 Hello, good afternoon, greeting from symptom Academy. So welcome to this YouTube live session on discussion of board level questions. So those who have joined in this session, I would request you to type in your names in the chat box so that I know who all are attending this session. So good afternoon, everyone. Good afternoon, Sai, Kushal. Kushal, long time. No see. So guys, let me set the agenda for today's session. In today's session, we are going to just take up board level questions that has been asked in the previous years. And we'll be also discussing upon the strategies that will help you to get more marks in your board exams. So before I start solving the problems, let me just brief you up about how the marking is done in board exams. So most of you would be already been briefed up by your school teachers. But to just recap the same things, first of all, try to understand that this exam is an exam of selection. Okay, where the examiner wants to allot you as many marks as possible. And for him to do that, you should have written some stuff in your paper, right? It is not an exam where you just write an objective answer and you move on, unlike you do it in your JEMN exam. So here you have to be as elaborative as possible as, you know, write it in as unambiguous way as possible so that it's very easy for the examiner to understand what you have written. All of you would be knowing that the marks allotted for any question are of three types. One is called the independent marks, okay? Independent marks is a marks where you actually write a formula which is pertinent to that particular topic. Let's say you have been given a question to find out the equation of tangent to an ellipse. Okay, let's say, okay? So if you just write down the equation of an ellipse, you will get some marks which we call as the independent marks. Means it is independent of whether you have solved the problem or not. You have written a formula which is pertinent to that particular topic and you get that marks, okay? Second type of marks which is allotted is called the method mark, okay? Method mark is a mark which actually is given if you write a formula which is helping you to solve that particular question. For example, in equation of a tangent, you know, you need a slope of the tangent and for finding the slope you need dy by dx at that point and you know the equation of a line which is y minus y1 equal to slope times x minus x1. So whether you have solved the question completely or not, if you have just written this method, you will be allotted some marks, okay? And it depends upon the chief examiner. And the third type of marks which is allotted is the answer mark, okay? Answer mark is when you finally obtain your answer. So I think maximum one or half marks is kept for the answer but the major part of the marking is being distributed among independent and method mark. So even if you don't get the answer right, you will be still getting the marks which are allotted for these two. Second important thing that you need to know is always put your best step forward. That means do not try to solve a question. Let's say if you have two different methods in mind, okay? Always write the best method first because many a times, not many a times always, your marks whenever they are allotted, it is actually posted in front of your answer script. So let's say I give you four marks on question number three. The moment I give you four marks, I will post that four in front of your answer script. So let's say if you have attempted that question number four once again later on, I would simply strike it out because I don't want to correct it again because I have already given marks for it. So first of all, don't try to attempt a question more than once because the second option will be definitely be striked out. And many a times what happens, let's say if you attempt it at the same place, many a times there are internal options. So if you attempt in CBC, there are three internal options for four marker question and three internal options for six marker. So many a times what happens if the examiner is lenient enough, what he will do is he will sit with a blue pen. Okay, this is something which is very internal and he will just strike out that part of this answer where you have got the least mark. But don't expect it to happen every time. So only when the examiner is very lenient, he may give you the best of the two attempts that you have actually taken on that question. But normally what happens? The moment they see you have attempted a question twice, the second option they will strike out. They will not bother to see whether you have got because they have a limited time to correct your paper. They are not in love with your paper. Okay, so these things you must keep in mind. Secondly, not secondly, thirdly in fact, whenever you are answering questions in physics and chemistry or for that matter in maths, try to understand the directive of the question. Okay, directive of the question is very important. Directive of the question means the first line of the question, explain what is, how, illustrate. These are called directive of the question. According to the directive of the question, try to answer it. For example, if they ask you what is this, that means they are asking you the definition of this. If they are saying illustrate this, means you have to use an example to bring out the concept. If they are asking you to describe it, that means you have to explain the concept along with an example. Probably you have to make a diagram as well. And always have a habit of underlining the keywords because that will make the life of the examiner easy. So always, after, I mean complete your paper 15-20 minutes early and sit and underline the important things with pencil so that he knows that you have understood the basic concepts and integrities of that question. Okay, and please do practice at least 5 mock papers before you write your board exams because it will give you an idea how to manage your time in the examination. Is that fine? Okay, in mathematics especially, please use mathematical statements wherever you can. Do not try to adopt any method which is not prescribed by your syllabus. Okay, for example, for evaluating determinant do not use SARAS method. Okay, because if you get a wrong answer you cannot demand or you cannot, let's say, claim a method mark or an independent mark because it is up to the examiner. So assume that the examiner doesn't know anything beyond the syllabus and you have to convince him with those answers which is within the syllabus prescribed to you by NCRT. Okay, anyways. So with this we start the session of today. First question here is find the distance between the two planes given by this vector form R dot 2i minus 3j plus 6k minus 4 equal to 0 and R dot 6i minus 9j plus 18k plus 30 equal to 0. Many times people ask me, sir, how much time should I devote to a particular question? Should I start with a six marker or should I start with a one marker? The idea is that the amount of time that you should dedicate to a given question, let's say having X mark, if a question is carrying X marks, the time allotted should be 1.5x. This is a simple rule that you should follow. So if it is a one marker, give 1.5 minutes. Okay, this many minutes. If it is a four marker, not more than six minutes. If it is a six marker, not more than nine minutes. Okay. So please ensure you are not devoting more than 1.5x to that particular question. Now, whether to start with a six marker or four marker, if you follow this time regime, you will never be following short of time. You will be always completing your paper much before time. Okay. Many people feel confident attempting six marker because they have ensured the heavy markers. Okay. Many people feel that if they do more questions, they will feel more confident. So they start with one marker. So that is up to your psychology. So I'm not giving any opinion on that. All right. So starting with this question. Yes. Anybody who has got the answer, please post it on the chat box. All right. So many of you are answering two units. Okay. So how do I solve this question? Very simple. So what I'm going to do is I'm going to first find out the distance of this plane. We know that both these planes