 Hello, good afternoon, welcome to the problem solving session on CET and CometK. So those who have joined in 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. Good afternoon, Sondarya. So those who are watching in this session, I would request you to type in your names in the chat box. And let me also tell you that this session would be on CET and Comet level problems. So there would be no J e main level of problem solved over here. That means it should not take you more than a minute to solve each of the problems that I'll be going to project on the screen before you. So altogether there would be 60 problems and it would be our effort to actually discuss it within the next three hours. So here is the first question in front of you. So put your timers on and let's start solving questions. So tan of x plus y is equal to 33 and x is given to you as tan inverse of 3. We need to find out what is our y. Okay, so I've started getting the response, so option 4 is what people are saying here. So it's a very easy question. Again all these questions are actually easy, it's just a test of your speed. So your y is going to be tan inverse of 33 minus tan inverse of 3. And we are going to use the formula x tan inverse x minus y by 1 plus xy. So that's going to be tan inverse of 30 by 100, that's going to be tan inverse of 3 by 10. Absolutely correct. So the first one to answer this correctly, Vaishnavi, good one. Moving on to the next question now, limit x tending to infinity under root of x square a square plus bx plus x and minus of x. So let me name these questions so that you can type your response by stating the question number. So this is your second question. One minute is the time that you should be taking to solve such questions, not more than that. All right. I need two more responses. Rithik, Sanjana, Sondarya, what do you think? So this is basically infinity minus infinity form of a problem, right? And for such cases, the first step that we follow is the step of rationalization, okay? So Ganesh thinks something else. All right, so let's just discuss this. So if you rationalize this, you will have a square x square plus bx plus x minus a square x square whole divided by a square x square plus bx plus x plus ax, okay? And you will cancel off the highest degree term from the numerator that is x square term. So it will be as good as b plus 1 times x, okay? And we can take an x common in the denominator as well that would give us a square and b plus 1 by x and an a over here. So this x and this x will get cancelled. And as the limit extends to infinity, we know that terms like 1 by x, et cetera, will start approaching zero. So this term will vanish over here, leaving us with the answer of b plus 1 by a plus a, that's b plus 1 by 2a. And therefore, option number one becomes the right option. And the first one to get this correct was, again, Vaishnavi, okay? Aaritwik, Sanjana, please check your working. I'm sure you must have done some cylinder mistakes here and there. So this was a pretty straightforward question. So without much waste of time, I'm taking you now to the third question for the day, a simple inequality question. Guys, let me tell you, exams like Comet K and CET are super simple exams. It's just testing your speed concept wise. It's a very simple exam. Okay. Two, two, two. Okay. Mostly people are going for option two. Again, this reminds you of your class 11th concept that is inequality. So first we'll make the half come on the other side. Okay. Always take the LCM. So 12x minus 4x plus 1 less than zero, that's going to be 8x plus 1 by 4x minus 1 less than zero. Remember two, you can always send out the other side because we know two is positive. And if you plot this on the real number line, you get minus 1, 8 and 1 fourth. So according to the baby curve sign scheme, this will be plus minus plus. And for your answer to be negative, it has to be in this interval. So it has to be minus 1 by 8 to 1 by 4, exclusive of minus 1 by 8 and 1 by 4. So option number two is correct. I think Sondarya was the first one to get this correct. Well done. Good. So without much waste of time, I'll take you to the third, sorry, I'm fourth question for the day. If I is a square matrix, we all know that it is a rotation matrix for us. And A transpose plus A is I2, I2 is an identity matrix of order two. Then which of the following option is correct? Okay. Vaishnavi says, Rithvik also says third. One more response I need. Awesome. So let's discuss this. So A transpose into A, we can do this mentally, but just to make things clear, I would write it down. So A transpose plus A would be this two matrices. And when you add them, you are going to get 2 cos theta 002 cos theta. And this is equated to an identity matrix, which is 1001. That means cos of theta is going to be half. So if you write the general solution, we know that cos theta equal to cos alpha. The general solution is theta equal to 2n pi plus minus alpha and belonging to integer. In this case, it will be 2n pi plus minus pi by 3 and belonging to an integer. Okay. So, of course, option number three becomes the right option in this case. Again, the fastest of you was Vaishnavi again, I'm saying fastest of you because this is just a test of your speed, nothing else. Next, if A and B are two matrices says that AB is equal to A and BA is equal to B, then B square is equal to which of the following options? This is your question number five for the day. And I'm repeating this. This session is helpful to those students who would be taking up even CT and Comet exams. We are not going to take up any J main level question here. All right. So I'm getting a response from Vaishnavi, Sanjana, Dupik, Sondarya and others who are attending the session but haven't typed in your names, I would request you to respond. Okay. So let's start with the fact that BA is equal to B. Okay. And when you multiply this with B post facto, I get BA is equal to B square. And we know that matrix multiplication is associative, so it becomes this. And since we know that AB itself is A, I can write this as A itself, so it becomes BA is equal to B square. And BA again is B, so it becomes B square. So B square is equal to B, and that's what we wanted to find out. So option number one becomes the right option in this case. Absolutely correct. Again, Vaishnavi was the first one to answer this. Let's move on to the next one. Given five line segments of lengths 2, 3, 4, 5 and 6 units, then the number of triangles that can be formed by joining these segments is. So you have to tell me how many triangles are possible with these segments as the length of the sites. This is your question number six, by the way. So if you are typing in your response, please mention this question number. Next 41, first option as per Svandarya, Vaishnavi also says first option. What about Sanjana Ritve, guys, please respond. Okay, first of all, it is clear that 5C3 will appear in one of your answers because you are choosing 3 from 5, so that's 5C3 obviously. Now these minus signify which of the cases cannot form a triangle. So as per my calculation, I can never form a triangle with this combination, no triangle is possible. Neither can I form a triangle with this combination for the simple reason that the sum of the two sites of a triangle must always be greater than the third. Neither can I form a triangle with 2, 4, 6 as well. Any other case that you take, you will actually find a triangle with that site. So these three cases must be removed. So your answer is going to be 5C3 minus 3, that's option number 1 is going to be correct. So the first one to answer this correctly was Svandarya. Great. So let's move on to the next one. That's question number 7 for the day. So guys, you are all maintaining a good speed. It's already 15 minutes down the session and we are into the seventh question. So