2023届高考数学专项练习极值点偏移一题30问含答案.pdf

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1、导 数 30 问导 数 30 问已知函数f(x)=lnxx ，若f(x)=a有两个不同的零点x已知函数f(x)=lnxx ，若f(x)=a有两个不同的零点x1 1，x，x2 2，且x，且x1 1x0)g【解析】令g(x)=lnx-ax(aR,x0)g(x)=1x-a(x0,aR)若a0,则g(x)=1x-a(x0,aR)若a0,则g(x)0恒成立,g(x)在(0,+)上单调递增,此时g(x)不可能有两个零点;若a0,则当x(0,1a)时,g(x)0恒成立,g(x)在(0,+)上单调递增,此时g(x)不可能有两个零点;若a0,则当x(0,1a)时,g(x)0,g(x)单调递增,当x(1a,+)时

2、,g(x)0,g(x)单调递增,当x(1a,+)时,g(x)0,g(x)单调递减;当x=1a时,g(x)取得最大值,且 g(x)(x)0即0a1e时,g(1)=-a0 x(0,+),lnx0即0a1e时,g(1)=-a0 x(0,+),lnxxlnx=2ln x 2 x2 x g(x)2 x g(x)2 x-ax=a x-ax=a x(2a-x(2a-x)g(4a)g(4a2 2)a2a(2a-2a)=0 x)a2a(2a-2a)=0 x1 1(1,e),x(1,e),x2 2(1a,4a(1a,4a2 2),g(x),g(x1 1)=g(x)=g(x2 2)=0,此时g(x)有两个零点,符合

3、题意综上所述,a的取值范围是0a1e,且1x)=0,此时g(x)有两个零点,符合题意综上所述,a的取值范围是0a1e,且1x1 1e1axe1a2a2a【解析】f(x)=a有两个不同的零点x【解析】f(x)=a有两个不同的零点x1 1,x,x2 2,则有a0且0a0且0a2 2x x1 1+x x2 2,由对均不等式易证由对均不等式易证:x x1 1+x x2 22 2a a3.求证求证:x x1 1+x x2 22e2e【解析】【解析】0 0a ae e,由由(1 1)得得x x1 1+x x2 22 2a a2e2e,x x1 1+x x2 22e2e4.求证求证:x x1 1+x x2

4、22 2a a【解析】【解析】lnlnx x1 1-axax1 1=lnlnx x2 2-axax2 2，a a(x x2 2-x x1 1)=()=(lnlnx x2 2-lnlnx x1 1)=)=lnlnx x2 2x x1 1=2ln2lnx x2 2x x1 14 4x x2 2x x1 1-1 1 x x2 2x x1 1+1 1=4 4x x2 2-x x1 1 x x2 2+x x1 1a a(x x2 2-x x1 1)()(x x2 2+x x1 1)4 4(x x2 2-x x1 1)x x2 2+x x1 1整理得整理得x x1 1+x x2 22 2a a5.求证求证

5、:1 1x x1 1+1 1x x2 22 2a a【解析】【解析】a a(x x2 2-x x1 1)=()=(lnlnx x2 2-lnlnx x1 1)=)=lnlnx x2 2x x1 12 2a a6.求证求证:e e2 2x x1 1x x2 2(反向对均不等式反向对均不等式).【解析】【解析】e e2 22 2,f f x x1 1 =a af f x x2 2 =a a lnlnx x1 1x x1 1=a alnlnx x2 2x x2 2=a a lnlnx x1 1-axax1 1=0 0lnlnx x2 2-axax2 2=0 0 lnlnx x1 1=axax1 1l

6、nlnx x2 2=axax2 2 即证即证a a(x x1 1+x x2 2)2 2,即证即证x x1 1+x x2 22 2a a,由由(2 2)知成立知成立7.求证求证:x x1 1x x2 21 1a a2 2(反向对均不等式反向对均不等式).【解析】【解析】x x1 1x x2 2x x1 1x x2 2,x x2 2-x x1 1a a(x x2 2-x x1 1)x x1 1x x2 2,x x1 1x x2 21 1a a得证得证8.求证求证:3 3a a-e ex x1 1+x x2 2.【解析】法一：因为【解析】法一：因为x x2 2-x x1 1lnlnx x2 2-ln

