 Now, let us take some questions which involve mod, can I move on to the next question anything that you want to copy from here please do that because I am going to clear the screen. Next question is find the area bounded by y is equal to 3 minus mod of 3 minus x and y is equal to 6 by mod x plus 1. Okay, now I will be drawing the curve but first of all you try on your respective notebooks and then we will be discussing it. Are you all done with the graph? Yes sir. Okay, so I was drawing on the screen the graph of this function first. So what I just now drew in front of you was the graph of y is equal to mod of 3 minus x. There is no doubt about this graph so please note this point, this point would be 3, 0 right? This is the normal modulus graph shifted 3 units to the right, yes? Now what is the effect of putting a minus sign over here? So what will happen if I put a minus sign over it? You would say the graph would become the mirror image of itself about the x axis, correct? Yes or no? Because here you are changing the sign of y right? This is like saying minus y is equal to mod of 3 minus x. So the graph would appear to be a reflection of the black one which is I am showing by a red one like this. And this point will be now 0, minus 3, correct? Now what is the effect of adding a 3 over here? What will happen if I add a 3 to this? Can I say the entire graph will now shift up by 3? Yeah. So can I say this entire scenario would appear to be like this? Yes or no? Where would this point go? This point will go at 3, 3 if I am not wrong. This point will start passing through origin if I am not wrong. Have I made myself clear here? So there is absolutely no doubt about the graph of y is equal to 3 minus mod of 3 minus x. Yes or no? Yes sir. Now what about the graph of this fellow? So first I made the graph of y is equal to 6 by x plus 1. Remember this that this is going to be a graph which is derived from the graph of a rectangular hyperbola which is actually of the nature y is equal to 1 by x. Okay. So how does the graph of y equal to 1 by x look like? It's a rectangular hyperbola like this, isn't it? Yes or no? Now if I multiply it with 6, it's just that it is going to change its curvature a bit. However the position of these two parts or these two arms of the rectangular hyperbola are not going to change. Now what is the effect of adding a 1 to the x? What is the effect of adding a 1 to the x? You would say the graph is going to shift to 1 units to the left, isn't it? So now this will become your x equal to minus 1 and it will be asymptotic to this line. Yes or no? Now the reason why I am not drawing the negative side of the graph is because what will happen is I am going to mod this first of all. That means I am modding the entire thing. Yes or no? It is going to give me another part of the graph which is of this nature which is not of requirement to me so I will not draw that. But I am definitely going to make this modulus function over here like this. So what I have done, I have drawn this graph now in the main graph. So which area am I looking at? I am looking at now this area. Yes or no? Yes or no? So now I would first require the coordinates of this point let me call it as p and this point let me call it as q. And this point coordinate is already known to us which is 3 comma 3. Okay. Now who will tell me the coordinates of p or at least tell me for solving the coordinates of p which two curves I need to solve simultaneously? Is it y is equal to 6 by x plus 1 and y is equal to x? Yeah. So what is this line? What is this line? Y is equal to x. Y is equal to x. No doubt about it. Correct. And what is this curve? What is this curve? Y is equal to? Y is equal to? 6 by x plus 1. 6 by x plus 1. Correct. So I am going to simultaneously solve both of them that means I am going to solve this equation which is going to give me x into x plus 1 equal to 6. Now I can be smart over here. Realizing that 6 is 2 into 3 my x will be 2. No doubt about it. Because this is just two such numbers which are separated by one from each other. So a good guess would be x is 2. Yes or no? So at this point my x is going to be 2. Right? Yes or no? Now what about the coordinates of point q? For getting q which two curves I need to solve? 2 6 minus x. Y is equal to? 6 minus x. Very good. And y is equal to 6 by x plus 1. So let us solve this simultaneously. So 6 minus x equated to 6 by x plus 1. Take it to the other side. So 6 minus x times x plus 1 is going to be 6. Correct. So if I open the brackets I am going to get 6x minus x which is 5x. Minus x square equal to 0 which is clearly giving me x as 0 and 5. Of course it will not be 0 but it can be 5. Now make no mistakes about getting the coordinates of these points because let me tell you friends they are going to decide your limits of integration. Is that clear? Now how many types of strips will I be requiring to cover up this entire area? How many types of strips will you be requiring to cover up this entire area? My personal call is I would need two types of strips. One is this type and another would be of this type. Yes or no? Why? Because see when I take this strip this strip upper end is on the line y equal to x and lower end is on the line on the curve 6 by x plus 1. But if I focus on the upper part of this strip it is lying on the line 6 minus x and the lower part is lying on 6 by x plus 1. So it is a different curve. Exactly so one type of curve is not going to suffice the entire area. Yes or no? So now if I have to write down the expression for that area correct me if I am wrong it would be integral from 2 to 3 x minus 6 by x plus 1 and this would give me the area of this first type of strip. Upper is x lower is 6 by x plus 1 upper minus lower, upper minus lower correct and I am going from 2 to 3 this is x equal to 3. Now from 3 to 5 it would be 6 minus x minus 6 by x plus 1. Yes or no? Now these are very simple integrals to evaluate. Just complete this and check whether you are getting this answer. I am just going to write down the answer for you. It should come out to be 13 by 2 minus 6 ln 2 square units. I will give you just 2 minutes quickly just check whether you are getting this answer or not. Are you getting that? Excellent. Can I move on to the next question? See here basically we are going to solve a lot and a lot of questions. So let me go to the next one. Sir before that can I have a small break? Yeah sure you can take a small break right now. Can we resume after just 5 minutes? Yeah okay sir. Okay. Yes sir. All of you are ready now? All right. Add a good break. Yeah. Okay all right fine. So now let us take up some questions which are based on different types of functions. We have mostly covered log function, exponential functions, polynomial functions.