dinesh8 wrote:

Is \(2x-3y\)<\(x^2\)?

1. \(2x-3y =-2\)

2. \(x> 2\) and \(y>0\)

How is it D?

I'm getting A

But in choice B there can be multiple cases

for example

X=3 and y=10 in this case its a Yes

but if x=3, y=1

the question becomes is \(X^2 > 3\) we can't answer that?

(1) \(2x-3y=-2\) --> question becomes: is \(-2<x^2\)? as square of a number is always non-negative (\(x^2\geq{0}\)) then \(x^2\geq{0}>-2\). Sufficient.

(2) \(x>2\) and \(y>0\) --> is \(2x-3y<x^2\) --> is \(x(x-2)+3y>0\) --> as \(x>2\) then \(x(x-2)\) is a positive number and as \(y>0\) then \(3y\) is also a positive number --> sum of two positive numbers is more than zero, hence \(x(x-2)+3y>0\) is true. Sufficient.

Answer: D.