 Yeah, can you see my screen now? Yeah, are you able to see the screen now? Okay, so having got the value of beta from here. Sorry the connection got lost Roughly, I'll put it over here. So r square would be 16 divided by a point two five four Okay, for practical purposes you can take this as one by four Okay, so r square could be taken as 16 divided by one by four, which is 64 Okay, okay, I understood I understood the problem here So basically you don't have to do this because as he rightly pointed out this will be 65 not 64 Okay, so we'll assume from it that makes our life very easy because Then it'll be a case of then we don't have to do all these things then it will be a case of a factorizable quadratic so you can write this as This minus 64 beta minus beta plus 16 is equal to 0 so 4 beta beta minus 16 minus 1 Beta minus 16 is equal to 0 so beta is one fourth. Okay, so it's exactly point two five Okay, so from here. I can get the value of r r will be nothing but under root of 64 which is going to be 8 So let's move on to the next question now Find the values of the following Kinematic ratios Evaluate number one cause for 95 degrees Can you see no? All right, sorry Again the net connection is not proper today. All right, so let's quickly discuss this the first one cause of 495 degrees Right, so 495 lies in which quadrant first of all, let's try to identify the quadrant in which 495 lies So 495 will lie in the second quadrant. Correct Correct in the second quadrant the value or the sign of cost is going to be negative correct next 495 could be written as 3 into 180 degrees minus 45 degrees Correct, so it's value is going to be cause 45. That's going to be negative 1 by root 2 Okay Quickly doing the next one also signed 1230 degrees So 1230 degrees would lie. Let's say I complete one around 360 to round 723 round 1080 1080 plus 90 is thousand so I'm here at 1080 so if I complete 90 more I will be 1 1 7 0 Okay, and then 60 more again So I'm I'm back to the second quadrant where sign is known to be positive Okay So I can write this as 1230 degree as 7 times 180 degrees So 7 times 180 degrees will make it 1260 minus 30 degrees So this will be as good as plus sign 30 degrees which is going to be plus half. Is that fine talking about caught minus 350 degrees is we know for that It's as good as saying minus caught 350 Yeah, can you see now? Okay, let's all the fourth one also fourth one. This is going to be minus tan of 1590 degrees So 1590 degrees lies in which quadrant? Let's figure that out first so 1590 degree means you have covered around three rotate four rotations, okay, and four rotation means you have covered around 1440 degrees. Correct. Okay, and this is 150 more than that. So you're back to the second quadrant Okay, so 150 degree more than that Will take you to the second quadrant in second quadrant again tan is known to be negative. So it's going to be negative tan Now this is 30 degrees. So you have tan 30 degree over here. So that is going to be plus 1 by root 3 Okay, that's going to be plus 1 by root 3 as your answer. Is that fine? next question if tan 35 degrees is equal to a Okay prove that tan 145 degrees Minus tan 125 degrees whole divided by one plus tan 145 degrees Tan 125 degrees is 1 minus a square by 2a. Is that done? All right, so let's discuss this now So we can write tan 145 degrees as tan of 180 degrees minus 35 degrees Correct, and we all know that tan 180 minus theta is minus tan theta. So it is minus 35 degrees Which is the negative Similarly a tan 125 degrees could be written as tan of 90 degrees plus 35 degrees Okay We know tan of 90 plus theta is going to be minus cot theta So it's minus cot 35 degrees, which is nothing but minus 1 by a Okay, so now putting it in into the left hand side of the expression Now substituting this in left hand side of the expression What do we get? minus a Minus of minus 1 by a Divided by 1 plus tan. That's nothing but 1 minus a square upon 2a. That's equal to your right-hand side Okay, so you see the screen now. Okay. All right. So let us move on to the next question now Next question is if cos of alpha plus beta is equal to 4 by 5 and sign Alpha minus beta is 5 by 13 Where alpha and beta Where alpha and beta both lie between 0 to pi by 4 find the value of tan of 2 alpha Find the value of tan of 2 alpha Then anyone please type in the response in the chat box once you're done So how do you get the value of tan 2 alpha given that cos of alpha plus beta and sign of alpha minus beta is given to you So let us understand this we can write tan of 2 alpha as tan of alpha plus beta Plus alpha minus beta correct now treat this as Your a angle Treat this as your b angle so which falls into the formula of tan a plus b which you can write it as tan a Plus tan b by 1 minus tan a tan b 1 minus tan a tan b correct But even to find this we need the value of tan alpha plus beta and tan alpha minus beta correct Now we know that since alpha and beta both line the interval 0 to pi by 4 Alpha plus beta will also lie in the same Alpha plus beta will also belong to the first quadrant Okay This is very important to know because it will decide what will be the sign of the trigonometric ratios that we are going to get Yeah, yes, exactly. That's