You have to keep both the theorems in your mind. What do you want to say? Such things are called the beauty of mathematics. It wants to say that A and 1 by B are not equal.
Now what to do? Now simplify everything. And throw the term there.
So the value of x, y, z will come to us from here. This is the question of IOQM 2021. If you have written all possible values of x1 and x2, then Hello everyone, how are you? I am with you again Today we are going to do third class of IUQM 2013 Lecture number 3 on Algebra topic Which class? Third class Again we are going to discuss many questions related to PRMO and RMO which have come in the past few years We will discuss with you many concepts related to PRMO we have to remember, which we have to prepare, we will talk about them together.
So, are you ready? You must be ready. Let's start. So, the first topic of today is Rational Root Theorem. What is Rational Root Theorem?
What does Rational Root Theorem say? It says a very simple thing. If, it says a simple thing, understand carefully, if you have any polynomial equation, if you have any polynomial equation, and p by q is a root of it so in p by q, the lower one will always be a factor of a n so q will be a factor of a n and p will always be a factor of a 0 this is For sure, this happens And this is the statement of rational root theorem Which you have to remember But the special thing is that The rational number in the form of p by q The rational root in the form of p by q Both p by q should be co-prime for each other That means the greatest common divisor of both of them GCD or HCF should be 1 That's it So for any polynomial of this polynomial For that we will for polynomial equation. Let's take a simple example. Try to understand.
I am writing a polynomial equation. You will think that it is a quadratic equation. Yes, it is the same thing.
Quadratic equation is also a polynomial equation. So, 6x square minus 5x plus 1 that is equal to 0. What is this? It is a polynomial equation.
If we factorize it, then it will have two factors. 2x minus 1 and 3x minus 1. 2x minus 1 that is equal to 0 and 3x minus 1 that is equal to 0. If we do this, separately we make these two equal to 0, then we will get two values of x, 1 by 2 and 1 by 3. Okay, sir. The roots that have come out here, the two roots that have come out of this polynomial equation, these are rational roots. Okay. What kind of roots are there?
Rational roots are there. I just want to tell you that the denominator, sir, 2 and 3, these two 2 and 3 are basically leading coefficient. What is leading coefficient?
Fact. leading coefficient factors and the numerator whose factors are they? the constant term's factors that's it this is what rational root theorem says let's move ahead to the next concept next concept is integer root theorem ok that means when are the roots of any polynomial equation integer? when are they?
last one in polynomial equation and there is a difference in this polynomial equation that in this leading coefficient how much is it? it is 1 in previous one leading coefficient was an here leading coefficient is 1 because of this in the form of p by q in our rational root in the previous slide Q was a factor of leading coefficient If leading coefficient is 1 Then in that particular situation If leading coefficient is 1 Then Q will be a factor of 1 Now if you write 1 in denominator of 1 Then definitely numerator part will be left, denominator part is 1, so we don't have to apply it so simple thing is, we will tell the roots in rational form because their denominator is 1, so these are integer roots how roots are there? they are integer roots, can you understand? simple point is this, if in any polynomial equation, the leading coefficient is 1 then understand, if the leading coefficient is 1, then understand how roots are going to come, integer roots are going to come The important thing is that all these coefficients should be integers. So this is integer root theorem.
If any rational root of this equation is there, then it will definitely be an integer because the leading coefficient is 1. In the previous one, I said that if any rational root of that equation was there, then that rational root was in the form of p by q. P will be the constant term and Q will be the leading coefficient You have to keep both the theorems in your mind What do you want to say? But now you also understand Without these theorems, the question can be solved You will say sir how? So let's discuss the question Let's hold hands on this question So how will we solve the question? What is the question?
The question is from our pre RMO 2013 It is given that The equation x square plus ax plus 20 is equal to 0. Equation x square plus ax plus 20 that is equal to 0 has integer roots. How are its roots? Its roots are integers. What is the sum of all possible values of a?
What will be the sum of all possible values of a? We have to tell this. You have to understand this.
The equation x square plus ax plus 20. that is equal to zero first thing is what kind of equation is this? this is a quadratic equation what kind of equation is this? it is a quadratic equation so you know that its roots are how many roots are there? two roots are there two roots are there so let's assume that its one root is x1 and the second root is x2 I assumed that its one root is x1 and the second root is x2 so how many of you remember that x1 plus x2 sum of roots is basically minus b by 1 equals if it AX square plus BX plus C So, minus B by A will be minus A upon 1 What will be?
Minus A upon 1 So, sum of roots is X1 plus X2 That will be equal to minus A And product of roots will be X1 into X2 That will be equal to C upon A, that is 20 So, I am writing important things here That X1 plus X2 is equal to minus A and x1 into x2, this is equal to 20. Clear? You must have understood till here. Now think, both the roots, that question has asked, how are they? They are integers.
How are both the roots? They are integers. Now I have to think, which are the two integers in the world, which can be 20 when multiplied?
20 can be made when multiplied. Okay, so we think, that if I, If I multiply 20 by 1, then it can be 20. First case. If I multiply minus 20 by minus 1, then also it can be 20. Plus or not?
Is this clear? So, in short, should I write both these things in short? If I take plus 20 and plus 1, then the product is 20. And if I take minus 20 and minus 1, then also the product is 20. Yes or no? You must have understood this well. If we talk about second case, what other values can I take of x1 and x2 to get product 20?
So, take 10 for one person and 2 for another person. So, 10 for plus minus and 2 for plus minus. If we take 10 for plus and 2 for plus, then product 20 will come. If we take 10 for minus and 2 for minus, then product 20 will come. Clear?
So, in both situations, we have this result. And if we talk about how can we get 20 more? Think more. So, think more.
multiply 4 5 and 4 multiply then also 20 so take plus minus 5 or 4 so product will be 20 so these 3 situations are possible in front of us 3 situations we have in which both the integer roots when we take the product 20 is coming basically 6 conditions plus 20 plus 1 minus 20 minus 1 plus 10 plus 2 minus 10 minus 2 plus 5 plus 4 minus 5 minus 4 6 situations are there now think what can be the value of minus a what can be the value of minus a so if I take minus 1 plus 20 and x2 is equal to plus 1 so x1 plus x2 is equal to minus 1 so we add both so if we take plus plus then 21 plus will come and if we take minus minus then minus 21 will come do you understand this? if you have written all possible values of x1 and x2 then 6 cases are being made in first case x1 value is 20 x2 value is plus 1 so from here I have kept the value x1 plus x2 x1 plus x2 value is 20, x2 value is plus 1, if we add both, then we get 21 plus. If you haven't clicked, then wait a minute, let's elaborate it a little more.
Let's elaborate it a little more, that I know minus a is equal to x1 plus x2. So, I am writing the different values of x1 plus x2 here. If I take x1 as 20, and x2 is 1, then the result will be 21. If I take x1, it will be minus 20. If I take x2, it will be minus 1, then the result will be minus 21. If I take x1, it will be 10. And if I take the other one, it will be 2, then it will be 12. If I take minus 10, minus 2, then the result will be minus 12. If I take plus 5, plus 4, then the result will be 9. If I take minus 5, minus 4, then the result will be minus 9. Clear?
So there are six possibilities of minus a. This is minus a's value. We have to tell a's value.
Okay sir, minus a's are possible values. So what will be a's possible values? Think.
Possible values of a. What can a's possible values be? Think, a's value is 21. So a's value is minus 21. Okay. And the other value is minus a's.
if minus a is equal to minus 21 then the second value will be plus 21 clear what will be the third value if minus a is equal to 12 then a value will be minus 12 and if minus a is equal to minus 12 then the next value will be plus 12 then if minus a is equal to 9 then a value will be minus 9 and if minus a is equal to minus 9 then it will be plus 9 these 6 possible values can be one what was the question what is the sum of all possible values So here I have written sum of all possible values of A Let's add all of them, so plus 21 minus 21 cancel, minus 12 plus 12 cancel, minus 9 plus 9 cancel, how much result did we get? Result came, sir, zero. How much result did we get?
Zero result came, the answer to this question is zero. Okay sir, how much is the answer to this question? It is zero and we solved this question very easily. What to keep in mind?
What to keep in mind? We have to keep in mind that if from these two places, constant term and the coefficient of x in both these places, if you know the value of one person if you have an integer number given on one person then you can think about all possibilities for the other you can think about all possibilities there is nothing special in this you just have to make possibilities and you have to sort them and according to your given condition you have to see that according to the requirement of the condition we will take the question further let's move ahead, what is its printed solution? you can see the printed solution now see the next question this question is from pre-RMO 2017 one more thing you will see here x square plus ax plus 20 was there in the previous question and it is also there in this question.
So, what did they do? In the 2013 question, they added some more items and gave us a new question. Okay, sir? They added some more, some extra and gave us a new question. And what do they say in the question?
Let a, b be integers such that all the roots of the equation, this one, are negative. Negative integers. All the roots of this equation, how are they all?
all the roots are negative integers What is the smallest possible value of a plus b? We have to tell the smallest possible value of a plus b. A plus b's smallest possible value.
Okay sir. Let's discuss this question. You understand it carefully.
Basically, here we have two equations given. How many equations are given? Two equations. And it has been said that this equation basically has negative integers roots. How are the roots?
They are negative integers. So the first thing to think about is from x square plus ax plus 20. From x square plus ax plus 20, the thing to think about starts here. That if we think about x square plus ax plus 20, and the second equation is x square plus 17x plus b, that is equal to 0. Let's think about the two roots of this, x1 and x2.
So let's think about them. x1 plus x2 will be... will be equal to minus 1 and if we do x1 times x2 then it will be equal to 20 and then we know that roots are negative integers so either the possibility will be minus 20 into minus 1 or possibility will be minus 10 into minus 2 or possibility will be minus 5 into minus 4 these three possibilities are required to get the product of root of 20 because integers are negative and negative integers are root so I have to think about negative so I will think when 20 product of root will come when I multiply two negative integers, then 20 can come on multiplying both negative Minus 20 and minus 1. Or minus 10 minus 2. Or minus 5 minus 4. There were three possibilities. We have written those three possibilities here.
Okay. Now I think, what will be the possible values of A? Possible values of A.