7、lnx x1 1x x1 1+x x2 22 2,令令x x1 1=e e得得x x2 2-e eaxax2 2-1 10 0,令令x x2 2=e e得得e e-x x1 11 1-axax1 10 0,+得得:a a(x x2 22 2-x x2 21 1)+()+(aeae-3 3)x x2 2-(-(aeae-3 3)x x1 10 0整理得整理得a a(x x1 1+x x2 2)()(x x2 2-x x1 1)()(aeae-3 3)()(x x1 1-x x2 2),),x x1 1+x x2 23 3-aeaea a法二法二:证明证明:左边左边:0 0 x x1 1e ex

8、x2 2,axax1 1=lnlnx x1 1=lnlnx x1 1e e+1 12 2(x x1 1e e-1 1)x x1 1e e+1 1+1 1 axax2 21 1+(+(aeae-3 3)x x1 1+e e0 0,-得得x x1 1+x x2 23 3-aeaea a.右边同法一右边同法一9.求证：求证：x x1 1+x x2 2-2ln2lna aa a【解析】要证【解析】要证x x1 1+x x2 2-2ln2lna aa a,lnlnx x1 1=axax1 1lnlnx x2 2=axax2 2 要证要证lnlnx x1 1+lnlnx x2 2a a-2ln2lna a

9、a a，即证即证lnlnx x1 1x x2 21 1nana-2 2,即证即证x x1 1x x2 21 1-lnlna aa a【解析】【解析】x x1 1+x x2 21 1-lnlna aa a=1 1a a-lnlna aa a ，x x2 2-x x1 1lnlnx x2 2-lnlnx x1 1x x1 1+x x2 22 2,先令先令x x1 1=1 1a a,x x2 2-1 1a aaxax2 2+lnlna ax x2 2+1 1a a2 2，再令再令x x2 2=1 1a a，1 1a a-x x1 1-lnlna a-axax1 10 0，-axax2 21 1-(-

10、(lnlna a-1 1)x x1 1-lnlna aa a -2 2a a0 0,+得得:x x1 1+x x2 21 1-lnlna aa a.11.求证求证:lnlnx x1 1+lnlnx x2 21 1-lnlna a.【解析】【解析】lnlnx x1 1=axax1 1lnlnx x2 2=axax axax1 1+axax2 21 1-lnlna a，x x1 1+x x2 21 1-lnlna aa a由由(7 7)得证得证.12.求证求证:x x1 1x x2 2e ea a【解析】【解析】x x1 1x x2 2e ea a,即证即证lnlnx x1 1x x2 2lnln

11、e ea a.即即lnlnx x1 1+lnlnx x2 21 1-lnlna a,由由(8 8)得证得证.13.求证：求证：1 1x x1 1+1 1x x2 22 2e e【解析】法一：要证【解析】法一：要证1 1x x1 1+1 1x x2 22 2e e,令令1 1x x1 1=t t1 1，1 1x x2 2=t t2 2，由题意得由题意得lnlnx x1 1=axax1 1，lnlnx x2 2=axax2 2，代入则有代入则有lnln1 1t t1 1=a a1 1t t1 1，lnln1 1t t2 2=a a1 1t t2 2即即0 0=lnlnt t1 1+a at t1

12、1，0 0=lnlnt t2 2+a at t2 2，两式作差有两式作差有lnlnt t2 2-IntInt1 1-a at t1 1+a at t2 2=0 0，lnlnt t2 2-lnlnt t1 1t t2 2-t t1 1=a at t1 1t t2 21 1t t1 1t t2 2，a a1 1e e2 2 t t1 1+t t2 22 2 t t1 1t t2 22 2e e，即，即1 1x x1 1+1 1x x2 22 2e e法二：另解法二：另解:令令t t=1 1x x,则则t t1 1=1 1x x1 1,t t2 2=1 1x x2 2,有有0 0t t1 11 1e

13、 et t2 21 1(t t1 1t t2 22 2e e;由题意由题意lnlnt t1 1=-a at t1 1,lnlnt t2 2=x x1 1t t2 2,lnlnt t1 1-lnlnt t2 2=-=-a a(1 1t t1 1-1 1t t2 2)=)=a at t1 1-t t2 2t t1 1+t t2 2,即即lnlnt t1 1t t2 2=a at t1 1-t t2 2t t1 1+t t2 2;lnlnt t1 1+lnlnt t2 2=-=-a at t1 1+t t2 2t t1 1 t t2 2;lnln(t t1 1t t2 2)lnlnt t1 1t t

