 Hello, good evening. Welcome to the YouTube live session on problem solving for class 11th So those who have joined in the session I would request you to type in your names in the chat box So that I know who all are attending this session and I also understand you have exams coming next week So don't worry. This session would be recorded. So those who will who are interested But still could not watch this session because of the exams coming next week Make sure you watch the recorded session. Okay, let me just brief you up about this session so this session will be solving some subjective problems and Yes, there will be no options given to you and you have to actually solve it and once at least three or four people have solved it I'll start giving you the solutions for it and I may ask you to send your solution To me personally on my whatsapp. Again, I'll repeat those who have joined in the session Please type in your names in the chat box This will be a school come slightly higher than school level questions There is a proper mix and match of the class 11 topics. That is tignometry quadratic equation Probability coordinate geometry binomial sequence series So I've I've tried to pick up one question each from every topic one from limits one from derivatives So here is the first question for the day So the question says if alpha and beta are the roots of the quadratic x square minus p x plus 1 minus c equal to 0 then show that alpha plus 1 times beta plus 1 is 1 minus c and Hence proof that this expression is true We often work. I'm not expecting many people for this session because I understand your exams in here And you want some time out to prepare for the exams Hey, Nikhil. Good afternoon. I would request you to set in your screens at at least 480p Or above. Good afternoon, Andrew. So in between you may hear my daughter's voice. She's with me. Actually, she's not well So please excuse that Yeah, so I think the first part is quite easy. I don't see a challenge doing the first part If you see this quadratic Can be written as x square minus p x Minus p plus c equal to 0 Okay, and if alpha and beta are the roots of this quadratic we can say that alpha plus beta is equal to p and Alpha into beta is going to be minus of p plus c okay Now alpha plus 1 times beta plus 1 is actually nothing but alpha beta plus alpha plus beta plus 1 and Since alpha plus beta is equal to minus of p plus c Alpha beta is minus p plus c alpha plus beta is p and this is 1 this cancels out and gives you the answer as 1 minus c Okay, so this proves the first part of this question. That is this part Okay, next is we have to prove that alpha this expression is equal to 1 How many of you have been able to do the second part? Hi Shishant, okay Great, so we'll discuss this so we can say that this expression over here can be written as actually alpha plus 1 the whole square Okay, and in the denominator actually we can write it as alpha plus 1 the whole square We can say plus c minus 1 Okay, plus c minus 1 I can write it as minus 1 minus c actually and Similarly here also we can write b beta plus 1 the whole square by Beta plus 1 the whole square Minus of 1 minus c now I am writing minus 1 minus c over here instead of writing c minus 1 is because I Know this expression right over here Right. I know 1 minus c could be written as alpha plus 1 times beta plus 1 So what I will do I will replace I'll replace 1 minus c with Alpha plus 1 beta plus 1 and when I do that This is what I will see coming in the expression okay, and Similarly here also. I'll see beta plus 1 the whole square and Beta plus 1 the whole square minus alpha plus 1 beta plus 1 Okay So alpha plus 1 I can cancel off From the numerator and the denominator Which leaves me with alpha minus beta in the denominator In a similar way I can have Beta minus alpha on the second part of the expression Now if you just see the denominator are same actually but with a negative sign So you can actually write this as negative beta plus 1 by alpha minus beta that gives you Alpha minus beta alpha plus 1 minus beta minus 1 so 1 and 1 gets cancelled So alpha minus beta by alpha minus beta of course is 1 which is your right hand side and hence proved Okay, and hence Okay, hence Is that fine guys quite easy problem? Hello, good afternoon everyone Okay, so Not a difficult problem to start with So let me now take you to the next one now. All right, so let's work on the second problem. Hope you all can read the problem Suppose a and b shoot independently until each hits his target And they have probabilities of three by five This is three divided by five And five five by seven of hitting the target at each shot find the probability that b will require More shots than a b will require more shots than a good afternoon pretty no worries So guys read the question very very carefully the probability of a hitting On each shot is three by five probability of Probability of b hitting is five by seven find the probability that b will require more shots than a three by seven No, no, that's not the right answer Neither is four by 35 Guys the total number of shots taken by them is not known to us It is just known that b has to take more shots than a Now how many more that is also not known to us So the question might appear to be an easy one or a easy sounding one, but it it requires a bit of thinking All right, so let me just give you some hint So let x be the number of times a shoots to hit the target for the first time so let x be the number of times That a shoots Okay a shoots To hit the target for the first time To hit the target for the first time Similarly, let y be the number of times Uh b shoots To hit the target For the first time Okay Now If let's say x is some particular value m Let's say x is some particular value m What does it mean? What is the probability that he will shoot the target on the mth shot? It means he must not have shot the target till m minus 1th shot that means 2 by 5 to the power of m minus 1 times 3 by 5 will be a probability because This is he has missed it for this means he has missed it Missed it m minus 1 Shots And the mth shot he has hit okay in a similar way if let's say If let's say b hits on the nth shot That means he was not successful till n minus 1th shot And the nth shot he was able to shoot correct Now since both are hitting independently Since both are hitting independently The probability that it will happen that x and y hit it in such a way that y is greater than x Will be nothing but First of all it will be the product of their respective shots Because we know independent in case of independent hitting Or when the when the when the two events are independent PA intersection b is PA into PV right right In your probability we have studied about independent events But this is subject to the fact that you are summing up You are summing up In such a