hello am I audible yes yes ma'am yes ma'am my screen isible yes ma'am' is breaking my voice is breaking no yes voice is breaking no it's fine most of the person says no so I think it's not that's fine now I'm audible clearly right yes now it's yes ma'am it's fine ma'am ma'am this is the revision s or so ma'am we'll start within two minutes uh yes someone wants to say something like this is the revision session right you're not audible clearly it is revision session the revision session yeah this is the rision session for week one and two we will go through some some of the concepts and uh we'll try to solve some of the practice questions which will help you in your quizes and uh yeah you will get some okay some benefits got it ma'am yeah ma'am actually I have a doubt regarding the quiz when exam yeah you can so my question is ma'am that is there going to be any negative marking in the quiz one exam no no no there won't be any negative marking uh so there is no negative marking yeah there is no negative marking okay thank you ma'am hello hello ma'am yes ma'am I have also a doubt regarding the qualify exam ma'am actually in the portal it is written that we have to score greater than or equal to 40% in the qualifi exam yeah ma'am so we have to score 40% in the four subjects separately or in the overall marks so I have not idea about that but I think you have yeah it depends on it depends depends category depends yeah depends on the Yes means in combination of the fourth subject that is CT math stats and English we have to score 40% overall no any [Music] overall 50% and 40% for each subject overall 50% individual 50% this is for General candidate and there is separate candidate for there are separate Mark for OBC and scst 35% for um uh OBC and 30% for scst okay ma'am thank you sir can you please tell me what is the timing for actually I'm a j level student so I will give I will give giving quiz one so can you please tell me what is the timings for maths and statistics paper as it is I know the exams will be between 2 to 6:00 p.m. but what is the actual timings for the maths and statistics purose actually I don't have any idea about that I think you can switch between uh it's up to you which fig you start first that's I'm sorry I can couldn't understand what you just said uh see actually I don't have that much idea about it that what's the particular timing but I think uh you will get overall four hours for four subjects so I think you can start any subject from your no I have I will be giving only two papers because I have chosen two papers uh I must add that I have directly qualified for the foundation course you will so you'll get two hours for all right uh all right so and then I will I think I'll can and leave right uh yeah fine ma'am can we switch the subject during the exam time uh I don't know I don't know about these technical ask I think these questions yeah I think these question should address to the co could we could we start the session yeah yeah okay so we will drop these all techical things we'll discuss it at the end of the session and I have request I have request you kindly mute everyone also close the chat can you mute everyone and also close the chat because we won't be able to take the session till that we can because otherwise we yes ma'am also please please do disable the don't unmute yourself this is someone share the WhatsApp chat fine so in this session I want to uh I want to discuss some of the topics uh from week one and two mainly the topic related to the relations definition of function domain and the the V diagram un intersection all those Concepts okay so let me start with the basic definition of relation my voice is clear to everyone right yes ma'am yes ma'am yes ma'am yes ma'am your voice is clear sorry what yes voice is clear reply the chat box only rather than yeah please reply in the chat box I'll mute you all but I mean can you make a full mute yeah now it's now you can't mute yourself uh you can ask me in the chat box I'm looking at chat box let's get started so we'll start with the basic definition relation I'll just quickly go through it so what do you we mean by relation so suppose we have a set a then a relation is nothing so R is said to be a relation R is a said to be a relation on a if R is a subset of a cross so by a cross a we mean that uh a cross a we mean that uh yeah you have to study up to four week this session will contain uh the uh this session Will based on week one and two and the next session next revision session will be based on week three and four so we plan two revision sessions for our so by a cartisian product a cross a we mean that this type of set that uh it will contain a pair a comma a Das type where a and a Das both are the elements of set a now in place of a we could take take two sets also that uh let's say A and B are two sets sets and R is a subset of A cross B then R is a relation On A and B which will take the elements of A and B both and by again A cross B is nothing but a set which contains a pair a comma B where a is an element of A and B is an element of now on relation we have some definition certain properties namely reflexivity symmetricity and so we will discuss uh each of the property uh here so let me once go through it quickly and then we'll solve our first push so if we have a relation R then this relation is said to be first thing is reflexive if a comma a belongs to R for all a in a that means every element should related to itself that means the sphere a Comm a should belongs to R for all a belongs to it then we we say that the relation is reflexive the second property is symmetricity symmetric so we say a relation is symmetric if a comma B belongs to R implies B comma a also belongs to so the hypothesis is a comma B should belongs to r that means a if a is related to B then B should related to a that's what we known as symmetricity what do you mean by Qs okay so now the third property is transitivity so by transitivity uh hypothesis is that if a b are related and BC are related then AC should be related so if a relation satisfy this condition we say that it's a transitivity I mean the relation is transitive okay so these three are the basic uh properties of a relation so a relation could have reflexivity symmetricity transitivity either of them and three of them also at the same time okay so now let we have a simple question here based on relation concept so is this question is visible to everyone yes so I'll give you five minutes to solve this question so just try it out and we will discuss it after if someone applied for two subject they'll get two hours for qualif okay so solve this uh and we'll discuss it at uh 825 yes set X is not visible now now is it visible yeah so by reflexivity so by reflex okay reflexivity is used only when the relation is based on the okay so here this option is not considered can this SL so but it's because reflexivity needs to uh needs to be defined only when the relation it's depend on only one set a cross then only we can check reflexivity because a a relation is said to be reflexive if a comma a belongs to R and for a comma a belongs to R we need a relation defined on a cross a only here relation R1 and R2 both are defined on X cross so that means reflexivity we don't need to consider sorry so R2 is all a comma B belongs to X cross Y in such a way that a + b mod 10 is equal to0 so you know modular arithmetic right what do we mean by a + b mod 10 is equal Z so in general If X Mod 10 is equals to0 means uh that 10 divides X that means 10 should divide X then uh it says that x 10 is equal to0 yeah the remainder is zero so that means the X should divide 10 should divide X and in a positive integer we don't have to consider zero zero is neither positive non negative so we don't consider positive in zero I mean zero to be positive integer if you want to write it if you want to consider zero then I should write it here non- negative integer in place of positive okay so by positive we mean strictly than Z not Z okay all POS set of OD positive integers OD posi Ines uh wait a minute what [Music] positive yeah so by mod mod is a something which represent the uh represent the remainder if we divide them so here here you just have to understand what does this uh this mean x 10 is equal to0 so if they written x m 10 is equal to 0 so that mean 10 should divide X that means the remainder is a remainder should be equals to zero when 10 divides X so here they say is R1 is symmetric is R2 is symmetric is R1 is transitive this is R2 is R2 is transitive and R1 is ultimately you have to say which of these R1 is symmetric or transitive R2 is symmetric or transitive I haven't got any of the answer so if you get the answer please answer in the chat box after that we'll discuss don't worry we can go up to 11 so we have plenty there's no such thing un symmetric okay so let's