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If x is not equal to 0, is |x|<a? 1. x^2<1 2.

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If x is not equal to 0, is |x|<a? 1. x^2<1 2. [#permalink] New post 24 Jan 2004, 05:17
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If x is not equal to 0, is |x|<a?
1. x^2<1
2. |x|<1/x
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 [#permalink] New post 24 Jan 2004, 05:43
Definetely E as we dont have the information of variable a.
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 [#permalink] New post 24 Jan 2004, 05:50
My apologies,,,the qn is

If x is not equal to 0, is |x|<1?
1. x^2<1
2. |x|<1/x
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 [#permalink] New post 24 Jan 2004, 15:45
Then A? becoz from (1) we can say that x is a fraction and so [x] < 1/x but from (2) we cant say
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 [#permalink] New post 24 Jan 2004, 16:46
I'm thinking that B is sufficient, too.

B) |x|<1/x

B tells us that for positive X values, any number between 0 and 1 works.
It does not work for any negative X values.

So b tells us that 0<x<1, thus making it sufficient.

I also agree that A is sufficient as well.
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 [#permalink] New post 24 Jan 2004, 17:12
I think B is ambigious and we need a definite one which can be said from A, So I think its A.
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 [#permalink] New post 24 Jan 2004, 17:26
Quote:
I think B is ambigious


It's not an opinion question.

Disprove my contention.
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 [#permalink] New post 24 Jan 2004, 17:42
Stoolfi u only proved that for +ve values it suffices but not for -ve values, so u can say that B is sufficient only if its true for both +ve and -ve values but in this case its not. So I think B is not sufficient. What do u say?
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 [#permalink] New post 24 Jan 2004, 17:56
is |x|<1 is the question.

B tells us, as I have shown, that x is a positive number that is less than 1.

1 will always be greater than all numbers less than 1, by definition.

You contend that I "only proved that for +ve values it suffices but not for -ve values", but B says that x is positive. In other words, there are no negative values of X that make B true.
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 [#permalink] New post 24 Jan 2004, 18:12
Ur rite Stoolfi, I agree now. :good
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 [#permalink] New post 24 Jan 2004, 19:45
U are right, Stoolfi

D is the answer. I had put A as my answer.

I just came up with one other method of answering, by drawing graphs.

the first eq, x^2<1, is a U shaped graph and for all values of x<1, y<1 too. So, A is correct.

For the second eq, |x|<1/x, draw two graphs, y=|x|, which is a 45 degree straight line hinged at (0.0) making a V.
y=1/x is an inverse of y=x.
For all values of graph 1 < graph 2, |x|<1. So, B is correct as well.

=>D is the answer
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 [#permalink] New post 24 Jan 2004, 20:18
Seems to be a nice method :wall
  [#permalink] 24 Jan 2004, 20:18
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