導(dǎo)數(shù)中的極值點(diǎn)偏移問(wèn)題（解析版）

1 1 1 1 5 8題型4指數(shù)型極值點(diǎn)偏移 題型5對(duì)數(shù)型極值點(diǎn)偏移 反饋訓(xùn)練 性。若函數(shù)f(x)在x=x0處取得極值，且函數(shù)y=f(x)與直線y=b交于A(x1,b),B(x2,b)兩點(diǎn)，則AB的x0(1)對(duì)稱化構(gòu)造法：構(gòu)造輔助函數(shù)：對(duì)結(jié)論x1+x2x1x2>(<)x型，構(gòu)造函數(shù)=f(x(-f(通過(guò)研究F(x)的單調(diào)性獲得不等式.數(shù)單調(diào)性證明.<1；【分析】(1)對(duì)函數(shù)f(x(=2xlnx-x2+1變形整理f(x(-1=2xlnx-x2=x(2lnx-x(，構(gòu)造函數(shù)結(jié)論成立.(2)對(duì)函數(shù)f(x(=2xlnx-x2+1進(jìn)行二次求導(dǎo)，從而可判斷函數(shù)f(x(單調(diào)性，要證x1+x2>2，只需證x2>2-x1，結(jié)合f(x(在(0,+∞(上單調(diào)遞減知只需證f(x2(<f(2-x1(，即證f(x1(+f(2-x1(>0，進(jìn)而構(gòu)造函數(shù)F(x(=f(x(+f(2-x(判斷其單調(diào)性即可證明.【詳解】(1)由題意，f(x(-1=2xlnx-x2=x(2lnx-x(，設(shè)g(x(=2lnx-x(x>0(，則g/(x(=x(<0，∴g(x(單調(diào)遞減，從而g(x(max=g(2(=2ln2-2=2(∴f(x(-1=xg(x(<0，故f(x(<1.x(=2lnx+2-2x，fⅡ(x(=-2=，x>0，∴f從而f/(x(在(0,1(上單調(diào)遞增，在(1,+∞(上單調(diào)遞減，故f/(x(≤f/(1(=0，若0<x1<x2≤1，則f(x1(+f(x2(>0，不合題意，若1≤x1<x2，則f(x1(+f(x2(<0，不合題意，∴0<x1<1<x2，要證x1+x2>2，只需證x2>2-x1，結(jié)合f(x(在(0,+∞(上單調(diào)遞減知只需證f(x2(<f(2-x1(，又f(x1(+f(x2(=0，∴f(x2(=-f(x1(，故只需證-f(x1(<f(2-x1(，即證f(x1(+f(2-x1(>0①,令F(x(=f(x(+f(2-x(，0<x<1，則F/(x(=f/(x(-f/(2-x(=2lnx+2-2x-[2ln(2-x(+2-2(2-x([=2ln-+4-4x，F(xiàn)，:F/(x(<0，從而F(x(在(0,1(上單調(diào)遞減，“F(1(=2f(1(=0，:F(x(>0，“0<x1<1，:F(x1(=f(x1(+f(2-x1(>0，即不等式①成立，故x1+x2>2.2.(2024·云南·二模)已知常數(shù)a>0，函數(shù)f(x)=x2-ax-2a2lnx.(1)若Vx>0,f(x)>-4a2，求a的取值范圍;x2>4a.(1)(0,(2)由(1)不妨設(shè)0<x1<2a<x2，設(shè)F(x)=f(x)-f(4a-x)，利用導(dǎo)數(shù)說(shuō)明函數(shù)的單調(diào)性，即可得到f(x1(>f(4a-x1(，結(jié)合f(x1(=f(x2(及f(x(的單調(diào)性，即可證明.(1)由已知得f(x)的定義域?yàn)閧x|x>0}，且f/(x)=x-a-==.“a>0，:當(dāng)x∈(0,2a)時(shí)，f/(x)<0，即f(x)在(0,2a)x)>0，即f(x)在(2a,+∞)上單調(diào)遞增.所以f(x(在x=2a處取得極小值即最小值，:f(x(min=f(2a(=-2a2ln(2a(，“Vx>0,f(x(>-4a2今f(x(min=-2a2ln(2a(>-4a2今ln(2a(<lne2，(2)由(1)知，f(x)的定義域?yàn)閧x|x>0}，“x1設(shè)F(x)=f(x)-f(4a-x)，則F/(x)=+=.x)<0,F/(2a)=0，即F(x)在(0,2a]上單調(diào)遞減.