 Hello all welcome to the YouTube live session on problem practice session This would be a problem practice session on important topics with respect to JEE main and JEE advance So request the people who have just joined in to type in your name in the live chat box so that I know who all are attending this session Good morning to all of you have joined the session So please type in your names so that I know who all are attending Hello, Aditi, Sai, Kushal, Ramcharan guys. How are you doing? Sure, without much I do we'll start with the very first problem of the day It's based on Koenig sections Koenig sections ellipse chapter So the question reads like this There is an ellipse with an eccentricity of half and one focus is at half comma one It's one directrix is the common tangent to the circle x square plus y square equal to 1 and X square minus y square is equal to 1 clearly. This is the rectangular hyperbola. This is going to be a rectangular hyperbola Then they're asking you the equation of the ellipse in the standard form Then they're asking you the equation of the ellipse in the standard form So again, the norm would be I would wait at least for five people to answer before I start discussing it Because I strongly believe that it's a problem solving session and everybody should put in heart and soul to solve this And try to be the first one to solve it. Good morning everybody who has joined the session So anybody who has managed to solve this I would request you to type in your response in the chat box as to which option is correct A, B, C or D The first one solving this would get a shout out for me so for the purpose of Realizing the situation geometrically I have drawn the graph for you so that you can refer to this So guys as you can clearly realize over here that the circle mentioned over here is the auxiliary Circle for this rectangular hyperbola, right? And that's why as you can see in the diagram I have drawn it in such a way that the circle is ending up touching the the vertices Right the circle is ending up touching the vertices. Let me call it as a and a dash Okay, so if you have to have a common tangent, it can be only at these two points Right So it's pretty obvious that the common tangents could be either of these two lines The common tangents could be either of these two lines and these lines are clearly x equal to One and x equal to minus one So now you have to figure out which one is nearer to this focus So which is the directrix that you're going to use we all know that when we know the focus and the equation of the directrix So let's say the focus is that alpha comma beta then we all know that the equation of the ellipse is given as e square Lx plus my plus n whole square by L square plus m square, right? This is the equation of the ellipse that we know. This is the most general form where your focus is that alpha comma beta and your directrix is And your directrix is Lx plus my plus n equal to zero So I'm sure the point the line the directrix which is nearer to half comma one would be x equal to one, right? So this would be your required this thing. By the way, you're a is one So I'll just replace this with one and one This will be one and this will be one. So our poor X is a okay So guys anybody else. So I've got the first response that is from poor Right poor it the directrix which is a closer to s closer to s will be x equal to one only Now everything is set. I don't see a Point not solving it right now once, you know, the Directrix is that one x equal to one and focus is that one comma. Sorry half comma one You can always write down the equation as you can always write down the equation of the ellipse as X minus half the whole square Plus y minus one the whole square is equal to e square is already given to us as half. So e square is one-fourth and X minus one square by one. Okay, so we just have to rearrange it So as to see which of the options is suiting the requirement of this So when you open this bracket, you get x square minus x plus one-fourth plus y square minus two y plus one is equal to one-fourth X square minus two x plus one So just let me simplify this further. So I'll have three x square by four and I will have minus x and plus x by two which is minus x by two and we'll have constants like One fourth and one fourth will get cancelled So we'll have y minus one the whole square is equal to zero Okay So just to further simplify I take three by four out common. So I'll have x square and When I take three by four common out from here, it becomes minus two by three X Okay, and for the completing this square over here without much ado. I'll have X minus one by three the whole square minus one by nine Plus y minus one the whole square equal to zero Which is clearly three-fourth X minus one-third the whole square Plus y minus one the whole square is equal to one by 12 is equal to one by 12 So according to the given options, I have to have a one on the other side Have to have a one on the other side. So I will divide throughout by one by 12 So if I divide throughout with one by 12 What I will see is nine x minus one-third the whole square Plus 12 y minus one the whole square is equal to one which I think Which I think is clearly your option number Option number a option number a by the way, I think there's a one missing out over it So yes, Purvik you are absolutely correct So the first one to solve it and the only one to solve it is Purvik Very good. Well done. Any question with respect to this? Please feel free to ask me right now So guys the moral of the story here is that make diagrams make diagrams without diagrams Never try to solve a coordinate geometry problem So I've been repeating this since the beginning of this course that please whenever there's a coordinate geometry problem Try to make a diagram that will always help you to find out the solution in a faster and easier way So with this note, I'll be moving on to the next problem. That's your again a question on coordinate geometry The second question so the question reads like this let there be a point P Whose coordinates are mentioned to you a seek theta comma beta and theta and there's a point Q also Where theta and Phi are complementary to each other where theta and pi are complementary to each other Okay on this hyperbola HK is the intersection point of the normals of P and Q Then find K Guys, let me tell you there is nothing important in the concept nothing rocket science in this question It's all about using your basics equation of a normal then the point of intersection of the lines and using the given condition in the problem That's it. There's nothing difficult about it and I'm expecting everybody to give me the right answer Okay, so Kushel is of the opinion. It's option D. Okay Rohan thinks it's option C Guys, they should not be anybody who is actually figuring out what is the equation of a normal at this stage of your journey So the equation of the normal at a point x1 comma y1 on a hyperbola Which is x square by a square minus y square by b square equal to one should be now clear The equation is a square x by x1 Plus b square y by y1 equal to a square plus b square Nobody should now be struggling. What is the equation of a normal? Okay, if you are means you have not Done enough practice on Koenig sections Now, let's say call this point Okay, so basically when you substitute x1 as a