 Hello, reading from Sendham Academy. Welcome all to the YouTube live session on problem solving. So this session will be covering up some important topics. Basically, it's a mix of class 11th and class 12th topics so that it is helpful not only in your pre-boards, but also in the upcoming JE exam. So guys, those who have joined in the session, I would request you to type your name in the chat box so that I know who are attending the session. So good afternoon to everybody who has joined in. Koshal, Saunanya, Sai, Sai Payamnath, Rithvik. So let's start guys, it's exactly three o'clock. And let's move on to the very first problem of the day. All right, so this problem that you see is a problem which has been picked up from the limits chapter. Okay, so very first problem. So you have already been given a limit expression over here and you have been asked to evaluate this term. Hello, Atmesh. Hi, Visist. So please feel free to type in your response in the chat box once you're done with the solving of the problem. Hello, Rohan, hello, Atmesh. So those who joined in late just to inform you, this session will be taking up a mixed-back problem from various topics which will be helpful also in your board exams as well as the main and advanced exams. I'll only start the discussion of the problem when I get at least three to four responses. Hello, Purvik. So guys, if you see this part clearly, even though it is not required just to inform you that this is nothing but it's integral of zero to one, log of one plus x dx. That's what they have evaluated over here. So even if this result was not known, you should have been able to find this anyways out. So please feel free to type in your response in the chat box once you're done with this problem. Okay, so I've already got one response, Sondarya, okay. I'm not commenting whether it's right or wrong. Kushal, not sure. Kushal, you should be very sure you're writing an exam where there's a negative marks. Sai has given a response. Hello, Rishal. Welcome to the live session. Okay, Visist also says see. Okay, guys, so let's start solving this problem. Let's call this as this limit as L, okay. Now, before I start solving this limit, I would like to introduce this n to the power k inside this one by nth power. Okay, when I do that, this is something which I experience. Limit n tending to infinity, n plus one to the power k, n plus two to the power k, and so on till n plus n to the power k, whole divided by, whole divided by, n to the power nk, n to the power nk, and this whole raise to the power of one by n. So what I did is, I just introduced this within the brackets, within the one by nth power. Okay. Now, I can individually give this n to the power nk, n times between each of these factors. So I can distribute it as limit, n tending to infinity, n plus one to the power k, n plus two to the power k, and so on till n plus n to the power k divided by n to the power k, n to the power k, and so on till n to the power k, whole raise to the power of one by n, whole raise to the power of one by n. Correct? Now, what I'll do here is, I will take log on both the sides. So I'll take log on both the sides. So log l to the base e, that will become one by n, and log of all these terms. So log of n plus one by n whole raise to the power k, plus log of n plus two by n whole raise to the power k, and so on till log of n plus n by n whole raise to the power k. Okay. And by the log properties, I can always pull a k out. So it becomes log n plus one by n plus log n plus two by n, and so on till log of n plus n by n. Okay? Guys, if you see at this expression carefully, you realize that this expression is nothing but k by n, summation of log n plus r, minus log n, okay? R varying from one to n. So this is my log l to the base e, right? And this expression is what we have been already given. We have been given that this expression, that is log of n plus r minus log of n, whole divided by n, this term as limit n tends to infinity, of course. So limit n tends to infinity. So this term here is known to us, right? So my answer is going to be k times the expression which was given to us, which is two log two minus half. So this k extra comes out. Rest everything is known to us. This is already given in the problem itself. So we can directly use it. And finally, just trying to simplify this expression, it becomes k times ln, this becomes two ln two minus one, which is k times ln four minus ln e, which is k times ln four by e, and k can be taken as the power on this expression. So finally, we have got ln l, that is the required limit is equal to ln four by e to the power of k, which means my l is going to be four by e to the power of k, which is clearly your option number c is correct. Exactly. So those who answer with option c, that option is correct, guys. Congratulations. So with this, we'll move on to the next problem. That's the problem number two for the day. This question comes to you from 3D Geometry, which is going to be immediately helpful for your second term exam. The question says, find the distance of the point one zero minus three from the plane x minus y minus z equal to nine, measured parallel to the line this. So we have to find the distance which is measured parallel to the line. Remember, we are not supposed to find out the perpendicular distance of this point from the plane, but only the distance which is parallel to this given line. Okay, Gushal says B, I need at least three to four responses so that I can start solving it out. So we have to find the distance p m. Now distance p m is the one which is measured parallel to the given line. All right, so most of you are giving the response as B. So let's check it out, guys. First of all, I need to write down the equation of the line through p m passing through one zero minus three. Right? So we can all use the direction ratios of the line as the direction ratio of the line passing through p m also because they are parallel. So I can say line, line passing through, line passing through p m would be x minus one by two, y minus zero by three, and z plus three by minus six. Okay, and then we'll make use of the parametric equation of a line. We'll say, let for a particular lambda, I get the point m or I get the coordinates of the point m. So x can be written as two lambda plus one, y could be written as three lambda, z could be written as minus six lambda minus three. Okay? And I could find the value of lambda by using the fact that this lies on the given plane. So this lies on the plane pi. So let's use the equation of a plane. So x minus y minus z equal to nine. So that will give you five lambda plus four equal to nine. So lambda is going to be one in this case. That means the point m is going to be three comma, three comma minus nine. Three comma, three comma minus nine. So the distance p m would be under root of three minus one square, three square, and minus nine plus three square. That gives you under root of four plus nine plus 36. And that's the seven units as your answer. Absolutely correct. So those who answer with option B, that's absolutely correct. So this was actually a school level question just to boost up your confidence in 3D. And 3D geometry and vectors. Now we'll move on to the problem number three for today which is on determinants. So the question says alpha and beta are the roots of this quadratic equation. X square plus BX plus C equal to zero. So we have to find that this determinant delta is equal to which of the following options. Yeah, any response guys from anyone? Okay, I'll give you a hint in this case. The hint is you have to use the concept of multiplication of determinants. You have to