way that n is always one more than m And it can go to infinity and m can be anywhere from 1 to infinity correct because You'd never know how many shots that they will take in order to finish the event So what i'm doing is i'm running two summations One on m going from one to infinity and other on n going from m plus one to infinity Okay, so now let me use the expressions that i have just now written So this is going to be two fifth m minus one three fifth And this is going to be two seventh n minus one five seventh Now how do i apply the summation process? Very simple. First of all, I will apply the summation on since they're independent m and n are independent I'll first apply the summation on n So I will just do it first Okay, so two fifth to the power m minus one This is going to be I think three seventh and two by seven to the power of m divided by One minus two by seven because it's a infinite gp infinite gp and the first term would be Uh, when you put your n as m plus one. So this is your first term And one minus r is in the denominator. Okay So it's summation from one to infinity, uh, two by five to the power of m minus one plus three seventh off This will become two sevenths to the power m and in the denominator you will have a five by seven So seven will go up So it will be left with three by five So three by five I'll bring out and it's a summation on I can also bring two by seven out and make it a four by 35 to the power of m minus one Yeah, so it's a gp again It's a gp again summation of four by 35 to the power of m minus one From one to infinity So this will give you one by because first term will be when you put your m as one. So one by one minus r Okay, so that's going to give you six by 35 And in the denominator, I will have a 35 by 32 if I'm not sorry 31 If I'm not wrong, yeah 31 So it'll get cancelled and your answer will be six by 31 So I think the process is clear. First of all, uh, I made use of the fact that Since they're independent events Since they're independent events the probability of them hitting is the product of their respective probabilities now I assume that uh The first person a takes m chances to hit it second person takes n chances to hit it and m Will go from one to infinity and n will always start from n plus one to infinity Okay, and after that I applied this summation process Thankfully m and n uh, in fact n was uh summed up first and then the summation of m happened And it was a simple gp to address gp to infinite terms Is that fine? Again, it's a higher than school level, but again a lot of learning came through this problem Now let's move on to the third problem So moving on to the third problem find the values of k for which this modulus is Less than two for all x belonging to r and once you're done, please type in your response in the chat box Or you can pm it to me directly And if you're done, please type done so that you know, I know who all are done One to infinity is uh Not correct So I'm letting you know if you're correct or not so that you can you know check you're working Minus infinity to zero four to infinity. No, Bharat. That is also not correct. Okay. Compliment of the oh, yeah Okay, how about others? All right. So basically First of all, we all know that whenever we have uh such kind of Uh inequality in modulus we all know that This is lying between minus a to a so in this case also I can say x square plus kx plus one divided by x square plus x plus one Will lie between Minus two and two Yeah, but that is correct. Now since x square plus x plus one is positive Why it is positive because it's the parabola opening upwards and its discriminant is negative Okay, so it says if this is positive I can multiply throughout by x square plus x plus one So it gives you this expression. Let me simplify this Now this amounts to two inequalities one is um three x square plus Two plus kx Okay, plus three is greater than zero. That's number one And the other one will be x square plus Uh two minus kx plus one Again greater than zero if you want if you want That this quadratic should always be positive That can only happen when it's a coefficient of x square is positive, which is already there So this is already positive And the discriminant of this should be negative because you want it to be a hanging parabola Okay, so in a case of a hanging parabola The discriminant would be negative because such a parabola will have no real root Right So Discriminant less than zero means Again b square that is k two plus k whole square Minus four ac minus four ac will be minus 36 less than zero Okay, uh, you can you can just uh factorize this as minus six square less than zero So it will give you If i'm not wrong, uh two plus k uh plus six And two plus k minus six less than zero So k plus eight and k minus four less than zero so k should lie between eight and four Okay So this is the response for the first one For the second one in a similar way, uh, if you want this quadratic to be always positive again the discriminant Since the coefficient of x square is one it should be always a hanging parabola Again the discriminant should be negative So b square minus four ac should be negative Which is two minus k the whole square minus two square is less than zero So I can just write it as this less than zero so this will be Minus k four minus k less than zero which implies k k minus four is Less than zero so k should lie between K should lie between zero and four Now these both the conditions must be two simultaneously. So the overlapping condition of this will be k lying between Zero to four that will be the intersection of These two is that fine So again, the only person to answer this correctly was Bharat But initially he gave it wrong and then he corrected himself Great So we now move on to the next problem. Now. This is a question coming to you from uh straight lines So read this question carefully. There's a line l one which is rotated About its point of intersection with the y-axis in the clockwise direction to make it l two Says that the area formed by l one l two x axis and this line is four by three square units and it's also given that The point of intersection with x equal to five lies below the x axis Find the equation of l two First I would request Uh, no, that's uh, that's slightly wrong Bharat. Just check x plus y is equal to two. Okay anybody else Okay, that that That makes it the same thing. I think Bharat others. Can you hear me? Okay, let me sketch this Oh, am I audible? Am I audible? Can you hear me? Oh, okay, great Apriti, I think just check your settings on your laptop or computer whichever you're using So let me first sketch this diagram. So here we have the Axises And we have this line three y minus two x minus six equal to zero and it's clear that when you draw this line it will be uh passing in such a way that It will cut the y axis. This