 Hello. Good morning. Welcome all to the YouTube live session on board level problem solving. Good morning to everyone who is present in the session. So I would request you to type your names in the live chat box so that I know who are attending the session. Hello. Good morning, Sondaria. All right. Yeah. So even before I start with the first problem. So yes, your answer is correct. So if a is a square matrix of order three says that a joint of a is 64 find the determinant of the transpose of a. So we all know that determinant of a joint of any square matrix a of order n is given by determinant a to the power n minus one, where n is the order of the square matrix, the square matrix a. Okay. So since it is of order three, so determinant a to the power of two is 64. Correct. Okay. So determinant a could have been plus or minus eight. So determinant a transpose also could have been plus or minus eight. Hello. Welcome all. Welcome all to the live session. So let's begin with the next question now question number two. So here is a square matrix satisfying a square is equal to I then what is the inverse of a again a super simple question. Just a one marker. Exactly. So basically this is a matrix. It is called an involuntary matrix where a is its own inverse. So if you just do a pre facto multiplication with a inverse on both the sides, you will get a is equal to a inverse. So inverse of a is a itself. Next, again, treat this as a rapid fire round. What are the degree of the following differential equation? What is the degree of the following differential equation? Past to absolutely correct. So basically this is the highest order. Okay. This is the highest order differential coefficient and the power highest power of the highest order differential coefficient defines the degree for you. So degree is going to be 2. Right. So answer is 2. Next, if A and B represent the two adjacent sites of a parallelogram, then write the area of the parallelogram in terms of A and B. Super simple. What is the area of the parallelogram? Just type it in the chat box. Mod of A cross B. Absolutely correct. Okay. Next, find the angle between the vectors A and B given that mod A is 3, mod B is 4 and mod of A cross B is equal to 6. It could be 150 degrees as well. Okay. So mod of A cross B is nothing but mod A mod B into sine of the angle between them. So this is 6. This is 3 into 4 into sine of theta. So sine theta is half. And this implies the angle could be 30 degrees or it could be 150 degrees. That's 5 by 6 or 5 by 6. So this is the last of the one marker. Find the direction cosines of a line passing through origin and lying in the first octant making equal angles with the three coordinate axis. Absolutely correct. So basically, let's say your direction cosines are L, M and N and each one of them is going to be cos of alpha because the angles made are all equal. And we also know that L square plus M square plus N square has to be 1 which implies 3 cos square alpha has to be 1 which implies cos of alpha has to be plus minus 1 by root 3. Okay. So either you can say this. Okay. Minus will be neglected because we are taking it into the first quadrant. Okay. So we are assuming the direction of the line to be in such a way that the angles made are all acute angles. Next. Show that the relation R in the set A given by the relation R is equal to defined as all ordered pairs A, B says that mod A minus B is 0 by 4 is an equivalent solution and find the set of all elements related to 1. So in relation questions, please ensure that you write all the steps properly without any kind of ambiguity. Try to show it's going to be all reflexive, symmetric and transitive. At least type in the answer for the last part of the question which says find the set of all elements related to 1. If done, do let me know. Just type done on the chat box. Okay. Great. So let's discuss this. So for all A belonging to the set A, okay, mod of A minus A is equal to 0 which is divisible by 4, which is divisible by 4, okay, implies A comma A belongs to the given relation, right. Therefore R is reflexive. Therefore R is reflexive. Okay. Now for symmetric, let A comma B belong to R. In fact, you can start by saying that let A and B belong to the set A and A comma B belong to R, okay. This implies mod of A minus B is divisible by, is divisible by 4, which also implies mod B minus A is divisible by 4, okay. And this implies B comma A will belong to R. Therefore R is symmetric. So for transitive, so you can write here reflexive, symmetric and for transitive, let A, B, C belong to A such that A comma B belongs to R and B comma C belongs to R, okay. This clearly implies mod of A minus B is divisible by 4 and mod of B minus C is divisible by 4. So I'll just make some space for myself by raising the first part. Let me raise this part. So this implies A minus B is divisible by 4 or you can say A minus B is equal to plus minus 4 lambda and B minus C is equal to plus minus 4 alpha, okay. So when you add it, so A minus C adding 1 and 2, adding 1 and 2. A minus C is equal to plus minus 4 lambda plus minus alpha, okay. Which clearly implies mod of A minus C is divisible by 4 and hence therefore R is transitive. Now regarding the set of all elements related to 1, so 1 itself is related to 1, 5 is related to 1, 9 is related to 1. After that it will go beyond the set A. So this will be your set which is related to 1. Next, 2 tan inverse of sin x is tan inverse of 2 Ck for x lying between 0 to pi by 2, solve for x. Please type in your response once you're done, okay. What about others? Alright. So let's discuss the first one here. So we all know that 2 tan inverse of x is tan inverse of 2x by 1 minus x square, okay. So applying the same onto the given problem, so 2 tan inverse sin of x will become tan inverse 2 sin x by 1 minus sin square x. And this is equal to tan inverse of 2 Ck, okay. So this means the inputs must be equal, okay, because