已知函数f(x)=2x^2+ax+b/ x^2+1的值域为〔1,3〕求a,b的值因为y=2x²+ax+b/ x²+1,所以(y-2)x²-ax+y-b=0(1)当y-2≠0时因为x∈R,Δ≥0,即a²-4(y-b)(y-2)≥0而4y²-4(2+b)y+8b-a²≤0又因为1≤y≤3
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