 Hello, good afternoon. Welcome to the YouTube live session. So those who are attending the session, I would request you to type in your names in the chat box so that I know who all are attending the session. So before we start, I would just like to set the agenda for today's session. It would be a mixed back problem solving from various topics, OK? And as we have been doing so far, I'll be waiting for a response to come from your side. At least three of you should respond before I start solving the problem. There may be some disturbance with respect to the sound because of some internet issue, but I'm sure you would be able to see the screen properly. That's not very convincing. Let's go ahead and get to them. Yeah. All right, so let's start with the first problem. f of x satisfies the requirement of Langranger's mean value theorem in the interval 0 to 2. And it's given that f0 is 0. And f dash x is less than or equal to half. For all x in the interval 0 to 2, then which of the following options are correct, OK? So Atme, Shohan, or it's mostly people are saying option B. So basically, this is the application of Langranger's mean value theorem. And let's say for a value of c lying between 0 to x, if I apply a Langranger's mean value theorem, then it'll be f dash c is equal to f of x minus f of 0 by x. Correct? Now since x is less than 2, in fact, c will also lie between 0 to 2, OK? So from this expression, I can say that f of x is equal to x times f dash c. So mod of f of x would be mod of x mod of f dash c, right? And it has been given that mod of f dash x for all x in the interval 0 to 2 is less than equal to half. So even this term would be less than equal to half, right? So I can say this entire expression will be less than equal to mod x into half. And the biggest value of x is going to be 2. So this expression will be further less than equal to 2 into half, which is equal to 1. So f dash x mod will be less than equal to 1. So option number b is the right option over here. And the first one to answer this correctly was Atmesh, OK? So now we'll move on to the next question. Hope there's no doubt regarding this. The value of f of 0 so that f of x given by this expression is continuous everywhere. And you may have seen this question in your preparation of board exams as well. It's a commonly asked question in continuity. Should I take more than one minute to solve this? Let's say this is question number 2. Please type the question number along with the response. Done? OK, so Amogh has given the response, OK? So if it has to be continuous everywhere, it has to be continuous at 0. So limit of this function as x tends to 0. So it should match with the value of the function at 0, OK? Now here I see a combination of three different types of limits. One is your limit of a to the power x minus 1 by x extending to 0. That's log of a to the base e. And you have limit of sine x by x extending to 0. That's going to be 1. And limit of log of 1 plus x by x that's going to be 1 again. So if I make these manipulations in this problem, I would get 4 to the power x minus 1 by x whole cube. In a similar way, I would divide this by x and I would divide this by x squared, OK? Now I need a 3 over here, so I'll multiply, divide with 3. I need a 4 over here, correct? So I'll multiply, divide with 4. Is that fine? OK, this 4 will go up, I guess. This 4 and this 3 will go up. Because you're dividing x by 4 to the denominator term, so this is as good as multiplying with 4 by x, OK? So this term here would become ln of 4 whole cube and is a factor of 12 coming up from here. And rest all these terms will become 1 and 1. That means your option number c will become the right option in this case. Is that fine? No question with respect to this? Let's move on to the next question. The next question says the number of ordered pairs alpha, comma, beta, where alpha and beta belong to the interval minus pi to pi. Satisfying these two equations, cos alpha minus beta is 1 and cos alpha plus beta is 1 by e, OK? How about others? So only Atmesh thinks it's slightly different. So if you look into this condition, first of all, cos of alpha minus beta equal to 1 implies only possibilities alpha minus beta has to be 0. That means alpha has to be equal to beta. And secondly, cos of alpha plus beta, which is nothing but cos of 2 alpha is equal to 1 by e. Now when you say cos of alpha is 1 by e, in what quadrant can your 2 alpha lie? Remember, your 2 alpha can actually lie between minus 2 pi to 2 pi, correct? If you're saying minus 2 pi to 2 pi, that means you have an option to take the first quadrant. Let's say I'm going from 0 to 2 pi. So these two are selected, correct? And if I'm going in the reverse direction also, I can select these two angles, right? That means there are four values of alpha that you can actually get, which implies there will be four values of 2 alpha, so there will be four values of alpha as well. Yeah, that's fine. It's in the first and the third quadrant. But your interval is from minus 2 pi to 2 pi. So altogether, you can actually take two complete terms. So you will get the same value for cos for different angles of alpha. And hence, the number of ordered pairs possible will be 4 in number. Is that fine? Now had alpha and beta been different quantities, then you would have got 8 as your answer. But in this case, alpha and beta are equal, so there's no point stopping the position. So whoever said d, I think the first person to answer this was Rohan. So d is the right answer in this case. Yeah, moving to the next question, which is from your mathematical reasoning, the statement negation of p or q or negation of p and q is logically equivalent to which of the following? All right, so I'm getting the response as b for most of you. Now this is as good as saying negation p and negation q. Or negation p and q. Now here I can write this as negation of p. We can take that as common. And negation q or q, correct? Now this is actually a tautology. This will always be giving you true as the response. So the response of this is going to be whatever is the response of negation p. So option b become the right option in this case. Absolutely correct. So most of you have answered this correctly. Shruti was the first person to answer this. Moving to the next question, if set a contains all those angles theta says that 2 cos square theta plus sine theta is less than equal to 2. And b contains all those values of theta, which is between 5 by 2 to 3, 5 by 2. Then we have to find the intersection of set a and b. This is your question number 5, let's say. OK, so I'm getting response from most of you as option c. All right, so let's check this out. So this I can, 2, 1 minus sine square theta plus sine theta is less than equal