 Hello, good morning. Welcome to the YouTube live session on board level problem solving So I would request you whoever has joined the session Please type in your names in the chat box so that I know who all are attending the session Okay, in this session, we're going to solve only board level questions Okay, and again the rule will be the same and until unless I get the response from at least four of you I Think the attendance is low today. So at least two of you I will not start solving the problem So let's start with the first question. This is a section a question. So we know section a has four question or one mark There's section B of two marks section C of four mark and section D of six markers So first question is a two and false question if the matrix is a diagonal matrix Having a diagonal elements as D1 D2 D3 then a inverse is the diagonal of D1 inverse D2 inverse and D3 inverse. So is this statement true or false? Now again, you must be knowing many formulas from your J But you know try to solve it in a proper way. Hello. Good morning, everyone So when I say a diagonal matrix means what the diagonal matrix means you have a Matrix whose diagonal elements are D1 D2 D3 So what will be the determinant of a determinant of a in this case? Let's see. I expand with respect to the row number one. It will be D1 D2 D3. Okay, Rohan says true for it What would be the adjoint of this matrix? Adjoint of the matrix is basically nothing but the transpose of the co-factor matrix of a So if I'm not wrong this co-factor of D1 will be D2 D3 Okay, co-factor of 0 will be 0 Again co-factor of this 0 will be 0 Again co-factor of this 0 will be again 0 Co-factor of D2 will be D1 D3. Okay, again co-factor of this 0 will be 0 Okay, co-factor of this 0 will be 0 again co-factor of this 0 will be 0 again co-factor of D3 will be D 1 D2 So we all know that inverse of a matrix is nothing but 1 by determinant a times adjoint of matrix So it's going to be 1 by D1 D2 D3 and we have a determinant over here. So that will give you 1 by D1 or D1 inverse 0 0 0 D2 inverse 0 0 0 D3 inverse So yes, this statement is true So the answer for this question is it's a true statement. Okay, simple one Moving on to the next question now Next is if abcr3 non-zero vector says that a dot b is equal to 0 and a plus B is equal to c then write the relationship between their magnitudes Then write the relationship between their magnitudes That means how is mod a mod b mod c related to each other So it's given that a dot b is 0 and a plus b is equal to c Again a very simple question. It should not take you more time. It's a one marker guys. Remember that One marker means less than 1.5 minutes. Okay done. So guys, it's very evident that when we talk about such a case Vector a and vector b are in such a way that they are perpendicular to each other because of this fact Okay, and resultant of a Resultant of a and b is equal to vector c Okay, so here b plus a is your vector c Okay, so it's very clear that mod of c square will be equal to mod of a square plus mod of b square But again, this is not the way you should be solving it. This is just for your own understanding When you're solving this in the exam since a plus b is equal to c take the dot product with The same vector on both the sides So it'll give you mod a square mod b square plus 2 a Dot b is equal to mod c square and this term here would be zero This term here would be zero. So it's mod a square plus mod b square is equal to mod of c square Okay, simple one Moving on to the next question now Next this is a question from binary operations Given that there is a set a which contains 1 2 3 4 5 0 1 2 3 4 5 and there's a binary operation star in Set 8 defined as a star b is equal to a b mod 6 a b modulo 6 means 1 when a star b is divided by this number 6 the remainder is a b So that's how they have defined it it represents the remainder obtained on dividing a b by 6 in the language of mathematics We pronounce this as a star b is congruent to a b modulo 6 is congruent to a b modulo 6 This is the concept which is actually Coming from the concept of number theory very simple. So if you have to find 3 star 5 means what is the remainder when you divide? a b by 6 so when you divide a b by 6 means you divide 15 by 6 your answer will be equal to 3 Okay, so this is equal to actually 15 mod 6 and 15 mod 6 is going to be 3 Okay, so answer is 3. Let's move on to the next one. A function is defined from r to r as f of x is equal to x cube plus 5 Find the inverse of this function. So this is I think Again last of the one marker after that will go on to the two markers In a one marker, you don't have to prove that it is actually one one and on two Okay, because it's just for one mark Right, so for finding the inverse we know the approach we first equate the function to y make x the subject of the formula So x will be y minus 5 to the power of one-third and this term is actually your f inverse y Right because if f of x is y x will actually be f inverse y remember this So f inverse y is going to be Y minus 5 to the power one-third which means f inverse x is going to be x minus 5 to the power one-third Okay, so this is how we finish the one marker now. We'll move on to the two markers Given that dy by dx is e to the power minus 2 y and y is 0 when x is 0 Find x when y is equal to 3 find x when y is equal to 3 You can type in your response once you're done Only you have to give the value. You have written in terms of y. That's fine. You have to give the value a Half of e to the power minus 6. Are you sure Vaishnavi? Check your answer Vaishnavi once again. I think yeah, I think you're correct. Let's let's check this So dx by dy is this So I can say dx is equal to e to the power minus 2 y dy Okay, let's integrate both sides. So it will be x plus c is equal to e to the power of minus 2 y by minus 2 Now in order to get the value of c we have been given this boundary condition that y is 0 when x is 5 so put 5 over here and Y is 0 means minus half that means c is going to be minus half minus 5 That's minus 11 by 2. Is that fine? So your function finally becomes x minus 11 by 2 is e to the power minus 2 y by 2 Now, let me put the value of y as 3 and see what's the value of x That becomes minus 6 by 2 so it's 11 There's a minus 2 over there. Yeah, okay So it becomes 11 minus e to the power minus 6 by 2. Yeah, now it's correct So this becomes your answer for this question moving on to the next question if b is collinear with a and a is given to you as 2 root 2i minus j plus 4k and mod of b is equal to 10 show that 2 times vector a plus minus b is equal to a 0 vector the small sign mistake It should be 11 minus e to the power minus 6 whole divided by 2 Yeah, take it easy. So basically your a is nothing but 2 root 2i minus j plus 4k Okay, and b is collinear to a and mod of b is given to us as 10 show that 2 a plus minus b is equal to a null vector guys you if you're done Just type done on your chat box so that I can start the discussion for this. All right, no issues Remember, it's a two marker not more than three minutes Okay, so on the area is done. All right So b is collinear to a means b is some lambda times a where lambda is some scalar quantity right lambda is some scalar quantity Okay, mod of b is equal to 10 mod of b is equal to 10 means a mod lambda under root of 2 root 2 square Minus 1 square 4 square is equal to 10. Okay, so this will be equal to mod lambda This will be 8 plus 1 plus 16 is equal to 10. That's going to be 5 So more lambda is going to be 10 upon 5 which is 2 implying lambda is plus minus 2 Okay, now, let's put it in 1. Let me