 Hello, welcome to Centrum Academy Problem Solving session. So those 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 this session. So this is going to be a problem solving session on mixed topics. And I've tried to pick up as many topics as possible, depending upon the time constraint that we have at hand. Hello guys, good afternoon, Vishesh, Kushal, Sai, Ritvik, Vaishnavi. So let me begin with the very first problem for the day. So this question has been picked up from the chapter quadratic equation. It's a more than one option correct question. So it says let 6a be less than 4a, 4a be less than 3c, then this equation which seems to be a quadratic equation has which of the following options? Options because more than one option may be correct in this case. Both roots real, both roots imaginary. One root lies between a by 2 comma b by 3. Other root lies between b by 3 comma c by 4. So again, the norm will be I will be giving you some time to solve this problem. So as in when I get around five responses, I will start with the discussion of the problem. Remember, more than one options are correct. So ensure that you are covering all the options without missing out on anyone. Hello, Sondarya, welcome to the session. Please feel free to type in your response in the chat box. Hello, Sai, welcome to the session. Hope your semester exams are going on well. All right, so I mean, since people are still joining in, this is the very first problem of the day. And it's a question on quadratic equation. And it's based on the position of the roots of the quadratic equation. All right, so let us just start the discussion for this. Is anybody near to the answer or do you want some time? OK, so OK, fine. So I'll give you a minute or two more. Sure, sure, sure. Take your time. All right, so option A, you are saying correct. But remember, there may be more than one option is correct. Anyone else? Purvik, Vaishnavi, Rithvayak, Sai. OK, so let us start the discussion for this. If I look at this question, I would first go through the options. In the options, I see these kind of terms, A by 2, B by 3, C by 4, et cetera. And I also see this inequality, which is between A, B, and C. OK, so what I'll do first, I will call this as your f of x function. So let's say this is your f of x function, which is 2x minus A times 3x minus B plus 3x minus B times 4x minus C plus 4x minus C times 2x minus A, 2x minus A. OK, now since I'm inquiring about the position of the roots, I would first see the sign of f of A by 2. See, if a root has to lie between A by 2 and B by 3, the sign of the function that I get when I substitute x as A by 2 should be opposite to the sign that I should get when I substitute x as B by 3. So when I substitute A by 2, the first term becomes 0. The second term becomes 3A by 2 minus B times 2A minus C. And I think third term also becomes 0. So I'll get 3A by 2 minus B times 2A minus C. Now, how would I know the sign of this? Which information will tell me how to find the sign of this? So look at this expression. This expression says that if you divide throughout by, let's say, 12, so A by 2 is less than B by 3 is less than C by 4. Why should be the sign negative? See, it's a very important concept that if a root lies between A and B, so let's say the root lies over here, the sign of the function at A should be opposite of the sign of F of A should be opposite to the sign of F of B. So if let's say this is positive, then this would be negative. Then we'll say one of the roots will lie between A and B. Is that fine, Vaishnavi? Is it clear why sign should be opposite? All right. Now here, how do I know the sign of this? Guys, let me ask you this question. Since you know A by 2 is less than B by 3, since you know A by 2 is less than B by 3, so can I say 3A by 2 will be less than B? That means 3A by 2 minus B would be negative, right? And since you know A by 2 is less than C by 4, it implies 2A is less than C, correct? Which implies 2A minus C will also be negative, yes? So when two negative quantities multiply, it will give you a positive result. So this sign would be positive. So F of A by 2 is positive. Let me now find B by 3. Just like I found F of A by 2, I will now find B by 3. So I think the first two terms will become 0, and the last term will become 4B by 3 minus C times. This will become 2B by 3 minus A, right? Again, ask yourself what would be the sign of 4B by 3 minus C? OK? Since you know A by 2 is less than B by 3 is less than C by 4, what would be the sign of 4B by 3 minus C? Will it be positive or will it be negative? Focus on this part. 4B by 3 minus C, will it be positive or negative? Just type in in the chat box. All right, it would be negative. What about 2B by 3 minus A? 2B by 3 minus A, will it be positive or negative? 2B by 3 minus A, will it be positive or negative? This would be positive, right? So this is a negative quantity, this is a positive quantity. So combine this result would give you a negative quantity, right? In a similar way, I would find out F of C by 4, right? So F of C by 4 will give you C by 2 minus A, 3C by 4 minus B, OK? So C by 2 minus A, that would be positive, right? And 3C by 4 minus B, that will also be positive. So this is a really positive, correct? Now, let us go back to the options. Let us go back to the options. If we see here, I'm getting the sine of F of A by 2N, sine of F of B by 3 as opposite, right? Look here, they're opposite in sign. That means one root has to lie between A by 2 and B by 3. And not only that, I'm also getting the sine of the function at B by 3 to be opposite to the sine of the function at C by 4. That means one root will definitely lie between B by 3 and C by 4. So option D is also correct. And since we have accounted for both the roots as lying between these two points, that means both the roots have to be real, right? So your options number A, C, and D would be correct in this case. Is that clear, guys? Hope you don't have any question with respect to it. Great. So let us move on to the second question. So read the question. If a line y equal to root of 3x cuts the curve, x cubed plus 3y square plus 4x plus 5y minus 1 equal to 0, add the points, add the points A, B, and C, then OA into OB into OC is equal to. Any idea how to proceed with this problem? I can give you a hint. Think in terms of distance form of a line. Think in terms of distance form of a line equation. OK, so Shondaria is getting none of these. Yeah, none of these could be the answer. I'm not denying it. So I'm just waiting for other people to respond in. Many people also know it as a parametric equation of a line. This is also known as the parametric equation of a line. Anyone else who is getting the answer as none of these, feel free to share it? All right, so let us start the discussion for this problem. I think only Shondaria has given the response so far, and her response is none of these. Is there anybody else who is getting none of these? All right, so let's look into this. The equation y equal to root of 3x, OK? Correct me if I'm wrong. If I try to apply this equation x minus x1 by cos theta equal to y minus y1 by sine theta equal to r, right? What does this equation actually represent? This equation represents the distance form of a line. It represents the distance form of a line where r represents the distance, or you can say directed distance, because it has a sign also. It represents the directed distance of x comma y from a known point x1, y1, where theta is the angle of inclination of the line, where theta is the angle of inclination of the line. This form is very less known to people because they hardly use it. So I can write this as y by root 3 is equal to x by 1, but I cannot write root 3 and 1 because it cannot correspond to any cos theta and sine theta, so I divided it by 2 as well. So y by root 3 by 2 is equal to x by half. This is the parametric form where the parameter here is r. So r is a parameter. So as r varies, you get different, different x, y points on the line. So I can say x is r by 2 and y is root 3 r by 2. Now such a point will satisfy this curve because it is cutting this curve. And coincidentally, r also represents the distance from any x, y point from the origin. That will also represent the distance of the points of intersection, points of intersection from the origin, from the origin. So it is solving two purpose for me. It is also giving me OA, OB, OC directly by the nature of the roots that we get from the equation. So let's see what do we get from this. So since this point satisfies this equation, I can always substitute x as r by 2 and y as root 3 by 2 r. So that will give me r cubed by 8 plus 3 y square. 