 Hello, welcome to Centrum Academy YouTube live session on theory of equations. So this is going to be a problem practice session in light of the upcoming exam of KVPY on Sunday. So in this chapter, I'm going to cover up various aspects of problem solving with respect to any equations and any equations which are primarily important from J advance point of view and KVPY point of view. So those who have joined this session, I would request you to please type in your names in the chat box so that I know who all are attending this session. All right, great. Hi all, hi vices, Khushal, Ritwik, Sai, Sondarya, Sanjana, Khushali. So without wasting much time, I'll directly take you to the first problem of this chapter. So I hope all of you can see your screen now. So we have a question over here. If f of x is equal to x square minus px plus q, p is an odd positive integer. So it's an odd positive integer. And the roots of the equation f of x equal to zero are two distinct prime numbers, two distinct prime numbers. And p plus q is given to you as 35. So the function is x square minus px plus q and the roots are prime numbers. The roots are distinct prime numbers. And we have been given an additional information that p plus q is 35 and p is odd. p belongs to an odd integer. So for every question, I'll give you around one, one-and-a-half minutes to think about it before I disclose the solution to you. And we have to find f10 summation of fr from one to 10. Hi Aditi, Nikhil, Shweta, Vashnavi, Shruti, good morning to you all. So we directly started with this first question. So all of these questions, you'll find them extremely challenging and they are very much oriented to your KVPY pattern of questioning. If you find the answer, please type in in the chat box. All right, so let's start the discussion for this. I know many of you would be in the process of solving it. So has anybody figured out what should be one of the roots of this function equal to zero must be? When you k2 to zero, what would be one of the roots of this? Has anybody figured out so far? So here is the catch actually in the problem. So let's say the roots are alpha and beta, which are distinct primes. And we know that in any quadratic equation, the sum of the roots is going to be minus B by A. So this is going to be playing the role of B. So minus B by A is going to be P. And this is an odd number. This is an odd number. Now, when the sum of two integers gives you an odd number, it cannot happen when both of them are even, right? And it also cannot happen when both of them are odd, right? That means one of these prime numbers have to be even, right? So one of these, one of them should be even, right? One of them should be even. And which is the only even prime number? The only even prime number is two right wishes. So only even prime number that we come across is two. So let's say your alpha is two, okay? Now that means two is going to satisfy this equation. Two is going to satisfy this equation which implies two square minus P into two plus Q should be equal to zero. In other words, two P minus Q should be equal to four, okay? Let's say I call this the first equation. And second equation is already in front of us. P plus Q should be equal to 35. Can we simultaneously solve it? Yes. Three P is equal to 39. So P becomes 13 for us. And as a result, Q becomes 22, I guess. Yes, 22, okay? So now we know the equation very well. So your equation x square minus Px plus Q equal to zero is finally x square minus 13x plus 22 equal to zero. So this is your f of x. And what does the question ask? The question asks you what is f of 10, summation f of r from 1 to 10? So first of all, f of 10. f of 10 would be 100 minus 130 plus 22, which is going to be minus eight. So they're asking you minus eight times. Summation from 1 to 10, f r, f r will be r square minus 13r plus 22. I think you can easily do this summation. Please hurry up and tell me the answer finally so that we can proceed to the next problem. Finish this off. So I'm giving you one minute to finish off the calculation. Hope you're all aware of summation r square, summation r. Just to recap, summation r square is from r equal to 1 till n is nn plus 1 to n plus 1 by 6. Summation r from r equal to 1 till n is nn plus 1 by 2. So guys, in KVPY chapters like theory of equation, chapters like inequality, properties of triangle, conic section, permutation combination, these are very, very important chapters. Yeah, I'm waiting for your response, guys. So just calculate this and tell me the answer. So it's minus eight times summation r square will be 10 into 11 into 21 by 6 minus 13 into 10 into 11 by 2 and you have plus 220. Plus 220. So that gives you 7 and 2 and it's 5. So it becomes minus 8 times 35 into 11. 35 into 11 is going to be 385. Minus and where you have minus 65 into 11. So you have 715 plus 210, 220. So how much does it give you? Minus 8 times 605 minus 715. That's minus 110. So your answer is 880. Your answer is going to be 880. So not the minus 880 rhetoric because you already have a minus sign in front of it. So this is going to be your answer for the very first question. So basically guys, we need to think through. It's not like you are solving some conventional approach. For example, here getting one of the roots as two was actually something which you need to think through. Alright, so well done to those who got the answer. Let's move on to the second question of the day. Alright, so here is your second question for the day. Find the number of ordered four tuple x, y, z and w where x, y, z and w belongs to the interval 0 to 10. So keep underlining these important information while you are solving problems in exam. With satisfied the inequality, 2 sin square x, 3 cos square y, 4 sin square z, 5 cos square w is greater than equal to 120. So here are your options, 0, 144, 81 or there are infinitely many. Again, primarily such questions have to be solved through good use of reasoning rather than going by a conventional approach. Whenever you are done, please feel free to type in in the chat box the answer that you get. Okay, Shrita has already come with the answer. She says it's going to be 0. Wish us to back it up. What about the others guys? Puruik, Aditi, Mahin, Vaishnavi, Shruti, Sondarya, please participate. Let's check it out. So when you say 2 sin square x, 3 cos square y, 4 sin square z, 5 cos square w, this is going to be greater than 120. Now 120, there is speciality about 120 that it can be written as 2 into 3 into 4 into 5. So 2 into 3 into 4 into 5. If you divide throughout by 2, 3, 4 and 5, let's say I divide this by 2, divide this by 3, divide this by 4, divide this by 5 and of course we are dividing it on the other side as well. What we are going to get is 2 to the power sin square x minus 1 and on this side we are going to get 2 cos square y minus 1, 4 sin square z minus 1 and 5 cos square w minus 1. Greater than or equal to 0. If I reciprocate everything, that means 1 minus sin square x 3 times 1 minus cos square y 4 times 1 minus sin square z and 5 times 1 minus cos square w that is going to be less than equal to 0. I am sorry, I will have a 1 on the other side. I will have a 1 on this other side. Thanks for pointing that out. Now when can this happen? Let's say I have cos square x 3 sin square y 4 cos square z and 5 sin square w less than or equal to 1. If you decide to take a log on the both sides you will get something like cos square x log 2 plus sin square y ln 3 plus cos square z ln 4 plus sin square w ln 5 less than or equal to 0. Now this is only possible when each one of these terms are equal to 0 because primarily all of them are supposed to be greater than or equal to 0. Because ln 2 is known to be greater than 0 and so is cos square x. Similarly ln 3 is known to be greater than 0 and so is sin square y. Similarly this also has greater than equal to 0 and this also is greater than equal to 0. They are saying it is less than equal to 0 that means the only possibility is that it would be equal to 0 and that will only happen when each of these terms are equal to 0. That means your cos square x should be 0 sin square y should be equal to 0 cos square z should be equal to 0 and sin square w should be equal to 0. Now think of the angles in the interval 0 to 10 radians. Treat this as in radians because if no unit is specified you have to treat them in radians. Think of angles in 0 to 10 radians where you find cos square x to be 0. So where all it can be 0? Where all it can be 0? It can be 0 