use the concept of multiplication of determinants. So if you look at this term three over here, it's actually one into one plus one into one plus one into one, okay? This term is one into one, one into alpha. One into beta. And this term over here is one into one, one into alpha square, plus one into beta square, right? In a similar way, this is nothing but one into one, one into alpha, one into beta. Again, this is one into one, one into alpha square, one into beta square. and this is 1 into 1 1 into or sorry alpha into alpha square beta into beta square and again the last is 1 into 1 ok 1 into alpha 1 into beta 1 into 1 alpha into alpha square beta into beta square and finally we have 1 into 1 alpha square into alpha square and beta square into beta square ok. Now the reason I have written it like this is for you to appreciate that this is nothing but you know a multiplication of two determinants which are as follows ok so when this multiplies to this you start realizing that your result is going to be the determinant which is given to you in the question. So guys how it works let me just tell you when this particular row multiplies with this particular row and you add it you get this element ok. Remember in determinants you can have the leeway to multiply row with row row with column column with row column with column so there are four types of multiplication you can perform and each one of them will give you the same result. However in case of matrices the limitation is you can only multiply row of the first matrix with the column of the second right. So the pre multiplier row is taken and the post multiplier column is taken but in determinant multiplication there is no such restrictions ok. So this element is obtained when you take the first row and the first row of both of them this element is obtained when you take the first row and the second column and this is obtained when you take the first row and the third row ok. So again I will repeat it this is the multiplication of row 1 and row 1 this is obtained from row 1 row 2 this is obtained from row 1 row 3 ok. In a similar way this is obtained when you multiply second row with the first row. So this is second row with the first row right and yeah by the way this is slightly this is nothing but alpha into alpha this was alpha into alpha and beta into beta. So this is taken when your row 2 from the first multiplies with row 2 of the second right and this is obtained when row 2 gets multiplied to row 3 row 2 with row 3 sorry this is alpha beta again yeah and this is obtained when row 3 multiplies with row 3 multiplies with row 1 that is this multiplies with this ok and this is obtained when row 3 is multiplied with row 2 and this is obtained when row 3 multiplies with row 3 correct. For the moment the moment you figure this out things will become very very easy for you I hope still this stage everything is clear in your mind any doubt so far guys please highlight it right now so guys this is nothing but this is nothing but 1111 alpha beta 1 alpha square beta square determinant square okay now evaluating this determinant is easy if you want to evaluate this determinant all we need to do is try to create maximum number of zeros in a given row or a given column for that we can take we can take this approach we can subtract we can subtract let's say the first row with the second row so I can do R2 as R2 minus R1 and we can do R3 as R3 minus R1 R3 minus R1 so by doing that it becomes 1111 I get a zero alpha minus one beta minus one and I get a zero alpha square minus one beta square minus one we can always pull out some factors common if you want to I think we don't need to pull out anything common we can directly expand with respect to the first column so alpha minus one beta square minus one minus alpha square minus one times beta minus one which I can write it as alpha minus one beta minus one beta plus one again alpha minus one beta minus one alpha plus one which gives you the result as which gives you the result as alpha minus one beta minus one and we'll have beta minus alpha left okay but remember we have to square this determinant so we have to square this determinant and when we square this determinant we end up getting alpha minus one square beta minus one square and of course we can write it as alpha minus beta square okay now comes the problem the problem says that this expression has to be expressed in terms of B and C where B and C are your coefficients of X and the constant term respectively right so here we can state that one thing since alpha beta are the roots of this equation since alpha beta are the roots of the equation this I can write it as X minus alpha X minus beta correct okay now these two terms I can easily obtain if I put X as one right if I put X as one I get something like this correct so if I square this up that means these two terms are obtained easily no problem with that which is one plus B plus C whole square correct now what about alpha minus beta square can I say alpha minus beta square is nothing but alpha plus beta square minus 4 alpha beta correct and alpha plus beta square is nothing but B square minus 4 alpha beta is nothing but minus 4 C which implies this result can be written as 1 plus B plus C the whole square times B square minus 4 C right exactly Purvik so our option number B becomes correct in this case so option number B becomes correct in this case is that clear guys so a very good question where you have to use your concept of multiplication of determinants okay so with this we'll move on to the next problem which is problem number four so we have been given a function f of X which satisfies this condition that 3 to the power X plus 3 to the power f of X is minimum of phi t is minimum of phi t where phi t is defined as the minimum of 2 t cube minus 15 t square plus 36 t minus 25 and 2 mod of sine t under the interval t lying between 2 to 4 to find the domain of this function any idea guys show show show take your time hello Rohit so this is the question from application of derivatives so this has been picked up from application of derivatives C okay so Purvik has given an answer which is C in this case anybody else guys time yourself for just four minutes not more than that four to five minutes should be sufficient enough to solve this problem okay wishes says D Purvik changes his answer to D now Vishal also says D what about others Rohan Akmesh, Ritvik, Sondarya, Sai, Nishal all right guys so let us check the minimum value you know that this guy will always be greater than equal to 2 right okay so 2 to 4 sorry it will be always be yeah it's sorry it's always be equal to 2 okay so always positive but less than equal to 2 now if you look at this expression this function let's say I call it as f of t let's find the derivative of this function with respect to t so derivative will be 60 square minus 30 t plus 36 which is nothing but 6 times t square minus 5 t plus 6 which is factorizable as t minus 2 t minus 3 and if you draw a wavy curve for this if you draw a wavy curve for this the sine scheme tells you this is the sine scheme that means the function is decreasing in the interval 2 to 3 so it is decreasing when t belongs to interval 2 to 3 and increasing from 3 to 4 and increasing from 3 to 4 okay that means the minimum value of this function will occur at 3 okay so minimum value of f of t will occur when t is equal to 3 isn't it right so now let us calculate what is f of 3 f of 3 is going to be a 2 times 3 cube minus 15 3 square plus 36 times 3 minus 25 which gives you a 54 minus 9 into 15 which is 135 plus 108 minus 25 which is a 160 to minus 160 that's going to be 2 right so of course