is your y axis x axis. It'll cut it at zero comma two Okay Now this line is rotated about the point of intersection in a clockwise sense To make it l two. So this is rotated in such a way that Yeah, let's say this is your l two This is l one. This is l two And again, there is a line x Equal to five line like this. So this is your x equal to five line. Okay, let me call this point as a Let's say this is the origin And let's say this is b This is c and this is d So let's find out the point of intersection. First of all, uh, the point b would be Five comma and if you put your Xs five, you will get 10 16 16 by three. So this will be 16 by three Uh c point will be five comma zero undoubtedly and uh d point would be again Not known but yeah one one thing I can do is I can assume this line itself to be the line l two itself to be Y is equal to uh mx plus two Right because slope is unknown, but I definitely know it's y intercept. It's y intercept will be two only So if I have to find out the intersection, I can write this point of intersection to be five comma five m plus two Just write it in a proper way five comma five m plus two Okay Now Let's say this point is e over here They're saying that the area of Uh the region formed by l one L two x axis and this line. So they're talking about this area. I think So this area is given to us as 49 by three square units The square units So can I say area of this unit will be let's say I call this area to be a Can I say area is nothing but area of the triangle Area of the triangle abd minus area of the triangle ecd Okay I can just drop a perpendicular over here and this will be of length five So area of the triangle abd will be half Uh base base would be the distance b so bd distance will be nothing but can I say, uh It would be the distance between the y coordinates of these two points So I can directly write 16 by three minus five m plus two Into height will be five minus area of ecd will be half Half base into height again Uh this length will be This length will be five m plus two in a negative sense Okay, and uh this point here this point here will be zero comma. Sorry, uh y will be zero in this case. So it will be minus two by m comma zero Okay, so this length here would be five plus two by Okay, and this a is given to us as 49 by three So let us try to simplify this so 49 by three Okay, let's simplify this This will be, uh Five by two And this will be Three three as the lcm That would be, uh 10 minus 15 m and plus one by two Five m plus two the whole square by m. This will be six m can be taken as lcm And here I can write five m Plus three times five n plus two whole square So, uh 75 and 75 m square will get cancelled Here I will get uh 50 m and from here I will get 10 20 60 50 plus 60 is 110 m Okay, and constants will be 12. Okay, just solve for m. Yeah, if I solve for m in this case I have to cross multiply here. So 49 uh into uh six will be uh 294 and this will be 330 m plus 36 Okay, so minus six m Sorry minus 36 m will be equal to 36 Which means m will be equal to minus of one if n is minus of one the equation of l2 will become This that means x plus y is equal to two would be the right answer The first one to answer this exactly correctly was Kavya and then of course Bharat also tried but He made some mistake In sign somewhere. Is that fine guys? Any question with respect to this? Oh, okay, never mind Bharat Let's look into the Fifth question for the day. So this question comes to you from the binomial theorem chapter This is uh filling the blanks actually the largest value of x for which the fourth term of this expansion is 336 Uh, no problem Ashutosh. You can just see whatever has been done so far. Yeah, so pretty light powers Just simplify the expression I think first before you start applying your Uh general formula of the rth r plus one-th term is not correct So first of all if you see this basically it's five to the power log of 4x plus 44 to the power of Uh one by five and in the same way this becomes five to the power log of Uh five Two to the power x minus one plus seven to the power of one-third. So this is straight away this Okay, after that Yeah, it's locked to the base five After this you can apply your concept uh that the fourth term is 336 So fourth term is t3 plus one so that's going to be eight c3 4x plus 44 to the power of Okay, one fifth to the power of five will be one itself and this will be two x minus one plus seven to the power of minus And this is given to us as 336 So we know eight c3 is going to be uh 56 if i'm not wrong and this is 4x plus 44 By this and this is equal to 336 given to us I think this will go by a factor of six itself You can actually assume your two to the power x as y So this will become y square by 44 divided by y by two plus seven is equal to six Which means y square plus 44 is equal to three y plus 42 So y square minus three y plus two is zero This is factorizable. So this gives you y as one Or y is two Now if it's y is one it implies y can be two also y can be sorry x can be two Sorry, i'm sorry y can be one that means two to the power x can be one or two to the power x can be So largest value will be x equal to one Is that fine? So I think it was a easy problem. I think people made mistake in the simplification process Okay, shava no no problem Let's look into the next one which is the sixth problem now So in a triangle abc this comes from the properties of triangle We have to show that a square sine s minus a Plus b square s minus b plus c square s minus c equal to this Once then, please let me know if you're done Yes, anyone hope you know the normal meaning of delta delta is the area of the triangle r is the Circum circle radius And small a small b small c are the sides of the triangle Okay, so let's discuss this if you take the left hand side s is half of a plus b plus c So s minus a would be half of b plus c minus a correct So I can write this as half b plus c minus a Okay Plus half b square c plus a minus b And half c square a plus b minus c So if you see you can actually take a common from everywhere So there's a a term over here a term over here a term over here. So if you take a common You end up getting uh c square Plus b square minus a square Okay, similarly you can take a b term common from here here and here And if you do that you get a square plus c square minus b square And similarly c term Can be taken common It will give you a square plus b square minus c square correct No This term is quite familiar to us. Isn't it? We have all learned the cosine formula So if I say cos of a we know that it's b square plus c square minus a square by 2 bc correct So I could write b square plus c square minus a square as 2 bc cos a Okay So what I will get over here is half 2 a b c cos a And this term likewise will become 2 a b c cos b And the third term will become 2 a b c cos c So 2 a b c