 Hello, good afternoon. Welcome you all to the YouTube live session on theory of equations So we still have a minute left Meanwhile those who have joined in the session I would request you to type in your names in the chat box So that I know who all are attending this session Hello, Niranjan, Andrew, Devdas, Sagoj, Ved, good afternoon. All right guys So we can get started with the session because I think it's already time now so we'll continue with the session on theory of equations and We'll continue with the method of change of variable of solving a Biquadratic equation Okay, so let us start with the next type. I think type one had already discussed in the class But I'll quickly repeat it again in the type one type of equation okay we had learned how to solve a Biquadratic equation of this nature right x minus a x minus b x minus c x minus d equal to some constant a where Where a b c Satisfied the condition that b minus a is equal to d minus c Right and if you would recall in the class the last problem that we had done was basically based on this type one Where the approach was we changed the variable involved So what do we did we change it to a new variable y which was actually the arithmetic mean of x minus a x minus b x minus c and x minus d Right, which is actually this that is we change our y to x minus a plus b plus c plus d by 4 correct Right So I had given a question on this in the class. So today. I'll be starting the session with the type 2 Questions okay. Good afternoon everybody who has joined in the session. So we have just begun So this was already we did in the class So we did in the last class, so I'm not going to repeat this again, okay in type 2 We have another Form of the equation by quadratic equation, which actually looks like this x minus a x minus b x minus c x minus d is equal to some constant times x square where where These constants a b c d are such that The product of Two of them is equal to the product of the other two. That means let's say a b is equal to cd Okay Now what do we do in such kind of bi quadratic equation? We actually reduce this to a collection of two quadratic equation by changing the variable to By changing the variable x plus a b by x as a new variable y Okay, so this is the approach that we normally take Let me explain you In fact, let me give you a question where you can actually try out and then I'll explain you how this particular method works Okay Alright, so moving on to the next page. I think you have copied till here. So let's take this question as an example solve the equation x plus 2 x plus 3 x plus 8 and x plus 12 is equal to 4 x square Okay, please note it resembles the format that we had discussed in the previous slide x plus a x plus b x plus c x Plus d is equal to a x square Okay, now guys, I would like you to spend some time solving this and Then I'll be telling you how to solve it by using the approach. I just now discussed with you Everyone's good afternoon Bharat Preeti Kavya Ashutosh Shri Ram Omkar Shri Ramya Do let me know your response on the chat box once you're done with this Hello, Nikhil. Welcome to the live session Sorry, we could not have the offline class today because of the difference of timings of comm science and bio students and moreover, right? You know my working hours And hence this YouTube live session Yes, guys, if you look at this expression, you realize that 2 into 12 that's 24 is same as 3 into 8 So it's something like AB is equal to CD, right? So what do you do here is that we try to club up? These two terms together and we try to club up these two terms together So if you expand this you're going to see x square plus 14x plus 24 And here also you'll see x square plus 11x plus 24 Okay now Divide throughout with x square But divide it in such a way that you give one x to this expression and one x to this expression So what I'm doing is I'm dividing by x square on both the sides of the equation Giving one x to this expression and another x to the other expression Okay, now the benefit over is that when you divide it it becomes x plus 24 by x plus 14 Okay, and again this becomes x plus 24 by x plus 11 is equal to 4 now try to recall In our approach I had told you that you have to substitute y is equal to x plus AB by x Right, AB here is 24. Okay, so this is AB and AB is same as CD So you will have to substitute this part as y Okay, so this part will be substituted with y in this equation So you'll end up getting a quadratic in y like this So y plus 14 times y plus 11 is equal to 4 Right now from here on it becomes super simple to solve this Right, so we can write this as y square plus 25 y This will become 154 okay, 154 minus 4 is 150 equal to 0 right and I'm pretty sure you can easily factorize this as y plus 15 Y plus 10 is equal to 0 So there are two possible values of y that we get y could be minus of 15 or minus of 10 Okay, now one by one you equate your x ultimately we have to solve for x not for y So we have to equate this to minus 15 and we have to equate this to minus of 10 Right, and from here you get the quadratic x square plus 15x plus 24 equal to 0 and From here you get the quadratic x square plus 10x plus 24 equal to 0, right? now This is pretty easy if you look at x square plus 10x plus 24 equal to 0 It could be easily factorized as x plus 4 times x plus 6 equal to 0 and hence your roots could be minus 4 and minus 6 right Absolutely, absolutely Ashutosh bang on No problem away. You're not miss a lot and When we talk about the second one, this is the first one and the second one is x square plus 15x plus 24 equal to 0 You can write your x to be minus b plus minus b square minus 4 ac So b square is 125 Okay 225 minus 4 ac is minus 96 by 2 a so your roots will be minus 15 Plus minus under root 129 by 2 Okay, so the four roots possible over here are x equal to minus 4 minus 6 minus 15 plus under root 129 by 2 and minus 15 minus under root 9 by 2 so these are the four roots possible and hence this is the answer So as you can see how the change of variable helps us to reduce a higher degree polynomial equation to a lower degree in fact It helps us to reduce a bi quadratic into a quadratic and get the solution for that Is it fine guys any question any concern regarding this please type it on the chat box So now we'll move on To the third type of bi quadratic equation that we normally come across Okay, that is of this form x minus a to the power 4 plus y minus b to the power 4 is equal to a constant Okay Here also the approach is more or less same as what we took for the very first type that we did We change the variable by making the substitution as y is equal to x minus a plus x minus b by 2 Right in other words we write it as x minus a plus b by 2 Okay, let's take a question based on this So let's take an example Solve for x minus This so find the real roots Find the real roots of this equation Okay x minus 4 6 minus x to the power 4 plus 8 minus x to the power 4 is equal to 16 Find the real roots of this equation. So you can treat this to be of the same nature I'm not sure about that Ashutosh, I'll have to solve it to see the number of real roots Okay, so I think did you guess that or did you find it or you found that great How about others? Plus minus root 7 Never mind away. Welcome to the live session. Oh, I'm sorry. I'm sorry. It's x actually that's a typo That's a typo. This is x Sorry, it's a bi quadratic in x instead of x. I wrote y by mistake. Thank you Niranjan for pointing that out Okay, guys, it's pretty simple again. The approach is same. We'll consider y to be We'll consider y to be x minus 6 plus x minus 8 By 2 okay, that's nothing but x minus 7 right? Now replace it over here. So we know x is equal to X is equal to y plus 7 So if you replace it over here you get 6 minus y minus 7 is minus 1 minus y to the power 4 and This will give you 1 minus y to the power 4 equal to 16 Which is as good as saying 1 1 plus y to the power 4 Plus 1 minus y to the power 4 is equal to 16 Okay Now if you expand this the good part about it is since they are binomial Expressions with a negative power alternate terms will start cancelling off So you'll be left with You'll be left with y to the power. Let's say I expand the first one. So you'll be left with 1 plus 4 y Plus 6 y square Plus 4 y cube plus y to the power 4 Okay, and again 1 minus 4 y plus 6 y square minus 4 y cube plus y to the power 4 is equal to 16 So plus 4 y plus 4 y mine Plus 4 y cube minus 4 y cube will get cancelled off Okay, leaving you with 2 y to the power 4 plus 12 y square Plus 2 is equal to 16 Which is as good as saying y to the power 4 Plus 6 y square minus 7 is equal to 0 I think it is easily factorizable as y square plus 7 and y square minus 1 equal to 0 correct Now it gives me two roots y could be plus minus 1 and This cannot give me any root because this cannot be 0. Okay, so this is ruled out So if I is equal to plus minus 1 your x which was actually I'm sorry x which was actually y plus 7 Okay, x was y plus 7 could be 1 plus 7 or minus 1 plus 7 So two possibilities are there 8 and 6 so these two are the Real roots that come from this equation. So yes 6 and 8 would be the right answer Absolutely correct. Ashutosh was the first one to answer this all the days So now we'll be moving across to the next concept of equations containing absolute value functions So we'll not be talking about equations containing containing modulus functions Equation containing the modulus functions, okay So guys, you all know the nature of mod functions, right? We have been all introduced to you know all types of mod functions and how they change their definition So we know that mod x behaves as x when x is greater than equal to 0 and minus x when x is less than 0, right? So if at all you have mod functions coming up, right? It's always advisable that you actually redefine the function according to different intervals of x and then solve for the value of x and Check whether it is complying with the restrictions of x that you have already imposed on that Let me just start with a simple example. Let's say I want to I Want to solve this equation. Okay a very simple one No, I'm just giving an example to