seek theta and y1 as b tan theta you get something like a cos theta x and b Caught Theta y is equal to a square plus b square So this would be the equation of the normal at p Similarly equation of the normal at q will be a cos phi x plus b Caught phi y is equal to a square plus b square Okay, so I is also of the opinion that it's D Purvik also says D Guys, I want the response from others. Ramcharan, Kondiniya, Atmaish, Shritvik, Naman, Sondarya, Shruti, Sanjana, Aditi So it's guys, it's very simple so the first equation that is the equation of the normal at p I would write it as x since I have to find K what I will do. I will try to eliminate x Correct, which is actually your H So let me divide throughout with a cos theta So if I divide by a cos theta, I'll get something like b by a Sin theta y is equal to a A square plus b square by a cos theta Similarly at q if you divide you get simply this So you don't have to do any hard work for the other one because it's just the replacement of theta with phi It's just the replacement of theta with phi Now just subtract these two. Let me call this as 1 and 2. So let's subtract 1 and 2 Let's subtract 1 and 2. So when you subtract 1 and 2 you get something like b by 1 by sin theta minus 1 by sin phi and On the right hand side, you'll end up getting a square plus b square by a 1 by cos theta minus 1 by cos phi There's a lot of things can be done over here So you can just get cancelled b and we can always write Sorry, this is y times so y is equal to a square plus b square by b and you will have cos phi minus cos theta by sin phi minus Sin phi minus sin theta and moreover you will have sin theta sin phi By cos theta cos phi now, of course this term would be tan theta tan phi Now guys by the way since they're complementary, this will become sin phi This will become sin theta. This is also becoming sin phi Okay, and this will become tan theta tan phi Now we know that since theta plus phi is equal to pi by 2 tan of theta plus phi will be undefined Okay, which means the expansion of this which normally has the tan theta tan phi will become Minus of 1 this should become minus of 1 Sorry, this become plus 1. Sorry. Yeah, because in the expansion you will have tan of theta plus tan of phi by 1 minus tan theta tan phi So your denominator should be equal to 0. This part should be equal to 0 So this part becomes 1 and I think this part will get cancelled off So I will get y as a square plus b square by b. Is it fine? Yes. So am I missing out anything? So, oh, I'm sorry. Yes. Yes. Yeah, this reverses Yeah, thank you for pointing that out. So this this becomes sin theta and this becomes sin phi, right? So there'll be a minus sign also left out from here. So yes, we'll get a negative sign Okay, so if this is a negative sign, of course option number D is going to be correct so I think the first one to answer this was Kushal Kushal was the first one to answer this well done Kushal very good Guys, I would rate this question as a typical JEE main question I don't think so. There was anything very difficult about this. It was all about knowing the equation of the normal and Finding out the point of intersection and this using this condition that theta plus phi is going to be 90 degree. Okay So moving on to the third question over here So this is a question which is based on functions. So there's a function from R to R and it's a periodic function says that F of t plus x is equal to 1 plus 1 minus 3 f of x plus 3 f of x whole square minus f of x whole cube So this this gap over here is just you know Unnecessary, but there was a gap in the question also. So So this gap, please ignore this gap. Okay So now t is a positive fixed positive number then the period of f of x then we have to find the period of f of x in terms of t only So I personally feel that this question is all about playing with the functional equation So on a roadmap, basically, I feel you just have to figure it out. What is what is this n over here? So when does your function become itself for a change of x by nt? So whether that n is 1 whether that n is 2 whether that n is 3 or none of these No problem. This is not the first time Aryan says B. Okay, Aryan, I have noted down your response How about others? So, Atmeshi is also getting B Sanjana, Naman, Ramcharan, Kondinia Where is Amog? Amog, Aditya, please respond Shwetha also is getting B. Okay, fine. You guys, it's very obvious that so don't consider t to be the period. Okay, see t is just a you know Alphabet which is used over here. So don't get confused. He is not a period of any function So if you see this part if you see this part of your problem, it is clearly One minus f of x whole cube, right and whole raise to the power of one third So that's going to be just one plus one minus f of x. That's actually Two minus f of x is equal to f of x plus t So f of x plus t plus f of x is Equal to two so now what I'm going to do is I'm just going to play with my expression see I already see a x plus t over here So what I will do I will replace I'll replace x with x plus t So in this functional equation, I'm going to replace x with x plus t. So that's going to be x plus t plus t plus f of x plus t is equal to 2 That means f of x plus 2t Plus f of x plus t is equal to 2 which clearly means that the position of f of x and f of x plus 2t are same because This is same as this Two is same as this. So these two have to be the same Which clearly implies f of x is equal to f of x plus 2t Clearly implying that the function is periodic with a period of 2t which is option number b is correct option number b is correct Okay, so the first one to answer this was a rn well done rn good Okay, I don't think it was a difficult problem. Well, you got carried away with the previous one Moving on to the fourth problem. The problem says GIF and fractional part are given in this function. So this is a function which is defined as to a to the power Two times gif of x plus fractional part of x minus 1 by 2 times gif of x plus fractional part of x when x is not equal to 1 And f of 0 is given to you as ln of a Okay, so we have to comment which of the following option is correct Okay, so our poor X is a C for this All right, so everybody is like okay for Vaishnavi who's saying for a everybody is replying C C C C All right, so let's have a look at it So well first we'll focus on finding the left hand limit That is a limit of this function when extends to a zero minus So a to the power two times Gif of x Plus fractional part of x minus one. I think minus one is down over here. Okay by two times fractional part of X So now when you're slightly less than zero, we all know that the gif of x will be minus of one So that will become minus two and we know that when you're slightly less than zero this fellow will actually become This fellow will actually become one minus x Right So just like you know, you can take a very simple value. Let's say X is 0.01 with a negative sign We know that fractional part of X in that case would be 0.99 Okay So it's just going to be it's going to be one plus X It's just going to be one plus X Okay minus one Similarly in the denominator will have a two into minus one and fractional part will be Will be One plus X You can look from the graph as well. So in the interval minus it was actually going to behave like this So this is a line Okay, whose slope is one so it'll