 Hello, greetings from Centrum Academy. Welcome to the YouTube live session on problem solving for board level. So those of you who have joined in the session, I would request you to type in your names in the chat box so that I know who all are attending the session. I know the attendance won't be too great because of the pre-boards going on in many of this many NPS schools. Good afternoon, Sai. Hi, good afternoon, shares. All right, so let's get started with this problem. First problem as you can see in front of you. Given a Binding operation, star is defined on R minus of minus one and is defined as A star B is equal to A by B plus one. Is star commutative? It's a one marker question. Clearly, it's not commutative because as you can see, A star B is A by B plus one and B star A will be equal to B by A plus one and this can only happen when A is equal to B or both will be equal only for certain conditions and the condition is A square plus A should be equal to B square plus B. That means A square minus B square is B minus A. Clearly, either A should be equal to B or A plus B should be equal to, you can just factorize this as A minus B, A plus B, okay, negative of A minus B or A plus B should be minus of one. So these are some restricted conditions under which they will be commutative. However, you don't have to show this in the exam. It's just for your own reference that under very special case, this will be commutative else it is not commutative for all values of A and B belonging to R minus of minus one. So you just have to write A by B plus one is not equal to B by A plus one for all A comma B belonging to R minus minus one, hence not commutative. Simple one to start with. So we'll now move on to the next question. Good afternoon everyone. Next is for what value of lambda are the vectors A, B and C coplanar? Again, a one marker. It should be like a rapid fire round. Okay. So for A, B, C to be coplanar, A, B, C vectors are coplanar implies the box product, this should be zero or the scalar triple product should be zero. Right? So box A, B, C will be nothing but a determinant formed by the coefficients of i, j, k. Okay. And this should be zero expanding with respect to the first row. So it becomes one minus two and minus of minus two will become zero plus one minus lambda plus three plus one to lambda minus six equal to zero. So this will give you the value as lambda minus three equal to zero. So lambda value is going to be three in this case. Is that fine? So i was at minus three or plus three. Okay. Great. Next, again a one marker. f of x is x square plus one, g of x is one by x minus five, f o g of five. That's absolutely correct. So f o g of five means you have to first find f of g of five. Right? So g of five is going to be nothing but one by five minus one, which is one fourth. And this you have to feed to the function f, which is going to be one fourth square plus one. That's going to be 17 by 16. That's going to be your answer. Right? So don't try to find out the formula for f o g of x and then find the value of f o g of five that would that would waste a lot of your time. Just find g of five and then whatever value you get feed to the function f that will give you f o g of five. Next, prove that cos inverse x minus cos inverse y is cos inverse x y plus under root of one minus x square under root one minus y square. If done, just type done on your chat box so that I can discuss it. Done? Okay. So let's discuss this. So we'll say let cos inverse x be a and cos inverse y be b. That is cos of a is x and cos of b is y. Now we know that cos of a plus b, sorry, cos of a minus b is cos a cos b plus sin a sin b. Okay. Now if cos of a is this, it implies sin of a would be under root of one minus x square. And if cos of b is y, sin of b would be under root of one minus y square. Okay. Correct? So cos of a minus b, I could write it as cos a, which is x cos b, which is y, sin a, which is under root of one minus x square, sin b, which is under root of one minus y square. And if you take cos inverse on both the sides, cos inverse of cos a minus b on both the sides. And normally in board exams, they assume that things are in the principal value branch. That's why they don't give any interval on x and y. Okay. So here you can easily assume cos inverse of cos a minus b is a minus b. But don't do that in J main and all because it all depends upon the range in which a minus b falls. Okay. So a minus b is going to be cos inverse xy plus under root one minus x square under root one minus y square. A can be replaced with cos inverse of x, b can be replaced with cos inverse of y. And finally this proves the result that was asked to be proven. Okay. Okay. Hence proved. Okay. Great. Normally such formulas are already there for ISC people, but not for CBC. For CBC, you need to prove them if you want to use them. And hence there was a question on proving it. Moving on to the next question. Next question, there are three coins in which one pair with probability half and two buyers with probability one third and two third for a head. One of the coins is tossed twice. If the head appears both times, what is the probability that the buyer's coin with probability two by third for the head was chosen? Please type in the answer in the chat box if you have found it. Okay. Size is 16 by 29. Anybody else who feels the same? Okay. So this is a case of Bayes theorem. Okay. Okay. So let basically, let even be the event that, let me call these coins by the name of, you know, coin C1, C2. No, sorry. This was C1 actually. Okay. Yeah. Let me call this as C1. Let me call this as C2. And let me call this as C3. Okay. So even is the event that coin C1 is chosen, E2 be the event that a coin C2 is chosen and E3 be the event that coin C3 is chosen. Okay. And A be the event that when you toss it two times, both head appears both the times. Okay. Now, what do we need? We need to find out, we need to find out that given head has appeared both the times, what is the probability that coin C3 was chosen? Okay. This is what we need. Correct. So according to conditional probability or you can say according to multiplication theorem, this is what we get. Isn't it? Okay. Which is nothing but PE1 into PA by even divided by PA. Right. Now, what is PA? Right. PA will be PE1 into PA by even PE2 into PA by two and PE3 into PA by three. Guys, let me tell you it's mandatory to write all these steps. Okay. You may be applying your shortcuts and writing the answer directly, but in order to get step marks and independent marks, method marks and independent marks, you need to write down these steps clearly. Okay. So this is going to be one third is the chance that you