for pi by 2. So it can be 0 for pi by 2. It can be 0 for 3 pi by 2. Please ensure you have to be within 0 to 10. It can be 0 for 5 pi by 2. Now if I go to 7 pi by 2 I am afraid I am going to exceed 10. So I will stop right here. When you think sin square y is equal to 0 sin square y is equal to 0 for pi 2 pi Can I take the risk of 3 pi? Yes I can because 3.14 times 3 is not going to exceed 10 and in similar way even z is going to show me the same set of values. Isn't it? I forgot 0. I forgot 0. This is also going to give me the same set of values that is 0 pi, 2 pi, 3 pi. Now when you want to find ordered tuplet means you can take you have 3 options for x you have 4 options for y you have again 3 options for z and you again have 4 options for w. So all together the total number of solutions number of total solutions is going to be how much? Do you want to change your answer? Guys do you want to change your answer? The answer is going to be 3 times 4 times 3 times 4 that's going to be right poovic so it's going to be 12 times 12 which is 144 so your option b is going to be the correct answer correct rationale So guys again here this is a very typical kvpy type of questions where your equation condition is met or inequality condition is met at the extreme cases so be careful about such cases again it's not a conventional way of problem solving so moving on to the next slide so here comes your question solve for x such that 1 factorial plus 2 factorial plus 3 factorial and so on till x factorial is k square where k is an integer so how many such x exist such that this condition is satisfied so psi says x equal to 3 psi if x were 3 your answer should have been saying that 1 plus 2 plus 6 would be a perfect square would be a perfect square is that correct? yeah it's correct so 9 so x can be equal to 3 is there any possible root? x equal to 1 that's fine that's also correct so x equal to 1 is another possibility because 1 is equal to 1 square so yes you have to mine out the roots for this particular problem so is there any other possibility? yeah we have only got x equal to 1 and x equal to 3 as the possibilities alright so moving on you're absolutely correct so guys you have to first see your square of any number can end in what? okay square of any integer square of any integer n sin it would end in either 0 okay 1 okay it will end in 4 and in 5 and in 6 and in and in 9 okay can any square end in 2? no I don't think so can it end in 3? no I don't think so can it end in 7? no can it end in 8? no right can it end in 3 also Vaishnavi? do you know any such integers whose square ends in 3? no great so these are the only possibilities in which square of any integer can end in right so guys when you start with 1 so of course 1 can be a possibility because you know the square of any integer can end in 1 when you add 2 it becomes 3 right so I cannot have any answer coming up for this when you add 6 it becomes 9 and yes we can have an answer which is 3 given by you okay the moment I add 24 which is 4 factorial I get the answer as 33 which ends in 3 okay the moment I add 120 the moment I add 120 you would realize that again it will give you something which is ending in 3 again it is ending in 3 the moment I add the moment I add 720 again I'll get something ending in 3 so if you see if you keep on going forward you would find that all the answers that you get from your left inside would be ending in 3 and hence there would not be any further solution from 4 onwards so Kushal was right so the only 2 possible solutions is x equal to 3 and x equal to 1 so these 2 can be the only possible answers so again this is how questions can be framed so it may look very impossible to solve these questions but as you reason it out you realize that you can exclude many such cases to reach to your final result yes for anything greater than 4 factorial it will end with a 0 and hence your remainder will always be 3 when you divide by 10 that means the last digit will always be 3 and no square or no perfect square can end with a 3 is that clear? we can move on to the next slide yes so question number 4 question number 4 so when x lies between 0 to 1000 x lies between 0 to 1000 gif greatest interior function of x by 2 plus greatest interior function of x by 3 plus greatest interior function of x by 5 is 31 x by 30 find the number of possible values of x try this out so your options are 34 33 32 none of these yes it's a simple question it's a very very simple question you can solve this within 10 seconds any response? 33 I will not reveal the answer right now because if I do so others will be influenced by this so kushal has said 33 and he's not sure let me add that as well yeah sure sure I'm not giving out the answer right now Shruti so Shweta says option D okay let's keep that also as one of our answers Nikhil says 33 not sure about that Purvik is getting none of these I think when you say x is 30 how about the others Sai says 33 Mahin is none of these Shruti is also none of these okay guys let's know reason this out do you see that the left hand side of this equation has to be an integer correct which implies right hand side should also be an integer right and if that has to be an integer that means whatever you get for x whatever you get for x since 31 is a prime number whatever you get for x must be divisible by 30 so x must be a multiple of multiple of 30 right and I want to ask you how many multiples of 30 do you see from 0 to 1000 so of course the first multiple that you see is going to be 30 itself 30, 60, 90 da da da da till it will go to 990 I guess yes or no yes or no so how many such numbers are there so it is going to be let's say the number of number is n so this is into 30 is going to be 990 so n minus 1 times 30 is going to be 33 sorry it's going to be 960 960 by 30 is going to be 32 so n is going to be 33 so 33 is going to be your answer and please note that whatever numbers you choose they are all divisible by 2, 3 and 5 as well right because LCM of 2, 3 and 5 is going to be 30 so all of these values are going to be a part of your answer right now think about a case where the LCM of 2, 3 and 5 were not were not 30 in that case will all my answers be there will all these values would be a part of my answer think about this and let me know on the group okay so meanwhile we will go to the next problem which is the problem number 5 so let's take up this problem problem number 5 so the question reads as if x square plus px plus 1 is a factor so this quadratic is a factor of this equation which seems to be a cubic in x then theta can be then theta can be which of the following options n pi n pi plus pi by 2 2n pi or n pi by 2 yeah your response is correct any idea how to go about with this for tomorrow's kvpo exam for bio students if you are good at bio even though I am a math teacher I am saying if you are not comfortable with maths choose physics chemistry bio alright so let's proceed towards solving this see if you are saying this is a factor of this that means you could express 2 cos square theta x cube as 2x this as this times something okay vashnevi is getting option a any other your response guys feel free to type in your response it doesn't matter even if it is not correct please just type in your response that's more important now of course this is a cubic right this thing is cubic and you already have a quadratic over here so you have to have a linear over here right now the best way to come to this linear is if there is an x square over here and there is 2 cos theta x cube that means it's obvious that 2 cos square theta x will come here okay sign here is saying option c let's check it out guys and second thing that you see since there is a sin 2 theta over here there has to be a sin 2 theta coming here as well so there is no option but to choose 2 cos square theta as the coefficient of x in the linear factor and the constant term as sin 2 theta right there is no other option right now if you are saying these 2 expressions are the same let us first find out the coefficient of let's find out the coefficient of x square on both the sides so on left hand side there is no x square so we can call it as 0 whereas on the right hand side the coefficient of x square will be cos square theta 2 p cos square theta and I will have sin 2 theta as well right in other words minus 2 p cos square theta is going to be sin theta sin 2 theta which is 2 sin theta cos theta right I should not have split it up anyways if I say cos theta is not equal to 0 I can say