to 2, OK? So this can be written as 2 sine square theta minus sine theta is greater than equal to 0. That means we can write this as sine theta times 2 sine theta minus 1 greater than equal to 0. So if I make my wavy curve for this, my sine theta can either lie in this interval or this interval. Or you can even take cases. For example, you can take a case where sine theta is positive and 2 sine theta minus 1 is also positive, right? So in this case, your sine theta will be greater than equal to half, which implies your theta should belong to the interval of greater than equal to half if you quickly make the sine curve graph. So greater than equal to half would be covering this zone from this value, pi by 6 to 5 pi by 6. But we are only allowed to start from pi by 2. So for this case, your answer will be from pi by 2 to 5 pi by 6. Second case is where you can take your sine theta to be negative and 2 sine theta minus 1 also to be negative. That is, sine theta is negative. And sine theta is less than equal to half. So that interval would be simply the interval between pi 2, 3 pi by 2, pi 2, 3 pi by 2. Because we cannot exceed 3 pi by 2 over here, OK? So from the second instance, we get the interval of theta to be between pi and 3 pi by 2, OK? Because in this interval, my sine theta is negative, that is less than equal to 0, as well as less than equal to half. So this basically becomes a redundant condition, OK? So this is the dominating condition in this, just like this was the dominating condition over here. So in these two conditions, I get these two as my interval. So it can either be this or this. So option C is the one which discusses about these two limits of theta. So C option is correct. Let's move on to the sixth question now. The coefficient of x to the power 50 in 1 plus x to the power 41 times 1 minus x plus x square to the power of 40, all right? So Vaishnavi has given a response, OK? Go ahead and back up with the same answer. All right, so mostly, Janta is going with option A. Let's check. So this term can actually be written as 1 plus x times 1 plus x to the power 40, and 1 minus x plus x square to the power 40. So this becomes 1 plus x times 1 minus x plus x square whole to the power of 40. This is actually the formula of A plus B times A square minus AB plus B square, which we know that it's going to be 1 plus x cubed, OK? Now when you expand such a binomial term, you would realize that you get terms of this nature, OK? So all the powers that you would see over here, they would be multiples of 3 over here. So this will all contain multiples of 3 power, multiples of 3, OK? And here you have 1 plus x. That means you can either get a multiple of, you can get an expression with a multiple of 3, or a term which is of the nature or power of x can be, let me write it in this way. So your terms could be of the nature x to the power 3n or x to the power 3n plus 1, right? And none of them can be equal to 50. This will not give me a whole number value of n. Neither this will give me a whole number value of n. That means there is no such term, OK? That would contain x to the power 50. So your answer is going to be 0 in this case. Is that fine? Good. So the first one to answer this correctly was Vaishnavi. Next, just for a change, a simple log question. So log to the base 7 of under root of 7, under root of 7, under root of 7 is. All right, so mostly people are going for option number C. Let's quickly check this. This is like saying log to the base 7 of log to the base 7 of 7 to the power half plus 1 4th plus 1 8th, right? And this becomes actually 7 to the power of. If you add this, you'll get 4 by 8 plus 2 by 8 plus 1 by 8. That's equal to 7 by 8. So 7 to the power of 7 by 8, which gives you log of 7. And this will become 7 by 8. So I can write this as log of 7 to the base 7 minus log of 8 to the base 7, which is nothing but 1 minus. And this 8 can be done as 2 to the power 3. So it's 1 minus 3 log 2 to the base 7. Option number C becomes correct. A very easy question just to break the monotony. Next is evaluate this integral, where this represents the GIF function. And this is a normal bracket. This is the normal bracket. OK, so mostly people are going for option 8A. So I can write this as minus half to half GIF of x plus minus half to half ln of 1 plus x by 1 minus x. And we all know that this function is actually an odd function. So if this is an odd function, this becomes a 0. And GIF function changes its definition about 0. So from minus half to 0, it behave as negative 1. And from 0 to half, it will behave as 0. So this is just about integrating negative x from 0 to negative half. That's going to be 0 negative of negative of negative half. So that's going to be plus half. So that's going to be negative half as your answer. So absolutely option number A becomes the right option. So we'll move on to the next question now. This question says this is square OABC and is formed by the line pairs x, y equal to 0. And x, y plus 1 is equal to x plus y, where o is the origin. A circle with center C1 inside the square is drawn to touch the pair of lines x, y equal to 0. And another circle with center C2 and radius twice that of C1 is drawn to touch the circle o and the other line pair. So find the radius of the circle with center as C1. This is your question number 9. OK, Rohan. OK, you're getting 1 by 3, which is not in the options. I would just suggest recheck your calculation once again. No one else? Sure, sure, take your time. No hurry. OK, I need two more people to respond before I start solving this. Let's try to sketch the situation over here. Let's say I have these two circles. And this is enclosed within pair of lines. Now x, y equal to 0 is basically nothing but it's a pair of line which signifies your coordinate axis. So x, y equal to 0 signifies your x equal to 0 and y equal to 0. That's nothing but your y-axis and the x-axis, respectively. If you look at the other line, that is this line, this is like saying x, y minus x minus y plus 1 is equal to 0. This is x minus 1, y minus 1 equal to 0, which signifies x can be 1, y can be 1. So x equal to 1, y equal to 1. According to the given situation, it should be like this. So according to the statement mentioned in the question, one circle with center at c1 touches x, y equal to 0 pair. And other circle, the center has c2. And twice the radius touches the other circle. One thing is for sure, if they are touching like this, then the diagonal should be passing through the centers of these two circles. So this is, let's say, r. Then this is, let's say, 2r. And if I drop a perpendicular, this will also be 2r. And if I drop a perpendicular here, that is also going to be r. So let's say I call this point as o. Now, what is the diagonal