So the same process, so I'll be giving you question one by one, I'll be giving you around one, one and a half minutes, two minutes depending upon the merit of the question and I'll be running a poll and you need to respond on that poll. In case you're not able to see the poll, you can give me a response on the chat box, okay? So a very easy question. Y is equal to seek of x degrees, find dy by dx. Very simple question, less than 30 seconds, it should be taking you to answer this, not even that. I knew somebody will be giving me that response which, you know, it should not be giving and I've already got response, that response from one of you. All right, 20 more seconds, 20 more seconds. So while people are joining in, okay, five, four, three, two, one, go. Okay, the answer to this question, as most of you have correctly answered it, is option number C, not B, okay? So when you write the derivative of C kicks as C kicks tan x, in that your x should be in radiance. But here in the question, it has categorically mentioned that it is in degrees. So before you differentiate it, you basically need to write it as seek of pi by 180 x, okay? So the same angle, okay? In radiance will be this, right? So looking at it, yeah. So when you write it like this, when you write it like this, your answer will be seek pi by 180 x into tan pi by 180 x into pi by 180. And now you can convert this back to degrees, okay? That clearly makes option number C the right option. Is this fine? Any questions? All right, next one. Paul is on. Why is under root of 1 plus cos 2 theta by 1 minus cos 2 theta, then dy by dx at 3 pi by 4, sorry, dy by d theta at 3 pi by 4, at theta equal to 3 pi by 4. Super easy question. Two different responses I'm getting. Seven of you have responded so far, one minute gone. Oh my goodness. So almost 50-50 to two of the options. Okay, should we stop in another 20 seconds? Five, four, three, two, one, okay, end of full, end of full. Okay, so let me ask one of you which option he or she has gone with. Anusha, Anusha you're there. Anusha, which option did you go for? Okay, Anusha has gone for option number A, okay. All right, and let me share the result also. Most of you have said A, but yes, equally same number of votes almost has gone to B as well. Okay, we'll check it out, which one is correct. See, first of all, I have been saying this so many times here that this is reciprocal of tan square theta. So it's under root of 1 by tan square theta. So it's under root of, under root of got square theta. Am I right? Yes or no? Is it fine? Any question so far?  It's a reciprocal. So this part is 1 by tan square theta. Correct. So won't it become mod of got theta? Correct. Now, this function in the neighborhood of 3 pi by 4 would be defined as negative cot theta. Right, because you're looking at an angle which is in the second quadrant. Correct. And second quadrant cot is negative. So now the derivative here would be cos 2 theta. Yes or no? Correct. So at theta equal to 3 pi by 4, it is going to be root to the whole square. So what's the answer then? What's the answer then? I was slightly shocked to see people giving A as the response, but it is actually option number B. Is it fine? Any questions? Any concerns here? So please, when you have a special function coming into picture, depending upon where are you carrying out the derivative, you need to redefine the function like that. So mod of cot theta, since you are finding the derivative at 3 pi by 4, you have to redefine this as negative cot theta. And negative cot theta derivative has to be performed. And then you have to put theta as 3 pi by 4. Is it fine? So Arusha, did you take care of that fact that it was negative cot theta? Yeah, that's the reason why it was wrong. Okay, anyways, next question. Next question. Poll is on. So x under root 1 plus y plus y under root 1 plus x is 0, then dy by dx could be which of the following option? And this is a commonly seen question. That's why many of you could answer it within 10 seconds. But I would request those who have not seen this question before to solve it properly. I'm giving you around two minutes time. Very good. Seven of you have already responded. Less than a minute. 30 more seconds. Okay, five, three, two, one, go. Okay, so now people have started voting. All right, almost 20 of you have voted. 20, 21 of you have voted. Most of you have gone with option C. But yes, few votes have also gone to BD as well. Okay, let's check. So first of all, in this type of a question, we need to do a bit of simplification before we start differentiating it. So first I will square both the sides over here. So that gives me x square 1 plus y. This is y square 1 plus x. Bring it to one side, you'll have x square minus y square plus x square y minus x y square is equal to zero. This is factorizable as x minus y, x plus y. And this is if you take x y common, this is again x minus y. Okay, so it's x minus y times 1 plus x plus y plus x y equal to zero. Okay. Now, if x is equal to y, then your dy by dx will be one, but that is not in my options. Okay, so I will not worry about it. So in this case, your dy by dx becomes a one. But if x plus y plus x y is equal to zero, then let's see what happens. So here y is nothing but negative x by one plus x. Okay, so let's find dy by dx for this case. So it'll be one plus x times negative one plus x plus x into one by one plus x the whole square. This is definitely negative one by one plus x the whole square. Okay, which is option number c. Is it fine? Any questions? Any concerns? Anybody? Any questions? Any concerns? Okay. Next question. Yeah. Derivative of sine inverse x under root one minus x minus root x under root one minus x square with respect to x. Again, if you would have solved inverse functions or if you would have solved the derivatives of inverse functions, you would have definitely come across this. This would be especially useful for the 12 ISC students because they have been taught these properties in their regular school exam, school curriculum. However, CVC people have not been taught this sine inverse x plus sine inverse y kind of a property. But nevertheless, you should be able to solve it anyhow. Almost a minute. Okay, one person has responded. Okay, three people have responded. Great. Two minutes gone. Only six people have responded. I can give 30 more seconds at the max. Okay, five, four, three, two, one, go. Okay, 14, only 15 people respond, 17 people by the way, by the time I stop the poll. So most of you have gone with option number C. Okay, see for tonight. Let's let's look into the solution for this. See, if you see this, try to compare this expression with something like this sine inverse of x under root one minus y square minus y under root one minus x square. Okay, so we all know that this comes from your simplification of this formula. Okay, I mean, I'm assuming things to be in the principal value branch only. So in this case, your x whole is being played by x only, but your y square rule is being played by x. So x is x, but your y is actually under root of x. Okay, if you see this, compare this. So in fact, this expression is sine inverse x minus sine inverse under root x. So this is your expression, which we need to differentiate. So when we do that, we end up getting one by under root one minus x square minus one by under root one minus under root x square is x. But you have to also follow chain rule, you have to do the derivative of root x, which is one by two root x. In that case, your option number C becomes the right option. Okay, is it fine? So well done people who gave answer as option number C. Any question, any concerns? Doing it by root force would be, I think it would be resulting into a very, very, you can say complicated expression, and you might not be able to simplify it, you know, so easily to option number C. Okay, so avoid using root force here. Let's take the next question. Let's take the next question. Okay, simple one, if for a function f of x f dash a is zero, f double dash a is also zero, but f triple dash a is greater than zero, then at x equal to a, what is happening? Is it a maxima? Is it a minima? It is not an extrema point, or it is some extrema point. What can you comment? Simple conceptual question. Double derivative, single derivative, double derivative, both are zero, but triple derivative is not zero. Okay, I'll be stopping this poll in the next 20 seconds. Okay, five, four, three, two, zero, go. Okay, most of you have got this right. So a clear cut example of something which is satisfying this is x cube function at zero. So if you talk about x cube function at zero, the single derivative at zero is zero, right, three x square, double derivative at zero, that is six x at zero is zero. Okay, but triple derivative is six, which is positive. And at zero, there is no minima, there's no maxima, there's actually a, there's actually a point of inflection, which is considered to be not an extrema point. So this is clear cut option number C. Okay, please note that they might not ask you to have the knowledge about the nth derivative, but it is always good to know your nth derivative test. Let us recall nth derivative test quickly. So what is this nth derivative test? nth derivative test says that, derivative test says that if the derivative of a function at a is zero, double derivative is also zero, I mean, dot, dot, dot. And you realize that your till nth derivative, it is zero, but n plus one-th derivative, n plus one-th derivative, this is the symbol for derivative I'm using, don't consider it to be a power. This is not zero. Let's say this is a value k, let's say k. Okay, then what is the conclusion? What is the conclusion? If n happens to be, if n happens to be an odd number, right, and k is a positive term, then at x equal to a, the function has local minima. And if k is negative, then at x equal to a, the function has a local maxima. But if n happens to be, if n happens to be an even number, okay, if n happens to be an even number, then at x equal to a, the function has neither maxima nor minima. So it could be a point of inflection. So in