LMVT SE KAR SKTE THE , ISKO EK NEW FUNCTION MAAN KE DIFFERENTIATE KARTE THE USSE EK F'(1)=F'(0) ayega that means equal slope or lmvt shows the point which have equal slope and f'(c) = f(1)-f(0)/1-0 Then we get f'(c) as 1 just put in given case integrate and solve
@@saswattangudu8634 why? Every extreme point is a minima or a maxima; as only one answer occurs we can safely assume it is minima;like a quadratic equation...
Can also be solved by putting f(x) = t, hence f'(x)dx = dt & x = f inverse t. The integrand becomes f'(f inverse t), which can also be written as d( f(f inverse t) /dt. Now change the limits, and get the answer as 1
Sir, yeh joh hain aapne prove kar diya kay >= 1 hain, lekin iskah matlabh yeh nahin kay answer bhi 1 hain. For example minimum value answer 2 bhi ho sakataa thaa. Phir bhi >= 1 satisfy ho jaataa hain. Haan 1 se kam jaroor nahin ho sakata. Iskay liye aapkoh minimum value at least 1 bhi prove karna padega. Iske liye ek example kaafi hain jiske liye yeh integral kaa value 1 hain. Aur yeh example hain f(x) = x f'(x) = 1 integral 0 --> 1 { [f'(x)]^2 dx } = integral 0 --> 1 { [1]^2 dx } = integral 0 --> 1 { 1 dx } = x | 0 --> 1 = 1 - 0 = 1
what if we assume a function f(x)=x as it satisfies the condition given and f'(X) will become 1 so integral of f'(X) will become 1 and by applying by parts in the asked integral we will easily get ans as 1
Yes, there is the logic behind your answer and f(x) = x is the only solution. From the inequality which sir has given above, there for integ((f'(x)) ^ 2))(x = 0 to 1) = 1 only in the equality case, i.e., integ((f'(x) - 1) ^ 2))(x = 0 to 1) = 0, which will imply that f'(x) - 1 = 0, because (f'(x) - 1) ^ 2 >= 0 and hence the antiderivative of (f'(x) - 1) ^ 2 will always be increasing and hence the definite integral over increasing range can be 0 only when the function itself is 0, therefore f'(x) = 1 and integrating both sides, we get f(x) = x + C. Substituting x = 0, we will get C = 0, therfore f(x) = x for the minimum value of integral.
Sir if we take [f'(x)-k]^2>=0 Solving by this method,we get I>=2k-k^2 As k approaches 0 from right side,it shows minimum value of I is 0 But from ur solution,I is minimum when f'(x)=1,i.e f(x)=x
You are a maths teacher. I have watched almost all your videos and your every video has amazing maths concepts. I study in class 10th and I am preparing for IIT JEE and I am able to understand.
Bro how you are managing!!! Initially I also have the enthusiasm but over the period of time managing time became difficult for me... I am stuck in continuity and also in physics I am stuck in electricity...Boards are also coming..So I have to read the class 10 books also ... please tell me your routine
[f'(x)-k]² leke dekho bhai given integral≥1-(k-1)² aaega To find the minimum value of lhs for all k satisfying this You must find maximum value of rhs So... As Max value of 1-(k-1)²=1 Minimum value of this integral=1 All f(x) in the form of x+c will satisfy this
sir aapne [f'(x)-1]^2 >=0 likha h ,laken agar ham general case ki baat karen like for some k, [f'(x)-k]^2 >=0 then ans kuch aur v aa sakta h yahn pe ye v mention nahi kiya h ki k +ve integer hona chaiye ya rational no then many possibilities are there
To find the minimum value of LHS for all k satisfying this You must find maximum value of RHS So... As Max value of 1-(k-1)²=1 Minimum value of this integral=1 All f(x) in the form of x+c will satisfy this
Why is it wrong to use leibniz theorem and equate it to zero to get min value through AOD? If we use leibniz theorem to differentiate then we get 0 that means from 0 to 1 it's a constant function...
sir aapne 1 hi kyu liya mene 1/2 leke solve kiya and min. value 3/4aaya and 2 lene par inc. kar raha hai . means as we go toward 0 iski min. value decrease kar rahi hai
Sir .... i think that just proving that the given integral is greater than 1 doesn't prove that 1 is the minimum value ... the mininum can be >1 as well. We can prove that at 1 , it will be minimum using Feynman's technique , where we assume a function integral 0 to 1 (f'x - a)^2 dx as a function of a .
