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This I missed it

Author Message
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VP
Joined: 28 Mar 2006
Posts: 1383
Followers: 2

Kudos [?]: 20 [0], given: 0

This I missed it [#permalink]  14 Feb 2007, 19:13
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
This I missed it
Attachments

File comment: Number Properties
CoOrdinate Geometry DS3.doc [55 KiB]

SVP
Joined: 01 May 2006
Posts: 1798
Followers: 8

Kudos [?]: 102 [0], given: 0

(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
Joined: 28 Mar 2006
Posts: 1383
Followers: 2

Kudos [?]: 20 [0], given: 0

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
Joined: 01 May 2006
Posts: 1798
Followers: 8

Kudos [?]: 102 [0], given: 0

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.
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