minus 2 p in fact 2 and 2 gets cancelled minus p cos theta is sin theta so I get tan theta as minus of p okay if I am assuming cos theta is not 0 there can be a possibility that cos theta can be 0 okay next is if I compare the coefficient of x on both the sides okay what do I get I get a 2 on the left hand side whereas on the right hand side I get 2 cos square theta that is when 1 multiplies to this and when px multiplies to this that is p sin 2 theta okay so again if you bring it to the other side will be 2 sin square theta is equal to 2p sin theta cos theta again if I am assuming sin theta is not 0 I can cancel them off okay and p comes out to be tan of theta okay now what conclusion can I draw from this what conclusion can I draw from this I cannot definitely have two values of one is giving me one is giving me p as minus tan theta and other is giving me p as tan theta okay when can possibly minus tan theta equal to tan theta when 2 tan theta is 0 right so when theta is a multiple of pi yes or no right so now can this be a possibility if theta is a multiple of pi if theta is a multiple of pi you know that cos of theta is not going to be 0 yes or no and this is the only possibility left from the first equation from the second equation both of them will be happy that tan theta will also be 0 and sin theta will also be 0 so in light of both these information that this is going to satisfy all the given conditions that I have in the problem the best possible value for theta is going to be n pi right and 2n pi whoever is getting an answer of 2n pi this is going to be a subset of the answer a so always in the exam try to give the biggest possible set that is going to satisfy the equation right so yes p has to be equal to 0 in that case p has to be equal to 0 in that case so option a becomes correct in this case so moving on to the next slide this is basically a paragraph based question let's call it question number 6 so consider a quadratic equation so you have been given a quadratic equation hope all of you can read the question so you have 1 plus mx square minus twice this x plus 1 plus 8m equal to 0 m should not be minus 1 on the basis of the above information answer the following questions the first question is the number of integral values of m such that the given quadratic has imaginary roots that is very simple to solve so this covers the various aspects of the positioning or the location of the roots of a quadratic equation which is again a very important concept from j point of view yes so the first question can be answered by simply using the fact that your discriminant has to be less than 0 correct so what is going to be a discriminant so b square minus 4 ac should be less than 0 you can drop the factor of 4 from everywhere so it's 1 plus 3m square minus 1 plus m 1 plus 8m less than 0 so if you simplify this you are going to end up getting 1 plus 6m minus 9m square minus m square plus 9m plus 1 less than 0 which is m square minus 3m less than 0 so m should lie between 0 and 3 m should lie between 0 and 3 so how many integers are possible not only 2 integers possible 1 and 2 so option c is absolutely correct so most of you have given this answer correctly good so in the meanwhile you can also find out when is the discriminant when is your discriminant going to be greater than equal to 0 when your m is going to lie between minus infinity to 0 so just keep this information so that you may need it elsewhere in the problem so just keep this information so that we can use it whenever it is required now let's talk about the second problem question 2 it says find the set of values of m such that the given quadratic has at least 1 root negative at least 1 root negative positioning of the roots location of the roots of a quadratic equation is one of the very important concept in quadratic equations so you have to be very very good with that especially how to deal with different situations question 2 option c that's ok I am not going to comment whether it is right or wrong so Vaishnavi just hold on before I disclose the answer any other response please resist, Purvik, Shweta Shruti Saundarya Aditi, Sai, Ritvik Kushal Prem guys back up Kushal also goes with option c for question 2 Sai also says the same Rishali alright so so for whatever response I have got for question 2 is c so let's check it out guys now let's say you consider your roots to be alpha and beta now when you want your at least one of the roots to be negative there are two possibilities one is one of them is negative exactly one of the roots is negative ok and the second possibility is both the roots are negative so both the roots are negative ok so when I say possibilities means this or this is possible right and this has to be ended with the situation that your roots should exist that means your discriminant has to be greater than equal to 0 this is going to be one of the very very important constraint because the roots don't exist there's no point saying it's negative or it's positive so your roots must exist and hence discriminant must be greater than equal to 0 and one of these option 1 or 2 can exist now when you say exactly one of the roots is negative when alpha into beta is negative right that means the product of the roots should be negative correct Vaishnavi when you say both the roots are negative when alpha plus beta is negative right that's the only condition when we can say both the roots are negative or means union or means union so basically you need to solve this inequality and take the union along with the intersection of the fact that discriminant should be greater than equal to 0 so this roadmap should be there in your mind when you're solving such problems so you must ensure that you're solving these three inequalities that is alpha beta less than 0 union alpha plus beta less than 0 whole intersection with discriminant greater than equal to 0 so discriminant greater than equal to 0 is already known to us so we don't need to slog it out alpha into beta is less than 0 so what is alpha into beta alpha into beta is going to be c by a c by a is going to be 1 plus 8m by 1 plus m okay so this guy has to be less than 0 so again let's quickly use wavy curve for this so minus 1 minus 1 by 8 plus minus plus so m has to lie between m has to lie between minus 1 and minus 1 by 8 now doesn't mean both a and b are less than 0 okay if it doesn't mean both of them are less than 0 then at least it means one of them has to be less than 0 shweta but it still meets my condition that at least one of the roots is 0 right so it incorporates both the conditions partially you are absolutely correct it doesn't mean that both of them have to be negative but it definitely says that it'll cover up at least one of the roots be negative so basically I want to say both the roots should not be positive that's what I want to say both the roots should not be positive okay great now alpha plus beta less than 0 means some of the roots which is minus b by a should be less than 0 okay and alpha plus beta is going to be 2 times 1 plus 3m by by 1 plus m correct if I'm not wrong that is my quadratic yes that's my quadratic and you want this to be less than 0 again plot it on the number line minus 1 minus 1 by 3 plus minus plus so m should belong to minus 1 to minus 1 by 3 now take the union of these two guys take the union of these two okay so when you take the union of these two let me plot all of them at the same time now so we have points like minus 1 we have points like minus 1 by 3 minus 1 by 8 we have 0 and we have 3 so first let's plot these 2 so minus 1 to minus 8 will be like this correct minus 1 to minus 1 by 3 is already covered under this okay and we finally have minus infinity to 0 so minus infinity to 0 so minus infinity to 0 union 3 to infinity so which covers the intersection zone is basically this zone intersection zone is basically this zone so the answer is going to be m belongs to minus 1 to minus 1 by 8 open interval that makes your option c is as your correct option so well done to those who answered option c sure Pushpinder I'll explain once again see the condition over here is you want at least one root to be negative right at least one root can be negative when exactly one of the root is negative or both the roots are negative okay when one of the roots is negative you actually use the condition the product of the roots is negative that's alpha into beta which is c by a should be negative and along with that I used the condition union alpha plus beta less than 