One correction on the DEC 2023 question : on 52:30 h(x) = x² is not uniform continuous on [0, ∞) hence the 4th option will be discarded. Sir, himself has mentioned it's not UC on the next question. Its not UC because we have two sequences aₙ=√(n+1) and bₙ= √n on [0,∞) such that |aₙ - bₙ|---> 0 as n--->∞ but |f(aₙ) - f(bₙ)|---> 1
Sir my another request, can you solve csir problems on singularity topic in complex analysis with short cuts. After completing differentiability topic in real analysis , as these series are very useful for our last minute preparation for csir exam.
Thank you so much sir ❤ you solve csir question like board exam questions you have great skills to solve problems and you teach us for free . thanks again ❤
Last question, option b is also correct because if I is compact ( closed and bounded) then f and g are both bounded and hence fg is uniformly continuous
Thankyou for this lecture Harish sir, with these videos we are able to analyse that csir is just repeating similar qns year after year. Your approach makes the csir look doable busting all the myths prevailing in Delhi based coachings. PS: sound issue is still prevailing, kindly look into it. As this problem ruins the quality content that you're delivering in lecture.
4.4k watch n why you are guys not liking the vedios ? I don't think anyone can explain here on yt history even i went to dips their questions discussions was in the air they didnt wait cz they taught iitians n DU students who were already toppers I never gets why they take choching even after coming from institutions like that. Well its humble request to all of you who are attending lectures series here to like the vedios n share too
Its my humble request to all the aspirants to kindly like the video too it would take just 2 seconds, but it would benefit the reach of this video thru YT algorithm. PS: despite 4.2k views only 200 something likes is non-reciprocation of sir's efforts!!
53:20 in this question it was asked that which of the following options implies that the function f is UC, it is not already given UC, which you told so :(
Thankyou sir. The session was awesome. I know you are proficient in this field. Sir please check it according to official answer key correct answers for the question (53:00) in 1,2 and 3, forth one is incorrect. Once again thank you.
Sir I have a doubt in 44:42 ... According to your shortcut, in option c, f(g(x)) g(x) is uniform continuous, but f(x) is un bounded so f(gx) cannot be uniformly continuous. How can I differentiate option 1 and 3?... Please clarify sir... Thankyou ❤
For the last question. Option a Since f,g are UC therefore both are bounded. So either f or g us bounded is wrong. So option a is wrong. Option b If we choose I as Real numbers then it is not compact. So option b is also wrong. Is it right sir??...
@@king-by1kh Bro h is defined from I to R ...so we need to consider an interval in which f and g are u.c that's why interval must be bounded for h(x) to be u.c.
One correction on the DEC 2023 question :
on 52:30 h(x) = x² is not uniform continuous on
[0, ∞) hence the 4th option will be discarded. Sir, himself has mentioned it's not UC on the next question.
Its not UC because we have two sequences aₙ=√(n+1) and bₙ= √n on [0,∞) such that |aₙ - bₙ|---> 0 as n--->∞ but |f(aₙ) - f(bₙ)|---> 1
Your lecture series is make us strong to fight any exam within suitable time
great lecture with very useful shortcut methods, thank you sir
Sir your approach to Csir net Questions is Always Unique...
❤❤❤
Thank you sir for providing this lecture
Thank you so much sir this is too much helpful and exam so near thanks again
Thnaku sir apke video se ese lgta h jse csir easy h
Very. Very helpful❤❤❤❤
very helpful👌
Thank you sir ❤
Thnku so much sir its really helpful 😊
Thanks a lot sir
very helpful
Very help full lecture for preparing net, I shared it among my students preparing net . thanks sir
Sir my another request, can you solve csir problems on singularity topic in complex analysis with short cuts. After completing differentiability topic in real analysis , as these series are very useful for our last minute preparation for csir exam.
Superb session.....❤
Thanks a lot sir ❤
Thank you so much sir 😊
Thank you so much sir ❤ you solve csir question like board exam questions you have great skills to solve problems and you teach us for free . thanks again ❤
Thank you so much sir 🙏🏻🙏🏻🙏🏻
🎉 thanks sir
Thanks Sir❤
Again thanks sir,
Thanks sir
Sir your video lectures are very helpful
I am really happy sir ... I understand all concepts easily
Thank you for more supporting sir 🙏🏻♥️
Your service for us will be remember for ever ❤❤
tysm
THANKS YOU SIR FOR THE LECTURE IN UNIFORM CONTINUITY 🙏🙏
Thanku you sir❣️🚩
I must say your analysis is so strong 🙌
nice sir
Differentiability ke baad complex analysis (bilinear transformation) ka topic Kara dijiye sir
Last question, option b is also correct because if I is compact ( closed and bounded) then f and g are both bounded and hence fg is uniformly continuous
50:05, last option doesn't seem correct because then f will also be uniformly continuous.