discuss this question see so here they say x is what so X is a set of all positive integers so that means since it is positive so that means we have to consider uh 0o one uh sorry 1 2 3 4 5 and so on and secondly it's an odd set so that means X is nothing but 1 3 5 7 9 11 and so on these type of elements are the members of s okay and the second set is y so Y is what the set of positive integer less than 30 and divisible by four by set of positive integer means again the integer of from starting from 1 2 3 4 and and we have to say that they are divisible by so the least positive integer which is divisible by four is four that means 4 8 12 16 and all multiples of four up to 30 so we know that 20 24 28 right these are the elements we have to take up to 28 so this is the set one now the relation R1 is defined to be that all a comma B belongs to X cross y so a comma B belongs to X cross y means a belongs to X and B belongs to Y in such a way that a is a factor of B so factor of b means B can be written as a multiple of something right so that that's the meaning of factor by a is a factor of b means B can be written as a multiplied by something so that means a should divide B you can understand from like this that if a divides B then we say that a is a factor of B right so R1 we have to take all such SP in where a should divides right so you know that one divides every integer so that means 1A 4 1 comma eight and so on up to 1A 28 are the elements of R1 now let's come to three now you know that three divides the uh three cannot divide any of the uh if three divides some four multiple of something uh because what we need that which are divisible five so that means if three divides four multiple of something there should be a three so it means three divides only 12 and uh three divides only now three divides 36 so that means three can divide only 12 so that means 3 comma 12 will be 11 of now let's go to five now we know that five if five divides four into something there should be a five so that means five divides 20 so 5 comma 20 will be there only so we are done with this now again if 7 divides 4 so there should be something is so that means 7 comma 28 is also an element of R so 7A 28 right similarly if 9 divides 4 into something 9 should be there so that means 36 but 36 is not an element of Y so that's why we cannot consider similarly if 11 divide 4 so there should be something but again 11 should be there so it's 44 again 44 is not an element of Y so it won't be there so that means these are the only elements belongs to R1 okay right so this is your R1 uh so R1 is clear to everyone is there any doubt or any missing element in R1 you find just reply know if there's someone raise hand uh D you can unmute yourself now ma'am you should include 3 comma 24 sorry what 3 comma 24 should be included 3 comma 24 3 divid 4 into something yeah there could be six so yeah correct T comma 20 thank you any other thing yeah you can uh check by this thing that if three divides four multiple of something that means three should present here now we can again take two because if three will get cancel out right so that means 24 should be similar thing can we do with five so five no five should be present there so 20 so if I take two here 40 will so that's means we cannot do this thing with print with five correct so these are the elements we get for R1 now similarly I'll try to find out all the elements which belongs to R right but R2 is little bit simpler to us because uh we have to take all the elements which divide divisible by 10 right so R2 is in R2 we have to take all such elements whose sum is divisible by 12 is divisible by 10 right so since if I take one then if I want to make it some multiple of 10 then there should be nine here then only it will become the addition will become 10 and 10 is divisible by similarly if I want one here but again uh we have to consider that this N9 should be an element of Y but we can easily check that 9 does not belongs to Y so that means one comma name won't be here so that means this element won't belong again so if I want one um plus something which is divisible by 10 or any odd number plus something which is divisible by 10 is it possible first thing we have to observe right so if A + B is equal to 10 multiple of something K now here you have an even number right here you have an odd number right and here you have some number which is divisible by four so that means B is some four multiple of something right so here you have even so is it possible one odd number and one even number add up and become an even number so is it possible right this kind uh this is this is not possible right you are discussing summary yeah I will discuss summary after these two questions okay so that mean we cannot make such type of combination that we cannot add one even and what one odd number to get an even number so that means R2 is simply F okay so here we get R1 and we get R now we have to answer the following question first thing says that R1 is symmetric so in R1 we observe that uh if uh one is here then one can't be here right because here we have to take all the uh multiple of four and multiple of four are always even so that means we cannot find any pair like this ab and B and ab uh yeah ab and ba a type of so if AB is there if AB is there then then B can't be there since B is some four multiple of something and four multiple of something is always an even right so if a is there then B can't be there so that means R1 is not symmetric so that means R1 is not set right this is pyq [Music] question yeah see in R2 what's happening here so in R2 we have to find out all those element a comma B belongs to X cross Y in such a way that a + b mod 10 is equal to0 right so what does it mean so it means that a + b is divisible by 10 right then what does it mean so it means that a plus b is 10 multiple of something now here we know that a is an odd number right B is an even number since why it is an even number because B is a multiple of four multiple of four means multiple of two also and any integer which is a multiple of two is a even number here we again have an even number because again it's a multiple of 10 and 10 contains two so it's a multiple of two so that means here we have an even number and you can observe that any if you add odd plus even you will again get an odd number why you will get again an odd number because even can be written suppose n is your even suppose this is M and this is n and n is even right so if n is even then n -1 + 1 you can do it right and here n minus 1 is an odd number right so that means this does not divisible by two this does not divisible by two and we have one here so you can easily observe here that you if you add one odd number and one even number the result will be an odd just basic observation nothing okay got it so that's why R2 is empty okay now we we observe that R1 is not symmetric why it is not symmetric because if any element a comma B belongs to R1 then uh a is an odd integer odd positive integer and uh B is in multiple of four that means B is an even so if I want if I want B comma a belongs to R1 then what I want I want B is a factor of a but before that I want B comma a is an element of X cross y so that means I want B belongs to X and A belongs to Y now you can easily check that X is a set of all odd positive integer so that means B cannot belongs to X and A cannot belong to Y so that means if a is there then B comma a can't be there so that's why we say R1 is not is it clear now because a is an element of X right and X is what x is the set of all positive odd positive integers yeah fine now let's move to another so again we have the transitivity property left so is R1 transitive so suppose that uh here from the before observation what I observe I observe that if if a comma B is a element of X cross y then uh B comma a would be an element of X cross right so for transitivity what I want I want this hypothesis that a comma B and B comma C should be an element of uh should belongs to R so that means this should be an element of X cross y this should be an element of X but if a Comm B is an element of X cross Y what does it implies so it implies that b belongs to Y and here it implies that b should belongs to X but if B is in y this will implies B is a multiple of four so that means B is uh that means B is a multiple of four means B is an even number right and B belongs to X will implies B is an odd number but this can't this can't POS so that means there does not exist any a comma B and B comma C uh belongs to uh R1 so trivially I meanly we can say R1 is Transit because we cannot find any pair which violate our property R so that's why R1 is Transit right so we have two questions after that we will discuss summary in I mean not only we'll try to do couple of questions also but uh first let me concentrate on main concept which