11 th maths , set theory

[00:00:00]No, don't do it. You do it. Done. Will be done. It happens.

[00:00:04]Then it will not be for more or less.

[00:00:08]No problem.

[00:00:12]Good afternoon students.

[00:00:15]Today, under 11th class, under 11 Maths, we have to do the next portion of the lesson of aggregate which we had studied yesterday.

[00:00:38]Today in this session of online class, let us discuss the second part of the set, its second relation and function, its second part was relation and function.

[00:01:04]In this context, as students, you understood in the context of set, it is a collection of well-defined objects.

[00:01:14]Looked at the type. I saw its construction form.

[00:01:18]In what form do you display it?

[00:01:20]We have also looked into that matter. Now let us start today. Let us look at its second portion in the context of the second portion.

[00:01:30]That is your connection and relationship. and function. So now let's see we read the most function from here. What is a function?

[00:01:52]So now let us define this function. The definition of function was the second part of the set. So we have defined the function in this way. What if, let's say, let's write down here how are a and b? are non-empty sets. Ok?

[00:02:18]It is done A and B are non-empty sets. A and B become non-empty sets.

[00:02:27]Now after this R will be a relation. R relationship is defined by what? From A to B. The relation from A to B will be defined as: What will be the factorial product of A and B? will be a subset. So A cross B is like this, this is the Cartesian product, right? R which will be the relation from A to B will be defined in such a way that R is a subset of A across B means that it is in the form of Cartesian product. Let us look at the definition once again that how can a and b be sets? are non-empty sets. So r is defined from a to b if a is a subset of b.

[00:03:20]What happens? It should become r. So this becomes your function. And we've represented this as f true that a cross b. Or we can say this as a function from A to B and this f is what it is doing to this?

[00:03:40]is defining. So this function is defined as Now if you want to write it, you can write it like this. r = x y, true, that is written from here, x belongs to a, y belongs to b, write it like this and here if you want, write x y belongs to r, this subset, sorry, how is it done in the form of a function, its definition is done, x and y can also be written from here, do it from here like this, this x r y, so this is what your function is, now after this, let us go to this function, I will tell you through a diagram how to take you through it, this is its definition, now after the definition, let us understand it through a diagram.

[00:04:39]Now what did we take to understand through the picture? Suppose a set is taken in this way. Its elements and taking another set which one? B Take its elements 1 2 3 4 taken like this. Now after this, let's say we take A, B, C and D from here.

[00:05:02]This function is defined by F as follows. From one to A, from two to B, from three to C and from four to D. So what will this one be?

[00:05:14]You are seeing that the elements of B which were being formed, this is called as image, this has become reflection and this has become pre-image, so in what form is the image of A B C D being obtained here, what is happening in the form of one, two, three and four, it is being obtained, so this will be yours from one to a, two to b, in this way one to b is, it has come with 3 C and four, D, so what has happened to you, it has become in the form of relation, now on this we define what is called domain, define domain or this is called province in Hindi. Ok? So what will the province be? We discuss in this context.

[00:06:07]How do you suppose the domains were two non-empty sets? There were A and B. Now these were being defined. From where? As A to B.

[00:06:14]So what will the first element of set A be called? It will be called a province. So this one that remains, should we see it again? What were A and B? empty sets are gone. A is defined after the empty set by whom? By R.

[00:06:30]In which was the function? from set A to B. So what will the first elements of the first set A be called? It will be called domain or we will call it province. Understand carefully. The first relation, the first set, its elements will be called domain. This became its province. Ok? If you want to write it, you can write it like this. r = From here we wrote x and then wrote x.

[00:07:05]After this, write it as x belongs to a, sometimes it is also written as x or you can write it like this also.

