解答題 10類導(dǎo)數(shù)答題模板（解析版）

11函數(shù)是高中數(shù)學(xué)主干知識(shí),單調(diào)性是函數(shù)的重要性質(zhì),用導(dǎo)數(shù)研究函數(shù)單調(diào)性是導(dǎo)數(shù)的一個(gè)主要應(yīng)1.(2024·全國(guó)·高考真題)已知函數(shù)f(x(=a(x-1(-lnx+1.(1)求f(x(的單調(diào)區(qū)間；f/(x)<02.(2023·全國(guó)·高考真題)已知函數(shù)f(x(=a(ex+(1)討論f(x(的單調(diào)性；思路詳解：(1)因?yàn)閒(x)=a(ex+a(-x所以f(x(在R上單調(diào)遞減；當(dāng)x<-lna時(shí)，f/(x(<0，則f(x(在(-∞,-lna(上單調(diào)遞減；當(dāng)x>-lna時(shí)，f/(x(>0，則f(x(在(-lna,+∞(上單調(diào)遞增；當(dāng)a>0時(shí)，f(x(在(-∞,-lna(上單調(diào)遞減，f(x(在(-lna,+∞(上單調(diào)遞增.(1)討論f(x)的單調(diào)性；f/(x(<0,f(x(單調(diào)遞減，22若x∈(0,+∞(，則f/(x(>0,f(x(單調(diào)遞增；當(dāng)0<a<時(shí)，若x∈(-∞,ln(2a((，則f/(x(>0,f(x(單調(diào)遞增，若x∈(ln(2a(,0(，則f/(x(<0,f(x(單調(diào)遞減，若x∈(0,+∞(，則f/(x(>0,f(x(單調(diào)遞增；x(≥0,f(x(在R上單調(diào)遞增；f/(x(>0,f(x(單調(diào)遞增，若x∈(0,ln(2a((，則f/(x(<0,f(x(單調(diào)遞減，若x∈(ln(2a(,+∞(，則f/(x(>0,f(x(單調(diào)遞增；(1)求函數(shù)f(x(的單調(diào)區(qū)間；思路詳解：(1)f(x)=ax-bx+e2,f/(x)=axlna-b，①若b≤0，則f/(x)=axlna-b≥0，所以f(x)在R上單調(diào)遞增；當(dāng)x∈(-∞,loga時(shí)，f'(x(<0,f(x(單調(diào)遞減，a,+∞(時(shí)，f'(x(>0,f(x(單調(diào)遞增.5.(2024·江西新余·模擬預(yù)測(cè))已知函數(shù)f(x)=-alnx+(2a+1)x-x2.(2)討論f(x)的單調(diào)性.所以函數(shù)f(x)的圖象在(1,f(1))處的切線方程為y-1=-(x-1)，即x+2y-3=0.(2)函數(shù)f(x)=-alnx+(2a+1)x-x2的定義域?yàn)?0,+∞)，求導(dǎo)得f/(x)=-+(2a+1)-2x=-，f/(x)<033(1)求函數(shù)f(x(在x=0處的切線方程；(2)討論函數(shù)f(x(的單調(diào)性；思路詳解：(1)因?yàn)閒(x(=e2x-2(a+1(ex+2ax+2a+1(所以函數(shù)f(x(在x=0處的切線方程為y=0；(2)函數(shù)f(x(=e2x-2(a+1(ex+2ax+2a+1(a>0(的定義域?yàn)镽，且f/(x(=2e2x-2(a+1(ex+2a=2(ex-a((ex-1(，x(=2(ex-1(2≥0恒成立，所以f(x(在R上單調(diào)遞增；所以f(x(在(-∞,0(，(lna,+∞(上單調(diào)遞增，在(0,lna(上單調(diào)遞減；所以f(x(在(-∞,lna(，(0,+∞(上單調(diào)遞增，在(lna,0(上單調(diào)遞減；當(dāng)a>1時(shí)，f(x(在(-∞,0(，(lna,+∞(上單調(diào)遞增，在(0,lna(上單調(diào)遞減；當(dāng)0<a<1時(shí)，f(x(在(-∞,lna(，(0,+∞(上單調(diào)遞增，在(lna,0(上單調(diào)遞減.