could be taken out. So it becomes a b c Cos a plus cos b plus cos c Okay Now, how many of you actually remember your conditional identities? How many of you remember that you could actually write this as 1 plus 4 sin a by 2 Sin b by 2 sin c by 2 If not, I could quickly derive this Okay, so derivation for cos a plus cos b plus cos c is 1 plus 4 sin a by 2 sin b by 2 sin c by 2. I'm going to do it briefly over here So this term I can write it as 2 cos a plus b by 2 Cos a minus b by 2 And this term over here, I can write it as 1 minus 2 sin square c by 2. Okay. Now, we know that in a triangle A plus b by 2 is pi by 2 minus c by 2. Correct So cos of that is going to be just sin of c by 2 Take one out and take minus 2 Sorry plus 2 sin c by 2 common You will get cos a minus b by 2 minus of sin c by 2 this term here would be Nothing but minus of cos a plus b by 2 again by the same property So this becomes 2 sin c by 2 and this becomes 2 sin a by 2 Sin b by 2 Which makes it 1 plus 4 sin a by 2 sin b by 2 sin c by 2 So placing it right over here, I can say the final result here can be written as a b c by 1 plus 4 sin a by 2 sin b by 2 sin c by 2 Okay, now still I'm not close to my result because my result says It should be 4r delta Okay, try to recall this formula Try to recall this formula That a b c by 4r is equal to delta So a b c will be equal to 4r delta Okay, so you could replace this with 4r delta 1 plus 4 sin a by 2 sin b by 2 sin c by 2 Which is nothing but your right hand side. Is that fine? So again, uh Simple problem but required you to understand or remember all your properties of triangle formula now Adi Sharma is not sufficient for that somebody was asking whether Adi Sharma is uh sufficient for knowing the formulas of properties of triangle I would say no because it just takes care of the basic sign rule cosine rule And tangent rule you may have to refer to standard j books. In fact, Cengage publishers would have all the formulas So let us now move on to the next problem, which is the seventh problem for the day So here we have a question coming from derivatives. So f of x is given as uh this Continued fraction compute the value of f 50 Into this dot is into f dash 50 Okay, so Bharat has given response Bharat, uh, that is not correct Check check once again. So same applies to you as Ashutosh. That is also not correct Okay, now somebody has changed his answer. So I'm not telling you this right or wrong. Let others try What will be answer 50? Okay. So most of you are saying 50 guys, uh, see here first of all Do you see that if you write this function like this x plus 1 by Now you realize that from this stage onwards Your function is getting repeated right, so you can say your f of x is equal to x plus 1 by x plus f of x So f of x minus x is 1 by f of x plus x Which means if you cross multiply f of x minus x times f of x plus x is equal to 1 Which means f square x minus x square is equal to 1 Okay Now differentiated both sides So if you differentiate this expression what you get on the left hand side you get two f of x into f dash x So this is basically application of chain rule Okay, so f of x square the derivative of it will be 2 f of x times f dash x And derivative of x square will be minus of 2x here and this is 0 Which clearly implies f of x f dash x is equal to x dropping the factor of 2 Now what do we need? We need the value of f 50 f dash 50 So we can put the value of x as 50 So when you put the value of x as 50 you get the desired result This is equal to 50 so your Answer for this problem is going to be 50 exactly 50. Okay Never mind happens So, uh, we'll move on to the next question which is uh The eighth question for the day So let a 1 a 2 etc till a n be real numbers such that under root a 1 plus under root a 2 minus 1 Plus under root a 3 minus 2 and so on till under root a n minus n minus 1 is equal to half a 1 plus a 2 till a n minus n n minus 3 by 4 compute The value of summation a i's Actually Okay, Bharat, uh, note it down your response. Let's see what others have to say Okay, so most of you are giving you the response as 5050. So let's see So what I'll do first is I'll assume the these terms under root a 1 to be let's say b 1 And under root of a 2 minus 1 to be b 2 And so on till under root of a n minus n minus 1 to be b n So of course a 1 will become b 1 square a 2 will become b 2 square plus 1 And so on till a n will become b n square plus n minus 1 Okay, so what is given to us in this problem is That b 1 plus b 2 At sector r till b n Sorry Is half of b 1 square b 2 square plus 1 And so on till b n square plus n minus 1 Minus of n n minus 3 by 4 Right, so here we can Actually write it as half You can send the 2 on the other side b 2 square Till b n square And you have 1 plus 2 plus 3 all the way till n minus 1 So half a b 1 square b 2 square up till b n square And this will be nothing but n minus 1 n by 4 let me write it as 4 itself here n n minus 3 by 4 Let me send the 2 on the other side b 1 b 2 till b n So here also I let me make it as 2 to each Yeah, so this will become b 1 square b 2 square Till b n square and this is going to be Minus 2 n by n so which is n Okay So overall I can write this expression as b 1 square b 2 square up till b n square minus twice of b 1 plus b 2 till b n And I can write this n as 1 plus 1 plus 1 n number of times. So this is n times The benefit of doing this is you would realize that you can actually make Perfect squares like this b 1 minus 1 whole square b 2 minus 1 the whole square Up till b n minus 1 the whole square And if you're having the sum of all the perfect squares equal to 0 the only possibility is that each one should be 0 So b 1 should be 1 b 2 should be 1 and so on till b n should be 1 So a 1 should be b 1 square So a 1 is b 1 square which is 1 a 2 will be b 2 square plus 1 which is 2 A 3 will be again 1 square plus 2 which is 3. So basically you realize that you are summing up When you are writing summation of ai From 1 to 100 you are actually summing up 1 plus 2 plus 3 all the way till 100 Which is nothing but 5050 Right. So almost everybody who answered it got it. Correct Where is the scamming I today? He's not attending the session today Okay So we'll move on to the next question now that comes to you from Coptic Summers chapter So if mod of z by Z conjugate mod minus z conjugate is equal to 1 plus mod z then prove that z is purely imaginary Then you have to prove that z is Purely imaginary If you're done, please let me know you're done and if possible send me your response on my whatsapp id Yes, anyone? Okay Ashutosh, can you just send me your working? Okay, so let's discuss this Please let me know once you're done See here whenever you have operations related to division and all it's advisable that you use the Euler form representation Okay So z conjugate will be r e to the power minus i theta So mod of z