roughly including your time and my time, we are taking around one minute to solve it. So that's what we expected from this session. How many numbers greater than 10 followed by 6 zeros, that's actually 1 million, can be formed from 2, 3, 0, 3, 4, 2, 3. Okay, Vaishnavi, need two more responses. Okay, Ritvik, Svandarya. All right. So it's very obvious that if this is going to form numbers, you have to use all the seven digits, right? And the only possibility you can end up a number getting, being less than 1 million if you are using all the seven digits is when your number is beginning with zero, correct? So if a number begins with zero, then that number would be lesser than 1 million, correct? So how many numbers are there which will be lesser than the given digit, okay? Let's say lesser than 1 million. So you have to exclude zero and see how many digits are there, 1, 2, 3, 4, 5, 6 digits are there. Out of that, I think 3 is repeated twice, so 3 factorial, 2 is repeated twice, so 2 factorial, okay? And the total number of numbers that you can form by using all the seven digits will be 7 factorial by 2 factorial, 3 factorial, correct? So your answer will be the latter minus the former. So 7 factorial minus 6 factorial can be written as 6 factorial into 6 and you have 2 and this is 6 again, so that is going to be 360. That's your option number 2 is going to be the right option in this case. Well done, Vaishnavi. Again, the first one to answer this. So with this, we'll move on to the eighth question for the day. I think with this rate, if we go on, we'll be able to finish our class today in two hours. Well done. I can see very good speed coming from few of you, in fact, some of you. So let's read this eighth question. An urn contains nine balls, two of which are white, three blue and four black. Three balls are drawn from the urn. The chance that two will be of the same color and the third of a different color. So basically, we have to find the probability that two balls are of the same color and the third one is of a different color. Okay, Vaishnavi, how about others? Okay. Fine. You change your answer. No problem. Ganesh, Sanjana, Sondarya, okay, all right. So let's discuss this now. First of all, how many equally likely cases would be there? That means how many elements would be there in the sample space? That means it is as good as drawing three balls from nine balls, which is nine C3, which is going to be nine into eight into seven by six. That's going to be 84. So this would be in your denominator. I think all of them have 84 in the denominator. Now let's take cases. Let's say I choose to draw two white and one black, let's say blue, or it can be two white, one black. So this would be nothing but two C2 into three C1 plus two C2 into four C1. In a similar way, I can have two blue, one white, or two blue, one black. So two blue will be three C2 into one white will be two C1, plus again three C2, one black will be four C1. And the last case is, let's say I pick up two black, one white, or two black, one blue. So two black I can get in four C2, one white I can get in two C1, again four C2 into three C1. So let's add all the cases over here. This is going to give me one into three, three, three plus four is going to be seven. This is going to be three, six, six, and twelve is going to be eighteen. And here I get twelve, twelve plus eighteen is going to be thirty. And all together if you add them, I think you should be getting fifty-five. That means fifty-five upon eighty-four is the desired probability. So fifty-five upon eighty-four is going to be your option number two. And again the first one to answer this was Vaishnavi. Let's move on to the ninth question for the day. So this is your ninth question. So the question reads six boys and six girls sit in a row at random, the probability that all the girls sit together. As it's all about speed, but remember one thing, it's equally important that you get the right answers, though there's no negative marks. But your aim should always be to score very, very high in these exams. Okay, Sondarya goes for the third option. How about others? I'm sure you must be taking some time doing the calculation because the concept is very easy. Sanjana, Ritvik, Vaishnavi, come on guys, it's already, we have spent around two minutes on this. So if you want all the girls to sit together, we have to use the string method where we tie all the girls and treat them as one entity. So six plus one, there will be seven entities and they can be arranged in seven factorial. But the girls themselves can sit in six factorial ways. So please remember, this would be the number of favorable events. The sample space, as you would all know, it's going to be 12 factorial. That's a simple way of arranging 12 people among themselves. So the probability would be nothing but seven factorial times six factorial by 12 factorial. 12 factorial, let's write after seven factorial, that's eight into nine into 10 into 11 into 12. And seven factorial, let's cancel it off. Seven 20 is the value of six factorial. So seven 20 would be obtained by the first three terms itself. So your answer will be one by 11 into 12, that's one by 132. That makes the third option as the right option. Absolutely. So Sondarya was the first one to answer this. Guys, many people may forget the internal or intrarangement of the girls. So this is important. Please do not miss out on this. So with this, we move on to the 10th question. Remember, we have to finish up 60 questions today. Three dice is, in fact, this is not a right English. Three dice are rolled once. The chance of getting a score of five is. The chance of getting a score of five is, OK? Hey, Saimeer, good afternoon. I think you joined us now. All right, Sanjana. I need one more response before I start solving this. Guys, just speed, nothing else. Great. So you can get a sum of five only when you have the numbers on the dice as 311 or various permutation of this, which is going to be actually three factorial by two factorial. Because you can have 131, 113 as well. Or you could have a 221 case. Again, that would be three factorial by two factorial. That's, again, three. So total number of favorable events will be 3 plus 3. That's going to be 6. And we know that the sample space would be 6 cubed because each die will show you six outputs. So 6 into 6 into 6 by a fundamental principle of multiplication. So your probability of the required event would be 6 upon 6 cubed. That's 1 by 36. Option number three is going to be the right option in this case. Since these events are countable, I would request you not to waste your time using multi-normal theorems to solve these kind of problems. That would be an overkill. So well done, Saimeer was the first one to get this right. Let's move on to the 11th question for the day. Again, a question from probability. I think back to back we have got questions from probability. A back contains three white, four black, two red. If two balls are drawn at random, find the probability that both the balls are white. This is just a five-second question. Absolutely. So that's going to be just 3C2 by 9C2, OK? So that's going to be 3 by 9 into 8, and 2 will go up. So 6, and this is going to be 12. 