call this as and 1 putting lambda in 1 We get b is equal to plus minus 2 a which means which means 2 a is equal to plus minus b which means 2 a plus minus b is equal to a null vector Okay, so that's what we had to prove over here simple So let's move on to the next one. So this is a probability question The question says a three dice Are thrown at the same time find the probability of getting 3 twos if it is known that the sum of the numbers on the three dice is six Do it accurately guys. No need to hurry up Okay, at least says one by 18 What about other guys? Again, it's a case of conditional probability, right? So first let's let's define the event always while in board exams always define the event. Let a be the event of of getting getting three twos okay be with the event of getting The sum on the three dice as six What do we need? We need to find probability of a by b. That means it's the conditional probability that that mean The probability of getting a given b has occurred. So according to the multiplication theorem Probability of a given b has occurred is p a intersection b by p b right Now what are the probability that you get a sum of six and you get three twos So that will always be one by two one six because only when you get two two two You are simultaneously getting a six also And you are getting all twos on this the die. So this is just one case like this total number of cases is Six into six into six because each die can show you six faces So here two one six is your sample space Now divided by probability of getting six so how many ways can you get six on? Three diced you can get it as one one four You can get it as one two three you can get it as one three two Then one four one Then two one three then two two two two three one You can also get three one two 2, you can also get 3, 2, 1, you can get 4, 1, 1. So these are the cases which are hominy number 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So this becomes 10 by 2, 1, 6 and your answer becomes 1 by 10. Absolutely correct. Guys, please do not start using multinomial theorem over here. That's another way to solve such kind of problems but it would be a overkill because it's just a board exams. Okay, correct. So the first one to answer this correctly was Rohan. Next, so next question is, find the point where the tangent to the curve y is equal to 2 to the, y is equal to e to the power 2x at 0, 1 meets the x axis. Minus half comma 0, Ali is saying to, Ali, I need coordinates. So do you mean 2 comma 0 like that? Okay, so 1 is also saying 2 comma 0. Okay, so let's see this. First of all, we need to find the equation of a tangent. So for that, we need the slope of the tangent. Slope of the tangent will be nothing but the derivative of the function at 0 comma 1, correct. So derivative of this function will be 2 times e to the power 2x at 0 comma 1. That is only going to be 2, okay. So equation would be, the equation of tangent would be y minus y1 is slope of the tangent x minus x1, correct. So y minus y1, 2 times x minus 0. So basically it will be y is equal to 2x plus 1. Now the point where it meets the x axis, your y should be 0. So on x axis, implies y should be equal to 0, which means 0 is equal to 2x plus 1, which means x is equal to negative half. And hence the point has to be negative half comma 0, absolutely correct. Guys, how come this is wrong? Sondarya only, Sondarya and Psi only got it right. Easy, go slow. Don't take bold exams lightly. Next, evaluate the integral sign inverse of x by 2 plus x. And it will be too long to type in. So if you're done, just type done or you can send me the snapshot of the working that you have done to my personal ID. Okay, done. Again, three minutes is the timeline for this. It's a two marker. Guys, Sondarya, Ali, Nishal, Rohan, whenever you get inverse functions to deal with, always remember substitution comes in very handy and useful. So try to attack such problems through substitutions. Okay, so it's more than three minutes now. Let me just intervene in this case. So in this case, I would like to substitute x as 2 tan square theta. Okay, so when x is 2 tan square theta, your dx is going to be 4 tan theta times secant square theta. Okay. So let me make the substitution over here. So it's sine inverse of under root 2 tan square theta by 2 plus 2 tan square theta. And of course dx will be 4 tan theta secant square theta. So sine inverse of this becomes 2 2 factor will get cancelled off. So tan square by 1 plus tan square is actually tan square by secant square. Okay, and let me copy rest of the terms as such. Okay, so we all know that this is going to be sine square under root of sine square is going to be sine theta. Okay, so it's just going to be 4 theta tan theta secant square theta. Now I know I have to use integration by parts over here, but I will take this function as u and I will take this function as v because this is easy to integrate. Because the presence of secant square along with tan x will make it very easy to integrate isn't it? Okay, so let me simplify this. So it's 4 times theta into integral of tan theta with respect to secant square theta. Guys, we all know that integral of tan theta with respect to secant square theta will be tan square theta divided by 2. It will be tan square theta divided by 2. Okay, minus, minus derivative of theta which is going to be 1. Again integral of tan theta secant square theta will be this. If you want you can pull out a factor of 2 giving you this term. Now this tan square theta you can actually write it as secant square theta minus 1. So this you can write it as secant square theta minus 1. That would be more convenient for us to deal with. So your answer will be 2 theta tan square theta minus tan theta plus theta plus 3. But remember you cannot leave your answer in terms of theta. So what I will do is I will just replace my theta back. So I will write it over here, my final answer. So theta from here can be written as x minus 2, sorry, x by 2, x by 2 under root tan inverse. So it will become 2 times tan inverse under root x by 2 into x by 2 because tan square theta will be x by 2 minus, again tan theta will be under root of x by 2 and theta will be nothing but tan inverse of under root of x by 2 plus c. So this becomes your final answer. So this becomes your final answer. See Ali we use signometric substitutions whenever you have inverse stick functions which are actually complicated. Okay. Is that fine? We will move on to the next one now. Find the value of sine inverse of cos 33 pi by 5. That's correct. That's absolutely correct Ali. It's minus pi by 2. That's correct. Yes, how about the rest one? So this one will be simplified to 33 pi by 5. You can actually simplify to cos of 6 pi plus 3 pi by 5 which is as good as cos 3 pi by 5. So think as if you are trying to simplify sine inverse of cos 3 pi by 5. Now we can write this as pi by 2 minus cos inverse of cos 3 pi by 5 by the supplementary angle property. Correct. Now since 3 pi by 5 belongs to the principal value branch that is 0 to pi, you can write this as 3 pi by 5 itself. So that's going to be 5 pi minus 6 pi by 10. That's going to be minus of pi by 10. That's going to be your answer. Is that fine guys? Shall we move on to the next one now? Show that a transpose and a transpose a are both symmetric matrices for any square matrix A for any matrix A. So if you have to show that A transpose is symmetric then we need to show that its transpose will give us the same matrix back. Isn't it? Okay. And similarly if we have to show that A transpose A is symmetric then we have to show A transpose A transpose is