0 alpha plus beta is basically the sum of the roots and when you say sum of the roots is negative means it also incorporates two cases when both are negative or one is positive and the other one is highly negative so both these conditions together take care of the fact that at least one root is insured to be insured to be negative right you could also deal with this by taking care of the fact that your both roots should be greater than equal to 0 that's another way to solve the problem now why intersection with this condition because this is the condition for your roots to exist this is the condition for your roots to exist if the roots do not exist there's no point talking about negative root or both the roots being negative so I have to take the intersection with this condition so that's what I did I found these three conditions individually and I performed this operation between them hope you all know how to take union for union you draw them on the same number line at the same level for intersection you draw it at a different level like this done over here it's at a different level so both of these are done at a different level and then we see where is the overlap happening where is the overlap happening that means both the conditions must be simultaneously be true right and I found this to be my answer for that I would suggest a reading on this on quadratic equation chapter that's very very important now guys moving on to the third question here find the values of m says that the given quadratic equation has both the roots as positive has both the roots as positive so visist has come up with the answer already he says d okay visist let's wait for others to come up with the answer sanjana also says d any other response guys back up not more than 4 minutes to give you a response rhetoric says d vashnavi says it will just be rest of the numbers satisfying the condition so option d yeah that's quite smart option d is going to be the answer for all of you right so this is going to be a correct response so as vashnavi rightly pointed out it's going to be the rest of the interval which is going to satisfy it so if this question comes as a stand alone question so here you could answer it because you knew the answer to the part 2 right if it comes as a stand alone question then what condition do you write for this what condition do you write for this if it comes as a stand alone question so if both the roots are positive if both the roots are positive so can I say has to be alpha beta should be positive and alpha plus beta should be positive correct and whole of this should be under the umbrella that your discriminant should be positive greater than 0 right so product should be positive some should be positive and discriminant should be positive so you are going to take the intersection of these conditions right that is c by a should be positive and minus b by a should be positive and your discriminant b square minus 4ac should also be positive so I am not going to solve this because here it is very easy from your option number 2 it becomes the answer becomes d is that fine right now moving on to the fourth part of the question find the set of values of m such that the given quadratic has at least one root positive at least one root positive again time yourself you should be able to solve it within 2 minutes now you know the approach any response guys sanjana says d hurvig will come to the fifth one later on so first let's solve the fourth one sai also says d vishas also says d nashal, shruti, aditi all say d so how do you solve this condition guys at least one root is positive at least one root is positive again can give you 2 situations one is exactly one root positive exactly one root is positive and we can also have a condition when both are positive both are positive okay now of course if exactly one is positive okay alpha into beta would be less than 0 right and if both are positive that means in combination we can say alpha plus beta would be positive now alpha plus beta is positive again can take 2 situations when both are positive or one is positive or one is negative of course it will not allow both of them to be negative so this guy will not allow it will not allow both to be negative okay so what should I do with these 2 should I take the union of these 2 or should I take the intersection of these 2 people who are not answering please respond should I take the union of these 2 conditions or intersection of these 2 conditions which just says union others please respond of course we should take union of these 2 conditions now the combined condition has to be intersected with the condition that your root must be real else there is no point talking about you know root being positive or negative at all so if we start solving this we have already solved for alpha beta curve okay so alpha beta less than 0 is when your m belongs to for this m belongs to minus 1 to minus 1 by 8 okay and when you say alpha plus beta greater than 0 for this your m should belong to I think we have already done less than 0 let's check it out yeah we have done less than 0 it has to be beyond minus 1 and minus 3 so it has to be minus infinity to minus 1 union minus 1 by 3 to infinity and of course the discriminant part is already there okay so let's collectively work it out on the number line so minus 1 minus 1 by 3 minus 1 by 8 0 3 minus 1 to minus 1 by 8 will be this interval minus infinity to minus 1 so basically it covers all the way minus infinity to minus 1 it covers all the way and minus 1 by 3 to infinity it also covers everything now okay and this has to be intersected with this has to be intersected with minus infinity to 0 minus infinity to 0 and 3 to infinity 3 to infinity okay so the combined condition is going to be the combined condition is going to be minus infinity to minus 1 okay of course minus 1 cannot be included because of this gap so it will be minus 1 all the way till we reach minus 1 by 8 right is that fine and ideally it should be minus 1 to minus 1 by 8 to 0 as well right but anyways let's go by the options option says 3 to infinity also so this part should also be there in your answer so we take the union of these these conditions that we get over here is that fine so option number D becomes your right option for this now going to the fifth question Purvik has already given me the answer what about the others let's work on the fifth question now the number of real values of M says that the given quadratic equation has roots in the ratio of 2 is to 3 any response guys any response okay so you can consider your roots to be 2 alpha and 3 alpha so if you want them to be in the ratio of 2 is to 3 so what will be the sum of the roots sum of the roots is going to be minus B by A which is 2 times 1 plus 3 M by 1 plus M which says 5 alpha is equal to this okay and what about the product of the roots that is 2 alpha I'm sorry this is going to be 3 alpha 2 alpha into 3 alpha is going to be C by A and C by A is going to be 1 plus 8 M by 1 plus M which is actually 6 alpha squared okay so what we can do is we can solve these two simultaneously for alpha that means 2 by 5 1 plus 3 M by 1 plus M which is going to be alpha substitute it over here substitute it over here so we are going to get 2 by 5 times 1 plus 3 M by 1 plus M square right into 6 is going to be 1 plus 8 M by 1 plus M so what I have done I have done 6 alpha squared is equal to 1 plus 8 M by 1 plus M right so let's quickly simplify this so that's going to give me 4 into 6 that is 24 upon 25 1 plus 3 M the square 1 plus M square is equal to 1 plus 8 M by 1 plus M and since M is not equal to 1 I can cancel them off which gives me the option which gives me the option to cross multiply so this square is equal to 1 plus 8 M times 1 plus M you can even take this 25 to the other side you can also take this 25 to the other side and when you simplify it you are going to get 9 into 24 minus 25 into 8 so that's going to give you 200 minus 24 into 9 that's going to be 216 16 M square and what about the M factor M is going to be 6 into 24 which is 144 and from here I will get 9 into 25 which is 225 which is going to be 81 so it's going to be minus 81 so it's going to be minus 81 and we will get 24 here and 25 which is going to be minus 1 ok now this is a quadratic in M this is a quadratic in M and we see that B square minus 4 AC would be always be positive B square minus 4 AC will also be always be positive so it implies we get 2 values for M 2 values for M which will satisfy this