be line wine equal to one plus X Okay So simplifying this I get a to the power of X minus one minus one by X minus one and as X tends to zero this is just going to become as Extends to zero minus it is just going to become a inverse minus one by minus one. That's going to be That's going to be one minus one by So this is your left hand limit Let us similarly find out the right-hand limit So for right-hand limit, I will write it as limit extending to zero plus a to the power two gif plus fractional part minus one by two gif plus fractional part so zero plus again this would become zero and Gif of X will become zero and fractional part will remain X So it'll become two into zero plus X minus one by two into zero plus X Which is actually a to the power X minus one by X which is log of X to the base log of a to the base E So what is happening is right-hand limit is matching with the value of the function at a but it is it may or may Not match with the value of the function. So these two are not matching for the same value of a Okay, so one minus a is not matching with ln of a Okay for all values of a Okay, it just means the functions limit does not exist the function limit does not exist So continuity there is no point of continuity in this case We cannot say removal discontinuity because it's the case of jump discontinuity because a values for the same a value Left-hand right-hand limit are not equal. So I cannot say this. So only option C is possible So the first one to answer this was again Purvik very good Purvik So guys any question with respect to this, please let me know this is just a basic question on continuity and differentiability Removal discontinuity is those cases where the limit exists But does that the value of the function at that point does not match with the limit? So we consider these two cases as your removal discontinuity where there is a missing point over here or where there is a Isolated point. So this is a case of removal discontinuity because you can always plug this value Always plug this value with the limit of the function at that point In case of jump or essential discount discontinuities, which we also call infinite discontinuities We cannot remove those kind of discontinuity So when this is a jump like this, we cannot do anything about it Or when these things go to infinity at this point, we cannot say we cannot remove these kind of discontinuities Exactly Sondarya when left-hand limit and right-hand limit are equal, but it is not equal to f of a those are removable discontinuities Again easy problem to score. So we'll now talk about The next problem, which is the fifth problem. So the question says g of x is defined as 1 4th f of 2x square minus 1 and half of f 1 minus x square and F dash x is an increasing function f dash x is an increasing function Then g of x is increasing in which of the following intervals then g of x is increasing in which of the following intervals Okay, again, I got a response from poorwick, which is option B Rithvik also feels the same. Okay, Krushal, Rohan Anyone from NPS Kaurangala? Atmeshe says 5d. Okay, so he differs from others. Okay. All right. So shares is supporting his Hsr fellow part. So he's also saying D Chalo, so I've got most of the responses Good to begin with So in order to know whether in which interval g is increasing the first obvious step is after differentiated, correct? And to differentiate it, I'll have to use the chain rule over here So that's going to be 4x and half We'll have f dash of 1 minus x squared times minus of 2x So 1 4th 1 4th goes off 2 and 2 goes off So g dash x will be x times I get f dash 2x square minus 1 and From here, I'll get minus f dash 1 minus x squared Now I can say that my function g of x is increasing when both of these terms are either positive or When both of these terms are both negative then only my g of x would be considered to be greater than 0 So let's take these as two situations. So we'll take case one case one where x is positive and f dash 2x square minus 1 Minus f dash 1 minus x square is also positive Remember f of x f dash x is an increasing function. So this has to be more than this That means even 2x square minus 1 should be more than 1 minus x squared That means 3x square should be greater than 2 So x square minus 2 3rd will be greater than 0 and that can only happen That can only happen when x is either less than minus of root of 2 by 3 or x is greater than root of 2 by 3 Right, so x can be less than minus of root 2 by 3 Or x could be greater than a root of 2 by 3 But this is not possible because of this restriction because they both have to be true simultaneously So from situation number one I get x should belong to a root of 2 by 3 comma infinity root of 2 by 3 comma infinity, okay Let's talk about situation number two in a similar way. So either x is negative and f dash 2x square Minus this should be negative So I don't have to you know solve again I can directly comment that x should belong to minus infinity to minus of root 2 by 3 Okay Now I have to take the cases cases means union of these two I have to take the union of these two that clearly implies option number B is correct Again the first one to answer this was Purik Bingo good guys Buck up other guys. I need to get the response from you as well So moving on to the sixth question over here. So you have been given a piecewise defined function f of x They're asking you set of real values of B such that f of x has the smallest value at x equal to 1 Again, it's a question from application of derivatives All right, so I've started getting options Ariane says B Aatmei C Kushal B Purik also B So mostly Janta is saying B or C B or C methora confusion. Hey The guys, let's try to solve this again. It's a piecewise function You know, it's a cubic equation right and cubic equation with a negative Leading coefficient. So it's graph is going to fall down Okay, this graph is going to always fall down So Let's say this is zero. This is one. This is three. So it's graph is really falling down So at zero, it'll be having some value. So it will be constantly falling Then you can also come to know from the derivative of the function So derivative of this will be minus of 3x squared from 0 to 1 which is always negative in nature Okay, and the derivative of this is always positive. So this is all is an increasing function So this is also an increasing function So guys the scene is something like this The other function is increasing this is decreasing and this is increasing Okay, now obviously if you want your Function to have the smallest value at one that means You can have these cases. Okay, so you can have these cases. It cannot go beyond this Okay, because it should have a lowest value at one so 2x minus 3 can actually You know come at these positions. In fact, it's the other way around 2x minus 3 will always be at minus 1 So this will always be at minus 1. Let's say something like this. So this should not fall below this line Right that means the limit of the function as x tends to 1 minus okay This should be Greater than equal to minus 1 right Then only the function will be set to have its minimum value at 1 Okay, so at 1 it will have a minimum value when the left hand limit of this function is Either greater or equal to the value of the function at 1 which is actually minus of 1 isn't it? So this means this means