(1)討論f(x)的單調(diào)性；詳解：(1)函數(shù)f(x)=ae2x-(ax+2-a)ex+x2的定義域?yàn)镽，求導(dǎo)得f/(x)=2ae2x-(ax+2)ex+x=(aex-1)(2ex-x)，函數(shù)φ(x)在(-∞,-ln2)上遞減，在(-ln2,+x-x>0，②當(dāng)a>0時(shí)，由f/(x)<0，得x<-lna，由f/(x)>0，得x>-lna，函數(shù)f(x)在(-∞,-lna)上單調(diào)遞減，在(-lna,+∞)上單調(diào)遞增，當(dāng)a>0時(shí)，f(x)在(-∞,-lna)上單調(diào)遞減，在(-lna,+∞)上單調(diào)遞增.44(1)討論f(x(的單調(diào)性；當(dāng)Δ=4-12a≤0,a≥時(shí)，f/(x(≥0,f(x(在R上單調(diào)遞增，,x2=1+1-3a1-1,x2=1+1-3a1-1-3a31--3a,1+-3a(時(shí)，f/(x(<0，f1-3a31-3a31-3a3,+∞1-3a31-3a31-3a3,+∞LL,1-3a3|2.(2024·青海海西·模擬預(yù)測(cè))已知函數(shù)f(x(=x3-x2+ax.(1)討論函數(shù)f(x(的單調(diào)性；，此時(shí)函數(shù)f(x(在R上單調(diào)遞增；2-2x+a=0的兩根為x=，此時(shí)函數(shù)f(x(的減區(qū)間為,，增區(qū)間為,+∞(；3.(2024·山東威?！ひ荒?已知函數(shù)f(x(=ln(ax(-x2+ax(a≠0(.(1)討論f(x(的單調(diào)性；55②當(dāng)a<0時(shí)，f(x(的定義域?yàn)?-∞,0(，令f/(x(<0,2x2-ax-1<0，解得：<(1)證明曲線y=f(x(在x=1處的切線過原點(diǎn)；(2)討論f(x(的單調(diào)性；所以f(x)在,上單調(diào)遞減；->1，665.函數(shù)f(x)=(x2+ax(ex(a∈R).(1)求f(x(的單調(diào)區(qū)間；f/(x)<0，f(x)單調(diào)遞減，f/(x(>0,f(x(在(0,+∞(上單調(diào)遞增2,x77(,g(x(<0所以f(x(在(0,x1(,(x2,+∞(單調(diào)遞增，在(x1,x2(單調(diào)遞減；故f(x(在(0,x2(單調(diào)遞減，(x2,+∞(單調(diào)遞增.綜上：a≥,f(x(在(0,+∞(上單調(diào)遞增；0<a<,f(x(在(0,1--2a(,(1+-2a,+∞(上單調(diào)遞增；7.已知函數(shù)f(x(=xlnx-x-a，g(x(=x2+lnx-ax.x(=2x+-a=2x2-+1，x>0，x(≥0恒成立，所以g(x(在(0,+∞(上單調(diào)遞增.2-ax+1=0的根為x=a±a2-8 定義1:若函數(shù)f(x)的導(dǎo)函數(shù)f/(x)在點(diǎn)x=x0處可導(dǎo),則稱f(x)在點(diǎn)x=x0的導(dǎo)數(shù)為f(x)在點(diǎn)x=x0的二階導(dǎo)數(shù),記作fⅡ(x0(,同時(shí)稱f(x)在點(diǎn)x=x0為二階可導(dǎo).88定義2:若f(x)在區(qū)間M上每一點(diǎn)都二階可導(dǎo),則得到一個(gè)定義在M上的二階可導(dǎo)函數(shù),記作fⅡ若f(x)在x=x0附近有連續(xù)的導(dǎo)函數(shù)fⅡ(x),且f/(x0(=0,fⅡ(x0(≠0(1)若fⅡ(x0(<0,則f(x)在點(diǎn)x0處取極大值;(2)若fⅡ(x0(>0,則f(x)在點(diǎn)x0處取極小值(1)討論f(x(的單調(diào)性；x(=a(eax-e-ax(令g(x(=a(eax-e-ax(，則g/(x(=a2(eax+e-ax(>0∴g(x(在R上單調(diào)遞增.f/(x(<0f/(x(>0∴f(x(在(-∞,0(上單調(diào)遞減，在(0,+∞(上單調(diào)遞增.