they are 1, 1 functions. So 2, 2 will get cancelled off, sorry, why did I write tan square, sin square, okay. So we can say that sin x by cos square x, so cos square x if you take on the other side it becomes cos square Ckx. So sin x is equal to cos x, that means tan of x is 1. Only one possible answer in this particular interval satisfies it and that answer is pi by 4. So this is going to be your solution for this inverse trigonometric equation. Now try the other part of it, let me know once you are done, okay. Alright, so here we can use our double angle formulas, sin inverse of 2x by 1 plus x square is equal to 2 tan inverse of x, okay. And this formula is valid for x lying between minus 1 to 1, correct. In a similar way cos inverse of 1 minus y square by 1 plus y square is equal to 2 tan inverse y, okay. And this formula is valid for y lying between 0 to infinity. Now all these conditions are being met over here, this condition is a subset of this condition, even this condition is met over here, correct. Now this can be written as, the left hand side can be written as tan inverse of, sorry, tan of half 2 tan inverse x plus 2 tan inverse y, which is actually tan of tan inverse x plus tan inverse y, okay. We also know that tan inverse x plus tan inverse y is going to be tan inverse x plus y by 1 minus xy. So this will become tan of tan inverse x plus y by 1 minus xy, okay. And this formula is valid when your xy is less than 1. And this condition is also met by this particular situation given over here. So tan of tan inverse x plus y will be nothing but x plus y by 1 minus xy, which is nothing but your right hand side of the equation, okay. Hence shown. Let's move on to the next question now. If none of a, b, c are is 0, using properties of determinant, prove that. So as you can see, you have to prove that the value of the determinant is bc plus ca plus ab the whole cube. Let me know once you're done. So remember I told you that there are three types of problems where you can lose a lot of your time if you do not take the right approach. One is integration, other is determinants, another is maxima and minima. So as you would find that many of you would be struggling in this problem. So these are the deciders, the ones who get 90% and the one who get 95% and the one who get 99% in maths. These are the deciders. Anybody please let me know if you're done. So Vaishnavi, Sondarya, Sanjina, others who have not disclosed their names. If at all you're done, do let me know. And if you need help, please do also type that. So you're not able to do it, then I can start solving it. I think 10 minutes is a good enough time to give it to this problem if you're not getting it. Should I start solving it? No problem Vaishnavi, I understand. Determine problem if you not take the right approach, it may kill a lot of your time. So let's try to tackle this problem now. So first of all, I need the terms a, b, b, c, c, etc. Now see what I will do here. I would try to get rid of the terms like c square, a square, b square and all because they are not important to me as the outcome of it doesn't contain those terms. So what I will do first, I will multiply row 1 with a, I will multiply row 2 with b and I will multiply row 3 with c. So when I do that, I get 1 by a, b, c. Of course, when you're multiplying, you have to divide it because if you're not doing any other operation after multiplication, you have to divide it also. So a, b, c and this will be a, b square plus b, c. And this will be a, c square plus b, c. And here I will get b times a square plus a, c. Again minus a, b, c. Again b times c square plus a, c. And this will be c times a square plus a, b. c times b square plus a, b minus a, b, c. Now start taking, take a common from c1, b common from c2 and c common from c3. So a, b, c will come out and a, b, c is already there in the denominator that will get cancelled. So if you take out this a, a, a from everywhere, you get a minus b, c. And this will become a, b plus b, c. Okay. Similarly, this will become a, c plus b, c. Here if you take a b common, it will become a, b plus a, c. Then minus of a, c. And then from here I will get b, c plus a, c. If you take a c common from column 3, you become, it becomes a, b plus a, c again. And a, b, a, b plus b, c. And minus of a, b. Okay. So far so good. No problem in identifying these two steps. Okay. Now what do you do is you do this operation. Add everything to the first row. That means R1 is transformed to R1 plus R2 plus R3. Okay. So I am writing the result over here in this amount of space. Okay. So when you add everything on this, you will realize that b, c will get cancelled and you will get a, b plus b, c plus a, c. Okay. And same will be true for the second element of the first row. A, b plus b, c plus a, c. Okay. And again, a, b plus b, c plus a, c. Okay. Rest of the elements will not touch it. So a, b plus b, c, copy it. A, c plus b, c, copy it. Minus a, c. B, c plus a, c. A, b plus b, c minus a, b. Now you can pull out an a, b plus b, c plus c, a or a, c. Common from the first row itself. Now I would have to make some space because I cannot write anymore. Okay. So let me just erase this part. Yeah. So when you take a, b plus b, c plus a, b plus b, c plus c, a common from the first row, you get 1, 1, 1. So a, b plus b, c minus a, c. A, b plus b, c. A, c plus b, c. B, c plus a, c minus a, b. Okay. Then you know that if you have all 1, 1, 1 generated in a given column or row, what to do next? The obvious operation is you can take one of the column, let's say c1 and start subtracting from the other column. So let's say c2 is c2 minus c1 and c3 is c3 minus c1. Okay. So after doing this operation, you would realize that we will be getting something like this. So you have taken this and subtract. So let's not disturb the first column because it is not transformed. This becomes 0. This becomes 0. This becomes negative