Question

已知函數(shù)f(x)=lnx-a²x²+ax(a∈R).

(1)當(dāng)a=1時(shí),證明函數(shù)f(x)只有一個(gè)零點(diǎn);

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

Answer 1

解:(1)證明:當(dāng)a=1時(shí),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(shí),f′(x)＞0;當(dāng)x＞1時(shí),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(shí),函數(shù)f(x)取得最大值,其值為f(1)=ln1-1²+1=0.

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

(2)∵f(x)=lnx-a²x²+ax,其定義域?yàn)?0,+∞),

∴f′(x)=-2a²x+a==.

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

此時(shí)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(shí),f′(x)＜0(x＞0)等價(jià)于(2ax+1)(ax-1)＞0(x＞0),即x＞-.

此時(shí)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(shí)數(shù)a的取值范圍是(-∞,］∪［1,+∞).