length in this case? Diagonal length in this case is root of 2, correct? And this length is also equal to 2r. And this length is also equal to r. So this is actually 2 root 2r. And this is actually root 2r. So the total length from o to b, which is root 2, is actually root 2r plus 3r plus 2 root 2r. So this implies root 2 is equal to 3 root 2 plus 3r. So r is going to be root 2 divided by 3 plus 3 root 2. And that is given by option number c in this case. Now, let's talk about the next question. OK, so 10a. So it's given that f of x has a second-order derivative, f double dash x, which is equal to 6x minus 1. OK, so f dash x will be what? f dash x will be 3 times x minus 1 whole square plus, let's say, a constant c1, OK? Now, it is given that the graph has a tangent of y equal to 3x minus 5 through this point. That means f dash 2 is equal to 3, correct? That is the slope of the tangent at 2 is equal to 3. So this implies 3 times 2 minus 1 square plus c1 is equal to 3. That means c1 is equal to 0. So f dash x is going to be 3 times x minus 1 whole square only, OK? So f of x is going to be x minus 1 whole cube, plus, let's say, another constant. But it has been given that the curve is passing through 2 comma 1, right? So that means f of 2 is going to be 1, right? So 2 minus 1 cube plus c2 is going to be 1, which means 1 plus c2 is going to be 1, which means c2 is also equal to 0. So putting it back over here, this implies f of x is going to be x minus 1 whole cube. So option number A becomes the right option in this case again. So let's now talk about the 11th question. A natural number x is chosen at random from the first 100 natural numbers. Find the probability that this expression is less than 0. None of these, OK? Only Shruti thinks 7 by 25. All right, so let's discuss this. So if I go by the wavey curve sine scheme, so this is 20, 30, 40. So it will be plus, minus, plus, minus. So it will be negative when x is less than 20, or x lies between 30 and 40, correct? So when you say x is less than 20, means you can only choose natural numbers from 1 to 19. So x would belong from 1 to 19, OK? And this means x should belong from 31 to 39, OK? So I think there are 19 such numbers over here. And there are nine such numbers over here. So altogether there are 28 numbers which are going to meet this given inequality. And this becomes your number of favorable events. And sample space is just going to be 100 because you're choosing one number from 100. So your probability of this event is going to be 28 upon 100, which is 7 upon 25. So option C becomes the right option in this case. Is that fine? So the only person to give right answer in this case, no, many people gave. And the first person was Shruti. Shruti, Aditya, Ramcharan. Again, a simple question, but again, a small counting error can lead to wrong results in this case. Again, a simple question. Just like we had log question a little while ago, a simple one just to break the monotony. Yeah, so mostly people are going with option number B. That's actually a very simple one. This is going to be just r square minus r minus one whole square. So when you start putting the value of r as one and all, you start getting terms like this. So this will become a 2010 square minus 2009 square. And terms, alternately, will start getting canceled. So this will get canceled with the next term which is mod A to C, which is three square minus two square. And this will lead to cancellation of all the terms, okay, which is 2010 whole square. So option B is correct in this case. Next is question number 13, if Z is a non-real, IE by IE, let's get this one by IE. So this is an angle of a triangle for which of the following up, okay? All right, so first of all, Z minus one by IE should be completely real, right? Because sine inverse domain is minus one to one, so this should be completely real in nature, okay? And if I assume Z to be X plus IY, Z minus one by I will be like X minus one plus IY by IE. Okay? Which is nothing but X minus one by I plus Y and this should belong to real. If this has to be real, it implies this term should have existed. This term should have zero. That means X should be one. So it is clear that real part of Z should be one. Now, option A and B clearly states that. So let's further look into this. Now, next we know that sine inverse domain is between minus one to one. And by the way, this term is going to be just sine inverse of Y because your Z is actually one plus IY. This is your Z. So one and one gets canceled, IE and IE gets canceled, leaving you with just with Y, right? So Y should lie between actually minus one to one. Okay, screen, okay? Now, if I scrutinize the question a little bit more, if it has to be an angle of a triangle, that means my sine inverse Y should only be between zero and pi by two, correct? That means Y should have been between zero and one only. Okay, so I think this question, so it may be challenged and this option, ideally should have been from zero to one. But anyways, if I have to mark the best of these options, I would go with option number B. All right, so we have the next question over here, a limit question. Find the limit theta tending to zero of four theta tan theta minus two theta tan theta by one minus cross two theta. Amesh, I'm believing that it has to be an angle of a triangle and we have angles between zero to one at 80 degree only excluding zero and 180 degree. So sitting in between zero to pi by two only. Okay, so mostly people are going with option D. So basically this becomes a very easy question. So I can write it as eight theta square tan theta divided by one minus cross two theta is two sine square theta and this is also two sine square theta. Okay, so in this case, in the first term over here, okay, we can divide the numerator by theta square and divide the denominator also by theta square. Okay, whereas in this case, what will happen since even if you divide by theta square in the numerator and denominator, you would realize that still the denominator, numerator will have an extra tan of theta, which will make things go zero in this case, okay? While in this case, it will become four by two because tan theta by theta is going to be one. Sine square theta by theta square is again going to be one. So your answer is going to be two in this case, which is option number D is correct. Absolutely, so the first one to answer this correctly was Sondarya. Let's move on to the next question, which is the 15th question. The ends A and B of a rod of length root five are sliding along the curve YZ for two X square. If XA and XB be the coordinates of the ends, then at the moment when A is zero comma zero and B is one comma two, find DXB by DXA. Yeah, yeah, sure, sure, take your time. Okay, what about