a, b plus b, c plus c, a. And this becomes 0 again. Here I will get, just in case I missed out anything, that's correct. Yeah, yeah, yeah, that's fine. Yeah. So here we will get again a 0 and this will become minus of a, b plus b, c plus c, a. Okay. Now from the first column, from the second and the third column, you can again pull out a a, b plus b, c plus c, a. So a, b plus b, c plus c, a whole cube will come out. Okay. And you get 1 a, b plus b, c, a, c plus b, c, 0 minus 1, 0, 0, 0, minus 1. Okay. Right. Now if you expand with respect to any one of the column number 2 or column number 3, let's say I expand with column number 3. So plus minus plus. So I'll get minus, minus of 1. Why am I getting a negative sign? Oh, sorry. Yeah. Minus of 1, but with again a minus, it will become plus. So it will become a, b plus b, c plus c, a whole cube. Because this is also minus only. Right. So minus of 1 minus 0 with a minus 1 sign will become plus 1. Okay. So this becomes your RHS and hence proved. Right. So these are the type of questions which can really kill a lot of your time. As you can see, we almost spent 18 minutes solving this problem. Right. Including the time I gave it to you. So be very, very careful of the approach and be watchful on what you want to prove. You cannot start doing any operation as you feel like. And always keep this multiplication of a row with ABC and taking then again ABC column, common in the columns. So this is also a very important kind of a property you can apply. Moving to the next question. Done. This is a simple problem. I guess. Do let me know once you're done. Okay. It's done. So under your Sanjana others. So for this kind of problem, it is best solved through substitutions. Right. Now the look and feel of these expressions under root of 1 minus x square under root of 1 minus y square gives you the feeling that we can actually substitute x as let's say sign alpha. Okay. And why are they saying sign of beta? Okay. So according to the problem statement, you can say cost of alpha plus cost of beta is equal to a sign alpha minus sign beta. Okay. Now we know the transformation formula cost alpha plus cost beta is to cost alpha plus beta by 2 into cost alpha minus beta by 2. And this will become to a cost alpha plus beta by 2 sign alpha minus beta by 2. Okay. So 2 can 2 can be removed. This can also be removed. Which means a cot of alpha minus beta by 2 is equal to a, which clearly implies alpha minus beta is 2 cot inverse a. Now, since a is a constant, a is a constant. This is also a constant. Right. But alpha is actually sign inverse of x and beta is actually sign inverse of y. Okay. So this is equal to a constant. Now differentiate both sides with respect to x. Differentiate both sides with respect to x. So that will give you under root 1 by under root 1 minus x square by under root 1 minus y square dy by dx is equal to 0. Which clearly implies dy by dx is nothing but under root of 1 minus y square by under root 1 minus x square is equal to your RHS. Right. Hence proved. So please keep in mind these kind of substitutions. Sometimes they complicate the problem by making it as under root 1 minus x to the power 6 plus 1 minus y to the power 6 is equal to 8 times x cube minus y cube. Right. And then here also they make this as to the power 6 and 6 and some terms will extra come over here. So in the same line you have to approach this problem as well by substituting x cube as sin alpha and y cube as sin beta. Okay. So be aware of the different versions of the same type. Okay. So we'll move on to the next question now. If y is equal to x plus under root x square minus 1 to the power of m then show that this differential equation is valid. Do let me know once done. Done. What about others? Done. Okay. So we'll start with the expression for dy by dx. So we'll be applying your chain rule on this. So m times this to the power of m minus 1 into derivative of this term itself which is actually 2x by 2 under root of x square minus 1. So this term gets cancelled off and if you try to take the LCM it becomes x plus under root x square minus 1 by under root of x square minus 1. So this becomes m times x plus under root x square minus 1 to the power of m by under root of x square minus 1. This is your dy by dx. So this expression is nothing but y itself. This expression is back to y. So you can write as my by under root of x square minus 1. Cross multiply. So take it on the left hand side. This is equal to this. Square both the sides. Square both the sides. Okay. When you square both the sides you get x square minus 1 dy by dx the whole square is equal to m square y square. Again differentiate with respect to x both the sides. I think there's a slight. Okay. This has to be minus I guess. So x square minus 1. This is 2 dy by dx into d2y by dx square. And you have 2x dy by dx the whole square. And here you will have m square 2y dy by dx. Now drop the factor of 2 dy by dx from everywhere. So 2 dy by dx gone, 2 dy by dx gone and 2 dy by dx gone. So you will get x square minus 1 d2y by dx square plus x dy by dx minus m square y is equal to 0. So this differential equation is valid. Let's move on to the next question. In all points of discontinuity of the function mod of x square. In fact it's not mod it is the greatest in the function. This has to be written in a proper way. Do let me know once you're done with the first part of the OR. Since the function definition is starting at 1 itself. It would not be a good idea to include 1 as your critical point. So you can do it like this. You can redefine the function as, you know, function which is going to be 1 for x lying between 1 to root 2. And your function will be defined as 2 when x lies between root 2 to root 3. And 3 when it lies between root 3 to 2. Okay, so this is how you can redefine the function. Now if you want to check the continuity of the function. You need to check it at the