1 by 12 is going to be the answer. So option number three is going to be the right option. Is that fine? Great. So the first one to get this correct was Sondarya. Let's move on to the next question. Question number 12. If two sets A and B have 99 elements in common, then the number of elements common to the sets A cross B and B cross A. OK, Vaishnavi says option one. Sondarya says option two. How about others? Ritwik, Saimeer, Sanjana. OK, let's discuss this now. So this property is basically that this question is based on the property that A cross B intersection C cross D is nothing but it's A intersection C cross B intersection D, OK? And if you take your C to be B and D to be A, OK? Then can I say A cross B intersection C cross D would be nothing but A intersection B cross B intersection A. That means the number of elements common to A cross B and C cross D would be the number of elements in A intersection B cross B intersection A. And we know that the number of elements in X cross Y is nothing but the number of elements in X times the number of elements in Y. That's going to give you the number of elements in A intersection B times the number of elements in B intersection A, which happens to be 99 in both of them because A intersection B is equal to B intersection A. So the answer is going to be 99 square. That's option number two, which is going to be correct. OK, Vaishnavi, is that fine? So we'll move on to the next one. If f of X is e to the power X and g of X is ln of e to the power X, this is your question number 13, then which of the following is true? One more response I need. All right, so let's discuss this. So as per the options, after first find f of g and g of F. So f of g would be, f of g of X would be e to the power log of e to the power X. That's going to be e to the power X itself. And g of F of X would be log of e to the power e to the power X, which is actually, again, you can take this down in front of this. So it would be e to the power X log of e, which is e to the power X again. And yes, they're equal. So option number two becomes the right option in this case. Well done, the first one to answer this, again, Psi. Next is your question number 14. The number of bijective functions from set A to A, where A contains 108 elements. So how many bijective functions are there from A to A? Bijective means both 1, 1 and onto. This should be pretty simple. It's just touch and go. Absolutely, absolutely. We know that if a set contains n elements, the number of 1, 1 functions that we can make is n factorial. And in order to be bijective, both the sets should have same elements, same number of elements, which is actually happening in this case. So your answer will be just n factorial, so it's option number two, which is going to be correct. The other touch and go questions, let's say if you lose out time in some question, these are the ones which will give you the cover up so that you can spend some decent amount of time in slightly difficult questions. Next, let's talk about question number 15 now. If phi is equal to mod of cos x plus mod of sin x, then find the value of dy by dx at 2 pi by 3. Koro will be having a session on J main problem solving. And again, it will be a mixed back topic. OK, Ashnabhi says 1, but Saim here says 3. I'm getting different answers surprisingly. Anybody else who can break this tie? Either it's 1 or it's 3. So I need one more response. OK, so Ritwik backs his friend, Saim. So guys, whenever you are met with such a situation where you have a mod function, always see where you are actually performing the derivative. This will help us to redefine the function. So at 2 pi by 3, that cost is going to be negative. So in the neighborhood of 2 pi by 3, this function is going to behave as negative cos x plus sin x. Because cos is a negative quantity there and sin is a positive quantity here. So just find dy by dx for this expression, which is going to be sin x plus cos x. And put the value of x as 2 pi by 3. That's going to give you root 3 by 2 minus half. So it's root 3 minus 1 by 2. That's going to be the right answer. So yes, option number 3 is going to be correct. Again, be careful. I'm repeating this. Whenever you are trying to find the derivative of mod function, redefine the mod function in the neighborhood of the point where you are carrying out that differentiation and then finally perform the derivative of that redefine function and then put the limits and then put the value of x. So why should we be careful with these questions? They are easy, but sometimes can be deceiving. So let's take the next question, which is question number 16. If ABC are in AP, then the value of the determinant is. All right, so I'm getting a response as 1. So basically, here all I have to do is do R2 as, you can do R2 as R2 first of all, OK? And multiply with a half. And then do this operation, R2 as R2 minus R1 plus R3. Or you can directly not do this operation and just say this. R2 as R2 minus R1 plus R3 by 2. So that will lead to 0, 0, 0 in the second row, giving you the answer as 0. So option number 1 is going to be correct. Next is question number 17. OK. So determinant of adjoint of a, we all know that determinant of adjoint of a is related to determinant of a by the formula, determinant a to the power n minus 1, n being the order of the matrix a. So in this case, it would be, first of all, determinant of a itself would be a cube in a diagonal matrix. We just multiply all the elements in the diagonal. So this will be a cube raised to the power of 3 minus 1. That's going to be a to the power 6. So option number 3, of course, becomes the right option. And again, psi is the first one to answer this correctly. Well done, Psi. Next is the differential equation of the family of parabolas. y squared is equal to 4ax, where a is a parameter. So this is your question number 18. So mostly people are with option number 1. So let's quickly do that. a is a parameter, so we need to get rid of a. So it's very obvious that if you differentiate it once with respect to x, we get 4a, correct? So we just have to replace it back over here. So y squared will be 4ax, 4ax will be 2xydy by dx. So you can cancel off 1y. So y by 2x is equal to dy by dx would be your required differential equation. That's option number 1 is correct. Absolutely. So the first one to answer this again is psi. All simple questions is just a matter of speed. So we'll now talk about the next one, which is question number 19. So this is a clear-cut case of a homogeneous differential equation. So since I can see terms like y by x appearing everywhere, you can write it as y by x plus tan of y by x. We can take our y to be vx. So dy by dx is going to be v plus x dv by dx. So when we replace over here, we get x dv by dx plus v is equal to v plus tan v. v and v gets canceled. So this becomes dx by x is equal to dv by tan v. Tan v is nothing but cot v. The integral of this is going to be lnx is equal to ln of sine x. And you can add a factor of c as lnc over here. So ln of cx is ln of sine y by x. So that implies sine y by x is going to be cx. So option number 2 is going to be the right option. And the first one to answer this was by Eshnavi. Very good. So we have a lot of grounds to cover. Let's move on to the next question. Question number 20. The general solution of the differential equation dy by dx plus y times g dash x is equal to g dash x times gx, where g is a given function of x, is which of the following options? So Sanjana thinks something else. So let's discuss this. So it's the clear-cut case of a linear differential equation. And in this case, our integrating factor is going to be e to the power integral of g dash x. That's going to be e to the power g of x. That becomes y into e to the power g of x is equal to integral of e to the power g of x times g of x, g dash x. So if I take g of x be, let's say, t, so I can say g dash x dx will be dt. So the right-hand side part, I can write it as integral of e to the power t into t dt. OK? So I can apply integration by parts, treating this as my second function, treating this as my first function. So that's going to be our t times e to the power t minus e to the power t plus c. So this is going to be y times e to the power g of x. So rephrasing it, so y times e to the power g of x is e to the power g of x, g of x minus 1 plus c. So let me bring every term to the one side. So let's bring every term to one side. So e to the power g of x times y plus 1 minus g