what we can take Venkat We can take sine alpha plus beta now to be 3 by 5 without any problem Okay, else there would be a doubt whether it would be 3 by 5 or a minus 3 by 5 Now that we know that it lies in the first quadrant We can take such calls very very easily in a similar way I can say cos alpha minus beta could be written as 12 by 13 Okay, and this implies my tan alpha plus beta, which is nothing but sine of alpha plus beta By cos alpha plus beta can be directly written as 3 by 4 Okay, and here I can write sine alpha minus beta by cos alpha minus beta that is going to be Tan alpha minus beta as directly I can write this as 5 by 12 Once I know these values, I can directly substitute over here So please substitute these formulas over here When I do that I get the value as tan alpha plus beta is going to be 3 upon 4 and This is going to be 5 upon 12 by 1 minus 3 upon 4 into 5 upon 12 Okay, which gives you 36 plus 20 which is 56 by 48 minus 15 which is 33 So your answer is going to be 56 upon 33. That's going to be your answer. Is that fine? Let's take the next question if a plus b is equal to pi by 4 Find the value of Find the value of 1 plus tan a Into 1 plus tan b If you're done, please type in the response in the chat box Okay, so let's quickly discuss this as well So I can always write angle b as pi by 4 minus angle a right So let us write this term as 1 plus tan a times 1 plus tan pi by 4 minus a Okay So this becomes 1 plus tan a and This becomes 1 plus. This is 1 minus tan a By 1 plus tan a correct Let me take an LCM over here. So this becomes 1 plus tan a and 1 minus tan a Okay, tan a and tan a gets cancelled in fact 1 plus tan a and 1 plus tan a also gets cancelled leaving you the answer as 2 So this becomes actually 2. Is that fine? Yes, correct Venkat answer is 2. Absolutely Let's take another one Find the value of Find the value of tan x into tan X plus pi by 3 plus tan x plus pi by 3 Into tan x minus pi by 3 plus tan x minus pi by 3 Into tan x Find the value of value of means you'll get a number you get a numeric answer for this particular expression any idea Okay, I'll just give you a hint here Just do this operation tan a tan b plus 1. Yeah. Yeah, take your time All right, take your time. No worries This is a very good problem actually see if you do this operation, which I gave you I give you a hint to do tan a tan b plus 1. Okay When you do that you realize you get sine a sine b by cos a cos b Plus one correct If you take the LCM of cos a cos b You get sine a sine b plus cos a cos b, right? This clearly the formula of cos a minus b by cos a cos b Now, let us try applying this to all these terms. Let's add a one one and One to each of these terms and subtract a three over here Okay, now this fits into this formula of 1 plus tan a tan b. Okay So I can write this as cos a Minus b which is pi by 3 divided by cos x Cos x plus pi by 3 Okay by 3 divided by cos x minus pi by 3 cos x plus pi by 3 Okay, and in a similar fashion this can be written as cos Pi by 3 divided by cos x cos x minus pi by 3 Okay, and you have a minus 3 outside over here, which is this minus 3 So now what I'm going to do is I'm going to club these two terms together Okay, and I'm going to first pull out a cos pi by 3 cos pi by 3 We all know is half so I don't have to write cos pi by 3 every time so 1 by 2 cos x I can pull out Correct, so it'll become 1 by cos x plus pi by 3 Plus 1 by cos x minus pi by 3 Okay This sum I can write it as Negative 2 Okay, now we all know the formula of cos a plus b into cos a minus b. Correct What are the formula? cos a plus b cos a minus b cos square a minus sine square b Or you can say cos square b minus sine square a correct So I can directly apply that over here So I can say this formula can be done as cos square pi by 3 minus sine square x Okay, minus a 3. Let's simplify this. So this becomes 1 by 2 cos x Okay, here I would get cos x plus pi by 3 plus cos x minus pi by 3 divided by cos x plus pi by 3 cos x minus pi by 3 Okay And this term will remain as such So this is going to be cos square pi by 3 minus sine square x Minus a 3. Now let us simplify this This term will become 2 cos x cos pi by 3 Okay, this will remain as such By the way, here there is no point to convert it like this. We can keep it as cos x plus pi by 3 into cos x minus pi by 3 Okay so This gets cancelled cos x cos x gets cancelled and we know that cos of pi by 3 again is a half So it's half of cos x plus pi by 3 into cos x minus pi by 3 and Here also I have 1 by 2 cos x plus pi by 3 cos x minus pi by 3 Okay These two terms will get cancelled off and I will be left with 3 minus 3 as the answer. So your answer is minus of 3 So this becomes your answer Okay Very good question can come in your uts, but it's very lengthy actually. Okay, so please make a note of this We'll move on to the next question Yes, it may be a six marker. You're correct. It can be given as a six marker if tan beta is n sine alpha cos alpha by 1 minus n sine square alpha show that tan of alpha minus beta Is 1 minus n tan alpha. Let me know once you're done. You didn't get it What about others? Okay, let's try it out. Let's start with tan alpha minus beta Tan alpha minus beta you can write it as tan alpha minus