critical points. So these are the critical points. So at x equal to root 2. If you see the left hand limit of the function as x tends to root 2 minus is actually going to be your 1. And the right hand limit as x tends to root 2 plus is going to be 2 which is not equal. So these two are not equal which implies that the function is discontinuous at x equal to root 2. Similarly for root 3 if you want to check its continuity. The left hand limit of the function at root 3 is going to be 2 itself. Okay, and this doesn't match which is actually equal to 2. Which doesn't match with the value of the right hand limit at root 3 where the function value is 3. Okay, so hence f of x, therefore f of x is discontinuous at x equal to root 3. So points of discontinuity is x equal to root 2 and root 3. Moving on to the next part of the question. Differentiate sin inverse 2x under root 1 minus x square with respect to cos inverse 1 minus x square by 1 plus x square. Just let me know once you're done. Okay, what about Vaishnavi, Sanjana? So again in case of let's say this expression is y. Okay, and this expression is let's say z. We actually need to find dy by dz, right? But y itself, we can first simplify this. But again do not use any readymade formula. We all know that this is the formula of 2 sin inverse of x. But do not use it directly. Make substitutions. So let x is sin of theta, right? So this implies y is sin inverse 2 sin theta into cos theta. Which is nothing but sin inverse of sin 2 theta. Which is nothing but 2 theta. And 2 theta will be 2 sin inverse of x. Okay. For z you can directly use the formula. z is actually 2 tan inverse of x. Okay. Now dy by dz will be nothing but dy by dx divided by dz by dx. So dy by dx is nothing but derivative of this with respect to x. So it's 2 by under root of 1 minus x square. Derivative of 2 tan inverse x is 2 by 1 plus of x square. So 2 and 2 gets cancelled. So your final answer will be 1 plus x square by under root of 1 minus x square. Okay. Sometimes they may also ask you a question like differentiate this with respect to this at x equal to let's say 0. Right. Something like this they can ask you. So that means after differentiating you have to put the value of x as 0. Next. Integrate 1 by cos x minus a cos x minus b. Okay. How about Sondarya, Sanjana? Done. So for the first type this is something which we have already done in our past as well. We need to multiply and divide with sin of b minus a. Okay. So whenever there are like terms in the denominator that is sin sin or cos cos. We have to do this same trick. And then we can write the numerator part as sin of x minus a minus x minus b. So it becomes 1 by sin b minus a. This will become sin of x minus a cos of x minus b minus cos of x minus a sin of x minus b. Divided by cos x minus a cos x minus b. Again cos x minus a cos x minus b. So this and this will get cancelled. This and this will get cancelled. So 1 by sin of b minus a integral this will become tan x minus a minus tan of x minus b. So our answer will now become 1 by sin b minus a. ln mod seek x minus a minus ln mod seek x minus b. Okay. Which you can combine and write it as 1 by sin b minus a ln. Seek x minus a by seek x minus b is as good as saying cos x minus b by cos x minus a. Please try out on the second one. That's also a very easy question. Do let me know once you're done. Done. Okay. Great. So here it's obvious that you have to apply integration by parts treating this as your first function, treating this as your second function. So applying integration by parts, which is nothing but integral of u v is u integral of v minus integral of du by dx into integral of v whole with respect to x. So this will become log of x square into x square by 2 minus derivative of log of x square will be 2 log of x into 1 by x and 1 by x multiplies with this will become x by 2. Okay. Now again here we will apply integration by parts treating this as our first function, treating this as our second function. So it becomes x square by 2 log of x whole square minus. Okay. Now this will become u times x square by 2 minus 1 by x into x square by 2 dx. So this and this gets cancelled. So it becomes x square by 2 log of x whole square minus x square by 2 log of x plus x this will become x to the power 2 by 4 plus c, x to the power 2 by 4 plus c. So this will become your answer for the second part. Next integration of x by x cube minus 1. Do let me know once you're done. Done. Okay. So here it's very obvious that you have to apply partial fractions. So this term x by x cube minus 1, you can write it as x by x minus 1 times x square plus x plus 1. Okay. And therefore we can split this as partial fractions in this fashion. Okay. Let us first figure out the value of a, b and c. So when you combine the term over here, you get this as x. Let us put the value of x as 1 first. Putting the value of x as 1, I get 1 is equal to 3a. So a is equal to one-third. a is equal to one-third. Okay. Put the value of x as 0. That means you're comparing the constant terms. Put the value of x as 0. You will get 0 is equal to a and minus of c. Okay. So c is equal to a. That means c is also one-third. Now let's compare the coefficient of x square. So coefficient of x square on the left-hand side is 0. And on the right-hand side it's a plus b. So b is equal to minus of a, which is minus one-third. So this entire expression, this entire integral, I can write it as one-third x minus 1 plus one-third. Okay. In fact, I can say one-third 1 minus x by x square plus x plus 1. Okay. Hold the x. Now this is a simple integration. This is one-third ln mod x minus 1. For this one, I need to use the fact that the numerator can be expressed as a times the derivative of the denominator. Okay. Plus b. So comparing the coefficient, a becomes negative of half. Correct. And one is equal to a plus b. a plus b. So b becomes 3 by 2. So just this integral, if I have to do, I