Question

已知函數(shù)f(x)=ln(x-2),其中a是不等于0的常數(shù),e為自然對數(shù)的底數(shù).

(1)當(dāng)a＞0時,求f(x)的單調(diào)區(qū)間;

(2)若f(x)在x₀處取得極值,且x₀［e+2,e²+2］,而f(x)≥0在［e+2,e²+2］上恒成立,求實數(shù)a的取值范圍.

Answer 1

解:由已知得函數(shù)f(x)=ln(x-2)的定義域為(2,+∞),f′(x)= ==［(x-1)²-(a+1)］.

(1)當(dāng)a＞0時,f(x)=(x-1+)(x-1-),∵x＞2,∴x-1+＞0,a(x-2)＞0.

①當(dāng)x＞1+時,f′(x)＜0,f(x)為減函數(shù).

②當(dāng)2＜x＜1+時,f′(x)＞0,f(x)為增函數(shù),

∴當(dāng)a＞0時,f(x)的單調(diào)遞增區(qū)間為(2,1+］,單調(diào)遞減區(qū)間為［1+,+∞).

(2)由(1)知當(dāng)a＜0時,f′(x)=＞0,f(x)遞增無極值,

∴由已知條件f(x)在x₀處有極值知a＞0,并且x₀=1+,∵x₀［e+2,e²+2］且e+2＞2,

∴f(x)在［e+2,e²+2］上單調(diào).

①當(dāng)［e+2,e²+2］為增區(qū)間時,f(x)≥0恒成立.

則有

②當(dāng)［e+2,e²+2］為減區(qū)間時,f(x)≥0恒成立.

則有

此時a.

綜上所述,實數(shù)a的取值范圍為a＞e⁴+2e².