I went to many institutions and invested a lot of money 💵 but no one ☝️ taught like you …..and you taught so very well 👌 without any fees thank you so much dear sir ❤
8:17 I have a doubt in selection of option c. We neglect [-1,1] because 0 belongs to it. Similarly option C also have 0. Then why we take as a correct answer?.... Please someone explain me 😢
The given option is set not interval because 1/n, n€N given .. So we need to find the function corresponding to this set and then check it continuity.... For example, for a and b option, function is given , but for c and d , function is not defined.... That's why, first defined the function and then check... Hope it help you.
Thank you sir🙏 Sir at 43:04 image of close interval under countinous function is always closed. and range of tan ^-1x is -pi/2 to pi/2 . So how option 1 is incorrect
@@Siyagupta52 ha to in this case you take A=R which is not closed But continuous function always map closed to closed Compact to compact Cauchy to cauchy So option 1 is true here
I went to many institutions and invested a lot of money 💵 but no one ☝️ taught like you …..and you taught so very well 👌 without any fees thank you so much dear sir ❤
Forget wrong things only take right and good things.
@@yogendravishwakarma5592yes ♥️
Because he is recognised by the top Indian scientist in the current time and recently included in the top 2% scientists in the world !!
@@aakashbhuse3170 absolutely 💯 right brother
@Dr.Harish garg , at 1:16:37 csir net June 2019 . Sir In this problem options (c) and (d) are also correct.
very helpful sir `thankyou so much sir .
Aisa content aaj tak nhi mila you tube pe... difficult topic hai isko banane ke liye bhot thankyou sir... kindly make topicwise aptitude pyq
Thankuuu so much sir.your teaching style is excellent.❤❤❤❤❤❤❤kash ap hmee phle
Very helpful sir . Please upload more videos.
Superb session.....❤
Really, you are teaching Real analysis, sir; thank you so much for sharing such videos.
at 35:40 how did we have discarded the unbounded option(d) .Please provide a valid justification for the same
1/1+e^x is cts, one one and bdd.
Take tan inverse x as example
Tku sir
Your videos very use full sir
My prayers and wishes are always for you sir 🙏
How to take f(x)=1, it is not one one ..at 35:01
Bahut jyada helpful h sir ji
@ dr harishgarg 41:00...a option kaise shi ho gya ...f(a ) bdd to a ko bdd kaise bol skte h ..inverse image thode hi bdd hoti h
Sir this is very helpful please upload next 2 lecture very soon please your trick very helpful in exam
at 16:11 how opt d is correct?
Thanks you sir aap great ho
sir 70min, lim x tends to zero [x] is not defined, so [x].sin(1/x) cannot be continuos at 0.
Very helpful ❤
34:24 you have not taking example f(x)=1 because one one function bola ha😢
Very nice explanation, thanks sir
8:17 I have a doubt in selection of option c. We neglect [-1,1] because 0 belongs to it. Similarly option C also have 0. Then why we take as a correct answer?.... Please someone explain me 😢
The given option is set not interval because 1/n, n€N given .. So we need to find the function corresponding to this set and then check it continuity.... For example, for a and b option, function is given , but for c and d , function is not defined.... That's why, first defined the function and then check... Hope it help you.
Yes sir thank you ❤
sir 44 min, option D is wrong. let f:(0,1) to R given by f(x)=1/x , then A=(0,1) is bounded but f(A)=(1, inf) is unbounded
37:10 how can we consider A=R in the (4) option sinceR is not compact
A={1} it is compact , f^-1({1})=R it is not compact
Thanks alot Sir 🙏
Thank you for your efforts sir 🙏
Very nice lecture sir....
@50:14 option (d) is incorrect as compact set will map to compact set.
Is this correct or not?
Can someone confirm?
Thank you sir🙏
Sir at 43:04 image of close interval under countinous function is always closed. and range of tan ^-1x is -pi/2 to pi/2 . So how option 1 is incorrect
Take f(x)=e^x , A=R then f(A)=(0,oo) which is not closed
@@Siyagupta52 ha to in this case you take A=R which is not closed
But continuous function always map closed to closed
Compact to compact
Cauchy to cauchy
So option 1 is true here
@@shefalithakur4580continuous function maps compact to compact
Closed to closed is not true .
As take the same example of e^x
@@Siyagupta52
I got it dear thanks
Continuous not necessarily map closed to closed
Continuity maps Closed inteval to closed interval...but not closed set to closed ..ie ex R
Thank you so much sir ❤
37:24 can anyone tell me..why sir had not taken compact subset A...why he has taken A=R
Yes A can't take R because A is compact
31:37 b option me continuous booa ha but greatest integer function always discontinuous at countable numbe of points how b option correct 😂
Really you are great sir ❤
Thank you sir 🙏
Thank you❤ sir
thanks sir
solid content sir
Thanks again ❤
You are genius sir
Thanks sir...❤
Sir ode or pde bhi topicwise krwa do
Sir... Please... Which software you are using for screen recording this video ?
Please explain in bilingual. Your lecture is very useful for me. Thanks sir
Sir please upload the vedios of linear algebra ... inner product spaces...and orthogonal...invariant subspcaes...
Linear algebra, inner product space etc already uploaded.. Explore at my youtube channel.
@@DrHarishGarg ok sir thanku
Very helpful 🎉🎉🎉
Thanks a lot sir jee .
Thanks sir ji
Please complete this topic with uniform continuity and all
Already uploaded.... See at my youtube channel.
Thank you sir
love you sir❤❤❤
Please upload the next lecture,
Thankyou sir
Thanku sir❤
Thankyou
Thanku sir
sir last request..... please upload a video on pyqs of ring theory.... 🙏🙏
At 7:50, c is not correct
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Thank you so much sir 🙏🏻🙏🏻🙏🏻
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Thank u sir ❤️❤️❤️❤️