 Hello, welcome to the live session greetings from Sendham Academy So those who have joined in the session I would request you to type in your names in the chat box So guys, let me tell you first thing before I start this session Congratulations to all of you. You have done us all proud with your performance in the JEMN exam and Please continue this good performance for the next phase two of the JEMN exam, which is supposed to be in April And in order to facilitate that we have actually come up with this fast track revision where we have shared the schedule of physics chemistry math So we are going to come up with some assorted questions on these topics, you know twice a week Right for math it will happen on Wednesday and Saturday Okay So these questions are a good mix of the entire concept So that the each and every subtopic is getting revised through problem practicing So good morning Amogh, Vishesh, Saritya, Ramcharanth So today we are going to do functions chapter and in this particular chapter We are going to cover all the smallest subtopics associated with functions Like finding the domain range The concept of even odd function the concept of inverse of a function Concept of types of functions many one on two into one one etc and Of course, we are going to talk about Mixed examples of all these concepts So without much ado, let's get started with this session So here comes the first problem for you. All right. So here's the problem number one I would request you all to keep it at a very high resolution so that you can read the question properly So the first question goes the number of integers in the domain of the function f of x Which is defined as log of 4 minus mod x to the base of gif of x square Okay, so you have a gif function over here. Okay Plus log of fractional part of fractional part of root x Okay, so be careful of these Brackets so we have a gif over here in the base of the log first We have a mod x over here and we have a fractional part of root x Again, all these questions are single option. Correct. So please feel free to type in your response once you're done with it Again, so let me tell you in the domain of the function. You just have to be very careful and identify The values of x for which the function is defined and real So when you have a function which is made up of some of two functions, we have to take the the case where The values of x make both the sub functions defined in real So we have to take the intersection of the solutions that we get from The domain of this function as well as this function All right So i'm only getting the answer the number of integer values uh purvik says it's d aditya also d Atmesh says c Atmesh says c What about others? Okay, so i would have Yeah, so many people are confused between c and d c and d except for vashnavi who says uh Option b All right, so i've got enough responses now so i can start the discussion of this question So guys first of all, uh, I would have to ensure that for this log the first log to be valid four minus in fact four minus the Mod of x this should be a positive term right that implies Mod of x would be uh less than four which clearly means x should lie between minus four to four x should lie between minus four to four right secondly The base of this log that is gif of x square should be positive and it should not be one Remember the restriction for any log function to be a real valued function The base should always be positive and should not be equal to one right now it means A fraction gif or integer is not One and it is greater than zero which means it has to be greater than equal to two. This has to be greater than equal to two Right, that means x square Should have been greater than equal to two and of course less than infinity right So which gives you a x This means x square is greater than equal to two. That means x square minus root two square is greater than equal to zero which implies Which implies x could be Greater than root two or x should be Less than equal to minus of root two okay Now this was about the first log function So these restrictions that we got restriction number one restriction number two we got it for the first log function How about the second log function the second log function the only restriction that we need to follow is uh gif of this Should be greater than zero now we know we all know that it's always greater than zero So it should not be equal to zero. That's it. That means root of x should not belong to an integer because of root of x belongs to an integer You know that its fractional part is going to be zero Right see guys, we all know that gif of any function gif of any function would always be Great lying between zero to one. Okay, so there's no problem in it being greater than zero Right the only thing that we need to watch out for that. It should not become equal to zero It should not become equal to zero that means We have to choose our values in such a way that x is not a perfect square of any integer that means x should not be a perfect square It should not be a perfect square Okay, so what are the possible values that we get over here? So let's take the intersection of all the conditions by drawing a number line so minus four so this is uh four and root two And minus root two Okay So our values of x would be in such a way that It should lie between this zone x should be greater than root two. So it should be lying in this zone Right Or it could lie in this zone Or this zone Yeah Now what are the possible values that x will satisfy? What are the possible values that x will satisfy over here? What are the possible integer values that will fall in this zone? Can I say x equal to two will be a possible value? Yes or no Can I say x equal to three is the possible value? Can I say minus two and minus three? No Minus two and minus three will not be included because you want This function to be a real valued function. So your x has to be positive for that Right So x has to be positive for that Let me just remove greater than equal to and just write greater than so be very very careful because Each and every Function should be a defined over here. So I cannot say minus two and minus three Right. So the only possibility is my x could be two or x could be three. That means Only two integer values of x will be a part of the domain of the function So option number c is correct Option number c is correct Is that fine So the first one to answer this was art mesh Art mesh was the first one to answer this correctly. Well done Okay Now guys moving on to the next question now Any question any any concern with respect to this solution? Please highlight it Okay, if not, we'll be moving on to the next question So guys the learning from this question was be very very careful while dealing with the domain Because you have to make each and every part of the function You know a valid and a real valued function So moving on to the second question of the day So we have to find the range of the function under root of cos inverse root of log x to the base 4 minus pi by 2 plus sin inverse of 1 plus x square by 4x remember the very first step before finding the range The very first step for finding the range is find the domain of the function because there is no existence of range without domain That's the biggest mistake people do while solving Range questions They don't bother to find out the domain which is wrong. Please find out the domain of the function first See for what values of x is this function valid And then go for the range of the function All right. So Vashtavi says 2d Okay, Ramchalan also backs it up 2d I need five responses for me to start solving this problem. So I'm waiting for three more people Okay So third in a row Purvik Again says option d all right Okay, so first find out the domain of the function for the domain of the function But let me redefine this function in a simpler format Uh, can I write this term as minus of sin inverse? uh under root of log x to the base of 4 and a sin inverse of 1 plus x square by 4x Okay now uh First of all if you want this A function to be a real valued function. That means sin inverse Under root of log x to the base 4 This should be a negative quantity. This should be a negative quantity Okay, right that means Under root of log x to the base 4 should be a negative