case you want to use this anywhere in solving the question that comes in your board exam, please feel free to use it. Nobody's going to ask you which method you adopted. So in this present example, you realize that n happens to be two, okay? So third derivative means n is two. So the moment you realize n is even, that point becomes neither a point of maxima nor minima. Let's move on to the next one now. The distance between one comma one and the tangent to the curve, this drawn at the point x equal to zero. Again, a simple question, just your basic tangent normal concept. Oh, sorry, the poll was being okay. I relaunched the poll. In case the poll is obstructing your view, that means if I forget to switch off the poll, please do highlight it on the chat. Extreme point is not a point of inflection. Point of inflection is considered to be neither maxima nor minima. Stress. Oh, two different options. The vote has been split between two different options. What is happening? I thought this was a straightforward question. The distance between one comma one and the tangent to this curve at x equal to zero. Okay, should we stop the poll now in next 15 seconds? Five, four, three, two, one, go. Oh, what happened? Nobody voted for B? Somebody please? How can you miss B? B is the most chosen one when you want to take a guess. Okay. Oh, now somebody has voted for me. No, people would. People may vote C. I got votes for B and D also. Okay. Yeah, so most of you have gone with C. So, guys, joking here is fine, but don't do that same joke with your actual paper. Anyways, when x is zero, y is going to be one. That is undoubted. So the point is zero comma one. So slope of the tangent is dy by dx calculated at zero comma one. Of course, you would not need a one to find the slope. So this will be two e to the power two x plus two x. When you put x as zero, this becomes a two. So y minus one is two x is the tangent. So y is minus two x minus one equal to zero. So this is the equation of the tangent and you want to know what is the distance of the point one comma one from this. So distance will be mod of one minus two minus one by under root of one square plus two square. Okay. Please let me know if I missed out anything that happens to be two by root five, which is option number C clearly. Okay. Simple touch and go question should not take more than one minute. Can we move on to the next one? Very much. Okay. My mistake. I think 3D question was seen. Let's take this one. A wire of 34 centimeter long is to be bent in the form of a quadrilateral of which each angle is 90 degree. Okay. Means to say a rectangle. What is the maximum area that can be enclosed within the quadrilateral? Okay. Let's do this. Tane and simple. Pole is on. Oh, I'm so sorry. Pole is on. Very good. Done. Should we close the pole in the next 30 seconds? Sir, 30 seconds is too much. 10 seconds only you close. Yes. Five, four, three, two, one, go. Okay. Well then I could see 23, 24 responses of which 23 of you have said option number D. It's obvious that that particular rectangle has to be a square to have a maximum area. So the perimeter is 4x. So let's say it is 34 by 4. 34 by 4 is going to be 8.5 ish. Correct me if I'm wrong. So area would be what? 8.5 square. This would be your area of the given figure. Okay. So what is 8.5 square? 8.5 square is 72, 25 put a dot. So option number D is correct. That's a go question. Next. The smallest value of this polynomial in the interval zero to nine, the smallest value of this polynomial in the interval zero to nine. And he launched the pole.  First numbers ending with five. I write a 25 and then I multiply that number before five with plus add a one and then multiply. Yes. I use also, I also use the same trick. 25 seconds and five people have already responded. Good. So it's a question of global minimum. Fast, fast, fast, fast, guys. Let me tell you, this is, these questions are not going to challenge you. It's all about being accurate. It's all about completing because I have been told. CBC has got 80 questions to answer, right? In one and a half hours. Is that right? CBC people? ISE, I know they have to answer 40 questions. Am I right? 50 questions. Oh, okay. So it's still more than Oh, Sumon, you're from which board? PUC board? PUC board? Okay, somebody's saying 60, somebody's saying 50. Why there's a confusion in that? Okay, let's take the low on 50. 50 also, like one and a half hours is like good enough speed. And there's one special person who has been given 22 and a half minutes extra. Okay. Okay, we'll stop in another 10, 15 seconds. Why? Because he has broken his hand. See, I keep telling no, be safe and sound during the exam time. Don't indulge in sports like football and all. Okay. So anyways, what has been done is done. Let's look forward. Five, four, three, two, one, go. Okay. So most of you have gone with option number, option number B, B for Bangaluru, which is zero. That is what you're saying, which is obvious also because from zero to nine, zero will have a least value. Okay, so option number. But I hope you all know the process involved process in this case, let's say this was not an obvious question. This was an obvious question. But if it was not an obvious question, you need to differentiate this, you need to differentiate this, you need to find out your critical points or stationary points, check the value of the function at zero, nine and all those critical points, whichever is the least, that is going to be your answer. So we'll have some statement, one statement, two questions. Two men A and B start with velocities V at the same time from a junction of two roads inclined at 45 degrees to each other. If they travel by different roads, then the rate at which they have been separated is under root of two minus root two V units. If they travel by different roads, then the rate at which they have been separated is two V sine pi by eight. Okay. Let me put the poll on. Yes. So I can see two minutes gone, only seven of you have responded. Okay. Can you close this poll in the next 15 seconds? Okay, rate measure is not there for, okay, fine. So we'll do one thing. We'll just cover this question up. Okay. Maybe if this question is not there for the board exams, don't worry too much. Okay. Five, four, three, two, one, go. All right. So this question basically is a simple, most of you are born with C by the way. Yeah. All right. So let's say this is the distance covered. Both of them start from the same point. A and B start from the same point. Okay. So what is this? This is going to be 45 degrees. So let's say this is X. So as per our cosine formula, cos 45 degree is VT square, VT square minus X square by two VT into VT. Okay. So just putting the values here, so it'll become two V square, T square, and this also becomes two V square, T square minus X square. Okay. So let's write X square as two minus root two V square, T square. So X is equal to under root of two minus root two VT. Okay. So DX by DT is the rate at which they're getting separated. So that is going to be this, which is obviously option number, sorry, statement number one is definitely true, but statement number two also, what is sign of pi by it? Pi by it means 22 and a half degrees. Isn't it? So what is sign 22 and a half degrees? So we can use the formula that sine square, 22 and a half degree. Okay. Two times this one minus this is cos 45. So cos of two theta is one minus two sine square theta. So it's one minus one by root two is equal to two sine square, 22 and a half. Okay. So that's going to be, oh, no problem, Rashmika. So we are just doing some problems from your board perspective. Yeah. Yeah. So this is going to be a root two minus one by two root two is equal to sine square, 22 and a half degree. Okay. So under root of this under root of this and multiply with the two V, multiply with the two V means multiply with the two here and let's say I introduce the two inside. So that is going to become four inside. So this into four whole under root into V. So this goes with the factor of two, this again goes with the factor of root two. So that clearly becomes two minus root two V. So both are same things. So A and B are ruled out. So option number C is correct. Both statements are true. Okay. C is the right option. If this, this topic was not there, rate majorer, then don't worry too much about it. But at least from J point of view, you should be aware of it. So let's move on. Okay. Let's take this question. Our window is in the form of a rectangle surmounted by a semi circular opening. The total perimeter of the window is 10 meters. Statement one says one of the dimension of the window to admit maximum light through the whole opening is 20 by pi plus four meter. Statement two says one of the dimension of the window to admit maximum light is 10 by pi by four. Okay. Fine. So let's put the poll on for this. One of the dimensions of the window, maybe they're talking about that one of the dimensions of the rectangle, the dia, it may represent a dia. Okay. One person has responded almost one and a half minutes gone. Slightly longer question. This is okay. So I'm giving you more time for this. Last 45 seconds I can give. Okay. Five, four, three, two, one. Okay. The responses have been all over the place now. Some of you have said A, B, C, D, two, two, three, four. Okay. So confusion related to which is the right option. Okay. Let's discuss this out. So I have assumed in this diagram that radius of this semi circular opening here is R. Okay. And these two sides are X each. So what are the perimeter here? Perimeter will be two R plus two X, two R plus two X and this part will be half of two pi R. Okay. This is 10. Okay. So two R plus two X plus pi R is equal to 10. Okay. So let's do one thing. Let's write our X in terms of R only. So we can write from here two X is equal to 10 minus pi plus two R. Okay. So what will be the area? Area has to be maximum for maximum light to come out of this window. So it's two R into two X, sorry, two R into X and we have half of pi R square. This is our area. Okay. So two X is already known to us. Two X is 10 minus pi plus two R and half of pi R square. Fine. Any question, any concerns in this so far? So for the maximum light, D A by D R must be zero. So let's differentiate this.   