So if the proposition p: function y=c2 is a true proposition, then 0

If the proposition q: when x∈, the function f (x) = x+1/x >; 1/c holds because the function f (x) = x+1/x >; =2, function f(x)=2 if and only if x= 1/x and x= 1, so when x∈, function f (x) ∈ [2,5/2] > 1/c, so 1/c < 2, so c> 1/2.

Because P or Q is a true proposition and P and Q are false propositions, one of P or Q is a true proposition and the other is a false proposition.

If p is true and q is false, then 0

If p is a false proposition and q is a true proposition, then c; = 1

Therefore, to sum up, the range of: is 0.

Quotation: 370 1 16