of x is equal to c. If you take log, so it becomes g of x plus log of y plus 1 minus g of x is equal to log c, which is another c. And I can clearly see that option number 2 is going to be the right answer in this case. So option 2 is the right option. And I think the first one to get this correct was, guys, can you see the screen? Yeah, the first one to get this correct was Sanjana. Sanjana is the only one who can get this correct. So option number 2 is going to be the right answer. Well done, Sanjana. So some questions may surprise you. So don't worry about that. Let's say if it takes more than one minute also, so don't get disheartened because you've already saved a couple of minutes in some of the questions. For example, this is one of the question which can save you a lot of time. By the way, there is a small typo here. This is plus sign over here. The question is the product of the degree and order of this differential equation. The product of the degree and order of this differential equation is question number 21. It has to be four because order is also two, degree is also two. So the product is going to be four. Option number one is going to be correct. So such questions which take around hardly three or four seconds can be used to balance those questions which are taking you more than a minute time. Next, integral of x to the power 27 times cos x plus e to the power x from minus one to one. Again, I don't think so. This question is more than 30 seconds worth. Question number 22, awesome. So now you see that whenever your limits are negative, the integral of odd terms become zero. So this is an odd term for us, so it will become zero. So it just converts it to this, right? Which is e to the power x itself from one to minus one. So it's e negative of e to the power minus one. So e minus one by e, which makes the option number three the right option. Pretty simple question, guys. Do not waste integrating minus one to one x to the power 27 cos x because this answer is going to be zero because it's an odd function. All right, so we'll move on to the 23rd one now. The value of this integral, okay? I'm getting two different answers from Sai and Vaishnavi. So let this be i. So according to King's property, I can write this as this 2n being an even number. It will not affect any of the sign terms, okay? This is by the use of King's property, okay? Let's add these two i's. So when I add these two i's, I will get pi times zero to pi, sine to the power 2nx by sine to the power 2nx plus cos to the power of 2nx dx. And now we can do one thing. I can half the limit because let's say I treat this as my function f of x. f of x and f of pi minus x are same. So I can half this and make this two. So this one goes off, okay? Again, if I use my King's property, I can write the same expression as zero to pi by two, cos to the power 2nx by cos to the power 2nx plus sine to the power 2nx, okay? Adding up these two will give me two i's equal to pi, zero to pi by two of one dx. So two i is equal to pi times, this is going to give you pi by two. So i is going to be pi squared by four, okay? So option number three is the right option in this case. Option number three becomes the right option in this case. So a psi has got the right answer in this case. I was slightly surprised because nobody else has responded to this. Was this a difficult question? By no means, okay? So we have to use our King's property twice. So here also I have made use of the King's property. Now move on to the 24th question. Integral of x cube minus one by x cube plus x. Guys, it's all about speed. I don't think so this is a difficult concept at all, okay? Mostly people are going for option two, okay? So let's discuss this. I can write this integral as x cube plus x minus x minus one by x cube plus x, right? So this gives me one minus one plus x by x cube plus x integral with respect to x, okay? So this one will integrate to x, no doubt about that. And let's spend some time integrating x cube plus x this term. Now here, I can again split this as integral of x by x cube plus x and integral of one by x cube plus x, okay? So let's split this as two separate integrals. So here x, x, x will get canceled. So I will have one by x square plus one dx. And here I can make use of, here I can make use of partial fractions, correct? So here I can write this as x into x square plus one. So at least I would generate a tan inverse x, okay? So let me save my time and see if I could find an expression with x and negative tan inverse x. And I think most of them have the same combination. So this is not going to help me out. So I'm going to work on this expression. So I can write that as one by x and I don't need to actually go into complete partial fraction because by observation also we can split it up like this, right? I just have to get x square canceled off, okay? So this term is going to be integral of this. So integral of one by x, x square plus one is just going to be integral of this term. That's going to be ln of x minus half ln of x square plus one. Okay? So my complete answer would look somewhat like this, x minus tan inverse x minus ln x plus half ln x square plus one, which clearly is option number two. So two is the right option in this case. So let me see who is the first one to answer this. The first one to answer this correctly was Vaishnavi again. Well done, very good. Is that fine? Again, I would categorize this problem as a time consuming problem. So we need to compensate for these lost time by the easier ones. 25th question. If integral of eight cos eight x plus one over tan two x minus cot two x is eight cos eight x plus c, find the value of a. Find the value of a, okay? All right, so let's discuss this. So the common sense is that I'm not going to find the value of a by integrating this. Actually, I'm going to take a shorter route. I'm going to differentiate both sides. This will give me this, okay? And on this side, I'll get minus eight a sine of eight x. Let's use some shorter tricks cos eight x plus one is going to be a two cos square four x. And here I'll get tan two x minus one by tan two x. And this side also I can write it as eight eight times. Two sine four x cos four x. So cos four x and one cos four x will be gone. This will become tan two x into cos four x by tan square two x minus one. That's minus eight a into two sine four x, correct? Now if you see this term carefully, this term is actually negative tan four x, right? This term here is negative tan four x. So negative tan four x and cos four x will give you negative sine four x, right? And on this side we have negative 16 a sine four x. So sine four x, sine four x is gone. Negative sine, negative sine can be dropped. And thereby a is going to be one by 16. That's option number three that is correct. Okay, right. So the first one to get this answer, correct was, most of you have actually changed your answer. Right, Rithik? It's option number three that's correct. Be careful, speed sometimes kills, okay? Next, question number 26 for the day. If integral of x times f of sine x from zero to pi is eight times integral zero to pi by two f of sine x, then find the value of a. A here is a constant. Okay, so mostly people are going for four. This is a very simple problem. You just have to let's say zero to pi x sine f of sine x. You just have to apply King's property. So that becomes zero to pi pi minus x f of sine x only. Okay, now this can be done as pi zero to pi f sine x dx minus zero to pi. So bring it on this side. It becomes two times zero to pi x times f of sine x is pi zero to pi f of sine x. Now you want the interval to be half. So put a half and two here because you can do that because this function will not change. Two, two factor drops. So this now becomes your a, which is going to be pi. So option number four is the right option. Well done. Psi