tan beta by 1 plus tan alpha tan beta Okay Now let tan alpha be as such and tan beta could be replaced by this value So let this value replace your tan beta So you'll get n sine alpha cos alpha by 1 minus n sine square alpha divided by 1 plus tan alpha times n sine alpha cos alpha by 1 minus n sine square alpha Okay write this tan alpha as sine alpha by cos alpha Sine alpha by cos alpha Okay in fact, we can write This tan alpha also as sine alpha by cos alpha cos alpha and this cos alpha will get cancelled off So let me simplify this so it becomes sine alpha 1 minus n sine square alpha Minus n sine alpha cos square alpha. Okay In the denominator, I will have 1 minus n sine square alpha plus sine You can write this as n sine square alpha. Is that fine? Just check once everything is fine so far This term gets cancelled with this and I can take a sine alpha common Okay, I get 1 minus n sine square alpha minus n cos square alpha Okay, and by the way, we'll have In the denominator 1 cos alpha coming up. So this I missed out So this will be cos alpha Because here we cancel out a cos alpha. So these two terms are gone right So when you take the LCM up in the up, you'll have this whole divided by Let me explain you what I did here this whole divided by cos alpha 1 minus n sine square alpha and Here in the denominator, I'll just have 1 minus n sine square alpha Correct. So what will happen is that this cos alpha will remain Okay, that's what I have written over here This cos alpha will remain. Okay So it'll be cos alpha into what? okay now from these two terms if you take a Negative n common you get sine square alpha plus cos square alpha. Okay, which is nothing but one So this will become tan alpha times 1 minus n Which is nothing but your RHS So this is what we wanted to prove over here. You wanted to prove that it is 1 minus n tan alpha Is that clear? Next question is 3 tan theta minus 15 degree Is equal to tan theta plus 15 degree where theta lies between 0 to 90 degree Find the value of theta find the value of theta 30 degrees. No, that's not the correct answer When are you watching in a lag because I'm already on the next page and I've not written anything on this page Just it is the top part of the page 45 degrees. That's correct. Did you guess it or did you solve it? Manav Was that a guess or did you solve it? Okay, so let's let's try to solve this now So we have been given that tan of 15 degree sorry tan of theta plus 15 degree by tan of theta minus 15 degree is equal to 3 Okay. Yeah So this as we were saying sine theta plus 15 by Cos theta minus 15 Sorry theta plus 15 and sine theta minus 15 Into cos theta minus 15 degree is 3 by 1 right So now we can do one thing we can apply component and dividend over here so I can do cos theta plus 15 Cos theta minus 15 Okay, plus cos theta plus 15 Sine theta plus minus 15 Whole divided by sine theta plus 15 Cos theta minus 15 minus Cos theta plus 15 This will be this correct Well, I'll take it to the previous slide when cut no worries. Okay. Let's first finish this off Now this is going to be sine of theta plus 15 plus theta minus 15 divided by Sine theta plus 15 minus theta minus 15 Okay, that's going to be to which implies sine to theta divided by Sine of 30 degrees is going to be to So sine of 2 theta is going to be 2 into sine 30 degree That's going to be 2 into half, which is nothing but a 1 correct Now sine of 2 theta is 1 implies To theta has to be 45 degrees. Sorry 2 theta has to be 90 degrees because it's given that your answer should lie between 0 to 90 only So theta has to be 45 degrees for this Is that fine any questions with respect to this? Next question When could I take you to the previous slide now if you want to copy something you can copy? So this is a previous slide to just understand the process what I did was I converted tan alpha to sine alpha by cos alpha and Tan beta substituted with whatever was given to me in the question And then if you simplify you'll automatically get the answer. Okay. Next question is x minus 5 by 6 is n tan x plus 2 pi by 3 Then prove that two times m minus n is m plus n seek to x Let me know once you're done. Just type done on your chat box Okay, only Venkat could do that What about others again same stuff just do m by n first m by n will be what tan x plus 2 pi by 3 divided by tan x minus pi by 6 Okay Just write this in terms of sine n cos So sine 2x x plus 2 pi by 3 by cos x plus 2 pi by 3 sine x minus pi by 6 and You'll have cos x minus pi by 6 Now again apply component and dividend oh so m plus n by m minus n Okay, so let's do that. So for the time being I'll consider this to be a this to be B So again, this is a and this is b just so that I don't have to write much So it's sine a cos b plus cos a sine b and And in the denominator, I'll have sine a cos b minus cos a sine b Which is clearly the formula for sine a plus b By sine a minus b, okay And sine a plus b means addition of these two terms that will give you sine x plus 2 pi by 3 minus pi by 6 Okay, 2 pi by 3 minus pi by 6 will be pi by 2 Divided by sine a minus b a minus b will be 2 pi by 3 plus pi by 6 which will be 5 pi by 6 So this will be m plus n by m minus n Okay, the sine 90 plus x is cos x Okay, okay, I'll have a 2 