e yeah welcome guys so please type in the chat box is the audio and video clear to you everyone of you so please type in the chat box is the audio and video clear to all of you everyone of you please type in the chat box yeah welcome all of you so please type in the chat box is the audio and video clear not coming a minute let me yeah good evening sagnika please confirm the audio and video Once is the audio and video clear to all of you or not it's clear right yeah thank God fine so uh let's start the session basically so we will continue our next lecture in this 7 p.m. series okay so of course we have started single varable calculus and we have finished couple of lectures okay in the last and the preceding classes and in today's class today we're going to look at lecture number three of the same module single variable calculus again okay so if you briefly go we have we are discussing the concepts of limits and derivatives of course now we will actually move you know a bit into this applications of differentiation okay like the mean value thems concepts of Maxima generation of the TA okay so in today's class the major Focus will be like applications of differentiation okay but anyway before that we'll again look at some standard limits and the ways in which we can apply the derivatives okay so but before going you know this is basically the codes for the updates on this channel okay so this is the telegram Channel link for the gate Wala English Channel so any update on this channel would be posted here so I recommend you to scan this and follow the group actually okay and this is my personal group of course so here you can see a lot of any update from me would be uh coming up on this uh Channel actually Okay and like if I I'm again practice are planning to post some practice sheet uh of uh you know basically linear algebra and also single variable calculus so these things we are going to post at this particular location okay and unity future is what we have recently launched so if you just type Unity PW and if you just you know join that then ultimately you can have a track cord of all your points the only thing you have to do is basically if you're asking a doubt then you can click exclamation doubt and then you can start your doubt similarly if you want to make a notes at one point you can type exclamation note and then keep the point then ultimately these get stored in a space okay so Unity space so again I recommend you to sign in for this Unity space it's absolutely free and once you sign in you'll have a record of all your interaction during the live classes okay at one particular place now coming to the last lecture in the last lecture we were mainly focusing on some nice uh standard graphs some standard graphs is what we have plotted okay and we have also seen some standard derivatives some standard derivatives okay so standard derivatives and you can also see apart from that we have also seen calculation of some limits okay so calculation of limits is what we have seen of course okay so calculation of limits so these are the things that we have discussed in a b uh things okay yes we want question yeah that's what okay I'll upload a practice sheet hopefully by uh Sunday uh sorry not Sunday by hopefully Monday evening or things like that so that you can have a practice okay so class H yeah I'm just coming up to the things so how to balance gate and semester yeah I'll tell you okay this is one of the most frequent question that people ask me how to manage your college time with your gate preparation of course I'll do tell you okay and in regarding tomorrow the block MX and properties yeah thanks seven I'll basically discuss this okay properties and also block matics and of course C Hamilton them yeah okay I'll discuss and and you know uh coming to class on tomorrow ASI basically we have a doubt solving session tomorrow okay so at 7:00 p.m. on the same channel of course we have a doubt solving and not only doubt solving if you have any queries okay like few days back we have conducted one on our telegram channel so something similar if you have any doubts or if you have any queries uh regarding anything you can just come up at 7: p.m. tomorrow and we will discuss the thing clear now let's see apart from this I have given you some homework okay so let's firstly quickly uh answer this homework and then we'll deal this C Hamilton the and block matrices of course a bit and then we'll go for you know today's uh Topic in general okay so I have given you this question limit X tends to Zer e^ 2x - 1 by sin 4X is equal to Dash Okay so this integer type and if you have solved please enter the answer in 30 seconds if you have solved definitely you can enter the answer so if you check this then this value of the limit is what okay that's all come on do this also tell us how to manage time for get preparation with job yeah definitely okay maybe in the last 20 minutes of today's session please or basically can we discuss these two things in tomorrow's class it's better right because tomorrow is like a generic session and of course you can ask your doubts very well I I have some telam queries which came in my telegram channel for a few days I'll answer them and also we'll answer a few other things okay so anyway the poll got finished so let's see how do we calculate this okay so solution limit x ts to 0 e^ 2x - 1 by sin 4X this is what you got okay so ultimately if you simplify this you can get a 0 by0 form okay because you know if you put x equal 0 this is E power 0 so 1 - 1 gives you 0 divided sin 0 is also 0 so this can be written as by applying l hospital rule you can say by L Hospital rule by L Hospital rule limit x ts to Z e^ 2x - 1 by sin 4X is equal to you know you can actually differentiate the function so functions individually so limit x ts to Z what would be the derivative of e^ 2x actually okay so the derivative of e^ 2x is 2 * e^ 2x daa of 1 is 0 divided by the derivative of sin 4X is nothing but 4 * cos 4X okay okay so this is what you'll get now if you replace x equal to 0 then ultimately this is 2 into e^ 0 is 1 divid 4 into cos 0 is again 1 this can cancel this two times so you have 1 by 2 which is equal to 0.5 okay so therefore limit x ts to 0 e^ 2x - 1 / sin 4X which is equal to 1x2 of course and this is 0.5 okay a simple l hospital rule is what you have to apply in order to solve this question and I hope many of you might have solved this because it's a very simple question based on simple logic that we have yesterday okay then then the look let's go to the next question so anyway 0.5 is the answer let's see if some of you have entered 0.5 show results yeah okay 100% one student answered correctly but who is it come on okay the box is not giving the name okay fine but still we can find if you go to leaderboard sagnika is the one who answered okay so sagnika good let's go to the next question quickly so this is one more question I've given you the value of limit x ts to 0 - sin x / 2 sin x + x cos x is what I have given you okay so again inte a type so 30 seconds if you have solved please enter the answer you can enter it in decimals whatever see if you have negative answer just enter the positive value okay because here there's no provision for entering negative so if let's say your answer is- 0.255 example okay then enter 0.25 okay Ash it's not 0.55 it's exactly 0.5 okay yeah good come on answer this quickly 0.33 is what it's coming maybe let's check it up okay so let's see okay maybe 0.33 I don't know let's solve and get to know solution limit x ts to z- sin x by 2 sin x plus X cos x okay so plus X cos x this is what you have now one thing that ultimately comes is if you keep applying the L Hospital rule it becomes a slightly lengthy problem cor so first problem I have given you using l hospital rule of course you can directly get the answer here if you try to apply the L Hospital rule of course the things slightly you know gets elaborated like I'll tell you two methods to solve this question so method one is what I'll just tell you maybe quickly method one and we'll also see the method two okay so method one the classical l hospital rule will apply and let's see because this ultimately gives you 0 by 0 form okay because if you put x equal to 0 sin 0 is 0 then 2 into 0 + 0 into 1 is again 0 so that's how we have so limit x ts to 0 minus derivative of sin x is cos x by l hospital is what I'm writing by L Hospital L by L Hospital rule let's see here so limit x ts to 0 limit X tends to 0 minus daa of sin x is cos x / then 2 into daa of sin is cos x plus da of X into cos x x into - sin x + cos x this is what you'll get actually here okay so this is what you'll get and ultimately if you put X = to 0 in this case we'll get - cos 0 1 / 2 into cos 0 1+ 0 into- sin 0 0 + cos 0 is 1 which is - 1 by 3 of course okay so - 1 by 3 which means this value is - 0.33 let's say if you own off to two decimal places so - 0.33 is what you have as the answer of course okay so- 1/3 which is- 0.33 of course okay now the second method which I would like to do is few times just because of adjusting the terms you'll get the answer very easily see you know limit x ts to Z minus okay this minus I'm pulling out for some time sin x by sin x no 2 * of sin x plus X into cos x x into cos x this is what you have okay now let's observe something can I I do a division with X all over okay so if I do a division with X all over this can be written as minus limit X tends to 0 sin x by X because ultimately when X is tending to when X is standing to Zer what does this total function tells you that means individually when X is standing to Z individually these functions will tend to some particular values and simplification of that okay so 2 is constant again limit x ts to Z sin x by X Plus then you know this x and x can be cancelled then limit x ts to Z cos x okay so this is what you have now ultimately you know if you just multiply this uh with X all over then you can understand this limit value in the numerator is one okay limit x ts to0 sin x by X is 1 similarly this value is also one and this value is also one because if you put 0 cos 0 is 1 so this is -1 by 2 into 1 + 1 and this value is same as this value okay so ultimately it is - 1/3 which is minus 0.33 that's it okay so by using some standard results you can sometimes just fit in some X maybe in the denominators numerators and you can convert the limits into some known functions okay is this clear to all of you yes or no this- 1X 3 is this clear to all of you or not okay aat sagnika ashini s Everyone of you okay now let's see so let's check the correct answer since I have asked you to enter uh this only with the Positive value right it's not showing you L board fine then okay so good hell let's move to the next question of course so in this question I've have given you uh again so I have given you some options as well so single Choice 30 seconds start the value of limit x ts to Infinity 1 + x² into E power - x come on do this limit x ts to Infinity 1 + x² into E power - x liit it is another indeterminant form okay it's like Infinity into zero things like that okay so quick done so let's see if you try to solve this then again it's a simple question limit x ts to Infinity 1 + x² into e power- x