[00:07:14]We have written x here x y belongs to r, this is your domain, what do we do with it, we define it, so now look at the domain here, which ones were there in it first, you will see the domain or the provinces, which ones are there, there is one, two, three and four, like this, okay, so this is your domain and province. Now just after this should we take it here? We read the range. The second thing would be range or let's call it campus in Hindi. This will be the campus. Now we will pay attention to the elements of B here, the components of B, what were they under some rule or instruction in some form or the other? The image was happening. Whose image was being captured? In which are the elements of A being imaged?

[00:08:13]on campus. So, that is, the elements of the second set of B, where were the images of its elements? It is going on at A.

[00:08:23]Look, relations are developing with the members of A.

[00:08:26]So this one will remain and this complex will be formed in this way.

[00:08:32]How do you write the premises? r is equal to as x written in it. Write here y x y belongs to r also. If you want, you can write it this way also. y y belongs to b and then x y is like this you have done it. You could have written it like x here. What could we do with this as x x belongs to a xy r? Let's define it.

[00:09:00]Now after that this became the province. This is the campus. If we understand in simple words, what will we say? If there is a relation between sets A and B, then the elements of the first set are called vital. What will be the elements of the second set which have relation with the first set? There is an image. There is a reflection.

[00:09:23]What will that be called? premises. Now what about this that happened after that? this here this walk walk like a b c d. This is its premises. Now after this comes the co-province. This is called co-domain or code domain. So what will be the co-province? This portion which will become co-province, whose image was there, whose image was coming, will become the range. And now let's take an element here, what are elements? Let there be one such element, say E. Let us assume that this is F. Now look, are there any images of E and F from here? Not there. Are E and F the reflections of the first element of the set A? Not there.

[00:10:11]And when the sets A and B are not reflections, this will become your what? Co- province. Are you getting the point? So what will you write under co-province?

[00:10:22]Tell me. Will the elements in this set come under the co-province? A will also come, B will also come, C will come, D will come, E will come, F will come. So who will be there in the campus in the same way? So can you tell me what all will come to the campus?

[00:10:36]A will come, B, C and D. So, what did we understand? The complex will be whatever the image of the first element is. And what will happen to those which remain as a whole, that is, each element of the second set? It becomes its co- province. So in this way, this province co-province and to understand it a little bit, we are using this word for the first time because in classes up to tenth, there are not many detailed discussions in the context of province complex and co-domain.

[00:11:07]So you should take it as a quick revision and yes, take it as a quick revision, what did it mean, just to say in clear words, in simple words, two sets A and B, each element of A will be the domain, each indeclinable of B will be the complex but it should be its image and all the sets of B which will be the second set will be called Sahap Prabandha, now on the basis of this we take up a problem related to a domain complex. Let's go with the question. Then your concept will become more clear on this.

[00:11:54]Suppose we are carrying it like this. Suppose, suppose, what are you assuming? Let's take a equal to like this.

[00:12:03]1 2 3 4 5 and 6 That's your elements of a. Then what is in it? R is defined. How is it defined? x y and then here I am giving y which means the image that will keep happening is defined such that define a relation from A to A by x + 1.

[00:12:46]It is telling you to define a relation and in this it is also saying that we have to explain it by drawing its diagram and after drawing the diagram, you will be asked what will be its second domain, what will be its complex and what will be its co-domain?

[00:13:06]We discuss in this context. Let's see, this question is being given in this manner.

[00:13:12]Now how will we understand this question? Let's go. First look at the question. How many elements does R have? 1 2 3 4 5 6. Now the elements of R are one two three 4 5 6.

[00:13:25]So first of all, if we were to multiply it by Cartesian factorial, we would do Cartesian multiplication. How would it have happened? One is with one, thus one is with one, one is with two, one is with three, one is with four, one is with five.

[00:13:40]Then one would be converted into six. Isn't it?

[00:13:42]Look, count the elements like 1 2 3 4 5 6.

[00:13:46]Square of 1 2 3 4 5 6 because it is done like this.

[00:13:50]It remains as 36 from here onwards and then it remains as two. As three as four. So in this way, if we take out the entire a cross n here, it would be 36. Now in which does this happen? Done as r.

