 Hello all welcome to the problem-solving session on complex numbers and other important topics today So those who are attending the session I would request you to type in your names in the chat box So that I know who all are attending this session But today we'll be covering up topics from complex numbers quadratic equations miscellaneous equations permutation combination binomial theorem sequence series straight lines pair of straight lines circles indefinite integrals area under curves differential equations and probability So let us get started with the first question, which is from the complex numbers Hello, all good morning. So those of you have joined in the sessions Please start working on the first problem And if you're done, please feel free to type in your response in the chat box So the person giving the first answer And the correct answer will get a shout out for me Alright, so Akshath has given the first response as B So I'll wait for at least five responses to come before I start solving it So here is a very beautiful application based question on complex numbers There's a curve defined for you over here and There are two spiders One male and another female. We're moving together along the curve The female spider suddenly realizes that the male spider is a rogue spider and Immediately tries to get away as far as possible like And hence it moved to another point on the curve What is the maximum distance that can be possible between the two? Given that they are on the same curve Guys, I need responses. I have just got one response and that too from Akshath Please respond before I start solving it I would require at least five people to respond on this Because it's a problem solving session where you have to solve the problems Okay, so Sundar also says B All right, so let's look into this question because It's already around seven to eight minutes and I've just got two responses So what we'll do here is again, we'll try to take that old approach old-school approach that we put Z as X plus I Y Okay, so if Z is X plus I Y, we all know that Z conjugate will be X minus I Y and Z plus Z conjugate will now become 2X What will happen to Z minus Z conjugate? Z minus Z conjugate will be 2 I Y So replacing it back in this into this equation that we have been provided 2 Z plus Z conjugate will be 2 mod 2 X and this will be 32 plus our 2 I Y the whole square So that's going to give me 4 mod X is equal to 32 plus 4 I square Y square, which is nothing but 4 mod X equal to 32 minus 4 Y square right Dropping the factor of dropping the factor of 4 throughout it will become Mod X is Equal to 8 minus Y square that is Y square is equal to 8 minus mod X Okay, now if you try to plot this function Y square You know that it's going to behave as 8 minus X when X is greater than equal to 0 and It'll behave as 8 plus X when X is less than 0, right? Now I can clearly see that these two curves are actually These two curves are actually the graphs of parabola Right, so let me just quickly sketch them So if you sketch this graph a Y square is equal to 8 minus X when X is greater than equal to 0 You would realize that it's going to be a parabola opening leftwards and Having its vertex at 8 comma 0 so it would be a graph like this It's going to be a graph like this correct So this point will be 8 comma 0 Right in a similar way if I plot Y square is equal to X plus 8 It's going to be a parabola opening rightwards and its vertex at minus 8 comma 0 so I'm drawing it in Orange Hope you can all see this Right, so this point will be minus 8 comma 0 Now from the graph itself, what do you realize? What could be the maximum distance possible between these two spiders? Let's say female is over here and male is over here Do you think that this distance is going to be the maximum distance? correct and hence the maximum distance is going to be the distance between 8 comma 0 and minus 8 comma 0 which is going to be 16 units and Akshath was the first one to answer it well than Akshath great I don't think so it was a difficult question guys I think you got scared just by the fact that complex numbers were involved Therefore a quick comment on this type of question is always resort to coordinate geometry method Complex number is such a chapter which he which uses vectors which uses coordinate geometry which uses of course your trigonometry And all those stuff is that fine. Can I move on any question with respect to this so option? Option B is correct. Let's begin with the second question now So, I hope the question is legible to all of you the question says Find the number of ordered pairs x comma y satisfying the equation 6 to the power x 2 by 3 to the power y minus 3 into 2 to the power x plus y Minus 8 into 3 to the power x minus y plus 24 equal to 0 And it's also given that x y is equal to 2 find the number of ordered pairs satisfying this condition Again the person answering this question correctly The first person will get a shout out from me All right, so Akshath again the first one to give a response He is not sure, but he thinks it's B Okay Koshal also backs up with B. Okay, Sanjana also B All right, so five people have already responded and all of them are saying B Okay, so let's let's look into this problem now Uh, so first of all what I'll do is I've been given two equations one and two okay, and I would try to first simplify this as 2 to the power x 8 to 3 to the power x and this is 2 to the power y by 3 to the power y and the rest of the terms I'm not going to disturb much Because I have a plan to factorize it. So let's see whether we can So this will become 2 to the power x plus y into 3 to the power x minus y minus 3 into 2 to the power x plus y Minus 8 into 3 to the power x minus y plus 24 equal to 0 right, so what I'm going to do is I'm going to Club these two terms together and I'm going to club these two terms together Okay So that's going to become 2 to the power x plus y times 3 to the power x minus y minus 8 times 3 to the power x minus y Minus 3 times 2 to the power x plus y plus 24 equal to 0. I Can always take a factor of 3 to the power x minus y out. So that's going to give me this Okay, loythia is saying C Psi is saying a okay. Let's check guys. Let's check. We are now very close to solving it So of course we can factorize this like this Right, so this was what I intended. I intended to factorize it So again, I moved on with a roadmap in my mind and I achieved what the Road map was intending to do so this can be giving you two possibilities whether this