Official Answer: С
Given that \(|x| = |y+1|\)
If \(x\) and \((y+1)\) both have same signs (either positive or negative), then \(x = y+1\) and \(x>y\), which makes \(x\) and \(y\) consecutive integers.
If \(x\) and \((y+1)\) both do not have the same signs (one is negative and the other is positive), then \(x = -(y+1)\) or \((x + y + 1) = 0\). In that case, either one could be greater in absolute value. S1 is not sufficient by itself.
If \(x^y = x! + |y|\), \(x\) and \(y\) both have to be positive because \(x! + |y|\) is always a positive integer. If \(y\) is negative, \(x^y\) can never be an integer. So for \(x^y\) to be an integer, \(y\) has to be positive. Let's see what values satisfy the equation:
If \(x = 1\), and \(y = 0\), \(x^y = x! + |y| = 1\).
If \(x = 2\), and \(y = 2\), \(x^y = x! + |y| = 4\).
So \(x\) and \(y\) have at least two possibilities. S2 is not sufficient by itself either.
Statement (1)+Statement (2):
Because of Statement (2), \(x\) and \(y\) have to be positive, in which case, by Statement (1) \(x > y\). Sufficient. Therefore, the answer is C.
I don't get it..Statement 2 gives you 2 solutions 1,0 and 2,2 so x and y are non-negative .. even combining with statement 2 you get 2 solutions.. so answer is E..