I have to clear okay so you got that R1 is not not transitive now similarly you can proceed with R2 again r R2 is what R2 is all a comma B belongs to X CR Y in such a way that a + b 10 is equals to Z so that means a + b should divides uh should be divisible by divisible by 10 but we are we got that R2 is five so whenever relation is an empty set this is whenever the relation is empty set then it is uh trivially trivially symmetric as well as transitive because we can't find any pair which will contradict our uh which will contradict our properties because we can't find any hypothesis condition so that's why it's trivially symmetric as well as Transit is it clear so r one is transitive vly because we can't find any pair which violates our property similar case happen with R2 R2 is again trivially symmetric and transitive as we cannot find any of pair which contradict our hypothesis okay is this clear so the question says uh this this is the question okay so let's quickly move to the next question where can I get previous year questions uh can someone uh share the links of previous year questions in the chat box okay so now I have this question here this question is based on the when diagram thing so first let me see transitive property says so transitor so what happen so if see hypothesis should be there if there is no there's no pair which satisfy our hypothesis we cannot we cannot contradict our conclusion so if hypothesis is there so that means if you could find two pairs a comma B and B comma C uh in r in your relationship then you have to check this problem then a comma c a comma C should belongs to r that means if you find two pairs of the form a comma B and B comma C in your relation then a comma C this pair should be in there in your relation if the hypothesis is not true then we can trivially say it we can vly say that okay my relation is satisfying that prop okay fine so now let's uh solve this before solving this question let me explain uh when diagram and set subsets and all those things so what does a set so a set is a well defined collection of OB right so there are two ways of writing set first is the write the elements of sets A1 A2 and so on and the second is the descriptive type that means A1 belongs to some other set with some certain properties some certain relations not particular relations some certain properties right these are the two uh ways to describe a set and after that uh we have some notion of subsets uh and the cardinality of a set and power set and okay so we say B is a subset of a if uh for all B belongs to B this this is a sign for belonging an if if for all B in B implies B is an also element of a so that means all all elements of B are in a then we say that b is a subset of a the most natural way to say that two sets are equal if and only if this is the way to write if and only if if and only by if and only means this way is also true and the conversely is also true so we say that two sets is equal if and only if a is a subset of B and B is a subset of a right okay someone raise hand no see now uh there's a what we call cardinality ofet so by cardinality we just mean the number of number of elements in that set this is a hash is denoted for a number on if I want to write the face number off I'll just write hash okay this is just a mathematical language to write Abate things okay now we'll move to vend so why when diagram is actually a pict pictural way to to demonstrate our sets and intersection and Union okay so let me Define what is the union so I'll say a union B is a union B is a set of all those element X such a way that X is in a or X is in B right this is the union the second thing is intersection a intersection b is all the sets X in such a way that uh either X is in a sorry sorry not either X is in a and X is in B so when we use or that means we are talking about Union and when we use end we're talking about intersection okay yeah pictorial way of pictorial representation of s is a vend so suppose you have three sets a b and c and the V diagram of those sets uh can be demonstrated using such type of pictures okay so this is your set your set B set C right here this uh this total Circle represent the whole sets and the intersection portion will represent the intersections right so here this the center one the center uh the center part of V diagram will actually represent a intersection B intersections right and uh and this part will uh this part with this part I mean you can easily check which part represent which set right so I'm just seeing this yellow part this yellow part is actually representing if you consider simply the this this yellow one not the this one simple this this y okay so it's actually representing the all elements which are a intersection C but not the elements of right so in this way you can interpret the V diagram common between these set yeah okay so now there is a formula for finding the uh we say the formula of inclusion exclusion to find the cardinality of Union of oets we know that the cardinality of a union B is actually equals to the cardinality of a plus cardinality of B minus cality of the intersection B this is one formula another formula is the cardinality of a union B Union C is actually the cardinality of a plus the cardinality of B plus the cardinality of C minus the cardinality of a intersection B minus the cardinality of B intersection C the cality of C intersection a and plus the cality of a intersection B intersection I'm not going to prove this these two formulas but I hope you can check on any of the videos video lecture there you will find the proof of these two very useful formulas for finding the union and all other things okay okay now let's move towards this question so this is your question I hope it's clearly visible I'll give you 10 minutes because it's a bit calculative so you can use the calculator and all and it's bit conceptual so just try to solve it observe which set which city we have to find and use when diagram to compute okay that's your head say 47 so let's meet at so Diva yes you want to say something you can meet yourself uh ma'am can you just read out the numbers that you have written uh for this particular question the numbers yeah all the numbers okay yeah so uh this is so there are, uh students and uh 570 for maths 390 for physics and 230 for chemistry 90 for physics maths and physics 140 for physics and chemistry 100 for chemistry and maths and 100 which does not like any of the we have to find the number of then this again the sign is actually denoting the number of so we have to find the number of students who likes only one of the okay for Union formula for three sets let me write it we have to find the number of students uh I mean the number of students uh who write only one of the subjects only one I mean uh any one of the subject you guys are understanding what I want to ask right either maths he likes either maths or physics or chemistry exactly one only one except only one means huh so a un B Union C it's simple just add all the sets add all the cardinality of single set subtract all the cardinality of one two sets intersection all possible two sets intersections C intersection a b b c and then C A and then add all the intersections the three sets interion it's called inclusion and exclusion formula yeah it's a e e yes sir do you want to say something please unmute your yes ma'am uh like ma'am the total sum of the marks is 1 11 190 and the total marks I'm not understanding what you okay ma'am wait a minute let me understand adiva you want to ask something oh yes m' am I audible yeah ma'am these two formulas that you just provided the then most of the questions asked like uh this can be solved from these two formulas right or the other formulas that we have studied in our respective classes will be used too uh see uh I don't know all about all the formulas but yeah these are the basic SE thetical formulas so they will will be used in many scenarios so it's better to learn much formulas any other some of you are getting correct answer but there are so so much variety of answer just recheck I can't tell you now let let try one again s are not right it's not the correct answer I think for okay so many of you got some answers so let's solve it so let's say uh cardal m is the set of students who like maths so the Cardinal of M is 570 570 the cardinality of p is 390 the cardinality of C the p means the set of students who likes physics okay similarly c means the set of students who likes chemistry cardinality of M intersection p is 90 cardinality of P intersection C is 14 and the cality of C intersection m is right and 100 students don't like any of the subject and 100 don't WR anything so in total so if I want to represent the the situation using when diagram how I do so in total my universe is of thousand student this is my universe and it's of thousand