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
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 🙃
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
Jee-IIST(trivandrum) 2028 Aspirant here I don't know Arithmetical progression no series and sequence all I know is mathematical reasoning -ve gives bigger 0 is smaller rational are not we are left with natural numbers smallest natural no. is 1.power of class 9th
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
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.
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👍🏻
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.
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
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
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
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 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.
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
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 \( 2023 \).
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
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
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
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.
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)
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
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
![KAYTRANADA - Witchy (feat. Childish Gambino) [Official Video]](http://i.ytimg.com/vi/jGK3YVmGZ3Y/mqdefault.jpg)
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?
badhiyaa🌻🌻
gajab
Very very nice bhaiya
❤❤mast
Bhai reddit aur youtube pe aapko kitna alag treat karte hai...
Anyways i like this channel, posts good questions
Question me creativity ✅ video me creativity ✅
Pls continue this method where u first think of probable methods ,it would give us a outlook on attacking questions the right way
your way of solving problem and make curiosity for a problem is amazing
grouping?
@@adityavikramsinha408 bro not a jee student lol
@@studyboy6335 ohk sorry
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
i did s(1+1/x) which works too
Did same
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
In the denominator of f(x) it should be (x-1)² and not (x+1)²
Thanks a lot for trying out the question. Loved the video
Kaha Mila tha bhai tumko yeh sawaal
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.
Approach 2 : Taylor series
Approach 3 binomial theorem application
Approach 4: Chod Do, move on
Approach 5: Try hi kyu krna hai
Approach 6 : Multiply both sides by 0
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
How do u know it increases- decreases and increases again between (1,2)?
Dayum, the new video style was great. Love the way you tried to think of the different possible methods to approach the question.
Full marks for creativity👍
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 liked the new way of video edting question 7-8 min mai solve hogaya tha...........😄😄
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
I knew the formula 1-2x+3x^2 🗿
Are u a ioqm student??, or did go for any math camp, cause ur math skills are really good, appreciated
Good one, using that memorised formula and replacing x by 2/x was a creative move
@@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
@@jeesimplified-subject🙏
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 🙃
Same maine bhi yahi kra
Its wrong can't it br anything between 0.9 and 1.1, not necessarily 1!
@@aaravarora2009 limit lagake kr skte h
@@aaravarora2009 bruh f'x is 0 at x =1 hence we verified that it is a point of extrema
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
Mains 2022 had some really good qs btw kitne aaye mains mai ???
ye sb dogla pan h
@@adityashaw4316 rank kya hai apka adv mei?
Jee-IIST(trivandrum) 2028 Aspirant here I don't know Arithmetical progression no series and sequence all I know is mathematical reasoning -ve gives bigger 0 is smaller rational are not we are left with natural numbers smallest natural no. is 1.power of class 9th
Bhai itna natak nahi karte hai 💀
@lkgnakengkjljkhaglnaw💀
@lkgnakengkjljkhaglnaw Chill bhai, ye neet waale h😂
When you don't know which RUclipsr you are talking about
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
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.
Ha but checking 1 is critical point is not very easy
I mean ek dam se strike na kare shyaad
[0,2] me sirf 1 hi ek critical point hai, are you sure about that?
If yes, then how?
Me still trying to figure out why didn't he choose x=1 earlier
So that I could explain you all the wrong steps one can think off
@@jeesimplified-subject mathematician for real🤝
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👍🏻
Voh fomula tha na binomial mein is series ka agar yaad hai toh yeh question kuch bhi nahi hai
That is for infinite series @@Ayush-mg6xw
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.
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
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
real
Bhaiya plzz we need maths content from you like eduniti for physics 😊
ig because the function is increasing in lhs and rhs of 1 so it will be global minima
U will certainly make me fall in love with maths...❤
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
Mujhe maths nhi aata phir bhi dekhne mein mza aaya🙂
It was dramatic but nice
Bhaiya agp series solve karke dekhenge toh( after differentiation) ek global minima aayega jo x=1 hai
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
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
Aur explanation de fir?
Kyu kiya x=1
Har cheez bas aise nhi hoti
Mai Nahi Bataunga
:In that sound:
ab koi ayega jo bolega bhaiya ye question toh maine 2 min me orally kar diya tha easy question tha
For real 😂😂
😂 vo log bhagwan hein
Bina question dekhe answer bta dete h aur bolte h ye to easy he tha
@@mohitpuri7238 iss baar koi 5th class ka aaya h
Bhai aagya h koi upar dekhna ek likh rha h pee dablu mein kra Diya tha 🤓....