is equal to 8 That means x minus y is 1 or x plus y is equal to 8 Sorry, I'm sorry x plus y is equal to 3 okay Now I have to use the other equation as well. That is my x y is equal to 2 right? So I have another possibility x y is equal to 2 So let's check if x minus y is 1 and x y is equal to 2. What are the possibilities that we get from here? So solving them simultaneously here. So Replacing y with 2 by x so y will be replaced with 2 by x equal to 1 so x square minus x minus 2 is equal to 0 and I can factorize this as x minus 2 x plus 1 equal to 0 so x could be 2 if x is 2 y is going to be 1 and If x is minus 1 y is going to be minus 2, right? So these are the two solutions already we got Let's take the other situation when x plus y is equal to 3 and When x plus y is equal to 3 y I can again replace with 2 by x that's going to be 3 So x square minus 3x plus 2 equal to 0 Again, this can be factorized as x minus 1 x minus 2 and This time it gives you x is 1 and x is 2 x equal to 2 I already took care of so this was already taken care of over here So I don't need to bother again. Let's see when x is 1. What is why why is also going to be 2? So this is the third possibility So the following solutions are possible x comma y could be 2 comma 1 1 comma 2 and It could also be Minus 1 comma minus 2 so guys unfortunately none of you were correct Because the option here is option number D, which is right All right, so don't be in a hurry to solve it the first it's important To just get it correct. That is more important any questions with respect to this guys All right, so moving on to the third question now Just let me name these questions. This was question number two for you and now let's move on to the third question Guys little more than two weeks remain For your a JMA exam, so you have to be very accurate accuracy should be a part and parcel of your problem solving Do not just blindly give any answers that you feel like you will be penalized for it I think I missed a question before this. No, that's fine. That's fine So this question is from binomial theorem. It says the coefficient of x to the power n in this expansion is non-zero Then n cannot be read the question carefully n cannot be of the form All right, so Siddhant has given the answer as a Rowan says C Akshath also says C. Okay. This is quite an easy question. I think many of you should give the answer for this Guys, I need two more response to start solving the problems again But don't hurry up. Take your time Okay, so Siddhan changes his answer to C Yes, guys, two more response two more response and I will be Starting to solve this problem for you Loitya, Brithvik, Saimeer, Kushal Vaishnavi, Shweta, Ramcharan, Kondiniya Sundar, Nidesh Okay, Shruti says option C. All right guys, so let's start solving this If you look at this term, if you look at this term carefully this term appears as you know One of the factors when you are using this formula, right? I'm just I'm just working on the roadmap that you could actually use correct So what I'll do since this is raised to a power of 100 I will break this 1 plus x to the power 101 as this Okay If I combine these two, I will get 1 plus x and I will get around 1 plus x 1 minus x plus x square to the power of 100 Which is actually 1 plus x cube to the power of 100 right Now when you expand this, you will always get terms which are of this nature 100 cr x to the power 3r Correct, and when you multiply with 1 plus x, you are going to get two types of summation that is 100 cr x to the power 3r and 100 cr x to the power 3r plus 1 because when this x multiplies with this term, it will generate 3r plus 1 Correct That means when you expand this, you will either get terms which are multiples of 3 Or they leave a remainder of 1 when you divide the The power with 3 right so you will get either terms like x cube x to the power 4 x 6 x to the power 7 like that right some terms like this will be there That means you will never ever get a term Which will have a power of 3r plus 2 on it correct That means Option number c is definitely correct. So there will be no terms with x raised to the power 3s plus 2 form I cannot say this because you can always have a term like x to the power 9 Which is actually 4 into 2 plus 1 so d is ruled out so A and b are definitely present. This is your b. This is your a so c cannot be there So rohan was the first person to answer this well done rohan good Rest others. I don't think so. It was a difficult problem and I think you just got scared Next question is coming to you from the sequence series chapters some of infinite terms of this series Again, let's see who is the first one to give the correct response Okay, kushal Aksha also says b same as what kushal says All right, so almost everybody says b option So guys just saw this first of all we'll write down the expression for the arith term for this Which if I'm not wrong is actually r by 1 into 3 into 5 up till 2r plus 1 right And now I will apply the vn method of summation on this By splitting tr as difference of two functions which are consecutive in nature So I can always break tr as The numerator r could be broken as 2r plus 1 minus 1 into half And rest of the terms you can copy as such Okay, which is clearly Which is clearly half And if you take 2r plus 1 together and once to one separately It will become 1 by 1 into 3 into 5 all the way till 2r minus 1 Minus 1 by 1 into 3 into 5 all the way till 2r plus 1 Okay This is called the vn method where you try to break the arith term as difference of two such functions Which are consecutive in terms of r. So if you're calling this function as vr function You will be calling this function as vr minus 1 function Now the benefit of breaking it like this is when you start putting your values of r as 1 2 and so on When you put 1 you get 1 minus 1 by 1 into 3 Correct second term would be half 1 into 3 minus 1 into 3 into 5 And so on till your nth term Which is going to be half 1 into 3 into 5 all the way till 2n minus 1 Minus 1 into 3 into 5 till 2n plus 1 Now when you add them all You get sn which is only going to be The first term Of the very first term and the second term of the last term. So this is what you are going to get And it's very obvious that if you're summing up till infinite terms If you're summing this till infinite terms This will tend to zero this expression will tend to zero So it'll become 1 into half into 1 minus zero. That's going to be half as the answer Exactly exactly. So basically Your option number b is