again this. Done easy? Okay. So let's look into this. So A A transpose transpose is basically following the law of reversal. Okay. It follows the law of reversal. Okay. Now A transpose transpose is A and this is A transpose giving you the same matrix back implies A transpose is symmetric. Okay. And in a similar way you can actually also do this problem. This is going to be A transpose A transpose transpose. That's going to be A transpose into A, which implies A transpose into A is symmetric. Now these properties can be used. You can also write that in the bracket because A B transpose is B transpose A transpose. This is an allowed property. You don't have to sit and prove it. Okay. So law of reversal can be used directly. Next. If X is A times theta minus sine theta, Y is A times 1 plus cos theta. Find the value of d2y by dx square when theta is equal to pi by 2. Once done, please type in the answer in the chat box. Minus 1 by minus 1 by 4a. 1 by 3 people have answered and all their answers are different. Okay. Let's check. So first let us find dy by dx. dy by dx will be nothing but dy by d theta divided by dx by d theta. So dy by d theta will be nothing but minus A sine theta and this will be 1 minus cos theta. So that's going to be minus sine theta by 1 minus cos theta. If you want you can break this into half angles. That will make your life a bit easy to deal with. Else you will have to use quotient tool once again. So 2 and 2 gone. One of the sine theta gone. So it will be minus of cot theta by 2. Okay. So dy by dx is negative of cot theta by 2. Now when you have to find the d2y by dx square, it is nothing but dy by dx of dy by dx, which is actually d by dx of minus of cot theta by 2, which is actually d by d theta of minus cot theta by 2 into d theta by dx. Okay. So minus of cot theta by 2 derivative will be cosecant square theta by 2 into half. Right. And d theta by dx will be the reciprocal of this term. So you can write 1 by A1 minus cos theta. So when you put pi by 2, dear students, we'll get secant square pi by 4. That's going to be, if I'm not wrong. And in the denominator, I will have 1 by 2A1 minus 0. So your answer is going to be 1 upon A. So yes, Sondarya was correct. Be careful. I think you're nervous. You've solved much more difficult problem than this. The idea here is do not miss out on this term. So please do not forget it. Very, very important term. Is that fine? Simple? Shallow. We'll move on to the next one. So x to the power m, y to the power n is equal to x plus y whole to the power of m plus n. Prove that d2y by dx square is equal to 0. If you're done, just type in done on your chat box. That's quite difficult. No worries. I'll just try, and just simplify it for you. Okay. So Arun says done. Arun says done. How about others? This shall sigh. Why should we? Done. Okay. So let me prove, solve this. So here, whenever you have these powers coming on these variables, the best approach is to take log on both the sides, right? Log to the base of E. So let's take log on this side. So log to the base of E on both the sides. Okay. So this becomes according to the log property m log x n log y. And this becomes m plus n log x plus y. Okay. For the purpose of brevity, I would use ln for log x to the base E. Okay. So it's m ln x n ln y and m plus n ln x plus y. Let me differentiate both sides with respect to x. So it becomes m by x n by y dy by dx. And this becomes m plus n 1 by x plus y times 1 plus dy by dx. I'll just try to collect dy by dx terms together. So dy by dx term would be present over here and coming from this side also. So it's n by y minus m plus n by x plus y. Right. And the other terms would be m plus n by x plus y minus of m by x. So far so good. So now I'm going to simplify this a bit. So dy by dx on the numerator, you would be getting n x plus y minus m plus n y. Okay. And the denominator would be y times x plus y. And on the right side, you'll get m plus n x minus m x plus y. Again, in the denominator, you'll get x by x plus y. So let's simplify it. I need some space. So I'll be just erasing the initial part of this. So let me erase this part. So I'll get dy by dx. And on the numerator, I would be getting n x minus m y n x minus m y denominator would be y times x plus y. Similarly, on the right hand side, m x will get cancelled. Again, I will get n x minus m y by x x plus y. So this and this gets cancelled x plus y x plus y will get cancelled. So this implies dy by dx is going to be y by x. Okay, let's say I call this as result one. Here is few more steps. Now d2 y by dx square would be nothing but applying quotient rule on y by x that will be x dy by dx minus y by x squared. Now from one, I can say that from this expression, I can say that dy by dx that is this term over here could be replaced again with y by x. So I will again write y by x over here. xx will get cancelled y minus y will become zero. So that implies d2 y by dx square is going to be zero for you. Okay, slightly lengthy. But again, if you know the approach, things are quite straightforward in this. Let's move on now to the next question. Find a, b and c says that f of x is continuous and differentiable at x equal to 1. I think you will get one in terms of the other probably. Okay, Rohan, without the real analysis, we just have to see where is the function continuous and differentiable. a is equal to b initial. Are you sure initial? Okay, for a function to be continuous at x equal to 1, the left hand derivative, sorry, the left hand limit, the limit of the function as x tends to 1 should be equal to the limit of the function as x tends to 1 plus and it should be equal to the value of the function at 1. Okay. So the left hand limit at 1 will be a plus b. Right hand limit will be b plus a plus c. Okay. And of course, the value itself is a plus b. So this is not giving me any new information. So we can directly equate these two and this gives me c as zero. Okay. Next is the function should be differentiable at 1. That means the left hand derivative and right hand derivative at 1 should be equal. Now in your board exams, please do not directly differentiate it. Use your first principles to find the derivative at 1 minus. For example, if I have to write derivative of the function at 1 minus, I will write it as limit 1 minus h minus f of 1 whole divided by minus h as h tends to 0. Right. So 1 minus h means a times 1 minus h square plus b. Minus f of 1 itself will be a plus b whole divided by minus of h. Yeah. So b and b gets cancelled. So you'll get a times 1 minus h whole square minus 1, which is actually minus 2h plus h square whole divided by minus it. Right. And this gives you the answer as 2a. This gives you the answer as 2a. Now in a similar way, let us find f dash 1 plus f dash 1 plus is limit h tending to zero f of 1 plus h minus f of 1 by positive h. F of 1 plus h means b 1 plus h the whole square a 1 plus h. You don't have to put a c because c is already zero. You can save your time on that. Minus f 1 will be minus b minus a whole divided by h. So this is going to be b 2h plus h square and this is going to be a h by h. If you divide by h, your answer will be 2b plus a. Now, since the function has to be differentiable at zero, 2b plus a should be equal to 2a. That means a should be equal to 2b. So this is one relation and c has to be zero, of course. So this will become your answer for this question. Is that fine? Clear. We'll move on to the next question now. That's the odd part of it. The question is prove that this function defined by f of x equal to, I'll rewrite this in a proper way, f of x equal to, it's a piecewise