 Hello everyone, welcome to the road-level problem-solving session. So those who have joined in this session, I would request you to type in your names in the chat box so that I know who all are attending this session. So let's get started with the road-level problem-solving. So here is the first question for you. Write down the value of the following integral of sin to the power 5x from minus pi by 2 to pi by 2. I would often say 0, absolutely correct because this is going to be an odd function. This is an odd function because sin of minus x to the power 5 is negative sin to the power 5x. And we know that for an odd function, integral from minus a to a is going to be 0 if f of x is an odd function. So this answer is going to be 0. Simple one to start with, so let's move on to the next question. Again, this is one marker. Find the distance of the following plane from the origin. Find the distance of the following plane from the origin. Absolutely correct, 1 by 3. So we know that in any plane Ax plus By plus Cz plus D equal to 0, the distance from the origin is going to be mod D by under root of A squared plus B squared plus C squared. So here mod of D is going to be mod of 1 which is 1 by under root of 2 squared minus 1 squared plus 2 squared. That's going to be 1 under root of 9. That's going to be 1 third unit. Let's move on to the next question. Again, a one marker. Find the value of lambda such that the cross product of these two vectors is a null vector. So I think Saturday onwards, your board exams are going to start right, Sai, if I'm not wrong. Saturday is your first paper on chemistry, sorry, English. So what are the timings of these exams? Will it be in the afternoon always? Okay, 1030 to 130. And what about the ISC, guys? ISC folks. Alright, so lambda is minus 3 as per Sai. So let's check quickly. So basically, if this is a null vector, then the cross product is given by the determinant ijk. 2, 6, 14, 1 minus lambda 7. This is going to be a null vector. So i times 42 plus 14 lambda minus j times 14 minus 14, that's going to be 0. K times minus 2 lambda minus 6 is equal to 0, means 0i, 0j, 0k. So you can compare any one of them. For example, if you compare this with 0, you get your answer as lambda is equal to minus 3. Even if you do 42 plus 14 lambda equal to 0, you get lambda same as minus of 3. So again, an easy problem. Let's move on to the next one. It's a question from Probability. A family of two children find the probability that both are boys. If it is known that at least one of the children is a boy. And second part is, if it is known that the elder's child is a boy, please type in the answer for the first part. So both Sai and Sundar are saying one-third. So let's check. Let A be the probability that both are boys. Let A be the event that both the children are boys. And B is the event that at least one of the children is a boy. What do we need to find out? We need to find out A given B. That means probability of A intersection B by probability of B. Now first of all, what is the event A? Event A when both children are boys. That can only be when both the children are boy-boy. What does the event B? At least one of the child is a boy. So it can be boy-girl. It can be boy-boy. It can be girl-boy. Now intersection of these two events is only boy-boy. So probability of N A intersection B will be 1 divided by the total number of cases. The total number of cases is the sample space would be boy-boy, boy-girl, girl-boy, girl-girl. So 1 by 4 is your numerator. Divided by B will be 3 by 4. So the answer will be one-third for the first part. Second part is, let's say I define another event C that the elder child is a boy. Elder child is a boy. So we need to find out probability of A given C. So again it's P A intersection C by P C. Now event A I have already written. Let's write event C now. Event C is when the elder child is a boy. So it can be boy-girl or it can be boy-boy. So A intersection again will be 1. So this will be 1 by 4. Divided by P C will be 2 by 4. So your answer is going to be 1 by 2 for the second part. So absolutely correct. So we'll move on to the next problem now. Move the following, tan inverse x plus tan inverse 2x by 1 minus x square is equal to tan inverse 3x minus x cube by 1 minus 3x square. Just try it done once you're done with the first part. Okay. So for the first part we just have to use the formula tan inverse A plus tan inverse B is tan inverse A plus B by 1 minus AB. So tan inverse x plus tan inverse 2x by 1 minus x square can be written as tan inverse x plus 2x by 1 minus x square divided by 1 minus 2x square by 1 minus x square. So that's nothing but tan inverse x minus x cube plus 2x divided by 1 minus 3x square. That's tan inverse of 3x minus x cube by 1 minus 3x square. That's equal to your right-hand side. Okay. Hence proof. Okay. Easy one. Now quickly do the other part of it. Done. Okay. So let's quickly discuss this as well. So here we need to make use of our reference triangle. So let's first take care of this part. So let cot inverse x be theta. So cot of theta is x. Okay. So let's say this is theta. So cot of theta is x means base by perpendicular is x by 1. So this is going to be 1 plus x square. Okay. So sine of theta is going to be 1 by under root of 1 plus x square. Okay. So now this reduces to cos of tan inverse 1 by under root of 1 plus x square. Okay. Again, let's take this as phi. So let's make a reference triangle. This is phi. So this is 1. This is under root of 1 plus x square. So this is going to be under root of 2 plus x square. So cos of phi is going to be, cos of phi is going to be base by hypotenuse. So base is under root 1 plus x square. Hypotenuse is under root 2 plus x square. And this is nothing but your right-hand side. Hence proof. Okay. Simple enough. So now let's move on to the next question. Express the following matrix