over here. So it's cos 2x Okay, and in the denominator, I'll get a half so to This can be written as 2 by seek 2x is m plus n by m minus n Okay, let's cross multiply When you cross multiply this This with this this with this you get 2 times m minus n as m plus n seek 2x And I think this is what we wanted to prove actually. Let's check up. Let's go up and check Yeah, this is what we wanted to prove over here. Is that fine, okay? Let's take one more question and then we'll move on to sets chapter prove that 1 plus tan x Tan x by 2 is seek x and This in turn is tan x Caught x by 2 minus 1 prove that tan x 1 plus tan x tan x by 2 is seek x And that in turn is also equal to tan x got x by 2 minus 1 Done. Yeah, hurry. I'll go back to the previous slide First let's do this question Done. Okay How about others? Done Adveta is also done. Okay So let us start with proving this is equal to this so 1 plus this I can write it as sine x By cos x and this becomes sine x by 2. Sorry sign x by 2 Cos x by 2 so I'm starting from my left-hand side. Okay Yeah, so this becomes cos x cos x by 2 and this is Cos x cos x by 2 plus sine x sine x by 2 Which is clearly the formula of cos a cos b plus sine a sine b which is cos a minus b divided by cos x Cos x by 2 Okay, that's nothing but cos x by 2 divided by cos x cos x by 2 These two gets cancelled and that is nothing but see Which is actually equal to your right-hand side Not to also apply the same so tan x by 2 tan x into cot x by 2. Let me write it as sine x cos x by 2 divided by cos x sine x by 2 minus 1 That will become sine x cos x by 2 minus cos x sine x by 2 divided by cos x sine x by 2 Okay, this is nothing but the formula of sine x minus x by 2 divided by cos x sine x by 2 correct and This is nothing but sine x by 2 divided by cos x sine x by 2 cancel off Sine x by 2 both from the numerator denominator. You will get c k x as your answer Right quite easy problem Now let us talk about sets for the timing in sets basically the Following concepts are important. First of all, you need to know what is a set. Okay Because they can give you a question as to whether that particular collection of object is a set or not All of you must know that the definition of a set is Or the definition of a set set is basically a collection of well defined It's the collection of well defined Objects, okay This well defined is very important because they should not be any ambiguity in the definition of a set So it's a well-defined collection of objects, which is called a set Okay So then I'll ask you this question. Tell me whether a collection of All beautiful girls in India All beautiful girls in India. Is this a set or not? Just type. Yes. If it is a set. No, if it is not. No, it is not a set Because beautiful is a very subjective term Right This is not well defined Who defines what is beautiful and what is not beautiful? Okay Next let's say all big cities of India all big cities of India. Is this a set or not? Okay, this is also not a set correct all intelligent all intelligent students of of NPS is this a set or not? No, again the word big the word intelligent. They are not well defined I'm getting this point So when such collection of objects are given to you you have to tell it is not a set Right, but if I say all colors of rainbow all colors Of the rainbow This is a set because there are seven defined colors with your inner rainbow. Is that fine? Next is the Tation of said which Presentations of one is called the roosa form and other is called the set middle of all Is it not visible? Can you see the screen everyone? Hello, I'm audible. Oh, okay Fine, so I'll just take a quick Question on converting Any of you know? No, so what's the answer? So you just have to put the values. So it's one by two two by three Three by four four by five five by six Six by seven seven by eight eight by nine nine by ten Oh, don't worry. It should be visible now Just once again, it would it should be visible now. Are you able to see the screen now? Okay? Let's take another question write the set builder form for set a defined as let's say Zero twenty four write a set builder form for this rotation N-Q-n absolutely correct. So basically in send middle of form you can write this as X where X In cube minus n We'll have done your mental ability in NTSC. We'll find this very easy Okay, next type of question that is framed on set is about subsets So all must be clear with the subsets when is a call a subset of B a is a subset of B if All elements of a also belong to be if set a has n elements The total number of subsets that can be formed is 2 to the power n How this it's very simple let's say a Has these elements a1 a2 a3 all the way till a n Okay Now if you want to form a subset of a Okay, let's say a subset of a has to be formed Let's say there is a subset of a Okay, if a subset of a has to be formed Every element here will have two options Correct. So a1 will have two options a2 will have two options a3 will have two options all the way till n correct, so the number of subsets of a can be obtained by Multiplying 2 into 2 into 2 n number of times Okay, so we say 2 to the power n number of subsets are possible for set a is that fine Okay, now there's something called proper