14、2 2=-=-t t1 1+t t2 2t t1 1-t t2 2即即lnln(t t1 1t t2 2)=-)=-t t2 2t t2 2+1 1t t1 1t t1 1-1 1 lnlnt t1 1t t2 2，又又lnlnt t1 1t t2 22 2(t t1 1-1 1t t2 2 -1 1)t t1 1t t2 2+1 1(0 0t t1 1t t2 21 1)lnln(t t1 1t t2 2)-)-2 2 t t1 1t t2 22 2 t t1 1t t2 2即即t t1 1+t t2 22 2e e,得证得证.14.求证：求证：2ln2lnx x1 1+lnlnx x2 2

15、e e【解析】易知【解析】易知x x2 2-x x1 1lnlnx x2 2-lnlnx x1 1=1 1a a，要证，要证2ln2lnx x1 1+lnlnx x2 2e e，即证即证2ln2lnx x1 1+lnlnx x2 2e e a a x x2 2-x x1 1 lnlnx x2 2-lnlnx x1 1，即即2 2axax1 1+axax2 2aeaex x2 2-x x1 1lnlnx x2 2-lnlnx x1 1，即证，即证2 2x x1 1+x x2 2e ex x2 2-x x1 1lnlnx x2 2-lnlnx x1 1，令令x x2 2x x1 1=t t(1 1

16、,+),+)，即只需证，即只需证lnlnt te e(t t-1 1)t t+2 2.令令g g(x x)=)=lnlnx x-3 3 x x2 2-1 1 x x2 2+4 4x x+1 1(x x1 1)，g g(x x)=)=(x x-1 1)4 4x x x x2 2+4 4x x+1 1 2 20 0，g g(x x)在在(1 1,+),+)单调递增，单调递增，g g(x x)g g(1 1)=)=0 0，lnlnt t-e e(t t-1 1)t t+2 23 3 t t2 2-1 1 t t2 2+4 4t t+1 1-e e(t t-1 1)t t+2 2=t t-1 1t t

17、2 2+4 4t t+1 1 (t t+2 2)(3 3-e e)t t2 2+(+(9 9-4e4e)t t+6 6-e e 0 0 原式得证原式得证.15.求证：求证：1 1lnlnx x1 1+1 1lnlnx x2 22 2aeae【解析】要证【解析】要证1 1lnlnx x1 1+1 1lnlnx x2 22 2aeae，证，证1 1x x1 1+1 1x x2 22 2a a2 2e e，由，由(4 4)得得1 1x x1 1+1 1x x2 22 2a a，又又aeae(0 0,1 1)故故1 1x x1 1+1 1x x2 22 2a a2 2e e，原式得证，原式得证.16.

18、求证：求证：x x1 1+1 1 x x2 2+1 1 3 3a a2 2-2 2a a+1 1【解析】要证【解析】要证 x x1 1+1 1 x x2 2+1 1 3 3a a2 2-2 2a a+1 1即证即证x x1 1+x x2 2+x x1 1x x2 23 3a a2 2-2 2a a，由由(7 7)知知x x1 1x x2 21 1a a2 2，由，由(9 9)知知x x1 1+x x2 2-2ln2lna aa a，-2ln2lna aa a1 1-1 1x x(x x1 1)-)-lnlnx x-1 1+1 1x x(x x1 1)-)-lnlna a-1 1+1 1a a(

19、a a1 1)得得-2ln2lna aa a2 2a a2 2-2 2a a，由，由+得得x x1 1+x x2 2+x x1 1x x2 22e2ea a2 2【解析】由【解析】由(2 2)知知x x1 1+x x2 22 2a a；由；由(1212)知知x x1 1x x2 2e ea a，则有则有x x2 21 1x x2 2+x x2 22 2x x1 1=x x1 1x x2 2x x1 1+x x2 2 2e2ea a2 218.求证：求证：x x1 1x x2 2aeae【解析】由题可知【解析】由题可知a a=lnlnx xx x ，易知，易知a a 0 0,1 1e e ，0

