Question

已知常數(shù),函數(shù).
(1)討論在區(qū)間上的單調(diào)性;
(2)若存在兩個(gè)極值點(diǎn),且,求的取值范圍.

Answer 1

(1)詳見解析  (2)

試題分析:(1)首先對函數(shù)求導(dǎo)并化簡得到導(dǎo)函數(shù),導(dǎo)函數(shù)的分母恒大于0,分子為含參的二次函數(shù),故討論分子的符號,確定導(dǎo)函數(shù)符號得到原函數(shù)的單調(diào)性,即分和得到導(dǎo)函數(shù)分子大于0和小于0的解集進(jìn)而得到函數(shù)的單調(diào)性.
(2)利用第(1)可得到當(dāng)時(shí),導(dǎo)數(shù)等于0有兩個(gè)根,根據(jù)題意即為兩個(gè)極值點(diǎn),首先導(dǎo)函數(shù)等于0的兩個(gè)根必須在原函數(shù)的可行域內(nèi),把關(guān)于的表達(dá)式帶入,得到關(guān)于的不等式,然后利用導(dǎo)函數(shù)討論的取值范圍使得成立.即可解決該問題.
(1)對函數(shù)求導(dǎo)可得
,因?yàn)?img src="http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053405250879.png" style="vertical-align:middle;" />,所以當(dāng)時(shí),即時(shí),恒成立,則函數(shù)在單調(diào)遞增,當(dāng)時(shí), ,則函數(shù)在區(qū)間單調(diào)遞減,在單調(diào)遞增的.
(2)解:(1)對函數(shù)求導(dǎo)可得,因?yàn)?img src="http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053405250879.png" style="vertical-align:middle;" />,所以當(dāng)時(shí),即時(shí),恒成立,則函數(shù)在單調(diào)遞增,當(dāng)時(shí), ,則函數(shù)在區(qū)間單調(diào)遞減,在單調(diào)遞增的.
(2)函數(shù)的定義域?yàn)?img src="http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053405811768.png" style="vertical-align:middle;" />,由(1)可得當(dāng)時(shí),,則 ,即,則為函數(shù)的兩個(gè)極值點(diǎn),代入可得
=
令,令,由知: 當(dāng)時(shí),, 當(dāng)時(shí),,
當(dāng)時(shí),,對求導(dǎo)可得,所以函數(shù)在上單調(diào)遞減,則,即不符合題意.
當(dāng)時(shí), ,對求導(dǎo)可得,所以函數(shù)在上單調(diào)遞減,則,即恒成立,
綜上的取值范圍為.