[00:14:04]This will be a subset of how it will happen.

[00:14:08]Delete it from here. r Sorry r it became A cross A as subset. Now it is 36 elements.

[00:14:16]We don't have to go into that.

[00:14:19]What is our R defined by here? You look at this. y is such that x is any element.

[00:14:26]If the first element of the first set is this, then if the element of a is such that it is 1 2 3 4 5 6, then what is the form of the second one which will be y?

[00:14:36]Like we also write in this form.

[00:14:38]f(x) = y Now look here y is defined as a function like this. Now we will keep it here. Now keep it going.

[00:14:48]We kept the value of x as one. If the value of x is kept as one, what will be the value of y? It's two. Now keep doing it.

[00:14:53]What value did you set for x? Two kept. If we keep two to two then what is happening corresponding to y? 2 + 1 = 3. Keep doing this. Everyone is continuously increasing this.

[00:15:11]This is three. Then after this we took x equal to two and went to three. The value of y becomes four. After this we took the value of x as four.

[00:15:20]So what is the value of y? According to this relation x + 1, then after this we take x = 5 and y = 6 and x = 6. y equal to is not here. Let it be.

[00:15:34]So what happened to it? The relationship continued to grow. Now after this relationship so many things are happening. Now what do we have to tell him? Its picture is to be shown in the form of a diagram which is called arrow diagram. Let us tell him. Let's go and see that.

[00:15:52]In the form of an arrow diagram. So now what did we get here? Took a set.

[00:15:57]Take the second set. So you take another set like this B this is A and A is A itself and what is it defined as fx here it is defined as f and what were its elements let's take a look at it let's take a look at it 1 2 3 4 5 6 so 1 2 3 4 5 6 now let's go next let's go to this what is happening?

[00:16:49]That step seems to have been done. Yes, okay.

[00:17:01]Drive in reverse. Moved ahead.

[00:17:06]Yes. This was it. After this yes.

[00:17:10]Let's write it here. a equal to we had elements 1 2 3 4 5 6.

[00:17:16]And how was r given? Ah sorry, here y was given as x + 1.

[00:17:23]So let's write down its elements. From here 1 then two then three then 4 then five and this is written six. Now after this, relations on this continue to develop. Now look here, we have written it here also.

[00:17:37]Let's call it 1, this, two, this, three, this, 4, this, 5, and this, six. Isn't it? So we wrote down 1 2 3 4 5 6. How was y equal to? was x + 1. That means we were keeping the first elements of the value of x in place of x. So what was the value of y corresponding to that? It kept on being received. So on this basis, if we look from here, then with whom is One now related? Two to two. Who did you get into? From three. Three ka to whom? From four.

[00:18:10]Four ka to whom? From five. Five of whom? From six. And this way this six is ​​being saved here. This is being missed. So this is the six that is missing. Now let it remain here like this and then what do we define on it now? So let's take this forward now.

[00:18:31]So here we will see which will be the first component. The first component is 1 2 3 4 5 so 1 2 3 4 5 and along with this this is like this 1 2 3 4 5 and from two this was five here and this is six yes the first component so this will be what will be its domain so what will be the components of the domain 1 2 3 4 5 in this way this will be your domain then after this this along with this what has happened its domain has become x belongs to r you can write this in the same way its range will come so what will be the range the range which will be the images that are coming with it, what will be its range it will keep becoming so see which range is left out here if one is left out then its range will be two then three then four then five and six. Many children ask, Sir, how did this get missed?

[00:19:41]What was the definition of Parivartanasar? That the element of the second set of b must have an image of the element of the first set. But you see the relation y = x + 1 was causing this? Was getting released. That means he will not come. Now coming to the co-province, what did we tell through the co-province definition?

[00:20:02]Co-province: He had told that all its components, then what will the components of its other set be called as a whole? will be called co-provinces. So what will happen to the forest here now? It will be included.

[00:20:15]1 2 3 4 5 and so on 4 5 and that's six. Will this happen? Sahap Province.