as a sum of symmetric and skew symmetric matrix and verify your result. Do let me know once you're done. So we know any matrix A could be written as half of A plus A transpose plus half of A minus A transpose, right? Where we know from the property of symmetric and skew symmetric matrices that this will always be a symmetric matrix and this will always be a skew symmetric matrix. So P will always be symmetric and Q will always be skew symmetric. No problem. Asunder, just see me doing it. Okay. So P is going to be half of A. A is nothing but 3 minus 2 minus 4, 3 minus 2 minus 5 minus 1, 1, 2 plus transpose of A. Transpose of A will be 3 minus 2 minus 4, 3 minus 2 minus 5, minus 1, 1, 2. Okay. So 3 plus 3, you can just add 3 plus 3 is 6 and this will be 1. This will be minus 5. This will be again 1, minus 4, minus 4, minus 5, minus 4, 4. Okay. So half of this is going to be 3, 1, half minus 5 by 2, half minus 2, minus 2, minus 5 by 2, minus 2. Okay. This is actually a symmetric matrix because if you transpose it, you would realize that you get the same matrix back. So this is going to be your P. Okay. In a similar way for Q, I'm just going to erase this up. Not everything. I'm just going to erase this part. For Q, this is going to be minus. So that's going to be half of 0, minus 5, minus 3, 5, 0, minus 6, 3, 6, 0. Okay. So half of it is going to be 0, minus 5 by 2, minus 3 by 2, 5 by 2, 0, minus 3, 3 by 2, 3, 0. Note the characteristic of a skew symmetric matrix that the leading diagonal will always be 0, 0, 0. And the elements which are symmetric about the leading diagonal will be such that they would be negatives of each other. Hi, Shantanu. Good afternoon. Okay. So we'll now move on to the next question. You have been given matrices. I'm sorry. You have been given vectors a, b and d. Find a vector of magnitude 6, which is parallel to the vector 2a minus 3b plus 2a minus b plus 3c. Yeah, sure. Sure. Once done, please type in your response in the chat box. Okay. Sai, what about Sundar, Shantanu? Oh, okay. No problem. Okay, it's quite simple because if you just find your 2a minus b plus 3c, you get this 2a minus b plus 3c. Okay. So I think that's going to give you i and 4 minus 6 is minus 2j and 2 minus 3 plus 2k. Okay. Now, there's a vector which is parallel to this vector and has a magnitude of 6. So let the vector be d. So d is lambda times this vector and magnitude of d is given as 6, which implies lambda square plus minus 2 lambda square and 2 lambda square under root is going to be 6, which implies 9 lambda square under root is going to be 6, which is nothing but 3 mod lambda is equal to 6. Okay. So here lambda value could be mod lambda is 2, so lambda value could be plus minus 2. Okay. So your possible vectors could be this or this. Okay. So both the answers are possible. So answers could be 2i minus 4j plus 4k or minus 2i plus 4j minus 4k. Now many people would have this opinion that this is parallel and this is anti-parallel. Correct? So even if you state this as your answer, it would be considered to be correct. Okay. And in fact, you can solve it by using the fact that you need a vector whose magnitude is 6. So it will be 6 times i minus 2j plus 2k by modulus of i minus 2j plus 2k. Okay. That will end up giving you the first answer. Okay. Let's do the second one, or part. If a, b and c are given, find a vector d which is perpendicular to both a and b and c dot d is equal to 18, none. So if a vector is parallel to a and b, you can safely say that the vector is along the direction of a cross b. So you can treat it as lambda times a cross b. Okay. So a cross b, you can find it by using your determinants. Okay. That's going to be 28 plus 4, which is 32 minus j, 7 minus 6 is 1 plus k, minus 2 minus 12 is minus of 14. Okay. So you can say your d vector is nothing but lambda times 32i cap minus j minus 14k cap. Now c dot d is given to us as 18. Okay. So this dot 2i minus j plus 4k is given to us as 18. So lambda 64 plus 1 minus 56 is going to be 18. Okay. So that's going to be 9. So lambda is going to be 2. So my d vector is going to be twice of this, which is 64i minus 2j minus 28k. This is going to be my answer. So I just check. I'm sure you would have gone wrong somewhere in calculation. What about Shantanu? Are you getting the same answer? Sundar? Okay. Let's move on to the next one now. Question is, find the points on the line at a distance of 5 units from the point 133. Once done, please type in the points in the chat box. Okay. There will be two such points. See, it is pretty obvious that let's say this is a line and the point p is here. So there will be two such points, which will be at the same distance of 5 from this. So let's say point A and point B. Absolutely correct. Absolutely correct, Sundar. So let's solve this, guys. So here, first of all, we'll assume that let lambda be the parameter, which is the parameter for the points on the line, which is at a distance of 5 from p. So equate this to lambda. So x will become 3 lambda minus 2, y will become 2 lambda minus 1, and z will become 2 lambda plus 3. Okay. So the distance of this point from, that means AP is equal to 5, which also means AP square is equal to 25. So let's use it. AP square will be under root AP square will be distance of this from 1 square, that is this minus 3 square, then 2 lambda minus 4 square, then 2 lambda square, that's equal to 25. So 4 plus 4 is 8, 8 plus 9 is 17 lambda square, and we will have the constants as minus 18 lambda minus 16 lambda and 25 is equal to 25. That clearly implies 17 lambda square minus 34 lambda is equal to 0. So lambda can be 0, lambda can be 2. If you put lambda as 0, you get the point as minus 2, minus 1, 3, that is this point itself, which is given to you in the equation. And when you put lambda as 2, you get the value as 4 comma 3 comma 7. So that's two possible points are there, so these will be your answer. Okay. Simple concept based on the parametric form of the line. Moving on to the other part, find the distance of the point P, which is 6 comma 5 comma 9 from the plane determined by the points A, B and C. First get the equation of a plane, which contains these three points and then use the distance of a point from a plane formula. 