“0<x1<2a，∴F(x1(>F(2a)=0，即f(x1(>f(4a-x1(，∵f(x1(=f(x2(=0，∴f(x2(>f(4a-x1(，∴4a-x1>2a，又∵x2>2a證明不等式.3.(2024·四川南充·一模)已知函數(shù)f(x)=x-lnx-a有兩個(gè)不同的零點(diǎn)x1,x2.x2>2.(1)f(x(的定義域?yàn)?0,+∞(，所以f(x(在(0,1(上單調(diào)遞減，在(1,+∞(上單調(diào)遞增，(2)不妨設(shè)x1<x2<1,x2>1又f(x(在(1,+∞(上單調(diào)遞增，所以只需證f(2-x1(<f(x2(，即證f(2-x1(<f(x1(.記g(x(=f(2-x(-f(x(=2-2x-ln(2-x(+lnx,x∈(0,1(，則g/(x(=+--2=-，x(>0，g(x(單調(diào)遞增，又g(1(=f(2-1(-f(1(=0，所以g(x1(=f(2-x1(-f(x1(<0，即f(2-x1(<f(x1(.4.(2024·安徽淮南·二模)已知函數(shù)f(x)=(1-x-k(x-1),x>-1,k∈R.(1)k=0時(shí)，函數(shù)f(x)=ex,x∈(-1,0)，則f/(x)=ex>0，f(x)在(-1,0)上單調(diào)遞增，所以f(x)=ex<f(0)=-1.設(shè)函數(shù)F(x)=-k，F(xiàn)/(x)=x)>0，故F(x)在(-1,0)上單調(diào)遞減，在(0,+∞)上單調(diào)遞增.設(shè)函數(shù)h(x)=F(x)-F(-x),x∈(-1,0)，則h(x)=+，/(x)<0，即函數(shù)h(x)在(-1,0)上單調(diào)遞減，所以h(x)>h(0)=0，即有F(x2(=F(x1(>F(-x1(，所以x2>-x1，5.(23-24高三下·天津·階段練習(xí))已知函數(shù)f(x)=x2-2ax+4lnx.(1)討論f(x)的單調(diào)區(qū)間；個(gè)公共點(diǎn)x1,x2,x3(x1<x2<x3(；②求證：x3-x1<47.(1)答案見解析2>2λ1-x1，只需證f(x1(<f(2λ1-x1(，構(gòu)造函數(shù)g(x(=f(x(-3-x1=(x2+x3(證.(1)f/(x(=2x-2a+=，x>0，其中t(x(=x2-ax+2，Δ=a2-8，函數(shù)f(x(在區(qū)間(0,+∞(單調(diào)遞增，f/(x(>0在區(qū)間(0,+∞(上恒成立，即函數(shù)f(x(在區(qū)間(0,+∞(上單調(diào)遞增，x1=所以函數(shù)f(x(的單調(diào)遞增區(qū)間是(0,和,+∞(，f(x(的單調(diào)遞增區(qū)間是(0,+∞(；當(dāng)a>2·2時(shí)，函數(shù)f(x(的單調(diào)遞增區(qū)間是(0,和,+∞(，又y=f(x)的圖象與y=b有三個(gè)公共點(diǎn)x1,x2,x3(x1<x2<x3(，故0<x1<λ1<x2<λ2<x3，則2λ1-x1>λ1，要證x1+x2>2λ1，即證x2>2λ1-x1，又2λ1-x1>λ1，且函數(shù)f(x(在(λ1,λ2(上單調(diào)遞減，即可證f(x2(<f(2λ1-x1(，又f(x1(=f(x2(=b，即可證f(x1(<f(2λ1-x1(，令g(x(=f(x(-f(2λ1-x(，x∈(0,λ1(，由f/(x(=2x-2a+==，則g/(x(=+x(2λ1-x(=2(x-λ1(?(x-λ2((2λ1-x(+x(x+x(2λ1-x(x(2λ1-x(=2(x-λ1(?-x2-2λ1λ2+2λ1x+λ2x+x2x(2λ1-x(故g/(x(在(0,λ1(上單調(diào)遞增，即g(x(<g(λ1(=f(λ1(-f(2λ1-λ1(=0，即f(x1(<f(2λ1-x1(恒成立，即得證；②由0<x1<λ1<x2<λ2<x3，則2λ2-x3<λ2，令h(x(=f(x(-f(2λ2-x(，x∈(λ2,+∞(，則h/(x(=+x(2λ2-x(=2(x-λ2(?