okay so and ultimately this gives you the form something like Infinity * real form okay so in last class I told you when you're getting Infinity into zero form but you know the technique only for Infinity by infinity or 0 by 0 by using using some fractions we can convert this into either Infinity by infinity or 0 by 0 okay so if you make this like simplification like slight adjustment 1 + x² / E power x and ultimately when X is very large both these functions tend to Infinities and so therefore limit X tends to Infinity by Al Hospital rule we can apply x ts to Infinity 1 + x² so derivative of 1 + x² is 2X by E power x this is also giving you again Infinity by Infinity so this is equal to limit x ts to Infinity 2 / differentiation of e^ X is again e^ X so this value is zero of course okay so therefore limit x ts to Infinity 1 + x² * E power - x is equal to Z okay you can actually understand this function if you slightly look at the mathematical reasoning of this the value why the reason why this function die down to zero as at very large x why this function goes to zero is because see it is product of two functions one thing is very clear it's product of two functions 1 + x² and e^ - x now when X is increasing compared to the rate of change okay means for example if you check when X is increasing 1 + x square keeps increasing and E power x keeps decreasing but I mean e powerus x keeps decreasing but the rate at which e- X decreases will be very fast compared to the increase in 1 plus X X squ okay means let's say at a particular value of x the product is some value this product is actually some some number okay so let's say l then if you slightly increase X the increase in 1 plus x² will be very much less as compared to the decrease in eus X okay so when you keep multiplying ultimately this e power- x is the strong function out of these two and this e power- x will try to bring down this 1 plus X squ drastically down and within some limit it will go to zero understood did you understand graphically what's happening in the background of course was doing limits and all it's fine but because when I started this question when I writed this I know the answer is going to be zero because I know mathematically in the background it happens is e powerus x is very fast decaying function so that will pull down this 1 plus x square very effectively okay if you see the graph you'll see even maybe I'll show you the graph actually oh sorry so maybe I'll plot this and show you 1 plus x² into e power- x okay so let's see 1 + x² 1 + x² into E power - x okay so you can see 1 + x² into e^ - x if you take this function then ultimately this function has got some shape of course but after you reach this point it dies down to zero you can see when X is increasing clear land X goes very large then ultimately this function die Downs because the reason is once it costes one that decay in the E power x is E power minus X is very fast okay as compared to 1 plus x² clear so anyway so the answer is going to be zero option A is the correct answer let's check it a none of you have answered okay fine so let's go to our next question I have given you one more question yeah this is a nice question again so if you have solved then enter the answer 30 seconds again I'm giving you the limiting value of the ratio of square of the sum of N natural numbers to the n * the square of the sum of squares of n natural numbers as n approaches Infinity is Dash so how do you know that e- X is dominant function after of the two functions because I know exponential decay is very fast as compared to the quadratic increase or quadratic decrease okay exponentially decaying you might have seen okay very drastic Decay within a very short span things die down or things climb up very fast okay especially if you're doing in vibrating strings or in vibrations you you'll come to know okay whenever you there's a wave which you start initially this wave immediately dice down when the when there's some overdamped oscillation or something okay obviously exponential functions are much very fast decaying F so what we're talking yeah this question will solve solution the limiting value of the ratio of square of sum of N natural numbers okay so first of all what is sum of N natural numbers I equal to 1 to n so Sigma I if you take sum of N natural numbers this becomes 1 + 2 and so on N you know this is n into n + 1 by 2 of course okay so n into n + 1 by 2 then summation then some sum of squares of n natural numbers okay so Sigma I = to 1 to n i sare is 1 square + 2 square and so on + n Square n into n + 1 into 2 n + 1 by 6 correct you know these are the two things that you should know in order to calculate this okay so sum of the first n natural numbers is n into n + 1 by 2 and sum of of first n natural numbers is basically n into n + 1 + 2 n + 1 by 6 okay clear to all of you okay so please type in the chat box is this clear do you know these things yes or no everyone please type in the the chat box are you you know are you aware of these two things actually if you know about those things then things is you know the solving this question is very easy of course okay yeah so let's see what we have written they have been asking you for when n tends to Infinity okay because they told as n approaches Infinity so for very large n let's see what happens okay so for very large n this value is so the value of ratio of of square of sum of the natural numbers so this is sum of N natural numbers so square of this limit n TS to Infinity n into n + 1 by 2 whole Square okay so this is square of some of the natural numbers this guy divided with n * the sum of squares of n natural numbers so n * sum of squares of n natural numbers is this guy so n * of n into n + 1 into 2 n + 1 / 6 is what you have so you have to simplify this limit that's it okay so if you simplify this limit let's see what you have limit n TS to Infinity then you know it's a fraction so you can simplify this n² into n + 1 s by 4 and you can flip this because if you multiplying this 6 by n² because n into n gives n² n + 1 into 2 n + one of course okay so this is what you have now ultimately you can cancel this n Square n Square this gives you two and this gives you three so 3x 2 * of limit n TS to Infinity if you expand this for example no need of even expansion you know if you expand this the highest order the highest degree term would be n Square okay so n s + 2 n + 1 divided by and if you expand this for example 2 n² + 2 n + n 3 n + one this is how you have okay now what would be the value of this limit that's the main uh you know interesting thing that you have to figure out now in last class I have given you one technique okay so when this n is standing to infinity and if powers or degrees of this both polinomial is same then ultimately ratio of the coefficients of leading degree is the answer okay you know this has the coefficient one and this has the coefficient 2 so ultimately this limit is going to be 1 / 2 that's what correct in last class we have seen of course so this value is 3x2 into 1X 2 which is 3x 4 of course and this is 0 75 clear so 0.75 is the actual correct answer in this 0.75 fine so please type in the chat box is this clear to all of you 0.75 is the correct answer in this case okay yeah so let's see 0.75 if someone has entered this 0.75 okay this is not showing but let's check here okay again uh questions sagnika good only looks like sagnika is doing homework what about others you guys don't want to do okay I'm not giving it just for fun okay so when I teach something you when you try to answer a few questions you'll get some doubts okay and that's how the concept Clarity actually builds up clear so anyway please spend time on solving the homeworks so anyway now coming to today's class we'll put our focus of course a bit on limits I'll have a few more points on limits and continuity okay and we have mean value thems then Taylor say is what we'll discuss and if time permits I'm not sure whether we can delete it today's class if not we'll delete in the next class so concept of Maxima and Minima is also what we'll try to focus on okay now let's see before uh we go for this mean value thems because I have a few things to talk in linear algebra the first thing that I would like to tell is K Hamilton them Hamilton here okay so which is again in syllabus of very few people and of course the application or number of questions which are coming on C hamton them is also very less so basically this themm tells you that every Matrix every Square Matrix in general of course every Square Matrix that is fice satisfies its own characteristic equ safies its own its own characteristic equation of course okay so every Square Matrix satisfies its own characteristic equation like you can see example I'll uh work out with an example so that you can understand it better if you have a matrix like let's say 1 2 3 - 1 let's say for example okay so 1 2 3 - 1 if you have this Matrix I'm explaining with 2x two but same thing can be extended for anything so you'll uh see one thing here if this is 1 2 3 - 1 then ultimately you know Lambda characteristic equation is something like characteristic equation is characteristic equation is so it's a 2x2 ultimately so Lambda Square minus Lambda Square minus trace of this Matrix okay so what is the trace 1 minus 1 so 0 0 is the trace clear so if this is Trace then we don't have Z element 0 into Lambda plus determinant what is determinant minus one - 6 so this is - 7 * is equal to 0 of course okay you know basically how to form this coefficients I have told you even for 3x3 case so if it's a 3 you have written Lambda CU minus t plus sum of the principle minus into Lambda plus this determinant is equal to sorry minus of determinant is equal to Z because you'll keep changing this plus minus signs alternatively okay so anyway now we can see we got Lambda Square - 7 is equal to0 this is the characteristic equation in this case okay if this is the characteristic equation now what C tells you is okay so this C Hamilton the Hamilton theem gives so if you replace Lambda with a a square so since this is a matx minus 7 should be repl with 7 I that's it is equal to Z so ultimately you know a square is equal to 7 * of I that's how okay so you can without actually calculating the actual a square you can tell a a square value okay similarly you can tell about higher powers of a in general if you know this C Hamilton theem and many times in matrices with large sizes you can calculate the inverse very easily without you know doing lot of stuff actually okay see how you can do that I'll just give you a hint obviously if you let's check check whether this a square will give us seven times Identity or not okay so a square is going to be 1 2 3 - 1 then 1 2 3 - 1 so if you simplify this let's say 1 + 2 3 are 6 so total 7 then 2 - 2 0 3 - 3 0 6- of - 1 into - 1 so sorry 3 into 2 6 Plus - 1 into - 1 is 1 again which is 7 so you can see this is 7 * I but you decided that this value is going to be 7 * I without actually calculating a square just by using this C hamon theem okay so that's how the when camon the satisfies things this zero is of course the tast of the M and obviously as you keep Maring forward you'll have some Trace then after