 Alright, so this is the first problem for the day. Let's start with this. We have to find the value of x which satisfies this particular equation. I'll be putting the poll on. Almost one and a half minutes gone. I have got one response so far. Good, good, good. Now responses are increasing. Six people. Almost two minutes gone. By the way, many of you are still joining in. So good evening, everyone. So we are up again with the problem solving sessions for your upcoming semester one exams. This is the first problem we have started with. Okay, so let's wrap this up in another 30 seconds. Five, four, three, one, go. Okay. Only 15 of you responded out of which 12 of you have to say option B. Okay, let's check whether B is right or not. So first of all, what's this cause of cot inverse x? So let's write it down in terms of x itself. So normally cot is treated as base by perpendicular, right? So it will become x by hypotenuse. Hypotenuse will be this. Okay, so the first term over here will become x square under root one plus x square under root. Okay, x plus under root of one plus x square. Next is sine of cot inverse x. So again, cot inverse x means x by one. This is your, this is your base. This is your perpendicular. So it'll be one by under root of one plus x square. So if you combine these two terms, so the given LHS will become, the given LHS will become, is my voice proper now, Shraddha? Yeah, in between, I think the internet was slightly weak. Now it is proper. Okay, thanks. So this is going to become x square plus one by under root of one plus x square whole square, which is 51 by 50. This clearly means this clearly means x square plus one is equal to 51 by 50. So x square is one by 50. So x is going to be plus minus, plus minus one by five root two. So any one of these two answers which is sitting over here that is good enough for us. I think one by five root two is sitting over here. Option number B is definitely right. Okay, well done. Most of you got this right. Any questions? It is still breaking. Let me check. I am on Geofiber. Yeah, that's fine. Should not be breaking actually. Yeah, sure. Fine. Any questions? Okay, I'll switch off my camera for something because bandwidth issues may be there. Okay. Now is it proper? All right. Should I move on to the next question now? Next. Sign of cot inverse one plus x is cos of tan inverse x. Then x is which of the following. Pole is being relaunched. Same strategy you have to adopt here as well. Use your reference triangle. One minute gone. Only two people so far. I can give 45 more seconds from here on. Five, four, three, one. All right. So this time participation has drastically improved. 51% of the people have replied and most of you have said option number D. D for Delhi. Okay, let's check. So here look at the first expression. Let's say I call this angle to be, I call this angle to be theta. Okay, so cot theta is given to us as one plus x. So if you make a reference triangle, if you make a reference triangle, cot theta is one plus x means base by perpendicular is this. Okay. So this is automatically under root of two plus two x plus x square. Am I right? Square of this plus one under root. So sign of that angle theta. So this is your theta angle. So sign of the angle theta that would become one by under root of two plus two x plus x square. Okay. In a similar way, let's say I call this angle to be five. Let's say I call this angle to be five. So tan five is x. So make a reference triangle. This approach is very well known to us. By the way, this was theta here. This was five. So this is your perpendicular by base. So this is x. This is one. So this is under root one plus x square. So cos of five, cos of five, that would become one by under root one plus x square. So as per the question, these both expressions are equal to each other. Okay. So from the question point of view, these two are equal. These two are equal. So one by under root of two plus two x plus x square is equal to one by under root of one plus x square. So that clearly means this is equal to one plus x square x square x square gone. So two x is equal to minus one. So x is equal to minus half, which is option number D. Okay. D is absolutely right. Well done. Any questions, any concerns here? Any doubts? Can I move on? Next question. Simple one. Poll starts now. Something wrong with the poll. Super easy. Idly should not be taking more than a minute for this. Excellent. Very good. Five, four, three, two, one, go. Again, in same lines. Okay. Most of you have voted for option number C, which is 15. Let's check. So on the same lines, tan inverse two is seek inverse. Can I say root five? Yes or no? So treat this as, treat this whole angle as theta. So again, when you make a reference triangle, tan theta is two. Tan theta is two means this is two. This is one. Okay. So this hypotenuse will become root five. So seek theta is hypotenuse by base. So it will become this. So seek of seek inverse root five square you have to do. So that's nothing but root five square. Okay. So that's five. That's five. Similarly, cot inverse three is cosec inverse root 10. Okay. Again, let's call this as some angle five. This is five. So cot five is equal to three. So this is three. This is one. So this is root 10. So cos cosec of five will be root 10 by one. So the same angle is cosec inverse root 10. So cosec of cosec inverse root 10, the whole square is going to be root 10, the whole square. That's nothing but a 10. So your answer is five plus 10, nothing but 50. Simple question. Option number C is correct. Any question anybody has? Should we move on? Okay. So let's take something from continuity differentiability chapter. The whole is on. That's a good question guys. Should not take so much time. I can only, I can only see five of you, six of you responded so far. 45 seconds more. Yes, it's done. Yes, yes. We can implement that also in the previous question. Correct. Okay. Let's start the countdown. Five, four, three, two, one, go. Okay. 17 of you have participated in this poll. Out of 17, 12 of you say option B. Let's check. See, look at each of the options. If you take a number less than zero, let's say I take a minus one. So your function will become cos of one by X divided by X when X is not equal to zero. Okay. Now, when you, when you're trying to find out, let's say the left hand limit, okay, what will happen? What will happen? This is a bounded quantity. Please remember, this is going to be a bounded quantity. So this is somewhere between minus one to one. Correct. And this is going to be a very, very small negative number. So it will be some, some fixed quantity divided by a very, very small negative number. So this result will go to minus infinity. And on the other hand, if you find the right hand limit, the same is going to give you something which is plus infinity. So of course, your upper and the lower limit, sorry, the left-handed limit and the right-hand limit will not be equal to each other and forget about it being equal to the value. They're not, they don't exist first of all. So this option cannot be my right option. Okay. If I take something greater than zero, let's say I take a one. So X cos one by X, I'd already discussed with you that this is an auxiliary function, but this guy arrests that oscillation in the neighborhood of zero. Okay. So do you remember this was done in our regular session of continuity and differentiability? So what happens that when you evaluate the limit of this as X tends to zero minus, again, this is a bounded function. This is somewhere between minus one to one. So this is of the nature bounded into tending to zero. So overall, this will be zero only. This is not infinity. If there was an infinity here, then I would say it is still an indeterminate form. Okay. And same will happen with your right-hand limit as well. So when you try to find the right-hand limit, that will also be a bounded function into something very close to zero. So that will still be a zero and that is equal to the value of the function also. Okay. So anything greater than zero is going to work. Even if it was X square, X cube or whatever number greater than zero, this is still going to be satisfying the fact that left-hand limit, right-hand limit both will be zero each and that will be equal to the value