Not a pro editor so bear up with that 😅 . . Bhai mai to normal video dalraha tha par pata ni kyun mera dimag satka aur storified format bana dala 😂 anyway how was it?
Another way of solving this que is by considering odd and even terms seperately. Then by S(1-1/x²) we simply get two gp from which we directly get (x-1)²x²⁰²⁰ -2/x²+2/x By this min value can be easily calculated by dy/dx
After summing up all the terms of f(x) , we get a closed form as f(x)=(x^2024+2024x+2023)/(x+1)² Clearly x=1 and x=-2024/2022 are the only critical points,(from the general expression of f(x) we see that f(x)>2023 for all x
I belive that f(1) will be the only minima because: 1)Negative numbers are eliminated 2)for numbers ranging from x=[0,1) f(x)~2022 [f(0)=2022] because for 0
( 7 minutes ) Approach 1 : Modular congruence, factor theorem, elementary group theory .. Let there be a group of all such polynomials .. Let Fi(x) be a polynomial belonging to the group. You notice : F0=1x^0 F2 = (x-1)^2+2 F4=(x-1)^2*q(x)+3 F2i gives remainder i+1 when divided by (x-1)^2.. Putting i = 1011 ,u get the mentioned polynomial and remainder is 1012 which obviously is the minimum value when x=1.
If you estimate f' from [0,1] then it will bear a negative value means f is decreasing function in [0,1] similarly, f is increasing in (1, 2] , f' will bear +ve value Hence we can say that 1 is global minima of the above function
There is some analysis to be done by checking what happens when numbers before and after 1 are put into the expression. If we first simplify the expression to x^2021(x-2) + x^2019(3x-4)+....x(2021x-2022) + 2023. When we put 1, we get 1012 as you have stated. Trying something less than 1 say 0.99 will essentially reduce the higher powers of x to near zero as such all the stuff that was going to be subtracted from 2023 will be reduced which will result in value higher than 1012 and if we were to put something greater say 1.01, then the lower powers will be reduced like (2021 -2022) which would be -1 will now be now around 19 as such this will also affect our answer. I know this is not a rigorous proof but i feel its a decent enough reason to assume that 1 will be the answer.
I was so immersed into the story that I almost forgot to pause 😂 Anyways heres my solution development: if we take x²⁰²²common we have (1- 2/x + 3/x² - 4/x³...2023/x²⁰²²) , In general do you know the formula for 1-2x+3x²...(-1)ⁿnxⁿ-¹? Me neither, but I do know the formula for x-x²+x³..x²⁰²³ and after I got the closed form easily I took the derivative and now we have formula for 1-2x+3x².. and then I put 1/x into the formula and now we have our function directly (after multiplying by x²⁰²²). I got {2023+2024x+x²⁰²³){1/(x+1)²} as f(x) , I took the derivative and tried putting x=1 in the solution which turned out to be 0 , now clearly for any number greater than 1 this would shoot up to a very big number because I had the remaining polynomial part of f’x as 2021x²⁰²³+2023x²⁰²²-2022x-2022 , these powers become useless between (0,1) but if you put something greater like x=2 the powers would be tremendously huge and hence f'(x) will continue to stay positive and also x cant be negative because for any -ve number its +ve counterpart is guaranteed to yield a lesser value , Therefore x=1 yileds the global minimum value being 1012 Edit: Your solution was so ingenious lol much better, in your solution in the form you wrote as f(x) putting x≥2 would give a +ve answer for all the individual brackets and + 2023 so they cant be minimum
@@annapurnasarangi9999I'm just a normal guy who likes maths )) sometimes I do solve ioqm Berkeley isi and research other interesting math related things or videos which are not related to jee for fun , I've diven too extensively into math and probably need to focus on my chemistry 😅 because if I bring it to the level of my maths and phy my rank might skyrocket, hopefully atleast
Brother agp method se we can directly deduce 1 as global minima…just x^n -1 ki identity use hori hai last me…2-3 lines of differentiation and the job is done👍🏻
Funtion zero se two mein increasing hai but 1 pe least value milti hai 0 pe 2023 and 2 pe more than 2023 but 1 pe least value aa rhi hai so function ka ek hi local minima and global minima hai i.e at 1. Infact ek rough graph bhi soch skte hai as a tick mark graph ✔️ where this down dip is at x equals 1 Question was awesome !! Editting too !