your minimum value would be obtained your minimum value will be obtained for those values where 2 sine 2 becomes minimum and the minimum of this will become 0 right because pi happens to lie between pi happens to lie between 0 and 4 sorry 2 and 4 because pi is approximately 3.14 which happens to lie between 3 and 4 so I can say that 3 to the power x plus 3 to the power f of x is equal to 0 correct so what should be the domain of this function so 3 to the power f of x is equal to negative 3 to the power f of x negative 3 to the power x this is not possible I guess why it is not possible because this will always be a positive quantity and this will always be a negative quantity isn't it yes or no so ideally the answer for this question should have been none of these now just now I checked into the question there has been a printing error over here this should have been 2 plus mod of sine t but even if you ignore this let's say if you don't consider this to be the new question that means let's say we consider it to be 2 mod sine t then the answer should have been none of these the answer should have been none of these okay guys if you take it to be 2 plus mod sine t then the answer would be different if you take it to be 2 plus mod sine t which is because of some printing error in the question then this will always be greater than 2 in that case the minimum value will be 2 in that case you have to find out the domain for this function right so that would have become 3 to the power f of x is equal to 2 minus 3 to the power x okay and since this function is always positive that it implies this should also be always positive that means 3 to the power x should always be less than 2 that means x should always be less than log of 2 to the base of 3 in that case your answer would have become x belongs to minus infinity to log 2 to the base 3 which is option D in this case okay but sorry it was my mistake there was a error in the question so ideally the answer should have been none of these but if you make this correction over here it becomes option number D in that case it becomes option number D okay let's move on to the next question now which is question number 5 the question says the adjacent points on the set of points x comma y where x and y both lie between where x and y both lie between 0 to 9 are joined with line segments parallel to either of the coordinate axis is to construct a path from 0 comma 0 to 9 comma 9 now if you choose the path what is the probability that the path has only one turning point so let me just explain the question to you the question is you have decided to choose a path from 0 comma 0 to 9 comma 9 by joining line segments parallel to the either of the coordinate axis is right for example you can go like this you can go like this you can go like this you can go like this you can go like this you can go like this in fact the question says adjacent points on the set of points x comma y are joined by the line segments parallel to either of the coordinate axis is to construct a path like this remember x and y here both are integers x and y both are integers so you can go only by integral distances like this let's say like this like this like this like this so this is one path okay like that you can make so many parts so what is the probability that a path chosen has only one turning point adjacent points which are like points next to each other for example these two are adjacent points then these two are adjacent points then these two are adjacent points then these two are adjacent points then these two are adjacent points like this points next to each other that's it so your path should have only one turning point first of all how many total parts are there can somebody find out what the total number of path this question is similar to somewhat like you know saying on a chess board you are moving from 0 0 to 9 9 and you can always move along the edges of your square boxes so it's very similar to that concept so total number of paths basically can be obtained if you see this figure clear clearly you have to move nine units horizontally correct and you have to move nine units vertically as well correct in order to reach from 0 comma 0 to 9 comma 9 you need to make nine units of horizontal displacements and nine units of vertical displacements right now think as if you have nine edges you have nine edges with you and you have nine v's with you you can arrange them in any fashion that you want right if you arrange them in any fashion that you want let's say I make one typical arrangement where you have H V V V H H H V V H 1 2 3 4 5 6 7 8 so let's say 1 2 3 4 5 6 7 8 9 and you have 1 2 3 4 5 6 7 8 9 what does it mean it means that you move one unit horizontally okay three units vertically then four units horizontally two units vertically one unit horizontally one vertically one horizontally two vertically like that okay that is how you reach the point yes or no so this is one way to arrange it so like that how many ways can I arrange it you will say it's very simple it's like arranging 18 objects of which nine are identical of type one nine are identical of type one and nine are identical of type two nine are identical of type two yes or no okay so it's like arranging the letters of a word where some letters are being repeated so if you recall the total number of ways would be nothing but total number of path would be nothing but 18 factorial by 9 factorial square correct absolutely correct perfect okay now out of this if you want to choose a path which has got only one turning do you agree with me that the path has to be either like this to this point right or it has to be from this point directly to this point yes or no so these two parts that is this pink part and this blue yellow part these are the only two parts where you can have exactly one turning right rest all the path if you take anywhere you have to take two turnings yes or no so in that case your answer will be what the probability of only picking up a path with two turnings with two turnings would become two divided by 18 factorial nine factorial nine factorial correct so I can simplify this right now is is there any option which is telling two factorial by nine factorial square no so let me simplify this just to check whether if some option matches with the answer so this will be two into nine factorial nine factorial 18 I can write it as one into two into three into four into five all the way till 18 and I can always take out the even terms put it separately and odd terms and keep it separately if I take two out from all the even terms I will end up getting two to the power nine into nine factorial and the odd terms would remain as such correct so cancel out nine factorial with nine factorial cancel out one of the twos so my answer is going to be nine factorial by two to the power eight one into three into five all the way till 17 so this one will become your answer do we have such an option I could see such an option over here yes in option numbers see we have the same expression so this is going to be your answer is that okay guys any question so far in this solution so let us move on to next question which is your question number six for the day this question has been picked up from area under curves application of integrals we have to find the area bounded by y equal to f of x x axis and the line y equal to one where your f of x is given to you in case in in in form of a integral okay it's given to you in a form of a equation like this so this is how typical g advance questions are framed where they would make your life no difficult to know the function itself so unless until you know the function you