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Is 1/p > r /r^2 + 2 1. p = r 2. r > 0

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Senior Manager
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Is 1/p > r /r^2 + 2 1. p = r 2. r > 0 [#permalink] New post 16 Dec 2006, 18:09
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A
B
C
D
E

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Is 1/p > r /r^2 + 2

1. p = r
2. r > 0
VP
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Joined: 28 Mar 2006
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Re: DS - inequality [#permalink] New post 16 Dec 2006, 18:39
gk3.14 wrote:
Is 1/p > r /r^2 + 2

1. p = r
2. r > 0


Assuming that the denominator is (r^2 +2)

from I we cannot get the solution 'cos if p=r=0 then the thing fails

so we need a condition >0 which is defined in II and this brings

C as the answer
Senior Manager
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 [#permalink] New post 16 Dec 2006, 19:01
1. for p=r=1
(1/p) > r/(r^2 +2)

for p=r= -1
(1/p) < r/(r^2 +2)

for p=r=0
1/p is not defined … insufficent

2. r >0 ….. we don’t know about p …. .unsufficient

from (1) and (2)

(1/p) > r/(r^2 +2)


Answer = C
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 [#permalink] New post 16 Dec 2006, 20:30
C)..

if 2) r<0 then also answer is C.
  [#permalink] 16 Dec 2006, 20:30
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Is 1/p > r /r^2 + 2 1. p = r 2. r > 0

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