# isi 2010 ME I

10 messages
Open this post in threaded view
|

## isi 2010 ME I

 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.
Open this post in threaded view
|

## Re: isi 2010 ME I

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

## Re: isi 2010 ME I

 I guess u mean f(x)= x-1/x Thank u!!
Open this post in threaded view
|

## Re: isi 2010 ME I

 In reply to this post by deepak I guess u mean f(x)= x-1/x Thank u!!
Open this post in threaded view
|

## Re: isi 2010 ME I

 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.
Open this post in threaded view
|

## Re: isi 2010 ME I

 Hmm..yes..thank u. but i'm unable to prove it. i've been trying for a long time..
Open this post in threaded view
|

## Re: isi 2010 ME I

 f(x)=e^x-e 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
Open this post in threaded view
|

## Re: isi 2010 ME I

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