Q25. Ma'am you used f(x)=x,which is neither constant nor bounded. You can't use it since option 3 says f is bounded. Your approach is wrong in this case,because epsilon must be independent of delta as per the given condition.
I think.. Onto wale me aapko problem hui h.. Yha aap esse krr skte ho.. f(m, n) =2^m (2n+1) , here 2^m not equal to 0 and 2n+1 is also not equal to 0 I.e. f(m, n) not equal to 0 But 0 belongs to Z So 0 does not have pre image Implies that not onto.
Very nice explanation madam it's great help thank you so much ....
Wlcm.. & thanks
Q 21 cardinality of domain is less than cardinality of codomain, kaise hua
Identity function is unbdd how option 3 correct mam ..question no 25
thanks mam
Wlcm
if we take 1/n then x tends to infinity exit but x tends to 0 not exist right?
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Your simplicity, normal and hand written notes allure me to watch more videos. Thank you so much. Sincer apperation and love.
If you speak in English it will be more helpful for many.
Q25. Ma'am you used f(x)=x,which is neither constant nor bounded. You can't use it since option 3 says f is bounded. Your approach is wrong in this case,because epsilon must be independent of delta as per the given condition.
1st question was solved in a wrong way... We cant conclude like that..
I think.. Onto wale me aapko problem hui h..
Yha aap esse krr skte ho.. f(m, n) =2^m (2n+1) , here 2^m not equal to 0 and 2n+1 is also not equal to 0
I.e. f(m, n) not equal to 0
But 0 belongs to Z
So 0 does not have pre image
Implies that not onto.
@@pmathematics ha ab thik hai.. ☺️
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