Question

函數f(x)=log_a(x²-ax+2)在區間(1,+∞)上恒為正值,則實數a的取值范圍為

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

Answer 1

答案:B

解析：令a=2,則f(x)=log₂(x²-2x+2)=log₂［(x-1)²+1］,當x∈(1,+∞)時f(x)＞log₂1=0滿足題意,排除A、C.假設a＞2,則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與假設a＞2矛盾.∴可排除選項D.因此選B.