JEE Advanced 2026: Functions Tricky Problem | How to improve rank in JEE Advanced?

[00:00:00]This function question asked in JE Advanced 2026 is not a new question. We have seen this type of questions before in JEE Mains and probably in Advanced as well.

[00:00:08]But the language has been modified in it in the way it is and things have not been given directly.

[00:00:15]So those things were a little new for you here. Or if you have practiced then you can recognize the thing here. So let us understand what kind of trap was hidden inside it. How did we have to catch him here? First of all, let n denote the set of all the positive integers. That means you have n natural numbers.

[00:00:34]Consider the sets a you're given a set that has five elements. B has given you a set which has a total of seven elements. So here we are told to let S be the set of all functions.

[00:00:46]Where you have been told that mapping has been done from A to B. Isn't it? Your mapping from A to B has been done here. So what is the total number of elements inside A? 1 2 3 4 and five and what are your total elements inside B? There is seven here. So 1 2 3 4 5 6 and seven, right? And what is said here is that F2 cannot be two.

[00:01:08]Meaning, this two of yours cannot be mapped with it here. Okay, right? F2 because what is here cannot be two.

[00:01:16]And also if I talk about F4 then it cannot be four here. Meaning four cannot be mapped with four.

[00:01:22]For the rest, anyone can get that number here.

[00:01:25]Consider the set T okay? Consider the set T f belonging to s, there exists a function g in which the domain here is called b, that is, these seven elements which can be inserted.

[00:01:36]And here the code domain is given by your natural numbers.

[00:01:38]True that gfx = 2 to the power x for all x belonging to a. Now here you might be thinking that this is the only limitation I have given regarding f.

[00:01:51]But then why do you need the g function which is explained separately here? So this should have flashed in your mind here that when you're asked about the function here, what's so special about g here? So if we look carefully here g(fx) = 2 to the power x, then we need to understand whether this is giving us any condition regarding fx or not. So how can we check that? So here I will give you a logic that if we take x = 1, is it okay?

[00:02:22]What value are we getting from that here? We are getting what is to value.

[00:02:25]Meaning that F1 is giving you pay to here.

[00:02:28]Similarly, if you put x = 2 then you can assume here that it is giving you the same value here. Okay, right?

[00:02:35]Meaning f2 is also giving you two. This thing is possible. It is absolutely possible.

[00:02:39]Because what will happen to us here? This type of function is many one function that one pays two and two pays two.

[00:02:45]This is absolutely possible. Isn't it? So what we have assumed is that in a way this is a many one function. Now if I put x = 1 here, what will this become? g of f1 means basically this is your g of 2. Ok? Do you understand this? And here the power of 2 is x, so instead of x, put one here. So how much will this come to? It will be equal to two. Just like that, I put g off here x = 2. So my f2 will be equal to whom, friend, this will also be equal to our g2 because f2 is also given as two and as soon as I put x = 2 here, then what happens is 2², now how is it possible that our g function gives two to two and also gives 2², brother, this will not remain a function, that is, it will not follow the definition of the function here, then g, but here you have been told that it is a function, so what we have accepted here is that this is a many one function.

[00:03:42]This is absolutely wrong. That means f has to be one. Are you understanding this thing? Ok? f has to be one one whatever it is.

[00:03:51]So we got limitations here related to the function f that our f can only be one. Second f2 cannot be equal to 2 and f4 which cannot be equal to 4.

[00:04:02]We have to take special care of these things here.

[00:04:04]Now you have to calculate the one by one function here.

[00:04:07]So for 1 minute, you have to keep in mind the limitations here also.

[00:04:11]So first of all let's do one thing, let's see the total number of here, our one one function.

[00:04:16]Total number off what can our one one function become? After that we will see what limitations we have to keep here. So what is the total number of one-one functions that can be created? So brother, one here can be mapped with one or it can be mapped with two or any seven.

[00:04:31]Now once he has mapped out what he has with one, then two is left with only six choices here.

[00:04:38]Ok? Now you will say that Yaar To cannot be mapped with Pay To here. We are not considering that thing right now.

[00:04:45]So here I am saying without any limitation. Ok? Without any limitation, you can remove the total one here.