[00:20:23]So this was a notation. This question can also be asked in another way. So we understood how to make it. Let's take a half of this and a half of this problem.

[00:20:36]Then it will become even more clear to you. Let's take that problem and another question from here.

[00:20:42]Now it was in the form y = x + 1. Let us take something else with us. Let's see this. Let's take another question.

[00:21:05]Let's say that or sorry proven yes yes.

[00:21:10]Suppose they are carrying the second question.

[00:21:12]Same is the case with beans. Let us understand this a little better. Let's take it one more time. Let a = 1 2 3 4 5 6 7 8 9 10 11 12 13 and 14 are given elements. One to 14 elements became A's. R is being defined.

[00:21:40]How is he telling a relation?

[00:21:45]How is X Y between x and y? So how is the difference between x and y defined?

[00:21:51]How is it defined as xy? 3y - x = 0 where these conditions are being given.

[00:22:03]x and y are the elements. Belongs to will come a.

[00:22:09]Ok? By A to B, right? It is from A to B, from A to A, sorry, we will do it here only.

[00:22:35]Write a relation R from A through A to A. Write a relation R.

[00:22:44]Write what it is. Also find its domain and range. Its province complex [ __ ] province also has to be known. Let's know so many things about it. Now this has to be found out. So how do we take it? First of all, understand it carefully. Elements of A Elements of A From where to where? 1 2 3 4 up to 14 like this. Components of what? A K.

[00:23:25]What defines a relationship? The relationship is defined. This equation becomes r = 3y - x = 0.

[00:23:36]You should implement this. How can I write this? You wrote 3y = x. Now 3y = x.

[00:23:44]When 3y = x is happening, then you can see here that I am writing it like this from here. And make this also. This first one from a to a. Now how many elements did a have? Write it down. 1 2 3 4 5 6 7 8 9 10 11 12 13 and 14 So far.

[00:24:15]Ok? Now these become the first elements of A.

[00:24:20]Now after the first element is there, look here R to R R which was A to M. So take it up on this as well. 1 2 3 4 5 6 7 8 done here 9 [nasal sound] Nine do it here. 9 10 11 12 13 & 14 What is the definition being given? Given that x, let's go over here and put x equal to what? If we put x equals one then yes x equals sorry you do one thing with this.

[00:24:58]3y = x continued to happen. So this is going to change with respect to x. Yes, okay.

[00:25:05]What is the value of x? It became a forest. So the value of x will become one. So what will y become? There will be three. Since 3y is here, you put x.

[00:25:15]What was the 3y relation given? Check this out. The relation had been given. 3x - y is correct? It was a bit of the opposite. Sorry, that's why I was asking how things were going. There is a typo here. This relation make it a. What was the question? The question was its 3x - y take it like this. 3x - y = 0 OK.

[00:25:46]Delete this. Yes. Now how was the relationship here? The relation was defined as 3x - y = 0 so let's transpose it. 3x is equal to how much? y is done. Now when 3x = y is happening then how much should we put here when 3x = y happens? What is the value of x? Forest kept. So what is y if the value of x is one? Look, 3 is becoming 1 three.

[00:26:18]Keep the value of x as 2. How much is y? It's six. The value of x becomes three.

[00:26:24]What is the value of y? It's nine. Then keep increasing it. What is the value of x? Four.

[00:26:29]So what is your value of y going to be? If you place it from here, it will become four. That means it became 12. Now did you put x?

[00:26:39]Kept five. So what will be the value of y?

[00:26:42]3 * 5 instead of x becomes 15. Stop here now.

[00:26:45]Now stop after making so many connections because now it was defined only till 14. There is no after this. So you see, if you were carrying one, then three came along with each one.

[00:26:59]These are its three diagrams. If x is set to two, what is the value of y? This is six. If you keep three, how much will it be? If you add three it becomes nine. Now after this, if we placed the four then we had to take it a little here.