Okay. See, the result of minus A is this. So if the value of A is asked, what will we do? Okay, hum.
we will add minus to the result of x1 plus x2 what will we add after that? minus so we will do x1 plus x2 so we add minus 20 and minus 1, minus 21 minus 20 and minus 1 x1 and x2's first possible value is minus 20 and minus 1 if we add these two, then minus 21 if we add another minus after that, then first possible value is plus 21 minus minus plus second point If we take the value of x1 and x2 as minus 10 and minus 2, then the addition of both will be minus 12. And if we add another minus outside, then minus of minus 12 will be plus 12. Clear? And let's talk about the third point, minus 5 and minus 4. Let's talk about minus 5 and minus 4. So, if we do x1 plus x2, then minus 5 and minus 4 will be minus 9. If we add another minus after minus 9, then what will happen? Plus 9 will come. So, possible values of a are 21, 12 and 9. Clear?
So, the first equation was already played. Now, we will talk about the second equation. The second equation is x square plus 17x plus b equal to 0. There was one more bracket after this. Now, we are talking about this. So, let's assume that its two roots are x3 and x4.
So, when we add x3 and x4, what will be? Minus b by a, that is, minus 17 upon 1. so it will be minus 17 and if we multiply x3 and x4 then the product of x3 and x4 will be b x3 and x4 product will be b now in the previous question we were given integer in product so we didn't have much difficulty in thinking but now the sum of both the roots is given as minus 17 and we know that the roots are negative integers so we have to break down minus 17 into two parts both should be negative so what are the possibilities? it is possible that x3 is minus 16 and x4 is minus 1 then minus 17 can be made first possibility if I do minus 15 instead of x3 then it will be minus 2 second possibility and if I write here minus 14 then it will be minus 3 minus 13 then it will be minus 4 minus 12 then it will be minus 5 means, the simple thing is write two negative integers like this which will be minus 17 when we add which will be minus 17 when we add our work is done so I wrote minus 16 minus 1, minus 15 minus 2 minus 14 minus 3, minus 13 minus 4, minus 12 minus 5 minus 11 minus 6 written, minus 10 minus 7 written, minus 9 minus 8 written.
If I go further, then here minus 8 and here minus 9 will be made, which are the same things, but the above case is repeated. What happened to the upper case? It is repeated, so we do not have to repeat the case.
So how many possibilities are made? 1, 2, 3, 4, 5, 6, 7 and 8 possibilities have been made. 8 ways are possible in which minus 17 sum of roots can be there and the value of both the roots will be negative integer so what value do I need?
possible values of b what are the possible values of b? if I take root of minus 16 and minus 1 then both will have product of b so 16 minus and minus 1 if both will be multiplied then minus 16 to minus 1 then plus 16 will be there and if I take minus 15 and minus 2 then the product will be 30 if I take minus 14 and minus 3 then the product will be 42 if I take minus 13 and minus 4 then the product will be 52 and if I take minus 12 and minus 5 then the product will be 60 if I take minus 11 and minus 6 then the product will be 66 if I take minus 10 and minus 7 then the product will be 70 and if I take minus 9 and minus 8 then the product will be 72 all these possibilities can be there A's possible values are these, B's possible values are these. Question has asked, what is the smallest possible value of A plus B? A plus B can have many possible values.
We have to tell smallest possible value of A plus B. Now think about a small thing. What should I do?
Which value of a and b should I take so that the result is small? So, you do one thing, take a as smallest and b as smallest. The smallest value in all the values of a will be taken as a value and the smallest value in all the values of b will be kept as b.
So, the smallest value of a is 9 and the smallest value of b is 16. So, 9 plus 16 is 25. This is the answer to this question I hope you understood this question What was the strategy of this question? It was the same strategy First assume the roots x1 and x2 Both will be seen as sum minus and both will be seen as product 20 Then what will you think? How can 20 be made? So 21 is 20, 10 to the 20 and 4 to 5 is 20 These are negative integers so I took all the negatives After that I thought what will be the value of a if we take x1-20 and x2-1 then what will be the result of this value so by doing this we got all possible values of x1 similarly we traced the second equation like this we wrote x3 and x4 as the roots both have sum-17 and both have product b so both have sum-17 how can it be possible if x3 and x4 are both negative integers so these were all possibilities out of all those possibilities, which one we got which gave us the smallest b value so that the smallest value is minus 16 and minus 1 so the product of minus 16 and minus 1 will be plus 16 so the value of 16 came from here and the value of 9 came from there 9 plus 16 is 25, this is its correct answer let's move ahead to printed solution so we get its printed solution here, 25 hope you understood this question well let's move ahead to the next point what is the next point, understand carefully ok Next question is from Priyarimo 2019. What does the question say?
Let f of x that is equal to x square plus a x plus b. x square plus a x plus b. If for all non-zero real x, if f of x plus 1 by x, f of x plus f of 1 by x is equal to and the roots of f of x equal to 0 are integers, what is the value of a square plus b square? what will be the value of a square plus b square let's think first of all how much is f x square plus ax plus b So what we will do is, how we will take this question? First of all, f is equal to x square plus ax plus b.
So I had told you something, just a little while ago, that if you get any question, then you have to focus on either constant term or x coefficient. If one of these integers is given, then you can take yourself a little further in the race of integer roots. Now in the race of integer roots, You can take yourself a little further, but here both are variables.
We don't know the value of both, a and b are unknown to us. So what we will do is, there are some more conditions given in this. f of x plus 1 by x that is equal to f of x plus f of 1 by x. This condition is given to us.
I write this condition here. After writing this condition, what do we have to do next? We have to think in this way that, Maybe we can get the value of one of a and b by using this.
What did we do in this search? f means that where x is written, where x is written in this function, f is written here, so here x, here x is written. If we want f, then what we will do is, from the given function, from the given function, in this given function, where x is written, we will put x plus 1 by x instead of x so what is the function?
x square plus ax plus b so it will be x square plus ax plus b there should be no problem but here x was there, then x was coming here at both the places here x plus 1 by x is there so instead of x, we will write x plus 1 by x here also and x plus 1 by x here also clear? we have written it that is equal to, then it is written f of x means this is as it is so write as it is, x square plus ax plus b clear after writing that plus f of 1 by x so f of 1 by x means in f of x, instead of x, put 1 by x instead of x, put 1 by x so here x square will be replaced by 1 by x square then here a into x so it will be a into 1 by x clear and what is written in plus, b so b as it is came there should be no problem till here What we did is very simple. We put values in the condition we had.
After putting values, what we did is, x plus 1 by x is whole square. So what we will do is expand it. So what will happen? x square plus 1 by x square. And plus 2 into x into 1 by x.
x into 1 by x is cut. You know, right? You should know, right?
It is very easy. If we multiply a by x, then it will be ax. If we multiply a by x, then it will be a by x. plus b that is equal to x square plus ax plus b as it is plus 1 by x whole square then 1 by x square will come plus a by x and plus b now those who are cutting will cut that is equal to both sides see equal to this is equal to sign okay equal to sign both sides all the things are same they will all be cut so a by x a by x ax to ax and x square to x square and 1 by x square to 1 by x square means everything is being cut B is cut from B On the left hand side 2 is left and on the right hand side B is left So you will get the value of B B value is 2 Sir this was the shortfall, there was no value of B So B value is there So basically this expression is complete f of x that is equal to x square plus ax and 2 in place of B Nothing was there Now we will think about it. Next he says, and the roots of f of x is equal to 0 are integers.
Question says, how are its roots? They are integers. So integer root, you should be able to see it. If roots are integers, then let's assume its two roots are x1 and x2. Okay sir.
So according to you, x1 plus x2, whose should be equal? x1 plus x2 should be equal to whose? and x1 times x2 should be 2 upon 1 which means 2k.
After this what to do? x1 into x2 is 2. So what to think? The thing to think is that what integer values can I keep instead of x1 and x2. So you will find that either you keep minus 1 into minus 2 then product 2 will come.
Or you keep 1 into 2 then product I don't see any other possibility So what we will do is If we ask what are the possible values of a? If I take minus 1 and minus 2 instead of x1 and x2, then it will be minus 1 and minus 2 will be minus 3. So if minus a is minus 3, then a will be 3. Minus minus is cut. And if I keep plus 1 and plus 2 instead of x1 and x2, then 1 plus 2 will be 3. 3 is equal to minus so the value of a will be minus 3 the value of a can be 3 or minus 3 according to this we think but no matter what, we don't have any problem we want the value of a square plus b square we want the value of a square plus b square you know, should I keep plus 3 or minus 3 in place of a if I square it, then 9 will be plus and if I square b, then how much will we keep in place of b? 2 so 2 square So, ultimately 9 plus 4 is 13. How many answers did we get?
13 answers. It was not a tough question. It was a very easy question. But yes, in the beginning, he showed some creativity. So, you need to understand this too.
The pattern of the question is the same. What is changing with time is the strategy. He is adding something to it.
we are putting conditions in it and the pattern will be same as the previous questions same to same question, I will not say that as it is question will come, as it is question will never come new questions will come but yes the pattern of their solution will be the questions of PYQs will come in them, they will put something new and give you and what are those new things or changes you have to face that but yes we will make you face so many problems here for the exam, we will face so many problems that you will definitely face the new problem there and you will solve the question. Believe me. Let's move ahead. Let's see the printed solution of this question.
What is the printed solution? That is 13. 13 answer. Okay.
Let's move ahead. Let's move on to the next thing. Okay. What does the next question say? Understand carefully.
What is the question saying? The question is simple. The question is saying, what is the smallest possible natural number n?
Okay. What is the smallest possible natural number n for which the equation has integer roots? What are the roots of the equation? Integer roots are the roots of the equation.
Pre-RMO 2014 question. Roots are integer roots. So, we have to tell the smallest possible value of n. What will be the smallest possible value of n? Let's understand this question.
x square minus nx plus 2014. because it is a quadratic equation so you will get 2 roots x1 and x2 and how are both the roots? both are integer roots so what will be x1 plus x2? x1 plus x2 minus n or n?
absolutely n minus b by a so if this quantity is negative from the beginning then it will be positive so basically sum of roots will be Minus of minus n upon 1. Okay, sir. So, what will come? n will come instead of x1 plus x2.
And x1 times x2. Right? We will do product of both. x1 times x2. So, what will come, sir?
If we do product, then I got result 2014. Right? 2014 came. Okay.