quantity right Now under root of something cannot be less than 0. However, we can say it can be equal to 0 I can say it can be equal to 0. That means x could be equal to 1. However, right In this case if you find we have to have 1 plus x square by 4x should lie between minus 1 to 1 that's what we know about the Domain of sin inverse functions Right, so if you take this condition 1 plus x square By 4x Uh, we already have a condition over here. So we can actually check whether this condition This condition is met when x is equal to 1 or not. So when you put x equal to 1 I get 2 by 4, which is actually half Yes, so half lies between minus 1 to 1 Correct. So this is the most restricting condition out of these two if I if I could say that So clearly the domain of the function is x should actually be 1 Right, it's it's a singleton set to be, you know, very precise So this is a singleton set. So x will only be 1 Right, so when x is only 1 and when you try to substitute the value of x as 1 in this function You would realize that this becomes 0 and this becomes sin inverse half sin inverse half is going to be pi by 6 So again, your range is going to be just one value pi by 6, which is option number d So the first one to answer this correctly was vaishnavi well done vaishnavi very good simple question this was Again, the moral of this story was to find the domain before you find the range Don't forget that All right, so quickly moving on to the next question question number three for the day Find the number of integers in the range of The function 1 by pi sin inverse x per stand inverse x Plus x plus 1 by x square plus 2 x plus 5 I'm sure the first sitting of j main would have come with a lot of learning to you So silly mistakes etc should be completely ruled out if you're sitting for a comparative exams. So try to be accurate It's good to be delta slow But it's very important to be exactly accurate I would request whenever you're typing in your response type the question number and then the option that you think is correct For example, in this case, if you think option a is correct type 3a again, since the range question go for domain first So the question is asking for the range. So go for the domain first Before anything else Okay, so ram chalan has come up with an answer Okay, purvay kathmesh noted down your response So I need two more people to respond the vaishnavi. Okay Ice buck up sigh sanjana sanjana also gave the response. Okay guys, so let's start with the discussion of this Or if you look at the first function, uh That is your sine inverse plus tan inverse x, you know that here x should lie between Minus one to one Here x could be infinity, correct? It can lie from anywhere from minus infinity to infinity So the restricting condition has been imposed by sine inverse over here. Okay Now this function this function can be Valid for any value of x because the denominator will never be zero because this can never be zero Because the discriminant of it is always negative Okay, that means it doesn't have it doesn't have any real root and it is always positive That is b square minus four ac is always negative. So it's always a positive function So in light of all this thing, we decide that your domain can only be The values of x lying between minus one to one Okay Now never forget the power of calculus, right calculus is a very potent tool, right? It tells us so many things. Let's find the derivative of this function Okay, when you find the derivative of this function, uh, we'll have one by under root of one minus x square Plus one by one plus x square And the derivative of this function would be x square plus two x plus five minus x plus one times two x plus two divided by x square plus two x plus five the whole square Okay Which is nothing but one by pi one by under root of one minus x square by one by one plus x square I'm just repeating this entire stuff over here and here. I'm going to just simplify it. It becomes a minus of x square And I will have two x plus two x four x two x minus four x is minus of two x And I'll have a plus three And I have a plus three. Okay Okay Now let us look at this expression in the numerator very very carefully. Let's let's scrutinize the expression in the numerator You know that uh minus of x square This I could write it as x square plus two x minus three which you can write it as x uh Plus three minus one okay Now if you draw the wavy curve of this wavy curve of this You know that It will be positive beyond one negative in the interval minus three to one And again positive over here But if I have to draw it for negative sign So this is just for wavy curve for wavy curve for x square plus two x minus three But if I have to draw the wavy curve for negative one what I will draw is I'll draw the opposite one That means for negative x Plus sorry minus two x Plus three I would draw this as minus plus minus Okay Now your domain actually falls in this interval minus one to one minus one to one falls in this interval Right, let's say there's minus one over here. So minus one to one falls in this interval Which means this expression is always going to be positive. This expression is always going to be positive Now in light of all these we can say that This entire derivative of this function is positive Which means f of x is an increasing function f of x is an increasing function If it is an increasing function its domain can easily be found out by substituting minus one And one into your function that will give you the maximum and the minimum value of the function Right, so when you put a minus one When you put a minus one you realize that uh, this entire thing will become So let me put it over here So when I put a minus one f of minus one gives you A one by pi this will become minus pi by two this will become pi by minus pi by four minus pi by four And this will become a zero Okay, so this becomes minus three by four if i'm not wrong Okay On the other hand if I put f value as one I will get one by pi Pi by two plus pi by four And this will become two by four plus four plus five So that's going to be uh Three by four Sorry, uh one plus two. I'm sorry. I'm sorry. This will become One plus two plus five That's going to be one by a two by eight, which is one by four. So that's going to be one again Okay, so the range of this function the range of this function Is Belongs to the interval Minus three by four to one minus three by four to one. So how many integers are possible The integers possible in this interval are zero and one only so there are only two integers possible So your option number c will become correct over here. So your option number c will become correct So the first one to answer this correctly was Ram Charan Well done Ram Charan very good. Is that fine guys? So hope there is no question with respect to this Great So now we can move on to the fourth question of the day again Be very very careful while reading the question. These are gif brackets. These are all greatest india function brackets So these are all gif Brackets these are fractional parts. These are fractional part So there's a function defined as this and it's an even function Then the sum of all the possible values of a is Again type in your response with the question number attached to it Let's see who is the first one to answer this All of you know the sgn function sgn function is the signum function So this is your signum function. So signum function gives you 1 0 or minus 1 depending when x is positive 0 or negative All right. So athmesh Purvik have given in the response I need three more people to respond before I start solving this All right. So moving on in this question See guys, if you see carefully this function has an x cube over here x and a tan x into signum x Okay Now both signum x and tan x are odd functions, isn't it? So I've already shown you that signum x and tan x are odd functions Now the product of an odd function with an odd function always gives you an even function So this is an even function However, we know x cube and x both are odd functions And since my entire function has to be an even function, can I say The x cube and x terms should not have been there The x cube and x terms should not have been there That means my term Gif of a square minus 5 gif of a plus 4 should have been 0 And so should