will not be able to find the area under the curve so the challenge here is to get the function okay we discussed that Purik you're getting area as e minus one okay so you want to say none of these let's wait for the response from others so that we can start discussing this sure sure check it out okay so I says one minus one by eight which is exactly half of this kushal gets option b which is at Mesh Rohan what are the answer you guys are getting Psi also says be at Mesh none of these alright guys so here we are going to use the concept of Leibniz rule so we are going to use a Leibniz rule to get our function right just to recall what's the Leibniz rule Leibniz rule is a method by use of which you can find the derivative of an integral so let's say you are integrating f of t with respect to t from psi of x to phi of x and you're differentiating it it is f of phi x times the derivative of the upper limit minus f of psi x times the derivative of the lower limit so in place of t you have to just put the upper limit differentiate the upper limit and multiply then in place of t you have to put the lower limit differentiate and you have to multiply chalo so with this let me just first multiply throughout with x so let me make it as x f of x equal to 1 plus sorry x plus integral of f of t from 1 to x now let us differentiate both sides by use of Leibniz rule so it becomes x times f dash x plus f of x equal to 1 plus f of x into 1 which is f of x and I don't need to put 1 because derivative of 1 will anyways be 0 so you get f dash x as 1 by x okay when you integrate this guys you get ln mod x and I'm sure Purvik missed out a C Purvik need to miss out a C very very important guys whenever you are integrating this is not to be missed out this is important because missing out the constant will change the function completely correct now how do I get this C how do I get this C if you see this expression can I say f of 1 will be equal to 1 1 by 1 integral from 1 to 1 f of t dt and this part anyways will be 0 so your f of 1 is actually 1 okay so your function f of x is equal to ln of mod x plus 1 right now we'll sketch this function first we'll sketch ln of mod x plus 1 so for that we sketch ln of x first and we shift it up by 1 so if you sketch ln of mod x and you shift it a ln of x and you shift it up by 1 that means your function will now cut the x-axis at a point where ln of x plus 1 becomes 0 which is x equal to 1 by e so at x equal to 1 by e it will cut the function it will cut the x-axis I'm sorry okay and the upper line is y equal to 1 so we have to just integrate till y equal to 1 so let's say y equal to 1 is like this let me further yeah so this is the graph bounded by x-axis bounded by x-axis and y equal to 0 to y equal to 1 but wait we have just plotted ln x plus 1 where it is modded it will become like this it'll become the mirror image of this like this in other words we have to we have to find this area as well but that is not a problem for us because I can say if I find this area it will be the same as this area both these areas would be same so let me find this one area out here I would prefer taking horizontal strips here I will prefer taking horizontal strips okay so what is x over here can I say x over here is going to be let me make x the subject of the formula so y is equal to ln x plus 1 so x is equal to e to the power y minus 1 so this will be e to the power y minus 1 right okay so ultimately the desired area that I am looking out for will be let's say two times the area which I am looking out for will be area from 0 to 1 of this expression e to the power y minus 1 dy okay so that becomes e to the power y minus 1 itself 1 and 0 so when you put 1 you get e to the power 0 and when you put 0 you get e to the power minus 1 so your answer would be 2 times 1 minus 1 over e 2 times 1 minus 1 over e which of the options are correct which of the option is correct right option number B is correct so I think Kushal had given a right answer for this well done Kushal Sai also gave a right answer rest of you guys underestimated the power of mod and you underestimated the power of the constant of integration guys do not miss out on these this is where you will lose your marks I mean you did absolutely well to you know use Leveny's rule but you did not take it to a conclusion correct so with this we'll move on to the next problem which is problem number 7 for you so read this question very very carefully it has been picked up from other chapter LCD limit continuity and differentiability more than one options may be correct in this case so the site of this question only makes you nervous because it's so long right but sometimes within these big questions very simple concepts are so first focus on your G guys G G is this term so first we need to find out what is this G then we can go on to find your F of X which is dependent on your G actually so G of X was actually two times sine X minus sine to the power NX plus mod of sine X minus sine to the power NX divided by two times sine X minus sine to the power NX minus mod of sine X minus sine to the power NX so size says C okay others please feel free to type in your response on the chat box okay so now this presence of this N makes a lot of difference this is quite important this is our decision maker in this problem why because that will decide the fortune of mod correct so how is this mod going to behave is what has been decided by the value of N right so guys if you look at these intervals which have been cited over here your X is between 0 to pi correct my X we are dealing with such X where we are between 0 to pi right and in 0 to pi interval we know sine X is positive no doubt about that okay now of course when it is positive I mean to say that it is between 0 and 1 right it cannot exceed that now if you talk about such values of N which are between 0 and 1 remember N is a positive real quantity okay if you talk about such values of N which is between 0 and 1 do you realize that N would be fractional in this case right N is a fraction in this case okay so which will be more sine X will be more or sine X to the power of N will be more which will be more in this case given N is a fraction please type it in the chat box which will be more sine X will be more or sine X to the power N will be more right Purvik sine X to the power N will be more so this expression will be greater than this expression in other words sine of X minus sine of X to the power N would be a negative term that means when you are between 0 and 1 my G of X is going to behave as 2 sine X minus sine to the power N X and this results into minus of sine X minus sine to the power N X why because mod mod will make it positive by putting an additional negative sign in front of it so the fate of G of X has now been decided when you are between when N is between 0 to 1 so let's see what is the answer that we get from this if I'm not wrong you get 1 by 3 as your answer correct so G of X is going to be 1 by 3 when N is between 0 and 1 right right exactly at 1 what happens exactly at 1 what happens exactly at 1 I think things will all become undefined right because G of X will become undefined and that happens okay and for that I have been given a situation that will happen when X now this will become 0 this will become 0 this will become 0 so it becomes 0 by 0 form so okay let's not talk about n equal to 1 now let us talk about n greater than 1 let's talk about n greater than 1 now when n is greater than 1 can I say in this case sine X would be greater than sine X to the power n right so your same G of X expression will now become 2 sine X minus sine to the power n X this will become plus