of the function. So B is the right option so far. Equal to zero, it is not going to be continuous at zero because it will be oscillating, which I already discussed to you. And because there's an equal to zero over here, this also cannot be your answer. Okay. It should be greater than zero strictly. It cannot be equal to zero also. So option B is right. And I think some of you got confused and mark D as well. What do we need to do if we want this? Okay. Very good question. So a follow-up question is what should be the value of K says that this function is differentiable at zero. So first of all, K should be greater than zero for it to be continuous. Okay. That is the necessary condition. Let us see what is the sufficient condition. So now if you want this function, so let's find out F dash zero minus. So that will be limit extending to zero plus F of zero plus H minus F of zero by H. Okay. So F of zero plus H means F of H that will be something like this. Okay. And very important thing is that this value should match with, oh, I'm so sorry. I think I had written zero plus here and this is zero minus. So for zero minus limit extending to zero plus F of zero minus H minus F of zero by minus H. So this is, this will be minus H to the power of K cos of one by H minus zero by minus H. So both should give you the same result. Both should give you finite and the same result. So here what will happen? It will become H to the power K minus one into cos one by H. Okay. Now see this guy should be greater than equal to zero. Yes or no? Greater than equal to greater than equal to zero or greater than equal to one. See H extending to zero plus if you want this answer to be finite. Okay. Aditya, what should be the value of K minus one? Greater than equal to zero. Can it be equal to zero? If it is equal to zero, it will become cos one by H. Correct. What are the limit of cos one by H as H tends to zero? It's oscillating, right? So this quantity should be a quantity which should be positive. Correct. Some positive value will be, should be there. Sorry. Greater than equal to zero. Am I right? Aditya, what is your route? Sir, it can't be equal to zero, right? Why? Any positive value that comes here, H will be on the numerator, right? So this will become zero. So K minus one, K should be greater than equal to one in this case. In fact, it should be, sorry, it should be purely greater than one. Yeah. Purely greater than one. Yeah. Got it. Because equal to one will make this get, getting, this will get cancelled and then the limit will not exist. Yeah. Same here also. You have minus, you have minus H to the power of K minus one, cos one by H. So K minus one should be greater than zero. So of course, the same criteria will hold true on this as well. Ideally, shouldn't it be infinity? What shouldn't be infinity? Kinshukh, unmute yourself and talk. What is your doubt? Power should be anywhere greater than one, which includes a very high power also. Why should the power be infinity? It should be a power greater than zero because you want this H to exist in some form in the numerator so that it arrests the oscillation of this fellow and make it as a finite limit. That is the concern here. So that can only happen when K minus one is greater than zero. Okay. If K minus one is equal to zero, that means K is equal to one, then what will happen? This will result into cancellation of H and H, giving you cos one by H. Limit of cos one by H as H tends to zero is an oscillating value. It's a value somewhere between minus one to one. This will also result into the same thing. So you cannot compare the two results. So you can compare the two results of your left hand limit and right hand limit only when you have certain trace of H left on the numerator to arrest the oscillation of this guy. Makes sense. Can we move on to the next one now? Sorry, I think poll was on. In case the poll is on and it is blocking your view, please bring it to my notice ASAP. This is an easy question. We have already done this type of question during our regular sessions. Last 30 seconds. So all of you have got your admit cards from the school. How is the exam happening? It is happening in the school premises only. Okay, great. And external examiners will be coming, right? Okay, all right. So five, four, three, two, one, go. Okay, so there has been a split of the votes between B and D, but B of course has got double votes than D. So there are people who think none of these is also an answer here. So first of all, let us understand here. What is the point of discontinuity of f of x itself? f of x is discontinuous, discontinuous at x equal to one. So this itself will also be the point of discontinuity of f of f of x also. Okay, now let's evaluate f of f of x that is one by one minus one by one minus x. So that will give you, if I'm not mistaken, one minus x by minus x or something, but x minus one by x. So this function will be discontinuous at x equal to zero. Okay, so this will also be the point of discontinuity for f of f of x as well. f of f of f of x as well. Okay, so these two points are already my points of discontinuity. Okay. So if I just have to take a call based on this, I would go for B, but let us also check out what is f of f of x equal to. So in this, of course, if you put your x as one by one minus x, let's see what do we get. So this is going to give you an x divided by, okay, they will all get cancelled. So this doesn't have any point of discontinuity. So overall, these are the two points where your composite function f of f of f of x will be discontinuous. So option number B is correct. Many of you who answered D, you probably would have taken a call just on the basis of this, which is not correct. Okay, because the function is not defined. The function is not defined at one. So you can't put a one in the function. The function is not defined at zero. Right. Yes or no. So you can't put a zero because f of f of x, that is not going to exist. Okay. So these are also included under the points of discontinuity of that final function as well. Is it fine? Any questions? Can we move on? Simple question. Something wrong with the poll relaunch. Want to continue? Can you see the poll now? Two minutes gone. I can get 30 more seconds. Okay, five, four, three, two, one. All right. Very less than 15 of you have only responded. Most of you have said option number C. Okay, let's check. So the first statement, dy by dx, it's dy by d theta divided by dx by d theta. So dy by d theta, that's very obvious. It's going to be minus a sin theta. And this is going to be a one minus cos theta. Okay. So this is broken down into half angles. Right. And this also can be broken down as two sin square theta by two, which is clearly, which is clearly nothing but negative cot theta by two. So first statement is definitely right. So B cannot be a wrong option. Sorry. Yeah, this cannot be the right option on because it says only two is true. And neither one or two is true. D is also gone. Okay. So people who have voted for a, sorry, B and D. Okay, they're wrong. Now let's check whether only one is true or both one and two is true. Let's check that. So for that, we need to differentiate this guy also second one. So dy by dt. So for dy by dt, we can use our quotient rule. So that will become, I'm just doing it directly over here into derivative of sine cube minus and derivative of this is going to be minus two sine two t upon two under root cos two t four divided by is it fine? This and this will get cancelled. So on simplification, this is going to give you three take multiple. Can you see my screen? I'm audible. Okay. Yeah. So this is going to be three cos two t. And this is anyways, sine square t cos t. Okay, here it will be plus and this will be sine two t into sine cube t. So you can write it as two sine to the power of four t into cos t. Is it fine? One more simplification I can do here is I will write this cos two t as one minus two sine square t. Are you able to see my screen now? Thank you. No, screen not visible. I'm audible. Hello. Just give me a second. I'll check my internet connection. Is it unstable? Can you see my screen? Is it fine? Okay. So if I check this out, this is going to be three sine square t cos t minus four sine to the power four t cos t upon this. So I think a bit