52) Which one is true

(x+y)² < x²
x and y have diff. sign

I) xy < 0 as xy will be -ve

(II) y ( 2x+y) < 0 or 2xy + y² < 0

(x+y)² = x²+(y²+2xy) < x²
x² is +ve so (y²+2xy) has to be < 0

I did not get it yet.

ern5231 wrote:I did not get it yet.

When (x + y) ² < x ², then x ²+ (y ²+ 2 x y) < x ², or (y ²+ 2 x y) < 0, or y ( 2 x + y) < 0, or II can be true.

Your example to deny II is perfect, but this is the trap in the wordings that are too crucial on all many occasions. The wording "can be true" extends more freedom to our imaginations than the wordings like "must be true" extend.