are... Let me write it as 2i minus 3j plus 6k equal to minus 30 by 3. Okay. So that's going to be minus of 10. Now you know that these two planes are actually parallel planes. These two are actually parallel planes. Okay. So when two planes are parallel, okay, you can actually see what is the distance of the origin from these two planes. Correct? Now, many times there is a confusion that how do I know that origin is stated between the two planes or it is outside the two planes? So how do you know whether it's at this position, that is position number one, or this position, position number two? Okay. It's very simple. The sign of this... If you keep these two terms same, then the sign of these two numbers actually tell you the relative position of these planes with respect to origin. If they are of opposite sign means the origin is between the planes. That means the planes are on the opposite side of the origin. If they are of the same sign, that means they are on the same side of the origin. Correct? So at this particular question, you will see that since they are of opposite sign, the planes are on the opposite sides of this particular origin. Okay? So now, what is the distance of origin from a given plane? It's nothing but mod of the D term divided by under root of 2 square, 3 square, 6 square. So let me call it as p1. Okay? So let me call this as pi1 plane. Let me call this as pi2 plane. So this is pi1 plane. This is pi2 plane. So distance of O from pi1 plane is let's say p1. Distance of pi2 plane from origin will be mod of minus 10 by again under root of 2 square minus 3 square and 6 square. So this is going to give you 4 by 7. This is going to give you 10 by 7 if I am not wrong. Correct? And since they are on the opposite side, we need to add these two values. So the distance between the two planes, let me call it as p, will be nothing but p1 plus p2 which is 4 by 7 plus 10 by 7 which is going to be 2 units. Okay? Always write 2 with a unit that you do not know, so you just write units. So the first one to answer this correctly was Vaishnavi. Well done Vaishnavi. Good. Let me now move on to the next question. Question number 2 now. If a and b are vectors such that mod a is half, mod b is 4 by root 3 and mod of a cross b is 1 by root 3, find mod of a dot b. Again, please solve it in a way you need to solve it in the exam. Don't treat it as a JEE main question. Be as descriptive, as elaborative as you can. Of course for the one marker and all, you just need to write your answer within one line or one word. But for 4 marker, 6 marker and 2 marker, you need to be elaborative. Little bit. One unit. So we know that mod of a cross b is mod a mod b into sin of the angle between them. Correct? Which implies 1 by root 3 is half into 4 by root 3 into sin of theta which we don't know. Okay? So from here we can say sin of theta is nothing but 1 by 2. Okay? Sin of theta is 1 by 2. Which clearly implies theta could either be 30 degrees or 150 degrees. Right? Now a dot b is given as mod a mod b into cos of the angle between them which is nothing but half into 4 by root 3 into let's say if I take cos 30 degree what will I get? Write it down over here. I will get half into 4 by root 3 into root 3 by 2. So this will give you okay, I will get 1. Correct? If I take cos of 150 degree I will get half 4 by root 3 into minus of root 3 by 2. That will give me minus of 1. Okay? So let me call it as 1 and let me call it as 2. So we can say from from 1 into 1 or 2 you can say mod of a dot b is going to be 1. Okay? Again first one to answer this was Vaishnavi again. Very good. Yeah. Let's move on to the next question now which is coming to you from inverse signometry chapter. First question is prove that 2 sin inverse 3 by 5 minus tan inverse 17 by 31 is pi by 4. Again remember one thing we have not been taught the formula for 2 sin inverse x. Correct? So we need to convert things in terms of tan inverse. Just write done if you are done with the proof for the first one. Hello wishes. Good afternoon. Welcome to the session. Again just to reiterate this is a problem solving session. Don't expect j e questions to be solved over here. Okay? Sai is done. Who else? So now we are just hold on we will do the second part. So let us finish the first part of this question. We have to prove this first. Now we all know that sin inverse of 3 by 5 you can actually take help of reference triangle over here. Let's say this is theta and you can show the examiner that this is going to be 3 and this is going to be 5 so this is going to be 4. Okay? So tan of theta you can say from reference triangle from reference triangle tan of theta is going to be 3 by 4. So theta is going to be tan inverse of 3 by 4. Okay? Now we know that we know that 2 tan inverse of x is tan inverse 2x by 1 minus x square but remember this is valid only when your x lies between x lies between minus 1 to 1. Okay? It's always important to state the complete formula. Okay? It's always important that you write the complete formula that you know many people will skip writing this. Okay? That is important. So the left hand side which is 2 sin inverse 3 by 5 could be written as 2 tan inverse 3 by 4 and could be written as tan inverse of twice of 3 by 4 by 1 minus 3 by 4 the whole square. Okay? Correct? And if I am not wrong this is going to give you tan inverse of 24 by 7. Right? Okay? Now I just write it above. Now we also know that tan inverse x minus tan inverse y what is it? It's tan inverse x minus y by 1 plus xy under what situation? Under what situation? What should be the interval of xy? xy should be greater than minus of 1. Correct? Correct? Now in this case when you are finding tan inverse of 24 by 7 minus tan inverse of 17 by 31 we can see that xy is greater than minus 1 of course. So I can use the formula that I just wrote down x minus y that is 24 by 7 minus 17 by 31 divided by 1 plus 24 by 7 into 17 by 31. Okay? That gives you the result as tan inverse I think this is a long calculation to be done. Okay? But again there are ways to do the calculation faster. Just do 24 into 31 minus 17 into 7 divided by 24 into 17 plus 7 into 31. Okay? Now all of you please focus on this approach which I am going to tell you here. You can see a factor of there is no common factor that actually you can share but okay let's calculate this this will come out to be 625 and this will also come out to be 625. So this will become 625 by 625 which is going to be tan inverse of 1 which is pi by 4. Okay? Now again many people ask me sir how elaborative I have to be? See guys, you know you don't need to show each and every step of the calculation but try to show the essential steps. There is something called omission of essential steps. So you may lose marks because of OEW. So don't do omission of essential steps. So wherever like these formulas are essential steps so if you don't write these formulas you may not get the desired independent and the method mark but don't try to be as elaborative as possible that means the examiner doesn't want to know how you have simplified it. Okay? It's okay to directly jump from this step to this step you will not lose marks for doing that. Is that fine? So this is how we solve this question. So I think no problem everybody could solve this. Let's now take the OR part of the question. So I am just clearing the screen so that you can work on the so that I can work on the OR part. Yeah. So for all this equation cos of tan inverse x is equal to sin of cot inverse 3 by 4. Guys normally two types of question come in inverse trigonometry. Please focus on them. One is the proof that question and other is inverse trigonometric equations. Okay? Please pay attention to these two types while you are dealing with inverse trigonometric functions and of course the formula has to be on your finger tips. So these type of questions normally come from inverse trigonometry. Yeah. How do I solve this? Very simple. First of all I can write this as this equal to sin of tan inverse of 4 by 3. Okay? So tan inverse of 4 by 3 I can write this as sorry cos of tan inverse of 4 by 3 I can write it as or this you can write it as cos of pi by 2 minus tan inverse of 4 by 3. Okay? That clearly implies tan inverse x is pi by 2 minus tan inverse 4 by 3. Right? Or tan inverse x plus tan