equation and hence the number of real values of M which satisfy this equation is going to be 2 please ensure many people after getting the quadratic they stop check whether these those quadratic have real roots or not because your possibility of 0 being the answer can also be there ok so it was a huge question it was covering 5 question in itself but we learnt a lot of concept about positioning of roots of quadratic equation so that's a good example let us now move on to the next problem which is your problem number 7 for the day hope all of you can see the question if alpha beta are the roots of the equation x square plus px plus q equal to 0 so alpha and beta are the roots for this and also x to the power alpha beta are also the roots for this so roots for this are alpha and beta as well and we have been given that alpha by beta and beta by alpha are the roots of this equation then they are asking you the value of N answer as and when you get it any response from anyone anyone any response alright let's work this out since alpha and beta are the roots of this equation I can say from using this fact that alpha plus beta is going to be minus p right alpha and beta is going to be minus p okay and alpha into beta is going to be q right now alpha and beta are also the roots of this equation correct right they are also the roots of this equation now if you treat this equation as x to the power 1950 square p to the power 1950 x to the power 1950 plus q to the power 1950 equal to 0 okay and I call y as x to the power 1950 for the time being then you would realize that I have written a quadratic like this I have written a quadratic in terms of y like this correct yes or no now if alpha beta are the roots of this equation what do you think will be the roots for this equation in terms of alpha and beta what would be the roots of this equation please type in the chat box guys all of you have suddenly gone quite that seems so weird to me so tell me if alpha beta are the roots of this equation what would be the roots of this y square plus p to the power 1950 times y plus q to the power 1950 this is the concept which is based on the transformation of equations or transformation of the roots of the equation so what value of y is going to satisfy it when alpha and beta is going to satisfy this equation is it a to the power 1 by 1950 or sorry alpha to the power 1 by 1950 or alpha to the power 1 by 1950 so there is a 2 answers coming up Purvik and Vaishnavi they have exactly given different answers see Purvik if you put alpha over here what do you see alpha to the power 1950 square plus p to the power 1950 okay alpha to the power 1950 plus q to the power 1950 equal to 0 right which clearly means this is playing the role of y isn't it this is playing the role of y so the root for this equation should be alpha to the power 1950 and beta to the power 1950 are you getting it Purvik right okay great so what can I say about the sum of the roots what can I say about the sum of the roots it's going to be minus b by a minus b by a that's going to be minus p times 1950 right and what is going to be the product of the roots the product of the roots is going to be my c by a which is q to the power 1950 which is q to the power 1950 now apart from this we have been given another information that alpha by beta and beta by alpha are the roots of this equation so I can say I can just simply substitute alpha by beta in place of x so alpha by beta to the power n plus 1 plus alpha by beta plus 1 to the power of n okay this should be equal to 0 so it is alpha to the power n plus beta to the power n okay and here also I have alpha plus beta to the power n that divided by b to the power n I have taken to the right side so it just becomes equal to 0 okay now we already know from our previous result that alpha plus beta is going to be minus p so we know alpha plus beta is going to be minus p so when you substitute it over here you get alpha to the power n plus beta to the power n and we get plus minus p to the power n equal to 0 okay so if you take this to the other side you get this expression sorry you get this expression okay now try to compare these two so if you compare these two you would realize that n is equal to 1950 is going to satisfy this particular condition okay so I am not saying that there cannot be any other value which will satisfy it but from the information that I have at hand I can only say my n can be equal to 1950 so this is going to be your answer does that make sense is the solution clear please type in clear on your screen if you think it's clear to you else please feel free to type in your question in the chat box so here what concept is being tested how do you transform your equation as per the roots requirement so this is a topic which you must all go through transformation of the equation as per the roots so guys moving on to the next question for the day which is your question number 8 so you have a function xq plus x square plus 100 times x plus 7 sin x okay and you have been given an equation which is this this equation is in terms of y and we need to comment upon the nature of the roots so no real roots one real root two real root more than two real root any progress guys anyone okay this seems to have made some progress kushal says B okay kushal let's wait and see what's the answer kushal if you see this this equation is actually a quadratic in y okay so it cannot have one real root either both the roots are real or both the roots are imaginary so this can never be the option got it kushal do you understand why it's a quadratic equation see let's say f1 is a f2 is B and f3 is C so it's something like y minus a 1 by y minus B 1 by y minus C equal to 0 correct sorry correct so that's going to give you y minus B y minus C twice of y minus A y minus C plus thrice of y minus A y minus B equal to 0 so this comes out to be a quadratic in y this is a quadratic in y so you can never have one real root it can either be no real root or two real roots okay so option B and D are anyways gone eliminated right Purvik this says A Sanjana says A Purvik says C let's see who is correct now guys these are the questions which like you know end up in a soup thinking you know where to start from okay so let's start with the finding the derivative of f of x now yes you must be surprised how come calculus so calculus can go anywhere it wants to so find the derivative of this function that gives you 3x square plus 2x 100 plus 7 cos x plus 7 cos x now if you try to solve this problem in this way that is 3x square plus 2x plus borrow a 7 from this so that you get 1 plus cos x then I'm pretty sure that this expression will always be positive right because 1 plus cos x will always be greater than equal to 0 and I see in front of me a quadratic in x whose coefficient of x square is positive and whose discriminant that is B square minus 4ac is negative in other words it's a quadratic which is always hanging in the air like this that means this quadratic will also be positive always right so here I have a function whose derivative is always positive and if derivative is always positive it implies my function is an increasing function it's an increasing function right so if it's an increasing function can I say that can I say that f of 1 will always be less than f of 2 and f of 2 will always be less than f of 3 yes or no that means according to my assumption of a b and c a should always be less than b and b should always be less than 3 correct yes or no now if you put the value of a into this function if you put y as a try to put y as a what do we get we get a minus b times a minus c right because these two terms will become 0 the last two terms will become 0 now a minus b is a negative term a minus c is a negative term that means when you put y equal to a this quadratic will be positive right so when you put y is equal to a you are getting a positive result from this expression let me call this as let me call this entire thing as let me not write it wrong let me call this entire thing as f of y okay so when you put f of a into this I will not call it f of y let's call g of 5 because g has already been f has already been used so when you put j into it you are getting a positive answer okay now put y as b when you put y as b you get what do you get you get two times b minus a b minus c okay now b minus a is positive but b minus c is negative why because b is lesser than c and a is greater a is lesser than b so this is positive this is negative so when you put b