Thankyou for this lecture Harish sir, with these videos we are able to analyse that csir is just repeating similar qns year after year. Your approach makes the csir look doable busting all the myths prevailing in Delhi based coachings.
PS: sound issue is still prevailing, kindly look into it. As this problem ruins the quality content that you're delivering in lecture.
Sir please make a video on Statistical Inference
4.4k watch n why you are guys not liking the vedios ? I don't think anyone can explain here on yt history even i went to dips their questions discussions was in the air they didnt wait cz they taught iitians n DU students who were already toppers I never gets why they take choching even after coming from institutions like that.
Well its humble request to all of you who are attending lectures series here to like the vedios n share too
Its my humble request to all the aspirants to kindly like the video too it would take just 2 seconds, but it would benefit the reach of this video thru YT algorithm.
PS: despite 4.2k views only 200 something likes is non-reciprocation of sir's efforts!!
For the last question, By Heine Borel theorem, I is not compact since option(d) is discarded.
53:20 in this question it was asked that which of the following options implies that the function f is UC, it is not already given UC, which you told so :(
Thankyou sir. The session was awesome. I know you are proficient in this field. Sir please check it according to official answer key correct answers for the question (53:00) in 1,2 and 3, forth one is incorrect. Once again thank you.
Sir tifr ke last two three years ke real and linear ke questions bhi krwa do if you have time tifr concept are repeate in csir I think . thanks ❤
Great 🙏 .. Some mistake I think in ques. Gate 2004?
🥀
Sir please upload sequence and series of function pyq video
Already uploaded.... Check my lecture "Uniform Convergence"
ruclips.net/video/Qi4cDKUCGmI/видео.html
This is the link
Okk sir . Thank you sir 🙏🙏
Sir I have a doubt in 44:42 ... According to your shortcut, in option c, f(g(x)) g(x) is uniform continuous, but f(x) is un bounded so f(gx) cannot be uniformly continuous.
How can I differentiate option 1 and 3?... Please clarify sir... Thankyou ❤
if that condition holds that only gurantees that..if it not then you can't say anything...apply derivative test
@@dhrubajotidas8276 Ok thank you
Kindly provide lecture on abstract algebra also.
Kitne bar bole.. Sit ka domain nahi hbye subject
52:00 Last option is incorrect as x^2 is not uc in R, its graph is parallel to Y axis
Ya I have the same confusion
25:38 que ka ans shi lga kya sbko ...?? f(2π) ki value galat h ..mere hisab s to c option shi hona chahiye ...anyone correct me if m wrong
galat hya woh
B and c
A,b,c
For the last question.
Option a
Since f,g are UC therefore both are bounded. So either f or g us bounded is wrong. So option a is wrong.
Option b
If we choose I as Real numbers then it is not compact. So option b is also wrong.
Is it right sir??...
real number is not an interval
@@dhrubajotidas8276 It is an interval
Sir f(x) =x and g(x) =1 then clearly h(x) is Uniform continuous on R clearly R is not bounded so how option c is true only option 1 is true sir
given I is an interval
@@dhrubajotidas8276 R is also an interval (-infinity, infinity)
(a, infinity) is also an interval
@@PrabhatTamrakar-yr2ew we have to discard the option so we take R other wise it always hold on finite interval
1:03, f is uniformly continuous doesn't imply f satisfies lipchitz condition
UC=> lip. But converse need not to be true
Converse is true only if derivative is bounded
ONLY ANSWER C IS CORRECT
Sound quality sahi nhi h😬....
Charan sparsh Gurudev....
ONLY ANSWE C IS CORRECT
In last pyq it is not compact since not closed and option a is true b is false
How option c is true bro if f(x) =x and g(x) =1 then clearly h(x) is Uniform continous On R but R is not bounded so how it is correct
@@king-by1kh Bro h is defined from I to R ...so we need to consider an interval in which f and g are u.c that's why interval must be bounded for h(x) to be u.c.
But it is not necessary because I is also consider as(-infinity, +infinity) f(x) =x is Uniform continuous on R but R=I is not bounded
Yes opt a is crt
I ask about option c brother
Sir make video on NBHM pyq
Thank you so much sir ❤
Thank you sir
Thanks sir ❤
Thank you Sir
Thank you sir
Thank you sir