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by ketkoag » Mon Aug 31, 2009 11:22 am
Please explain the answer.. IMO E..
OA given : c
second statement is "x is less than equal to 1"
in the figure the second line after 1 is the cursor..please ignore it.
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by gmatmachoman » Mon Aug 31, 2009 12:10 pm
When X lies between 0 & 1,we can certainly say that the exp is equal to 1

ex: let x= 0.2
x-1= -0.8
|x-1|=0.8

|x| +|x-1|=0.2+|-0.8|
=1.

So it proves that any value between 0& 1 will be eual to 1 for |x| +|x-1|

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by ketkoag » Thu Sep 03, 2009 10:54 am
thanks for pointing it out..
kinda silly mistake.. isn't it.

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Re: Mod

by qwe12 » Fri Sep 04, 2009 1:42 am
ketkoag wrote:Please explain the answer.. IMO E..
OA given : c
second statement is "x is less than equal to 1"
in the figure the second line after 1 is the cursor..please ignore it.
where did you get this question?