akhilsuhag wrote:Is |x²+y²| > |x²-y²| ?
(1) x > y
(2) x > 0
Alternate approach:
When both sides of an inequality are enclosed in absolute value symbols, we can SQUARE the inequality.
|x²+y²|² > |x²-y²|²
x� + y� + 2x²y² > x� + y� - 2x²y²
4x²y² > 0
(xy)² > 0.
In the resulting inequality, the left side will always be positive as long as xy≠0.
Question stem, rephrased:
Is xy ≠0?
Statement 1: x > y
Case 1: x=1 and y=0
In this case, xy=0, and the answer to the rephrased question stem is NO.
Case 2: x=2 and y=1
In this case, xy≠0, and the answer to the rephrased question stem is YES.
Since the answer is NO in Case 1 but YES in Case 2, INSUFFICIENT.
Cases 1 and 2 satisfy BOTH STATEMENTS.
Since the answer is NO in Case 1 but YES in Case 2, the two statements combined are INSUFFICIENT.
The correct answer is
E.
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