inverse 4 by 3 is going to be pi by 2. Okay? Which clearly implies that if you write this as tan inverse x plus cot inverse 3 by 4 equal to pi by 2 from the complementary angle properties that is tan inverse x plus cot inverse x is equal to pi by 2. Okay? We can claim that x is equal to 3 by 4. Is that fine? Now some people have written plus minus 3 by 4. Okay? Just verify whether plus minus 3 by 4 also will meet your requirement or not. Now 3 by 4 definitely will meet. So let's check minus 3 by 4 is meeting the requirement or not. So if you write cos of tan inverse of minus 3 by 4. Correct? Now this will actually give you pi minus cos inverse sorry cos of this will actually give you cos of minus of tan inverse 3 by 4. Correct? Which is as good as saying cos of tan inverse 3 by 4. So in this case yes both of these solutions can be possible. Yeah another method is you can draw a triangle and solve it. You can draw a triangle and solve it. So plus minus 3 by 4 both are possible. So another method to solve this is you take your tan inverse x to be angle theta. Okay? So tan of theta is going to be your x. So this is going to be x. This is going to be 1. So this is going to be 1 plus x square. Okay? So cos of theta is 1 by under root 1 plus x square. So left hand side will become this. Okay? Similarly this let's say cot inverse 3 by 4 is 5. So in the triangle if you know this is angle 5 then this is going to be 4 by 3. So this is going to be 5. So sin of 5 is going to be 4 by 5. So 1 by under root of 1 plus x square equated to 4 by 5. Okay? So from here you can so from here you can say 1 plus x square is going to be 25 by 16 which means x square is going to be 25 by 16 minus 1 that is going to be 9 by 9 by 16. So x is going to be plus minus 3 by 4. Okay? So both the possibilities are there. Great. So now we'll move on to the next question. This is from the integration chapter. Question number 4. First one is evaluate integral of sin square x by sin x plus cos x from 0 to pi by 2. Guys can you see the screen? Is the screen visible? Okay, done. People are done with the first part can start working on the second part. Sai can you type the answer that you have got for the first part? Just to be sure that you have done it correctly. Yeah, you can just PM me your solution on the WhatsApp. Please type done if you are done with the integration of the first integral. Alright, so since 3 people have responded, let me start discussing this. So for the first one, you will first call this as let's say integral i. So by King's property by King's property i could also be written as 0 to pi by 2 sin square of pi by 2 minus x by sin pi by 2 minus x plus cos pi by 2 minus x. So which is going to be integral 0 to pi by 2 cos square x by sin x plus cos x. Okay. So adding both the i's so 2i is going to be 0 to pi by 2, you are going to get sin square x plus cos square x by sin x plus cos x which is going to give you 2i is equal to 0 to pi by 2 1 by sin x plus cos x. Okay. Now there are various ways to solve this. One is by splitting up into half angles of tan okay. Or you could just multiply and divide with root 2. Okay. So let me take this approach where i 1 by let me take root 2 over here so 1 by 1 by root 2 sin x plus 1 by root 2 cos x. Okay. So i is going to be 1 by 2 root 2 0 to pi by 2 this is going to be sin of cos of because you can write any of the formula that suits you cos of pi by 4 minus x okay. Which is 1 by 2 root 2 0 to pi by 2 seek of pi by 4 minus x dx okay. And we all know the integral of that that's going to be you can do one more thing over here you can actually take pi by 4 minus x to be t. So dx will become negative dt and your limit of integration will be from pi by 4 to minus pi by 4 so you can always reverse it you can always reverse it and write it as pi by 4 to minus pi by 4 seek of t with a minus of dt and you can always write it as root 2 from pi by minus pi by 4 to pi by 4 seek of t dt okay. Now integration of seek is known to all of us it is ln of mod seek x or seek t plus tan t and from minus of pi by 4 to pi by 4. So I will make some space for myself by raising the initial part of the solution so let me raise this part. This gives me the answer as 1 by 2 root 2 and this will become log of seek pi by 4 is going to be root 2 tan pi by 4 is going to be 1 okay minus log of seek this. So it's 1 by 2 root 2 log of root 2 plus 1 by root 2 minus 1 okay. If you want you can rationalize it over here and write it as 1 by 2 root 2 log of root 2 plus 1 the whole square by 1. So you can always bring the 2 in front of it so it becomes 1 by root 2 ln of root 2 plus 1. So this becomes your answer. No need to put a mod over here because you know that root 2 plus 1 is always a positive quantity. So if you put a mod nobody is going to mark it wrong so this becomes your final answer for your i. Okay so most of you have got this answer try the second part or the other part of this question 0 to 1 integral of cot inverse 1 minus x plus x square guys let me just warn you against 2 type of questions which can kill lot of your time and another is determinants and one more type is maxima and minima right. If you don't take the correct approach to solve these 3 types of questions it may kill lot of your time right. So always be very very careful while attempting these 3 type of questions what are they integration, determinants and maxima minima questions second one okay tapas note down your answer let's see what others have to say so for the second one you first have to write it as tan inverse of 1 by 1 minus x plus x square guys remember one thing that tan inverse 1 by x is cot inverse x only when x is positive okay this formula is to be used only when x is positive and vice versa okay write it as tan inverse x is equal to cot inverse 1 by x when x is positive now remember this term this term will always be positive why? because it's a quadratic whose discriminant will be negative right if you see this is a quadratic expression okay and the coefficient of x square is positive and its discriminant delta will be b square minus 4ac which is negative so it will always be a hanging parabola so it will never cut the x axis so it will always be positive and therefore I could use this formula with surety so here I can write that 1 as first of all in the denominator I can write it as 1 minus x 1 minus x in fact you can write it as let me just write it as x x minus 1 okay so in the numerator also we can make this adjustment I can write 1 as this so this gives me the value of i as 0 to 1 tan inverse x minus 0 to 1 tan inverse x minus 1 correct now how do I integrate tan inverse x integrate tan inverse x I will have to use integration by parts okay so I will take this as my first function I will take this as my second function so it will become x times tan inverse x minus 1 by 1 plus x square into x so put a 2 over here make a half so this becomes x tan inverse x minus half ln of 1 plus x square now this part over here by king's property you can write it as integration 0 to 1 1 minus x minus 1 dx which actually becomes minus of 0 to 1 tan inverse x correct so your i is actually nothing but 2 times integral 0 to 1 of tan inverse x itself okay here we already know the formula so I can write it as 2 x tan inverse x minus ln 1 plus x square and just put the limit of integration 0 to 1 so let me make some space right here at top so that I can write down the final result for it so when you put 1 you get 2 pi by 4 minus ln 2 and when you put 0 you will definitely get a 0 everywhere so answer is going to be pi by 2 minus ln 2 and the first one to answer this was tapas awesome guys good so this question is actually a j e question which came long ago in j e now they have started asking them in c b s c exams guys one suggestion to the is c folks over here let me tell you that since is c is going the c b s c way you should not only attempt solving your is c previous year papers but also c b s c previous year papers because I have a feeling that they would be actually taking up questions from c b s c directly next question number 5 ensure in definite integrals if you are using a property do mention that property in the previous question we had used the kings property right so just don't write kings property because many examiner would not be knowing