going to be correct And the first one to answer this was kushal Well done kushal great. All right. So moving on to the fifth question This question comes to you from straight lines chapter again a class 11th topic of coordinate geometry And this question read like this a rare flight incident at the point minus 2 comma minus 1 gets reflected From the tangent at zero comma minus 1 to the circle The reflected rate touches the circle Find the equation of a line along which the incident ray moved Again, no need to hurry. Just take your time. Accuracy is always important Speed comes with time. Speed cannot be developed overnight But accuracy is something which you can always take care of All right. So Rohan is saying b siddhan is also backing up with c. Sorry Rohan is saying b siddhan is also saying b Anybody else? Okay, so laudhitya also says b akshet, saimeer, pratik shutti Vaishnavi, shweta So guys, let's draw the diagram for this from the diagram. Many things will be very very clear So we have a situation like this where You're drawing a tangent at a point 0 comma minus 1 Okay, so let's say So this is 0 comma minus 1 Okay, and a rare flight is incident at minus 2 comma minus 1 on this tangent So let's say a rare flight comes it gets incident on this point and the reflected ray touches the circle So let's say this is the reflected ray The green one And it touches the circle. So the rare flight comes like this Hits the tangent at minus 2 minus 1 And goes grazing the circle Okay And the circle is given to us as x square plus y square is equal to 1 now. Let us first do one thing Let us find the equation of the reflected ray in terms of a given slope. Let's say Let the slope of Let the slope of Reflected ray Reflected ray b m Okay Now if this ray has a slope of m and it is passing through minus 2 minus 1 I can always write down the equation as y plus 1 is equal to mx plus 2 Right, which is nothing but y is equal to mx plus 2m minus 1 And we know that the distance of this reflected ray from the origin will be equal to the radius of the circle correct So I can say mod of 2n minus 1 by under root of 1 plus m square is going to be the radius of the circle Okay, so let's solve for this Let's take it to the other side Let's square both the sides So I get 4m square minus 4m plus 1 is m square plus 1 that is 3m square Minus 4m is equal to 0 that clearly gives you two values of m that is 0 and 4 by 3 Now remember 0 will be the one which is corresponding to this line because this is parallel to the x axis this line is parallel to the x axis So I need the other one. So this has to be a line with a slope of 4 by 3 So the other one has to be a line with a slope of 4 by 3 right So I can write down the equation of the reflected ray as y plus 1 is equal to 4 third x plus 2 That's going to be 3 y plus 3 is equal to 4x plus 8 That means 3 y 4x minus 3 y Uh plus 5 is equal to 0 So this is the equation of the reflected ray. This is the equation of the reflected ray Okay Now guys another important thing that I would like to highlight over here If you draw a normal at this point If you draw a normal at this point You would realize that the normal is basically a line y is equal to minus 1 I'm sorry x is equal to minus 1 x equal to minus 2. I'm sorry. Yeah, because it's going to be Perpendicular to this line right and this is parallel to the x axis. So this has to be parallel to the y axis parallel to the y axis means No, shweta. No, no, no Tangent from an external point x1 y1 is never that you are actually confusing with the chord of contact equation If x1 y1 were to lie on the circle. Yes, what you stated would be the equation of the tangent Okay, so a clever way to solve this problem is to guess any point on this line Guess any point on this reflected line and take the mirror image about this line For example, I can always guess a point on the reflected ray. Let's do one thing. Let's put x as minus 5 Let's put x as minus 5. I'm guessing a point on the reflected ray So minus 20 minus 3 way plus 5 is equal to 0 So 3 y is equal to minus 15. So y is also equal to minus 5 So I can always guess a point over here Uh minus 5 minus 5 comma minus 5 So what I did is I guessed a point on the reflected ray. Okay It's actually not here. It's it would be actually somewhere over here This will be minus 5 comma minus 5 What I want to do is I want to guess a mirror image of that on the incident ray So can I say mirror image of that would be this point would be Let's not draw it over here. Would be confusing Yeah This mirror image would be can can somebody tell me what would be this mirror image if this is point minus 5 comma minus 5 See the distance from this the distance from this is going to be minus 3 So if you increase a minus 3 on plus 3 on this side, what will happen? So this is this distance is plus 3 Lohetia are you sure? right It's going to be 1 comma minus 5 Okay Now all I need to do is write the equation of a line connecting Write a equation of a line joining. So the incident ray would be a line joining minus 2 comma minus 1 and 1 comma minus 5 Right. So it's very simple. You can use the two point formula So y minus y1 Is equal to slope x minus x1 So that's always going to be 3 y plus 1 minus 4 x plus 2 So that's going to be 3 y plus 3 minus 4 x minus 8 That's going to be 3 y plus 4 x plus 11 equal to 0. So which of the options is going to be correct 4 x plus 3 y equal to 11 is option number b right And I think the first one to answer this was Again Rohan was the first one to answer this Well done Rohan Any questions with respect to this guys? Now just a quick suggestion whenever you are solving problems of this nature Diagram is always going to help you out But how can you say that the incident ray in this case? Okay, it happened to be the incident ray is perpendicular to the reflected ray But can you generalize like this? Shweta, can you generalize it like this? All right. So moving to the sixth question now So this question is being taken up from a pair of straight lines So you have been given that there is a pair of straight lines represented by u as a short form Then the equation of the third pair of straight line Through the points where these meet the axis is No, I don't think so Shweta that statement is correct Take extreme cases like this Take a case like this. So where this is the This is a reflected ray and this is the incident ray. I cannot say this angle is always 90 degree No, I think Rohan and Siddhant answered the previous question Simultaneously But in my list