk and work with a lot of customers. And by implementing a platform that supports, you know, Docker's native orchestration swarm, as well as Masos and Kubernetes, we really appeal to a lot of larger organizations that really like the direction of containers are going. But they don't want to necessarily pick one tool to try and use across all the different projects in their company. So what's the lens strategies? You guys look at that and say, hey, you know what? We're not going to have dogma with any one platform. And it's more than just that because you look at a tool like Kubernetes, it's a great tool. It's a really powerful platform. But if you're an organization that's putting it into production, you don't want to be building these Kubernetes clusters over and over again for each team. So delivering, using something like Rancher to build a container service that can support, you know, one click deployment of Kubernetes infrastructure anywhere. And then hand that off to your app development team. That's really appealing. So a lot of the demand we're getting from the enterprise is, you know, my developers want to use containers. I want to have some sanity to that. And a nice open source tool like Rancher that provides a full platform for running containers. It just fits in really well with the strategy. You see yourself as a hedge or almost an insurance policy with a customer. In a way, that sounds what it sounds like. Is that accurate? I think it's more of an enabler, but yeah, I agree. I think there is, you know, one of the problems with any early technology is there's a bit of analysis paralysis. People are afraid to bet on a tool and especially big bets. Always, you know, there's always some leaders who will take those big bets and say, you know, we believe in some tool early on, but usually what you get is companies saying like, you know, team over here is telling us how great Kubernetes is. These guys over here love Swarm. We've got some other team who's been lobbying us to put in a Mesa's cluster for the last year. I don't know which one to do. And so a tool like Rancher that supports them all and sort of says, look, orchestration is, you know, these schedulers are somewhat swappable. They all have value. They have different case probabilities. They're all moving at their own pace. Don't pick one, put in a platform, let people use the tool that works best for them.  They're like, yeah, we've used schedulers for a long time. Schedulers by themselves aren't that novel. Schedulers have been part of IT for 20 years. There's probably three today. There might be 10 in five years. Let's just, we'll build out the infrastructure. I can put it before you get at least this question is that Martin Casada had a slide up on the keynote. Developers don't care about the old way Gartner report. They care about, you know, other developers, but the key is try and buy. So your point about these big bets, and that one's, I mean, they're making a big bet in general, but they're not making big bets at once. Yeah, we break the world into, there's two types of projects we see. So it's interesting, we're totally open source. And so very often we don't know somebody's using us until they call us. And so we'll get a call and they'll be like, hey, we really like Rancher. We'd love you guys to come out and do an analysis of our implementation or we want support. But there's two approaches. One of them is it's a team. They've got an app. They've decided they want to run this on Kubernetes. They're using our distribution of Kubernetes and Rancher and they're going live. And they don't care what else is happening in the organization. This app is going live in two months and they've already made their decisions. They're tried, they're using it, it's baked in. They're now just trying. It's a production track. They're just getting sign off by everyone and part of sign off might be having a support contract or having some support. There's another side that is more strategic that's saying we want to build a container platform for our company. We're a Fortune 2000 company and we want to understand containers across the board and set global policy. But those guys, they're actually moving much slower. That's where you get a lot of, well, let's try and do a lot of RFPs. Let's get some analysis. And it's surprising that those projects, not surprising, they'll take a lot longer to code anywhere. But they're both happening. I mean, I see it both ways. I see some organizations that sort of have a lot of central control and they're sort of pushing a central vision and they're going to get there as here's our container platform, use that on whatever clouds you want to. And other ones where they've been given enough rope to run and they're taking tools, picking them and going live. And in a lot of ways, those projects are turning out to be quite successful because as always, if the team believes in the tools they're using, then they'll defend them and make them work. And if they own the decision to choose them, they're a lot more likely to, you know, fight to keep them all working and aligned and upgraded and everything else you need to do. So pros and cons of different models, I think it's going to be a mix, but I definitely agree with the sentiments that it's bottoms up. People try, they test, they use open source tools. They talk to each other in the communities. I'm sorry. Right, no, no, great point. So you've been in business for a couple of years now as open source software. Talk to us about some of the customers, some of the use cases that are really driving the evolution and the maturation of Ranger. Yeah, there's a couple of big, I'd say buckets. One of the things we're seeing is early on in the first year, the people who were really taking containers into production were startups. They were SaaS companies, they were web services companies, they were online gaming companies, and they're just all sorts of types of companies, but they were generally, you know, very, very knowledgeable DevOps teams at small startup B type organizations that were writing cloud native applications and they were moving quickly. And they were just taking our stuff raw, even during the beta, going into production, and we'd find out about it later. But towards the end of last year, we saw a start to see a shift. We saw government agencies, healthcare companies, insurance companies, a lot of telcos, a lot of companies and financial services that were at an application level taking tools like Kubernetes and implementing them with Rancher and allowing their teams or for their own team to put apps in production. So a shift towards more, I wouldn't say necessarily yet we were taking the legacy job application, that's probably next year. This year, what we're seeing a lot of is new applications, microservices driven applications, sort of mode two applications in Gartner speak that are being written and being deployed using containers, predominantly onto VMware and Amazon. I would say that if it's running on-prem, it seems to be running still in VMs