2011年—2019年高考全国卷(1卷、2卷、3卷)理科数学试题分类汇编——8.函数与导数

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(1)求a;

(2)证明:f(x)存在唯一的极大值点x0,且e?2?f(x0)?2?2.

解析:(1)法一:由题知:f(x)?x?ax?a?lnx??x?0?,且f(x)?0 , 所以a?x?1??lnx?0,

lnxlnx;当x??1,???时,a?;当x?1时,a?x?1??lnx?0成立. x?1x?11x?1令g?x??x?1?lnx,g'?x??1??,

xxlnx当x??0,1?时,g'?x??0,g?x?递减,g?x??g?1??0,所以:x?1?lnx,即:所以a?1; ?1,

x?1lnx当x??1,???时,g'?x??0,g?x?递增,g?x??g?1??0,所以:x?1?lnx,即:?1.

x?1即当x??0,1?时,a?所以,a?1. 综上,a?1.

法二:洛必达法则:由题知:f(x)?x?ax?a?lnx??x?0?,且f(x)?0 ,所以:a?x?1??lnx?0. 即当x??0,1?时,a?lnxlnx;当x??1,???时,a?; x?1x?1当x?1时,a?x?1??lnx?0成立.

11x?1??lnx1??lnx?lnxx令g?x??,g'?x??x. ?22x?1?x?1??x?1?令h?x??1?1111?x?lnx,h'?x??2??2. xxxx当x??0,1?时,h'?x??0,h?x?递增,h?x??h?1??0;

lnx?'?lnx1所以g'?x??0,g?x?递减,g?x??lim?lim?lim?1,所以:a?1;

x?1x?1x?1?x?1?'x?1x当x??1,???时,h'?x??0,h?x?递减,h?x??h?1??0; 所以g'?x??0,g?x?递减,g?x??limx?1?lnx?'?lim1?1,所以:a?1. lnx?limx?1x?1?x?1?'x?1x1故a?1.

(2)由(1)知:f(x)?x?x?1?lnx?,f'?x??2x?2?lnx,x?2?lnx,设??x??2则?'?x??2?x.

?1??1?x?0,x?,???'x?0??????时,?'?x??0. 当时,;当

?2??2?广东省中山一中,朱欢收集整理,欢迎交流

所以??x?在?0,2?递减,在?2,???递增.

??1???1????2?e???0,??2??0,??1??0,所以??x?在?0,2?有唯一零点x0,在?2,???有唯一零点1, 又

?1?????1???1???且当x??0,x0?时,??x??0;当x??x0,1?时,??x??0; 当x??1,???时,??x??0.

又f'?x????x?,所以x?x0是f(x)的唯一极大值点. 由f'?x0??0得lnx0?2?x0?1?,故f?x0??x0?1?x0?. 由x0??0,1?得f?x0??14.

?1?1?2?1因为x?x0是f(x)在?0,1?的唯一极大值点,由e??0,1?,fe?0得f?x0??fe?e

????所以e

?2?f(x0)?2?2.

(2017·新课标Ⅲ,)21.已知函数f?x??x?1?alnx. (1)若f?x?…0 ,求a的值;

(2)设m为整数,且对于任意正整数n,?1+??1??1???1+?2??22??1??1+n??m,求m最小值. ?2?解析:(1)f(x)?x?1?alnx,x?0,则f?(x)?1?ax?a?,且f(1)?0, xx当a?0时,f??x??0,f?x?在?0,???上单调递增,所以0?x?1时,f?x??0,不满足题意; 当a?0时,

当0?x?a时,f?(x)?0,则f(x)在(0,a)上单调递减; 当x?a时,f?(x)?0,则f(x)在(a,??)上单调递增.

①若a?1,f(x)在(a,1)上单调递增所以当x?(a,1)时f(x)?f(1)?0矛盾; ②若a?1,f(x)在(1,a)上单调递减所以当x?(1,a)时f(x)?f(1)?0矛盾;

③若a?1,f(x)在(0,1)上单调递减,在(1,??)上单调递增所以f(x)…f(1)?0满足题意. 综上所述a?1.