So, what to do? So, what to do? Now, we will think that how can we break 2014? How can we break 2014?
and we can check the value of the values according to our conditions and what are the possibilities of x1 and x2 we have to take such possibilities in which n is the value n is the value of natural number and n is the value of smallest first if n is the value of natural number and it is equal to the sum of two integers and the product of both integers is positive then what is the product? x1 and x2 are multiplied and the result is positive integer and if two integers are multiplied and the result is positive integer then there are only two possibilities either both are positive or both are negative either both are positive or both are negative now I will think one more thing n is a natural number and you know that x1 and x2 are either positive or negative If both are positive, then the natural number will be added. I am sure about this. But if both are negative integers, x1 is also a negative integer, x2 is also a negative integer, then the sum of two negative integers will be negative integer. Right?
If the result is negative integer, then the value of n will not remain the natural number. So, this is sure that, I got to know from observation that x1 and x2 are positive integers. Basically, these two are positive integers.
If you understand the story up to here, that x1 and x2 are both positive integers, then it is very good. Before this, just consider the question to be over. Now, we have to think how to break 2014 into two parts.
So, what will we do first? 2014 is prime factorization, so divide by 2 so it will be 1007 what we will do after that what we will divide by we can divide by 19 19 times 5 is 95, so 5 is left so 19357 and then divide by 53 one time ok, so I got to know that 2014 2014 can be written 2 multiply 19 multiply 53. Okay, sir. Okay.
What to do after this? What to think after this? After this, we can think, kids, that I have to make these three numbers, so there are only two numbers. So, how can we write 2014? If X1 into X2 is 2014, then I am writing its possibilities.
I am understanding it carefully, son. If I write X1 as 2014, then how much will X2 be? It will be 1 and product... product will be 2014. If I write 1007 instead of X1, then it will be 2 instead of X2.
Even then the product will not be 2014. If I do what after making 1 and 2. Now if I do what, multiply 2 with this, then it will be 106 times 19. So this is also 2014. Ok, and after that what will be the next possibility? 2 is given to 53, now 2 is given to 19 so if 2 is given to 19, then 1 will be 38 and 1 will be 53 these are the possibilities, there are 4 possibilities because 3 factors are being made so the total number of factors that can be made will be 2 x 2 x 2, means 6 factors will be made 1 is 1 and 1 is itself, 1 is 1007 and 1 is 2 and 1 is 106 and 19, so 53 and this Okay, okay, okay, very good. Right? So, we have reached here.
Absolutely right. Okay? We have made four possibilities. Now, we think that what value of x1 and x2 should we take?
Right? Because we ultimately need n value. Which value do we need? We need n value. Okay?
And how did I want n value? Smallest possible value. Right? Smallest possible value of n. Okay?
So, you tell me. This value is of x1 and this value is of x2. Okay, sir?
The one below. Okay? So, which value of x1 and x2 should be taken?
You think for yourself. Which value of x1 and x2 should I take? So that x1 plus x2, which is n, becomes smallest.
What should that be? Smallest. If you want the sum of two numbers to be smallest, then take both the numbers as small as possible. So, if we see x1 and x2, then x2 is the smallest here at 1. But in that condition, x1 is 2014. It is a very big number.
So, we will not take this possibility. And in the second case, x2 is 2. but X1 is still not a very small number it is 1007 so we will cut these possibilities also because their sum is more in third one, if you see their sum is 125 and in fourth one, both numbers 53 and 38 their sum is 91 the lowest sum of both numbers whose product is 2014 the lowest sum of both numbers is 53 and 38 so definitely we can say that x1 is 53 and x2 is 38 and in that particular situation n will be 53 plus 38 if we do 53 plus 38 then what will be 91? n will be the value and it was saying what is the smallest possible natural number value so natural number value is 91 our answer is 91 for this question let's move ahead and see the printed solution of this question ok, so in printed solution also 91. Okay.
Did you understand the question? Let's move on to the next question. No, there is no next question.
Next is concept. Okay. What does this concept say?
Basically, transformation of roots. Assume, in any situation, small things. If alpha and beta are roots of an equation of ax squared plus bx plus c that is equal to 0. Okay.
There is an equation of ax squared plus bx plus c that is equal to 0. And its roots are alpha and beta. Who are the roots? alpha and beta so if I Instead of alpha and beta, I want to have such an equation A dream equation Such an equation whose root is minus alpha and minus beta Minus alpha and minus beta So to solve this, to make such an equation, there is no need to worry Just in the given equation Replace x by minus x In the given equation, if you replace x with minus x Then the equation will look something like this and its roots will be minus alpha and minus beta instead of alpha and beta this happens basically what we did in this we transformed the roots this is called transformation of roots pattern is fixed if you want roots 1 by alpha and 1 by beta the above equation ax square plus bx plus c its roots are alpha and beta if you want such an equation whose roots are 1 by alpha and 1 by beta then what you do so replace x with 1 by x so the equation will be like this and its roots will be 1 by alpha and 1 by beta if I want root alpha par n comma beta par n means we don't get satisfaction from alpha and beta roots we want to do alpha par n and beta par n so what transformation we will do is we will replace x with x power 1 by n where x will be there we will replace x with x power 1 by n so the equation will be like this and its roots will be alpha power n comma beta power n now let's talk about if I want to have a constant in roots suppose roots were 2 and 3 and I want roots to be 8 and 12 so I multiplied 4 in roots multiplied 4 in this, it became 8 and multiplied 4 in this, it became 12 so I want roots to be 4 times or I am saying that the roots should be multiplied by k so what you should do is, if you want to multiply k in the roots then instead of x, you should keep x by k and divide the k you want to multiply by x that means you should replace x by k so the roots of the equation will be k alpha and k beta If there are alpha and beta roots and I want to add a constant k to them So remove the constant from x and replace x with x-k So the roots of the equation will be k plus alpha and k plus beta Now let's talk about the next point If I want to divide k in roots So multiply k in x and it will be kx and replace x with kx so the new equation will be of this type now let's talk about roots, I want alpha to be 1 by n and beta to be 1 by n instead of nth root so what we will do is we will replace x with x to the power of n so it will be like this hope you understood this story wait for 5 seconds ok, let's move on to the next point ok, so what is the question next? understand it carefully understand it very very very carefully What is the question?
It is a very good question. It says, suppose that A and B are the real numbers. What are A and B?
They are real numbers. Such that A and B's product is not 1. A and B's product is not 1. If something like this is written, then I get a doubt. That A and B is not equal to 1, this is not its purpose.
It does not want to say that A and B's value is not 1. It wants to say that If you divide b, then a is not equal to 1 by b. You are saying that a is not equal to 1 by b. You are saying that a and 1 by b are not equal.
And the equation is this one. 120 a square minus 120 a plus 1. And b square minus 120 b plus 120. Hold. a is not equal to 1 by b. and both equations hold then find the value of 1 plus b plus ab upon may a think in reverse How to think? Think a little backwards.
How to think? How to think? Understand carefully sir. The thing to understand carefully is that I have the first equation.
120a square. Okay sir. Minus 120a plus 1. That is equal to 0. Okay.
And the second equation is b square minus 120b plus 120. That is equal to 0. Okay sir. Okay. What is special about both? What is special about both?
Understand carefully. If you have a question. If I put 1 by b in the same place as a, no, I can't.
It is written there that a is not equal to b. So, this means a is not equal to 1 by b. This means this.
But still, let's think for once. What do we do for once? Let's think that if I put 1 by b in the same place as a, what will happen? So, 120 as it is.
Instead of a square, 1 by b square will come. So, 120 upon b square will be done. Minus 120, if you put 1 by b in the same place as a, then 120 upon b will come.
and plus is 1, that is equal to 0 and the LSM of all will be B square so here will be 120 minus 120B and plus B square that is equal to 0 so B square is thrown in 0 so the one who fell in 0 will fall, no one will come back so here comes B square minus 120B B square minus 120B plus 120 that is equal to 0 there should be no problem here see this and this equation is exactly the same it is exactly the same sir It is the same, it is a fraud. It has kept 1 by b in the place of a, but now it doesn't want to accept it, so it has written a is not equal to 1 by b. The equations are the same. Originally, in one equation, if we keep 1 by x in the place of x, or 1 by b in the place of a, then the other equation is formed. Here, the transformation of roots has happened.
What is it? What happened to the roots? Transformation has happened. Okay, so, sir, if we...
and replace any one of the variables with its reciprocal then what is the relation between the new equation and the old equation's roots? so if I say that its roots were x1 and x2 if its roots were x1 and x2 then its roots will be 1 by x1 and 1 by x2 what will happen? 1 by x1 and 1 by x2 will happen there should be no problem till here let's move ahead so why did we write this now? This is just to show that the transformation of roots is happening.
So if I am writing its roots as x1 and x2, then I will write its roots as 1 by x1 and 1 by x2. You should not have any problem in this. Now you think that you know that a is not equal to 1 by b.
Okay? This means that x1 is not equal to 1 by x1. Okay sir?
x1 will not be equal to 1 by x1. Or, x2 will not be equal to 1 by x2. Okay, sir?
These two things are true. Okay? But, but, but, if I say that x1 is equal to, x1 is equal to 1 by x2, then this thing can be possible.
That one of its roots, what happens to the other root, it becomes equal. This can happen. Clear? And if this can happen, then understand, our work is done. How?
That x1, So, if I ask you the sum of roots of x1 plus x2, then it will be minus b upon a. So, minus 120 upon a is 120. So, minus minus plus will be 1. So, x1 plus x2 will be 1. And x1 times x2 is the product of both roots. So, it will be c upon a, i.e. 1 upon 120. Clear? Do you know x1 plus x2? Do you know x1 into x2?
Do you know? and I have accepted that x1 is equal to 1 by x2 because I know that roots have been transferred if you are not ready to accept that x1 has 1 by x1 then you have to accept that its reciprocal is this you have to accept this and if you have accepted this then x1 stands for a and 1 by x2 stands for b for which x1 stands? and for 1 by x2, we are standing for b and we need the value of this expression expression is 1 plus b plus ab upon a so what we do is, 1 as it is, in place of b, 1 by x2 and in place of a, x1 and in place of a, b is also there, so in place of b, 1 by x2 That is divided by x1 in the same place. Okay sir. I have put all the values.
Now what to do? I took LCM above x2 and here x2 is x1 and divided by x1 Main divide is this, the denominator above it will come down and multiply So basically it will be x1 plus x2 plus 1 upon x1 into x2 After this, what did we do? How much did we keep in place of x1 plus x2? How much did we keep in place of x1 plus x2?