6 by root 34. Okay. What about others? Sundar? First of all, the equation of a plane that passes through these three points, we all know the equation of a plane formula when it passes through three points x minus x1, x minus, sorry, y minus y1, z minus z1, x2 minus x1, y2 minus y1, z2 minus z1 and x3 minus x1, y3 minus y1 and z3 minus z1 equal to 0. This is the equation of a plane, which passes through x1, y1, z1, x2, y2, z2 and x3, y3, z3. So we can take the first point itself as x1, y1, z1, so x minus 3, y plus 1, z minus 2 and then 5 minus 3 will be 2, 3, 2, minus 4, 0, 4. Okay. Take this as 0. Okay. Let's expand this. So x minus 3 times 12 minus y plus 1 times 8 minus minus 8 is 16, okay, and z minus 2 times. You may drop the factor of 4 from everywhere. So 3 times x minus 3 minus 4 times y plus 1 plus 3 times z minus 2 equal to 0. That gives you the equation of the plane as 3x minus 4y plus 3z and constant terms will be minus 9, minus 13, minus 6 is equal to minus 19. So this will be the equation of the plane which contains ABC. Now we all know the distance formula. The distance formula is mod of ax1, by1, cz1 plus d by under root of a square plus b square plus c square, correct, okay. So the point given is 659. So the distance can be written as 3 into 6, 18 minus 20 plus 27 minus 19 mod by under root of 9 plus 16 plus 9, okay. So if I am not mistaken, you will get 7 minus 1, 6 by root 34 units. 6 by root 34 is absolutely correct, okay. So Shantanu, after you have got the equation of a plane, we all know that this is the formula for distance of a point, distance of x1, y1, z1 from the plane ax plus by plus cz plus d equal to 0, okay. So you can use this formula directly to get the distance of a point from a plane, okay. Great. So let's move on now to the next question. Sundar, I hope it is clear to you, okay. So next is our differential equation. Solve the differential equation x square minus 1 dy by dx plus 2xy is equal to 1 by x square minus 1. Just type in done if you are done with the first part so that I can start the discussion for the same. Okay. Yeah, it's a linear differential equation. So for the first part, we can solve it by first converting it to this form. So it will be dy by dx plus 2x by x square minus 1. Y is equal to 1 by x square minus 1 the whole square. So integrating factor will be e to the power 2x by x square minus 1 dx which is e to the power ln of x square minus 1, okay. And this will be nothing but x square minus 1 itself, okay. So it becomes a super simple question. So answer will be y into x square minus 1 integration of 1 by x square minus 1 into x square minus 1. So this will be gone. So your answer will be y times x square minus 1 is equal to x plus c. So the first part is quite simple. Now try out the or part of it. Oh, sorry, sorry, sorry, sorry. Yeah, there is a x square whole square over it. I'm sorry. Yeah. Sorry about that. So you'll have this is equal to integration of this. Okay. So this is 1 by 2a ln of mod x minus a by x plus a, okay, okay. So your final result will be y times x square minus 1 is 1 by 2 ln mod x minus 1 by x plus 1 plus c, okay. So let's focus on the second part of the problem, find the solution or solve the following differential equation under root of 1 plus x square plus y square plus x square y square plus x y dy by dx equal to 0. Let me know once you're done. Is that done? So you would have all figured out that this term is actually factorizable like this. And then you can actually separate the variables out easily from here. Okay. So you can separate the variables out like this dy is equal to negative. So once you have separated the variables, let's integrate. This part is quite easy. Here you can actually directly take 1 plus y square as let's say t square. So automatically y dy will become t dt. Okay. So let me call this as I1 by the way. So I1 can be easily evaluated by t dt by t. So that will give you t is nothing but under root of 1 plus y square. Okay. Let me call this as I2. So in I2, what we can do is, yeah, I don't type it just I wanted to know whether you are done, whether you could do it or not. So let's multiply with x both in the numerator and denominator. Okay. Now what do you do is take 1 plus x square as let's say z square. So x dx will be equal to z dz. So x dx will become z dz. So it will become z. This term will become z. Again, x dx will become z dz. So z square dz and in the denominator, you have z square minus 1. Okay. Now let me just erase the initial part of the problem so that I have some space to write on. I'll just erase this part. So integration of z square by z square minus 1. You can write it as integration of z square minus 1 plus 1 by z square minus 1 z square minus 1. So that's going to be z. And this is going to be again 1 by 2 a ln of mod z minus 1 by z plus 1. So your final result is going to be, I'm just writing the final result over here. The final result is going to be under root of 1 plus y square is equal to, since there was a negative sign over here, this entire thing will be multiplied with a negative sign. Negative of under root 1 plus x square minus half ln, z is going to be 1 plus x square under root. So it is going to be this as your final result. Okay. So this is going to be your answer. Integration of part is not required over here. It's not required. Method of substitution is sufficient enough to deal with i2. So don't go for the overkill. So let's move on to the next one. So solve the differential equation x minus y dy by dx is equal