已知函數(shù)f(x)=lnx+x2.
(Ⅰ)若函數(shù)g(x)=f(x)-ax在其定義域內(nèi)為增函數(shù),求實(shí)數(shù)a的取值范圍;
(Ⅱ)在(Ⅰ)的條件下,若a>1,h(x)=e3x-3aexx∈[0,ln2],求h(x)的極小值;
(Ⅲ)設(shè)F(x)=2f(x)-3x2-kx(k∈R),若函數(shù)F(x)存在兩個(gè)零點(diǎn)m,n(0<m<n),且2x=m+n.問:函數(shù)F(x)在點(diǎn)(x,F(xiàn)(x))處的切線能否平行于x軸?若能,求出該切線方程;若不能,請說明理由.
【答案】
分析:(Ⅰ)先根據(jù)題意寫出:g(x)再求導(dǎo)數(shù),由題意知,g′(x)≥0,x∈(0,+∞)恒成立,即
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由此即可求得實(shí)數(shù)a的取值范圍;
(Ⅱ)由(Ⅰ)知
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,利用換元法令t=e
x,則t∈[1,2],則h(t)=t
3-3at,接下來利用導(dǎo)數(shù)研究此函數(shù)的單調(diào)性,從而得出h(x)的極小值;
(Ⅲ)對于能否問題,可先假設(shè)能,即設(shè)F(x)在(x
,F(xiàn)(x
))的切線平行于x軸,其中F(x)=2lnx-x
2-kx結(jié)合題意,列出方程組,證得函數(shù)
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在(0,1)上單調(diào)遞增,最后出現(xiàn)矛盾,說明假設(shè)不成立,即切線不能否平行于x軸.
解答:解:(Ⅰ)g(x)=f(x)-ax=lnx+x
2-ax,
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由題意知,g′(x)≥0,對任意的x∈(0,+∞)恒成立,即
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又∵x>0,
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,當(dāng)且僅當(dāng)
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時(shí)等號(hào)成立
∴
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,可得
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(Ⅱ)由(Ⅰ)知,
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,令t=e
x,則t∈[1,2],則
h(t)=t
3-3at,
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由h′(t)=0,得
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或
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(舍去),
∵
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,∴
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若
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,則h′(t)<0,h(t)單調(diào)遞減;若
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,則h′(t)>0,h(t)單調(diào)遞增
∴當(dāng)
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時(shí),h(t)取得極小值,極小值為
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(Ⅲ)設(shè)F(x)在(x
,F(xiàn)(x
))的切線平行于x軸,其中F(x)=2lnx-x
2-kx
結(jié)合題意,有
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①-②得
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所以
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,由④得
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所以
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設(shè)
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,⑤式變?yōu)?img src="http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181447942685081/SYS201310241814479426850021_DA/25.png">
設(shè)
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,
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所以函數(shù)
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在(0,1)上單調(diào)遞增,
因此,y<y|
u=1=0,即
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,也就是
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此式與⑤矛盾
所以F(x)在(x
,F(xiàn)(x
))的切線不能平行于x軸
點(diǎn)評(píng):此題是個(gè)難題.本題主要考查用導(dǎo)數(shù)法研究函數(shù)的單調(diào)性,基本思路是:當(dāng)函數(shù)為增函數(shù)時(shí),導(dǎo)數(shù)大于等于零;當(dāng)函數(shù)為減函數(shù)時(shí),導(dǎo)數(shù)小于等于零,根據(jù)解題要求選擇是否分離變量,體現(xiàn)了轉(zhuǎn)化的思想和分類討論以及數(shù)形結(jié)合的思想方法,同時(shí)考查了學(xué)生的靈活應(yīng)用知識(shí)分析解決問題的能力和計(jì)算能力.