 Now I will be introducing some typical questions, typical functions which are very commonly asked in JEE as well, okay. So let us begin with this question, I am going to just clear the screen off, okay. So let's start with this question, let f of x, let f of x be a function defined as max, max of 4 sin x and 4 minus 2 sin x, okay, for x lying between 0 to 2 pi. Now the question is find the area, find the area which is bounded by f of x graph, the y axis graph, the x axis graph and x equal to 2 pi lying, find the area bounded between these four functions y equal to f of x, y axis means x equal to 0, x axis means y is equal to 0 and x equal to 2 pi. So I am sure you would have heard of max function and all, yes or no? So I will quickly recap also for those who are not aware of max function, max and min functions are special functions which are actually graphically understood very well as compared to mathematical way of understanding them. So I will just quickly tell you what is max function, meanwhile those who are solving it please avoid looking at the screen and do it on your notebook. So when I say, let me just draw these two functions on the GeoGebra graph that is 4 sin x and 4 minus 2 sin x, so y is equal to 4 sin x, so here is the graph in front of you and y is equal to 4 minus 2 sin x, 4 minus 2 sin x and I am just going to draw the graph between the y axis and x equal to 2 pi. So I just quickly mark the graph which I am most interested in in this particular question. I am interested only in this part of the graph which is from x equal to 0 till x equal to 2 pi, this is your x equal to 2 pi. Now what is the meaning of max? See when I say max means you need to choose the higher of the two for a given value of x, let's say I say x is 0, so this will give me 4 into sin 0 and this will give me 4 minus 2 sin 0, so this will obviously give me 0 and this will give me 4. So higher of the two is 4, so the value that the function will take at 0 will be 4. If I say what is the value of the function at pi by 2, so first you will evaluate both of them and take the maximum of the two, for example 4 sin pi by 2 and 4 minus 2 sin pi by 2. Now clearly it will give you 4 over here and 2 over here which is the maximum of the two, you will say 4, so value of the function at pi by 2 is going to be 4. So when we talk about the same thing from a graphical perspective, we normally take that part, so if you see this graph we are going to actually talk about the part of the graph which is the highest part of the graph, which is this part and we are going to find out the area under this curve. Now first of all I would request all of you to find out the coordinates of this point, so that we are able to see what is my limit of integration. So all of you try this out, then I will discuss with you the points, so what is going to be the coordinate of this point? Sin inverse 2 by 3. Right, so this point will have x coordinate of sin inverse 2 by 3. What about the x coordinate of this point? You can say from the symmetry of the figure it will be pi minus sin inverse of 2 by 3, so what is the area expression that you can write? Anybody area expression, what should I write? So 4 minus 2 sin x from 0 to sin inverse 2 by 3 plus 4 sin x from sin inverse 2 by 3 to pi minus sin inverse 2 by 3 plus 4 sin x from sin inverse 2 by 3 to pi minus sin inverse 2 by 3, sin plus of 4 minus 2 sin x from pi minus sin inverse 2 by 3 to pi. Alright, now the expression that you would be writing here would be integration from 0 to sin inverse 2 by 3 of which curve, of which curve, which is this curve? 4 minus sin 2 x. 4 minus 2 sin x correct, so it is 4 minus 2 sin x. Yeah, 2 sin x sorry. Okay, then which is this curve? It is 4 sin x, yes or no? So there the expression would now become sin inverse 2 by 3 to pi minus sin inverse 2 by 3 integral of 4 sin x plus what is this curve now? Again, we are back to 4 minus 2 sin x from pi minus sin inverse 2 by 3 till 2 pi correct. So let's write that expression now, so pi minus sin inverse 2 by 3 to 2 pi of again 4 minus sin 2 x. Okay, so if you evaluate this expression, your answer would come out to be 4 pi plus 8 sin inverse 2 by 3 plus 4 root 5. I am assuming you would have already evaluated this expression. Yeah, so just draw GIF of x plus GIF of y equal to 1 first. We will take care of the mod later on. So how would GIF of x plus GIF of y equal to 1 look like just in the first quadrant? So just draw this in the first quadrant. So when you think will this, these are all integers, right? This will be an integer. This will be an integer. So what is the possibility of you getting an integer plus an integer equal to 1 in the first quadrant? What are the various options I have? This can be 0, this can be 1, correct? Or 1 or 0. And can be 1 or 0. There is no other possibility. So now if I draw this in the first quadrant, x is 0 and y is 1, when can that happen? 1, 2, 3, 1, 2, 3. x is 0, y is 1. Greatest integer x is 0, y is 1 can happen when you are in this box. Yes or no? Right? In this box, your GIF of x will be 0 and GIF of y would be equal to 1. And also in this block where your GIF of x would be 1 and GIF of y would be 0. Is that fine? What would be the effect of modding x and y? So if I mod it now, what will happen? Can I say modding x will just reflect the same thing about the y axis? So this area is also included, this area is also included. And modding y will include, reflect everything about the x axis. So this is also included, this is also included. Similarly, this would also be included and this would also be included. Yes or no? So what would be the total area? Sir, can you explain it again? How does it happen? The box and all? We were first drawing it only in the first quadrant. So in the first quadrant, we figured out in which box my x coordinate GIF of x will give me 0 and GIF of y will give me 1. Okay. So that happened in this box. So I am just shading the box for you. So that happened in this box. Now, in which square box do you see mod x becoming 1 and mod y, sorry, GIF of x becoming 1 and GIF of y becoming 0? That happens in this box. These two are the boxes that I obtained in the first quadrant. Okay. Now I have modded my x first. Modding x means when you do, when you replace your x in a function with mod x, what happens? What happens? What is the step which we follow while making this transformation? It's reflected about an inverse. Right. So we reflect it about y axis. Yeah. So these two got reflected and we got these two. We got these two. Correct? Now, what happens when you mod y? Whatever was above will get reflected below. Yeah. So we got this box, this box, this box and this box. Correct? So altogether, my total area would be the sum of these boxes which are nothing but 8 in number. So 8 square units. Is it clear? Is that clear? Is the approach clear?