Functional Equations Masterclass 1 | The Ultimate FE Toolkit | Setup, Substitutions & 40+ SE

[00:00:00]Today we'll study functional equations.

[00:00:04]Now functional equations they have been fundamental blocks of any competitive examinations including J advanced, ISI and CMI as well as mathematics olumpi.

[00:00:16]Now the first question here itself is what is a functional equation? So basically a functional equation is an equation whose unknown is a function and not a number.

[00:00:28]Say for example we are given f_sub_x + y = f_sub_x + f_sub_y for all xy belongs to r.

[00:00:43]Now answer to this equation will be a certain class of functions and that's why it's a functional equation. So when we solve a functional equation our answer is not a number but every function which satisfies this given relation. So when it says find all f it means first writing every solution. So we must write every solution to this given condition and second prove this list is complete.

[00:01:25]And there's nothing we have left behind.

[00:01:29]Now before studying functional equations, we'll study some prerequisites which is fundamental to the study of functional equations. Now the first thing is before solving any functional equation you must answer these three questions very clearly and the questions are what is the domain of this problem.

[00:01:52]Second is what is the co-domain of the problem and third is which condition of regularity is imposed.

[00:02:04]So what is the regularity? And when we say regularity we talk about continuity, differentiability, boundedness or monotonicity and such things.

[00:02:26]So before solving any problem, these are the first three questions that you should answer for yourself. Now the question comes why do we need to do this? Essentially these three things they determine which techniques we are going to use. Now a small example for this is this very important functional equation f_sub_x + y equals f_sub_x + f_sub_y for all xy belongs to r.

[00:02:57]This is the most important equation in entire functional equation and it is called as coochi equation.

[00:03:07]Now for this function if the function is defined on natural numbers to natural numbers the result of this function is f(x)= ax or some a belongs to natural numbers.

[00:03:25]If this function is defined from rational to real and there's no regularity given then this function is defined as fq equals aq where a belongs to r.

[00:03:47]Now if the same function is defined from R to R and we are given some regularity condition. Say we are given that the function is continuous. Then in that case this function f(x) lb ax where a belongs to r. But if we are given this condition f is defined on r2 r but there is no regularity given in the question then this equation will have uncountably many b solutions and we call these solutions as either wild solutions or ham bases and that is the reason for every functional equation we must start by noticing its domain, co-domain and conditions of regularity.

[00:04:45]Now functional equation is well defined if for the given setup there is at least one solution and ideally infinitely many or a parameterized family of solution.

[00:04:58]So a functional equation is well defined if it has at least one ideally finite solutions or some family of parametric equations and basically there are three failure modes where we say this setup is not well defined.

[00:05:21]First is when no solution exist.

[00:05:25]So if we have a functional equation where no solution is possible, it's a valid question but it's generally not a well- definfined functional equation.

[00:05:38]Second could be if the equation have trivally infinitely many solutions.

[00:05:56]And the third failure mode is when there are bite solutions or hammers bases which are hidden due to missing regularity condition.

[00:06:07]Suppose we have this functional equation f_sub_x minus f minus x it is equal to 1 for all x belongs to r. Now if you put x as 0 we'll get 0 = 1 which is impossible. So this equation it does not have any solution or suppose we are given fx= fx + 1. Now basically this condition it is true for any periodic function whose period is 1. So it'll give us trivally infinitely many solutions. And this wild solution case we have already discussed gochi's equation when a function is defined on r but no regularity is defined.

[00:06:59]Now before starting functional equations we'll study some prerequisites that we'll need. The first one is a function is injective or a function is a one one function. So we know that a function f will be injective if from f a= fb the only condition we get is a= b.

[00:07:28]Second condition is your subjective function or we say it is an onto function.

[00:07:38]So a function is an onto function if its co-domain is equal to its range or in simple words for every y in co- doain there is some x such that fx= Why?

[00:08:08]Now this third class is bjective functions.

[00:08:15]Now bjective is basically both injective and subjective. So when a function is both injective and subjective we say it is a bjective function.

[00:08:27]Now the fourth term is involution.

[00:08:35]If a function is such that f of fx is equal to x we say it is an involution.

[00:08:44]Now here we need to keep in mind every involution is a bjective function. So every involution is a bjection.

[00:09:00]And for such functions f_sub_x is equal to f inverse x. Now the fifth condition is it important.

[00:09:16]A function is it important if f of f_sub_x is f_sub_x itself. that is f is the identity on its own image.

[00:09:32]Now the sixth one is periodic function with period t which is greater than zero. So if a function is such that f(x) + t equals f_sub_x for all x in the domain then we say this function is periodic with period t and if this t is the smallest then we say it is its fundamental period.

[00:10:08]Now the seventh one is odd and even function. So odd functions and even functions.

[00:10:22]So if the function is such that f of min - x is equal to minus f_sub_x we say it's an odd function. But if f of minus x is equal to f_sub_x we say it's an even function.

[00:10:37]And finally we have fixed points.

[00:10:46]So fixed point of any function f is any value of x say x not such that f x is equal to x note. Another prerequisite that I recommend before studying functional equations is recurrence.

[00:11:13]So if you understand reoccurrence and the techniques required, it'll be of great help when you'll be solving functions questions because at times you will see your functional equation can be easily solved by converting it to a recurrence. We have already studied recurrence in sequences and series. So what I'll do is I'll put the link to that recurrence relations video in the description below.

[00:11:41]And I strongly recommend completing recurrence relation before studying functional equations. So far we have studied. First three conditions that we must check while solving a functional equation. First is its domain, then its co- doain. Third conditions of regularity and then we must know all these eight conditions. Now here the question is for these four equations we need to state the domain co-domain and any regularity hypothesis in each of the following problems without solving them.

[00:12:15]So we need not solve we just need to find domain and co-domain. Now for this first function domain is n co- doain is also n. We are given f_sub_2= to now f mn is equal to fm into fn where gcd of mn it is equal to 1 which is this function is multiplicative on co-prime arguments.

[00:12:47]So when gcd of mn is one then we can always write fm into n as fm into fn say f_sub_6 we can write this is f_sub_2 into f_sub_3 so it'll be simply f_sub_2 into f_sub_3 and then we have this regularity condition that if m is less than n then fn is less than fn so it is strictly increasing so we have regularity condition that it is strictly monotonic.

[00:13:20]Now for the second question, we are given this function f from r to r such that f_sub_x + y= f_sub_x into f_sub_y where f_sub_x is differentiable at x= 1.

[00:13:30]Now domain r co- doain r this is the functional equation and the regularity condition we are given here is it is differentiable at one point x= a.

[00:13:42]Now in this question C domain is R except 0 and 1 co- doain is R. We are given this condition and there is no regularity defined in this question. And this fourth one is domain integers co-domain integers.

[00:14:07]These are the condition. These are values given and again in this there is no regularity.

[00:14:15]So for these questions we are not supposed to solve them. We just need to find their domain co-omain and any regularity hypothesis. Now guys the question is a function is defined from r positive to r positive such that x + y f of f_sub_x into y is equal to x² f of f_sub_x + f_sub_y for all xy = to0. Now we need to show there is no function which satisfies this given condition.