Sir Humble request to you please. Sir please can you make a video of class 9 and 10 important geometry and mensuration chapters which help in physics please sir I am not studying any geometry and mensuration chapter I'n class 9 and 10 due to my state board school please sir all topics of geometry and mensuration which use in physics. Please sir I really need them I am in class 11. 🙏
2:13:31 Type 5️⃣ Eqn of the form: a sinx + b cosx
Since f'(x) is always increasing function, it's gonna yield the minimum value at the lower limit, i.e. 0, pretty simple.
LMVT SE KAR SKTE THE , ISKO EK NEW FUNCTION MAAN KE DIFFERENTIATE KARTE THE USSE EK F'(1)=F'(0) ayega that means equal slope or lmvt shows the point which have equal slope and f'(c) = f(1)-f(0)/1-0
Then we get f'(c) as 1 just put in given case integrate and solve
Sir,(f'(x)-n)^2>=0 for any n which is a real number,so I think that this is not a rigorous way of proving the theorem...........
one another way in term of thinking in lmvt beacuse of f is differentiable s that f^,x=1 in 0,1 interval ans comes in one line
ha bhai waise hi kiya; f0 and f1 ki values bhi given hai na..
@@rugvedkshirsagar4917 BUT THERE ARE PROBLEM IN DOING THESE . BY GIVEN INFO WE CANT SAY ABOT MIN. VALUE ATTAIN AT X=c
That can be a critical point but not min value
@@saswattangudu8634 why? Every extreme point is a minima or a maxima; as only one answer occurs we can safely assume it is minima;like a quadratic equation...
@@hindrajmali-b2h h pr answer to aa gaya na,lmao
Can also be solved by putting f(x) = t, hence f'(x)dx = dt & x = f inverse t.
The integrand becomes f'(f inverse t), which can also be written as d( f(f inverse t) /dt. Now change the limits, and get the answer as 1
How though the integrand is f'(f inverse(t)) not f'(f inverse(t))* derivative of f inverse(t) you have apply chain rule
∫(f'(x))^2dx bhi toh greater than or equal to zero hoga toh minimum value to aise hi 0 likh skte hn tb. 👀👀
Agar upper limit (ye +ive)- lower limit(ye bhi + ive) kre
Toh minus me bhi ans aa skta h
Nhi ho skta
u mean f'(x) =0 everywhere if the integral =0
But acc to q, it cant
0 is the greatest lower bound but you can't say it is the min value
Sir, yeh joh hain aapne prove kar diya kay >= 1 hain, lekin iskah matlabh yeh nahin kay answer bhi 1 hain. For example minimum value answer 2 bhi ho sakataa thaa. Phir bhi >= 1 satisfy ho jaataa hain. Haan 1 se kam jaroor nahin ho sakata. Iskay liye aapkoh minimum value at least 1 bhi prove karna padega. Iske liye ek example kaafi hain jiske liye yeh integral kaa value 1 hain.
Aur yeh example hain
f(x) = x
f'(x) = 1
integral 0 --> 1 { [f'(x)]^2 dx } = integral 0 --> 1 { [1]^2 dx } = integral 0 --> 1 { 1 dx } = x | 0 --> 1 = 1 - 0 = 1
🤝👍
what if we assume a function f(x)=x as it satisfies the condition given and f'(X) will become 1 so integral of f'(X) will become 1 and by applying by parts in the asked integral we will easily get ans as 1
Thanks sir for this amazing questions
For functions belonging to the family f(x) =x^n/n have the minimum value of the asked integral n belongs to natural number
We can suppose the function as f(x)= x , because it supports all the three conditions and then q. Is dolved!
for f(x)=x
Sir f(x)=x ke liye. Mene sir phle y=x wala hi socha tha. Phir mene f(x)=x^2 ke liye bhi socha usse answer 4/3 aaya Jo ki 1 se bada tha...