of simplification is, you know, involved over it. Okay. So let's say this is my result for dy by dt. For dx by dt, let's do the same expression. So for dx by dt, dx by dt is we are differentiating this expression. So that is going to become or I think I did dx by dt. Sorry. So I was thinking how the same expression is done. Yeah. So let's do dy by dt now, dy by dt. So dy by dt will be, okay. This also we have to simplify in a similar way. So I'm just writing the simplified step version of this. So this will be minus three cos two t cos square t sine t plus plus cos cube t. And this is going to be again two sine t cos t by cos two t under root of cos two t. Okay. Now this time here you write this fellow as two cos square t minus one. So when I do that, I'm going to make this replacement over here itself so that we save a bit of our simplification time because the problem is not difficult, but it is taking some time to simplify. So when I do that, I will end up getting plus three cos square t sine t and minus six plus two will give you minus four, minus four cos to the power 40 sine t. Okay. Yeah. So this term is your dx by dt and this term is your dy by dt. So what is the ratio of, let me call this as one and let me call this as two. So two divided by one. Two divided by one will only leave you with, just make a barricade here. Yeah. Two divided by one will leave you with three cos square t sine t minus cos to the power 40 sine t upon three sine square t cos t minus four cos to the power, sine to the power 40 cos t. Now we can cancel off, we can cancel off one sine from both numerator and denominator. Okay. We can cancel one cos from both numerator and denominator. Okay. Now what is this formula? The numerator formula is clearly the formula of negative cos three t, isn't it? Because cos three t formula is four cos to t minus three cos t and denominator is the formula of sine three t, which means it is negative cot three t. So is the question set us saying that this is negative cot three t? Yes. Means option or statement number two is also correct. So both one and two are going to be the right options. Okay. But yes, this question comes with a bit of simplification involved and is slightly time consuming. Okay. So try to write a cos in sine in abbreviated format. Don't write it like SIN, COS, etc. That will take a lot of your time. So try to write in an abbreviated format. Okay. But again, don't do it if you have not practiced it so far. Don't start experimenting things in the examination part. Can we move on to the next question? All is on. Simple question. Okay. Can you check that out? Done? Okay. Most of you have gone with option number A. Okay. Let's check. Let's check. Now, this question, if you differentiate it once, you end up getting negative a sine x minus b cos x. If you differentiate it twice, it becomes negative a cos x plus b sine x. Yes. So n could be two. So a option is definitely correct. But some of you are also claiming that even six is going to work. Okay. We will verify that as well. So triple derivative is going to give you negative, sorry, triple derivative is going to give you a sine x plus b cos x. Fourth derivative is going to give you a cos x minus b sine x. Fifth derivative is going to give you negative a sine x minus b cos x. And sixth derivative is going to give you negative a cos x plus b sine x. Yes. Or even option c is correct. Okay. Now, here, many times, normally we discuss the nth derivative formula with the students. So nth derivative of sine of a x plus b. Okay. I'm not forcing anything, anybody to remember anything. But if you could just keep this in your mind, it might make your life simpler in many situations. Okay. And nth derivative of cos of a x plus b is a to the power n. Same expression, but with terms of costs. Okay. So this is something that I would like you to note down and keep it may save your time provided you are differentiating multiple times. Okay. So in our present situation, in our present situation, what is happening is you want this cost to become negative costs. And you want this minus sign to become plus sign. So n equal to two ways basically works fine over here. Provided you have put your a as a one and b as a zero. So do that. So a to the power n will become a one. Okay. So this formula, in case you need it, I mean, I'm not saying that it is absolutely important to know these formulas. In case you need it at some places where you are performing a higher order, a very high order derivative, then you can keep this formula in your mind. Okay. But yes, in the present example, both a and c options are right. Is it fine? Can I move to the next question? Let's take this one. Oh, a function is defined from R to R as this. Find the value of a so that this function is continuous at zero. Definitely I'll share this recording. No issues. I've been already doing it. Okay. Five, four, three, two, one, go. Okay. Most of you have gone with option number, option number D, D for Delhi. So it's just a question of limits actually. So for the function to be continuous, your value of the function at zero, okay, that is your a must be equal to the limit of the function as x tends to zero. So your a value will be limit extending to zero to sign x. And this is just to sign x cos x by two x cos x. So this two will go off. This two will go off. Take a sign x common one minus cos x upon cos x and x. Okay. So this is as good as a tan x by x. This is as good as this is as good as tan x by x. And you can just split this limit as two separate limits. So this is anyways one, but this is a zero. So your overall answer is zero. So option number D is going to be right. Is it okay? Any questions? Any questions? Any concerns? Next question. Oh, yeah, Prakul, I remember you having, you know, doing this question exactly. Now, I remember that. So how are your exams going on? ISC people? So far so good. Paper has been coming easy. Okay, good as of now. No problem with the use of scribe. I mean, I hope there's no delay and all time till four. Till what time to what time is the writing time for ISC people? Kinshuk, here M is an integer. M is any integer. Okay, I can give you 30 more seconds. Simple question. Okay, five, four, three, two, one, go. Awesome. 100% people have said option number C. Okay, let's check. So it is very obvious that this term should be, this term should be of the nature. Correct? Right? A multiple of pi plus pi by two, because at pi by two, if this becomes discontinuous, then it is periodic. The whole function is periodic with pi. So at every m pi plus pi by two, the function will start facing infinite discontinuity. In other words, x by x plus one is m plus half. Right? You just take a reciprocal of it. So one plus one by x is equal to two by two m plus one. Subtract a one. Okay. So if you subtract a one, it becomes one minus two m by two m plus one. Take the reciprocal, it becomes option number C. Oh, sorry. Option number C. Is it fine? Oh, good. Good, Prakul. Should we move on to the next question? It's a one-minute question actually, lesser than that. If you know your properties of differentiability well, basics, basics, Prakul, paper will come easy. Okay, you just have to control your emotions. So practicing harder problems will not, what you need to practice is calming down your nerves and trying to read the question properly before answering. Okay. It is applicable to all of you here. Don't go for tough problems. You're not preparing for J advance here. Just keep your basic understanding clear. So last minute, I would just request you to just look through your inverse trig formulae because that is something which is slightly property heavy. And application of derivatives concepts, you can just practice a bit because that is the heavier of all the chapters. And just a bit of formula for a number of functions, bijective functions, how to find domain, range, etc. Beyond that, you don't have to worry too hard. Practice too hard. Okay, five, four, three, two, one, go. All right. Most of you have said option number eight. Now see here, this is a function which is like this actually mod x square is as good as x square. So this is a function which is continuous and differentiable. This is a function which is continuous but not differentiable at x equal to zero because this guy