[00:14:43]Now domain is r positive co- doain is r positive and there is no regularity to this equation. Now in this what we'll do is we'll first check if the function is injective that is we'll start with let f a equals fb. Now we can write a + y f of f a y will be equal to a² f of fa + f_sub_y.

[00:15:29]And if we put B in this equation, X as B, it'll be B + Y f of FB into Y, it'll be B² F of FB + F_sub_Y. Now since fa= fb f of f a y will be equal to f of fb y and also f of fa + f_sub_y will be equal to f of fb + f_sub_y. So we can write a + y upon a square will be equal to b + y upon b² and we can simplify this into y into b² - a square will be equal to a² b minus a b square. Now we can write this as y into b minus a b + a will be equal to a b into a minus b.

[00:16:43]Now this should be true for all y because this functional equation it is true for all xy belongs to r positive.

[00:16:51]That means it must be an identity and it'll be an identity if and only if both sides are zero and the only factor which is common to both is b minus a. So from here we can say a must be equal to b. So this function f it must be a one one function or an injective function. So that's our first part of the solution.

[00:17:16]Now in the second part what we'll do is we'll try and cancel out x + y with x² and for that what we'll do is we'll say there is some alpha such that 1 + alpha it is equal to alpha².

[00:17:38]Now in this equation we'll put x is alpha and y is 1 then we'll get alpha + 1 f of f alpha will be equal to alpha² f alpha plus f_sub_1. Now we have chosen alpha such that 1 + alpha is equal to alpha square and f is a 1 function. So we'll get f alpha equals f alpha + f_sub_1.

[00:18:11]Basically we'll get f_sub_1 equals zero.

[00:18:15]But looking at its co- doain its co- doain is r positive positive real numbers. So f_sub_1 it cannot be zero and therefore there is no solution to this equation.

[00:18:28]So this question is little bit advanced for an early discussion but it clearly shows how conditions of domain co-omain regularity and the eight terms that we have studied they are going to be used while studying functional equations. So as of now if this question is not clear to you it's perfectly all right.

[00:18:51]Now the question is a student claims f_sub_x + 1 equals fx + 1 for all x belongs to r and it also says f_sub_1 is equal to 1 which forces fx= x then we have to find an explicit counter example and identify the missing hypothesis that would make the claim correct.

[00:19:20]Now basically this functional equation it's a simple variation of coochi's equation. So it is f_sub_x + y equals f_sub_x + f_sub_y where they have just put y is 1 and is defined on r and since kom is not mentioned we'll assume it is r. So this is coochi's equation but without regularity. Answer it is without regularity and we know that without regularity this equation will have uncountably many wild solutions.

[00:19:54]We just need to provide a counter example in this and a simple counter example would be we'll take f(x) as x plus some other function whose period is one. Say fx is x + fractional part of x. We know that fractional part of x it's a period function whose period is one. Now we put it in this equation. Then fx + 1 it will be x + 1 plus fractional part of x + 1. Now we know that fraction part of x + n is simply fractional part of x. So it will be x + 1 plus fractional part of x. Now x plus fractional x is f_sub_x. So it'll be this f(x) + 1. So this function it satisfies this given equation. So the given conditions they do not force f(x)= x as only solution. There will be infinitely many y solutions. Now the missing hypothesis in this is hypothesis of regularity. So if we apply any regularity say continuity then we can say the only solution forced here is f(x)= x.

[00:21:13]Now we come to the toolkit that we have for solving a function equation and the most important toolkit that you'll ever need is the substitution toolkit.

[00:21:37]Now most of the functional equations for J advanced even ISI and CMI they'll be solved using substitution and particular for J advanced almost 80% problems they require substitution and for ISI and CMI it is usually the first dominant move. There are seven fundamental substitutions that are generally required to solve a functional equation. Now before studying those seven substitutions, we'll answer why are we actually doing a substitution and also what forces our choice of substitution. So, so there are three underlying questions for which we are trying to find an answer. And the first question is can I make an equation simpler by eliminating a term?

[00:22:35]So is there a substitution which is going to vanish the part of an expression?

[00:22:41]Second question is can I make the equation from two variables to one variable. So second could be reduction of variables.

[00:22:52]If the equation involves x and y and by putting y equals 0 our function equation reduces to one variable equation then nothing like it.

[00:23:02]And the third question is can I have simultaneous equations?

[00:23:10]So is it possible by making a substitution that we get two equations with two unknowns? So basically it'll become a simultaneous equation problem which we can easily solve. Now generally there is no order in which we should apply the substitution toolkit but it is preferred that you start with this first one and this usually works in most of the questions where 0 is in the domain of the functional equation and that is to put x as 0 and y as 0 and using this we generally find f0. Now in some cases rather than finding zero we'll find one.

[00:23:50]So it'll be either x as one, y as one or could be one of the variable is one and the other variable is zero to find the value of f_sub_1. So this is your first and the basic substitution that you'll need.

[00:24:04]Now the question is let f from r to r be a function satisfying fx = f_sub_1 - x and fx + 1 = fx + 1. Now this if we put x as0 we'll get f_sub_0= f_sub_1 and if we put x0 here we'll get f_sub_1= f0 + 1 which will give us 1=0 which is impossible that means there is no such function.

[00:24:46]So by simple substitutions we have proved that this functional equation has no solution. Second substitution is y = x. If the equation is in two variables by putting y = x it'll reduce to one variable equation.

[00:25:11]In some cases rather than putting y = x we'll put y = -x or generally y = a - x. Now what we'll do is it'll put x + y as zero. So from here generally we get the condition that a function is odd or even. So it expresses odd or even structure of the problem. Now the fourth substitution is swapping x and y. So there'll be some questions where if you just swap x and y you may get a simultaneous equation or you'll get a simpler condition. So by swapping and the swapping it won't work for symmetric equations. It'll work for unymmetric equations and these unymmetric equations they can give you some clean identity or some reduced form or maybe simultaneous equations also. Now the fifth substitution type is we replace x with some function of x.

[00:26:18]Now we do this generally to get simultaneous equations.

[00:26:27]Say for example suppose we are given a f_sub_x plus b f 1 upon x equals suppose x². Now we need to find f_sub_x.

[00:26:41]Now we know that if we replace x with 1x or if we use this function 1x then here this x it'll change to 1x and 1 upon 1x it will change to x. So it'll become a simultaneous equation. So if we replace x with 1x we can write a f 1x plus b fx it is equal to 1x². Now it has become a simultaneous equation in f_sub_x and f_sub_1 byx. Now you can eliminate f_sub_1 byx to find this function f_sub_x. So there'll be questions where the equation will be converted into simultaneous equations maybe in two cycles, maybe in three cycles, maybe in more cycles. There you need to determine this substitution that we need to apply again and again.

[00:27:38]Now the sixth substitution is when we substitute the unknown onto itself. So suppose we substitute y as fx or x as fx.

[00:27:56]So generally what it does is it gives us this condition of injectivity or sugjectivity.