the left hand limit of this function as x tends to 1 will be minus 1 plus b cube minus b square plus b minus 1 By b square plus 3b plus 2 this should be greater than equal to minus of 1 This should be greater than Minus of 1 so let's solve this inequality. By the way, this is minus 1 and minus 1 will go off from here and you will get something like This is factorizable and I can factorize this as b square plus 1 times B minus 1 and Denominator is also factorizable b plus 1 b plus 2 this should be greater than equal to 0 So I will use the wavy curve for this. This is always positive. This is always positive term So I just have to plot the others which is minus 2 minus 1 and 1 So it'll be positive negative positive negative. So when you want positive it has to be minus 2 to minus 1 Minus 2 minus 1 both will be exclusive Okay, union 1 to infinity 1 should be included because it can be 0 at 1 So guys clearly option number b is correct over here So the first one to answer this was Arian again Arian was the first one to answer this Okay, so guys, please be accurate. Don't be in a hurry things are easy This was an easy question. I had discussed many of such questions in the class as well So can we move on to the next one? Let's move on to the seventh question now So I think this question comes to you from integral calculus indefinite integrals Good to see many people are responding many people have responded for the previous questions So I would expect the same going forward as well Anybody any success? Okay, so at Mesh says option a okay So does Sanjana and Saimi here others please respond if you're done. Okay, so on there. You also says option a So almost everybody who has replied so far is going for option a others. What do you think Purvik Shweta? Okay, Shweta also says a just Arian has a different answer to give Yes, everybody is saying a okay guys. Let's let's look at this Now let's figure out first. What is f dash x so I'll just use this particular information I'll differentiate both the sides and For that I will use the product rule. I Will use the product rules. So it's going to be 6 f of x f dash x Okay So if I take f dash x common, I will get x minus 6 f of x is Equal to f of x in fact minus of f of x so f dash x is going to be f of x by 6 f of x minus x 6 f of x minus x Okay Now I can see this term Actually appearing over here Okay, and not only that I can also see that if I take a 2x common from these two terms I Could see 6 6 f of x minus x appearing So from the numerator what I'll do is I'll first take 2x common so that will become x minus 6 f of x plus f of x divided by 6 f of x minus x Times and we have x square minus f of x the whole square So let me take x minus 6 f of x common from both the numerator and denominator So when I take the common x minus 6 f of x from both numerator and denominator, I Would be left with this. Oh, I'm sorry. This is not the square of this term. This is the square only outside Yeah, I'm sorry. So I'll get something like this Okay Now this term is clearly your f dash x this term is clearly your f dash x so it becomes 2x plus f dash x By x square minus f of x the whole square now what I get Can see clearly is that the derivative of the denominator is present in your numerator So I just have to use a simple substitution. Let me call this as t So 2x minus f dash x DX is going to be your dt That's just going to be dt by t square. That's going to be negative of 1 by t Okay, and that's going to be minus 1 by x square minus f of x is an x minus So there's a minus sign on top. Okay. I'm sorry. Yeah, there's a minus sign over here Yeah, there's a minus sign over here. Okay, so yeah, I can take a minus out common first So I'll take this minus sign common out first So yes, it's going to become this Yeah, so it's this which is going to be your answer or you can just see there's this Yes, absolutely correct guys. So option number a becomes correct in this case. So option a is correct so the first one to answer this was Arthmesh Very good Arthmesh again not a tricky question not a difficult question. It just made you think in a different way Okay, so it was just a case of a simple substitution But they complicated it by introducing a lot of functions and a function equation like this Any questions? Can we move ahead? Please feel free to stop me if you have any questions so moving on to the Eighth question for the day, which is on definite integrals So we have a function over here x by 1 plus ln x ln x all the way till infinity x belonging to the interval 1 to infinity Then find the Find the Limit of sorry integral of f of x from 1 to 2e It's a very very simple question very very simple question Okay, so shares says Option a okay Anybody else apart from shares? Yes guys anybody else So we are only interested in evaluating this function from 1 to 2e Okay, so let's say We know that ln x becomes equal to 1 exactly at x equal to e So let's see what happens when I take x between 1 and e So between 1 and e I can say that ln of x In this interval ln of x would actually be a quantity which is between 0 and 1 okay And if you see this term, it's actually ln x multiplied to itself infinitely many number of times so what will happen to Such a scenario when ln x is multiplied to itself infinitely many number of times and your ln x happens to be between 0 and 1 It's obvious that this is going to become 0 right Which means your function is going to behave as Just x when you are between 1 and e Because this part is going to become 0 So can you take a hint now and and try to complete it from here on? Uh, sure. I mean you can take a minute. No worries. So all right. Now people have started uh answering this Okay So let's see whether it's a or no Now what happens when you go beyond e when you go beyond e what happens Anyways, I'm not bothered exactly at e because that's going to create a spike and that is going to be x by 2 spike So I'm not going to be worried about it So when it is greater than e We know that ln of x is going to be greater than 1 if ln sx is going to be greater than 1 this will become infinitely big correct And if this becomes infinitely big that means the entire function will collapse That means the entire function will collapse and become 0 correct Because ln of x will be greater than 1 when x is between e to 2 e So ln x to the power n will actually go to infinity if n is going to infinity Okay So your function is going to become just 0 Now I just have to use the Definition of this function. So from 1 to e you are just integrating x And from e to 2 e you're just integrating 0 So that's as good as saying x square by 2 from e to 1 that's going to be e square by 2 minus half That's e square minus 1 by 2 and clearly option number a is correct, right So well done shares shares bhaktram was the first one to answer this and that too correctly Well done So I think guys this is very simple You just had to redefine the function and once you redefine the function according to the intervals Split the limits of integration and then just carry out the process. So I I'm sorry So next question will take up on area under curves and after this question We can expect a small break of five to ten minutes. So this is your question number nine Hope the question is legible The area bounded by the curves given by So there are three curves over here. One is this one. In fact, it's a