bigtreezl wrote:

is 2x-3y<x^2 ?

1) 2x-3y = -2

2) x>2 and y>0

1) 2x-3y =-2. This implies that -2 < x^2 . Sufficient since -2 is negative and x^2 is always positive.

2) x>2 and y>0.

2x-3y<x^2 will always hold good. We have to first understand that for +ve numbers ( apart from those between 0 and 1) x^2 is always bigger than 2x. And here in our equation, we are subtracting a part of 2x in the form of 3y where y is a positive number. So 2x-3y will definitely be smaller than x^2

Some examples:

if x=3 and y=1 => 2(3)-3(1)= 3 which is less than x^2 ie 9

if x=3 and y=3 => 2(3)-3(3)=-3 which is less than x^2 ie 9

if x=1000 y=100 => 2(1000)-3(100)=1700 again less than x^2 which is 1000000

2 also sufficient.

answer is D?

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