choose your coin C1 and half is the probability that you'll get a head once and again half is the probability that you get head once more. Okay. Then one third is the probability of getting coin C2 and one third into one third is getting two heads directly and again one third is the probability of choosing coin C3 and a two third into two third is getting the heads. Okay. Now, out of this, this term has to be in the, sorry, E3 will be there. Sorry. This term has to be on the numerator. So one third into two third into two third will be in the numerator and rest entire thing will be in the denominator. So one third, half into half plus one third, one third, one third plus one third, two third, two third. Now, at least one third factor will go off from everywhere. Okay. So it'll become four by nine divided by one by four plus one by nine plus four by nine. Okay. So that's going to be four by nine divided by one fourth plus five by nine. So let me find some place I'm not able to yeah, let me do it over here. So end result would be four by nine into you'll get 29 by 36 36 will go up. So this will become 16 by 29. So this will be your final answer. And I think Sai was the first person to answer this correctly. And of course everybody answered this correctly. Well done. Okay. So there was this was a simple problem to solve. But what I wanted to highlight was the defining of events which is important and using the using the correct formula for the Bayes theorem. Next, if vectors A, B, C are such that A plus B plus C is equal to a null vector. And vector A has magnitude three vector B has magnitude five vector C has magnitude seven find the angle between A and B. Okay. 60 degrees. Okay. Rohan also feels the same. Shallow will discuss it. So A plus B plus C is negative of C vector because they add up to a null vector. Let's take the, in fact, you can take mod of this. Okay, let's square it. So this will become a A plus B dot A plus B. And this will become minus C dot of minus C. That's going to be mod of C square. And here I will get mod of A square mod of B square plus two a dot B. Okay. So, mod of A square is going to be nine mod of B square is going to be 25. And this will be two mod A mod B into cos of the angle between them. Let's say that's theta is equal to mod C square which is seven square which is 49. So you will get a 30 cos theta is equal to 15. That means cos of theta is going to be half. That means theta is going to be pi by three or 60 degrees. Okay. Easy question again. Write down the integrating factor for solving the differential equation. One minus y square dx by dy plus xy is equal to ay given that mod of y is less than one. These are all two markers. Okay. So first of all, when you realize this differential equation is a linear differential equation. And this is exact form how it should be looking like. That is it should be appearing to be in this form dx by dy plus px equal to q where p and q are functions of x, sorry, functions of y or constants. Yeah. So in this case, your p is going to be this, q is going to be this. We know that integrating factor is integral of e to the power integral of p dy which is e to the power integral of y one minus y square dy. Right. Now you can just put a factor of minus half and minus two over here. So your denominators derivative is minus two y which is present on the top. So it becomes e to the power minus half ln mod one minus y square. Okay. Now there is no need to put a mod because you know that mod of y square will be less than one. So one minus y square would be greater than one, sorry, greater than zero. Okay. So this can be written as simply e to the power minus half ln one minus y square which is nothing but e to the power ln one by under root one minus y square. So you can write it as one by a root one minus y square. This will be your integrating factor. Okay. Absolutely correct, Sai Sondarya. Next. Again, a two marker. Integrate five to the power five x into five x, sorry, five to the power five to the power x into five to the power x dx. If done, let me know. Five to the power five to the power x plus c. Okay. Let's check. So the questions can be of various format. They can also ask you, you know, five to the power five to the power five to the power x into five to the power five to the power x into five to the power x dx. Okay. So but both the cases, the approaches same. What do we do is we take the highest of them, that is five to the power five to the power x as t. Okay. Now differentiate both sides with respect to x. Now, when you're differentiating the left sides with respect to x, ensure that you're using chain rule properly. Okay. So five to the power something derivative will be five to the power something ln five into derivative of something that is going to be again five to the power x ln five. Okay. So this dx will be equal to dt. So five to the power five to the power x into five to the power x times ln five square dx will be equal to dt. That means five to the power five to the power x into five to the power x dx will be dt by ln five square. Okay. So this entire thing can be replaced with integration of dt by ln five the whole square ln five the whole square being a constant can be pulled out of the integration process and that gives you t by ln five the whole square plus c. So your answer is going to be five to the power five to the power x by ln five the whole square plus c. Okay. So whatever you take, I think this is going to be the result. So even try this out. Try this one out guys and tell me what is the result that you find over here. Got it Sai, where you went wrong? Great. Be careful. These are hot problems. They can come any time. They can be repeated also. Yes done this one. Again the approach is same. You just have to take this one as your u or t or k or y whatever you want. Okay. Do I need to discuss this or can you move ahead? So I think in this case your answer will come out to be five to the power five to the power x by ln five whole to the power three plus c. Yes. Are you getting that? Okay. Great. Next again a two marker. If y is equal to sign inverse of five x plus 12 under root one minus x square by 13 find dy by dx. Okay. Guys, whenever inverse trigonometric function comes, please make sure you use proper substitutions. Don't try brute force over here. Don't try to apply chain rule on this. Okay. Just use proper substitutions. For example, if you see this term itself, we can simplify this like five by 13 x plus 12 by 13 under root of one minus x square. Okay. And if you see this clearly, you can actually assume your, let sign of let's say some angle five is five by 13 then automatically your cos of five will become 