20、0 x x1 1e ex x2 2法一：由题可知法一：由题可知x x1 1x x2 2aeaelnlnx x1 1-lnlnx x2 2lnlna a+1 1a a x x1 1-x x2 2 lnlna a+1 1x x1 1-x x2 21 1a a+lnlna aa a 易知易知x x1 1-lnlna aa a ,x x1 1-x x2 21 1a a+lnlna aa a 法二：由题可得法二：由题可得x x1 1x x2 2aeae=lnlnx x2 2x x2 2 e ex x1 1e elnlnx x2 2,由由0 0 x x1 1e ee ex x1 1,满足满足.19.求证求

21、证:x x1 11 1-1 1-aeaea a【解析】法一：因为【解析】法一：因为x x1 1,x x2 2是是lnlnx x=axax的两个根的两个根,且且1 1x x1 1e ex x2 2,且且lnlnx x1 1=axax1 1lnlnx x2 2=axax2 2 只涉及到变量只涉及到变量x x1 1,故只用故只用lnlnx x1 1=axax1 1 a a=lnlnx x1 1x x1 1,要证要证x x1 11 1-axax axax2 21 1-2 2x x1 1+e e0 0 x x1 1lnlnx x1 1-2 2x x1 1+e e0 0,构造函数构造函数h h(x x)=

22、)=x xlnlnx x-2 2x x+e e,x x(1 1,e e),),则则h h(x x)=)=lnlnx x-1 10 0,h h(x x)在在(1 1,e e)上递减上递减,h h(x x)=)=h h(e e)=)=0 0,原不等式成立原不等式成立.法二法二:因为因为x xlnlnx xx x-1 1当且仅当当且仅当x x=1 1时等号成立时等号成立.x xe elnlnx xe ex xe e-1 1,当且仅当当且仅当x xe e=1 1时等号成立，时等号成立，即即x x(lnlnx x-1 1)x x-e e当且仅当当且仅当x x=e e时等号成立时等号成立,即即x xlnl

23、nx x2 2x x-e e当且仅当当且仅当x x=e e时等号成立时等号成立,1 1x x1 10 0,原不等式成立原不等式成立.20.求证求证:x x2 21 1+1 1-aeaea a;【解析】要证【解析】要证x x2 21 1+1 1-aeaea a，即证，即证axax2 2-1 11 1-aeae.即证即证axax2 22 2-2 2x x2 2+e e0 0 只需证只需证axax2 22 2=x x2 2lnlnx x2 22 2x x2 2-e e,f f(x x)=)=lnlnx xx x f f(x x)=)=1 1-lnlnx xx x2 2,f f(x x)在在(0 0,

24、e e)上单调递增上单调递增,在在(e e,+),+)上单调递减上单调递减.f f(x x)maxmax=f f(e e)=)=1 1e e,0 0a a1 1e e,且且1 1x x1 1e ee e,x x2 2lnlnx x2 22 2x x2 2-e e得证得证21.求证求证:x x2 2-x x1 12 2 1 1-aeaea a;【解析】法一：【解析】法一：f f(x x)=)=lnlnx xx x f f(x x)=)=1 1-lnlnx xx x2 2 f f(x x)在在(0 0,e e)上单调递增上单调递增,在在(e e,+),+)上单调递减上单调递减.f f(x x)ma

25、xmax=f f(e e)=)=1 1e e,0 0a a1 1e e,且且1 1x x1 1e ex x2 2,由由lnlnx xx x-1 1时时x x(0 0,1 1)()(1 1,+),+)恒成立恒成立,lnln1 1x x1 1x x-1 1,-,-lnlnx x1 1-1 1x x,由由axax=lnlnx x得得axax-1 1=lnlnx x-1 1=lnlnx xe e1 1-e ex x,x x1 10 0,上式化简为上式化简为axax2 21 1-2 2x x1 1+e e0 0,又又 a a0 0,=4 4-4 4aeae0 0,x x1 1=2 2 4 4-4 4ae

26、ae2 2a a x x1 1=1 1+1 1-aeaea a或或x x2 2=1 1-1 1-aeaea a,0 0 x x1 1e e,且且0 0a a2 2a a x x1 1+x x2 22 21 1a a.x x1 1+x x2 22 2-x x1 11 1a a-1 1-1 1-eaeaa a,x x1 1+x x2 2-2 2x x1 12 21 1-eaeaa a x x2 2-x x1 12 21 1-eaeaa a,即即x x2 2-x x1 12 2 1 1-aeaea a法二法二:若证若证x x2 2-x x1 12 2 1 1-aeaea a只需证只需证x x2 2-

