 All right, so now I'm assuming the screen is visible to all of you. Hello. Now it should be visible. Yeah, so here's the next question So you have a function which is defined as a limit of a summation X by r x plus one by r plus one x plus one Now we have to comment about the continuity and differentiability of this function at zero So I think this is a 21st question Yeah You guys first focus on getting the function then we'll talk about its continuity and differentiability So first ensure that you have found out the function f of x I think it looks to be a clear-cut case of The VN method of summation if I'm not wrong Because I can get a feel from r x plus one and r plus one x plus one So try it out in a similar line all right, so if you look at this term x by Rx plus one and R plus one x plus one You can actually write your numerator itself as r plus one x plus one minus Rx plus one which actually becomes one by rx plus one minus one by R plus one x plus one So this is you can call it as your rth term in fact r plus one-th term and now starts Applying the VN method on it So you realize when you apply the VN method on it the summation on it from r equal to zero till n minus one will end up giving you when this is Zero this will be one and this will become one by nx plus one Okay, so this will be a summation result. I should not write this over here anymore Okay, so this will be a summation. So this will be your function. This will be your f of x Okay, now still the limit is operating on it as n tends to infinity Now if you're trying to identify whether the function is continuous and differentiable at zero Let's first find what is f of zero So f of zero will be one no doubt Right because the moment you put this zero it becomes one Sorry it becomes zero Let's become zero. Okay, so f of zero is zero now. Let us find out the right hand limit So let us find limit of the function as x tends to zero plus When x tends to zero plus basically you are trying to find out the limit of a limit of One minus nx plus one correct so one minus one by nx plus one as n tends to infinity what will happen this term will become huge correct so limit of limit of f of x as x tends to zero plus Right, what will happen this term? This term will become huge now many people say this term is actually Indeterminate so how do you say that it's actually huge? What we do is we take a fixed value of x If it is not tending zero plus means you are taking a very fixed value of x and n is tending to infinity Okay, so this is a fixed small value of x so this term will become very very large So your answer is going to be one minus zero which is going to be one so be careful about these kind of problems because When n is tending to infinity and x is tending to zero plus There's always a element of confusion That nx is actually tending to zero into infinity so this should be an indeterminate form Right, so how are we sure sure that this term will be going to infinity? So when you are taking a very small value to the right hand of x and your n is tending to infinity I'm considering x to be a fixed small value not a value which is tending to okay, and therefore this Expression is going to tend to infinity and so this is going to become zero the entire thing is going to become zero So this is going to become a zero value And your answer is going to be one minus zero which is one So right-hand limit itself doesn't match with the value of the function at zero Okay, hence the function is not continuous and if it is not continuous it cannot be differentiable also So option number B is sorry option number three is going to be correct It is neither continuous nor differentiable at x equal to zero So let's move on to the last problem of the day Which is your question number 22? So it's given that the function is differentiable at x equal to one Find the value of mod a plus p again. Look at the function. How it is defined here You have the n value tending to zero Sorry, it's tending to infinity make it this change. This is tending to infinity. This should not be zero This should be infinity guys. Please make this change This should be infinity over here. Okay, please be careful Okay, so in the interest of time, I'll be solving this because we are almost about to Finish the class. So remember This function is defined for two intervals of x, right? So if I talk about this interval when x is between zero and one, let's say I talk about this interval, okay? If I evaluate this limit guys, I think there is a small mistake in the problem Just give me one second The brackets here are not written properly Let me just re-correct the question here Because if this is tending to n what will happen? It'll create a problem Okay, so this is whole raise to the power of n. Just forget this question. I'm just rewriting it limit n tending to infinity ax by x minus 1 And this is multiplied to cot Pi x by 4 whole raise to the power n plus Px square plus 2 divided by Cot pi x by 4 whole to the power n plus 1 Okay, this is just multiplied to this and there's a power of n over here So just forget about this question. There has been a big typo error in this Right now when I'm talking about the interval zero to one If I talk about this interval x lying between zero to one, you know that cot Pi by 4 cot x pi by 4 Okay This expression when you're slightly less than one this expression will be greater than one Right, so this term will become huge this term and this term will become huge correct Right, so if I divide both the numerator and denominator by cot pi x by 4 to the power of n What will happen or I can take this as common for both numerator and denominator So in the interval zero to one, I'm just going to Take not get to see the screen. Yeah So if I take this term as common Okay, so this term will become zero This term will become zero Okay, and ultimately what will happen is this will get cancelled with the denominator term Now is it visible? Yeah, so you'll be just getting a x by x minus one So moral of the story is when you're between zero to one your function will just give you a x by x minus one Okay In a similar way when you are between one and two, so let me just erase this off Okay, so the idea is your function will start behaving as a x x minus one when your x is between zero and one Okay Similarly, when your x is between one and two your this term will become zero correct So your function will start behaving as px square plus two Okay, and exactly at one it is zero Now we need to work with this function now this function is differentiable at one Which means f of x must be continuous at one so f of x is Continuous at x equal to one if it is continuous at x equal to one That means left-hand limit and right-hand limit should be equal to zero So left-hand limit is going to be zero itself. That is fine a right-hand limit is going to be p plus two So p plus two is going to be zero Which means p is going to be minus of two correct Now they also say that it is differentiable at one. So let us find f dash x Okay, so f dash x if I'm not wrong It'll give you a times two x minus one for x lying between zero to one and two px when x lies between one and two and Zero when x is one In fact, I don't need the value of the function at zero okay So differentiable at one means left-hand derivative and right-hand derivative at once should match so two into one A into two minus one should be equal to two p So a should be equal to two p if a is equal to two p means a has to be minus of four So a plus p mod will be a plus p mod will be mod of minus of six which is going to be Six itself, which is option number two is correct. All right guys We have 34 questions in this particular slide So I'll be sending the rest of you as homework along with the Assignments on this particular chapter. So over and out from Centrum Academy. Thank you so much for coming live today Bye. Bye. Have a good night