Thank you for your unique and engaging teaching methods. Your guidance has helped me develop innovative ways to solve problems. I'm truly grateful for your support and encouragement. never stop this series sir 🙂🙂🙂🙂. And yes sir I follow your videos from your "Stolz-Cezaro theorem" in the limits video
It's really heartening to see genuine Math lovers following our videos. Thanks a lot. If you can share it on your social media, or circulate on relevant groups then it would be massive help to our small channel.
Amazing approach sir. Op series Sir aap please ek do video probability pe bhi bana denge flow chart wale method se conditional wale pnc mixed jaisa ..... plz ek SKT episode probability pe bhi sir apke unique method se 🙏🙏
Sabhi channel par hoti hai...Hum actually padhne waale bachho ka channel banana chah rahe hain...isliye bheed kam hogi but content ekdum honest gem milega. We are bringing chem and phy SKT very soon.
Sir what I did is this : As f(0) =0 so there must be only terms of x in the function so f'(x) must have 1 as it's minimum value as the minimum coefficient of x in the function will be 1 so after that I created f'(x) and squared it and integrated it and the answer came 1 as the minimum value of f
sir 7:42 p aapne bola ki I=1 should be achievable...voh toh hme pta tha soln se... pr hm kaise yeah deduce krenge ki f'x should be a constant...pls explain
Woh last mein bataya hai...That ultimately hum f'x ko ek constant K banana chah rahe hain so that woh integral zero ho jaaye. Constant banana hai to slope 1 hi Lena padega due to (0,0) and (1,1)
Sir sayad ham agar f dash square ke graph ko evaluate kren to aisa feel aa jata h ki integral ki value 1 hogi i have tried this in some other qns also pata nhi kaise ye work krr jata h hamesa square wala area sab variations ko dominate krr deta h plzz think this and try to see ki agar iss info pe koi inference milta h to btw enjoyed the video ❤❤
apply by parts and then its just 1- {integral f''(x)f(x)dx} from 0 to 1 and for some minimum value of function lets say m so f'(x)-m > (equal to zero) so for this equality to hold f'(x) must be constant so f''(x) must be 0 so ans is 1 infact we can do this with Variational Calculus becomes very easy
Sir but when you say k=1 in the end , that means f dash(x) >=1 for all x.know But since f can be anything, f dash can also be anything? Like even -1000?
Nahi...Dekho humein woh (f'x -k) square ko universally zero karna hoga so that uska integral also becomes equal to zero. Because when that integral becomes zero tabhi question waale integral ka min value achieve hoga. Ab socho ki agar f'x -k ka square universally zero hai to matlab f'x universally k hai. Matlab woh st line hai jo 0,0 and 1,1 se pass karega. So yes waise fx bahut saare ho sakte hain but minimum value dene wala ek hi hoga jo straight line hoga jiska slope 1 aaya
Sir bhut hii acha video hain. Sir mera ek question tha kii kyaa aane bhi bsc and msc maths kiye hua h. BTW I am in bsc 5th sem and preparing for IIT JAM Mathematics 2025
Nahi, B.Sc nahi, I have done B.Tech from IIT B. That's why my math is good for JEE only. I do have decent knowledge of Math beyond JEE as well but that's a result of my curiosity and interest.
Sir according to chatgpt this is a oneliner with Calculus of variations This seems to be useful for certain types of Qs for JEE like this one Can you take a session on this topic
@anshul8330 use logic of symmetry, if there is a positive slope deviation from 1 then there has to be an equivalent negative deviation and vice versa and the area will always be equal to 1/2 or greater, equal to 1/2 when the respective deviation is greater than its conjugate. And equal to 1/2 when conjugate deviation is equal to original deviation. Hence when you superimpose both conditions then answer wil be 1
Sorry for thr confusion i thought this was another q reply @anshul8330 This is what I am talking about 1. Imagine you have a f'x^2 graph, keep a 'base' graph as constant 1. Now imagine the curve of fx,our base f'x graph represents a straight line y=x. Now if you deviate about fx then you can plot the respective deviation about the f'x ^2 curve as well, if you plot several deviations you will always see the graph of (f'x)^2 will increase beyond 1. Hence minimum area is 1
No. LMVT use karna galat hai. LMVT se ye nahi milega ki har point par derivative 1 chahiye. Wahan milega ki kisi ek jagah koi random c hai wahan derivative 1 hai.
Wrong idea hai beta. Rolle's Theorem se ye prove ho raha hai ki atleast ek jagah derivative 0 hai. Humein derivative ko kisi ek jagah zero nahi...Us region mein universally zero karna hai.
@@vinodkumar-py1rk To ensure that the function is f'x - k is not just ≥0 but ALWAYS zero. Inequality ke andar Jo equal to wala part hai, woh prove karna hai ki ho sakta hai for ALL x in domain. Kyonki minimum value us equality wale part se hi aayegi.
@@anshul8330 and sir why are we doing that? to ensure I =1 is achievable? sorry if its a dumb doubt, i got sort of confused during certain parts of the video. PS: what if we had just proceeded with (f'(x)-k)^2?
