 Hello all green from Centrum Academy. So today is going to be a problem-solving session on binomial theorem and This is a very important chapter for you in class 11th And of course one question comes every year in J main on binomial theorem So in light of that We'll be having this session today. So meanwhile those who have joined us now Please type in your name in the chat box so that I know who are attending the session All right, so let us start with this question on your screen over here. That's the first question for the day in the expansion of cube root of 4 plus 4th root of 1 by 6 whole to the part 20 We have to answer which of the following options are correct. Remember here more than one option may be correct Because we are also targeting J advance a pattern of questioning So the number of irrational terms and rational terms is what we need to figure out in this question So as a routine, I'll be giving you around two to two and a half minutes three minutes to think about it and then we'll be discussing the solution and Please feel free to type in your response in the chat box Whichever option you feel is correct. Please feel free to type in Hello all. Hello, we just with the kushal Nikhil Vaishnavi Purvik Good afternoon. Hello, Lalitha Okay, so I have started getting the response number of rational terms Okay, hello, Sonia Yeah, so as I already told you more than one option may be correct in this case So be very very careful in answering it in 2015 J in more than one option correct If you don't mark all the options you may end up getting a negative marks Okay, others I Can see the responses from only a few of you Hi, Shruti guys to begin with I am I'm sure all of you know the expression for the R plus one-th term R plus one-th term is Ncr Okay The first expression to the power of n minus R and the second expression to the power of R right Just for you to recall. This is the expression for Any R plus one-th term in an binomial expression. So guys, let me tell you The solution for this up till now I can see only one answer to be correct So one of you has given the right answer Just one of you has given the right answer so far. Okay. Now now I've started getting the response. Okay, great So guys, let us begin by solving this problem So NCR of course first of all the total number of terms so the number of total terms is going to be 21 in this case right because the number of terms is going to be always one more than the exponent of the binomial expansion correct and now when we write the R plus one-th term R plus one-th term can be written as 20 CR 4 to the power 20 minus R by 3 and 6 to the power of Minus R by 4 Okay But it will try to express it in terms of powers of 2 and 3 because 4 and 6 are not prime numbers So what we'll do is we try to express this in powers of 2 and 3 So when I do that I get 20 CR Okay, 2 to the power of 2 times 20 minus R by 3 and again here I'll get 2 to the power minus R by 4 into 3 to the power minus R by 4 Okay Hi Amog Yeah, now collating all the powers we get 20 CR 2 to the power of 20 minus R by 3 minus R by 4 and 3 to the power minus R by 4 okay So collating the power on 2 it gives you 2 to the power of If I'm not wrong, it'll give you 8 into 160 minus 8 minus 3 is 11 R by 12 Okay, and 3 to the power minus R by 4 Now for you to have a rational term For you to have a rational term the powers on these prime numbers must be whole numbers that means 16 Must be integers that means 160 minus 11 R by 12 This must be an integer and so must be R by 12 Sorry R by 4 right Now from here, I have to pick up such ours which lies between 0 to 20 because as you can see power is 20 My R cannot exceed this interval So 0 to 20 is the value of R that I should I should be looking out for so I can actually count them It will be 0 4 8 12 16 and 20 Okay Now out of these I need to check which one of them will also make This fellow as integers. So we have to account for that also So I'm sure 0 is not going to help us Connect 4 44 and this node is not going to help us What about 8? What about 8? 160 minus 88 That's 72. Yes. So 8 is something which I'll be choosing So a is fine What about our 12 12 no, it will be 160 minus 132 which is 28. No What about 16? What about 16? No, I think 160 minus 16 will give you 176 which is minus 16. No, this is not an integer as well What about 20? What about 20? So 160 minus 220 by 12 minus 60 by 12. Yes, it's an integer, right? So in this case In this case, I would realize that only 8 and 20 meets the requirement of both of them being integers So both the requirements are met only by 8 and 20 Therefore, therefore, there will be only two rational terms Which implies There will be 19 irrational terms There will be 19 irrational terms Okay Now remember here The middle term Since n is n is equal to 20 Right, the middle term would be obtained by The middle term is your n plus 1 by 2th term n plus 2 by 1th term is your middle term n by 2 Plus 1th term is your middle term, right? This is your middle term, correct? That means out of 20 21 terms that we have the 11th term is the middle term And the 11th term is obtained by putting r as 10 And r equal to 10 does not appear in our rational term So middle term has to be irrational Okay Now number of irrational terms is 19. That is correct. Number of rational term is 2. That is also correct Now what about the ninth term ninth term is obtained when you put r as 8 correct And r equal to 8 is appearing in my rational terms. So basically the ninth term is also Rational in nature. So all the options a, b, c, d are correct So I could see Kushal, Sai and Purvik only answering these questions. Hello, Nitya Is that fine guys? Okay, a good problem to start with Now let us move on to the second problem of the day Guys try to be as accurate as possible. Don't be in a hurry to answer You will not be awarded marks for being fast Okay Next question is based on multinomial theorem find the coefficient of x to the power 8 b to the power 4 c to the power 9 d to the power 9 in the following expression In this expression, we have to find the coefficient of this term No problem. Hello, Vaishnavi So just to quickly recall what's a multinomial theorem for those who have already forgotten it Let's say if you have more than one variables, let's say x1 plus x2 plus x3 all the way till xr raised to the power n Then this expansion is written as summation of n factorial by alpha 1 factorial alpha 2 factorial all the way till alpha r factorial Where alpha 1 is the exponent of x1 Alpha 2 is the exponent of x2 and so on till alpha r is the exponent of xr Where all your x1 x2 etc. They are whole numbers And their sum has to be equal to n Right just a quick recall for you guys if in case you have forgotten the multinomial expression i minus v Okay, so purvik has given an answer Okay, fine. I'm not saying it's right or wrong. Let's wait for the others to chip in This is a j e main level problem So I'm expecting all of you to give me the right answer in this case. It's not a difficult problem Okay, which is also says the same kushal Others fast. I want to see your response Doesn't matter Don't shy away from answering However, don't hurry up because accuracy is very important. I keep on repeating this In j exam. It's very important to be accurate Just solve 60 65 problems, but it should be absolutely correct Okay, shruti says d nikhil again says c nitya minus v rohan amog shweta nishil lalita Manasvi also says d. Okay. Okay. So almost everybody has started responding now So guys, let's discuss this There are two ways to solve this problem one is, you know, if I have to Okay, so I would say let Let a term like this that is n factorial by 10 factorial by alpha 1 factorial alpha 2 factorial alpha 3 factorial alpha 4 factorial determine Determine the coefficient of such a term which simplifies to give you The required expression so let's say alpha 1 alpha 2 alpha 3 and alpha 4 are such powers that you raise on these four terms, okay Such that it collectively gives you It collectively gives you something like This that means I get the desired expression By expanding this term right So guys, if you collate the powers of a you realize that a would be subjected to power of alpha 1 plus alpha 2 plus alpha 3 B will be subjected to the power of alpha 1 plus alpha 2 plus alpha 4 And c will be having a power of alpha 1 alpha 3 alpha 4 And d will be having a power of alpha 2 alpha 3 and alpha 4 Okay And of course this term I can only evaluate when I know my alpha 1 alpha 2 alpha 3. So let us just for the time being write it like this Okay, now these powers we are These powers we are Comparing with the powers given to me in the question so Apart from this We also should be knowing