was the first one to get this correct. So we'll move on to the 27th question. Okay, Psi and Vaishnavi both go for option number three. What about others? Okay, so let's discuss this. I can write this as tan inverse of pi by two minus tan inverse of 21. This I can write it as pi by two minus tan inverse of 13. And this I can write it as pi by two minus tan inverse of negative eight, right? So this becomes plus over here. You can drop this and make a plus sign over here. Okay. So we have three pi by two minus tan inverse of 21 plus tan inverse of 13 minus tan inverse of eight. Is that fine? We all know that tan inverse of X plus tan inverse of Y. If X and Y both are positive and XY is greater than one, we follow the formula of pi plus tan inverse X plus Y by one minus XY. Okay. So what I'll do is I'll apply the same to these two. So three pi by two minus pi plus tan inverse of, this will become 34 by 21 times 13 is going to be 63, 21, 273 minus 272. And I think that's going to be negative one by eight. Okay. And we already have tan inverse eight waiting over here. When you open the bracket, it becomes negative pi by two, negative tan inverse one by eight. Okay. And negative tan inverse eight. So I can write this as minus cot inverse eight. And again, there's a minus tan inverse eight. So you can take a minus sign here. This will become pi by two again. So if I'm not mistaken, your answer should come out to be minus pi by two minus pi by two. Let's check our working if everything is fine. So this I wrote it as pi by two minus tan inverse 21. That's absolutely correct. This I wrote as pi by two minus tan inverse 13. That's absolutely correct. This is pi by two minus tan inverse negative eight. So that's becomes pi by two plus tan inverse eight. Let me just clarify this. Pi minus tan inverse cot inverse of this and pi by two and pi by two minus. Yeah, so that's absolutely correct. So pi by two, pi by two will combine to give you three pi by two and I just took care of this and I use the formula. Tan inverse X plus tan inverse Y is pi plus tan inverse X minus Y by one minus X Y. So this will give me minus one by eight. And if I open the brackets, I get negative pi by two. Oh, oh, oh, oh, oh, oh, my bad. That's positive pi by two. That's positive pi by two. Because none of these is a very rare chance that it should appear as your answer. So my mistake. So answer here is going to be zero, my dear friends. Option one is going to be correct, right? And many of you have claimed the answer to be pi, which is slightly alarming to me because then you are not out of your school mode. This formula has to be used. I think many of you are still continuing with the school mode that tan inverse X plus tan inverse Y is always tan inverse X plus Y by one minus X Y. So please revisit the formulas. Revisit the formulas. This is a mistake which many people would make, okay? That formula which you have learnt in school works only when your XY is less than one. In this case, my X into Y was absolutely greater than one. And hence, the school formula does not work in that case. Is that fine? It happens, it happens. Since you're all out of the board exams, it is a natural tendency to take these formulas slightly, lightly. So this was an eye opener, okay? Great. So let's move on to the next one. Again, from inverse trick, this is your question number 28. All right, so people are going for option number four. Let's check this out. So first of all, I need to convert these cos inverses and sine inverses to tan inverse, right? So for that, I will quickly make use of the reference triangle. So let's say this angle is theta and I take this as theta. So this will be one. This will be five root two, right? So this obviously would be seven, right? So theta is nothing but tan inverse seven. Let this angle be five, okay? And again, I'll make use of a reference triangle with this as five. So four by root 17. So this has to be one. So five is going to be tan inverse of four, okay? So it's tan of tan inverse seven minus tan inverse four. So that's going to be tan of... Now, again, remember your formulas correctly. That's most important part in inverse trigonometry. Tan inverse X minus tan inverse Y. You can write this as tan inverse X minus Y by one plus X, Y. Only when your X, Y is greater than minus one. Only when your X, Y is greater than minus one. Which happens in this case because 28 is greater than minus one. So I can use seven minus four by one plus 28. And that's going to give you three by 29. That's option number four. That's going to be correct. Absolutely. So the first one to get this correct is Psi Mir, right? So let's move on to the 29th question. Psi says three while Sondarya and Ethvik are of the opinion that is going to be two. Okay, so this is pretty simple. So basically you have to do just one cube, two cube all the way till 20 cube minus one cube, two cube all the way till 10 cube. Okay? So that's going to be n into n plus one by two. So this is going to be 20 into 21 by two whole cube. And this is going to be 10 into 11 by two whole cube. So this is going to be, I'm sorry, not whole square. Yeah, I've already used the formula. So that will be whole square, yeah. So this is going to be 210. That is four, four, one, double zero. And this is going to be 55 square. 55 square is three, zero, two, five. So the difference is going to be four, one, zero, seven, five. So this is clearly an odd number divisible by five. That means option number two is going to be correct in this case. So the first one to get this correct was Sondarya. Absolutely. So let's now move on to the 30th question. We are halfway through. So we have 30 more questions to go. So we have to be pretty fast. We have just around one hour, 25 minutes more. Okay. So the class of 175 students, the following data shows the number of students opting one or more subjects. Mathematics is 100, physics 70, chemistry 40. Maths and physics is 30. Maths and chemistry is 28. Physics and chemistry is 23. All of them is going to be 18. So whenever such a problem comes, it is best solved by making Venn diagrams. And we normally start it, and we normally start filling from that position where there's a maximum number of overlap. Okay. So let us consider this to be your maths, physics, chemistry. Maximum overlap happens here, 18. Okay. Now maths and physics was 30. So this has to be 12. Maths and chemistry was 28. So this has to be 10. And physics and chemistry was 23. So this has to be five. All together in maths, they were 100. So we have already accounted for 40. So that's going to be 60 over here. Okay. Now this actually answers the question because the question is asking the number of students who have opted for maths alone. So maths alone has to be 60. Option number three is correct. So well within one minute, absolutely correct. The first one to give this answer was Sanjana. Oh, I'm so sorry. Sondarya. Great. So guys, moving on to the next one, which is question number 31. And again, this is a super simple question. All right. So total number of proper subsets, we know that it's going to be two to the power, the number of elements in the set minus one. So two to the power four minus one, that's 15. So no brainer questions. So absolutely correct. So option number three is the right option in this case. Let's move on to the next one, question number 32. Guys, remember we have 28 questions more to go. This is not multiplication sign. This is 26.25. Read it like that. Yes, anyone? Okay. So mean is quite clear. So mean is going to be six plus seven plus 10 plus 12 plus 13 plus four plus eight plus 12 divided by total number of data, which is going to be eight in number. So if you add this up, it's going to be 13, 23, 35, 48, 48 and P8, 68, 72, 72 by eight, which is going to be nine. So these two are ruled out. Okay. Now variance is nothing but summation of mod of xi minus x bar square by total number of data. Right? So let's start with the six minus nine. Six minus nine is three. Three whole square is going to be nine. And this is going to be four. This is going to be one. Again, this is going to be nine. This is going to be 16 from a five. This is going to be 25. Again, from eight is going to be one. And again, it's going to be nine. So whole divided by eight. Okay, let's add this up. So 13, 14 and 14 plus nine, 23. 