divided by this Will be r e to the power i theta by r minus r e to the power minus i theta mod is equal to 1 plus r So this is saying that e to the power i theta minus r e to the power minus i theta Is equal to 1 plus r Now just replace your e to the power i theta plus with cos theta plus i sin theta minus r cos theta minus i sin theta Okay, this is equal to 1 plus r now. Can you simplify it? If you write it it'll become uh cos theta 1 minus r plus i Sin theta 1 plus r. So this is a complex number Now the modulus of this complex number will be simply the square of the real part plus of the part under root So which you can actually write it as square of this part Now here you have cos square theta uh 1 minus r square you can write it as this And here you have sin square theta 1 plus 2 r plus r square And this is 1 plus r the whole square So cos square theta plus sin square theta you can collect it that's becomes that becomes 1 and you will have 1 minus Sorry, you have 1 plus r square as well, right? You will have 1 plus r square as well and then you'll have uh Plus 2 r sin square theta minus cos square theta This will be this So when you bring it to the other side you get 2 r Sin square theta minus cos square theta minus 1 is equal to 0 That's that's nothing but minus 4 r cos square theta equal to 0 Now since r cannot be zero because then your complex number will be zero And that's not possible because in that case The kosher itself will not be a valid one because this will become zero over here So this only implies that your cos theta can be zero If cos theta is zero means the real part of that complex number has to be zero And if real part is zero, which clearly implies z is purely imaginary Okay So partially I use the Euler form and partially I use the polar form representation And the moment I get cos theta is zero means your Theta is some odd multiple of 5 by 2 and any odd multiple of 5 by 2 argument will actually lie on the imaginary axis Correct? So it makes it purely imaginary in character without Euler you can do it But again, it's a matter of time also like you can you can have 100 of ways to do it But what takes you the least time is what we'll prefer All right, so moving on to the 10th question for the day This equation has no solution when it belongs to this interval So it's pick up to a this is a mixed and Tignometry and calculus In fact, I can say it's a mixed and match of tignometry calculus. Of course having functions element Bharath, can you send me the solution? Great great, okay That's correct actually. Yeah Okay guys, this problem is best solved by using graphs See, uh treat is treat this as if you have two curves one is two sine x And other is mod x plus a and you want to see that they have no solution Okay We all know the graph of Two sine x it's like this Okay It's just that its amplitude will be This Now mod of x Mod of x plus a is basically mod x graph Shifted up or down by a depending upon what is the magnitude of a correct So I'll just make it in different color So I'll just make it in different color. So this is uh mod x and can shift up and down It can shift up and down depending upon a Now it is clear that since the both the branches are facing upwards So these branches of mod x are facing upwards No matter how down you take it will always cut the graphs But you don't don't want any solution any no solution means no cutting No cutting slash touching Okay, so it can only take it upwards Now I can take it upwards only till the point where I realize that it is just managing to touch the graph correct And after this if you take it up you would realize that they will have no solution at all So I need to first find out what is this x point where they will both touch each other Now if they both touch each other the tangent to the sine x curve should match with the slope of the line Remember the slope of this line is one right? So this is actually x plus a line So tangent to two sine x will be nothing but the derivative of the function which is two cos x And this should match with the slope of the line this will match with the slope of the line So two cos x should match with one that means cos x should be half So x should be pi by three right so if you're If your system of equation like this has no solution It means that Pi by three plus a should be greater than Two sine pi by three okay So that means a should be greater than two sine pi by three Minus pi by three So a should be greater than three root three minus pi by three which clearly implies A should belong to the interval three root three minus pi by three comma infinity Is that fine All right, so we'll move on to the next question now Question number 11 for the day Uh, I would call the previous question J main level question Yeah, yeah, take your time 121 Okay Why do you think it's wrong? That doesn't mean it's right Okay, so why do you think I should we're making silly mistakes check it out Okay, so this is a circle concentric with a ellipse And it passes through the foresight. So this is f1 and this is f2 So let me draw it like this. Okay, and it is also Intersecting it at four points Let's say this is f1 f2 and one of the points intersecting is let's say p Now major axis is 15th and the area of this triangle p f1 f2 is 26 Now major axis is 15 Which means p f1 plus p f2 is equal to 15 correct because we know in an ellipse The length of the major axis is basically The sum of the distance of a point from the focus of foresight. Okay And of course, this is going to be 90 degree because it has been drawn on a circle So I can say half pf1 into pf2 Is going to be 26 Okay, so let me call this For the time being as x. Let me call this as y Because we can easily deal with x and y so 15 And this makes x and y as 52 So if I want to find out, uh, let's say f1 f2, let me call it as z. Can I say z is nothing but Following a Pythagoras theorem like this x z square is x square plus y square Which is actually nothing but x plus y the whole square minus 2xy Why I'm converting it to x plus y and 2xy is because I know these information with me So x plus y the whole square will be 225 minus 2xy will be 104. So that will give me, uh, 121 Okay, so z square will be 121 Okay Which actually means 4a square e square is actually 121 Which actually means 4 a square minus b square is 121 and I think this is what we needed. So your answer is going to be 121 So ashutosh you are correct surprisingly Even though you are not confident about your answer. That is also bad. You should be confident about your answer Because if you are not confident either you will not mark it Or you will waste your time confirming your answer that both of them is hazardous for your, uh, you know competitive exams Guys all of you know that a square E square is equal to a square minus b square because you can write e square as 1 minus b square by a square So