others? All right, so mostly people are going with option D. See, first of all, let's say this is the rod AB, whose length is given to us as root five. Let's say this coordinate is XA, YA, this coordinate is XB, YB. So I can say the distance between A and B is root of five. So XA minus XB whole square plus YA minus YB whole square and the root is root five, right? So XA minus XB square plus YA minus YB square is equal to five. Now, we know that they are lying on this parabola Y is equal to two X square. So I can say YA is two XA square and YB is two XB square, okay? So I can write this as XA minus XB square and this I can write it as two XA square minus two XB square square, okay? So that's given to us as five, okay? Now, let me differentiate with respect to XA, both the sides. So this will become two times XA minus XB into a derivative of this is one minus DXB by DXA. In a similar way, this can be done as two times, two XA square, two XB square into derivative of this term, which is going to be four XA, minus four XB DXB by DXA. And this is equal to zero, okay? Now, I want the value of DXB by DXA at the instant when XA is zero and XB is one. So what I'll do is wherever there is an XA, I will put zero and wherever there is an XB, I'll put a one. That becomes two, zero minus one, one minus, let me call this quantity as K for the timing. So let me call this quantity as K for the timing. So two times, again, it will be zero minus two, four into zero, which is again zero, minus four K. And this is equal to zero, right? So if you simplify this, it will give me two K minus two, minus 16 K is equal to zero. That means 18 K is equal to two. So K is going to be one by nine. That means DXB by DXA is going to be one by nine. So option number D becomes the right option. And the first one to answer this was Amok. So now let's move on to the next question. That's question number 16. So a couple of things I want to tell you guys. So if you think for one, one and a half minutes of trying, you are going nowhere, then please abandon the question. Do not invest any more time. Because if you have not been able to make any progress in one and a half minutes, it's like, the question may be beyond your scope or things may not be clicking to you at that moment of time. So don't unnecessarily waste time. I think max a question should take two and a half minutes or three minutes more, max to max to solve. If you're taking more than that time, that means you need to work out on your speech. Of course, without compromising on the accuracy. Okay, Aditya. So I've got three responses so far. Two of them are in favor of option B. Koshal says option A. Okay, so let's say this is our G of X function. So when you multiply, it becomes this term. Now what I'm doing is I'm just focusing on writing terms which are having even powers of X. I will not bother to write terms which are containing odd power of X because we know that for those cases, your answer is going to be zero because the limits of this integral are opposite in sign. So just evaluate this integral. So this is going to be P times two to the power, sorry, X to the power two N plus three by two N plus three. And this is going to be X to the power three by three. And this is going to be BNQ times X, okay? Or you can do two times this from zero to one, right? So this will give you two P by two N plus one, sorry, two N plus three. And you have two ANP by three and you have two BNQ. Now for all linear functions, PX plus Q, this integral has to be zero. That means it is independent of your PNQ. In other words, the coefficient of P and coefficient of Q should be zero each. So this should be zero and this should be zero, which clearly implies B has to be zero or BN has to be zero and AN has to be negative three by two N plus three. AN has to be negative three by two N plus three. So option number B says that, so B is the right option. Okay, so first one to answer this correctly was Aditya, great, well done. Let's go on to the next question. Find the, or match the conditions of column one with column two. Yeah, it's a fractional part. No, this entire thing is not a fractional part. This is a fractional part. Yeah, so let me tell you this bracket is your GIF, but this is the ordinary bracket, normal. Normal brackets. So basically it's GIF of X over here and this is going to be a fractional part of X. Okay, so mostly people are saying 17A. Okay, let's discuss. Sign of pi GIF of X is basically nothing but always a zero value, right? Because you're dealing with an integral value of pi. So this is always an integer, okay? So this has to be differentiable everywhere. So A will map with first option. Whereas if you see this function, we know that this function is going to behave as sine pi X when X is between zero and one, right? So sine pi X would be a graph which is like this from zero to one. And after that, it will be periodic with one. That means it'll keep repeating itself like this. So this would be your graph of sine of pi times fractional part of X. As you can see, this function will start having kinks or you can say cuffs. So these all points are like cuffs. And at cuffs, you know that the function is not going to be differentiable. So only at integer values. So this function will be not differentiable at one and minus one to state just two of the points. So B will map to option three. So this is given by option number A in this case. Next, the two vertices of a triangle are given to us four comma minus three and minus two comma five. If the ortho center of the triangle is at one comma two, find the third vertex. This is your 18th question. Okay, so 18th. So this is a very simple problem to solve because you can just plug in the points. For example, I know that this is going to be perpendicular, right? So k plus three by h minus four should be negative reciprocal of the slope of let's say this point BH. So it should be one by slope of BH. Slope of BH is going to be five minus two by minus two minus one, which is going to be three by minus three. That's going to be negative one. So that's going to be one, okay? So k plus three is going to be h minus four. So k plus three is equal to h minus four is only satisfied by option number B in this case, correct? Because 26 plus three is equal to 33 minus four, right? None of the other options satisfied. So option B is the right option. However, if you want to be sure that it is not none of these, you can again do the slope of h is negative reciprocal of the slope of BC but it will end up giving you the same answer. So it's an easy question. Let's not spend too much time on this. Next, normal AO, A1, A2 are drawn to the parabola y square is equal to 8x from the point h comma zero. If triangle OA1, A2 is equilateral, then the