ISI UGB PYQs | Must solve | Most Important Questions | Part 03 | Anirudha sir

Añadido: 2026-05-09

[00:00:01]Hello everyone, welcome to this video. I hope you're all doing well. As usual, uh we will continue our preparation journey towards ISI and CMI. We'll be doing PYQs because that is the best way to prepare for an examination. And uh that's all I got for now. And another thing important information, we have started uh IOQM preparation batch. And if you are willing to prepare for aim and if you don't find time to prepare it while preparing for J ne these things please consider joining uh the course because we will be conducting it three times a week in the night during the time of 9:00 p.m. to 10:30 p.m. So you won't find it intersecting with other uh preparation timings. You'll find it very very helpful. So yeah, the links will be in the description and please uh go to the channel and watch the announcement video for to find the details about the batch. All right, that's it. So we'll start with this without much uh delay.

[00:01:14]So yeah find with proof all the possible values of t such that this happens this is more of a j advanced type of a question isn't it? So I hope you can figure it out immediately.

[00:01:34]Uh I hope you have paused the video and tried it for yourself. So let's just begin. So now the important thing which we will be using is the fact that if you have a limit of this type limit n ts to infinity uh 1 by n summation f of k by n k is varying from 1 to n to n 0 to n.

[00:02:07]This will be equal to integral from 0 to 1 f of x t.

[00:02:17]All right, this is the we uh I explained why this happens in the uh you know what is it called? Um definite integration video, right? Uh I made a whole playlist on calculus covering calculus. If you have not watched it, please go and watch in the is my preparation. Uh there is I don't know what the playlist is called. Please go and check it out. So yeah. So in that I did cover that this sum happens to be 0 to 1 this and it is totally a mathematical statement to be used. You can use this statement directly. Okay.

[00:02:55]So now we have something like this. So let's rewrite the question. So we have limit n ts to infinity 1 by n^ t uh summation k power 1 by 3 k is varying from 1 to n okay so this is not like that but we can convert it into that form so and this is equal to c and what is the value of c that's the And for which value of k okay so this is limit n ts to infinity if I want to have k by n k by n whole power something k by n function of k by n that's what I want to write it as so I want to have this basically k by n^ 1x 3 so what do I need to get here I need to get 1x n^ t - 1x 3 because one of the 1x3's goes here right Okay. And I need to get a 1 by n outside. So I can write this again as uh okay rewrite limit n ts to infinity. I need to get 1x n summation k by n^ 1 by t right. So this is this whole thing. We know what this is. So we will have n power t minus 1x 3 - 1. So 4x3 right this is what we have.

[00:04:34]So if t is exactly 4x3 that would imply that this value is simply you know c this is c right. So c is equal to integral 0 to 1 x^ 1x 3 dx.

[00:04:53]So that will be x^ 4x 3 by 4x 3 0 to 1. that is simply 3x4 right so C will be 3x4 if t was 4x3 but if what happens if t is let's say greater than 4x3 if it is greater than 4x3 you'll have limit n ts to infinity 1x n summation k by n^ 4x 3 uh 1x3 K from 1 to n whatever all divided by n power - 4x 3 and this is a totally you know there it's there this is tending to some constant this is tending to the integral but this is tending to infinity so overall this will be zero so if t is greater than this c is zero what happens if t is less than 4x3 then our C will be limit n ts to less than 4x 3 in the sense what 4x3 minus t is positive so instead of in the denominator I'll write it in the numerator n ts to infinity n power 4x 3 - t into the whole thing 1 by n summation k by n whole q now this is finite or tending to finite quantity but this is tending to infinity so this will be so yeah so you know for which case what will be happening it's a very easy question I did it under five minutes you can also do it probably okay uh all right so next question try to do it this is somewhat uh trickier then that okay so this function is supposed to have only two uh real roots.

[00:07:20]Okay, how do we prove that?

[00:07:26]First of all, how if you are given a polomial right now, this happened to me in my ISI entrance exam. If I I was given a polomial actually so and I was asked to find out what happens uh like how many roots does it have? They didn't even ask how many roots they asked what are the roots of this polinomial.