from this you can actually come to this is it So we'll move on to the next question question number 12 Or if you know the meaning of summation x plus y by x minus y sin Square alpha minus beta If you want I can write down this in detail. It actually means x plus y by x minus y sin square alpha minus beta plus uh y plus z by y minus z Sin square beta minus gamma and z plus x by z minus x Sin square gamma minus alpha. So you have to prove that We have to prove that this term is equal to zero again. If you're done, please, uh, let me know you're done And if possible send me the solution on my id Which paper do you have on monday? chem Yeah, of course take your time Till at least four people have said I'm done. I'm not going to discuss this I'll send it to my whatsapp Okay, so Ashutosh, Omkar and Bharata are done. Okay, so let's let's discuss this now because since some of you have done So actually look and feel of this tells you that first of all I need to do something about this expression x plus y by x minus y okay So from this expression I can say x by y could be written as and uh theta plus alpha by tan of theta plus beta Okay, and let us apply componento and dividendo componento and dividendo because It's a easier way to obtain something like x plus y by x minus y. So that's going to be tan of theta plus alpha plus tan of theta plus beta by tan of theta plus alpha minus of tan theta plus beta Okay So basically I'm not going to write their expansions But I can directly say on the numerator over here You will end up getting something like sin of theta plus alpha plus theta plus beta And in the denominator you're going to get theta plus alpha minus theta minus beta. Okay You can actually expand it at your end and check whether you're getting this So this is going to give you this If anybody knows wants to know how we get this term from here Please let me know then I'll explain it If you're fine, I can proceed Is everybody fine? So let me take the denominator on the other side. So x plus y by x minus y times a sin alpha minus beta is going to be Sin of 2 theta plus alpha plus beta Now since you want sin square over here, I have to additionally multiply with sin alpha minus beta Okay, now I can actually use my transformation formula. I can write this as twice This is to half cos a minus b cos a minus b That's going to give you I think 2 theta Plus 2 beta Okay, and minus cos Plus b Okay Similarly without wasting much time I can claim that y plus z by y minus z sin square beta minus gamma This can be written as half cos 2 theta plus 2 gamma minus cos 2 theta Plus 2 beta and not only that we can also write z plus x by z minus x sin square Gamma minus alpha as half Cos 2 theta plus 2 alpha Minus cos 2 theta plus 2 gamma And when you add these three when you add these three what happens So when you add these three expressions, you realize that This term gets cancelled with this this term gets cancelled with this this term It's cancelled with this getting you zero as the final answer, which is your RHS And hence proved So moving on to the next question now question number 13 This is coming from the chapter parabola The line ab makes an intercept of a and b on the coordinate axis is find the equation of a parabola passing through a and b And the origin if ab is the shortest vocal chord. Okay, just let me know when you're done So knowing to simplify you just express your answer in a crude way. Whatever you You can actually best find by the way, how many possible parabola would be there? Oh, no, that's not correct. Omkar All right, there would be two answers Absolutely correct Show us your take your time. I'm I'm going to give you another five minutes to do it. Don't worry So let's say this is your line a comma zero. This is zero comma b Ab the shortest vocal chord. So this should be your latter sector Okay, so one can be possible like this another can be possible like this No, but it has to pass through the origin as well, right If it is like this, the other one cannot pass through origin So let's say this is point a this is point b It is clear that the midpoint will be your focus. Let's say a by two comma b by two Yeah, I have a feeling that they can be only one such parabola Because if it is passing through origin, I cannot draw the other one, right? Just correct me if I'm wrong The origin is the vertex No, even if the origin is the vertex It will not work, right? Okay. Anyway, let's let's figure out the equation of this parabola at least Okay, so basically This would be your axis and we can have a Or Directrix like this Let me call this as d Let me call this as m Now first of all this line will have an equation line ab will have an equation of x by a plus y by b equal to 1 Right, so if directrix is parallel to the lattice rectum This this directrix will have an equation something like this x by a plus y by b is equal to some lambda Because they are parallel lines They will only differ in terms of their constant terms Okay Now what is the length bd? We can know that bd and bm are equal right because of the definition of the parabola itself So bd length is going to be half of ba which is half of uh under root of a square plus b square And this is equal to bm bm would be if I'm not wrong 0 by a Plus b by b minus lambda mod by under root of 1 by a square plus 1 by b square So if I take it up here, it will become half of a square plus b square by ab and here I will have uh mod of 1 minus lambda now from here We can note that lambda value has to be Will lambda be positive or negative? Please note that origin lies here So the point where it cuts the x-axis should be negative, right? So lambda should be negative So here if I get rid of lambda, I'll get a square. Sorry if I get rid of mod I'll get the two possibilities. What is plus minus this? So lambda would be 1 plus minus a square plus b square by 2a b Which gives me 1 2a b plus minus a square plus b square by 2a b okay So the only possibility is that it can be negative of a minus b whole square by 2a b. This can be a lambda Because this is only the negative quantity. So this entire thing is a negative quantity Right, so once I've got that means I have actually got my directrix This is my directrix And once you know the equation of a directrix and your focus, let's say, you know your focus is h comma k and directrix is Lx plus my plus n equal to zero, you know the equation of the parabola. What is that? x minus h the whole square plus y minus k the whole square is equal to lx plus my plus n the whole square by l square plus m square, right? Just try to recall this the general form so here my h was a by 2 k was b by 2 And the directrix equation was this ugly equation whole square By 1 by a square plus 1 by b square. Okay, it's fine. We can stop over here. So this will become our answer But why did you assume the vertex to pass through it? Where did