sine X minus sine to the power n X and in the denominator and in the denominator I will get a minus sine over n and correct me if I'm wrong this is going to give you actually 3 this is going to give you actually 3 correct so G of X behaves as 1 by 3 when your N is between 0 and 1 and it behaves as 3 where n is greater than 1 now I don't know what will happen to the function when it is equal to 1 for that we have to evaluate that that's actually become 0 by 0 but that is not important to us because in your options they only talk about 0 to 1 greater than 1 0 to 1 greater than 1 none of the places they bother about what is X equal to 1 so you can say at n equal to 1 it is undefined okay now what is my f of X f of X is G of X greatest integer function when you are between 0 to pi union pi by 2 to pi yes or no okay now what happens to the greatest integer function of 1 by 3 when n lies between 0 and 1 you will say this function will start behaving as when you are between 0 and 1 your function will start behaving as 0 3 clearly indicating that the function has a discontinuity and hence non-differentiability at X equal to pi so option a cannot be correct option a cannot be correct right but what happens when n is greater than 1 when n is greater than 1 both will be 3 3 each this will also be 3 this is also 3 that means the function will be a constant function thereby the function will have a continuity and differentiability at pi by 2 so option B is correct option B is correct option C again is wrong because it is saying function is continuous at pi by 2 which was not right so option C is also wrong since it is not continuous option C is also wrong and it says that option function is continuous but not differentiable this is also wrong because it is continuous and differentiable both at pi by 2 so guys your option number B is correct your option number B is correct in this case I think Rohit gave nearly correct answer Atmesh gave a correct answer okay great guys so a good question to have at this point any question with respect to this so with this we move on to the next question which is your question number 9 sorry 8 for the day consider a real-valued function f of x satisfying this functional equation satisfying this functional equation and f of 1 is equal to a where a is not equal to 1 then a minus 1 times summation of f i from i equal to 1 to n so treat this as 1 so psi a sine to the power nx means you are rooting it so when you take a square root of an expression or take the square root of a number which is a fractional you always get a higher value for example if you take square root of 1 by 4 you will get 1 by 2 1 by 2 is higher than 1 by 4 got it psi with which says D atmash also says D which is also says D okay so most of you have been able to solve this so let's solve this question first of all I'll put the y value as 1 in order to know the function so it gives you 2 f of x equal to f of x plus f of 1 to the power of x and f of 1 is only given to us as a so it becomes f of x is equal to a to the power x that's quite simple all it is required to do the summation of a to the power i from i equal to 1 to n that means a to the power 1 a to the power 2 all the way till a to the power n I think this is a simple geometric progression with the first term as a common ratio as a as well so this is what is going to happen correct these two terms get cancelled and when you expand it you get a to the power n plus 1 minus a which is option number D is correct so few simple problems here and there just to boost up your confidence so now we'll move on to the 10th question sorry 9th question so after this question we can take a break guys which exam do you have on Monday okay physics great so guys please write a mock paper tomorrow early in the morning the same time when you're writing an exam time yourself for three hours check out for the shortcomings that you experience while writing the test whether it's time mismanagement whether it's poor presentation whether you are missing out on the question right whether it's the handwriting right everything matters so keep practicing mock papers of board level full length before going to any exam or com practicals so again it's a locus based question from coordinate geometry with a mix of trigonometry in it so b and c are fixed that's 2 comma 0 and 8 comma 0 and a is free to move a is free to move guys take a clue from the options take a clue from the options do not start blindfolded all the options suggest that it has to be an ellipse that's the biggest hint the option gives you never ever start attempting any problem without giving a glance at the options options are the best guide that you have when I wrote my J in 2003 and 2004 we didn't have any options now you guys are fortunate that options are given to you option makes the effort almost one one-third we had to start with a blindfold but now you have been guided that it has to be an ellipse anyhow yes or no so if I were you I would have considered my B and C to be the foresight I would have considered these to be the foci of the ellipse okay so center will be exactly at 5 comma 0 right and as you can see in all the options you have 5 comma 0 correct now guys we need to know here we note note 2a e 2a e is known to us but we don't know what is 2a so for that this condition is mentioned to you now when I say 2a I'm not referring to the sides of a triangle I'm referring to that x square by a square plus y square by b square that a I'm referring to I'm referring to this a okay so in this case your 2a will be actually b plus c right this 2a will be actually b plus c isn't it that is the sum of the distance of a from the two fixed point is equal to the length of the major axis right so 2a is b plus c so let's see whether you're able to get b plus c some somewhere so wishes already has given his response which is b okay let's see whether it's right or wrong others just to tell you one thing since your origin lies on the let's say you're considering b and c they are on the x-axis right so b and c are on x-axis why because you have considered that them to be the foci foci always lies on the major axis right so x-axis is basically containing your major axis okay when x-axis becomes your major axis that means your a has to be greater than b so these two cannot be your options anyways so we just be it's we can be solely rule out b cannot be your answer you can either have a or you can either have c because foci are always on the major axis and in this case the major axis with the x-axis means it has to be an ellipse of this nature not the standing one got it we just chill out so we'll work this out guys let's not spend too much time because we have seven more problems to go after this now guys here I'll use the half angle formulas of tan right how many of you remember tan a by 2 tan b by 2 tan c by 2 formula do you remember the half angle formulas properties of triangles correct so tan b by 2 would be under root of s minus a s minus c by s s minus b and tan of c by 2 would be under root of s minus a s minus b by s s minus c right so I'll directly use this formula where s is the semi-perimeter of the triangle s is the semi-perimeter of the triangle so 4 times tan b by 2 tan c by 2 is 1 so when you multiply these two if I have not mistaken you will get 4 times s minus a square s minus c s s minus c s s minus b s minus b right so this goes off this goes off you get under root of s minus a square by s so which means 4 times s minus a is equal to is equal to s hope I'm not missing out on anything so from here I will have a plus b plus so from here I will have 3s is equal to 4a right is it fine guys so we'll have 3 