0 se 2 ke beech 1 ke agal bagal kahi global minima nhi ayega ye kaise keh skte ho?? wohi pucha gya h,.. 1 per aa rha wo tum kewal 3 digit ke al bagl check krke bol rhe ho
The only reason i hate mains 🙃cauz mains mostly ask for going towards calculation side but yar dimaag bolta hai ki kuch intuitive type accha kuch solution hoga and ultimately time waste ho jata hai wo sochne ke chakkar mai
So as we know that it will ocuur in the interval [0,2] and 1 is the critical point. So, f(0)=2023 f(2)= greater than 2023 f(1)=1012 And hence we can say that at x=1 global minima is present.
Putting 1.1 in f'x will give us a positive number i.e slope positive however putting 0.9 will give us a negative number hence negative slope we can infer that x=1 is global minima 🙃
Bhaiya, after AGP and differentiation, The condition for f'(x)=0 comes 【(1011x+1012)(x^2023-1)=0】 I.E ,one (-ve solution)(rejected), and (all 2023th roots of unity) So only feasible value for f'(x)=0 is "1(one)" And that by putting values is selected
We can just write the derivative in the form of sigma (2023-r)(r)(x^(r-1)-1)= sigma(2023-r)(r)(x-1)(1+x+x^2+x^3+.....+x^(r-1)) and so the only values of x for which derivative is 0 are -1 and 1 but as we identified that it is between 0 and 2 so the answer is therefore 1.
Itna bhi creative question nahi tha bhaiya, 2min mai orally solve hogaya(mai abhi 5th class mai hu) And iske jesa similar question ashish sir(PW) ne class mai karaya tha
My intution and working Sbse phle fn ko dekh kr muje lga ki ye mai generalisation se jaunga like y=Sum (-1)^(k-1)kx^(n-k) so yha mene phle x^n nikala to x^(-k) ka issue ban rha tha so y=x^(n+1)Sum(-1)^k *k/x^(k+1) Ab andr wala differential h y=x^(n+1)d/dx(Sum (-1/x)^k) y=x^(n+1)d/dx((-1/x*(1-(-1/x)^n)/(1-(-1/x) y=(x^(n+1)-(-1)^n(nx+x+n))/(x+1)^2 dy/dx iska ab nikala and y' ek fn aay lenghty sa Y'=(x-1)((-1)^n-x^n)+n(x+1)(x^n+(-1)^n)/(x+1)^3 Isse y'(x)=0 ke liye roots nikala for every n=even ie here 2022 x=1 is a root and iske aage wale pe x>0 hence inc and peeche wale x
bhaiya i used binomial , this series is the expansion of the differentiation of the term (x-y)^2023 wrt y where i assumed x as a constant and later put y=1 , usse bhi min value 1 pr ayi
x=1 is the global min because for x2 f(x ) is greater than 2023 and therefore if we consider points around x= 1 we get that the min value of fxn will be at x=1 , i.e it is the global minima ... Awesome qn btw
I Solved it without differentiation or AM-GM My method involves modifying the function by finding patterns. But comments me explain karna muskil hai So, I have sent you the solution in email and attached the solution pdf.
