STATEMENT I

(x+y)^2>6
take (i) x=1 and y=2
(ii) x2 and y=2
not sufficient

statement II
xy=2

take (i)x=1 and y=2
(ii) x=1/2 and y=4
so not sufficient

combining
take x=1/2 and y=4
(ii) x=-1 and y=-2
so not sufficient

hence ans is E.

angxif wrote:Is x^2+ y^2> 6?
(1) (x+y)^2>6
(2) xy=2

I think the answer is B.
1) (x+y)^2- x^2+y^2+2xy 2xy could be negative or positive hence insuff
2)for xy to be 2 it can either be -1x-2 or -2x-1 or 2x1 or 1x2.
In all the above cases x^2+y^2will be less than 6 so B is suff.

OA[spoiler]E

Help!

Your reasoning in support of (2) is flawed. You have considered only integers. Take for example x=4 & y = ½. Here xy = 2 and x^2 + y^2 = 16.25 > 6.
Hence (2) alone is not sufficient.

When (1) & (2) are combined.
(x+y)^2 > 6 and xy = 2.
X^2 + y^2 = (x+y)^2 - 2xy = [>6] - 4.

The first part of this equation (even though > 6) is not sufficiently known. It could be 7 or even 12 giving two different values of x^2 + y^2. Hence E is the correct answer.