running on VMware. If it's being deployed in the cloud, it's going to Amazon, Azure, Google, IBM or other cloud providers that people are consuming. But generally, you're not quite to the point yet. I think if you look down the road, what you'll see more is containers running on bare metal and legacy applications moving to containers. We're already seeing the beginnings of that. A lot of the questions that are coming in right now from users are, how do I run persistent workloads? And surprisingly, it's actually quite straightforward. I don't think, I think a lot of people are, the nature of a container is you have to attach a disk to it if you want to have persistence, but that's not that different than anything else. And so once people kind of understand it, they start moving. And so I expect next year to start seeing a lot of migration of existing workloads into containers to get better density and to get portability. So. All right. Just really quickly, John, thank you. Just one of the things I saw on your website was can running Docker really be that easy? Talk just about the feedback loop that you're getting with your customers to drive the evolution of your software and to really become a leader, maybe a sneaky leader in that space. Yeah, you know, everything in open source comes down to word of mouth. And you know, the first thing you find when you put some software out there and you know, kind of put it up on GitHub as you start to get, people start to find, hey, this doesn't work, that doesn't work. And you know, pretty quickly you get forums up and you start letting people talk to each other. And one of the things we saw really quickly as people started using Rancher, was they liked that it had a really clean interface, that it had support for different orchestration platforms, that it was supporting, plugging into your existing AD and your existing ops tools. And so about three months before we GA'd, we started reaching out to some of the more vocal people in our community and saying, hey, would you mind sharing your stories and sharing a quote and kind of using it, being a case study for us? And I was amazed. I mean, I've been working in software for almost 20 years now. And pulling case studies out of customers is pretty much the hardest thing to do. But in open source, it was just like, I mean, everyone from large insurance, they were like, yeah, we'll put a quote up, we'll write a blog, Sony came on one of our meetups and talked about how the PlayStation group was using containers. And people like Orange gave us quotes for our website. And we've had a ton of companies create their own videos where they talk about how they're using this or blog about it. That's great product validation too. I mean, that's what open source does for it. It is, it's a feedback loop. I think you said it perfectly where they just, they want to share their use case. And one of the cool things we've seen, if you go to our YouTube channel, we've got these videos that are being created where users are just telling their story. But it's not like the old classic VMware, Cisco case study where you bring in a big team. It's like two guys sitting in front of a webcam saying, hey, I'm DevOps Joe one, and this is DevOps Joe two. And we just built our whole CI CD platform on containers. This is how we're doing it. And it's very granular. It's not, I mean, they go through all the problems. Oh, this stuff isn't that good about this, but it's really good here. And so you get a very honest feedback that you just don't get in general and technology. Right, best validation of what you're delivering through the voices of your customers, absolutely. So what's the business model now? Obviously you got great traction on the customer front with the open source. So that's a great testament. Is it just kind of like giving enough cash in the bank? You're not trying to force it? Or is it standard support? How are you guys planning on making money? And not that it's a pressure point, but. No, it is a pressure point actually. It's always a pressure point. Because the reality is, yeah, you have to build a business, I think. And open source is starting, I think there's still a lot of questions in the investment community and in the general community is how are open source companies making money and how will you make money? We've been very focused on building a good open source project to begin with and not taking a closed core or open core closed function approach. So pure open source. Just pure open source being open because what we find is that we get a ton of contributions and we get a lot of attention. What we've been lucky and I think is still very true is when people find value and they start to run real applications on it, we have no problem selling them supported versions of our product that they can count on running at enterprise grade, you know, five, nine level support. And so they want 24 by seven support. They want to know that the people who wrote the software can come and provide, you know. So this is classic open source, great software in production, client will see value in support contract and they're not going to break the bank but enough that's significant. It's not a tip jar but it's big. It's real because for them, I mean the key is they want support. They also want to work with us. I think it's like anything. As you make a tool strategic, you want to have an engagement with the people working on that platform and supporting it. So we found a lot of success in that approach. We're also seeing a big growth among managed services companies who are using Rancher as the basis for their container service. Where they're rolling out a container service to their customer base and Rancher gives them a lot of flexibility. It gives them a multi-tenant platform and we've built a model that allows them to charge for the software as they use it. So that all has been really appealing to companies that are using OpenStack maybe building OpenStack based public clouds or hosted managed services. A lot of VSPP type of companies that want to add containers to the portfolio. And that's another great one. Thanks for coming on theCUBE and sharing the insight. Final word I'll give to you. What's next for you guys? What's your priorities? What are you going to be doing? More open source, more all the time, all the time. What's going on? I think right now we're heading towards, there's a lot of automation that's getting built around this stuff to solve even the more tricky problems. How to do really more complex failover, better storage management. But I think we're still in such an early growth phase. I think the focus is getting people successful and getting them talking about what they're doing. All right Rancher Labs from herding cats to herding cattle. And we've got other new opportunities. Congratulations on your success. Young startup kicking ass, taking names here in the open source community. I'm John Furrier, Lisa Martin. We'll be right back with more after this short break. You're watching theCUBE. Thanks for having us.