be 6 times fractional part of a square minus 5 times fractional part of a plus 1 should have been right Right. So only when these two terms are not there My function will clearly be an even function It cannot happen that I have an odd function plus an even function giving you as an even function Remember odd plus even is neither Okay, only even plus even can give you an even function So in that case these two terms should not have been there Right, which implies if you treat this as a quadratic in gif of a you would get something like this gif of a minus 1 times gif of a minus 4 Equal to 0 That means gif a of a could be 1 or it could be 4 and similarly this can also be factorized This is also factorizable as 3 times Uh fractional part of a minus 1 times 2 times fractional part of a minus 1 Equal to 0 which implies fractional part of a could either be 1 third or half Okay, so what are the permutation combinations possible over here? So a which we all know is nothing but gif plus fractional part could be what are the possibilities The possibilities are 1 plus 1 by 3. It is 1 plus half It is 1 4 plus 1 by 3 and it is 4 plus half Right in other words your a could be some of all the a could be some of a could be a 1 plus 4 twice And half plus 1 third twice That's going to be 10 this is 5 by 6 twice will be 5 by 3. So your answer will be 35 by 3 Which is your option number d option number d is correct in this case Okay, option number d is correct in this case So the first one to answer this question correctly was atmesh maha patra very good Well done So guys any questions any concern with this this was a simple question Right just a use of the even odd concept over here that The sum of two even functions will be even sum of two odd will be odd But if you add an even to an odd it will result into neither So that was the concept which was getting tested over here Okay So without much waste of time we'll move on to the fifth question of the day So read this question carefully and be our natural numbers and we have been given a function Sine plus cos and it is periodic with a finite fundamental period Then what is the period of this function? Then what is the period of this function? Take your time. No need to hurry All right. So sondarya is the first one to answer. So She says five c Okay, saimir says this five c atmesh five b Okay, so shares backs is a cool fellow He also says five b So Okay I need one more response. I need one more response All right. So pati kondanya also says five b. Okay So guys, let me just start this conversation the solution for this Now guys remember here that under root of this and under root of this right If at all this is a third that means in irrational quantity, they will never be periodic that means if Under root of a square minus three doesn't belong to a rational number and this doesn't belong to a rational number Then your function Then your function f of x can never be a periodic function. That's something that we need to understand first of all correct which means Under root of a square minus three is a rational number That is a square minus three should be a perfect square It must be a perfect square. Okay, and so should be b square plus seven Okay, so can you suggest me some natural numbers? For which a square minus three is a perfect square Okay, I can think of as two I Can think of as two correct Is there any other possibility? Yeah minus cannot be accepted because we are talking about natural numbers over here Okay, now there's no other possibility right because between let's say I have the perfect squares as one Four and then it's nine. So this gap will continuously keep on increasing. So I cannot have you know A gap of lesser than three between two perfect squares. So the only possibility is a is two and here the only possibility is B is going to be three because nine plus seven is going to be sixteen. That's going to be a perfect square Right, so you'll not find any other case where you would realize that a square minus three and b square plus seven Are perfect squares apart from this scenario correct So that means your function itself is actually sine of Okay x And cos of Forex Isn't it now Sine of x is periodic with a period of two pi and cos of four x is periodic with a period of pi by two So we have to take the lcm of two pi and pi by two which is clearly going to be two pi That means your option number b is correct in this case Your option number b is correct in this case So, uh, well done. Uh, the first one to answer this question correctly was atmesh maha patra again Okay bingo Is that fine guys any question with respect to this? Again, this is not a difficult question. It just required a bit of analysis. That's it So now we'll move on to the sixth question of the day So here we have a question where there is a function defined from r to one to infinity And is defined as log of under root of 3x square minus 4x plus k plus 1 plus 10 whole base to the whole to the base 10 This function is subjective. This function is subjective. Then what is the value of k? Or in which interval does k lie Okay, so I've started getting the response b b b b b. Okay, everybody is going for b So, uh, we know that for the function to be surjective That means the function to be onto the range should be equal to its codomain Right, so range must be equal to its codomain for the function to be surjective That means the range of this function should be from one to infinity Right, that is log of this quantity 3x square minus 4x plus 1 plus 10 To the base 10 should be belonging to the interval One to infinity Right, which means If I remove the log if I remove the log under root of 3x square minus 4x plus k plus 1 Plus 10 should belong to the interval 10 to infinity Right, that means under root of 3x square minus 4x plus k plus 1 should belong to Again zero to infinity right Yes or no Which again means Which again means 3x square minus 4x plus k plus 1 should belong to zero to infinity Right That means this function Which is an upward looking which is an upward opening parabola It's it opens upwards Right, it's an upward opening parabola It should Be such that it should touch the function like this touch the x-axis like this. That's why its range is from zero to infinity Yes, guys, is the screen visible? Okay, can you hear me? Am I audible? All right, great Okay, sorry for the technical glitch. We'll move on to the next question. Which is the question number seven All right So we have the next question. Which of the following function is a one one function? Which of the following functions is a one one function? Yeah guys, uh In case you have missed out the explanation for the previous one I will definitely explain you after the end of the session So wishes, uh, I'll explain you towards the end of the session. Don't worry about it The seventh question. Please, uh, type in your response along with the question number 7 c Okay All right, so most people are going for option c c c c at me sondaria prathik Sanjana One more I need guys to start discussing this problem All right, so let's try to analyze each of these functions one by one Okay So this function The only point of contention is whether x is positive or x is negative. Okay So when x is exactly zero If I talk about the first function that is e to the power of a signum x Plus e to the power x square uh, when your x is equal to Now this function behaves as Uh e plus e to the power x square when x is positive Okay, this behaves as uh When x is zero And it behaves as one by e plus e to the power x square When x is negative, okay So again, when you draw the graph for such a function When you draw the graph for such a function, it's going to pass through zero comma two And at the same value zero, this will be at Uh one plus e so it will be going in this manner Okay And the last one will be one by e so it'll start with one plus this is actually one plus e zero comma one plus e and this is actually Zero comma one plus one by e and this will start moving in this direction This will start moving up correct Now clearly you will see that at a certain stage this function will not be a one one function because it will be cut by a horizontal line at more than one place It will be cut by a horizontal line at more than one place And if it is cut by a horizontal line at