function, x by mod x plus 2x square when x is not equal to zero and k when x is equal to zero. This function is discontinuous. I mean, they wanted to say add x equal to zero regardless of choice of k. So you have to prove that whatever is your k, this function will not be continuous at x equal to zero. That means left hand limit itself will not match with the right hand limit indirectly. Same. Okay. So here we need to just redefine the function. We need to redefine this function. So the function would be redefined as x by x plus 2x square when x is positive. It would be defined as minus x plus 2x square when x is negative and of course k when x is zero. Okay. So let us find the left hand limit at one limit of the function as x tends to zero minus. So for that, I will use this definition x by minus x plus 2x square, take x common and cancel from numerator and denominator. When you put x as zero, you get the answer as negative of one. So this is your left hand limit. Let's find the right hand limit at zero. That means f of x as limit x tends to zero plus. So this will be x by x plus 2x square again cancel out an x that will become one by one plus 2x put x as zero. It will become one. So this implies left hand limit does not match with the right hand limit at x tends to zero. It implies f of x is having which type of discontinuity. Can anybody tell me which kind of discontinuity is shown by this particular question? What should I write? Left hand limit is minus one. Right hand limit is one. So what type of discontinuity is there in this problem? Jump discontinuity. Right. It is undergoing jump discontinuity or we call it as finite discontinuity. And this is actually non-removable. It is non-removable. Non-removable means you cannot correct such kind of discontinuities. So probably in this case, the function at zero is having such kind of discontinuity. Left hand limit is minus one. Okay. Right hand limit is one. So there is a jump in the function. And the value of the function at zero is this. So there is a hole. There is a hole. So this is the jump. Cannot be corrected. Jump discontinuities cannot be corrected. Next. So a school wants to award its student for the values of honesty say x, regularity say y, hard work say z. With a total cash award of 6000 bucks, three times the award money for hard work added to that given for honesty amounts to 11,000. The award money given for honesty and hard work together is double the one given for regularity. Represent the above information. Algebraically, there's first question. Second is as a matrix equation and third part, find the award money for each value using matrices method and suggest one value with the student should be awarded. One more value that the student should be awarded. Sorry guys, there was a power cut. At least let me know the values for x, y and z that you get. Okay, so let's discuss this now. So the total x is the amount of money given for honesty, y is the amount of money given for regularity and z is the amount of money given for hard work. So this should amount to 6000 bucks. That is also given to us that three times the award money for hard work added to that given for honesty amounts to 11,000 rupees. So x plus three y or you can say x plus sorry three z is equal to 11,000 rupees. So you can write it as zero y if you want just to make it complete. And it's also given that the award given for honesty and hard work together is double the one given for regularity. Okay, so x plus z is equal to two y, which is actually nothing but x minus two y plus z equal to zero. So finally we have three equations coming up. One is x plus y plus z is equal to 6000, x plus zero y plus three z is equal to 11,000 and x minus two y plus z is equal to zero. Okay, if you write this in terms of matrices, this is nothing but 1111, 103, 1 minus 2, 1 times x, y, z is equal to 6000, 11,000 and zero. Now in order to find out, let's say this is your x, a, x, b. So in order to find your x, you have to do a inverse b. So we can find a inverse by using the formula one by determinant of a adjoint a. Okay, so let's find determinant of a. Okay, so determinant of a will become, let me expand with respect to the first two. So it's one times zero minus minus six, which is seven minus one minus two and we'll have minus two. So that's going to be six. So determinant is equal to six. Okay, let's find adjoint of a. Again, for adjoint of a, I have told you the shortcut method. You can use your shortcut method in your exam as well, because they are not going to really check upon how you're going to do adjoint. They just bothered about the result of it. So if you recall, I had given a shortcut method for finding the adjoint, which is first write it down like this, then repeat the first and the second row after this and repeat the first and the second column like this. Okay, now focus on this area. Now start cross multiplying these two, these two, these two, these two, these two, these two. Again, these two, these two, these two. So it'll give me one minus six is six. This is three minus one, three minus one is going to be two. Okay, and this is going to be minus two. So I'm just writing it in a already transpose manner so that I don't have to waste time doing transpose. Then this will be minus two minus one, which is minus three. Then one minus one, which is zero. Then again, one minus two, which is going to be minus or minus two, which is three. Again, this is going to be three. This is going to be minus of two. And this is going to be negative one. So this is your adjoint of a. Okay. Now, once we have got a joint of a, we can use this formula x is equal to a inverse b. Let me just make some space for myself. So x is going to be one sixth of six to minus two, minus three, zero, three, three, minus two, minus one. This is going to be 6,000, 11,000, and zero. So that's going to be one sixth. Yeah. Let's do the multiplication of these two matrices. So 36 minus 33. That's going to be 1000. And this is going to be a 12,000. This is going to be minus 12 plus 33, which is going to be 21,000. Okay. So one sixth of that is going to be 500. One sixth of this is going to be 2000. And 21 by six is going to be 3500. So the award given for honesty is 500. The award given for regularity is 2000 and the award given for hard work is the most. Okay. Now, it's a value-based question. So which other quality do you think a student should have? Any suggestion? According to the, according to me, the other quality a student must have is it should be helpful. It should be ready to help other students. Okay. You can write anything you want, but okay, this is what I feel. Okay, good. So we'll move on to the next question now. If the straight line x cos alpha plus y sin alpha is equal to p touches the curve x square by a square plus y square by v square equal to one, show that a square cos square alpha plus b square sin square alpha is equal to p square. Scroll up. That is gone now, right? I've erased it. It's actually I'm writing it on a PDF. You can go back on the video and check what I written. Again, avoid using any kind of parametric representation for the ellipse because again, if the examiner doesn't appreciate it, he may not award you marks. Run. Okay. Great. So first of all, we need to find out the equation of a tangent at any point x1 y1. So for equation of a tangent at x1 y1, you first have to differentiate the function. So dy by dx is going to be zero. So this implies dy by dx is going to be minus b square x by a square y and you have to find it at a point x1 y1. Okay. So at x1 y1, it will be minus b square x1 by a square