the question say that the vertex will pass through it? That's a I would say loss of generalization Okay, that would be a loss of generalization actually Let the situation of the question say that it's your vertex. You cannot assume origin to be vertex Because the question never says that question says it is passing through origin I think the main purpose of giving the concept that it is passing through origin is to you know, ensure that you choose only one lambda I mean we started thinking that there are two lambda possible But later on we realized that the question purpose was to you know Make us choose only one of the lambda It can be but I mean How how are you sure that it will be at the origin? Is there anything from the expression that tells you it is the vertex is at origin? I mean, I'm just asking you a question. I'm not challenging you I'm just saying is there anything in the question that tells you that origin can be your vertex Right, if it is it should come out from the question itself So we'll move on to the next question that is your 14th question again coming from tignometry I don't think so you should make such assumption because it's a loss of generalization Or you're saying if they had made equal vertex equal intercept Of course, it will pass through origin because see then what will happen Chauhan you can see from our answer itself. Okay, I'll go back to the previous question See what will happen over here? If you're saying a is equal to b Chauhan, what will happen? This will automatically become zero If it automatically becomes zero Then direct tricks will be passing through origin that I can say I cannot say about the Vertex because then this quantity will become zero here You can say direct tricks is passing through origin, but I cannot still say the The Vertex will pass through origin Okay All right, so let me move on to the 14th question I'm sorry. I will not be giving you any break because we have three questions to be covered and we have exactly 30 minutes So I'm assuming 10 minutes per question will it will take Okay, 79 by 18 That's not correct As you told you are very close, but that's not the correct answer Oh, you're getting the same answer. Okay So then let's wait for this to Respond others Gaurav Omkar Bharat Chauhan Kavya Shishant Sreeti Extremely weird answer way too high. Okay Okay, first of all, uh, if you see x plus y Can I say x plus y will be sine to the power 5p Cost to the power 5p by Sine square p Cos square p cat Now somehow focus on getting these two terms Now denominator is quite simple because if you square this up You get sine 2p as sorry sine p cos p as 1 4th minus 1 By half That's minus 3 by 8 Okay, so denominator will be 9 by 64 Now, what about the numerator? Now think as if you have been given a plus b And you have to find a to the power 5b to the power 5 Okay, so, uh, what I can do is first I can just do a long division So a to the power 4 will give me a to the power 5 plus a to the power 4b Which is minus a to the power 4 And You can put minus a cube b So minus a to the power 4b And you will have minus a cube b square. So that will give you a cube b square Now we can have plus a square b So a cube b square and plus a square b cube That will give you minus a square b cube And I can see a pattern building up minus a b cube Minus a b to the power 4 a b to the power 4 plus b to the power 5 and you'll have finally b to the power Okay, so this will be a quotient. So I can write Correct me if I'm wrong. I can write sine to the power 5p plus cos to the power 5p as sine p cos p Times Sine to the power 4p I'll group this also Cos to the power 4p. So this is your these two terms are taken care of okay, here I'll get minus If I take a b common that is sine p cos p I will get a square plus b square a square plus b square is going to be 1 and I will get sine square p cos square p Now here also I can say this term over here. I can write it as Sine square p plus cos square p the whole square Minus 2 sine square p Cos square p So this minus will make it one less Now this is half. This is going to be 1 This is uh If I'm not wrong Minus 3 by 8 whole square which is 9 by 64 And this is going to be plus 3 by 8 This is going to be a sine To the power 5p cos to the power 5p now if you're dividing it by sine square p cos square p means you're dividing by 9 by 64 so Just to simplify it. I'm assuming a factor of 64 As the lcm So the answer will be 18 And you'll have 83 minus sorry 88 minus 9 which is 79 Oh, that's correct. Actually Ashutosh. I'm sorry Yeah, that's correct. Let's move on to the next question question number 50 No, I'm not expecting it to solve within two to three minutes. Yes, they are Five minute questions. I think it'll take it easily five minutes. This is a super easy question I think you should solve this within two minutes Is it half? Are you sure? Okay, chalo we'll check Anybody else who wants to answer I got one from Omkar one from Priti one fourth From Ashutosh again one from Kabya If not, I can start solving it This I can write it as 1 minus cos x square by 2 and this is 1 minus cos x square by 4 Right, that is the expansion. In fact, that is the factorization of this Now this expression is as good as saying two sin square x square by two And this is two sin square x square by four right Now in order to neutralize this In order to neutralize this I would need an x square by x x square by two whole square, right And to this neutralize this I need x square by four whole square So what I will do is I will just Write this as x square by two whole square and make a four And x square by four whole square and make a 16 Okay Limit extending to zero. So this will be neutralized To one This will be neutralized to one So I think two two will get cancelled with this This this will cancel. I think half is the answer One by 32 Let me check So so so so so so I got I got the problem. This is actually Four and this is eight Okay. Okay. Okay. Now probably yeah, you may get the difference answer. Sorry So this is going to be Four and this is going to be 16 in that case And this is going to be eight and this is going to be 64 in that case Okay, so this will be 16 and this will be 64 Yeah, now the answer is going to be one by 32 One by 32 will be the answer So the first one to got this right was surprisingly Trideev Trideev is from class 11th Hsr Great. So we'll now move on without wasting much time To the last question which is on hyperbola So slowly I'll draw the diagram for you as well so that you know, you can actually connect things So we have One line y equal to mx minus one. So mx minus one would be Passing through this. So let's say It has the slope and we have y x equal to 2 y x equal to 2 y will be passing through or then like this And we have y is equal to minus of 2 x y is equal to minus of 2 x somewhat like this So this is y is equal to mx