the name of that property okay so instead of writing kings property always write the property itself second thing is always try to substantiate your answer with graphs if at all you are if you have time but don't try to solve it only by using graphs graph should be used as a supporting tool rather than the initial tool okay many people think I can solve this very easily by graph and all because of their j e aptitude but don't do that in your board exams many examiners will not be aware of the graphical way of solving questions not to underestimate anyone but you know there are some there will be some you know examiner who will not be aware of that concept because they have not gone beyond NCRT syllabus okay so kushal has given the answer I mean he has given his response I am not commenting whether it is right or wrong as of now okay I need one more response to get started with this okay great so here what you do is first you take log x as t that means x will become e to the power t which implies dx is becoming e to the power t dt okay so this integral will convert to e to the power t log of t plus 1 by t square dt okay and since log t derivative is 1 by t I will introduce 1 by t and subtract 1 by t like this which will convert it to two separate integrals log t plus 1 by t dt and the other being e to the power t minus 1 by t plus 1 by t square dt okay so now it resembles so from the formula e to the power x f of x plus f dash x is equal to e to the power x f of x plus c we can write this as e to the power t log t plus e to the power t into minus 1 by t okay plus c e to the power t is x log of t will be log of log x and this will be minus e to the power t is again x and t is log of x don't forget to put this c towards the end you may lose out some marks if you don't put c so this is going to be your answer for the question now people ask me said do we need to derive this formula is this directly given in your ncrt textbook the answer is yes so no need to derive it derivation of those formula should be done which is not mentioned in your ncrt syllabus moving on to the next integration problem that's the sixth problem for the day integral of 1 minus sin x by sin x 1 plus sin x okay okay the bus registered your answer anybody else even if you're done just type done so don't have to write down the answer always but I need at least three response for me to start solving it okay so let's start the discussion of this problem so what we'll do is we'll multiply first the numerator and denominator by 1 minus sin x okay so I'll multiply the numerator with 1 minus sin x and in the denominator if I do that I will get 1 minus sin square x and this gives you 1 minus sin x the whole square you can actually expand it here itself rather than writing repetitive steps is that fine now here we can split the integral into the following integrals one is this other would be integral of sin x by cos square x and other would be minus 2 by you can say 1 by cos square which is actually secant square okay let me call this as integral i1 i2 okay now i3 is pretty easy i3 is integral of secant square which is going to be minus 2 tan x okay i2 in i2 you can actually take cos x as t so take cos x as t so minus of sin x dx will become dt okay so this becomes minus dt by t square and the integration is actually 1 by t okay so this will become ckx now problem comes with i1 so let us solve i1 separately just demarcate it for i1 again let me multiply and divide the numerator and denominator with sin x so it becomes sin x okay cos square x okay which you can actually also write it as sin x by 1 minus cos square x into cos square x now take cos x as t so again sin x dx will become negative of dt so it will become negative dt by 1 minus t square into t square okay which is nothing but dt by t square minus 1 into t square which you can actually also write it as t square minus 1 minus of t square now this integration is 1 by 2 ln of t minus 1 by t plus 1 and integration of this is 1 by t right so i will place back my t as cos x over here again and for that i need some space so i will just erase this part over here and i will directly write down the entire result so i1 will become half let us say i this is called i integral so i will become half log of cos x minus 1 by cos x plus 1 and there is a 1 by cos over here and there is a seek over here so it will become 2 seek and minus 2 tan so you can take 2 common this will become your final answer now this is subjected to further simplification but even if you leave it at this stage you will not lose marks okay so this can be accepted as your final result is that fine okay so i think tuppas should also get the same result if you simplify this further because see these trigonometric ratios can be intraconvertible okay so you can play with them to leave your answer in any form there is no end to that so tuppas was the only one to answer this well done let's move on to the next question this is question number 7 for the day find the equation of a normal to the curve ay square equal to x cube at a point whose x coordinate is am square yeah yeah yeah that's absolutely correct you can use partial fractions as well is it done so please type done if you are done with it so that we can start the discussion for this okay okay so i start getting the response so first of all since this is not a linear in y we have to know the y coordinate at the point of normal seek so to know the y coordinate we have to substitute x value as okay so y square is going to be a square m to the power of 6 so y can be both plus minus am cube okay now let us first consider so let the point be let the point be am square comma am cube there can be two possible points like that so we will take the positive one first now to find out the slope of the normal we first need to find out the slope of the tangent at that point that means dy by dx at that point and then negative reciprocate the answer so it will be a times 2y dy by dx is equal to 3x square so dy by dx is going to be 3x square by 2 ay okay so this gives you the slope of the tangent when you substitute the point x1 y1 so let this be x1 y1 okay so here you have to substitute x1 y1 so you get 3x square will be going to be am square whole square by 2 ay am cube so if I am not wrong this is going to leave you with 3 by 2 so slope of the normal is going to be minus 2 by 3m so equation of normal am square comma am cube is going to be y minus y1 so y minus am cube is equal to slope times x minus x1 so if I simplify this it will become 3my minus 3 am to the power 4 plus 2x minus 2 am square equal to 0 okay but remember this is the equation of a normal at this point in a similar way you need to find out the equation of normal at am square comma am cube but even if you don't do that you will not lose marks because the question never said equation of normals okay there is just one normal that they ask so any one of the answer is acceptable is that fine great so I think the first people to answer this wishes to got 3 y I think there is something wrong with your answer wishes just check so on area is correct again it's factor of 3 is missing okay anyways so we will move on to the next question question number 8 okay so mostly I am getting the oh yeah I am sorry for the type over here it's given that this function is continuous at x equal to 0 alright so if it is continuous at x equal to 0 implies limit of the function as x tends to 0 minus should be equal to limit of the function as x tends to 0 plus should be equal to the value of the function at 0 right so limit of the function as x tends to 0 minus you will follow this definition which is going to be k times sin 5 by 2 x plus 1 remember it's the straight forward case of substitution so when you substitute you get sin 5 by 2 into 0 plus 1 that's going to be k itself limit of the function as x tends to 0 plus would be tan of x minus sin of x by x cube now ISC people have the liberty to use lopital lopital indeterminate forms are there in your course but not the cbsc people okay so for cbsc guys you have to solve it the regular way that means you have to break up this as sin x 1 minus cos x by cos x times x cube okay limit x tending to 0 plus which you can write it as limit x tending to 0 plus sin x by x and 1 minus cos x by x square and 