subset. What's the proper subset? Proper subset is a subset where at least one element of that set B should not belong to that subset Okay, well Let me write it in terms of language If a is a proper subset of B if a is a proper subset of B okay, then at least one element of a Does not belong to Sorry one element of B does not belong to a Does not belong to a are you getting it? So, let me take an example So, let's say a is Let's say B is not a let's say B is One two three Okay Now the subset of a can be the subset can be one Okay, it can be two It can be three It can be one two It can be one three It can be two three. It can be one two three Correct. It can also be a null set So altogether, there will be eight elements one two three four five six seven eight out of this This is called the improper subset Because it has got all the elements of set B and all the remaining ones which are marking with a tick mark They are proper subsets. So these all are proper subsets so as a rule we say that if The number of elements in a set a is n Then the number of proper subsets Then the number of proper subsets is equal to 2 to the power n minus 2 to the power n minus 1 Now questions that can come on proper subsets and subsets. Let's take that. Let a and B be two finite sets Let a and B be two finite sets such that a has m elements and B has k elements Okay, the total number of subsets of A is 56 the total number of subsets of a is 56 more than The total number of subsets of B. Okay, find the value of 64 What is that out? Mac? What is your doubt? Okay, so you're correct. So basically let's solve this so total number of subsets of Total number of subsets of a would be two to the power m Okay, and total number of subsets of B will be 2 to the power k Okay So 2 to the power m will be 56 plus 2 to the power k So 2 to the power m minus 2 to the power k will be 56 That's very obvious that m is greater than k, right? Oh that we converted to a rooster form Or that we converted to a set builder form how how does x lie between 1 to 9 not x? It's n that lies between 1 to 9 Sorry for the misprint there Not as clear now from this it is very clear that this is an even number and This is an odd number and this even number some power of 2 So basically it's very obvious that this has to be split up as 8 into 7 Which clearly implies 2 to the power k is 8 and 2 to the power m minus k minus 1 is 7 which means k is 3 and 2 to the power m minus 3 is equal to 8 so m minus 3 is also equal to 3 Because it's can be done as 2 to the power 3 so m is equal to 6 So the answer is k is 3 and m is 6 Next important concept is the concept of power sets power set power set of a set of a set a is basically nothing but it's a set which contains all subsets of a Okay, I'll give you an example. Let's say a is 1 comma 2 Okay, I want to write the power set of a this is how we represent the power set of a So what I'll do is I will write all the subsets of a possible which is null set Set containing one set containing two set containing one and two Mac, you know guy because n will go till 1 to 9 right including 9 so 9 by 10 would be there What is wrong in that? Now just a quick question for you if a is a set containing only one write down The power set of power set of a Okay, now let's let's solve this question. Now power set of a will contain Only two elements null set and the set of one itself correct Now when it comes to power set of a power set of a we have to start writing the Subsets of this now subset of this will contain Null set a set containing null set a set containing one and A set containing null set and one itself This is very important many people have a confusion regarding this. They say sir Is it these two the same thing? Please note that null set and a set containing null set are not the same things. They are different things This is like a empty box and this is like a box containing an empty box Are you getting it? There's a difference between these two are you getting it? Please note couple of things Set Containing one and a set containing the set one. Please note. They are two different things They're not the same Here this is a set which contains one as the element and this is a set which contains set one as its element, okay? Are you getting this point? Okay, let's let's take a very interesting exercise Okay, let a be a null set Let a be a null set. Okay Find power set of power set of power set of a Okay, let's discuss So first of all What will be power set of a power set of a will basically contain a set? which contains the Null set itself Okay Now what will be power set of power set of a will contain a set which contains the null set and the set of null set Are you getting this? Now what will be power set of power set of power set of a? So it'll be a set containing null set Okay, a set containing null set a set containing the this Okay, and a set containing These two elements is that fine? So please note. They're all different things. They may all look very similar especially these three looks similar But they are not they're different things this is a null set element This is an element which is a set of a null set Are you getting this point? Very important now next is the concept of union intersection which