设函数f(x)=(x+a)lnx-x+a. (Ⅰ)设g(x)=f'(x),求g(x)函数的单调区间; (Ⅱ)若a≥
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(Ⅰ)g(x)的定义域是(0,+∞)∵g(x)=f'(x)=
+lnx,a x
∴g'(x)=-
+a x2
=1 x
,…(2分)x-a x2
(1)当a≤0时,g'(x)>0,∴g(x)在(0,+∞)上单调递增,
故g(x)单调区间是(0,+∞)…(4分)
(2)当a>0时,g'(x)>0,∴g(x)在(a,+∞)上单调递增,
再由g'(x)<0得g(x)在(0,a)上单调递减.
g(x)的单调区间是(0,a)与(a,+∞)…(7分)
(Ⅱ)由题(Ⅰ)知,g(x)在x=a时取到最小值,且为g(a)=
+lna=1+lna.…(9分)a a
∵a≥
,∴lna≥-1,∴g(a)≥0.1 e
∴f'(x)≥g(a)≥0.f(x)在(0,+∞)上单调递增,…(11分)
∵f(e)=(e+a)lne-e+a=2a>0,f(
)=(1 e
+a)ln1 e
-1 e
+a=-1 e
<0,,2 e
∴f(x)在(
,e)内有零点.…(13分)1 e
故函数f(x)=(x+a)lnx-x+a的零点个数为1.…(14分)