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(1)a=e,或a=3e(2)
(1)求导得f′(x)=2(x﹣a)lnx+
![](https://img.yulucn.com/upload/9/19/919e4ffb06182e182036c62c7444744f_thumb.jpg)
=(x﹣a)(2lnx+1﹣
![](https://img.yulucn.com/upload/7/72/77235f1c24c6eed51ebc99b0f18b2150_thumb.jpg)
),
因为x=e是f(x)的极值点,
所以f′(e)=0
解得a=e或a=3e.
经检验,a=e或a=3e符合题意,
所以a=e,或a=3e
(2)①当0<x≤1时,对于任意的实数a,恒有f(x)≤0<4e
2 成立
②当1<x≤3e时,,由题意,首先有f(3e)=(3e﹣a)
2 ln3e≤4e
2 ,
解得
由(1)知f′(x)=2(x﹣a)lnx+
![](https://img.yulucn.com/upload/4/ad/4ad9513a3f2de6727cc0fb05bf222f89_thumb.jpg)
=(x﹣a)(2lnx+1﹣
![](https://img.yulucn.com/upload/f/c2/fc2cea8e31acebf2eb4bcb5cc44fc531_thumb.jpg)
),
令h(x)=2lnx+1﹣
![](https://img.yulucn.com/upload/f/c2/fc2cea8e31acebf2eb4bcb5cc44fc531_thumb.jpg)
,则h(1)=1﹣a<0,
h(a)=2lna>0且h(3e)=2ln3e+1﹣
![](https://img.yulucn.com/upload/e/fe/efed9ca7876077d1b4c5c47bc7cc7dab_thumb.jpg)
≥2ln3e+1﹣
![](https://img.yulucn.com/upload/5/17/517b9b4eecbd268c7a16660fd5f40ca0_thumb.jpg)
=2(ln3e﹣
![](https://img.yulucn.com/upload/d/9a/d9a1d79a75a8ae4004e00a9f2b5f341c_thumb.jpg)
)>0
又h(x)在(0,+∞)内单调递增,所以函数h(x)在在(0,+∞)内有唯一零点,记此零点为x
0 则1<x
0 <3e,1<x
0 <a,从而,当x∈(0,x
0 )时,f′(x)>0,
当x∈(x
0 ,a)时,f′(x)<0,
当x∈(a,+∞)时,f′(x)>0,即f(x)在(0,x
0 )内是增函数,
在(x
0 ,a)内是减函数,在(a,+∞)内是增函数
所以要使得对任意的x∈(0,3e],恒有f(x)≤4e
2 成立只要有
有h(x
0 )=2lnx
0 +1﹣
![](https://img.yulucn.com/upload/e/1e/e1e1a85433771a79b495f68bd1d836cc_thumb.jpg)
=0得a=2x
0 lnx
0 +x
0 ,将它代入
![](https://img.yulucn.com/upload/d/4c/d4cf1450ec10748b452024434672660c_thumb.jpg)
得4x
0 2 ln
3 x
0 ≤4e
2 又x
0 >1,注意到函数4x
2 ln
3 x在(1,+∞)上是增函数故1<x
0 ≤e
再由a=2x
0 lnx
0 +x
0 ,及函数2xlnx+x在(1,+∞)上是增函数,可得1<a≤3e
由f(3e)=(3e﹣a)
2 ln3e≤4e
2 解得
![](https://img.yulucn.com/upload/b/8f/b8f9195de1079f39abde55ccd42e47ac_thumb.jpg)
,
所以得
综上,a的取值范围为
1年前
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