to x plus 2y. First of all, show that it's a homogeneous equation and then solve it. Okay. So for homogeneous differential equation, it is just sufficient to show that your dy by dx is coming out to be a function of y by x. Okay. If you're able to show that, that means you have been able to show it's a homogeneous differential equation. And also please solve it also. Done. So dy by dx here, you can write it as 1 plus, sorry, x plus 2y by x minus y, which is nothing but 1 plus 2y by x by 1 minus y by x. Okay. Which is clearly a function of y by x. So it's a homogeneous differential equation. Now put y is equal to vx. So dy by dx is going to be v plus x dv by dx. Okay. So it's v plus x dv by dx is equal to 1 plus 2v by 1 minus v. Let's take the v on the other side. So x dv by dx, x dv by dx is going to be 1 plus 2v by 1 minus v minus of v. Okay. So that's going to become v square plus v plus 1 by negative 1 minus v. So 1 minus v dv by v square plus v plus 1 is equal to dx by x. Now we all know how to deal with these kind of problems. Here we try to write the numerator as a times the derivative of the denominator plus b. So clearly your a will be negative of half. Okay. And a plus b is going to be 1. So negative of half plus b is going to be 1. So b is going to be 3 upon 2. Okay. So I can write this as minus half of 2v plus 1 by v square plus v plus 1 plus 3 by 2 times dv by v square plus v plus 1 is equal to dx by x. Let's integrate now. So this will simply become half ln v square plus v plus 1. This will become dv by v plus half the whole square plus root 3 by 2 the whole square. Okay. ln x plus c. Just let me make some space for myself over here. Let me just raise this part. Okay. So here my answer is going to be finally minus half ln v square plus v plus 1 plus 3 by 2. And this is going to be 1 by a which is 2 by root 3 tan inverse of x by tan inverse of x by will be this. And this will be equal to ln x plus c. Okay. You will combine the ln's together. For example, you may take it on the other side. First of all, let's multiply it throughout with 2. So I'm just simplifying a bit. Okay. So let's say this 2 is gone and we have ln x square over here. Okay. And you can write y as this. So you can write it as ln x square y square x square plus y by x plus 1 is equal to this will be root 3, 2 root 3 tan inverse of 2y plus x by root 3x. x square will get multiplied over here. So your final result will look like this ln y square plus xy plus x square is equal to 2 root 3 tan inverse 2y plus x by root 3x. So this will be your final result. Is that fine? Excellent. Correct. So now let's talk about the next question. Evaluate the integral x by 2 divided by under root of x minus 2x minus 3. Done? Okay. So again here, the approach is same. As what we did for the previous problem. So this is going to be x square minus 5x plus 6. Okay. And again, we have to express your x plus 2 as a times the derivative of this expression, which is 2x minus 5 plus b. If you compare the coefficients, a will become half. And minus 5a plus b is equal to 2. So b is going to be 2 plus 5 by 2, which is 9 by 2. Okay. So you can write it as half 2x minus 5 dx under root of x square minus 5x plus 6 plus 9 by 2 dx by under root of x square minus 5x plus 6. Okay. Now this is pretty easy to integrate i1. So let's say i1 we have to integrate. So for i1, you can take x square minus 5x plus 6 as t square. So 2x minus 5 dx will become 2t dt. So it's half 2t dt by t dt gone. 2t also gone. So it's going to be t. That's nothing but under root of x square minus 5x plus 6. For this, we have to complete the square. For the i2 integration, we have to complete the square. And we can complete the square by under root of x minus 5 by 2 the whole square. And extra you will have to put. So it's 1 by 2 the whole square. So this will become 9 by 2 into ln x minus 5 by 2 plus x square minus 5x plus 6. So your final result is going to be final answer is going to be under root of x square minus 5x plus 6. So this will become 9 by 2 into ln x minus 5 by 2 plus x square minus 5x plus 6. So this will be under root of x square minus 5x plus 6 plus 9 upon 2 ln of x minus 5 by 2 plus under root of x square minus 5x plus 6. You can further factorize it like this. So this is going to be your final result. Excellent. Correct. I think you can skip this problem. You can start with this one. Shantanu has already done it. I guess. So let's do this first find dy by dx. dy by dx will be e to the power a sin inverse x into a by under root of 1 minus x square. You can just replace your e to the power a sin inverse x with y again. So it's y a under root of 1 minus x square. So that's going to be under root of 1 minus x square. dy by dx is equal to ay. Let's square both the sides. Let's square both the sides. So when you square both the sides, you get 1 minus x square dy by dx whole square is equal to a square y square. Now we can apply here differentiation with respect to x on both the sides. So we'll use product rule on the left side. So 1 minus x square 2 dy by dx into d2y by dx square. And then you'll have a minus 2x dy by dx whole square is equal to a square into 2y dy by dx. So drop the factor of 2 dy by dx from all the sides. So 2 dy by dx, 2 dy by dx, 2 dy by dx. That will end up giving you 1 minus x square d2y by dx square minus x dy by dx is equal to a square y. And hence shown. So please note that do not try to forcefully substitute these expressions. And try to show that it's coming out to be 0 instead be smart in your approach wherever you can put y, you put that. So that you keep getting the answer as you simplify the steps. So please do not try to forcibly put into this expression. Next, use the properties of determinants to prove that this is equal to this. Yeah, Shantanu, they are various ways to do it. So one of them is what you are suggesting. Okay, so you take