you all know I don't have to tell you all that things One important thing is a difference of sets. This is important difference of Of sets which is written as a minus b Right, it is to be read as all elements x such that x belongs to a and X does not belong to b Okay, please note that a minus b is same as a Intersection b complement. This is a very important relation Now most probably the questions that you will get would be practical problems on sets Okay, or we say problems based on cardinal number Problems based on Cardinality of sets So all of you must be familiar with these formulas and a union b That's na Plus nb Minus and a intersection b Okay And a union b union c is nothing, but which what do you want to do in a reliable way? And a minus b would be nothing, but n a Union b minus nb, right and a minus b is basically this part This is set a and this is set b. This is called a minus b. This part is called a minus b Okay, okay, and this part is called the yellow part is called nb minus a you can also write this as na minus na intersection b similarly, you can write nb minus a as na union b minus na Or you can write it as nb minus na intersection b No for power said you just have to write down the subset of that set. Okay, and just Put all those elements in a curly bracket God, Venkat now basically all these theorems you can directly prove from when diagrams Okay, so when you say na union b is equal to na plus nb Minus na intersection b you can prove this by when diagrams So let's say we have this as set a This has set b Okay, let's say this part let me show this by Yellow part. Let's say this is m Let's say this part is n and Let's say I change the color. This part here is oh Okay Now see it when you say and a union b. It's actually nothing but m plus o plus n Okay, now according to this formula and a is nothing but m plus o correct, this is na and B is nothing but M plus o Right and a intersection b is nothing but oh correct, so if you do n a plus nb minus na intersection b You end up getting m plus o plus n plus o minus o Which gives you nothing but m plus o plus n which is what had written over here. Okay So all of these questions you need not do you do not remember these properties You can directly do it Okay, I'll go back to the previous slide her in one second. Is that fine? Let's take practical problems, which you can solve by using when diagrams. Have you done these type of questions in your school? Practical problems on when diagram They say it's so many number of students Take physics so many number of students take chemistry like that kind of problem. Have you done that in school? Okay, so most likely those problems are going to be asked for tomorrow's test Okay, so I just run you through a few of them. Okay. Let's start with this question. Are you sure? Then what type of problems have been done in your school for sets? So what is left in sets then have it in the properties? Prove that questions. Oh Okay, only problems related to that. Okay, fine. Then we'll do a problem said it to that only rather than you know doing anything extra Okay, fine. So I'll give you problems related to a rooster set set build a form Etc. If let's say set a is basically the set of All alphabets of the word it on the power set of a Write down the power set of a Power sets are basically a set of subsets of a. Okay. So first write down a A will be nothing but P OR please remember in a set you cannot write one element more than once so even if there are two os You have to only write one oh, okay Yeah, that's correct. That's correct minus Manav, sorry So what are the subsets of a? Subsets of a would be a null set a Set containing P a set containing O a set containing are a set containing PO a set containing PR a Set containing are oh and a set containing PO are so one two three four five six seven eight Correct all you need to do is you have to put them all in a curly bracket And this would become the power set of a So please note the number of elements in the power set of a Is same as the total number of subsets of a so basically this is very important formula for you. Is that fine? next question a is one two three five six and be is two three four seven eight Write down the set a minus b none Yes, so basically from a remove the elements which are also there in be so two and three are in B So your answer will be one five six Okay easy one Let's take another question If a is a set which contains one two three four Five six. Okay, then state true or false Say to our false statement number one a statement number a One is a subset of a statement number B This is a subset of a statement number C this belongs to a statement number D one two three Is a subset of a Okay When you're writing your response write it like this a is to B is to C is false like this Please state your response like this. Okay, so that I know which which Statement you are writing as true or false. I can see a lot of response coming But you are writing it separately. I would request you to write in one go a true B true C true D false like this, please write the response like this don't give a Individually one-one response in one go you write this one go in your in your box. You write this in one go So with it says a