by 2 a plus b plus c is equal to 4a so a plus b plus c is equal to 6 sorry 8a by 3 so b plus c is equal to 5a by 3 right and 8 self is the distance between which is between b and c right a is your bc which is equal to 6 I guess so that is 5 by 3 into 6 which is equal to 10 that means 2a is equal to 10 which means a is equal to 5 and 2ae is equal to 6 which means e is equal to 6 over 10 which is 3 by 5 right now what is b square so we know that 1 minus b square by a square is your e square 1 minus b square by a square is equal to e square so I can say 1 minus e square is your b square by a square which means b square is a square times 1 minus e square which means it is 25 times 1 minus 9 by 25 which is going to be 16 correct so now you have to draw an ellipse or you have to write the equation of an ellipse whose center is at whose center is at 5 comma 0 and whose a square is 25 and b square is 16 which is clearly which is clearly the ellipse equation would be x minus 5 square by 25 plus y square by 16 equal to 1 that makes your option number a as the correct answer which makes your option number a as the correct answer so guys I think you need to revisit your properties of triangles and of course basics of ellipse nothing very difficult about it so now we are going to have a small break of let's say 10 minutes we'll assume at 5 0 5 so we'll have a break okay we'll assume at 5 0 5 p.m. sorry we'll assume at 5 o'clock we'll assume at 5 o'clock well so those who are back from the break so here's the problem in front of you you can start solving it so again this is a question on indefinite integrals right where you have to integrate x times f of x where f of x is not given to you directly but in the form of a limit you guys here the challenge is to evaluate the limit and hence to get your function I think integration will not be that much of a challenge for you so again a typical j question where they make things very very they ask you to mine out some information in order to reach the end of the problem anybody who could get the function please share with me if you have got the function also so guys here please remember this limit is on n so your answer will be in terms of x and that is what is going to be a ref of x okay so first who will tell me which form of indeterminacy is theirs which this limit is trying to address what is the indeterminate form over here exactly so it's tending to 0 this part is tending to 0 and this part is infinity so it's tending to 0 times infinity form of indeterminacy okay so now in order to address this first of all I'll convert it to 0 by 0 form so which I'll do it like this so x to the power 1 by n minus x to the power 1 by n plus 1 divided by 1 by n square now it has become 0 by 0 form so that I can apply lopital over here okay further to just make my life easy what I'll do is I'll pull out an x to the power 1 by n plus 1 common that means I will write it like this okay this is not a mandatory step but it is just to make my life simple because now I will have this term as x to the power of almost tending to 0 quantity which is going to be 1 anyways correct so ultimately what I get is limit and tending to infinity x to the power 1 by n minus 1 by n plus 1 minus 1 by 1 over n square correct remember this is still 0 by 0 this is still 0 by 0 form of indeterminacy okay now I'll apply lopital over here applying lh rule now tell me when you're applying lopital rule will you differentiate with respect to x or will you differentiate with respect to n will you differentiate this with respect to x or will you differentiate this with respect to n n exactly because the limit is applied on n correct very basic questions because many people will make mistake in these small basic concept because they're so used to differentiating with respect to x so guys it is like a constant raise to a power variable isn't it so it'll be x raise to the power 1 by n minus 1 by n plus 1 okay ln x into derivative of this term which is actually this and the denominator it will be minus 2 by n cube correct so now let us simplify this further so I can write it as x to the power n n plus 1 and on the numerator I will have 1 ln x here I will have minus n plus 1 whole square plus n square by n square n plus 1 whole square and in the denominator I have a minus 2 by n cube so minus 2 and n cube I can take it on top okay remember your limit here is n tending to infinity now can I say this term here we'll have n raise to a power 0 we'll have x raise to the power 0 which will approximately be 1 again correct ln x is a constant so is minus 2 I can take it out so I can write it as I can write it as minus half ln x and we have something like n square minus n plus 1 square by n square n plus 1 whole square into n cube limit and tending to infinity so which is minus half ln x this will be if I'm not wrong minus 2n minus 1 by n square and this will go off and you'll have a minus n over here and n plus 1 whole square over here limit and tending to infinity which is nothing but minus half ln x and you have limit and tending to infinity minus 2n square minus n by n square plus 2n plus 1 now this is infinity by infinity form of indeterminacy where you know the answer is the coefficient of n square in both numerator and denominator so your answer will become this into minus of 2 so minus 2 and minus 2 will get cancelled off that means your function was actually ln of x your function was actually ln of x is so much of hard work j made you to do in order to get the function itself now post that I think the process will be pretty much simple for you x ln x you'll follow integration by parts treating ln x as you and treating x as we so it becomes x square by 2 ln x minus 1 by x x square by 2 dx xn x gets cancelled so it becomes x square by 2 ln x minus x square by 4 plus c so which option is correct x square ln x minus x to the power 4 plus c oh yes option number d is correct in this case is that fine guys any question with respect to this all right so moving on to the 11th question now so here we have a question where we have been given a parabola y square is equal to 4ax so let me start sketching along with the question okay so we have a parabola y square is equal to 4ax and there's a focal cord and tangent is drawn at the focal cord okay so let's say at a focal cord here a tangent is drawn here now two circles are drawn in which one circle is drawn taking focal cord as diameter and the other is on by taking intercept of tangent this should be tangent between point t and the directrix as diameter so let's say this is your directrix let's say this is your directrix so with these two points as the extremities you have drawn a circle and let me draw another circle in the red in yellow now we have to find the locus of the midpoint we have to find the locus of the midpoint of the common cord of these two circles so pink circle let's say s dash equal to 0 and yellow circle s double dash equal to 0 we have to find the locus of the midpoint of the common cord locus of the midpoint of the common cord so pink circle has been drawn with pt as the diameter and the yellow circle has been drawn even though it doesn't look like that pq as the diameter so the one is with pq as diameter pq as the diameter and the pink one is with pt as the diameter anybody has any idea how to how to oppose this problem sure sure take your time not getting no worries kushal we'll discuss this how many of you don't have any idea so that I can start discussing it so okay no worries so guys all of you please pay attention over here recall some of the properties which you have studied if I connect t2q if I connect t2q can I say this angle here would