more than one place the first answer Okay, so option number a cannot be correct What about option number b? Option number b again, let me redefine the function first So your function f of x which is e to the power x square plus mod of x This function behaves as in fact, you can clearly see that when you put x value as Let's say one or minus one you get the same answer, right? Here we can directly check by putting some special values when you put one you get e to the power one Plus one which is e square Right and the same value you get when you put minus one Correct, which is actually e to the power one plus one again, right? So of course since you are getting Same value of the function for two different inputs second function cannot be a one one function So this cannot be a one one function for sure Okay What about the last one? So here this function Is defined from three to four and this function is defined like this Now again when you're defining it from three to four, we can actually redefine this function We can actually choose to redefine this function So in the interval three to four, this will behave as x minus one This will behave as x minus two. This will behave as x minus three But this will behave as four minus x right, which means your function will be actually One and one will get cancelled two x minus two So two x minus two has to be a one one function because it is increasing And you can also see that when x is three Your value of the function is four and when x is four the value of the function is six So basically this function is a one one function This function is a one one function And if it is a one one function option number c will become correct in this case Okay So those who have answered it with c The answer is correct Okay, so the first one To answer this was arthnes now arthnes is saying even d I think as there is only one element in the domain of the function So let's check about d. What is there in d function? Uh, if you look in the d functions Yes, this function will be valid Your function d will be valid when ln of cos of sin x Would be a positive function, right? Which means cos of sin x should be greater than equal to one However, cos of any value cannot be greater equal than one. That means cos of sin x should only be equal to one That means sin x could be zero And actually that can happen for many values arthnes. It can happen for all multiples of n pi, right? Right, so there are so many inputs which will give you the same output So it cannot be a it cannot be a one one function. So option number d is a many one function again Right. In fact, you can call it as a constant function Right. So only option c can be correct in this case Not even option d Is that fine Okay, so well done those who have answered you have answered it absolutely correct And the first one to answer was arthmesh Good So let us now move on to the eighth question of the day So again, we have a function here from defined from a positive real numbers to a set of inputs minus 101 Of course, it's a signum function Okay, signum of this input x minus x to the power four plus x to the power seven minus x to the power eight minus one Then we have to comment on the nature of this function Many one on two many one into one one on two one one into All right. So amok says Option b. Okay Note it down amok Oh my god now Sad getting b b b b from everyone. Okay Well done. That was a good turnaround time. Let's try to solve this now Now, uh, since we're talking about r plus, okay, and since we're talking about the values of x I'll be actually trying to test the waters for different smaller smaller intervals of x. So first I'll start with Let's say my x belongs to zero to one Okay When max belongs to zero to one The few things which I can observe here is that First my x minus one would be negative. That means this minus this term would be negative correct Can I say x to the past seven minus x to the power four will also be negative Because x to the past seven will have a lesser value than x to the power four. This will also be negative Right, and of course minus x to the power eight will be negative. Right So when I add them all When I add them all that means I make x minus x to the power four plus x to the past seven minus x to the power eight Minus one. This would be a negative term Okay, so this will give me a negative value right So if this gives me a negative value, that means signum of this function Signum of this function is going to end up giving me negative of one Okay Now let's analyze what happens when your x is Greater than one. Let's say one to infinity. Okay. Let's see what happens Now when x is from one to infinity, uh, we can conclude that Of course x will be less than x to the power four Right, so this will be less than this and similarly x to the past seven will be lesser than x to the power eight Yes, I know That means x minus x to the power four would be negative And so would be x to the past seven minus x to the power eight And of course minus one is already negative So when you add them all once again When you add them all once again Again, you realize x minus x to the power four plus x to the power seven minus x to the power eight minus one Is a negative quantity That means signum of this function would be minus of one Signum of this function would be minus of one Right Okay, now let us see what happens when I put x as exactly one When I put x as exactly one what happens So it becomes one minus one plus one minus one minus one That's again minus one So a signum of minus one is going to be minus one again Okay So what do I see for many inputs? I am getting the same I'm getting the same output. So it's actually a constant function So it's actually a constant function And if it is a constant function, it has to be a Many one function It has to be a many one function and moreover I see the range of this function is always minus one the range of this function is always minus one That means the codomain is not equal to the range And hence it cannot be an onto function And if it is not an onto function, it has to be an into function Which means the function is many one and into it is many one and into Again, let's see who's the first one to answer the first one to answer this question was Amok Bharadwaj. Amok Bharadwaj Well done Amok Is that fine guys? So you need to put in some analysis You just need to put in some analysis while dealing with these special types of functions Right. Well, great. So now moving on to the ninth question of the day So here we have a function g of x defined as four cos to the power four x Minus two cos to x minus half cos four x minus x to the power seven whole to the power of one by seven So find g of g of 100. So basically it's based on composition of functions It's based on the composition of functions So let's see who is the first one to get this right Here I want to response from other other people as well. I can see only few people answering it Aditya Sondarya Ramcharan Lalitha Please back up guys Rithvik. Okay. So wishes gives a response Need four more responses Okay, so Purvik backs his fellow nafler Okay, Sai Need two more response All right. So mostly Janta is going with option number D. Okay So we'll do one thing. We'll try to first Simplify the function g of x. So let me just simplify this function g of x Okay, especially this expression for cos to the power four x minus two cos two x minus half cos four x So I can write this as four cos to the power four x minus two Uh, this will become our two cos square x minus one and minus half Uh, two cos square two x minus one Okay, which again can be uh For the simplified minus four cos square x plus two Uh, and this becomes minus cos square two x Uh plus half and cos square two x I can further simplify this to be Two cos square x minus one Okay, uh once again, I think add to Two cos square two x minus one the whole square. Okay, let me just correct this up So it'll be Yeah, it will be this now when I open the brackets. I get four cos to the power four x minus Four cos square x plus two and I'll have a minus four cos to the power four x Plus four cos square x. Uh minus one plus half Okay, so this term and this term gets cancelled this term and this term gets cancelled and I will be left with uh One plus Half which is three by two So this