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VP
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New post 14 Feb 2007, 20:13
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SVP
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New post 15 Feb 2007, 00:06
(A) for me :)

n < 0 and n^2 < 1/100...

Where is 1/n ?

n^2 < 1/100
<=> (-n)^2 < 1/100
<=> -n < 1/10 as the function sqr() increases when x increases on x >= 0
<=> (0 >) n > -1/10
<=> 1/n < -10 as n<0 and the function 1/x decreases when x increases on x < 0.
VP
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Joined: 28 Mar 2006
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Kudos [?]: 28 [0], given: 0

 [#permalink]

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New post 15 Feb 2007, 05:33
Fig wrote:
(A) for me :)

n < 0 and n^2 < 1/100...

Where is 1/n ?

n^2 < 1/100
<=> (-n)^2 < 1/100
<=> -n <1>= 0
<0>) n > -1/10
<=> 1/n < -10 as n<0 and the function 1/x decreases when x increases on x < 0.


Fig thanks a bunch for your response

This is how I have proceeded

Since n is -ve

n^2 < 1/100 (given)

means n <1/-10 ( But the 1st choice is n<-10) this is where I got confused

Can you explain what is wrong with my approach?
SVP
SVP
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Joined: 01 May 2006
Posts: 1798
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Kudos [?]: 137 [0], given: 0

 [#permalink]

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New post 15 Feb 2007, 09:18
trivikram wrote:
Fig wrote:
(A) for me :)

n < 0 and n^2 < 1/100...

Where is 1/n ?

n^2 < 1/100
<=> (-n)^2 < 1/100
<=> -n <1>= 0
<0>) n > -1/10
<=> 1/n < -10 as n<0 and the function 1/x decreases when x increases on x < 0.


Fig thanks a bunch for your response

This is how I have proceeded

Since n is -ve

n^2 < 1/100 (given)

means n <1/-10 ( But the 1st choice is n<-10) this is where I got confused

Can you explain what is wrong with my approach?


The bold part is not equivalent to the original inequation. :)

U should look at the equation n^2 < 1/100 (without focussing on n> 0 or n<0).

n^2 < 1/100
<=> sqrt(n^2) < sqrt(1/100) (which is actually |n| < 1/10)
<=> -1/10 < n < 1/10

Then,
as n < 0, we have -1/10 < n < 0.
  [#permalink] 15 Feb 2007, 09:18
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