y1. Okay. Then the equation of the tangent can be written as y minus y1 is equal to slope. Guys, now we don't have the liberty to do whatever we learned in JEE. That means we cannot do this. We cannot say replace your x square with xx1, replace your y square with yy1. Now these things cannot be done in board exams. Okay. So whatever I said in the first day, assume that your examiner knows nothing beyond the ncrt prescribed textbook. Okay. Yeah. So let us simplify this. So it's a square yy1 minus a square y1 square is equal to minus b square xx1 plus b square x1 square. Okay. So which is going to be b square xx1 plus a square yy1 is equal to b square x1 square a square y1 square. Okay. Divide both sides with a square b square a square b square a square b square a square b square. Okay. So this gives me xx1 by a square yy1 by b square is equal to x1 square by a square plus y1 square by b square. Now we know that since x1 y1 belongs to the ellipse, this term is going to be one actually. Correct. So this becomes your equation of the tangent. So as you can see to get the equation of the tangent itself, you had to do so many steps. Now the question is claiming that this is also a tangent at the same point. Yes or no? That means these two equations are same as per the given question. So we can compare the coefficients. So first we can compare the coefficient of x which is x1 by a square divided by cos alpha should be equal to y1 by b square sin alpha and this should be equal to 1 by p. Okay. So this gives me x1 as a square cos alpha by p and y1 as b square sin alpha by p. And since they both satisfy, they both satisfy the equation of the parabola, sorry, the equation of the ellipse, we can substitute this in this equation. So I will end up getting a to the power 4 cos square alpha by p square divided by a square b4 sin square alpha by p square divided by b square equal to 1. a square will get cancelled. b square will get cancelled. So it will give me a square cos square alpha plus b square sin square alpha equal to p square. Hence proved. So guys, so many things have to be done because you have to solve under the constraint of ncrt prescribed syllabus. So sometimes working under constraints can be very painful like this. Is that fine? Next, next question is it's a odd part of the previous one. The volume of the cube is increasing at a constant rate prove that the increase of its surface area varies inversely as the length of the side. Let me know once you are done. Okay, great. Anyone else? Okay, cool. So we have been given that the rate at which the volume is changing is a constant. Okay, and we know that volume is x cube for a cube. So this implies a 3x square dx by dt is equal to k. Okay, we can keep your dx by dt as k by 3x square. Now we know that surface area of a cube is equal to 6x square. So ds by dt is going to be 12x dx by dt. And if you replace your dx by dt with k by 3x square, it becomes 4k by x. Which implies ds by dt is directly proportional to 1 by x. Or you can say ds by dt is inversely proportional to the length of the cube, which implies ds by dt is inversely proportional to x. Next question. If a, b and c are three mutually perpendicular vectors of equal magnitude, then show that a plus b plus c is equally inclined to a, b and c at an angle of cos inverse 1 by root 3. So a couple of things here, they are mutually perpendicular. All the three are mutually perpendicular. And all of them are equal in magnitude. Done. Okay, so let's discuss this. So let mod a is equal to mod c. And let's say each of them is equal to k. Correct? Now, let's say alpha is the angle, let alpha be the angle between a plus b plus c and the vector a. Then we can say that by dot product, cos of alpha is a dot a plus b plus c by mod of a, mod of a plus b plus c. So this is going to give you mod of a dot a will be mod of a square and rest all of them would be zero. In fact, I can get this, which is mod a by a plus b plus c. Now, I do not know this value. This value is not known to me. So how will I find that out? Again, very simple. We can say that mod a plus b plus c square is nothing but a plus b plus c dot a plus b plus c. So that's going to give me mod a square, mod b square, mod c square and rest all the products will be zero, zero, zero. So that's going to give me three k square. So mod of a plus b plus c is going to be a root of 3k. So substituting over here, so substituting over here, this will become k by root 3k. So k and k gets cancelled. So alpha is going to be cos inverse of 1 by root 3. And similarly, you don't have to do all the things similarly. We can say if beta is the angle between b and a plus b plus c, beta will also be cos inverse 1 by root 3. And let's say gamma is the angle between c and a plus b b plus c, then gamma will also be cos inverse 1 by root 3. Next, solve the differential equation sine 2x dy by dx equal to y plus tan x or show that this differential equation will do the second part later on. First focus on the first one. Let me know if you're done with the first one. All right, so what is done. So first one, what kind of a differential equation is this? So what kind of a differential equation is this? This is an LDE, right? It's a linear differential equation. Now we need to identify what is my PNQ in this case. So first of all, I would just reframe this as minus y by sine 2x is equal to tan x by sine 2x, correct? So my P is going to be negative 1 by sine 2x, which is actually negative cos c 2x, okay? And Q is going to be, if I'm not wrong, this can be done as sine x by cos x. And this is 2 sine x cos square x. So that's going to be half of secant square x, okay? So integrating factor in this case will be integral of e to the power integral of P dx, which is going to be integral of minus of cos cos c 2x. So that's going to be e to the power integral of cos c 2x is ln cos c 2x minus cot 2x by half. So it becomes minus half ln cos c 2x minus cot 2x, okay? Which you can actually write it as 1 by root of cos c 2x minus cot 2x, okay? So this is your if. Is this fine? Now I cannot work with this complicated form. Let me simplify it a bit. So I can write it as 1 under root of, I can write it as 1 minus cos 2x by sine 2x, which is actually under root of 1 minus 2 sine square x by 2 sine x cos x. So 2 2 gets cancelled and sine x gets cancelled. So it becomes 1 by root of tan x, 1 by root of tan x. Now, once I've got my if, I can directly use the formula for the linear differential equation. So I'm going to just erase this off. So my formula is going to be y into if, y into if is equal to integral of cube, which is half secant square x into if, okay? Now things become quite simple over here. If you take your tan x to be t, then your secant square x dx will start behaving as dt. So this will start behaving as your dt. So it's actually 1 by 2 root t dt. So integral of that will be what? Integral of that will be a root of 2 root of t. So it's half of 2 root t. And here I will have y by and a root of tan x. Let me clear off some mess, which has been created here. Plus c, which is going to be y by root of tan x is equal to 2 2 gets cancelled. T is again root of tan x plus c. If you want, you can write it as y is equal to tan x plus c root tan x. Now, even if you leave it to this form, nobody is going to cut marks for that. So this is your answer. Let me tell you two types of differential equation are very, very important for your board exams. One is your linear differential equation and other is your homogeneous differential equation. Please practice a lot of problems on these two forms. Oh, I'm sorry, we have to do the second