Observing the question we find that this eqation never be negative and sum of last rth term and first rth tearm s sahgunak is 2024 and sign become opposite when we move to center then we put 1 the we get 0 which is least value of function .. we easely solve by obsevation
To find the minimum value of the function \( F(x) = x^{2022} - 2x^{2021} + 3x^{2020} + \ldots + 2023 \), we need to find the critical points by taking the derivative of the function with respect to \( x \) and then setting it equal to zero. First, let's find the derivative of the function: \[ F'(x) = 2022x^{2021} - 2 \cdot 2021x^{2020} + 3 \cdot 2020x^{2019} + \ldots + 0 \] \[ F'(x) = 2022x^{2021} - 2 \cdot 2021x^{2020} + 3 \cdot 2020x^{2019} + \ldots \] Now, to find the critical points, we set \( F'(x) \) equal to zero: \[ 2022x^{2021} - 2 \cdot 2021x^{2020} + 3 \cdot 2020x^{2019} + \ldots = 0 \] Factoring out the common term \( x^{2019} \): \[ x^{2019}(2022x^2 - 2 \cdot 2021x + 3 \cdot 2020) = 0 \] \[ x^{2019}(4044x^2 - 4042x + 6060) = 0 \] Now, to find the critical points, we need to solve the quadratic equation \( 4044x^2 - 4042x + 6060 = 0 \). After finding the roots, we need to check which one minimizes the function. However, due to the very high degree of the polynomial and the specific values involved, it might not be practical to solve it algebraically. Instead, considering the leading term \( x^{2022} \), which has a positive coefficient, we can infer that as \( x \) approaches positive or negative infinity, \( F(x) \) also approaches positive infinity. Hence, there is no minimum value for this function. . Let's continue with a different approach to find the minimum value of the function. Given the function: \[ F(x) = x^{2022} - 2x^{2021} + 3x^{2020} + \ldots + 2023 \] Notice that each term has a decreasing exponent but an increasing coefficient. To find the minimum value of the function, we can consider the term with the lowest exponent, which is the constant term, \( 2023 \). Since all other terms involve \( x \) raised to a higher power, they will dominate the function as \( x \) approaches positive or negative infinity. Therefore, we can conclude that the minimum value of the function occurs when \( x = 0 \). Substituting \( x = 0 \) into the function, we get: \[ F(0) = 2023 \] So, the minimum value of the function \( F(x) \) is \( 2023 \).
this was the hardest problem ? and this youtuber, who seems to consider himself a pro problem solver did so much melodrama to solve " this " question ??!! Its pathetic.... isse acha bhai primary school ki mental maths solve karta... udhar melodrama accha bhi lagta.......
@@jeesimplified-subject Tu Jee Mains me 1 lakh rank lane wala 6kka hai jo idhar baccho ko JEE Advanced me AIR 1 lane ke sapne dikhata hai aisi chindi questions ko god level hype karke. 12th ke bad college to mila nahi tujhe, padhai to hui nahi tujhse, to aur kya krta, baccho ko chutya hi bnaega tu.
agar ham f(x) ke do consecutive terms ko uthaye to ham unhe x^r ( nx - n - 1 ) likh skte h, now for x>2, nx - ( n+1 ) > 0 to isse ye pta chlta h ki fx ke do consecutive terms ka sum positive h aur badi value h due to x^r, to usse intuitively pta chlta h ki fx aage jake increasing hi hoga
8:59, yaha pe ham dy/dx 1+ aur 1- pe dhek skte hai ki kaha -ve aa rha hai kaha positive, agar 1+ pe dy/dx positive hai to matlab funtion increasing aur 1- pe negative hai matlab function decreasing, iska matlab 0 se 2 ke beech me 1 minimum hai
Put x=1 your question is solved 😂😂😂 solved in 10 sec by seeing question try my method you will get correct Answer...😊😊 Edit: I not seen any video or full soln
I search this type of question's solution in web and found they solved by using higher mathematics ( not included in IIT syllabus ) but that is not point that it can't not solved by the chapters we learnt in class 11 and 12, yes it can be solved
Bhaiya honestly bolu toh iss video mein question kab start hua kab khtm idea hi nhi laga. Jiske kaaran khud solve karne ka mauka hi nhi mila XD Purane waala style preferable tha jisme aap question solve karne ka time dete the. IF possible usko iske saath integrate karlo. Also thanks to comments section jisse mereko idea laga why x=1 is global minima.(Khud nhi soch paaya tha)
Orginal series ko dekhe to power 2022 se start Hui thi aur end hui power 0 ( constant term par ) mtlb 2023 terms in total sab dy/dx or differentiation kiya constant term ka differentiation 0 hogya aur no of terms 1decrease Hui toh total no of terms then 2022 hogayi so we can no of terms are even after dy/dx Or simply you can think har term cancel hogi 1st and last jaise toh no of terms even honi padegi tabhi isa possible ha
wahi sahi hai bhai me bhi tha woh extraordinary me viswas rakhne wala aur no aate the 150/300 and all that shit away stick to basic and boom i landed on 227/300 99.42 percentile jm24@@jeesimplified-subject