(2)由(1)知当x??1,???时,x?1?lnx?0, 令x?1?11?ln1?得,?n2n?2?1??n, ?2广东省中山一中,朱欢收集整理,欢迎交流

1?1?1111?1???从而ln?1???ln?1?2??...?ln?1?n???2?...?n?1?n?1,

22?2??2??2?221??1??1??故?1???1?2?...?1?n??e. ?2??2??2?1??1??1??而?1???1?2??1?3??2,所以所以m的最小值为3. ?2??2??2?

(2016·新课标Ⅰ,12)已知函数f(x)?(x?2)e?a(x?1)有两个零点.

(Ⅰ)求a的取值范围;(Ⅱ)设x1,x2是f(x)的两个零点,证明:x1?x2?2.

由已知得:f'?x???x?1?ex?2a?x?1???x?1?ex?2a

x2解析:⑴

??① 若a?0,那么f?x??0??x?2?ex?0?x?2,f?x?只有唯一的零点x?2,不合题意; ② 若a?0,那么ex?2a?ex?0,

所以当x?1时,f'?x??0,f?x?单调递增;当x?1时,f'?x??0,f?x?单调递减; 即:

x

???,1? ? 1 ?1,??? ? f'?x? f?x? 0 极小值 ↓ ↑ 故f?x?在?1,???上至多一个零点,在???,1?上至多一个零点 由于f?2??a?0,f?1???e?0,则f?2?f?1??0, 根据零点存在性定理,f?x?在?1,2?上有且仅有一个零点. 而当x?1时,ex?e,x?2??1?0,

故f?x???x?2?ex?a?x?1??e?x?2??a?x?1??a?x?1??e?x?1??e

222?e?e2?4ae?e?e2?4ae?1,t2??1,则f?x??0的两根t1? t1?t2,因为a?0,故当x?t12a2a或x?t2时,a?x?1??e?x?1??e?0 因此,当x?1且x?t1时,f?x??0

又f?1???e?0,根据零点存在性定理,f?x?在???,1?有且只有一个零点. 此时,f?x?在R上有且只有两个零点,满足题意.

2e③ 若??a?0,则ln??2a??lne?1,

2当x?ln??2a?时,x?1?ln??2a??1?0,ex?2a?e即f'?x???x?1?ex?2a?0,f?x?单调递增; 当ln??2a??x?1时,x?1?0,ex?2a?e调递减;

ln??2a?ln??2a??2a?0,

???2a?0,即f'?x??xa???1?e?x?2?0,f?x?单

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当x?1时,x?1?0,ex?2a?e即:

x ln??2a??2a?0,即f'?x??0,f?x?单调递增.

???,ln??2a?? + ↑ ln??2a? 0 极大值 ?ln??2a?,1? - ↓ 1 0 极小值 ?1,??? + ↑ f'?x? f?x? 而极大值 f??ln??2a?????2a??ln??2a??2???a??ln??2a??1???a??ln??2a??2???1?0

2?2?故当x≤那么f?x?≤f?1时,f?x?在x?ln??2a?处取到最大值f??ln??2a????0恒成立,?ln??2a???,即f?x??0无解

而当x?1时,f?x?单调递增,至多一个零点 此时f?x?在R上至多一个零点,不合题意.

e④ 若a??,那么ln??2a??1

2当x?1?ln??2a?时,x?1?0,ex?2a?eln??2a?ln??2a??2a?0,即f'?x??0,f?x?单调递增 ?2a?0,即f'?x??0,f?x?单调递增

当x?1?ln??2a?时,x?1?0,ex?2a?e又f?x?在x?1处有意义,故f?x?在R上单调递增,此时至多一个零点,不合题意.