part as well, right? Yeah, second part. Show that this differential equation is a homogeneous and find the particular solution for this differential equation where y is 0 and x is e. Done? Okay. So how do you show that a particular differential equation is a homogeneous one? So guys, the characteristic of a homogeneous differential equation is it should give you something of this nature. D y by dx should be something of this nature where this is a homogeneous function, where this should be a homogeneous function. Now, how do you identify a homogeneous function? If any function f of x, if x comma y is a homogeneous function of degree n, then it implies that f of x, y, sorry, lambda x, lambda y should be lambda to the power of n x, y. This is the characteristic of a homogeneous function, correct? So now let's check whether this particular differential equation meets this criteria. So here, dy by dx would be, dy by dx can be written as, I'm just writing it one shot. It will be y divided by x plus x. Oh, there's no, I'm sorry. This is one dy by dx only. So it's going to be y minus x e to the power y by x divided by x. So that's going to be y by x minus e to the power y by x. Right? So this is your function. This is your function. Now here, it should be a homogeneous function of degrees zero. Okay? It should be a homogeneous function of degrees zero. That means this term should be same as this. Why? Because ideally, whenever we study homogeneous differential equation, it has to be of this, where both these two functions are homogeneous function of same degree. So overall, the effective degree of this should be zero. So effectively, the degree of this should be equal to zero. Okay. So this is your function. And if you try to replace your x with lambda x and y with lambda y, nothing is going to change. Nothing is going to change. It's just going to be the same function back again. Okay. So this is the proof for that. The equation here is a homogeneous differential equation. Right? Now, let's solve it. For homogeneous differential equation, we use the substitution y is equal to vx or x equal to vy as necessary. So in this case, since we have y by x in all term, it is advisable that you substitute y is equal to vx. So dy by dx will be v plus x dv by dy. x dv by dx. Okay. And replace it over here. So v plus x dv by dx is equal to v minus e to the power v. So v and v can get cancelled off. And you can write this as minus dx by x is equal to dv by e to the power minus ln x. And this will be e to the power minus v by minus one plus c. So e to the power minus y by x minus ln mod x is equal to c is your general solution. But I want a particular solution. So what is the particular solution given that y is zero when x is e? So y is zero means it will become one x is e means it will become one. So this will become c will become zero. The answer is e to the power minus y by x. Okay, is equal to ln mod x, you can say. That means you can say minus y by x is log of log of x. So you can say y is equal to minus x log of log of mod x. This is going to be a particular solution. Is that fine? So we'll now move on to the next question. So here we have been given up a question. So just a minute. This is a zero to a under root of x dx integral is two a zero to pi by two sine cube x. Then find the value of a to a plus one one minus two x the whole square. So let us focus on the first part first and then we'll come to the next part. Yeah, you can do that. But again, c will anyways be zero, right? No, no, see, if you're doing that, c will be coming out in such a way that it will become one. Yeah, in that case, c will become one. You can do that. So please type in the answer for this. This is a numerical answer question. Yes, done with the first one. What are you getting as the value of this integral? All right, so let's start the discussion. So root of x integral from zero to a, we know that root of x integral will be two third x to the power three by two, okay, from a to zero. And on this side, I can do this integral by first expressing sine cube x as three sine x minus sine three x by four. Okay, so I've used the identity sine of three x is three sine x minus four sine cube x. So I've used this identity over here. Okay. Okay. Yeah, so let's integrate this. So this side we'll get two by three a to the power three by two. On this side, we'll get integral of, in fact, by four, we can divide outside. So this will give me three cos x minus three cos x. And this will become plus cos three x by three from pi by two to zero. Correct. So this we can cancel off two. And when you put pi by two, this becomes zero. And cos of three pi by two, that is also zero. And when you put zero, we get three minus one by three. So three minus one by three a by two is equal to two third a to the power three by two. Okay. So two third a to the power three by two is equal to this is going to be eight by three. So it's becoming four a by three. This gives me two possibilities. Either a could be zero. Or if you cancel off a, make it half, three, three also cancels. So a to the power half is two. So a could be four. So there are two possibilities of a. Now, when a is zero, your integral will be from zero to one of one minus two x squared ex. Correct. So that will give you one minus two x cube by three into minus two. And when you put one and zero, you get the answer as one by minus six plus, sorry, minus one by minus six plus plus one by six again. Okay. So the answer is one by three in this case. Now what happens when a is four, when a is four, you need to integrate it from four to five. Okay. So again, the result is the same one minus two x whole cube by minus six. Okay. Let me make some space for myself over here. Oh, I'm sorry. Yeah. So when you write over here, it becomes one sixth of minus seven twenty nine, sorry, minus one sixth of minus seven twenty nine plus plus three forty three. Okay. Which is a three eighty six, three eighty six by six. That's one ninety three by three, one ninety three by three. So two possibilities are there, one by three and one ninety three by three. So, Rohan, you just got one of the answers. There are two possible answers in this case. Please do not cancel out a blindly because a could have been zero from here because if you integrate it from zero to zero, this will be zero also. And if you keep a as zero, this will be also zero. So a equal to zero is a possible value. Don't cancel it out. Okay. So we'll move on to the second part of this question. Evaluate integral from zero to pi by two. This is pi by two. Let me know once you're done so that I can start the discussion for it. All right. Pi by two minus half pi by two minus one. Okay. Let's check. So here first of all, we'll take since I can see tan inverse of sin x also and sin x can also be obtained from this term. So I can write it as two times zero to pi by two. Sin x cos x tan inverse of sin x. Right. So let sin x be t. Okay. So that implies cos x dx is going to behave as dt. So this becomes integral zero to one t tan inverse t. Right. Now here we need to use integration by parts treating this as my first function and treating this as my second function. So it's two times tan inverse t integration of t is going to be a t square by two minus again t square by two into derivative of this term will be one by t square. Okay. If you want, you can cancel off this two from everywhere to two and this two. So tan inverse of t times t square one to zero. And yeah, what about this? For this case, I have to write it as t square plus one minus one t square plus one. Okay. So when you substitute one over