3 y square means 3 r square by 4 plus 4 r by 2 plus 5 y, which is root 3 r by 2, minus 1 equal to 0, correct? Which clearly is giving you a cubic equation in r. So it's giving me a cubic equation in r. So 9 by 4 r square. And this gives me r times 2 plus 5 root 3 by 2 minus 1 equal to 0, right? Now let's say the roots of this equation is r1, r2, r3, where clearly r1, r2, and r3 would signify the distances of the point of intersection from the origin. So let's say this is OA, this is OB, this is OC, correct? So what do I need from the problem? I need from the problem the product of OA, OB, OC, which is r1, r2, r3. That means I need the product of the roots from this, correct, right? So minus 9 by 4, minus b by a gives you the sum of the roots. C by a gives you product of the roots 2 at a time. And again, minus d by a will give you, so minus d by a would give you the product of the roots. So minus d by a will give you 8 as your answer, OK? So OA into OB into OC should be 8. So Sondarya, you were correct. The answer for this is none of these. Excellent. Good, but surprisingly only one of you could solve it. Guys, never ignore this form. This is very, very important. However, it is very less known and less used, right? There can be so many methods, right? Was your method smaller than this, shorter than this? You would have found out the points themselves, and then you would have found OA, OB, OC, and multiplied the result. Did you do that? All right, so moving on to the third problem of the day. In a triangle, A, B, C, A is greater than B. B is greater than equal to C. And it's given that A cube plus B cube plus C cube by sine cube A sine cube B plus sine cube C is equal to 7. Then they're asking you the maximum possible side of A, maximum possible value of A. I think this should be a very simple question for you to answer. I'm expecting answers from everybody in this case. Others, Sai Premnath, Vaishnavi, Kushal, Vishisht. Which exam do you have on Monday, guys? R&R people, which exam do you have on Monday? Achha, Nafal has physics, Sai Meer and BioComp. Okay, so Rajavi Nagar people are at peace. Oh, it's on Tuesday. Okay, chill, say a lot of days. Please keep writing mock tests and evaluate your performance, see where you're going wrong, work on your weaknesses. You can only work on them for the next two weeks. The last two weeks, you cannot do anything new. You can just write tests. And that would be your, you know, score in the JEE exam as well. Okay, so Purveik, Vishisht and Ritvik all have given the response as 5C. Not 5, it's question number three. Okay, anybody else? Aditya also says C. This is very simple question, guys. You just have to use your sign rule over here, right? Sign rule has to be used, which says A by sign A is equal to B by sign B is equal to C by sign C is equal to the diameter or the twice of the circum-circle radius, okay? Of the triangle. So let's say this is your circum-circle and you're drawing a triangle like this, A, B, B, C. So this is your A, this is your B, and this is your C, okay? Now, if I use this result, I can say A is two R, A is two R sign A, B is two R sign B, and C is two R sign C, right? And if I substitute them in the numerator over here, I will get two R cube sign cube A plus sign cube B plus sign cube C, okay? And of course, in the denominator, I have sign cube A summation as well. So these two will cancel off giving you our two R cube is this. So two R is cube root of seven, cube root of seven. Now, remember, this is the diameter of the circle and under no circumstance can any side exceed the diameter of the circle, right? So no side can exceed the diameter of the circum-circle in which it is inscribed, okay? That clearly means, that clearly means your A, which is the greatest of all sides, can never exceed seven, cube root of seven. So this is your maximum possible value. So absolutely correct, option C is the right option, all right? So next problem is basically a problem on trigonometry. This is A times tan A, okay? So there's unnecessary gap in between. So it's A times tan A plus under root of A square minus one tan B plus under root A square plus one tan gamma is equal to two A, where A is a constant and alpha, beta, gamma are variable angles. Find the least value of this expression. Of course, it is an integer type question. Please type in your response in the chat box. So it's A tan alpha plus under root A square minus one tan beta plus under root A square plus one tan gamma is equal to two A, then you have to find the least value of this expression. Any HSR, Indira Nagar, Kaur Mangala students attending this session, please type in your name. Yeah, anybody, any success? All right, so let us discuss this. So guys, let me start with a very obvious result. For A tan, A tan alpha minus under root of A square minus one tan beta, right? Can I say the square of this term will always be greater than equal to zero, right? In a similar way, yeah, so let me do this. A tan alpha, okay. Now I can say under root of A square minus one tan beta minus under root of A square plus one tan gamma. Whole square will also be greater than equal to zero. And so would be A tan alpha minus under root of A square plus one tan gamma square that will also be greater than equal to zero. Now I'm just taking a roadmap into our consideration because I wanted to generate tan square alpha tan square beta tan square gamma into my expression. So I took this route, but let me tell you whatever route you may plan out that may or may not work, okay? So I'll never be under the impression that whatever approach you are taking that will always lead to the desired result. So you may have to change your strategy sometime. So be prepared to have some leeway or some spare time in your hand so that even if you have to change your strategy you should be able to do that, okay? So why I use this approach is because if I say these terms are all positive that means there's some will also be positive, right? So if I add them all, right? Then the left-hand side will be added up to give you a positive term, isn't it? Okay. So basically what I'm doing, I'm doing something like this. I'm doing L minus M whole square, M minus N whole square, N minus, let's say, L whole square is greater than equal to zero. This is what I'm doing, right? So it'll give me twice of L square plus M square plus N square minus twice of LM, MN and NL greater than equal to zero, right? Which means L square plus M square plus N square minus LM plus NM or MN plus NL is greater than equal to zero, okay? Now this term, this term that you see over here, if I want I can write this term as L plus M plus N whole square, okay? So when I do the whole square, what do I get? I get L