this that the sum of all alpha 1 alpha 2 alpha 3 and alpha 4 should add up to give you 10 because We are raising this multinomial to the power of 10 Right Now this let's say this is our first equation Second equation is when you start comparing the power of a on both the sides. So this power is going to be 8 Okay in a similar way Alpha 1 plus alpha 2 plus alpha 4 is going to be 4 Alpha 1 plus alpha 3 plus alpha 4 is going to be 9 And alpha 2 plus alpha 3 plus alpha 4 is again going to be 9 Now because of this first expression It becomes super super simple to solve for alpha 1 alpha 2 alpha 3 from these five equations So I can clearly see from these two that alpha 4 is 2 Correct from these two I can figure out that alpha 3 is 6 Correct again from these two I can figure out that alpha 2 is 1 And again from the last and the first expression I can figure out that alpha 1 itself is 1 right So that makes our life so simple. So our answer would be 10 factorial by 2 factorial 6 factorial 1 factorial 1 factorial right that's 10 into 9 into 8 into 7 And I'll have two factorial down over here. So that's going to give you 5 That's 360 into 7. That's going to be 2520 Okay, so that makes your option number option number C correct Well done to those guys who answered this correctly However, there was an easier method to do this whenever you get this in a shortcut Where you can take a b c d to the power of 10 out Right that leaves you with a inverse b inverse c inverse d inverse to the power of 10 right Now guys, you know that you already have a to the power 10 b to the power 10 c to the power 10 d to the power 10 out right Yes or no So in this multinomial expression in this multinomial expression, you just desire a to the power minus 2 Right, which is actually this You desire b to the power of minus 6 which is actually same as this and you desire c to the power of Of Of 1 I guess that is minus 1 to the power 1 and again d to the power of minus 1 to the power 1 right So so this term will occur along with 10 factorial by 2 factorial 6 factorial 1 factorial 1 factorial So directly this becomes your answer in a much faster way. That's again going to give you 2520 So I would recommend this method Okay, so this method is suggested for you to solve the problem. The other method becomes slightly lengthy Yeah, yeah, same thing same thing. I'm okay. So that becomes 2520 Well done to the guys who answered this now guys moving on to the next problem Oh, I'm sorry. I think I missed one slide. Let me go back. Okay So here we have a very big problem in front of you It's a match the column problem And of course, let me tell you this is a je advanced level of problem But still let's give it is give it a full try It's equivalent to solving four questions So as you can read The party of the question says column one party says The last non-zero digit in the sum two times Summation The double summation one acting on i another acting on j Remember i is always less than equal to j and both are going from 0 to 10 And you have i into j i into j 10 c 1 10 c j Hope you can all see the problem clearly on your screen Hello, Sai. That is a good problem. Just take your time. Don't rush through it Right now. I'm not disclosing any answer for which so you can try it out And then we can we can discuss it. We'll go step by step I'm giving you some time to do the part a okay, and then we'll discuss a then you can start attempting part b Guys, everybody is clear with what does this mean? What does double summation mean? One summation is working on i another summation is working on j right People who have been into coding will understand an nested for loop is something like that, okay Where i is going from 0 to 10 and j is going from i to Whatever number it's going to okay So i starts from 0 to 10 and j starts from whatever is the position of i till it goes to 10 Okay, so at least for the first one, let us start solving the first one because It's going to be a long problem because it has four problems in itself so uh Guys, I would just like you to understand this simple thing Uh, if your i and j are operating If your i and j are operating independently Let's say i also goes from 0 to 10 and j also goes from 0 to 10 Okay Of any expression of any expression Okay, then this expression would contain those cases where First i and j are equal Right, it would consist of those cases where i and j are equal So i is equal to j and both are going from 0 to 10 correct Plus it would consist of those cases where i is Less than j and it is less than 10 Correct plus it would consist of those cases also Where j is less than i and both are less than 10 correct So these are the three possibilities that come when i and j are operating independently Is it fine? Is there any question with respect to this that this actually summation would be equivalent to saying When i and j are equal correct Okay, then Summation when i is less than j Plus summation when j is less than i so only there are three possibilities either they're equal or i is greater Or j is greater right Now in those type of functions In those type of functions of i and j Where you realize that this function is symmetrical with respect to i and j This function is symmetric with respect to i and j. What does the meaning of symmetric with respect to i and j means If you swap the positions of i and j it doesn't make any difference to the function Then in those situations guys listen to this very very carefully These two would be same These two would be same These two will be identical in values. Okay If this is understood then we can start solving the actual problem Anybody having any doubt with respect to this argument Whatever argument I have given is there anybody who is not convinced with that Please let me know because this is going to be the basis of solving Problems where double summation is involved Yeah, I'm talking about the problem in general Okay Now what the problem that is given to me here is finding this fellow right so we have to find s So let's say I call this as I call this as s Okay So can I say this will also be my s Right So what i'm trying to say here is that What i'm trying to say here is that summation from 0 to 10 on i summation from 0 to 10 on j of this expression i dot j n c i n c j will be equal to Will be equal to Summation i equal to 0 to 10 When i is equal to j when i is equal to j. Can I say the expression will become i square n c i square Because i and j are equal So i into j will become i square n c i and n c j are equal. So they'll become n c i square And I will have this as 2 s. So let me write it as 2 s Okay Now guys here This is super simple to evaluate. It is like finding summation of i n c i From i equal to 0 to 10 and multiplying it with summation j n c j from 0 to 10 Okay, so both these results I just have to multiply in this case Whereas this is also easy to evaluate which we'll be evaluating now Okay So once we get these two Once we get these two We can always make s the subject of the formula and get our answer. Okay Now let us go step by step How do I find this expression? What is summation? R n c r What is that anybody knows the result for this? Are you guys aware of the result for this? By the way n is also 10. So I can you know use that 10, but I'll use it later on Okay, as of now i'm just summing up Uh, I'm just planning to sum up r into n c r From r equal to 0 till n. Okay later on I'll put the value of n as 10. That's later Not to worry right now So does anybody know this summation? If you don't know this summation, I would like you to recall that uh If you all recall this expansion of 1 plus x to the power n It is c 0 c 1 x c 2 x square c 3 x cube and so on till c n x to the power n correct By the way, I'm not writing n over here, but if you want you can write it Okay Now differentiate with respect to x. So when you differentiate with respect to x, this is what we see So I'll get n c 1 Okay, or you can say 1 into n c 1 Then 2 into n c 2 x Then 3 into n c 3 x square And so on till n into n c n x to the power n minus 1 Okay Now put x as 2 If you put x as 2 We get n into 2 to the power n minus 1 as this which is nothing but summation r n c r Okay from r equal to 0 to n remember 0 term would be 0. So don't need to bother about it So guys here. I know the answer for this which is n 2 to the power n minus 1 into n into 2 to the power n minus 1 right no doubt So far in evaluating the left-hand side of the equal to Okay Now guys, let us evaluate this now Let us evaluate this now that is summation of r square n c r square from r equal to 0 to n Okay, does anybody have any idea how to do this? Please type. Yes, if you have an idea how to do it So this one problem is sufficient enough