So this should be zero. Any questions so far? On expansion, I can see this is going to be 10. We'll have minus four R and you'll have a minus pi R minus pi R. Yeah. That means R is equal to 10 by four plus pi. So one of the dimensions of the window will be 10 by a four plus pi because one of the dimensions is two R. Now we cannot straight away dismiss the other statement other than without checking what is my X value. Maybe this dimension may correspond to X. So let's first check that and confirm whether that is X or not. So if I put over here, two X is equal to 10 minus pi plus two times R. Okay. This is my R. Okay. So let's take the LCM and all. So pi plus four, this is 10 pi plus 40 minus 10 pi minus 20 by two X. Is it fine? Anything that I missed, please let me know. So this is 20 by pi plus four. So X is going to be 10 by pi plus four. That means this statement is also correct. That means one of the dimensions is this, another dimension will be this. So both the statements are true in this case. Is it not? But only two people went for option C. Okay. So this is something which I would say it was just a fact that you did not think of checking the other dimension of the window. Okay. So please whenever you see like options, both of them are true. So please give it a thought that I have to check the other one as well. Didn't we find two X? No, two X is 20 by pi plus four. X is 10 by pi plus four. You didn't understand why the second statement is correct because both of them are talking about one of the dimensions. Okay. So one dimension is two R, which happens to be 20 by pi plus four. Okay. So this is one of the dimension. Another dimension is X, which is 10 by pi plus four. So both the options are right in their own sense. One of them is 20 by pi plus four. Two R is the dimension of the window. Got it? Yeah. Okay. I know everybody's in a rush to finish the question. So I know you may miss out checking the options, which might be correct. So be careful with respect to that. Next. Let's take this question. The largest distance of A comma zero from this curve. Fractual, I know you have seen this question. Oh, sorry. Oh, poll. Let me put the poll on. Okay. I have got a response from one of you in almost one minute. That was a good speed actually. Two minutes almost. I can give one more minute. Okay. Just five, six responses so far. Should we close the poll? Three minutes have already gone. Okay. Let me close this at the countdown of five, five, four, three, two, one. Please, please, please, please vote. Okay. After so much of request, nine people have only voted. Let me close it now. Almost equal votes have gone to option B and D. Okay. So let me call up on a person. RAB, which option did you go for? If at all you went for any. Akash, did you answer? No? Went for B. Okay. Akash went for B. Let's check whether B is correct or D is correct or none of the two is correct. See, you take any point X comma Y on this curve. Okay. So let X comma Y be at largest distance, at largest distance from A comma zero. So let's say the distance is S. Okay. So if the distance is largest, even S square will be largest. This term, this term, I will write it like this. I'll convert it to a single variable by using the fact that Y square. So look at this expression. Look at this expression. From here you can say Y square is 2X minus 2X square. Okay. So can I say derivative of, derivative of S, oh, I'm so sorry, derivative of S square with respect to X, that should be zero for S square to be a point of, or you can say a stationary point, maxima or minima. So if you differentiate this, you end up getting, I think first of all, let's try to simplify this term. So S square is going to be negative X square and you will have, let it be 2X plus 2X, no issues. So now derivative of S square with respect to X will give you negative 2X, negative 2A plus 2. Okay. This should be equal to zero. In other words, 2X is equal to 2 minus 2A. That means X is equal to 1 minus A. Okay. Now 1 minus A, if I put it in this expression. Okay. Let's say I put X as 1 minus A in this expression. Let's see what do we get.  Okay. So let's see what happens when we simplify it.   So this and this will go for a toss. This and this will go for a toss. Anything else to be taken care of? Okay. 2A square and minus 2A and I think plus 1. Okay. So this is your distance square. So distance is under root of this, which clearly is option number D. Option number D is right. Okay. So Akash, B was not correct, D was right. Okay. Prakul also, I think you said A initially, right? And then you later change it to D. Okay. So option D was the right option. Is it fine? Any questions, any concerns here? So maxima minima question. Is it fine? Any questions, any concerns? And you would all see that the double derivative is negative. So, you know, I've not done that test, but the double derivative of this distance with respect to X is a negative, which clearly signifies that at X equal to 1 minus A, it's a S square is a maxima. Okay. Just too early. Yeah. Please, you know, you may find questions which are familiar to you because you all have practiced so many questions that it is, you know, quite possible that you may see a question which may look very familiar. But even though in that case, don't rely on your memory. Okay. Solve it. I'm sure you all have good speed that you will be able to complete the question paper in the stipulated time, other than I think Prakul. Okay. And Prakul, one more feedback I would like to give you. You make the person write everything. No, some things you have to do in your mind. Are you getting my point? So only ask him to write that thing which you may forget to write, which you may forget.  Don't ask him to write every statement. Like I could see you that you were asking him, asking the person to write, you know, something which, which was an obvious calculation, cancel this out, cancel this out. Oh, you can do it on your mind and you can just tell him to jot down a particular thing. Okay. So try practicing in that sense. If you want to take my suggestion, because if you make him write too much, you will waste time dictating him. Okay. And that is not required. Right. It's like what you are telling yourself to write, you should be telling him to write. Are you getting a point? So I don't write obvious things. Right. I only write things which I feel I have to see in order to make a judgment. That thing only you make him write. Okay. It's very much pertinent to Prakul because he will be writing the exam with a scribe next. The shortest distance between this line and the curve, the shortest distance between this line, y minus x equal to one and this curve. All is on. So we have done this question a lot of times while our regular chapter of AOD was going on. So I think everybody should get this right. Oh, it came for school, perhaps also this time. Okay. Six of you, seven of you have voted so far, almost two minutes gone. You could solve this question by using your concept of tangents and normals as well as maxima minima. So both the options are open. You may choose whichever way is more convenient to you. Okay. We'll be closing the poll in the next 15 seconds. Five, four, three, two, one. Most of you have gone with option number A. 11 vote has gone to BCND. Okay. 18 of you say option A. Let's check. See, if I have to solve this question by using maxima minima, okay, let's say this is your y minus x minus one equal to zero. So I would take the point on this curve to be t square comma t. Okay. Just to rhyme with this particular curve, x is equal to y square. Correct? So t square is square of t. Isn't it? And I would use my regular formula that is the distance of a point from a line. Let me call that as a D. So it would be t square minus t minus one mod by under root of two. Now, in order to make it maximum, this fellow has to be least or this fellow has to be greatest, right? That means if you differentiate this fellow, you may square it also doesn't make a much of a difference. If you square it also, it will not make much of a difference. So if you want to make this maximum, you need to have t as a half. That means this point is one fourth. Oh, I'm so sorry. This is going to be, yeah, sorry. This is t minus t square. Yeah. So it is going to be same thing. I mean, one minus two t equal to zero. So this is one fourth comma half. Okay. So what are the least distance of one fourth comma half from it? Again, you can put your value half minus one fourth minus one mod by under root of two. That's going to be minus half minus one fourth by root two, which is clearly, which is clearly three by four root two. Three by four root two, just multiply and divide with a root two. That will become three root two by eight. Option number A is correct. This is by the use of maxima minima to solve this question. If you want to use your concept of tangent and normal, that means you need to first ensure that that tangent drawn at this point, I mean, it may not look from the diagram should be parallel to this line. Right. So that value of t will be chosen for which the tangent drawn at this point, t square minus t should have the same slope as the given line because the shortest distance would be along the line, which is perpendicular to the tangent at that point and basically the given line. So it should have a common normal. Okay. So if you differentiate this derivative here will be dy by dx, dy by dx at that point t will be nothing but one by two t and they should match with the slope of that given line. So here also you get t as a half. Same thing, you will get the point as one fourth, one fourth comma half and you can find the distance of the point from that line using this formula. So that gives you again three root two by eight. So both the approaches can be used. Is it fine? Any questions? Any concerns? Anybody? Let's move on to the next question. Okay. This might not be an important question for the ISC students, but