Good Questions to feel the real JEE Advanced - 25

追加: 2026-05-11

[00:00:00]So hi people and welcome back once again to my channel and I'm really back after a very long time because I didn't have the scope of making videos because I was really busy in something else. So yeah we are back and J advanced is right around the corner. So let's keep it to some really good questions. I guess I will be regular from now on. So I guess you will be watching my videos and I hope you are all preparing well.

[00:00:28]Anyways, let's start off with the question. So, today's question is once again from our maths series. This question is from the maths group. So, it's a really nice question I guess worth to be taking up. Okay. And maybe uh by now you guys might have read the question as well and you can understand that we have to find two things, right?

[00:00:48]Uh BD are actually one question and AC are another question. These are two questions.

[00:00:55]So, yeah. Okay. So without further ado, let's start off with the question. So basically there are two functions f ixx and g jx. Okay. Defined in such a way that for every different value of i and j the dependency of the function varies respectively as f i of x they have given as 1 - roo<unk> under of 1 - x² for i= 1 and i is equal to 2. It is maximum value of f_sub_1 of t when 0 is less than equal to t is less than equal to x.

[00:01:27]Okay. And gi of x is also something like <unk>2 - 1x2 into roo<unk> under 1 - x square. And for f_subi of x they have said that it is will be the maximum value of f_sub_1. And for gi of x it is said that it will be the minimum value of g1. So basically we have to find out the area bound between the curves f_sub_2 and g2. And we have to compare between F2 and G2. Okay, you can see. So that's it I guess. So starting off with the question, I guess this is very clear. So I will move my screen a little bit downwards.

[00:01:59]Okay. So starting off with the question we have been given first that f 1 of x is equal to okay sorry it is 1 minus of roo<unk> under 1 - x² right. So in order to determine f_sub_2 of x okay because you can see it is given as the maximum value. So first thing is we have to check whether this function is increasing or decreasing right. So very specifically we have to differentiate.

[00:02:29]So f_sub_1 dash of x that will be equal to ddx of this thing. So one is a constant that will be zero obviously.

[00:02:36]And so I guess no more difficulties.

[00:02:38]This will come out to be min - x by roo<unk> under 1 - x square right because you will have 2x here and there will be a two here. So this two and this two will get cancelled. I just skipped one step. Okay. So this thing is your f_sub_1 dash of x. Now if you see for 0 less than equal to x less than equal to 1, this particular quantity is actually greater than or equal to zero which proves that the function is actually an increasing function. f_sub_1 of x is increasing right.

[00:03:18]Okay. So because f_sub_1 of x is increasing. Therefore, if we as per the question is telling if we consider the interval 0 less than equal to t less than equal to x then the maximum value of f1 of t will obviously occur at x is equal to t or you can say uh rather I should say t is equal to x right so therefore I guess it's very easy to say that your f_sub_2 of x will be equal to 1 - roo<unk> under 1 - x square itself.

[00:03:58]Okay.

[00:04:01]Okay. So moving on to your g1 of x it is given to be equal to <unk>2 -/ [clears throat] into root under 1 - x square.

[00:04:15]So basically from this thing once again we have to like uh do it like f right in order to find out g2 of x we need to differentiate and check whether this is increasing or decreasing. I hope you can understand okay so g1 dash of x <unk>2 -/ is obviously a constant so no issues. So <unk>2 -/ remains as it is and for root under of 1 - x square the differentiation will be as similar as we did it last time right. So I guess no confusions in this. It will be min - x by roo<unk> under 1 - x^ 2. Okay. And uh I guess no more doubts in that that this particular quantity is positive. 1.414 - 0.5.

[00:04:58]That's a positive quantity and this has a minus and obviously root under 1 - x square is positive. And even x is also positive here because uh the interval we are considering here is 0 less than x less than one. Right? That means g1 - x is actually less than zero.

[00:05:20]Which means that g1 of x is decreasing.

[00:05:26]Right?

[00:05:29]So because your g1x is decreasing uh very simply we can conclude from here that minimum value I hope you remember we are talking about minimum value in g and maximum value in f. So minimum value of g1 of t in the interval 0 less than equal to t less than equal to x in will be at t = to x. So once again g2 of x also becomes <unk>2 -/ into roo<unk> 1 - x². Okay.