[00:07:48]What are the roots? Like it was a degree seven polinomial or something. I I was I tried to put one two and it didn't work and it was getting a messy and and later I realized that it had terms which had only odd powers. It didn't have any term with even power right that means if I if I uh differentiated it the differential will have all the even powers only and it also had a constant term outside. So that means that it will be strictly increasing.

[00:08:26]It was strictly greater than zero. Fdash was strictly greater than zero. So it was a strictly increasing function. And I immediately said told them that it'll have exactly one root because it is strictly increasing Z and it'll keep on increasing. So it'll intersect the x-axis only once. And they said okay very good. And they did not even ask what are the roots. So they were just trying to trick me probably into uh doing some mistake uh doing some extra overthinking and uh you know spoil the paper but it is the presence of mind which they are expecting from you. That's what I got from that particular question. I was asked four uh I was asked three questions. There was another logarithm question and one of the questions was I think uh permutation combination. Uh but it was totally fun. I thoroughly enjoyed after the three questions were done. It was done within 20 to 25 minutes. I was jumping up and I just went and hugged my father and uh I cracked it and my father also said it'll happen. Don't worry. So yeah uh yeah that was a good memory.

[00:09:35]Anyways let's get back to this question.

[00:09:37]So here we are supposed to find only two real roots namely zero and one. Okay.

[00:09:45]Uh so for two roots how would a function look like if if it was only one root I said that if you differentiate it it will strictly increase and whatever for two roots exactly two roots for every n so let's see what happens again by differentiating. So what is the differentiation q n dash of n.

[00:10:09]So this is obviously 0 - n x^ n -1 minus uh n into 1 - x^ n and because I'm also differentiating chain rule the inside thing this is minus one so I will have - n x^ n -1 this - this - + n 1 - x^ So this is q n dash of n.

[00:10:45]So when is qnd0?

[00:10:49]qn dash of x =0 implies - n x^ n -1 + n into 1 - x^ n - 1 0 and we are talking about real numbers only. So obviously we won't consider the complex roots or anything. So if I cancel out n so I basically have 1 - x^ n - 1 = x^ n and yeah as I said we were talking about real numbers so it has to happen that 1 - x= x so x = 1x2 right so it q dash has only one root so what does that immediately tell you for every value of n qn dash has only one root. So that means for every value of n it it it will be something like this or something like this.

[00:11:46]That means that uh you know this is the point. This is 1x2. That means that it will intersect x-axis only two times. It will intersect x-axis only two times. So how do we prove it exactly? How do we prove it?

[00:12:02]So what happens when x is less than 1x2?

[00:12:06]x is less than 1x2 implies x^ n is less than 1x 2 power n.

[00:12:14]And what is 1x nx? 1 - x will be greater than 1x. That means 1 - x^ n will be greater than 1x 2 power.

[00:12:26]Right? So what do we have? QN of X is 1 - X^ N - 1 - X^ N that is 1 - of X^ n + 1 - X so if you h how do I add these two I don't think they're getting added so easily Is this what I had? Is that what I had?

[00:13:14]Can I I I want to show oh okay.

[00:13:26]What happens if x is uh less than okay okay okay okay okay let's not talk that let's not let's not talk about qn right let's talk about what happens to q dash nuh all right so I have q dash n of x to be 1 - x power n - 1 - x^ n minus one right and uh if you see the difference okay - 1 - 1 even if I take that if I take the difference this is going to be positive this is going to be positive for this for x less than 1 by 2 that means the graph of qn of x will be increasing when x is greater than 1x2 And by similar calculations if you see similarly qndash of x will be less than zero for x greater than 1x2 that means it is something like this right at 1x 2 it is zero for less than 1x2 it is increasing increasing here it is decreasing. So it has to intersect only at two places and we know that 0 and one are the roots of this. So yeah it's very simple very very easy question but you have to write the steps properly. I think that's what was the important part in this question. So I hope you found it helpful guys. Uh and of course I want to remind you again about IOM preparation course. If you are interested please join. So meet you there. Take care. Good night.