[00:28:05]And seventh one is using any substitution which supports regularity hypothesis.

[00:28:17]Say it is given that function is differentiable. So we can find f d-0 and it may convert a functional equation to ordinary differential equations. So these are seven important substitutions.

[00:28:35]that we need to solve majority of our questions. Now we take examples of these. So for this first one, suppose we are given a function f which is defined on R2R such that f_sub_x + y equals f_sub_x + f_sub_y plus xy for all xy belongs to r.

[00:29:03]Now we put x as 0 and y as 0 we'll get f0 and it will be equal to f0 + f0 + 0 which will give us f0 = 0. So from here we get this condition that r f0 is zero and this f_sub_0 it is very important because it helps solving other conditions and when the equation is multiplicative say for example if it has x y then you'll try both x0 y0 or x = 1 and y = 1. Now we'll come to the second substitution type.

[00:29:47]Suppose in the same problem if we put y as x we can write f2x it is equal to 2 fx + x². Now it has reduced to a one variable equation and basically this is a reoccurrence relation in disguise. So this is so you can solve it using recurrence or or we can solve this question by converting it to coochi's form.

[00:30:30]So we know that f_sub_x + y = f_sub_x + f_sub_y. This is coochi's equation. We just have this extra term xy. So if we can change the functions which cancels xy it'll solve our problem and from this one we know this function f_sub_x or f_sub_2x it contains x² so it forces this substitution let f_sub_x equ= gx + x² by 2 because in x + y whole square we'll get 2xy and because we want to cancel xy we'll take this x² by 2. So f_sub_x + y will be g x + y + x + y whole square by 2. Now this is f_sub_x which is gx + x² by 2 g y + y² by 2 + xy. Now we open this cancel.

[00:31:34]So we'll get this equation which is gx + y = gx + g y which is coochi's equation and we have a standard solution of coochi's equation which we are not discussing now but now this problem it becomes solvable.

[00:32:00]Now the third substitution type is setting y = -x.

[00:32:06]Now in coochi's equation f_sub_x + y equals f_sub_x + f_sub_y. If we put y as min - x then it'll be this f0 and this is f_sub_x plus f minus x. and f0 we can find out by plugging in values x=0 and y=0.

[00:32:31]So from here we get f_sub_0=0.

[00:32:33]So we can write this as f of min - x= - f_sub_x that is this f_sub_x it is an odd function. So this substitution it has revealed the nature of this function. Now this fourth substitution was swapping x and y. Now for swapping y and x we can take a simple example. Say for example we are given this equation f x + g y it is 2x + y + 5.

[00:33:12]Now if we swap x and y we can write f y + gx it'll be equal to 2 y + x + 5. So if it is unymmetric structure by swapping x and y you can get another equation which may be used to solve the functional equation.

[00:33:40]Now fifth one was replacing x with some transformation of x. So we have this question where fx + 2 f 1 - x it is equal to 3x. Now we know that 1 - of 1 - x is simply x. So if we take this transformation 1 - x we'll get this another equation which is f_sub_1 - x plus 2 fx that'll be this 3 1 - x. Now we can multiply the second equation with 2 and we can subtract this first one. So this f 1 - x will cancel.

[00:34:23]Then we'll get 3 f(x). I'll be this 6 - 6x - 3x or your function f(x) will be 2 - 3x.

[00:34:38]Now your sixth type where you change a variable to the function itself. Say for example you're given a function defined on R tor and it says f of f_sub_x is equal to x + 1 and suppose it is given that f is continuous.

[00:35:01]Now if we replace x with f_sub_x we can write f of f of f_sub_x it will be fx + 1 and f of fx is x + 1. So we can write fx + 1 and it is equal to fx + 1.

[00:35:22]Now this is that question where it has that missing regularity condition. But here since we have given f is continuous then it forces fx= x and finally the seventh which uses regularity hypothesis. Suppose we are given this equation f defined on R2R such that f_sub_x + y = f_sub_x into f_sub_y and suppose this function is differentiable in r and fd0= k. Now since this is differentiable what we'll do is we'll partially differentiate it with respect to y. we'll get fdash x + y into f_sub_x fd y. Now we'll put y as zero. We get fdx equals f_sub_x into fd0 which is k fx. So clearly this function f_sub_x will be a kx.

[00:36:34]Basically we get this by solving this simple differential equation. fdx is dy by dx and f_sub_x is y. So this is dy by dx= ky which is a simple first order differential equation.

[00:36:50]Let's take another example. Suppose we are given a differentiable function f defined on r. So it is differentiable in R such that f_sub_x + y = f_sub_x + f_sub_y plus x² y + y² x. Now this if we put x and y as both zero we'll get f0 = 0. And if we partially differentiate this with respect to y, we'll get fd - x + y.

[00:37:28]Now this is zero. This is fdy.

[00:37:32]This is x² + 2 yx. Now again we'll put y as 0. We'll get fdx will be equal to f0 + x². Now this again is simple differential equation. So if we integrate this we'll get f_sub_x it'll be f0x plus xq by 3 + c and since f0 is 0 this c is zero. So your function fx will be xq by 3 + fd0 into x. Now here the question is in a function we are given that 2 f_sub_x plus x f 1x - 2 f and then we have mod <unk>2 sin<unk> into x + 1x 4 and this is equal to 4 cos² px xx2 plus x cos by xx.

[00:38:49]Now first thing is it says prove that f_sub_2 plus f_sub_1x2 is 1.

[00:38:55]So for the first part we need to find the value of f_sub_2 + f 1 by2.

[00:39:02]Now for this what we'll do is we'll first find the value of f_sub_1 which should be simple. So we'll put x as one.

[00:39:08]we'll get 2 f_sub_1 + f_sub_1 - 2 f_sub and here it will be sin<unk> +<unk> by 4. So it'll be 1x <unk>2. So this also will be f_sub_1.

[00:39:29]Now this is 4 cos²<unk> by2 cos 90 is 0 and this is 1 cos p<unk> - 1. So from here we can say value of f_sub_1 is minus1. So we have the value of f_sub_1. Now we need to find the value of f_sub_2 + f_sub_1 by2. And for this what we clearly see is if we put x as2 we'll get this as f_sub_2 + f_sub_1x2.

[00:39:56]So we'll put x as2 we'll write 2 f_sub_2 + 2 f 1 by2 and this is 2 f and this is <unk>2 sin pi 2 + 1x4 and it'll be equal to 4 cos².

[00:40:24]Now p<unk> xx2 it'll be simply pi and plus 2 cos by 2. Now this is 2 which we take common and then it will be f_sub_2 plus f_sub_1 by2. Now here it will be sin 2 pi + theta and sin 2 pi + theta is sin theta which is sin p<unk> 4. So it be - 2 f_sub_1 and cos pi is - 1 but it is cos square pi. So it will be this +4 and cos 90 is 0. Now we know that f_sub_1 is -1.