12 by 13. Okay. And you can say let sign of theta be x then automatically your cos of theta will become under root of one minus x square. Unfortunately, we cannot use the formula of sign inverse x plus sign inverse y over here because it's not a part of our CVSE syllabus. So we have to use such substitutions very strategically. So sign inverse, this will become a sign five. In fact, you can take this as cos x because that will make things more simple. So let's say this is cos theta and this is sign of theta. Okay. So sign five cos theta plus this will become a cos five sign theta and this is nothing but sign inverse of sin theta plus five. That's going to be simply theta plus five. Okay. So y is theta theta is nothing but cos inverse of cos inverse of x and a five is nothing but sign inverse of five by 13, which anyways is a constant for me. This is a constant for me. So when you do dy by dx, we end up getting negative one by under root one minus x square as our answer. Okay. So why there's a confusion between plus and minus half the people have said minus one by under root one minus x square, half the people have said one by under root one minus x square. See ultimately it's theta plus five only, right? Oh, okay, okay, I got your, I got your concern. I got your concern. Yes. So yeah. So one party is getting a sin inverse x plus cos inverse five by 13. So both the answers are acceptable in this case because, yeah, because they are not specified on any limitation on x actually. So both answers are acceptable in this case. So let's move on to the next one. If matrices a and b are commutative, then prove that a b to the power n is b to the power for all n belonging to natural number. Sure, sure, sure, take your time. Right. Okay. So this is going to be a simple problem. See when you're writing a b to the power n, it actually means it actually means a multiplied with b, b, b and so on, where all the b's are written n number of times. Correct. Now, when you say a b, a b actually is equal to b a because they are commutative. So I can write this as b a b b like this. Okay. Now this is n minus one times actually. Okay. Because a b is equal to b a. Okay. Again, this a b again can be swapped as b a b b b written. This is written n minus two times. Okay. And if this trend continues, we will realize that it becomes something like this towards the last step. And again, this can be swapped as b a n number of times. Okay. So this is nothing but b to the power n a. Okay. Which is your RHS hence proved. Is that fine? Let's move on to the next one now. Now I think we started with the four marker. So the first question says prove that all the points on this curve y square is equal to two a times x plus a sin x by at which the tangent is parallel to the x axis lie on the parabola. They should say lie on a parabola. We'll take up the other part after we have done the first one. If at all you found the equation of the parabola, do share. Okay. So Rohan says y square is equal to two x. How about the others? Psi, shares, Sondarya. Okay. Still trying. Now from here onwards, you will start realizing that the questions will become slightly more difficult and challenging. And you have to be very, very careful while solving them. So let's say the point which you are drawing the tangent on is h comma k point. Okay. So let the tangent be drawn at h comma k point. Now the slope of the tangent at this point is parallel to the x axis. That means dy by dx calculated at this point should be equal to zero. Okay. So let's find the expression for dy by dx first of all. So 2y dy by dx is equal to 2a 1 plus, if I'm not wrong, this will be 2 cos x by a into 1 by, so a and a will get cancelled off. Substituting the value, so 2k dy by dx at that point, 2a 1 plus cos h by a is going to be such a way that this is going to be zero. That means 1 plus cos h by a is going to be zero. Right? It means cos h by a is going to be negative of 1. Correct? Now cos of any angle is negative of 1. That means that angle has to be an odd multiple of pi. Correct? So h by a has to be an odd multiple of pi. Now, if I use this point back again into the expression of the curve of the parabola, let's see what do we obtain. Let me just quickly clear off the initial part of this particular expression. Okay. So let me just clear this off. So we'll get k square is equal to 2a h plus a sin of h by a. Now h by a is nothing but 2n plus 1 pi. Correct? And we know that sin for any multiple of pi is going to be zero. Whether it's odd multiple or even multiple, it doesn't care about that. It's going to be zero. So it's going to be satisfying this. And if you generalize this, if you generalize this, that means you replace your k with y and h with x, you get y square is equal to 2ax which represents a parabola, which clearly is a parabola. And Rohan was exactly correct. This is representing a parabola. Again, not a difficult question, but I think you got carried away with the concept of parabola being introduced. Anyways, so let's move on to the next question. This says that if a particle is moving in a straight line such that t is equal to a square plus bs plus c a positive, s is the distance travelled in time t show that the particle undergoes a retardation which at any instant is proportional to the cube of the velocity. So we have to show that the retardation is proportional to the cube of the velocity where ds by dt represents velocity and d2 by ds d2s by dt square represents retardation, retardation, acceleration, whatever you want to name it. There's nothing in the name at least in months. Let me know if you're done with this. Okay, done. So let's take this up. So t is equal to a square plus bs plus c. Let's differentiate both sides with respect to t. So this will become 2as ds by dt plus bds by dt. So this implies 1 is equal to 2as plus bds by dt. That means you can say 1 by 2as plus b is equal to ds by dt. Now, again this expression differentiating once more with respect to t here. So it will become 2a. You have to follow product rule over here. Okay, so it will becomes ds by dt into ds by dt which is ds by dt whole square then 2as d2s by dt square and this will become again bd2s by dt square. So what happens here is that we get 0 is equal to 2a ds by dt whole square plus 2as plus bd2s by dt square. Now, this expression 2as plus b, this expression is well known to me. That means it's reciprocal. That is 2as plus b is equal to 1 divided by ds by dt. So I can write this as let me make some space for myself. So I can write 0 is equal to 2a ds by dt whole square and this is nothing but d2s by dt square divided by ds by dt ds by dt. So if you take it on the other side it means minus 