Question

【題目】函數(shù).

(1)討論函數(shù)的單調(diào)性;

(2)當(dāng)時,方程在區(qū)間內(nèi)有唯一實數(shù)解,求實數(shù)的取值范圍.

Answer 1

【答案】(1)答案見解析；(2) 或.

【解析】試題分析：

(1)求導(dǎo),分類討論得f(x)的單調(diào)區(qū)間即可；

(2)問題轉(zhuǎn)化為有唯一實數(shù)解；構(gòu)造函數(shù),求導(dǎo)得或.

試題解析：

(1) ,

(i)當(dāng)時, ,令,得,令,得,

函數(shù)f(x)在上單調(diào)遞增, 上單調(diào)遞減;

(ⅱ)當(dāng)時,令,得

令,得,令,得,

函數(shù)f(x)在和上單調(diào)遞增, 上單調(diào)遞減;

(ⅲ)當(dāng)時, ,函數(shù)f(x)在上單調(diào)遞增;

(ⅳ)當(dāng)時,

令,得,令,得,

函數(shù)f(x)在和上單調(diào)遞增, 上單調(diào)遞減;

綜上所述:當(dāng)時,函數(shù)f(x)的單調(diào)遞增區(qū)間為,單調(diào)遞減區(qū)間為;

當(dāng)時,函數(shù)f(x)的單調(diào)遞增區(qū)間為和,單調(diào)遞減區(qū)間為;

當(dāng)時,函數(shù)f(x)的單調(diào)遞增區(qū)間為;

當(dāng)時,函數(shù)f(x)的單調(diào)遞增區(qū)間為和,單調(diào)遞減區(qū)間為

(2)當(dāng)時, ,由,得,又,所以,

要使方程在區(qū)間上有唯一實數(shù)解,只需有唯一實數(shù)解；

令,∴,

由得得,

∴在區(qū)間上是增函數(shù),在區(qū)間上是減函數(shù).

,

故或