[00:40:58]So we'll get 2 * f_sub_2 + f_sub_1 by2 will be equal to 2 or we can say f_sub_2 + f_sub_1 by2 this is 1 which is what we need to prove in this first part. Now we'll come to the second part where we need to prove f_sub_2 + f_sub_1 equals 0. And for that we have to find the value of f_sub_2. Since we already have one relation between f_sub_2 and f_sub_1x2, we'll try and find another relationship. And for this what we'll do is we'll substitute x= 1x2 in the given equation.

[00:41:45]So if we substitute x = 1 by 2 we can simply write 2 f_sub_1x2 + 1x2 f_sub_2 - 2 f <unk>2 this mod sin Pi 1x2 + 1x 4 and it'll be equal to 4 cos²<unk> by 4 + 1 by 2 cos 2 pi.

[00:42:38]Now this is 2 f_sub_1x2 plus 1x2 f_sub_2 - 2 f. Now this is sin 90 + theta that'll be cos by 4 which is 1x <unk>2.

[00:42:51]So this again will be f_sub_1. Now cos 4 is 1x <unk>2. So it'll be this 2 and then plus cos 2 pi is 1. It'll be this 1 by 2.

[00:43:04]Now f_sub_1 is minus 1.

[00:43:07]So this 2 and 2 will cancel. If we multiply everything with 2, we can write f_sub_1 by 2 into 4 plus f_sub_2. This is 1. So now we have two equations in f_sub_2 and f_sub_1 by2.

[00:43:26]So we can write this first one as 4 * f_sub_2 + 4 * f_sub_1 by2 is equal to 4 and this is 4 f_sub_1x2 + f_sub_2 this is equal to 1 and if we subtract this f_sub_1x2 will cancel so we'll get 3 f_sub_2 = 3 or value of f_sub_2 is simply 1. So we have f_sub_2 and we also have f_sub_1.

[00:44:03]So this value f_sub_2 plus f_sub_1 will be 1 + -1 and it will be zero which is what we need to prove in this second part.

[00:44:14]Now the question is a function f is defined for all xy belongs to r such that f_sub_1 is 2 and f_sub_2 is 8 and we're given fx + y - kxy is f_sub_x + 2 y square where k is some constant.

[00:44:43]We need to first find f_sub_x and then show that f_sub_x + y into f_sub_1 upon x + y is equal to k for x + y and equal to zero. Now in this what we'll do is we'll put x as 0 and input x as 0 we'll get fy equals f0 + 2 y². Now we need to find the value of f_sub_0 and we already know we are given two conditions f_sub_1 is 2 and f_sub_2 is 8. So if you put f_sub_1 f_sub_1 is 2 and it'll be f_sub_0 + 2. So from here we can say f_sub_0 is 0. We can also check if the system is consistent. So f_sub_2 is 8 will be this f0 + 8 again we'll get f0 as 0.

[00:45:46]So basically our function fx it is 2x².

[00:45:52]Now first we'll find the value of k. So we'll substitute this function in this given equation.

[00:46:01]Then we can write 2x + y² - kxy.

[00:46:12]It'll be equal to 2x² + 2 y². So it'll be this 2x 2 + 2 y 2 + 4xy - kxy = 2x² + 2 y². that means value of k is four. So from here we'll get the value of k as four.

[00:46:37]So this is your first part where you need to find the function. And for the second part we need to prove that fx + y into f 1 upon x + y it is equal to k. Now if you look at the left hand side f_sub_x + y will be 2 into x + y² and f 1 upon x + y will be 2 into 1 upon x + y square cancel. So this left hand side l will be equal to 4 which is nothing but k. So it means we have proved this second one also.

[00:47:25]Now the question is consider a real valued function f_sub_x satisfying 2 f_sub_xy and this is equal to f_sub_x ^ y + f y ^ x for all xy belongs to r and it says f_sub_1 is equal to a where a is unal then we need to prove that a minus 1 into this sum ation I varies from 1 to n f_sub_i will be equal to a n + 1 minus a now since we are given the value of f_sub_1 for f_sub_x we'll simply put y is 1 we'll get this as 2 f_sub_x and then it will be f_sub_x ^ 1 which again is f_sub_x plus f_sub_1 ^x so from here we can simply write this f_sub_x is nothing but a ^ x. So your function f(x) is a ^ x.

[00:48:30]Now we need to prove this result. So we need to find this summation.

[00:48:35]I varies from 1 to n f_subi. So this is a minus one and then this summation i varies from 1 to n and this is a ^ i. Now this is nothing but a GP whose first term is a and common ratio is also a and number of terms is n. So it'll be this a minus one and then a a ^ n - 1 upon a minus one. This a - 1 will cancel.

[00:49:09]It'll be this a n + 1 minus a which is what we need to prove in this question. Now the question is if for all real values of u and v we given 2 fu cos v is equal to f u + v + f u minus v. Prove that for all real values of x first one is fx + f - x this is 2 a cos x. Now clearly we can see f + v and f minus v. So all we need to do is we'll put u as zero and v as x. So you can write 2 f0 cos x and it will be this fx + f - x. So from here we can write f(x) + f - x lb 2 f_sub0 cos x and if we consider f_s0 as a we can write this as 2 a cos x. So that's your first part. Now in the second part we need to prove f<unk> - x + f - x is zero.

[00:50:44]Now the second part we need to choose u and v such that it becomes<unk> - x and - x. One way of doing it is you can choose u + v as p<unk> - x and u - v as minus x. If you'll add them, you'll get u as by 2 - x and v as by 2.

[00:51:10]So we'll choose u as by 2 - x and we'll take v as by 2. So it'll be this 2 f by 2 cos by 2. Now this is f u + v. So it will be simply<unk> - x plus f u - v and that'll be f - x and cos 90 is 0. So from here we can write f<unk> - x + f - x will be zero which is your second part.

[00:51:48]Now this third part is we need to prove f_sub_<unk> - x + f_sub_x it is - 2 b sin x. Now here rather than having - x we are having this as x. So we take this as x here. So we'll get u as by 2 now and v as by 2 - x. So for this third part we'll take u as by 2 and v as<unk> by 2 - x. So it'll be 2 f by 2 cos<unk> by 2 - x and it will be f u + v which is f - x plus f u - b and that'll be your f_sub_x which we can write f - x + f_sub_x will be 2 f by 2 cos 90 - x is simply sin x and since we need to prove it this is - 2 b sin x so we'll take b as - f<unk> by 2 so we can write f<unk> - x + f_sub_x lb - 2 b sin x. Now we need to prove f_sub_x = a cos x - b sin x. So we already have three equations. First equation is f_sub_x + f - x.

[00:53:37]It is 2 a cos x.

[00:53:43]Second equation is f<unk> - x + f - x it is zero. And third equation is f<unk> - x plus f_sub_x that is - 2 b sin x. Now what we'll do is we'll add first and third and we'll subtract this second. So we'll subtract second and we'll add first and third. So this is going to cancel. It'll cancel. We'll get this as 2 f_sub_x and it'll be 2 a cos x minus 2 b sin x.