[00:06:11]uh like uh you guys can worthwhile question me that uh why was this stuff given then if both of them are same. Okay. So yeah question manipulation is done like that they will have lots of things to confuse but sure top property is right it could be something different question be manipulated in lot of this but if you know the concepts clearly obviously just do it okay so moving on uh I have already told you that we have two basic questions right one is pairing between f_sub_2 and g2 and another is fine.

[00:06:52]Let's do one thing first.

[00:06:55]Let's compare between your integral to as a function of tdt and 0 to x g of two as a function of dt.

[00:07:13]We have to compare between these two right as said in the options obviously.

[00:07:18]So let's define one function like uh I obviously thisly 5,000 let's say let's take let us take a function capital f of x which is basically g of2 sorry not g of2 g2 of e dt okay that's write dt in the end of it difference. So f_sub_2 of t als also comes within the bracket itself and then outside we have so yeah now if we try to find capital fdash x from here that obviously comes g2 of x minus f_sub_2 of x as per limit right so if we write down the values this will be <unk>2 -/<unk> - x² minus of 1 minus so if you have to find this I guess finding this will be really easy to distribute the terms I guess <unk>2 remains within a bracket minus/ this will be -1 + 1 so -/ here and + one here will make plus/ roo<unk> 1 - x square and one independent minus one will be left Cool. So, uh I hope you have noticed in the question that they have given us a condition that x is less than equal to 1x2 in both the option. Okay.

[00:09:00]So, owing to this condition. From here we can say x² is less than equal to half. So 1 - x² will be greater than equal to half. If we take the root root under of 1 - x² will be greater than equals to 1x <unk>2. So from here we can easily conclude that g2 of x minus f_sub_2 of x is greater than or equal to <unk>2 +/ into 1x <unk>2 - 1. So simplifying this thing I guess this will be 1 + 1 by 2<unk>2 - 1 - 1 by 2<unk>2. Am I correct? Let's check once. Okay. 1 + 1 by 2<unk>2 - 1.

[00:09:53]Uh okay, this thing is actually extra uh sorry for this. Sorry for this mistake. So obviously + 1 and minus 1 get cancelled and so we get this to be greater than equals to 1 by 2<unk>2 which is greater than zero. So yeah that's a nice conclusion actually. So we get that your f dash x is greater than equals to zero. Also one more thing if we find out f of0 okay you will understand that f of0 is also actually equal to zero. So from here we get integrating this or taking the integral on both sides.

[00:10:35]Okay we get f ofx is greater than equals to z.

[00:10:43]So from here manipulating the difference like we wrote this is a difference right manipulating the difference and taking either the f I guess we will take the f term to the other side we will get integral 0 tox f_sub_2 of t ddt is less than or equal to integral 0 tox g2 of t dt. Okay. And therefore we get option number B correct.

[00:11:13]Obviously option number D is discarded.

[00:11:16]Okay. So that's like half 50% of the question done. Okay. Moving on to the next one which is your option number A and C.

[00:11:29]Okay. So here I guess let's check the option once.

[00:11:35]Okay.

[00:11:37]I mean say it tell us that we have to find out the areas between f_sub_2 and perfect. Okay cool move on to it.

[00:11:47]Okay so moving back uh let's give a small reference because we have changed a lot of screens. So 2 of x is equal to 1 -/ roo<unk> under 1 - x² and your g2 of x is equal to <unk>2 -/ by<unk> 1 - x² okay so because we are obviously solving for this thing the area so we have to check whether there is an intersection point right and I guess there will be an intersection point so we have to solve these two and it will be I guess a really hefty calculation Okay, this part is a bit lengthy and maybe I will prefer skipping a few steps.

[00:12:31]I'm I will try to skip on the calculation on the sum side from now because obviously you know it wastes.

[00:12:39]So if we move the square root terms further I guess just like the last part we did this will be 1=<unk>2 +/ into<unk> 1 1 - x square. So this is 1 - x² = 1 by <unk>2 +/ okay. So this comes out to be 2 by 2<unk> 2 + 1. And now if we rationalize this thing to something better I guess uh I'm moving the screen now.