(x-λ1((2λ2-x(+x(x+x(2λ2-x(=2(x-λ2(?-故h/(x(在(λ2,+∞(上單調(diào)遞增，即h(x(>h(λ2(=f(λ2(-f(2λ2-λ2(=0，時(shí)，f(x(>f(2λ2-x(，由x3>λ2，故f(x3(>f(2λ2-x3(，又f(x3(=f(x2(，故f(x2(>f(2λ2-x3(，2-x3<λ2，λ1<x2<λ2，函數(shù)f(x(在(λ1,λ2(上單調(diào)遞減，故x2<2λ2-x3，即x2+x3<2λ2，又由①知x1+x2>2λ1，故x3-x1=(x2+x3(-(x2+x1(<2λ2-2λ1，2-2λ1=2(λ1+λ2(2-4λ1λ2=2a2-8≤262-8=47，故x3-x1<4、7.-x1=(x2+x3(-(x2+x1(<2λ2-2λ1.6.已知函數(shù)f(x(=3lnx+ax2-4x(a>0).(1)由題意可知：f(x(的定義域?yàn)?0,+∞(，f/(x(=+2ax-4=，即f/(x(>0在(0,+∞(內(nèi)恒成立，所以f(x(在(0,+∞(內(nèi)單調(diào)遞增.f/(x(=+x-4=，1<x<3,f/(x(<0;0<x<1,或x>3,f/(x(>0;故f(x(在(0,1(,(3,+∞(內(nèi)單調(diào)遞增，在(1,3(內(nèi)單調(diào)遞減，由題意可得：0<x1<1<x2<3<x3，因?yàn)閒(x1(=f(x2(=f(x3(=b，令g(x(=f(x(-f(2-x(,0<x<1，則g/(x(=f/(x(+f/(2-x(=+x-4(+-+2-x-4(=>0，可知g(x(在(0,1(內(nèi)單調(diào)遞增，則g(x(<g(1(=0，可得f(x(<f(2-x(在(0,1(內(nèi)恒成立，，則f(x1(=f(x2(<f(2-x1(，且1<2-x1<2,1<x2<3,f(x(在(1,3(內(nèi)單調(diào)遞減令h(x(=f(x(-f(6-x(,1<x<3，則h/(x(=f/(x(+f/(6-x(=+x-4(+-+6-x-4(=>0，可知h(x(在(1,3(內(nèi)單調(diào)遞增，則h(x(<h(3(=0，可得f(x(<f(6-x(在(1,3(內(nèi)恒成立，，則f(x2(=f(x3(<f(6-x2(，且3<6-x2<5,x3>3,f(x(在(3,+∞(內(nèi)單調(diào)遞增，由x1+x2>2和x2+x3<6可得x3-x1<4.x2>1<a<1；由=,設(shè)h(x(=g(x(-g，對(duì)h(x(求導(dǎo)可得g(x1(<g，又g(x1(=g(x2(，再由g(x(在f(x(=(1+lnx(eln=又f/(x(=1+lnx(?ax2=-lnx，由f/(x(=0，得xax2當(dāng)0<x<1時(shí)，f/(x(>0，則f(x(在(0,1(上單當(dāng)x>1時(shí)，f/(x(<0，則f(x(在(1,+∞(上單調(diào)遞減，<x1<1<x2，=， 設(shè)h(x(=g(x(-g=-x(1-lnx(，x(=+lnx=lnx≥0，所以h(x(在(0,+∞(上單調(diào)遞增，又h(1(=0，所以h(x1(=g(x1(-g<0，即g(x1(<g，又g(x1(=g(x2(，所以g(x2(<g，又x2>1，>1故x1x2>1.=g(x(-g，對(duì)h(x(求導(dǎo)，得到h(x(的單調(diào)性和最值可證得g(x1(<g，即可證明x1x2>1.故f/(x)=+x-2==≥0，令g/(x)=0//即當(dāng)-1<a<-1時(shí)，f(x)=ax2有兩個(gè)不同實(shí)根x1，x2，上述兩式相除得ln(x1x2)=x1+x2，lnx2-x1不妨設(shè)x1<x22>e2，設(shè)t=>1，令F(t)=lnt-=lnt+-2，則F′(t)=-=>0，2>e2.1x2<4.