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by rohangupta83 » Thu Dec 04, 2008 4:29 am
If x &#8800; 0, is x^2/|x| < 1?
(1) x < 1
(2) x > &#8722;1

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by scoobydooby » Thu Dec 04, 2008 5:35 am
C?

Statement: x<1
say x=1/2
or (1/4)/(1/2)=1/2 <1 yes

say x=-1
or 1/1=1<1 no not sufficient

Statement: x>-1
say x=-1/2
or (1/4)/(1/2)=1/2<1 yes

say x=1
or 1/1=1<1 no not sufficient

combining 1 and 2 , -1<x<1 or x must be some fraction either positive or negative (x cannot be 0)

x=-1/2 or x=1/2 gives (1/4)/(1/2)=1/2<1
hence C

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by cramya » Thu Dec 04, 2008 6:03 am
Question stem:

x^2 / |x| < 1 can be rewritten as x^2 < |x| since we know |x| is always positive (so dont have to worry about affecting the inequality)

This is true is for fractions between -1 and +1

Stmt I

x<1

If x=-2 for eg x^2>|x|
If x=1/2 x^2<|x|

INSUFF

Stmt II

x>-1
If x=2 for eg x^2>|x|
If x=1/2 then x^2<|x|

INSUFF

Combining stmt I and II we get

-1<x<1

Therefore for all values within this range x^2<|x|

SUFF
C)

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by rohangupta83 » Thu Dec 04, 2008 6:06 am
scoobydooby wrote:C?

Statement: x<1
say x=1/2
or (1/4)/(1/2)=1/2 <1 yes

say x=-1
or 1/1=1<1 no not sufficient
This is what I did -

As the x in the denominator is |x|

so I assumed 2 cases

case 1: x>0
here your example is valid where you took x = 1/2

therefore, x^2/|x| < 1

But for x < 0

i.e. x is negative

if I take x = -1

I get

x^2/x = (-1)^2 / - (-1) = 1 ----- (ohh! I realize my mistake now. I didn't account for the negative sign of the negative number and took (-1)^2/(-1) instead)
scoobydooby wrote: Statement: x>-1
say x=-1/2
or (1/4)/(1/2)=1/2<1 yes

say x=1
or 1/1=1<1 no not sufficient

combining 1 and 2 , -1<x<1 or x must be some fraction either positive or negative (x cannot be 0)

x=-1/2 or x=1/2 gives (1/4)/(1/2)=1/2<1
hence C
Thanks Scoobydoo - sometimes another perspective is what you need to understand your own mistakes. :)[/quote]

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by cramya » Thu Dec 04, 2008 4:13 pm
Rohan,
A good one. Whats the source?

Regards,
Cramya

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by raajan_p » Thu Dec 04, 2008 5:23 pm
Ok, for this question, I dint substitute any numbers..

so guys, please check and let me know if there's any flaw in my approach..

X^2/|X| < 1 can be written as

X^2/X < 1 and X^2/(-X) < 1

Or X < 1 and X > -1

So the given question can be re-written as if - 1 < x < 1

From A we know that X < 1, but it can be any value till - infinity.

From B we know that X > -1, similarly any value till infinity.

Combining both we surely know that - 1 < X < 1

Hence C is the answer.