square plus M square plus N square plus two times LM, MN plus NL, right? Right? So I can say this term over here, I can replace it with L square plus M square plus N square minus L plus M plus N the whole square, N the whole square. At least I can write this term, I can write this term that is minus two LM plus MN plus NL, okay? So why I'm doing this is because I can generate L plus M plus N which is given to you in the question so that I can generate this term, I can generate L plus M plus N term, right? So let me do one thing, I will not erase the two factors, so let it be there, okay? So now I can replace, now I can replace this term as L square plus M square plus N square minus L plus M plus N whole square, okay? Which gives me three times L square plus M square plus N square minus L plus M plus N whole square greater than equal to zero, right? Now let us figure out what is this term, what is this term and what is this term? So this term is L square, L square is A square tan square alpha, M square which is A square minus one tan square beta and N square which is A square plus one tan square gamma, correct? And this side I get two A square, so this will be greater than equal to zero. So far so good guys, now let us see whether I'm able to generate that term which I need, I want the value of, least value of tan square alpha plus tan square beta plus tan square gamma, okay? But I don't think so that term is getting generated over here, okay? So I don't think that term is getting generated over here. So this approach is not yielding the results. So what I will do is I will modify my approach. See this is what I'd say, I mean you may not be able to get the answer in the first go, okay? So you have to keep playing with the expressions. So what I'll do next is I'll make a small change in this approach. So what I'll do is I'll just clear this up so that I can start with the other approach because of the paucity of space, okay? So I will change this approach. And don't get disheartened even if you have to cross the problem because let me tell you, J problems cannot be solved or sometimes it may happen, you may not be able to solve in one go, okay? So don't have high hopes that in one go only I will be able to solve all the problems. You may have to take multiple iterations, okay? So now I'll change my approach. What I'll do is I'll take the same expression but slightly in a different way because ultimately I realize that my tan square alpha tan square beta tan square gamma are not appearing. So I will now start with this approach. Yeah, but Simeer, you should not get anything apart from tan square alpha tan square beta tan square gamma. Let us see, I mean I have another approach in my mind so I'll start with this fact. Let's say this is greater than equal to zero, okay? Now see how I will distribute the term so that they all get the same coefficient. Just carefully watch this selection of terms which I am making. Tan square beta minus into tan square gamma and I'll take a tan square gamma, a tan gamma minus under root a square plus one tan alpha, okay? Now let's try this expression. So note that I have made some small changes. I have made some small changes in the expression. So again, this is my L, this is my M, okay? In fact, they're all different different terms. So let us try expanding them. So now when you expand it, you realize that you will get a square tan square beta from here, from here you will get a square plus one tan square beta, right? Tan alpha terms would be tan square alpha terms would be obtained from a square plus one and it would be obtained from here as well, a square minus one. And tan square gamma term would be obtained from here and here, which is again, a square plus a square minus one. But again, the problem is the same. I'm getting two a square term from here. I'm getting, I should get the same coefficient of all the tan square terms, right? So this should also give me two a square. This should also give me, so I think all of them are giving me different different expressions. Okay, nevermind. So we will approach with this problem. So we'll do one thing. Can we write this as, can we write this as, let's say I get a square for tan square alpha, a square plus one from here. I get a square minus one from here, okay? So a square plus one from here, a square minus one from here. Let's introduce a square also, okay? So let's do this thing, okay? Similarly for tan square B, I will get a square plus one from here, a square from here, correct? And let's introduce a square minus one also from our side. Just remember what terms I'm introducing, okay? Plus I will get for tan square gamma, I'll get a square from here. I'll get a square minus one from here. And from my side, again, I'll introduce a square plus one tan square gamma, okay? Now remember the terms which we introduced, we introduced minus a square tan square alpha, so I'll subtract it, okay? I introduced minus a square minus one tan square beta, so I'll subtract it. And I introduce minus a square plus one tan square gamma, a square plus one tan square gamma, so I'll subtract it. And the rest of the terms that you see over here would be minus two a tan beta a tan beta under root a square minus one tan alpha minus two a square plus one tan beta under root a square minus one tan gamma, and minus two a tan gamma under root a square plus one tan alpha, okay? This is all greater than equal to zero, remember that. Now you would be wondering why I introduced these terms because now you would see that these all terms over here are same, this is also three a square, this is also three a square, this is also three a square, right? So I'll get three a square tan square alpha tan square beta tan square gamma, okay? And the rest of the terms, if you take a minus sign common, right? So if you take a minus sign common in these terms, you would realize that you actually get something like this, a tan alpha under root a square minus one tan beta under root a square plus one tan gamma whole square, right? And this entire expression is greater than equal to zero. This entire expression is greater than equal to zero. Now you see guys, I've got this term also and I've got this term also. Correct. So three a square tan square alpha plus tan square beta plus tan square gamma minus this is two a square, which is greater than equal to zero. That means, that means tan square alpha plus tan square beta plus tan square gamma would be greater than equal to four a square by three a square, okay? So ultimately, what did I get? Ultimately, I got tan square alpha plus tan square beta plus tan square gamma is greater than equal to four by three. Correct. Now look at the desired question. The desired question was the least value of three times this, so three is also there. So bring this three up. So just remove this three from here, okay? And bring this three up. So that clearly means that this expression will always be greater than equal to four. So the least value will be equal to four. So four is your answer. Two a is because it is given in the question, no? Swetha, this is two a. Read the question. A tan alpha plus under root a square minus one tan beta plus under root a square plus one tan gamma is two a. Now this is a very, very long question, but I would say it is a J advance question. But again, it teaches you how to fiddle with the terms. How to fiddle with the terms, okay? Great. So moving on to the next question now. Question number five. So it's a question from Circle. Tangents are drawn from origin to touch the circle x square plus y square plus two g x plus two a five plus c at points p and q. Find the equation of the circumcircles, circumscribing triangle