possible values of h. Okay, 19s. Anybody else? Okay, but that's not in the options for it. Possible values. That means one of the values can be equal to 8x from the point h comma zero. Okay, atmesh gets 19a. Okay, so let's discuss this. So OA is a normal, so is A1 and A2, correct? Okay, so basically these three points are actually called the co-normal points. So we have discussed them in our parabola chapter that these are called the co-normal points. And if you connect them, as per the question, this triangle is an equilateral triangle. This triangle is an equilateral triangle. So this angle has to be, correct? So what is the slope of OA1? Slope of OA1, let's say I call this point to be the point with a parameter t1. That means let's say this point is two t1 square comma four t1. And this point A2 is two t2 square comma four t2. So slope of OA1 is going to be four t1 by two t1 square. That's going to be two by t1. And this is given to us as tan of pi by six. So this slope is nothing but tan of the angle of inclination. That's going to be one by root three. Which means t1 could be safely written as two root three. In other words, what I'm trying to claim is that if I have a parametric equation of the normal, okay, which we all know is this, this is the equation of the normal at any parameter or any parametric point t, then one of the roots, then t1 is a root of this equation. t1 is a root of this equation, correct? Similarly, you can have this slope, let's say t2, then t2 is going to be negative of two by root three. So this is also going to be a root of this equation, okay? Anyways, so what am I claim is here that if let's say this normal is passing through h comma zero, so let me put y as zero and this will become our th, this will become two into two into t plus two into tq, which will actually give me, I'm writing down over here, th is equal to four t plus two tq, okay? Canceling the factor of th is going to be four plus two t square, right? Now, irrespective of whether you put t as t1 or t2, your answer will become four in plus two into two root three square. That's going to be four plus 12 into 224, that's going to be 28. So 28 is going to be your answer. That is nothing but option number C is correct. Now, we all know that the algebraic sum of the T's at the co-normal point is zero. So this is corresponding to your t equal to zero. So you can call this as t3 equal to zero point, okay? So that is automatically coming from here when you were cancelling the factor of t from both the sides, it was actually saying that t equal to zero could be a possible value of the root. All right, so let's move on to the next question, which is question number 20. Okay, Vaishnavi. So alpha, beta, gamma here are scalar quantities. A, B, C are unit vectors. Okay, Sai also says option B. So it's given that A dot B is zero and A dot C and B dot C is going to be cos of theta. And of course they are all unit vectors, so their moduli are equal to one each. So now I have been given that C is alpha A plus beta B plus gamma of A cross B. Okay, so let us take the dot product with A first. So C dot A is going to be alpha A dot A is going to be one, this is going to be zero and this is again going to be zero. That means alpha is going to be cos of theta. If you take dot product with B, it's going to be alpha cos theta. Sorry, it will be zero again because A dot B is zero. This will be beta and this again will be zero because A cross B dot A and A cross B dot B, they'll both will be zero. Okay, so even beta is going to be cos of theta. So it's very clear that option number B is not true. So you can just mark option B. Okay, but let's say we want to ascertain whether option A, C and D, none of them are incorrect. So let's check them out quickly. So if I now take C dot C, I'll get alpha A dot C plus beta B dot C, okay? Or you can do one thing, you can take the dot product of this vector with itself. So this dot product with itself. So this into this will give you alpha square, beta square and gamma square mod A cross B square. Rest, this into B will give you zero, this into this will give you zero, B into A would give you zero and this into this will give you zero. Similarly, this into this and this into this will give you zero again. So this is the only term that I would get. And mod C square is going to be one. So one is going to be alpha square plus beta square plus gamma square. And this is also going to be one. Now since alpha is equal to beta, two alpha square plus gamma square is going to be one. That means gamma square is one minus two alpha square, which is one minus two cos square theta. So that's going to be negative cos two theta. So this option is correct. So this cannot be my answer. This option is also correct. So this cannot be my answer. Okay. And from here I can write alpha square as one minus gamma square by two, okay? Which means beta square is going to be, so one minus gamma square is two cos square theta by two. Okay. And two cos square theta is going to be one plus cos two theta by two. So even option number D is saying the right thing. So this cannot be your answer. So only possibility is option number B. Okay, so had the question been a multiple option correct question and they would ask which of the options are true, then obviously A, C, D would have been the answer. Now let's talk about question number 21. Okay, Homo, where should I be? Easy question. So this becomes sine square theta plus 81 by 81 to the past sine square theta and that's equal to 30. So if you call this term as Y, it will become 81 by Y is equal to 30, which means Y square minus 30 Y. So this can be factorized as 27 and minus three. So Y could be three, Y could be 27. So 81 sine square theta if it is three, that means sine square theta has to be one fourth, correct? So sine theta has to be plus minus half and if sine, if 81 to the past sine square theta is equal to 27, that means sine square theta has to be three by fourth, which implies sine theta can be plus minus root three by two. So my possible values here could be, yeah, I think 120 can also be a correct answer. So of course 30 is a correct answer and root three by two could be obtained for 120 degree as well. So in this present case, both these options are correct. So option A and C both are correct. How is B, how is D correct? Yeah, D is also correct. D is also correct. So, okay, so they should have asked this question. Theta cannot be, I think that would be a right question to us. Theta cannot be then option number B is correct. And then in this case, in the present form A, C, and D will be correct. Next, find the sum of N terms of the following series. So here you can easily put some special values like N as two or something and you can get your answer, but I would