1 by cos x so you can break this up as separate limits if you want that is sin x by x is separate into limit x tending to 0 plus 1 minus cos x by x square is separate and limit x tending to 0 plus 1 by cos x is separate which is going to give you half 1 into half into 1 that is going to be half okay and that is also the value of the function at 0 so k should be equal to half for the function to be continuous so k is half okay next differentiate sin 2x to the power x plus sin inverse root of 3x with respect to x and there is an odd part of the question as well to this this is question number 9 for you both the parts Vaishnavi those who have done the first part can attempt the odd part of it yeah yeah I know that no need to type the answer if you are done it is fine okay first one is huge fine so let y is equal to u plus v where u is sin 2x to the power of x and v is sin inverse of root of 3x correct so this implies dy by dx is equal to du by dx plus dv by dx okay now u is equal to sin of 2x to the power x so ln of u will be x ln of sin 2x okay differentiating with respect to x on both the sides we get 1 by u du by dx is equal to x times derivative of ln of sin x will be 1 by sin 2x into cos of 2x into 2 okay plus ln of sin 2x so this implies du by dx is going to be u which is sin of 2x to the power of x times 2x cot 2x plus ln sin 2x now since v is taken as sin inverse of dv by dx will be 1 by under root of 1 minus square of this into root 3 so this implies your final result is going to be dy by dx is equal to sin 2x to the power of x 2x cot 2x plus ln sin 2x plus root 3 by 1 minus under root 1 minus 3x square okay this is going to be your answer okay please for God's sake do not start taking log with a plus sign in between right many people are misusing this property log of a plus b log a plus log b don't do all these things despite several repetitions people are making this mistake fine so assuming you all have got this answer let's move on to the other part now other part of the question need it very carefully tan inverse this we have to differentiate with respect to this okay now even though it is not mentioned in the question we know that for this function to exist x should lie between minus 1 to 1 right because if x doesn't lie between minus 1 to 1 what will happen 1 minus x square will become negative and under root of it will start giving you non real values okay so now whenever you have to differentiate with respect to some function so let's say y is equal to this we will say let theta be in fact let y be this let theta be cos inverse of x square right so what do we need right we need to find dy by d theta right this is what we need to find okay correct now can I say dy by d theta can also be written as in fact I can make this substitution over here first so this means cos of theta is x square so if you make that substitution over here tan inverse of this will become under root of 1 plus cos theta minus under root of 1 minus cos theta by under root of 1 plus cos theta plus under root of 1 minus cos theta yeah now this result could be written as many people can rationalize it also but normally you may also write it as 2 cos square theta by 2 minus under root 2 sin square theta by 2 2 cos square theta by 2 plus under root 2 sin square theta by 2 okay since x lies between minus 1 to 1 we know that x square will lie between 0 and 1 correct so if x square lies between 0 and 1 theta will belong to the interval of 0 to pi by 2 correct so cos theta this actually is nothing but tan inverse of mod cos theta by 2 minus mod sin theta by 2 by again mod cos theta by 2 plus mod sin theta by 2 and since sin theta by 2 and cos theta by 2 both will be positive because your angles are lying in the first quadrant they belong to the first quadrant so you can actually write it as tan inverse of yeah so you can write this as cos theta by 2 minus sin theta by 2 divided by cos theta by 2 minus sin plus sin theta by 2 tan inverse of this is clearly tan inverse of tan pi by 4 minus theta by 2 again I am skipping some step but you can show that step in the paper if at all you get this question so let me just clear off a bit of the solution part so that I can fit so tan inverse of tan pi by 4 minus theta by 2 is going to be pi by 4 minus theta by 2 correct so this is your y so this implies dy by d theta is going to be minus of half now remember I did not make the check over here but if you want you can always make a check that this term should lie in the principal value branch so we already know theta lies in the interval 0 to pi by 2 right so theta by 2 will lie in the interval 0 to pi by 4 the minus of theta by 2 will lie in the interval minus pi by 4 to 0 okay and pi by 4 minus sorry pi by 4 minus theta by 2 will lie in the interval 0 to pi by 4 so it is very much in the principal value branch so this very much is in the principal value branch and you can actually write down the result to be this directly however let me tell you in board level in board level questions they will always give you such situation that tan inverse of tan pi will always become pi because they always deal in the principal value branch okay so there you don't have to bother that okay let me first verify whether it is in the principal value branch or not take it for granted that it will be in the principal value branch okay is that fine so answer is going to be minus half so most of you who have answered this you were correct in fact all of you are correct so we will now move on to the next question okay and what should be lambda tapas okay so I have noted down your response let's see if others also get the same answer okay Vaishnavi lambda also same or different lambda okay great so you can directly use the formula equation of a plane passing through x1 y1 x2 y2 and x3 y3 is given by x minus x1 y minus y1 z minus z1 x2 minus x1 y2 minus y1 z2 minus z1 x3 minus x1 y3 minus y1 and z3 minus z1 equal to 0 okay and you can choose any one of the points to be your x1 y1 not a problem so let me choose this as your x1 y1 z1 point okay so I will write down the equation as x minus 3 y minus 2 z minus 1 and x2 minus x1 is going to be 1 y2 minus y1 is going to be 0 minus 3 okay and this is going to be 3 minus of 2 let me expand this with respect to the first row itself x minus 3 times 0 plus 9 which is 9 minus y minus 2 times minus 2 plus 9 which is going to be 7 and z minus 1 times 3 this is equal to 0 so if you expand it you are going to get 9x minus 7y plus z constants will be 27 minus 27 okay so your final answer will be this is equal to 16 okay so this is the equation of the desired thing now this point must satisfy this plane to be coplanar now lambda 5 5 let me call this as 1 must satisfy the first equation in order to be coplanar with the other 3 points so 9 lambda minus 35 plus 5 is equal to 16 so 9 lambda is equal to I am sorry I think it is 3z over here yeah sorry that is going to be 15 so this is going to be 36 so lambda value is going to be 4 so great the first person to answer this correctly was again tapas 7th question for the day find the coordinates of the point where the line meets the plane which is perpendicular to the vector and at a distance of 4 by root 11 from the origin so there is a line given by this equation so this line meets the plane whose perpendicular is this and this plane is at a distance of 4 by root 11 from the origin okay 2 2 0 let's see what others have to say I am not saying it is right or wrong let's wait for others to give in their response so kushal is getting some different response different answer I need a third person to break this tie who is correct tapas or kushal you know to be very frank there can actually be 2 answers for this question because when you say there is a plane the perpendicular to which is this and at a distance of 4 root 11 from the origin so let's say this is your origin you can have 2 possible planes like this both parallel to each other both being at a distance of 4 by root 11 so this can also be 4 by root 11 this can also be 4 by root 11 correct and if you take these 2 different situations your answers will differ so this is something which I call as a controversial question and in that case what happens the examiner actually chief examiner actually tells the examiner to give marks for both