27、x x1 1+x x1 1+x x2 2=2 2x x2 22 2+2 2 1 1-aeaea a即证即证axax2 2-1 11 1-aeae 即证即证axax2 22 2-2 2x x2 2+e e0 0 只需证只需证axax2 22 2=x x2 2lnlnx x2 22 2x x2 2-e e,由法一可知由法一可知1 1x x1 1e ee e,x x2 2lnlnx x2 22 2x x2 2-e e得证得证22.求证求证:x x2 2-x x1 11 1-aeaea a;【解析】由于【解析】由于2 2 1 1-aeaea a1 1-aeaea a,由由(2121)得证得证.23.求

28、证求证:x x2 2-x x1 11 1-aeae;【解析】因为【解析】因为0 0a ae e1 1,1 1a a1 1-aeae 1 1-aeae,由由(2222)得证得证.(从从2222问到问到2424问是一串连续放缩问是一串连续放缩,只要证明第一步放缩成立只要证明第一步放缩成立,即后面的放缩全部成立即后面的放缩全部成立,连续连续放缩也为后续题目提供了很好的思路放缩也为后续题目提供了很好的思路).24.求证求证:x x2 2-x x1 12 2e ea a-e e2 2【解析】放缩设【解析】放缩设(x x)=)=lnlnx x-2 2(x x-1 1)x x+1 1(x x1 1),),(

29、x x)=)=1 1x x-4 4(x x+1 1)2 2=(x x-1 1)2 2x x(x x+1 1)2 20 0恒成立恒成立,(x x)在在(1 1,+),+)单调递增单调递增,(x x)(1 1)=)=0 0,即即lnlnx x2 2(x x-1 1)x x+1 10 0(x x1 1)由由lnlnx x1 1=axax1 1lnlnx x2 2=axax2 2 ，可得，可得a a=lnlnx x2 2-lnlnx x1 1x x2 2-x x1 1=lnlnx x2 2x x1 1x x2 2-x x1 11 1x x2 2-x x1 1 2 2(x x2 2x x1 1-1 1)

30、x x2 2x x1 1+1 1=2 2x x2 2+x x1 1，从而从而x x2 22 2a a-x x1 1则则x x2 2-x x1 12 21 1a a-x x1 11 1-aeaea a要证明要证明x x2 2-x x1 12 2e ea a-e e2 2,只需证只需证x x2 2-x x1 12 2e ea a-e e2 2只需证只需证1 1-aeaea ae ea a-e e2 21 1a ae ea aaeae)0 0【解析】【解析】f f(x x)=)=1 1-lnlnx xx x2 2即证即证f f(x x1 1)+)+f f(x x2 2)=)=1 1-lnlnx x1

31、 1x x2 21 1+1 1-lnlnx x2 2x x2 22 20 0即即x x2 22 2x x2 21 11 1-lnlnx x1 1 +1 1-lnlnx x2 2 0 0设设x x2 2x x1 1=t t1 1,x x1 1=lnlnt ta a(t t-1 1),x x2 2=t tlnlnt ta a(t t-1 1),要证要证f f(x x1 1)+)+f f(x x2 2)=)=1 1-lnlnx x1 1x x2 21 1+1 1-lnlnx x2 2x x2 22 20 0,由由x x2 22 2x x2 21 11 1-lnlnx x1 1 +1 1-lnlnx

32、x2 2 0 0即即t t2 21 1-lnlnt tt t-1 1 +1 1-t tlnlnt tt t-1 1 0 0即即lnlnt t2 2即即lnlnt t2 2(t t-1 1)t t+1 1(此时飘带函数与所求相反此时飘带函数与所求相反)，故求导，故求导g g(t t)=)=lnlnt t-(t t2 2+1 1)()(t t-1 1)t t(t t+1 1),则则g g(t t)=)=-t t(t t-1 1)()(t t2 2+3 3t t-1 1)-)-1 1t t(t t+1 1)2 21 1故故g g(x x2 2x x1 1)(e e2 2-2 2)()(1 1-aea