some of minus then after determinant with plus minus alterations okay I have told you how to generate a characteristic equation okay now let's say we'll take one more matx and we'll inspect like I'll show you the advantage of calculating inverse example a is equal to say for example 1 2 3 7 3 - 4 maybe some numbers whatever you like 2 - 1 2 Okay so let's say this is your Matrix then if I ask you what is the way to calculate it a inverse okay so how to calculate this a inverse then you know the classical position you should calculate 9 minus total okay then after calculating this 9 minus then you have to find co-actors then calculate transpose it you'll get a joint okay then divide it with determinant of a you have to do all these things of course okay but let's see without doing all these things how you can calculate a inverse first of all if you add the characteristic equation I've told you just by looking at the matx how to add the characteristic equation for 3x3 characteristic equation is so it's 3x3 ultimately so Lambda Cube minus because it's plus so next would be minus so what is T 1 + 3 4 4 + 2 6 so 6 Lambda squ then here it is minus so next it's going to be plus sum of principle minus so first let us calculate the minor of this element okay so if you delete this row and this column and row so 6 - 4 so 6 - 4 is 2 then after minor of this element 2 - 6 - 4 plus the last element 3 - 14 - 11 * Lambda then you have used plus then again minus now determinant value is equal to 0 so what is the determinant let's check so determinant is 1 into 6 uh - 4 so 2 - 2 into 14 + 8 22 + 3 into - 7 + 6 - 1 okay so ultimately this is 2 uh this is 44 of course 44 and 3 44 and 3 is 47 47 + 2 so - 45 so determinant is - 45 you know that a is equal to - 45 this is what you have so - 45 is equal to 0 so which means you know Lambda Cub - 6 Lambda squ then this is -5 + 2 -3 Lambda + 45 is equal to 0 of course Okay so this is the characteristic equation I've told you just by looking at the matx and doing few calculations how to write this normally you need not expand the total 3x3 determinant and then multiplying the terms factorizing you need not do all these things of course to calculate this equation now as per C Hamilton theem it tells you something like C Hamilton DM this tells you ultimately a cub - 6 a² - 13 * a + 45 * identity Matrix is equal to 0 okay so this is plus 45 into I is equal to 0 okay now if you multiply it with a inverse you know multiplying with a inverse on both sides of course you'll see you can decide whether you can multiply whether a inverse exist or not if a inverse exist then ultimately you have this constant term okay this 45 if a inverse do not exist then ultimately this determinant is zero you won't have the constant term then C Hamilton doesn't work of course okay so if you multiply with a inverse a s - 6 a - 13 * a into a inverse is identity Matrix plus 45 * a inverse is equal to0 so what is a inverse from this equation you know 6 1X 45 * of 6 a + 13 * i - a² is what you're going to get okay and you have generated an expression for a squ you can see you know it's you feel that this calculation is a bit lengthy but actually whenever you go for matx Beyond 5x5 and 6x6 this equation is actually very helpful okay means this C Hamilton the is actually very helpful because at such matrices rather than calculating minus of let's say for example if you have 5x5 Matrix you have to calculate 25 minus okay to calculate the inverse okay but if you know this technique then ultimately calculating a square is much simple as calculating compared to calculating minus of 25 elements okay so that's why this K Hamilton them is very useful in general clear to all of you yes all no please type in the chat box did you understand how basically what camilon can actually help you help you I'm afraid of calculus I failed it several times it gives me a panic attack oh my God so I need to avoid this do you have any solution for this problem just attend the lectures okay and whenever we are starting from the base please do follow that's all okay see anything can bring discomfort in your life only once or maximum twice that's it okay that's all once you face the problem next time when it comes you are not afed okay let's say for example if G qualifying gate is one of your you know issues okay then you have qualifi if you it seems impossible only you know it's done for the first time okay then after it's like a very common thing clear it becomes a very common thing like first time when I have written gate examination backs down the line you know uh that's the first time I was in bch final year then so I felt like very you know uh you know a bit of I don't have anxiety but a bit of you know less confident okay feeling less confident is what I have then but once I have cleared it then after every I'm writing every I'm qualifying and I've been doing well in get examination so it it it's like a matter of first time that's it okay so anyway this C ter actually helps you up now coming to one uh thing actually the partition matrices which people have asked so partition matri or partition Matrix or sometimes it's also called as block matx so let's see what's a partition matx basically what does a block matx can actually help you okay now if you have a system okay you'll have lot of systems which keep developing now if you have this system okay then if you take any matx for example say matx m is what you have for example so there is a matx m and there is a system such that m x is equal to or maybe I'll take some other variable M you know Alpha is equal to Beta let's say this is what you have and you have to solve for Alpha for example okay then when you have to solve for this Alpha few times what happens here is this matx m will not be full occupied okay many of the elements of M would be zeros or some part of M could be identity Matrix and like for example I'll show you if you maybe I'll show you one equation so diagonal in fluid mechanics in cfd is what one matx I'll see so tagal matrices in cfd we'll experience a lot of times okay so T diagonal system for elliptic equations what we can anyway we'll right now don't worry about what is antic equation what is a differential equation and all but my main intention is to tell you if you check sometimes you can see this Matrix okay so there's something called thas algorithm which will maybe we'll see numerical methods but right now don't worry about all these things right now just look at this story as this is a matrix and this is a vector and this is also one more Vector okay now if you for example identify this matx one thing that you can figure out is only these three diagonals okay so the principal diagonal then the below diagonal and the diagonal which is above so only these three diagonals contains elements and the rest of all elements are zeros okay so whenever you try to calculate this values of X1 X2 and so on till xn few times what happen the computer has to unnecessarily perform lot of operations okay so to eliminate this we will actually split this Matrix into blocks okay so let's say for example I have taken a square Matrix here then maybe some part is here there's one part here and last part is here okay so this is what we actually divide so whenever you have this Matrix this Matrix a can be partitioned like this okay so when you whenever you have this Matrix you can partition this Matrix so this is a maybe b c and d of course okay so these are called parts of course okay so whenever you have a very big matx you can actually split it up into some blocks okay maybe they can be uh you know Square matrices or rectangular matrices anything like let's say if this is off size M by M for example because I'm using M then let's say a is of size n byn okay then B is of size M minus n n CR M minus n of course okay so M minus n is what you have right so this is of size n by n n by n so this is off size n by m minus n and what about c c will be off size so since you have covered M rows so M minus n cross you already have filled this n columns here so c will also have n Columns of course and D would have same thing like m- n CR m- n so these are basically the matrices okay so you can split up this big Matrix of size M bym into four maybe you can split into any number depending on your comfort so you have this ABCD that's what we call partitioning now the advantage of this partitioning is few times especially when you dealing with triagonal systems or systems where we have majority of the limits as zeros let's say for example when you're doing this performance this a is identity Matrix for example okay or let's say the partition is such that these two are equal so that this partition is symmetric okay so you have very special cases where the solution the value of this Alpha can be calculated with very easy calculations actually okay so anyway let's see what we can do is this Alpha can also be written as some vector xar and some Vector Y Bar where this let's say for example Alpha is something like alpha 1 Alpha 2 and so on alpha n then ultimately this x is of size uh you know it it has some elements here this can be written as X1 X2 and so on you have how many elements do you have here so a is Multiplied to this X for example okay so you have M by n element so it should have n by1 so n by1 this is xn and similarly you have maybe y1 Y2 and so on y m minus n this is what you can have of course okay so that you have total M elements here so total you'll have n number of this is m alpha M because we have taken this as n m bym Matrix m is of size M by m so this total consist of M elements that M elements we are splitting it into two vectors okay so this is the first vector and this is the second Vector of course okay and we will actually calculate the things similarly this beta can be split into maybe C1 and D1 similarly this beta can be split into two vectors maybe first Vector is C1 let's say it has some columns I mean some rows which which is of size a column matx of course it has some R and a vector D1 of course okay so this is what you can do the advantage of doing this kind of splitting is you need not perform individual calculations for individual elements in fact using matx operations you can solve the system at a single go I'll show you how we can do this let's say m alpha is equal to B for example and this m can be treated as a b c d it's a partition Matrix of course okay so it's a partition matx and this multiplied with X and Y okay so these are the vectors which we have done so this Alpha is split into two parts and ultimately C1 and what we have taken C1 and D1 okay fine so let's say this is D1 clear so this is C D1 now exactly we have split this according to the sizes okay so we have split this xar Y Bar C1 D1 depending on the size of this n byn if this this is of size n byn then ultimately this X1 has got n elements so that this matx multiplication a with X is possible okay similarly this B is of size n by m- n so ultimately this has M minus n columns and this has M minus n rows then ultimately this B can be multiplied to this Vector y okay so correspondingly we'll generate X C1 and D1 where C1 is of size n by 1 and this is M - n by 1 of course okay so these are the two vectors which when you