here, this becomes pi by four undoubtedly and here I will have integration of one minus one by t square plus one. So that becomes t minus tan inverse of t. Okay. And when you put one, you get one minus pi by four and when you put zero, you get a zero. So your answer is going to be pi by two minus one. So pi by two minus one is absolutely correct. And the first one to answer this was Arun Rohan Arun. Absolutely correct. So let's now move on to the next question. Find the smallest and the largest value of tan inverse of one minus x by one plus x for x lying between zero and one. Find the maximum and the minimum value of tan inverse one minus x by one plus x for x lying between zero and one. Yes, what are the smallest and what is the greatest value, largest value in that interval? Here's a very simple question actually. I don't know why it's taking so much time. First of all, I can say smallest is zero, largest is pi by two. Not completely correct, Sondarya. Okay, let's take this. See here, tan inverse, I can write this one as tan of pi by four and I can write this x as tan of tan inverse of x. Again, one plus tan pi by four and tan of tan inverse x. So this is tan inverse of tan pi by four minus tan inverse x. So till this step, everybody would have got, so it's pi by four minus tan inverse x. Now tan inverse x is an increasing function. Tan inverse x is an increasing function. Its graph is like this, if you would recall. Okay. So it's at pi by two and this is at minus of pi by two. But you're only concerned with this interval zero to one. So zero and at one, its value is pi by four. So its minimum can, this can be minimum when this is maximum. So its maximum is pi by four in this interval. So the minimum or the least value will be pi by four minus tan inverse one. Okay. And this is going to be zero. And the greatest value would be when this is zero itself. So pi by four minus tan inverse zero, which is going to be pi by four. So zero is the smallest value and pi by four is the greatest value. Is that fine guys? Show that if a coin is tossed n times, then the probability of getting three or four heads is given by n plus one c four, half to the power of n, where n is greater than four. Where n is greater than four. Simple. Okay. Most of you are done. Okay. So the probability of getting three heads or four heads is basically you're trying to find out the sum of the probability of getting three or four. So plus of this. So when a coin is tossed n times, we know that the probability of getting our heads is given as ncr half to the power of n minus r into half to the power of r, which is actually ncr half to the power of n. Okay. So this implies getting three heads will be nc three, half to the power of n and getting four heads will be again nc four, half to the power of n. So when you add these two, when you add these two, you get half to the power of n common and you have nc three plus nc four. Okay. Now try to recall Pascal's identity or Pascal's law. Okay. Pascal's rule ncr plus ncr minus one is n plus one cr. So you can write this result as half to the power n times n plus one c four and hence proved and hence shown. Simple one. Next. So this is again a question based on probability and anti aircraft gun can take a maximum of four shots at an enemy plane moving away from it. The probability of hitting the plane at the first second turn in the four shots are 0.4, 0.3, 0.2 and 0.1. What are the probabilities that the gun hits the plane, the gun hits the plane? Pretty simple question. Okay. That's absolutely correct. So it's going to be the probability of one minus that a be the event that he hits it in the first shot. So hits in the first shot, be with the probability, be with the event that it's hits in the second shot, see hits in the third shot, D hits in the fourth shot. So the probability that he will actually end up hitting in at least one of the shots is hitting in at least one of the shots is one minus none of the shots hit the plane. Okay. That's going to be one minus. Now, since each of these shots are independent, you have to do P of A complement, P of B complement, P of C complement, P of D complement. So that's going to be one minus, P of A complement is going to be 0.6, P of B complement 0.7, 0.8 and 0.9. If I'm not wrong, 42 into 72 will be 84, 14, 29. So it's going to be a 3024. So one minus 0.3024. That's going to be 0.6976. 0.6976 is absolutely correct. So why should we do the first one to answer this? Okay, easy one. So let's move on to the next. So next question is, find the equation of the right bisector plane of the line segment joining these two points. So basically, it's something like this. There's a point A, A, A minus A, minus A, minus A. And you have to find a plane which exactly bisects right bisector mean this has to be 90 degrees. Again, an easy one. Please tell me the answer quickly. Okay. So easy one. So first of all, the normal direction will be nothing but 2Ai, 2Aj and 2Ak. Okay. It's just the difference between the coordinates of these two points, correct? Okay. And the point over here that satisfies the equation of a plane will be 0, 0, 0. So you can also say these are your direction ratios. So direction ratios will be 2A, 2A, 2A. You can also use 1, 1, 1 if you want because they're all same. So your equation of the plane would be, equation of the plane will be 1 times x minus 0, 1 times y minus 0, 1 times z minus 0, equal to 0. Remember, I'm using the formula A times x minus x1, B times y minus y1 and C times z minus z1 equal to 0. So I'm using this formula to get the equation of the plane. So this has been used over here. So your final answer is going to be x plus y plus z equal to 0. So it will not depend on A at all. So almost everybody is correct. Okay. So next we have a question from a chapter on relations. So there is a relation S in the set of all the numbers and defined as ASB is equivalent to 1 plus AB greater than 0. Is S an equivalence relation? Is S an equivalence relation? Okay. So Rohan says it is not an equivalence relation. So which property it fails to satisfy? Yeah, A would be origin. Yes, because you can actually cancel off. Now the point A is the position vector over here. So A is the position vector of the midpoint of the given two points. This is not the A that you are talking about. Are you getting its size? The equation that you have written R minus A dot N, that A is ocular origin. So it will be just R dot N equal to 0. And N will be 2Ai plus 2Aj plus 2Ak and you can drop the factor of 2A from everywhere. So Sondarya is saying yes, it is an equivalence relation. So let us check for each one of them reflexive. So it has to be reflexive. It means A should be related to itself by S. Is it true? Yes, because A into A is always positive. It implies A is related to A by S. Hence, S is reflexive. Next, is it symmetric? Let us check. So let A comma B belong to R such that A is related to B, which implies 1 plus AB is greater than 0. Or 1 plus BA will also be greater than 0, which means B is also related to S. And hence, S is symmetric. Now the last property check is the transitive property check. Now let me take a scenario where let AB1, let's say B be half. So A will be related to S because 1 plus 1 into half is greater than 0. Now let C be minus of 3 by 2. So B will be related to C because 1 plus half into minus 3 by 2 is 1 minus 3 by 4, which is 1 by 4, which is again greater than 0. But A is not related to S, right? A is not related to S because 1 plus AC, that is 1 into minus of 3 by 2 is negative, which means S is not transitive. If F is not transitive, it implies S is not