[00:27:15]Let's improve it a little bit. This is how it became five. After this, here six 7 8 9 10 11 12 yes this 14 13 and 14 yes complete it. Now look, the image of one has become three here. What will happen to the image of two? The image of two became six.

[00:27:53]Do this two's image with six. Then what will you do with the image of three?

[00:28:05]Let's take a slightly larger diagram again. It was getting smaller.

[00:28:11]You will not have any problem in understanding it all at once.

[00:28:14]Now let's do this and make it. A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 That's it A.

[00:28:30]Make another one too. Now after creating A, on this also 1 2 3 4 5 6 7 8 9 10 11 12 13 and 14, now see that the relation that was obtained with x became one, its image became three. The image of two became six.

[00:28:53]Who got the image of three? On the nine.

[00:28:56]On whom will the image of four be formed? At 12. And if I had kept five then how much would five be? 15 is not here, so you leave it, this is what its picture became, it became a diagram, this was the picture in the form of a diagram, like if I had said arrow diagram, then it is obtained in the form of arrow diagram like this, so now let's go, the first part of this was done, what was its domain, what was the definition of domain, the definition of domain was that if two sets are given, then the elements of the first, so the elements of the first, now see here, not the entire elements of the first, what is it by some relation? are defined. So who all are there?

[00:29:39]1 2 3 4 So how much is it here? Its province became 1 2 3 4 Now after this came the matter of campus. The matter of the campus came up. So look, this should be the image of the first element. There must be reflection. So the image of one was three.

[00:29:56]Two had six, three had nine and four had 12. So keep writing here. One became three. Then after three it became six.

[00:30:06]After six came nine and after nine came 12. Now what will happen to the remaining co-province? The definition of co-domain is that it is said that all the elements of the second set are complete. So where will this co-province come from? The components from 1 to 14 will become co-provinces. This way it will be your province.

[00:30:37]This will be the province. After this it will become co- province.

[00:30:42]And what happens to the images that are coming? It becomes a complex. 3 6 9 12 So this is the question. Please understand this well. Examples of this are given in many other exercises as well.

[00:31:06]You will learn to make those questions based on those relationships and once you learn, you will remember this in the context of the province and campus. The elements of the first will be the domain whose relation is defined by any set with the mean r. The image in the second set will be its attendant and what will be all the elements of B? There are co- provinces. Let's move on to the next one.

[00:31:38][nasal sound] It 's only 15.

[00:32:04]Let's move on to another question after this. The domain and range belong to the same.

[00:32:19]We take that question forward as an important question in our syllabus and it is a very good question. Now let us take this question forward.

[00:32:40]This question is function The question is function fx = 9 - x² defined by defined by real function f Find the domain and range of the real function f.

[00:33:40]This question has been given to us.

[00:33:45]Now try to understand this question.

[00:33:49]Such questions are very important. In many ways, it will also give 16 minus x² in a book.

[00:33:57]25 in a book will give a loan x².

[00:34:04]4 minus x² will also give. But the method of making all the questions will be the same.

[00:34:11]Now the function was defined by 9 - x.

[00:34:15]Read the question once.

[00:34:18]And how is it when defined?

[00:34:21]is a real function. The best word here is real and what is this real word doing in itself?

[00:34:30]is defining. You will know that whenever there is a real function, it will be a real function, so if we take the root in this way, it is in the form of a real function.

[00:34:49]This is a real function like this, if x happens, then always remember it will be real.

[00:34:55]This means what will be its value? It will be equal to or greater than zero. Will you pay attention to what will remain? Zero or greater than zero. Suppose what happens if this happens? If it becomes less than 0, then will this number here become infinite or complex number or which number will it become? will become a complex number. Isn't it brother? So it becomes a complex number. So, will you ever notice how this set of real numbers will always be? It will be in plus, it will be zero or it will be greater than zero. This is the first concept you should learn that how real is this function, it is a real function, meaning the value of this 9 - x², what will it always be like? Will it remain zero or greater than zero? And if it becomes less than zero, then it will become a complex number and when it becomes a complex number, then in this situation it will not be a real function. The issue arises as to what becomes the root of an irrational number. So we don't go into that much detail. So first of all this is a real number. It is a real number, meaning what will be its value?

