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Source: — Data Sufficiency |
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by clock60 » Sat Apr 16, 2011 12:39 am
here i got A
is a>0
(1)x^2-2x+a>0 for all x. this is quadratic function, with parabola
according to the st any x resulted in +ve value for function, it means that graph lies above x-axis
and the value of 2^2-4a<0-discriminant
(y=ax^2+bx+c D=b^2-4ac)
4-4a<0.1-a<0, a>1 the answer is yes and it is suff
(2)ax^2+1>0
it is easy here to find both +ve and -ve values for a
say a=1
then x^2+1>0 any time yes
a=-0,5, x=1. the -0,5+1=0,5 again yes but a<0
to me insuff
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