to teach us a lot of concepts Okay, if nobody has any idea about how to do this very simple. Let us Look in the look here We all know that let's now be derived that n into 1 plus x to the power n minus 1 was actually n c 0 Oh, sorry my bad. It was n c 1 Then uh 2 into n c 2 x 3 into n c 3 x square And so on till n into n c n x to the power n minus 1 right Now what I'll do now I'll replace my x with 1 by x In this expansion, I'll replace my x with 1 by x so of course this becomes n c 2 into 1 by x this becomes 3 into n c 3 1 by x square And so on till n into n c n 1 by x to the power n minus 1 correct Now guys In the multiplication of these entire terms, let us try to multiply these terms Okay, if you multiply these terms you will get n square 1 plus x to the power n minus 1 into 1 plus 1 by x to the power n minus 1 correct And you'll get lot of terms Among those terms there would be a term Which would be obtained when you multiply these two together These two together these two together And so on till your last term And that would be a term which is basically a constant term right So your n c 1 square And 2 square n c 2 square etc Would be coming along with A term which is actually independent of x correct So apart from so many terms there will be one term which is going to be independent of x Okay, in other words in other words if I look for The coefficient of that term which is independent of x from here Then I'll be able to get the desired series So I have to find a term which is independent of x from here Right So let us now find the term which is independent of x from here So when you simplify this it actually becomes 1 plus x fold to the power 2 n minus 2 By x to the power n minus 1 correct So if you want your term to be independent of x can I say I'm looking for I'm looking for the coefficient of x to the power n minus 1 in 1 plus x to the power 2 n minus 2 correct Which is what? Which is what 2 n minus 2 c n minus 1 correct Yes or no So the term which is independent of x in this term would be The term which is independent of x would be finally n square Into 2 n minus 2 c n minus 1 are you getting this point? Right So now that we have found your Answer for this also So let's replace it back So we have found the answer for this also which is n square into Into 2 n minus 2 c n minus 1 Now we can go for finding the value of s Correct Meanwhile, you can also substitute your n as 10 back in this Okay So your 2 s expression will become Will become n square 2 to the power 2 n minus 2 minus 2 n minus 2 c n minus 1 So this is going to be your 2 s Okay Now put your n value as 10 So let's put your n value as s n value as 10 So 2 s will become 100 times 2 to the power 20 minus 2 which is 18 minus 18 c 9 Correct Yes, I know So basically it will become 50 times 2 to the power 18 minus 18 c 9 Okay Now guys the question says what the question says I'll repeat it once again. Don't worry as of now Let me just complete the problem Shweta and Nitya I'll come back to you guys Sorry, I'll come back to you girls. So, uh, let me just finish it off as of now Now 2 to the power 18 if you see See this will be a number which is multiplied to 10, right? Yes or no So last digit has to be last digit has to be zero Correct Now I want to see what is the last digit of these numbers. So last digit of these numbers Will actually determine what is going to be the first non zero last term, right? So last digit of 2 to the power 18 if you see 2 to the power 18 follows this pattern Uh, 2 to the power 18 if you divide it by 4 you get a remainder of Reminder of this is going to be 2, right? So your last term of this will be 2 to the power 2 So 2 to the power 2 will be the last term which is actually 4 Okay, so from this the last term will be 4 But 18 see 9 If you see 18 see 9 You are basically writing 18 factorial by 10 factorial in sorry 9 factorial into 9 factorial so you are writing 9 factorial by 9 factorial Okay, and this will definitely end with a zero this will end with a zero Why it will end with a zero because If you write this 18 to 10 You will have something like this Right Yes or no, so this 10 once it makes an appearance it will end up with a zero correct So basically the last digit of these two guys will be zero plus four which is going to be four And hence the last digit of this entire expression will be four. So this is going to be the answer for The first part which is part a Now i'm going to explain this again So nitya doubt was how did I get the term independent of x see nitya We already have x to the power n minus 1 over here, right? So if I get x to the power n minus 1 from here These both will cancel each other out Yes or no so the term x to the power n minus 1 or the term containing x to the power n minus 1 from this term Is going to be the term which is independent will be independent of x in this entire expression Okay, and that term would be 2n minus 2cn minus 1 And finally when you multiply n square you get this entire expression You end up getting the entire expression that I wrote it over here Okay from here. I found your 2s And then I put the value of ns10 Shweta, is your doubt clear or is it still there n square to the power of n square to the power of n minus 1 I didn't get that term If possible, can you write it and send me the snapshot on the whatsapp personal whatsapp? Okay Yeah, so this was a huge question But again, uh, it took some time because we were not We did not recall the formula for these type of series which you should actually remember Okay Good Now moving on to part b So a is mapped to q So this is mapped to q B anybody has been able to do b part Anybody who could do the b part Lawyer says b is r Okay What about others? Okay, let's attempt the b question anybody else who wants to give the answer for the b You have please feel free to give the answer for b part Just actually taking the entire screen Okay, so I just attempt the b part over here Now for the b part, uh, when we see this expression like cos theta plus nc1 cos 2 theta etc Till ncn cos 3 theta. Sorry ncn cos n theta Now this actually gives me a feeling about, uh, the demoverage theorem in complex numbers, right? So can I say this expression is actually the real part of this expression is actually the real part of real part of Um cos theta plus i sin theta Okay, plus plus nc1 cos 2 theta plus i sin 2 theta Plus nc2 cos 3 theta Okay, my mistake this n plus 1 should be here There should be n plus 1 here. Is that what the question says? Correct? Yeah, it goes till cos n plus 1 theta, okay So this is going to be And this sizzle will go all the way till you reach ncn Cos n plus 1 theta plus i n plus 1 theta So the real part of this whole giant expression would be the required series. Isn't it? Yes or no Now if you look at it carefully, it's actually the real part of this term is e to the power i theta correct This term is nc1 e to the power i2 theta Euler form now i'm converting it to Euler form And so on till ncn e to the power i n plus 1 theta If you take e to the power i theta common You'll end up getting 1 plus nc1 e to the power i theta all the way till ncn e to the power i n theta correct Which is clearly the real part of the expansion of e to the power i theta 1 plus e to the power i theta to the power of n Yes or no Okay Now once you have got till this stage It should become a simple thing for you. It should not be very difficult for you to solve at this stage Now you may use your polar form if you want Let's use the polar form I'm looking out for space now. Yeah, let's go on to this end I'm sorry. I'm juggling around the pages because It's totally filled up Yeah, so I want the real part of e to the power i theta 1 plus e to the power i theta to the power of n Okay, so which is cos theta plus i sin theta And here I have 1 plus cos 2 theta plus i sin 2 theta whole to the power of n Okay, okay Yeah Now me you may use your half angles formula over here and half angle formula over here So this is 2 cos square theta by 2 Plus i into 2 sin theta by 2 cos theta by 2 Okay, and this term remains as such cos theta plus i sin Theta so take 2 cos theta by 2 out so it'll be 2 to the power n cos n theta by 2 And we have cos theta plus i sin theta And we'll be left with cos theta by 2 plus i sin theta by 2 whole to the power of n And we know from de Moivre from de Moivre's theorem This will become Cos n theta by 2 plus i sin n theta by 2 Right. I can use de Moivre over here Okay Now just find the real part of it out because real part can be obtained just by multiplying these two And just by multiplying these two Okay So that will give me we have to find real part everywhere. Please do not forget that real part of this Okay So the real part will be 2 