Question

已知函数f(x)=1/2ax^2+2x,g(x)lnx

Answer 1

h(x)=xg(x)-2x=xln(x)-2x,x>0.
h'(x)=ln(x)+1-2=ln(x)-1,
0<x<e时,h'(x)<0,h(x)单调递减.
x>e时,h'(x)>0,h(x)单调递增.
f(x)=ax^2/2+2x,
x>=1时,f'(x)=ax+2>=0.
x>=1,a>=0时显然满足要求.
x>=1,a<0时,
f'(x)=ax+2>=0,
ax>=-2,
ax>=-2>=-2x,a>=-2.
a的取值范围是a>=-2.
g(x)/x=ln(x)/x=f'(x)-(2a+1)=ax+2-(2a+1)=ax-2a+1,x>0.
ln(x)=ax^2+(1-2a)x,
s(x)=ln(x)-ax^2+(2a-1)x,
1/e<x<e,a>0.
s'(x)=1/x-2ax+2a-1=[-2ax^2+(2a-1)x+1]/x=[-2ax-1][x-1]/x=(2ax+1)(1-x)/x,
1/e<x<1时,s'(x)>0,s(x)单调递增.s(1/e)<s(x)<s(1).s(x)在1/e<x<1上至多有1个实根.
e>x>1时,s'(x)<0,s(x)单调递减.s(1)>s(x)>s(e).s(x)在e>x>1上至多有1个实根.
s(1/e)=-1-a/e^2+(2a-1)/e=[(2a-1)e-a-e^2]/e^2=[a^2-(a-e)^2-a-e]/e^2.
s(e)=1-ae^2+(2a-1)e=1-e+ae(2-e)<0.
s(1)=a-1.
要使得s(x)在1/e<x<e上有2个不同的实根,则必须,s(1)>0,s(1/e)<0.
也即,
a>1,
0>(2a-1)e-a-e^2=a(2e-1)-e-e^2,
e+e^2>a(2e-1),
a<(e+e^2)/(2e-1).
(e+e^2)/(2e-1)>(e+e)/(2e-1)>(2e-1)/(2e-1)=1.
1<a<(e+e^2)/(2e-1)时,方程g(x)/x=f'(x)-(2a+1)在区间(1/e,e)内有且只有2个不相等的实数根

Answer 2

(1)y=xlnx-2x
y'=lnx+1-2=lnx-1
令y'=0
x=e
0
=0在[1,+无穷）上恒成立
1/2ax^2+2x>=0
1/2ax^2>=-2x
a>=-4/x
所以
a>=0
(3)lnx/x=ax+2-2a-1
lnx=ax^2-(1-2a)x
ax^2+(1-2a)x-lnx=0
令F(x)=
ax^2+(1-2a)x-lnx
F'(x)=2ax+1-2a-1/x
令F'(x)=0
x=2a-1/2a
因为x>0
所以2a>1
a>1/2
可知x=2a-1/2a是F(x)的极小值点
若有且只有2个不相等的实数根
只需
F（2a-1/2a）<0
后面自己解吧
打这么多
累死了

Answer 3

h(x)=xg(x)-2x=xln(x)-2x,x>0.
h'(x)=ln(x)
1-2=ln(x)-1,
0
e时,h'(x)>0,h(x)单调递增.
f(x)=ax^2/2
2x,
x>=1时,f'(x)=ax
2>=0.
x>=1,a>=0时显然满足要求.
x>=1,a<0时,
f'(x)=ax
2>=0,
ax>=-2,
ax>=-2>=-2x,
a>=-2.
a的取值范围是a>=-2.
g(x)/x=ln(x)/x
=
f'(x)-(2a
1)=ax
2-(2a
1)=ax-2a
1,
x>0.
ln(x)=ax^2
(1-2a)x,
s(x)=ln(x)
-
ax^2
(2a
-
1)x,
1/e
0.
s'(x)=1/x
-
2ax
2a-1
=
[-2ax^2
(2a-1)x
1]/x
=
[-2ax-1][x-1]/x
=
(2ax
1)(1-x)/x,
1/e
0,
s(x)单调递增.
s(1/e)
x>1时,s'(x)<0,
s(x)单调递减.
s(1)>s(x)>s(e).s(x)在e>x>1上至多有1个实根.
s(1/e)=-1-a/e^2
(2a-1)/e
=
[(2a-1)e-a-e^2]/e^2
=
[a^2
-
(a-e)^2
-
a
-
e]/e^2
.
s(e)=1-ae^2
(2a-1)e=1-e
ae(2-e)<0.
s(1)=a-1.
要使得s(x)在1/e
0,
s(1/e)<0.
也即,
a>1,
0>(2a-1)e-a-e^2=a(2e-1)-e-e^2,
e
e^2>a(2e-1),
a<(e
e^2)/(2e-1).
(e
e^2)/(2e-1)>(e
e)/(2e-1)>(2e-1)/(2e-1)=1.
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