start with after that. I'll give you a follow-up question So here you start with this parent condition. Let x is greater than equal to 0. So this is your parent condition Now when you have assumed x is greater than equal to 0 you start getting This as so this is redefined. This is redefined as this the word here is redefined Okay, so x square minus 5x plus 6 will become the new redefinition of this particular polynomial so this roots are minus x minus 2x minus 3 equal to 0 So here you find that your x comes out to be 2 and 3 now both of them should comply with the parent condition Right, so if they comply as the parent with the parent condition, then these are acceptable values of the root But if they don't you have to reject the root Okay, similarly I can say x could be less than 0 and If you take x is less than 0 you get x square plus 5x plus 6 because your modulus of x will become negative of x and when you solve this You get the value of x as minus 2 and minus 3 and yes both of them are less than 0 So basically this will also be accepted just like these were accepted So your final answer would be your roots will be 2 3 minus 2 minus 3 Okay Graphically speaking guys, how do we actually plot this? Graphically speaking, you know that if I have to plot y is equal to x square minus 5 mod x first I will plot this function Okay, so this function will be a parabola which is cutting the x-axis at 2 and 3 Correct now when I'm modding x basically what am I doing? I'm doing this transformation on the function I'm doing mod x square minus 5 mod x plus 6 remember modding x Okay, modding x. What does it do? It actually? Reflects that part of the graph which is on the positive x-axis about the y-axis. Isn't it? So what I'll do is I will remove this part and I will make the reflection of this like this About the y-axis right as you can see it will generate two more roots Which is minus 2 and minus 3 and therefore? The solution for this equation is wherever this graph is meeting the x-axis that happens to be at these four points minus 2 minus 3 2 and 3 respectively Is this approach clear? How do we actually deal with equations containing modulus? Okay Now let me give you more examples to work on solve the equation Mod of x minus mod of 4 minus x Minus 2x is equal to 4. So it's mod of x minus mod of 4 minus x Minus 2x is equal to 4 Guys don't be the hurry please focus on accuracy Normally Two minutes will be allotted for you to actually work on every question two to three minutes maximum Okay, so some of you have already started giving your response I need at least five people to respond before I start giving you the solution Okay, so yeah, okay, so mostly I think I'm getting the response as x equal to zero so let's try to equate let's try to solve this question, okay? Now if you see this mod of 4 minus x now we know that When x is less than equal to 4 When x is less than equal to 4 this equation will behave as This modulus 4 minus x will behave as 4 minus x correct So let me just redefine this function first Let me call this as f of x and let me redefine this function first Okay, so when x is less than equal to 4 I will get mod of x minus This will become 4 minus x Okay Mod Minus 2x Correct, which is actually mod of 2x minus 4 Minus 2x So let me just write that up. So here I will write it as mod of 2x minus 4 Minus of 2x Is that fine and when you take x greater than 4 It's just going to become mod of x minus x minus 4 and let me again confirm this this will actually become 4 minus 2x This will become 4 minus 2x So mod 4 minus 2x is just like 4 minus 2x So I'll be just changing this to 4 minus 2x now if you look at The second system this equation and you try to equate it to 4 So if you take this condition and try to solve this equation Subject to the condition that x should be 4 you realize that you end up getting 2x equal to 0 That means x should be 0 But remember this is your parent condition Right, that means this solution will be rejected because 0 is not greater than 4 Right, so now I'll do what I will try to focus my attention on this particular function Okay, so now mod of 2x minus 4 Will again change its definition about 2 so when x is greater than equal to 2 It will behave as 2x minus 4 minus 2x Okay, but remember it is greater than equal to 2 but less than equal to 4 So let me just make this correction over here greater than equal to 2 but less than equal to 4 and when x is less than 2 What happens this becomes 4 minus 2x minus 2x Correct Yes, or no, so this is your second. This is your redefinition of this Now, let us take this condition. Let's take the first one if you take the first one It gives you minus 4 is equal to 4, which is actually not possible okay Right and second situation gives you 4 minus 4x is equal to 4 which means again x is 0 and Remember we have a parent condition that x should be less than 2 So does the child satisfy the parent condition or is there any overlap between the child and the parent condition? Yes, that means x equal to 0 is the only solution possible for this particular Modulus equation. So well done guys those who gave the answer as x equal to 0 Good evening, Sukheer then Lakshya. Welcome to the live session Is that fine guys any question with respect to this? So the keywords over here is redefining the function and Seeing the overlap between the parent and the child conditions on x that you have imposed Okay, under no situation you should actually exceed or go beyond the limitations which are imposed by the parent condition Is that clear can we move on to the next question now? So next question is solve mod x minus 1 plus mod 7 minus x plus 2 mod x minus 2 is equal to 4 Yes, but guys first redefine the word here is redefine your function first the critical points are one two and seven and then start One by one solving the redefined version with the right-hand side and see whether the solution obtained is Overlapping with the parent condition that you have actually decided for that interval. Okay, so Ashutosh has given his response Ashutosh, I will not be commenting on anything right now because that will influence others so Let them first work out. I should get at least five response before I start Solving this for you. Okay, so Shiram also thinks the same as what Ashutosh does. Oh, I see a solution coming from Omkar. Okay Omkar says There is a solution existing So Keith so Keith also says the same no solution. Okay guys, so we have got enough responses now so what I'll do is I'll first make a number line and Plot the critical points that we see over here. Okay, so one two and seven are the critical points So now these critical points divides the number line into four zones first second third and fourth Now guys, this is the reason why this method is sometimes also called method of intervals Method of intervals that's a very very important method when it comes to solving of modulus equations Now if you look at the first interval where your x is less than equal to 1 You realize that these equations change their definition. So first one will become Minus of x minus 1 Right, this will become 7 minus x Okay, and this will again become minus 2 x minus 2 okay, and this is equated to 4 correct Let's see we do we whether we get a solution from here or not So minus x minus x will get a minus 4x and we'll get 1 plus 7 8 8 plus 4 Which is 12 is equal to 4 That means you get minus 4x is equal to minus 8 means x is equal to 2 Which is definitely not a solution because it does not obey its parent parent is x is less than equal to 1 So no overlap is happening between these two Let's look at the second interval where your x is between 1 and 2 Guys, you can actually include one at both the places. It will not make any difference to your result Okay, so let's take the Next condition x lies between 1 and 2 Let us redefine the function again in this interval So the first one will become x minus 1 second one will remain 7 minus x Third one will remain minus 2x minus 2 equal to 4 Correct. So let us try to simplify this so it will give you xx will get cancelled So minus 2x and you'll get 6 plus 4 10 10 is equal to 4 which implies x is equal to 3 Again, it does not obey its parent. So this will be rejected once again Taking the third situation where your x lies between 2 and 7 Okay, let's say x lying between 2 and 7 Again, let me redefine the function. Okay, so when you redefine the function, this becomes x minus 2 Again 7 minus x and this time it will become plus 2x minus 2 is equal to 4 Right, which is actually xx will get cancelled. So 2x here you'll have 5 5 minus 4 which is plus 1 is equal to 4 Correct. Oh, I'm sorry. I'm sorry. I'm sorry So this is x minus 1 Yeah, I'm just Okay, so let me just change this this will become xx will get cancelled so 6 minus 4 which is 2 Okay, so this gives you x as 1 Again, this will be rejected because it doesn't adhere to the parent condition Now going to the fourth situation where your x is greater than equal to 7 x is greater than equal to 7 as the fourth condition So here what we'll realize that Only 7 minus x will become negative Rest all will be positive Okay, so it becomes 2x 2x which is 4x and I will have a minus 8 Minus 12 is equal to 4 So this implies x is equal to 4 Okay, again, this will be rejected. So you'll see that there is no solution existing for this system of equation right So what do we realize is that this modulus equation has no solution existing So guys after so much of hard work, we realize that there isn't a solution existing Is that fine guys? Any questions any concerns? Please do right in the chat box Sir, how do we solve with complex numbers are allowed? Okay, we'll take that in complex number equations, but here right now. We are dealing with the real roots Okay, so whatever roots that we are expecting from these equations I'm expecting them to be real in nature all right now let us now I would like you to Understand this particular form which we normally see in modulus equations Whenever we have equations of the form, let me write it as an important form Okay mod of f of x plus g of x Is mod f of x plus mod g of x Okay, now these type of questions come a lot Can somebody