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We have \(2\sqrt{x}+y^2<5\)

    Best way here could be to take scenarios using some calculations.

    First of all, x has to be non-negative and the same for \(y^2\). However, y can be negative here.
    If x = 0, y can be 0, 1, 2, - 1, - 2. However xy = 0 for all the cases.
    If x = 1, y can be 0, 1, -1 so possible value of xy = 0, 1, - 1
    If x = 2, y can be 0, 1, - 1 so xy = 0, 2, - 2
    If x= 3, y can be 0, 1, - 1 so xy = 0, 3, - 3
    If x = 4, y can be 0 xy =0

Thus possible values are -3, -2, - 1, 0, 1, 2, 3

The answer is option D.

But it is stated at many places that in GAMT unless specifically stated negative integers are not taken in to consideration, So should we follow the no negative numbers method ?

Hey, there is nothing like that. We can and in fact we should consider negative numbers if it is feasible and in line with the constraints given in the question. Could you please share the source of the above info. There could be some misinterpretation and that's why we would prefer to know the source.
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Asked: If x and y are integers and \(2\sqrt{x} + y^2 < 5\), then x*y can take how many different values ?

\(2\sqrt{x} + y^2 < 5\)
\(2\sqrt{x} < 5-y^2\)

Since
\(2\sqrt{x} > 0\)

\(0<=2\sqrt{x} < 5-y^2\)

If x=0; 5-y^2>0; y={-2,-1,0,1,2}; x*y=0
If x=1; 5-y^2 > 2; y^2<3; y={-1,0,1}; x*y={-1,0,1}
If x=2; ; y={-1,0,1}; x*y={-2,0,2}
If x=3; ; y={-1,0,1}; x*y={-3,0,3}
If x=4; ; 5-y^2> 4; y^2< 1; y=0; x*y=0

x*y = {-3,-2,-1,0,1,2,3} : 7 values

IMO D
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