1 So, I have kept 1 in place of x1 plus x2. In the next place, I have kept 1 and upon x1 into x2. The product of x1 and x2 is 120. So, I have kept 120 in place of x1.
What do you say? Will we keep this much? No, not 120. 1 by 120. 1 by 120. So, the one who divides by the main will come up. 1 plus 1 is 2. And this 1 by 120 will come up and be reciprocal. So, 120 will come up and be multiplied.
2 will be 240 so the answer to this question should be 240 is it clear? hope so, you must have understood this pre-RMO 2016 question the only special thing was that the roots have been transformed what are the roots? transform if its roots are alpha and beta then the other one will be 1 by alpha and 1 by beta but it said that a and 1 by are not equal so it means that x1 and 1 by x1 are not equal so x1 will be equal to 1 by x2 and x2 will be equal to 1 by x1 so I have finished this question by getting the value from here let's move on to the next point what does the printed solution say? the printed solution says that the answer will be 240 let's move on to the next question a 5-digit number in base 10 you understand base 10, right? So, the number system is 163 So, 3 is a unit digit in 163 Plus 6 is a tenth digit So, 6 is 10 to the power of 1 And 1 is a hundredth digit So, 1 is 100 to the power of 2 Clear?
So, you should understand that Based on means that if a number is written in the form of 10's powers in the expanded form, then it is in base 10. Clear? Okay. So, there are some other things in base as well, right?
So, when will we discuss those things? We can discuss those things in the number system. So, a 5-digit number in base 10 has digits k, k plus 1, k plus 2, 3k, k plus 3. Clear sir?
Is it clear? You think that the question is saying something simple. that there is a 5 digit number and the digits are in the order of first digit is k, then k plus 1, then k plus 2, then 3k, then k plus 3 in that order from left to right, there is a number if there is a 5 digit number, suppose the number of 5 digit is 15625 then whenever I write its expanded form, then the leftmost digit the leftmost desert, how much power does it have in the multiplier? here it can be 0, 0, 0, 0, 4, 0, so it can be 10 to the power 4 then what was in the multiplier of 5?
in the multiplier of 5, it was 10 to the power 3 then what was in the multiplier of 6? in the multiplier of 6, it was 10 to the power 2 then in the multiplier of 2, it was 10 to the power 1 and finally in the last, it was 5 if I write this number in its expanded form, it will be something like this and here the 1, 5, 6, 2, 5 In the same order, these digits are also written. So, can we frame the number with the help of these digits?
Yes, we can. So, our number will be, it is saying that if this number is m square, so our number is m square, and what will it be called? The first left digit will be multiplied by 10 to the power 4. Okay, sir.
Then, the next digit is k plus 1, how will it be multiplied? 10 to the power 3. Okay. The next digit is k plus 2, from which it will be multiplied? It will be multiplied by 10 to the power 2. Then the next digit is 3k, from which it will be multiplied?
It will be multiplied by 10 to the power 1. And the next digit is k plus 3, it will be multiplied by 1 or as it is. So what I did is, m square, if this number is m square, then the expended form of m square will be this. we have written this for some natural number m find the find that sum of the digits of m Basically, we have to tell the sum of the deserts of the value of m. So, let's think about it. See, we have made the expanded form of m square that is equal to a.
And who is doing the work of deserts? All these things written here, k, k plus 1, k plus 2, 3k and k plus 3, what are all these? They are deserts. Now you have to understand carefully, listen to me very carefully.
The digits are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. These are the digits in the market. We keep making numbers by rotating them and putting them up and down. We don't have anything else to take with us.
So these are the digits and you know that if you see the biggest one among all these, if you see the biggest one among all these, In this market, it is 3k. 3k is the highest. What value can it take?
It can take 9. What value can 3k take? It can take 9. 3k value, i.e. the digit of 3k. Instead of this digit.
Instead of this digit. Because K is also a digit. So, K, K, 1, 2, 3, 4, 5, 6, 7, 8, 9. Is one of them.
0, 1, 2, 3, 4, 5, 6, 7, 8, 9 is one of K and 3K is also one of these so 3K can take maximum value of 9 so from here K value is 3 you should know that K value can be 3 or 2 or 1 not 0, why not 0 we will talk about that too but K value can be maximum 3, there is no possibility beyond 3 if you understood this then we will move ahead with this question so m square as it is lets simplify it a bit we can get some direction from here 10 to the power 4 is 10,000 so it will be 10,000 times k ok sir then 10 to the power 3 is 1000 1000 multiplied by k is 1000k plus 10 to the power 3 is 1000 plus then 10 to the power 4 is 100 100 multiplied by k is 100k plus 100 multiplied by 2 is 200 clear 3k is 30k and plus is k plus 3 in expanded form now add the numbers separately this is the number and this is the number now add the numbers m square and you know that m can be 0,1,2 or 3 it can't be more than 3, we have done this now I said why 0 can't be, listen to that if I keep k value 0, then this part will be 0 only 1203 will be left, 1203 which is not a 5 digit number so we need a 5 digit number If we get the value of k, then we will not get the number of 5 digits. If we do not get the number of 5 digits, then the value of k cannot be zero. This thing also clicked. So what we did is that we narrow down the options.
So now I know that either the value of k will be 1 or k will be 2. Or the value of k will be 3. The value of k is either 1 or 2 or 3. No other value of k can be. Okay. So.
So, there is no value other than k. No problem. Let's move ahead.
If we keep k as 1, then what will be m squared? Let's talk about this. So, if we keep 1, then as it is, this number will be multiplied by 1. So, as it is, we will add 1, 2, 0, 3 in it.
So, the total number will be 3 plus 1, 4. 0 plus 3, 3. And 2 plus 1, 3. 1 plus 1, 2. And 1 ahead. so it will be 12334 what will happen? now this is not a perfect square so check it with long division method 12334 ok sir what to do?
we will make pair of 2 digits from last first date 1 1 1 1 here minus here plus here minus here 2 will be here 23 forward carry forward 1 1 1 here 21 will be minus here plus here 22 will be here 2 will be 3 4 forward as it is then then then then then then then then then if I take 1 here then 2 21 will come something will be saved so it means this is not a perfect square this is a perfect square and this is not perfect square so we will cut this option because this is not our option now if we take k value 2 then m square will be m square will be 2 if we multiply by 2 then 2 2 2 6 2 plus 1 2 0 3 ok we doubled all we doubled all and now we will add them 3 plus 2 will be 5 0 plus 6 will be 6 2 plus 2 is 4 and 23. 23465 will be the number. Now let's check if this is a perfect square. 23465, made a pair of 2 desserts. 111 minus 1 will be left.
34, carry forward. No, not carry forward, we took them down. Here 2 is done.
Then what to do? We will take 5. so 1,2,5 will be 9 then 65 is copied and here 30 is done what to do? if we take 3 here then it will be 3309, 0, 3309 perfect square is not here now this is not perfect square, which option is left?
only k equal to 3 so we select k equal to 3 option we put k value 3, so m square will be we will multiply by 3 3, 3, 9, 3 plus Okay 1, 2, 0, 3 If we add both, then what will be the value of m square? Let's think about the value of m square Okay 3 plus 3 is 6 Okay 0 plus 9 is 9 Okay 2 plus 3 is 5 1 plus 3 is 4 And 3 3, 4, 5, 9, 6 Right? Let's check this Whose square is this? Because ultimately the value of m We need digits of m value. 3, 4, 5, 9, 6. We made pair of 2 digits.
If we take 1, 1, 1, then 1 will be here. 2 will be here in remaining part and 45 as it is. Here 2 is made. If I take 8, then how much will it be?
No, it won't work if I take 8. Let's take 9. 9, 9 is 81. 1, 8. and 9 to the 18, 18 and 8, 26, this is too much, we were taking 8 correctly, right? Okay, so here we will take 8, here also we will write 8, here also we will write 8, okay sir? 8 times 8, 64, 4, 6, 8 times 2, 16, 16 and 6, 22, okay? So here I have 21, right? You are understanding this much, right?
We have even brought this question here, that means we have solved this question, okay? This was one of the good questions, okay? In IUQM 2021, okay? So... 21 left here, then the next pair is 96 here 28 and 8, 36 is done so after 36, which digit should I take here that 6 comes here ok, so either 4 comes in 4, 4, 16 or 6 comes in 8, 8, 9, 9, 6, 6, 36, 6 so either I have to take 4 here or I have to take 6 here so 4, 3, 12, 100 is being made but there is 21, 100 so don't take 4, so you have to take 6 only so multiplied by 6, 6, 6, 36, 6, 3 6x36 and 3x39, 9 will come, here 3 will come, 6x38 and 3x21, absolutely correct.
Okay, remainder 0 came, means here 186 came. So, the m is basically, m will come, sir, 186, and sum of digits, sum of digits, let me talk about sum of digits of m, right, sum of m's digits, so 1 plus 8 plus 6, sir. 8 and 6, 14 and 1, 15. So, the sum of digits is 15. What will happen?
The answer of this particular question will be given. Okay, sir. 15 answers.
Okay, sir. Let's move ahead. Hope you understood the question.
Very well. Let's move ahead. Okay.
This is its printed solution. 186M value was coming. You can calculate the digital sum.
You can do it. Let's go. Now, let's talk about the next pattern of questions. Next pattern of questions.
Next pattern. This is the question of IOCUM 2021. I.e. 2021 is not too old but not the latest either. What does the equation say?
Positive integers A, B, C satisfying A upon A minus B. A upon A minus B that is equal to C. What is the largest possible value of A? The largest possible value of A. What is it?
No, no, the largest possible value of A is not A. It is A plus B plus C. How many do you have?
A plus B plus C its largest value what is it? its largest value which is not exceeding 99 its value is always smaller than 99 99 is equal to or smaller than 99 it can't be bigger than 99 not exceeding 99 so we have to calculate the largest possible value of A plus B plus C which is 99 or smaller than that what will be that value? and we don't know anything else Apart from this, we don't know anything.
So, let's think what can be done in this. First of all, some students will get this jump that we cross multiply this. So, what will be A equals to? A minus B will be multiplied by C.
So, it will be A minus B. A minus B will be formed. After that, we will throw minus B there. So, A plus B is equal to A. There should be no problem here.