has an issue. Okay. So overall, a sum of a differentiable and non-differentiable function is non-differentiable only. But if you still want to make it differentiable, that is what the question is saying, I want to make it differentiable, then I don't want this guy to be there. So if this guy is not to be there, I have to, I should choose my a value to be zero. Okay. b could be anything. So a should be zero, b could be any real number. So option number a is going to be correct. You don't have to worry about doing anything else. Is it fine? Any questions? Should we go to the next one? Poll is on. Oh yes, these special brackets are gif's. This bracket is the gif function. Okay. If you feel multiple options are right, please give me a response on the chat box as well. Okay. Should we discuss it now? Five, four, three, two, one. Okay. I think if more than one option is right, then this poll has no meaning. Okay. So see, they're all talking about neighborhood of two. So let us redefine the function close to two. So if you talk about very value, which is very close to two, let's say less than two, I would say, I would not say zero to two. I'll just say less than two, slightly lesser than two. So lesser than two, gif of this will become a one. So that will be as good as an x. Okay. Equal to two, if you check this function value will be, or let's say greater than equal to two, also you can take simultaneously. This function value will be just x, x minus one. Okay. Now, is this function continuous at x equal to two? If you check the left hand limit is two, right hand limit is also two, right? Because this, when you put, it'll get a two. This also, when you put, you get a two into two minus one. So this is definitely continuous at x equal to two. Is it differentiable at two? So for finding the left hand derivative at two, you just have to differentiate this and put x as two. By the way, there's no scope of putting anything. They'll get a one. And if you differentiate this, you will get two x minus one. And here if you put x as two, you will end up getting a three. Okay. That means it is not differentiable at two. So option c is correct and option b is correct. So bc is going to be the right option. Simple. Oh, I'm so sorry. Anusha, any special reason for finding it differentiable at x equal to two? Then what happened? x equal to one? Okay. Maybe this kind of functions will not be asked in board exam, but just wanted to check your understanding with respect to min function, min max function. I'm putting the poll on, but if you feel that there are multiple options, right? You can give me a response on the chat box. See, let me tell you again, my questions will be of a slightly higher level than what you will be getting. Okay. So this is because we are practicing it. We can't practice with easier questions. Have you seen player practicing for their cricket match? They will not practice from 22 yards. The bowlers will bowl from lesser than 22 yards for the batsman at least. Okay, Aishwetish. Okay, Aditya. Aishwetish wants to change his answer. Okay, five, four, three, two, one, go. Okay. 10 of you have only voted out of seven of you say option a. Okay. Fine. See, first of all, min function or max function, whatever special functions you're given with respect to min max, you make both the graphs. So mod x, x plus one is like this. Okay. And mod of x plus one is going to be a graph like this. So I've already told you for mod of x plus one, you make the graph of, you make the graph of x plus one first, which is going to be like this and remove the part of the graph, which is on the negative side of the x axis and reflect the graph, which is on the positive side like this. Okay. Now our graph will be the minimum of the two functions. So minimum of this blue and green one is the green one till you reach zero. And after that it is all, both of them are same actually. So it is final graph is actually a straight line y equal to x plus one. So it is differentiable everywhere option number a is correct. b is not right. Is it always greater than one? I don't think so. It is going down in the negative direction as well. Okay. So option only a is correct. Is it fine? Any questions? Any questions, any concerns? Are there any special reason for marking b and c? Oh, you plotted max by mistake. That's the thing I'm worried. Many of you will make a mistake because you read the question incorrectly, not because you don't know it. Yes, all is on. Okay. I've got one response so far. Okay. Last 30 seconds, just two of you have voted so far. Okay. Five, four, three, two, one go. Okay. So C and D have got the vote so far. And most of you say option number C, C for Chennai. Let's check this out. See, first of all, we need to know what is the function actually. Okay. For that, there is a simple approach here, which probably you would have used several times to replace your x with pi by two minus x. So that sign and cost interchange their positions. So when I do that, this is what I get. Okay. So on replacing x with pi by two minus x, we get that. Okay. Now think as if you have been given, you have been given something like this. Two a plus b is x and two b plus a is pi by two minus x. Okay. You have to find your a and b. Okay. Let's say I want to find my a. So what I'll do here, I'll multiply this with a two. Okay. Whole thing. And then we'll subtract the result. Okay. So when I do that, I get three a, two b, two b will get cancelled and this will become, if I'm not mistaken, it will become three x minus pi by two. Correct. That means three a, a is nothing but sin inverse x, sorry, sin x. Okay. Which means f of sin x is x minus pi by six. Now in order to make this, in order to make this x, you can change your x with sin inverse x. Okay. So again, replace your x with sin inverse x. So that will become f of sin of sin inverse x equal to sin inverse x minus pi by six. In short, you have got your function. This is your function. Okay. Now the question is asking us for the derivative that is clearly one by under root one minus x square. Is it fine? So option number c is going to be the right option. Any questions? Okay. Now try your hand on this. It's of the similar type. So those who could not get the first one, try your hand at this problem. One response so far. Okay. Last 30 seconds. Five, four, three, two, one. Go. Okay. b is the most chosen response. Most of you say it's half. Let's check. Again, the same story will change our replace your x with one by x. Okay. So when I do that, it becomes this, this becomes f of x and this becomes this. Okay. Again, think as if you have written three a minus two b is x and three b minus two a. So you can just write it minus two a plus three b is one by x. You have to solve for a thing like that. So multiply this with three, multiply this with or two. Correct. And add them. So when you add them, this will become five a minus six, six will get canceled. So this will become three x plus two upon x. So your a is one fifth of three x plus two upon x. Okay. And a is nothing but f of x. Okay. Now go for the derivative. Find the derivative at two. So the derivative here is going to become one fifth, three minus two by x square put x as two. So that is one fifth, three minus two by four. So that is one fifth, three minus half. That's nothing but one fifth of five by two. That's a half option number. Option number b is definitely right. Well, is it fine? Any questions? Easy question. Find the maximum value, find the maximum value of this expression. Very good. This question came in your pre words. Okay. Okay, five, four, three, two, one, two. Okay. And a four will only 13 people responded. And most of you have got option number c. Let's check c is correct or not. So let's take all this as y. So you have to take a log on both sides to the base of e. Okay. You can write it as negative x ln x. Okay. Differentiate both sides with respect to x. So it's this. Okay. Now put it to zero. Putting it to zero means putting this as zero, which clearly means ln x is equal to minus one. So x is one by e. Okay. Since this is the only thing and they're asking you which of there is the maximum value and none of these is not there, then probably putting this will give you a maximum value only. Don't waste your time checking the double derivative