an important question for the CVSE students. Linear programming problems. So minus x one plus x two is less than equal to one and minus x one plus three x two is less than equal to nine x one x two both are positive. Define a bounded feasible space, unbounded feasible space, both or none of these. This is from the linear programming problems, which is not a part of your ISC curriculum. So don't worry about it. You can just relax. We call this as an LPP problem, LPP question, linear programming problem. So CVSE people, please give a response. See, this is just an inequality question in two variables. You can think it like this. You are done in equality in class 11, right? So try to make a space. You can treat your x one to be like your x and x two to be like your y. So try to see which space does it, what space does it actually show? What is the space shown by these inequalities? Is it a bounded space or is it an unbounded space? That is what the question is asking. I was also surprised to see C as an option, both bounded and bounded. It's like, you know, it's an oxymoron. Cannot be true simultaneously. Okay, should we discuss it? Five, three, two, one. Oh my God. A and B almost neck to neck. Nine people have voted for A. Eleven people have voted for B. Let's check, which is the right option. See, the first one is like minus x plus y less than equal to one. Okay, so if I have to draw this graph first, normally I will draw this graph, which is going to be a straight line like this. Now, which part of the region I have to shade? Origin. Does it, does origin satisfy this inequality? Zero, yes. So towards the origin. And since I have to be in the first quadrant, I have to make sure that I'm below this line. Okay, and I need not go into this quadrant. Yeah. Next one is minus x plus three y equal to nine. So I may write it as three y is equal to x plus nine y is equal to x by three plus three. Okay, so x by three has a slope of one third and it is cutting at this. So it will be mostly like a line, which is flat out like this. Okay, and cutting it out. Right. Now, I have to only bother about my first quadrant points because x one and x two both are greater than equal to zero. So again, does origin satisfy it? Does origin satisfy this inequality? Zero is lesser than nine. Yes. So below it. Okay. So below this. So below this means you are going to shade this part. Let me use a blue color marker below this. So I can see the overlapping zone. The overlapping zone will be this part. Okay. And this is clearly an unbounded region. So option number B is correct. Can you, can you special reason for changing it to A? Why did you change your answer to A? Meanwhile, you got confused. Okay. Yeah, that's a very good question. If a feasible reason doesn't exist, will you call it as a bounded, unbounded or neither? I would call it as a neither actually. Call it as a neither. Yeah. Next. Oh, sorry, poll. I keep forgetting. Find the height of the cylinder of maximum volume that can be inscribed in a sphere of radius A. So find the height of the cylinder of maximum volume that can be inscribed in a sphere of radius A. Again, a familiar question. You might have done this before during your practice sessions. Okay. Unbounded region is out of the syllabus. Arjun is telling me. See, since I have not checked into this topic of linear programming problem and normally we don't teach it for J also. So you, you're the best person to tell me whether it is there or not. So is Arjun correct? He says unbounded is not a part of the syllabus. Okay. Thank you. Thank you for confirming that. All right. So almost two minutes gone. Nine people have voted. I can wait for 30 more seconds. Okay. Okay. All right. So five, four, three, two, one. Okay. So two options have got the maximum response 11 for B and five for A. Let's check, which is correct. See, in this case, I would take a, I would take a parameter to solve this question. So I definitely know that this is A and this is A. And let's say this angle is theta. Okay. So this cylinder, the volume in terms of theta would be pi R square.  And height of the cylinder would be two A sin theta. Okay. So volume will become two pi A cube cos square theta into sin theta. And if you want the volume to be maximum, dv by d theta must be a zero. So if you differentiate it, this is to be treated like a constant. So k times derivative of this will be cos square theta into cos theta and minus sin theta into two cos theta sin theta. So this will be sin square. Okay. So cos cube theta is two cos theta sin square theta. So take off one of this because theta cannot be a 90 degree. So cos square theta is equal to two sin square theta. In short, one minus sin square is equal to two sin square. So sin theta is going to be one by root three. So this length is going to be two by root three option number B is correct. Clear any questions? Yes, yes, you can go by the pathological theorem. But in case you feel that assuming a parameter to relate the two unknowns, for example, here the two unknowns is your radius of the cylinder and the height of the cylinder. So now in this case, what I have used is I have related them to a parameter theta. Okay. So both R and H could be written in terms of theta, thereby converting the entire expression as a single variable expression facilitating the differentiation process. But you can always use the process of writing it as a Pythagoras theorem also and doing it depends upon question to question also. Would you like to take this question? Because somebody told me that the optimization problem has been taken away. The maximum value, sorry for that spelling, the maximum value of P which is your objective function in this case, x plus three y says that two x plus y is lesser than equal to 20, x plus two y is lesser than equal to 20 and x and y both are positive. Is which of the following? I relaunch. Okay. Fine. Agreed. Chalo. Okay. Seven people have voted so far. One and a half minutes gone. I'll stop this poll in the next 30 seconds. So please give forth your response within that time period. Not really Karthik, but yes, I thought I would make my diagrams. Okay. Five, four, three, two, one, go. All right. I have got almost 21 people voting and 12 of you say option C. Okay. Let's check. So as you can see here, first of all, I have made the two lines here, two x plus y equal to 20, the white line and x plus two y is equal to 20, the blue line. So the region which is going to be common to this will be this part. So this is the convex polygon that you get. So as you know, the optimum values occur at the corner positions. Okay. So we'll see the corner positions of this. So this is zero, zero. This is already 10 comma zero. And this point, we need to figure it out by simultaneously solving these two equations. So let me write it as two x plus two y equal to 40. So y is equal to 20 and oh, sorry, this is forward. So if you subtract it, three y is going to be 20. So y is going to be 20 by three. And if y is going to be 20 by three, x is also going to be 20 by three. Okay. So now what do we do? We make a table. We make a table. So this is my value of the objective function for this following x comma y. So this is my objective function x plus three y. So the corner points are zero 10. Zero 10 will give this as 30. Okay. Zero zero will anyways give you a zero. Okay. 10 zero will give you a 10 and 20 by three comma 20 by three. That will give you a 20 by three plus plus plus plus plus 20, which is going to be 80 by three. Am I right? Out of these four, it is clear that 30 is the maximum because this is 26.66. So the maximum value is obtained at zero comma 10 position. So that is going to be your maximum value. And the answer is option C. Okay. So this is the optimization. But of course, my feasible solution is bounded feasible solution zone is a bounded convex polygon over here. So this is a part of your syllabus. Is it fine? Any question? Any questions? Any concerns? Okay. Now let me give you a vector question because then the other people, the ISC people will feel that I'm only taking CVSE questions. Okay. Let's take this one. By the way, you can also, CVSE people can also try thinking as if it's a, you know, practice for your comparative exams. Let A and B be two equal vectors inclined at an angle theta. Then which of the following option gives you sine theta by two? Which of the following option gives you sine theta by two? Easy question. Last 30 seconds. 5, 4, 3, 2, 1, go. Okay. I've got almost 15 people responding and 14 people responding. Most of you say option A. Okay. Let's check it out. See here, let's say mod A and mod B both are equal to lambda H. Okay. So let's start with this expression.  