考點:分段函數(shù)的應(yīng)用
專題:函數(shù)的性質(zhì)及應(yīng)用
分析:若函數(shù)f(x)=
,若存在兩個不相等的實數(shù)x
1,x
2,使得f(x
1)=f(x
2),則函數(shù)不為單調(diào)函數(shù),進(jìn)而根據(jù)二次函數(shù)和一次函數(shù)的圖象和性質(zhì),可得實數(shù)a的取值范圍.
解答:
解:若存在兩個不相等的實數(shù)x
1,x
2,使得f(x
1)=f(x
2),
則函數(shù)f(x)函數(shù)不為單調(diào)函數(shù),
由y=-x
2+2ax在(a,+∞)上為減函數(shù),
故函數(shù)f(x)=
為單調(diào)函數(shù)時只能是減函數(shù),
此時a<0,
故函數(shù)f(x)函數(shù)不為單調(diào)函數(shù)時,a≥0,
即實數(shù)a的取值范圍為:[0,+∞),
故答案為:[0,+∞)
點評:本題考查的知識點是分段函數(shù)的應(yīng)用,其中根據(jù)已知分析出函數(shù)f(x)函數(shù)不為單調(diào)函數(shù),是解答的關(guān)鍵.