求導(dǎo)得f/(x)=ax-(2a+1)+=，若0<a<，當(dāng)0<x<2或x>時(shí)，f/(x)>0，當(dāng)2<x<時(shí)，f/(x)<0，若a>，當(dāng)0<x<或x>2時(shí)，f/(x)>0，當(dāng)<x<2時(shí)，f/(x)<0，所以a的取值范圍是a≤0.0<x1<2<x2，要證x1x2<4，只證x1<，即證f(x1(<f，就證f(x2(-f<0，令g(x)=f(x)-fx>2，求導(dǎo)得g/(x)=f/(x)-f/-=(ax-1)(x-2)+-1-2(?4=(ax-1)(x-2)(x-4a)xx2x=ax-1+=?)<0，即f(x2)-f<0?f(x2)<f，又f(x1)=f(x2)，因此f(x1)<f，顯然0<x1<2,0<<2，又函數(shù)f(x)在(0,2)上單調(diào)遞增，則有x1<，所以x1x2<4.a+eb<1.f(x(min<0，求出m>-，結(jié)合題目條件，得到當(dāng)-<m<0時(shí)，f>0，根據(jù)零點(diǎn)存在性定理得到f(x(在a-b=x(≤0，g(x(在定義域內(nèi)單調(diào)遞減，故g(t+1(<g(t(，即ln(1+t(+-<0，證明出結(jié)論.(x)=(x+1(ex，故f(x(在(-∞,-1(內(nèi)單調(diào)遞減，在(-1,+∞(單調(diào)遞增，故要使f(x(有兩個(gè)零點(diǎn)，則需f(x(min=f(-1(=-e-1-m<0，故m>-，由題目條件m<0，可得-<m<0，當(dāng)-<m<0時(shí)，因?yàn)閒=e>m-m=0，又<-e<-1，故f(x(在(-∞,-1(內(nèi)存在唯一零點(diǎn)，又f(0(=-m>0，故f(x)在(-1,0(內(nèi)存在唯一零點(diǎn)，則f(x(在R上存在兩個(gè)零點(diǎn)，故滿足題意的實(shí)數(shù)m的取值范圍為(-,0(；a-b=，-at=t，故a=，要證ea+eb<1，即證e+e<1?elnt+1(<1?e+1(<1?eeln(t+1(<1?e+ln(t+1(<1，即證ln(1+t(+-<0，u(>0，故h(u(在(0,1(內(nèi)單調(diào)遞減，在(1,+∞(單調(diào)遞增，所以h(u(≥h(1(=0，所以u(píng)-1-lnu≥0，令u=得-1-ln≥0，x(在定義域內(nèi)單調(diào)遞減，11.(2023·湖北武漢·三模)已知函數(shù)f(x(=ax+(a-1(lnx+，a∈R.(1)討論函數(shù)f(x(的單調(diào)性；(2)若關(guān)于x的方程f(x(=xex-lnx+有兩個(gè)不相等的實(shí)數(shù)根x1、x2，(1)答案見解析出函數(shù)f(x(的增區(qū)間和減區(qū)間；(2)(i)將方程變形為ex+lnx=a(x+lnx(，令t(x(=x+lnx，令g(t(=，可知直線y=a與函數(shù)g(t((1)解：因?yàn)閒(x(=ax+(a-1(lnx+，所以f,(x(=a+-==，其中x>0.①當(dāng)a≤0時(shí)，f,(x(<0，所以函②當(dāng)a>0時(shí)，由f,(x(>0得x>，由f,(x(<0可得0<x<.所以函數(shù)f(x(的增區(qū)間為,+∞(，減區(qū)間為(0,.當(dāng)a>0時(shí)，函數(shù)f(x(的增區(qū)間為,+∞(，減(2)解：(i)方程f(x(=xex-lnx+可化為xex=ax+alnx，即ex+lnx=a(x+lnx(.令t(x(=x+lnx，因?yàn)楹瘮?shù)t(x(在(0,+∞(上單調(diào)遞增，易知函數(shù)t(x(=x+lnx的值域?yàn)镽，t=at(*)有兩個(gè)不等的實(shí)根.由g,(t(<0可得t<0或0<t<1，由g,(t(>0可得t>1，所以，函數(shù)g(t(在(-∞,0(和(0,1(上單調(diào)遞減，在(1,+∞(上單調(diào)遞增.t(=>0.作出函數(shù)g(t(和y=a的圖象如下圖所示：t=at，所以只需證t1+t2>2.t1<1<t2.令p=(t>1(，只需證lnp>.