M02-19

Official Solution:


Is \(x \gt y^2\)?

(1) \(x \gt y+5\).

Rewrite as \(x-y \gt 5\). This is clearly insufficient, for example: if \(x=1\) and \(y=-10\), then the answer is NO, but if \(x=10\) and \(y=1\), then the answer is YES. We have two different answers, so this is not sufficient.

(2) \(x^2-y^2=0\).

Rewrite as \((x-y)(x+y)=0\): so either \(x-y=0\) or \(x+y=0\). This is also insufficient: if \(x=1\) and \(y=1\), then the answer is NO, but if \(x=\frac{1}{2}\) and \(y=\frac{1}{2}\), then the answer is YES. We have two different answers, so this is not sufficient.

(1)+(2) Since we know from (1) that \(x-y \gt 5\), we can conclude that \(x-y \neq 0\). Therefore, based on (2), we must have \(x+y=0\). This means that \(y=-x\), and we can rewrite the question as: is \(x \gt x^2\)?

Now, let's substitute \(y=-x\) in (1): \(x \gt -x + 5\). From this inequality, we can infer that \(x \gt \frac{5}{2}\). Since \(x \gt \frac{5}{2}\), it implies that \(x^2 \gt x\) (because for any \(x\) greater than 1, \(x^2 \gt x\)). Therefore, the answer to the question of whether \(x \gt y^2\) is NO. Sufficient.


Answer: C