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If |x| < 1, so, -1 < x < 1

Stat1: x> 0.5∗x2 or, x^2 - 2*x< 0 or, x*(x-2) < 0 or, 0<x<2 (but, given, -1< x <1), so, 0<x<1. We can say, x won't be less than Zero. Sufficient

Stat2: x/|x|=1; denominator is positive and right side is positive. So, x will be positive too. Sufficient

So, I think D. :)
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TarunKumar1234

Quote:
Stat1: x> 0.5∗x2 or, x^2 - 2*x< 0 or, x*(x-2) < 0 or, 0<x<2 (but, given, -1< x <1), so, 0<x<1. We can say, x won't be less than Zero. Sufficient

x*(x-2) < 0

Why aren't we taking 2 cases

where

Case 1:- The one u took

Case 2:- x<0 OR x-2>0 or x>2
x cannot be greater than 2 therefore -1<x<0

I took this case also which led me to choose option B and not D. Or am i missing something
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TarunKumar1234

Quote:
Stat1: x> 0.5∗x2 or, x^2 - 2*x< 0 or, x*(x-2) < 0 or, 0<x<2 (but, given, -1< x <1), so, 0<x<1. We can say, x won't be less than Zero. Sufficient

x*(x-2) < 0

Why aren't we taking 2 cases

where

Case 1:- The one u took

Case 2:- x<0 OR x-2>0 or x>2
x cannot be greater than 2 therefore -1<x<0

I took this case also which led me to choose option B and not D. Or am i missing something

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rsrighosh!

Please refer below explanation, both cases are considered. You are right! we have to consider both cases along with given condition. :)
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TarunKumar1234
Thanks it is clear now.. I totally forgot to use this method.. :facepalm:
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