33、e)【解析】由题已知【解析】由题已知:x x0 0,f f(x x)=)=1 1-lnlnx xx x2 2,令令f f(x x)=)=0 0时时,x x=e e,f f(x x)在在(0 0,e e)上单调递增上单调递增,在在(e e,+),+)上单调递减上单调递减,f f(e e)=)=1 1e e.f f(x x)=)=a a时有两个不同的零点时有两个不同的零点,则则a a0 0过极值点作与过极值点作与(2 2,0 0),(),(e e2 2,0 0)之间的直线之间的直线,g g1 1(x x)=)=1 1e e2 2-2e2e(x x-2 2)()(2 2x xe e),),g g2

34、2(x x)=)=1 1e e2 2-e e3 3(x x-e e2 2)()(e ex xe e2 2),),当当g g1 1(x x)=)=a a时时,x x3 3=aeae2 2-2 2aeae+2 2,当当g g2 2(x x)=)=a a时时,x x4 4=aeae2 2-aeae3 3+e e2 2,h h1 1(x x)=)=f f(x x)-)-g g1 1(x x)=)=lnlnx xx x -1 1e e2 2-2e2e(x x-2 2)=)=lnlnx x-1 1e e2 2-2e2e(x x2 2-2 2x x)x x(1 1x xe e)令令 1 1(x x)=)=l

35、nlnx x-1 1e e2 2-2e2e(x x2 2-2 2x x),),1 1(x x)=)=1 1x x-2 2e e2 2-2e2ex x+2 2e e2 2-2e2e=e e2 2-2 2e e-2 2x x2 2+2 2x xx x e e2 2-2e2e =-2 2x x2 2+2 2x x+e e2 2-2e2ex x e e2 2-2e2e y y=-=-2 2x x2 2+2 2x x+(+(e e2 2-2e2e)，对称轴，对称轴x x=-=-b b2 2a a=1 12 2,y ymaxmax=e e2 2-2 2e e-4 4(e e)=)=0 0,h h1 1(x

36、 x)0 0恒成立恒成立.f f(x x)g g1 1(x x)在在(1 1,e e)上恒成立上恒成立.f f(x x3 3)g g1 1(x x3 3)=)=f f(x x1 1),),由于由于f f(x x)在在(0 0,e e)上单调递增上单调递增,故故x x1 1x x3 3恒成立恒成立.令令h h2 2(x x)=)=f f(x x)-)-g g2 2(x x)=)=lnlnx xx x -1 1e e2 2-e e3 3(x x-e e2 2)=)=lnlnx x-1 1e e2 2-e e3 3x x2 2-xexe2 2 x x(e ex x)令令 2 2(x x)=)=lnl

37、nx x-1 1e e2 2-e e3 3(x x2 2-xexe2 2),),2 2(x x)=)=1 1x x+1 1e e3 3-e e2 2(2 2x x-e e2 2)=)=2 2x x2 2-e e2 2x x+e e3 3-e e2 2x x e e3 3-e e2 2 ，y y=2 2x x2 2-e e2 2x x+e e3 3-e e2 2,对称轴对称轴x x=-=-b b2 2a a=e e2 24 4,1 1e e2 24 40 0,2 2(x x)在在(e e,+),+)上单调递增上单调递增,2 2(x x)minmin=2 2(e e)=)=1 1+1 1e e3

38、3-e e2 2(e e2 2-e e3 3)=)=0 0,h h2 2(x x)0 0恒成立恒成立.f f(x x)g g2 2(x x)在在(e e,+),+)上恒成立上恒成立.f f(x x4 4)g g1 1(x x4 4)=)=f f(x x2 2),),f f(x x)在在(e e,+),+)上单调递减上单调递减,x x4 4()(1 1-aeae)()(e e2 2-2 2),),x x2 2-x x1 1x x3 3-x x4 4(1 1-aeae)()(e e2 2-2 2)27.求证求证:x x2 2-x x1 12 2(e e-2 2)1 1-aeae【解析】以【解析】以

39、(e e,1 1e e)为顶点为顶点,(,(2 2,0 0)和和(2 2e e-2 2,0 0)为函数与为函数与x x轴的交点轴的交点,拟合二次函数拟合二次函数y y=a a1 1(x x-2 2)()(x x-2e2e+2 2)代入代入(e e,1 1e e)有有a a1 1=-=-1 1e e(e e-1 1)2 2,y y=-=-1 1e e(e e-1 1)2 2(x x-2 2)()(x x-2e2e+2 2)=)=a a x x2 2-2 2exex+4 4e e-4 4=-=-aeae(e e-2 2)2 2,x x2 2-2 2exex+4 4e e-4 4+aeae(e e-

