2?x231?x??2,??? , h?(x)??0 ?h(x)?h(2)?ln2??0 即lnx?(x2?1)2x44
ab?ax227. (Ⅰ)f??x??2 2x?b??由f??1??0及f?1??2得,a?4,b?1
?2k?f??x0??4?2?1?x0???2?? ?21?x0??1设
11?x02?1??t,t??0,1?得k???,4?
?2?4?4x2(Ⅱ)f??x??,令f??x??0?x???1,1?
1?x22??f?x?的增区间为??1,1?,故当0?x1?x2?1时,
f?x2??f?x1??0. x2?x1即k?0,故x0???1,1? (法一)由于f?(x0)?f?(?x0）,故只需要证明x0??0,1?时结论成立
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(法二) 由已知:
?x1?x02022?1x0?12????x?1?x1x221?1x2?1??2? ①
下面以反证法证明结论:
假设x0?x2?x1,则x0?x1x2,
因为x0???1,1?,x1,x2??0,1?,所以0?1?x0?1?x1x2,
2又0??x120?1??x2?21?1x2?1??12??x22,故
1?x02022?1x0?1????x?1?x1x221?1x2?1??2?
与①式矛盾
假设x0?x1?x2,同理可得与①式矛盾
综上,有x1?x0?x2成立
?x1?x020?1x0?1????x?1?x1x221?1x2?1??2?
12x2?1?28.解:(1)??(x)??
x(x?1)2x(x?1)2?x?0,x?1,???(x)?0,增区间为(0,1)和(1,+?)
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(2)?f?(x)?111,?f?(x0)?,切线方程为y?lnx0?(x?x0)① xx0x0x设l与y?g(x)切于点(x1,e1),?g?(x)?ex,?ex1?1,?x1??lnx0, x0?l方程y?lnx011x??,② x0x0x0lnx01x?1?,?lnx0?0, x0x0x0?1x?1在区间(1,??)上单调递增, x?1由①②可得lnx0?1?由(1)知,?(x)?lnx?e2?1e2?3e?1?222又?(e)?lne??2?0, ??0,?(e)?lne?2e?1e?1e?1e?1由零点存在性定理,知方程?(x)?0必在区间(e,e)上有唯一的根,这个根就是x0,故在区间(1,??)上存在唯一的x0,使得直线l与曲线y?g(x)相切
21?x??1?lnx??1lnx'29. (1)函数f(x)定义域为?0,???,f?x??x, ??22xx由f'?x??0?x?1,当0?x?1时,f'?x??0,当x?1时,f'?x??0,
则f(x)在?0,1?上单增,在?1,???上单减,函数f(x)在x?1处取得唯一的极值.
?a?02??2?由题意得?1??a?1,故所求实数a的取值范围为?,1?
3a?1?a??3??3?x?1??1?lnx??k1?lnxk(2) 当x?1时,不等式f(x)?. ???k?x?1xx?1x令g(x)??x?1??1?lnx?,x'?x?1?,由题意,k?g(x)在?1,???恒成立.
?x??x?1??1?lnx??x'x?lnx??x?1??1?lnx????' g(x)??22xx令h?x??x?lnx?x?1?,则h'?x??1?1?0,当且仅当x?1时取等号. x所以h?x??x?lnx在?1,???上单调递增,h?x??h?1??1?0
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因此g(x)?'x?lnxh?x??2?0,则g(x)在?1,???上单调递增,g?x?min?g?1??2 2xx所以k?2,即实数k的取值范围为???,2?
2恒成立, x?11?lnx22x22即??lnx??1?1??1?,
xx?1x?1x?1x(3)由(2)知,当x?1时,不等式f(x)?令x?k?k?1?,k?N,则有ln??k?k?1????1??21??1?1?2???.
k?k?1??kk?1???1??11?,ln2?3?1?2?????? 2??23?分别令k?1,2,3,?,n,n?N?则有ln?1?2??1?2?1?1??1,?,ln?nn?1?1?2??????将这n个不等式左右两边分别相加,则得 ???nn?1?1?2?222? ln?1?2?3???nn?1?n?21??n?2???????n?1?n?1?故1?2?3???n222?n?1??en?2?2n?1,从而?(n?1)!??(n?1)e2n?2?2n?1.n?N?
30.解:(1)由题意知,f(x)的定义域为(0,??),
112(x?)2?b?b2x?2x?b22 (x?0) f'(x)?2x?2???xxx1?当b?时, f?(x)?0,函数f(x)在定义域(0,??)上单调递增
22(2)当b?0时f?(x)?0有两个不同解,
x1?11?2b11?2b ?, x2??222211?2b??0?(0,??)，舍去, 2211?2b??1?(0,??), 22x1? 而x2?此时 f?(x),f(x)随x在定义域上的变化情况如下表:
x (0,x2) x2 第24页，共29页
(x2，??) f?(x) f(x) ? 减 0 极小值 ? 增 由此表可知:?b?0时,f(x)有惟一极小值点, x?11?2b, ?22(3)由(2)可知当b??1时,函数f(x)?(x?1)2?lnx,
此时f(x)有惟一极小值点 x?11?2b1?3 ??222且x?(0,1?31?3)时，f'(x)?0, f(x)在(0,)为减函数 22141?3??，n32111
?恒有 f(1)?f(1?)，即恒有 0?2?ln(1?)nnn1?当 n?3 时恒有ln(n?1)?lnn ?2 成立n?当 n?3 时， 0? 1?1?令函数h(x)?(x?1)?lnx （x?0)
则 h'(x)?1?1x?1 ?xx?x?1 时，h'(x)?0 ，又h(x)在x?1处连续?x?[1,??)时h(x)为增函数1111?n?3 时 1?1? ?h(1?)?h(1) 即 ?ln(1?)?0nnnn 11?ln(n?1)?lnn?ln(1?)?nn11综上述可知 n?3 时恒有 ?ln(n?1)?lnn?2nn14分 31. (Ⅰ) 由原式?1?1lnx??b, ················ xx1lnx令g(x)?1??,可得g(x)在?0,1?上递减,
xx在?1,???上递增,所以g(x)min?g(1)?0
即b?0 ···············3分 (Ⅱ)f?(x)?2ax?lnx,(x?0)
令f?(x)?0,得2a?lnxlnx1当x?e时设h(x)?h(x)?maxx,x,e
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