tell me when can this situation hold true when will mod of f of x plus g of x will be equal to the sum of their individual mods Guys, please type in a type in your response in the chat box Okay Okay Same science exactly guys remember for this to happen f of x and g of x must be of the same sign Okay f of x and g of x must be of the same sign must be of same sign Which means This system is equivalent to saying you are trying to solve this inequality Are you getting this point? Okay So this is equivalent to saying that you are trying to solve this inequality So even though it looks in as an equation to start with or prima facie It looks like a equation but ultimately it converts itself to an in equation like this And I'm sure all of you are exposed to solving of quadratic inequalities Remember wavy curve. I think everybody remembers wavy curve. Let's take a question based on this Solve the equation mod x by x minus one Plus mod x is equal to x square by mod of x minus one Yes, so I have got a response from sukeet gaurav Shavan niranjan Okay guys, so if you look at this equation very closely You would realize that This expression that you have over here You can actually write it as mod of x square by x minus one Right, it would not be wrong to include x square within the mod because we know it's always greater than equal to zero And x by x minus one plus x actually you realize would be x square by x minus one So here is the situation which is arising from this condition that we had discussed a little while ago Right now The conclusion that we had drawn from the theory was that this is equivalent to let me write it here equivalent This is like solving this inequality Okay, that means f of x into g of x that is x square by x minus one would be greater than equal to zero Again, you can make the wavy curve for this. So zero and one Right, so this will be uh plus Minus and this will again be minus Now they're asking you where it is greater than equal to zero greater than equal to zero will happen when x will be greater than equal It cannot be equal to one because then the denominator will become undefined So x should be greater than one Or it could be zero Or x could be zero Okay, so you can say the solution of this would be x should belong to the interval one to infinity Union with the element zero Is that fine? So this is the answer for this modulus equation Is that fine guys? So you would realize in certain situations that even though you are solving an equation You may not get Roots for it, but you may get intervals for it as your answer Let me show you another example where if you start with an equation You will end up getting a interval as the answer not a particular value or values of x So let me give you this question mod of x square minus 8x plus 12 By x square minus 10x plus 21 Is equal to negative of x square minus 8x plus 12 by x square minus 10x plus 21 Pretty simple guys Please focus on accuracy. I should Get right answers from each one of you The prerequisite is you should all be knowing your wavy curve very well Okay, so got a response from Bharath Bharath Can you tell me which values are inclusive which are exclusive? You have not used any bracket Two did or two excluded three included or three excluded So I don't see a right response with respect to the brackets used Right, so yeah So one of you has given the right answer. I'm not disclose who Wow, Shavan curly brackets. Okay. Okay. Great. So mostly, you know the Core principle behind solving this equation So if modulus of something is giving you a negative response That means the input to this modulus must be a negative quantity But remember many of you have taken it to be less than zero It can be actually less than equal to zero Right because zero whether it doesn't matter whether you put a negative sign before zero or not. Okay Now, I think this is a rational A function which can be factorized so you can factorize the numerator as x minus two x minus six Denominator could be factorized as x minus three x minus seven less than equal to zero Let us plot it on the wavy curve So the zeros for these factors. Let us plot them first. So two three six and seven Right mode right most interval will start with the plus and then keep alternating the sign because or each of these factors have been raised to odd powers So less than equal to zero means the intervals where you have written a negative sign Which is two comma three and six comma seven Be careful about the brackets Square brackets will be Only along that numbers which are in the numerator. So two is in the numerator. You can put a square bracket Three is not in the numerator. So you have to put a round bracket Union again six is in the numerator put a square bracket seven is in the denominator put a round bracket, right So put a square bracket around that number which is in the numerator and round bracket around that number Which is present in the denominator in this rational function right so this becomes your answer is that fine any question any concerns please let me know so guys I will quickly take you through modulus inequality questions as well because you have already done this in your inequalities chapter that's why quickly take you through in fact I can directly start with a question solve the in equation solve the in equation 1 minus mod x by 1 plus mod x whole mod greater than equal to half greater than equal to half so just a quick since we have already started discussing modulus based functions so I thought I would give you a question on modulus based inequality question guys the idea here is to get the redefined version of these any equations right we cannot solve questions with mod in place we have to redefine it for a particular interval or intervals of x and then try to address the inequality involved okay so I got a response from Sukirt that was fast good I'm not commenting whether it's right or wrong Sukirt right now I'll keep you in suspense okay Ashutosh backs Sukirt up okay how about other Shaban Gaurav Abhay Kavya Niranjan Omkar Bacabh guys I would love to see response from each one of you okay Shaban has got a slightly different answer well great so I've got response from most of the people so let me start solving this problem so guys the first thing that we see is we have you know mod of x involved within this bigger mod right so what I will do is I'll try to redefine this expression so solving this inequality is like solving this inequality subject to the condition that x is greater than equal to 0 or solving this inequality feel free to correct me wherever you feel so when x is less than 0 so solving this inequality is equivalent to solving this system of inequalities correct when I say solving the system of inequalities means simultaneously solving them that means you have to take an overlap of the answers that you get from them okay now let me take a bit of time to simplify this even further so if I'm not wrong the first one just gives you 1 by a mod of 1 by a mod of 1 plus x is greater than equal to half okay if your x is greater than equal to 0 and the second one gives you 1 by a mod of this greater than equal to half if x is less than 0 right is that fine now guys remember if x is greater than 0 1 plus x is also greater than 0 correct so I can say that this is as good as saying 1 by 1 plus x is greater than equal to half correct remember x is greater than equal to 0 now assuming x is greater than equal to 0 let us try to address this so this will become 2 is greater than equal to 1 plus x so x is less than equal to 1 that means combining these two conditions I can say your x lies between 0 and 1 guys remember I could have the courage to take it on the other side because I knew x was greater than equal to 0 and hence 1 plus x was positive right is that fine now what about the next one when x is less than 0 again we can say convincingly that this expression this expression would be as good as saying 1 by 1 minus x is greater than equal to half right since x is less than 0 I know 1 minus x is positive and hence I can write it like this that means x is greater than equal to minus 1 so combining these two I can say that your x is lying between minus 1 to 0 correct is that fine now these two situations are like the possibilities so these two are cases so we'll take the union of these two conditions so when I take the union of these two conditions I end up getting this as my answer right or you can also say x belongs to the interval minus 1 to 1 right so the first one to answer this correctly and was Sukirt yes Sukirt was the first one to answer this correctly I think Sukirt and Omkar were the first people to answer this correctly well done guys now guys we'll start discussing equations involving equations involving greatest integer functions least integer functions and fractional part functions let me tell you these functions are very very frequently used in the J main and J advance exams apart from your kbpy exams where they try to test how you actually deal with such kind of special functions so these are called special functions because they have some rules which describe them for example if I talk about gif function that is greatest integer function we all know that greatest integer function gives you or returns to you the greatest integer less than or equal to x so when we say greatest integer of x it basically returns to you the greatest integer greatest integer less than or equal to x less than or equal to x okay we all know this as an example I can cite some situations like 3.55 greatest integer will be 3 correct 0.89 gif will be 0 but if you put something negative like 8.34 the greatest integer for this will be minus 9 so we all know the we all know the definition of greatest integer correct right now in a nutshell if your x happens to lie between two integers n and n plus 1 your gif will actually be n right this is how we actually know our greatest integer functions now there are some properties that we are going to discuss for greatest integer functions which are very very critical in solving questions properties of gif the first property that I'm going to talk about is if you find the gif of x