2as ds by dt whole cube will become equal to d2s by dt square. Clearly implying that your retardation will be directly proportional to cube of the velocity. Okay, hence shown. Okay, simple. Good. So let's move on to the next question. Prove that the derivative of tan inverse of under root 1 plus ax whole square minus 1 whole by ax with respect to this expression, tan inverse 2x under root 1 minus x square by 1 minus 2x square at x equal to 0 is a by 4. How about others? So far I've got the response only from one of you and wait for at least three of you to respond before I start solving this. Okay, psi is also done. Sondarya also done. There are more people attending this session. I think they are not introduced themselves. They have not written their name in the chat box. I request them to do so so that I know who are attending. Okay, anyhow, let's discuss this. So if you want to find the derivative of this expression with respect to this expression, it's very important that we simplify these expressions. Okay, so if I talk about this expression tan inverse of under root 1 plus ax whole square minus 1 by ax. Guys, is the screen visible to you? Am I audible? Okay, fine. So in this case, what I will do is I will put my ax as let's say tan theta. Okay, so when I do that, I get the expression as tan inverse of 1 plus tan square theta under root which is secant theta minus 1 by tan theta, right, which we can write it as tan inverse 1 minus cos theta by sin theta, which is nothing but can be further reduced to tan inverse of 2 sin square theta by 2 divided by 2 sin theta by 2 cos theta by 2, which is going to be giving you tan inverse of tan theta by 2. That's nothing but theta by 2 itself. Correct. In a similar way with the other expression as well, we can do the same similar kind of a substitution. So if I talk about tan inverse of 2x under root 1 minus x square by 1 minus 2x square, I can substitute here x as let's say sin phi. Okay, so that gives me tan inverse 2 sin phi cos phi and denominator gives you 1 minus 2 sin square phi, which is nothing but tan inverse of sin 2 phi by cos 2 phi. Okay. That's nothing but tan inverse of tan 2 phi which is 2 phi itself. Okay. Now, ultimately what do I need? Ultimately, I need the derivative of this with respect to this, with respect to this. That means I need 1 fourth of d theta by d phi. Okay. Try to be smart. Write less as less as possible. You have to find the derivative of theta by 2 expression, which was actually the reduced form of this. This was actually reduced to theta by 2 and this was reduced to 2 phi. So I am trying to find out derivative of theta by 2 with respect to 2 phi. That means I am trying to find out 1 fourth of d theta by d phi. Correct. Now, x is sin of phi here and Ax is equal to tan of theta over here. We can do one small activity. I need some space over here. So I would just clear off a bit of this portion. Okay. I will do a simple activity. From this expression, I can say adx is equal to secant square theta d theta and from here, I can say dx is equal to cos phi d phi. Correct. Now, when you know x is 0, when you know x is 0, sin phi and theta both would be multiples of phi. Right? You can assume very simple case to be 0 itself. So theta is 0 and phi is 0. If you put that over here and divide it, let's divide 1 by 2. So a will be equal to secant square 0 by cos 0 d theta by d phi. So ultimately, you will get d theta by d phi as your a because this is going to be 1 and this is also going to be 1. So this expression is going to become one fourth of a that is nothing but a by 4 and that is what is required from the problem. Right? So hence proved. Okay. So not a difficult problem, but of course you have to use some smart substitutions. Okay. And you can and you may not be able to use the same substitution for x at both the places. That's also a catch in this problem. Is that fine? Any question with respect to this? Great. So we'll move on to the next question now. Discuss the continuity of this function e to the power 1 by x minus 1 by e to the power 1 by x plus 1, if x is not 0 and 0 if x is equal to 0 at x equal to 0. Okay? So psi has given a response. What about shares? Sound area? Others? Okay. So let's discuss this. First let us find out the limit of this function as x tends to 0 plus. Okay? We know that if it has to be continuous, the left hand limit, right hand limit and the value of the function at 0 must all be equal to each other. So at 0 plus we can write a limit extending to 0 plus e to the power 1 by x minus 1 by e to the power 1 by x plus 1. Now let 1 by x be equal to let's say h. So as x tends to 0 plus, h will tend to plus infinity, right? Okay? So we can convert this problem as a limit on h. So limit h tending to infinity e to the power h minus 1 by e to the power h plus 1. So let's say e to the power h common from the numerator and denominator. Okay? So this is what we'll get, right? Now please remember e to the power minus h as h tends to infinity will be 0. So this term over here, so recall this. So this term over here will become 1 minus 0 by 1 plus 0 which is equal to 1 which actually doesn't match with the value of the function at 0. Okay? So you don't have to do a further test. Since the right hand limit itself doesn't match with the value of the function at 0, it implies f of x is discontinuous at x equal to 0. However, let's say if the situation comes in that you have to find out the left hand limit as well, then you can also find out the left hand limit very easily. So limit as x tends to 0 minus would be limit x tending to 0 minus e to the power 1 by x minus 1 by e to the power 1 by x plus 1. Here again, you can take x as or 1 by x as minus of h. So as x tends to 0 minus, h will again tend to plus infinity. So you have to write this as limit h tending to infinity e to the power minus h minus 1 by e to the power minus h plus 1. So this will automatically become 0 0 each, giving you the answer as negative of 1. So again, it also doesn't match with the left hand limit which is negative of 1. So further reinforces the fact that the function is discontinuous at x equal to 0. Is that fine? This was a simple test of how you evaluate your limit of those kinds. Next is the all part of the same question. Show that the function f of x is equal to mod x minus mod of x plus 1 is not differentiable at x equal to 1. Yeah, yeah, you can stop at right hand limit side. Great. So here first, we need to redefine the function in the neighborhood of minus 1 before we start working on the differentiability part. So we will first write down the status of the function when x is less than minus 1 and when x is greater than equal to minus 