students here some students uh like Smiths some likes maths some likes physics and s lik right and uh we have to find we have to find number of students uh whose likes exactly one subject right so from this uh this line that 100 student don't like any of the sub so that mean 100 students are lying in this area this area outside these three circles so from this line we can imply that the cardinality of M intersection P intersection C is nothing but sorry not intersection Union that because 100 student are outside of our outside of our circles so that means the cardinality of Union is nothing but 1,00 minus 100 right because 100 student does not belong any of this so that's why the cardinality of Union is 900 and we have to find the number of students which likes exactly one subject so that means the number of students which present in the S either they could like maths or they could like chemistry or they could like physics right so that means I have to find the number of students which lies in this black sheded in right is this clear to everyone what I want to find right I want to find the number of students uh which are sitting in this black shaded so for finding this black shaded area I get some intuition okay if I find the whole Union and this this this portion this intersection portion this portion this portion and this portion if I find the union and find the uh find the area which uh which presented in this yellow region and I subtract the this yellow region from my union then I'll get the required number which I want to find right so now what's our goal our goal is to find out the cardinality of this yellow shaded so how could you find this yellow shaded reg right so this yellow shaded region so the number of uh student lies in a yellow region is equals to so what we have to do we have to consider the cardinality of Math's intersection with chemistry and we have to add up with the cardinality of chemistry intersection with physics again if we add the cardinality of math intersection physics then I can find the area of this reg but you can observe that we calculated this middle part tce right we calculated this intersection that M intersection P intersection C twice three times right so that means I have to subtract it two times so that I can calculate it exactly once right because I calculate it three times because in this region also that the whole intersection is contained in this region the whole intersection is also contained in this region the whole intersection is also contain in this so that means if I uh and I want to count that region but exactly once so that's why I have to subtract that twice right so as intersection C intersection B the thing is if I compute this area and subtract this area from my union I'll get the required uh the required number right so for that what I want I want the cardinality of all those sets but here I don't have the cardinality of intersection so for finding the cardinality of intersection I'll try to use the cardinality of U okay but we have the formula that the cardinality of M un P un C is nothing but the cardinality of M plus the cardinality of B plus the cardinality of C minus the cardinality of M intersection B- B inter cus C inter M then plus M intersection P intersection right so using this formula I can calculate what is the cality of m intersection P intersection C what is it and some quickly answer 490 630 I 40 50 4 4 okay so let's assume let 4 I mean not let's assume you guys calculated already so it's fo so that means the intersection is fo so that means the the number of students belongs to shaded area R is equals to uh we have to add up these three intersection that means so we have to add 90 + uh 140 + 100 and then I have to subtract two times of 40 so that means I have to subtract 8 so answer is 250 so that means from the above discussion we'll get the number of student lies in yellow region is equals to 250 so that means the number of student uh likes exactly one subject is nothing but the union and the union is M Union P Union C minus this yellow shaded region so that means 900 minus 250 which is nothing but 6 right so the answer is 64 yeah so many of you got the correct answers so why did we multiplied by two see uh in this for counting the number of student lies in lell region we counted this area right because this area is also part of M intersection c c intersection p and p intersection so that means we are counting this area three times that's why we have to SP how can you use the formula because the formula will contain either the cardinality of m p or C we have to find this region this region plus this region plus this that means the formula is not there you have to use the intuition okay if I find the uh the number of student lies in the union and subtract this area I'll get my the the intersection is 40 then we are just mtip let me explain once more yeah calculations is correct so see here we have to find the number of uh students lies in Black shaded are right we have to find the number of students which lies in this black shaded if I want to find the number of student lies in Black shaded area I just have to do that Union this is the total Union and then you have to subtract this part then only you will get the black shaded areas right so we can find out we have already the UN because Union is computed by the fact that 100 student does not like any of the so that means 1, minus 100 that's why 9 that means we have 900 student present in this Union now we have to find out the students which lies in this yellow part so this yellow part this Wing this swing this swing is actually represent the cardinality of M intersections this Wing will represent the total the whole Wing this type of this Wing will represent the cardinality of C intersection and similarly this swing will represent the cardinality of M intersection so if you add these all intersections you will counting this centered regions thce because this region is again a part of the Swing right part of the Swing as well and part of the Swing as so that's why we have to subtract this and after we'll do you can uh try another way if you want but uh that's a most most General way anyone good okay so I'll unmute you guys one by one so that's start with Sur Sur you want to see something ask something yeah that how did we calculate M intersection C intersection that's not I have we just use the formula we have all all the values for substituting in this formula we have cardinality of m p c intersections and Union also we just computed Union right how - 100 yeah can can you scroll down once please yes you mentioned two Cal of M right so what will we substitute there no no no not don't use this formula for un for finding the intersection you have to use this formula the second one okay huh here you have all the values we just don't know the union so that's true understood M thank you okay uh you can unmute your uh' yes yes ma'am ma'am we found maths intersection physics intersection chemistry that is the middle part of it is 40 huh so and uh maths intersection physics is 90 minus 40 50 uh okay so you're Computing in little different way but that's also correct you're is getting indu from V diagram that if you subtract that area so ma'am the yellow colored part according to the diagram V diagram is coming out to be 250 yeah right and so ma'am then only maths or only physics or only chemistry will be 650 right huh right so 650 is on we oh got it thank you sorry now let's come to uh next person granel yes ma'am m'am I had a doubt like uh how did we get 40 I didn't quite get it you're not audible that clear uh hello I'm audible no your voice is little bit low I can't uh understand you can message in chat box I'm looking at the so now U yes ma'am what's your ma'am I have a doubt in the question ma'am the total of students number is 1,000 take and the number of student in mathematics is 570 and physics would be 390 and chemistry 230 uh when we add the numbers the total would be 1190 yeah but there's overlapping of uh students now that's right uh if there is no overlapping then obviously this sum will be more than the number of student but there are some students who like maths and physics boths so they only counted once that's oh okay ma'am huh okay so you guys can lower your hands uh yes sir do you want to say something ma'am uh I have a question in this like you use ma' two formulas can't we just use the one formula which you told like the long one no see we cannot use that formula directly because what region we have to compute is not present in that form here you have to compute exactly once so exactly once means not entirely cardinality of and not enely once we will find the like the intersection like a un sorry your voice is too low I can't understand ma'am I'm I'm can you understand can you tell me why ma'am I'm not getting you can you can message