We have already discussed in our vector chapter. So this is mod A square, mod B square minus two A dot B. Now mod A mod B both are lambda H. So we can write it like this. And here also we can say two mod A mod B, cos of the angle between them. Okay. So this becomes two lambda square cos theta. Okay. This is mod A minus B the whole square. So this is two lambda square minus this. So if you take common, it becomes one minus cos theta. And one minus cos theta clearly is two sine square theta by two. So sine square theta by two is going to be mod of A minus B the whole square by four lambda square. Four lambda square you can write it as four mod A square. So that clearly implies sine theta by two would be mod A minus B by two mod A, which is your option number A, which is your option number. If you would have started with mod of A plus B, then you would end up getting one plus cos theta and that would not lead to sine theta by two. That is why I started with mod of A minus B. Is it fine? IST students especially. Any questions here? Any questions? Any concerns? I could have started only with two expressions. Either I could have started with mod of A minus B or mod of A plus B. But as I told you, I predicted that if I do mod of A plus B square, I will end up getting one plus cos theta kind of an expression. So of course a bit of reverse engineering I did in order to start with. Fine. All right. So let the poll be on for this. Easy question. People making mistakes here also. Okay. We'll stop the poll in the next 20 seconds. You can write your answer on the chat box if you want to change it. Okay. Fine. All right. So five, four, three, two, one. Okay. So again, again, again, very few people polling in. 16 of you voted and out of which nine of them say, nine of you say option C is correct. Okay. Let's check. See AP and BQ are two vertical polls. Let me name them. One is of length 16. Another is of length 22. This length is 20. Okay. Then the distance of a point R on AB says that RP square plus RQ square is equal to minimum. Okay. So what is happening? Let's say there is a point that you choose here. Call it R. So let's say this is X and this is 20 minus X. So you want RP square. Oh, in the same way, RP square. RP square will be 16 square plus X square. And RQ square. RQ square is going to be 22 square, 20 minus X the whole square. And you want this to be summed to be minimum. This derivative should be zero. That means if you differentiate it, you get 2X minus 2, 20 minus X. This should be equal to zero, which clearly means X is equal to 10. Okay. So then the distance of a point R on AB from the point A, this is going to be 10. If you double differentiate it, if you double differentiate it, you will end up getting 4X minus 40. So derivative of this will be equal to four, which is a positive term. That means at this point, this expression, let me call this expression to be E. So E will be minimum in this value of X. Okay. Simple question. Which of the function defined below is a 1, 1 function? No, not necessary. Second test, you did not do it. Easy question again. Six people and all the four options have been touched upon. Okay. Almost two minutes. I can give you 30 more seconds at the max. Five, three, two, one. Cool. Okay. 25 of you voted out of which 10 of you say option B, but 14 of you say non-B also. So that is more concerning. That means more people are wrong than maybe right. I'm not sure whether B is right or not. Okay. We'll check it out. Okay. Let's discuss. See, first of all, the function mentioned in option A domain is zero to infinity. Please note that here. The minimum point of this will come at X equal to two. That means at X equal to two. Okay. Let's say, I'm just assuming this function is going to be negative there. Yeah. So at two, this function is going to take a turn, something like this. So it has to be a many one function because this function will be cut by a horizontal line at more than one point. So this A cannot be my option. Okay. But the second fellow, the minimum point will be, or you can say the turning point for it would be at minus two. Right. So let it turn at a minus two. Okay. So at minus two, its value will be some negative number. So let it turn at a minus two. But my analysis only starts from zero onwards. So I will not be bothering about what is the function to the left side of zero. So this is clear cut a one-one function because it will be cut by a horizontal line only at one point. Okay. So option B is the right option for sure. Now, how do I know C and D are not? C will definitely be not the right option because F1 and F minus 1 both give you the same answer, isn't it? That is E plus 1 by E. So this cannot be my answer. How do I know D cannot be my answer? Okay. Let's do a small test over it. Yeah. If you want this guy to become a zero, then let's see for how many values it becomes zero. For it to become zero, this guy should become a one, isn't it? So for ln of x square plus x was going to be zero, this guy should become a one, which clearly happens for two values. One is zero and other is minus one. That means two inputs zero and one gave you the same output. Hence, it cannot be a one-one function. So only option B is correct. And more or less, it's only single option correct. Oh yeah. You can always perform the calculus test as well. Yes. Is it fine? Any questions? Any concerns? Anybody? All fine? Okay. Let's take another one. Yes, yes, yes. You see the stationary point and then you see what happens to the sign change in the derivative. Is it showing you both the signs, positive, negative, both in and around that stationary point? If that happens, that means the function will not be a one-one. So in the mentioned domain, it should either be always positive or always negative, not both. Yes, Fakul, definitely Saturday, Sunday. Please focus on English. Okay. Don't do any math, physics, chemistry during that thing. See, these classes are only revision classes. Okay.  Don't, no worries. Sorry, I didn't put the poll on, my bad. Tomorrow, you normally have a physics class, right? Which class you have? I mean, depending upon the reminder, CBC people, you please keep attending the classes until next week also. No harm. ISC people exam is starting this Monday itself. So they can, they can afford not to attend the class. Okay. Because boards are there, nothing more important than boards now. So that's the reason why we have stopped the monthly tests also for you all, so that all of you can focus on the board. So CBC people, you can attend. ISC people, you can take an off. Please prepare for your boards. No comments on that as of now. Okay. Read the question properly. It says symmetric only, reflexive only, transitive only and equivalence. It just, it did not just say symmetric. It is a single option, correct? Okay. Very easy question, but I've got only 12 responses. Why is that so? I can give 30 more seconds. Okay. Five, four, three, two, one, go. All right. So 21 people have responded out of which 15 say option D is correct. Okay. Let's check. See here. Clearly A, B will be related to A, B. Why? Why? Because ABB plus A is same as BA, A plus B. Am I right? Because this is also commutative and this is also commutative. So it is reflexive, but is it only reflexive? Let's check. Let's say if this is related to C comma D. Okay. In short, I'm saying ADB plus C is equal to BCA plus D. Then let's check what happens whether this is related. So let's say this is a correct statement assuming that this is a correct statement. So if I do this, it will be CB D plus A is equal to DAC plus B, which is actually correct because this AD has become DA and this B plus C has become C plus B.  So this term is this term. So this statement is true because of the assumption that one is true. So assuming A comma B is related to C comma D, it implies that C comma D is also related to A comma B. So it has to be symmetric as well. So what happens is option A and B both go for a toss straight away. Both these options go for a toss straight away. Okay. And since it is reflexive and symmetric both, only transitive will also go for a toss. So only legitimate option is equivalent solution. But let us prove its transitivity also. So for transitive relation, we'll say let A comma B be related to C comma D, which means AD B plus C is equal to BC, BC A plus D. In short, you're trying to say one by B plus one by C is one by A plus one by D. And let C comma D be related to E comma F. That means in short, you're trying to say one by D plus one by E is equal to one by C plus one by F. Okay. Just add them. So one by B plus one by C, one by D plus one by E is equal to one by A plus one by D plus one by C plus one by F. One by C, one by C goes one by D plus one by D goes. So it basically says B plus E by B E is equal to A plus F by A F, which means, which means A F B plus E is B E A plus F. Isn't this saying that A comma B is related to E comma F, because by the relation that we had learned B E A plus F will be equal to A F B plus C according to this, which is basically true from here. So it is transitive as well. So it is transitive as well. So it is an equivalent solution, but you don't waste doing this test because you already know that it was both symmetric and reflexive. So it cannot be transitive only. So ABC are gone. D is the only possible option. Okay, don't waste doing tests which are not required. Save your time and energy for other questions. Okay, next one. No, no, no, it's not my favorite question. It's just that this question is seen at so many places. My favorite question. Two minutes gone. Two and a half minutes gone. Almost 18 of you have responded. Should we discuss it now? Five, four, three, two, one, go. Okay, so D is the most chosen response. Let's check. See, this function is defined from a set of natural numbers to integers. Okay, now looking at the option, they are basically interested in knowing whether they are 1, 1 onto or not. Okay, now let's check. Is this function a 1, 1 function? Is this function a 1, 1 function? So 1, 1 function, yeah, we'll check. 1, 1 function basically will assume that let X and Y be two natural numbers, okay, such that it is giving you the same output. Okay, so now there are various possibilities. If X happens, X and Y both happen to be odd, then definitely X minus 1 by 2 equal to Y minus 1 by 2 implies X equal to Y. No doubt about it. If X minus Y both happens to be even, then minus X by 2 is equal to minus 5 by 2 again implies X equal to Y. That means only two same inputs can give you the same output. Okay. But if one happens to be odd and other happens to be even, okay, then what will happen? Then what will happen? So if you compare this result, what will happen? X minus 1 by 2 is minus 5 by 2. Am I right? In such cases, you would realize that yeah, this is not possible only. Why? X and Y are both natural numbers. So even if you take the minimum natural number, 1 for X, Y will become a 0. So it will become unnatural. Okay, so this situation cannot arise. So it has to be 1, 1 function. So option number, this is ruled out. This is ruled out. Okay. Now whether this is on to or not, let's check that as well. So on to means any integer you take, it should be coming from some or the other number. So normally integers, we can classify them to be even or odd. Okay. So if you take any integer, let's say I'm taking some examples. Let's say I take a 5. Okay. So n minus 1 by 2, if you put it as 5, that means 11 gives you 5. Yes or no? If you take 4, if you take 4, n minus 1 by 2 equal to 4, n is equal to 9. So that is fine.  