OPQ. Okay, Purvik says B, okay? I'm not commenting right or wrong, okay? So most of you are going for option number B. What about others? Vishist? Vandarya, Swetha, Aditi, Kushal, Shashank, Sai Premnath, Rithvik, Vaishnavi, please share your answer as well. All right, so let's say I'm drawing tangents from the origin and I'm trying to draw a circle like this. So this is my OPQ. OPQ. So I know the equation of this circle, which is x square plus y square plus two g x plus two a five plus c equal to zero. And I have to find this out. I have to find this circle out. So Vishist and Sai also getting option B. Guys, what about the others? I don't see any participation from others. Rithvik, where are you? Aditi? All right, guys. So I have a very, very small question for you. Do you realize that PQ? Do you realize that PQ is actually the chord of contact? Right? Chord of contact drawn to the circle from the origin, right? So can you tell me the equation of PQ? Can you tell me the equation of PQ? Can I say equation of PQ will be nothing but t equal to zero? Right, t equal to zero is the equation of the chord of contact, correct? So t equal to zero means x, x one, y, y one, g x plus x one, f y plus y one plus c equal to zero, right? Now, remember your x one, y one here is origin because you're drawing the chord of contact from origin, correct? So when x one, y one is origin, this will become g x, f y plus c equal to zero. So this is your equation of chord of contact, okay? The equation of chord of contact is this. Now, do you realize that the circum circle which you are looking for, it's actually belonging to a family. So this belongs to a family, to a family which is created by the intersection of the given circle and chord of contact, right? Do you agree with me on this? So I know from the concept of family of circles that a circle passing through intersection of a circle and a line is given by s plus lambda l equal to zero. So I can clearly say that the desired equation which I'm looking out for is x square plus y square plus two g x plus two f y plus c plus lambda times chord of contact equal to zero, correct? Yes or no? And now how do I get this lambda? How do I get this lambda, guys? Please type in in your chat box. How do I get this lambda? Because if I get this lambda, my job will be done. Excellent, Sondarya. Absolutely correct, absolutely correct, Shweta. So by the fact that this will pass through origin, this passes through origin. So that would be a useful information for me to get the value of lambda. So put x as zero, y as zero, you get c plus lambda c equal to zero, which means lambda is going to be minus one, right? So when lambda is minus one, this equation would reduce to x square plus y square plus two g x plus two f y plus c minus g x plus f y plus c equal to zero, thereby giving you the answer as x square plus y square plus g x plus f y, this. And I think that is option number B. So well done. I think Purvik was the first one to answer this. Yeah, Purvik was the first one to answer this. Very good, Purvik. Give it up. Is that fine? Any question with respect to this? Okay, let's move on. Question number six for the day. The question reads as the number of phase in which four distinct balls can be put into four boxes, labelled as A, B, C, D, so that exactly one box remains empty. Once done, please type in your response in the chat box so that we can start the discussion. Okay, so Tridhi was saying option D. Okay, Kushal is also saying D. Vishis is saying A. Sai is also saying D. Aditi, okay. All right, so I've got the required number of response. So guys, let's try to solve this problem. Now you want exactly one box to remain empty, right? You want exactly one box to remain empty. So out of four box, exactly one can be chosen. Exactly one box can be chosen from four box in four C one way. Okay, this can be chosen. In four C one way. Okay, let's say you decide that, you know, a particular box C, let's say this will be empty. Okay, so A, B, D should be filled up by four balls in such a way that none of them should be empty. Correct. Now you will have following options. You can only have one, one, two as the option, correct. And of course, you can also have a two, one, one, or one, two, one like that, right? Yes or no? So now if I have to give one ball to box number A, I can do it in four C one way, right? One ball to box number B can be done in three C one way. And the remaining two will definitely go in box number D, right? But there can be three factorial by two factorial combinations of these as well because it can be one, two, one, two, one, one. So three combinations like this can happen. So that's why this three has come because of that, okay? So it will be 12 into three, which is 36. And here one empty box could be any one of the four boxes. So in four ways. So the total number of ways of doing this task would be four into 36, that's 144 ways. Excellent. Nice to see our eleventhers solving this problem before anybody else. Well done, Tridhi. Good. So option D is correct. Or you could also do the same task by using the concept of distribution of, distribution of our distinct objects, distribution of, in the class I've told you, and distinct objects, so I'll keep it in, okay? Distribution of and distinct objects in our different groups, groups such that no group is empty, such that no group is empty. So here you are applying that funder, right? No group is empty here, no group is empty over here. The one which is empty is already chosen before the process has started, okay? So you can also do this problem by using this concept. Remember the formula for that was R to the power n minus RC1, R minus one to the power n plus RC2, R minus two to the power n, and so on, right? Now here, R is going to be three because you have to give it to three non-empty groups, and n is going to be four. So it's three to the power four minus RC1, three minus one to the power four, then three C2, three minus two to the power four. The rest of the terms would be zero, so I would not even bother writing it. So it's 81 minus three times two to the power four, which is 16, and this will be three, right? So it will be 81 minus 48 plus three, which is 84 minus 48, which will automatically give you 36, okay? And multiply this 36 with four C1, that would be your answer, four into 36, which is 144. So so many ways you can do it, but I would suggest you stick to your common sense way of this because it is quite possible that you may forget this formula in your exam. Okay, so let's move on to the seventh question. So this is a question which has been picked up from a series sequence in progression chapter. So find the sum of this series to infinite terms. Guys, some of you brought this point to my notice that when you are writing a mock test, the evaluation part or checking the, what is the correct way to solve every problem is taking you a longer time, right? Somebody has reported that I'm taking four hours to see the solution and understand what they have done. See, initially it will happen. So it's a normal thing and try to reduce it to two hours, right? Two or less than two hours. The process of looking at the solution and trying to understand is as good as, you know, studying the theory once again. So 90 questions you are looking at the solution of that, at least you're looking at the solution of those questions which you could not solve at all. So it's a learning for you and it takes in some time to understand it. So it's very normal to take, you know, around two hours or two and a half hours to scan through the entire paper solution and see where you went wrong. That's fine. But whatever is your score in the last two weeks of