minus one. This is x is equal to 2 y this is y is equal to negative of 2 x So This is a point and this is b point We have to find the locus of the centroid of oav and prove that it is a hyperbola passing through the origin Yeah, i'm just drawing the curve graph Seven i'm not Two to three minutes at least Can you see the screen visible? Okay, now let's just come back Is anybody done this type in done so that we can start discussing this That was an honest confession. I did not get it. Okay. Shoshan thinks he is at the last step Okay, others Gaurav, Bharat, Preeti, Kavya, Ravan Chalo guys Time to discuss So let's say this is origin and let's say this is a point x1 y1 and this is the point x2 y2 First of all x1 y1 and x2 y2 both lie on Uh y equal to mx minus one so I can say first of all They will satisfy the equation of y equal to mx minus one, okay Let me call this as condition number one correct now, uh Can I say x1 y1 is obtained from the intersection of these two lines? So I'll solve them simultaneously so, uh x by two is equal to mx minus one so at x1 So I can say x1 Half minus m is equal to minus one. So x1 is equal to one by In fact two by Two m minus one, okay x2 is again obtained by intersection of Uh this line y equal to mx minus one and y equal to minus of two x So minus of two x2 is mx2 minus one that means one is equal to m plus two x2 That means x2 is one by m plus two This is also important information for me because Let me call it as two rather than three Because I would need x zero x1 x2 to get the x coordinate of the centroid Okay, so let the centroid be h comma k So let centroid be H comma k so I can say H will be nothing but x1 plus x2 by three and k will be y1 plus y2 by three But I don't have to work hard for y1 and y2 because I know that y1 and y2 can be obtained from here Right if you add y1 and y2 here You will get m x1 plus x2 minus two by three Which actually means that I can write x1 plus x2 as three h and replace it over here And replace it over here So it implies k is going to be Three h m minus two by three Which means I can get m as three k plus two by three h We also know x1 and x2 in terms of m So here I can use Three h is equal to x1 plus x2 So x1 is this term and x2 is this term correct And the m over here I can use it here So it's two times m minus one plus one by m Plus two see what is the agenda behind doing all these things see neither. I know x1 neither. I know x2 neither. I know m So I want an answer which is independent of these. I just want a relationship between h and k. That's it That should be my agenda Are you getting this point? So in order to get the relationship between h and k I am doing all these things So I can say here three h is going to be uh, correct me if I'm wrong. Uh, it'll give me six h by In the denominator, I will get Uh six k Plus four minus three h And here also I'll get three h and in the denominator. I'll get three k plus two plus six h In fact three h and all I can cancel off from both the sides So let me now simplify it In fact, I can replace I can generalize here By replacing h with x and k with y And when I do that I get one is equal to two by Six y plus four minus three x And one by three y plus two plus six x Okay, so let's let's multiply First of all, is it passing through origin? Let me check that because uh, then only it will be a curve passing through origin Yes, I can see that because when you put a x zero and y as zero you'll get one is equal to two by four plus one by two That's that's actually true. Okay Now, let us see whether it's a hyperbola or not In case of a hyperbola you have to simplify it my god, I'm Too scared to multiply this because it's a huge expression anyways, let me just try to be as Brief as possible So this is going to be two times three y plus Two plus six x plus six y minus three x plus four okay Yeah, so let me expand terms in such a way that I pinpointedly get terms so x square terms So x square terms will be minus 18 x square from here Okay, y square terms y square terms will be 18 y square okay Will I get an x y term? Let's check x y term will be 36 36 minus nine. That's it. Which is 27 x y Okay, will I get an x term x term will be uh minus uh six x from here If I'm not wrong And from here I will again get uh 12 12 x minus three x which is nine x And 24 so I'll get a 24 from here And minus of 12 x plus three x So how much is this? 27 minus 18 plus nine x Okay, y term will be 12 y Plus 12 minus six minus six so I'll get 24 minus 12 24 minus 12 That's going to be 12 y equal to zero Oh wonderful. I can drop a factor of minus three throughout so I'll get six x square minus six y square minus nine x y minus three x minus four y equal to zero now for it to represent For it to represent a hyperbola Oh God have you sent something to me on whatsapp? Now for it to represent a hyperbola. I should just ensure that Delta is not equal to zero Okay, and h square should be greater than ab Now at least this part I can verify h square will be nine by two whole square Okay, is it greater than ab? Is it greater than six into six? I don't think so, right? Oh, sorry six into minus six Okay, so of course it is going to be true And delta of course you can verify so please try this on your own I'm sure it is going to represent it hence the question Okay So guys we have already exceeded eight minutes anyways Thank you so much for uh coming online for this youtube live session and all the best for your upcoming semester exams Please be confident. I'll be sending some more papers on the group Okay, at least three papers I'll send please time yourself And solve them If possible try to send me a solution also so that I can give you some feedback A couple of tips, please do not try to go for a semester exam in j mode. Okay Try to be as descriptive and elaborative as possible Use proper mathematical notations. Don't jump essential steps. Okay Try to ensure you are getting the independent and the method marks even if you're not getting the answer Independent is something where you get a marks just for writing a formula related to that topic Method is where you get a marks writing a formula which is helping you to solve that question So normally answer carries just one or a half a mark and rest of the marks are allotted between independent and method marks Okay, so, uh, please ensure you are writing all the steps so that you you get both these marks and please Do not spend more than 1.5 minutes per marks So that's that's actually the analogy that we normally keep 180 minutes 100 marks means 1.8 minutes per mark So reduce it by a factor of let's say, you know 90 percent because of course you'll be distracted and here and there So 1.5 minutes per marks. So if it's a two marker, don't spend more than three minutes So guys, thank you so much all the best Make Center Academy proud over and out from our side. Bye. Bye