unless and till there is a none of these or unless and till you get one more option or one more value of x. So your answer is one by one by e to the power of one by, which is e to the power one by definitely option number c is right option. Is it fine? Any questions? Can you comment on the minimum value? See a function need not have a minimum actually. It's not necessary that there has to be a minimum existing. See this function, this function x to the power one by x. I don't know whether you have seen the graph of it or not. Okay. The graph of this function, if you draw, it basically starts with one over here goes high and then after that it starts diminishing. Okay. So when it is still falling, it doesn't have a minimum. Are you getting my point? It's like something which is rising. What is the greatest value? It is still rising. Infinity is not the greatest value as I already told you. The global maxima or the global minima, they are fixed values. You cannot claim infinity and minus infinity is to be global maxima or global minima. So it is still falling. So global minima doesn't exist. No tending to zero is not a global minima. That's the problem. Why should one be the minima? It has a lot of values lesser than one also. But this value is still dropping. That's what global minima is not defined. It is still falling down, down, down, down, down, down. It is approaching zero. Agreed. But you can't claim that the global minima is equal to zero. Exactly. Global maxima and global minima concepts when we were talking about global maxima and global minima, if the functions, if the function limiting value is beating the maximum or the minimum value occurring in it, then that particular absolute extrema is not defined. So let's say x square function. I give you a simpler example, x square function. It's global minima occurs at x equal to zero. But what is the global maxima? It's not defined. It does not exist because it is still rising. Okay. And there's no fixed value that you say, okay, this is the global maxima it attains. Same is happening with the case of this function one by x to the power x as well. Next one, if the parabola y is equal to f of x having axis parallel to the y axis touches the line y equal to x at one comma one, then which of the options is correct? Two minutes, two and a half minutes gone. I've only got two responses so far. Five responses I can give maximum 45 seconds more. Okay. Five, four, three, two, one. Very, very few people have only seven of you have responded to this. And out of that four of you say option B. Okay, let's check. See, first of all, a parabola whose axis is parallel to y axis means it's an upward opening parabola. So you can definitely take the parabola or your f of x to be your normal quadratic expression like this. Okay. So what are the things that is given to us in the option f0 and f dash 0? First of all, f0 everybody agrees will be c and f dash 0. f dash 0 will be nothing but b. So at least they're trying to say something like 2b plus c is one. This is trying to say 2c plus b is one. Okay. This fellow is trying to say again, 2c minus b is one. And this fellow is trying to say, let me not write here, let me write above the option, 2c minus b is one. And this fellow is trying to say 2b minus c is one. Okay, let's see which of them is correct. Now, a couple of things to be utilized here. First of all, the function touches this line at one comma one. That means the derivative of this function when x is one is actually one. Right. Which clearly means 2a plus b is one. Okay, let's call it as first equation. Second thing is the function will pass through one comma one. That means f of one is one. That means a plus b plus c is one. Okay. Let's call this as two. Now let's subtract one and two. If you subtract one and two, a minus c becomes zero. That means a is equal to c. If you put this back in the first one, you get 2c plus b equal to one. So which option says 2c plus b equal to one? 2c plus b equal to one. Option b says it. I hope you can read it. I've written it in yellow. So option b is right. Okay. Any questions? Any concerns? Can I move on to the next question? It's opening upward, downward doesn't matter. It's just a value is going to decide it, but it is of this nature. That is what I can say. So Aditya, it could open downwards also because I've never claimed a is positive. The equation of one of the tangents to the curve y is equal to cos x plus y from x line between minus 2pi to 2pi inclusive. That is parallel to this line x plus 2y equal to zero. Is which of the following poll is on test for the options. Okay. Please act as per the options because they say equation of one of the tangents to this curve. Okay. 45 more seconds I can give for this. 5, 4, 3, 2, 1, go. Okay. Very less people have voted this time also. 11 people only. Okay. 11 people and out of that seven say option number b. Okay. Let's check. Okay. First of all, a line parallel to this should off be, should be of this nature, right? Of course, all of them satisfy the same criteria. Now you want that line to be parallel, tangent to be parallel to this means having a slope of minus half. So first of all, let's find dy by dx. So let's differentiate both sides. So when you differentiate both sides, okay, this is what you end up getting. And now you have to put your dy by dx as a minus half because this is the slope of this line. And this is going to be half. That clearly means sin x plus y is one. Okay. So you can also say that cos x plus y is zero. That means x plus y is a or multiple of pi by two. Okay. So let's say I take a, yes, that's what. So this is going to give you y as zero, right? That means your x is off this nature. Now take any x, which as per the option is going to work out. I think let's try taking pi by two first. So if I take x as pi by two and y as zero, that means I want to find the tangent at this point. And let's say this is the tangent. It must satisfy pi by two comma zero. That means C is pi by two. So it falls in line with our option number B. Okay. So you could take any x of this nature. There could be so many x depending upon your n value, but I decided to look at the option B and decide to take n as zero. Okay. If you take n as one also, you'll get pi by two comma zero. In that case, your answer will become x plus two y equal to five by two. So there could be so many tangents. We want one of the tangents. That is what the question says. Okay. So option number B is going to be the correct option. I think we have done this question before. Okay. We'll skip it. This function is increasing in which of the following intervals. This question can be solved in less than 30 seconds actually. Okay. Should we stop in another 30 seconds? Five, four, three, two, one, go. Okay. End of poll. Most of you have gone with option number B. See, I would like you to recall the concept of increasing decreasing functions where we had talked about how do we decide the increasing decreasing nature for composite functions. Okay. So let us say I have a function F. I have a function G and I have a function F or G. Okay. Let's try to decide if the function F is increasing and G is also increasing. What could you comment about F or G? Will it be increasing or will it be decreasing? Increasing? Correct. Good. If this function is increasing and this function is decreasing, then what can you comment about F or G? See, G is a decreasing function. So if I put, let's say X1. Okay. And X1 is supposed to be lesser than X2. Let's say. Right. Then what will happen? This guy will have, if this G is a decreasing function, this guy will have a higher value. So this guy will have a higher value because F is an increasing function. G is a decreasing function. So for lower inputs, G will show higher values, isn't it? So if F gets a higher value, so this will be G of X1 will be more than G of X2, isn't it? So F of G of X1 will be more than F of G of X2 because F is an increasing function. So if you increase the output, sorry, if you increase the input, what is going to happen? Your output is going to fall down finally. Yes or no? So this is a decreasing function, no? How you people are saying increasing function? Shittich, any special reason for calling it as increasing function? Sinchen? Anusha? Okay. If this is a decreasing