add the combination you'll again get the beta clear first of all the splitting up is it clear then we'll see how to solve the systems okay but first of all please tell me is the splitting up clear to all of you how we we can actually divide or basically we can part partition a big matx is this clear to all of you we'll see how to solve but first of all the partition is understandable or not everyone of you please type in the chat box M minus N means number of uh you know totally have M RS and M columns in this matrix it's of size M by m so when you have partitioned it we are taking Yos in the A and ultimately n and n columns in a so ultimately we should have n minus m m minus n columns okay see whenever you at a side by side you'll have total n columns okay but a consisting n columns it's of size n by n so out of total M columns if you have n columns in a then ultimately number of columns here is M minus n that's how okay did you understand okay see basically what I'm telling is look let's say I have a 4x4 matx for example okay 1 2 3 4 1 2 3 4 You know just some 3x4 Matrix you take okay if I take a 3x4 matrix if I can partition this into this situation then ultimately this is of size 2x3 and this total is of size 3x 4 so total this matx completely should have four columns so if I take this part as one Matrix a which is of size 2x3 and if I call this as Matrix B then ultimately number of columns that I'm going to come here is 1 because this 3 + one column will actually give me total four columns that's why if this matrix consist of n columns ultimately B should consist of M minus n columns if you add these two then ultimately total number of columns is M again okay similar if a consists of n then B should also consist of n because this is a horizontal line so like you can see here it is 2x3 so here it's also 2 by one okay so this two is actually same in this case that's how in general we are doing the partition okay clear did you all get this yes or no please type in the chat box is this clear to each and every one of you or not now did you all understand how we have done the partition partition just a minute I'll have water okay Prat everyone is this clear how we are doing the partitioning okay now yeah now we'll see how this partitioning is effective for us okay how we can help us in solving the a system okay so let's see how we can uh do this so ultimately we got this system as a b c d we have partitioned this then this Vector also we have partitioned into two vectors and this also we have partitioned D1 D1 but of course the partitioning is exactly the in sync with this matx m partitioning now let's see if you simplify this you you have a into xar + B into Y Bar is equal to C1 bar of course okay and this C into xar plus d into Y Bar is equal to D1 bar exactly okay so this is D1 bar is what you can write okay now let's see if this is C xar + d y bar equal to D1 bar then on this equation let's say say if you want to calculate X for example okay how do we calculate you can use elimination procedure but it's not like you multiply c b and CD then cancel and all okay because whatever you do this multiplication doesn't matter so we'll go for elimination Y is equal to if you calculate Y how we will calculate y this Y is actually D1 minus C * X this multipli with d inverse okay because these are matrices you should not write 1 by D these are matrices everywhere we have Matrix so ultimate this is the expression for the Y Vector actually clear so y can be calculated by using D inverse * D1 - CX of course okay now if you replace this in first equation therefore a X plus b y so what is y d inverse D1 minus D inverse TX is equal to T1 of course okay so this is T1 now if you take the coefficients which have a inside so this gives up something like a minus D inverse sorry A minus b d inverse C this complete thing multiplied with X operated on X plus b t inverse D1 is equal to constant 1 you know this okay so this is constant one so you can ultimately write this x is equal to a minus b d inverse C whole inverse this total inverse multiplied with C1 minus b t inverse D1 this is what you'll get correct so this is how you can calculate what is X completely here okay but the main interesting thing in this uh cold story is of course see this is what you got expression and here you have to calculate this D inverse only one inverse is what you have to calculate see in actual case if you want to solve this system okay if you want to solve m alpha equal to B inverse m alpha is equal to B inverse you can you have to calculate inverse of this Matrix M okay and let's say if this is of size M by m total there will be M Square elements in this and to calculate the inverse of this you'll definitely always calculate this total M Square yes or no you have to calculate minus of all those M squ elements right correct will if I want to calculate that inverse M inverse ultimately I'll have to calculate M you know basically I'll have to calculate all M by m minus okay means total M Square minus is what I have to calculate which is very big operation but in this case if you want to calculate this x inverse you have to calculate this D inverse okay so only D inverse and of course the inverse of this Matrix as well but you can observe one thing the size of any of these matrices would be definitely less than the size of M bym okay and also this partitioning will actually make one thing very simple let's say this complete Matrix M you have some M Square elements let's say the total Matrix m is singular for example okay then ultimately you cannot solve Alpha equal to M inverse B Because M inverse does not exist but when this complete Matrix is singular there can be a chances where this part if you take this small part this small part may not be singular yes or no correct and for calculation of X and Y you need only this D inverse so definitely if this is non- singular then still you can solve this system cor did you understand the advantage of partitioning the things yes so please type in the chat box is this clear to all of you did you understand why we are interested in the partitioning business especially when the mat is consist of lot of uh you know things okay fine okay now let's see so this is how you'll actually calculate X and here you'll observe one thing this Matrix because a minus b d inverse C is what we have this a minus sry a minus B D inverse C this Matrix is called sh complement sh complement of D of course okay so this is called T complement of D of course okay so can we solve 4x4 Matrix from this yeah you can solve even th by th000 also okay just pick 999 by 991 as one and 1 by one as last or maybe 2x2 as the last small part so just calculate the inverse of the 2x2 that's very simple and you can use this technique okay now if you replace this in that we'll get the value of y also okay so this is called the scuss complement or sh complement basically of this D inverse of D of course okay so sh complement of D so just by calculating this D inverse and putting it here you can calculate this x because everything is rest all is matrix multiplication that's it okay so multiplications are very simple to calculate the solute you can calculate of any size okay and in fact you can use this technique for non Square matri also okay let's say for example I have a matx of size 4X 5 okay whenever I have a matrix of size 4X 5 see for example if I uh have one more column and one more Row for example so 1 one one 1 1 1 1 1 okay so clearly it's of size 4X 5 now in this case this is of size 4X 5 of course okay so let's say maybe for the comfort I'm telling this uh this is two different elements we are having now if this is 4X 5 then you can it's a nonlinear system of I mean sorry it's a linear system but it's of nons square Matrix then when it is nons square matx by Framing a 2x2 here you can calculate what is D inverse and you can keep performing this operation okay so that's how this Vector can be completely calculated using some Matrix multiplications and calculating inverse of D and once you get this you can also calculate this very easily okay it is almost like substitution okay so this calculating X inverse is the main stuff once you get this x inverse just replacing this x inverse here will give you this complete y okay which means these two are connected of course and identifying one will just give the other in very few steps okay that's how this complete block of elements with the help of matrices this complete block is calculated in a single go and once this is known this can also be calculated in very few operations okay that's why we have the advantage of dealing with partitioning M now one important stuff that comes here is few times what happens this B comes out to be a symmetric Matrix okay B comes out to be a Sy matx which means B can be written as B transpose now when you're using some properties of inverse and all things get simple easily and few times this B can also be an sorry D okay the last Matrix D can also be an orthogonal Matrix okay so whenever it's an orthogonal Matrix this D inverse is just D transpose and things would be very simple okay you need not even calculate the inverse then in that case of D you just transpose it so if D is you know orthogonal Matrix D is same as D transpose and if you just put D transpose and D transpose and simplify you can solve okay fine so is this clear to each and every one of you why we are dealing all this partition matrices or you know other things basically okay yeah correct or wrong please type in the chat box is this true is this clear to all of you okay he pleas please type in the chat box here you're understanding the things or not yeah examples this okay this is what I have told you just now okay so let's say you have a matrix like this then you can take out some part out of that okay and calculate the inverse and then apply these rules which can give you let me write for y also and Y Bar is equal to D inverse times okay D inverse times D1 minus C * okay so I think this I have written C and this is C1 and D1 okay fine you can uh recognize so D1 into C into X so what is X a minus b d inverse C whole inverse multiplied with C1 minus b d inverse D1 this is how you have the elements of okay and of course normally when a matx SI goes beyond a very large value we'll just use programming using this simple uh identification of partitioning how do we calculate the determinants of the partition Matrix see basically you can calculate this whole determinant is what you have okay so a see for example if this is a non-square matrix first of all for calculating the determinants we'll check the determinance matx and once you have the determinant matx you can apply any or form or anything to make the to calculate the debt very easily okay so let's go to the applications okay we have few more things to deal okay so this is basically about the partic and of course last year is the first thing first time when they have introduced data science paper and last year there's no question on par matics of course so we'll see what happens this okay hello let's before we go for mean value thems I would like to give few points actually okay some standard limits is what you can do some stand limits which would help you okay so