into the expression you are getting a negative answer what does it mean if a polynomial changes its sign for two values a and b that means there has to be at least one root between a and b okay at least one root between a and b that means there has to be two roots between a and b there has to be at least exactly two real roots because we cannot have one root real and one root imaginary so here is my verdict that for this equation you will have exactly two real roots coming up is that fine is there any question with respect to that so again here the concept of location of the roots is being used along with the fact of increasing decreasing nature so guys whenever you are solving any kind of a polynomial equation or for that matter any equation always keep this angle of increasing decreasing nature always in your mind okay a question may be also asked to you how many real roots does this equation have f of x yeah sure I will repeat it when you put y as a into this function you will realize that you get y- sorry a-b times a-c right because all the other terms like this will become 0 0 right so when you put y as a that means in this function which I am calling as g of y this entire thing I am calling as g of y so when you put y as a you get the answer as positive okay when you put y as b you get the answer as negative that means between a and b g of y should have one root at least so this implies g of y has at least at least one root in the interval a to b right if it has at least one root in the interval a to b means one root has to be real and since it is a quadratic the other also has to be real so I am not saying I am putting y as a I am just trying to check the value of the sign of this expression so the actual value would be somewhere between a and b are you getting it yeah I understand that we should so I am not saying my root will be a I am saying the root will be somewhere between a and b so I am just trying to check the sign of the numerator expression that's it the value is slightly heavier than a you will get the same result because anyways the denominator is going to go to the right side and become 0 is that fine great with this we will move on to the next question question number 9 if alpha beta gamma be the roots of the equation this x minus a times x minus b times x minus c equal to dd not being equal to 0 then the roots of the equation x minus alpha x minus beta x minus gamma plus dr yes sure after this question Shweta will take a break this is a very easy question you should be able to solve this within one minute any response from anyone Kushal says c okay Sandhariya also c Nikhil also c others Rushali is Vishak there Purvik Aditya Rithvik Vaishnavi is also getting c alright guys Aditya is also getting c so if alpha beta gamma are the roots of this equation I can easily comment that this is fact-prisable as x minus alpha x minus beta x minus gamma normally should always write a k over here but we know that the coefficient of x cube over here coefficient of x cube is equal to 1 so my k actually becomes 1 okay say k actually becomes 1 so I don't have to write the value of k over here so I can write it as x minus alpha x minus beta x minus gamma okay now if I send the d to the other side if I send the d to the other side it straight away shows that this expression can be factorized as this this expression can be factorized into this right which clearly implies that the roots of this equation are going to be a b and c so your option c becomes correct it was a super simple question just in between so those difficult questions so guys we'll take a break right now we'll take a break and we'll resume at we'll resume at let's say 11 30 would be fine thank you alright so let us resume after the break hope everybody is back so here is the next question question number 10 the solution for this equation x minus 1 equal to x minus gif of x times x minus fractional part of x okay this should be an easy problem and always remember when you're solving such problem there is a cycle so remember this cycle always first treat mod wherever there is a mod treat that first that means take situations so that the mod acquires definition according to that interval of x so first take the interval and break mod x that is the first step second step is always break x as gif plus fractional part okay third step is make make fractional part the subject of the formula okay why do we make it the subject of the formula because we all know that fractional part can lie between 0 and 1 so that gives you fractional part in terms of greatest interior function and greatest interior function is very very easy to solve because you are dealing with integer values of the expression so this is the approach that you should be following whenever you are dealing with these three functions that is mod x x fractional part and greatest interior function so this is a pretty easy one so let's solve this quickly so this is the cycle that you should be following psi says option C okay Purvik also says C others please pitch in okay Kushal also says C alright so I think most of you are under the favor that option C is correct so let's solve this out as I already discussed with you first of all first of all this is nothing but fractional part of x and this is nothing but gif of x and x change it with change it with fractional part plus integer part like that okay right now here if you see clearly it's actually gif fractional part minus gif fractional part plus 1 equal to 0 which is actually of the form a b a b minus a minus b plus 1 right which can easily be factorized as a minus 1 times b minus 1 okay so I can write this as fraction greatest interior minus 1 times fractional part minus 1 0 which implies either this can be 1 or this can be 1 but we know fractional part can never be 1 because it has to lie between 0 and 1 it has to be lesser than 1 and between 0 so the only possibility is your greatest interior x is 1 for that your x has to lie between 1 to 2 2 not included so absolutely correct most of you given this answer correct well done guys this was slightly easier question let's go to the next question which is question number 11 for you so here in this question we have alpha beta gamma such that they satisfy these 3 conditions and you have to find the gif of alpha to the power 4 plus beta to the power 4 plus gamma to the power 4 minutes so think on this guys any progress feel free to type in your answer in the chat box sure sure kushal take your time no worries purvig says 18 all I can say purvig is no so that you can try once again any other answer please kushal no that is not the right answer yeah so what is the greatest integer of that the question says find the greatest integer of that expression so let me tell this now this is your correct so guys the answer to this question is 7 so now let us discuss this question let us discuss this question now when it comes to the fact that I have to find alpha to the power 4 beta to the power 4 gamma to the power 4 it's actually alpha square beta square gamma square it is actually influenced by this square expression right so we can find this expression within let me write it within ok we can find this in this expansion square ok so when we actually do this that is alpha square beta square gamma square whole square I am going to get alpha to the power 4 beta to the power 4 gamma to the power 4 apart from that I get 2 alpha square beta square 2 beta square gamma square and 2 gamma square alpha square correct now this expression is known to me which is 6 square this is something which is not known to me so x so 2 times alpha beta gamma square I need ok now this expression is influenced by alpha beta beta gamma gamma alpha how do I get this how do I get this we can get this from the fact that I know these 2 expressions that is if I use this identity alpha plus beta plus gamma whole square I am going to give me alpha square beta square plus gamma square plus 2 times alpha beta beta gamma gamma alpha right so this is 4 square which is 16 this is going to be 6 so alpha beta beta gamma gamma alpha is obtained from this expression which is 10 by 2 which is going to be 5 so this is going to be 5 ok now again if I square this expression up if I square this expression up alpha beta beta gamma gamma alpha square which is going to be 5 square I am going to end up getting alpha square beta square beta square gamma square gamma square alpha square apart from that I am going to get to alpha beta square gamma then I will get alpha