Question

已知函數(shù)f(x)=lnx-a²x²+ax(a∈R).

(1)當(dāng)a=1時,證明函數(shù)f(x)只有一個零點;

(2)若函數(shù)f(x)在區(qū)間(1,+∞)上是減函數(shù),求實數(shù)a的取值范圍.

Answer 1

解:(1)證明:當(dāng)a=1時,f(x)=lnx-x²+x,其定義域是(0,+∞),

∴f′(x)=-2x+1=.

令f′(x)=0,即=0,解得x=或x=1.

∵x＞0,∴x=舍去.當(dāng)0＜x＜1時,f′(x)＞0;當(dāng)x＞1時,f′(x)＜0.

∴函數(shù)f(x)在區(qū)間(0,1)上單調(diào)遞增,在區(qū)間(1,+∞)上單調(diào)遞減.

∴當(dāng)x=1時,函數(shù)f(x)取得最大值,其值為f(1)=ln1-1²+1=0.

當(dāng)x≠1時,f(x)＜f(1),即f(x)＜0.∴函數(shù)f(x)只有一個零點.

(2)∵f(x)=lnx-a²x²+ax,其定義域為(0,+∞),

∴f′(x)=-2a²x+a==.

①當(dāng)a=0時,f′(x)=＞0,

∴f(x)在區(qū)間(0,+∞)上為增函數(shù),不合題意.

②當(dāng)a＞0時,f′(x)＜0(x＞0)等價于(2ax+1)(ax-1)＞0(x＞0),即x＞.

此時f(x)的單調(diào)遞減區(qū)間為(,+∞).

依題意,得解之,得a≥1.

③當(dāng)a＜0時,f′(x)＜0(x＞0)等價于(2ax+1)(ax-1)＞0(x＞0),即x＞-.

此時f(x)的單調(diào)遞減區(qū)間為(,+∞).

依題意,得解之,得a≤.

綜上所述,實數(shù)a的取值范圍是(-∞,］∪［1,+∞).