If you take a negative number, let's say minus 3, then minus n by 2 equal to this, that means n is equal to 6. So this guy takes care of all the negative numbers. This output takes care of all the negative numbers and this output takes care of all the positive numbers. Okay. So whether positive even or positive odd, there will be some odd number for which you will get that answer. Okay. And this is going to take care of all the negative integers. Is it fine? So it has to be on to as well. So this option B is also ruled out. Hence option number D is the only option. By the way, many students have done this question so many times that they remember this result that it is a 1-1 and on to function. That is why I think Fakuru was suggesting that it is my favorite question. Is it fine? Any question, any concerns? All is on. I think Sir Raghav seems familiar to me as well. Very good. Can we stop the poll in the next 10 seconds? 5, 4, 3, 2, 1. Okay. So A is the most chosen response. Let's check whether A is right or not. See, if X and Y are equal, then definitely this will be root 2. That means you will get only root 2 which is a irrational number. Okay. So X is related to X for sure. It is a reflexive relation. So you can stop there. But is it a symmetric relation? Let's check it out. See, root 2 will be related to 1 in this case because if you do root 2 minus 1 plus root 2, this is definitely an irrational number. But is 1 related to root 2? No, because in this case you realize you get a rational number. So it is not going to be symmetric. Is it transitive? Give me an example. Okay. Let's take this example. Same thing. Root 2 is related to 1. Okay. 1 will be related to, let's say, 2 root 2. So 1 minus 2 root 2 plus root 2. This is definitely an irrational number. But is root 2 related to 2 root 2? Let's check. So root 2 minus 2 root 2 plus root 2. This is actually 0. 0 is a rational number. So this relation doesn't hold true. Okay. So it is not a transitive as well. Now I'm solving this question from general perspective, not from objective level question. Of course, objective level if I have to solve, the moment I realize it is reflexive and I'll stop there. Okay. But if they say reflexive only, then there would be an issue because none of these is also sitting. In that situation you need to take care of that. Are you getting my point? So watch out for options like reflexive, reflexive only. Because reflexive only, you have to ensure that it is reflexive only, not just reflexive. Is it fine? All right. We'll take one more question and then we'll probably, you know, go for a break or something. Small break. Did we do this question yesterday? I didn't, I don't think so. Okay. Let's do this domain of this function. Okay. Very good. Eight of you have responded. We'll close it in the next 15 seconds. Five, four, three, two, one, go. Okay. This is a rare occasion when 100 out of 100, 100% vote goes to one of the options. Okay. So log of x square by two should be between minus one to one. So x square by two should be between half to two. That means x square should be between. Okay. So x square is between one to four. This only happens in option number C. They all look very similar. So it's only that their brackets are different. So one minus one doesn't create an issue because there's an x square sitting over here. So there's no point not including minus one or something like that. Okay. So option number C is the right option. So we'll take a break over here. Six, one PM. Let's meet at what time? How much 15 minutes you want or 10 minutes you want? Okay. Fine. Let's meet at six 15 with more questions. All right. I hope people are back. So let's get started with this question. This function for sine cube x minus six sine square x plus 12 sine x plus 100 is which of the following options? Just two people responded almost one and a half minutes gone. Okay. We'll close this in another 15 seconds. Five, four, three, two, one. Okay. I can see only 11 of you have responded to this and out of 11, eight of you say option B. Okay. Let's discuss this question out. See, first of all, what's the derivative of this function? Derivative of this function will give you 12 sine square x into cos x, 12 sine x into cos x plus 12 cos x. So you can take cos x common. Okay. Yes or no? If you take 12 common, you have sine square x minus sine x plus one into cos x. Now, this sum that you have over here is actually sine x minus half the whole square plus three by four. Okay. That means this sum will always be positive no matter what. So everything is decided by this guy cos x. Okay. So your cos x will be the whole answer decision-making over here. So in pi cos x graph, we all know is this. Okay. I'm just making, I'm just making a, of course, we're not going to pi it and all. So pi two, three pi by two cos x is negative. So this derivative will be negative in that case. So the function would be decreasing in that interval. So increasing in pi two, three pi by two is not correct. Decreasing from pi by two to pi, that is correct because it is negative option B is right. Decreasing from, yeah, decreasing from minus pi by two to pi by two. That is also not correct because from pi by minus pi by two to pi by two cos x is positive. Decreasing from zero to pi by two, that is also not correct because it is positive in that interval. So only option number B is the right option. Is it fine? Any questions? Any questions? Any concerns? Let's move on to the next one. Yeah, let's move on to the next one. Most of these questions we have already taken in the last determinant question. A minus B, B plus C, A, this is the determinant. Okay, very good. This is equal to which of the following options. Okay. So here, none of these basically comes as a problem factor for us, right? If that none of these were not there, maybe a special substitution would have definitely helped us out. Last 20 seconds, 5, 4, 3, 2, 1, go. Okay. So 11, 13 of you responded and 8 of you out of 13 say option number C. C for tonight. Let's check. See, if I were you first, I would have done this operation C2 transform to C2 plus C3. Okay.  So I'm not writing obvious steps. And this is what I was telling Prakul a little while ago. So let's say if you know that you can do that step in your mind, don't ask the person to write it. Are you getting Prakul what I'm trying to say? Okay. So things that you can do in your head, no need to ask the person to write that. Now, from here, I can, I mean, it's easy to move from here, I can do this operation R2 transform to R2 minus R1, R3 transform to R3 minus R1. So take the first row and subtract it from the other two rows. So that gives us A minus B1A. This will give you, if I'm not mistaken, 2B minus 2A, 0B minus A. And this will, this will give you, yeah, it'll become C plus B minus 2A, 0 and this is C minus A. Correct me if I'm wrong. Okay. Now let's expand this with respect to the second column. So minus one and you already have A plus B plus C. So minus one times, this is gone, this is gone. And you can do B minus A also common you can take, isn't it? So I can take B minus A common. So two and one, B minus A is gone. So it's minus one, 2C minus 2A, minus C minus B plus 2A. So 2A, 2A gone. So it's going to be negative one, C minus B, so that will become B minus C. That will become B minus C. Am I right? Anything that I'm missing out, do let me know. Okay. Convinced. So does this match with any one of the options? Does it match with any one of the options other than none of these? Let's verify also. So does it match with AQ plus BQ plus CQ? I don't think so. 3BC definitely not because a lot of terms will be left. And this is anyways A plus B plus C times A square plus B square plus C square minus AB minus BC. That is not going to come from here. So it's none of these. It's none of these. Yes, in original C1, okay. So you want to do that operation also that you can do. If you do that, you will get minus B minus A minus A. And then you could take B plus C plus A common also 111 ABC. So does it help us? Yes, of course, we can do that step as well. C1 minus C3. Okay. C1 minus C3 and then C2 plus C1. So that gives you CAB will come in between. Okay. It'll give you a standard result. Okay. So anyways, in this case, option D is going to be the right option. Oh, you are getting that once again. Let's do the operation which she is suggesting. So first you're saying C1 minus C3, which I already got it. Okay. You want the minus sign to be here itself? Then you add it. What does it become? Minus B minus A minus A. You're saying add C2 and C1. C1 transfer to C2 C1. Okay. So that will give you CBA. Then what do you do? Oh, you read it by mistake. Okay, fine. So anyway, so since you realize your mistake, we will not be wasting time doing this any further. Next question. We had mostly covered all these questions yesterday. Okay. Find the angle of intersection between y square is equal to x and x square is equal to y. And I didn't set the question. It was taken up from some of the resource material. Okay. Most of you have said option number B. Should we stop now? Okay. Let's discuss this out. So what is the point of intersection? It is very obvious that the point of intersection is either the origin or 0, 1 comma 1. Okay. So dy by dx for this. So 2y dy by dx equal to 1. So dy by dx at 1 comma 1 will be half. So for the curve 1, it will be half. Let's call it as m1. And dy by dx for the other one is going to be 2. Let's call it as m2. So tan of theta is mod m1 minus m2 by 1 plus m1. So that's 2 minus half by 1 plus 2 into half. That's nothing but 3 by 4. Yeah. So tan theta is 3 by 4. So tan inverse 3 by 4, option number B is correct. Clear? Next. Quick fire. The slope of the tangent to the curve x equal to t square plus 3t minus 8 and y is equal to 2t square minus 2t minus 5 at the point 2 comma minus 1 is, which of the following? Very good. Six of you responded already. Seven of you responded already. Very good. Last 20 seconds. What kind of a curve is this that can only be known once you eliminate t? t can be easily eliminated. You just eliminate that the second degree term. You'll get a t from there and just substitute it in, substitute it back in one of the curves. Doesn't appear to be the ones which are known to us maybe. Okay. Meanwhile, let's close the poll. 5, 4, 3, 2, 1. Okay. 17 of you responded. 