Rohan name comes first So I shouted out his name. No problem Siddhant always the next time Yes guys any response from anyone? Guys look at the options the options are giving you the hint All the options are of the form U and is an x y always over here with a constant So all these options that you see over here are of this form It's actually a hint for you which tells that you have to use the concept of family Now can you think in this line x y if I'm not wrong is basically the combined equation of the x axis and the y axis this is the combined equation of The x axis and the y axis Does it does this ring a bell? Okay, so Akshath says B. Kondinia says C even Sundar says C Guys I need two more response Always whenever you're trying to solve an objective based problem look at the options See whether options are trying to give you a clue. Is there any pattern hidden in the question? You guys are lucky to have an option based question We never had in our days when I wrote my IIT in 2004 I never had an option given to me in the subjective paper So it made us much more in fact double difficult to crack those exams as compared to the present form of IITJ So guys what it's trying to say is that you need a pair of lines which passes through We all we need a pair of lines which passes through U and x y So of course the the required pair of straight lines will be this Okay, now I need to get the value of lambda now. How will I get the value of lambda? guys always remember if A x square plus 2 h x y plus by square plus 2 g x plus 2 f y plus c equal to 0 represents a pair of straight lines Then delta which is actually given by abc plus 2 f gh Minus af square minus bg square minus cs square will always be zero Please remember this the condition for any general second degree equation to represent a pair of straight lines Is this determinant should be zero? Hope all of you are aware of this the same as what we call as delta This is the expanded version of this determinant Now since this represents a pair of straight lines, I can clearly say that for this abc plus 2 f gh Minus af square minus bg square minus cs square will be equal to zero Okay, now I want this also to represent a pair of straight lines Right, so when you actually write it you get ax square plus 2 h x y plus by square plus 2 g x plus 2 f y plus c plus lambda x y is equal to zero In other words, you get ax square plus 2 h plus lambda x y By square 2 g x 2 f y plus c is equal to zero Now this has to represent a pair of straight lines. So if this has to represent a pair of straight lines The same condition that is abc plus 2 f gh minus af square minus bg square minus cs square should also be zero right So abc will not change 2 h x y Sorry, uh 2 f gh will become now 2 f gh will become this term Okay Minus af square minus bg square minus c h square So h if I am not wrong will be this New h square this should be equal to zero. Is that fine? Now Let us open the brackets over here If I open the brackets, uh, I'll get Uh, one mistake. This will be h Just a second. This will be h plus lambda by 2 Yeah, same here as well In fact, it will be here because we have abc plus 2 f gh So 2 h is already given to us as this Okay, yeah So now I can just expand it up. So it'll be abc plus 2 f gh Minus af square minus bg square minus c h square And apart from that I'll get lambda f g Okay, lambda f g and I'll get minus lambda I'll get minus lambda c h Okay, and I will get minus c lambda square by 4 equal to zero So what I did I just expanded this And this last term that is 2 f F g into lambda I took separately And I simplified this Now we know from the previous fact that this is going to be zero because It comes from the condition of pair of state lines So this should also be zero So lambda f g minus c h should be equal to c lambda square by 4 Lambda cannot be zero. So lambda has to be 4 f g minus c h By c correct So if I place it back If I place it back over here, I get u plus lambda x y is equal to zero Okay, multiply throughout with c you'll get cu is equal to 4 f g minus c h x y equal to zero. So this would be your desired Equation of the line So yes Option number c becomes correct And the first one to answer that was Pratik Kondiniya Well done Pratik very good This was not a easy question this involved our understanding of the family family of pair of state lines Okay, and of course using the concepts as discussed. So guys, let's move on to the next question Question number seven So the question read like this two points p and q are taken on the line joining the points 0 0 and 3 comma 3 a comma 0 Such that a p is equal to p q is equal to q b And circles have drawn on a p q b and p a p p q and q b as diameters The locus of the point s says that the sum of the squares of the tangents from which To the three circles is equal to b square. So some of the squares of the you can read it as length of the tangents Some of the squares of the length of the tangents from which to this three circle is equal to b square Okay, so swetha has said option b What about others? Need four more responses Okay, so one says c Okay, so all the three people who responded has given different different answers a from saimir Akshatsina also says c One more response I need guys All right So this is let's say a zero comma zero point and this is let's say three a comma zero point And you know, this is going to be a comma zero. This is going to be uh to a comma zero And there's a point h comma y sorry h comma k From where the length square of the length of the tangent drawn to these circles is equal to b square okay first of all Is everybody aware of the fact that if s equal to zero is the equation of the circle Then the length of the tangent Then the length of the tangent Drawn from x1 y1 Drawn from x1 y1 is given by Under root of s1 Is given by under root of s1 s1 means you substitute the point in the s part of the circle and take the root of it correct So, uh, let's first find out the equation of these three circles. So if I'm not wrong this circle Would have the equation of x minus a by two whole square plus y square minus a square by two equal to zero This circle will have the equation of x minus three by two a square plus y square minus a square by two equal to zero And again this equation will have this circle will have equation x minus five a by two whole square Plus y square minus a square by two equal to zero correct So this is like your s. Let me call it as s dash Let me call this as s double dash And let me call this as s triple dash