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If x is not equal to 0, is |x|<a? 1. x^2<1 2. [#permalink]
 24 Jan 2004, 06:17
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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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Manager
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Definetely E as we dont have the information of variable a.
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Manager
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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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Manager
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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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Director
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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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Manager
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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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Director
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Quote: I think B is ambigious
It's not an opinion question.
Disprove my contention.
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Manager
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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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Director
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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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Ur rite Stoolfi, I agree now. 
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Manager
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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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Manager
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Seems to be a nice method 
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