1. So when it is less than minus 1, it behave as negative x. And this will behave as plus x plus 1. And when x is greater than equal to minus 1. Now, again, we have to be careful. Just greater than minus 1 is what we are looking at. We are not looking at, you know, it is not going to pass 0. Just greater than minus 1 we are looking at. So this will again be a negative quantity, right? And this will be negative of x plus 1. You can be very precise that this should be less than 0. Okay, let me write it in a proper way. So minus 1 less than 0. Okay. So your function is going to be just 1 when x is less than minus 1 and minus 2x minus 1 when x is between minus 1 and 0. Correct? So left hand derivative at x equal to minus 1 is nothing but f dash minus 1 with a negative sign, which is nothing but limit h tending to 0 f of minus 1 minus h minus f of minus 1 divided by minus h. So just prior to this, it will be 1 minus 1 by minus h, which is going to be 0. And right hand derivative at minus 1 will be nothing but f dash minus 1 plus will be equal to limit f of minus 1 plus h minus f of minus 1 by h. f of minus 1 plus h will become minus 2 times minus 1 plus h minus 1. And when you subtract the value of f of minus 1, that's going to be minus 1 again by h. Okay? So here in the numerator part, you'll realize you get 2 minus 2h minus 2 by h. That gives you minus of minus of 2 as your right hand derivative as limit h tends to 0. So right hand derivative and left hand derivative do not match and hence the function is not differentiable at x equal to minus 1. So right hand derivative at minus 1 does not match with left hand derivative at minus 1. Please ensure you are using first principles to evaluate your left hand right hand derivative, guys. Please do not jump. Please do not use direct methods of differentiation. Okay? Don't differentiate this and say it is 0 and differentiate this and minus 2. Since they're not equal, it is not differentiable at x equal to minus 1. You may lose marks. Please, please, please in board exams, man. Please ensure you are as elaborative as possible. Use first principles. Of course, if the function becomes super difficult to do it by first principles, I can understand. But such will not be the case at least in the board exams. Okay? So chalo. Next is find the equation of the line through the point 3 comma 4 which cuts from the first quadrant a triangle of minimum area. So a maximum minima question for you. So please type in the equation of the line once you're done with it on the chat box. Okay? So Vaishnavi has given her response. How about others? y is equal to minus 4 by 3x plus 8. Okay? I think it means the same thing as what Vaishnavi amends. Okay? 4x minus 3y plus 9 is equal to 0. Sondarya has to be a negative slope line. Your slope is coming out to be positive. I mean logically thinking your answer cannot be correct because it's in the first quadrant, right? The triangle is being generated in the first quadrant. It has to be from a negatively slope line. Okay? So something like this is happening that the line is, you know, generating. So let's say this point is a comma 0. This point is 0 comma b. Okay? So the line equation can be safely assumed to be this. Okay? Now it is also given that this line passes through 3 comma 4. So 3 by a plus 4 by b is equal to 1. And area of the triangle is nothing but half mod a b. Okay? And you want this to be minimum. Correct? Now remember that if area of the triangle has to be minimum, its square also has to be minimum. That means this has to be minimum. Okay? Okay? Now you can replace your, in fact you can replace a square or b square with each other. So from this equation I can write 3 by a is equal to 1 minus 4 by b which clearly implies a is equal to 3b by b minus 4. Correct? So I can write this a square as 9b square by b minus 4 the whole square into b square. So the constants are immaterial to me. What I am most interested is in the expression b to the power 4 this part. Okay? Now if you want the area to be minimum the derivative of this term with respect to b should be 0. Right? That means if I apply a quotient rule over here into 4b cube minus b to the power 4 to b minus 4 whole divided by b minus 4 to the power 4 this should be equal to 0. Correct? Okay? So a lot of terms can be removed. So here we can say b minus 4 whole square into 4b cube should be equal to 2b to the power 4b minus 4. Now of course b minus 4 cannot be 0 so you can cancel it off. Okay? I am just erasing it off because if b minus 4 is 0 which means b is equal to 4 it will lead to the denominator being 0. So b minus 4 can be safely removed. b cube can be removed. Okay? A factor of 2 can be removed. A factor of 2 can be removed from here and you can make a 2 over here. So definitely b is going to be 8. b is going to be 8. If b is going to be 8, a is going to be 3 into 8 by 4 which is going to be 6. So your answer is going to be x by 6 plus y by 8 equal to 1. So Vaishnavi Deshpande absolutely correct. Well done. I think Rohan also got the same stuff. Yeah. Rohan also has got the same stuff. Yeah. Rohan also is correct. It is just the y equal to mxc form of this intercept form. Okay guys. So just remember that you can always take the square of the objective function also if you want to find your maxima minima. But one thing which I have not done over here is the double derivative test. That you have to do because that will carry one or two marks. So don't miss this out. I am writing here. Okay. Please, please do, please perform double derivative test also. Please double, do double derivative test. Okay. So double derivative of delta square with respect to b should be positive for this value of b. Okay. So don't miss out on that one. So guys, we will take a break here for 5 minutes. Okay. So let's have a break. We will be back by 5, 10 p.m. All right, guys. Welcome back. So here is the next question. Show that the general solution of the differential equation, this is given by this where a is a parameter. So let's focus on the first part. We will come to the other part after doing the first part. Let me know once you are done. Great, great. How about others? Okay. I need two more people to respond before I start solving this. Okay. Great. So we will divide both sides by these two terms. That means we will separate out the variables like this. So it will become dy by y square plus y plus 1 plus dx by x square plus x plus 1. Okay. Equal to 0. Since it is a variable separable, we can integrate both sides. And here I can write it as dy y plus half the whole square plus root 3 by 2 the whole square. And same goes with dx as well because the terms are exactly same. So we don't have to, waste