[00:13:14]This is I guess coming out to be 4 <unk>2 - 2 and this thing is 8 - 1 7.

[00:13:23]So this is your roo<unk> 1 - x square.

[00:13:28]So I guess after this there will be a lot of manipulation and we have to square this up and all. Uh so I got the value of x to be something something big like this is coming but don't worry this will simplify might be a but I guess it will simplify as per okay so now the thing is we have to like if we are finding out the area we have to take the difference between right so we need to find out which is the above so either two options Either you can do it using drawing the graphs which is not very difficult I guess but yeah we have another option is by substituting x equal to0 so at x= 0 f_sub_2 of 0 is equ= to 1 - 1 which is 0 and g2 of 0 is equals to -/ which is obviously a positive quantity from here we can conclude that G2 is above.

[00:14:36]So your estimated difference will be integral 0 to g of x - f of x dx. Okay. And alpha is obviously the intersection point which we hunted for here.

[00:14:56]So cool. So solving this up I guess not a very difficult.

[00:15:02]This will again like we have done this difference multiple times. Okay. So I'm directly writing the thing. This will come out to be <unk>2 + half x. Okay. So if we split the integral of integral 0 to alpha <unk>2 +/ into<unk> 1 by dx - root gives us 1 dx. So this thing will come out to be the root 2 plus/ outside and 0 to alpha integral roo<unk> 1 - x² dx is a very very common rule that we learn which is x by 2<unk> - x² + half sin x from 0 to alpha minus uh okay this is I guess x right So uh this will be x from 0 to alpha. So I guess we can directly write it as alpha.

[00:16:10]Cool. Okay. So I guess expanding this a little bit.

[00:16:17]This is coming out to be <unk>2 +<unk> 1 - alpha² +/ <unk>2 + 1 by 2 sin inverse alpha - alpha. Okay.

[00:16:40]Cool.

[00:16:43]Okay. Um so before proceeding actually there is uh one thing I would like to tell so guys this calculation is going to be a bit long okay so to make it short I figured out something basically where we were finding out the value of 1 - x² as your 2 by 2 <unk>2 + 1 you can take this to be some cos theta KN. Okay.

[00:17:16]So your x basically becomes sin theta kn. After this taking this thing after taking this thing you can put that into the integral and then evaluate the integral. Okay. In that case your limits will actually change from 0 to theta and here you will have everything in terms of theta. Okay. If you try to solve that actually I guess your uh integral is getting a bit easier because see the calculation is really uh I tried to calculate this the calculation is really hefty okay so somewhere around I I guess I am like can approximate somewhere around because the value that we are getting is very very close to 5 by 4 <unk>2 okay it's very very close to<unk> by 4 <unk>2 but it is not exactly 5 by 4<unk>2 and even I tried to graph these two like the curves and so that you also the curve is something like this wait the curve is something like it's actually a symmetric curve okay the curve something like this. And uh our area is basically this part.

[00:18:40]So actually there is one more thing to be added here.

[00:18:46]I'm not messing up the upper part. Okay?

[00:18:48]Cuz I have already done that. Just wherever you start off with a multiply a two. If you are taking your limits from zero to this particular point uh whatever it will be it if it is alpha very well and if it is theta kn then start off from zero and till alpha or till theta kn okay it's an even curve as you can see I mean an even bounded area okay so that's it and uh I don't know somewhere in the end we are forced to make an approximation okayision is going to be really hefty so I would advise if you go to decimals. You can find out the values of decimals. <unk>2 is 1.414. The calculation will be a bit hefty but the area is finally coming out to be by 4.

[00:19:36]Okay, it was very confusing. I can't uh at the timing see any errors in my solution. If you guys see, you can obviously write down in the comments. That's always open I have said. So yeah uh that's it for option number A.

[00:19:54]So guys, our correct options for this question are option number A and your option book. Okay.

[00:20:02]So that's it for today guys. I guess you have liked this video and uh regarding my regularity, I will try to be regular from now on. I have been really busy in something else. So yeah and that's it for today. Uh and keep preparing well and if you have liked my channel do like, share and subscribe. Obviously any doubts come out in the comment section and also if you find out any errors in the calculation you can also tell me in the comments. Okay. So that's it for today guys and I hope you enjoyed this video. Thanks for watching.