[00:54:24]So we can write fx is of the form a cos x minus b sin x where a and b are arbitrary constants. Now here we are given a function which is f x + a y comma x - a y is axy. We need to find fxy. So this what we'll simply do is we'll take this as sum u and we'll take this as v. So we'll get x + a y = u and x - a y = v. If we add them, we'll get x as u + v by 2. And if we subtract them, we'll get by as u - v upon 2 a. Now we'll substitute these in this given equation. Then we can write f u v will be equal to a. Now x is u + v by 2 and this is u - v by 2 a. Now we'll replace uv with xy. So we can write fxy will be equal to x² - y² by 4. And that's your option b.

[00:55:48]Now the question is let f be a real valued function such that f_sub_x + 2f 202x it is equal to 3x. Now clearly this is reciprocal substitution. So if we replace x with 202x then 22x will become 202 by 202x which is simply x. So we'll replace x with 22x. So it'll be this f 22x plus now this is 2 fx and it'll be this 3 into 202x.

[00:56:42]Now it's a simultaneous equation in f_sub_x and f202 by x. So what we'll do is we'll multiply this second with two.

[00:56:52]So it will be this two and here will be this four and it'll be this into two and then we'll subtract first from it. So this 2 f 202x will cancel. So we'll get this as 3 fx and it'll be this 3 into 44 by x minus 3x this 3 will also cancel. So we'll get our function f_sub_x as 44 by x - x. Now we need to find the value of f_sub_2. So this f_sub_2 will be 44 / 2.

[00:57:32]So it will be this 202 minus 2. So f_sub_2 is simply 20,000. And that's your option b. Now here we are given a function f(x) satisfying f_sub_xy equals fx by y and we are given that f_sub_30 is 20 and we need to find the value of f40.

[00:57:57]Now what we know is for a product f_sub_y we just need to find f_sub_x because we'll get y simply by substitution. Now what we'll do is we'll write this f 30 as f f 10 into 3 because we need 10 for 40 also. So it'll be this f10x3 and we given that this is 20. So value of f10 it is 60. Now we can write f40 as f 10 into 4. So it will be this f10 / 4 and f_sub_10 is 60. So 60 by 4 is simply 15.

[00:58:40]So value of f14 is 15 and that's your option a. The question is let f(x) and g(x) be functions which take integers as arguments. So domain is integers and we are given let f_sub_x + y = f_sub_x + g y + 8 for all integers x y and says let f_sub_x = x for all negative integers. And it also says g8 is 17. We need to find the value of f0.

[00:59:24]Since we need to find the value of f0 and we know the value of fx when x is negative, we'll take x as minus 8 and y as 8 because we know g8. So we'll get this as f0 and it will be this f minus 8 plus g8 + 8. Now fx is x when x is negative integer. So it'll be this minus 8 g8 is 17 and then + 8. So value of f0 is simply 17. And that's your option a.

[01:00:05]Now the question is a real valued function f_sub_x satisfies the functional equation f_sub_x minus y and it is equal to f_sub_x into f_sub_y - f a - x into f a + y where a is a given constant and f0 is Now we need to find the value of f 2 a - x.

[01:00:44]Now this if we put both x and y as zero we'll get this as f0 it'll be this f_sub_0² minus fa1 square. Now f_sub_0 is 1. So this is one and this is one. So from here we can write f a it'll be simply zero.

[01:01:11]Now we need to find value of f 2 a minus x.

[01:01:17]Now we'll use this.

[01:01:19]So what we'll do is we'll put x as a. So we'll get f a minus y. It'll be this f a into f_sub_y minus f_sub_0 f a + y. Now fa is 0 and f_sub_0 is 1.

[01:01:39]So we can write f a minus y it will be equal to now fa is 0 and f0 is 1. So it'll be this minus f a + y. Now we need f 2 a minus x. So we'll replace y with a min - x. So it'll be this f a minus a min - x and it'll be this f and this is 2 a - x. So from here a and a will cancel. So we'll get f 2 a - x - f_sub_x and that's your option b. Now the question is if a function f satisfy the relation fx + y + fx - y equals 2 f_sub_x f_sub_y and f_sub_0 is unal then which of the following is or are true.

[01:02:47]Now first we'll find the value of f_sub_0. If we put x and y both s0 we'll get 2 f0 equals 2 f_sub_0². So from here either f_s0 will be zero or f_sub_0 is 1. But in the question it says f0 are equal to 0. So here we'll have this result that value of f0 is 1. Now we need to find whether fx is an odd function or an even function. So for this what we'll do is we'll simply put x as zero. We'll get f_sub_y + f minus y. It'll be 2 f0 f_sub_y. Now f_sub_0 is 1.

[01:03:33]So from here we'll get f - y will be equal to f_sub_y. That is f_sub_x is an even function. So option a is correct.

[01:03:44]And if f is even then f_sub_2 equals f -2 and if f_sub_2 is a f -2 is also a and f_sub_4 will be equal to f -4 and if f_sub_4 is b f -4 will also be b. So the correct options are only a and c. Now the question is let f_sub_x + f_sub_y = f x into under root of 1 - y² + y into under root of 1 - x² and f_sub_x is not a constant function then which of the following is or are true? Now if you look at it, it clearly looks like sin inverse function because we know sin inverse x + sin inverse y is sin inverse x into under root 1 - y square + y into under root 1 - x². So one way is just by guessing the function f_sub_x is sin inverse x. But here we are not going to use this. We'll try and solve it some other way. So what we'll do is first we'll find the value of f0 and for it we'll let both x and y is 0. So it'll be 2 * f0.

[01:05:10]This is equal to f0 which implies f_sub_0 is simply zero.

[01:05:17]Now first we'll find if this is an even function or an odd function. And for it what we'll do is we'll replace y with minus x. So we'll get f_sub_x plus f_sub - x will be f x into under root of 1 - x square and then - x into under root of 1 - x square which is f_sub_0 f_sub_0 is 0. So from here we can say f_sub_x + f - x = to0 that is f_sub_x is an odd function and we clearly know sin inverse x is an odd function. Now first we'll try and figure out f 2x into under root 1 - x² and in this we see if we replace y with x we'll get 2x into under root of 1 - x². So for this f_sub_2x into under roo<unk> 1 - x square we'll take y as x. So it'll be f_sub_x + f_sub_x and it'll be this f x into under root 1 - x² + x into under root 1 - x². So we'll get 2 * f_sub_x = f 2x into under root of 1 - x² and that's your option D. So one of the option is this option D.

[01:07:00]Now since we have under root 1 - x square and 1 - y square we know both x and y they will lie between -1 and + one. So what we'll do is we'll let x sin theta. So we can write 2 f sin theta will be equal to f sin 2 theta. Now this looks like sin 3 theta. So what we'll do is in this first equation we'll put x as sin theta and y as sin 2 theta. So we can write f sin theta plus f sin 2 theta will be f sin theta cos 2 theta plus cos theta sin 2 theta or we can write f sin 3 theta will be equal to fs sin theta + f sin 2 theta. Now sin 3 theta is 3 sin theta - 4 sin cube theta and this is f sin theta and f_s sin 2 theta is 2 f sin theta which is 3 f sin theta and sin theta is x. So we can write f 3x - 4x cq will be equal to 3 fx but in the option we are given 4xq - 3x and we know that fx is an odd function. So from here we can say f 4xq - 3x will be equal to minus 3 fx.