region The overlapping of these three region is what is Required by this question Again, please do not try to solve it without sketching it I want everybody to sketch these three regions and then realize which region they are looking for Nine a okay. So athmesh says option a But not solved rigorously Purvik also says a okay anybody else So guys so far. I've just got the response from two students Okay, so arian is another one to respond as a I need at least two more responses So these two are pretty simple, right? This means x should lie between minus half to half. So And y should also lie between minus half to half So minus half to half. Let me just you know, sketch this over here. Let's say this is minus half And let's say this is plus half So x is equal to half. This is x equal to minus half Okay, and in a similar way, this is let's say y is equal to half and y is equal to minus half y is equal to half y is equal to minus half Right yeah No, I think most of the people are you know Trapped in this graph because of too many mods involved guys. Just think without mod think as if you are in the first quadrant X and y both are positive. So get rid of mod first. You cannot think with mod in place Okay So think as if there is no mod think there's no inequality also, right make your life as simple as you can Think there is no inequality also. Let's say this is equal to one. Okay. Now. How would you plot? So it's pretty obvious It's pretty obvious that when you make y the subject of the formula it becomes 2 to the power minus x and minus half Right, so just plot this graph first without any any problem first try to plot this graph Okay, so I'm giving you 30 seconds to just plot this graph not 30 second 10 seconds to plot this graph So go step by step first plot First plot y is equal to 2 to the power of x then see what will happen if you put a minus x And then see what happens when you Put a minus half So y to the power minus x is just going to be a parabola like this. Sorry. It's going to be a Exponential curve like this cutting the y at zero. So it's going to fall down like this Okay, so it's going to be an exponential function falling like this Y is equal to 2 to the power minus of x right cutting at 0 comma 1 right And at half it will be at When x is half it will sorry when x is this is x equal to half so it becomes 2 to the power of minus half Okay Now when you subtract the half from it you just have to pull this half down So this part will actually end up coming over here Okay, so this will go off This will go off and this will come down like this Remember this value will be 2 to the power minus half minus half Now when you're modding when you're saying y is less than this Or when you're saying this is greater than equal to 1 Okay This is greater than Sorry, this is less than equal to 1. I'm sorry. This is less than equal to 1 So what will happen? This inequality will also become less than equal to 1 This will also become less than equal to so less than means it is has to lie Below this graph. It has to lie in this zone Let me change the pen It will lie in this zone now when you're modding Now you're modding x and y what will happen modding x just reflects this graph about the y axis So when you mod x what happens this goes this side as well Okay, and when you mod y it just gets reflected down as well So if you would realize from the graph it would be of this nature So all you need to do is find out this area Now can you try it out? Psi is also saying a So you just have to take one of the Quadrants and find the area in that quadrant and just do four times So can I say it's four times zero to half The area under two to the power of Minus x Minus half dx So that's going to be four times zero to one Let me just write down the result over here. So two to the power minus x By l n two with a minus sign and you'll have an x by two over here Okay, so zero to half So when you put a half it becomes four times Minus two to the power of minus half by l n two Minus one by four and minus when you put a zero you get minus one by l n two And this will be zero. So let me simplify this a little bit. So it becomes four times This becomes minus one by root two ln x Sorry ln two not ln x ln two Minus one fourth plus half ln two. So it's just going to be uh One by ln two you can take common. So one minus one by root two into four minus one Okay Which is going to be four minus two root two by ln two minus one. I think None of these options are correct in this case. So option number d is correct So guys, I'm sorry. None of you are correct Okay So option number d is correct in this case If you still feel that, you know, I've done some mistake anywhere or some correction needs to be done Please highlight it right now. But according to me the answer is none of these Is that fine? Did you do you realize your mistakes? So guys, let's take a break now and uh Let's take a break and we'll resume at 11 Let's say 11 18 am All right guys, so welcome back Welcome back after the break So those who are already back can start thinking on this problem, which has been taken from our differential equations Okay Again a very very important concept for your school exams as well So please start working on this and Do let me know your response in the chat box Yes guys any response? Okay, so Purvik says option a others Atmesh, Kushal Aryan Ritvik, Ramcharan, Kondiniya Sondarya, Sanjana Gaurav, Shweta Rohan Apurva K. Sai wants to go with option c Okay, Apurva has other plans. He is saying option b Atmesh says a How about others Kushal also says a All right guys, let's try to see how we can approach this problem Again, not a variable separable neither it's homogeneous. So let's see if things work around linear differential equation form So first of all, I'll reciprocate this entire term sign of y square plus one So dx by dy is equal to a x cube y sign of y square plus xy Okay Let me bring this so y xy on the other side and we have x cube y sign of y square Now because of this problem because of this problem Because of this x x cube term over here things are not in the exact linear differential equation form So what I'll do is First I will divide throughout with x cube So let me divide throughout with x cube When I do that I get something like this Right now guys, do you identify which is this form? If I take one by x square If I take my in fact, you can take minus one by x square as let's say you Do identify that this is actually going to give you something like this correct So this term here will be half of du by dy and this term over here will be Uy is equal to y sign of y square Okay, multiply throughout with two. So when you multiply throughout with two It becomes like this. So basically if you would recall, this is actually called the Bernoulli's form Okay, this is actually the Bernoulli's form It is one of the reducible to linear differential equation form. Okay Now it becomes easy to solve this. I can always take my 2 y to be my p. So integrating factor will be e to the power integral of 2 y dy That's going to be e to the power y square So that's going to be u times e to the power integral of 2 y dy which is y square into 2 y sign y square e to the power y square dy So in this you can always put your y square as t So this will behave as your dt term. So it'll be u e to the power y square That's going to be e to the power t sign