too much time on completing the square. So this will become 1 by a tan inverse x by a formula, right? So it's 1 by a, this is 1 by a tan inverse x by a, correct? This will also become 1 by a tan inverse x by, that's going to be c, correct? Now what I'm going to do is I'm going to take this root 3 by 2 on the other side. So it's again another constant c. I don't care what is that c. Okay. Then apply the formula of tan inverse x plus tan inverse y, right? Which is actually tan inverse of x plus y by 1 minus xy. Okay. And this c, if you take the tan on the other side, that will be another c which you can call it as c dash. Okay. Now on the numerator side, if you simplify, on the numerator side, if you simplify, you will get 2x plus 2y plus 2 by root 3. And in the denominator, you'll end up getting 3 minus 2x plus 2y plus 1 plus 4xy. Okay. And 3 will go up. This is again equal to c dash. I need some space. So I'll be erasing the initial part of the solution. So please bear with me. So I'll just erase this part. Okay. All right. So numerator will leave me with 2 root 3 x plus y plus 1. And denominator will leave me with 2 minus 2x minus 2y minus 4xy is equal to c dash. A factor of 2 can be removed easily from numerator and denominator. So I'll just subtly drop this 2 from everywhere. So there was a 1 over here. There was a 2 over here. And this root 3 also can be brought down. This root 3 also can be brought down over here. Correct. Now you can call this as your A, right, which actually becomes the general solution for this differential equation. So this is how we solve the first part of this OR. Now please work on the second part of the OR, find the differential equation of the family of curves represented by y equal to c times x plus c the whole square. Again, do let me know once you're done. So remember the c both sides are same. So this c and this c are same. There's no difference between these two. And hence it has only one arbitrary constant. So it has to be an order one differential equation. Okay. Why double lash? Again, I told you why double lash cannot appear because it's order one differential equation. There's just one arbitrary constant. So it has to be an order one differential equation. So d2y by dx square cannot appear. All right. So let's discuss this now. So first of all, if you differentiate this with respect to x, you actually get 2cx plus c. Correct. Okay. Let's square this up. Let's square this up. This is fine, but you cannot differentiate it once more. Squaring is allowed, but differentiating it once more is not allowed. Okay. So it becomes x plus c the whole square. Let me call it as 1. Let me call it as 2. Okay. Now if you divide 2 by 1, what do we get? We get 4c. 4c is equal to y divided by y dash square. Correct. That means c for me can be written as y by 4 y dash square. Okay. Let me put in the original equation itself. Right. Put this in one. When you put this in one, we get y is equal to y by 4 y dash square into x plus y by 4 y dash square whole square. Now you can further simplify it to any extent that you want. You can actually take the LCM in this case and I think 16 y dash to the power 4 will come and 64 y dash to the power 6. So it's 64 y dash to the power 6 will come over here. Okay. Yy can get cancelled off and this is x 4 y dash square plus y the whole square. So you may end up writing it as y dash cube 16 y dash square x plus y the whole square. Is that fine? It's even fine to leave the answer at this place also. There's no issue with respect to that. Other option would be to divide these two. When you divide these two, you get y by y dash is equal to x plus c by 2. Correct. So from here you can get c as 2 y by y dash minus x. Okay. And then you can also substitute it back in the original equation. So let's check what do we get from there. So I'll just erase this part. So you get y is equal to 2 y by y dash minus x, x plus c is 2 y by y dash square. Okay. Both will simplify out to give the same expression. So either ways you can adopt. So you can divide one by this expression, get your c, put it back in the first equation. Let's move on to the next question now. Let's do the first integral. Evaluate x by x square minus 1 times under root of 1 plus x square. Okay. You may simplify the final expression if you have time, but they won't direct marks for that. But as far as possible, my recommendation would be to simplify it till the extent you can. Oh, okay. Great. Done. First one. Yeah. For the first one, done. Anyone? Okay. So Vaishnavi is done. How about others? Done. Okay. So let's discuss this quickly. This wasn't a great one though. I don't know why you took so much of time. You should have just taken this as t square. Okay. So this substitution would have just given you x dx as t dt. And when you substitute it over here, your numerator will become t dt. The denominator would have become t square minus 4 into t over here. So this t and this t would have got cancelled. So it just becomes, you know, t square minus 2 square, which you can write it as 1 by 2 a. Okay. ln t minus t by t plus 2. Okay. Which is 1 fourth ln of instead of t, you can write it under root of 1 plus x square minus 2 by under root of 1 plus x square plus 2 plus c. Okay. Again, that's what I told you. There are three types of problems which will kill a lot of your time. One is integration. Other is determinants. Other is maximum, minima. If you take a wrong approach. So right substitution is very, very important for such kind of problems. Is that fine, guys? Okay. Now please try out the second one. Evaluate integral of x plus sin x by 1 plus cos x from 0 to pi by 2. Yeah, guys. Any idea? Still trying? Any success? How many of you are wondering whether I have flipped the numerator with the denominator? Let me tell you, that's not the case. Had that been the case, it would have been a one marker. Sure, sure, sure. Take your time. Take your time. My shoes. I'm really sorry, Vaishnavi. That's not the case. But I don't think so. It's difficult. It may be a bit lengthy, but it's not difficult, I think. Okay, okay. So let's explore this. So first of all, we can split this as 0 to pi by 2 x by 1 plus cos x and 0 to pi by 2 sin x by 1 plus cos x. Okay. Now let me call this as i1 and let me call this as i2. Now let's take care of i2 first. i2 is I think a simpler one to integrate because if you can take your cos x as t, then minus of sin x dx will become your dt. So your numerator will become minus dt by 1 plus t. And of course, when x is 0, t is going to be 1 and this is going to be 0. So I'm going to flip it and get rid of this negative sign. Okay. So that's going to be ln of 1 plus t from 0 to 1. That's