[01:09:03]That means the correct option is this option A. So the correct options are A and D.

[01:09:09]Now the question is determine all functions from R to R such that f_sub_x into f_sub_y minus f_sub_xy it is equal to x + y for all xy belongs to r.

[01:09:27]Now we put x is zero and y is zero. We can write f0² minus f_sub_0 it will be equal to 0. So from here we can write either f_sub_0 is 0 or f0 is 1. But if 0 equals 0 it will give us y =0 if we put x as 0 for all y which is not true. So the only condition possible here is f0 = 1. Now we'll simply put y is 0 in this given equation. So it will be this f_sub_x into f_sub_0 minus f_sub_0 and this is equal to x + 0. Now f_sub_0 is 1. So f_sub_x -1 will be x or your function f_sub_x will be simply x + 1.

[01:10:30]Next question is find all the functions g defined on r satisfying gx + y plus gx - y it is equal to 2x² + 2 y² for all xy belongs to r. If we put simply y as zero, we get 2 gx = 2x² or simply gx it'll be x² and if we put gx is x² it'll be x + y square + x - y square it'll be simply 2x2 + 2 y². Now the question is we have defined a function from r to r such that f x² + x + 3 + 2 * f x² - 3x + 5 it is 6 x² - 10 x + 17 for all x belongs to r we need to find the value F5.

[01:11:46]Now we can see 1 + 1 + 3 is 5. But here it will be 6 - 3 3. So it'll become F_sub_3.

[01:11:55]So we get an equation in F_sub_5 and F_sub_3. And if we put zero, we'll get relation between 3 and 5. So it'll become simultaneous equation in F_sub_3 and F5. So first we'll put X as 0. We can write f_sub_3 + 2 * f_sub_5 it'll be equal to 17.

[01:12:15]And if we put x as 1, we can write f_sub_5 plus 2 * f_sub_3 will be 13. From here we can find the value of f_sub_5.

[01:12:35]So we'll multiply this first one with two and we'll subtract second from it.

[01:12:40]So you'll get 3 f5 and there'll be 34 - 13 which is 21. So value of f5 is simply 7.

[01:12:53]So answer to this first question is 7. Now we'll come to the second question.

[01:13:02]f_sub_x is an odd function that means f of min - x is minus f_sub_x and we are given that f_sub_1 is 3 and f_sub_x + 2 equals f_sub_x + f_sub_2. We need to find the value of f_sub_3.

[01:13:30]If we put x as one, we can write f_sub_3 as f_sub_1 + f_sub_2. f_sub_1 is three. So it'll be this 3 + f_sub_2. And if f_sub_x is odd, we know f_sub_0 is zero. If we put x as 0 here, we'll get f_sub_2 as f_sub_0= f_sub_2. No result.

[01:13:58]If we put x as minus2 we can write f_sub_0 f -2 + f_sub_2 again no result. Now we need to find f_sub_2 and what we have is f_sub_0 f_sub_1 and also f_sub_us1.

[01:14:16]What we'll do is in this we'll put x is minus1 we'll get f_sub_1 = f -1 plus f_sub_2 now f_sub_1 is 3 f -1 is - f_sub_1 so that'll be -3 plus f_sub_2 that means value of f_sub_2 is 6 so clearly f_sub_3 is 3 + 6 9 so answer So this second question is 9.

[01:14:50]Now it says f(x) is a continuous onto function satisfying f_sub_x + f minus x= to0. That means f(x) is odd and also we know f0 will be zero.

[01:15:07]And it says f - 3 is 2 and f_sub_5 is 4 and - 5 and + 5. The minimum number of roots of this equation fx=0 is. Now since we know it is an odd function we know f_sub_3 will be -2 and f -5 will be -4 and also f_sub_0 is zero. So f -5 into f -3 will be this -8 less than zero.

[01:15:53]also f_sub_3 into f_sub_5 it is minus a less than z then from bolzaros theorem we can say there is at least one root in -5 to - 3 and at least one root in 3 and 5 and then we have another root which is at x=0 so this equation will have at least three roots so the minimum number of roots is three. The other question is a function f is continuous and has the property f of f_sub_x is 1 - x. Now we'll replace x with f_sub_x.

[01:16:38]We can write f of f of f_sub_x will be 1 - f_sub_x. And we know that f of f_sub_x is simply 1 - x. So we can write f 1 - x lb 1 - f_sub_x or f_sub_x + f 1 - x will be 1 and if we put x as 1x4 we can write f_sub_1x4 plus f_sub_3x4 it'll be equal to 1. So answer to this question is simply 1.

[01:17:15]Now it says f(x) is an even function and g(x) is an odd function satisfying this relation. So if f_sub_x is even we know that f of min - x is equal to f_sub_x and g of minus x= minus g(x). And we are given this relation x² f_sub_x - 2 f 1x it is equal to gx.

[01:17:49]Now this equation we replace x with minus x. We'll write x² f - x - 2 f 1 upon - x and this is g - x. Now f - x is f_sub_x. So it'll be this x² f_sub_x - 2 f 1x and this is equal to g minus x.

[01:18:18]Now both these left hand side they're same. So from here we can write g minus x= gx. So gx is both it is an odd function and it's also an even function. So that means this can only be true when gx is zero. So gx is identically zero. So it's a constant function whose value is zero.

[01:18:40]So that means your option d is correct.

[01:18:46]So now we can write this equation as x² f_sub_x - 2f 1x it'll be zero. And if we replace x with 1 upon x we'll get 1 upon x² f_sub_x - 2 fx it'll be zero.

[01:19:14]And if we solve this we'll get fx= 0 where x is unequal to zero.

[01:19:26]B 2009 is going to be zero and nothing is said about f(x) when the value of x is zero. So f0 could be anything. So it may not be a constant function and it may not be zero for all x belongs to r. So the only options correct are b and d. Now the question is for this a we are given a function g from r to q and f from r to q. So co- doain is set of rational numbers and f and g are continuous functions such that roo<unk>3 fx plus gx = 3. Now both f_sub_x and gx their co- doain is rational number. So an irrational can never be equal to rational. So from here you can clearly see this f(x) it must be zero and gx it must be three for all x belongs to r. So both the functions they are constant and they are uniquely defined. Now we need to find the value of 1 - fxq plus gx - 3q.

[01:20:40]Now fx is zero. So it'll be this 1 cube and gx - 3 is zero. So this is equal to 1. So this a it matches with b.

[01:20:56]Now for this b we are given three functions fg and h which are continuous and positive. And it says f + g + h. It is under root f_sg plus under root gh plus under root hf.

[01:21:15]Now this is basically a square + b² + c²= a b + b c + ca and we know that in this what we have is a= b= c or we can multiply it with 2 and we can write this as f + g minus 2<unk> fg plus g + h minus 2<unk> gh plus h + f - 2<unk> hf and it will be equal to 0. Now this is a square + b square - 2 a b which is a minus b whole square.