of these possibilities so if you consider the equation of a plane to be r dot n cap is equal to d ok now d is the distance right which actually we write it as mod of this so mod of this is given to you as 4 by root 11 so there are 2 possibilities one is your equation could either be r dot i plus j plus 4 by root 11 is equal to 4 by root 11 or there could be possibility of r dot i plus j plus 3k by root 11 equal to minus 4 by root 11 both are possible both are equation of the planes which is at a distance of 4 by root 11 and having the same normal right so if you just simplify it you get the plane equation to be r dot i plus j plus 3k equal to 4 and this gives you r dot i plus j plus 3k equal to minus of 4 ok so both could be your possibility let me take the positive one first now remember if a line is cutting through a plane remember when a line is cutting through the plane let's say i see the point of intersection to be point m so from the equation of the line itself you can predict the position vector of m so this thing is actually your position vector of m itself so position vector of m you can write it as 3lambda minus 1i plus 4lambda minus 2j plus 3lambda minus 3k now m should satisfy the equation of a plane so this should satisfy this so r is this term itself so replace this in term place of r so this r will be this entire thing so dot i will become 3i minus 1 plus 4lambda minus 2 plus 3 times 3lambda minus 3 equal to 0 let us simplify this so 9, 12, 16lambda and this will become minus 9, minus 10 minus 12 so this is equal to 12 so lambda is equal to 3 by 4 putting a lambda back in om so 3 by 4 will be let me just quickly verify if everything is correct 13, 16, minus 3 is 12 it's not equal to 0, it's equal to 4 a small mistake so 16lambda is equal to 16 so lambda will become 1 so your point will become 2i plus 2j plus 0k that is one possibility so of course your point is 2, 2, 0 that's what tapas said now let us check with the other equation what I am getting so now I will substitute this point in the other plane so 3lambda minus 1 4lambda minus 2 plus 3 times 3lambda minus 3 is equal to minus of 4 so again we will get 16lambda minus 12 is equal to minus of 4 so lambda could be half so if you put lambda as half the other possibility of the position vector of m is half i plus 0j minus half k okay so 3 by 2 minus 3 is minus 3 by 2 sorry this is minus 3 by 2 yeah so the other answer possible is half 0 minus 3 by 2 so even kushal is correct both of kushal you are correct so there are two possibilities one is your lambda could be one another is lambda could be half so both the answers have to be accepted by the board is it fine guys? next given that vectors a,b,c form a triangle such that a is b plus c find pqrs such that the area of this triangle is 5 root 6 where a,b and c are given to you as follows is it done guys okay good so tapas has given the response so first of all we can just compare these two vectors this is given as b plus c now since i,j and k are linearly independent which implies you can compare the components that means p is equal to s plus 3 q is equal to 4 and r is equal to 2 okay now we still haven't found p and s for that I will make use of the other information given to us that is the area of the triangle is 5 root 6 so if you have a triangle like this a,b,c where a is the sum of b and c the area of the triangle is given by mod of cross product of a and b yeah so first of all a cross b if you have to find out I can make use of the determinant the outer mod is the modulus and the inner mod is the determinant symbol so i,j,k p,q,r s,3 4 this is given to us as 5 root 6 okay or you can just take this and write it as 10 root 6 so expanding this i component will be 4q minus 3r j will be 4p minus sr it would be a good idea actually to take half of b cross c because c doesn't have any variable in it so it's good idea to take b cross c rather than this so let me just make that quick change just to simplify my efforts so this I can write it as s 3,4 3,1,2 so it's going to be i times minus 6 minus 4 is minus 10 minus j times minus 2s minus 12 and plus k times s minus 9 so minus 10 i plus i can say j times 2s plus 12 plus k times s minus 9 modulus is going to be this 10 root 6 so now let me just make some space for myself so 100 plus you can say 4 times s plus 6 square s minus 9 square is equal to 100 into 6 that is 600 so this and this will become 500 over here if you expand it it becomes 5 of s square and you get 12,48 48 minus 18 is 30s and constants will be 36 into 4 which is 144 plus 81 which is 225 so you can actually write here as minus 225 sorry minus 275 equal to 0 drop a factor of 5 so we will get 6s minus 55 equal to 0 yeah that is clearly factorisable as plus 11 and s minus 5 so s could be 5 or s could be minus 11 and if s is 5 so when s is 5 p will be 8 and s is minus 11 p will be minus of 8 little long question I think it should be a 6 marker so tapas is the only one who got the answer right well done tapas is that fine so we will move on to the 13th question there are 2 bags A and B A contains 3 white and 4 red balls whereas B contains 4 white and 3 red balls 3 balls are drawn at random without replacement from one of the bags and are found to be 2 white and 1 red find the probability that these were drawn from back B and A is saying half Ramcharan is saying 3 by 5 so the response is 50-50 between half and 3 by 5 almost okay now let even be the event so let even be the event that bag A was chosen E2 be the event that back B was chosen and A be the event that 2 white and 1 red is drawn correct now what do we need to find out we need to find out given that A has occurred what is the probability that it came from back B right so it is E2 okay which is actually P E2 intersection A by PA which is nothing but PE2 into PA by E2 by PA okay now PA is nothing but PE1 into PA by E1 and PE2 into PA by E2 okay you can say law of total probability okay so let's evaluate this first so what is the chance that bag A is chosen that's going to be half itself and given bag A is chosen you draw 2 white and 1 red from it now bag A has 3 white and 4 red so 2 white and 1 red can come out from it in 3C2 into 4C1 by 7C3 okay again what is the probability that back B is chosen again half and given that bag B is chosen what is the probability that you get 2 white and 1 red which will be 4C2 into 3C1 by again 7C3 okay if I am not wrong this is going to give you half I will not calculate half and 7C3 because I know it's going to get cancelled so this is going to be 4 into 3 which is 12 and this is going to be 6 into 4 which is going to be 18 okay so putting over here your answer will be the second term which is half into 18 by 7C3 divided by half into 12 plus 18 so factor of half will get cancelled off you will be left with 18 by let me write it over here for the want of space 18 by 30 will be your answer which is 3 by 5 so answer is going to be 3 by 5 and not half I could see many people answering me with half answer is 3 by 5 okay so a classical question on base theorem but make sure whenever you are solving such questions always define your events this is important many people they ignore these things these are like the small differences that can create between you getting 92% and getting 99% okay moving on to the next question which is question number 14 for the day so there is a person Ishaan super generous person and he wants to donate a rectangular plot of land for a school in his village when he was asked to give the dimensions of the plot he told that if its length is decreased by 10 meters and breadth is also decreased by 20 meters then the area will be decreased by 5300 okay so he was a mathematical person so he was in terms of equations using matrices find the dimensions of the plot and also give reason why he wants to donate the plot donate the plot for a village now for value based question please write good answers okay so don't say that he probably had lot of black money that's why he wanted to make a school and all second part is in those who are challenging the completeness of the question let me just quickly verify okay so apart from this one more part of the question is there if the length is decreased by 50 if the length is decreased by 50 50 and breadth is increased by 50 and breadth