t dt right So let's follow simplify So if I'm not mistaken, it's just going to become u into e to the power y square. This is already a ready made formula So it'll become 1 by a square plus b square e to the power y square sign of t minus cos of t plus c right So let us bring this to on the other side to u and divide by e to the power y square So it's going to be sine of y square minus cos of y square Plus c times e to the power minus y square and we already know y is 1 by x square So 2 is equal to x square times x square times the entire expression which is going to be So this will this is minus 1 by x square, right? So we'll have a negative sign coming over here and this will become sine y square. This will become Minus c e to the power minus y square Yeah So I can write it this and down brackets So this will become your answer. So which option is matching with this? So 2 is equal to x square cos of y square minus sin of y square minus 2 c to the power minus y square. So it's fine even if they write c as 2c doesn't make much of a difference Okay So I think option number a is the closest Option number a is the closest to my answer So your option a will be correct in this case Option a will be correct in this case All right. So the first one to answer this was Purik again The purik was the first one to answer this Well done purik. So moving on to question number 11 for the day So this is a question which is based on vectors Let pqr be three mutually perpendicular vectors with the same magnitude and x be satisfying this equation Then x is which of the following 11 b That's what sai says Kondanya also backs him up So does koshal I need two more responses Ramcharan Sure. Sure. Take your time. Take your time You guys whenever there is a case of a vector equation I always told you try to express the unknown vectors in terms of non-coplanar vectors Right and three mutually perpendicular vectors will always be non-coplanar They will always be non-coplanar. So non-coplanar vectors are actually linearly independent of each other So whenever you get a unknown vector try to express it in terms of The non-coplanar vectors in this case. They are these mutually perpendicular vectors pqnr So I would always start my problem by assuming that Let my Unknown vector be this You can actually take it as ijk in your mind because ijk also forms a system of Non-coplanar vectors Okay, and hence they're linearly independent of each other So you can always start your assumption that your unknown vector is this And not only that we have been given that mod of p mod of q and mod of r are all the same So I may call this as you know k for the timing Now let us see what this equation has to say the equation says Now these are all symmetrical terms right so p P and q are involved over here q and r are involved over here r and p is involved over here So what I'll do is I'll just take one of the terms Let me take the very first term and try to simplify it and I can replicate the results for the others So let's say p cross x minus q Cross p So I'm applying the vector triple product over here. We all know vector triple product a cross b cross c That's going to be a dot c times b minus a dot b times c correct So I'll be applying the same formula over here as well So it'll be p dot p times that's mod p square mod p square. I can take it as k square so k square x minus q minus minus p dot x minus q times times p Okay, okay, so let me expand this up So it becomes so if I'm not wrong it just becomes p dot x p dot q will anyways be zero Okay, so this times p it will become is that fine which is actually k square x vector minus k square q vector and p dot x p dot x from here I can say will be clearly x1 into k square okay, so it'll be p into a p dot x1 p which is going to be p mod p square x1 rest all will give you zero because they're perpendicular So this will give you x1 p k square Right. This is your first term So this is your first term So similarly your second term. I can also write it as k square x minus k square r minus And I will get uh Since there's a p there'll be x2 q k square And the third term will be similarly k square x minus k square p minus x3 r k square, okay If you add the three If you add the three terms that is your first second and third term in the given expression you'll get three k square x So that's the sum of all these three terms Minus k square You'll get p plus q plus r and you get minus k square x1 p x2 q x3 r which is actually again your vector x itself correct Right, and this is given to you as a null vector So you get two k square x because three k square x is over here and minus k square x is over here So you get two k square x vector minus k square is e Is equal to zero Which clearly implies x is going to be p plus q plus r by two p plus q plus r by two, okay So I think option number b, yes option number b is correct in this case The first one to answer this was Sai Meer Sai Meer was the first one to answer this very good so guys the Critical point over here was this assumption. This assumption is something which was important That means any vector in space could be expressed as a linear combination of three non-coplanar vectors And after that it was just a mere application of the vector triple product and you know Following the symmetry of the figure and trying to see what which terms You know combine and which terms cancel out So guys any question with respect to this, please let me know right now All right, so moving on to the 12th question for the day The question says decimal parts of the logarithms of two numbers taken at random are found to six places Okay, so I think it's just the decimal places which are six in numbers The probability that the second can be subtracted from the first without borrowing All right, so Rohan says c is correct. Okay I have registered your response Rohan Let's wait for the others again probability and permutation combination very heavily linked chapters But if at all you have not You're not very confident about the use of permutation combination and probability Uh, don't try to slog at the 11th are okay It's fine. It's not necessary that these topics would be strong for everybody Because it's more of an art. I believe So don't waste too much time doing it Yes, of course some basic things like base theorem law of total probability probability distribution conditional probability You can focus on all these concepts Atmesh backs it up with c Shweta says a okay Sai also says c So does a purva. Okay, fine. So I have got enough response now for me to start solving this So guys again focus on one number at a time. So let's say our uh Decimal places the number after the decimal places that is the six places are x 1 x 2 x 3 x 4 x 5 x 6 For the first number and y 1 y 2 y 3 y 4 y 5 y 6 for the second number, okay So what i'm trying to say is that if you're trying to subtract the second from the first Without borrowing that means each xi should be greater than yi Okay, in fact greater than equal to I can say Okay, so let's focus on any one of them. Let's say I focus on this one x 1 and y 1 okay Now it's clear that x 1 and y 1 both can take values You know from 0 to 9 correct 0 to 9 that means they both have 10 10 options each So the sample space for choosing x 1 y 1 would be Uh, nothing but 10 into 10 Nothing but 10 into 10 that is 10 options for x 1 And 10 options for y 