[01:21:58]So we can write this as under root f minus under root g square under root g minus under root h² + under root h minus under root f whole square equals to 0. Now all of them they are real numbers and sum of squares of real numbers can be zero if and only if each of the square is zero. So from here we'll get f will be equal to g and it will be equal to h.

[01:22:31]So in that case fx + gx - 2 hx will always be zero.

[01:22:38]So this b it matches with q. So this B it matches with Q.

[01:22:47]Now this question C it says we are given this function Y= FX satisfying this equation then Y-1 + Y1 would be equal to So here we are given YQ - 2 Y² X + 1 + 4 XY + X² - 2 into y - 2 = 0.

[01:23:17]Now we'll take these two terms together and these two terms together.

[01:23:23]So if we take y square common we'll get y - 2.

[01:23:28]And here if we take 2x y common we'll get y - 2 and + x² - 2 y - 2 = 0. We can take y - 2 common y² - 2x y + x² - 2 = 0.

[01:23:55]Here this is x² - 1. So it'll be this one here. So it be x square - 1.

[01:24:02]Now this is a - b whole square. So it'll be this y - 2 and then y - x² - 1 = 0 which is a + b a - b. So it'll be y - 2 y - x + 1 and y - x -1 =0. So either y is 2 or y is x -1 or y is x + 1.

[01:24:31]Now we need to find y-1 + y1. For this first one yd1 is 0 and y1 is 2. So in that case this value will be equal to 2.

[01:24:49]In this second case y1 is 0 and y-1 is 1. So this value it'll be 1 and in this third case y1 is 2 and y dash1 is one. So this value it'll be3. So this c it matches with p r and s.

[01:25:22]Now in this d part it says we have this gp. So we can solve this GP. So it'll be this a and then f(x) ^ 100 - 1 upon f(x) f(x) - 1 = 0.

[01:25:47]So from here we can write x fx ^ 100 lb 1 and this x fx it should not be equal to 1. So from here we can write x fx equals plus or minus one. Now since it cannot be + one so f_sub_x will be min - 1 by x.

[01:26:20]Now f_sub_1 is minus1 so 1 + f_sub_1 will be simply zero. So this d it matches with q.

[01:26:32]The question is a real valid function satisfies the relation f_sub_x into f_sub_y and it is equal to f 2xy + 3 + 3 f x + y - 3 f_y + 6 y For all real numbers x and y, we need to find the value of f8.

[01:27:09]Now since we have 6 y here, what we'll do is we'll try and put x as zero. And if we put x as zero, we'll get f_sub_0 f_sub_y will be equal to f_sub_3 + 3 f_sub_y - 3 f_sub_y plus 6 y.

[01:27:38]So essentially we'll get this f_sub_y as 6 y + f_sub_3 upon f_sub_0.

[01:27:54]Now using this we'll try and find the value of f_sub_0 and f_sub_3. So if we put y as 0 we'll get f_sub_0² equals f_sub_3 and if we put y as 3 we'll get f_sub_0 f_sub_3 lb 18 + f_sub_3. Now we'll put f_sub_3 as f0 squared and let f_sub_0 be a. So it'll be this a cube - a² - 18 = 0 or we can write a² into a - 1 and this is 3² into 2. So clearly value of a is 3.

[01:28:50]So f_sub_0 is three and f_sub_3 is 9. And once we have f_sub_0 and f_sub_3, we can write f_sub_y. So f_sub_y it will be 6 y plus f_sub_3.

[01:29:06]f_sub_3 is 9 upon f_sub_0. It'll be 3.

[01:29:10]So it'll be simply 2 y + 3. And we need to find f_sub_8. So f_sub_8 will be 16 + 3 19. So answer to this question is 19.

[01:29:23]19. The question is we need to prove that there do not exist functions f and g with either of the following properties. So we'll start with the first one and we're given that f_sub_x + g y it is x into y for all xy.

[01:29:45]Now if we put x as zero we can write f0 plus g y =0 that means gy is minus f_sub_0 a constant.

[01:30:05]So your f_sub_x will be xy + f_sub_0. But a function which takes only one argument x. So it cannot depend on y. So such a function is not possible. And for the second case we are given fx into g y is x + y.

[01:30:30]Now we'll replace y with minus x. We can write f_sub_x into g minus x. It'll be equal to z.

[01:30:42]So either f(x) is zero or g(x) is zero. And if we put x is zero, we'll get gy as y upon f_sub_0 and g y is defined if f_sub_0 is unequal to zero. And the same way if we put y is zero we'll get f_sub_x is x upon g0 which is defined when g 0 is unequal to zero. So such a function is also not possible.

[01:31:27]Let f(x + 1 upon y + f x - 1 upon y equals 2 f_sub_x f_sub_1 upon y for all xy belongs to r and y is unequal to zero.

[01:31:52]And it is given that if f_sub_0 is zero then show that f_sub_1 equals f_sub_2 equals z.

[01:32:06]Now if we put x as zero we'll get f 1 upon y + f - 1 upon y and it'll be 2 * f0 and f0 0. So that means this function f_sub_x it is an odd function.

[01:32:29]Now if we put x is 1 and y is 1 we'll get f_sub_2 plus f0 it'll be 2 * f_sub_1² and if we put x as -1 and y is -1 we can write f -2 2 + f will be 2 * f - 1 square. Now we are given f_sub_0 is 0. So from here f_sub_2 will be 2 f_sub_1² and it'll be greater than equal to 0 and f - 2 it'll be 2 f_sub_1 square again greater than equal to 0. But since f_sub_x is an odd function f -2 it should be minus f_sub_2 and it should be less than equal to zero. So both the conditions simultaneously can only be true if f_sub_2 is zero. So from here we can say f_sub_2 is zero and since f_sub_2 is 2 f_sub_1² it also implies f_sub_1= z. Now we are given a question where a function f is defined on r and says f_sub_x + y - 2 f_sub_x - y plus f_sub_x - f_sub_2 - 2 f_sub_y = y - 2 we are defined f_sub_x. Now we put both x and y is 0.

[01:34:24]We'll get f_sub_0 - 2 f_sub_0 + f_sub_0 - 2 f_sub_0 = -2 that is value of f_sub_0 it must be 1.

[01:34:43]And in this equation if we put x as 0 we'll get this f_sub_y - 2 f - y + f_sub_0 f_sub_0 is 1 - 2 f_sub_y = y - 2 or we can write - f_sub_y - 2f - y will be equal y - 3 or f_sub_y + 2 f - y will be 3 - y. Now in this we'll replace y with minus y to get a simultaneous equation.

[01:35:34]So we get f - y + 2 f_sub y and it will be equal to 3 + y and we'll multiply this second equation with two and we'll subtract first from second.

[01:35:52]So 2 f - y will cancel here will be this 3 f_sub y and this will be 6 + 2 y - 3 + y or simply f_sub_y is y + 1. So for this question your function f(x) is x + 1.

[01:36:23]Now the question is f is a function on r2r such that f_sub_x + y + 2 fx - y + fx + 2 f_sub_y = 4x + y. First we'll find the value of f0. So we'll put both x and y as zero. So here on the left hand side we'll get this as 6 f0=0.