beta gamma square then I will get alpha square beta gamma which is nothing but let's say I call this quantity as y so 5 square is going to give me y plus 2 alpha beta gamma into alpha plus beta plus gamma so what I am doing is I am taking alpha beta gamma common from this so what I am doing from this I am taking alpha beta gamma as common ok now the problem is I know alpha into beta into gamma which is I think 4 right which is 4 but I don't know what is alpha beta gamma ok so basically I am stuck with this problem now what is alpha beta gamma ok let me call this as z for the time being ok so see one is leading to another now how do I get z how do I get z if you would recall you would have learned this expansion that alpha cube plus beta cube plus gamma cube minus 3 alpha beta gamma is actually alpha plus beta plus gamma times alpha square beta square gamma square minus alpha beta minus beta gamma minus alpha gamma correct now are these values known to me yes I know this value is going to be 4 this value let me check what was that value 6 this is going to be 6 right and this value is going to be 5 which I already found out earlier right so alpha cube beta cube plus gamma cube whose value itself is known to us that value is 8 so this value is 8 and this is my z so 8 minus 3 z 8 minus 3 z is equal to 4 so z is equal to 4 by 3 ok if you place this z over here I get y as 25 minus 32 by 3 right which is going to be 75 minus 32 75 minus 32 is 43 by 3 ok and now when I place this value back over here I get 36 is equal to x plus 2 times y which is 43 by 3 am I correct ok so x value from here I can get as 36 minus 86 by 3 which is an 108 minus 86 by 3 which is going to be 22 by 3 so the required expression which is the expression within the greatest integer sign that is 22 by 3 so gif of 22 by 3 is going to be 7 and that's the answer for this question right so is that clear is the step clears so again you have to have a road map in your mind and accordingly act towards what is required so here I needed alpha square alpha beta whole square plus beta gamma whole square plus gamma alpha whole square so I proceeded towards finding that so as and when I needed a term I needed an expression I found a formula to derive it good with this we can move on to the next problem question number 12 so the question says this is a big equation of degree 9 and they are asking you what are the nature of the roots so today just one paper of kvpy completely time yourself and just revise a bit of your three subjects that you have finally made up your mind to attempt don't try to study too many things and sleep early if possible so that you can get up early so nature of the roots is going to be a very important type of questions asked in these exams so you have to be very very careful what approaches you can take and as I have already discussed calculus method is a good approach to solve such kind of problems and I have not seen anybody using it so far now so that was a hint actually for you see in real roots okay so Mahin and Kushal are saying option C okay guys let's look into this so let us call the left hand side of the expression as a function this let's differentiate it with respect to X so that's going to give you 9 times X minus 3 to the power 8 9 times X minus 3 square to the power 8 and 9 times 3 cube to the power of 8 and so on till 9 times X minus 3 to the power 9 to the power 8 now you would see yourself that each of these terms are positive each of these terms are positive because they are subjected to even power so f dash X is positive it means it's an function it implies it implies f of X f of X is an increasing function it's an increasing function okay if f of X is an increasing function and it's a degree 9 it means it's an odd degree polynomial it's an odd degree polynomial now remember any odd degree polynomial graph if you sketch if you sketch the graph of any odd degree polynomial whose leading coefficient is positive leading coefficient means leading coefficient means coefficient of highest power of X whose leading coefficient is positive then the graph of such functions will always be of this nature so they will go something will happen and then they will end up up this is already discussed in the class so the ending branches will be oriented in this way okay something will happen in between which I am not able to comment right now okay but it will always be like this now the function is an increasing function function means once it cuts the X axis it will never cut it back again because it's increasing isn't it if an increasing function once crosses X axis it will never come back again because it always has to increase it's a monotonic function a monotonic function has to be a 1-1 function now do you want to change your answer guys now would you like to change your answer guys yeah now you started changing your answer so it's going to be option B so it has to only have one real root and rest of the roots will be imaginary in character because it will be all in the air so it will not be cutting X axis anymore guys you have to understand this fact how graph is going to help you solve this problem along with functions you just can't rely completely on algebra right so there are so many things which comes into action while you are solving questions like this right great so with this we will go to the next question that's question number 13 for you this is quite similar to what we have already done so you should be able to solve this easily so the number of solutions of the following inequality 2 to the power 1 by sine square X2 3 to the power 1 minus sine square X3 and so on this guys I am expecting you to give me the answer I have already solved a similar looking problem for you Vashtavi says option B Shweta says A Shweta seems to like option A a lot Purik says A okay Pusha, Nishal all go with option A so only Vaishnavi says B okay others please pitch in Sanjana where are you Mahen Vishest Aditi, Sondarya where is Ritwik gone so guys let's work on this so basically this is 2 to the power cosecant square X2 3 to the power cosecant square X3 4 to the power cosecant square X4 and so on N to the power cosecant square Xn on the right hand side we can write it as 2 into 3 into 4 all the way till we reach N so just like we did the last time we divided by 2 3 4 till N so that's 2 3 4 till N so that gives me 2 to the power cosecant square minus 1 which is cot square X2 3 to the power cot square X3 4 to the power cot square X4 and so on till N to the power cot square Xn less than equal to 1 less than equal to 1 again when you take log on the both sides you realize we get cot square X2 ln 2 plus cot square X3 ln 2 sorry ln 3 and so on till cot square Xn ln N okay less than equal to 0 now this option will only be true when it is exactly equal to 0 because all these quantities I will not say positive first say I will say they are greater than equal to 0 they are greater than equal to 0 okay and since it is less than equal to 0 the only possibility is each of them should be equal to 0 that means every cot X2 should be 0 cot X3 should be 0 and so on till cot Xn should be 0 okay now the interval 0 to pi by 2 when is these values going to give you 0 when is cot 0 in the interval 0 to pi by 2 sorry 0 to 2 pi it is going to be at pi by 2 and 3 pi by 2 right only 2 possibilities for each one of them so X3 can also be pi by 2 3 pi by 2 like that all these angles can only be pi by 2 and 3 pi by 2 right so from X2 to X1 from X2 to Xn each one has 2 2 options each so according to fundamental principle of multiplication my answer would be 2 to the power n minus 1 that's option B is correct so only Vaishnavi was correct well done Vaishnavi good guys I think you copied the answer which others told in majority okay great so let's move on let's move on to the 14th problem for the day so the question says this and this denote the fractional part and GIF respectively solve this equation solve the equation mod of 2x minus 1 is equal to 3 GIF plus 2 fractional part explain the last part again okay sure let others copy this question then I'll switch to the next question let's just hold on for a minute so that others can copy this question and then I can explain you the last part so it's mod of 2x minus 1 is equal to 3 GIF of X plus 2 times fractional part of X yeah all of you look here so basically ln 2 is the positive term okay and this is also a positive term in fact when I say positive means greater than equal to 0 so left hand side all quantities are greater than