function and this is an increasing function, then what will happen? Let me make some space over here. If F is a decreasing function and G is an increasing function, what will happen to F of G? Decreasing only? I don't know if you agree with this. Okay. If this is a decreasing function and this is also a decreasing function, then what will happen to F of G? Check it out. So let's say X1 is less than X2. Okay. So G of X1 will be more than G of X2 because G is a decreasing function. And F of G of X1 will be lesser than F of G of X2 because even F is a decreasing function. So what is the nature here? It becomes increasing. It becomes increasing. Okay. Now, we will use a part of it in our problem solving. If you see this function, it is made up of this as your F and this as your G, where I definitely know that tan inverse is an increasing function. And the question setter is also asking the interval where this entire thing is an increasing function. So basically, we are catering to this need. So if your F is increasing and F of G is also increasing, that means we need to tell the examiner that even G should be increasing in that interval. So in which interval do you think sin X plus cos X is increasing? Now, this function is written in a harmonic form like this. Right. So this is a function which will be graph of sin X shifted pi by 4 to the left. Okay. So pi by 4 to the left. So this graph looks like this. At 0, it becomes 1. It goes surpasses. So this is your minus pi by 4. This is your pi by 4. Okay. And this is going to be 3 pi by 4. Okay. And this is going to be a pi by 2. Add a pi by 2 here. It will become 5 pi by 4. Anyways, so out of these intervals, out of these intervals, which is the interval where the function is increasing, is it pi by 4 to pi by 2, pi by 4 to pi by 2? No, it is falling. A cannot be my option. Answer. Okay. Minus pi by 2. Okay. This is going to be minus 3 minus 3 pi by 4 if I'm not mistaken. Correct. Yeah. So minus pi by 2 will fall somewhere in between. So minus pi by 2 to pi by 4. Is it increasing? Yes, it is increasing. Option B is definitely correct. 0 to pi by 2. Pi by 2 will be somewhere here in between. Okay. So 0 to pi by 2. No, it is showing both the nature. It cannot be my option. Minus pi by 2 to pi by 2. Again, it is increasing and decreasing both. It cannot be my option. So simple. Only by looking at the graph of root 2 sine x plus pi by 4, you can answer your question. Is it fine? No need to go for any kind of a derivative or not. So we'll take a small break over here. 601. We will meet exactly at 6.15 pm with further more questions. I think we have to touch upon matrix determinants and your concept of linear programming problems. So we'll take that up on the other side of the break. So see you in 10, 15 minutes. All right. So let's take this next question. Suppose alpha, beta, gamma are real numbers such that sine alpha, sine beta, sine gamma, none of them is equal to 0. Then delta, this determinant cannot exceed which of the following values. I'm relaunching the poll. So two responses. So far almost two minutes gone. Okay. Last 30 seconds. Five, four, three, two, one. Go. Okay. So I can see a confused response between A and B. Both of the options have got equal number of votes and one person has responded with option C. See, first of all, what we can do is we can take, we can take sine square alpha common from the first row, sine square beta common from the second row, sine square gamma common from the second row that will leave you with a standard determinant which you have actually seen before. Okay. So if you recall, you have seen something like one x, x square, one y, y square, one z, z square kind of a determinant. Am I right? And for such cases, we have also learned that I'm just recalling that concept for you. So for one, one, one, x, y, z, x square, y square, z square, the value of this determinant becomes x minus y, y minus z, z minus x. Okay. So you can use same thing over here. So it is going to become, it is going to become cot alpha minus cot beta, cot beta minus cot gamma and cot gamma minus cot alpha. Okay. Now, just a quick recall of trigonometry, cot A minus cot B is actually sine B minus A. Okay. Let me just write it down in an expanded way. Oh, sorry. Cos A sine B, cos A sine A minus cos B sine B. So this becomes sine B cos A minus cos B sine A upon sine A sine B upon sine A sine B, which is the formula of sine of B minus A by sine A sine B. Okay. So why I decide to, you know, take up this particular identity because each of these terms here, for example, this term, you can write it as sine beta minus alpha by sine alpha sine beta. This will become sine gamma minus beta again, sine beta sine gamma. And this becomes sine alpha minus beta by sine alpha sine gamma. Okay. So sine square terms will start getting cancelled. This gets cancelled. This gets cancelled. This gets cancelled. Okay. So basically, you're left with a product of three sine terms. Okay. You can, yeah. Okay. So you can write it as product of sine beta minus alphas. Okay. So this will always be less than or equal to one. It cannot exceed. It cannot exceed one. Okay. So option number one is correct. Or option number A is correct. Is it fine? Any questions? Any questions, any concerns? Next one. If A square plus B square plus C square is minus two, then the polynomial f of x is off which degree? Yes. Two and a half minutes gone. Last 30 seconds. Okay. Five, four, three, two, one. Okay. Take a call. Even if you're not able to solve it, take a call, take a guess. Okay. 10, 11 people have responded. Most of you say option number D. D means degree two. See here, first of all, in order to utilize this information, let's do this information, let's do this operation, add every column to the first column. So that clearly gives us, if I'm not mistaken, one plus take x common x square plus B square plus C square plus two. Okay. Yes, I know. Same here also. It will give you one plus, again, A square plus B square plus C square plus two. Okay. And one plus x again here, A square plus B square plus C square plus two. Okay. And let this be as it is. Be careful about the use of brackets here. Okay. Now, this is anyways going to be zero because you have been already provided that this is negative two. Okay. So this entire first column will reduce to 111. I'm making that change here itself. I don't want to write this again. Takes a lot of time to replicate the same data. Now you can use the fact that all the elements of one of the columns have become a one. So you can start doing this operation R1 as take any one of the rows. Let's say I take the third one and subtract it from all of them. Okay. So this will become a zero. This will become a zero. This will become, if I'm not mistaken, x minus one. Okay. Zero. This will also become a one minus x. Okay. And this also x minus one. Now expand with respect to the first column. So you will only get this into this and this into this, which is actually a second degree equation. Okay. So you'll end up getting negative of negative that is this whole square. So this is your function. So degree two, absolutely right. Option number D is correct. Any questions? Any questions? Any concerns? Yeah. Next question. X and Y are real numbers. Then this determinant lies in which interval? Almost two minutes. Okay. Last 45 seconds. Last 30 seconds. I can give max to max because already three minutes have gone. Five, three, two, one. Okay. A split between almost A and C options. Both of them have got almost equal number of votes. Okay. Let's check. See, let us do this operation here. Since there's a cos x plus y sitting over here. Okay. Cos x plus y is what? Cos x plus y is cos x cos y minus sin x sin y. Correct. Okay. Cos x is always sitting here. Sin x is always sitting here. Okay. So let me do this event. R3 is R3 minus minus cos y R1 and plus sin y R2. Okay. So what will this do to this expression? This will create a zero here. And thankfully, if you see this expression, if you multiply this with negative cos y, it will cancel out the negative sin x cos y here also. So it will create a zero here as well. Okay. Check it out. Okay. And this will create a sin y minus cos y here. If you want, you can expand it. If you think it is a shorter way to expand it and do it. But this