would write it as integral of minus half 2x plus 1 plus 3 by 2, divided by x square plus x plus 1, 1 by x square plus x plus 1. Okay. So this will become half ln x square plus x plus 1. And here I have to complete this square. So if you complete this square, here it will become x plus half the whole square. Guys, can you see the screen now? Girls, so once you've done this, this part answer will become 3 by 2 into 1 by a tan inverse of x by a. Okay. So collating the final answer, your final answer will now become, I'll just erase this part. Your final answer will become 1 third ln x minus 1. And since you've got half and 1 third over here, you'll end up getting, you'll end up getting minus 1 sixth ln x square plus x plus 1. Okay. And you have got root 3 over here and 1 by 3 over here. So that will give you plus 1 by root 3 tan inverse 2x plus 1 by root 3 plus c. So this will become your final answer for this question. Please check your results. Great. So now we'll move on to the next question. Using properties of definite integral, evaluate integral of x by 4 minus cos square x from 0 to pi. Let me know once you're done with this and also type in your response on the chat box. Sure, sure. Take the time. Pi square by 3. Okay. Anybody else who's getting the same answers? Sondarya, Sanjana. Okay. So we'll first start with the king's property. Okay. It says that integral 0 to a f of x can be written as integral 0 to a f of a minus x. Okay. So let's say I call this as i. So i will be equal to 0 to pi, pi minus x dx by 4 minus cos square x. Okay. So add the 2 i's. So it will be pi dx. So you can write pi by 2, 0 to pi. You can write this as since let's say this is your f of x. Since f of x is same as f of 2 a minus x. That means f of x is same as f of pi minus x. Nothing is going to change. You can write i as, sorry, you can write i as pi by 2 into 2, 0 to pi by 2 dx by 4 minus cos square x. So this gets cancelled. Now divide the numerator and denominator by cos square. So this will become c square on top. And here it becomes 4 secant square minus 1. Now in the denominator, change that secant square to tan 1 plus tan square. So this will become 3 plus 4 tan square x. Now take tan x as t, which implies secant square x dx is going to be your dt. So i will be pi times 0 to infinity dt by 3 plus 4 t square. Okay. Take 4 common. So pi by 4, 0 to infinity dt by t square by 3 by 4 will be root 3 by 2 the whole square, which is pi by 4, 2 by root 3 tan inverse of 2t plus root 3 by root 3. Okay. Now when you put your t as infinity, it becomes pi by 2. And when you put it as 0, it becomes 0. So the answer will be pi square by 4 root 3. So i will be pi square by 4 root 3. Vaishnavi, that's not correct. Something went wrong. Please check your calculations. Is that fine? Everything fine with the solution? No questions, no concerns? Now we'll move on to the next one. Okay. Now this appears to be a very easy problem. I don't know how it figured it in between these problems. This is super simple. So let the vector be xi plus yj plus zk. So dot product with the first one means x minus 3z is equal to 0. Okay. Dot product with the second one means x minus 2y is equal to 0. And sorry, is equal to 5, not 0. And the dot product with the third one is 8. That means x plus y plus 4z is equal to 8. This is also my bad. Okay. So subtracting, you get z is equal to 5. That's for sure. If z is 5, x is going to be 15. And if this is 15, this is y. This is going to be 20. So y is going to be 8 minus 35, which is minus of 27. So your required vector will be 15i minus 27j plus 5k. Okay. That's an easy problem. So let's not waste too much time on this. Yeah. 3D problem. Let's take a 3D problem. Find the equation of a plane passing through the point 1 to 1. And perpendicular to the line joining the points, these two. Also find the perpendicular distance of the plane from the origin. Okay. What about others? So there's a plane which contains 1 to 1. And it's perpendicular to the line joining the two points. So this is a line joining the two points. You have to find the equation of this plane. Guys, can you see the screen? Can you see the screen? Yeah. So basically the normal vector direction can be known by the 3 minus 4j. And 5 minus 2k. Okay. That's going to be i minus j plus 3k. This will be a normal vector. And we all know that the equation of a plane is r minus a dot n is equal to 0. So r dot n is equal to a dot n. So r dot i minus j plus 3k. And this is your a vector. So a vector is the position vector of this point i plus 2j plus k. So the dot product would be nothing but 1 minus 2, 1 minus 2 plus 3. Okay. So that is going to give you 2. Hence the final answer, if you want to write it in the Cartesian form, you can write down the equation to be x minus y plus 3z is equal to 2. This is the equation of a plane in the Cartesian form. Now the next part of the equation is also find the perpendicular distance of the plane from the origin. Now we all know if you want to find the distance of the origin, we have to take mod d by under root of a square plus b square plus c square. Okay. This is the distance from the origin. This is the perpendicular distance from the origin. So that is going to be 2 by under root of 1 plus 1 plus 9, which is 2 by root 11. Okay. Now let us do the all part of the equation. Find the equation of the perpendicular drawn from the point p to the line. Okay. Okay. Fine. No issues. Others, are you getting the same stuff? So this is our line and from a point 2, 4 minus 1, you are drawing a perpendicular on this particular line. So we have to find the equation of this line. Okay. So first of all, we will consider this point to be a point m. Okay. Whose coordinates can be obtained by using the parametric representation for this line? Let us say, call it as lambda. So x will be lambda minus 5. y will be 4 lambda minus 3 and z will be minus 9 lambda plus 6. So this point is