isi 2010 ME I

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isi 2010 ME I

Vasudha
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.
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Re: isi 2010 ME I

deepak
+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.
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Re: isi 2010 ME I

Vasudha
I guess u mean f(x)= x-1/x
Thank u!!
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Re: isi 2010 ME I

Vasudha
In reply to this post by deepak
I guess u mean f(x)= x-1/x
Thank u!!
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Re: isi 2010 ME I

Amit Goyal
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.
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Re: isi 2010 ME I

Vasudha
Hmm..yes..thank u. but i'm unable to prove it. i've been trying for a long time..
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Re: isi 2010 ME I

archita
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
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Re: isi 2010 ME I

Vasudha
yea archita but how do u prove it for the general case?
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Re: isi 2010 ME I

komal
In reply to this post by Vasudha
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
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Re: isi 2010 ME I

Amit Goyal
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.