令h(p(=lnp-，其中p>1，則h/(p(=-=>0，所以h(p(在(1,+∞(上單調(diào)遞增，故h(p所以原不等式得證.(1)直接構(gòu)造函數(shù)法：證明不等式f(x(>g(x((或f(x(<g(x()轉(zhuǎn)化為證明f(x(-g(x(>0(或f(x(-g(x(<0)，進(jìn)而構(gòu)造輔助函數(shù)h(x(=f(x(-g(x(；(2)若函數(shù)g(x(=xf(x(-x+a有兩個(gè)不同的極值點(diǎn)，記作x1,x2【分析】(1)對(duì)函數(shù)f(x(=lnx-3x求導(dǎo)后得f/(x(=,x>0，分別求出f/(x(>0和f/(x(<0的(2)由g(x(=xf(x(-x+a有兩個(gè)極值點(diǎn)x1,x2?lnx1=2ax1,lnx2=2ax2，從而要證lnx1+2lnx2>3?2ax1+4ax2>3?a>?->，令t=,t>1，構(gòu)建函數(shù)h(t(=lnt-【詳解】(1)由題意得f(x(=lnx-3x，則f/(x(=,x令f/(x(>0，解得0<x<；令f/(x(<0，解得x>，∴f(x(在(0,上單調(diào)遞增，在,+∞(上單調(diào)遞減，∴f(x)max=f=ln-3×=-ln3-1，∴f(x(無(wú)最小值，最大值為-ln3-1.(2)∵g(x(=xf(x(-x+a=xlnx-ax2-x+a，則g/(x(=lnx-2ax，又g(x(有兩個(gè)不同的極值點(diǎn)x1,x2,∴l(xiāng)nx1=2ax1,lnx2=2ax2，∵0<x1<x2,∴原式等價(jià)于證明a>①.x2-x1x1+2x2，x2-x1x1+2x2，即證ln>=.令h(t(=lnt-，則h/(t(=-=，∵t>1,∴h/(t(>0恒成立，∴h(t(在(1,+∞(上單調(diào)遞增，t(>h(1(=0，即lnt>，13.(2024高三下·全國(guó)·專題練習(xí))已知函數(shù)g(x(=lnx-ax2+(2-a(x(a∈R).(1)求g(x(的單調(diào)區(qū)間；(2)若函數(shù)f(x(=g(x(+(a+1(x2-2x，x1,x2(0<x1<x2(是函數(shù)f(x(的兩個(gè)零點(diǎn)，證明：f/<(1)答案見解析h(t(=(1+t(lnt-2t+2，利用導(dǎo)數(shù)研究函數(shù)h(t(單調(diào)性，確定其最值，得到h(t(<h(1(=0，即可證得結(jié)論.(1)函數(shù)g(x(=lnx-ax2+(2-a(x的定義域?yàn)?0,+∞(，g/(x(=-2ax+2-a=-，，則g(x(在(0,+∞(上單調(diào)遞增；x(<0，則g(x(在(0,上單調(diào)遞增，在,+∞(上單調(diào)遞減.(2)因?yàn)閤1,x2(0<x1<x2(是f(x(=lnx+x2-ax的兩個(gè)零點(diǎn)，所以lnx1+x-ax1=0，lnx2+x-ax2=0，將兩式作差可得a=+(x1+x2(，又f/(x(=+2x-a，所以f/=+x1+x2-a=-，所以要證f/<0，只須證明-<0，即證明>lnx1-lnx2，即證明>ln，令=t∈(0,1(，即證>lnt，2t-2>(1+t(lnt，令h(t(=(1+t(lnt-2t+2，則h/(t(=lnt+-1，令u(t(=h/(t(=lnt+-1，則u/(t(=-=<0在(0,1(上恒成立，∴h/(t(在(0,1(上遞減，又h/(t(>h/(1(=0，∴h(t(在(0,1(上遞增，則h(t(<h(1(=0，即(1+t(lnt<2t-2， 行求解.f(x(=圖象上(ii)證明：x+x>.ax-ex+ax2-1=lnx，進(jìn)行同構(gòu)eax+ax2=elnex+lnex，將問(wèn)題轉(zhuǎn)化為方程ax2值范圍.e(ii)由2(x+x(>(x1+x2(2知，先證x1+x2>.2，即極值點(diǎn)偏移問(wèn)題，構(gòu)造函數(shù)F(x(=h(x(-e得h(x1(<h-x1(即h(x2(<h-x1(，再由設(shè)h(x(=(x>1)，則h/(x(=，由h/(x(>0?2lnx-ln2x>0?lnx(2-lnx(>0.