23 is 39, 39 and this is going to be 64, 64 plus 10 is going to be 74, 74 by eight. Is that what you're getting? 74 by eight, 74 by eight should be approximately 9.25. That means none of the options are coming out to be correct, right? None of the options are there. Yeah, this rule should not have been there actually. Yeah, yeah, yeah. Then option number two would be correct. Anyway, this is a doubtful question. So I'll move on to the next one. Question number 33. You know, they clearly asked about variance, not standard deviation. So they must have been a typing error. Don't worry much about that, okay? So your A is clearly on the Y axis. B is your origin and C is A comma zero. So the medians AD, this is your median AD, they are perpendicular to each other, right? So the slope of AD, it's very clear from the diagram that the slope of AD will be minus B by, minus two B by A, okay? And this point is going to be A by two comma B by two. So slope of BE is going to be B by A. So minus two B square is equal to negative A square. So two B square is equal to A square, which means A is equal to plus minus root two B. Option number four becomes the right option. An easy question. Let's move on to the 34th question. Let A and B be non-zero reals such that A is not equal to B, then the equation of a line passing through the origin and the point of intersection of these two lines. All right, so if you have to find a line which is passing through the intersection of these two lines, it's better to start with the, I think this is Y, by mistake it has X. It's better to take the concept of family of lines in this case, okay? And now it is passing through origin means, origin would satisfy this. So if you put the value of X and Y both as zero, you get lambda value as negative one, okay? So the moment you put lambda value as negative one, you get something like this equal to zero, which is X times one by X times A minus B plus Y times one by B minus one by A is going to be zero. So X minus Y is going to be zero, which is going to be option number three, that's correct. Okay, so guys, whenever they talk about point of intersection of lines, it's better to take the approach of family of lines that would be very, very helpful in this regard. Let's move on to the 35th question. A straight line meets the coordinate accesses at A and B so that the centroid of the triangle OAB is one comma two, then the equation of the line AB is, okay? So that's a pretty, pretty simple question. So OAB, okay, this is the triangle and the centroid is one comma two. So let's say this coordinate is zero comma B, this A comma zero. So A by three is going to be one, so A is going to be three and B by three is going to be two, so B is going to be six. So X by three plus Y by six is going to be the one, that's two X plus Y is equal to six. Option number two is correct, absolutely. Again, Vaishnavi was the first one to get this right. Great, good speed. Let's now talk about the next question, that's question number 36. The direction issue of which of the line is perpendicular to the two lines, okay? So the best way to solve this is start checking your points. Else you can say we have to solve these two equations, okay? And if you use your cross multiplication method, that'll give you six minus two, which is going to be four, and this is going to be one plus four, which is going to be five, and this is going to be four plus, four minus minus three, which is going to be seven. So four, five, seven is the one, which is going to be your right answer, option number one, well done. Next is your question number 37. A line makes an angle of 45, 60 with the positive direction of X and Y axis. With the positive direction of Z, the line makes which of the following angles? These are all time saving questions. So let's be very fast, try to solve this within 10 to 15 seconds, not more than that. Okay, so we know L squared plus M squared plus N squared has to be one. So cos squared 45 is half, cos squared 60 is going to be one fourth, right? So N squared is going to be one minus three fourth, that's going to be one fourth again. So N is plus minus half, that means if let's say the angle is gamma, gamma is plus minus half, so gamma may be 60 degrees and 120 degrees. Okay, so option number three becomes the right option. But wait a second, they're asking with the positive direction of Z axis, so it has to be 60 degrees. So option number one is correct, so this has to be dropped off. This has to be dropped off, yes, yes, yes. So now let's move on to the next one, which is question number 38. And after this question, we'll take a five minutes of break and resume with the remaining 22 questions. Find the number of solutions for this equation, okay? Three solutions, okay, so let's discuss this. Now, if I say my quarter fix is positive, that means my X should lie either in the first quadrant or in the third quadrant, right? If that happens, I would be seeing an equation of this nature, okay? Which clearly implies that I would get something like this which is not possible, okay? Because we know that cosec X belongs to the interval R minus one to one, right? In other words, my cortex cannot be positive, it has to be negative, right? If it has to be negative, that means my cortex is negative, means X has to be either in the second quadrant or in the fourth quadrant, right? So in that case, I would get something like this, which means minus two cortex is equal to one by sine X, which means minus two cosec by sine X is equal to one by sine X. And of course, we know sine X cannot be zero because if it does, the equation should not hold to itself. So cosec should be negative half. Now, in the second and the fourth quadrant, only possibility that you can get as a negative half when your angle is two pi by three. So this is the only solution. So you'll have one solution as your answer, which is option number two is correct. Okay, is that fine? So guys, here we'll take a break for five minutes, okay? And we'll resume with more problems. Is that fine? All right, thank you. We'll exactly meet at 525. All right, so welcome back. So let's move on to the next question, which is your question number 39 for the day. So it's a pretty simple question based on trigonometry. And you can always choose a special value of X to solve such questions. Yeah, so here you can actually simplify as well. For example, numerator, you can simplify it as two cos two X into sine X divided by negative cos two X. So basically it gives you negative two sine X. That's option number one, that's correct, okay? All right, next moving on to the question number 40. In a triangle, A, B, C, find the value of cosec A times sine B cos C plus cos B sine C. And again, I think this is not more than a 10 second question. Absolutely, so this is going to be cosec of sine B plus C. And we know in a triangle, sine B plus C, sine B plus C is going to be sine A itself and you get the answer as one. Again, these are just 10, five seconds questions, not more than that. So let's move on to the next one. That's question number 41. I don't think it would be negative two sine X because sine square X minus cos square X is negative cos two X. So given two vectors, i cap minus j cap and i cap minus two j cap, the unit vector coplanar with the two given vectors and perpendicular to i minus j, okay? Again, the best way to solve this is