1 no doubt about that. So that's 100 correct Now if you want x 1 to be greater than y 1 Okay, then let's say if you choose x 1 as some a Okay, let's say some a which where a belongs to Uh 0 to 9 Okay, a belongs to 0 to 9 Okay Then y 1 can only take values from a plus 1 onwards till 10 Right not 10 9 I'm sorry. Yeah 9 Okay, that means for every option a y has how many number of ways It's a number of ways in which you can choose y 1 would be nothing but 9 minus a plus 1 plus 1 No, it can be a also it can be a also right. So I'm sorry. So it can be a also So it will be a over here Because boring you can put pickup a also Yeah correct Right, so it will be 10 minus a Okay, in this case, you will always realize Oh, I'm sorry. You have Let's do the other way around. Let's say this is your uh This is your let's say Y i then this is your x a yeah, no, it's fine. Is that okay? So very for every y 1 y i as a x i can become a a plus 1 all the way till 9 Okay, so the number of ways in which you can choose your x i is going to be 10 minus is going to be 10 minus a okay So can I say the total number of sample space in the favorable event would be nothing but Summation of 10 minus a all the way from 0 till 9 Okay, so that's going to be 10 plus 9 plus 8 all the way till you reach 1 That's going to be some of this which is going to be 10 into 11 by 2. That's going to be 55 Okay So your probability that your x 1 would be greater than y 1 would be 55 by 100 That's going to be 11 by 20 And there's six such places the six such places. So your answer is just going to become 11 by 20 11 by 20 11 by 20 Okay, hold to the power of 6 which is clearly option number C is correct So I think Rohan was the first one to answer this well done one So with this we move on to the next question over here Question number 13 Yes guys anyone any success? all right It cannot be too for sure It'll either be unique or no solutions or infinitely many solutions Okay, so mostly people are going for one 13a Rohan says 13c a Purva says a Sai also says a okay All right. So can we start the discussion for this? So first of all if t is real and we had we have a equation over here So let me do one thing. Let me generate a quadratic in terms of T So lambda t square plus 3t plus 4 will be a t square minus 3t plus 4 That means a t square lambda minus 1 plus 3 times lambda plus 1 times t Plus 4 times lambda minus 1 equal to 0 Now since t belongs to real numbers, I can say the discriminant would be greater than equal to 0 So the discriminant Will be this minus 4ac will be 16 lambda minus 1 the whole square would be greater than equal to 0 So, I'm sorry, why did I write 14 it should be 16? Yeah So, let me write this as a square minus b square formula. So it'll become 3 lambda plus 1 Plus 4 lambda minus 1 And 3 lambda plus 1 minus 4 lambda minus 1. This should be greater than equal to 0. That means 7 lambda minus 1 and I have 7 minus lambda greater than equal to 0 Okay Which clearly makes me feel that I can write this as 7 lambda minus 1 times lambda minus 7 is less than equal to 0 So lambda has to be a value between 1 by 7 to 7 So far so good Now let me Write down the determinant for this system of linear equations So let me write down the determinant for Or the determinant of the coefficient of the variables in this equation Okay, let's see. What does the value of this come out to be? Let me expand with respect to the first row itself So 2 lambda plus 15 plus 1 lambda plus 18 minus 4 Uh 5 minus 12 5 minus 12 is minus 7 So that would be 7 lambda 45 plus 28 Plus 18. So that's going to be 45 plus 46. That's going to be 91 Why did I write minus over here minus 28? Yeah, so that's going to be 35. That's going to be 35 Okay So now as your lambda lies in this interval We can see that this expression This expression Cannot be 0 Because for it to be 0 lambda has to be minus of 5 Which is which does not belong to this interval which does not belong to this interval Okay, which means the determinant of this cannot be 0 If determinant is not 0, there is no chance of getting no solution and infinitely many solution That means it can only have a unique solution. So option number a is correct So option number a is going to be correct in this case Is that fine guys? All right. So moving on to the 14th question now So this is a question which is Picked up from 3d geometry if p be any point on the plane lx plus ny plus nz equal to small p q be a point on the line op Such that op is equal to op into oq is p square Then the locus of the point q is Then the locus of the point q is Yeah, so these are planes These are planes If they intersect at more than one point that means they would intersect on a line of intersection So they will have infinitely many solution or just one solution just like it happens in case of x y y z and zx planes Or it can have no solution Okay, so rohan has already come up with the answer rohan says It's c Prati kondinia also says c Okay kushal also goes for c I need two more response So i'm very also goes for c kushal also. Okay So most of you are giving the c as the answer. So let's see whether c is the right answer or not Okay, so uh Let the point p Be x1 y1 z1. Okay So since it lies on the plane, I can say the equation of this plane would be So this point will be satisfying this plane. So l x1 My one plus n z1 is equal to p. Okay Now let us write down the equation of The line op so Let's say this is your plane. This is your point p. This is your origin And there is a point q in this line. There's a point q on this line. Okay, so first of all I can always write down Let's say this is alpha beta gamma And this is x1 y1 z1. In fact, I don't have to write the equation of op I can say their direction ratios will be proportional correct So direction ratios of Direction ratios of op Would be proportional to the direction ratios of oq That means x by alpha Will be y by beta is equal to z by gamma Okay, let me call this as k Now the third condition says op into oq is p square op will be x1 square Okay y1 square z1 square this will be op Oq will be under root alpha square beta square gamma square Okay Now since I have to write The locus of alpha beta gamma I can do one thing I can substitute x1 as alpha k y1 as beta k And z1 as gamma k square Under root So this will become k alpha square plus beta square plus gamma square is equal to p square Okay Now k also should not be a part of my answer K also should not be a part of my answer So I can do one more thing I can substitute x as alpha k y1 as beta k And z1 as gamma k In this equation Okay Which clearly means k can also be replaced with p by l alpha m beta n gamma. So let us put it back over here Let's put it back over here. So when we do that we get p by l alpha m beta n gamma x square plus sorry alpha square plus beta square plus gamma square is equal to p square So one of the p's will get cancelled which clearly means Alpha square plus beta square plus gamma square is p times l alpha m beta plus n gamma And we can all always generalize this We can always generalize this by replacing our alpha with x beta with y and gamma with z So it gives you the equation of the desired locus as x square plus y square plus z square is p times lx plus my plus n z So according to me, this should be your answer. Let's see which of the option matches I think option number b matches No, no, sorry option number c