let's see some standard limits so first let us see limit x ts to a x^ N - a^ n / x - A this would be n into a power n minus1 so similarly limit x ts to a x^ m - a^ M / x^ N - a^ n this gives you m by n * a^ m minus n of course okay so M by n into a^ M minus n then let's see limit x ts to 0 sin a x by X is equal to limit x ts to Z tan a x by X this is also equal to a of course okay so this is a then 4 limit x ts to 0 1 - cos x by x² is = to a² by 2 guys please remember many times in get to confuse you they'll don't give x square here they'll just give you only 1 x if they give you only 1 x the answer is zero okay because ultimately you know if you have limit X tends to Z 1 - cos a x by X okay okay then you can do this as a multiplication with X and write it X squ so this can be written as limit x ts to 0 1 - cos x by x² this guy into limit x ts to 0 x is what you have okay so ultimately you know this gives you a s by 2 but this gives you zero so this is zero so please make sure whether you have X or X squ here when you calculating this limits okay because many times people are habituated to write this shortcuts and they don't actually remember the question okay so please keep this in mind then we have few more things which we can write 5 limit x ts to z e power a x - 1 / X okay so if you uh try to simplify this then we have E power a x - 1 / X is log a to the base e correct so this is log a uh just a minute we'll calculate this okay so log a^ X - 1 Let's Take here a^ x - 1 is equal to log a to the base e of course okay so log a to the base e because if you apply l hospital a x into log a to the base e so we have okay so this is log a to the B then you can write few more limits see all this can be calculated using l hospital okay till here it can be calculated using L hospitals rule of course then six limit X tends to 0 a^ x + b^ x - 1 by 2^ 1 by X is equal to so let's say you have this limit for example okay so such that it gives you the form 1 to the^ Infinity because when you put x equal 0 a^ 0 is 1 b^ 0 is 1 and minus 1 so this total becomes 1 1 by 2 power uh you know basically we have a^ X Plus okay just a minute I think I have made some small mistake when writing this a^ X Plus b^ X by 2 - 1 okay so a x + b^ x by 2 - 1^ 1 by X okay so you can see when you put x equal to 0 this is 1 1 so 1 by 2 sorry 2 by 2 is 1 1 - 1 0 no just a minute then this minus one is not there maybe okay yeah Power 1 by x correct so this gives you 1 Point Infinity so this is square root of ab okay so this is square root of a you can actually do that the reason for that is why you can do this is I'll just show you how to get this result limit X tends to0 a^ x + b^ x by 2 ^ 1X X this gives you something like 1 to the^ Infinity correct because if you put x equal to Z then ultimately that gives you 1 to the^ Infinity then you can see in last class I have given one expression which could help you if limit X tends to Infinity F ofx power G of X is 1 power Infinity then okay so then limit X tends to Infinity f of x power G of X is equal to e to the power limit X tends to Infinity in fact X tends to any limit a you know you can keep a to generalize the things x ts to a is 1 infity then x ts to a is equal to E power limit x ts to a p of X into F ofx minus 1 this is what we have seen okay so this is the expression that I have given you in the last lecture so since if you put x equal to Z this gives you the form 1 power Infinity so what you can do is you can apply this and let's try to get this down therefore limit X tends to infinity a^ x + b^ x by 2^ 1X X is equal to E power limit x ts to Infinity G of X so what is g of X of course 1X X into f of x - 1 a^ x + b^ x by 2 - 1 this is what you have okay so if you simplify this let's see what we'll get this is equal to E power limit X tends to Infinity you can write this as if you take LCM you have a^ x + b^ x - 2 / 2 we can write that as a^ X - 1 + b^ x - 1 / x / by X and this 1x2 which is a constant you can bring it outside the limit okay so 1 by2 is constant you can bring it outside the limit then ultimately since limit can be split up at plus sign so in fact at any sign of course so limit X tends to Infinity this a^ X - 1 by X this term will actually give you log a to the base C okay this is the limit what I've given you if you apply Hospital you'll get the answer because if you put x equal to 0 this gives you 0 by 0 form so upon applying H hospital it gives you log a to the base e and here if you pull out this 1 by2 then E power limit extends to Infinity you have sum of two limits so as per this logic you'll have E power half * L A plus lwn b log B to the base e of course okay so this is what you have now when you have sum of you know two logs you have log a plus log B then ultimately log AB is what you have so e to the power half L AB so this half which is in the multiplication of log can be thrown into the power so E power L root of ab which is root a because you know log basically you know a power log a log something to the base is that something so therefore this value is root a okay yeah this is correct fine so you can check this are the things the properties that we have used here is this half when it goes to this power basically you know the two logarithm properties which you can use is n * l x to the base a can be written as lwn x^ n the base a this is one thing and uh you know one more thing is a power log X to the base a is equal to X that's it okay so these two properties is what I have used in simplifying this uh things of course okay is this clear to all of you yes or no do you know this two properties do you know these two properties do anyone of you know if some of you don't know I'll explain you this logs of course it's like some nth class or 10th class stuff but still I'll explain you if some of you don't know please say no then I'll explain how these things arec only one of them you know okay which one the first one or the second one top or the bottom no idea if some of you don't know please ask me I'll definitely clarify okay or if you know then it's fine okay everybody knows I guess none of you is typing okay good okay so doesn't know the second okay second one you don't know fine maybe if you have studied this you know aptitude basically school math you can understand the definition of logarithm is if a^ x is equal to n this implies log n to the base a is equal to X this is how the log is defined Cor yeah will you AG this is how the log is defined now you can consider one thing this implies a power log n to the base a is equal to a power x because these two are equal so this will be equal to this but what is a^ X N so this implies a log n the base a on the power is equal to n okay clear okay we are keeping the base same okay so ultimately this ax turns out to be n and this complete expression gives you n fine hello let's uh okay I hope this is clear so in general if you have a limit like this similarly one General limit which I would like to give is six so 7 limit X tends to zero 1^ x + 2^ x and so on + n to the^ x / n^ 1xx which is of the form 1 power Infinity this is of the form y root of n factorial is what we have okay so this is n of n factorial clear so like for example you see when you have only two numbers two root of product of these two which is square Ro of a so let's see this is one again important thing which you have of course so yeah tan we have taught and this kind of things we have taught then 8 you know in last class I told you if limit x ts to a f ofx is equal to 0 then limit X tends to a f of x power f ofx this is always equal to one okay in bracket please write this is ^ 0 in general is not equal to one okay so this ^0 is not equal to one but this one comes up because of some logic what I Told You So based on this there are a lot of gate questions which came now I would like to ask you one limit I hope you'll all will answer me limit x ts to 0 oh Infinity let's say infinity sin x by x what is this value can anyone tell me limit x ts to Infinity 1 + x sorry limit x ts to Infinity sin x by x what is the value of this what is the value of this limit X tends to Infinity sin x by x z why it should be zero it could be something else right limit x ts to Infinity s x by 0 x by x z two yeah of course it is zero okay because in the graph yesterday I have given you as X Ines the sin x byx value keeps dying down and it goes to zero and here you can see the reason for this is whatever you keep in place of X you'll always have have a finite value in -1 to 1 of course so this divided by a very large number okay so this TS to zero of course a finite value in minus 1 to 1 divided by very large number gives you a very number the number which is very close to Z of course okay so this is one thing then we can see 10th question 10th limit X tends to Z 1 + a x^ B by X is same as limit X tends to Infinity 1 + a by X power x okay so power BX let's say BX in both these cases the value is E power a okay you can see they give you the form one 1. Infinity so if you do the classical procedure you will see basically this gives you 1. infinity form and if you apply this uh uh result then ultimately this gives you E power AB of course okay this also very important uh things which save you some time okay so these are some standard limits which could keep rolling out again and again get examination okay so this it could you know it could slightly save time but of course these are something which you need to mandat know even if you don't know if you know the techniques you can D at any point okay so let's solve one question actually this question okay so let's solve the following inequality is two for all xclose to Z what is the value of limit X tends to infinity x into sin x by 1 - cos x I'll give you some time okay so maybe some uh 60 seconds is what I'll give you come on solve this question it's a nice question actually okay there's something called sandwich them in maths which you can see now of course but if some of you have any idea you can try this question come on to it one maybe okay time got finished but anyway let's check okay so let's check it up what is the answer the first thing which I would like to tell you is first of all what is the sandwich theorem is is what you have to know let's say if you have 2 less than x less than 2 for example okay so let's say you have a number like this the value of x is trapped between two okay means this is two and this is also two that means X should stay at two it cannot move anywhere correct yes or no if you see on an aial line if you have aial line of course you have a line and let's say this is the value X is equal to 2 and let's say the upper limit and lower limit of this x is only two okay so that means this value of x has to be at two doesn't matter it should it cannot move anywhere okay so therefore in such cases X is equal to 2 that's what you call sandwich the means whenever it's fixed by it's locked by two limits here but if both the limits are same then ultimately the value of this function is the value of the limit itself okay so let's see what we can uh calculate of course if this is two for all X close to 0 limit x ts to 0 2 - x² by 3 less than limit x ts to Z x sin x by 1 - cos x less than limit x ts to 0 2 this is what we have okay so