plus minus n right where n belongs to some integer it is as good as saying gif of x plus minus n so this n will come out of the gif bracket right is it fine take for example let's say I say 2.3 plus 5 okay we all know that if you evaluate it it is be it'll be gif of 7.3 which is actually 7 right it is as good as saying gif of 2.3 plus 5 which is actually 2 plus 5 which is 7 so we get the same output you may even try a negative number let's say I take minus 2.3 plus 5 right that's going to be 2.7 gif which is actually 2 right we can also write this as gif of minus 2.3 plus 5 which is minus 3 plus 5 which is again 2 so we get the same result irrespective of whether our x was positive or negative and hence this property is that clear okay moving on to the next property guys gif of minus x is just minus gif of x if x happens to be an integer in fact you can just write it as minus x as well because if x is an integer gif of x is as good as x okay now this property doesn't need an example to support it is too obvious but what is important is when x is not an integer okay when x is not an integer what happens to gif of negative x when x is not an integer gif of negative x becomes minus 1 minus gif of x so this is to be followed when x is not an integer okay let me take an example to support the second one let me make this bracket proper way okay yeah so let's say I want to find out gif of minus 3.7 right we all know the answer for this the answer for this is minus of 4 okay now try to verify whether this is correct or not so it's minus 1 minus gif of 3.7 right gif of 3.7 will be just 3 so minus 1 minus 3 gives you minus 4 yes both these answers match right and hence the property as you can see in front of you okay moving on to the next property now property number three property number three this is a property which you will find as direct question in many regional entrance exams gif of x plus gif of x plus 1 by n plus gif of x plus 2 by n and so on till gif of x plus n minus 1 by n where n belongs to a natural number is actually gif of nx is actually gif of nx okay let me try to illustrate this property through an example let's say I have x as 1 by 3 let's say okay and let us say my n is a natural number which is let's say 10 fine now I want to find gif of 1 by 3 plus gif of 1 by 3 plus 1 by 10 gif of 1 by 3 plus 2 by 10 and so on till I reach gif of x plus 9 by 10 by the way it was 1 by 3 so let me just write 1 by 3 over here now guys how do we solve this problem if you just apply common sense you are not going to see one unless until you realize that you reach 1 by 3 plus 2 by 3 right okay right so unless until you see reach 1 by 3 plus 2 by 3 its gif is not going to give you 1 correct so when is the first time that you will realize that you are you have started getting 1 as your answer what should ever add what should what what value of k should I keep over here says that I start realizing that this comes one which is the least possible value of k so till 6 by 10 you guys are saying it will not become 1 let's check so 6 by 10 is as good as 3 by 5 so that'll give you 15 5 plus 9 okay yes you're correct so till 6 by 5 it will not give you 1 the moment you step to 7 by 10 start giving you 1 isn't it so can I say that from this value all the way that are till 1 by 3 plus 6 by 10 I will get zilch I'll get nothing from it correct after that I'll start getting 1 okay this will be one so will be this okay so will be this okay so 1 1 1 is what we get from these three so we get 3 as our answer okay so we get 3 as our answer which actually maps on to this property where we have gif of 10 into 1 by 3 10 into 1 by 3 is approximately 3.33 gif of it is going to be 3 okay so this is how we can illustrate this particular property moving on to the fourth properties in fact this is a property which is based on inequalities when you say gif of x is greater than equal to n let's say n is some integer right remember it is as good as saying x is greater than equal to n is that fine so I will number it as a secondly when you say gif of x of x is greater than n it is actually as good as saying x is greater than equal to n plus 1 right again n is an integer over here so when we say gif of x is greater than n and you know gif of x will always be an integer okay so the next integer possible after n is n plus 1 isn't it so x should be greater than equal to n plus 1 for this particular inequality to be satisfied is this clear any questions if you say gif of x is less than equal to n okay it is as good as saying x is less than n plus 1 again n here is integer guys please remember these inequalities are very very important they have been asked n number of times in the comparative exams similarly when you say gif of x is less than n it is as good as saying x is less than n again n is an integer is that fine so these four inequalities are very important okay moving on to the fifth property always remember gif of x plus y will always be greater than gif of x plus gif of y please do not please do not open this and say gif of x plus y is equal to gif of x plus gif of y normally this will always be greater than equal to gif of x plus gif of y now let us try our hand at solving a problem based on gif x and y satisfy let x and y satisfy the equation y is equal to 2 gif of x plus 3 and y is equal to 3 times gif of x minus 2 okay simultaneously okay so these two conditions are satisfied by x and y simultaneously determine the value of determine the value of gif of x plus y okay some people are asking me the proof why don't you come with the proof and let me know so whatever other properties some of them are very intuitive okay and especially that natural number thing i would like all of you to prove that and send me the response on the group simple problem to start with okay so nirajan has already given a response anybody else who feels the same okay shawan also backs it up less than 30 suki says equal to 30 some of you are saying equal to 30 some of you are saying less than equal to 30 okay let me just clarify the confusion over here so first of all we can equate the y on both the sides when we equate the y on both the sides remember since 2 is an integer we can actually pull it out by using our property so it will become gif of x minus 2 right so if i solve this i get 9 is equal to gif of x right undoubtedly 9 will be the gif of x now substitute this substitute this in y is equal to 2 gif of x plus 3 that will give you y as 2 into 9 plus 3 which is going to be 21 correct right now the requirement of the question is gif of x plus y which is actually gif of x plus 21 so far i don't know x i only know gif of x right but thankfully because of the property that one of them is an integer i can write it as gif of x plus 21 which is going to be 9 plus 21 so this answer will be exactly 30 this will be your answer it will not be less than or something like that it will be exactly 30 so the first one to answer this correctly was shravan shravan was the first one i would not consider niranjan to be the first one because he himself asked me a question so instead of giving the answer he asked whether 30 was correct or not so he was not confident so shravan was confident great so without wasting much time we'll move on to the second special function which we call as the least integer function okay so let's we have already seen the greatest integer function let's talk about the least integer function now greatest integer function is called as the floor function least integer function is actually known by the name of ceiling function okay i'm sure you guys are aware of the graph of this function by the way let me define this first least integer function is basically it gives you when i say least integer x by the way the symbol for least integer is like this or many books will write it as this so least integer function is defined as the least integer least integer greater than or equal to greater than or equal to x right take some example let's say i want the least integer for 3.57 that will be 4 okay least integer function of minus 8.29 that will be minus 8 least integer of minus 0.7 that will be 0 okay and if you plot the graph for such a function your graph would look like this so there's slight shift from the graph of gif now guys in a nutshell if your x lies between two integers n and n plus 1 your ceiling function x is defined as n plus 1 simple as that just to summarize this in few words i can say if your x lies between n and n plus 1 where n of course is an integer okay then the least integer function of x will be n plus 1 right what is important for us is to understand the relationship between gif and lif so relation between gif and lif it's a very simple relation guys lif is basically equal to gif when x belongs to integer in fact then all of them will belong to x actually and it is one more than gif if x is not an integer please try to understand this simple logic okay so i'll repeat this again lif is equal to gif is equal to x if x belongs to integers and lif is equal to one more than gif if x belongs to if the x doesn't belong to integers right so many a times we use this kind of relation in order to solve equations involving lif and gif so what do we do wherever i see an lif i try to express it in terms of gif by bifurcating it into two such conditions okay let me take you through some examples to illustrate what i mean to say let's say i want to solve a simple equation like this lif of x into gif of x equal to one okay let me tell you whenever as such questions will be given they will explicitly mention the statement where this represents gif and this represents lif okay unless until stated please do not assume any bracket to be a special function bracket okay treat them as ordinary brackets now guys i would like to see if somebody can solve this please solve it quickly and send me a response or type in your response in the chat box very simple question to start with okay so i've started getting response okay so mostly janta is saying one and minus one okay i understand you can actually verify by using the simple substitution in this particular question but let's say i have to solve it in