like you to do it in a rigorous way. Hey. Let's go. Shit. Okay, all right. So basically what I'm going to do is I'm going to first multiply with one minus X and divide with one minus X. So if you do that, you get one minus X squared. This becomes one minus XQ and this term will actually become one minus X to the power four and so on. So it's going to form something like this N minus X plus X square plus X cube all the way till X to the power N, okay? Which again is a GP, okay? So this gives you the result like this. Let's take the LCM of one minus X inside. So it becomes one minus X square and one minus X minus X times one minus X to the power N. So option number C is the right option in this case. So this is a function from R to R defined as two X plus sine X, then F of X is, which of the following? One, one and onto, one, one, but not onto, onto, but not one, one, neither one, one, nor onto. Okay, Purvig says option B. Okay. Purvig wants to change his answer to A. What about others? Okay, it's very obvious that the derivative of this function is going to be two plus cos X. And we all know that cos X belongs to the interval minus one to one, right? So this will always be a positive term. That means your function is monotonically increasing. And if it is monotonically increasing function, it has to be a one, one function, okay? And if you take any value of X, let's say if X goes to infinity, even F of X goes to infinity. And if X goes to negative infinity, even F of X goes to negative infinity. And not only that, all the values of the function are going to be addressed by this function because it's a continuous function. Two X is a continuous function, sine X is a continuous function. So it has to be one, one and onto both. So option A is going to be correct. Next question is from statistics for this arithmetic progression, A, A plus D, A plus two D, till A plus two and D, the mean deviation from mean is. So we have to find out the mean deviation from mean. So Ramcharan has given the response, okay? Okay, anybody else? Okay, so mostly people are going with option number C. So we'll discuss this. First of all, we know that the mean deviation from the mean position is going to be one by N, summation of mod of Xi minus X bar. Okay, so the mean of the data itself would be the first data and the last data divided by two. So that is A plus N D will be your mean, okay? Undoubtedly. So the total number of data present over here is two N plus one. So I have to do summation of, okay? Each of the data, let's say I generalize the data with A plus R D and let's say from zero to two N minus A plus N D, okay? So basically we have to do one by two N plus ones, summation zero to two N, R equal to zero to two N, mod of R minus N times D. So this is nothing but one plus two plus three all the way till N, twice stop. Isn't it? Correct? So this is going to be D into two N plus one. This is going to be N into N plus one by two into two. So two and two gets canceled. So your answer will be N into N plus one D whole divided by two N plus one. So in that case, your option number B will become the right answer and not C. So none of you got it correct. Any question with respect to this? Did you get your mistake? Where you all went wrong? All right, so moving on to the next question. A dictionary is printed consisting of seven letter words. Only that can be made with the letters of the word cricket. If the words are printed in an alphabetical order, what is the number of words before the word cricket? So a very simple question that you would have done in your class 11 permutation combination. So you just have to see how many words appear before the word cricket itself. Vashtami, the sum from N to zero would be N into N plus one by two only, right? Because zero is as good as starting from one itself. So you're going from one to N. Going from one to N means N into N plus one by two. So the first word will be CC. So let me call this question number as question number 25. So Aditya says, okay, option C. Okay, let's check. So first, all the words starting with CC will be before cricket, right? So that word will be five factorial. Then words starting with CE, okay? That will also be before cricket. Then again, will be five factorial. Then similarly, CI will be five factorial, okay? CK will also be five factorial. Then CR, when it comes to CR, so the first word will be CRC, okay? So CRC, it will be one, two, three, four. That's going to be four factorial, okay? Then CRE, that's again, will be four factorial. Then CRI, now cricket also start with CRI. So the next word will be CRIC. So cricket also starts with CRIC, correct? So CRIC. So the next word will be E. Then these two words can be filled in two factorial ways, okay? So it's E, KR, sorry, CRIC, KT. So next word will be CRIC after the next word will be K. So if K comes, the next word has to be ENT. So this is the word exactly after this. So let's count how many of them are there. 120 into four will be 480. This will be 16 into 16, 32 plus two. That is going to be 24 into two. That is 48, 48 plus two is going to be 50 again. So total words before cricket going to be 530, that's option number C is going to be correct. Absolutely, so first one to answer this correctly was Aditya. Okay, so next is a two and false question. So this is the direction for the two and false questions. All of you would have already gone through it. Statement one is true, two is true, two is the correct explanation. One is true, two is true, two is not the correct explanation. One is true, two is false. One is false, two is false, okay? So this is the first question. Consider a function g of x as f of x minus one and another function f of x plus f of one minus x is two. Statement one claims that g of x is symmetrical about half comma zero. Statement two says, if g of a minus x is negative g of a plus x, then g of x is symmetrical about a comma zero. So which of the options a, b, c and d are correct in this case? Let's say this is 26th question. Okay, 26th question, Amogha is giving the response. Okay, let's wait for one more person to respond. Okay, so let's discuss this. So if you look at the second statement, can I write this as f of x minus one plus f of one minus x minus one equal to zero, correct? This can be written in this way. Now this itself is g of x plus g of one minus x equal to zero, correct? That means g of x is negative of g of one minus x, okay? So it tries to say that whatever is the value of the function at x, it has the same value with a negative at one minus x. Let's say the value at zero is this. Then the value at one would be this. Correct? Now here I think one thing that they should have mentioned that f of x is a continuous function. Because if I do that, the value let's