Cauchy integral inequality direct question
Substitute Fx = x
idk if there is a logic but only f(x)=x satisfies the conditions and also gives the same ans...
Yes, there is the logic behind your answer and f(x) = x is the only solution. From the inequality which sir has given above, there for integ((f'(x)) ^ 2))(x = 0 to 1) = 1 only in the equality case, i.e., integ((f'(x) - 1) ^ 2))(x = 0 to 1) = 0, which will imply that f'(x) - 1 = 0, because (f'(x) - 1) ^ 2 >= 0 and hence the antiderivative of (f'(x) - 1) ^ 2 will always be increasing and hence the definite integral over increasing range can be 0 only when the function itself is 0, therefore f'(x) = 1 and integrating both sides, we get f(x) = x + C. Substituting x = 0, we will get C = 0, therfore f(x) = x for the minimum value of integral.
No , actually f(x) = x is the only function bounding minimum area Varna f(x) toh kitne bhi hogl sakte hai jo iss condition ko satisfy kar sate hai
Sir if we take [f'(x)-k]^2>=0
Solving by this method,we get
I>=2k-k^2
As k approaches 0 from right side,it shows minimum value of I is 0
But from ur solution,I is minimum when f'(x)=1,i.e f(x)=x
no,the max value of 2k-k^2 is 1(at 1), so I >=1,, also for minimum value of int. k=1
Thank you sir😊❤😊
You are a maths teacher. I have watched almost all your videos and your every video has amazing maths concepts.
I study in class 10th and I am preparing for IIT JEE and I am able to understand.
Bro how you are managing!!! Initially I also have the enthusiasm but over the period of time managing time became difficult for me... I am stuck in continuity and also in physics I am stuck in electricity...Boards are also coming..So I have to read the class 10 books also ... please tell me your routine
yes bro tell me how can i start
I have done 11th physics how much course have u done
Bro enthusiasm is nice but if you dont do the problem urself and just understand it by watching it you wont be able to solve it urself bro
@@seemonverma963 whole 11 maths alongwith class 10 science and maths
Fx is a polynomial function i.e x^n... therefore the min fx value is 'x' only
But It is not given in question that fx is a polynomial function
@@bhavyasoni7425 i assumed it to be..
@@bhavyasoni7425 as f0=0 and f1=1
@@sairajtripathy4752 f(x) = sin(πx/2) bhi 0 aur 1 pe same value deta hai
We need more like that.❤ You
Good evening sir …sir I think y=x is the only function which satisfy this .
for the family of curves y=x+c this is true
just use rms >= am
exactly
?
kaise
RMS>AM>GM>HM , if all no. r distinct positive values❤
Isse solution kaise nikalege??
Sir -1 ki jagah -2 bhi le sakte the then answer kuch aur ata. So how did you decide that youll take -1 only.
@@kushrungta4025 💀
[f'(x)-k]² leke dekho bhai given integral≥1-(k-1)² aaega
To find the minimum value of lhs for all k satisfying this
You must find maximum value of rhs
So...
As Max value of 1-(k-1)²=1
Minimum value of this integral=1
All f(x) in the form of x+c will satisfy this
Kyunki f(x) [0,1] me hi differentiable ha.
@@YashRohra-i4h abe bhai kisi chis ki min value chahiye to tu use kis chis mai max value ke equal kaise kar raha hai
@@YashRohra-i4hhan bhai sahi h
K lekr solve krna best option tha
Pr sir ne seedha 1 he kya liya ?
Sir I think function can be f(x)=x
sir aapne [f'(x)-1]^2 >=0 likha h ,laken agar ham general case ki baat karen like for some k, [f'(x)-k]^2 >=0 then ans kuch aur v aa sakta h yahn pe ye v mention nahi kiya h ki k +ve integer hona chaiye ya rational no then many possibilities are there
Then f'(x) dx ²≥ 1- (k-1)²
Sir ne k=1 wala hi soln dekha hoga
To find the minimum value of LHS for all k satisfying this
You must find maximum value of RHS
So...