[01:36:47]So value of f0 is simply zero. Now in this equation if we simply put y as zero we'll get this as f_sub_x plus f_sub_x plus f_sub_x. So that'll be 4 fx plus 2 f0 and that'll be 4 x + 0. Now f0 is 0. So clearly f_sub_x will be equal to x. So we'll get this function f_sub_x as x.

[01:37:23]Now the question is we given this function f again on r2r f_sub_x into f x + y it is equal to f_sub_y² into fx - y² into e^ y + 4.

[01:37:39]Now again we'll find f0.

[01:37:41]So for that we'll again put x and y is 0. So it'll be this f_sub_0² and here it will be f0 ^ 4 into e ^ 4. So clearly f_sub_0² it will be e^ -4 and f_sub_0 it will be + - e^ -2 and since I can see this variable y here so I'll put x as zero. So I put x as0 it will be this f0 into f_sub_y f_sub_y² f - y² and it will be e power y + 4. Now f_sub_y will cancel once.

[01:38:51]So we can write f_sub_y into f - y² it will be f_sub_0 upon e ^ y + 4. And if we replace y with minus y we'll get a simultaneous equation. So it will be this f - 5 into f_sub_y² and it'll be this f_sub_0 upon e^ - y + 4.

[01:39:28]Now we'll substitute f minus y in this first equation.

[01:39:37]Then it'll become f_sub_y. And here will be this f_sub_0 upon f_sub_y² e power - y + 4² and it'll be equal to f_sub_0 upon e^ y + 4.

[01:40:07]Now it'll be f_sub_y cq. So we can write f_sub_y cq it will be equal to f_sub_0 into e ^ y + 4 upon e ^ 8 - 2 y. So we can write f_sub_y cq as f_sub_0 which is plus - e^ -2 and this is e power 3 y - 4. So f_sub yq will be + - e ^ 3 y - 6 or simply f_sub_y will be + - e ^ y - 2.

[01:41:03]So in this case this function f_sub_x will be plus minus e^ x - 2. And then the fourth question is fx + y + fx - y - y + 2 f_sub_x + y into x² - 2 y = 0.

[01:41:46]Now there's some symmetry with these two terms. So if we replace y with minus y, it'll be same. So what we'll do is we'll replace y with minus y. Then it'll be this f x - y + f x + y. And here will be this minus - y + 2 fx and then - y x² - 2 y = 0. Now we equate these two we'll get y + 2 fx - y x² - 2 y and it'll be equal to - y + 2 fx + y x² and here we replace y with - 1. So it will be x square + 2 y. So yx² + 2 y. Now here 2 fx will cancel. So we'll get 2 y fx and here 2 y square will cancel we'll get 2 yx². So our function f_sub_x will be simply x². And that is the answer to this question.

[01:43:09]Now the question is we are given a function f defined as f_sub_xy = x fx plus y f_sub_y for all xy.

[01:43:32]Now if we put x and y both as zero we'll get f_sub_0 as zero and if we put xy is one we'll get f_sub_1 as zero. So if we put y as 1, we'll get fx equals now f_sub_1 is zero. So f_sub_x should be x into f_sub_x for all f_sub_x. So the only solution possible in this case is f(x) must be zero.

[01:44:20]Now the question is let f and g be two functions defined on r2r such that f x + g y is equal to 2x + y + y. Then we need to find expression for g x + f_sub_y.

[01:44:43]Now what we'll do is we can write f_sub_x as f x - g 0 plus g 0. So in this given equation we'll put x as x - g 0 and we'll put y as 0. We'll get this as 2 * x - g 0 + 0 + 5 or your function f(x) it will be simply 2x - 2 g 0 + 5.

[01:45:22]Now we'll put it in this given equation to find gy.

[01:45:26]So f_sub_x + d y will be 2 x + g y - 2 g 0 + 5 and it will be equal to 2x + y + 5. Now this five will cancel 2x will also cancel.

[01:45:47]we'll get gy = y + 2 g 0 upon 2. Now we need to find g of x + f_sub_y. So this g of x + f_sub_y it will be x + f_sub_y + 2 g0 upon 2 and f_sub_y + 2 g0 it will be 2 y + 5.

[01:46:25]So we can write this as x + 2 y + 5 on 2. So this function g(x) + f_sub_y will be x + 2 y + 5 / 2.

[01:46:36]Now the question is domain of f is set of positive integers and we are given that f_sub_xy it is equal to f_sub_x + f_sub_y for all xy. Then answer the independent questions below.

[01:46:58]Now this first question is f 2025 is zero, f20 is 10 and f25 is equal to 20. What is the smallest n for which fn is not uniquely determined?

[01:47:24]and write values of f(x) for each positive integer x less than n.

[01:47:30]Now we can write f25 as f5 into 5 and that'll be f_sub_5 + f_sub_5. So there'll be 2 * f_sub_5 and it is 20. So from here we can say value of f_sub_5 is 10 and from here we can say this is f 5 5 into 2² so it'll be this f_sub_5 plus f_sub_4 and f_sub_4 will be f_sub_2 + f_sub_2. So there'll be f_sub_5 + 2 f_sub_2 this is 10. So from here we can say value of f_sub_2 is zero and 2025 is 5 into 5 into 3 ^ 4. So f 5 5² 3 ^ 4 it is zero. So it'll be 2 f_sub_5 + 4 f_sub_3.

[01:48:43]So f_sub_3 will be - f_sub_5 by 2. So this is - 5. So we have f_sub_5, f_sub_2 and f_sub_3.

[01:49:03]We can work out f_sub_1 from this equation itself. So if we put x as 1, we'll get f_sub_1=0. So f_sub_1 is 0, f_sub_2 is 0, f_sub_3 is - 5.

[01:49:22]F_sub_4 is 2 F2 0.

[01:49:28]F_sub_5 is 10.

[01:49:32]F_sub_6 is f_sub_2 + f_sub_3 - 5 and f_sub_7. Now there is no information given about this seven because that's the next prime in the sequence 3 2 and five. So this f7 it is not uniquely determined.

[01:49:52]So the smallest value of n for which this function is not defined is n= 7.

[01:49:58]Now this b part it says is there a function f for which fx is equal to0 for all positive inteious less than 225 to the^ 225 but f is not identically zero. show how to define such f and or show that it is not possible. Now we know that we can define any number n as p1 ^ a1 p2 to the power a2 and pk to the power a k where p1 p2 and pk are all primes.

[01:50:38]So we know that for natural numbers we can write them as prime factors.

[01:50:44]Now if we define a function such that f p0 fpi is zero for all pi less than 2025 to the^ 2025 then in that case this function fx will be zero for all x less than 2025 to the^ 2025 because any Any number less than 2025 to ^ 2025 can be expressed by these primes for which we have taken value of f as zero. So we can define a function fx in which f pi is equal to 0 whenever the prime is less than 225 to the^ 2025 and then we can extend this definition to make it not constant function. So take any prime number say Q greater than 225 ^ 225 at which value of this function is non zero say one. So this is one way of defining such functions.