a rigorous way how will i solve it my first thing is i would first consider x to be integer so this is my parent condition so when x is an integer we know that gif and lif both will be x each correct so this equation will convert itself to saying x square is equal to one which clearly implies x equal to plus minus one and undoubtedly both of them belong to integer and hence they are acceptable right but let's say if i take the parent condition as x not belonging to integer it would be beneficial for us to convert lif to gif now we know that lif is one more than gif so i can say lifx is gif plus one times this equal to one which clearly actually brings us to a quadratic in gif right now using sridhar acharya formula we can solve it as minus one plus minus b square minus 4ac by 2a and let me tell you that this is not an integer this will never give me an integer right ideally this should have been an integer right so there will be no solution existing in terms of no solution existing in x belonging x not belonging to integer so guys you all are correct the only possible solution is this so this will become your answer that is x is plus or minus of one well done again sukit was the first one to answer this correctly well done sukit sure we'll take some few more cases just to give confidence to you so next is solve solve lifx square plus lifx plus one square is equal to 25 where this bracket denotes the least integer function least integer function guys take your time do not be in a hurry to solve it i'll appreciate if you can give me the right answer rather than giving me a wrong answer quickly it's good again to make two situations for x belonging to integer x not belonging to integer for x belonging to integer it is super simple right it's no brainer but x not belonging to integer i would request you to convert lif to gif okay so i've been getting answers from people omkar has given the answer two two three two excluded minus three to four minus four excluded uh okay so let me tell you that one of you has given the right answer now who i will not tell but okay son giving me the right answer rather than guessing who gave the right answer sure sure sure take your time i'll give you a minute or so guys let me tell you there would be intervals that would be satisfying this particular solution of the problem time up now let me solve this so first thing is we'll take a case one where i'm considering my x to belong to integers now when x belongs to integers it is as good as saying i'm trying to solve this quadratic equation right which i'm sure none of you would have any problem or inhibitions solving it it will be simply x square plus x minus 12 equal to zero and this can be factorized as x plus four x minus three equal to zero giving you the result as minus four and three undoubtedly right considering case two where i'm taking x as not integers okay when x is not an integer we have seen that i could write lif as gif of x plus one so this can be written as this square okay and here i can write it as gif of x plus one plus one whole square now don't read these curly brackets as factional part because i have not defined a curly bracket like that yet okay now we know the property just now we learned that gif of x plus an integer could be written as gif of x plus one so this one could be taken out of the brackets okay correct and this is equal to 25 right now treat your gif of x as some k okay just for the purpose of writing less i am writing it as k for the time being so this is k plus one k plus two the whole square is equal to 25 and if you simplify this you'll get 2k square or 2k plus 4k which will give you 6k and i will have five minus 20 equal to zero which is actually k square plus 3k minus 10 equal to zero and if i'm not wrong it is factorizable as x plus five times x minus two equal to zero correct this clearly implies gif of x plus five times gif of x minus two is equal to zero which means gif of x can be equal to minus five or gif of x can be equal to two right okay so remember my x is not integer over here so you saying that gif of x equal gif of x equal to five actually implies that your x could be any value any value between minus five to minus four isn't it correct around brackets and you saying gif of x is equal to two can be any value between two to three right remember two and three should not be included because of this parent condition where you have said that x cannot be an integer right but now we'll try to combine these i'm going to combine these two so let's combine these two so when you combine these two your end result would be x belonging to minus five now minus four can be included because minus four here is allowed union two to three three can be allowed right so be very very careful while you are choosing your brackets around these numbers because your options will have all permutation combination of these brackets and here and there if something goes wrong you'll end up getting a negative marks right so guys this becomes your answer for this question right so hope you guys are now a little bit confident how to deal with at least gif and lif functions so now we'll be talking about another special function which we call as the fractional part function again not a new thing to you when we say fractional part of x it basically means this right and we define it as nothing but x minus gif of x so this is how it is defined i think in the bridge course we had also seen the graph for fractional part of x right it's a periodic function it is one of those algebraic functions which are periodic okay of this nature correct so zero one two three minus one minus two so it keeps on going in both the directions remember it can never become one so essentially speaking gif sorry a fractional part of x should lie between zero and one including zero excluding one okay one of the very important properties that actually result from this definition is that if you since it is periodic if you want to find the gif of sorry a fractional part of x plus n where n is an integer it is as good as fractional part of x okay this is a very important property which we normally make use of in solving questions okay right let me begin this with a question over here so all of you please attempt this question now question is find or simplify rather should not say find i should say simplify i yeah simplify gif of x plus summation of fractional part of x plus r by 2000 where r is taking integral values from 1 to 2000 simplify this i'll just give you 30 seconds for this super simple okay so i started getting the response guys you're too fast good right so most of you have answered this correctly guys the only thing that you need to keep in your mind is that you realize that gi fractional part of x plus r where r is an integer will always be fractional part of x so you actually end up summing fractional part of x 2000 times and you have of course 1 by 2000 waiting down over here okay so it's just going to be gif plus 2000 by 2000 fractional part of x which is actually going to be x itself so x is going to be the answer very good let me slowly take you to the next question next question is solve x minus 1 equal to x minus gif of x times x minus fractional part of x where this represents gif and this represents fractional part again an easy question guys okay solution may not be a single value of x it may be a range as well so far i have not got the right solution in fact i should say i have not got the complete solution from any one of you all right so i have got a right answer right now that's one to two inclusive one exclusive two guys i don't know why why you got this question as wrong let's see first this x i could write it as gif plus fractional part minus 1 and we all know that this is fractional part this part is actually fractional part times x minus fractional part okay x minus fractional part is as good as saying gif okay so i can write this as fractional part times gif okay now bring all the terms to one side so we can say gif fractional part minus gif minus fractional part plus 1 is equal to 0 now guys if you see this very very carefully you would realize that it's actually factorizable like this okay which implies that either gif of x could be 1 or fractional part of x could be 1 but the latter is not possible because we know that your fractional part of x will always lie between 0 to 1 is that fine so you can only have one possibility that your gif of x is equal to 1 which gives me a range of value of x that is one to two two exclusive one inclusive so this will be your answer okay so yes sukih was the only person to have solved this question correctly now guys let me tell you some important tip over here if you have been given a problem that's what i call the problem solving cycle problem solving cycle if you have been given a problem which involves x mod x gif of x li f of x fractional part of x correct then what do we do we follow a particular cycle we follow a particular convention in order to minimize our effort in solving the problem the first thing is we first try to redefine mod that means this should be redefined according to the interval that means you can you convert everything to this okay that means there should not be any mod mod has to be redefined correct next is you redefine your li f li f should be written completely in terms of gif right so see we are dropping one by one so first of all this got dropped second this got dropped dropped means you're you're redefining it okay now third step is your x should not be there so this will x should be converted completely as gif and fractional part okay so now this is dropped okay and finally when you get this you try to express you try to express this in terms of the greatest integer function okay and use the condition that your fractional part always is sandwiched between 0 and 1 okay to get the values of gif of x let me give you a simple example to illustrate this let's say I want to solve this question for fractional part is equal to x plus gif of x okay now as I told you we'll first since there is no mod involved over here I have to first make sure that I write this x as gif plus fractional