term is actually three by two So your entire function g of x could actually be written as three by two minus x to the power seven hold to the power of one by seven Okay, so this was just uh Um inflated version of this, you know simple looking function over here. Okay Now we have to find g of g of hundred Okay, so g of g of hundred first of all, let's find g of g of x So g of g of x means you are substituting this in place of x over here So it's going to be three by two minus in place of x will have three by two minus x to the power seven Hold to the power seven whole raise to the power of seven Whole raise to the power of one by seven Okay, so this and this gets cancelled So when this and this gets cancelled you have three by two minus three by two plus x to the power seven Whole raise to the power of one by seven Which is actually nothing but x That means g of g of hundred will actually just be Just be hundred This will just be hundred. So of course option number d becomes uh, your right option So the first one to answer this was Vishishthra, Vishishthra is the first one to answer this well done Vishisht Okay, again Simple-looking question, but again would eat up a lot of your time in simplification if you are not fast enough So that that's what happened in the previous j e main sitting Uh people who are used to using calculator are not too fast with their calculations lost a lot of marks So let's move on to the next question, which is your 10th question for today So this question is based on inverse As I can see from the options So a is a positive number greater than one And f of x is defined as log of x square to the base of a x greater than zero if f inverse is the inverse of the function b and c are real numbers What is f inverse b plus c f inverse b plus c? Oh, I'm always already up at the answer So amore has given the response for this question All right, so mostly people are bombarding with option a a a So time to discuss this quickly So this is the function. We know that if this is the function The inverse of this function is very easy to find out make x the subject of the formula So a to the power y is x square. That means x is a to the power y by two correct so f inverse f inverse x is going to be a to the power x by two Okay So when I say f inverse b plus c, I mean a to the power b plus c by two Which is nothing but a to the power b by two into a to the power c by two Which is nothing but f inverse b Into f inverse c Okay, so easy question over here option number a is correct So without much waste of time we'll move on to the 11th question now So in the 11th question, uh, there is a piecewise function f of x Okay Defined as minus of x plus one when x is less than equal to zero and minus of x minus one whole square when x is greater than one Then the number of solutions of f of x minus f inverse x equal to zero is Again another question based on your understanding of inverse But this time the function is a piecewise defined function So be very very careful while dealing with the inverse of such functions Okay, atmesh Says a option. Okay Note it down Okay, so how you backs it up Same as purvik I need two more response to start discussing this problem Okay, so mostly janta is saying option a a a a a All right, so let's discuss this question now, uh So we have been given a function f of x. Let's find out the inverse of this first Let me first begin with the first part of the function minus of x plus one So y is equal to minus of x plus one when x is less than equal to zero So if I make x the subject of the formula, so x is nothing but one minus y Okay And now when you say x is less than equal to zero, which means one minus y should be less than equal to zero Which means y is greater than equal to one Okay So in the inverse of this function, let me call it as y I can write this as one minus y for y greater than equal to one In a similar way, uh When you call y as negative x minus one whole square for x greater than equal to one So I could say minus y is x minus y whole square So under root of minus y One plus Okay Uh, this is going to be your x Okay, since x is greater than equal to zero. I'm not taking a negative sign over here Please note that I could have written plus minus over here But I'm not writing negative sign because I have a restriction on x that is x is greater than equal to one correct So this is possible when x is greater than equal to one that means One plus under root minus one is greater than equal to one. That means under root of This should be greater than equal to zero. Which means first of all y should be negative That means y should be negative So I can redefine the function as One plus under root negative y when y is less than equal to zero. So this is your f inverse y So from here, I can always get f inverse x So f inverse x would be nothing but One minus x when x is greater than equal to one And one plus root of negative x when x is less than equal to zero okay Now in order to solve this equation, it's like saying that where do f of x And f inverse x meet right where does the graph of f of x and f inverse x actually meet each other Okay, so let me plot these two graphs and see where do they actually meet Okay So let me plot these two graphs Let me just take up this space to plot it Okay So minus of x plus one is going to be just a line like this which is passing through Zero comma one But remember I have to only plot this graph when x is greater than Sorry when x is less than equal to zero x is less than equal to zero means I can only plot this graph in this part Okay, so this is your one minus x line y is equal to one minus x line. Okay Now, uh, the second part of the graph which is actually x minus one the whole square Okay, minus x minus one the whole square Minus x minus one the whole square when x is greater than equal to one So let me just draw that as well X is greater than equal to one it will be behaving as a downward opening parabola like this So this is your function correct Now even Even if you don't decide to find this function, you can actually plot the graph because you know that Your f inverse x graph is going to be the mirror image About this line. It's going to be the mirror image about this line Okay, so if you're drawing going to draw the mirror image about this line, let me draw that in uh, yellow color Let me draw that in a blue color If you're going to draw the mirror image of Let's say this parabola Right, so the mirror image of this parabola would appear to be This line Correct yes or no And the mirror image of this line is going to be This line right, which is again going to be the same line just a Part of that line when x is greater than equal to one. So this is The blue one is The blue graph is the graph of f inverse x The white one is the graph of f of x now see where do they intersect? Where do these Two graphs intersect I can see them intersecting at zero Right, that's one place Okay, I can see them intersecting here Again, I can see them intersecting over here And over here So I can see four points where these graphs are actually intersecting each other Okay, so the graphs are intersecting each other at four points means there are four possible solutions Of f of x equal to f inverse x that means option number d is correct So I think ram charan is the only person who gave this answer as the right answer followed by purvik Is that fine guys to be very very careful? However, this was not required this would this was not required Had you played just with the graph you could have solved the problem But just for you know better understanding and clarity of the concept I I gave you the function also if you know f inverse function also Okay Great. So now we can move on to the next problem That's problem number 12 Okay, so problem reads like this The graph of the function y equal to g of x is shown in the figure and there is a function f of x Defined as minus 3x square minus kx minus 12 k belonging to real number If fog is positive for all x belonging to real numbers, then the least integral value of k Then you have to find out the least integral value of Okay Yes guys any response anyone Okay, so ram charan aatmesh they go for option a So does resist Shares also goes for option a all right. So mostly people go for option a Okay, so let's discuss this now Now if you