e⑤ 若a??,则ln??2a??1

2当x?1时,x?1?0,ex?2a?e1?2a?eln??2a??2a?0,即f'?x??0,f?x?单调递增

当1?x?ln??2a?时,x?1?0,ex?2a?eln??2a??2a?0,即f'?x??0,f?x?单调递减

ln??2a?当x?ln??2a?时,x?1?ln??2a??1?0,ex?2a?e?2a?0,即f'?x??0,

f?x?单调递增 即:

x ???,1? + ↑ 1 0 极大值 ?1,ln??2a?? - ↓ ln??2a? 0 极小值 ?ln??2a?,??? + ↑ f'?x? f?x? 故当x≤ln??2a?时,f?x?在x?1处取到最大值f?1???e,那么f?x?≤?e?0恒成立,即

f?x??0无解

当x?ln??2a?时,f?x?单调递增,至多一个零点,此时f?x?在R上至多一个零点,不合题意. 综上所述,当且仅当a?0时符合题意,即a的取值范围为?0,???. ⑵ 由已知得:f?x1??f?x2??0,不难发现x1?1,x2?1,

x1?2?ex?故可整理得:?a?2?x1?1?1?x2?2?ex?2?x2?1?2,

?x?2?exg?x??,则g?x1??g?x2? 2?x?1?广东省中山一中,朱欢收集整理,欢迎交流

?x?2??1x,当x?1时,g'x?0,gx单调递减;当x?1时,g'x?0,gx单调递增.

g'?x??e????????3x?1??设m?0,构造代数式:

2g?1?m??g?1?m??m?11?m?m?11?m1?m1?m?m?12m?e?e?2e?e?1? m2m2m?m?1?m?12m2m2e2m?0,故h?m?单调递增,有h?m??h?0??0. 设h?m??e?1,m?0,则h'?m??2m?1?m?1?因此,对于任意的m?0,g?1?m??g?1?m?.

由g?x1??g?x2?可知x1、x2不可能在g?x?的同一个单调区间上,不妨设x1?x2,则必有x1?1?x2 令m?1?x1?0,则有g??1??1?x1????g??1??1?x1????g?2?x1??g?x1??g?x2? 而2?x1?1,x2?1,g?x?在?1,???上单调递增,因此:g?2?x1??g?x2??2?x1?x2 整理得:x1?x2?2.

(2016·新课标Ⅱ,21)(Ⅰ)讨论函数f(x)?x?2xx e 的单调性,并证明当x>0时,(x?2)e?x?2?0;

x?2ex?ax?a (Ⅱ)证明:当a?[0,1)时,函数g(x)=求函数h(a)(x?0)有最小值.设g (x)的最小值为h(a),

x2的值域. ?x?24?x2ex????2???2,???时,f??x??0,∴f?x?在证明:⑴f??x??e?,∵当x????,?x?2?x?2?2??x?2?2??x?2xe?f?0?=?1,∴?x?2?ex?x?2?0. ???,?2?和??2,???上单调递增,∴x?0时,

x?2x?2xx2xxx(x?2)(?e?a)e?a?x?2x?e?ax?a?x?xe?2e?ax?2a??x?2?1?,由(1)⑵ g??x??,a??0,?x3x4x4x?2xt?2t知,当x?0时,f?x???e的值域为??1,???,只有一解.使得?e??a,t??0,2?,当

x?2t?2,g(x)单调减;当x?(t,??)时g?(x)?0,g(x)单调增,x?(0,t)时,g?(x)?0t?2ttte?t?1?e??et?t?1?e?a?t?1?etett?2?0,,记k?t??,在t??0,2?时,k??t??h?a????2t2t2t?2t?2?t?2?x?1e2?∴k?t?单调递增,∴h?a??k?t???,?.

?24?(2015·新课标Ⅰ,12)已知函数f(x)?x3?ax?1,g(x)??lnx. 4(Ⅰ)当a为何值时,x轴为曲线y?f(x)的切线;

(Ⅱ)用min{m,n}表示m,n中的最小值,设函数h(x)?min{f(x),g(x)}(x?0),讨论h(x)零点的个数.

2解:(Ⅰ)f?(x)?3x?a,若x轴为曲线y?f(x)的切线,则切点(x0,0)满足f?(x0)?0,f(x0)?0,

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