Aastha Choudhary

Give me some time.I am not getting things clearly.

For u''(x)<=0 implies u'(x) is strictly decreasing and in an alter way u'(x) not strictly decreasing implies u''(x) is not <=0.

@DevanshBhardwaj

Now for u′(x)≤0 to be true for some x, u"(x)<0 for some x∈[0,∞) as u′(0)>0. How you concluded this line?

GoodBye!

yes but this is for x=0 only :/

for u(x)<0 i.e. negative!

For u'(x)<0 we can claim u(x) is strictly decreasing.

@DevanshBhardwaj no I meant things were unclear, not useless :)

oho!

Sorry for?

I am not getting you right?

The quest is for [0,inf)

Since this is only for a specific x=0.

Seems unclear @DevanshBhardwaj

There seems some hollowness in your solution!

It would be possible only if u''(x)>0 for all x in R.

@DevanshBhardwaj How did you claimed the convexity?

Now as u′(0)>0, it is clear that the function will be convex throughout the interval [0,∞), thus u(x),u′(x),u"(x) will all be positive in the interval. Both statements follow.

I am exploring Kaczor to get a better solution for the same, will revert beck soon@DevanshBhardwaj

Had found it being a really very interesting problem.

yes @DevanshBhardwaj

yeah! not promoted on stack.

Asking for sources on internet for mcqs?

:)

Bye!

undoubtedly IISc is,let's see!

Hopefully!

Thanks again :)

Institute of Advanced Study!

I am not aware of advanced study

yeah!

That's great!

Wow IAS lectures!

Sure! Good luck!

@DevanshBhardwaj undoubtedly its better.

No! they are not taught!

Sadly, I am not shortlisted for ISI :(

yeah! Thanks for your efforts btw what exams are you preparing for?

okay! So this was the usage of the given condition again and again!

yes..

yes basically like continued fraction.

mathjax is time consuming.

simple typing is fine

okay!

This is kinda continued fraction but not actually continued. Isn't?

sorry?

yes

yes upto that only!

yes, that way I also observed.