

If f(1) = 0, f'(x) > f(x) for all x > 1, then f(x) is
(a) positive valued for all x > 1,
(b) negative valued for all x > 1,
(c) positive valued on (1, 2) but negative valued on [2,∞).
(d) None of these.


+vely valued, i would think.
For eg., if the function is assumed to be f(x) = 11/x,
it satisfies f(1) = 0, f'(1) = 1+1/x^2 > f(x) for x>1. So if the observation of the function 11/x is considered, you will see it increasing in the positive direction.


I guess u mean f(x)= x1/x
Thank u!!


I guess u mean f(x)= x1/x
Thank u!!

Administrator

Check that neither x  1/x not 1  1/x works because none of them satisfies f'(x) > f(x) for all x > 1. Anyways the following example will work: f(x) = e^x  e. This will help you rule out (b) and (c) but is not the proof that its (a). Try to prove it.


Hmm..yes..thank u. but i'm unable to prove it. i've been trying for a long time..


f(x)=e^xe
f'(x)=e^x
e^x>0
therefore f(x) is increasing.
for x=1,f(1)=o
therefore for x>1,f(x)>f(1) (since f(x) is increasing)
therefore for x>1,f(x)>0,
hence prooved


yea archita but how do u prove it for the general case?


f'(x)>f(x) means that
lim h tending to zero 0{ f(x+h)f(x) / h} > f(x)
Rearranging we will get f(x+h)> f(x ) {1 + h) [limit h is tending to zero]
Consider case when h is positive then, f(x+h)> f(x) , h>0
i.e. f is increasing.
function is contionous( as it is differential), so f(x) when x is tending to 1 = 0
and as function is increasing so f(x) >0 for every x >1

Administrator

Komal, Its a good effort. But there is some problem. This is what you did:
Correct steps:
We are given that f'(x) > f(x) ∀ x > 1.
That is, lim_{h goes to 0} (f(x+h)f(x))/h > f(x) ∀ x > 1.
This gives us: f(x+h) > (1+h)f(x) for h positive and small enough, ∀ x > 1..
Wrong step: This implies f(x+h) > f(x) and hence f is increasing.
Note: Above does not imply f(x+h) > f(x), because this step requires f(x) > 0 and its not shown yet that f(x) > 0.

