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# If x≠0, is |x| <1? (1) x2<1 (2) |x| < 1/x

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If x≠0, is |x| <1? (1) x2<1 (2) |x| < 1/x [#permalink]

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26 Jul 2009, 23:46
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. If x≠0, is |x| <1?
(1) x2<1
(2) |x| < 1/x

Thanks
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27 Jul 2009, 00:35
IMO D

1) is sufficient
If x² <1 it means that x is between -1 and 1.
So |x| < 1

In statement 2, we can notice that x can not be negative. So for any positive x, |x| = x and then x<1/x means that (x²-1 )/x <0 ==> x²-1 < 0 same as statement 1) sufficient.

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27 Jul 2009, 00:52
nitishmahajan wrote:
. If x≠0, is |x| <1?
(1) x2<1
(2) |x| < 1/x

D.
1) x^2<1 implies that -1<x<1 which is nothing but |x| < 1.
SUFF.
2)
If x>0, then the inequality becomes, x < 1/x. Is this possible for integers >= 1 ? No, because for integers >=1 the reciprocal is always smaller than the integer itself.

So we have to think about fractions between 0 and 1. Reciprocal of this fraction will an integer greater than 1. So all the fractions between 0 and 1 will satisfy the equality.

If x<0, then LHS becomes +ve and RHS becomes negative. Is this possible ? No, so anything <0 will not satisfy the inequality.

Hence, we can say for sure that if |x|<1, then |x|<1/x.
SUFF.
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27 Jul 2009, 01:06
Economist wrote:
nitishmahajan wrote:
. If x≠0, is |x| <1?
(1) x2<1
(2) |x| < 1/x

D.
1) x^2<1 implies that -1<x<1 which is nothing but |x| < 1.
SUFF.
2)
If x>0, then the inequality becomes, x < 1/x. Is this possible for integers >= 1 ? No, because for integers >=1 the reciprocal is always smaller than the integer itself.

So we have to think about fractions between 0 and 1. Reciprocal of this fraction will an integer greater than 1. So all the fractions between 0 and 1 will satisfy the equality.

If x<0, then LHS becomes +ve and RHS becomes negative. Is this possible ? No, so anything <0 will not satisfy the inequality.

Hence, we can say for sure that if |x|<1, then |x|<1/x.
SUFF.

Thanks for nice explanation ! It helped !
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27 Jul 2009, 01:07
IMO D

1) is sufficient
If x² <1 it means that x is between -1 and 1.
So |x| < 1

In statement 2, we can notice that x can not be negative. So for any positive x, |x| = x and then x<1/x means that (x²-1 )/x <0 ==> x²-1 < 0 same as statement 1) sufficient.

OA is D only, thanks for he explanation !:)
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27 Jul 2009, 12:32
hey , I did not get this can you please elobarate ?

Quote:
If x<0, then LHS becomes +ve and RHS becomes negative. Is this possible ? No, so anything <0 will not satisfy the inequality.

Hence, we can say for sure that if |x|<1, then |x|<1/x.
SUFF.

if x < 0

then the inequation becomes -x < 1/x ==> $$-x^2 < 1$$==> $$x^2 > -1$$ how can we rule this out ?

please throw some light on this..
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27 Jul 2009, 14:03
skpMatcha wrote:
hey , I did not get this can you please elobarate ?
if x < 0

then the inequation becomes -x < 1/x ==> $$-x^2 < 1$$==> $$x^2 > -1$$ how can we rule this out ?

please throw some light on this..

There are two problems in what you have derived:

1. if x<0, then value of |x| is always x. Note that |x| is ALWAYS positive no matter whatever is the value of x. That is why we can say that when x<0, |x| = -x....because -(-x) will be positive x.

2. This is not related to your question because if x<0 your derivation -x < 1/x is wrong(as I explained above)..it should be x < -1/x.

Now, just analyzing your statement, If x<0 and -x < 1/x ==> [m]-x^2 < 1.
Here, if -x < 1/x then we can derive -x^2 < 1 only if x > 0. If x<0, then it becomes -x^2 > 1.
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27 Jul 2009, 15:20
Yep D.

Good explaination econ.
Re: DS Question   [#permalink] 27 Jul 2009, 15:20
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