look at the graph the graph actually tells you the range of g of x Right the g of x happens to lie between minus 2 to minus 1 Right, isn't it As you can see the function becomes asymptotic It becomes asymptotic At minus 1 Okay, so it is between minus 2 to 1 so square brackets at minus 2 round brackets at minus 1, okay now When you say f of g of x is positive that means you're trying to say f of minus 2 That's also positive and you're trying to say f of minus 1 Okay, now minus 1 is not a part of it. So minus 1 could actually be Uh greater than equal to 0 now If you look at this quadratic expression If you look at this quadratic expression f of 0 is going to be minus of 12 Right f of 0 is going to be a minus of 12 So all these actually tells us about the nature of the quadratic expression that we'll be getting So f of 0 is minus 12 means the graph is cutting at 0 comma minus 12 correct and since coefficient of x square is negative It's going to be a downward opening parabola, right? And let us say minus 1 is here and minus 2 is over here. Okay Okay, now the value of the function before this is positive Okay, so At minus 1 it could actually become zero or it could actually be greater than zero So there could be two situations It could be a case like this Okay Or it could be a case like this. So I'm just redrawing it in a dotted fashion Or it could be a case like this Okay, so there are two possibilities of your function f of x So if these are the two cases I can say I can use these two conditions f of 2 minus 2 is greater than 0 and f of minus 1 is greater than equal to 0 Okay So f of minus 2 greater than 0 if I use what will I get? So let me put the value over here. So I'll get a minus 12 Plus 2k Minus 12 greater than 0 That means 2k is greater than 24. So k should be greater than 12 Okay On the other hand, if I use f of minus 1 greater than equal to 0, I would get Minus 3 plus k minus 12 greater than equal to 0. So k should be greater than equal to 15 Now both these conditions should be simultaneously true this and this should happen So this is simultaneously true. That means the intersection point is this point k should be actually be greater than equal to 15 So your option number c becomes correct So your option number c becomes correct not option number a so almost all of you have answered this incorrectly Okay So now let us talk about The next question which is question number 14 Oh, I think this is question number 13 by the way My mistake. I wrote a 14 over here. Yes anyone Any idea how to do this Okay, so size says Option c Okay So this is the question which is based on the functional equation which is again a very important part of the function chapter So I need four more response. Okay. Amog has also come up with an response Okay, atmesh goes for c Come on guys. I want to see a response coming from almost everybody Aditya, Ramcharan, Kondinia, Vishis One more response guys before I start solving this All right. So what we'll do is first We'll work out for some special values So we have been given this function Okay, this is f of f of y Plus x f of y Plus f of x Minus one for all x and y belonging to real numbers and find f of two So let us first do one thing. Let's put x and f of y both as zero zero each Let's put this and this as zero zero each Let's see. What do we get the value of f of one? Sorry f of zero as So when we do that we get f of zero minus zero, which is zero And this will give you f of zero again. This will give you zero and this will give you again f of zero minus one Okay So obviously f of zero will now become One f of zero will now become one Now what I'm going to do next is I am again going to put x is equal to f of y So let us now put x equal to f of y in this particular expression Okay, so first thing was this expression and second is my this substitution x is equal to f of y So when I do that I get f of zero So Okay, this will become f of x correct, this will become x square And again, I have an f of x minus one Now f of one value. We already know it's one Here I get two f of x plus x square minus one That means two f of x can be written as two minus x square Which clearly implies your function f of x is one minus x square by two So this is your final function that you are actually going to use here So the question demands f of two So just put x as two it becomes one minus four by two which is nothing but one minus two which is minus one That means option number c becomes correct in this case. Let's see who was the first one to answer this correctly So amog was the first one to answer this correctly as 13 option No, I'm sorry Simear was the first one to answer this so 13c that is minus one. That's correct So well done Simear Good, so with this we move on to the 14th question now so This is a 14th question. There is a function f of x defined from r to rational numbers And there's another function g also which is defined from r to rational numbers and they are both continuous functions their continuous functions Such that under root three f of x plus g of x is four And one minus f of x cube Plus g of x minus three whole cube is equal to which of the following options So guys, I'm not going to give you a break today. I'm just going to finish it up 15 minutes early So rather than giving you a break, uh, let's get the other session over a little bit early Okay, so poor wick has given a response Okay, Simear Kondinya, Ramcharan, Aditya Naman, I haven't seen Naman answering it. Naman, where are you? Are you there in the session? Okay, so Aditya also goes for b option. I need one more response Prisist Atmesh Cheers sanjana Sondarya Vaishnavi Okay Still solving guys. I think this is a pretty straightforward question, but slightly, you know It's a tricky question because here you have to just do some analysis You know, the function output is always rational, right? It'll always going to give you a rational output. Correct Now root 3 is an irrational number Correct. So unless and till f of x becomes irrational again, you cannot have You cannot have this sum as a rational quantity So when I say root of 3 into a rational number Okay, plus a rational number Is giving you a rational number Right Isn't it because 4 is rational, g of x will always give our rational output f of x will also give a rational output And this root 3 is actually irrational This is actually irrational, right Now that can only happen if this rational quantity is actually zero Right, if it is not zero, that means some element of irrationality will exist over here and your answer can never be a rational quantity So the only possibility is your f of x can be a function which is zero. So it's a constant function So it is a constant function because it has to be a continuous function So no matter whatever is the input your f of x will always be zero Right Okay Now Which clearly implies g of x will also be a constant function and that will always be four because then only zero plus Four will be giving you four. So this is also a constant function right So one minus f of x would be like one minus zero cube And g of x minus three whole cube will be four minus three cube Which is actually one cube plus one cube that's going to be two That means your option number b is going to be correct Right, wasn't it a super simple question? Right and the first one to answer this was Purvik Well done Purvik very good So let us start with the next question which is question number 15 for the day Again, this is a functional equation that I have given you So there's a function from r to r satisfying f of x into f of y minus f of x y is equal to x plus y And f of one is always greater than zero then which of the following option is correct Okay, so our page Has come up with a response Okay, Purvik also All right, sorry Guys these these sessions are best to be you know attended life So people who are missing out on this session thinking that you will watch the replay of this session on youtube Let me know it will not be that effective So the efficacy of this pitaya process lies when you are participating and solving it real time All right, so mostly people are saying c c c c So let us discuss this So first of all when I'm dealing with a functional equation, I'll try to figure out f one because f one is something whose information is given to us directly or indirectly