New SKT Video agaya hai
Amazing sir
Glad you liked this video. There are many more videos in both Funda and SKT series. Have a look!
Pls share to help our little channel.
Thank you for your unique and engaging teaching methods. Your guidance has helped me develop innovative ways to solve problems. I'm truly grateful for your support and encouragement. never stop this series sir 🙂🙂🙂🙂. And yes sir I follow your videos from your "Stolz-Cezaro theorem" in the limits video
It's really heartening to see genuine Math lovers following our videos. Thanks a lot. If you can share it on your social media, or circulate on relevant groups then it would be massive help to our small channel.
Amazing approach sir. Op series
Sir aap please ek do video probability pe bhi bana denge flow chart wale method se conditional wale pnc mixed jaisa ..... plz ek SKT episode probability pe bhi sir apke unique method se 🙏🙏
Done!
Please share this video so that it helps our little channel.
@@anshul8330 definitely sir. Thank you so much
Thanks sir pls continue this series
❤ Ofcourse. Aap bhi ise share kar dijiye. Jitna support milega utna hi hum energy mein rahenge
Thanks sir
Aise hi padhte raho!❤
Damn! I explored a great channel, teaching us on how to think. Great work,
❤ thanks Krish. Please share our video with others to help our channel. We need your support.
Ushrahar
@piratedwell4608 Iska kya matlab hota hai?
You are unparalleled, sir.
Thanks, I am just trying to help.
Please do share the video so that it helps our little channel.
Yes sir very nice thought
sir what are your thoughts on using the sum of squares of deviations?
we know that sum[(xi-A)^2] is minimum when A = x bar (mean)
mean(f'(x)) =1
Yes. I realised it was a good approach
And thankyou sir idea click series lane k liye
Glad you liked it! Please share this video with others to help the channel.
aapke channel pe hi asli padhai hoti hai!!
Sabhi channel par hoti hai...Hum actually padhne waale bachho ka channel banana chah rahe hain...isliye bheed kam hogi but content ekdum honest gem milega.
We are bringing chem and phy SKT very soon.
@@anshul8330 yes sir you are right!! bas jan session se pehle jitne funda videos ho sake utna please daal dena. aap sacchi mein amazing padhate ho!
Sandaar❤❤
🎉Thanks for support
Sir what I did is this :
As f(0) =0 so there must be only terms of x in the function so f'(x) must have 1 as it's minimum value as the minimum coefficient of x in the function will be 1 so after that I created f'(x) and squared it and integrated it and the answer came 1 as the minimum value of f
Samjha nahi...Are you using expansion?
Brilliant solution sir ❤
May I suggest one thing from student point of view -
Pl use 2 playlist - one for class 11th and one for class 12th .
Complicated nahi ho jaayega?
Har series ke 2 playlist, 11th and 12th.
Thanks for this feedback. Please share the video so that it helps our team.
@anshul8330 You are right sir. I am a parent, just for your info. And, definitely I share this with my kid n his friends.
thanks sir
i solved it using cov but thanks for this watched full video
Haan alternate methods bhi hai...humne sirf sochne ka tareeka par focus kiy
@depressedguy9467 what is Cov?
@@_parallax_0 I think ye variance ke baare mein baat kar rahe hain
@@_parallax_0 New video is out. Check
sir 7:42 p aapne bola ki I=1 should be achievable...voh toh hme pta tha soln se... pr hm kaise yeah deduce krenge ki f'x should be a constant...pls explain
Woh last mein bataya hai...That ultimately hum f'x ko ek constant K banana chah rahe hain so that woh integral zero ho jaaye. Constant banana hai to slope 1 hi Lena padega due to (0,0) and (1,1)
@@anshul8330 understood sir...thanx
@MadhavSah007 Share the video to support our small channel!
Sir sayad ham agar f dash square ke graph ko evaluate kren to aisa feel aa jata h ki integral ki value 1 hogi i have tried this in some other qns also pata nhi kaise ye work krr jata h hamesa square wala area sab variations ko dominate krr deta h plzz think this and try to see ki agar iss info pe koi inference milta h to btw enjoyed the video ❤❤
Sir ji jee advanced ki four months weekwise strategy par video bnaiye please sir vote for It everyone
apply by parts and then its just 1- {integral f''(x)f(x)dx} from 0 to 1 and for some minimum value of function lets say m so f'(x)-m > (equal to zero) so for this equality to hold f'(x) must be constant so f''(x) must be 0 so ans is 1 infact we can do this with Variational Calculus becomes very easy
Half of your comment I agree with.
Sir but when you say k=1 in the end , that means f dash(x) >=1 for all x.know
But since f can be anything, f dash can also be anything?
Like even -1000?
Nahi...Dekho humein woh (f'x -k) square ko universally zero karna hoga so that uska integral also becomes equal to zero. Because when that integral becomes zero tabhi question waale integral ka min value achieve hoga.