testing, please let me tell you, that would be your score even in the JMAIN exam. Unless until JMAIN exam becomes, you know, easy for everybody. So then everybody's marks would increase. Okay, so Kushal is getting A, Sai is getting C. Anybody else? I need at least three more responses. Okay, minute one. Fine, so I'll wait for one minute. Sai, Premnath, Vishisth, Aditi, Hithvik, Shweta, Sondarya, please respond, guys. Okay, yeah, that is the main problem. TR, anybody can find out. Okay, so Vishis says C. All right, so let's find this guys, find the sum out, guys. So can I say, I can write the RIT term as four R plus one, five to the power R into R, R minus one, where R is greater than equal to two. So basically I'm looking for the sum where we are trying to sum up TR from R equal to T, two till infinity. Okay, now all of you please watch over here. So TR is equal to four R plus one, five to the power R, R minus one. Can I write this four R plus one as five R minus R minus one? Okay, now individually divide five R by five to the power R, R, R minus one, and minus R minus one by five to the power R, R, R minus one. Okay, so this R and this R gets canceled and one of the five gets canceled. Okay, so it leaves you with, and this also gets canceled, so it leaves you with TR expression as one by five to the power R minus one, R minus one, minus one by five to the power R into R. Now do you all realize that it has not converted into something like VR minus VR, VR minus one minus VR, right? So basically now we can solve this. Now we can solve this by using your VN method of summation. Now we can solve this by using the VN method of summation. So yeah, start with R equal to two. So T two will be one by five to the power one into one minus five to the power two into two. T three will be one by five to the power two into two, minus one by five to the power three into three. T four will be one by five to the power three into three minus one by five to the power four into four, and it will continue all the way till infinity. So you would realize terms like this starts cancelling off and since it goes to infinity only this thing will be left off, okay. So some till infinite terms would be nothing but 1 by 5. So the first one to give this answer was Saim here, well done Saim here, good. Is that fine guys? Any question with respect to VN method over here, alright. So next moving on to a question which is coming to you from the binomial theorem chapter. You have to find the coefficient of x to the power n in this polynomial. I think this is going to be an easy question, nota, okay. Okay Vishis says C, Shweta backs Purik but Purik itself backs out, okay. I also say C, these are all JEE main level question guys, you should not be making any mistake in them. Yeah, I do admit some of the questions with JEE advanced level but this particular question is a JEE main level question, Khushar also says C, alright. I think this is very simple, when you are finding coefficient, remember there are n plus 1 terms over here. So altogether there are n plus 1 terms are there, right. So when you are finding the coefficient of x to the power n, okay. So coefficient of x to the power n is basically nothing but it would be the sum of the roots. So treat them as factors, correct. So in any polynomial you know that, any polynomial of this form, this will be actually you can say minus, it will be like this, right. Now here it comes the sum of the roots, sum of the roots, right. So the coefficient of x to the power n term would be, what are the roots of this? The roots of this is minus this, minus 2n plus 1c1, etc., till minus 2n plus 1cn. We have to find this term, coefficient of x to the power n which will be nothing but sum of this, very simple. Sum of such term still 2n plus 1cn, okay. Now you would realize that had this series continued, you would have got the next term as n plus 1c, n plus 1, alright. Next term would have been n plus 1cn plus 2 and all the way till 2n plus 1c2n plus 1. Now notice that this term and this term are actually the same, they are the same because of the property ncr is equal to ncn minus r, alright. So what is happening is, they will be pairing up happening between every term. For example, the term just before this would be 2n plus 1cn minus 1 and this will get paired up with this term, right. So since there are even number of terms over here, since there are even number of terms over here, what you will get is the whole sum, the entire sum will be actually 2 times the required sum. So let's say I call this as s. So this entire term will also be s because they are same just now I proved them. So 2s will be equal to sum of 2n plus 1cr from r equal to 0 till 2n plus 1 which we all know is 2 to the power 2n plus 1. So s will be 2 to the power 2n and your option number c therefore becomes the right option. So it was Vishisht who answered it correctly for the first time, well done. So we will take one more problem guys and then we will take a small break. So this is a problem which has been picked up from complex numbers. So 1 by a plus omega, 1 by b plus omega, 1 by c plus omega plus 1 by d plus omega is 1 by omega where a, b, c, d are real numbers and omega is a cube root of unity. Then find summation, guys the meaning of this is clear meaning of this is 1 by a square minus a plus 1, 1 by b square minus b plus 1, 1 by c square minus c plus 1, 1 by d square minus d plus 1. That is the meaning of this summation term. Good to see Tridheb attempting these questions. Keep it up. Okay. I am not saying right or wrong. Let's wait and see what others have to say. Anybody else? So far only Sai has responded to it. Guys need your response. So far only Sai Meer has responded. What about others? Sritvek, Kushal, Shweta, Purvek, Vishisht, Aditi, Sai, alright see guys here one important thing that I would like to ask you. Can you factorize this term? Can you factorize this? Can you break up into product of two factors? I am sure if I ask you this you would be able to do it, right? You would say a minus omega, a minus omega square, right? In a similar way, can you factorize this term? So say yeah, it can be factorized a plus omega, a plus omega square, right? Remember these are important factorizations which normally we do not study because the zeros of these polynomials are complex numbers, right? So now take a clue. So when it's very important to identify the clue from the question, the art of problem solving is hidden within identifying the clue from the problem. You cannot literally crack all the problems by using conventional methods, okay? Now I can clearly see at least 1 by a plus omega over here, okay? Can I do one thing? How would I generate 1 by a plus omega square? How would I generate, how would I generate 1 by a plus omega square? Think about that. If any idea please type it in the chat box. So I have an idea. So we have been given this, right? Plus 1 by d plus omega is 1 by omega, right? So what I'll do, I will take conjugate on both the sides. I will take conjugate on both the sides, okay? When I take conjugate on both the sides, it becomes 1 by a plus omega conjugate, 1 by b plus omega conjugate, 1 by c plus omega conjugate and 1 by d plus omega conjugate. This is equal to 1 by omega conjugate, right? Now 1 plus a is already a real number, so it'll be a plus conjugate of omega, 1 by b plus conjugate of omega, 1 by c plus conjugate of omega and 1 by d plus conjugate of omega. And this will be, yeah. And we all know that, we know that conjugate of omega is what? Conjugate of omega is what? Type in, what is the conjugate of omega? Everybody, omega square, right? Wished, correct. So do you see, now I'm able to generate the term which