Okay So the square of the length of the tangent means s1 dash s2 double dash I should not write s2. I should write s1 only s1 double dash and s1 triple dash is adding up to b square according to the given question Okay So what will be s1 dash? What will be s1 dash? So in place of in place of x and y you just have to put k h and k So this plus this So if you simplify this it becomes uh a square by four. Oh, yes. Yes. I'm sorry. I'm sorry Thank you for correcting So this square plus I will have three times k square minus a square by four equal to zero Sorry is equal to not zero is equal to b square. I'm sorry is equal to b square. Yeah Now again, let's simplify this I will get h square h square h square from there. So I'll get uh three h square I will get uh minus a h minus three h Minus five h that means minus nine a h I will get And apart from that I'll get a square by four nine a square by four and 25 a square by four Plus three k square minus three a square by four is equal to b square. So three h square plus k square Minus nine h and I think I'll have uh 10 25 minus 124 plus eight a square minus b square equal to zero Now just generalizing it Generalizing it my answer will become three times x square plus y square minus nine a x Plus eight square eight a square minus b square equal to zero So let's see which of the options talks about this All right, so clearly you can see option b talks about it So shweta was the first one to answer this well done shweta very good So this question basically tested your understanding whether you knew the length of a tangent from an external point Rest everything was pretty simple. I did not find this question at all challenging Any questions with respect to this? Please do ask So next we'll take a question on integration and after that I'll give you a break for 10 minutes So we have to find the integral of this term seek seek alpha by coseek to alpha plus seek 18 alpha by Coseek seek alpha plus seek 54 alpha by coseek 18 alpha with respect to alpha Yes, guys waiting for your response Yeah, nobody's able to crack this Ramchandra, sorry Ramcharan Ramcharan nidheesh Rithvik lauhtya Vaishnavi Okay, so lauhtya is saying b Sanjana is also saying b nidheesh Where are you? What's your response nidheesh? Yeah, Ramcharan says a Sundar says d all right guys, so Let's just discuss this You would all realize that these all terms are somewhere symmetric in nature, right? They show some kind of symmetric nature So let us pick one of the terms Let's take the first term That is seek 6 alpha by coseek to alpha Which is actually this Right Now I'll do one thing we'll multiply and divide with cos 2 alpha So that becomes sine 4 alpha by 2 cos 2 alpha cos 6 alpha Now sine of 4 alpha is sine 6 alpha minus 2 alpha By 2 cos 2 alpha cos 6 alpha So now you can break sine 6 minus 2 alpha by using your compound angle formula sine 6 alpha cos 2 alpha minus cos 6 alpha sine 2 alpha by 2 cos 2 alpha cos 6 alpha, okay Now when you individually divide you get half cos 2 alpha cos 2 alpha gets cancelled so it becomes tan 6 alpha minus tan 2 alpha So this is your finally sine of 6 alpha by coseek by coseek 2 alpha So can I similarly comment that if I have to find seek seek 18 alpha by coseek 6 alpha I just have to replace my alpha with 3 alpha So that's going to be 18 alpha minus tan 6 alpha And finally when you have to find tan 54 alpha by coseek by coseek 18 alpha That would become half tan 54 alpha minus tan 18 alpha Okay, just add these three Right, so when you're adding these three you're adding the right-hand side as well So you can clearly understand over here that sine 6 alpha by coseek 2 alpha plus It's not sine it's seek. This is seek So, yeah It's going to be half when you add these two these terms will get cancelled So we'll be left with tan 54 alpha minus tan 2 alpha It's very similar to the type of questions we have been doing in tignometry also. Remember last year we did So you are integrating this Now we know that integration of tan is log of seek So it's going to be half log of seek 54 alpha by 54 minus log seek 2 alpha by 2 So that's going to become your ln seek 54 alpha by 108 minus ln seek 2 alpha by 4 plus c So let's see which option is correct Yeah, I think option number a becomes correct in this case So well done Ramcharan. So Ramcharan was the first one to answer this correctly So see you are not sure but your answer was correct, right? So you should take some chances You should take some chances as well Okay So guys we'll take a break right now and Will return at exactly 11 55 So we'll take a break Okay, and we'll resume at 11 55 a.m All right, guys, welcome back after the break So, uh, let us begin with question number nine Which is actually based on area under curves So you have been given a question find the area bounded by y is equal to sine inverse mod sine x And y is equal to sine inverse mod sine x whole square from x lying between 0 to 2 pi All right, so Sundar says option a Okay, I think b Okay, so Rohan says d Okay option d That they also backs it up option d All right, so almost most of you have answered this Guys try to identify this In this particular question sine inverse Mod sine x this is appearing at both the functions Right Okay Now we know that sine inverse is an increasing function, right? So Whatever is the period of mod sine x that will also be the period of y We all know that we all know that mod of sine x is periodic with Is periodic with pi right So if I just concentrate on framing finding the graph of this function From x lying between 0 to pi Okay Now if you try to recall the graph of sine inverse sine x Please note that the graph of it was like this From 0 to pi it was it goes like this and it comes back like this just from 0 to pi So if you extend it further it would go to minus pi by 2 So it is like this from minus pi by 2 to pi by 2 And this is i'll pi by 2 and this is pi So I only need this part of the graph This part of the graph is only what I need The rest part is not required because I will only draw it between 0 to pi and we know it's going to be periodic after that So it's going to repeat anyways Okay, so the next part of the graph would be like this Hope all of you recall these graphs. Nobody has any Question regarding how these graphs come up Okay Now This graph is done. Now. What about this graph? So y is equal to sine inverse mod sine x Whole square again, what I will do is I would first redefine sine inverse mod sine x So, you know that in the interval 0 to pi by 2 it's going