to the power n cos to the power n theta by 2 And if you see cos theta cos n theta by 2 minus sin theta n theta by 2 that will be cos cos n theta by 2 plus theta Correct Which is going to be 2 to the power n cos n theta by 2 cos n plus 2 by 2 theta right Now guys the question says you have to find the value of You have to find out the value at an angle where theta is pi n plus 2 Okay, so if theta is Pi by n plus 2 If theta is pi by n plus 2 it means n plus 2 theta is going to be pi And divided by 2 is going to pi by 2 So this is going to be pi by 2 thereby making everything a big zero Thereby making everything a big zero So lohitya Well done your answer was correct good. So basically this is Is Mapped to r this is mapped to r Okay, great anybody else Who has tried out Who has tried out uh part c? Anybody who has tried out part c? Yeah, see my purpose right now is to you know revise the theory with you I know such a difficult question may not come as a the question for je advance Even it's difficult for je advance level But through this question my purpose was to you know, give you different dimensions of this chapter Okay, so don't worry about the time right now. I know it's going to be taking more than 20 minutes of time Anyone who has got the answer for c part? All right, so for c let us start the discussion and I'll discuss it on a separate page so Yeah, so c says those who have Not written down the question the question says in the expansion of 1 plus x square to the power of n Which can be written as a 0 a 1 x a 2 x square And so on till a 2 n x to the power 2 n find the value of a 0 a 1 a 3 a 4 A 6 a 7 and so on Okay Now first of all You need to ask yourself. What is the pattern that you see in the listing out of these numbers? What is the pattern over here? What is the pattern in which it has been written? I can see a pattern a 0 a 3 a 6 the next term would be a 9 etc Okay And a 1 a 4 a 7 next term would be like something like a 10 Okay, and so on. This is the pattern Denominator the pattern is quite obvious a 2 a 5 a 8 Okay So basically they are jumping by a difference of three each Correct. Yes or no right now Sai has given an answer Sai and lohitia Both of you say c is s, right? Let's check guys. Okay. Now guys in this case we have to put x value Since they are they are being added in a gap of three You basically would realize that such substitutions will be helpful Right Where omega is basically the complex cube root of unity. We all know that one plus We know that one plus omega plus omega square etc is giving you zero, right? So what will happen? This will end up eating up the terms Which are at a difference of three three each? Okay, so see how it helps So when we put x as one we get three to the power n as a 0 a 1 a 2 a 3 a 4 a 5 a 6 etc a 2 the power 2 n Okay When you put omega you get One plus omega plus omega square to the power n which is zero to the power n as a 0 a 1 omega a 2 omega square Again, then you get a 3 a 4 omega a 5 omega square and then again a 6 like that will continue Right. This is when you put x as omega When you put x as omega square When you put x as omega square We'll again get a zero because now it will become one plus omega square omega to the power four which is again omega So this will again be zero to the power n So we'll have something like this Omega to the power four will be back to omega again Okay, a 3 Again, it will be a 4 omega square and again it will be a 5 omega a 6 etc Okay When you add it up You get 3 to the power n is equal to 3 a 0 will come Right, but this entire row This entire row or column you can say this will become zero Right because one plus omega plus omega square can be It tend to be as zero by this particular property Similarly, this will also vanish Correct, so I'll get 3 a 3 3 a 6 etc Correct, which implies a 0 plus a 3 plus a 6 etc Will be 3 to the power n by 3 which is 3 to the power n minus 1 Correct now How do I get a 1 a 4 a 7 etc? How do I get the other series? Now all of you please pay attention to this If you want to get the other series You multiply the entire expression with x square Okay, when you multiply with the entire expression with x square, this is what you get correct And again start putting the value of x as 1 Omega and omega square one by one So when you put one you will again get 3 to the power n which is a 0 a 1 a 2 Da da da da da when you put omega you again get zero But this time you end up getting omega square over here Okay This one will become omega cube means omega cube means again a 1 This one will become a 2 omega to the power 4 which is omega and so on Again when you put omega square you get zero a 0 omega again a 1 because omega Square whole cube is omega to the power 6 and omega to the power 6 will again be 1 Correct and this again will be omega square and so on So now when you add it you get 3 to the power n when you add all these three You get 3 to the power n as 3 times a 1 plus a 4 plus a 7 etc Which is nothing but a 1 plus a 4 plus a 7 is also Dot dot dot is also equal to 3 to the power n minus 1 And you visualize that same would be true even for a 2 plus a 5 plus a 8 etc So that will also be 3 to the power n minus 1 correct so ultimately Ultimately, what do I desire ultimately? I desire 3 to the power n minus 1 plus 3 to the power n minus 1 by 3 to the power n minus 1 isn't it because this expression on the numerator is made up of a 0 a 3 etc plus a 1 a 4 a 7 etc And in the denominator it is a 2 a 5 is 8 etc So both are all of them are 3 to the power n minus 1 So 3 to the power n minus 1 3 to the power n minus 1 by 3 to the power n minus 1 So that will give you the answer as 2 So that will give you the answer as 2 So let us go back to our slide I think it was the third question Yeah, so c will be mapping to s That clearly implies d will be mapping to p but not necessarily guys it may happen that uh Two of them may map to the same But nevertheless, this is a simple problem which you can anyway solve Okay So how do you solve this by using which rule? Which rule will be helpful for this case? Anybody has any idea Pascal's exactly that's the case of a pascal identity So guys, let me solve this hope you have copied this expression down because I'll be changing the slide now Okay, so let me go on to the next slide So we have to find 47 c 4 by 57 c 4 plus Since 57 c 53 is constant it can be pulled out And you have summation of 50 minus j c 3 From j equal to 0 to 3 plus 1 by 57 c 4 Summation of 56 minus k c 53 minus k from k equal to 0 to 5 Now Since all these terms are same 57 c 4 is same as 57 c 53 is same as 57 c 4. They all are same They all are same Okay, so what we can do is we can take it common outside. So 1 by 57 c 4 or 57 c 3 Whatever you want to write it you can take it outside So you have 57 c 4 And if you start putting the value of z over here j over here This will become a 50 c 3 Okay, and 50 c 3 plus 50 Sorry 49 c 3 48 c 3 And 47 c 3 Okay The other term will be and this is as good as saying 56 minus k c 3 So which will be 56 c 3 55 c 3 54 c 3 53 c 3 52 c 3 And 51 c 3 Okay now guys Watch here Pascal's identity. We all know that n c r plus n c r minus 1 is n plus 1 c r. This is called the Pascal's identity Pascal's identity Okay Now if you group up these two terms 47 c 4 and 47 c 3 Can I say it will become 48 c 4? Correct So this will become 48 c 4 And 48 c 4 and 48 c 3 will become 49 c 4 49 c 4 and 49 c 3 will become 50 c 4 and 50 c 4 and 50 c 3 will become 51 c 4 correct 51 c 4 and 51 c 3 will become 52 c 4 correct Then 53 c 4 54 c 4 55 c 4 56 c 4 57 c 4. So your answer will be 57 c 4 And outside 1 by 57 c 4 is waiting So they will get cancelled giving you the answer as one So basically As I already discussed with you For option d P will be correct Okay, so p will be correct for option d So it was a huge problem, but again we learned a lot of things from this So in case you have not understood anything Make sure you revisit this Problem once again And do feel free to ask me offline as well Okay So moving on to the next question guys So in this question we have to Sum up the series c 0 by n n plus 1 minus c 1 by n plus 1 n plus 2 etc Note there is an alternating plus minus sign here minus plus like that To n plus 1 terms Okay, of course there will be From c 0 to c n there will be n plus 1 terms Now this answer has to be expressed in terms of an integral This answer has to be expressed in the form of an integral Now I know you can always take up a special value of n and you know quickly evaluate these four integrals But I would request you here to solve it by the regular method Yeah, yeah, the usual meaning is nc 1, nc 2, nc 3 and so on like that. Yeah Yeah, this looks like a beta function, but beta