As Max value of 1-(k-1)²=1
Minimum value of this integral=1
All f(x) in the form of x+c will satisfy this
Plz continue this series sir❤
Sir please upload lectures of series "love with graphs"
Why is it wrong to use leibniz theorem and equate it to zero to get min value through AOD? If we use leibniz theorem to differentiate then we get 0 that means from 0 to 1 it's a constant function...
Yeah you're absolutely right ✅️
kyuki yeh integral ko as constant treat karenge iski limit ka diff. hamesa zero hoga agar limit x ke terms me hoti tab wo kar sakte the
could have taken f(x) = x by observation and verified very simple
We can use euler's formula [calculus of variation] too here
Y=x
why did you do (f¹(x) - 1)²?
f(x) = x pe minm value aa rhi hai
F(x) = x ke liye minima
I think for y=x
Please upload class 10th Olympiad video
sir aapne 1 hi kyu liya mene 1/2 leke solve kiya and min. value 3/4aaya and 2 lene par inc. kar raha hai . means as we go toward 0 iski min. value decrease kar rahi hai
Yes ... i think we have to prove that using differentiation under the integral sign , assuming a function integral 0 to 1 (f'x - a)^2 dx
i dont get it . its defnite intergral we cannot use newton lebinitz since limits are constant.
Same doubt also came on my mind
AM ≥ GM
Star problem series kab ayega?
Sir .... i think that just proving that the given integral is greater than 1 doesn't prove that 1 is the minimum value ... the mininum can be >1 as well. We can prove that at 1 , it will be minimum using Feynman's technique , where we assume a function integral 0 to 1 (f'x - a)^2 dx as a function of a .
value cant exceed 1 for minium val
Hello sir golden ratio se hp sequence ka sum kaise derive karte hai please sir ispe video bana do please
Sir pure youTube pe dhundliya but nahi mil raha hai
Please bana do
You are new ramanujan sir ❤❤❤😊😊😊😊
Kuch jyada ho gya bhai😂
bhai aap pehle padh lo ki ramunujan kon the ..no disrespect to sir but sir ka kaam in mathematics is not even 1/infinity bit of his work......
Can we take f(x)=x
X=1/2
6:16 fx =1
Sir Humble request to you please. Sir please can you make a video of class 9 and 10 important geometry and mensuration chapters which help in physics please sir I am not studying any geometry and mensuration chapter I'n class 9 and 10 due to my state board school please sir all topics of geometry and mensuration which use in physics. Please sir I really need them I am in class 11. 🙏
Sri chaitanya ka tag kyun use kar rahe ho
i also am from shree chaitanya
It's Sri not shree😏
😢
Which branch bro?
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Sir 11th 12th all chapter one shot upload kar di jiye plz
I did it with help of Lmvt no one is taking about that guse i am wrong though the answer is correct but method should be proper
F'(x) - 1 he kyu liya, minus 2 kyu nhi kar skte??? 2:07
Kyunki f(x) [0,1] me hi differentiable ha.
Sir what approach aapne 1 hi kyo liya koi logic to diya hi nahi
I applied LMVT
sir pls FAT kum karo
sir plz make videos on itt proff notes q plzzz
Sir mote hote ja rahe ho aap..❤
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I didn't get why sir took f'(x)-1 ? Can anyone please explain ??why didn't you too -2 instead of any other number?
Because we cant directly determine the integral of (f`(x)^2) without knowing the function but can write ∫f`(x)dx as f(x) as it has linear power
@@Love_Coding so can we also consider f'(x)-2 ???
@@kotakikahanik Yes but then you get ∫f`(x)dx >= 0 so it doesnt give any answer
@@kotakikahanik yes
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sir this is wrong solution
Jai mahaprabhu jai bangla
1st view 🎉
Disclaimer:- nishant jindal is not harmed in this video .💀☠️☠️
how ?
Context?
th-cam.com/video/_GxkVbv6siw/w-d-xo.html
Yeah he is really not harmed in the video😂😂😂
I think that because this problem is from “NOTES”
🫵🐅💚🚩
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Any body from 11th😅
F(x)=X