Engineering Maths-02 Endsem Paper Solution - Dec 2025 | Ep.01 | FE QP Solution Series

Added: 2026-05-26

[00:00:05][Music] [Music] Friends, I, Professor Mangesh Kaire, welcome everyone to this paper solution series of the First Year 2024 pattern. Friends, here we are going to see the paper solution of our paper which was held in December 2025.

[00:00:25]Friends, here I have taken the solution of the paper that is worth 60 marks. Now the kids with 70 marks will say sir what is our paper solution? So friends, whether it is 60 marks or 70 marks, the syllabus is the same, only the marks have been changed, that is, today the question that was for four marks has come for three marks, so whether it is a 60 mark paper or a 70 mark paper, everyone has to refer to the same paper solution and score their subject.

[00:00:48]Here we are giving one paper solution.

[00:00:50]All the remaining 24 pattern paper solutions with detail solutions and detail explanations are available on the Purple Eye app. As you can see on the side, in this way, you can download the Purple application and subscribe to each paper solution. You can access each paper solution and score your subject.

[00:01:10]So let's enjoy this subject.

[00:01:13]Hello friends, today we are going to see the M2 2025 paper and its solution. In detail, this is the M2 2025 paper which was held in January. It has an OK pattern of 70 marks. The first question in it is MCQ.

[00:01:37]Write the correct option. The first MCQ in it is the reduction formula. The value of zero to pai ba sa re fo xd x Now there is a ready-made formula to get the same solution. The reduction formula is called, let's see how to get the value of sa re fo x on by. If this is the power, then the answer is to multiply pai ba, take it down, reduce one above and subtract two from two and stop until one and two come. That is, continue until then. Mass 3-two fo matu 2 stop one and two came stop pa ba multiplied because it is ok. Then this reduction formula is used sa fo ch integral between the limit zero to pa ok.

[00:02:43]Then the first loop of the is sa along is clearly visible here. If you mean the first al is your zero, the second will be pa ba si. Now the first al will come to zero, but if you put pa And the year is so and that means what is the first t sila your first loop comes then area bound by kar now here for area bound by kar th formula used common point found out and solved using area formula one by si unit option number one then asked integrating factor of differential equation now this differential equation is your linear then linear formula here integrating factor is taken out ok then maximum now next two one by one marks answer is directly in it ok maximum current i max gun by ba and orthogonal trajectory of the family ok intersect of the family al 90 when it intersects 90 we call it tra ok that's it your six mcqs of the marks then come questions t question beta gamma and rower has come question number two question two a So we put the question to the question but here you can also convert the beta function without putting it.

[00:04:20]Here you can solve it by putting the information x and slats. We can solve it by bringing it into the definition of beta.

[00:04:27]But here I used an alternative method by using the trigonometric function to put x and s so that here you get the reduction formula. See what you got here. Reduction formula and how to use the reduction formula. See what it means. Then what does it mean? This is the bottom of the itz and the top of the sen and keep doing minus two minus two until one and two come.

[00:04:51]See the top one came to stop. Below two came to stop and the foot batu multiplied for this because it is OK. Here we have solved the first question using the trigonometric function. The second question is clearly the gamma function. If the function has information exponent here, we put the difference and differentiate it by differentiating the limit. To change and bring it into the form of the given integral, convert it into the definition of gamma and after converting it into the definition of gamma, the value is easily obtained there. Okay, this is your gamma function example.

[00:05:30]After that, your D is Rudy.

[00:05:34]We will do this example in four steps. First of all, what do we do? We apply R. Let's see what we have to apply. Let's call it the integral. Why do you have one variable and another? Let's call it the integral.

[00:05:49]Then we use R here.

[00:05:53]Since both the sides are connected. After entering the integral, we get the partial of it. With respect to x, zero is constant. Then the cut is made.

[00:06:11]X comes. Okay, now after using R, here we integrate x using the formula x n, that is, n is equal to a n is equal to n and then by putting the upper limit and lower limit, we get 1 a a p now.

[00:06:26]Whose play is it because we put the d body, what is the d body of?

[00:06:34]After getting the value of the soi, we have to take it out and integrate it here to find out 1 a ps integration lag psv p now why have we taken c here because there is no limit to the integral here so we took c here and there was a limit so there is no need to take th ok then to find the value of c we have to put a suitable value of a in the given integral so that this integral should be zero then if we put zero in place of a here then what will be zero so we put zero here a and find out c and put c in that one that is the required thing to show us [clearing throat] then question two came now question number three so in question number three also here we have used trigonometric function But here we can solve it without using trigonometric functions. How do I put x² as t here so that it comes in the form of the beta function and can be solved using the definition of the beta function?

[00:07:38]But what I did was put x as tn θ so that you can easily bring this trigonometric term into the form of sa and co so that if sa and co come in the form, here you will use the reduction formula.

[00:07:50]How to use it is very simple.

[00:07:53]Here, start with one, start with two, and then the division of the two, start with f, and keep doing it between the two until one or one comes, then stop when one and comes, and this is what you have done.

[00:08:09]Now here, you have not taken any steps and the answer has come.

[00:08:15]Now, you have left this very simple question.

[00:08:18]They did not want you to worry about anything here. Here, the definition of the beta function is directly visible. Look, what is the mass of To take and get the answer, it is very easy to put the sum on the beta function.

[00:08:40]You can also do the sum of f and b and you can also solve this with beta.

[00:08:45]Then C asked again, " Look, here first put the body on both sides, then break the body inside, after the body is formed, we take the partial derivative with respect to a, the partial date is here, after this, we integrate it using the formula. We saw this formula in the four series and if coi and sa are, then we use the formula here. What is the formula? ax a sp b s co bx p b sa bx. We solved that formula here. Let's say a sp spsv now a sp spsv is the value of whose value is it and what do we need to do to find out what we need. We also integrated this formula with respect to this integration. Since the derivative of psv and its is not there, what did we do? We took ba down and adjusted it here so that the lower term is presented above the derivative and the integration lag term is ok. Then to get c, we put the value of a one because to make this integral zero, if you put a one here, both will be equal and the integral will become zero. So, what will you put a one to make it equal? ​​We put one one, the integral will become zero and you can easily find the value of c.

[00:10:13]Look, put e so that the value of i of v will be zero and after it is zero, you can find c. Okay, by putting the value of c, you can easily find the search you asked for.

[00:10:24]In this way, we have solved question number three, a, b and c.

[00:10:29]Okay, so friends, the solution that your teachers have explained to you, I am sure you will like it and you will be confident, but you will also want the complete solution. Don't try it, just go to Purple Power, every course has its paper solution folder, there you have this paper, there is a PDF of it, there is a detailed explanation, plus we have solved all the papers after this, 60 marks, 70 marks, all papers, yes so both types, 60 and 70, both types of students should refer all the papers because the syllabus is the same, yes so go to Purple Power, download Purple Power, every course has paper solutions, so go to the course you want the paper solution for, subscribe to the course, refer the paper solutions and just do your engineering scoring, best of luck [Music]