care, all the best. Yeah, done with the first part. So for the first part, we just have to split the determinant as two determinants, 1, 1, 1. And we have xx square px cube yy square py cube zz square pz cube. Okay, now you can take p common from the third column as well as x, y and z common from the all the rows. Okay, so when we do that, this will be simplified to pxyz 1, 1, 1. xyz x square y square z square. Okay, right. And we can actually also make the shifting of or interchanging of columns over here. First of all, this and then whatever comes over here, we can again shift with this. So this part can also be rewritten as 1, 1, 1, xyz x square y square z square. Okay, so as a result, I'm just making some space, making some space over here. So it's 1 plus pxyz this. Okay, and now for this part of the determinant, you can easily simplify by doing the operation r1 as r1 minus r2. Okay, and you can do r2 as r2 minus r3. Okay, and this result is basically known to many of us because we have done it so many times. So I'm not going to waste time doing it again. So that is going to be x minus y, y minus z and z minus x that you can easily figure out. Once you subtract, you can take x minus y, y minus z, etc. common and then you can get this result. Now, try the second part out. So elementary operation basically has to be performed with very lot of precaution because we are dealing with a lot of numbers. So it's very obvious that things may go wrong. Okay, so whenever you're applying elementary row operations, again, I'll tell you the process. First of all, you have to create a 1 over here, which is already there. Then you have to make this 0, 0. Then again, you have to make this as 1 and again 0, 0. Again, this has 1 and 0, 0. So this is the order in which we follow. So you go column wise making one in the column first and then the two subsequent zeros. Okay, so this is one, this two zero, this is one, these two zeros. Again, this is one, these two zeros. And when you're performing elementary row transformation, start with the statement a is equal to ia. Okay, so a is equal to ia. Okay, and whatever operations you do on this, you have to do the same thing on this as well. So first operation that I can do is r2 is r2 plus r1. Okay, so when you do it at both the places, you will end up getting, I'm just making changes here itself because I cannot afford to write too much because of the scarcity of place. So this will become 0, 5 minus 2. And this will end up becoming 1, 1, 0. Once you have created 0, 0, now time is to make this as 1. So you do r2 as 1 fifth of r2. Okay, and again when you do that, you will end up making a 1 over here and minus 2 by 5 over here. And this will become 1 by 5, 1 by 5, 0. Okay, now we have to make 0, 0 over here. So you can do another operation r1 as r1 plus r3. Okay, so when you do r1 as r1 plus r3, this will become 0, this will become minus 1. Okay, and here also you will get 1, 0, 1. Then do r3 as r3 plus 2 r2. So when you do that, this becomes a 0, this becomes a 1 fifth. Okay, and here you get 2 fifth, 2 fifth, 1. Okay, then do r3 as 5 r3 to make this as 1, correct. So this will become 1, okay, and this will become 2, 2, 5. Now let's do, to make this 0, let's do r1 as r1 plus r3. Okay, so r1 plus r3, this will become 0 and this will become 3, 2, 6. Okay, now let's do r2 as r2 plus 2 fifth r3. So I have to make this as 0 over here. So this will become 0 and here I will get 1, 1 and 2, correct. Now I have reached a stage where I have got this as I, so this becomes your A inverse. This becomes your A inverse. Is that fine? So if you follow these procedure, you will not require more than 9 set of operations. Here I could manage just with 7 because, you know, I did not do the first operation because there's only a 1 created over here, so less than number of operations were required. Is that fine? So we'll move on to the next one. Question is, if a machine correctly set up, it produces 90% acceptable items. If it is incorrectly set up, it produces only 40% acceptable items. Experience shows that 80% of the setups are correctly done. If after a certain setup, the machine produces 2 acceptable items, find the probability that the machine is correctly set up. Any idea, Sai? Okay, so let me just do this. So let A be the event that the machine produces 2 acceptable items. Okay, and let even be the event that the machine has been correctly set up. Okay, and E to be the event that the machine is incorrectly set up. Okay, so first focus on how when, under what situation will your machine give you 2 acceptable items. So if it has been correctly set up, whose chances is 0.8. And if it is correctly set up, 0.9 is the probability that will give you a correct acceptable item once and 0.9 again that will give you twice. So 0.9 into 0.9. Or the machine has been incorrectly set up, whose probability is 0.2. And then 0.4 is the first acceptable item, 0.4 is the second acceptable item. Okay, now when you're finding out the probability that, given that the machine has produced 2 acceptable items, what is the probability that it was correctly set up will be this. So you'll have to keep this in your numerator and the entire thing in the denominator. So the answer will be 0.8 into 0.9 into 0.9 divided by 0.8 into 0.9 into 0.9 plus 0.2 into 0.4 into 0.4. So this is going to give you 81 into 8 which is 648 by 648 plus 32. So the answer will be 648 by 680. That's approximately 0.95. Is that fine? Next, linear programming problem. One kind of a cake requires 300 grams of flour and 15 grams of fat. Another kind of cake requires 150 grams of flour and 30 grams of fat. Find the maximum number of cakes which can be made from 7.5 kgs of flour and 600 grams of fat. Assuming that there is no shortage of the other integrant used in making the cakes. Make it as a linear programming problem and solve it graphically. So