part okay which means three fractional part can be written as three fractional part can be written as 2 gif of x now make gif fraction part the subject of the formula okay and since we know that this lies between 0 and 1 it means 2 gif of x by 3 lies between 0 and 1 correct which means 2 gif of x lies between 0 and 3 correct which means gif of x lies between 0 and 3 by 2 now the only integer that is possible between 0 and 3 by 2 including 0 is 0 itself and 1 correct so these are the only possibility right now if this is possible let's go back to this situation over here so we can say fractional part could be either 2 by 3 into 0 or it could be 2 by 3 into 1 so both are possible you can have fractional part as 0 and 2 by 3 so your possible values of x could be 0 plus 0 which is 0 or it could be 1 plus 2 by 3 which is 5 by 3 so these are the two solutions your x could be 0 or your x could be 5 by 3 okay sure Pithi you can join so do you understand how this cycle works so just remember the order in which you have to redefine functions so first mod x then comes lif okay then comes x right so finally you get everything in terms of gif and fraction part then make the fraction part the subject of the formula and subject it to the interval that it lies between 0 to 1 0 inclusive 1 exclusive and try to identify what integer values of gif can be possible and then try to construct your x back hope this example is clear and with this I would like you all to work on this question solve mod 2x minus 1 is equal to 3 gif of x plus 2 fractional part I'm sorry there's a 1 over here I think I missed out a 1 yeah it's mod of 2x minus 1 correct okay so I've got a response from Sukirt, Kavya and Bharat so far okay so even Ved got the same so let us quickly discuss it now first thing is you have to redefine this function so we know that mod of 2x minus 1 will behave as 2x minus 1 when x is greater than equal to half and behave as 1 minus 2x when x is less than half correct so we'll start with the case one where I purposely choose my x to be greater than equal to half okay so this is my parent condition to begin with right so under this situation your problem is going to be simplified to this correct and as I told you according to the problem solving cycle x could be written as gif plus fractional part minus 1 plus 3 gif plus 2 fractional part of x okay so if I simplify this I get gif minus 1 is 3 gif that means gif is equal to minus 1 correct right which means your x could actually lie anywhere between minus 1 to 0 right but unfortunately we need to reject this because it is not complying with the parent parent says x should be greater than equal to half and here I am getting x somewhere lying between minus 1 to 0 so I have to reject this solution okay now let's let's look into the second case case number 2 where I am using the interval x is less than half when x is less than half I get the expression as 1 minus 2x as 3 gif of x plus 2 fractional part of x which means 1 minus 2 is this so this will become let me just simplify it to one test so 5 gif plus 4 this is equal to 1 so 4 fractional part of x can be written as 1 minus 5 gif of x so fractional part could be written as 1 minus 5 gif of x by 4 okay now use the restriction that your gif of x will always lie between 0 and 1 so 1 minus 5 this by 4 will always lie between 0 and 1 that means 1 minus 5 gif of x will always lie between 0 and 4 so minus this will always lie between 3 so divide by minus 5 so it will become minus 3 by 5 gif of x will less than 1 by 5 now the only integer which happens to lie between this interval is 0 correct and if you put x gif x as 0 over here you get the fractional part as 1 by 4 so finally when you construct your x x is nothing but gif of x plus fractional part you'll end up getting 0 plus 1 by 4 which is 1 by 4 as the answer which is 1 by 4 as the answer right so guys now we'll go for a small break okay you can just have a break we'll be back at 5 13 let's be back at 5 13 or 5 15 so we'll resume at 5 15 p.m just around eight minutes of break so i'm gonna read you know that i got to going out each and every test to see if it falls from my face and that in that course i used to waste half an hour time and i did that to the boss everywhere no i can't wait for them to say oh i can't just use what i'm gonna call and manji and the crazy so why is it so scary that i don't want to let anxious more under us that tomorrow i'm gonna bring it over and you're not gonna do that so this psychology is happening because you watch the video so this is where he was coming from he doesn't want to come here he wants to take part because he wants to be here i don't want to do this i don't want to do this i don't want to do this i don't want to do this i don't want to do this yes things are very high there but he knows that ha ha ha ha things How many times do you do it all? And then I thought it was funny. That's the case. I have my reality. You do it all the time. I don't know if you're going to do that. I think, um, maybe you can actually do it. You don't go on for a long time. You may have to go and have fun. I think it's already very much. I think it's going to be tomorrow. Why is that? I told you that. Okay. All right, guys, welcome back. So let us start with a problem itself. So here we have a system of equations in X, Y, Z. Okay. And we need to solve for X, Y, and Z. Okay. So note these symbols are usual symbols. That means this represents GIF. Okay. And this represents the fractional part of X. Or let me just write dot over here. Just like I have written for GIF. So this represents the fractional part. So guys, take your time. Don't be in a hurry to solve it. It's a simple question. Let me tell you even though there are three variables, three unknowns involved, it's a simple question. Okay. So I see a solution coming from, uh, Sukirt. Okay. 2, 1.3, 3.1. Well, I'm not saying it's right or wrong. Let's wait for others to reply. Yeah, got it. Got it. Sukirt, I understood. I understood. Okay. So Bharat also thinks the same. All right. So I've been getting a response from all mostly all of you. So let's try to discuss this. So first thing is I will add all of them. Right. So let me call it as first, second, and third. Okay. Let me add all of them. When you add all of them, remember, so you have these two terms which will combine with this X to give you 2X. Okay. Then Y will combine with these two to give you 2Y and similarly we'll have 2Z equal to this will be 7.4 and 5.4 will be 12.8. That means your X plus Y plus Z will be equal to 6.4. Okay. Now just add first and second. If you add first and second, remember you will be getting X, Y and Z. So X, Y and Z will be obtained when you add these two along with fractional part of X and greatest integer function or GIF of Y. So this will be equal to 7.4. Correct. Now we already know this result. We already know X plus Y plus Z is 6.4. So substituting over here I get this is equal to 1. Right. Now when this is equal to 1 the only possibility that can happen is your greatest integer is 1 and your fractional part is 0. Right. There's no other possibility apart from this. Now we will add the second and let's say the third equation. So when you add the second and the third equation what do we get? We get plus Y plus Z plus GIF of Y fractional part of Y plus GIF of Z. Right. This is equal to 8 9.7. This is equal to 9.7. Now this is already 6.4 over here. So this will actually give you 3.3 which clearly implies which clearly implies the GIF of Z should be 3 and fractional part of Y should be 0.3. Correct. Similarly adding 1 and 3 adding 1 and 3 what do we get? We will get X plus Y plus Z plus GIF of X plus fractional part of Z equal to 8.5. And again this is 6.4 So I can say GIF of X plus fractional part of Z is going to be 2.1 which clearly implies that this is going to be 2 and this is going to be 0.1. Now we have got everything required for us to get our X. So X will be nothing but GIF plus fractional part which is going to be 2 plus 0 Y is again going to be GIF plus fractional part of Y which is going to be 1 plus 0.3 which is 1.3 Z is going to be GIF of Z plus fractional part of Z that is going to be 3 plus 0.1 which is 3.1. So this will be your result. The first person to answer this correctly was Sukirt. So now let us take a last question on this. Solve GIF of X whole square is equal to GIF of X whole square plus 2X. Which class you are talking about? 11? 9. Just one second. I have 10. Only 9. We will have to wait for a few more environments to happen and then we can keep you updated. We could not get that strength. Yes guys. I have got the response for most of you. So let's try to solve this now. So again, let's say I consider 2 cases when X is an integer as I already told you we have 2 cases X as integer and X not as integer. So if X is integer all of them will be X itself. So basically 2X will be 0 that means X is equal to 0. Case 2 is when X is not an integer. When X is not an integer we have already discussed that your LIF can be written as GIF plus 1. So I can write it as GIF of X plus 1 the whole square is equal to GIF of X whole square 2X. If you simplify this by opening the brackets this can be cancelled off and here itself you can actually write it as 2GIF plus 1 and this will be again 2GIF plus 2 fractional part. So from here we get fractional part is equal to 1 by 2. Fraction part is equal to 1 by 2. Now when I say fraction part is equal to 1 by 2 your GIF can be any integer. So your X could be any number of this type where n is an integer. So this will be your solution. So it's not necessary that you should have your n as 0 it could be any integer plus half. So this condition must be met. Is that fine? So combining all the solutions you will have X equal to 0 as well along with n plus half. So for example let's say I take n as 3 by 2. So I take n as 1. So my X will become 3 by 2. Let's say 3 by 2 satisfies this or not. So 3 by 2 LIF will be 2 square 1 square plus 2 into 3 by 2. Yes it is satisfying 4 is equal to 4. So it can be any X value of the order n plus half n being integer or n could be 0. X equal to 0 also holds true. Now we are going to