lambda minus 5, 4 lambda minus 3, minus 9 lambda plus 6. Now the direction of this line is going to be the difference of the coordinates. So lambda minus 7, 4 lambda minus 7 again and then minus 9 lambda plus 7. Okay. This should be perpendicular to the given line. So 1 into lambda minus 7, 4 into 4 lambda minus 7 and minus 9 into minus 9 lambda plus 7 will be equal to 0. So this will be 81, 81 plus 16 plus 1, 81 plus 16 plus 1 lambda. Okay. So we will get minus 7, minus 28, minus 63 equal to 0. So that will become 98 lambda and here also I will get 98 itself. So lambda is going to be 1. So if lambda is 1, the point is going to be minus 6, minus 3, minus 2. I am sorry. The point is going to be minus 4 comma 1 comma minus 3. Okay. So this is going to be your point. So once you know a point, we can directly find also the direction from here. So direction of the line will be, direction ratio of the line will be minus 6, minus 3 and minus 2. Is that fine? So the equation of the desired line will be, I am just going to erase this. So the equation of the desired line will be x minus 2 by minus 6, y minus 4 by minus 3 and z plus 1 by minus of 2. Okay. In the vector form, it will become r is equal to 2y plus 4j minus k plus some lambda times 6i minus 3j minus 2k. Is that fine? Sondarya, just check why your answer is not matching with mine. 1, i plus 2j plus k is not proportional to what I have got which is 6 is to 3 is to 2. But 1 is to 2 is to 1. Okay. Just check out your working. Meanwhile, we will move on to the next problem. A bias dice is twice as likely to show an even number as an odd number. The dice is rolled three times. If the occurrence of an even number is considered a success, then write the probability distribution of successes. Find the mean number of successes. So let me know once you are done. At least you can type in the mean number of successes. How? How much should we? Alright. So in this case, we know that your p is going to be 2 by 3 and your q is going to be 1 third. Okay. So it's actually mean or the expected value can directly write over here itself as np. Okay. And it's known that there are the dice are rolled three times. So it's going to be 3 into 2 by 3. That's going to be 2 itself. Okay. Now regarding the probability distribution, you can actually expand this. Okay. And you can just make a column like this. So probability could be 0, 1, 2 or 3. So for 0, you just have the first term, which is 1 third cube. Second term will be 3c1, 1 third square into 2 third and 3c2, 1 third into 2 third square and finally 2 third cube. So these values will be coming over here. So 1 by 27 is this value and this will be 3 into 6, 6 by 27. This will be 12 by 27 and this will be 8 by 27. This is going to be the probability distribution and the mean number of successes will be nothing but 2. Using matrices, solve the following system of equations. 1 by x minus 1 by y plus 1 by z is 4. 2 by x plus 1 by y minus 3 by z equal to 0. 1 by x plus 1 by y plus 1 by z is equal to 2. Okay. The problem becomes very simple once you start taking your 1 by x as capital X, 1 by y as capital Y and 1 by z as capital Z. Okay. So it's actually x minus y plus z is equal to 4. 2x plus y minus 3z is equal to 0. Yeah. So first we can convert it to a matrix form. So 1 minus 1, 1, 2, 1 minus 3, 1, 1, 1, x, y, z, 4, 0, 2. Okay. So this is of the form ax is equal to b. So let's find out the inverse of a. First of all determinant of a. Let's find it out which is going to be 1, 4 plus 1, 5 plus 1, again a 5. So again, this will become 1. That's going to be 10. Correct. Let's find out the adjoint matrix. So I will use my method here. So 1 minus 1, 1, 2, 1 minus 3, 1, 1, 1. Again, copy the first row, second row, first column, second column. Okay. So let's cross multiply this 2, this 2, this 2, this 2, this 2, this 2, this 2, this 2, this 2, this 2, and this 2. So that will become 1 minus 1, which is 4, minus 3, minus 3, minus 2 is minus 5, then it will become 1. 1 plus 1 is 2, then this will be 0, then this will be minus of 2, then this will be 3 minus 1, which is 2, then this will be 5, and this will be 3. Okay. So this into 4, 0, 2 multiplied with 1 by 10. Okay. So I can clearly see that X is going to be 16 plus 4, 20, 20 by 10, which is 2. Y is going to be minus 20 plus 10, which is minus 10, minus 1, and Z is going to be 4 plus 6, 10, that is 1. So X is going to be half, Y is going to be minus 1, Z is going to be 1 again. Yeah, so almost everybody is correct. Again, the call is yours for finding out the inverse of A, you may use elementary transformation or you may use adjoint, but adjoint is always a faster method. Show that the volume of the greatest cylinder that can be inscribed in a cone of height H and semi-vertical angle alpha is 4 by 27 pi H cubed tan square alpha. Done. Okay. So let's discuss the first one. So we have a cylinder inscribed within a cone. So let's say this is a cone and there is a cylinder which is inscribed within a cone like this. Okay. The height of this cone is small h and the semi-vertical angle here is alpha. This is alpha. Okay. Right. Now, see, let's say the height of the cylinder itself is capital H to the Y, R square into capital H where R is the radius of the cylinder. Correct. Now, this part, this part is H minus H. Okay. So what will be the radius? Right. So radius by H minus H is equal to tan of alpha. So radius is going to be H minus H tan alpha. Remember, alpha is fixed. Alpha is a constant. Okay. Okay. And capital H is a variable. Small h is also a constant, but capital H is a variable. Okay. Now, the expression for the volume will be pi R square, which is tan square alpha, which is H minus H whole square into H. Now, if you want it to be maximum, the derivative of the volume with respect to capital H should be 0, which means the derivative of this expression, which is H minus H the whole square plus 2H H minus H with a negative sign. So