I was missing? Again, see, understand the roadmap here. You cannot memorize such problems. You have to see what does it take for me to generate this fellow, right? Now tell me how would I generate this entire term? How would I generate this entire term? Tell me that. Now I have these two terms with me, 1 plus a by omega, 1 plus b by omega, 1 by c plus omega square, 1 by d plus omega square. I'm sorry, I'm sorry, I'm sorry. Omega, omega, equal to 1 by omega, okay? How do I generate 1 by a plus omega, a plus omega square? From where will I get this term? Right, can I say this term will, yeah, absolutely, Purveks. Subtract these two. When you subtract these two, you get something like this, okay? By the way, this is omega. So when you subtract these two, in fact, you'll get all summations of this kind. And this will be 1 by omega minus 1 by omega square, correct? Now, it's as good as summing up, omega square minus omega by a plus omega, a plus omega square. And here I will also get omega square minus omega, omega cube, correct? Now this is a constant because summation is applied on a. So this will be canceled off. So it implies summation 1 by a plus omega, a plus omega square is going to be 1 by omega cube, which means summation 1 by a square minus a plus 1 will be 1 by 1. So your answer for this question would be option number a. So Psi, sorry, option a was the answer for this. Again, try to understand the roadmap, guys. If you're not coming up with the roadmap to solve the problem, right? How will you be getting your answer? And let it fail. It's fine that if you pick up a roadmap and it doesn't work. Remember in the technology problem, I picked up a roadmap and it did not work, right? So be prepared to fail, no worries. If you don't fail, you'll not succeed. Failure is a stepping stone towards success. It is not opposite of success. So with this gun, I am going to take a break, okay? So let's take a break, guys. And we'll return on the other side of the break with five more problems. So let's take a break. We'll resume at 540 sharp, okay? We'll resume at 540, okay? Sharp. All right, guys, welcome back to the live session again. So we are at our 10th question. So here we have another question here. Find the maximum value of x square y subject to the constraints given to you as in the problem. X and Y both are positive. Hope everybody is back from the break. So just a small clue that I may give over here that when you see this AM greater than GM, this will be helpful. Okay, try this. I mean, again, it may or may not work. Yeah, this has been picked up from, again, series sequence and progressions. I think I'm yet to do this chapter in HSR class 11th. Okay, so let me give you a clue. Try to produce this term in GM from this term, okay? It may not happen that collectively that will help you, but you can do piecewise also. Try to generate from x and y this term x square y, right? So try to generate this term from x and y. So of course, this will come in your geometric mean part, okay? And this will come in your arithmetic mean part, right? To give you just an idea, I can break x as x by two, x by two and keep y as such. So I can say this by three will always be greater than x by two, x by two into y whole to the power of one by three. So am greater than equal to GM, right? So x plus y by three will always be greater than x square y by four cube root. I mean, I'm just giving you one of the ways to do it, right? So you can say that x plus y, x plus y is greater than equal to three times x square y by four whole cube, okay? Now, in a similar way, try to see whether this term, whether from this term, similarly, whether from this term under root, just forget about under root. Under root we can take care later on also. Whether from this term, can I generate, let's say this term in a arithmetic mean part, can I generate x square y term from here? I think this is the most strictest part of the problem. Can I take a clue from the answer? Okay, so in the answers I see root 15, root 15, root 15, somewhere 15 has to appear. Three minimum will appear from here, so 12. You guys, again, you have to just take a clue from wherever you can because again, brute force will not work on such problems. So just let me try this. I don't know whether this will work or not. Let me break two x to the power, two x square into, y is already broken into three parts, let's say. So three y square I break up as y square plus y square plus y square, okay? So six, y to the power six is already there. So I should have minimum x to the power 12 generated, minimum x to the power 12 generated. So let me try this, let me break two x square into eight parts. So one, two, three, four, five, six, seven, eight. And let me break two x, y into four parts. Let me break two x, y into four parts. Okay, now if you see the powers, so how many terms I will have? You'll see I am now generating 15 terms. One, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15. So basically I took a clue from this. Maximum, minimum would not be easy either, Kushal. You may try that approach as well, okay? Now let us see whether my powers of x and y come in the ratio of two is to one or not. So I'll have two plus two, 16 power of x over here. And I have one, two, three, four. So x to the power four. And y will be, yes, y will be 10 power, okay? So basically I'm successful in writing it like this, okay? Whole raise to the power of 15th root, okay? This will always be less than equal to two x square plus two xy plus three y square by 15. Is that fine? Let me simplify this quickly without wasting much time. So this will be one fourth. This will be half. So it's going to be half to the power of 24, x to the power 20, y to the power 10. Whole raise to the power of one by 15 is always less than equal to two x square, two xy, three y square by 15. So let's see what do we get from here. So half to the power 24 by 15. First of all, I can take out a factor of, I can take out a factor of 10 at least, right? So I can take this as one by two and one by two to the power of, sorry, one by four and this to the power of 20. X to the power 20, y to the power 10, whole raise to the power of one by 15 is less than equal to this term. So if I'm not wrong, it becomes x square y and I would be, what is the term that would be left inside? Hope I have written the terms correctly, one by 20. Oh, why did I write a power of 10 over here? This was power of eight. So power of eight means this term would become 20. This term will become 20. So this term will not appear at all. Yeah, I'm sorry, I'm sorry for that mistake. So that's going to give you four inside the bracket. Okay, and outside you'll have 10 by 15, which is two by three. So this term would always be greater than equal to 15 times this. So under root of this, under root of this would be greater than root 15. Now you see root 15 has appeared, correct? Now what I will do next is I will combine, I will combine this expression, combine expression one and two. So from one and two, I can say x plus y plus under root of two x square plus two x y plus three y square would be greater than equal to three plus root 15. And I'm exactly getting x square y by four to the power of one by three. Right? And if I'm not wrong, this term is given to us as k. So k will be greater than equal to three plus root 15 x square by y, x square y by four to the power of one by three. So it'll become k by three plus root 15, okay, whole cube, whole cube into four is greater than equal to x square y. Which if I'm not mistaken, sorry, which if I'm not mistaken means the greatest value, the greatest value of x square y will be four k cube divided by three plus root 15, whole cube. Is