be 90 degrees because we know the property that tangents drawn at the end of the focal cord meet the directrix at right angles correct and that is precisely the reason why I had drawn the structure in such a way that I my circle should pass through t okay right so I purposely made this yellow circle pass through t because I know that the tangents drawn from t to the ends of the focal cord are at 90 degree that means the direct trick behave direct trick behaves as the director circle for the parabola it's a director circle for the parabola is that fine now another important point that I would like you to recall is that the intercept of the intercept of any tangent the intercept of any tangent between the point of tangency between the point of tangency and the directrix subtends 90 degree at the focus this problem I use it as an opportunity to revise these two properties which many people forget especially the second one first one people know because of the concept that the directrix is the director circle of a parabola second one most of you forget because this is very less used intercept of any tangent between the point of tangency and the directrix subtends a 90 degree at the focus so that is the reason why I made this also pass through the focus so this is your focus so this angle will be 90 degree this angle will also be 90 degree are you getting this point so guys what what can you conclude in this case can I conclude that the common tangent the common cord for the pink circle and the yellow circle right can I say that it is nothing but the diameter PT itself so can I conclude that the common cord is PT itself that is the diameter of the pink circle is that fine now I have to find out the center of this circle I have to find out the locus of the center of this circle H comma K right now equation of the tangent we all know for a parabola is T y is equal to x plus a T square right please recall the parametric form of the equation of a tangent to a parabola correct now let's say P is a T square comma 2 a T what will be T point T point is where this particular tangent will meet the directrix so directrix means x equal to a so I'll put T y so for T point will find T y so T y is equal to minus a plus a T square so y is equal to a T minus a by T so T point will be minus a comma a T minus a by T correct so midpoint midpoint H will be a T square minus a by 2 and K will be K will be 2 a T plus a T minus a by T by 2 now from both of these equations I have to get rid of I have to eliminate I have to eliminate T because T is the parameter which I have chosen right so that is very easy to do that because from first let me call this as the first equation let me call this as a second equation from the first equation I can write T square minus 1 is 2 h by a so T square is equal to 2 h by a plus 1 correct from the second equation I can say K is equal to 3 a T minus a by T right correct so K by a is equal to 3 T minus a by T that means 3 T square minus a is equal to K T by a right now let's square both the sides let's square both the sides so you get 3 T square minus a whole square and K square by a square T square now put your T square as 2 h by a plus 1 minus a whole square equal to K square by a square and in place of T square you put 2 h by a plus 1 correct now time to expand this so we'll write it as 3 times 2 h plus a by a minus a whole square is equal to K square by a square 2 h plus a by a so this will become a cube right so it becomes 6 h plus 3 a minus a square times a is equal to K square 2 h plus a well we have to now square both the sides we have to square both the sides sorry we have to find the square of this term yeah 2 K by a is 3 T minus 1 by T correct oh 2 K by oh yeah thanks thanks for reminding that so 2 K by a will be this I think I did a small mistake here so this will be one let me just redo this part yeah sorry so I get this term which means which means 2 K by a T is 3 T square minus 1 the 4 K square by a square T square is going to be square of this term which is okay so 3 T square will be 6 h by a 6 h by a plus 3 minus 1 and here I'll get 4 a K square by a square T square is 2 h by a plus 1 yeah now I can take a 4 common so 3 h by a plus 1 the whole square and this 4 and this 4 will get cancelled so you'll have K square a square 2 h plus a by a so that will give you 3 h plus 1 3 h plus a whole square as K square 2 h plus a by a so now you can expand it so you have 9 h square plus 6 a h plus a square times a is equal to 2 K square h plus 2 K square h plus a K square which is 9 x square 9 a x 9 a square x plus a cube is equal to 2 x y square plus a y square there's any of the option match there's any one of the option match so we have 9 a x square I'll just write all the terms properly minus a y square then you have minus 2 x y square minus 6 a square x plus a cube equal to 0 so I think option number b matches yes or no is that fine guys any question with respect to this is it clear any idea any any doubt or any concern that you have no questions great so we'll move on to the next slide now which is question number 12 so the question says find the value of alpha for which the points alpha comma alpha plus 2 is an interior point of the smaller segment of this circle made by the chord whose equation is this best would always be to make a diagram first this would be a line where it is meeting the y at 3 and x at 4 okay and this is interrupt 2 so this is 2 comma 0 point this is 2 comma 0 point so this line will pass through I can write this as 3x plus 4 y equal to minus 12 so x by minus 4 plus y by minus 3 equal to 1 okay so it is making an intercept of minus 4 and minus 3 so minus 3 is somewhere over here minus 3 is minus 4 is here minus 3 is the interior point of the smaller segment of the circle made by the chord whose equation is this guys I have a feeling that this chord this chord is not even touching the circle will it test the circle let's find out the distance from the center yeah distance from the center is 2.4 right so 3x plus 4 y plus 12 comma 0 center distance will be mod by 5 right so this will be 2.4 which is greater than the radius of the circle oh yeah so this will be a chord it'll not even touch the circle right so problem is done straight away none of these yes yes hello so it was just a question whether you knew the concept whether it is going to test the circle or not so sometimes questions which look very very ugly from outside if you try to solve them they become so so simple okay so none of these will become the answer in this case so let's move on to the 13th question now find the value of the integral minus pi by 2 to pi by 2 pi to the past sine x by 1 plus pi to the past sine x with respect to x okay so Purvik has already given his response options C psi also says C others Atmesh okay so guys just recall this property which we had studied in definite integral that this can be written as 0 to a f of x plus f of minus x dx right and from here we had drawn two conclusions that if the function was odd it as answer would be 0 because both these functions will become negative of each other and if the function was even the answer will be twice 0 to a f of x because both of them will become the same right so here we have to use the parent property to solve this so from minus pi by 2 this can be a potential school level question for you many people just remember the result of the conclusion drawn from it but they don't result they don't remember the original result so please remember this so 0 to pi by 2 it becomes pi to the power sine x by 1 plus pi to the power sine x plus pi to the power minus of sine x by 1 plus pi to the power minus of sine x okay which becomes 