in chat box please okay so let's move forward don't question I want to take so this is your question so in week one and week two we have two topics first majorly two topics first is a relation when diagram set sets subsets cardinality of sets and all and second thing is the equation of line the equation of intersection of two lines the point of intersection the section formula and all those things but those things are much more what we can say computational part there you have particular formulas they derive formulas in lectures so I assume that you guys know the formulas I'll just explain how the equation of line what is the equation of line what is this slow and all those things and after that we can jump to some practice questions okay the formula and from there I'm finding the intersection and then subtracting it from the subject and then adding sub with so you can do in this way also but I don't think there is any uh error but uh we have to check okay if you try in this way is your answer is correct 650 then it's okay function domain okay function domain can cover cover yes gurra you want to say something no oh hi yes I'm still confused with this cardinality thing so my main question is like when we are supposed to minus 2 and I have posted similar question in the chat as well and in that question uh the answer is 110 it's a part of a mock question and in that question we are subtracting it three times so I'm sort of confused when we are supposed to use the uh exact formula and when we are supposed to minus two and we supposed to uh by intuition you will you'll get some idea that if I right so by intu you get at some idea that this intersection part is a part of this swing as well as uh this wing and as well as this wi right so here I want this intersection part to be to be counted in my yellow region right that's why I take one part and subract it from others that's why I subtract I do minus two if you don't want this region then you will subtract it minus three times right so by using when diagram take some intution that uh you know how does area puzzle look like you have to take that area so that means you have to take exactly one so am I audible hello 40 we get just by using the formula okay so let's move toward the domain of function and after that I'll move towards the which week revision is going on is uh yeah one and to both I'm not dis I'm not discussing each and every topic I'm just discussing the major revision portion okay so now what is the function okay so function is actually a map map means what again uh for for defining function we need two type of sets so a map from A to B or instead of MTH just say relation so relation from A to B is said to be a function function if for uh for every a after 4 what I did you will look at so let's get back to the function so function is what so relation from A to B set to be a function if for every a in a uh there is a there is a unique B in B in such a way that uh this function F will maap a to B now in the definition of function there are two important things you have to observe first thing is every so that means for every element of a there should be an output we can consider function as a machine which will take inputs and give you outputs but it should take every input I mean every member of a to be as an input and give a unique output not several outputs unique output so in the definition of function there's only two keywords first thing is every and the second one is uni so if every element of a you are getting unique output for B then that relation is said to be a function right and here the set a is said to be the domain of function so that means all elements which can be taken as unique uh inputs of the function right those elements are set to be domain the set B is set to be Cod and if you consider this set this set that all elements of B in such a way that there is an there is a in a say that F Max a to so that mean if you collect all those elements from B which has preimage by pre image which has an input then such type of then this set is known to be the range of by unique you mean singular yeah singular means yeah unique singular because unique is one to one function no no no no unique is not one to one function uh by one to one we means unique output unique uh sorry unique input unique output that's called one to one function if we have an input uh which can we can have two inputs which can correspond to single output numers yeah sure I'll give you simple example very simple example to understand okay let's answer just answer my questions so this is your a b b and this is a contains one and B contains a b c only uh now F Maps one to a 2 to a that's it so is f Define a function Robin why Define a function why it defines a function just you have two keywords every and unique no no no not about the same output you can have same output for different inputs there is no constraint by unique means yeah that's now I understood by unique mean singular only one output the same output can correspond to two different inputs okay yeah correct the domain is not ma precisely so that means this every word this every word get violated that's why this is not a function okay now let's consider another example suppose now three will get mapped to B and four will get mapped to B as well as C now is this a function now is this a function no why who say yes no this is not a function right why it is not a function because for a single output sing single output sorry for a single input we are getting two different output distinct outputs that means the uniqueness uniqueness is got viated that's why it's not a function right now let's consider this example this is same outut so is this a function yes it is a function because it is satisfying both of the property every element of a domain get ma to a unique output right right so this is a function now the other properties of function will be discussed in next rision session because I think that's a part of week three and week four okay now we have to go for equation of line so I'll just briefly Define equation of right so if you have one point so Point XY not not XY ab and uh you have a slope slope is what slope means uh if we change X if we change X how does y will change at what rate does y change so that means if you have two points X1 y1 X2 Y2 then uh M denotes the slope of line slope of line passes through A and B so it's simply Y2 - y1 that means change in y upon X2 - X1 change in X this is the slope of line passing through two points and the equation of line is simply y = mx plus c c is the Y intercept okay H that's in we five one one on two yeah that's not a I don't think that's a part of of this discussion okay so this is a equation of line General equation of line where i'mot the slope slope means if we change X how does y will change and what rate y will change right so if we have two points we can easily compute this slope using the formula this is a general equation of line and if you have two point the equation of line This is three points equation of line is y is equal to the slow Y2 - y1 upon X2 - X1 x - X1 and + y this is the equation of passing through theun function yeah function is mentioned in week one because we need functions but the precise definition of domain range I don't think that's a part of our discussion but although I stated the domain range both of so this is the equation of line now uh if you have uh if you have two lines let's say y1 is = m1x sorry y = m1x + C1 and y = m2x + 6 right so there's a formula for finding the point of intersection right how does we find the point of intersection we will just take X1 comma y1 to be a point of intersection substitute the value here and solve the equations that's it yeah we it is mentioned in week one but the only that definition domain codomain range I already stated right about one I'm not sure but I'll state it okay okay so the point of intersection you guys know the section formulas section formulas right if we have a line segment AB with the some points X1 y1 X2 Y2 and if some point C this line segment in M ratio n then you have the formula how to compute the uh cardinalities of this points okay so yeah correct elimination method we can use for finding the point of intersection and similarly there's formula for finding the coordinate of Point C if the line segment is got got sectioned internally using M ratios I don't remember exact formula so something m is there this is something like this see for finding point of intersection you have two equation of line right so let's suppose this is your point of intersection P now substitute the value of p in these two equations and eliminate the M1 and C1 M1 M2 C and you'll get the answer okay fine so let's bring a couple of questions uh you guys want to solve some questions or you want to discuss more Theory because I think I discussed major of the