directly gives you, if you expand with respect to R3, this will give you sin y minus cos y times plus minus plus minus plus. Yeah. Cos square minus sin square, which is minus minus sin square, which is one. So now if you have a function like this, the max value of this function is root two. The min value is negative root two. We have already done this in our trigonometry. So sin y minus cos y max value is root two. Min value is negative root two. So the interval in which your delta will lie is minus root two to root, which is clearly option number A. Is it fine? Any questions? Should we move on to the next one now? Yeah. If one plus A111 is equal to zero, A is not zero, B is not zero, C is not zero, then this is equal to. Pole is on. Can you all see the pole? Now it is on. There is a small lag in the appearance of the pole. Don't you think this should be 2C actually? I feel like this should be 2C. Oh no. That's fine. Okay. Last 30 seconds. Just four of you have responded. Five, four, three, two, one, six of you out of which, sorry, seven of you have responded out of which four of you say option C. Let's check whether C was correct or not. See, take A common from row number one, B common from row number two and C common from row number three. That will leave you with this expression. Yeah. Now do this activity R1 transform to this. Okay. So when you do that, you will end up seeing one by A, one by B, one by C plus three appearing in the first row. Correct me if I'm wrong. Okay. The rest of the elements will be as they are. Now take this guy, the first row guy common from the entire row number one. So this whole thing I'll take common out. Okay. So you'll have one by A, one by B, one by C plus three. Now, since you've created one, one, one in row number one, you can always go for the operation C1 is C1 minus C3, C2 is C2 minus C3, because we want to create enough number of zeros before we start expanding it. So this will be zero, zero, one, one, two, one by B minus two, minus two, three plus one by C. So when you expand this, you'll end up getting minus two plus four, which is two. So it makes it two ABC, one by A, one by B, one by C plus three. And this entire result is given to you as a zero. That means this whole thing is a zero. But ABC, oh, I'm so sorry. ABC are non-zero quantities, if I remember. So it is only possible to get a zero with these guys not being equal to zero when this is equal to zero. So one by A, one by B, one by C is negative three. Okay. So Janta, you were wrong. Option number B was correct. Most of you, I'll share the result again. Most of you went for option number C. Okay. Aditya, which option you got for this one? Okay, let's take some relation functions question. Simple one. May have seen this question a number of times while practicing relation functions of class 11th, maybe 12th also. Okay. Let's stop this in another 30 seconds. Five, three, two, one. Most of you have chosen A, but there is a strong conflict between A and D both. Okay. Now, first of all, is it a reflexive relation? Because yes, I feel that this will be a part of, sorry, this will be the part of the relation for all A belonging to real numbers. So it has to be reflexive. So reflexive, undoubtedly. What about symmetric? If this is belonging to the relation, then it obviously implies this will also be true, which means B comma A will also belong to that relation. So it means it has to be symmetric as well. It has to be symmetric as well. But is it transitive? That is something that we need to figure out. Now I would request all of you to think of an example of such number A, B, C. And I'm sure you can do it where A comma B, or you can say A is related to B by that relation. B is also related to C. But A is not related to C. Think of an example. Aditya, your A is also related to C only. Okay. Yes, a lot of examples are coming up, by the way. You can think of an endless number of examples. Okay. So one example, which I liked from Bever, he said, let A be minus three. So minus C is related to minus one because yes, one plus A B will be positive. Minus one will be related to half because one minus half is positive. But minus three will not be related to half because one minus three by two is negative. So it is clearly not transitive. So there is no, there is no scope of option D being correct. So option A only is correct. Okay. This anyways was ruled out because it was symmetric and it was reflexive. So A and D, the confusion was there. Now D is wrong. A is the right option. Is it fine? Any questions? Another easy question. I want the answer for this in 30 seconds. Poll is on. Simple. Okay. People have already responded. So let's stop. Five, four, three, two, one, go. Okay. F O F of this means you're first putting one minus root three into the function. When you're putting one minus root three into the function, the function is going to see any rational input coming. So I'll throw one at you. Okay. Now you're showing one which is a rational input. So this will give you a negative one. Option B is definitely right. Am I audible? Okay. Okay. Some issue with the net today. I don't know what is wrong. Anyways, we'll move on to the next question. If the function GF is defined and is one one, then which of the following option is correct? This is a theory based question. I think we had already discussed this in our functions chapter early in this year. Okay. Five, four, three, two, one, go. Oh, see the response everybody. Confusion confusion everywhere. Four people saying B, four people saying C, four people saying D. Okay. A quick recap of that theory, even though we're not going to the details of it, if there is a composite function, okay, and this happens to be one one. Okay. If this function is a one one, then this guy must be one one. This guy must be one one. Okay. And if a composite function is on two, then this guy must be on two. Okay. Just remember this theory. Very, very important. So if GF is one one, then F this guy has to be one one. So option number D is correct. Okay. So this question can be asked in either format. They can say FOG is one one. Okay. Then which of them should be one one? This guy inside function, the function which is fed inside. If that is not one one, there is no chance FOG can become one one. Because for two different inputs, G can give you the same output. So F will also give you the same output. Right. So it cannot be a one one in that case. Similarly, if FOG is on two, then F must be on two. That's very important. Could you give example where GF is one one, but G is not one one. GF is one one. G, O, F, this is one one. And you want okay. And you want G to be not one one. G is not one one. Not one one. Okay. What example we can have for this? No, Pradhin, his question is G, O, F. G, O, F is one one. Okay. And G is not one one. G, O, F, G, O, F is one one. Okay. Fractional part you want to use. You can make, you can make a function. I mean, not a, maybe not a well defined function, but you can just try to make a function. For example, let's say G is a function which follows this kind of a mapping. Okay. So let's say I just take some dummy numbers and let's say this is A, B, C. Okay. And let's say the mapping is F is a function. F is a function which gives you one and three. Well, let's say alpha beta. So in this case, in this case, if I do G of F of X, okay. So if I put an alpha G of F of alpha, it will become G of one. So that will give you an A. And if I put beta, it'll give me G of three, which will only give me a B. So this becomes a one one function. Will this example suffice? I mean, you need not always have an example coming from your daily life functions. What if beta gave two in F of X? Now, if beta were, if beta were giving you a two, let's say, then what will happen? This will become two. And G of two and G of one, both will give you A. Okay. So G of one and G of two will both give you A each. In that case, F of G will not remain a one one. Right? That's it. Okay. So I think you're trying to, you know, you're getting confused with here. If this is a one one, then F has to be one one. Please understand it. So I'm not trying to say if F is one one, then G of F will definitely be one one. I'm not trying to say that. No, that is not a, yeah. So Siddhij has another example. G of X is X square. And