going to be ln of 2. So this part will become i2 will become ln of 2. Now for i1, I can write this 1 plus cos 2x as 2 cos square x by 2. So it becomes half x secant square x by 2 dx. Correct? Now I can apply integration by parts on this. Since secant square x by 2 is an easy function to integrate, we can take that as v. And the algebraic function which is x, I can take that as u. So that becomes half x times integral of secant square x by 2 is going to be tan x by 2 divided by half. Okay? So let me not behave like Gandhi over here. Let me just write it over here directly. Right? Minus derivative of this is 1 again integration of tan x by 2. Okay? So that gives you tan x by 2 minus half. Integral of this will be divided by half to log of sec x by 2. Is that fine? So 2, 2 will get cancelled off. Let me just remove this mess from here. And your integral is from 0 to pi by 2. So when you put pi by 2, you get pi by 2 tan pi by 4 which is 1 minus log sec pi by 4 which is root 2. Okay? And when you put 0, it will get 0 everywhere. Correct? So it gives you actually pi by 2 minus half ln 2. And there's already an ln 2 over here. So that will give you pi by 2 plus half ln 2. Psi, why are you not getting an extra half with ln 2? Check. Okay? Chalo, great. Not a difficult one. Of course, a little bit lengthy. Of course, these kind of problems are a bit lengthy. We'll move on to the next question now. Six dyes are thrown 729 times. How many times do you expect at least three dyes to show five or a six? So Rohan says 233. Okay? Okay, so Meher also backs Rohan up. So he also thinks the answer should be 233. What about others? Vaishnavi, Sondarya, Shes. See, first focus on this word expect. That actually gives you the idea about the expectation. Expectation means the mean value. Okay? Expectation means the mean value. Okay? Now, if you take one dye, if you take a six dye. Okay? And throw it once. What is the chance that at least three of them will show five or six? So first of all, getting a five or a six probability itself is two by six, which is one by three. Okay? So that means not getting five or six. Let me call this as event E. So not getting five by six will be two by three. Right? Now, when you're tossing it one-third and there are six such dyes and you want to get at least three dyes which show you five and six. Correct? So you'll be using binomial distribution over here, right? Right? It's better to find out those cases where, so you want three, four, five or six dyes. So it's one minus, no dyes shows you, no dye shows you five or six. So that probability will be six C0, two by three to the power of six. And one dye shows you five or six will be equal to this. And a two dye show you five by five or six will be this. Okay? So this will be a simpler way to calculate it. Okay? Now, this is becoming your new P. You can call it as P dash. And this is going to be your N. So your expected value will be nothing but N P dash. So that is going to be 729 times this entire expression. So one minus, by the way, I can just quickly write the simplified form of it. Two to the power six will be around 64 by 729. This will be around six into 32. Six into 32 will be around 192 by 729. And this will be around 15 into 16. 15 into 16 is I think 240. Yeah, 240. Okay? So when you multiply throughout, you get something like 729 minus 64 minus 192 minus 240, which if I'm not mistaken is going to be 256 minus 240, which is 729 minus 496, which is 233. So absolutely correct. Rohan was the first one to answer this. So 233 will be the number of times when you will see at least three dice showing a five or six. So it's a binomial distribution within a binomial distribution actually. Is that fine? All right. So we'll move on to the next one. Three rotten apples are mixed with seven fresh apples. Find the probability distribution of the number of rotten apples. If three apples are drawn one by one with replacement. Okay. If you're done, let me discuss this. Let me know it so that I can discuss it. Done. Okay. Great. Great. So three rotten are mixed with seven ones. So the probability of a particular apple to be rotten will be three by 10. And probability that it will not be rotten will be seven by 10. Correct? Three apples are drawn one by one. Okay. Find the probability distribution of the rotten apples. So basically what we are doing is we are writing a probability distribution where we can use binomial theorem. Okay. So here are the possible values of X. So your X could be zero, one, two or three maximum. Correct? So if you don't want any rotten apples, it will be seven by 10 cubed. If you exactly want one rotten apple, that will be three C1 seven by 10 whole square into three by 10. Correct? If you want exactly two rotten apples, three C2 seven by 10 into three by 10 square. And if you want all the three to be rotten, then it will be three by 10 whole cube. Okay. Now I would suggest you to actually simplify this and write in the exam, let's say 343 by 1000. Again, this will be 49 into 9. 49 into 9. This will be 441. Okay. This will be 27 by 1000. Okay. So I think the remaining will be, if I'm not wrong, 27 will be 370, 441, which will be 811. It will be 190, 189. Okay. So this will be the probability distribution for the number of rotten apples. Next, if a line makes some angle theta with each of X, X's, Z axis, and if the angle alpha, which it makes with the Y axis is such that sine alpha is three sine square theta. Find the value of cos square theta. Okay. What about others? Okay. That's quite easy now. We all know that cos of these angles are going to be the direction for science. So cos square theta, cos square theta, cos square alpha is going to be one. So2 cos square theta is going to be sine square alpha, and it's actually given to us as three sine square theta as well. It's very obvious that tan square theta is going to be two by three. So one plus tan square theta is going to be five by three. That is nothing, but six square theta is going to be five by three. that means cos square theta is going to be 3 by 5 so absolutely correct easy question i don't know what's doing in the six marker next for any vector a prove that mod of a cross i whole square mod of a cross j whole square plus mod of a cross k whole square is twice mod a square run because there are many ways to do it normally you can use your uh languages identity for this which says that mod of a cross b square is actually a dot a a dot b b dot a b dot b so if you replace your b with icap it will become now let let a be a1 i a2 j a3 k so a dot a would be mod a square only right a dot b will become a1 this will also become a1 and b dot b will become one so if you evaluate it it