16 of you say option B. Here, first of all, it is very important that if you find your dy by dx, you'll use this formula dy by dt divided by dx by dt. Okay. So, dy by dt will be 4t minus 2. Okay. And dx by dt is 2t plus 3. But in order to know what is the slope, you need to substitute the t value. What t value will you substitute? For that, you need to see which t simultaneously satisfies this. Okay. So, for that, you solve only one of them. Let's say this is easier for us to solve. Okay. So, this is nothing but t plus 5 times t minus 2. So, does 2 satisfy? Let's check. So, when you put a 2 here, you get an 8 minus 4 minus 5. Does it give you a minus 1? Yes. So, 2 is the one which you have to put here as well. So, this question mark will be 2. So, answer will be 8 minus 2 by 4 plus 3, 6 by 7. Option number B is right. Okay. Please ensure that when you're getting the two different values of t, one of them might not satisfy. One of them will not satisfy because for a particular point, there can only be one t possible. So, t is like, I always keep on saying t is like that Aadhaar card number for that point. You cannot have multiple Aadhaar cards for the same person. So, the one which satisfies both the condition that is this and this, that only has to be put in your expression for dy by dx. Is it fine? Let's take the next question. We'll take some from, should I take some from 3D geometry? Give me a second. How many of you have done 3D geometry in school? I know I have not done it in symptom. In school, how many of you have done it? Okay, not in YPR, not in many of the places. Okay. Anyways, we'll take some from vectors first, then we'll go back to the, let's take this question. If A is 4i plus 6j and B is 3j plus 4k, then the component of A along B, then the component of A along B is which of the following? Poll is on. Maybe 775, I will start taking some 3D questions for some of you. Others, if you want to solve, you can solve, else you may drop the session, drop off the session. Not an issue. See, I hope you understand the meaning of a component of a vector in the direction of another vector. Okay, so let's say this is your vector A. Okay, so what do you have to find out? You have to find out this blue vector. Okay, this blue vector is the component of A in the direction of B. Okay, so you need to tell which vector is going to be this vector, not the length. Length is the projection. The vector itself is the component. I think I've given you enough hint. It's very simple. That is the projection length, propyl. You need to find out the component of A along B. I've already drawn the diagram also. Refer to the diagram and find this vector, which I've shown by this blue vector. Should we solve the poll now? 5, 4, 3, 2, 1, go. Right, correct in shook. So a confused response. A and B almost equal number of votes. Let's check. So see, what is this length? This length is A dot B cap and the unit vector along B is B cap.  That will be your answer. So A dot B cap, what is, by the way, this is B vector. They should not write a cap over it. This is a B vector. So B cap will be 3J plus 4K by 5. So dot product of this will give you 12 plus 24 by 5. So this will be A dot B cap. And again, a B cap means scalar. Yes, sir. Sir, A has INJ. It will only be 60. I'm so sorry. Thank you for reminding me that. So this will only be 18. Okay, 18 by 5. And B cap will be 3I plus 4J by 5. So answer would be 8 by 25, 3I plus 4J. Option number B is correct. Option number B is correct. We got most number of votes, by the way. 3J plus 4K. Is it fine? Any questions? I think we had done all these questions. If the function F is given by XQ minus 3X plus 3, there are four statements given to you. One is X equal to plus minus 2 are the only critical points for local maxima or local minima. X equal to 1 is a point of local minima. Local minimum value is 2. Local maximum value is 5. Then which of the following statements are correct? These are the most dangerous of objective type question when there is options within options. Very good. 5, 4, 3, 2, 1, go. So 18 of you have responded to this question. So 14 of you say option number D. Let's check. So F dash X is equal to 3X square minus 3. So the critical points will only be plus minus 1. So this option, wherever there is a 1, you cross it out. A and C gone. Now, double derivative will give you 6X, add X equal to 1, it gives you positive. So 1 is a point of minimum and minimum value will be 1 minus 3 plus 3, which is going to be 1. So 3 is wrong. So wherever there is a 3, so B is gone, only D is left. So D has to be the right option, but let's check what is the local maximum.  So this 4th, 4 is correct and 2 is correct. So 2 and 4 are true statements. So option number D was right. All right. So I'll take one 3D geometry question. Let's see how much does it interest other people. I'll give you an idea how 3D geometry questions will normally look like. Whether you try it or not, it's a different thing. Okay. For the ISC students, one 3D geometry question, find the point of intersection of these two lines. Apply your common sense. It looks like six lines. This is basically how a 3D symmetric form of the equation of a line looks. Don't worry. We'll be doing it very soon. But just to keep everybody interested that, okay, this is how 3D questions look like. We'll check Kinshukh. I'm not sure of the answer. Even I have to solve it along with you. Nafla, at least I would request you to give your response on the chat box also, because I'll get all answers which like CVC people will be giving who are already not aware of this topic. Okay, Raghav. Anybody who wishes to put forth their response? Okay. Anusha wants to really give it a try. Go for it. Okay. Arjun, good try. We'll discuss it. I know you have not done it in school also. You have not done it in center also, but I would really appreciate if somebody gives it a try and tries to solve it. Right, Shatish. Okay. So what do we do? Okay. We'll stop the poll here. Most of you have gone with option B. Okay. Let's check. So we normally take this to be some parameter. Let's say R and S. I'll not take S. I'll take a T because S and 5 looks very similar. So normally the point which you'll get from here is X is 3R minus 1. Okay. Y will be 5R minus 3 and Z will be 7R minus 5. And from here you'll get T plus 2. Sorry. T plus 2. 3T plus 4 and 5T plus 6. So if they really meet, this should be equal to this. This should be equal to this and this should be equal to this. Correct. So from these two equations, let's see what do we get? 3R minus 1 is T plus 2 and 5R minus 3 is 3T plus 4. Okay. Let's solve for T and R over here. So this equation gives you 3R minus T is equal to 3 and 5R minus 3T is equal to 7. Multiply this with a 3. So this will give you 9R minus 3T equal to 9 and 5R minus 3T equal to 7. Subtract it. 4R is equal to 2. So R is half. So if R is half, so 3 by 2 minus 1 is T plus 2. So this is going to be half minus 2 is equal to T. So T is going to be minus 3 by 2. But the point is, does this satisfy the other condition? First of all, let's check that. So 7 by 2 minus 5. Does it satisfy minus 15 by 2 plus 6? Yes, it satisfies. That means, yes, intersection is happening. And at what point are they intersecting? You can obtain it by putting T as a minus 3 by 2 here or R as a half over here. Doesn't matter. So if you put R as a half over here, it gives you a half.  So the point of intersection is half minus half minus 3 by 2 and indeed it is option number B. All those stories will take up when I do the question, when I do the topic with you. If I did not get the third condition, very good question, if my third situation did not get fulfilled by the values of R and T that I have got over here, means those two lines will be like this. Not intersecting like a road below a flyover. We'll talk about this topic. I just thought I would just give you an exposure about 3D lines also so that mentally you are prepared that these kind of questions you will be seeing in your later on concepts. Next one. An isosceles triangle of vertical angle 2 theta is inscribed in a circle of radius A. The area of the triangle is maximum when theta is 5 by 6. The area of the triangle is minimum when theta is 5 by 6. Let's talk about maximum in my head. Vertical angle is this. Let's make a different color. Why does option number C even exist? All right. Three people responded one and a half minutes already gone. See, I mean just to give you this thing, either both could be false or only one could be true. Only A and D have to choose because it cannot be minimum. Minimum will be zero. So either it will be maximum at 5 by 6 or this 5 by 6 figure itself is wrong. I mean, I'm just taking a devil's advocate. I'm just thinking out of this thing. But if I have to mark this particular problem, I would only focus on this figure 5 by 6. Whether it is coming or not for maximum area. Other options are just bogus options. Okay. Both cannot be true that you already know. Cannot be minimum. Minimum is zero. But yes, both could be false if this figure 5 by 6 doesn't come, which also is very less likely. Okay. Let's check. I've given you enough hints. It's already three minutes gone and only 11 of you have responded. Yeah, you can use that. But again, that length x and y that you're talking about, A and B you're talking about, that themselves will be variables as of now. Okay. And there's already assigned two Theta sitting over there. Sorry. There will be two Theta sitting over there. So most of you have gone with A. Let's check. See, area of this triangle is half base into height. Okay. Since it is an isosceles triangle, this will also be Theta. Correct. And if this is Theta, this will also be Theta. Correct. Right. So what will be this angle? 