this is root X. Okay. X square root X that will give you a one one function and yes. Correct. So that's another example, which Siddhij figured out. He said X square for this and root X for this. Okay. So G of F is going to give you an X, which happens to be one one. Okay. But G of X happens to be X square. Okay. Now here also we have to be very careful. When you talk about F of G of X, your domain is decided by this guy. So X should be greater than equal to zero. Correct. So by definition of the function X square seems to be positive. I mean, not a one one function. So G, I didn't get that. What is this? G of F you wanted to write? Hmm. Technically, this is a one one function. And as per him, this is a many one function. See, I mean, you can always define, you can always define G to be from R to R. Let's say. Okay. And this function is like this. So in this case, it becomes a many one function. Okay. And he's defining F from R plus to R as root X. Okay. Now the question center asks you to find out what is this? So this function will be G of F will be X, but for X belonging to R plus. So the actual definition of G was from R to R. So it is a many one function. Correct. So this criteria is fulfilled. Whatever you are saying, it should be not one one, which is fulfilled. All right. So there could be so many examples. Let's move on with more problem solving. F of X is sin X plus cos X G of X is X square minus one then G of F of X is invertible in which domain. Okay. Let's close this in another 30 seconds. Okay. Five, four, three, two, one, go. Okay. Most of you have gone with option number B. Let's check whether B is right or not. See G of F of X assuming that G of F of X exists. So that will become sin plus cos square minus one, which is actually sin two X. If this function is invertible, right? It can only happen when two X lies between minus five by two to five by two, then only it will be one one. And of course you have to, you know, check for the arranged part of it. So X has to be between minus five by four to five by four. In fact, in reality, it could be any, an interval where the function is not allowed to take a turn back. So as per the options, I basically found out that this is the most exhaustive answer, which happens to be option number B. Okay. Next. Let's take some questions from simple question minus five by two to five by two will also work. Okay. Let's check. So this is your graph off. Okay. Sine two X. Okay. This is zero. This is five by two. And this is minus five by two. Right. How come you're saying it is going to work? It is not one one here. It is not one one here. Sine two X graph. You'll get almost zero marks if you do that. Yes. Yes. Should we discuss it now? Guys, this is again a simple question. You just have to find which region is the, see this line is obviously given to you two X, a four X minus two Y equal to minus three. Now just check origin. Does it satisfy this condition? A zero minus zero is less than equal to three. Does it satisfy it for that matter? Minus three. So this is not a part of your question also. Yeah. Does it satisfy this condition? Zero minus zero is less than equal to minus three. No. Right. You only have to look at the inequalities which are based on this line. Okay. Don't entertain A and C at all. B and D, one of them will be your answer. So does B, you know, entertain? No. Only D entertains that inequality. I'm so sorry. They're saying that a region which is shaded, right? So zero. Yeah. It should be the opposite. Yes. So once again, small. Yeah. So zero satisfies this inequality. D one. So this should not be your answer. So the answer should be B actually. Correct? D, right? The inequality which is not satisfied by zero, that is going to be your answer. Correct? So zero is less than equal to minus three. Correct? This inequality is not satisfied by zero. So B should be your answer. Right? So B is the right option. Yeah. B is only correct. B is only correct. Simple question. Okay. Let me solve the poll. I think most of you have got it. Okay. Five, four, three, two, one, go. Okay. So most of you, in fact, all of you have said option number B. Let's check it out. See, this is going to be a simple question. All you need to do is put these points in your objective function. Okay. So for this point, for zero comma eight, you end up getting minus 32. For four comma 10, you end up getting minus 28. For six comma eight, you will end up getting minus 14. For six comma five will get minus two. For five comma zero, you'll get 15. And for zero comma zero, you'll get a zero. Okay. Out of which the minimum is occurring at zero comma eight. This is your minimum value. Option B is correct. Okay. So these are all easy questions. Poll is on. Yes. Two people have responded so far. Okay. Five, two, one. Okay. End of poll. Most of you have said option number C. Most of you have said option number C. Okay. Now see, x plus y less than equal to seven, it has to be lesser than this line. Okay. And x plus two y is lesser than equal to 10. It has to be within this line. Okay. So obviously our desired zone is going to be desired zone is going to be this zone. What are the corner points? Zero, zero, zero, five, seven, zero. And this will be the meeting point of these two, which is going to be, if I'm not mistaken, four comma three. So we'll have to check what is the value of the objective function at these points. So p value at the points. So I think the minimum of them will come out to be zero, zero will give you a zero. Zero, five will give you a 10 and seven, zero will give you a 35 and four, three will give you a 20 plus six, 26. So this guy is the maximum of all 35 is the answer. Oh, Kiran. Okay. Most of you are going to lose marks because you will be in a hurry to just zip past the question. So read the question properly. That's very, very important. Okay. Let's take this question. The parameter on which the value of this determinant does not depend. You can do an expansion here because I think you should be getting technometric identities when you are expanding it at every stage. None. Just five of you have responded. Okay. Five, four, three, two, one, go. Okay. So seven of you responded out of which four of you say option B. All right. So let's discuss it. So if you expand with respect to, expand with respect to row number one, so it'll give you one, this into this minus this into this. If I'm not mistaken, that is nothing but it is sine P plus DX minus PX. So one into this, okay, minus A. Now, this into this minus this into this, that will give you sine P plus DX minus P minus DX. Okay. And last but not the least, this into this will give you sine PX minus P minus DX. So overall, if you see the first expression is just sine DX. Second expression is sine 2 DX and third expression is again sine DX. So there is no, there is no, you know, P seen anywhere. But yes, you can see A, you can see D and you can of course have X. So only it is independent of P. So it does not depend on P. Does this find any questions, any questions, any concerns? Last, find the value of K. I've already told you how to solve these kinds of problems. So it should not take you much time. Five, three, one. See to solve this kind of a question, I've already told you don't have to, you know, go beyond a simple act over here, just put your values of the ABC as something, for example, put something which will make one of the, you know, values easy to calculate. Don't put such a value which will make, you know, both sides become zero. So this will give you minus one, two, two, two, I think again a minus one, two, two, two, minus one. And this is going to be 8K. So expand it. So this will be one minus four, which is minus three. So minus two, minus two, minus four, which is minus six and plus two, four plus two, six. So that's going to be 20, how much? 1224, 24, 27. Have I done some calculation error? Just check. Minus two. This is minus two, minus four. Oh, sorry. This is 27K. So this is equal to 27K. So 27 is equal to 27K. K has to be one option number B. Is this fine? Any questions? Thank you. So with this, we come to an end. Tomorrow we'll have the last round of revision. Post that. We are going to meet only in December last week. Okay. Keep practicing. We'll meet again tomorrow. Thank you. Bye-bye. Good night.