this in the options? Yeah, option number B is correct. Again, very good problem. It made us think a lot how to break the terms at least. Again, you may have to look at permutation combinations over here. Let's move on to question number 11. Again, this is a question based on continuity and differentiability. This is power of t. So I'll write it down again here. f of x is limit t tending to infinity, one plus sine pi x whole a to the power t minus one by one plus sine pi x whole a to the power t plus one. Yeah, Shweta, I mean, I needed the total of 15 terms, okay. And I had to break x square into even number of terms. Okay. And I needed to break, what was the other term? Two x, y. That also had to be broken down into even number of terms because x square y had to be generated, okay. So, and that to the ratio of the power of x and y should be in the ratio of one is to two. So three terms are already broken up. So I had to break up 12 more terms. So I started with eight and four combination. If that would have not worked, I would have gone into six and six combination like that. So it's basically trial and error in breaking the terms till you get x square y terms throughout the expression. Or x and y should be in the ratio of power should be in the ratio of two is to one. May not strike in the exam, I'm telling you. So these are advanced level questions. It may not strike in the exam. That's why in advance, you don't have to attempt all the questions. Just pick up 60% or 65% questions and solve them all correctly. In these kind of questions, question number 11, try to redefine the function. Try to redefine f of x in simpler words by choosing some intervals of x, okay. In various intervals of x. Because then only the question about continuity or discontinuity at certain values of x will arise. Okay, so this is the hint. I mean, it was needless to give this, but if anybody doesn't know, then you should be following this approach. Redefine your function in various intervals of x. Ah, no, no, no, no. It is a normal bracket, Purvik. Unless until mentioned, don't treat any bracket to be a special function. Okay, so how do I know in which interval to redefine the function? Again, this question will be arising in your mind, right? So what are these various intervals? How would one come to know? Okay, now see here, the only thing that is dictating is you see this number is raised to a power of infinity, right? Both the places. Here also it is raised to a power of infinity. So only thing that will determine your interval prediction or the choice of interval is when is this greater than one and when is this less than one, correct? And when it is equal to one. That's also a point of contention for us, correct? Now, we all know, okay, Vishesh has already given an answer. Two, two, three, it will be positive again like this. So this is the graph of sine pi x, okay? Now, we know that is going to be, this term is going to be greater than one when it is in the interval zero to one, correct? Or let's say two, two, three, correct? Or let's say from four to five like that, correct? So correct me if I'm wrong. Can I say one plus sine pi x would be greater than one if I take all values of x belonging to two n pi two, two n plus one pi, correct? And one plus sine pi x would be less than one if I take any value of x belonging to two n plus one pi two, two n plus two pi. Sorry, why I'm writing pi? Sorry, no pi, right? And it will be exactly one, that is one plus sine pi x will be exactly one if x is some integer, right? Zero, one, two, three like that. Is everybody agreeing with me on this? Is this okay with you? Yes or no, please type in okay or not okay on your chat box, okay? So all of you please listen to this very, very carefully. So when you're applying the function on these intervals, let's see what is the fate of the function in these intervals. So since one plus sine pi x is greater than one, is greater than one in the interval x belonging to two n, two n plus one, okay? What I will do is I'll take this term as common or I will divide by this term. So divide numerator and denominator by this term, okay? So when you do this, you are going to get one minus one by one plus sine pi x to the power of t by one plus one plus one by one plus sine pi x whole to the power of t. Now, since you know this term is huge, right? So one plus sine pi x to the power of t will be almost tending to infinity, right? So this will become zero, this will become zero, giving you the answer as one by one, okay? In a similar way, can you tell me what will happen if x were to lie in the interval two n plus one, two n plus two. What will happen to this term? What will happen to these two terms which I'm circling out? What will happen to them? Now remember one plus sine pi x is less than one over here. So if I'm taking a power of t which is huge, what will happen to this term? Something which is less than one, what will happen to that? It will become zero, right? It will tend to zero, okay? So since it is less than one, this power infinity will tend to zero. So what will your answer be? Minus one, correct? Yes or no? Correct? So do I see a discontinuity at two n plus one? I do. That means that every odd point it will be discontinuous. What about every even point? Let's check. Okay, now when one plus sine pi x is exactly one, it will become one minus one by one plus one, which is anyway zero, right? So for every zero, one, two, three and so on it will become zero, exactly. So the function is behaving like this. So the function is behaving like this. It is one between zero and one, okay? Exactly at this point it becomes zero. Then it is becoming minus one here. Then again it is becoming zero. Again it is following this trend. Yes or no? Yes or no? So I can clearly say that at every integer points, my function is showing discontinuity. So at every integer points, my function is showing discontinuity. So your option number B is correct. So wishes was the first one to answer this. Well done, very good. Guys, a general feedback. How are you finding these questions? Thought intensive or okay, can we manage with little bit of more practice? Or is too too difficult? I just want to take a poll over here. How do you feel it? How do you feel these questions are? Okay, wishes things, manage with practice. Thought intensive, intensive. It is moderate level, okay? Honest feedback, okay. Are they very conceptual or not? Detest your concepts in and out? Great, great. So here is your 12th question. This question is basically on pair of straight lines. So if A and B are positive numbers, A less than B, then the range of values of K for which real lambda can be found such that the equation represents a pair of straight lines. So look at the question, observe this part, range of K for real lambda, it represents pair of straight lines. If possible, now of course in your online, you cannot underline in the question paper, but keep an eye on the keywords. Real lambda, whenever this comes, I think about discriminant somehow, I don't know. Whenever range comes, of course, discriminant is associated with greater than or less than sign, so range also comes in my mind. So this is the clue, okay, that I'm trying to generate from this question. It may or may not work. So the other day, somebody was asking me, sir, how many of these problems or how many problems of such types should be solved every day? See, I would say if you're solving it completely on your own without taking any help from the solution or hint, just five to 10 questions like this in each physics chemistry maths, if you're solving on your own, it is sufficient. There's no need to do 50 and 100 problems per day. So every