if you simplify this this implies if you put x equal to 0 this limit gives you 2 less than limit X tends to Z x sin x by 1 - cos x less than limit x ts to 0 2 is 2 again okay because you don't have X to substitute Z so that means this limit will lie between two as lower and two as upper limit therefore this implies limit x ts to Z x * sin x by 1 - cos x this value is definitely two in this case Okay so this is of course two clear that's how even without knowing this knowledge if you try to calculate this limit still you will get two okay which I can show you easily I I think many of you might have forgot one thing many of you might have forgotten one thing so that's why you entering one but let's see if you want to uh you know basically Sol this limit X tends to Z x sin x by 1- cos x okay so let's say you have this you can write this as limit X tends to Zer X sin x by 1 - cos x you divide with x² on both numerator and denominator so if you denote ultimately this limit if you split okay so this is of the form limit x ts to Z sin x by X ID by limit x ts to Z 1 - cos x ided by x² okay so ultimately you know this value is 1 this value is 1 by 2 so the total answer is two that's it okay so even if you don't know this logic still by just calculating this you can say the answer is two or you can apply lpal whatever you have a lot of things to do so whatever you do you'll can clearly figure out this value is two of course okay but to make you uh eliminate this process they have already given you this inequality so they they want to tell you that as X tends to zero ultimately this function okay this function will lie at two because the lower and upper bounds are two so ultimately this function has to be two so if you know this logic you need not do this calculation clear okay sagnika hash Alex aay gami everyone of you yeah so please type in the chat box is this clear to all of you yes or no type a bit fast quick did you all understand why it's true because since you guys have given one so I know there's some you know we dra back so that's all now I just want to solve few questions before we uh okay solve this question once again so with a one unit change in B so if you're changing B by one unit what is the change in X in the solution of the system equations okay so which means Y is fixed so if B is changed by some amount what is the change in X that's what okay so what is a change in X that could give a change of uh two units a change of one unit in B come on think anyway for this question the answer is two I think none of you have answered fine so for this question single Choice I'm giving you 120 seconds or maybe yeah 50 seconds you take it's okay all isue so there are two equations which are connected linear equations and if your B has to change by one unit what should be the change in X that's the question okay solve it's again a previous year question Sol many times of course come on solve this quickly B is coming two unit maybe okay let's see everyone should try it could be B maybe it's possible because B is in one of the option fast come on quick come on guys quick see what some of you feeling see I gave you time for thinking okay actually if you know the concept to solve this you need only 30 seconds okay but few who want to think they have to think okay okay so that's why I have given some time okay yeah C is what Ashman is feeling okay so let's see if you solve this question so first of all they're asking you with a one unit change in B what is the change in X this is what you want correct so which means you see you see here they want you to calculate DX by DB this is what for a one unit change in b means if DB equal to 1 what is DX that's what which means you have to calculate what is DX by DB that's what they asking cor from the definition of differentiation when B changes by one unit what is the corresponding change in X that's what they asking the first of all what is the connectivity between X and B let us figure out okay what is the connectivity so let's see let's eliminate y from these two equations so that you can have connection between X and V directly okay so let's see if you uh maybe if you multiply point 99 okay so this implies 99 x +99 y is equal to 99 into 2 1.98 of course okay so 1.98 is what we have and this equation gives you 1.01 X +99 Y because I have intentionally multiplied 99 because I want to eliminate y so this is uh you know B of course now this is equation one maybe and this guy is equation two now two - 1 gives you something like 1.01 - 0.09 gives you 0.02 x is equal to this gets canceled of course B minus 1.98 so this is the connectivity between X and B directly now what they asking you when B changes by one unit what is the change in X which means you have to figure out what is the value of DX by DB okay so let's see if you differentiate this equation we get2 DX by DB is equal to 1 and what is DX by DB because if you're differentiating with respect to B then this x becomes D of X becomes DX by DB D of B becomes 1 and D of constant is zero so 1 by 0.02 so this is 50 of course no 50 1 50 50 units what you have okay so therefore change in X is equal to 50 three times change in B this is what you can clearly understand so for unit change in B ultimately change in X is 50 units simple definition of differentiation it's even a previous year get question okay so 50 units is the correct answer clear did you all understand why it's C so let's check it up in the poll so C so 50% of Alex has answered it correct in one not one seconds maybe but I think few more CES okay but uh I think uh they are after the time okay as I think because of typing Capital C your answer was not taken I think okay so so anyway let's see this is a one thing again then let's talk about few uh nice points here so you know ultimately uh first limit X tends to a f of x plus or minus maybe G of x if you have something like this this can be return into two separate limits limit X tends to a f of x plus or minus limit X tends to a g of X is what we have okay so that's something then second limit x ts to a f of x into G of x if you have the product of f of x and G of X then ultimately you can write this limit x ts to a f of x is what we have multiplied with limit X tends to a z of X okay so that's what we have then the third thing is I just want to give you and of course if you have a division here same thing holds of course limit X tends to a f of x by G of x if you have for example then this can be written as two separate limits XTS to a because ultimately it's the ratio of the functions at that value of x so G of X this is what these three points are very much common known to you of course there's nothing big deal but the fourth point is really a bit nice point if limit X tends to a f of x does not exist and limit X tends to a g of X does not exist okay maybe or you can say and or or and this does not exist then first point limit x ts to a f ofx into G of X may exist this product may exist may exist okay means if individually one of the limits is not existing but if you take the product of the two functions then that product will have may have some value actually okay I'll show you quickly example you all know limit X tends to Z sinx which is a very quite common example you know this is zero it exist okay the answer is zero then second you know limit x ts to 0 1 by X I have shown you clearly this does not exist okay so x ts to 0 1X X does not exist because I've sh you using the graph of yal 1 byx in the right neighborhood the value is plus infinity and in the left neighborhood the value is minus infinity the function discontinues at that point and ultimately this limit does not exist okay but if you check the product of these two functions but limit x ts to Z sin x into 1X X so if you take this then this is nothing but limit x ts to zero sin x by X and of course this value is one okay so that exist okay so that's why even individually some of the limits are creating problem or issue but when you multiply them or when you divide them then the problem gets resolved few times okay so that's how this is a very important clear and and coming to the concepts of differentiability I'd like to just give one point which we have dealt in last class of course the points of the points where mod f ofx for some polinomial function f ofx or some polinomial function you know basically some function f ofx you can say it can be polom it can be technomic anything so the points where mod f ofx is not differentiable is or basically are not differentiable are not differentiable are differentiable are the points are the points where f ofx is equal to Z where f ofx is equal to Z the reason why at all these points where f of x equal Z the function is not differentiable is because at the points where f ofx is equal to Z at all these points they generate sh pages okay and at any Shar page function is not differentiable you know because why at a Shar page function is not differentiable let's see if you have a function like this that's that it's generating a sharp page okay and after this it's changing and here also there's some change but exactly at this point it is generating a sharp so when when it's generating a sharped the left neighborhood if you take the tangent to this function so the tangent appears something like this okay so in the left neighborhood the tangent appears something like this so if this is my x- Axis okay so let's say this is my x axis in general then ultimately this tangent is making an angle more than 90 so slope is negative and whenever you go for the right neighborhood of course so in the right neighborhood what happens is this value is actually making a positive slope clear and ultimately these two slopes are not equal because one is negative and one is positive and of course they're non zero so clearly this at this sharp point in both the neighborhoods slopes will be different okay because of the orientation of the tangents so therefore at all these points because at all points where f of x is equal to Z the function produces tages produces edges of course okay so the function produces T Pages clear so this is a very important thing like you see mod x for example okay you take this function modex then ultimately f of x is equal to X okay so the point where F ofx equal to0 is nothing but x equal to0 so you know at mod x you know at x equal to Z mod X is not differentiable clear so example you know I like to write few functions example so mod X is not differentiable is not differentiable mod X is not differentiable at this function so the function inside the modulus if you call that F ofx so this is X is my f of x so F ofx equal to 0 that means X is equal to 0 and obviously we have proven this many times so second mod sinx if you take for example this is not differentiable not differentiable at the points at the points where sin x is equal to 0 okay so sin x is equal to 0 that is at x equal to 0 + - pi plus orus 2 pi okay and so on of course of course at all these points the function is not okay fine like you can see like if you for example plot this sin x sin mod x I'll just show you it's very easy you can interpret okay you know sin x always lies between minus 1 to 1 so mod sin x always lies between 0 to 1 okay so if you like if