function normally has this expression Beta mn is 0 to 1 x to the power m minus 1 1 minus x to the power n minus 1 Anyone any success so far trying? So because of the alternate sign you can see that 1 minus x would be involved, right? So this can never be your answer d okay A purvik what if I say more than one options are correct Yeah, Rohan same to you also more than one options may be correct in this case So purvik according to purvik it's c and d then right Only see sai only see kushal only see is correct Exactly. So c and d would be the same if you apply the king's property, right? So wishes also says c and d Rohan also says c and d, okay So many of you are in the favor of c and d because c and d matches All right guys, so let us just start the discussion of this, uh, you know, it's a very interesting problem So we'll start with the expansion of 1 minus x to the power of n Okay It's going to be c0 minus c1 x plus c2 x square minus c3 x cube and so on, okay Now because I have to generate n and n plus 1 Can I first multiply this with x to the power n minus 1 because I want to integrate this So when I multiply this I end up getting This expression like this Okay Okay, now when you integrate it Now when you integrate it you definitely end up getting Let's say you integrate it between 0 to 1 Okay, you end up getting c0 by n x to the power n minus c1 x to the power n plus 1 by n plus 1 Okay, and c2 x n to the power n plus 2 by n plus 2 and so on correct Now the problem is Now the problem is when you integrate this You need to integrate it once more, right? And in order to integrate once more We should know how to integrate this first of all, correct Yes or no And do you think it's simple to integrate? Can I easily integrate this? Is it easy to integrate? Is this fellow easy to integrate? Even if I apply King's property and all can I easily integrate it? Yes or no Can I integrate this term easily? Especially I'm asking this question to those who gave the answer right now If yes, what substitution should I use? Of course, I may go into a beta function But assuming that I don't want to use beta function So of course, you would realize it is not easy to integrate, correct? So the approach which I have taken right now Right will not work that means This step of integration may not help me Okay Because of course I have to integrate it once again So what is happening? This is a problem where J is restricting me from integrating it once more Right So J is restricting me for from integrating it once more Okay, but nevertheless I'll play a trick over here. I'll now again multiply with 1 minus x So I'll again multiply with 1 minus x So it's c0 x to the power n minus 1 into 1 minus x again c1 x to the power n into 1 minus x Plus c2 x to the power n plus 1 Which is actually this Okay And on this side, I'll get x to the power n minus 1 minus x to the power n minus c1 x to the power n minus x to the power n plus 1 c2 x to the power n plus 1 minus x to the power n plus 2 and so on Okay Now if you try integrating it from 0 to 1 Now if you try integrating this from 0 to 1 Okay, see what happens The first term will become c0 x to the power n by n minus x to the power n plus 1 by n plus 1 correct If you put your limits 1 to 0 see what comes out Similarly second term will become c1 times x to the power n plus 1 by n plus 1 minus x to the power n plus 2 by n plus 2 Again put your limits see what comes right and so on and so forth Okay If you see this clearly the first term gives you c0 1 by n minus 1 by n plus 1 Second term gives you 1 by n plus 1 minus 1 by n plus 2 Similarly third term will give you 1 by n plus 2 minus 1 by n plus 3 And so on Which is nothing but c0 By n n plus 1 minus c1 by n plus 1 n plus 2 Plus c3 by n plus 2 n plus 3 etc Correct Which means the answer that you should get will be integral of 0 to 1 x to the power n minus 1 1 minus x to the power n plus 1 dx So of course When you said it was option d sorry it was option c you were correct But apart from that also you will see if you apply king's property If you apply the king's property of integration This says integral 0 to a f of x is integral of f of a minus x Then you realize integral 0 to 1 x to the power n minus 1 1 minus x to the power n plus 1 could also be written as integral 1 minus x to the power n minus 1 And this will become 1 minus 1 minus x 1 minus 1 minus x to the power n plus 1 Which is 0 to 1 1 minus x to the power n minus 1 Times x to the power n plus 1 dx So your option number d also is correct in this case Your option number d also will be correct in this case So both these options are correct Again, it's a typical j advance problem Yeah, I'll show it. I'll repeat this once again First I have multiplied this expression by n minus 1 so it became this Correct And later on I realized that I cannot integrate once more because I am not able to integrate the result that I get over here So what I did again, I multiplied with 1 minus x. So here I multiplied with x to the power n minus 1 Here I multiplied with 1 minus x Is that fine swantharya? Now when I get the expression and I integrate it from 0 to 1 Both sides you'll start realizing that the required series starts appearing on the right hand side So you'll see the required series coming up over here Yeah, you can use beta function, but guys Then the problem is you you cannot integrate once more because the moment you put the value of x as 0 to 1 x will be lost, right? So you have to integrate doubly And double integration will will be like, okay, almost, you know, very difficult for you at this stage, right? Once you put the substitute once you substitute the limits of integration in that Your answer will come in terms of n, right? So x will vanish off, right? So you wanted to integrate once more so that n plus 1 also comes down, right? So that will not be able that you'll not be able to do That means you had to do something like this integral of of this You had to do this then Okay Yeah, and that would become very difficult. This would be very difficult to integrate Okay So that is why you have to use this approach Okay Guys, can you move on to the next problem now? We have only been able to do four problems so far In one and a half hours Okay, so we are into the fifth problem now I'll give you a break at around five o'clock If that is fine with you So this question reads if ac is greater than b square then some of the coefficients in the expansion Then some of the coefficients in the expansion of a alpha square x square plus 2b alpha x whole to the power n Where all a b c alpha are real quantities and n is a natural number Is positive if a is positive positive is c is positive Negative if a is negative and n is odd Positive if c is negative and n is even So please try this out and feel free to type in your response in the chat box So Gushal says c Yeah, multiple may be correct in this case as well Okay, so wishes says a and c With big says a okay So I'm not commenting upon right or wrong as of now So I'm here also Nishal also a Amog also a okay, Gushal Yeah, we'll discuss that we'll discuss about the discriminant Yeah, that's that's rightly pointed out by you discriminant is negative I'm just waiting in for a few more responses. Then I'll start the discussion Shweta Sondarya Aditi Loetia Manasvi Tapas Nitya Okay, so loetia also says c and c and d. Okay guys, let's look here. Uh, some of the coefficient is basically nothing but The same binomial expansion when you put x as one right when you put x as one This term itself becomes the sum of all your sum of all your Coefficients right Now this term is clearly a quadratic In alpha, so let's call it as f f of alpha Okay, so this will this is clearly a quadratic in alpha And if you see the discriminant of this expression that is b square minus four ac That means four times b square minus ac and since ac is since ac is Greater than b square this term would be negative in nature correct right Now we have known that in any quadratic expression This expression is always positive if discriminant is negative and a is positive correct And this expression is always negative If discriminant is negative and a is negative that means it has the same sign as a if the discriminant is negative correct Now Since the discriminant is negative and in the option number one it says a is greater than zero That means you will have a positive number. That means f of alpha will be a positive number So it raised to any power natural number will also be positive Correct yes or no, so let me call it as G alpha So g alpha is always positive that G of alpha will always be positive So option a is correct. No doubt Okay And if a is negative If a is negative Of course when a is negative