[00:04:51]So in that case, how many choices do you have left for the second? Six survived.

[00:04:55]Similarly, the third one will be left with five. Fourth will have four left and Five will have three left. So if we understand this better from the concept of PAC, then there are total five elements here and seven elements here.

[00:05:06]So the five want any five out of these seven.

[00:05:09]So how can that be yours simply 7C5 and as soon as he has selected those five elements.

[00:05:15]Now we can arrange them also. Isn't it? Anyone can find value here. So you have to convert what you have into 5 forial. Now if we calculate this here, how much will it simply come to? 7 * 6 / 2 right? 7orial / 5orial, this is your value and how much will this 5orial be? 120 So what is the total here?

[00:05:39]21 * 120 means this will be your 2520.

[00:05:43]Ok? So this is your total number of one one function. Next, what you said here was that we have to take care of these conditions.

[00:05:51]So now in this we use the inclusion exclusion principle. Isn't it? If I try to explain it to you through sets, it will be easier for you.

[00:06:00]What do you have to do? Here we have to assume that we take case one, let's say our P1 is defined as when our F2 = 2. Ok?

[00:06:14]Our F2 here should become equal to 2. And what is happening in our case of P2 is that our F4 here becomes equal to 4.

[00:06:21]Now what exactly do we want?

[00:06:24]We want this to be this is also not equal to this is also not equal to 4.

[00:06:28]So whatever their common intersection is here, that is our solution here.

[00:06:33]So basically here you want that first of all you take the complement of P1 because we do not want that condition and similarly take the complement of P2 here and whatever will be their intersection, whatever will be the common solution here because both these conditions should be valid here at the same time.

[00:06:48]So here we are taking the common intersection. Now here if we apply DeMorgan's law which we see here at the time of set.

[00:06:56]So what will happen here according to that? We can simply write this, change the sign in the middle and you can put the complement here above. Isn't it? And if we look at this one carefully, then apart from P1 union P2, we will have to consider all the parts. So basically, from the universal set, subtract the part of P1 union P2 here and we also see its formula inside the set. That is equal to NP1 + NP2 - of N P1 intersection P2, so this is our principle of inclusion and exclusion here. Okay, right? If you know it directly then you can do it very easily. Right here, I tried to explain to you exactly what we do.

[00:07:43]Now what does np1 mean, that is, when brother, your f2 also becomes equal to pe two here. Now as soon as you make f2 = 2, meaning you have fixed one value, then you are left with four values ​​here and six values ​​here, so what will be their value? 6C4 and you have taken total four values ​​here, so you will have to arrange it with 4orial.

[00:08:05]Similarly, if I talk about np2, it will also give us the same value.

[00:08:10]So make it into 2 because what will happen in it? f4 = 4. Meaning a value has been fixed here. It's fixed here.

[00:08:16]So there are four left here and six left there too.

[00:08:18]Then the same condition will occur. Minus what do you have to do with this? np1 intersection p2. So now what will be the value of this one? P1 intersection P2 means both two and four have been assigned to you. So now there are three values ​​left here and how many are left here? Five values ​​left. So this will be basically 5C3 and into what we have here, how much will it be?

[00:08:41]So 3orial because we will arrange those three elements here.

[00:08:45]So if I calculate the value here, how much will it be? 6 * 5/ 2 means 15 * 24 * 2 means 30 * 24, so this is how much you have? 720 this is coming up what is your 720 minus 5C3? 10 10 * 3 means 6 is 60, so this value is coming to you as 720 - of 60, okay, so what will be yours, here it will be 660, so your universal set, that is, the total number of which we had derived the one by one function, because that will be your complete set, from that we are subtracting this.

[00:09:21]So subtract 660 from 2520 here.

[00:09:24]That will be your final answer. So your final answer here will be 1860, which means your correct answer was 1860. So I understood the logic here, it took me time for the explanation but if you have practiced some questions of this type then you can do it very easily.

[00:09:41]Then you will cook all these things there like khichdi. Isn't it? Then we will simply apply the concept and find the answer. Well, hopefully things have become clear to you here.

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