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

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Senior Manager
Joined: 14 Aug 2006
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Is 1/p > r /r^2 + 2 1. p = r 2. r > 0 [#permalink]

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16 Dec 2006, 19:09
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Is 1/p > r /r^2 + 2

1. p = r
2. r > 0

Kudos [?]: 49 [0], given: 0

VP
Joined: 28 Mar 2006
Posts: 1367

Kudos [?]: 38 [0], given: 0

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16 Dec 2006, 19:39
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

Kudos [?]: 38 [0], given: 0

Senior Manager
Joined: 19 Jul 2006
Posts: 358

Kudos [?]: 8 [0], given: 0

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16 Dec 2006, 20:01
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 â€¦ insufficent

2. r >0 â€¦.. we donâ€™t know about p â€¦. .unsufficient

from (1) and (2)

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

Kudos [?]: 8 [0], given: 0

Manager
Joined: 18 Nov 2006
Posts: 123

Kudos [?]: 2 [0], given: 0

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16 Dec 2006, 21:30
C)..

if 2) r<0 then also answer is C.

Kudos [?]: 2 [0], given: 0

16 Dec 2006, 21:30
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