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
Very helpful sir . Please upload more videos.
Bahut jyada helpful h sir ji
Superb session.....❤
Thankuuu so much sir.your teaching style is excellent.❤❤❤❤❤❤❤kash ap hmee phle
Tku sir
Your videos very use full sir
How to take f(x)=1, it is not one one ..at 35:01
Really, you are teaching Real analysis, sir; thank you so much for sharing such videos.
Very helpful ❤
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
My prayers and wishes are always for you sir 🙏
Thanks you sir aap great ho
Sir this is very helpful please upload next 2 lecture very soon please your trick very helpful in exam
Thanks alot Sir 🙏
Very nice explanation, thanks sir
Very nice lecture sir....
@ 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
at 16:11 how opt d is correct?
Thank you so much sir ❤
Thank you for your efforts sir 🙏
Thank you❤ sir
Really you are great sir ❤
Thank you sir 🙏
Thanks again ❤
You are genius sir
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 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 70min, lim x tends to zero [x] is not defined, so [x].sin(1/x) cannot be continuos at 0.
solid content sir
Thanks a lot sir jee .
Very helpful 🎉🎉🎉
@50:14 option (d) is incorrect as compact set will map to compact set.
Is this correct or not?
Can someone confirm?
thanks 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
Please explain in bilingual. Your lecture is very useful for me. Thanks sir
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
Thanks sir ji
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
Thank you sir
love you sir❤❤❤
Sir ode or pde bhi topicwise krwa do
34:24 you have not taking example f(x)=1 because one one function bola ha😢
Sir... Please... Which software you are using for screen recording this video ?
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
Thanku sir❤
Thankyou sir
Please complete this topic with uniform continuity and all
Already uploaded.... See at my youtube channel.
Please upload the next lecture,
Thankyou
31:37 b option me continuous booa ha but greatest integer function always discontinuous at countable numbe of points how b option correct 😂
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 u sir ❤️❤️❤️❤️