tell me how many cakes of the first type and how many cakes of the second type need to be produced to maximize the total number of cakes. Alright, so in this case, first of all, we'll write the constraints. So let the first type cake produced will be X and second type cake is let's say Y. Now we have a constraint that the number of the amount of flour used will be this much and it should be less than equal to 7500. So this is your first constraint. Similarly, the fat used will be this, this should be less than 600. And of course, X and Y both have to be positive. So these are our constraints and our objective function is we have to maximize X plus Y. So this equation, first of all, you can divide it throughout with 150 and if you divide it by 150, you get 2X plus Y less than equal to 50. Here if you divide by 15, you get X plus 2Y is less than equal to 40. Okay, so let me just calibrate this. So 10, 20, 30, 40, 50 and this is 10, 20, 30, 40, 50. Okay, so first line is 2X plus Y is equal to 50. Now first you divide by 50, it will become X by 25 plus Y by 50. So X by 25, Y by 50 means you have to connect like this. Okay, second if you divide by 40, X by 40 plus Y by 20. So X by 40, Y by 20 will be like this. Okay, now the feasible solution zone will be less than that means whichever site contains the origin. So that will be a feasible solution zone. So this will be a feasible solution zone. So let us see the corner points occurring over here in this convex polygon. So this is 0, 20. Okay, and here if you subtract 2 times, this will become 2X plus Y is equal to 50 and 2X plus 4Y is equal to 80. So 3Y is 30, so Y is 10. So Y is 10 means X is 20. So this is 20 comma 10 and this is 25 comma 0. So let's check which gives you the maximum value of X plus Y. So X plus Y will be maximum for this case it will be 20. For the first point, let's say point A, point B, point C. So X plus Y for point A, it will give you 20. For point B, it will give you 30 and point C, it will give you 25. So this is the max. That means 20 cakes of type 1 and 10 cakes of type 2 must be baked for you to have maximum number of cakes within the given resource constraints. Next, find the coordinates of the foot of the perpendicular and the perpendicular distance from 3 comma 2 comma 1 from the plane 2X minus Y plus Z plus 1 equal to 0. Also find the image of the point in the plane. Alright, so let's look into this problem. So we have a plane and from a point 321, you're dropping a perpendicular. So we have to find the foot of the perpendicular, let's call P and not only that, we have to find the image of P on this plane. Let's say I call it as P dash. Okay, now first of all, we'll write down the equation of the line perpendicular to the plane and passing through P. So that can be easily written X minus 3 by coefficient of X Y minus 2 by coefficient of Y and Z minus 1 by coefficient of 1. And let's say for this point the parameter involved is lambda. So let's say this is lambda. So this point M could be written as 2 lambda plus 3 minus lambda plus 2 and lambda plus 1. Now this point should satisfy the equation of the plane. So let me call this plane as pi. So it should satisfy the equation of the plane. So 2X minus Y plus Z plus 1 should be equal to 0. Okay, so let's simplify this. So 6 lambda and from here I will get plus 6 is equal to 0. So lambda is negative 1. So if lambda is negative 1, your point M is nothing but 1, 3, 0. 1, 3, 0. So this is the foot of the perpendicular. So perpendicular distance, let me call this distance as small d. So perpendicular distance would be simply the distance between these two points which is nothing but 2 square plus 1 square plus 1 square which is under root of 6. Okay, under root of 6 units. Now let's say the image of this point is alpha, beta, gamma. So we can say that the midpoint of P and P dash will be your M. Alpha plus 3 by 2 will be equal to the X coordinate. So this means alpha is going to be negative 1. Okay, beta plus 2 by 2 is going to be 3. That means beta is going to be 4. And gamma plus 1 by 2 is going to be 0. That means gamma is going to be minus of 1. So the image point will be minus 1, 4, minus 1. Okay, so let's move on to the next question. So we have to find the area of the circle which is interior to this parabola. So this is your circle and your parabola is the one which is opening upwards. So we have to find out this area. So first of all you have to see the point of intersection. Let me call it as P and Q. At least let's find out the X coordinate of the point of intersection. So we'll simultaneously solve it. So let me replace my X square with 4Y over here. So this will become 16Y plus 4Y square is equal to 9. So we have a quadratic over here, 16Y minus 9 equal to 0. So 36 you can factorize it as 18 and 2. So minus 2Y plus 18Y minus 9 equal to 0. So 2Y, 2Y minus 1 plus 9, 2Y minus 1 equal to 0. So Y cannot be negative of 9 by 2. So if Y is half, X is going to be plus minus root 2. So this is going to be root 2. This is going to be negative of root 2. Now we are going to just focus on one of the parts because it is symmetrical. So this area would be, area for the one part would be integral from 0 say root 2. The Y of the upper curve which is the circle which is going to be 9 by 4 minus X square minus Y of the lower curve which is X square by 4. So we have to integrate this and double up this to get our answer. We will just evaluate it, evaluate this. So the integral of this part will be twice X by 2 under root of 9 by 4 minus X square plus 9 by 4, sorry 9 by 8 it will be A square by 2. The inverse of X by, X by will be 2X by 3 and this is going to be minus X cube by 12 from 0 to root 2. So you can cancel the factor of 2 from everywhere and when you put root 2 you will get the answer as root 2 by 6 plus 9 by 4 sin inverse of 2 root 2 by 3 plus C. The next question, using integration