becomes mod mod a square minus a1 square similarly i can say mod of a cross j square would become mod a square minus a2 square and mod of a cross k square will become mod a square minus a3 square so let's say one two three so adding them all it will become three mod a square minus a1 square a2 square a3 square which itself is mod a square so three mod a square minus mod a square will be two mod a square that's going to be your right hand side hence proved even if they don't allow it it's you can actually prove it's quite simple to be proven it's just based on the use that this could be actually written as mod a square mod b square 1 minus cos square theta so it's mod a square mod b square minus a dot b the whole square at least this they can allow is just that i'm writing in terms of determinant it may look a different one but this will be sufficient or alternately you can if you want you can actually take the literal cross product okay and you find the mod of each one square it up and you'll get the answer next so here we have a question given that a non-empty set x and a binary operation start defined on the power set of x cross the power set of x giving you a power set of x as this a minus b union b minus a for all a and b belonging to the power set of x find the identity operation for star okay please ignore this part this is just an extra information that i don't know why they're written and find the inverse of a yeah is it done see if if some set has to be an identity element then a star let's say the set is called i then this has to be equal to a correct so a minus i union i minus a has to be equal to a itself it's very obvious that i has to be a null set over here find the inverse of a so again let b be the inverse of a so a star b should be equal to the null set that means a minus b union b minus a should be a null set that can only happen when individually each of them has to be a null set itself correct that means a should be equal to b that means a is its own inverse okay so a inverse is a itself is that fine so let's move on to the next question now yeah this is a question on determinants using the properties of determinants prove that this determinant is equal to this oh really okay okay fine fine fine then we can just skip this no issues find the area of this region so there's a circle and there's a line so I've drawn both of them for you this is as good as coming from this inequality equal to half so I think the required region that they want is this one so you have to use integration remember you cannot use direct formula like area of the sector minus area of the triangle that would not be accepted because the question specifically mentions using integration and it should not be like that you find the area of a triangle by integration and subtract it from the area of the sector that will also be not accepted the entire area has to be found out by using integration that's the catch over here that's it I know that's not a difficult one to do so first let us find out the points so this point definitely is going to be half x equal to half half comma zero okay now till this very point till this very point your rectangle your thin vertical strips will go from the circle to the straight line okay and after that your vertical strip will go from circle to the x axis so for the first part you will say integration from zero to half the y of the circle would be under root one minus x square okay minus half minus x dx and from half to this point is going to be one comma zero so half to one it's going to be just the circle part now if you want you can club these two so you can club these two and write it as integral of zero to one under root of one minus x square okay minus half minus x dx from zero to half so ultimately it becomes the area of the sector and this becomes the area of the triangle okay but you have to show it by this manner so without wasting much time I'm moving on to the other problem this is simple you can easily calculate it so there's no point I know solving it yeah let's do a question from 3d find the distance of this point minus 2 comma 3 comma minus 4 from this line measured parallel to the plane 4x minus 12y minus 3z equal to 0 find the distance of the point measured parallel to the plane so I hope the situation is clear so there's a line and there is a point like this minus 2 comma 3 comma minus 4 okay and there is a plane let's say it's a plane like this you have to make measure the distance which is parallel to this plane which is parallel to this plane so this distance is required this d is required so guys again in the interest of time again let's say I call this point as m right so you can obtain this m by assuming a certain parameter lambda so I can say coordinates of m can be chosen as 3 lambda minus 2 comma 4 lambda minus 3 by 2 comma 5 lambda minus 4 by 3 okay so this point here can be chosen as 3 lambda minus 2 4 lambda minus 3 by 2 5 lambda minus 4 by 3 now if you see clearly the perpendicular from the plane the perpendicular from the plane will be also perpendicular to this line correct so let me call this as p for the time being so p m will be perpendicular to the normal of the plane yes or no right so what is the direction ratios of what is the direction ratios of p m so if you subtract it it becomes 3 lambda and here if you subtract it becomes minus 9 by 2 and here it becomes 5 lambda plus 8 by 3 now this multiplied with minus 2 sorry 4 this multiplied with 4 this multiplied with 12 this multiplied with minus 3 should add up to 0 so 12 lambda plus 6 times 4 lambda minus 9 and minus 5 lambda plus 8 should become 0 okay so let's open the brackets so it will become 36 minus 1 which is 31 lambda and minus 54 minus 8 is minus 62 minus 54 minus 8 is minus 62 is equal to 0 so lambda comes out to be 2 for us so put your lambda as 2 over here and your m point can be found out as 4 comma this will be 5 by 2 okay and this will be 10 minus 4 6 by 3 which is 2 okay so 4 comma 5 by 2 comma 2 I can easily calculate the distance by using the distance formula it will be under root of 6 square 36 and half square which is 1 fourth and again 6 square 36 so it's under root of 72 plus 1 fourth which is 288 289 under root 289 by 2 okay 289 is actually 7 root 289 is actually 17 by 2 so you can write this as 17 by 2 as well is it fine guys do you have any exam tomorrow any pre-votes schedule for tomorrow if yes then all the best for the same okay no exam tomorrow so please work hard I'll send you this pdf on the whatsapp so whatever questions you couldn't solve please solve them and try them out so we're now from Centrum Academy thank you so much for coming online bye bye have a good night