1) x=y+1, so x^x is (y+1)^(y+1). In this case, x^x will clearly be bigger than y^y for y>0. For example, if y=4, then x=5 and 5^5 is bigger than 4^4. But if y is negative and even, y^y will be bigger. For example, if y=-2, then x=-1 x^x=-1 and y^y=1/4 and y^y is bigger. INSUFFICIENT.
2) x^y>x and x is positive. x is positive so we can divide both sides by x to get x^(y-1)>1. This is true for x=5, y=2 in which case x^x is bigger. And also true for y=5 x=2, in which case y^y is bigger.
1&2 together: The fact that x is positive and x^(y-1)>1 implies that x>1. If x=1, 1^(y-1) would be 1 for all values of y and hence not greater than 1. However, x cannot be 2 either, because then y would be 1, and 2^1 is not greater than 2. Thus the lowest value x could have is 3. If x=y+1, the lowest value y could have is 2. As discussed in the analysis of statement 1 above, x^x is always bigger than y^y if x=y+1 and y>0. SUFFICIENT
Ans: C
Large X/Y
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