see which of the options are going to meet your requirement, right? So the first one, of course, meets your requirement, but the second one doesn't. But there's a possibility of plus minus both meeting the requirement over here. So we can say option number three will be the best option to go for because there can be two possible vectors which are coplanar to this vector and perpendicular to this one, okay? Other way to solve this problem is we have to consider a vector which is of the form lambda times i minus j plus mu times i minus two j, okay? Use the fact that, let's say I take i common, so I have lambda plus mu plus j minus lambda minus two mu. So first of all, we can use the fact that lambda plus mu plus lambda minus two mu will be equal to zero because the dot product of this with this should give you zero. So that brings you to the fact that two lambda minus mu is equal to zero. Oh, sorry, there's a negative sign here as well. Something once again, plus. So this will be plus sign if I'm not mistaken, yeah, yeah. Okay, so in this case, and this would be j is negative lambda, yeah. So this will give me two lambda plus three mu equal to zero. And next thing is that we know that this vector has to be a unit vector. So lambda plus mu square, okay? And lambda plus two mu square should be equal to one, right? Again, this will lead me to two different equations, but this would be a lengthy approach because it's much faster to actually take up your vectors and check the dot product with i minus j to be zero. And of course it should have a unit of, I should have a magnitude of unity. So third is the most exhaustive option among all the three options. In fact, two is not correct. So I would go for option number three. Next, if A, B, C are three non-zero vectors such that each one of them are perpendicular to the sum of the other two, then what is the value of mod A plus B plus C square? So this is question number 42 for you. Okay, so if each one is, guys, can you see my screen? Yeah. If each is perpendicular to the sum of the other two, that means this is zero. That means this is zero and that means this is zero. Okay, if you add them, you get two A dot B, B dot C, C dot A to be zero, right? And we all know that mod of A plus B plus C square is going to be mod A square, mod B square, mod C square plus two times this, which is going to be zero itself. And hence the option number A exactly matches with our requirement. Yes, so absolutely correct. Vaishnavi, Sanjana, Vandarya, option A is correct. Option one is correct. In fact, let's move on quickly without wasting much time to question number 43. The slant height of a cone is fixed at seven centimeter. If the rate of increase of its height is 0.3 centimeter per second, then the rate of increase of its volume when its height is four centimeter is. Okay, let's discuss this quickly. So volume of this cone would be one third pi R square H. Okay, and I can actually write my, we know that slant height, H and R, we can write R square as L square minus H square. So I can write this as L square minus H square times H. Okay, and the slant height is fixed. Slant height is going to be seven. So volume is going to be one third. This is going to be seven square, which is 49 minus H square into H, which is clearly 49 H minus H cube. Now if you want your volume rate of increase, we just have to differentiate both sides with respect to T. That's going to be one third pi 49 D H by DT minus three H square D H by DT. Okay, so your D H by DT is 0.3. That's the rate at which the height is increasing. And you want it at a height of four centimeter. So you can just use H as four. Okay, so that's going to be 0.1 pi. And this is 49 minus 48. That's going to be pi by 10 centimeter cube per second. So I think option number four is correct in this case. Absolutely. Let's talk about the next question, which is your question number 44. If S square is AT square plus two BT plus C, then the acceleration is, okay, psi has given the response. All right, so let's discuss this. So we have been given that S square is this. So the first operation that we can do is we can differentiate both sides with respect to time. So that will give you this, right? In other words, you can drop the factor of two from both the sides. You'll get this expression, right? So DS by DT is going to be AT plus B by S. Let's find the double derivative of S with respect to T. And that would be S times A minus AT plus B, DS by DT divided by S square. Now DS by DT itself is, DS by DT itself is AT plus B by S. So in that case, if I, okay, we have an S square down as well. I'm sorry, S square, yeah. So in that case, if I simplify this, I will get A S square minus AT plus B the whole square by S cube, correct? And A square will be nothing but A square T square plus two ABT plus AC. And when you subtract the square of this term, that means you're subtracting this. So this will go off, this will go off, leaving you with AC minus B square by S cube, which clearly states that it is proportional to one by S cube. So it is independent. This is a constant independent of time. So it is going to be proportional to one by S cube, inversely proportional to one by S cube. So option number four is going to be the right option. Absolutely. So the first one to answer this question correctly is PSI, PSIMEAR. Let's continue with this momentum and try to solve the next one, which is question number 45. Zero typical question from rate measure. So we have a situation like this where there's a ladder of five meter length leaning against and this bottom of the ladder is slipping away at the rate of two meters per second. That means dx by dt is two, okay? So we know this x square plus y square is going to be 25. So two x dx by dt plus two y dy by dt is going to be zero. So they're asking you what is the decrease of the height at the instant when the ladder is four meters. So at the instant when this is four, this has to be three. So we'll put the values over here. So two into four dx by dt is two. In fact, I could have easily dropped the factor of two unnecessarily we are writing it. And y is going to be three dy by dt. So dy by dt is going to be negative of eight by three meters per second, suggesting that there is a fall in the value. And hence, since there's already a word decreasing mentioned over here, option number two is going to be the right option. Well done. Next, let's talk about question number 46. Yes, guys. Any response for question number 46? Okay, Sondarya, anybody else? So let's quickly solve this. First of all, we like to solve these two curves simultaneously to see where do they intersect. So y square could be replaced with two x square by a whole square is equal to four x, which means four x cube, four x to the power four is equal to four a cube x, right? Which clearly implies x could be zero or x could be a. If x is zero, that means they both meet either at zero comma zero or at a comma. So let's see what happens at zero comma zero. So when you find the dy by dx for both the curves, so two y dy by dx is equal to four a. So for this point, dy by dx becomes four a by two y. Correct? Similarly, in this case, dy by dx becomes four x by a. Now in this case, you'll realize that this goes to infinity, right? And as x tends to zero and this tends to zero, correct? So this point, the curve is meeting at, in fact, we can check from the diagram as well. At this point, both the curves are meeting at right angles to each other. As you can see, this is the tangent to the curve. Ay is equal to two x square. And this is the tangent to the curve. This is the tangent to the curve. Y square is equal to four x. So this has to be 90 degrees. Okay, so this cannot be your, any one of the options. So we have to check at the other point, a comma two a. So at a comma two a, ma will be nothing but four a by four a, which