find the area of the triangle ABC. Now you have to use integration, cannot use any direct formula. The coordinates of whose vertices are 4 comma 1, 6 comma 6 and 8 comma 4. Okay so let's first make the diagram for this. So 4 comma 1 I can take roughly over here. Let me change the color. So 4 comma 1, 6 comma 6, 8 comma 4. Okay so this is 4 comma 1, 6 comma 6, 8 comma 4. So let's write down the equation of the line first. So this line will be Y is equal to 12 minus X. Okay this line will be Y is equal to 5X by 2 minus 9 and this line will be Y is equal to 3X by 4 minus 2. Okay this point here is 6, this point here is 4 and this point here is 8. Okay so let us first take a vertical strip in this zone. So in the region 4 to 6 you are taking the difference between 5X by 2 minus 9 and 3X by 4 minus 2. Okay and from the region 6 to 8 you are taking the difference between 12 minus X and 3X by 4 minus 2. So you can simplify it further and write it as from 4 to 6 it is 5X by 2 minus 9. From 6 to 8 it is 12 minus X dx and negative from 4 to 8 you have 3X by 4 minus 2. If you integrate this your result will come out to be 7 square units. Okay so guys here we will take a small break. Okay we will resume at 550. Alright so welcome back. So let us start with the problem on maxima and minima. If the length of the 3 sides of a trapezium other than the base is 10 centimeter each, find the area of the trapezium when it is maximum. So this is the scenario so this is 10, this is 10, this is also 10. So it is a kind of an isosceles trapezium and you have to find out the area of the trapezium when it is maximum. Okay so here if you look at the construction this will be 10 itself and let us say this is X and X each. Okay so we know that the area of the trapezium is half into the height into sum of parallel sides. So height of it will be under root of 100 minus X square into sum of the parallel sides. Sum of the parallel sides will be 2X plus 20. Okay now because of this under root figure many people find it inconvenient to differentiate. Let me first divide it by half. You can square both the sides. So if A is maximum even A square will be maximum. So if you square both the sides you will get something like this. Okay now for area to be maximum the derivative of A square with respect to X should be 0. Okay so let us differentiate this so 10 plus X into 2 into 10 minus X and we will have 10 plus X whole square into minus 1 is equal to 0. So twice of 10 plus X into 10 minus X is equal to 10 plus X the whole square. Now X cannot be 10 okay so we can cancel off an X plus 10 factor for both the sides. So 20 minus 2X is 10 plus X or we have a square of this term as well just a second. So we have 10 minus X and 10 plus X yeah so this is going to be A square will be 10 plus X the whole square and 100 minus X square. So that is 10 plus X the whole square and this is 10 minus X 10 plus X. So it is going to be 10 plus X whole cube into 10 minus X. Now we can do the derivative of A square with respect to X as 0. So that is going to be 3 times 10 plus X the whole square into 10 minus X is equal to 10 plus X whole cube okay. So it will be 3 times 10 minus X is equal to 10 plus X. So 4X is equal to 20 so X is going to be 5 okay. So this X has to be 5 for the area to be maximum and that value of the area will be half under root of 75 into. In fact you can get rid of this 2 X plus 10 is going to be 15. So 15 root 75 will be around 75 root 3 75 root 3 centimeter square. So with this we move on to the next question. Find the intervals in which the following function is strictly increasing and strictly decreasing. Okay so the function is increasing where the derivative of the function is greater than 0. Strictly increasing where the derivative of this function is greater than 0 f of X is strictly increasing. Okay and where the derivative of the function is less than 0 f of X is strictly decreasing. So when you differentiate it f dash X is going to be minus 9 plus 12X minus 3X square. So take a minus 3 common so it is X square minus 4X plus 3. So that is minus 3X minus 1X minus 3 okay. Now minus 3X minus 1X minus 3 will be greater than 0 where X minus 1X minus 3 will be less than 0. Correct and that will happen between 1 and 3. So when the X value is between 1 and 3 the function is strictly increasing. So we can say X belonging to the open interval 1 comma 3 the function is strictly increasing. And the remaining intervals that is minus infinity to 1 union 3 to infinity the function is strictly decreasing. So this will be your interval for strictly decreasing. This will be your interval for strictly increasing. Next integral of X cos pi X from 0 to 3 by 2. Okay so first we need to redefine this function mod of X cos pi X okay. This function is going to be behaving as X cos pi X when your X is between 0 and half. Because exactly at half it becomes 0 okay. And post that it is going to behave as negative X cos pi X. So from half to 3 by 2 it is going to behave as negative cos pi X. So when you are integrating this you are going to integrate from 0 to half X cos pi X. And from half to 3 by 2 you are going to integrate negative cos pi X. Now here you have to use your integration by parts taking this as U taking this as V. So it is X cos pi X in fact X sin pi X by pi minus integration of sin pi X by pi okay. X sin pi X by pi plus cos pi X by pi square okay. So first let us put half. When you put half it becomes half into 1 by pi and this will become 0 okay. And when you put a 0 this will become 0 and this will become 1 by pi square. In a similar way let us evaluate this one also. For this one your answer is going to be negative sin negative X sin pi X by pi negative cos pi X by pi square okay. So from 3 by 2 to half. So when you put 3 by 2 you get negative 3 by 2 in fact plus 2 by pi okay. And this will become 0 negative. And when you put a half