oquiella wrote:2. Is (x+y)^2 < x^2+y^2?
(1) x^2 < y^2
(2) x<0<y
Please explain why not 1? throughly
(x+y)² < x² + y²
x² + y² + 2xy < x² + y²
2xy < 0
xy < 0.
xy<0 only if x and y have DIFFERENT SIGNS.
Question stem, rephrased:
Do x and y have different signs?
Statement 1: x² < y²
It's possible that x=1 and y=2, in which case x and y have the SAME SIGN.
It's possible that x=-1 and y=2, in which case x and y have DIFFERENT SIGNS.
INSUFFICIENT.
Statement 2: x < 0 < y
Since x is negative and y is positive, x and y have different signs.
SUFFICIENT.
The correct answer is
B.
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