this lie between 0 to 1 then ultimately the graph comes like a loop that's it okay such that this value is y equal to 1 this value is y is equal to 1 X Y of course and clearly you can see at all these points this is pi this is 2 pi okay this is 0er then minus Pi - 2 pi at all these points it generates Shar page okay which is similar to this so if you just plot if I can maybe show in decimos because you have decimos so mod sinx if you plot you'll see sorry okay so mod sinx is what I'm plotting of course basically so you'll see this is the mod sinx function I have plotted so you can check it comes up like a bridge like this okay and all these Loops are bounded by this minus one and if you take any Loop for example you see here it's a sharp edge okay it's similar to the modx uh the loop what you have okay so in fact if you go to any Shar page you go to any Z it again generates a sharp edge so at all these sharp edges ultimately this function doesn't become defensible clear to all of you yeah is it understandable to all of you or not so please type in the chat box clear this point is very important because many times few questions actually come up again what about mod tanx mod tanx is like again yeah correct at the points where at X is equal to Z it generates sh page okay so if you want to see because in mod tanx what happens if you take the interval - P by2 to P by2 fromus by2 to 0 it is actually negative okay so if it gets flipped it's like symmetric about Y axis okay so it becomes symmetric about Y axis because obviously tan - x is- Tan x so the graph becomes symmetric about Y axis I'll show you maybe if you're interested because we have plotter so let's see okay so you can see this is mod tanx of course and if you make a close view so this is symmetric of course they go to Infinity but within the range if you see at all points where the value of function is zero then ultimately there's a sharp edge again okay so you can zoom in much more to see the clear sharped this part is almost like modx okay so that's what keep zooming in the things you'll keep understanding lot of uh nice things actually clear so we'll solve few questions again before we go to the mean value the or whatever so let's see come on solve this question at x equal to0 the function f x equal to mod sin 2 pix by L where X goes fromus infity to Infinity L is some constant which is positive is Dash continuous differentiable not continuous not definable come on think single Choice maybe some 90 seconds is what I'm giving faster a bit fast quick continuous and defensible not continuous and not defensible b d about a and C hello fine I ended the poll now let's see the correct answer is of course continuous but not defen continuous but not differentiable okay so you can see simple like you can I think many of you might be calculating the continuity of this function and all when we have this mod function mod function functions are always continuous function okay and the only place where it's not differentiable is basically the points where sin 2x is equal to 0 so at x equal to0 ultimately the function inside the modulus is z okay so that's why a is the correct answer so at you know basically if you take this function f of x is equal to mod sin 2 pix by L if you take this function this is continuous always because why it's continuous is continuous the reason why it's continuous is because whenever you take sign function sign function is always continuous okay the only thing it has is plusus sign variations now when you put a modulus it comes to be positive for all values of X that's what right there's no variation in the means there's no point weight baks actually but the function is smooth but instead of having negative and positive Alternatives it has continuous positives which I have just shown okay in Desmos now similarly if you can see this it is not differentiable at the points it is not differentiable at the points it is not differentiable at the points where at the points where sign 2 pi x by L is equal to 0 this implies 2 pi x by L should be equal to n Pi well of course n is equal to 0 Plus or - 1 plus or - 2 what whatever you have lot of integers then if you simplify this then pi and Pi gets cancelled so X is equal to NL by 2 these are the points if you if L is defined then at all these points where n is equal to again 0 + -1 plus -2 at all these points it is not differentiable okay and at n equal to 0 ultimately X is also zero so this is one point where this function is not differentiable clearly okay so therefore at X x equal to 0 this n uh at n equal to 0 there comes a point x equal to 0 and at that x equal to 0 this function is not differentiable clear so option A is the correct answer got it so yeah D would be reasonably the correct answer let's see 50% of you Pat has answered it on time and correctly I think yeah Pat has typed D yeah as you have typed maybe after the after the time so that's why you don't have we'll solve one more question actually so okay I'll give you this question homework so you figure out out of these functions which function is not continuous at xal to 3 okay so this question I'm giving you homework you please try it now we'll quickly move on to the mean value thems of okay so let's go to mean value theems so the first mean value theem which means basically what does the mean value tells you of course okay so if you look at a mean value theem it tells about the value of the function at some point okay the name itself mean value the value of x which is somewhere in between the limits can be calculated okay using this mean value theorems so let's see what is this me Val the and physically I'll explain you weight is much useful of course you have lot of instances first is L just mean value the likees main value DM okay so it is L in value here so let's see you calculate if you see what is like if f ofx is continuous if f ofx is continuous in a comma B closed interval and differentiable in open interval a comma B then there exist at least one value of x of X such that okay so there exist at least one value of x such that X belongs to open interval a comma B and f- of C is equal to F of B minus F of a / B minus a of course okay so this is what we have so f- of C is given by this so what is the physical meaning of this okay so fine you you know the formula good are happy but what does this formula actually tells you okay so that's what you have to understand see because Frankly Speaking formula wise it's only up to this point but if you know only this point there are questions in Gate examination where you cannot answer okay and even in many textbooks unfortunately you'll find only this but nothing else you'll find okay but let's understand what this actually tells you geometrically if you take this two if you take this curve okay so let's say you have this xaxis and you also have y axis okay I'm taking first quadrant so that uh General so X and Y let's say if you take this function so let's say you have some function like this a random function okay so whenever have this y equal to F ofx of course this is y is equal to F ofx now let's take two points of X such that it all have this function so let's say this is my X for example in fact you can take somewhere anywhere in between doesn't matter I'm taking the ends so that I have some space to write the things so let's say this is X is equal to a and this is X is equal to B of course okay so this is xal to B and this is xal to a now if this is X is equal to a then ultimately this point which is here will give us F of a correct to this value is f of a similarly if this is B then ultimately this value is f of B just a minute maybe I'll okay so let's check it so we have this is the function y equal to F ofx of course so this is the function y is equal to f of x okay so this limit here is f of B actually now if that is f of B and this function is yal f of x this is xal a f of a xal b f of B simultaneously then if you call this as point a for example if you call this point a and if you call this point B General then if I want to write the coordinates of this point a ultimately the x coordinate of point a is a and y coordinate is f of a correct similarly if I want to add the coordinates here B comma F of B will you all agree yes or no please type in the chat box will you all agree with this if this point is capital A a comma F of a and that point is capital B then the coordinates of that point B is small B comma F of small B right because this is the function yal f of x two yes or no please type in the chat box correct now if I connect this two with the help of a straight line let's say I'm connecting these two guys with the help of a straight line okay so I've have connected this as with this the help of a straight line actually so can someone tell me if I calculate slope of ab what would be the value of slope of ab actually what is the value of slope of ab you know a a b is the line segment passing through the points a which has this coordinates small a comma F of small A and B has the coordinates small B comma F of small B so if there's a line segment which is passing through A and B then ultimately slope of this line AB is nothing but the right hand side of this equation correct yes because slope of a is Y2 - y1 y X2 minus X1 is what you have correct so the slope of this line is nothing but the right hand side of this equation two good all of you so please type in the chat box is this clear to all of you yes or no cor now let's see what is f- of C of course you know I have given you one point if you have any function f ofx and if you calculate f- of a this f- of a is nothing but slope of tangent to the fun function at the point xal to a correct okay so let's see if you calculate f- of a of course then T slope of tangent to the curve slope of tangent to curve to the curve at X is equal to C is f Das of C of course okay so this is is f- of C because ultimately if you take any point C then ultimately the slope of the tangent to the function at that point C is nothing but f- of C of course okay now if these two are equal when slopes of two lines are equal what does the let's say for example this is the tangent and a is a line segment you know if slope of ab and slope of tangent are equal that means what is the relation between those two what's the relation between those two if slope of tangent and slope of the line segment are equal that means what's the connection between those two line connection between the line segment and the tangent connection of connection between these two both lines are parallel yes correct okay they are parallel to each other which means whenever you have a curve which is continuous and defensible like this at least at one point at least at least at one point you will have a point where the tangent will be parallel to the line maybe you can have two you can have two you can have three as many you can have but at least there will be one point like for example you see here here at this point the tangent is parallel okay this line ab and this tangent both are parallel and this point is C of course okay so this point is of course C is what clear so this is two fine okay so let's see this point is xal to a and this is xal to B and this point is C so that's what mean value