theory part there's only this formula part is there so you can uh go through the theory questions right so we have half an hour more than half an hour so let's solve some good questions they can come I don't know yeah so here the answer is already given but uh let's forget about this question okay let's solve this question okay it's a good question this second one this one I'll give you uh 10 minutes to solve it sit back the leg yeah it is compuls to Sol I think you got if you score less than 50 H you can apply for re qualifier also there's one option uh it depends on your preparation how you will prepare the question will M up or easier into that so just solve this question and answer in the chat box e e e you can uh no you can just tell tell the number person not this question I think based on when diagram e e e e e e hello I'm audible audible hello this 650 okay let's start [Music] no I think this is the only find the number of people who like at least two of the G that's the second part that's the third part and the second part is find the number of people who likes only of so let's try to solve first so they say we have to find out how many people like all the three moves okay so that means if I consider so in a survey 500 peoples are there so total number of people is 500 okay now 49% like to watch Comedy movs okay so that means 49% for comedy so let's say cardinality of C is putting C is note the number of person people like the comedy BS okay 53 lik movies so cardal of p is 53 and uh 62 likes romantic movies so let's say cardinality of R to be 62 and also 27 like to watch both comedy and cality of C intersection T is 27 uh 29% like to watch both Thriller and romantic so T intersection R is 29 now 28% like to watch both form and fin of C intersection R is 28 and 5% like none of the oppos so that means uh 5% does not like anything anything so that means 95% will like at least one right 95% likes at least one so that means the cardinality of C intersection P intersection R is 91 in terms of percentage I'm not calculating it in terms of number of people we can do it at the end or you can first convert this percentage into the number of people and then proceed both the ways are correct so I'm just uh trying to leave percentage as it is and at the end I'll compute the number of just by just solving percentage yes FAL take inter because they say end here right they like comedy and film whenever we have end in our phrase we'll take it to be an intersection whenever we have or in our phrases we'll take to be un huh we need to find all of three so that means we need to find oh sorry sorry this is not in it's a huh 5% do not like anything so that means 95% will like at least one of them and whenever we have such at least one of them kind of thing we use Union right so now we have to find out how many people like all the three movies so that means we have to find out uh people who like comedy as well as Thriller well as r that means we have to find out the cardinality of C intersection T intersection R right what you'll do you'll just use our formula so by formula I'm Computing it uh directly so you can uh write the formula try to try to solve the calculations and just try to do so it's 49 27 2 28 so the answer is for this intersection C intersection T intersection R is 15% of people will like all of the three movies so if you count the number then it's 15 multiplied by 500 upon 100 right so that means the answer is 43 75 right so got it is there any doubt we'll just apply the formula and do the calculation we will come to the 15% or 7 okay so is this question clear to everyone rise see what happened we just uh uh defined some sets according to our question here I'm not converting percentage into people you can do both ways you can convert percentage into people first and after that you can apply the formula the basic formula for inclusion exclusion was this form that c un T un R is to add all things first add one first then subtract one intersections I mean two set intersections then add your now here all other quantities are known right and this quantity this quantity is so we have to find out this so just take all these into your left hand side and you 15 aners so yeah here I am working with the percentage that's why I got 15 using just by the formula just by the nothing else just do the computation you'll find it the answer is 75 okay tell move forward or you want to spend more time on this question is there any doubt still okay this is all about thetic question thir question okay so I'll give you the this the little bit intution here so in the second question they say that find the number of people who likes only one of the three JS only one we have to find so again that this you have to find out that uh that regon so this is your V diagram only one me either this one either this one or this so again you have to go by that uh that subtraction method find the union and then subtract the yellow region and find the answer that will be your only one uh number and in the third subard they say find the number of people who like at least two of the given at least two means either M either this intersection that means in the third region you have to find out in the third question you have to find out at least two that mean it could be three but at least two there should be so that means you have to just find out the this reion I mean you just have to do Union minus intersection that will be your answer no no no wait wait wait I explain it WR for at least three you have to do something else this way see you have to find out the number of people who like at least least two of the given okay so this is your V diagram then you have to find out uh people with at least two so if people belong to this region this is a ctm CTR so if people belong to this region then they will like R and C if they belong to this region they will like R and D if they belong to this region they will like CN so that means here we are satisfying this condition at least but if you also consider the intersection region here the people like three movies so that means here also people sittings which like at least so that means you have to find out the find out the people who are lying in this uh this vend this this region that's the answer for third one right let's uh try to solve this thir and SS SS this is one ss based question cul yeah although SS is not part of equal thank you sol that like at least one yeah for at least one you have to just calculate this region that means for calculating this region you have to find out this expression that all the two terms intersections that means C intersection t plus T intersection R plus r intersection C and then since we are counting this intersection region thce we have to subtract it twice because we want is exactly so two times C intersection T intersection that would be your answer for at least three you're getting 70 say yeah 270 is the correct answer for third question there will be uh there will be a session day after tomorrow tomorrow I don't know but this this question has answer okay so let's uh solve this question it's a bit easy question based on this information answer the answer question huh just match the following type of thing so now I just explain what do we mean by anti no uh in only one subject we have to subtract at least two from the union at least two we have different formula for only one subject we have to subtract the number which we'll get from at least two situation from the union situation that's then only we can find the so here M Sy the relation is said to be anti symmetric if AB and ba belongs to R this is your hypothesis for anti sricity you will imply AAL to so that means a is related to B and B is related to a then AAL okay so just uh tell which type of relation uh this one is just match it over here let's say this is 1 2 3 4 five yeah five six already given let's say this is a b c d e f match the columns according yeah symmetric and anti symmetric have oh all are symmetric then we have to go for sub question okay so this is the four sub part just answer accordingly but how could you see all the now the question is totally visible right just answer these four sub parts so symmet so symmetric we have hypothesis that if a belongs to R sure and the conclusion is then should belongs to R this is called symmetry R means relation okay uh now anti symmetric says that if uh a b and b a belongs to R will emply a that means if there is two pairs a and ba which are in your relationship then it will imply is to okay now we have to State true or false R6 does not match with any of the for for e e okay so let's discuss this question so here uh First Option says that uh R six does not match with the any type of relations given in column Bel so in column we have symmetric anti symmetric identity transitivity reflexive and equivalence Rel so