and discriminant is negative this expression will also be negative So this will be negative If a is negative And if I raise it to a power of n or 2 n plus 1 whatever an odd natural number then of course this will be negative So Option c will also be correct Option c will also be correct Okay now guys Remember one thing over here if c is positive If c is positive Can I say a will automatically become positive because this is this is greater than b square Correct Isn't it if c is positive And since everything is greater than b square, which is anyways a positive number Can I say a will automatically become positive? So indirectly it is saying a is positive Correct So that means if c is positive implies a is positive and hence the entire expression will be positive. So b is also correct Uh, no Russian. Unfortunately, we are not providing any pendrive courses, but we are working towards it Definitely by next year you should see one coming out in the market Okay Now guys in a similar way Let us look at option number d If c is negative If c is negative What will happen? If c is negative, then can I say If c is negative, can I say a also has to be negative? Correct And if a is negative that means your f of alpha will be negative But if you raise it to an even power But if you raise it to an even power That is power of 2 m Can I say it will still become positive in nature? Correct Which means option number d is also correct so guys Because of rushing through solving the problem all of you gave the wrong answer to this So in this case the answer is all option a b c d are correct. Is that fine? Any question with respect to this? Great Let's move on to the sixth question before we all take a small break So here is the question in front of you c0 c1 c2 till cn be the coefficient of various terms in the expansion of 1 plus x to the power n in ascending power of x And summation of k plus 1 whole square ck From 0 to n is 2 to the power n minus 2 f of n Then the product of the roots of the equation f of x equal to 0 is This should be a very simple question All of you should be able to do it So basically this is the series that is given to you Please feel free to type in your response guys The answer is a single integer type So it's an answer between 0 to 9 So please feel free to type in the answer So if you see this summation we need to find we need to find 1 square c0 c0 then we'll have 2 square c1 3 square c2 and so on till n plus 1 square cn This is what we need to find out Now go for calculus approach the calculus method approach Yeah, yeah, I'm not solving it. I'm not solving it. I'm just writing out the series in a proper way I'm sure all of you would be able to solve it. It's pretty easy. Pretty easy any response guys Okay Shweta says zero Sure, sure Okay. Amog got one awaiting responses of others Till I get five answers. I will not start the discussion Nitya got four Okay Kushal says two Amog now changes this answer to four. Sondarya says two Kushal, okay guys again, uh This is one of the standard problems that we have been solving involving series of binomial coefficients, correct So in order to create these one square two square three square You know, it's it's going to take a multiple number of differentiations, right? But we can we cannot start differentiating right now because we will lose c zero, right? So in order to protect c zero and order to have two in front of c one, I'll have to multiply this with x correct So multiply this with x. So this makes x x square x cube x to the power four And ultimately this will be x to the power n plus one, correct now when you differentiate this Now when you differentiate this Okay This is what you see So on the right side you will get c zero two c one x three c two x square Four c three x cube Till n plus one c n x to the power n right And since I have to produce two square and one square and three square and four square like that I have to again multiply this with x correct So again multiply and x to this entire thing. So x times Okay, it will become x times one plus x to the power n plus n x square One plus x to the power n minus one equal to c zero x two c one x square three c two x cube Four c three x four and so on Till n plus one c n x to the power n plus one Now let's differentiate it once again Okay, so when you differentiate it once again, it'll become one plus x to the power n And again n x one plus x to the power n minus one Here I'll have n two x one plus x to the power n minus one And then n n minus one x square one plus x This is going to be your c zero or two square c one x three square c two x square and so on now put x as one put x as one Okay, when you put x as one you get two to the power n This becomes n into two to the power n minus one this becomes two and Two and two to the power n minus one This becomes n n minus one into uh two Right, which is actually two to the power n n into two to the power n minus one and this is n into Two to the power n again And this is two times. I missed out n minus one here. Okay now take Two to the power n minus two common. Okay, so take this term common from the entire series So what will you end up getting? You'll end up getting uh four you'll end up getting two n Again, you'll end up getting four n and you'll end up getting Two n n minus one Right. Is that correct guys? Everything is fine so far. Let me quickly cross check This is n minus two Ministic so this will become n minus two. So this will not give you a two over here. Right Now having got this If you simplify it it'll become n square plus five n plus four Right And this is what the equation is calling as f n. This is what your question is calling as f n So if f n is this f x will be simply replace your n with x That's going to be x square plus five x plus four And of course the product of the roots will be c by a in this case, which is going to be four by one. So answer is four So absolutely correct So nitya was the first one to answer it Well done. Very good At this juncture guys, we are going to take a break for around 505 so let's take a break Let's take a break and we'll resume at We'll resume at 505 p.m. All right, so welcome back After the break So this is the first question after the break So those who have come back early can start thinking on it the units digit in the Expression 17 to the power 2009 plus 11 to the power 2009 minus 7 to the power 2009 Okay So you have to find the units digit of this giant expression 17 to the power 2009 plus 11 to the power 2009 minus 7 to the power 2009 There are many methods to do it actually Okay, so vicious has already come with the answer. Amog also has given the answer Okay, let's see whether your answer is correct or not. Okay, nickel as well Okay, so almost all of you have answered this and that too correctly. Okay So yes, the answer is going to be one. That's very simple to solve So guys if you see here This particular question can be solved easily by using binomial theorem by knowing the fact that We know that x to the power n minus y to the power of n is definitely divisible by x minus y correct So I can say 17 to the power 2009 minus 7 to the power 2009 Will definitely be divisible by 17 minus 7 Right and apart from this we'll get something which I don't care what it is, right? So obviously it will be 10 into something Okay, if 10 into something is there means it should end with This should definitely have the last digit as 0 Okay So these two will have combined will have a last digit as 0 Now what about 11 to the power 2009 11 to the power 2009 Right, which we know that we can write it as 10 plus 1 to the power of 2009 Which again if you start writing the binomial expression for this This is how it goes on Right and it'll go all the way till the last number 2009 c 2008 10 And we'll have 2009 c 2009 into 1 So obviously Before this everything will have at least 1 0 at the end and this will be equal to 1 So the last digit for this will be 1 Okay, so last date is 1 So last date of these two combined is 0 last date of this is 1 so total the last date is going to be The last the final last digit is going to be 1 only which is option a Okay So good problem to begin with after the break Let's move on to the next problem This was your problem number 7 and next is your problem number 8 again, this is An integer type question For all x belonging to 10 by 11 and 11 by 10 Open interval the greatest term in the expansion of this has the greatest coefficient The greatest term has the greatest coefficient Where n is 4 times the fourth term in the expansion of this Then find the value of mk So you have to find the value of mk It's a single integer type question. Please feel free to type in your response in the chat box