因?yàn)閤>1，所以2-lnx>0?x<e2，所以h(x(在(1,e2(上單調(diào)遞增，在(e2,+∞(上單調(diào)遞減，所以h(x((2)(i)因?yàn)閒(x(=g(x(，即eax-ex+ax2-1=令u(t)=et+t，所以u(píng)(ax2(=u(lnex(，令h(x(=，所以h(x(=h(x(max=h=e.令F(x(=h(x(-h-x(，x∈(0,，因?yàn)镕(x(=-，所以F(x(=--，F(xiàn)=0，所以h(x(<h-x(.<h-x1(，又因?yàn)閔(x1(=h(x2(，所以h(x2(<h-x1(，2(x+x(>(x1+x2(2>所以x+x>.極值點(diǎn)偏移問(wèn)題的一般方法--對(duì)稱化構(gòu)造的步驟如下：(1)求極值點(diǎn)x0：求出函數(shù)f(x(的極值點(diǎn)x0，結(jié)合函數(shù)f(x(的圖像，由f(x1(=f(x2(得出x2,x1的取值+x1>2x0的情況，構(gòu)造函數(shù)F(x(=f(x(-f(2x0-x(；①F/(x(=f/(x(+f/(2x0-x(>0，則F(x(單調(diào)遞增；②注意到F(x0(=0，則F(x1(=f(x1(-f(2x0-x1(<0即f(x1(<f(2x0-x1(；③f(x2(=f(x1(<f(2x0-x1(，根據(jù)f(x(在(x0,+∞(單調(diào)減，則x2>2x0-x1④得到結(jié)論x2+x1>2x0.-x1=-x1=轉(zhuǎn)化為f(x1(=f(x2(，再設(shè)x1<x2，數(shù)形結(jié)合得到<x1<1<x2，接著構(gòu)造函數(shù)，利用函數(shù)的單調(diào)性得到x1+x2>2，最后利用放縮法證明不等式.-x1=得ex1-x1得x2(lnx1+1(=x1(lnx2+1(，即=，即f(x1(=f(x2(.而f=0，當(dāng)x>1時(shí)，f(x)>0恒成立，不妨設(shè)x1<x2，則<x1<1<x2.記h(x)=f(x)-f(2-x)，x∈,1(，則h/(x)=f/(x)+f/(2-x)=--> --=->0，所以函數(shù)h(x)在,1(上單調(diào)遞增，所以h(x)<f(1)-f(2-1)=0，即f(x)<f(2-x)，x∈,1(，于是f(x2(=f(x1(<f(2-x1(，2-x1∈(1,2-，所以x+x>x+(2-x1(3=8-12x1+6x=6(x1-1(2+2>2.(2)構(gòu)造輔助函數(shù)(若要證x1+x2>(<)2x0，則構(gòu)造函數(shù)g(x)=f(x)-f(2x0-x(；若要證x1x2>(<)(3)代入x1,x2，利用f(x2(=f(x1(及f(x)的單調(diào)性即得所證結(jié)論.1.(2021·全國(guó)·統(tǒng)考高考真題)已知函數(shù)f(x(=x(1-lnx(.(1)討論f(x(的單調(diào)性；(1)f(x(的遞增區(qū)間為(0,1(，遞減區(qū)間為(1,+∞(；(2)證明見解析.調(diào)性.(1)f(x(的定義域?yàn)?0,+∞(.由f(x(=x(1-lnx(得，f/(x(=-lnx，故f(x(在區(qū)間(0,1[內(nèi)為增函數(shù)，在區(qū)間[1,+∞(內(nèi)為減函數(shù)，由blna-alnb=a-b得1-ln=1-ln，即f=f.①令g(x(=f(2-x(-f(x(，則g/(x)=ln(2-x)+lnx=ln(2x-x2)=ln[1-(x-1)2]，從而f(2-x(>f(x(，所以f(2->f=f，由(1)得2-<即2<+.①令h(x(=x+f(x(，則h'(x(=1+f/(x(=1-lnx，從而x+f(x(<e，所以+f<e.又由∈(0,1)，可得<1-ln=f=f，所以+<f+=e.②由①②得2<+<e.令=m,=n．則上式變?yōu)閙(1-lnm(=n(1-lnn(，于是命題轉(zhuǎn)換為證明：2<m+n<e.