equivalence relation means it's a combination of reflexive plus uh symmetric and plus transitive that's the relation is called equivalent okay and by identity relation means uh it maps to every uh to every a Comm this type of relations are ident but here uh a no so the identity relation is this type of relation that every every element should every other element so now R1 is uh given to uh this only one Comm right so that means R1 is uh uh symmetric relation obviously because it satisfy the V condition that a comma B is in R1 so since 1 comma 1 is only in R1 so that means B comma a which is again one comma this is 1A 1 then B comma is again 1A so it's also an element of R so it means R1 is a symmetric relation so this option is false because it's corresponding to corresponds to symmetric okay the second option says R1 matches with all type of relations except anti symmetric relation given in the column B that option is also false because for anti symmetricity we have a we want some hypothesis right for anti symmetricity we want the hypothesis of a comma B and B A belongs to R then this will implies AAL so here only one element is there 1A 1 1A and 1 is equal so it's it's anti symmetric an so that means this option is also not true R1 matches with all type of relation except anti but one is anti so now the question is in total how many relations given in column A matches with transitive relation [Music] so because it matches to they say it does not match with any type of D but R1 is symmetric so that's why uh this option is false right R six oh sorry sorry sorry sorry R six they I'm so sorry and this so let's check the first option again because I computed for R six so R six is uh 1A 1 1A 2 and 2 comma 1 and 2A 3 okay so this is our R six now the first option says that R six does not match with any type of relation so here we have a first thing symmetricity so since you know that 2 comma 3 is an element of R six but uh 3A 2 does not belongs to R six so that's why it's not symmetric now anti symmetricity so again 1A 2 and 2 comma 1 are the elements of R six but uh two is not but two is not equals to 1 so again it's not anti symmetric obviously it is not identity since 2 comma 2 and 3 comma 3 is not an element of R six so again not identity uh for transitivity again it's not transitive because 1 comma 2 and 2 comma 3 is there but 3 1 comma 3 is not there so that's why it's again not transitive and again not reflexive because again 2 comma 2 and 3A 3 are not the IND of our so obviously not any so that's why first option is I'm so sorry I got so one is true yeah one is true and the second one is false right because R1 is anti symmetric since it is regly because there is no two PA to compare we want ab and ba both now go to option number so option number three says that in total how many relations given in column A matches with transitive relation okay so R1 is transitive backwardly transitive about what about R2 so R2 is again vly transitive because there is no two pairs A and B C type of thing if we want to say it is transitive we need some PS yes you want to say something uh ma'am I have a doubt in that first uh uh option R six does not match with any type of relations given in column B uh just check if I'm correct uh my understanding uh what I I I feel or I thought that R6 satisfies transitive relation U why because that is this is my um explanation just let me know if I'm wrong oh R six is transitive no R six is not transitive because yeah that is one one comma one is there uh a comma B so then 1 comma 2 is there that is B comma a and then uh a comma B that is one one is there H right but uh we have one pair this right that uh 1 comma 2 is there and 2 comma 3 is there but uh 1 comma 3 is not there yeah but uh this is not there but then I I think in the previous sessions what I understood was even if it is not there since one comma 2 is there it is transitive no no for all pairs we have to uh uh we have have to check our conditions for all pairs if you find such pairs any number of pairs here we got two pairs but there could be any number of them so you have to check for every pairs so you have to say you have to check for every element that is there in second huh right every element should satisfy the transitive uh relation only then we can say that relation is transitive yeah correct okay okay thank you so this is one thing what we have R2 is transitive Again by vectors choices uh now what about R3 many so R3 we have one one and one two so again we have one two there so that's again R R3 trans now about R4 again R4 is transited C for transitivity there is no two pairs to consider a comma b b comma C right so that means hypothesis is uh uh already I mean there is no there is no two pairs using which we can contradict our statement of transitivity if you find two pairs ab and BC such a way that AC is not there then that will imply that it's not transitive here uh we cannot find any two pairs which will contradict that's why it's Transit okay so symmetricity simply says that if a symmetric means what symmetric means if a comma B is there and then B comma should be right and uh anti symmetric means that if a comma B and B comma C is there in R then only uh then a comma C should be there so let's uh go through options one again so first option says that R six does not match any of the relations R so for R six uh we have uh um for symmetricity we have that 2 comma 3 is there in R six but 3A 2 is not there right so this will imply okay now for anti symmetricity for anti symmetricity we have 1 comma 2 and 2 comma 1 is there in R six but uh two is not equal that means a is not equals to B condition so that's why it's a not anticip now for identity relation for again that 2 comma 2 is not there in R six so that's why not Iden or we have one comma two type of PA also there so one comma two is also there so not identic in identity we want all a comma a type of now transitivity so again this 1A 2 and 2 comma 3 is there but uh 1a3 is not there in R six so that's why not see transitivity in transitivity we have this type of B comma c will imply a comma C is this is in transitivity okay in anti symmetricity we have this kind of thing A and B will imply there's a much there's a significant difference between okay yeah like no EVC and no VC then there's no AC yeah sply Transit we can no see uh this type of pair need not always exist a B and B if you have both of them then only a is equals to that's why it cannot be totally identic okay so I think you guys can do this question now okay let's uh do this question in full detail so that no one will get confused so this is my question here I'll get some of the things for R six that R6 is uh neither I mean it's not satisfying any condition that's why option one is true okay now let's move to second option okay so second option says that R1 matches with the all type of relation except antisymmetric relation given in column B okay so what is your R1 so R1 is basically R1 is this onea one right and what does anti symmetricity say so anti symmetric says that if a comma B and B comma a is there in R1 then it should imply a isal right but here only 1A 1 is the El so that means 1A 1 is there 1A 1 is there and 1 is equals to so that means this condition is triv so that's why R1 is antis and that's why this option is not correct second one okay now let's go to third option so this third one says that uh total how many relation given in column A matches with the transit so we get that R so that means there is no pair to compare a and BC we want such type of pair to compare transitivity so that means R1 is trivially transitive now R2 again R2 we don't have such pair so that means R2 is also D transitive now R3 again R3 we don't have such pair I mean we have such pair that we have 1 comma 1 and 1 comma 2 but again 1 comma 2 is a element of one so that's why R3 is transitive now in R4 also we don't have such pair so that's why R4 is again transitive and in R5 Let's see R5 so in R5 One Two is there and 22 is there but again one two is there so yeah we don't have any such pair we have this thing that 1 one and 1 two is there but this is implying that 1 two is there so this is correct and again we have two two and one two right one two and two plus there but again one two is there so that's why RI is also transitive but R6 is not transitive we just saw that uh what is this 2A 1A 2 is there and 2 comma 3 is there but 1 comma 3 is not there in R six so that's why the option for the answer for fifth one is five okay the last one says in total how many relation given column matches with reflexive relation reflexive so what does reflexive says okay so reflexive reflexive means for all a in a a comma a should be there in R right okay so first one is not reflexive because 2 comma 2 and 3 comma 3 not there right the second that means R1 R2 okay R2 is reflexive what about