So the greatest term in the expansion has the greatest coefficient This is a very very critical word for you. This gives you a clue Amog, Atmesh, Loitya, Vishis, Purvik, Koshal Sai and response Not getting it. Okay. Purvik is still trying 6 I can just say No The answer is not 6 Shweta, Sondarya, Shruti Sai says 3 So let us start the discussion for this question I think I've got few responses from Purvik, Sai Now guys the greatest term in this expansion has the greatest coefficient. That's actually a hint given to you You know that the greatest coefficient in this is the basically the greatest binomial coefficient right because all the coefficients other than One and one they're all right. They're all one in this expression. So binomial coefficient is the highest coefficient in this case So what is the highest binomial coefficient? When your power is an even number So you'll say 2ncn is the highest Binomial coefficient right In this case, it is the highest coefficient as well So 2ncn comes from n plus one-th term right n plus one-th term Of this expansion Will be 2ncn Right into x to the power n right. So this is the highest coefficient or greatest coefficient occurring Okay now This term happens to be The greatest term as well This term happens to be the greatest term as well that means Tn plus one is greater than tn And at the same time it is also greater than Tn plus two That means 2ncn x to the power n is greater than 2ncn minus one x to the power n minus one And simultaneously 2ncn x to the power n is greater than 2ncn plus one x to the power n plus one right So from this we get 2ncn by 2ncn minus one greater than Or this into x let me write it in the other way around I can say x is greater than 2ncn minus one by 2ncn And from here we get x is greater than sorry x is less than x is less than 2ncn by 2ncn plus one correct That means x lies between 2ncn minus one by 2ncn And 2ncn by 2ncn plus one correct Now what is 2nc? We all know this formula from our permutation combination chapter that ncr by ncr minus one Is what n minus r plus one by r Right, let's apply this formula over here Let's apply this to This situation So this will become r which is n by n minus r plus one Okay lies between And here it becomes Yeah in a similar way here it becomes n plus one by 2n minus n 2n minus n plus one plus one right So in short my n by n plus one Is lesser than x and it is lesser than n plus one by n Right now try to compare this with 10 by 11 less than 11 by 10 Okay, try to compare these two Try to compare these two So when you compare these two automatically you realize that n comes out to be 10 Okay, n comes out to be 10 for you Now read the next part of the question the next part of the question says n is four times the fourth term in this expansion So n is Four times the fourth term in this expansion. What will the fourth term? For t4 I have to put nrs3 so it will be mc3 kx to the power m minus three one by x to the power three Right, which is four times mc3 k to the power m minus three x to the power m minus six right Now it is very obvious that this doesn't have any x term in it That means this term should have been ideally been zero Right because 10 doesn't have any x term in it correct Which clearly implies m should be six Now when you know m is equal to six I have to find k because ultimately the question asks me what is the value of mk Right So now let's compare this part With 10 So 10 by 4 is mc3 k to the power 3 So 5 by 2 is equal to If I'm not wrong, it's 20 k to the power 3 So k to the power 3 is 1 5 by 40, which is 1 by 8 So k has to be half correct Which ultimately gives me mk value as 6 into half, which is going to be 3 in this case So your answer to this integer type question is mk value is mk value is equal to 3 Is that clear guys any question with respect to this? Good So let's move on to our ninth question So again, it's a column match question for you Yeah, because I have to ensure that from both the sides shruti it should be the greatest term Okay, so it should be greatest then n term. It should be greater than nth term Also, it should be greater than n plus 2 th term also. All right, so hope you all can see the question clearly In case some things are not legible. Let me say it's a plus sign Okay, so it's x plus 1 x plus 2 x plus 3 all the way till x plus n And this is a 0 plus a 1 x a 2 x square and so on again will go Column wise so answer for a I'm first looking out for answer for a I'm sure that is going to be the most easiest of all Yeah, it's x plus 1 it's x plus 1 So for a which is the answer guys Exactly a will map to p obvious put put x as 1 When you put x as 1 you get a 0 a 1 plus a 2 etc till a n And that's going to be a 2 into 3 into 4 all the way till n plus 1 that is nothing but n plus 1 factorial So for a option p is correct This is easy problem. This should not be a problem for you. Yeah, b is also done Anybody done with b part q for b. Okay, no idea anybody else So let there says for a b. It is q Okay. Nikhil also says the same Good week also Okay, so most of you have given the response It's very easy now because you have already dealt with so many problems So multiply with x throughout first So we'll multiply with x that's going to be a 0x a 1 x square And so on till a n x to the power n plus 1 Now differentiate with respect to x. Okay, so when you differentiate with respect to x again, you'll apply product rule over here one by one So first you'll write this and then you will write this Okay, we'll keep on going. So I'm not going to write all the terms So it's a naught plus 2 a 1 x right Then we'll have 3 a 2 x square and so on Till n plus 1 a n x to the power n Okay Now put x as 1 Okay, when you're putting x as 1 you'd realize, uh, here it comes out to be uh n Plus 1 factorial Okay, and just because your 1 is missing you can just put a 1 and divide by 1 or divide by yeah and divide by 1. Okay Second will be 1 into 3 2 n plus 1 so you can say it's n plus 1 factorial only but you have to compensate by dividing it by 2 correct So this trend continues Right, that means you can take n plus 1 factorial And you'll have 1 by 1 plus 1 by 2 plus 1 by 3 all the way till All the way till you reach You reach n Right, so we'll have 1 by n plus 1 down over here Okay, so from So from n factorial if you take n plus 1 common that is what you are going to get 1 by n plus 1 So it is clearly mapping to your Q of column 2 Yeah, you can take log as well Yeah, sure see uh So when you differentiate x when you differentiate this term When you differentiate this term You have to apply integration by Sorry multiplication by using uh differentiation by using product rule So first one gets differentiated Other terms remain as such correct then x plus 1 gets differentiated and other terms remain as such then x plus 2 gets differentiated And other terms remain as such are you getting this nitya? So one at a time you're differentiating keeping the other same Now when you put x as 1 you get n plus 1 factorial from here You get n plus 1 factorial by By 2 here You get n plus 1 factorial by 3 here and so on till you get n plus 1 factorial by n plus 1 So if you take n plus 1 factorial common you are left with 1 plus half plus one third up till 1 by n plus 1 1 by n plus 1 Is that fine it? So product tool you have to remember this u v w dash Is u dash v w? u v dash w? u v w dash uh C is s okay C is s Purik also C is s Khushal also C is s loitia also C is s So many people are saying C is s, okay So let's talk about C first So in case of C what I'll do is If we just open the question So in case of C what I'll do is Since I want n a naught n minus 1 a1 plus n minus 2 a2 And so on till a n minus 1 correct So x x plus 1 x plus 2 all the way till x plus n Which is a0 a1 All the way till a n x to the power n. So what I'll do now is I will take x to the power I like take x to the power n common Okay So I'll take I'll do one thing. I'll I'll divide everything by Or I'll replace let's talk this I'll replace x with 1 by x I'll replace x with 1 by x Okay So, uh, when you do that If you replace x with 1 by x What do you get on the Right side on the right side I get a1 by x a2 by x square and so on till a n by x to the power n On the left side I get 1 by x I get 1 plus 1 by x I get 1 plus 2 by x And so on till 1 plus n by x Okay Now take x as the lcm So on the left side if you take x as the lcm I'm sorry. I'm sorry. Uh, I think it's the wrong way. I wrote it. This will become 1 by x plus 2 1 by x percent. Yeah So this will become x plus 1 by x x here I'll get 1 plus 2x by x And so on till 1 plus nx by x Here also I can write it as a0 x to the power n a1 x to the power n minus 1 and so on till Till a n All divided by x to the power