令f(x(=x(1-lnx(，則有f(m(=f(n(，不妨設(shè)m<n.由(1)知0<m<1,1<n<e，先證m+n>2.要證：m+n>2?n>2-m?f(n(<f(2-m(?f(m)<f(2-m(?f(m(-f(2-m(<0.令g(x(=f(x(-f(2-x(,x∈(0,1(，則g′(x(=-lnx-ln(2-x(=-ln[x(2-x([≥-ln1=0，再證m+n<e.因?yàn)閙(1-lnm(=n?(1-lnn(>m，所以需證n(1-lnn(+n<e?m+n<e.令h(x(=x(1-lnx(+x,x∈(1,e(，所以h(x(<h(e(=e．故h(n(<e，即m+n<e.綜合可知2<+<e.證明+>2同證法2．以下證明x1+x2<e.不妨設(shè)x2=tx1，則t=>1，1-lnx1)=x2(1-lnx2)得x1(1-lnx1)=tx1[1-ln(tx1)]，lnx1=1--，t(x1<e，兩邊取對(duì)數(shù)得ln(1+t)+lnx1<1，記h(s)=-ln(1+s)，則h′(s)=-<0，所以g(s(在區(qū)間(0,+∞(內(nèi)單調(diào)遞減.由t∈(1,+∞(得t-1∈(0,+∞(，所以g(t(<g(t-1(，由已知得-=-，令=x1,=x2，不妨設(shè)x1<x2，所以f(x1(=f(x2(.令φ(x)=lnx+-2(0<x<e)，則φ′(x)=-=<0.所以φ(x(>φ(e(=0,h′(x(>0，h(x(在區(qū)間(0,e(內(nèi)單調(diào)遞增.又因?yàn)閒(x1(=f(x2(，所以=,>，即x-ex2<x-ex1,(x1-x2((x1+x2-e(>0.x2<e證明題中的不等式即可.在.2.設(shè)函數(shù)f(x(=lnx-ax(a∈R(.(2)若函數(shù)g(x(=xf(x(-x+a有兩個(gè)不同的極值點(diǎn)，記作x1,x2【分析】(1)對(duì)函數(shù)f(x(=lnx-3x求導(dǎo)后得f/(x(=,x>0，分別求出f/(x(>0和f/(x(<0的(2)由g(x(=xf(x(-x+a有兩個(gè)極值點(diǎn)x1,x2?lnx1=2ax1,lnx2=2ax2，從而要證lnx1+2lnx2>3?2ax1+4ax2>3?a>?->，令t=,t>1，構(gòu)建函數(shù)h(t(=lnt-【詳解】(1)由題意得f(x(=lnx-3x，則f/(x(=,x令f/(x(>0，解得0<x<；令f/(x(<0，解得x>，∴f(x)max=f=ln-3×=-ln3-1，=xf(x(-x+a=xlnx-ax2-x+a，則g/(x(=lnx-2ax，又g(x(有兩個(gè)不同的極值點(diǎn)x1,x2,∴l(xiāng)nx1=2ax1,lnx2=2ax2，欲證lnx1+2lnx2>3，即證2ax1+4ax2>3，∵0<x1<x2,∴原式等價(jià)于證明a>①.x2-x1x1+2x2，x2-x1x1+2x2，即證ln>=.令t=，則t>1，上式等價(jià)于求證lnt>.令h(t(=lnt-，則h/(t(=-=，t(>h(1(=0，即lnt>，t=，t=x2+x1(；x2>1<a<1；由=,設(shè)h(x(=g(x(-g，對(duì)h(x(求導(dǎo)可得g(x1(<g，又g(x1(=g(x2(，再由g(x(在f(x(=(1+lnx(eln=又f/(x(=1+lnx(?ax2=-lnx，由f/(x(=0，得xax2當(dāng)0<x<1時(shí)，f/(x(>0，則f(x(在(0,1(上單當(dāng)x>1時(shí)，f/(x(<0，則f(x(在(1,+∞(上單調(diào)遞減，設(shè)h(x(=g(x(-g=-x(1-lnx(，h/(x(=+lnx=lnx≥0，所以h(x(在(0,+∞(上單調(diào)遞增，又h(1(=0，所以h(x1(=g(x1(-g<0，即g(x1(<g，又g(x1(=g(x2(，所以g(x2(<g，(x(在(1,+∞(上單調(diào)遞減，所以x2>，=g(x(-g，對(duì)h(x(求導(dǎo)，得到h(x(的單調(diào)性和最值可證得g(x1(<g，即可證明x1x2>1.4.