is increased by 50 meters then the area will remain the same then area will remain the same okay so I am getting the answers for most of you okay so let a be length into breadth so in the first expression when length is decreased by 10 and the breadth is also decreased by 10 sorry 20 then the area gets decreased by 5300 you can replace this a with lb itself so if you expand this you get minus 20 l minus 10 b plus 200 is equal to lb minus 5300 lb will also be not there from both the sides which means 20 l plus 10 b is equal to 550 0 you can drop the factor of 2 also so 2l plus b is equal to 550 first equation second equation is if the length is decreased by 50 and the breadth is increased by 50 area remains the same correct which is 50 l minus 50 b is equal to 2500 which means l minus b is equal to 50 this is a second equation now we have to use matrices to find the dimension of the plot so you are trying to solve for l and b so you will have equation 1 2 1 1 minus 1 lb this will be 550 and 50 so let this be ax equal to b so x will be a inverse b okay now for a 2 by 2 matrix finding adjoint is very simple so how will you find the adjoint just swap the position of this so it becomes 1 and 2 and reverse the sign of this determinant of a would be minus of 3 so I am just making some space for myself so x will be a inverse which is minus 1 by 3 minus 1 minus 1 minus 1 2 and this will be 550 50 so that will give you minus of 600 and this will give you minus 550 which is 100 which is minus of 450 into minus 1 third which is going to give you 250 that means your length is 200 and your breadth is 150 absolutely correct please do not forget to write the units Ramcharan that is a question which you can actually ask your teachers as well because normally they don't focus much on how you have found adjoint but let's say if they have kept a half a mark for it let's not take chances you can actually write the normal way of finding the adjoint but nobody generally makes a mistake in finding the adjoint of a 2 by 2 so we know that the diagonal positions are switched and the sign of the other diagonal is reversed next question number 15 yeah shortcut is given I think in ncrt textbook just type done if you are done with it okay just one more response I need guys before I start solving this alright since x by y terms are mentioned it would be advisable to write your dx by dy we all know that it's a homogeneous equation right so dx by dy is going to be 2x e to the power x by y minus y by 2y e to the power x by y let x be equal to okay so dx by dy is going to be v plus y dv by dy so replacing it over here it becomes v plus y dv by dy this would become 2vy e to the power v minus y by 2y e to the power b so y could be cancelled so if you cancel out y you get 2e to the power v minus 1 by 2e to the power v so bring this v on the other side that would give you minus 1 by 2e to the power v so this is your y dv by dy okay so this clearly implies 2e to the power v dv is minus dy by y which gives you 2e to the power v is equal to minus of ln y plus c 2 times e to the power v plus ln y is equal to c which implies 2e to the power x by y plus ln y is equal to c so absolutely correct tapas again tapas was the first one to answer this so let us now move on to the next question again the differential equation question in differential equation please pay more attention to linear differential equation and homogeneous differential equation so these are the two topics in differential equation that mostly questions are framed in board level questions in board exams do let me know once you are done just mention if you have done or not no need to type in the answer okay tapas has typed in the answer great what about others okay done so let us discuss this so it is a clear cut case of a linear differential equation where the dependent variable is y and independent variable is x okay so if you try to compare this you can compare this with dy by dx plus py is equal to q where q p and q are functions of x or constants okay so p is going to be minus of 1 plus x and q is actually e to the power 3x 1 plus or x plus 1 the whole square so integrating factor is going to be e to the power integral of minus 1 plus x dx is going to be e to the power minus ln 1 plus x which is again going to be 1 plus x okay so your answer is going to be y into integrating factor is equal to e to the power 3x 1 plus x the whole square into 1 plus x integral with respect to x so it is like integrating e to the power x plus x e to the power 3x so this will be e to the power 3x by 3 and for this you have to use integration by parts treating your x to be your first function and this to be your second function so x e to the power 3x by 3 minus integral of e to the power 3x by 3 so just collating the entire result over here in yellow so y by x plus 1 that is going to be e to the power 3x by 3 plus x by 3 e to the power 3x and this is going to be 1 by 9 e to the power 3x plus c absolutely correct so tapas absolutely correct again the first one to answer this moving on to the 16th question for the day using integration find the area of the region given by all such points x comma y where is less than 6ax and x square plus y square is less than equal to 16a square so always in such kind of a question please please catch the graph because without graph you cannot even think of solving questions like these so what is desired is this zone so y square is less than 6ax means it has to lie within the arms of the parabola and x square plus y square less than equal to 16 means it has to lie within the circle so the intersection zone is this area now these kind of integrals are best solved by using horizontal strips and if you are able to find out one horizontal strip I will be automatically be able to find out the other one because both are symmetrical halves okay first of all point of intersection for point of intersection I need to simultaneously solve these two equations so forget about the inequality we will have to solve the equations so this will become x square plus 6ax minus 16a square I think it is factorizable as 8 and 2 if I am not mistaken so taking x common it's x plus 8a minus 2a common again x plus 8a equal to 0 so x has to be 2a so this has to be x equal to 2a point so if x is 2a point y is going to be 12a square under root which is 2 root 3a now I have to take the area of this horizontal strip so I will take the difference of x over here let's say x2 and let's say this is x1 so my area will be difference of x2 minus x1 dy where dy is the thickness of this strip and I have to integrate it from y equal to 0 till 2 root 3a and double up my answer to get the entire area now for x2 expression I have to use the circle equation so that's going to be 16a square minus y2 under root now why not minus because here you know that on the right hand side of the x axis your x will always be positive and x1 will be y2 by 6a so the desired area will be twice 2 root 3a 16a square minus y2 minus y2 by 6a is that fine ok so let us evaluate these two integrals so this will become twice of it follows the formula of under root of a2 minus x2 form so it's x by 2 under root of 16a square minus y2 ok plus 16a square by 2 and you will have sine inverse y by 4a ok and this will be minus y3 by 18a this will be minus y3 by 18a from 0 to 2 root 3a ok guys when you put in the value I am just giving you the final result final result should come out to be 4a square by root 3 plus let me write it up over here 4a square by root 3 plus 16a square pi by 3 is this the answer that you all are getting this many square units this many square units ok those who have solved it are you getting this answer Vaishnavi, Lalitha, Suresh ok great so we will now move on to the next one so the question is in which interval is this function strictly increasing or strictly decreasing this is your question number 18 for the day let me know once you are done with the first part so that we can discuss the first part quickly ok alright so let's discuss this so if you want to find out the interval in which it is strictly increasing your derivative of the function should be positive right for f of x to be strictly increasing right so its derivative will be 4x cube minus 24x square plus 44x minus 24 equal to 0 if you drop a factor of 