40、2 2)2 2=0 0解得解得x x3 3=2 2e e-2 2，x x4 4=2e2e+2 2 x x2 2-x x1 1x x4 4-x x3 3=2 2(e e-2 2)1 1-aeae28.求证求证:x x2 2-x x1 14e4e3 32 2-1 1-(-(2e2e2 2+1 1)a a,(,(3 32e2e3 32 2 a a1 1e e)【解析】由【解析】由f f(x x)=)=1 1-lnlnx xx x2 2,易知易知f f(x x)在在(1 1,0 0)和和(e e3 32 2,3 32e2e3 32 2 )处处切线切线l l1 1:y y=x x-1 1,l l2 2:

41、y y=-=-1 12e2e3 3(x x-1 1)+)+3 32e2e3 32 2 易知易知f f(x x)-)-l l1 10 0,f f(x x)-)-l l2 2x x3 3=a a+1 1和和x x2 2x x4 4=4e4e3 32 2-2 2aeae,x x2 2-x x1 1x x4 4-x x3 3e emm+1 1;【解析】证明【解析】证明:f f(x x)=)=a a有两个不同的零点有两个不同的零点x x1 1,x x2 2,则有则有a a0 0且且0 0a ae emm+1 1即证即证axax1 1+maxmax2 2mm+1 1由由lnlnx x1 1-axax1 1

42、=0 0lnlnx x2 2-axax2 2=0 0 作差得作差得 lnlnx x1 1-lnlnx x2 2 -a a x x1 1-x x2 2 =0 0,lnlnx x1 1-lnlnx x2 2x x1 1-x x2 2=a a,代入上式化简得代入上式化简得:(:(x x1 1+mxmx2 2)lnlnx x1 1-lnlnx x2 2x x1 1-x x2 2mm+1 1,令令x x1 1x x2 2=t t(0 0t t1 1+mm,即证即证t tlnlnt t+mmlnlnt tmtmt+t t-mm-1 1,令令h h(t t)=)=t tlnlnt t+mmlnlnt t-m

43、tmt-t t+mm+1 1,则则h h(t t)=)=lnlnt t+mm-mtmtt tF F(t t)=)=t tlnlnt t+mm-mtmt,则则F F(t t)=)=1 1+lnlnt t-mm，0 0t t1 1且且mm1 1,F F(t t)F F(1 1)=)=0 0,即即h h(t t)0 0恒成立恒成立,h h(t t)在定义域内单调递增在定义域内单调递增,即即h h(t t)h h(1 1)=)=0 0恒成立恒成立,t tlnlnt t+mmlnlnt tmtmt+t t-mm-1 1,原式得证原式得证.30.若若1 1a a(1 1-mm)x x1 1+mxmx2 2

44、恒成立恒成立,求求mm的取值范围的取值范围.【解析】易知【解析】易知x x2 2-x x1 1lnlnx x2 2-lnlnx x1 1=1 1a a，原式，原式x x2 2-x x1 1lnlnx x2 2-lnlnx x1 1(0 0,x x2 2x x1 1-1 1lnlnx x2 2x x1 11 1,则则t t-1 1lnlnt t0 0,lnlnt t0 0,即即lnlnt t-t t-1 11 1-mm+mtmt0 0令令g g(t t)=)=lnlnt t-t t-1 11 1-mm+mtmt，则则g g(t t)=)=(t t-1 1)mm2 2t t-(-(mm-1 1)2

45、 2 t t(1 1-mm+mtmt)2 2=mm2 2(t t-1 1)t t-(mm-1 1)2 2mm2 2t t(1 1-mm+mtmt)2 2当当(mm-1 1)2 2mm2 21 1时时,即即mm1 12 2时时,g g(t t)0 0,g g(x x)在在(1 1,+),+)递增递增,g g(1 1)=)=0 0 g g(1 1)g g(0 0)=)=0 0,满足满足.当当(mm-1 1)2 2mm2 21 1时时,即即0 0mm1 12 2时时,则则t t(1 1,(1 1-mm)2 2mm2 2)时时,g g(t t)0 0,g g(x x)在在(1 1,(1 1-mm)2 2mm2 2)单调递减，单调递减，g g(t t)g g(1 1)=)=0 0,矛盾矛盾综上综上:mm 1 12 2,+,+