0 to pi by 2 pi to the power sine x by 1 plus pi to the power sine x plus 1 because here I can in this case I can actually write this as 1 by pi to the power sine x plus 1 it actually becomes 1 which is going to be pi by 2 is your answer simple question moving to the next question now question number 14 if the normal to the curve y equal to f of x at x equal to 0 be given by this equation 3x minus y plus 3 equal to 0 then we have to find the limit of x square by this as x tends to 0 this should be a simple problem guys see what is the slope of this line slope of this normal is going to be 3 right so slope of the tangent is minus 1 by 3 correct which is nothing but f dash 0 right which is nothing but f dash 0 okay so Purvik has given his response let's check now clearly if you see this is going to be 0 by f 0 minus 5 f 0 plus 4 f 0 which is going to be 0 by 0 form correct so I can apply Lopital I can apply LH rule so it will become this 2x what's the derivative of f of x square derivative of f of x square will be 2x f dash x square similarly here I will get minus 5 into 4 dash f dash 4x square into 8x and 4 times f dash 7x square into 14x correct so I can cancel off x x x and x from here and put x as 0 put x as 0 and you put x as 0 we get the limit as 2 by 2 times f dash 0 minus 5 times f dash 0 into 8 which makes this 40 which makes this as 40 plus 56 times f dash 0 so your answer will be 2 times 58 minus 40 which is 18 times f dash 0 which is 2 times 18 into minus 1 by 3 right which is going to be minus of 1 by 3 so absolutely correct those who answered 1 minus 1 by 3 that is the right answer so with this we move on to the second last problem of the day which is based on differential equations question number 15 for the day which is based on differential equation any idea guys how to do this doesn't look like a variable separable doesn't look like a homogeneous either may be exact may be exact because I can see like terms on both the sides 1 by x 1 by y x y square this like terms on both the sides so may be exact that's a leap of faith I mean it may work it may not work yeah sure sure sure work towards it so at least let me separate the variables out just to see whether something happens okay I get this so these two are fine no problem with these two so only we have to see what happens to this so this part if I divide both numerator and denominator by x square y square do you realize I get something like this yes now it is going to be exact differential equation so if you see in this differential equation if you take 1 by y minus 1 by x as t then minus dy by y square plus dx by x square is going to be dt so it's going to be dx by x minus dy by y plus plus it becomes minus dt by t square right so this is the exact differential of so this part can be exactly differentiated as ln of mod x minus ln of mod y and this can be written as plus 1 by t right equal to some constant okay so it becomes ln mod x by y plus 1 by t now t here will be x minus y by x y so it will become x y by x minus y equal to c do we have any options which talk about this yeah option number c is correct okay so yes indeed it was a case of an exact differential equation any question with respect to this guys so alright it's in the interest of time we'll move on to the last question of the day which is the 16th question then we'll call it a day so this question is basically from yeah it has been picked up from series sequence and progressions the question says the number of values of x for which x gi f and fractional part are in HP are in harmonic progression previous question okay let me just go back to the previous slide oh I'm sorry I'm sorry yeah yeah you're correct you're correct my mistake this is not option C this is option A thank you thank you Sai for pointing that out yeah I'm sorry by mistake I mark C anyways moving on to the next question so we all know that when three terms are in HP ABC are in HP then B is equal to 2 AC by A plus C right so everybody is aware quick revision of class 11th progressions so as per the question this is equal to 2 x fractional part of x by x plus fractional part of x and in the class I have told you how to deal when you have x mod x fractional part x gi f of x first which first you'll always break up mod x okay under a given interval then you'll break x as gi f plus fraction part then you'll make fractional part the subject of the formula so this is the approach that we normally follow when we have all these mod x fractional part get us into a coming in the same problem so please follow that approach I'm sure you'll be able to find the answer to this so you have to tell me the number of values number of values is it one okay we'll check it out what about other guys any other response so as I discussed with you I would break x as gi f plus fraction part times fraction part by gi f plus two fraction part let's cross multiply cross multiply it'll give you a fraction gi f square plus two gi f fraction part is equal to two gi f fraction part plus twice fraction part square cancel this off so you get a fractional part twice a fraction part square is equal to gi f square that means fractional part is plus minus gi f of x by root of two okay now we all know that fractional part happens to lie between zero and one that means plus minus by root two will lie between zero and one correct now what I'll do is I'll take two different inequalities from here one is with a plus sign this lies between zero and one and other with a minus sign okay now this will give you a fractional part lies between zero and root two and this says fractional part lies between minus root two and zero minus root two and zero okay now when you say fractional part lies between zero and two the only one possibility is it can be zero or it can be one correct and from here the only possibility is it can be minus of one correct let's check let's check it outside now when you take fractional gi f is zero you realize that fractional part will also come out to be zero right from this relation so this is the relation which connects the fractional part to the gi f which is not possible because x will become zero and you know that in an hp no term can be zero in an harmonic progression there'll be no zero term never there cannot be any term which is zero so the only two possibility left with me is one and minus one okay so now when x is one sorry gi f of x is one your fractional part could be now it could be plus minus one by root two but we know that fractional part is always positive so that means it will always be plus one by root two that means my x will be my x will be one plus one by root two it can never be one minus one by root two remember fractional part is always positive it lies between zero and one and when you take minus one then in that case the fractional part still will be one by root two that means your x could be minus one plus one by root two so we have only two answers coming from here which is this one and this one so the number of values of x for which these three terms would be in an harmonic progression would be two in number will be equal to two is that fine Saimir right all right guys so we'll end the session right now thank you all for coming online so on behalf of Centrum Academy I would like to wish you all of you best of luck for your upcoming exams do well keep your calm and remember to practice a paper a day before the exam full length paper just check you know your time management stress management mistakes handwriting everything more than two paper like two or three paper would be good enough okay so thank you so much so we'll call it a day thank you for coming online bye bye and best of luck