matches option number c matches So this will be your desired answer So the first one to answer this was rohan So rohan was the first one to answer this well done rohan. So some of you are constantly replying and you're replying correctly, especially I've seen some of yours Strike rate is very very high So guys, uh, moving on To the last question of the day So we have a question over here picked up from properties of triangles It's a column match question. I'm sorry. This is happens to be your 15th question So match the column one with column two should actually draw it in a very So let's say this point is e Yeah, so this is 90 degree, right? This is 90 Okay, now they're asking you the minimum value of cot b and cot c Okay, how do I find that out? So, uh, I can always find this angle. Let's say this is theta and let's say this is phi So tan of b is tan of theta plus Phi which is actually tan theta plus tan phi By 1 minus tan theta tan phi Okay Now from the figure itself, I can clearly say tan of theta will be x by 2 y And tan of phi will be tan of phi will be x by y and here I will have 1 minus x square by 2 y square Okay So it implies It implies that tan of beta will be nothing but Uh x y x y 3 x y and in the denominator, I'll have 2 y square minus x square So cot of b will be 2 y square minus x square by 3 x y, okay I can similarly say Cot of c would be 2 x square minus y square by 3 x y, okay So let's add these two. Let's add these two. So cot b plus cot c Will be nothing but x square plus y square by 3 x y Now here, I will use am greater than gm since both are positive quantity. I can say this will always be greater than root x y Correct. So x square plus y square will always be greater than equal to 2 x y So this quantity will always be greater than 2 x y by 3 x y That means it will always be greater than 2 by 3 So minimum value will be 2 by 3 minimum value 2 by 3 is option number q So a will be mapped to q. So a answer will be q Let's see anybody has been able to give a s q No, Rohan says a is r sai also says a is r Okay, so a is q my dear students Right, let's talk about the b part if capital r is 13 by 2 small r is 2 And small r1 is 3 find the area of the triangle in square units area of the triangle in square units Okay, so how do I do this problem? So the circum circle radius is given to you The in circle radius is given to you and one of the x circles Radius is given to you. So whoever are saying b is equal to s. Did you actually solve it or Sure, there's a mixed bag. I keep on picking up from different different books Okay, I would like to know whether how many of you are aware of these basic formulas which connects R small r r1 r2 r3. Okay. I wanted to know properties of r1 r2 Are you all aware that the sum of all x centers radii is actually r plus 4 r. Are you aware of this? How many of you are not Then let me tell you You need to keep these formulas in your mind Another important property was one by r is one by r1 plus one by r2 plus one by r3, right Now why I'm using this is because I've been given such information. I've been given I've been given. What is the value of capital r? I've been given. What is the value of small r and small r1? Isn't it? So r1 is given to us So what is the value of r1 given to us? Let's check r1 is 3 So let's use this so r1 is 3 r2 and r3 is not known And r is given to us as 2 And capital r is 13 by 2 So 2 and 4 into 13 by 2. So r2 plus r3 is going to be 26 plus 220 a 28 minus 3 which is 25. So that's your first equation Now using the second one. This is by using the first one using the second one. I can get 1 by r is 2 r is 2 Is equal to 1 by r1 which is 1 by 3 And 1 by r2 and 1 by r3. I don't know right So I can write this as 1 6th as r r2 plus r3 by r2 r3 That means r2 r3 is going to be 6 times r2 plus r3 which is 6 times 25 which is going to 150 Which is going to be 150 which clearly gives me These two equations Which everybody can guess that r2 will be r2 or r3. One of them would be 10 and another one of them would be 15 correct Now still how does it help me to get the area of the triangle? Okay Now if I know r2 r3 I can use yet another property which is which says that S square that is semi perimeter square is r1 r2 r3 by r. How many of you are aware of this formula? Semi perimeter square is r1 r2 r3 by r So using this formula I can say s square is r1 r2 r3 which is going to be 3 into 10 into 15 By r r which is 2 that's going to be around 75 to 25 so s is going to be 15 Now area of the triangle is nothing but r s That's going to be 2 into 15. That's going to be 30 square units 30 square units So yes those over those whoever said that your b will be s you are correct b will be s you are correct Okay c what do you think will be the answer for c anybody who has given the answer for c c is q C is not q So those those ever who are saying c is q that's wrong I think working on d is easier. So let's work on d first So let me let me work on d first okay So I have been given that uh sine 4a plus sine 4b plus sine 4c plus 8 cos a Is equal to zero so first of all I can Club up these terms So let me take sine 4a plus sine 4b together And then sine 4c. I will see what to do with that Okay, so I can write this as 2 sine 2a plus 2b into cos 2a minus 2b now since we know a plus b plus c is Pi minus c 2a plus 2b will be uh 2 pi minus 2c So this is going to be minus sine 2c Cos 2a minus 2b and this I can write it as 2 sine 2c cos 2c So take uh minus 2 sine 2c common it will become cos 2a minus 2b minus cos of 2c and again 2 pi minus 2a plus 2b will be 2c So cos of 2c Will be least affected that will just become cos of 2a plus 2b All right, so when you expand it you get minus 4 sine 2c sine 2a sine 2b Okay, now don't forget 8 cos a on at the at the end So that's going to become I can again split this up as minus 8 sine a cos a sine 2b sine 2c And I have a 8 cos a as well So I can take 8 cos a common and I'll get 1 minus Sine a sine 2b sine 2c Okay This is equal to zero given to me. This is equal to zero given to me. Okay, so there is a possibility that your a could be 90 degree Your a could be 90 degree And sine a sine 2b sine 2c is equal to 1 Which clearly implies if a is 90 degree Then Sine 2b into sine 2c will be 1 which clearly implies each one of them has to be 1 Because they cannot exceed anyone cannot exceed 1 right Which means b is equal to 45 degrees c is also equal to 45 degree which clearly satisfies this case And hence we can say that Hence we can say that cos of a is going to be zero correct Cos of a is equal to zero. So this is going to map to p So I think This is the only one which is left to map to r. So I think this is map to r But I would leave this as a homework for you to prove that cos a is equal to half in this case Okay, so please do get back on this Okay, so guys we have already exceeded The time limit that we had set for today. So anyways, it was a great having the session with you all So hope you learned many things new especially in the last problem Make sure you know your weaknesses and keep revising it through these problems Thank you everybody for coming online and all the best for your exams which are which you have when your school reopens And of course a very very happy new year to you. So may the new year bring a lot of good marks good ranks and happy faces Okay So enjoy bye bye Take care over and out from syndrome academy