Ab socho ki agar f'x -k ka square universally zero hai to matlab f'x universally k hai. Matlab woh st line hai jo 0,0 and 1,1 se pass karega.
So yes waise fx bahut saare ho sakte hain but minimum value dene wala ek hi hoga jo straight line hoga jiska slope 1 aaya
Sir bhut hii acha video hain. Sir mera ek question tha kii kyaa aane bhi bsc and msc maths kiye hua h. BTW I am in bsc 5th sem and preparing for IIT JAM Mathematics 2025
Nahi, B.Sc nahi, I have done B.Tech from IIT B. That's why my math is good for JEE only.
I do have decent knowledge of Math beyond JEE as well but that's a result of my curiosity and interest.
@anshul8330 okay sir 🙏
Sir according to chatgpt this is a oneliner with Calculus of variations
This seems to be useful for certain types of Qs for JEE like this one
Can you take a session on this topic
That's a great point! I'll definitely consider this soon
👏👏👏
🎉🎉🎉
Sir acadza app pe revision dost aur formula sheet dost free hai ya paid
Acadza is only paid. You can ask for a discount to support@acadza.com
Sir can we start from Lagrange theorem??
LMVT se nahi aayega
Sir alok sir aapke colleague haina? You both are best
Yes Alok sir is my dear friend and an investor in Acadza. He is exceptional in Maths.
@anshul8330 sir luv you both
😊 Share the video to help our channel @@Cooososoo
@@anshul8330 yes sir I will , sir can you Bring some sessions for good level mains questions
@Cooososoo will do so
We can think f'(x) =1
Using LMVT
We get easily....
No. Incorrect logic.
I send soon why taking 1
If we proceed as [f'(x)-k]²>=0
We get I >= 2k-k²>=1 .....
@@originmathematicsbyer.deep2099 But what's the guarantee that 1 is achieved. Also this is not LMVT.
@@anshul8330 minimum value of 2k-k² is 1 at k=1
Sir aap cute ho
Meri ex ko aisa nahi lagta sadly!
@@anshul8330 sir uska taste kharab hai xd aap cute ho
@@unnati5385Woh IIT Bombay se thi! Akalmand to hogi.
Chalo ye chhodo...Aap padh lo.
@@anshul8330 sir konse semester wali thi aapne to Fibonacci series banaya tha gf ka cllg me xdxdxd
@@unnati5385Padh lo! 😂
Orally kardia lmao
Kaise?
@anshul8330 use logic of symmetry, if there is a positive slope deviation from 1 then there has to be an equivalent negative deviation and vice versa and the area will always be equal to 1/2 or greater, equal to 1/2 when the respective deviation is greater than its conjugate. And equal to 1/2 when conjugate deviation is equal to original deviation. Hence when you superimpose both conditions then answer wil be 1
@@doge_the_cat
1. Which area are you talking about? Area under f(x) graph? That can be less than 1/2.
2. How does superposition give minimum value?
Sorry for thr confusion i thought this was another q reply @anshul8330
This is what I am talking about
1. Imagine you have a f'x^2 graph, keep a 'base' graph as constant 1. Now imagine the curve of fx,our base f'x graph represents a straight line y=x. Now if you deviate about fx then you can plot the respective deviation about the f'x ^2 curve as well, if you plot several deviations you will always see the graph of (f'x)^2 will increase beyond 1. Hence minimum area is 1
@@doge_the_catI feel you are right. We are just not able to communicate effectively in YT comments.
Use LMVT
No. LMVT use karna galat hai. LMVT se ye nahi milega ki har point par derivative 1 chahiye. Wahan milega ki kisi ek jagah koi random c hai wahan derivative 1 hai.
@@anshul8330bhai LMVT se bhi answer 1 aayega aur inhone bhi to greater than equal to 1 kiya hai
@@shivanshupadhyay1874 bro vo anshul sir hai ,sir bat karre the tere sath
Let g(x)=f(x)-x g is diff. g(0)=0& g(1)=0
So we can say’s g’(c)=0 for at least one c€(0,1)
And f’(x)-1=g’(x)
Or {f’(x)-1}^2=(g’(x))^2
Wrong idea hai beta. Rolle's Theorem se ye prove ho raha hai ki atleast ek jagah derivative 0 hai. Humein derivative ko kisi ek jagah zero nahi...Us region mein universally zero karna hai.
Why are u forcing to make identical 1
@@vinodkumar-py1rk To ensure that the function is f'x - k is not just ≥0 but ALWAYS zero.
Inequality ke andar Jo equal to wala part hai, woh prove karna hai ki ho sakta hai for ALL x in domain. Kyonki minimum value us equality wale part se hi aayegi.
@@anshul8330 and sir why are we doing that? to ensure I =1 is achievable? sorry if its a dumb doubt, i got sort of confused during certain parts of the video.
PS: what if we had just proceeded with (f'(x)-k)^2?
@@_parallax_0 then we would have proved that integral Ī ≥1 but we wouldn't have proved that min value of ī=1