an equivalence relation. Is that fine? Yeah, you can take many examples. Okay, I have just taken one example. Let's move on to the next question. Let the binary operation star B from n cross n to n cross n. This is actually not a right way of writing. This is actually n cross n cross n cross n because it is operating on two ordered pairs and giving you one ordered pair. Okay, anyways, this is just a typo. So this is a binary operation which gives you this result for all ABCD belonging to natural number. Show that this operation is commutative and associative and find its identity element if any. Okay, identity doesn't exist. So have you been able to prove that it is commutative first? In commutative, you have to just show that A comma B star C comma D is same as C comma D star A comma B. So for ABCD belonging to the natural numbers. Okay, so according to this definition, this is actually AD plus BC comma BD. What it does is AD plus BC comma BD. So here also it will do CB plus DA and BD here will be DB. Since this is same as AD plus BC comma BD. So since this is true, yes, it is commutative implies that SAR is commutative. Next is associative. In associative, we need to show that A comma B star C comma D. This operation done with E comma F is same as A comma B star C comma D. Now this operation has to be done first. Correct. So let me make some space for myself so that I have enough space to work on. So first let us evaluate the left hand side. So A comma B star C comma D is going to be AD plus BC comma BD star with EF. Okay, I think that's going to give you ADF plus BCF plus BDE comma BDF. Okay, now right hand side will be A comma B star CF plus DE comma DF. So let me simplify this. So this will become again ADF BCF BDE plus BDF. Okay, hence both are same LHS and RHS match. So LHS is equal to RHS implies star is associative. Now they want to find out the identity element if any. Okay, let's go on to identity element. So we'll say let E1 comma E2 be the identity element. E1 comma E2 be the identity element. So A comma B star E1 comma E2 should be same as A comma B itself. Right, so according to the definition E E2 plus BE1 comma BE2 should be equal to A comma B, which implies E E2 plus BE1 is equal to A and BE2 is equal to B. This implies E2 is going to be one. Okay, if E2 is going to be one, replace it over here. So A plus BE1 is equal to A. So this implies BE1 is equal to zero. This implies E1 could be zero. So identity element happens to be zero comma one, which actually doesn't belong to n cross n. So it should belong to the same set from where you are picking up your A and B, A comma B, right? So the answer here is absolutely correct Arun, no identity element exists. No identity element exists. Is that fine? So we'll move on to the next one now. So alpha gamma, alpha beta gamma are the roots of the quadratic sorry, of the root of the equation x square px plus q is equal to rx plus 1. Prove that this determinant is equal to zero. Prove that this determinant is equal to z. So first of all, this equation is like this. Please let me know if you're done. I would recommend use properties wherever you can. I would suggest use properties because you never know that they may direct marks for anything. Yeah, yeah, yeah. That's within the scope. Okay, done. Alright, so let's let's do that. First of all, let me expand this determinant. So I can write I can do these operations. Let me do C1 as C1 minus C3 and C2 as C2 minus C3. So I will generate an alpha zero minus gamma zero beta minus gamma 111 plus gamma. Now let's expand this. So we'll expand with respect to the first row. So it's alpha beta plus beta gamma. Guys, is this free invisible? Is this free invisible? Then why is it not visible to me? Okay, yeah. So expanding this will give me alpha beta beta gamma alpha gamma plus alpha beta gamma. Okay, so this is basically nothing but product of root second two at a time which is actually given by just one second. Now can you see? Can you see this free now? So this term here would be C by which is minus r by p and this term will be minus of r by p. So this will give you zero. Is that fine? So now moving on to the last question for the day. Again, a determinant question if x plus y plus these zero show that this determinant is equal to this. See guys, this is a slightly tricky problem if you try to see properties will not be working over here, right? If you try to simplify this, you'll not be able to apply properties. So I think you have to expand this by brute force. So first of all, let me expand this by brute force. So it becomes, let's say I expand it with respect to the first row. So it becomes x a times y z a square minus x square b c minus y b y square a c minus x z b square and plus z c x y c square minus a b z square. Okay. So it becomes x y z a cube and we will have x cube a b c as well, right? Similarly, we have y cube a b c plus x y z b cube, then again x y z c cube, then minus a b c z cube. Which actually clearly you can see x y z could be taken common. Okay. And you get this term. Similarly, if you take a b c as common, you get this term. Now, what I'll do is I will introduce subtly this term over here minus three a b c and here also I'll subtly introduce minus three x y z because whatever you introduce will get cancelled from this. Okay. So this pair will get cancelled with this pair. So if you see this, it is just x y z a cube plus b cube plus c cube minus three a b c and this term over here will become zero, right? Why it will become zero because we all know this formula. Everybody knows this formula. I'm just writing it in yellow. x cube plus y cube plus z cube minus three x y z is x plus y plus z x square plus y square plus z square minus x y minus y z minus z x. Correct? So that's why it becomes zero. Okay. So now I have to show that my other expression is going to be this. In other words, x y z is already there, right? So I could just show that this part comes out to be a cube plus b cube plus c cube minus three a b c. Correct? So let's try to show that. So far, so good. There's no problem with till this stage. So now I'm going to erase this and I'm going to just prove that determinant of this is going to give you a cube plus b cube plus c cube minus three a b c. That's quite easy. You can actually add everything to the first column. So a b c a b c and a b c. So b c a b c a a plus b plus c common from the first column. It becomes one one one b a c c b a. Then just do this operation r one as r one minus r three and r two as r two minus r three. Okay. So this will become a plus b plus c. And here if you do the operation, you'll get a zero zero one b minus c a minus c c zero b minus a, sorry, a minus c is c minus a over a. I'm sorry. Correct? Now if you expand it, now if you expand it, you get a plus b plus c times b minus c b minus a and you'll get plus a minus c square. Okay. Which actually gives you a plus b plus c. And you get a square plus b square plus c square minus a b minus b c minus c a. Okay. Which is nothing but a cube plus b cube plus c cube minus three a b c, which is your right hand side. There are four more problems left, which I'll be sending across in the group. So please ensure you solve it. And I'll be solving I'll be sending you some few more papers of the same level. Do you have any exam on Monday? Nafal guys? NPS? Do you have any exam on Monday? Oh, board exam, which paper? Is it board or pre-board? I think it's going to be board only. Oh, pre-board exam. Okay, fine, fine, fine. All right, guys. Give you a best. Okay. So signing off over and out from syndrome Academy. Bye bye. Have a good day.