Question 296658
First, we need to see if statement I is true or not. Assume that {{{y=x}}}, if this is the case, then {{{x+1/x=x}}}. Subtract 'x' from both sides to get {{{1/x=0}}}. Since this equation is NEVER true for any 'x' values (let alone positive 'x' values), we can say that {{{y=x}}} is false where {{{y=x+1/x}}}. So this means that {{{y<>x}}} and that statement I is true.

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If {{{x>1}}}, then {{{1/x}}} is NOT an integer. So this consequently means that {{{x+1/x}}} is also NOT an integer. So statement II is false meaning that choices C and E can be ignored.

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Finally, because 'x' is an integer and 'x' is positive, we also know that {{{x<x+1/x}}} or that {{{x<y}}} (do a simple substitution here). Multiply both sides by 'x' to get {{{x^2<xy}}} (note: the sign doesn't flip because 'x' is positive).


So we've shown that if 'x' is an integer and 'x' is positive, then {{{x^2<xy}}} where {{{y=x+1/x}}}. So statement III is true.


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So because statements I and III are both true, we can say that the answer is choice D.