Now we want that A, B and C are basically positive integers. What are positive integers? So we can think about them when they are seen separately. What to do?
They are seen separately, single single. Because when two integers are in the product and both are unknown, then it becomes very difficult to think about them. So we want that both of these separate integers come single single. So what to do to make them single single single?
Key. divide A here also divide A here also and divide A here also no problem here, my coding should not be done why did you do this sir? when you will divide A here also then 1 by C will come here when you will divide B to B then 1 by A will come here and when you will divide A to A then 1 by B will come here so basically you got this correlation of three integers we got a relationship in between and this relationship is of this type that all three are alone in it now I have to think which integers should I keep that this integer's reciprocal plus this integer's reciprocal should be added and what should I get?
I should get a new third integer's reciprocal okay, let's think so the first thing I thought I am taking some examples because it is new for me too so what I did here if we keep the value of c as 1 and a as 1 so 1 by 1 plus 1 by 1 so how much result will it give? so basically if it becomes 1 plus 1, it will become 2 so this cannot be written in 1 upon something it cannot be written in the form of 1 upon integer so we cannot take 1 as 1 first thing, it was clicked well ok sir ok, now if I take the value of c as 2 in 1 upon I take the value of c as 2 and here if I take 1 only 1 upon 1 so this half plus 1 will be 3 by 2 if I want to write it as numerator 1 then it will be 1 upon 2 by 3 so the value of b is 2 by 3 so the value of b is 2 by 3 so this is very strange b was an integer so how can it be 2 by 3 so lets do one thing to sort out this situation lets assume that instead of approaching them directly if we assume that that A is K and C is KN and B is KR I thought about M, N and R and when we multiply K in them, A comes I thought that A comes So, here 1 by C will be 1 by KN plus 1 by A will be 1 by K equal to 1 by B will be 1 by KR Now, if you take 1 by K common and cut it If you take 1 by K common and cut it Then, here in the bracket, 1 by N plus 1 by M That is equal to Let's write this situation on the side A plus B plus C less than equal to 99 Okay Sir So, if we take 1 by K common, then we get 1 by KN, no, KR. After that I have removed k from k. Now I have the same form again. 1 by n plus 1 by m that is equal to 1 by r.
Clear? You should not have any problem till here. Now think, think a little. If I write 1 instead of m and n, it will not work. If I write 1 instead of either one, it will not work either.
Why it will not work? Because here I have to bring 1 by integer. So basically my attempt should be that whatever the sum of these two fractions is, that sum of fractions, because 1 upon integer and 1 upon integer means both are fractions, both are fractions, and if it is a positive integer, then both are positive fractions, both positive fractions will merge and a fraction will be formed which is 1 upon something, what will happen? 1 upon something is there.
Okay, so what to do? Now I think that we will do one thing in this, because I want 1 upon integer, so I thought that in this particular situation, 1 by If I add 1 by 6 to 3, then the result is 1 by 2. Now you will think how did you think this? I thought of this type of situation.
a, b, plus b, c is equal to a, c. So you have to take the numbers from this type. that if you see the integers in this way that if you multiply these two and add these two then they should be equal to the product of A and C what should be equal to the product of A and C so if I directly compare it with this then the value of C is 3 and the value of A is 6 what is the value of C?
3 what is the value of A? 6 and what is the value of B? 2 now think about it how much is 6th Risa?
18 and basically it is written as 6x2 and 2x3 so 6x12 plus 6 is equal to 6x18 so we thought of this pattern that all three numbers are used and what happens is that all three numbers have one numerator and below are integers and for smaller numbers than this if apart from 2, 3 and 6 we can think of some other small integer for which this is possible then it is not at all possible because the smallest integer is 1 positive integer 1 we don't have to take it because if we take 1 then the overall fraction instead of being a unit fraction it will become a complete whole which we don't need we need this unit fraction, this unit fraction and this unit fraction so for this one should not be 1 out of all three so after 1 I put an option 2 if we keep all three 2,2 then also it will not be 1 If we keep all three as 3, then also it will not be made. So, after evaluating, we get a result. 1 by 3 plus 1 by 6, that is equal to 1 by 2. If you add, if you take 6 LCM, then 2 plus 1, that is equal to 3 by 6, which is simplified and we will get 1 by 2. So, it happens that these two things make this.
After this, it is understood that if we compare, then the value of n is 3. How much is the value of n? 3. If we compare the value of m, then how much is it? 3. 6 and R is 2 and when these three values come and I have assumed them for A, K was there what was the value of A? K times M if we do K times M, then K times 6 which is 6K A's value will be 6 ok what was the value of B?
KR so K into 2 which is 2K and the value of C is that will be Kn, so we will do Kx3 so basically the result will be 3k I got the value of a, b, c in terms of k 6k, 2k and 3k, now why did we do this? the only thing to think about is that if we take any of the multiples of these three numbers that is, the multiple of 2, 3, 4 of all of them then for all of them this property will always be the same, it will be followed follow how it will be followed? you check by taking 2 multiple if 1 by 3, 1 by 6 and 1 by 2 multiply 2 in all denominator then it will be 1 by 6 plus 1 by 12 that is equal to 1 by 4 you check by taking LCM if you take LCM 12 6 2 is 12 so 2 plus 12 1 is 12 is 1 so 2 plus 1 3, if you do 3 by 12 then if you cancel then it will be 1 by 4 so this property is always following in multiples of these numbers and it is called Horiya So what we did is, A is a multiple of 6k, B is a multiple of 2k and C is a multiple of 3k. Okay sir? so we need the maximum value of a plus b plus c which is not more than 99 so we calculated a plus b plus c a plus b plus c is 6k plus 2k plus 3k 6k plus 2k plus 3k we added all three 6k plus 2k is 11k so all three of them are 11k and i don't want the value greater than 99 So what values of k do I keep?
I can keep 1, 2, 3, 4, 5, 6, 7, 8 and maximum 9 Maximum is 9 because if I keep 9 then what will be the value of a plus b plus c? At k equal to 99, what will be the value of a plus b plus c? 99k So I was asking what is the largest possible value of a plus b plus c which is not exceeding 99 which is not more than 99, what is the possible value of A plus B plus C?
so the possible value of A plus B plus C is 99 what is it? 99 is its possible value that's it I hope you understood this question I won't say it's easy, it was a good question but it had a property following it so these things are called beauty of mathematics beauty of mathematics Beauty of Mathematics So basically this is the beauty You can see it in the questions You can see it somewhere Because one of the best Maths exam is there So you can see something like this Let's move on to the next point So this is its printed solution You can see from here The maximum value it can take is 99 Let's move on Okay next question IUQM 2022 question or IUQM 2022 question because of which we were getting to know that the paper was very tough. How was the paper? It was very tough.
Because of which we were getting to know that. One such question from those questions. And what does the question say?
Let X, Y, Z be complex numbers. What is this? Complex numbers. Sir, what is a complex number? Bitta.
I am telling you for 10th class students, 11th class students know this well 11th and 12th class students know complex numbers well but the students who are below 10th standard I will tell them that complex numbers are that big world whose small world is real numbers and there is a big world outside this real numbers world actually this real numbers world is in that world so complex numbers means the father of all these numbers let's go So, x, y, z are also complex numbers such that these three equations are given to us. x upon y plus z plus y upon z plus x plus z upon x plus y that is equal to 9. And x square upon y plus z plus y square upon z plus x plus z square upon x plus y that is equal to 64. And x cube upon y plus z plus y cube upon z plus x plus z cube upon x plus y that is equal to 488. So, if we know this, x upon yz plus y upon zx plus z upon xy that is equal to m by n where m and n are positive integers with greatest common divisor. So, find m plus n. Who to find? Find m plus n.
Okay. So, it's a big question. I got tired of reading.
Okay. So, let's solve this question. Okay. First thing. You have to understand this question very carefully.
Whenever you get this type of format, you have to think about how to apply something in it that it can be used and it can be used. Okay? Meaning?
Meaning? Meaning? What you have to do is, you have to get your work done with this.
If you go to solve this, you will reach a very very painful situation. Right? So, you will just get your work done with this. nothing more so in the first situation, we know that x upon y plus z y upon z plus x, z upon x plus y the sum of these three is equal to 9 if I multiply x in this whole equation once multiply x in both sides so what will happen? what will happen?
let's think about that first so as soon as I multiply x in x upon y plus z then it will become x square upon y plus z okay sir and If I multiply x in this, it will be xy. on z plus x ok why not write xy on z plus x in the side because we are making the format so make the correct format so here it will be written xy on z plus x ok sir and if we multiply x in this then it will be xz on x plus y clear and what will be in equal to? in equal to, as soon as x is multiplied by 9, then it will be 9x how much will it be?
it will be 9x, clear? x is multiplied in the first equation first equation is this, in this first x is multiplied so this, 1, 2, 3, 3 expressions are this and in equal to 9x is there, clear? now what I did, in this first equation now I am multiplying, whose child is it? now I am multiplying y in this, I am multiplying y if I multiply y, then When I multiply y with the term y, what will it be?
First of all, multiply y with y, y square. How much will it be in upon? z plus x. Clear? This term will come here.
And if we multiply y with the other two terms, it will be something like this. xy upon y plus z plus yz upon x plus y. And when we multiply y in front with 9, it will be 9y. Clear? Hope so.
Now in this equation we are going to multiply z. So in the third term when we multiply z, then z into z will be z square upon x. z square upon x plus y.
And then we will multiply that z term in the other two. So here we will get y plus z in zx upon and If we multiply z in this term, then it will be yz upon z plus x. Clear?
Okay. And if we multiply z in the 9 in equal to, then it will be 9z. No problem here. No problem.
Now what to do after this? Think. What is the sum of these three terms? What is the sum of these three terms? x square upon y plus z, y square upon z plus x, z square upon x plus y.
Basically, this second equation is ready in your hand. Right? and the sum of these three terms is 64. So, instead of these three terms, I wrote 64. How much did I write?
I wrote 64, sir. And we will listen to the next part now. The remaining six terms are left.
What we have to see in these is whose denominator is z plus x. This denominator is z plus x and this denominator is z plus x. These two have denominator z plus x.
Okay? Did you see? So, what you do is, z plus x, whose denominator is z plus x, those two have denominator same because then numerators will be added. So here is xy and here is yz.