is false B is false C is true D is true. Okay D and D. Okay, let us discuss this Let us discuss this a Says one is a subset of a this is a false statement You can say one belongs to a one is not a set Right if you want to write it you have to write it as this This is a subset of a but one belongs to a Okay, so this statement is a wrong statement. Is that fine? So those who are saying a is true I'm so sorry. It is not true. It is false Second is This is a subset of a this is also false Because this is an element. So basically this will belong to set a Right, but if you want to write it a subset you have to say Set of this will be a subset of a Okay, so this statement is also false This is true. This is also false Because This is not a subset. So it should you should write it like this. This is A subset of a It is not an element. There is no element like this present in the set Okay, if you want to write it as an element, you should just state 5 comma 6 is An element of a like this. So there's a difference between these two. Okay Last is 1 2 3 a subset. This is true So the answer will be F F F T So a is F B is F C is F and D is T Nobody got this right guys try to understand the difference between subset and Element of this isn't this means it's an element of or belongs this means it's a subset of So this must be a subset. This must be a set next question is similar looking question 1 3 4 in a set 5 Okay state true or false state true or false first statement is 3 comma 4 is a subset of a second one 3 comma 4 belongs to a third one Set off set off 3 comma 4 is a subset of a option 4 1 2 5 set is a subset of a 5 1 2 5 set belongs to a 6 1 1 belongs to a 7th one One is a subset of a 8th one Null set is a subset of a and 9th one Null set is a subset of Okay, let's go step by step All of you, please answer first one. Is it true or false? just write I True or I false. What do you what do you think? Yes? That would be considered a single element. This would be a single element When you're writing the power set is this true or false It's false It's an element of set a so it should ideally be This belongs to a why am I getting a mixed response? I just did an exercise right? It's false This is an element of the set You can't write it as a subset subset. It is not a set itself. I mean, it's it should be like This to be called a subset of a this would be a subset of it Okay, wait wait with your tough first. Let me clarify the first one. Why people are making mistake with the first one Okay Next one This is true Yes, this is true. This is correct, but this is not correct first one was not correct second one is correct Third one third one is true. Correct as I only wrote here third one Set of a set containing three by four is a subset of a Okay, fourth one fourth one is also true. Correct So now one two three if you put this in a set it will become a subset of a but this is false Fifth one is false Okay, six one one is an element of set a this is true But one is a subset of a this is false seventh one Null set is a subset of a this is true But a set of null set is a subset of a this is false I think you interchange the answer 8th is true and 9th is false Let's move on to the next question write the set builder form for Set a which contains one No set of a null set is Not a subset of it because null set itself is a set So you can say this is a subset of a that's fine But when you're saying this is a subset of a that means a should contain Null set as its element then only you can write this Else you cannot write it that in words Venkat Yeah, and I understand you have identified the pattern the pattern is the difference of consecutive term is natural numbers. That is fine So you have identified that one two four seven eleven sixteen twenty two This difference is one This difference is to this difference is C four five six, but how do you write this as a formula? I cannot be descriptive right after write it in as Simplistic formula pattern as possible any idea. I'll show you something very interesting. Let's say I decide to add these numbers Okay, let's say I decide to add these numbers up till the last term. Let's say the net term Let me call this as s Let's write the same term one shifted to the right. Let us subtract these two So this will become zero. This will become one. This will become one. This will become two This will become three. This will become four. This will become five. This will become six Up till Tn minus Tn minus one and a minus Tn correct Let's bring this Tn to the left side So it'll become Tn is equal to one plus one plus two plus three all the way till Tn minus Tn minus one, okay Now here if you see there are n minus one natural numbers added correct Who some we know that we all know from our junior classes that? The sum till n terms is nn plus one by two, right? If you add one plus two plus three till n terms We know that this sum is going to be n into n plus one by two correct If I sum the very same thing till n minus one instead my sum will be n minus one into n by two correct So from here I can say that Tn would be one plus n minus one into n by two Okay, take an LCM. So it'll be two plus n square minus n So this is basically nothing but n square minus n plus