So I'll put x and y both as one in this case. Let's see what happens So f of one into f of one, which is f of one square Minus f of one is equal to one plus one Okay So f of one square Minus f of one is equal to two So you can treat it as if you are solving a quadratic k minus k is equal to two Right, which is actually factorizable as a k plus two, sorry k minus two into k plus one So this gives you two values of k one is minus one and other is Two that means f of one can either be minus one or two, but I've been given that it is always positive So this is rejected. So f of one is going to be your two. No doubt about that Okay Now what I'm going to do next is I'm going to put y value as one So these hidden trials you have to keep on doing while solving questions So I'm going to put y value as One and keep x as x. So I'm just going to put y value as one keep x as x itself So I will get f of x into f of one which is actually two minus f of x is equal to x plus one That clearly gives you the function f of x as x plus one Right If f of x is x plus one its inverse Will also be x plus one only Sorry, it will be x minus one Correct you can always take this as y make x the subject of the formula So y is equal to x plus one means x is equal to y minus one So f inverse y is equal to y minus one. So f inverse x is going to be x minus one Right now we have to see in light of these two information that we have f of x equal to x plus one and f inverse x is equal to x minus one Which of the following options are correct Of course a is not correct b is not correct. Yes c is correct. So option number c becomes the right answer in this case and the first one to answer this was Atmesh Mahapatra Well done Atmesh Okay, any question any concern with this question was a simple question though So now let us move on to the 16th question So this question again says f of x square minus six x plus six plus f of x square minus four x plus four is two x Then f of minus three plus f of nine minus five times f of one is equal to Okay, rationally says option c Hey guys, can you see this screen? Yeah, sorry, there was a power cut now. Can you see the screen now? Can you see it? Okay Yeah, so i'm getting the responses as option c from most of you So shreyas, uh, sorry sigh Aditya vishast Have given this have given the responses. What about others guys purvik? Okay. I need one more response now Getting the answer as four four is in not in option a b c or d. Okay Check out page once again your calculations Getting the answer is one All right. So guys, let's let's discuss this let's discuss this. Okay Uh, first of all When we look at this function I need to get one minus three And nine, okay So first of all what I'll do is I'll try to put x square minus six six plus six as one And when I do that I get x square minus six x plus five is equal to zero Which is nothing but x minus two times x minus Sorry, uh, it's factorizable as x minus one and x minus five Okay, so that will give you the root as x equal to one or x equal to five In a similar way if I try to make this expression as one Okay, that implies x square minus four x plus three equal to zero Which is nothing but x minus one times x minus three equal to zero So I get the value of x as one and three in this case now So if I put the value of x as one If I put x as one, what do I get this becomes f of one This also becomes f of one And this becomes two into one Which clearly implies f of one is actually one Okay, let me call this as my first you know equation Now let us try putting x as three When I put x as three This expression becomes nine minus eighteen Nine minus eighteen plus six, which is actually f of minus three Okay This becomes f of one as we have already found out that the root of this equation is three So this will become f of one And two into three, which is going to be six Okay, so this implies f of minus three plus one is equal to six, which means F of minus three is equal to five. Let me call it as the second equation So I've been able to get f of one. I've been able to get f of minus three Okay, let's try putting five. Let's see what happens when we put five When we put five I get f of 25 minus 30 plus six that is actually f of one And this will become 25 minus 20 Plus four, which is actually nine And we'll have two into five, which is 10 So this implies f of nine could be written as 10 minus f of one, which is actually nine itself Okay, let me call this as three Now I've got all the required Values that I need to solve this problem f of minus three. I have got which is actually two Plus f of nine. I've got which is actually nine minus Five f of one five f of one Sorry, f of minus three is five. I'm sorry This is five And f of one is one. So this is minus five. So five and five gets cancelled leaving you the answer as nine So option number c becomes correct in this case Option number c becomes correct in this case Um, the first one to answer this was Vaishnavi Deshpande Okay Right well done Vaishnavi very good Again the concept was very easy. You just had to try out the values of x Moving on to the next question Question number 18 Sorry question number 17 for the day so f of x A function continuous even periodic function f of x with a period of eight Okay And f of zero is zero So it's an even periodic function with a period of eight. We have been given f of zero is zero f of one is negative two f of two is one f of three is two f of four is three and we have to find the value of tan inverse Of tan of f of minus five plus f of 20 plus cos inverse of f of minus 10 plus f of 17 Okay, so I've got a response now Okay, so atmesh has also given a response Chotas Lalitha All right, so let us start the discussion for this problem Now we have been asked to find out the value of this expression now We can see that since the function is periodic, right So you know that the function is periodic So can I say the value of f of minus five Would be the same as the value of the function at f of minus five plus eight Right because it's periodic with a period of eight Right So we'll have the same value as f of three f of three is actually two Okay, so this value over here. We can write this value as two Right In a similar way f of 20 f of 20 would be the same as f of four plus 16 Right, so it'll be same as four because f of four plus eight plus eight It's going to be same as f of four And f of four is given to us as three. So this value will be three right f of minus 10 would be Same as f of minus two Correct and f of minus two would be same as f of two Because it is also mentioned that the function is even in nature That means my answer is going to be One for this Yes or no, okay What about f of 17? f of 17 would be same as f of one Right because 17 can be obtained by adding 16 to 1 And f of one is given to us as minus of two Okay, so these values now are available to us. This is three This is cos inverse of cos inverse of Yeah, this is the bracket is over here. It closes over here And this is cos inverse of zero. Sorry cos inverse of one, which is going to be zero and f of 17 is minus two That means this will actually lead to This will actually lead to tan inverse of tan three This will actually lead to tan inverse of tan three when I say tan three means three radians This three is in radians Okay, and if you try to recall your tan inverse tan theta graph Right. Hope most of you remember the graph of this It's like this So this is going to be pi by two. This is going to be minus pi by two Uh, this is going to be pi. So basically you are in this branch. You are in this branch Okay And the equation of this branch will be uh, y is equal to Uh, pi minus x sorry y is equal to x minus pi y is equal to x minus pi Right So tan inverse of tan three is going to be three minus pi. That means option number d is going to be the correct answer in this case So the person who got this right the first one to answer this right was amog Amog bardwaj again some of you are Getting the answer right every time others. I need response from you as well So moving on to the next question, which is the last question for this particular session That's question number 18 for you So we have again a piecewise defined function f of r defined as 2x plus, uh, alpha square minus x is greater than equal to two and uh alpha x by 2 plus 10 