If we take common y from both, xy and yz, if we take common y from both, then z plus x will be in bracket. Think carefully, think carefully. Plus, in terms upon which x plus y is there, in terms upon which x plus y is there, in terms upon which x plus y is there, so here is xz and here is yz. If we take common z from both, so in bracket we have x plus y clear? what will be left after taking common z?
x plus y in bracket and next thing we will do write the terms which have y plus z in them write them too y plus z in the denominator and xy and zx xy and zx if we take common x from both of them then in bracket we will have y plus z clear? so what do you want to say sir? you want to write common x?
plus 9y plus 9z some students said that they are saying that if we take 9 common then x plus y plus z will be left in bracket there is no problem in this sir, from z plus x z plus x is cut from x plus y, x plus y is cut and from y plus z, y plus z is cut so here also x plus y plus z is left so here basically it is written 64 plus x plus y plus z ok sir that is equal to 9 times x plus y plus z ok sir ok, that is equal to how much? no, it comes equal only once then after that this x plus y plus z comes here and minus 9 times x plus y plus z out of 9 times something, 1 times something will be minus so 8 times something will be left so it will be 8 times x plus y plus z that is equal to how much? 64 and if this is the case, then how much will x plus y plus z be sir? 64 upon 8, right? so the answer will be 8 x plus y plus z value is 8. There should be no problem till here.
Right? Do you understand till here? I have calculated 8 for x plus y plus z value.
Now what will we do? Now what will we do? Now we will play again. Right?
Now what will we do? We will play again. We will play the same thing again.
This is equation 1. This is equation 2. And this is equation 3. Now I say that if I multiply x once in the second equation, multiply y once and multiply z once in the second equation, multiply x once, multiply y once and multiply z once. If I multiply z, what will be the result? So, what we will do is, on the next slide we will write the second equation and in that, the second equation will be x square upon y plus z plus y square upon z plus x plus z square upon x plus y that is equal to 64 given this is the second equation you can check, this is the equation so, we have written the equation now what we will do? now we can do a small work first multiply x with x square then it will be x cube x cube upon y plus z first multiply x with x x cube upon y plus z then instead of writing 3 things if i multiply x with x square then it will be xy square upon z plus x and if I multiply x with this, then xz square upon x plus y and what will be equal to?
If we multiply x with 64, then what will be 64x? This is the multiplication of x Now what we have to do is, we have to multiply y with both sides The whole equation is from y So when I multiply y with y square terms, then y cube upon z plus x and if I multiply y with x square then it will be x square y upon y plus z and if I multiply y with z square then it will be y z square upon x plus y and if I multiply y with 64 then it will be 64y now we will multiply z in this equation and add it to this equation means I have written x,y,z and multiplied them means in this equation I have written x,y and z and added all three equations separately we did the same thing on the previous slide we are doing the same thing here we multiplied with power 1 and now we are multiplying with power 2 so if we multiply z then what will be here x square z upon y plus z we will multiply z with y square, how much will it be No, no, no, we wrote something wrong. We have to multiply z here.
So multiply z with the last term. How much will it be if we multiply z with z square? How much will it be in z cube upon?
x plus y. And multiply z with the rest of the terms and write here. So we multiplied z with x square. x square z upon is y plus z. And if we multiply z with y square, then it becomes y square z upon is z plus x.
Clear? Okay. After playing this much, after playing this much, so we multiply z with 64, 64z now what we do is we take 64 common here, so we have x plus y plus z and these three terms which are developed, you already know the value of these three terms how much value is there of these three terms? 488 how much value is there?
488 so we will put this value there so here instead of these three terms I wrote 488 Now, which terms are in the denominator of z plus x? In the first and last term. So, take z plus x as common. No, no, no. Take z plus x as it is.
Take LCM of both these terms. Okay? LCM of both these terms of z plus x is here. So, you know numerators will be added. That is, xy square and y square z.
What will happen to both? They will be added. If both are added, then what is common between both?
I see y square as common and in bracket, x is left in z and plus. So, as soon as you take common in both, y square, then you will see z plus x in bracket. In plus, what will you do after this, kids?
After this, x plus y, look at the terms in the denominator. And LCM will be x, y of those two terms. And what common do you see in both of these? Z square common, bracket has x plus y.
So, take z square common, bracket has x plus y. Clear? Okay. then after that in both terms we will focus both have denominator y plus z so y plus z here and x square common in bracket will be y plus z that is equal to 64 in bracket x plus y plus z we will write here Now what to do next?
Z plus X cancel, Z plus X cancel, X plus Y cancel, Y plus Z cancel. So here 488 is remaining. Plus X square plus Y square plus Z square is remaining. What else to do?
We can do this. In the multiplier of 64, the value of X plus Y plus Z is known. What else to do? x plus y plus z is 8. So we will put 8 here in the 64th multiplier.
64 into 8 is 512. So 512 will remain as it is. 488 will be minus here. Okay sir. As soon as it was minus, we came to know that what is the value of x square plus y square plus z square?
The value of x square plus y square plus z square came out. We have 24. Okay sir. The value is 24. Now what do we do?
you have x plus y plus z you have x square plus y square plus z square now we will do one thing you all know the formula of x plus y plus z whole square in the first day of algebraic identities I told you the formula of a plus b plus c whole square we will use that formula we will expand it so here we will have x square plus y square plus z square plus 2xy plus 2yz plus 2zx this is the formula, ok, write it down now what to do sir? how much is the value of x plus y plus z? 8, so 8 value is kept, square above how much is the value of x square plus y square plus z square? 24, keep it keep it sir then we will take two common from these three terms, what will be left? xy plus yz plus zx, ok sir ok, keep it how much is the square of 8 sir?
64 and if it goes there plus 24, then it will be minus 24 that is equal to 2 in bracket xy plus yz plus zx. Okay, sir. I wrote it.
Okay. What should be done after that? What should be done after that? So, let's do one thing.
Let's do 24 out of 64. How much will it be? 40 will be left. We will throw 2 in its divide.
So, we will keep 2 in 40's divide. So, this value will come. xy plus yz plus zx. Okay. Okay.
This value came. We have xy plus yz plus zx. How much value we got?
This value is 20. We have 3 values. x plus y plus z is 8. x square plus y square plus z square is 24. And x y plus y z plus z x is 20. Now we need last value which is x into y into z. We need value of x into y into z.
So what we have to do? The most initial equation we had is this one. In this equation, because we know that x plus y plus z is 8 so why not write 8 minus x instead of y plus z if we ask the value of y plus z then what will you say? if you throw x there, it will be 8 minus x y plus z value is 8 minus x similarly x plus y value is 8 minus z similarly x plus z value is 8 minus y we know these values so we substitute all three places here and then try to solve this So what we did is, I had written x upon y plus z, so instead of that, x upon 8 minus x and y upon 8 minus y and z upon 8 minus z. Okay sir, that is equal to how much will it be?
Think, how much will it be? If I think what to do in this, if I... I am coming to put the value instead of all the terms. So, instead of y plus z, I have kept 8 minus x.
And instead of z plus x, I have put 8 minus y. And instead of x plus y, I have put 8 minus z. Clear? I have put all three values. And what is written in equal to?
9. So, I have written 9 in equal to. Now, let's simplify this a little. So, what should we do?
Let's take LCM of all three. I have taken LCM of all three. How much will it be? 8 minus x, 8 minus y.
and 8-z ok now let's do one thing in this x's multiplier these two denominators are coming so write both denominators as 8-y and 8-z ok in this y's multiplier the other two denominators 8-x and 8-z ok then in this z's multiplier ok both have a denominator 8-x 8-y ok then after that, that is equal to 9 Now simplify the term and get the value of x,y,z x,z is multiplied by both brackets 8x is 64 minus 8 common bracket y plus z plus yz this will be the first one if both brackets are multiplied and plus y is multiplied by 64 minus 8 times x plus z and in plus, xz then, keep the same z out and multiply both brackets 64 as it is, 8 to the 64 then, minus 8 will be x plus y in the bracket and in plus, xy in plus, xy bracket close that is equal to In the 9th multiplication, all these 3 things are gone. So, when all these 3 brackets are multiplied in the 9th multiplication, what will happen? I will tell you. 8 x 8 x 8, this is the first term which will be 512. Then, multiply 8 of 2 brackets and take the variable of 3rd bracket.
So, 8x8 is 64. So, in minus, x plus y plus z will come in 64. Then, if we take 1x8 and 2 variables, then here it will be 8 in plus bracket, and the last term will be xyz in plus bracket this is the term this much clear, let's move ahead so what we do after this if I multiply x by 64 and y by 64 and z by 64 so basically in last write in starting that common of all three is 64 in bracket x plus y plus z will be made ok, we have taken first term of all and if x will be multiplied in this bracket 8 minus 8 times y plus z and y will be multiplied in this bracket and z will be multiplied in this bracket so in all three terms take common of minus 8 so here xy plus y will be xz here x when multiplied by these two will be xy plus xz and here y will be xy plus yz and here if z is multiplied then xz plus yz so basically 2 times in bracket xy plus yz plus zx will be made xy plus yz plus zx will be made so what we did is we took 2 more common and wrote it like this then if x is multiplied by yz then here xyz will be made here also xyz will be made and here also xyz will be made so basically it will be 3 times xyz that is equal to after this we will think that 9 will be multiplied by 512 9 will be multiplied by minus 64 and we will add the value of x plus y plus z which is 8 then 9 will be multiplied by 8 so 9 into 8 and the value of x plus y plus z is 20 so we will add this value 20 and in plus will be 9 is multiplied by xyz so 9xyz we need the value of xyz you know it so 64 as it is x plus y plus z value will be 8 minus 8 times 2 is 16 xy plus yz plus zx how much value will we put in this term? 20 and plus 3xyz which was written, we will throw it there it will be minus 9xyz so how much will be left there? 6xyz will be saved there. And when you multiply 64 by 8, 512 is being made.
How much is being made? 512. 64 times 8 is 512. So this is 9 times 512 plus and this is 9 times 512 minus. Both are cut.
Okay? As it is, cancelled. And more and more.