Mai to 1 sec me kr diya tha
😂
Bro please let me know how did you studied maths and become so much better
If anybody has purchased Set of 60 please share your review
🎉🎉 kya baat h bhaiya...progress kr rhe ho...😊 ..mja aaya .😂
Bhai ab toh ham bhaiya ki videos download bhi nahi kar sakte
Bro editing ke liye 2 se 3 ghante lag gaye itna time nahi lagta hai bro 😢😢
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.
Amazing video bro ❤
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
Kyuki aapne keh diya isliye😅
Please give ❤
Love you brother
Bahuut cool ban rhe bhaiya kisko patana hai 😂😂😂
integral from 0 to 1 of x*[1/x]*{1/x}
Where [.] Is gif and {.} Is fif and it is given that summation of n from (1 to inf) of. 1/n² is pi²/6.
I guess the integral that u have given has the value is equal to N?, also, are the terms in power or multiplied?
@@annapurnasarangi9999 what's N?
5 elder stars !!
I multiplied f(x) with x getting
xf(x) added with f(x) and get an equation that was easy to differentiate and when checked for
f'(x) = 0 , at x = 1
creativity
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
mitochondria is the a²+b²+2ab of the motion
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 \( 2023 \).
May we put 1 x =omega
2. X =omega ² in the expansion of (1+ix)²⁰²³
May question form a gp
Sorry for all things
All people who gave reason for x=1 to be min, all are blatantly flawed.
yea, most of them 😂
U should take help of sandal bhaiya
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
Isko kehte hai choti si baat ka batangar banana
Bhai agp sum me Sx-S karke S nikal jaiga usko d/dx =0 karke x=1 dekhte hi pata lag jaiga
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
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
Bhai aap bahut achhi video banate ho
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.......
Bhai I am not a pro problem solver, auto assume karlete ho.
I just enjoy solving questions. Cuz its fun. besides, It ain’t the hardest but its fun 😍
@@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.
Abeyyyy ye tohh mere coaching ke mock test ka question h😂
x=1 pe har do terms se -1 ban rha tha so 2022 se -1011, 2023-1011= 1012
Bhai tune khud 12th kbaad padhai ki nhi h aur bakchodi kr rha h 😢😂😂
Very simple question
x^2022 common lekar agp lagado
1 is only solution as f'(x) is zero only at x=1 in [0,2]
( mujhe solve karna nhi aya per desmos par check karke pata chala 😢)
Respect the honesty bro
Isko binomial se bhi kar sakte the btw.. zyada asan parta
Beautiful bhaiya beautiful ❤️
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)
True, Its a different style. rather more entertaining 😂
christopher nolan real id se aao
mast
naya kuch try kiya
badiya 👍
ese video aur cha
hiyee ab toh
Bhiaya 11 th me hu I dont know what is global minima but I😊 will try to find it
Bhaiya question me bata Dena kya 11th wale ye questions try kar sakte hai ya nahi?
kya ye ek mission tha ......................
3:40 ke baad
bgm itna loud hai aapki avaaz samjh hi ni aarhi
bhaiya you have sticker on ur lptop,screen damage hogyi thi meri,just a cautionn🙌,although great ques
Aisa ek quesn mathongo ke cc adv wle me tha jo burnout hua tha 8 baar repeat kra tha video ko tb jake smj aya
Michael penn problem
kya hi mast lag raha hai face cam video aag laga dii
Bhaiya bg music low rakho
1012?
5 min problem
zoro lost again
As usual 😕
See my question also i sent it was based on complex no 😁
Bhai merko bhi bhejiyo
Telegram
@@yatharth2463 @Jeeadv202477
Mereko bhi
Op fan behind Zoro
bhai 1:49 am ko upload kar rhe ho, so jayo 🗿
Priyank Nolan
Bahut achha sawaal tha ye
Lmvt use karna tha naa
This video wasted my 8 minutes
only zoro can solve this monster question
but he’ll certainly get lost a million times before getting to the answer 😂
I choose ans jisse sahi ans aye
😂 voi to nai pata bhai
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
7:03 bhaiya yaha pe app kaise dekthe hi bataya ki vo even no.of terms hai
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
Which company chair you use. Is it comfortable
Local hai bhai par mast hai kafi
Bhai prep band kardo PYQ nehi karoge toh kya karoge