say at a point we can take one sixth is same as negative of the value at let's say five sixth. So this trend continues, you would realize that the graph will be in such a way that it will cut at half comma zero, okay? Because at half it will become g of half is equal to negative g of half. That means g of half should be equal to zero. I'm assuming in this case that my function is continuous function. So this function, if g of f of x is continuous will be symmetrical about half comma zero. So this is correct statement. And here if you replace your, now this means g of a minus x is negative of g of a plus x. See what does it mean? It means that, let me just make some space for myself. Let's say this is a, if you go x units left of a and x units right of a, they are negative of each other. So let's say I move x units to the left of a and I move a x units to the right of a then the graph would be negative of each other. That means even in this case you would realize that the graph will come out to be symmetrical about a comma zero. So this statement is also correct, okay? So this statement and this statement is also, both are correct. And two is helping me to, or two is explaining my statement number one. So option number a become the right option in this case. Is that fine? Okay, so one more thing you should remember that if g of a minus x is equal to g of a plus x, right? Then the graph will be symmetrical about x equal to a line. Okay? And if g of a minus x is equal to negative g of a plus x, the graph will be symmetrical about a comma zero point. Provided it's a continuous function. Yeah, so are the given lines coplanar? So this line is passing through one comma minus three comma two and this line is passing through two comma one comma minus three. Okay, so if these three lines are coplanar, we know that x one minus x two, let's say I call these as points x one, x two y, x one, y one, z one, x two, y two, z two. So, and let's say I call this as a one, b one, c one, a two, b two, c two. So this determinant should be zero if they are coplanar. So I can write this as a minus one comma minus four comma five, two one minus three one minus three two. So let's find out the value of this determinant. This will be negative one, two minus nine, which is going to be negative seven. By the way, if you look at statement two, it doesn't talk about all these things, right? So I'm just doing it just to make a confirmation whether these two lines are coplanar or not. Okay, so four plus three and five minus six. Minus one. So that's going to be minus plus seven and this is going to be 28 minus 35. Yes, they're coming out to be zero. But this is not the intended way that in which we are actually supposed to solve this question. If they're coplanar and you know that these are not proportional, these terms are not proportional, that means these two lines should actually meet, correct? Or they should actually intersect, right? So if they're not proportional, two of the conditions can happen. Either they are skew lines, right? Then only they will not meet, but skew lines are actually non-coplanar, okay? And if they're coplanar and they are parallel, then also they will not meet. But we know they are not parallel because two one and minus three is not proportional to minus one, sorry, one minus three and two, okay? So that means if these two lines are coplanar, they should actually meet. So let's say I call this as, okay, I call this as some value lambda. Let's say they meet at x one, y one and z one, okay? That means x one, y one, z one should be satisfying both these lines, right? So x one is going to be two lambda plus one, y one is going to be lambda minus three, z one is going to be minus three lambda plus two. Now if you subtract these two equations, let's say I do two x one minus y one, or you can do one thing, you can just take according to the requirement of the question, you can call this as x one point itself and you can call all these points as y one itself, okay? x one, y one are some parameters. So here x will be two x one plus one, y will be, let me clear off everything else which is not required. So x one will be two x one plus one, sorry, x will be two x one plus one, y will be x one minus three, okay? And z will be negative three x one minus two. Similarly for the other one, it will be x equal to y one plus two, y will be equal to minus three y one plus one, and z will be equal to two y one minus three, okay? Now why I took this situation is because these two should be equal, these two should be equal and these two should be equal. That is two x one plus one should be equal to y one plus two, x one minus three should be minus three y one plus one, and minus three x one minus two will be two y one minus three, okay? So this will result into your first equation. So this will be your first equation. This would result into your second equation and this would result into your third equation. Now here your x one and y one value should be existing. That means you should be able to find the values for x and y one and that's why they should be consistent. So if you are able to find x one y one values, that means these two lines are intersecting and since they are not parallel, they have to be co-planar, okay? So statement one and statement two both are correct and statement two is still explaining statement one. So again in this case, option number A will be correct. Okay, so ignore this, this is just a repetition. Ignore this part, just do the question itself. Okay, so Amogh, see to answer your question, how is statement two explaining it? If you go back to the previous question, x one and y one are like your parameters over here. So if these values of x one and y one are found out, that means there exists some value of x one and y one. Okay, that means these two intersect. The moment any two lines intersect, right? We can always pass a plane through them and call it co-planar, right Amogh? So this statement two is actually trying to say that these two lines actually intersect and hence they are co-planar. Is that fine Amogh? If you have any other question, follow up question. Yeah, we realize that using the co-planar question which was not the intention of the question, the question was actually meant to be solved by using the fact that these two lines intersect, that's it. What will make these two lines intersect? That is statement number two is saying that. So they will intersect only when there is a solution for x one and y one. It should not happen that the answer from these two is not matching with the third one. If there is an answer, that means they intersect and hence they are co-planar. So we started by using the determinant methods, but that was not the intention of the question. So even