n Okay so The left hand side expression will look like this now and so on till this All divided by x to the power n And here also I'll get a0 this a1 x to the power n minus 1 a2 x to the power n minus 2 All the way till a n whole divided by x to the power n These two gets cancelled Correct these two gets cancelled Now when you differentiate both sides When you differentiate both sides with respect to x What do we get we get 1 plus 2x 1 plus 3x dot dot dot 1 plus nx again one at a time will differentiate So now you get uh x plus 1 into 2 1 plus 3x and so on Okay, then you get x plus 1 1 plus 2x into 3 Into 1 plus 4x All the way till 1 plus nx This will continue and on the left on the right side we get n times a0 x to the power n minus 1 n minus 1 times a1 x to the power n minus 2 Dot dot dot dot till a n minus 1 Okay Now when you put x as 1 Now when you put x as 1 Let's see what happens When you put x as 1 On the left side you get 3 into 4 till You reach n plus 1 factorial Which is actually n plus 1 factorial by 2 Right Similarly here you get 2 into 2 into 4 Etc till this one will be what? 2 into 2 into 4 into 5 till you reach n plus 1 Can I say it is as good as 2 times n plus 1 factorial By 3 Correct right Similarly the next term would be what next term will be uh 2 into 3 into 3 into 5 Into 6 till n plus 1 Which is as good as saying n plus 1 factorial Into 3 by 4 Because 3 is extra and 4 is missing Right So if the trend continues can I say the last term will be n n plus 1 factorial by n plus 1? correct So if you take n plus 1 factorial common you will get You will get Half plus 2 by 3 plus 3 by 4 and so on till n by n plus 1 Does this match with any option? Does it match with any option? It matches with option number s It matches with s isn't it Yes or no guys Isn't it divided by x to the power n plus 1? x to the power n plus 1 Why? How many terms I have the first term was x plus 1 itself right? Oh one small error this term was not there This term was not there. Yeah Is that fine so that 1 by x was by a mistake taken it should be of degree n only right So there'll be equal number of x's both in the denominator and numerator So there'll be n x's over here n times Which gets cancelled with this Yeah, yeah, vashnavi it was not n plus 1 times but the first term by mistake I wrote 1 by x This term is actually not there This term is actually not there so What was the response that I got a c is s So c is s is absolutely correct So this maps on to s And of course then d will map to r but let us still find it out Oh, this is simple. You just have to differentiate And start putting x value as 1 Okay earlier we in the case of The b part we have to multiply with x and then differentiate here You don't you just have to differentiate from the word go itself So this will map to your r Okay So let's move on to now question number 10 Again, this is a j-email level question If 1 plus x to the power 10 is given as a 0 a 1 x a 2 x square and so on till a 10 x to the power 10 Then you have to find the value of the expression a 0 minus a 2 plus a 4 This plus this Whole square this whole square plus this whole square You've done please feel free to type in your response in the chat box Yeah, anybody Okay Oh gaurav is also joined in Gaurav says d Purvik says one minute Amok says c lauhtla says b. It's only a is missing I have got all the options Shweta also backs option d none of these All right guys I'll give one minute to those who are trying so that you know, I can start the discussion Sondarya also says d purvik b Sorry purvik c size b Khushal is d Okay, so most most of you are saying option d that is none of these. Okay, so Let's do one thing in this expansion of 1 plus x to the power 10 Which is a 0 a 1 x a 2 x square Let's put x as i Okay, when you put i We get a 0 Plus a 1 i Minus a 2 Minus a 3 i Plus a 4 Plus a 5 i Correct minus a 6 Minus a 7 i and so on correct If you take all the real terms together Your real terms will be This Correct And if you take your imaginary terms together Which i'm writing in orange So it'll be a 1 minus a 3 Plus a 5 minus a 7 dot dot dot right Okay Now isn't the question asking you the modulus of this? Right if you try to recall modulus of any complex number is going to be Under root of x square plus y square. So it's asking you the modulus of this square Correct Okay, so it's asking you the modulus of this square Which is actually going to be a root 2 to the power 10 into 10 square which is going to be your required result Which is going to be a required result So the answer is going to be 2 to the power 10 is your answer which is option number b Right correct psi wishes atmesh Correct So the answer is going to be 2 to the power 10 Is that clear guys? All right, so let's move on to the 11th question This again is an easy question If curly bracket x represent the fractional part of x then What is 5 to the power 200 by 8 fractional part? Okay, gaurav says b Porek also backs it up Okay, now I've started getting the answers b b b. Okay guys in the uh interest of time. Let's let me do this problem So 5 to the power 200 when it is divided by 8 Okay, what is the remainder? What is the remainder that I need to find out? Okay So very simple. I'll take 5 to the power 200 As 25 to the power 100, which is actually 24 plus 1 to the power of 100 Right now if I use binomial expansion for this, I'll get 100 24 to the power 100 plus 100 c1 24 to the power 99 And it keeps on going till the last term is going to be 1 Just before that I'll get 100 c 99 into 24. Okay Now when you finally divided by 8 When you finally divided by 8 you would realize that Till this term Till this term All the terms would be divisible by 8 correct So when you divide you'll get an integer till you divide it and the last term will be 1 by 8 Correct So the fractional part would be your 1 by 8 term This would be your answer for the fractional part of this So absolutely correct option b is right answer moving on to the 12th question This is the second last question After that we have we have one more question coming up. Oh, sorry Yeah, this is the question question number 12 You have to find b 10 The function f of x is given f of x by 1 minus x is given a 0 is 1 b 1 is 3 Then b 10 given a 1 a 2 etc are in gp Yeah, yeah, yeah, both the series go to infinity Answer is an integer but not a single integer It's not a single integer type, but it's an integer answer No, it's a four digit number. It's a four digit number not a five digit anyone Okay, so let's let's try to solve this problem So if you see this term Can I write it as f of x into 1 minus x to the power minus 1? So basically this is the problem where a binomial expansion involving Radical powers involving negative integer powers is being used Okay So this is equal to b 0 b 1 x b 2 x square Till b and x to the power n and so on So we all know that we all know that 1 minus x to the power minus 1 can be written as this It's basically a infinite gp with first term as 1 And common ratio as x, okay And the right hand side i'm copying it as such Okay Now if you compare the coefficients, you would realize that your a 0 is actually b 0 Right, no doubt about it Okay and your a 0 plus a 1 is actually your b 1 Okay And if you compare your x square coefficient Oh, no, no, no, I'll just solve it Vishishtha and Purvik. Just watch it over here. Okay If you compare your coefficient of x square, you would realize x square can be a 0 A 1 and a 2 and this is actually your b 2 Right, so if you see the trend actually your b n is actually a 0 plus a 1 plus a 2 Till a to the power Till a n right, this is your trend that is going on Isn't it Okay Now According to the given question b 0 is what b a 0 is 1 so a 0 is 1 that means b 0 is also 1 Okay, and a 0 plus a 1 is equal to 3 And a 0 is already 1 So a 1 has to be 2 correct Which implies a 0 a 1 etc R in g p R in g p With common ratio as with common ratio as 2 Invisible just a second Yeah, now it should be visible Yeah, now it is visible now. Okay So basically, uh, you have these terms. So basically it's going to be a g p Whose first term is 1 common ratio is 2 all together. There will be 11 terms So your answer will be 2 to the power 11 which is going to be uh 2 0 4 8 minus 1 Which is 2 0 4 7 No, no, no psi From 1 you from 0 to 10 there are 11 terms from a 0 to 8 10 there are 11 terms Okay, so first term is 1 So 11th term would be 2 to the power 10. So basically there are 11 terms over here Okay, so it will be 2 to the power 11 minus 1 by 2 minus 1. So answer is this Okay So guys, thank you for coming on live youtube. So we'll be calling off now. Thank you very much For joining in the session Okay over and out from centerm academy. Bye. Bye. Have a good night