4 it becomes x cube minus 66 square plus 11x minus 4 equal to 0 sorry minus 6 equal to 0 ok so for such cases we basically have this as your roots so if you make a wavy curve sign scheme for this 1, 2, 3 it will be plus, minus, plus, minus so from 1, 2, 2 union 3 to infinity the function is positive so in this interval it is strictly increasing and minus infinity to 1 union 2, 2, 3 the function is strictly decreasing absolutely correct so first one to answer this is again tapas next is find the or part of it find the maximum and minimum values of these functions once done please type in you are done remember they are asking you the maximum and minimum values they are not asking the points where they are maximum or minimum of course you have to go through that root only but remember to mention that is what is the directive of the question right don't answer anything which is not asked to you or don't conclude your answer at that stage which is not required in the answer so they are asking you the maximum and minimum values of this function itself maximum is 2 minus 2 log 2, minimum is minus 1 let's check so for extima points for maxima slash or you can say for extima points you first need to find the derivative of the function and put it to 0 which means ckx tanx and this will become 2 into minus of sinx by cosx that's equal to 0 which means ckx tanx is equal to 2 tanx so one possibility is your tanx is 0 that can happen when x is pi and the other possibility is your ckx is equal to 2 which means cosx is equal to half which can happen at 60 degree pi by 3 or 2 pi minus pi by 3 correct? so 2 pi minus pi by 3 is 5 pi by 3 okay now ensure that whatever points you are getting it the function must be defined for these points so never include those values where the function itself will become undefined in nature or the derivative will become undefined because it will not be there now how do I know which is my point of maxima or which is my point of minima okay for that you don't have to actually sit and put the values find the double derivative you can actually check at 0, pi by 3 pi pi by 3 in fact you don't have to check at 0 because 0 is not included and let me just first write so at pi by 3 this value will become 2 and this will become 1 by 4 which is minus 2 log 2 okay at pi it will become 1 by cos pi which is minus 1 plus 0 which is minus of 1 and again at 5 pi by 3 this will become 2 and this will again become minus 2 log 2 so this is your maximum value this is your minimum value but guys there is a catch in this question catch in the question is since it is an open interval since it is an open interval you need to see whether the max is greater than the limit of the function as x tends to 2 pi and limit of the function or you can say 2 pi minus or limit of the function as x tends to 0 minus okay similarly you need to check whether the min of the function is less than in fact less than equal to can also be possible or limit f of x as x tends to 0 plus but normally in board level they would actually not go into such complexities normally so I am talking with respect to your concept of clarity point of view when open interval comes whenever interval comes it is not a question of local maxima or local minima it is actually a question of global maxima or global minima okay so here double derivative test is not important but what is important is what is the maximum that it attains at the stationary point and is this maximum greater than the value it can attain at let's say this is a and b whether your maximum is greater than limit of the function as x tends to b minus or limit of the function as x tends to a plus but I am sure in case of board exams they don't mean to do these tests because you know this is meant for people who are not preparing for je as well so even though the question is not correctly framed if you have to solve this please do not take chances do your double derivative test as well okay so if I have to do this I will do the double derivative test as well so double derivative test at pi, double derivative test at pi by 3, double derivative test at pi by 3 so please do these tests and realize that at pi it will be positive at pi by 3 and pi pi by 3 it will be negative in nature is that clear guys, is the concept clear even though the intention may not be to solve it by using these methods but you should be clear in your mind about the approach okay next is a determinant question so prove by using the properties this word is very important use of properties now when I say use of properties means which are helping you to solve the problem so do not use any arbitrary property that you want and just justify that you are using property it should actually help you to solve the problem once you are done please type done so that we can discuss this okay size is done so is Sondarya this is your 19th question of the day alright so let's discuss this so first operation that I can do is column 1 as column 1 minus 2 column 3 guys let me tell you the other day one of you was using this property c1 as c2 minus c1 can you ever use such kind of expression also you are saying c1 is transformed as c2 minus c1 won't this be an incorrect property so please do not use such properties if you are using such properties first you have to make c1 as minus c1 for if you are making this you have to multiply and divide with minus outside okay so please do not misuse such properties only operations of this nature that is ri is equal to ri plus krj okay so such kind of properties you can use if you are using ri is equal to kri ensure that you divide with divide with k outside the determinant okay anyways so when I use this property you can see here that the first column will become b2 plus c2 this is c2 plus a2 this is a2 plus b2 and you will have this okay then we can do c1 as c1 plus c2 so this will become a2 b2 c2 a2 b2 c2 a2 b2 c2 a2 b2 c2 bc caab and you can all pull out a2 plus b2 plus c2 common you will get 111 a2 b2 c2 bc caab okay and the moment you have generated 111 you can do r1 as r1 minus r3 and r2 also as r2 minus r3 so you get a2 plus b2 plus c2 times 0 0 a2 minus b2 sorry a2 minus c2 this will become b2 minus c2 this will become bc minus ab and ca minus ab and 1c2 ab okay so you can always pull out c minus a common from the first row and c minus b common from the second row let me make some place for myself so a2 b2 c2 and I am taking let's say a minus c and b minus c common from the first row I will get 0 a plus c and here I will get a minus of b this will be 0 b plus c and here if you take a common minus of a okay now you can expand it with respect to the third row so this will give you a2 b2 c2 a minus c b minus c and this will become minus a2 minus ac plus b2 plus bc now club these two terms you will get this desired result okay so here you can write it as b2 minus a2 and cb minus a so you can always take b minus a common you get a plus b plus c and hence the result now there are more questions left guys but I will leave them as homework for you okay so this next question I will just quickly discuss the format when you are asked to use elementary row transformations to find the inverse we always go in this direction we always go column wise so first we create a 1 then 00 then 100 then 100 in order to get the inverse remember in your row operations to find out the inverse we follow this formula a is equal to ia please use this for elementary row operations use a equal to ai if you are using elementary column transformations okay in elementary column transformation the approach should be always create a 100 then 100 00 okay is that fine so guys I am going to stop over here and I am going to send this pdf version to you so that you can solve the remaining problems I think 4 more problems are left off okay and all the best for your upcoming exams so this entire week will be focusing only on board level preparation now you would have got a better idea how to present your answers and what may lead to some method marks and independent marks so only answer is not important in board they also see what is the methodology that you adopt so write everything that you can under the time constraint given to you so thank you so much for coming online today over and out from Centrum Academy bye bye have a good night