statement number two could be used to prove that the lines were co-planar. Determinant method is another method to prove that the lines were co-planar. Yeah, so this question is easy. So let's take this up first. So the volume of the parallel pipette formed by this is maximum, okay? Okay, so this statement two is definitely correct. And the volume of the parallel pipette is given as modulus of the determinant formed by one a zero in this case and a one one and zero one a. Okay, so if you simplify this, it becomes a minus one and minus of a q. So db by da should be zero. That means one minus three a square should be zero. So a should be equal to plus minus one by root three. So double derivative of this is going to be minus six a. Which by the way, when I say determinant, so when a is equal to plus one by root three, this is going to be negative, right? When a is going to be minus one by root three, then double derivative is going to be positive. So this statement one is wrong. Statement two is correct. So d will be your right answer. That doesn't matter, Amok. Statement one is false, that's it. Whether it's explaining or not, it's immaterial because one is false. Oh, you were asking about the previous one. Okay, so this is clear. Oh, sorry, I meant I thought that it was for this question. So again, a question from differential equations. Okay, Aditya, Sondarya, all right. So if I talk about the first equation, this is actually a linear differential equation, right? So here the integrating factor is going to be e to the power integral of 2x dx. That's going to be e to the power x squared. So your solution is going to be y into e to the power x squared. So integral of 2 to the power, 2 into e to the power x squared, x squared. That's going to give you y into e to the power x squared is equal to 2x plus c. So statement one is correct, that's for sure. Now, statement two says that if y is equal to this, then dx, that means dx by dy have to show that it is e to the power minus x squared minus 2xy. So let's take this expression, differentiate both sides with respect to y. So it will be e to the power x squared plus y into 2x e to the power x squared dx by dy is equal to zero, okay? Unfortunately, I don't see this coming. I don't see this equivalent to this. So statement number two is going to be wrong. Statement number one is right, so option C is going to be correct. All right, so last question that is the 30th question for the day and we are done with the session then. Okay, so the distance of origin, this distance is given to us as, and from here, there's a point Q which is at a distance of one by P from this. And I have to find out the locus of Q, okay? So first of all, small P is going to be, if I'm not wrong, one by under root of cos square theta by a square plus sine square theta by b square. That's going to be a b under root of b square cos square theta plus a square sine square theta. This is going to be your P. Now, in order to find out the coordinates of Q, I would use the distance form of the equation of a line. I'm sure all of you are aware of the distance form that is x minus x one by cos theta equal to y minus y one by sine theta is equal to r. Where r is the distance of the unknown point x one y one from the known point x one y one. So Q is my unknown point, so let it be x. This point is a cos theta, okay? Now, the slope of the line, the slope of this tangent is nothing but b cos theta by minus a sine theta, right? So I can do one thing. I can write this minus a sine theta, okay? Or you can say, we can find, this is your tan of theta. This is your tan of, or you can say tan of phi. So cos of phi is going to be minus a sine theta by under root of a square sine square theta plus b square cos square theta, okay? So this expression I'm going to write over here. Minus a sine theta by under root of a square sine square theta plus b square cos square theta. This is equal to y minus b sine theta by sine phi. Now, let me call this as phi because I've started calling it phi over here. Theta has already been used. That's why I'm calling it as phi. So sine of phi will be b cos theta by again under root of a square sine square theta plus b square cos square theta. And this expression is one by p. This expression is one by p, right? Now let me rewrite this so that we can understand this. So we have x minus a cos theta divided by minus a sine theta by under root of a square sine square theta plus b square cos square theta. And this is equal to y minus b sine theta divided by b cos theta by under root of a square sine square theta plus b square cos square theta. And this term is one by p. One by p means under root of a square sine square theta plus b square cos square theta by a b. So from here I can say that x minus a cos theta by minus a sine theta is going to be one by a b. That means x is going to be a and a gets canceled. Or let it be at the place. So x is going to be a cos theta minus a sine theta by a b. Similarly, y is going to be b sine theta plus b cos theta by a b. Okay, now I have to eliminate theta from here. I have to eliminate theta from these two expressions. Okay, so I can see one thing that if I do x by a square and y by b square, I get something cos square theta plus sine square theta by a square b square minus two. So a and a will get canceled minus two. a by b sine theta cos theta. And the other term gives me b square sine square theta plus this b and this b gets canceled. This is going to be just b square. This is going to be cos square theta by a square and again minus two b by a sine theta cos theta. Just a minute. I think I had already taken a comment down, right? This is going to be your a, this is going to be your b and b. So x by a is going to be a cos theta. Okay, so I should not have this thing over here. And this is a square b square, that's correct. Okay, and let me change here itself. So a, a and I'll bring a a over here. This also b, b and let me bring a b over here. Yeah, so I won't even have this b square term over here. And again, I'll have a square b square and this will be two a by b. So when adding these two will get canceled, these two will add up to give me one and these two will add up to give me one by a square b square. Okay, so my final locus will be this expression, which I believe is your option number a. So a becomes the right option in this case. Okay, this is slightly lengthy question, but the process was simple. You had to use a distance form and of course you should be knowing the equation of a tangent in parametric form. Okay, so guys, thank you so much for coming online. Okay, so over and out from my side and all the best for the exam, computer science exam, which is left off. We'll meet again with the CET session on Tuesday, next Tuesday.