when x is less than 2 if f of x is onto function then alpha belongs to which interval All right. So almost everybody is giving the response as option number b I think one person gives the answer and other other people follow after that Okay All right, so let us discuss this question. Uh, if you look at the let's let's do a graphical analysis of this question Okay So, uh, let's do a graphical analysis of this question Now 2x plus alpha square when x is greater than equal to two is definitely going to be a straight line Right. So let's say this is x equal to two And we'll have a straight line having a slope of 2x Okay, and having a positive intercepts because alpha square will be positive. This will be a positive term Okay, now what about the other line? Alpha x by 2 plus 10 Now Since my alpha is not known to me. It can be positive. It can be negative. I'm not sure about the slope of this line Right. I'm not sure about What is the slope of this line whether it's positive or negative is not known to me, right? Now can alpha be negative? Can alpha be negative Can alpha be negative? Just try to answer this question If it is a negative Then what will happen? The graph of the function of course this value of the function at zero should be 10 Right. So when x is zero it should be 10. So it should be cutting it over here Okay, we'll be cutting it over here at 10 And if it is negative what will happen the function will Go up. Okay, so it can be either like this Either it can go like this Or it can go like this Correct. Yes or no Right. So will it cover all the real numbers? Because it has to be onto if it is onto that means your core domain should be all real range should all be all real numbers So is it possible that your alpha is negative? The answer to this is no it is not possible Because then the entire real values will not be covered up Are you getting it point? Are you getting this point because when x is less than 2 What will happen? Right, it starts from some infinite finite value Okay Right and go upwards Yes or no So those who have said answer option b Be straight away rule out b cannot be your answer Okay, so do you want to change your response? I'm still giving you some time if you want to change your response Okay, it can start at the same ordinate also. I'm not denying it. It can start at the same ordinate It can be like this also, but how does it? So all real numbers are not covered right Okay, so alpha has to be positive in that case And if alpha is positive And if alpha is positive Again, I can have multiple scenarios. I can have My line starting like this and going down Or my line starting from here Or going down Or my line starting from here and going down Now again, which of these three possibilities a b and c cannot happen Which of these three possibilities a b or c cannot happen Remember your range has to be all real numbers. This has to be your range Okay, so art may say c cannot happen absolutely if c happens. That means this part will be left unmapped No, b can happen. Why not b can happen So what I'm trying to say that The value of the function at 2 that is 2 into 2 plus alpha square And the value of this function at 2 that is alpha by 2 into 2 plus 10 This should always be Greater than equal to this value because c is not possible because I'm ruling out the Possibility of c because it will leave to unmapping of certain values of y So your domain range cannot be all real numbers correct So which clearly means alpha square plus 4 should be less than equal to alpha plus 10 That means alpha square minus alpha Minus 6 should be less than equal to 0 Which is clearly factorizable as alpha minus 3 times Alpha plus 2 is less than equal to 0 That means alpha should belong to the interval minus 2 to 3 But as I told you it can never be negative Right, it can never be 0 also Right, so this possibility is ruled out. So you have to redefine it You have to retune it from 0 to 3. That means option number c can be correct Option number c can be correct So none of you got this problem, right? So again, these are some conceptual questions looks very simple, but you know, it may be very very tricky at times Okay, so guys These were a few questions. I would send you some more questions as assignments on this Okay, and do complete it and send me across on the group or personally to me So We have different different sessions lined up for every day. I think you have been already been given the schedule Oh, okay. Fine. I'll I'll stay in this problem once again So Purvik, is it clear why alpha can never be less than equal to 0? If alpha is equal to 0, the other function would be a straight line at 10 So it will never cover all the real values if it is less than 0 it will go up like this So the values below it will be left off That's why it cannot be less than equal to 0. So this is clear to you Next is if it is positive, they can be again three situations. It can go down like this It can go down like this it can go down like this again In situation number c what will happens if it is below this function That means there would be a gap left in the values of y. That means all real numbers will not be covered That means this situation c cannot happen So your other line that is alpha x by 2 plus 10 can either either be positioned in this fashion or been positioned in this fashion So only these two are the possible positionings that means your initial value that is the value of the function at 2 for the Both the function should be such that This function should have a higher value or at least the equal value as the function this at 2 So 2 into 2 plus alpha square should be lesser than equal to 2 into alpha by 2 plus 10 Okay, so this is 10 over here Then we solved it and we get an inequality over here and from here I got the this condition and again I ruled out this condition that x cannot be Sorry alpha cannot be less than equal to 0 and hence this answer comes out. So, uh As per the schedule the next class which is going to be on saturday We are going to do uh saturdays Sorry today is saturday next class is going to be on uh 24th of jan which is going to be on trigonometry and properties of triangles Okay, so request you guys to study it. Meanwhile, I will send you more problems on this chapter to practice Okay So over and out from my side Thank you so much for coming online Next class would be a slightly longer session. So we'll compensate for the time lost Okay Thank you guys. I'll I'll share this pdf with you all Oh, yeah css wishes to I'll explain you the sixth question once again Yes, as per the schedule if there's a test you'll be sent a test across All right, so this question is what wishes was asking Sorry sixth question. Yeah, so this function has to be surjective when you say function is surjective means range should be equal to codomain correct so This is your codomain and your function should belong to this interval That means this should belong to tend to infinity That means this should belong to zero to infinity Now this belongs to zero to infinity means x square minus four x plus k plus one should also belong to zero to infinity Now the range of the function is zero and above that means it is just touching your x axis So this is y equal to zero So it is just managing to touch your x axis touching your x axis means real and equal roots So for real and equal roots your discriminant should be greater than equal to zero Okay, so this should be greater than equal to zero right Okay, so if it is greater than equal to zero, this should be also be greater than equal to zero Okay, okay. I understood. So this would be k should be greater than So k here, uh, 12 k should be less than equal to four. So k should be less than one by three Less than equal to one by three Okay, now less than equal to one by three. It cannot include Any one of these situations it cannot include b Okay, because equal to equal to uh, one by three will be missed out This will also be not the answer. This will also be not the answer. So It has to be exactly one by three Because if you say less than less than one by three you are going to miss out on the value Okay, so it will not be a subjective function So great guys, so we'll uh, call up this session right now. Thank you so much for coming online Over and out from Centrum Academy Bye. Bye. Have a good day