now begin with solution of irrational equations. Solution of irrational equations. Guys what is irrational equations? Irrational equations are those equations which are subjected to radical sign. Radical sign means power of half, power of one third, power of any kind of power where the power is not an integer or a negative integer. That is called a radical power. The function which is subjected to radical power it is called a radicand. Example is like function having root X function having X to the power 1 by 3. So if you find these kind of expressions appearing in any equation those will be called as irrational equations. Now we have to be very very careful while we are solving irrational equations. So before I start your irrational equations can be broadly classified as irrational equations containing even roots. Containing even roots like X to the power half X to the power one fourth you know or any kind of function let me not write X it could be any function to the power half okay. Any kind of function to the power of fourth root sixth root etc. Now remember if your radicand here f of X is called a radicand f of X will be called as a radicand okay. If your radicand is negative if your radicand is negative then such equations will have their roots as imaginary okay. If your radicand is zero okay then your roots will also contain the roots of f of X equal to zero right so let's say we are trying to solve such kind of equations where under root of f of X is equal to zero and you find your f of X to be you know a negative or you choose such an interval where it is negative so you will have imaginary roots. If your radicand is positive okay then only you can expect your roots to come out to be positive is that fine. And the second situation that let me not write f of X over here it will become slightly complicated so let me let me write it just as X X X X so root will be zero in this case okay. So second type of irrational equations you can have is one with odd roots okay like X to the power one by three X to the power one by five etc okay here if your radicand here if your radicand is positive roots will be positive roots will be positive if your radicand is zero right you will have roots as zero okay. And if your radicand is negative you will have real roots as negative okay. The real roots I am talking about the real roots only I am not commenting about the complex or imaginary roots for such rational irrational equations. Now guys one very important thing which I would like to highlight over here whenever you are trying to solve questions involving irrational functions ensure that your root always lies under the domain of that given expression so always check the domain because you cannot work with under root of a negative quantity and expect a real root to come from it okay. So let me try to explain this with some examples let's say I want to solve this question solve under root of two X plus seven plus under root of X plus four equal to zero I am talking about real roots itself real roots can be rational roots also okay. In this entire chapter we are worried about real roots when we are saying solving of the equation we want real roots only so I have already started getting a response now before you start solving the equation as I told you you should must watch out for the domain of the function. This function will be defined only when you have these two criteria simultaneously met that means X should be greater than equal to minus 7 by 2 and X should be greater than equal to minus 4 the intersection of these two is X greater than equal to minus 7 by 2 right X should be greater than equal to minus 7 by 2 so if you take any value of X greater than equal to minus 7 by 2 you would realize that four plus X will always be greater than half So under root of 4 plus x or x plus 4 whatever you can call it will always be a quantity which will be greater than equal to 1 by root 2. So this quantity can never be equal to 0 and therefore when you rightly said there is no solution you were correct. So when you said there was no solution you were correct because this quantity will always be positive in this case simultaneously they cannot become 0. Are you getting it? Similarly you can also get you know questions which look very weird for example under root x minus 4 is minus 5 can we have a solution for this right? We cannot we cannot have a solution for this because our under root of x minus 4 whatever is our x or whatever is the permissible value of x will always be greater than equal to 0. It cannot be a negative quantity. So here also no solution alright so let me give you few more cases solve this under root of 6 minus x minus under root of x minus 8 equal to 2 guys remember only when you have an expression raised to an even root then only you will be bothered to check whether the expression or the radicand is positive or not. If you have an expression subjected to an odd root we do not care to check the sign of the radicand because you can find out the root of that particular radicand even if it is negative positive or 0 okay so again there is no solution because as you can see here the radicand are subjected to even roots so I should have I should ensure first of all that these two terms must be positive simultaneously correct. So this says x should be less than equal to 6 and this fellow says x should be greater than equal to 8 right and simultaneously they must be true so you have to take the overlap of this condition which is not possible right so there is no there is no value of x for which both of these radicands would be positive simultaneously and hence there is no point solving this equation at all that means your domain itself is a null set right so we cannot expect to get a solution for such kind of equation right so the idea here is to first find the domain and then start working with the solution now try this one out under root of minus 2 minus x is equal to 5th root of 5th root of x minus 7 now again do not think like minus 2 minus x will always be negative it depends on x it's a big question is it really Ashutosh what do others think again Shilam says no solution alright so guys this is subjected to a odd root so I don't care about the sign of this but I'm definitely worried about this term being positive that means x should be less than minus of 2 okay so it should be less than equal to minus of 2 now when x is less than minus of 2 we know that x minus 7 will be less than equal to minus 9 correct so 5th root of this would be a negative quantity so this would be a negative quantity for sure correct now right hand side is a positive quantity left hand side is a negative quantity is it possible is it possible even when you say both as 0 you realize 1 becomes 0 at minus 2 other becomes 0 at 7 so again my dear friends there will be no solution for this equation right so you must be wondering that you know will there ever be a solution for such kind of equations so now we'll start solving some questions where you will see some solutions happening okay so let us begin with this question I'll again start with a very simple one root of x is equal to x minus 2 guys before we start solving this just a word of caution here whenever we get terms which are having square square roots and all of course we are I know more inclined towards squaring it and trying to solve it there's nothing wrong in doing that but remember when you square an equation on both the sides you introduce extra roots into the system which we call as the extraneous roots and make sure when you've got extraneous roots you check all the roots with the original equation given to you to reject the roots which are extra okay so just like see I'm getting a solution 1 comma 4 but you see one will not satisfy this so one is an extra root are you getting that on car so when you said 1 comma 4 just try to see whether one satisfies your equation or not it will not satisfy it because root of 1 is equal to 1 minus 2 is a fallacy right so it's okay to square it and when you square it you get x is equal to x minus 2 the whole square which means x square minus 4x plus 4 is equal to x which means x square minus 5x plus 4 is equal to 0 which is factorizable as x minus 1 x minus 4 equal to 0 so x could be 1 or x equal to 4 but this one if you see under root of 1 is not equal to 1 minus 2 so this root will be rejected only this root will be accepted so x equal to 4 is the only solution so be careful about squaring it is that fine let's take another example solve solve 3 times x plus 3 to the power half minus x minus 2 to the power half is equal to 7 yes anyone okay Ashutosh says no solution and he thinks he has made a mistake okay so Kavya says okay six okay six plus six only okay great so let us discuss this so first of all we'll write this as 3 times x plus 3 to the power half is equal to 7 plus x minus 2 to the power half again in order to solve it I'm going to square both the sides so square both the sides so it becomes 9 times x plus 3 is equal to let me simplify this side x plus 2 plus 49 plus 14 under root of x minus 2 okay so simplifying this will give me 8x and we have minus 47 27 which means minus 20 is equal to 14x minus 2 drop the factor of 2 from everywhere it becomes this now again square again square both the sides so squaring both the sides will give you 16x square plus 100 minus 80x this will give you 49x minus 98 so simplifying this as a simple quadratic in x you get minus 129x and plus 198 equal to 0 okay so you can write this as x by 16 plus 198 by 16 equal to 0 so I think you can break 198 by 16 as 6 into 33 by 16 and they will also add up to give you 129 so this can be factorized as x minus 6 into x minus 33 by 16 equal to 0 so we have two possibilities here x could be 6 or x could be 33 by 16 okay now when you check with 33 by 16 when you check with 6 you get 3 times 6 plus 3 minus under root of 6 minus 2 which we get as 3 to 3 minus 2 which is 7 so that's correct so 6 is a possible root so this is correct now let's check with under root 33 by 16 so 3 plus under root of 33 by 16 minus under root of 33 by 16 minus 2 that is going to give you 48 plus this which is going to