[00:36:16]Tell me whether it will be 0 or greater than 0. Now look here this function fx is defined. Let's solve this. Let's make a solution from here. The solution is defined by fx = 9 - x² and what is this thing? It is real. is a real function.

[00:36:40]Now the actual function is done. Meaning its value which will be 9 - x² is always >= 0, always equal than or greater will always be equal to or greater than, can be written as we have written it like this, 3² minus from here, x² has been written like this. Now you must have read this as square equation. If a and b are in the form of two quantities a and b. You've written it as 3 + x.

[00:37:19]From here I wrote 3 - x. Is >= 0 Now look at this, this is 3 + x written as 3 - x written.

[00:37:28]I kind of plotted a line here.

[00:37:31]After plotting it, let's say it is 0 and here we took it +3, let's take it here, what? -3 And there will be a lot of what comes here between -3 to +3? The numbers will come.

[00:37:43]Now a lot of numbers will come.

[00:37:46]Now look here, we have not studied yet.

[00:37:49]But I want to tell you that when you study linear inequalities under linear inequalities, then you keep putting the values ​​here and see whether this equation is getting satisfied or not. Now how much smaller value did you put here than -3? If you put -4 then you will make 3 -4 and here three and then minus minus will become plus.

[00:38:13]Look, it became -1 and this was how much, from here 3 + 4 became 7, it came to -7. If you put this here -7, what is happening? Look, he is saying -7. Greater than or equal to 0. Is this even possible? Not there. It will not be possible at all. Now if you keep the second value bigger instead of smaller.

[00:38:37]If we had put four here, you would not have to write x = 4. I am telling this to explain. If we put x = 4 then what is 12 here now? 4 and + 3 is going to be 7.

[00:38:49]And after this, if we had done three minus this here, then this -1 becomes 12 and it becomes -1 again. The minus amount arrived. Look, keep the value bigger than this. How to keep it big?

[00:39:01]We did three to four. I had done four here.

[00:39:03]Here it was changed to -4. So if you keep its value as -4 then it will come to minus. If four is kept greater than three then what is this also happening? Again it is going minus. And when this minus is happening, that means what does our function here say? Greater than or equal to 0 means it must be a real number. What is happening? It is not happening. From where? We will study linear inequalities further. It will be on that. I would like to discuss it in a little detail, otherwise we will see later. But I am telling you how to explain this.

[00:39:36]So this line was -3 after 0 and three, if you keep even a very small number or a big number and on keeping its value, it again becomes negative. Meaning this relationship cannot be defined. I am not satisfied. And when it is not satisfied, that means this x here will be between close interval three and -3.

[00:39:58]Now this close interval is happening between -3 and three here. So pay attention to x belongs to.

[00:40:06]x Belongs to Ya.

[00:40:09]Now after that the question arose again, Sir, how did you take it? So come on brother, if you place this here between these three. Keep it between -3 to 0.

[00:40:17]Keep it between 0 to 3.

[00:40:20]We put 0 in place of x. x = here let's make a little bit right here. Suppose we had put x = 0, what would have happened if we had put 0? Three and then how much would it be from here? 3 How much is it? It's nine. What do you do here x equal to? Keep the forest. If you include one also then 3 + 1 will become 4. Four and one goes from here so 2 becomes 4 and 2 becomes 8. I mean, what is this thing coming? Positive amount is coming. Between whom?

[00:40:47]From here, whatever amount you are keeping between -3 to 3, whatever amount you are keeping between these, you are getting positive. I mean, what was this equation going on? I was getting satisfied.

[00:40:57]Satisfied Being satisfied took close interval -3 and three.

[00:41:02]Now one thing came up here. What did this turn out to be? Domain or what did it come to? Yes, the province has arrived.