5 by 2. Okay. So that angle will be 5 by 2 minus Theta. Correct. Now, once again, why did I choose this to be give me a second? Yeah. If this is 2 Theta, this is pi minus 2 Theta by 2. So this angle is this whole angle is 5 by 2 minus Theta. All right. Yes or no? Okay. Now, so for area of the triangle, I would need its base and height. So this part would be 5 by 2 minus 2 Theta. Okay. Any questions? Any concerns? Now, so if this is A, this base will be A cos 5 by 2 minus 2 Theta and height will be A plus A sin 5 by 2 minus 2 Theta. In short, you'll end up getting half A square sin 2 Theta 1 plus cos 2 Theta. So in order to have a maxima or a minima, the derivative of A with respect to Theta should be 0, which means constant.  Okay. Anything that I missed out, do let me know. In other words, you're signed to say that sin square 2 Theta is cos 2 Theta 1 plus cos 2 Theta. So here, this could be written as 1 plus cos 2 Theta times 1 minus cos 2 Theta. This and this will get cancelled off, cannot be 0. That means 2 cos 2 Theta is equal to 1. That means cos 2 Theta is equal to half, which means 2 Theta is 5 by 3, which means 5 by 6 figure is correct. So it has to be option A, nothing else. Is it fine? Any questions? Any concerns? That's a very good question you asked. How to maintain calm? See, many times what happens is we check the entire question paper and that disturbs us mentally. This is an important thing that I would like to discuss with you. See, this is the difference between board exam and competitive exam. In competitive exam, you have to scan the entire paper to see where are the easy questions lying. And you have to solve them first correctly in order to get confidence. What is the difference in a board paper like this? Is there anybody who is going to leave some questions? No, you have to attempt all the questions at the end of the day. So I would not suggest you to look at the entire question paper. I would suggest you to start from the first question onwards because any ways you are going to attempt that question. There's no preference of one question over the other, which normally happens in a competitive exam where you will find easier questions and try to do them. Tougher questions you may not even attempt to them because there is a negative mark involved. So don't look at the entire paper because if the paper happens to be normal, board papers are not difficult. If it happens to be difficult, you will get demotivated. You will be scared of some question which is about to come and you will be thinking about that question. Oh, I saw that question and I don't know how to solve it. Oh my God, what to do? And that will affect your present question also. So treat the question as per the merit as they come. If you feel you can solve it, go for it. If you feel you are not able to solve it, but still you have to take a guess because there is no negative mark in your board exam. So many of you get demotivated if at all you happen to see a question which you feel is difficult. Many of you get over enthusiastic. Paper is easy. Like that. And you start making mistakes. Both the situations are bad. And yes, so for 12 ISC people, you have some reading time. So I would say, I mean, read the first few questions and start mentally solving it maybe. Ten minutes you can mentally solve three, four questions. Especially, you have to do that exercise only. So this is what I would suggest. But again, it is up to you how you manage your, this is not applicable to everybody. It is only applicable to those people who get overexcited. I mean they are having some kind of an anxiety issue that they become very happy to see a easy question or very slightly sad to see a difficult question. But in most probable scenario, you will be getting very easy question. Okay. So don't worry too much about it. So my problem is the second scenario. That means you should not get over happy and enthusiastic and overlook some tricky aspects of the question. Are you getting my one? So this is what I would suggest. I mean, up to you to make some changes to this approach. Okay. So we'll take some more AOD questions. This would be the last question and other people, non-CVAC people, you can stay back. We'll do some 3D questions. If other wants to do 3D question, you can stay back else you can leave the session. After this, we will be taking some 3D geometry questions, 3D geometry and vector questions. A right circular cylinder which is open at the top and has a given surface area will have the greatest volume if its height and radius are related by option is all. I take rest for cool. Relax. Take a breather. Go take a walk. Okay. Relax. Keep telling yourself. I am important. Right? No exam. Nothing is more important than me myself. Keep telling yourself like this. Anybody who has got this anxiety issue. Keep reminding you are the most important thing for yourself. No other thing is no other thing is as important as you are. Okay. Let's discuss this. Most of you have said option number D. So here the cylinder is open from the top. So surface area of the cylinder will be 2 pi Rh and pi R square. Okay. And you want the cylinder to have the greatest volume. Now see in this type of problem, what is happening is that you have a 2 variable involved and those 2 variables are linked to each other by this relationship. Okay. So try to convert it to a single variable function. So how do we do that? Very simple. I can get the edge from here as s minus pi R square by 2 pi R. Okay. So pi R square H. Okay. This will be pi R square H. Is it fine? Okay. So this will get cancelled and this will be R. Right. So it is R s by 2 minus pi R q by 2. That will be your volume. Anything that I have missed out, do let me know. So dv by dr should be 0. That means s by 2 minus 3 pi R square by 2 should be 0. Okay. In other words, s is 3 pi R square. Okay. If s is 3 pi R square and if you put it over here, you will end up getting you will end up getting 2 pi R square as 2 pi Rh. Clearly meaning H is equal to R. Option D is correct. If they just mention about surface area in general, you have to take everything, top, bottom, everything. Okay. All right. So people, CBC people can leave, ISC people can stay back if you want and we'll do some questions on vectors and 3D as well. Slide 31 if you want to see. Give me a second. Hariharan, anything that you would like to ask? This will be theta. Okay. So a few questions from vectors first. A, B, C are 3 vectors says that A plus B plus C is a null vector and mod A is 2, mod B is 3, mod C is 5. Find A dot B plus B dot C, C dot A. Yes. Should we discuss it? A plus B plus C. A plus B plus C is adding up to null vector. Now see, if you see the dimensions, that should give you some clue. I can see few of you are here, plenty of you are here. Okay. Yeah. Let's close this poll in another 30 seconds. Okay. Let's discuss this 5, 4, 3, 2, 1. Okay. So this scenario, this scenario will arise only when something like this happens. A is 2. Okay. B is here. Let's say B is 3. And C is exactly like this. Then only A plus B plus C will be 0. So if I say A dot B, if I do A dot B, A dot B will be minus 10 because they are oppositely directed. Angle between them is 5. Okay. B dot C, B dot C will be minus 15. Correct. This is A, B, C. Sorry. A dot B will be plus 6. I'm sorry. A dot B will be plus 6. B dot C will be minus 15. And C dot A will be minus 10. So overall, this answer becomes minus 19, which is option number C. Is it fine? Any questions? Okay. These two lines are co-planar for what values of K? Okay. Raghav, which option did you go for? Okay. Anybody else who wants to give it a try? Should we discuss it now? Okay. Raghav wants to change the answer. Okay. Fine. So six of you have responded. You can see that most of you are confused here, 2, 2, 2 here. Okay. Now see, if these three lines are co-planar, please note that the vector connecting these two points, which I've shown with the yellow dotted line and the vector formed by 1, 1 minus K, okay. And the vector formed by K, 2 and 1 must be co-planar. Okay. In other words, IE plus J, I should not use IJK here because K has already been used over here. I will use, okay. Let me use it for my shoes. Minus KK. This and the difference between them, which is nothing but IE minus J minus K. These three vectors must be co-planar. So if you remember, we had done the STP formula. STP formula says that if three vectors are co-planar, the volume of the pallet of pipette will be zero, right? So 1, 1K, K21 and 1 minus 1 minus 1. This determinant should be zero. Okay. So it's as soon as solving this determinant. So let's solve this. So this will be minus 2, minus 2 plus 1, which is minus 1, minus K minus 1 and plus K minus K minus 2. This should be zero. So let's simplify this. So it's minus 1 plus K plus 1 minus K square minus 2K should be zero. In short, you'll end up getting K square plus K is zero. So K is zero or K is minus 1. Let me know if I missed out anything. Oh, this is minus. Sorry about that. Sorry. So this will become a minus K. I'm sorry. So it's minus 1 plus K plus 1 and plus K square plus 2K is equal to zero. So it's K square plus 3K is zero. So K could be zero or K could be minus three. Is that fine? Any questions, any concerns? So we'll take one more question. Last question for the day. Simple conceptual question. Oh, yes. People choose all the options, but for the right option. Correct. Oh, yes. Sorry. Simple question. Okay. Five, four, three, two, one, go. So A cross B is a vector. So they should not write as zero. They should write as zero vector. Okay. So these are some small glitches in this. Most of you have said option number C. See, if A and B are non-zero vectors, then this says A and B are perpendicular. Okay. If at all you are considering A and B to be non-zero vectors, then this means A and B is perpendicular. And this means A and B are collinear. Okay. But both of them cannot happen simultaneously. That means any one of them has to be a null vector. So either A or B has to be a null vector. So option C is correct. All right. So ISC students, all the best for your upcoming exam next week.  But CVSE people, one more week we can spend together solving more and more questions. Okay.