R3 R3 is not reflexive but because again 2 comma 2 is not there R4 R4 is again not reflexive R5 is reflexive because our a is only 1 23 2 right and R six is again not reflexes that means R2 and R5 is are reflexes yes R5 is reflex you can consider R5 as an relation which is reflexive but not identical because identity relation will contain R1 is not reflexive because 2 comma 2 and 3 comma 3 is one is not reflexive because 2 comma 2 and 3 comma 3 is not there right reflexive means reflexive means that for all a in a a comma a should be there in means for all element of a they should be related to itself that's the meaning of reference uh yes haridas you can uh raise your hand yes haridas want to say ' a comma B belongs to uh some relation but B comma C doesn't belong is it still transitive H it will be still transitive because we don't have pairs to contradict our uh definition yeah but in in the first question you asked in this uh program you took R2 as non-transitive R2 start of this lecture no no no I yeah I think by mistake and I take R2 not transitive but R2 is transitive okay thank you okay so with this uh I'll share this PDF with you guys so you guys can uh look at these questions uh in this PDF there are many questions related to every I mean all other weeks also so I'll share this PDF in chat box you can uh try to solve it by yourself okay so I shared uh the PDF here please uh have a look at this uh and try to solve the remaining questions there are 15 question related to every other week also so you guys have any other doubts uh let me un first yes yes I doubt regarding the whole Ms I have cleared my all the concepts but when I am trying to solve the question I have some difficulties to face solving the questions same problem ma'am I mean uh the concepts are clear but you are not able to get how to solve the question right yes just a slight of the answer a slight of uh the working will make us write the whole question properly whole answer properly but you know that exor is missing in every question Sor what hello yes ma'am when when I am trying to solve the Activity questions that are given in the dashboard then those those questions are solving very easily but when I am trying to solve the graded assignment assign solution solution ass are a bit tough but uh we gave it a bit tough because you have so many resources right you have internet you have chat GPT and all things so you can use those into to solve it so that's why we want you to yes first the level of qualif my question my question was that not that my question is that when I am solve the qualifier then the difficulty is same for like graded assignment or decrease uh what can I say about this see uh in exam there will be some questions uh which will map to graded assignment difficulty but there will be some questions which are easy I mean it's a combination exam is a combination of all tough and easy question so we won't ask anything out of syllabus so yeah PDF only one and week one on two only right sorry what the PDF is week one and two or because it's having only week on one and two it revision of one and two but the PDF contain much more questions I think I I can see only one and two week syllabus only not two and three sry three and four in the PDF yes I are you only one PDF right huh only one p so I only one week one and two two questions there is some SS question s is not in week one or twoce SS week two SS is a part of week two okay so then it's only week two week one and two uh H SS is a part of it because it's based on line equations yes so ma'am please give me some tips to solve the questions and your voice is too low I can't hear it so ma'am please give me some tips regarding the solving questions hello speak a little bit just practice for practice you want questions there are plenty of question you have activity practice this one previous he asking he's asking for tips to clear exam that's why I said practice tip to clear exam is study as much as you can ma' ma'am I want I want a suggestion uh regarding yeah regarding the word problems that we get in uh please uh speak a bit louder I can't hear okay okay uh I need a suggestion uh regarding the word problems that we get for week two and all how to uh um re mean how to arrive at the uh equation or what they are asking uh can you simply can you just can you just explain like how to break that question and uh uh get what they want so that after that we'll be able to solve uh with the uh formulas that we know and everything like everywh get stuck everywhere get stuck in the equation only see justable and then put the relations on that variable that's the only thing I mean there's some unknown quantity right so assume that quantity to be X and then put all the conditions which uh involve that particular quantity uh that's the only uh way to solve such equations problem assume whatever unknown quantity is there to be X and then put the conditions there that they say twice multiple of this is this so it means 2x plus this is this so something like this just take first consider your particular thing that whatever the number of age or person AG is there or some weight is there some whatever the numerical quantity is there take that quantity to be X and then try to apply all the relations on that unknown variable X then only you will find the relations and because yeah first step is to find out the correct relation right then only you can apply the formulas which you whichever you know then yeah yeah yeah yeah exactly exactly okay yeah it will require a little bit practice I think but uh when once you get a clear intuition what you have to do with the text then I think it's okay okay is there any higher weightage for any of the week is there any weightage for the weeks like the week three got higher weightage or week four got high so the weightage I think sir will discuss in next revision session I think uh because he knows the weightage that's why I don't have that clear idea about weightage hello ma'am I have a doubt it should be equal weightage right it should be equal weightage but when when will the next session but I hear from others that only week three and four will be given important in other session they normally emphasize on week third and fourth because of the complexity being higher on the side to week one and two three and four a bit complex but hello ma'am I have a doubt what uh uh I am unable to create polinomial graph like direct I am confused in creating the direction of uh so you can ask these doubt to the sir I mean he will explain from week three and four whatever guy whatever doubts you have just collect it and ask to they will answer when is coming I think on this Thursday maybe you have a next revision s session tomorrow I think thday means H Thursday is tomorrow so I think tomorrow okay we are waiting for tomorrow you have a session okay Friday morning Friday I don't think because we are missing him link I don't get I okay guys I think uh we are done with the session hello ma'am I don't get a direct link I get from someone others group you don't get the direct what direct link of the sessions why there is a calendar I think my calendar not work properly you to gole I message but they didn't get any any solution of that okay so just share your email ID I'll check once if I can do something uh I okay I have this thing today today there is an update about the examination centers that's been out so the exams normally will be held in the morning or both the session will be there uh you got any admit card no they say the Hall ticket will come three days before but the exam Center is out today they have sent the exam updated the examination centers morning where did I share the link uh no just share your email ID to me so that I can check your profile that your calendar where I share huh in the chat box okay hello ma'am yes uh ma'am where did you share the uh problem PDF in the chat box itself you can scroll ma'am after so many messages are come is there on the top in between I just got disconnected so would you ma'am please share it again yeah yeah sure M I shared the link okay guys if you sorry I I Shar the email please check uh where is your email ID this uh RPL yes ma' yes share the PDF first okay guys you guys uh you can leave now thank you for joining all the best for your exam ma' please do something for my calendar ma'am sorry what for my calendar and update of session so the exam been held in uh morning as well as evening uh it depends on the student uh we have two phases for that so if you got selected for morning session then we'll have exam at morning the PDF is coming I mean loing very slowly thank you yeah it's yes got it ma'am thank you yes okay thank you for joining you guys can leave now good night good night thank you thank you no e e e hello ma'am is this someone can someone take my out e e e e e e e e e e e e