let me make a negative over here. So this will be negative and this would be equal to 0. So from here we get H minus H square is 2H H minus H since H small h cannot be equal to capital H. That means you cannot have a cylinder with the same height as that of the cone. So here H is equal to 2H plus H. That means capital H is H by 3, small h by 3. And R will be nothing but 2H by 3 tan of alpha. Right. Now, always do the double derivative test. D to V by D H square. Okay. So D to V by D H square will be pi tan square alpha. Okay. So this will be 2 times H minus H with a negative sign and this will be minus 2H minus H. And again you will have, this will be positive I guess. Okay. And yeah, minus 2 times H minus H and then we will have minus 2H into minus 1. Okay. So if you look at the entire expression, you would get the answer as negative of 2H, pi tan square negative of 2H, pi tan square negative of 2H which is actually a negative expression. That means the volume will be maximum, the volume of the cylinder will be maximum. Now having found out the radius and the height capital H, the volume can be found out by using pi R square. R square will be 4H square by 9 tan square alpha and capital H is nothing but small H by 3. So that will be 4 by 27 pi H cube tan square alpha. So hence proved. Okay. Now let's talk about the next part of the question. Show that the normal at any point on the curve is a constant distance from the origin. Let me know once done and also let me know what is the shortest distance, what is the distance from the origin that you are getting. Done. Okay. The distance is A. So let me quickly discuss this also. We have to find the equation of the normal to the curve. Okay. So for normal to the curve, we need the slope of the normal which is nothing but minus dx by dy. So which is minus dx by d theta divided by dy by d theta. So minus dx by d theta will be negative of negative A sin theta. Okay. Plus A sin theta plus A theta cos theta divided by dy by d theta which is A cos theta minus A cos theta plus A theta sin theta. So this two will get cancelled. This two will get cancelled and you will get expression as minus of cot theta. So the equation of the given normal would be y minus y1 is equal to slope times x minus x1. So let's write this cot theta as cos theta by sin theta itself. Let's cross multiply. So y sin theta minus A sin square theta plus A theta sin theta cos theta is equal to minus x cos theta plus A cos square theta and plus A theta sin theta cos theta. This and this gets cancelled. So ultimately you get the equation as y sin theta plus x cos theta is equal to A. Okay. And we know that this line is basically nothing but at a constant distance of this is nothing but the normal form of a line right which is always at a distance of A from the origin. Okay. But you can also use the distance of a origin from this line to get the value and this constant distance is going to be mod of A. And the constant distance is going to be mod of A. Next, find the area of the region bounded by this particular set of points which satisfies these inequalities. When is your first board exams? And what which subject you have? I think fifth is physics sir. But I think when is your other subjects like English or electives? Okay. Second march is English. Once done, please type in your answer in the chat box. Is it done? Okay. 43 by 6 units. How about others? So let's discuss this. First of all, let us find out this point of p of intersection. So we'll simultaneously solve y is equal to x square and y is equal to x plus 2. So if you just compare their y's. Okay. It's obvious that x has to be 2 in this case. So this point is x equal to 2. Okay. So x equal to 3 has to be somewhere over here. So this is x equal to 3. Now the inequality zone suggests that you have to be y has to be lesser than x square. So you have to be below this part. And y also has to be lesser than x plus 2. So basically the region that they're looking out for is this region. Okay. But in this region also when you're trying to find out the area, you have to follow a different type of vertical strip in this zone. And a different type of vertical strip in this zone. So from 0 to 2, you will be integrating the area below x square. And from 2 to 3, you will be integrating the area within x plus 2. Correct. So your answer will be x cube by 3, 2 to 0. And again, x square by 2 plus 2x from 3 to 2. So this is going to be 8 by 3. This is going to be 9 by 2 plus 9 by 2 plus 6 minus 2 plus 2. That's going to be 8 by 3 plus 9 by 2 plus 2. So 6, this will be 16, this will be 27, this will be 12. So just let me, this will be 4, I guess. Small mistake here. This will be 4. Yeah. So 6 and 6 will get cancelled. So this term will not be appearing. Yeah. That's correct. Yeah. So now your answer is going to be 43 by 6 square units. So please mention square units in your answer. So absolutely correct. So all of you have done it. Got the right answer for this. So let's move on to the next one. Find the particular solution for the differential equation. Given that y is pi when x is 3, just a minor correction over here. Make this as plus. Okay. How about others? Okay. So this is clearly a case of homogeneous differential equation, right? This is clearly a case of a homogeneous differential equation. And since you see terms like y by x appearing, you can always put y is equal to vx, which means dy by dx can be written as v plus x dv by dx. So this will become x v plus x dv by dx minus vx into vx into sin v is equal to vx plus x dv by dx x cos v. Okay. You can drop the factor of x from all the places. Okay. So you may subtly drop x from every one, every place. This x will go off. This x will go off. This x will go off. This x will go off. This x will go off. This x will go off. Okay. And if you expand it over here, you will get vx dv by dx into sin v. And here you will get 2v plus x dv by dx into cos v. So