[00:41:11]Now what do we come up with after this? Let's find out its range, find out its range. Now if we discuss it in its context then it becomes province -3 and three. Now after this delete it from here. Yes. Now let us find its range.

[00:41:31]Range. This is a very good question. It is very important. Children, did everyone understand till here? Let me repeat it again. What was it? f(x) = 9 - x². How was it? It was real. It was real, meaning what would its value be? It will either be zero or greater than zero.

[00:41:47]So we took this taking zero or greater than zero. 9 is squared to 3. x is converted to x² two times a² - b² in the formula. 3 + x is done. 3 - x is done. Now is greater > equal to 0 we plotted the line between -3 and 3.

[00:42:06]Put an amount smaller than -3. Put an amount greater than three.

[00:42:09]So what do we get when we put the value of this in the linear infinities under these conditions? This comes out negative. Whereas it should be zero or greater than zero. So this equation is not satisfied. And what did we do when we were not satisfied?

[00:42:26]Put any number between -3 and 0 and between 0 and 3 then this positive value comes on putting the number. Positive value came.

[00:42:30]That means it became 0 or greater than 0.

[00:42:33]So, how much of this close interval do we get? -3 and three are being received. Now just after this we find its range. So to find the range, look, this range is taken as y = fx. Isn't it? So here I have written y as 9 - x². I would like to explain one thing to the students. Like we took x equal to how much? I took 16 and left. Ok? Now if we take x = 16 and go under the root, what will be its value here? x = only and only 4 will come.

[00:43:13]What do many students do? Let's write this here as plus minus 4. Do not write plus minus 4 at all.

[00:43:18]Why won't you write? Because what kind of equation is this? It is a linear equation. It is a linear equation. So due to the linear equation it will have one value only and only. Now here if we write this x² - 16 = 0 then x² = 16 or x = here it will be plus minus your 4, please pay attention.

[00:43:40]Why? Because this is a quadratic equation. is a quadratic equation. So x² - 16 is put then this x sorry it has equal sign.

[00:43:49]This equals sign will become x² = 16. Let's delete it. x x² = 16 then x = + -4 what is the basic difference between the two? Here, when there is an under route, you are seeing only one and only one value here. What is its value?

[00:44:14]Two values ​​are being obtained. Why is this happening?

[00:44:16]Because why did this quadratic equation become single value? Linear equations. Now look here it's coming out as y = 9 - x². So what should we do with it now? Now look here. Square it from both sides. So did you understand this?

[00:44:34]Look here, its y equal to is given in this form. So if we square this, then after squaring it, we wrote how much? Let's write y² = 9 - x². Now what do we do with it? We transpose it.

[00:44:48]What will you get if you transpose? Let's go now. x² equals two.

[00:44:55]What do you get if you move this around? Let's erase this a little bit from here.

[00:45:01]y² plus x² = 9, so what did we write x² equal to? 9 - y².

[00:45:11]Now you write x² = 9 - y². So look from here, what have you done? x = 9 - y². Now it is written in the form 9 - y².

[00:45:25]So what do we do now in these conditions? Greater than or equal to then make it 0. So how much did this cost? 9 - y² > equal to 0 Same procedure we were doing.

[00:45:39]When nine is instead of nine then as it is. x is replaced by y. Now if we do this also from here, how much would it bring? 9 will be written as 3² minus, write it as y² and after this what will be obtained here, it will become one 3 + y, okay and it will become one, from here how much 3 - y > = 0 again y belongs to here, it comes in this form, in which form it is obtained in the form of -3 and 3, in the form of -3 and 3, okay now when it is obtained in the form of -3 and 3, then a little contradiction is coming here. What kind of contradiction is coming? Look, what was its interval?

[00:46:26]-3 and 3 did we go with this here? We left with y being >= 0. fx = y or it was taken like this fx = y. Is > = 0 Now what will this be -3 and 3? Will not done.

