Question

函数f(x)=log_a(x²-ax+2)在区间(1,+∞)上恒为正值,则实数a的取值范围为

A.(1,2)                 B.(1,2］                  C.(0,1)∪(1,2)              D.(1,)

Answer 1

答案:B

解析：令a=2,则f(x)=log₂(x²-2x+2)=log₂［(x-1)²+1］,当x∈(1,+∞)时f(x)＞log₂1=0满足题意,排除A、C.假设a＞2,则x²-ax+2＞1在(1,+∞)上恒成立,即x²-ax+1＞0.令g(x)=x²-ax+1,∵＞1,

∴g(x)≥g()=1.故1＞0.解得-2＜a＜2与假设a＞2矛盾.∴可排除选项D.因此选B.