is going to be one. And here it will be four a by a, which is going to be four. So M two is going to be four. So the angle between the tangents at the point a comma two a would be M one minus M two by one plus M one M two. And that is going to be tan inverse of, right? You can take the modulus of this angle of intersection because we normally take the acute angle of the point of intersection. So this is going to be one minus four, that's three by five. Okay. So tan inverse three by five is going to be the answer in this case. So which is option number two. Is that fine? Clear? Let's move on to the next one, which is question number 47. The area bounded by the curve y equal to x cube, the x axis and the ordinates at the point x equal to minus two and x equal to one. Yep, pretty simple question. So x cube, this is the graph. And you want to find out the area trapped between minus two and one. Okay. So in case of minus two to zero, you would actually be integrating negative x cube. And from zero to one, you will be integrating x cube. So this will give you negative x to the power four by four and x to the power four by four. So this is going to be one fourth. And this is going to be plus 16 by four. So that's, so answer will be 17 by four, which is option number D or option number four. When they say area, you have to be calculating the absolute value of this integrals. Next question number 48. The area in square units bounded by the normal at one comma two to the parabola y square is equal to four x, the x axis and the curve. So let's quickly draw the diagram for this particular situation. So we have a parabola y square is equal to four x and we are drawing a normal at a point one comma two like this. So they're mainly concerned with the area of this part. So this is the area that we are actually looking out for. So at one comma two, if I draw the equation of a normal, we know that one comma two is basically corresponding to a point t equal to one, correct? So I'll use the readymade formula that we know y plus tx is equal to two at plus at cube and a is one over here. So that'll give me x plus y is equal to three. So this is going to be my normal, correct? So it's better to solve this problem by choosing horizontal strips. So x for this point will be three minus y, x for this point will be y square by four. So the difference in their x would be this and you just need to integrate this from zero to two. So that would give me three y minus y square by two minus y cube by 12. Okay, from zero to two, that gives you six minus two minus eight by 12. That's a four minus two by three. And if I'm not wrong, that's 10 by three units. 10 by three is going to be option number one. Exactly, option number one. All right, so let's move on to the 49th question. The area bounded by cos x and sin x between zero to pi by two is. So we need to figure out this area. So as you can see, the area would be nothing but the area from zero to pi by four of cos x minus sin x cube because cos x would be higher in value than sin x in the interval zero to pi by four, correct? So integral of cos x is going to be sin x and integral of sin x is going to be negative cos x. So it's going to be this from pi by four to zero. So that's going to give me root two, root two minus zero plus one. That's going to be root two minus one square units. And that's option number one that is correct. So never hesitate in making the curve because that actually gives you a clear cut idea which curve is above the other, which curve is below the other. And then you can make a differential strip and find out the area in the desired interval, okay? So let's make a half century now. Let's move on to the question number 50. The area in square units of the region bounded between y square is equal to nine x and y is equal to three x. Yeah, okay? All right, so let's check this. So you just have to integrate a differential strip chosen over here. So y at the top is three root x. y at the bottom is three x. We just need to integrate this from zero to one. So three times, this will be x to the power three by two into two by three minus x square by two from zero to one. That gives you two by three minus half. That's four minus three, one sixth. So your answer is going to be half square units, half square units, okay? Okay, so this is a case which comes very, very regularly. So normally there is a readymade formula which says that area between y square is equal to four x and y is equal to mx is eight a square by three m square square units. So if you remember this formula, you really save a lot of time in this question. So this is a very, very specific formula, only meant to find the area between the curve. y square is equal to four x and y is equal to mx. So in the present question, my a was actually nine by four. So eight into nine by four whole square by three into m square. m square is going to be three square, correct? So if you solve this, it is going to be eight into nine into nine by four into four into three square into three. So this is going to get canceled off. This is going to be two. And this is going to be again a two. Let me just rewrite this again. I think I've messed this up, yeah. Yeah. So it's eight into 81 by 16 into 27. So that's going to be a two. And this is going to be, no, this is already going to be three m square. No, I'm sorry, this is going to be three m cube, not three m square, three m cube. So three m cube will give me 81. So that would get canceled, giving you the answer as half. Yeah, so small change. The formula is eight a square by three m cube. Anyways, this is a very, very specific formula. So even if you don't know it, that's fine. That's only meant for the case of a parabola, y square is equal to four a x and the line y is equal to mx. Next is the question on binomial theorem. This is question number 51. Hello, Shantanu, how are you? Hope your exams went well, the board exams. Yes, so in this case, just to save our efforts, first of all, we'll pull out a two to the power seven out and we'll start referring to this expression as a, okay? So it's one by one plus a to the power of seven minus one minus a to the power of seven. So if you expand this expression, we get one plus, okay. Now, let's not write everything because some of the terms are going to get canceled off. So one plus seven a, then seven c two, that's going to be 21 a square, then a seven c three, which is 35 a cube, okay? Seven c four, which is 35 a to the power four, then 21 a to the power five, then seven a to the power six plus a to the power seven. And this term will give me one minus seven a plus 21 a square minus 35 a cube plus 35 a to the power four, minus 21 a to the power five, minus seven a to the power six, plus a to the power seven. And when you finally subtract it, you're going to get seven a 35 a cube, 21 a to the power five, and seven, oh sorry, this is plus and this is minus, a to the power seven twice stop. And outside we already have two to the power seven and a. So one of the a's will lose off. So we'll have a to the power six as the highest degree term, a to the power six will be the highest degree term. That means three a plus one to the power three will come up. And this will lead to a polynomial of degree three. This will lead me to a polynomial of degree three. So we don't have to write everything here. We have to be carefully selecting out the terms which we need to write. And of course use of this made our task easier. Is that fine? So option number four becomes the right option in this case. We could still not cover up nine more questions and I'll be sending you on the group to solve it. Okay. So we'll end the session now. I hope to see you again tomorrow with the session on JEE main problem solving. Thank you everyone for coming online over and out from syndrome Academy. Have a good day. Bye bye, take care.