20 multiplied by 8 is 160. 20 multiplied by 8 is 160 and 9 multiplied by 8 is 1440 1440 is minus 1440 this is the value of 6xyz so we will get the value of xyz from here see, let's do one thing let's write this 1440 without multiplying this was 9x8x20 so 9x8x20 will come here Now I took 8 as common from these 3 numbers. If I take 8 as common from these 3 numbers, then I will get 64 here. Minus 2 here, so 2 times 20 will be 40. And here 20 times 9 will be 180. Okay, minus 180. Okay, we are coming to one more mistake.
when all three will be multiplied, then it will be minus xyz minus xyz will be minus 9xyz and when 3xyz will go there then it will be minus 12xyz because that term will be negative 1 plus minus is fine so here we have taken out the value of minus 12xyz so 12xyz is negative divide by this, then the product of x, y, z will be 64 minus 40, so 24 will be left, from 180, 24 will be less, how much will be left? 156 minus, so 8 multiplied by minus 156 divided by minus 12, so I will divide 12 by 156 so I will get 13 so 13 into 8 will be xyz value 104 xyz value is 13 times 8 which is 104 after getting xyz value understand our work is done why our work is done? let me tell you ok we are asked in this question if x upon yz plus y upon zx plus z upon xy whose value is equal to this? M upon N is equal to where M are positive integers and M and N have greatest common divisor 1 so we have to tell the value of M plus N if I take LCM of these three terms then you will know that LCM will be XYZ what will be LCM?
XYZ what will be above? it has YZ so if we multiply X, then it will be X square here it will be Y square and here it will be Z square Now you know the value of x square plus y square plus z square. You calculated the value of x square plus y square plus z square here. How much is it?
24. So, I put 24 here. I put 24 above. In upon, I calculated the value of x into y into z.
How much did it come? 104. So, you will put 104 there in divide. So, 24 and 104. Cancel it. So, 8 3s are 24 and 8 1s are 8 and 8 3s are 24. So the answer is 3 by 13. But he was saying that the result of this is in the form of m upon n. So it is coming in the form of m upon n.
So I got the value of m as 3 and the value of n as 13. So he asked the result of m plus n. So we solve this on a new page. Basically, that question needs the value of x upon yz plus y upon zx plus z upon xy. And this is equal to m by n.
So I took LCM of all three, made XYZ, and made X square plus Y square plus Z square. I have value of X square plus Y square plus Z square is 24, and value of XYZ is 104. That is equal to M upon N, right? So 8 is 3 and 8 is 13. So 3 by 13 will be the value of M upon N.
Okay sir? So the question is to ask the value of M plus N, if they don't have any common factor. So they don't have any common factor other than 1, both are co-prime. So adding 3 plus 13, the answer is 16. this is the question i think this is the question increase the value of m plus n so we got m plus n value 16 hope you understood this question it was a very heavy question but let's assume you are preparing for olympiad you are not preparing for exam you are preparing for the toughest exam in India so you will get good questions what will you do with good questions? you will get this type of question if you change these numbers then also the approach will be same numbers will change but what will be the approach?
So whenever we see a question in this pattern, we have to keep the same approach. So let's see its printed solution. Don't think that the printed solution will come in one page. 1, 2, clear.
Then 3 and then 4. They also took 4 slides. Let's move on to the next question. What is the next question?
Let's understand it carefully. Let be real numbers such that x into y is 1. The product of x and y is 1. In the previous question, we had come to know that the product of a and b is not 1. Here, the product of x and y is 1. Further, he said, let T and t be the largest and smallest value of the expression this one. The largest value of this expression is T. and the smallest value is t and what we want to say is if T can be expressed in the form of M by N where M and N are non-zero integers with greatest common divisor MN that is equal to 1 then we want the value of M plus N, I.Q.M 2022 let's see the question capital T and small t, the largest value and the smallest value in both values, you have to add both values and represent it in the form of m by n and m and n should be co-prime, both should have gcd1 and after that tell me the value of m plus n, how much is it?
IUQM 2022 again another great question, algebra's great question understand that this is the question because of which this is eaten that the IUQM 2022 paper was very tough very tough so these are the same questions let's solve them come let's see let's use our brains the question said we need the minimum and maximum value of this expression so look, the way to calculate minimum and maximum value is very limited today we are looking at one of them the expression is written x plus y whole square minus x minus y minus 2 upon is x plus y whole square plus is x minus y and minus is 2 clear this is written in our hand ok now what to do think first thing is this is written x plus y and this is written x minus y looking at this there is a little problem either do these two x minus y or do these two x plus y see this minus will not accept this will not accept this can accept so what to do for once we will expand it, x plus y whole square, so it will be x square plus y square plus 2xy ok, so instead of 2xy, we will write 2xy here and add another minus 2xy here and add another plus 2xy here, till here there will be no difference we have added 2xy and subtracted it, and write 2xy here then we will add minus x minus y and minus 2 divided by x plus y whole square we will expand it x square plus y square plus 2xy we will write it here and here we will subtract 2xy and add 2xy no problem, it should not be here and then we will add x minus y and minus 2 ok, now what we will do ok, what will happen by doing this By doing this, x square plus y square minus 2xy, these three are made of x minus y whole square. And here, 2xy plus 2xy, how much is it? 4xy.
So, I wrote 4 in plus. And what is the value of xy? 1. So, I disappeared x and y because instead of x and y, I will write 1. 4 times 1 is 4. And in minus, x minus y is there.
And in minus, 2 is there. Clear? Divided by.
Below, there are three terms. I clubbed these three terms and wrote x minus y whole square. Okay, sir? plus 2xy plus 2xy plus 4xy and xy is gone because xy is 1 and plus x-y as it is and minus 2 what we do after this write x-y whole square as it is and 4-2 will be 2 so we will write it at last minus x-y and plus 2 divided by whole square of x minus y plus x minus y and 4 minus 2 so it will be 2. This is the expression after making it. Now here also x minus y, here also x minus y, here also x minus y and here also x minus y.
I have done all four things the same. And why did I do all four things the same? There was a strategy behind it. What was it? The strategy was that we will let x minus y some new term.
So, let's assume a new term x minus y. What should we assume? Let's assume a.
Don't assume that the variable is constant. I have a left, t is less here. So, I have a left instead of x-y. So, this will be a2 divided by a2.
This is the expression I have. And this expression's key value. I assume its value is k What is its value?
I assume its value is k I let its value be k So what we will do? We will throw this term there Whenever you have quadratic upon quadratic Write the concept here Quadratic upon quadratic Or linear upon quadratic Or quadratic upon linear or linear upon linear these four things and we have to find their range range means how much value will it take in the fall and how much value will it take in the increase minimum and maximum and all the intervals between them if we want to find the range then what we will do quadratic upon quadratic, linear upon quadratic, quadratic upon linear and linear upon linear first of all take the expression kk equal take kk equal and after taking kk equal cross multiply and make a single quadratic equation you will say yes, we will make it so what we did, a square minus a plus 2 that is equal to ka square plus ak and plus 2k now we will take all these terms so ka square a square will be minus so a square common will be k minus 1 this is a k, minus a goes there plus a, if we take a common from both then k plus 1 will be in bracket and if plus 2 goes there then minus 2 will come in plus, 2k minus 2, clear? 2k minus 2, that is equal to how much?
0, this is a quadratic equation, what is it? quadratic equation is made in terms of a, as a variable we are getting a quadratic equation in terms of k variable is like a here, so what to think? because So, what was a? It was x minus y.
And the question has said something. x and y are real. So, if x and y are real, then if I subtract 2 reals, then only real number will come.
So, what is a? It is real. Here, a is real.
So, a is a real number. And what is a? It is the root of this equation.
And if a is real and root, then definitely the roots of this equation should be real. The roots of this equation should be real, which means the discriminant of this quadratic equation should be greater than or equal to 0. So how will we find the discriminant of this equation? If we do b square minus 4ac, then who is sitting in the place of b? k plus 1. So k plus 1 whole square minus 4 into k minus 1 into 2 times, we took 2 common from both of them, in the bracket k minus 1. Okay sir, that is greater than or equal to 0. Okay?
Now what we will do is expand the whole square of k plus 1 So it will be k square plus 2k plus 1 And k minus 1 into k minus 1 will be k square minus 2k plus 1 So what we will do is 4 to the 8 outside and k square minus 2k plus 1 Will be greater than or equal to 0 Because k minus 1 into k minus 1 is a whole square Now what we will do is k square plus 2k plus 1 and minus 8k square minus minus plus 8 to the 16k and minus 8 that is greater than equal to 0 so subtraction made minus 7k square and plus 18k and minus 7 that is greater than equal to 0 ok sir ok now what we can do next ok what we can do next sir minus 7k square plus 18k plus or minus 7 we will take all the things in the direction of inequality so 7k square will be plus and minus 18k and plus 7 that is greater than equal to 0 ok, after this what we will do this is again quadratic so if we want we can take the roots so if we take the root of this then the formula to take the root is minus b plus minus under root b square minus 4ac and upon will be 2a This is the formula for the root of quadratic equation. So, if we take the root here, it will be minus of minus 18 plus minus under root 18 square root of 3 is 24 minus 4 into a into c so 4 times 7 is 28 28 into 7 is 190 Wait a minute, 49 double 98 double 196 divided by 29 Sorry, 2a, so 2 into 7 will be So, minus minus plus will be 18 In the plus minus under root, we will subtract 196 from 324. So, we subtract 196 from 324, so we will get 128. How much will we get? We will get 128 divided by 14. So, the square root of 128 is 8 root 2. So, what will happen here? 18 plus minus will be 8 root 2 divided by 14. Divide 2 in both. So, you will get 9 plus minus 4 root 2. divided by 7 clear?
from here, whose value is left? K's value is left for K K is left from here K's value we can take either plus or minus now let's talk about the next this expression we had this expression a square minus a plus 2 upon me a square plus a plus 2 whose value we took as K k is equal to 7k square minus 18k plus 7 which is less than or equal to 0 if I want to factorize it in terms of k then it will be k minus 9 minus 4 root 2 upon 7 plus 4 root 2 upon 7 and 9 minus 4 root 2 upon 7 9 plus 4 root 2 by 7 and k minus 9 minus 4 root 2 by 7 these values are there, basically these two factors will be made after removing roots, k minus first root and k minus second root now if we look at it according to inequalities so what we will do according to inequalities, we will represent both on number line so here 9 minus 4 root 2 upon 7 will be upon 7 will be here and here 9 plus divided by 7 will be here. And we want less equal. So what will be our attempt?
Whenever you know, there are two critical points on the number line. And we want positive. If we want result greater than 0, then we take the outer one. And if we want less than, then we take the inner one.