(23-24高二下·云南·期中)已知函數(shù)f(x(=3lnx+ax2-4x(a>0).(1)在(0,+∞(上單調(diào)遞增f(x(在(0,+∞(內(nèi)單調(diào)遞增；(2)f(x(=3lnx+x2-4x，求導(dǎo)，得到函數(shù)單調(diào)性，得到0<x1<1<x2<3<x3，構(gòu)造g(x(=f(x(-f(2-x(,0<x<1，求導(dǎo)得到函數(shù)單調(diào)性，得到x1+x2>2，再構(gòu)造h(x(=f(x(-f(6-x(,1<x<3，(1)由題意可知：f(x(的定義域?yàn)?0,+∞(，f/(x(=+2ax-4=，Δ=16-24=-8<0，可知2x2-4x+3>0在(0,+∞(上恒成立，即f/(x(>0在(0,+∞(上恒成立，所以f(x(在(0,+∞(上單調(diào)遞增.f/(x(=+x-4=，1<x<3,f/(x(<0;0<x<1,或x>3,f/(x(>0;故f(x(在(0,1(,(3,+∞(上單調(diào)遞增，在(1,3(上單調(diào)遞減，由題意可得：0<x1<1<x2<3<x3，因?yàn)閒(x1(=f(x2(=f(x3(=b，令g(x(=f(x(-f(2-x(,0<x<1，則g/(x(=f/(x(+f/(2-x(=+x-4(+-+2-x-4(=>0，可知g(x(在(0,1(上單調(diào)遞增，則g(x(<g(1(=0，可得f(x(<f(2-x(在(0,1(上恒成立，，則f(x1(=f(x2(<f(2-x1(，且1<2-x1<2,1<x2<3,f(x(在(1,3(上單調(diào)遞減令h(x(=f(x(-f(6-x(,1<x<3，則h/(x(=f/(x(+f/(6-x(=+x-4(+-+6-x-4(=>0，可知h(x(在(1,3(上單調(diào)遞增，則h(x(<h(3(=0，可得f(x(<f(6-x(在(1,3(上恒成立，，則f(x2(=f(x3(<f(6-x2(，〈3,f(x(在(3,+∞(上單調(diào)遞增，由x1+x2>2和x2+x3<6可得x3-x1<4.【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：構(gòu)造兩次差函數(shù)，解決極值點(diǎn)偏移問(wèn)題，即構(gòu)造g(x(=f(x(-f(2-x(,0<x<1，求導(dǎo)得到函數(shù)單調(diào)性，得到x1+x2>2，再構(gòu)造h(x(=f(x(-f(6-x(,1<x<3，求導(dǎo)得到函數(shù)單調(diào)2+x3<6.5.(2024·遼寧·模擬預(yù)測(cè))已知函數(shù)f(x(=ex-ax2(a>0).(2)若f(x(有三個(gè)零點(diǎn)x1,x2,x3，且x1<x2<x3.(1)f(x(在(1,2(上單調(diào)遞減，在(2,+∞(上單調(diào)遞增將三個(gè)變量的問(wèn)題轉(zhuǎn)化為單變量問(wèn)題，即可構(gòu)造函數(shù)μ(x(=lnx-(x>1(，證明其在(1)當(dāng)a=時(shí)，f(x(=ex-x2，f/(x(=ex-x，令g(x(=f/(x(=ex-x，g/(x(=ex-，令g/(x(=ex-=0，可得x=ln=2-ln2，-ln2,+∞(時(shí)，g/(x(>0，即g(x(在(1,2-ln2(上單調(diào)遞減，在(2-ln2,+∞(上單調(diào)遞增，又g(1(=f/(1(=e-<0，g(2(=f/(2(=e2-e2=0，f/(x(>0，故f(x(在(1,2(上單調(diào)遞減，在(2,+∞(上單調(diào)遞增；(2)(i)f(x(有三個(gè)零點(diǎn)，即ex-ax2=0有三個(gè)根，令h(x(=，h/(x(=，x(<0，即h(x(在(-∞,0(、(2,+∞(上單調(diào)遞增，在(0,2(上單調(diào)遞減，-+(ii)由h(x(在(-∞,0(上單調(diào)遞增，h(-