equal to 0 on the right hand side you have less than 0 so the only situation where both the left and the right hand side will agree to each other will be when they are equal to 0 each so that will happen only when each of these terms here is 0 0 0 so cot x2 should be 0 cot x3 should be 0 all the way till cos xn should be 0 no no no it is from 2 to n 2 to n how many terms are there 2 to n there is n minus 1 terms only right so that's why it is 2 to the power n minus 1 okay yeah now see here a cot x2 would be 0 only for these 2 angles pi by 2 and 3 pi by 2 in the interval so in the interval 0 to 2 pi cot will be 0 at pi by 2 and 3 pi by 2 so every angle has 2 options so x2 has 2 options that is pi by 2 and 3 pi by 2 similarly x3 has also 2 options pi by 2 and 3 pi by 2 like the angle has got 2 2 options okay so how many how many possible solution sets can be there the possible solution set can be 2 into 2 into 2 into 2 n minus 1 times okay that makes your answer 2 to the power n minus 1 Rithvik is that fine with you now it's all this question now question number 14 Kushal says 1 by 4 okay Kushal let's wait for this to answer Shruti also backs him up says 1 by 4 Shrita says any value that belongs to minus 1 to minus 2 okay so she has given a range of values of x guys I am waiting for others Sai also says 1 by 4 alright so majority of you are in the favor of saying 1 by 4 Shrita adds and x equal to 1 by 4 okay recall the cycle recall the cycle which I have discussed always treat mod first so I know mod of 2x minus 1 is going to behave in 2 ways it's going to behave as 2x minus 1 when x is greater than equal to half and it's going to behave as 1 minus 2x when x is less than half right so let's take the parent condition let's say I will work under this condition x greater than equal to half so under that condition I will have the left hand side which is the mod operation as 2x minus 1 here I will have 2gif plus 2 fractional part then remembering the next step of the cycle we need to change our x with gif plus fractional part so which boils down to this expression which is 2gif 2 fractional part minus 1 is equal to 3gif 2 fractional part 2 fractional part 2 fractional part 1 okay which gives you gif of x equal to minus 1 now this is 2 when your x belongs to the interval minus 2 to minus 1 now which is not satisfying this criteria okay so under this criteria when x is greater than half you cannot say x lies between minus 2 and minus 1 right in fact you should say 0 to minus 1 not that you should say minus 1 to 0 okay minus 1 to 0 even this cannot be the answer this cannot be the answer because it violates this condition yes or no so now I will move on to the next condition when x is less than half so this is my initial condition or parent condition under that circumstance I will get this expression equal to this expression okay again break x as gif plus fractional part okay expand things up so this is going to be this so now if you look at this closely you are going to get 1 minus 4 fractional part of x is equal to or you can say 5 gif of x 1 minus this is 4 fractional part now make fractional part the subject of the formula which gives you 1 minus 5 greatest integer by 4 and we all know that we know that fractional part lies between 0 and 1 so as a result 1 minus 5 gif of x by 4 will lie between 0 and 1 that means 1 minus 5 gif of x will lie between 0 and 4 which means if you divide by minus 5 it actually switches the sign like this should not be equal to 1 here I will stop this 1 so less than equal to 1 by 5 now which is the integer lying between minus 3 to 1 by 5 1 possible integer and that is 0 that is going to be only 0 ok so if gif of x is 0 what is going to be the fractional part if you substitute it over here you get fractional part as 1 minus 0 by 4 which is 1 by 4 so as a result your x will come out to be gif plus fractional part 0 plus 1 by 4 so this is going to be your value of x does it satisfy the condition that it has to be less than half yes it very much satisfies the condition that it should be less than half so my answer is going to be x equal to 1 by 4 so well done to those who solved this great going guys so coming to the last question for the day that's question number 15 solve the system of equation solve the system of equation in x, y and z satisfying the following condition satisfying the following equations so these are greatest integer functions these are fractional parts so please work this out x, y, z all values I need purvik is 3.1 0.1 not 3 into 1 this is 3.1 4.3 5.4 it's decimal not multiplication shruti says x equal to 2 okay I'm not commenting whether it's right or wrong need all the 3 values y is 1.3 okay no matrix method won't work because there are different natures right action part x one is gif of x so matrix method will not work and shruti says z is 1.1 sai also gives a similar answer just that sai says z is 3.1 okay shruti changes it to 3.1 now guys how about others nikhil purvik okay where is the rest gone shweta get a good sleep today okay today your answer is that off the track I don't know why normally you are very accurate for tomorrow you should be very very fresh so sai is saying y is 1.3 z is 3.1 what about x sai aditi saundarya sai is saying x is 2 fine so we have just one minute left so we are going to solve this quickly first of all add all these three equations together okay if you are going to add it up you are going to see that fractional part and gif is going to become x correct here these two are going to become y and here these two are going to become z and you already have an x y and g separate so correct me if I am wrong if you add them if you add one two and three you are going to get twice x y z is equal to the sum over a that is 7.4 7.4 and this is going to be 12.8 so x y x plus y plus d is going to be 6.4 that's your first equation okay now add one and two when you add one and two you will get you will get x plus fractional part of x y plus gif of y and gif of z plus fractional part of z that's going to be z anyways so that's going to be 7.4 okay now out of this we already know x plus y plus z that's 6.4 so 6.4 plus fractional part of x plus gif of y is going to be 7.4 that means fractional part of x plus gif of y is going to be 1 which clearly implies that fractional part of x is going to be 0 and gif of y is going to be 1 okay so let's say this is our first information that we obtained so far now let's add 2 and 3 let's add 2 and 3 so I'll get x plus y plus fractional part of y plus z plus gif of z is going to be 9.7 is going to be 9.7 out of that I already know the value of these x plus y plus z which is 6.4 so 6.4 plus fractional part of y plus integral part of z is going to be 9.7 which means integral part of gif of z plus fractional part of y is going to be 3.3 okay which clearly implies gif of z is 3 and fractional part of y is 0.3 so this is our next finding now let's add 1 and 3 let's add 1 and 3 so adding 1 and 3 we get x plus gif of x plus y plus z plus fractional part of z is equal to 8.5 out of that we only know these 3 values which is 6.4 plus gif of x plus fractional part of z is going to be 8.5 so this is going to give me 2.1 so greatest integer of z is 2 and fractional part of z is going to be 0.1 so that's another achievement for me now let's start finding in the values of x, x will be gif that is 2 plus 0 that's going to be 2 y is going to be gif plus fractional part which is 1 plus 0.3 which is 1.3 and z is going to be gif plus fractional part which is going to be 3 plus 0.1 which is going to be 3.1 so this becomes your answer guys whenever you are finding the answer to any equation just always verify whether it satisfies any one of the equation don't get complacent that oh I've got the right answer so always verify with one of the equations so guys hope this session was very useful to you in understanding different natures of the problem we went through fractional part, greatest integer function mod function, we looked into calculus aspect, we looked into team cases of inequality some problems can also require the good use of graph, we looked into transformation of quadratic equations so all these aspects I hope was very very insightful for you to solve further problems, all the best for tomorrow, have a good sleep today don't try to study a lot of things just put away your books by 9.30 and just go off to sleep and do well in tomorrow's exam ok, so thank you very much for my side over and out from Centrum Academy thank you for coming live on youtube