problem that you attempt, you should try to do it on your own. Take it to completion. Yeah, any idea, guys? So just try to recall a couple of things. First of all, if your second degree equation like this, a x square plus two h x y plus b y square plus two g x plus two f y plus c equal to zero represents a pair of straight lines, then this determinant a b c, a g h g h f f should be equal to zero, right? This is something which we all know, correct? If you try to expand it, it just becomes a b c plus two f g h minus a f square minus b g square minus c h square equal to zero, correct? Now, let us see what do we get from this expression in this given problem? a b c. So what would be a b c in this case? a is a, okay? b is b, okay? What is c? c is 2k, c is 2k, and two g is 2k, two f is also 2k, okay? And two h is two lambda, okay? So if I use in this formula, a b c will be a b into 2k two f g h. So two f g h minus a f square, a f square is minus a k square minus b g square minus b again g square, okay? Minus c h square. So minus 2k h square will be lambda square, okay? This is equal to zero. Now, let us simplify this. If you simplify this, you would realize you end up seeing a quadratic in lambda. So you get minus 2k a lambda square. So this term, then you have plus 2k square lambda, okay? And apart from that, we get 2abk minus a plus bk square, okay? So far, so good? So now the problem over here is that since this is a quadratic in lambda and it has been mentioned in the question that lambda has to be real, right? So when lambda is real means the discriminant. So as predicted from the roadmap, right? So I'm going to get real lambda so I convert it into a quadratic in lambda. And if it is real, my discriminant should be greater than equal to zero. My discriminant should be greater than equal to zero, okay? Which implies b square minus 4ac should be greater than equal to zero, correct? So what is going to be b square over here? b square is 4k to the power 4 minus 4ac. So minus 4ac, correct me? This will be minus 4ac. This should be greater than equal to zero, correct? Let's take a k square out over here. Let's drop a factor of 4, 4 first. 4, 4 is not required. 4, 4 is dropped off. And if I take a k common from the bracket over here, this k and this k, I can write a k square over here. And you can also drop a k square as well. So it becomes k square two times 2ab minus a plus bk is greater than equal to zero. Now this gives me a perfect ground to find the range of k, correct? So see how things are automatically unfolding themselves. Correct? So k square minus 2a plus bk plus 4ab greater than equal to zero. Now I'm sure this is factorizable as this, okay? So take k common, k minus 2a, take minus 2b common, k minus 2a is greater than equal to zero. So k minus 2a, k minus 2b is greater than equal to zero, which implies, now which of them is lesser, which of them is more? For that, is there any condition given to you in the problem? Yes, it is given that a and b are positive numbers where a is less than b. So if a is less than b means 2a is less than 2b, right? So 2a is the smaller of the roots. So I can say k has to be less than equal to smaller of the roots or k has to be greater than equal to greater of the roots. Hope you remember how to solve an inequality, right? At this juncture, it should not be like I should revise inequality. So which option is correct, guys? Which option is correct? Option number D is correct in this case. So moving on to the next problem. So we'll keep it as a last problem because we have one more problem to go which I will give you as a homework. So please solve it and send it to me on the group right away. Before 7.30, I want the answer to be on the group. So this is an integration problem. Again, there's an element of always doubt created by ITG exams where they do not explicitly mention everything. They make you work hard to get the required parameters of the function. So here f of x is unknown to you unless and until you evaluate that limit which I think you should be able to. I mean, I don't see a point not being able to evaluate that. First tell me what is f of x? Everybody, before you solve the integral also. Exactly, Vaishnavi, correct. Oh, Shruti is also there. I thought she is not in the session today. Yes, correct. f of x is going to be one. Never mind Shruti, no problem. It's already recorded so you can always access it. But if you're solving it along with the class, actually it keeps you, you know, it lets you know what is your time management with respect to the problem. Never see a video on solution of problems because unless until you solve it on your own, you'll not get that hang. Again, it's not about number of problems that you see me solving it. It's about how many you solve it. Normally, this is a method used by IIT so that people do not scam. People will start differentiating the options and see which of the, which is matching with the question. So this is to make your question scam proof. Okay, so in the interest of time, let me just begin solving the problem. Undoubtedly, f of x equal to one. So for those who could not figure that out, since x is greater than one, you know x to the power n will tend to infinity, right? So it's always a practice that we take that term or we divide by that term throughout, correct? So if you divide by, if you divide by x to the power n throughout, you are going to get one minus x to the power minus two n by one plus x to the power minus two n, correct? And no doubt it, no doubt these two terms will become zero so where f of x is one, okay? So now you are left with integration of this term, okay? So whenever this term comes in my mind, I actually see that this is integrable, right? This is easy to integrate, correct? Because if I have two x by one plus x square under root, I can easily integrate it by choosing, by choosing one plus x square as t square. So x dx, x dx becomes, sorry, two x dx becomes dt. So it becomes dt by, or let me choose t square that will make my life easy. So if I choose t square, I get x dx as t dt. Then in that case, I don't need a this, okay? So x dx becomes t dt. So it becomes integration of t dt by t, which is under root of one plus x square, okay? So this is easy to integrate. So I will take integration by parts where I will call this as v and I would call this as u, okay? So I will do ibp integration by parts. So let me just copy down this term. So it will become log of x plus under root one plus x square times x by under root one plus x square dx. So it becomes log x plus under root one plus x square. Integral of this, we already know under root of one plus x square minus derivative of this term. Derivative of this term would be following chain rule, correct? One plus x by under root one plus x square, okay? And of course under root one plus x square will get multiplied. Now if you simplify this, you would realize that lot of terms would get canceled off. So this will become one by x under root one plus x square times x plus under root one plus x square y under root one plus x square into under root one plus x square dx. So this, this, this, this cancels. So you are left with integration of one which is going to be x itself. So is there any option which is talking about this as your answer? I think option number D, option number D. So D is the correct option. Okay. So guys, we'll stop over here. One more problem is left off which anyways I'll be sending it on the group. I'll send the entire working on the group itself. Please make sure you solve that and post the answer on the group, okay? Thank you so much over and out from Centrum Academy. Have a good night and best of luck for your upcoming exams. Bye-bye.