it becomes minus 1 by 2 pi and this will again become 0. So all together it becomes 4 by 2 pi 4 by 2 pi. Now add these two results so when you add the two results your final answer is going to become 5 by 2 pi minus 1 by pi square. So be careful to split the limits of integration as per the change of the definition interval of this function. Next integrate X square e to the power X by X plus 2 the whole square alright. So for the first integral you can write this as X square plus 4 minus 4 e to the power X okay. In fact you can reverse the sign over here X minus 4 plus 4 divided by X plus 2 the whole square. So this becomes X minus 2 X plus 2 by X plus 2 the whole square plus 4 divided by X plus 2 the whole square e to the power X. This will get cancelled off. So X minus 2 by X plus 2 plus 4 by X plus 2 the whole square e to the power X. Now let us take e to the power X as let's sorry let's take X minus 2 by X plus 2 as some function of X. And let's try to find out its derivative. So we'll use quotient rule over here so X plus 2 the whole square. So X plus 2 minus X minus 2 that's going to be 4 by X plus 2 the whole square. So I realize over here that if this is your function this will become your derivative of the function right. Which clearly implies your answer will be e to the power X f of X plus C which is nothing but e to the power X X minus 2 by X plus 2 plus C. So this is based on the standard result this integral is based on the standard result e to the power X f of X plus f dash X. Integral is e to the power X f of X plus C. Now let's work on the other part of the problem integral of X square plus 1 by X square minus 5 X plus 6. Okay so in this case we'll first have to write this as X square minus 5 X plus 6 plus 5 X minus 5 and you divide it by the denominator. Okay so this simply becomes a 1 and here you can in this part that is 5 X minus 5 divided by the denominator is factorizable. You can apply partial fractions and if you apply partial fractions a is going to be minus 5 and b is going to be a 10. So you can write this integral as minus 5 by X minus 2 plus 10 by X minus 3 dx. So your result will be X plus in fact X minus 5 ln X minus 2 plus 10 ln X minus 3. Okay this is going to be the answer for the second integral. Next consider the experiment of throwing a dice if a multiple of 3 comes up throw the dice again and if any other number comes toss a coin. Find the condition probability of the event that the coin shows a tail given that at least one dice shows a 3. Yes the answer for the first part it's very simple the answer of the first part is going to be 0. Because there is no overlap of you getting 3 and the coin showing a tail. Okay now second part of the question how many times must a man toss a fair coin so that the probability of having at least one tail is more than 80%. So at least one tail means 1 minus no tail at all. Okay this should be greater than 80% 80% means 4 by 5. So 1 minus 4 by 5 should be greater than 1 by 2 to the power n. That means 1 by 5 should be greater than 1 by 2 to the power n. Now for n equal to 1 it is not true because half is greater than 1 by 5. For n equal to 2 also it is not true. For n equal to 3 it is true. So he must toss the coin 3 times so that the probability of having at least one tail is more than 80%. Let's move on to the next question. A, B and C are 3 unit vectors such that A dot B is equal to A dot C equal to 0. And the angle between B and C is pi by 6 prove that vector A is equal to plus minus 2 times B cross C. Okay again this is a quite simple problem. The fact that A dot B is equal to A dot C equal to 0 means A is in the direction of B cross C. So you can write A as lambda B cross C correct. Now the modulus of this vector is going to be modulus this, modulus delta. Sorry modulus lambda times modulus B cross C. So this is going to be 1, this is going to be mod B mod C into sine of the angle between them which is going to be half. And since B and C both are unit vectors mod lambda is going to be 2 that means lambda is going to be plus minus 2. And if you put it back over here your vector A becomes plus minus 2 times B cross C. So moving on to the next question, find the equation of the perpendicular drawn from the point 2 comma 4 comma minus 1 to this line. So basically this is your line and from a point 2, 4 minus 1 you have to draw a perpendicular to this line. So again let's say this point is lambda, this point is corresponding to lambda. So equate this to lambda so this point will be lambda minus 5, 4 lambda minus 3 and minus 9 lambda plus 6. Now you can use the fact that direction cosines of PM, direction ratios of PM is such that it is perpendicular to the given line L. So direction ratios of PM will be lambda minus 7, 4 lambda minus 7, minus 9 lambda plus 7. The dot product of this with 1, 4 and minus 9 which is lambda minus 7, 4 times 4 lambda minus 7, minus 9 times minus 9 lambda plus 7 should be equal to 0. So 81 plus 16, 97, 97, 98 lambda and here we will have minus 7, minus 28 which is minus 35, minus 35, minus 63 is again minus 98 is equal to 0. So lambda is going to be 1. So direction ratios of PM can be written as minus 6, minus 3 and minus 2. So equation of PM can be written as x minus 2 by 6 is equal to y minus 4 by 3 is equal to z minus 6, sorry z plus 1 by 2. So this becomes the equation of the line PM. Now the other part I have already done a similar question so you can treat this as your homework question. And there are few more problems left over so I will be sending this sheet on the group so that you can complete it. So guys over and out there will be a Google form send out on the group asking you whether you want classes in the boards because I can see very less attendance today. Anyways all the best for your exam starting on Saturday. Over and out from Centrum Academy. Thank you so much. Have a good day. Bye-bye.