分析:(Ⅰ)討論(1)當(dāng)a=0時(shí),(2)當(dāng)a<0時(shí)(3)當(dāng)a>0時(shí),①當(dāng)x∈(0,
)時(shí),②當(dāng)x∈(
,+∞)時(shí)的情況,從而得出當(dāng)a>0時(shí),函數(shù)f(x)的單調(diào)減區(qū)間是(0,
),單調(diào)增區(qū)間為(
,+∞).
(Ⅱ)討論(1)當(dāng)a≤0時(shí),(2)當(dāng)a>0時(shí),①當(dāng)
≤1,②當(dāng)1<
<e,③當(dāng)
≥e情況,從而得出a=2.
解答:
解:函數(shù)f(x)的定義域是(0,+∞),f′(x)=
,
(Ⅰ)(1)當(dāng)a=0時(shí),f′(x)=-
<0,
故函數(shù)f(x)在(0,+∞)上單調(diào)遞減.
(2)當(dāng)a<0時(shí),f′(x)<0恒成立,
所以函數(shù)f(x)在(0,+∞)上單調(diào)遞減.
(3)當(dāng)a>0時(shí),令f′(x)=0,又因?yàn)閤>0,解得x=
.
①當(dāng)x∈(0,
)時(shí),f′(x)<0,
所以函數(shù)f(x)在(0,
)單調(diào)遞減.
②當(dāng)x∈(
,+∞)時(shí),f′(x)>0,
所以函數(shù)f(x)在(
,+∞)單調(diào)遞增.
綜上所述,當(dāng)a≤0時(shí),函數(shù)f(x)的單調(diào)減區(qū)間是(0,+∞),
當(dāng)a>0時(shí),函數(shù)f(x)的單調(diào)減區(qū)間是(0,
),單調(diào)增區(qū)間為(
,+∞).
(Ⅱ)(1)當(dāng)a≤0時(shí),由(Ⅰ)可知,f(x)在[1,e]上單調(diào)遞減,
所以f(x)的最小值為f(e)=1,解得a=
>0,舍去.
(2)當(dāng)a>0時(shí),由(Ⅰ)可知,
①當(dāng)
≤1,即a≥1時(shí),函數(shù)f(x)在[1,e]上單調(diào)遞增,
所以函數(shù)f(x)的最小值為f(1)=
a=1,解得a=2.
②當(dāng)1<
<e,即
<a<1時(shí),
函數(shù)f(x)在(1,
)上單調(diào)遞減,在(
,e)上單調(diào)遞增,
所以函數(shù)f(x)的最小值為f(
)=
+
lna=1,
解得a=e,舍去.
③當(dāng)
≥e,即0<a≤
時(shí),函數(shù)f(x)在[1,e]上單調(diào)遞減,
所以函數(shù)f(x)的最小值為f(e)=
ae
2-1=1,
得a=
,舍去.
綜上所述,a=2.