be 81 by 16 minus under root 1 by 16 that's going to be 3 times 9 by 4 minus 1 by 4 so that's going to give you 26 by 4 which is not equal to 7 so this root will be rejected this root will be rejected so yes the only possible solution is so your answer is going to be x can only be 6 x can only be 6 absolutely correct okay so moving on to the next question solve this 6 minus 4x minus x square is equal to x plus 4 okay so Nikhil says minus of 1 Bharat says 1 oh minus 1 I'm sorry yeah so guys whenever you're trying to solve such questions always ensure that you're taken care of this let's say x plus 4 should always be greater than equal to 0 okay so this is equivalent to saying that this should be greater than 0 and this is equal to x plus 4 whole square so solving this is equivalent to saying this equivalent to solving these two conditions right so now this condition says x should be greater than equal to minus 4 and from here I get 6 minus 4x minus x square is equal to x square plus 8x plus 16 which gives you 2x square plus 12x plus 12 is equal to 0 sorry plus 10 equal to 0 which is x square plus 5x plus 6x plus 5 is equal to 0 which is factorizable as x plus 1 x plus 5 equal to 0 so one value is minus 1 and another value is minus 5 now minus 5 cannot be the answer because this restriction has to be met that is x should be greater than equal to minus 4 right minus 5 is lesser than minus 4 so this is an extra root this is something which we call as the extraneous roots okay so the only possibility is your x could be equal to minus 1 is that fine guys okay I'll give another opportunity those who could not get it right so 2x minus 1 to the power of 1 3 1 3rd plus x minus 1 to the power of 1 3rd is equal to 1 okay so people are getting only one as the answer so we can quickly check this so if we cube both the sides if we cube both the sides you get 2x minus 1 x minus 1 plus 3 times cube root of 2x minus 1 x minus 1 times the sum of this 2x minus 1 to the power 1 3rd x minus 1 to the power 1 3rd equal to 1 cube remember I have used the formula a plus b whole cube is a cube plus b cube plus 3 a b a plus b okay right so this becomes 3x minus 2 plus 3 times cube root of 2x minus 1 x minus 1 and this will actually become 1 this term over here is actually become 1 okay so I'm not writing it so equal to 1 so 3x minus 3 or you can say 3 minus 3x could be written as 3 times this so you cube both the sides so if you cube both the sides you get 27 times 2x minus 1 x minus 1 okay you can drop the factor of 27 also from both the places so you'll get something like this okay now let us take some terms common from both the sides for example I can always say x could be equal to 1 that's a solution and cancel out x minus 1 please note that you can cancel but you should always register this as one of your roots so I can say 1 minus x whole square is 2x minus 1 so I can say x square minus 2x plus 1 is 2x minus 1 so x square minus 4x plus 2 equal to 0 now here you would realize that b square minus 4ac that is 16 I hope I have not made any mistake oh there's a minus sign over it I'm sorry so there's a minus sign left off yeah so minus plus 1 then in that case you don't have to do all this thing so you'll have 2x 2x cutoffs so x square is equal to 0 so this will give you x as 0 right now but when you put x is 0 this is going to give you minus 1 and this is also going to give you minus 1 so minus 1 plus minus 1 cannot be 1 so this is rejected because this is an extraneous roots this is an extraneous root is that fine so only one possibility is there your x is equal to 1 so most of you have answered this question correctly next question a lot more is left astutosh we haven't yet started solving exponential equations logarithmic equations and then there are in equations also involved so I think we'll require more classes yes we will need another four class I'll let you know when we can have that so Sukir says no solution Barat also says no solution okay now if you want to solve such questions you can see some similar looking terms for example if I start calling this term as let's say u square let's say u or let's say under root of this as u so you can say 2x square plus 5x minus 2 is equal to u square by the way all of them will have roots so there's no point checking the domain so this function will be valid for x belonging to r okay now so if you're calling this as u square you must call this as u square minus 7 now what I'll do is I'll again call this term as let's say v so this term is actually v square so v square is equal to u square minus 7 okay that's our first equation let me call this as first equation and I get u plus v is equal to 1 okay so from first equation I can say u square minus v square is 7 and from the second equation I have u plus v is equal to 1 so if you divide it you get u minus v is equal to 7 so u minus v is 7 u plus v is 1 so if you add them you get 2 u is equal to 8 so u is equal to 4 and if you say u is equal to 4 v is going to become minus of 3 which is not possible because this expression cannot become minus of 3 so here the answer is no solution see why I'm saying the domain of the function can be any real number the domain can be any real number because b square minus 4 ac this is always greater than equal to 0 this will also be greater than equal to 0 that's what I'm trying to say which is not necessary but you should always check the domain all right so guys we'll now spend some time talking about irrational inequalities so before we wrap up the chapter of before we wrap up the concept of irrational equations we should also look at irrational in equations okay so there are few things that we shouldn't all you know keep in our mind if you have so I'll be writing various types okay so let's say type 1 if you get an in equation of this nature f of x raised to power even root is less than g of x raised to the power of even root remember remember solving this inequality is equivalent to this is equivalent to solving f of x is greater than equal to 0 and g of x is greater than f of x okay so if you solve these two simultaneously then it is as good as saying you're solving this inequality what if x is minus of 1 let me check that so oh this is oh yeah yeah that's correct that's correct it's not cover all real numbers because your discriminant is I'm sorry your discriminant is not coming negative so I was under the impression your discriminant is coming negative yes your domain will not be all real numbers your correct Sukirt okay but again this is not important for us because we'll be only checking the domain once we have got a solution okay all right so moving on to this condition let me just give you a question on this let's say solve the inequality solve the inequality under root of x plus 14 is less than x plus 2 I'm sorry I think I have forgotten no it's correct yeah or inclusive minus 14 to minus 5 okay Sukirt just a question to you if it is let's say minus 10 can a positive quantity be lesser than so let's say you take x as minus 10 according to your answer let's say x is minus 10 so can under root of 4 be less than minus 10 plus 2 is it possible so Bharat and Sukirt please note your answers are not correct now I am sure I'm sure you must have thought that this question is actually based on this condition right actually not this is not based on this condition so guys in such kind of questions I'll discuss a separate theory this is actually type two type of problem where you have and even through top f of x is less than g of x where let's say n belongs to a natural number in this kind of problem the system is equivalent to okay these three inequalities first of all your f of x should be greater than equal to 0 correct right secondly your g of x should also be greater than 0 right and thirdly your f of x should be less than g of x raised to the power of 2n so all these things you have to solve simultaneously right so don't get confused this is not based on this form so I think most of you ignored this condition g of x is positive so when you say f of x is positive you get x plus 14 greater than equal to 0 that means x should be greater than equal to minus 14 when you say g of x is 0 that means this should be greater than 0 that means x should be greater than minus 2 right and once you've got that you square both the sides that means you take the power of 2 on both the sides so x plus 4 is less than x plus 2 the whole square which is saying x plus 4 is less than x square plus 4x plus 4 yeah 4 4 gets cancelled so x square plus 3x sorry it's 14 not 4 I'm sorry so this will not get cancelled so x plus 14 is less than x square plus 4x plus 4 so x square plus 3x minus 10 is greater than 0 so I think it is factorizable as x plus 5 x minus 2 greater than 0 so x could be greater than 2 or x could be less than minus 5 okay so now these three conditions have to be overlapped so we have to see the overlapping of these three conditions so let us make a number line so you'll have minus 14 minus 2 in fact you have a minus 5 over here and 2 so this says x should be greater than equal to 14 this says x should be greater than equal to minus 2 this says x should be less than 2 this says x should be greater than 5 greater than 2 and x greater less than minus 5 okay so overlap occurs in this zone this is the overlapping zone so your x should belong to 2 to infinity so this is going to be your final answer okay guys there's much more to cover but before I wrap up in condition number one I wanted to take two situations a and b b is where you are given something like an odd power of f of x is less than an odd power of same odd power of g of x now guys here it is very simple it is as good as saying or this is equivalent to saying f of x is less than g of x so here you don't test for f of x being positive okay because of an odd power involved we don't test whether f of x is positive so solving this is equivalent to solving this okay so guys there are much more concepts to be covered I think we would require one more class okay so I'll update you on the timings of another class so as of now we're in out from my side thank you so much for coming online bye bye take care