tells you whenever you have a continuous curve such that it's continuous and Def whenever you have a curve it's continuous and defensible within two limits at least between those two points at least there will be one point such that the slope of the tangent will be will be parall to the line connecting both the points actually and this you can actually you know use it many places like I'll give you one physical example okay so let's say you have a road for example okay so you have road I'll just tell you one thing nowadays you know there are something called speed guns okay you know there's a speed gun of course which measures the speed of the vehicle of course so what technology that our government uses in general is they'll put one speed gun here and maybe speed guns here they'll put individual speed guns okay and they'll check let's say this is the road and vehicle is moving and let's say on this road the speed limit is 80 km/ hour okay let's say this is 80 km per hour then if or 60 km per hour is the speed limit let's take the calculations is easy so 60 km/ hour is the speed limit then what they do they'll keep a speed gun here and they'll observe whether any vehicle is coming with a speed more than 60 okay and people citizens are good enough they are clever Enough by the time they they know there's a speed gun here by the time they come here they'll slow the vehicle and they'll make sure the speed of the vehicle is below 60 so that this gun do not capture this vehicle right similarly in between they drive it very fast again and by the time they come here again they'll drive slow so that they don't get caught between these two speed guns yes or no this is what people do generally okay so if you have a speed gun here then ultimately when you come close to the speed gun then ultimately you decrease the speed of the vehicle so that you don't get caught in the gun similarly the same thing happens here instead of this is the technique which is followed by you know in general traffic police and all but in general rather than doing this technique if they follow a different technique 90% of the population get fined okay you know what's the technique I'll tell you look if this is 60 km per hour and let's say the distance between these two is 3 km let's say in a 3 km Gap they have put a speed gun okay two speed guns they have put instead of putting the speed guns forget about the speed gun business just put two cameras okay now let's say you track a vehicle number here for for example let's say one vehicle is coming so you track a vehicle number so let's say for example the vehicle number is you know something which you know you have some vehicle number say let's say for example okay I love K so much so a some XX g s XX XX so let's say there's some number like this now whenever you have this number you know the classical technique they initially check only at this points okay they check the speeds at this point and also they check the speed at this point and if if at both the points your speed is slow they'll leave you but instead of doing this if they do this let's say this is 3 km this is 60 km/ hour then ultimately the minimum time minimum time you need to cross from this point A to point B if you follow the speed limit is 3 minutes correct means you cannot once let's say for example by the time you cross this if your time t equal to Z then the minimum possible time that you need is to come here if you follow the speed limit is 3 minutes correct because the distance is 3 km but if you calculate this thing like at this point t equal to 0 and at this point x is equal to 0 and let's say x is equal to 3 km at this point and T is equal to maybe some T seconds or t minutes let's say for example you calculate in minutes no so let's say t minutes if you plot distance versus time graph okay you will find out one thing average velocity you should follow in order to be within the speed limit is 1 kilm per hour okay 1 km per hour now let's say instead of 3 minutes you have reached this point in 2.5 minutes for example okay so let's say in reality you have reached much faster than the minimum time 2.5 minutes is what you have done okay in 2.5 minutes you reached this point but at both these points your speed is less than 60 let's say okay at both these points your speed is less than 60 so that you have actually not you know find because you didn't get caught in both of the speed guns but you have reached from this point a to this point B at 2.5 minutes okay which means definitely you might have done one thing if you have reached this in 2.5 minutes so you know this is X and this is T of course so if you plot this point initially at time t equal to Z your distance is z at time T = to 2.5 your distance is 3 km so 2.5 comma 3 so these are the two points definitely on your traveling because initially at time t equal Z let's say that's starting point within 2.5 minutes you reach 3 km if you connect these two points okay then slope is nothing but 3 / 2.5 so kilom divided by minutes so if you convert this 30 or 3 by 2.5 if you convert this to minutes so 1 hour is 60 minutes so 120 180 by 2.5 of course okay so 180 by 2.5 rly 75 I guess yeah 72 okay so this is 72 which means this is 72 km per hour means if you have definitely made these two points definitely at least at one point in your trajectory your speed is 72 km/ hour yes or no correct definitely would have traveled somewhere in between at a speed more than 60 only because of that you have reached at some Less Than 3 minutes only 2.5 minutes okay if you apply this mean value theorem because since your vehicle is satisfying these two points and you know your vehicle has traveled obviously at every time let's say then ultimately at least at one point you have crossed the speed limit of 60 km/ hour on this limit yes or no correct did you get this at least at one point between those two points your speed of the vehicle is 72 km/ hour 100% at least at one point Definitely Maybe you have traveled at many other points also at 72 km Maybe 72 then 30 72 30 you might have done this also but definitely at least at one point you might have traveled at a speed which is equal to 72 km/ hour which is more than 60 that means you have violated the traffic C clear but people don't use luckily they use the spotting Just Between the speed gun okay so if you don't have the speed guns then ultimately still you can catch lot of people because you T the vehicle number here and you taste the vehicle number here the same vehicle and you see the time difference between these two spots distance is fixed of course you know let's say if I keep my cameras between say 5 km down the line then ultimately you'll measure the you'll look the vehicle here and you'll also look for the same vehicle number here and if this time defense is less than this value then ultimately you can find the vehicle undoubtedly okay because you can show them the proof which means if you have crossed these two points within a very less time which means you have definitely viated the speed on the RO okay this is what so over speeding can be find understood and velocity is nothing but rate of change of displacement so V is equal to d s by DT which is nothing but f- of X at any given time if you it is as X DX by DT and derivative with respect to time is what you're taking so f- of T is what you're looking at Clear did you understand this simple story yes or no did you understand what we mean by this leg mean value the clear so basically the main is at least at one point the rate of change is equal to the slope of the tangent okay fine so is this clear to all of you okay are you all understanding this or not okay fine so uh what we have done so we have done this uh leg me value them of course that's what we have seen then uh okay maybe the time is up for today only 5 minutes is left okay we'll continue with the Min Val thems in tomorrow's class no issue but I have given you one question homework today so please work out this question okay we'll deal in maybe Monday's class okay so which of the following function is continuous at xal to 3 so this is what we have so has anyone this practically no luckily none of them has done this okay that's why all of us are safe many of them many of us are safe but if you do that the fins will take a huge boost the economy of the traffic okay but anyway H so uh but not totally see just go through these things once again okay it gives you much more clarity okay and uh coming to tomorrow at 7:00 p.m. tomorrow we have the doubt revision class okay so on Sunday at 7:00 p.m. we have the doubt revision class okay doubts and division class of course you can ask all your doubts whatever you have maybe in the maths or whatever whatever have taught or maybe in some other generalized doubts regarding gate examination or things like some of you have asked how to balance this college and G exam preparation both simultaneously or maybe college or maybe job and this gate examination so all these questions will answer in a much detailed form okay so tomorrow at 7: p.m. that's on Sunday we'll have a doubt solving okay so tomorrow so anyway so let's quickly get into the things what we have done so we talked about you know limits again we give a bit of introduction about the mean value the okay lmvt of course L man just mean value them is what we have done okay and I have given you one question homework of course today so please solve this question as your homework and then yeah we'll again continue on Monday of course okay and of course tomorrow also at 7: p.m. we have the doubt solving session so please do attend without any fail and thank you for today you have submitted your Google form d right yeah it's on YouTube out class is on YouTube okay on YouTube the same space the same location so next time please add space for uploading images as it will be easier that way yeah sure see you can actually add images you know basically if you click that you're going to a batch okay so in that batch if you have any doubt the slides are divided so you can click on the particular slide and you can ask the doubt the class was awesome thank you fine so is each and everything clear to all of you okay so if you have any doubts you can ask me or else it's fine we'll call it today and we'll close thanks for hearts yeah thank you all then okay we'll meet tomorrow yeah of course we'll meet talk about some doubts okay yeah yeah please Alex please tell me I'm always open to sessions okay you can always tell me because ultimately the main purpose of me teaching is to transfer some knowledge and that has to be done at the maximum possible best level okay Max possible efficient yeah good night Alex has some doubt so after that we'll clear okay so rest of you can leave but please practice as question concept wise yeah that's what I told you basically we'll I'm planning to post one document for practice on my telegram Channel very soon okay so you can practice that I'll give you questions and just answer not the solution but I'll just give the answer so that you can try and Sol if you cannot solve you can discuss in the gole okay that's a doubt right yeah no we'll update so then thank you all we'll meet tomorrow at 7 p.m. for the doubt solving session thank you