to behave as A straight line y equal to x correct like this Right from pi by 2 to pi It's going to behave as y equal to minus x and it'll also say in the show the same nature That'll start also start repeating the same thing over and over again. Okay So when you square it When you square it I'm sorry. This will be y equal to minus x minus Minus x plus pi. I'm sorry. Yeah, when you square it It actually means the square of this term Right, so again, let us draw it from 0 to pi. So 0 to pi it's going to be First 0 to pi by 2 it's going to follow y equal to x square line So I'm going to draw y equal to x square line y equal to x square line will be like this Okay, remember it will go up at pi by 2 it'll become pi square by 4 and From pi by 2 to pi I will draw the graph of y is equal to x minus pi the whole square means the same shift same thing Same thing shifted Same thing shifted to the right. So it will be again a curve like this Okay, okay And for pi to 2 pi also the same thing will be repeated Now, what am I looking for? I'm looking for this area I'm looking for the area in In yellow Okay, so hope you can see whatever I'm shading in yellow That's what I'm looking out for now from the symmetry of the figure. I can say whatever is this area Let's say I call it as area a then this area will also be area a this also will also be area a and this will also be area a correct So let us find that out first That area is simply the area between the line y equal to x and the parabola y equal to x square Okay, what is the meeting point of these two meeting point of these two is at one Okay So for area a I can say a will be nothing but integral of 0 to 1 x minus x square So that's going to be x square by 2 minus x cube by 3 from 0 to 1 that's going to be half minus one third That's going to be one sixth Okay, so total area here of a's will be 4a so 4a will be 2 by 3 square units So this part is that let's say the area of this part Is b now here also I'll break this into two zones Here also I'll make a line like this And I'll break this into This is b This is also b This is also b and this is also b again from the symmetry of the graph So how would I find this area? Again same thing area between x square And x This time from 0 to 5 by 2 This time from 0 to pi by 2 This will give you the area of the Part b Yes or no, so I'm just zooming it up. It's something like this There's a line and there is a curve And you're finding this area you're finding this area That's what I've shown as b So area between x square and x from 0 to sorry from 1 to pi by 2 I'm absolutely sorry From 1 to pi by 2 because it's from 1 over here right from 1 to pi by 2 So let's solve this so it'll be x cube by 3 minus x square by 2 So if you substitute value you will get something like Pi cube by 24 Minus pi square by 8 Minus 1 third minus half is that fine So when you finally add these two When you finally add This is only b so 4b will be pi cube by Let's simplify this so it'll become So this will become minus pi cube by 12 This will become your 4b. So this will become 2 third minus pi cube by 12 So your total answer will be Your total answer will be 1 6th. Sorry 2 3rd So total area will be 2 3rd plus 2 3rd Minus pi cube by 12 Which is going to be 4 3rd minus pi cube by 12 Oh, I have not done minus x square by 2 minus x square by 2 if I put Uh, I will get Sorry, there's more term over here That's why I was wondering something is missing. Yeah, this time I was missing. Is it fine guys? So I'll just Do this part again because I think Some Calculation error has happened. I'll just I'll just redo this part again So x cube by 3 minus x square by 2 and when you put pi by 2 you get pi cube by by 24 minus pi square by 8 Minus 1 by 3 minus 1 by 2 Yeah So this is a b part So which is going to be a pi cube by 24 minus pi square by 8 minus Plus 1 by 6 So 4b will be a pi cube by 4 minus 2 pi square plus 2 by 3 So you can take pi square Yeah So total area will be Total area will be 4b plus 4a Total area will be 4b plus 4a. That's going to pi square Pi by 4 minus 2 plus 4 by 3 And I don't think so it matches with any of the options So option number d is correct So I think Rohan was the first one to answer this You guys move let's move on here. What was tested was whether you are aware of the graph of sine inverse sine x And how to use the graph to find out the desired area Next is your question number 10 for the day If f of x be a positive continuous and differentiable function on the interval a to b and if limit of f of x as extends to a plus Is 1 and limit of f of x as extends to b minus is 3 to the power 1 by 4 And f dash x is greater than equal to f cube plus 1 by f of x then which of the following option is correct Okay Laugthia says a All right, so let's discuss this question. How to proceed with this. So first of all you take this condition F dash x is greater than equal to uh f cube x plus 1 by f of x So you just take the LCM over here f of x 1 plus f to the power 4x is f dash x Okay, and since it is a positive function I can always Switch the sides of it without affecting the inequality sign This will always be less than equal to 1 Sorry greater than equal to 1 Now let's integrate both sides with respect to da from a to b If you see clearly in this function if you take your f to the power 4, sorry f square x as some t You can always write Derivative of it as 2 f of x into f dash x dx is equal to dt So it's like f square a f square b And you have a half dt by 1 plus t square This is always greater than b minus a This is always greater than b minus a a right So, uh, if you simplify this it's going to become half of tan inverse tan inverse t and put f square b there Minus tan inverse F square a there Okay, and this will always be greater than b minus a Now guys, let's take a limiting case of this because your function is not defined at a and b Your function is defined in an open interval a and b. So we'll have to take the limiting case over here That means you cannot find exactly f of b square. You can always find the limit of tan inverse f square b as beat as x tends to b minus So you can always find this as x tends to b minus Okay, and here you can always find tan inverse f square x as x tends to a plus So, uh, let's use the result. We already know the value of the question only provides me the value of tan inverse Value of f of x as x tends to b minus which is 3 to the power 1 fourth So here it will become tan inverse 3 to the power 1 fourth square which is going to be half Okay minus tan