[00:46:45]Why not? Because what did we do with Assumption?

[00:46:49]What should be the value of y? It should be zero.

[00:46:52]So now look, from here if we want to understand this on the basis of this line then see this -3 has become one zero. When? When the value of y here was zero. Look at how many things are available for this y. We took the value of y to be 0 or greater than 0. If you solve it here, how much will it be? -3 and three.

[00:47:10]So here we have -3 and 0, one more one will come, now from here, this one from here to three, so find the intersection of both, so what did we get, the first one was from -3 to 0 and just after this we took it here, let us take it a little bit like this -3, this 0 and 3, the value of y should be 0 or greater than 0, this became one value and the second value from here to where? -3 to three. So when you find its intersection or this is what is called its common, what will be its common? 0 and three. What will happen to it? It will be a campus.

[00:47:56][Applause] Do you understand? Look again once. What was it? The range y = fx happened. We took y - x².

[00:48:03]This was Given. How to carry it?

[00:48:06]Because the value of fx = y is kept here, we defined the value of fx as 9 - x², took the square root and got this, now let us proceed with this, its value should be 0 or greater than 0, -3 and 3 came, but we have assumed that its value cannot be 0, so it cannot be 0, so we plotted a line between -3 and 0 and 0 and 3, so see on this line one is becoming minus -3 to 0 and one from -3 to 3, so we will take its intersection, so in this way what we are seeing there is 03, this zero zero 3 which we are getting, what will happen to it, it will become its range, so write it like this, y y belongs to 03, what will happen to it, it will become its range, so we will understand this question a little better, just remember this concept and after this concept, also remember this one concept that the value is satisfied due to the presence of linear equal inequalities. It does n't happen. Now let's take up another small problem. Let's move on to the next problem. Let's move on to the next problem. Do it quickly. It's almost time. Yes.

[00:49:39]Now let's move on to a small question related to the algebra of real functions. Isn't it? Algebra of real functions So under the algebra of real functions there was a problem, suppose the question was, suppose [sound of clearing throat] fx = x and g(x) is defined by g(x) = x.

[00:50:46]If f + g x f - g x f gx is defined then what does f gx have to do with it? Have to find out. Let's take this question that has been given to us, this last portion of it is coming up here, related to the algebra of real functions. So it has the form fx = x under the algebra of real functions.

[00:51:30]g(x) = So let's write this as can we? f here you have done it like this. Do it like this.

[00:51:36]This is the operation fx plus sign on x plus sign g(x) fx is defined by?

[00:51:46]What is fx defined by?

[00:51:48]Put a plus sign between x. What is g(x) defined by? By x.

[00:51:53]Similarly you can find this also.

[00:51:56]How? f off x done.

[00:52:00]What is the value of g ofx fx minus? x minus f.

[00:52:03]Then come f * g you'll turn this into fx turn g(x) what is it defined as?

[00:52:10]Yes fx, look, you don't have to do anything. Well, it has to be done in the same way as we do normal algebra because see, it is clear from the name itself that it is the algebra of real fruits, so the value of fx is kept as x and the value of x is like this, now after this comes f / gx, so you will write it as fx by gx, what is the value of fx here, the value of fx is x and what is this x, so pay attention, it will go up to x or you can write this, how much 1 x can be written. All you have to do is pay a little attention here that the value of x should not be zero here. Will pay attention. Otherwise it will be undef meaning it will not be defined. Why? Because if it is made 1 / 0 then it will become indefinite. So in this way, in the algebra of real functions, you have many problems in the next classes also, in 12th also by doing fx of g(x), then keep on finding the value of f and g(x) immediately on the basis of the definition which was defined.

[00:53:18]Different people can put different things here. Let's put 2x - 1 there. Let us substitute x² here. Even then you don't have to worry. The sum of two functions in general algebra is simply the difference of two functions. These are the product of two functions and the division of two functions, thus it becomes this. Today's time ends here. Come on students, we will meet again tomorrow. Thank you.