If you put any value between these two numbers, 0 is coming. If you put 0, then how will the result be coming? The result will be negative.
How will it be coming? Negative result will be coming. That's why, that's why, that's why, what are we going to do? That's why, what are we going to do?
We are going to take this result in the middle. The value of this expression will be from here to here. From here to here means 9 minus 4 root 2 upon 7 to 9 plus 4 root 2 upon 7. Whose value?
This expression's value, that is, K's value. Okay, sir. Okay.
So, the question said that the minimum value of this expression is t is small t and the maximum value is capital T so the minimum value of this is this so this is small t and this is capital T so what he said is add capital T and small t so add capital T and small t so we get 9 plus 4 root 2 divided by 7 plus 9 minus 4 root 2 divided by 7 so we add both 7 LCM comes plus 4 root 2 minus 4 root 2 will be deducted 9 plus 9 is 18 so the result will be in the form of m upon n he said it will be in the form of m upon n it will be in the form of m upon n now what we have to do after coming in the form of m upon n that m and n should be co-prime there should be no common factor other than 1 i.e. what should be their greatest common divisor 1 should be there question said sir i am not saying it from my heart what should be the GCD 1 should be there so we have taken GCD 1 in 18 and 7 no other than 1 common factor divide does not go so we have taken GCD 1 so we need the result of m plus n m plus n means 18 plus 7 if you add 18 and 7, what will be the result? 18 plus 7 will be 25 so 25 was the answer to that question which question? the answer to this question was 25 whenever you get such question with such algebraic expression and you are asked about its value if it is related to minimum or maximum value then you should understand that we need the range of that expression we need a range which we are going to study in domain and range function What was the use of quadratic in this? It was full of quadratic So what we did first? We converted it in a single variable term We converted it in this term After converting it in this term What was our work?
After converting it in this term, we assumed it We assumed its value is k Simplified it and made it quadratic Its roots should be real We forced it on it We did its discriminant greater than or equal to 0 From there discriminant told us Using its maximum value capital T plus small t solved and we got the answer 25 this is the summary of this question let's move ahead, you must have understood this question let's see the printed solution of this question so this is the solution of this question, they have done it with some other method, I don't know ok sir, they have used this method ok sir, this is the printed solution ok sir, so the answer is 25, clear sir? ok, let's move ahead next question, ok sir One more great question to start with. It's a question from IOQM 2022 again. It's a question from IOQM 2022 again.
What does the question say? What does it want to say? It wants to say that A is real numbers. And what is the first equation saying?
The first equation is 3A plus 2 that is equal to 6B. What is the second equation? The second equation is 3B.
plus 2 equal to 5C ok sir and third thing you know what he is saying 3CA plus 2 that is equal to 4A ok sir as he said so many things I got doubt what happened to me I got doubt on this what doubt always here 3 and here 2 here 3 and here 2 here 3 and here 2 what do you want what do you want to say let's ask him so whenever such situation comes When you see all the things are same, then you should understand that we have to multiply all these things. If we multiply all these things, then we will get something which we require. But, but, but, and it is very important to understand this carefully. After giving three equations, the question asked us, let Q denote the set of all rational numbers, given that the product A can take two values, R upon S, which belongs to rational, and t upon u which belongs to rational. Q means rational.
In lowest form, then find r, s, t, u, 4's sum. Okay. So what to do?
What to do? What to do? Let's do one thing.
This b, here b is here, here c is here, here a is here. Let's remove these. What do we do? Let's remove these.
Let's not have variables. Then let's multiply all three equations. Because in all three, we can see the same thing.
In left, 3, 2, 3, 2, 3, 2. We can see the same thing in all three. so if we can see the same thing in all three, then what we are going to do? first thing is, divide b here and here similarly, divide c here and here and here divide a here and here after dividing here b will be divided and here 3a will be left plus 2 by b, that is equal to 6 b will be cancelled here c will be cancelled and what will happen? 3b plus 2 by c that is equal to 5 and what will be here?
here 3c because a is a cancel plus 2 upon a that is equal to 4 because a is a cancel ok then what to do? I mean from abc, I want value of abc I mean from abc, abc can take two values he is saying so what to do? multiply these three equations 1, 2 and 3, multiply all three multiply all three so left portion will be multiplied here left left portion will be multiplied here so here we multiplied 3a plus 2 by b and 3b plus 2 by c and 3c plus 2 by a that is equal to 6 into 5 into 4 6, 5, 30, 30, 4, 120 120 is here we have multiplied all three brackets now what we have to do is multiply all three brackets so first of all the best way to multiply first multiply all three terms so 3a, 3b and 3c multiplied together, so how much is it? so this becomes 3c, 9c, 27 27abc, ok sir then from here we took 3a, 3b but from there we don't take 3c So, we will do this once, and this once, and this once. So, in short, 3a multiplied by 3b, how much?
9ab? 9ab multiplied by 2 by c. So, 9 to the 18, will be ab upon c. What will happen?
18ab upon c. So, once 18ab upon c will be made. Then once 18bc upon a will be made. bc upon a will be made.
Right? No, we are making some mistake. Wait, let's think a little. 3 3s are 9ab and we will multiply it by 2 by a so in 9ab, 2 is multiplied by 18ab and we will divide by a so it will be 18 by b so the first term will be 18 by b if we multiply 9bc by 2 by b then it will be 18c so here it will be 18 18C, 18B, 18A these 3 terms will be made after that, we will take 3A once and the remaining 2 times we will take these terms so multiply 3A or 3C with the products of these terms so how much is 2 by B and 2 by C multiplied? 4 upon B, if we multiply it with 3C, then C will be cut from C 12 upon B will be left Once 12 is followed by B, then C and then A.
These are the terms. And do you know what will be the last term? When 2 by B, 2 by C and 2 by A will be multiplied, then it will be 42008 upon A. After that, it will be 120. What I did is that I multiplied it.
When 2, 2, 2 and 3 are multiplied, then 8 terms will be made. 1, 2, 3, 4, 5, 6, 7, 8. 8 terms are there. Now what we can do is write this 27abc as a dis. And take 6 common from these 3 terms. And write 3c as a dis.
3b and the next place will be 3a. And if we take 6 commonly from here, then 2 by b, 2 by c, 2 by a will be left. So, I wrote 2 by b with 3a. And 2 by c with 3b. Or a.
2 by 1 with this, 2 by a with this, with 3c, so 2 by a with this, why did you do this? Sir, why did you do this? Why are we doing this specifically? So, son, we are doing this because a format is being made, what is it showing? That the value of 3c plus 2 by a is known from before, how much is it?
It is 4, so we will replace it, what will we do? We will replace it, sir, okay? So, for the purpose of replacing it, 27 abc as it is, okay sir?
Plus 6 as it is? 3C plus 2 by A will be 4 3B plus 2 by C will be 5 So we will write 5 in place of 3A 3A plus 2 by B will be 6 So we will write 6 in place of 3A So we will write 6 in place of 3A So we will write 6 in place of 3A So we will write 6 in place of 3A So we will add page 27 A as it is in plus 4 plus 5 is 9 9 plus 6 is 15 15 times 6 is 15 times 6 is 90 so 90 will be minus so 90 will be minus so how much will be left from 120 30 will be left so 8 upon A is equal to 30 we need the value of A, consider A as K so this will be 27K plus 8 by K that is equal to 30 K is LCM, K into K will be 27K square and plus 8 that is equal to 30, we will throw K here so 27K square plus 8 that is equal to 30k clear? no problem we will throw 30k there 27k square minus 30k and plus 8, that is equal to what?
0 now we will factorise this or we will directly calculate the roots so k is equal to minus of minus 30 minus b plus minus under root b square minus 4ac minus 30 square 900 minus 4 into a into c so 4 times 8 is 32, 32 times 27 so 27 x 32 is 54 5 81 5 86 864 will be minus and divide will be 2 x 27 2a clear then what will we do? minus minus plus 30 both will be minus plus then under root of plus minus 900 minus 864 is 36 you know it and I know it too so 27x2 is 54 then k will be 30 plus minus 36 square root 6 30 plus minus 6 is 36 30 minus 6 is 24 and both of them have 54 so 18x2 is 36 and 18x3 is 54 so 2x3 is 1 and 1 value is 6x4 is 24 and 6x9 is 54 so 4x9 So what was the question saying? Let's ask the question now. He was talking a lot.
What was he saying? What was he saying? The question was saying A can take the two values r upon s and t upon u in lowest form.
So we have to tell r plus s plus t plus u. So one value is r upon s and one value is t upon u. Okay sir. So what is the value of r? r's value is 2 and s's value is 3. Okay sir.
r's value is 2 and s's value is 3. t's value is 4. and u is 9 all are in simplest form so i wrote it so we need result of r plus s plus t plus u so how much will be? 2 plus 3 plus 4 plus 9 so r plus s plus t plus u will be 2 plus 3 is 5 5 plus 4 is 9 9 and 9 is 18 so here result is 18 18 is the answer it was very easy question not too tough But because IOCM was of 2022, in which PRMO and RMO were merged in some stages, so it was a good paper. Don't consider it as a PRMO paper, it had a good level. So today we have asked a tough paper, which we were going to discuss.
So 18 is the answer, let's see its printed solution. So we have this printed solution and here its answer is 18. I hope you understood today's story DPP and notes are shared with you You will get DPPs on the app Solve their questions Starting questions are easy I am telling you every time Do them They are tough questions You have to give them some time If you give them some time, they will be questions It is not that the questions are so tough that they are not made by you. It is not like that at all. If all the questions are easy, then what was that exam, sir? If we are talking about a good exam, then definitely the effort should be that we give at least one question, which is of RMO level, 30 minutes.
Give 30 minutes to one question. Because look, you are getting 30 minutes for one question in the exam as well. At an average. in RMO, not in PRMO in PRMO, the level of IOQM which is stage 1 you are not getting much time in that but in RMO, you are getting 30 minutes for each question at an average so you should consider this if you are preparing for IOQM level then you should merge first and second stage you have to practice good questions so the last question in DPP is RMO level question there are some questions of INMO level also, so you try it and if you face any problem, then if you focus then the questions will be solved and most of the questions are like this, the PYQs we are discussing here so you will get a lot of help from them okay, that's all for today, thanks to all of you, thank you bye bye, have a nice day, we will meet in the next class, thank you