inverse This will become Uh, one to the power. So this will become one. So this always be greater than b minus a So half tan inverse root 3 is pi by 3 Minus pi by 4. This will always be greater than b minus a So if I'm not wrong, this will give you pi by 12 So b minus a will always be less than pi by 24 b minus a will always be less than equal to Option c is correct guys Option c is correct So, yeah So gaurav is correct. So the first one to answer this correctly is gaurav itself Great Let's move on to the next question guys This is a question which have been picked from probability So the question says the length Find the probability that the length of a randomly chosen chord of a circle lies between 2 3rd and 5 6th of its diameter Gaurav says uh c All right, let's wait for others to respond gaurav So I need at least four more responses for me to start discussing this Yeah, so let's say this is a randomly chosen chord Its length should be somewhere between 2 3rd and 5 6th of the diameter So the length of ab should be somewhere between Let's say this is a circle of radius a So it should be somewhere between 2 3rd of a and Sorry 2 3rd of 2 a and 5 6th of 2 a All right. So the purpose of convenience. I can assume this to be angle theta Right So I know the length of ab is going to be 2 a sin theta, right? So this is going to be 2 a sin theta That means uh 4 a by 3 This should be the condition So drop a factor of 2 so 2 a by drop a factor of 2 a in fact So I can say 2 by 3 Should be less than sin theta should be less than 5 by 3 Now, let's say I call this as r Can I say r is now going to be a cos theta? R is going to be a cos of theta Okay, and if sin theta lies between 2 by 3 and 5 by 6. I'm sorry. This has to be 5 by 6 5 by 6 cos theta will lie between cos theta will lie between So, uh 5 5 by 6 1 by 25 by 36 and the root which is 11 by root 6. I'm sorry root 11 by 6 and 1 minus 4 by 9 and the root which is going to be 5 root 5 by 3 So it's going to lie between root 5 by 3. Okay Now what it is trying to say that it's trying to say that If your chord is in such a way then r has to vary That's sorry cos theta has to vary between root 11 by 6 to root 5 by 3 So if you attach an a to this It means r should vary It means your r should vary From root 11 by 6 a to root 5 by 3 a Correct Right, so can I say Your chord is going to lie in the region Your chord is going to lie in the region Which is trapped between two concentric circles Which is trapped between two concentric circles Which is trapped between two concentric circles Whose r can vary from r can vary from root 11 a by 6 to root 5 a by 3 Okay, remember this is still within the outside circle It is still within the circle a so i'm not drawing the outer circle So let me draw the outer circle as well It is still within that outer circle Whose radius is a so this is a circle whose radius is a So this entire thing is your a So your chord can only happen to lie in this annular zone Your chord can actually happen to lie in this annular zone That means it is a case of geometrical probability where Your probability of your chord to be between Two third and five six of the diameter will be nothing but The area of this region which is pi r square minus pi root 11 by 6 a square Whole divided by the total area which is pi a square So that's going to become pi pi will get cancelled in fact a will also get cancels will be 5 by 9 minus 11 by 36 Which if i'm not wrong it's going to be 20 by 36 minus 11 by 36 which is 9 by 36 Which is going to be 1 fourth Okay, swetha. I'll repeat this once again nischel also. See what is happening is This chord A focus on this circle the pink one In the pink circle First of all, I realize that This chord This chord Is going to be between the circle whose radius happens to lie between Root 11 by 6 a and root 5 by 3 a It can't be above it can't be below it correct So there are two circles Make such two circles Your two circles It will be like this that your chord That will trace out only this annular zone So either your chord can be like this Chord or it can be like this So now if I trace this chord, if I start moving this chord, it will basically cover an annular zone This will cover an annular zone And this is this annular zone area which I have shown to you by the shaded area This is this annular zone Correct, so it will be trapped within two concentric circles The chord should be such that The area traced by that chord is equivalent to the area between these two concentric circles Whose radius is under root 11 by 6 a and under root 5 by 3 a Okay So this area shade, this area will cover up It will be basically equivalent when you rotate it About the whole circle, so basically this area will come Are you getting this point? And that area I found out to be this expression Divided by the total area which is pi a square of the entire circle because r can go from 0 to a So total area is pi a square But for r lying between these two values only your length of the chord will be meeting the required Inequality and hence total area will be So a total probability will be the area of the annular region divided by area of the total Circle which is going to be one by four Next is a question which comes from permutation combination So number of ways to choose an ordered pair a comma b from the set 1 2 3 till 10 says that mod of a minus b is less than equal to 5 So gaurav says 100 Navin says a sondarya says a sai may also says a Any other response guys? All right, so let's try to solve this now So a minus b mod is less than equal to 5 Now they're asking you ordered pair a and b. Okay So what I'll do first I'll first find out When is this equal to 0? That means when a is equal to b The way is equal to b You can only form 10 ordered pairs by this right Only 10 ordered pairs can be formed That is 1 comma 1 2 comma 2 3 comma 3 all the way till 10 comma 10 right So only 1 comma 1 2 comma 2 3 comma 3 can exist till 10 comma 10 okay Now if I consider mod a minus b to be any number Other than 0 let's say this is somewhere between 1 and 5 This is somewhere between 1 and 5 Okay, now what I'll do is From 1 to 10 I will choose a somewhere And b somewhere Okay Now let's say x numbers are coming between 1 and a y numbers are coming between a and b and z numbers are coming between b and 10 So can I say x and y x and z the restriction will be they should be just greater than equal to 0 But for why can I say this restriction has to be between 0 to 4 Now why 0 to 4? They cannot be any number between a and b. Let's