x dv by dx will be nothing but v sin v minus cos v. And this will be equal to 2v cos v. So dx by x will be, dx by x will be v sin v minus cos v. In fact, you can take the two on the other side by v sin v. Sorry, cos v cos v. I hope there is no sin mistake done anywhere. No. That's correct. Now, if you look at the right-hand side integral, if you take v cos v as t, then minus v sin v plus cos v dv will become dt. That means it is as we were saying, negative dt by t. Okay. Let me just clear some space on top. So this will become 2 ln x is equal to minus ln t plus ln c. That means ln x square into t is equal to ln c. So that means x square into t. t is nothing but v cos v is equal to c. And v cos v is nothing but y by x cos y by x is equal to c. So this will be clubbed as xy. This will be clubbed as xy. Now, what is given to us? Boundary condition is y is pi when x is 3. So 3 pi cos of pi by 3 is half. So c is pi by 3 pi by 2. So your final answer will be xy cos y by x is equal to 3 pi by 2. This will become your answer. Yeah, I think it's the same thing as what Sondaria also got. Okay. Find the equation of a plane passing through the point 111 and containing the line. Also show that the plane contains this line as well. Once you're done, let me know. Okay. So let's discuss this also. Containing the line this is containing the line means you have been given two information that you know a vector on this plane. That is 3i minus j plus 5k. Okay. And you know two points 111 and the other is minus 315. So you can also get this vector. So two vectors can be known on the plane. So if two vectors are known, let me call them as a and b. I can always get the direction of the normal as well. Okay. So a vector will be minus 4i plus 0j plus 4k. Okay. b vector is already known to you 3i minus j plus 5k. So n will be in the direction of their cross product. So let's write their cross product. Okay. So this will give you 4 minus j minus 20 minus 12 is minus 32 and k is going to be 4. So you can write it as i plus 8j plus k also. Okay. So your equation of the line will be r dot n. r dot n is equal to a dot n. And a dot n can be written as n. So it's x plus a to y plus z is equal to n. Absolutely correct, Sondarya. Next is show that this plane also contains this line. Now in order to show that the plane contains the line, you also have to show that the line is perpendicular to n. That means 3 into 1 minus 1 into 8 plus 5 into 1 should be 0. That's true. And not only that, this point must satisfy it. So minus 1 plus 16 plus 5 is it equal to 10? Yes, it is equal to 10. Hence this plane contains this line. Okay. So please do not just show that the direction of the line is perpendicular to the normal. You also have to show that the point also satisfies the plane. This is an LPP question. You can just skip it as take it as a homework. Let's do this probability question. Hope you have understood the question. So bag A and bag B. A contains four white and three black. B contains two white and two black. So two walls are transformed from this back to this back. And then a ball is drawn from B. And it is found out to be black. What is the probability that one white and one black was transferred? 3 by 5 is absolutely correct. Yeah, that's absolutely correct. Again here, approach the problem by defining the events. So let even be the event that two white balls were transferred. Two white balls were transferred from A to B. Okay. E2 is the event that two black balls were transferred from A to B. And E3 be the event that one white and one black were transferred from A to B. So what do we have to find out? We have to find out probability of E3 given B has occurred, right? So according to the conditional probability, this formula is going to be P E3 intersection B by PB. Which is nothing but P E3 into P B by E3 by PB. Okay. Now, first of all, PB. How many ways can PB happen? PB can happen with P E1 into P B by E1 plus P E2 into P B by E2 plus P E3 into P B by E3. Right? Now, even if the event that you're transferring two white balls from bag 1 to bag B. So the probability of getting two white balls will be 4C2 by 7C2. Okay. And then getting a black ball given that two white has been transferred. So there will be two black and four white. So its probability will be 2 by 6. Okay. Then transferring two black balls from bag A to bag B will be 3C2 by 7C2. And then getting a black ball will be 4 by 6. And transferring one white and one black will be 4C1, 3C1 by 7C2. And then transferring a black ball will be 3 by 6. And then getting a black ball will be 3 by 6. Okay. So out of this, the third expression is in the numerator. As you can see in the formula, third expression is in the numerator. Rest everything is in the denominator. So your answer is going to be, I'm just going to erase this part. So your answer is going to be 4C1, 3C1 into 3 by 6 divided by 7C2 divided by the entire thing. 4C2 by 7C2 into 2 by 6. 3C2 by 7C2 into 4 by 6. 4C1 into 3C1 by 7C2 into 3 by 6. So 6 and 7C2 can be cancelled out. Okay. So if I'm not wrong, the numerator will become 36. And denominator will become 6 into 2, 12. 12 and this is going to be 3 into 4 again, 12. And this is again going to be 36. So it's 36 by 60. That is going to be 3 by 5. So answer is going to be 3 by 5 in this question. Okay. So guys, we're going to stop over here. And there are a few more questions. I think one more question after this, an LPP question. You can always try that as homework. So over and out from Centrum Academy. Thank you and have a good day and keep practicing. Keep writing mock tests. Okay. So the next one week is going to be very, very critical. Make your hand, eyes, everything. Tune to your board exams. Okay. So think as if you have to write board exams day in and day out. So accordingly, you should start practicing. Every day you should be writing at least one or two mock tests. Thank you so much. Bye-bye. Take care.