two by two Okay, so your nth term is going to be this Now you can try out put n as one put n as one you will get t one as One minus one plus two by two, which is one. So this is the first term put n as two you get put n as two You get second term as one plus you can put it over here also if you want Two minus one into two by two, which is again a two correct Put n as three you will get one plus three minus one, which is two two into three by two, which is four Put n as four you will get one plus three into four by two Which is going to be seven like that. Okay So what I can do is while writing the set builder form I can say it contains x such that x is one plus n n minus one by two For n belonging to natural number and n less than equal to how many terms are there by the way One two three four five six seven terms. So I can write it as n less than equal to seven Yeah, I subtracted both the series. I Shifted it one to the right and I subtracted them. Is that fine? I'm not going to ask such complicated questions. It's just a slightly more complicated version of writing the set builder form Okay, let's question Again the two sets a and b a has m elements B has k elements Okay the number of subsets of a Is hundred and twelve more than the number of subsets of b? find m and k The number of subsets of a is hundred and twelve more than the number of subsets of b Find m and k seven and four. Okay. Let's check Again, two to the power m is hundred and twelve plus two to the power k same approach Two to the power m minus two to the power k is hundred and twelve So two to the power k into two to the power m minus k minus one is hundred and twelve. Okay Now this has to be an even number and a power of two. This has to be an odd number The only possibility is sixteen into seven So this implies two to the power k is sixteenth That means two to the power k is two to the power four that means鍵 is four and To do the power M minus four minus one is seven that means you do the power M minus four is equal to eight that is Two to the power M minus four is two to the power three that means M minus four is three So M is equal to seven So K is 4, M is 7, absolutely correct, okay. Next question is, set A is 1, 2, 3, set B is 2, 3, 4. Find intersection of power set of A and power set of B. Null set, why? Why do you have to be a null set blanket? Okay. Yeah, so basically if you try to do it by a regress method, power set of A would contain null set 1, 2, 3, 1 and 2, 2 and 3, 1 and 3 and 1, 2, 3, correct. Power set of B will contain null set 2, 3, 4, 2 and 3, 3 and 4, 2 and 4 and 2, 3, 4. Okay. Out of these common, commoner as these, these two are common, these two are common, these two are common, these two are common, that's it, correct. So this is going to be nothing but null set 2, 3, 2 and 3, right. Now if you see this very closely, if you write the power set of this, what's A intersection B? A intersection B from here would be just 2 and 3, right. If you write the power set of this, you will get null set 2, 3 and 2 and 3. What does signify? These two are same, these two are same or not? So this brings a very important property, please note this down. Power set of A intersection B is same as power set of A intersection power set of B. Is that fine? Fine. So I'll give you a question containing a null set also. Let's say there is set which contains 1, 2 and a null set and a set of null set, okay. Write down the power set of, let me just remove one element here. Yeah. Write down the power set of A, okay. Let's quickly look into this. So first write down all subsets of A. So by default you have a null set, correct. You'll have a set of 1, you'll have a set of null set, you have a set of set of null set, correct. You'll have 1 and null set combined as a set. You have a set of 1 and the null set as a set and finally you'll have a set of null set and a set of null set as a set. So if you put a curly bracket around this, this will become the power set of A. So altogether 8 answers should be there. 1, 2, 3, 4, 5, 6, 7. I think I missed out a set containing all of them. Yeah, 8 answers should be there, okay. Is that fine? So guys, all the best for tomorrow. Don't panic out, especially trigonometry question. Be very careful in choosing the means depending upon what you want to prove. It will be a very easy paper. Don't worry about it. It's a unit test only, okay. Just don't do silly mistakes. I don't want to hear I did a silly mistake. Silly mistake cannot be corrected, okay. And do send me the paper after the school is over. Please send me a, on the group itself we can post the paper so that I can share it with other people also who, why yet to write their UTs, okay. So all the best. Over and out from my side, practice from NCRT exemplar, practice from NCRT exemplar and RD Sharma, okay. These two books are sufficient enough, okay. Mostly they'll pick up questions from these two books itself. No Venkat, I'll not be taking KVPY today because tomorrow you have a test, I guess, right? Tomorrow which paper you have? English you have, okay. Anyways, we'll take up KVPY session again. It's not a very big topic, okay. So as of now, over and out from my side, all the best for English as well. Bye bye, all the best.