mean a and b can be consecutive also If they're consecutive then the difference between them will be 1 Right, so maximum four numbers can come between them and zero numbers can minimum is the zero number that can come between them So think as if you are solving this inequality x plus y plus z is equal to 8 Under the restriction that x and z are greater than equal to 0 and y should lie between y should lie between 0 to 4 So think as if you are solving this under this constraint Under these these two constraints this and this Correct So can I use multinomial theorem over here? Can I make use of multinomial theorem over here? So for x and z There is no constraint so it can take value from x to the power 0 x to the power 1 x to the power 2 all the way till x to the power 8 Whereas for why the constraint is it can take x to the power 0 x to the power 1 x to the power 2 x to the power 3 x to the power 4 So in this I have to find coefficient of from this expression. I have to find coefficient of x to the power 8 That is this Okay So guys, let us simplify this this is going to be a gp pure simple gp with one term and 1 minus x to the power 9 by 1 minus x Whole square and this is again going to be a 1 minus x to the power 5 By 1 minus x So this is going to be 1 minus 9 whole square Into 1 minus x to the power 5 into 1 minus x to the power minus 3 Now this term is never going to contribute an x to the power 8 So it's only going to be from these two terms that I'm going to get coefficient of x to the power 8 Again, we don't have to go very far over here. We can just In pick up terms which are coefficient giving you coefficient of x to the power 8 So since there is a 1 over here, I need coefficient of x to the power 8 from here so I need x to the power 8 from 1 minus x to the power minus 3 And since there is an x to the power 5 over here, I need x cube from 1 minus x to the power minus 3 Okay So let's find these two out and let's see what is the answer By the way, we all are aware of what is the coefficient of x to the power r in this How to find coefficient of x to the power r in this So it's going to be something like this Okay, like this. So if you want to get x to the power r It will have n n plus 1 all the way till n plus r minus 1 by R factorial this will be coefficient of r which you can write in a very simple word as n plus r minus 1 c r So here I'll get 3 plus 8 minus 1 c 8 which is nothing but 10 c 2 10 c 2 is going to be 45 And this will be 3 plus 3 minus 1 c 3 that is going to be 5 c 3 which is going to be 10 So 1 into this 45 minus 1 into coefficient of this will be minus 10 So 35 will be the coefficient of x to the power 8 Now remember for each of these 35 terms For each of these 35 terms since they're asking you ordered pairs Since they're asking you ordered pairs So for each of these 35 Terms you can have Two possibilities that means a b and b a that means 70 such terms would be coming from there So your total answer will be 70 plus 80, which is option c is correct So I think Sondaria is the first one to answer this Option c all right guys. So the last question of the day, which is the 13 question So it's given that integral of a mod of sine x From a to b is 8 integral of mod of cos x from 0 to a plus b is 9 Then you have to find integral of x sine x from a to b So guys if I just ask you what is the integral of mod of sine x from 0 to pi by 2 What is the answer for this? We know mod sine x is going to be just sine x. So it's going to be Minus cos x from pi by 2 to 0 that's going to be 1 Now imagine that Each of these areas 1 1 1 Okay, it starts from anywhere Start from a and go till 8 and you have covered 8 of those areas That means what can I comment about? Sorry the total area will be from 0 to pi. This is this is going to be 2 then yeah, this is going to be 2 So from 0 to pi it's this 0 to pi by 2 it's 1 so from 0 to pi it will be 0 to pi this will be which is actually the period of this function that's going to be 2 yeah So think as if you have covered 8 of such areas it means what can I say about the difference between these two Can I say for sure that the difference between b and a Has to be 4 pi similarly Cos x is also a periodic function. Correct. So let's find out the area of mod of cos x From 0 to Let's say first pi by 2 we'll find it out. That's going to be 0 to pi by 2 This is the graph of Okay mod of cos x So it becomes cos x its integration is sin x 5 by 2 to 0 its area will be 1 correct That means if it is covering this area is 1 So this area has to be 2 This area also has to be 2 This area also has to be 2 And you will have one more area This area also has to be 2 and this area has to be 1 so then only we'll get a total of area of 9 So this only will become your 9. This is your 9 right So it has to start from 0. This is pi by 2 if I'm not wrong Okay, 0 pi by 2 then 3 pi by 2 then 5 pi by 2 Then 7 pi by 2 9 pi by 2 and this will be 11 Sorry plus pi by 2 which is going to be 5 pi. I think am I correct? Is that fine guys? So can I clearly say that This point Will be a plus b. So a plus b will be 5 pi Yes or no Is that clear guys? Can I say a plus b is 5 pi? Now B minus a is 4 pi and a plus b is 5 pi. What does it mean? What is a and b respectively? Just find a and b respectively So when you add it 2 b is going to be 9 pi So b is equal to 9 pi by 2 And when you subtract it 2 a is going to be pi So a is going to be pi by 2 Correct Exactly So now what is the question asking you finally the question is asking you to integrate x sin x The question is asking you to integrate x sin x From pi by 2 to 9 pi by 2 correct So please integrate this You may call this i and you can use the kinks property over here as well if you want Anyways, you can you apply partial you can apply integration by parts also add them If you add this one and this one you get this So please complete this and tell me the answer So this will be minus sin x Indication of sin x is minus cos x I think it's going to become zero right? What are the answer that you're getting? Is it coming out to be zero? So I think None of these is going to be your answer. Is it fine any question guys? So all right with this I'll end the session. I'll share this pdf with you on the group as well Okay Thank you all for coming live today for the session And all the best for your semester exams or both practicals pre-vote practicals I think from january 2nd you have another preparatory exam starting up. So all the best for that And Merry Christmas in advance. Thank you so much. Bye. Bye