解法二：f/(x(=a(eax-e-ax(=①當(dāng)a>0時(shí)，由f/(x(<0得x<0，由f/(x(>0得x>0∴f(x(在(-∞,0(上單調(diào)遞減，在(0,+∞(上單調(diào)遞增②當(dāng)a<0時(shí)，同理可得f(x(在(-∞,0(上單調(diào)遞減，在(0,+∞(上單調(diào)遞增.2.已知函數(shù)f(x(=x2-axlnx-1(a∈R(.-2xlnx-1，函數(shù)f(x(的定義域?yàn)?0,+∞(，f/(x(=2x-2(lnx+1(=2(x-lnx-1(.令g(x)=x-lnx-1，有g(shù)/(x)=，99可得函數(shù)f(x(單調(diào)遞增，3.已知函數(shù)f(x(=(ex-a-a(lnx+x+.思路詳解：(1)解法一：因?yàn)閒(x(=(ex-a-a(lnx+x+，所以f/(x)=ex-alnx+(ex-a-a(+1-=ex-axlnx+(ex-a-a(x?+x-x=ex+lnx-alnx+ex+lnx-a?-a+x-x=+lnx((ex+lnx-a-1)+x+lnx-ax易知+lnx≥+1-=1>0，設(shè)lnx0+x0=a，則當(dāng)0<x<x0時(shí)，x+lnx-a<0，ex+lnx-a-1<0，所以f/(x)<0，則f(x(在(0,x0(單調(diào)遞減；當(dāng)x>x0時(shí)，x+lnx-a>0，ex+lnx-a-1>0，所以f/(x)>0，則f(x(在(x0,+∞(單調(diào)遞增；當(dāng)a=1時(shí)f(x)=(ex-1-1)lnx+x+，則f/(x(=ex-1lnx++1-=ex-1(lnx++1--=-=>0x-1關(guān)于x單調(diào)遞增且ex-1>0，所以f/(x(單調(diào)遞增，則f/(x(>f/(1(=0，所以f(x(在(1,+∞(單調(diào)遞增.當(dāng)0<x<1時(shí)，ex-1<e1-1=1，f/(x)=ex-1(lnx++1--<lnx++1--=lnx+1-，所以f/(x)<0，所以f(x(在(0,1(單調(diào)遞減.4.已知函數(shù)f(x(滿足f(x(=ex-x2.x(>0，所以m(x(在(-∞,ln2(上單調(diào)遞減，在(ln2,+∞(上單調(diào)遞增，所以m(x)min=m(ln2(=2(1-ln2(>0，即f/(x(>0恒成立，5.已知函數(shù)f(x(=x(alnx-x-1(，其中a∈R.則f/(x)=lnx-2x(x>0)，設(shè)g(x)=lnx-2x(x>0)，(x)=-2=，得g(x)max=g=ln-1=-ln2-1<0，即f/(x)=g(x)<0，6.已知函數(shù)f(x(=ex+a-ln(x+1(-a(a∈R(.令h(x(=f/(x(，則h/(x(=ex+>0在x∈(-1,+∞(上恒成立，所以h(x(在(-1,+∞(上單調(diào)遞增，即f/(x(在(-1,+∞(上單調(diào)遞增.1.(2024·全國(guó)·高考真題)已知函數(shù)f(x(=a(x-1(-lnx+1.(1)求f(x(的單調(diào)區(qū)間；-1恒成立.f/(x)<0x-1-f(x)=ex-1-a(x-1)+lnx-1≥ex-1-2x+1+lnx，令g(x)=ex-1-2x+1+lnx(x>1)，下證g(x)>0即可.(1)求曲線y=f(x(在x=2處的切線斜率；(2)要證x>0時(shí)f(x(=+ln(x+1(>1，即證ln(x+1(>，令g(x)=ln(x+1(-且x>0，則g/(x)=-=>0，所以x>0時(shí)f(x(>1.=<1，∵x>0,ln(1-x(<0，∴xln(1-x(<0，即證x+ln(1-x(>xln(1-x(，化簡(jiǎn)得x+(1-x(ln(1-x(>0；=<1，∵x<0,ln(1-x(>0，∴xln(1-x(<0，即證x+ln(1-x(>xln(1-x(，化簡(jiǎn)得x+(1-x(ln(1-x(>0；令φ(t(=1-t+tlnt，φ/(t(=-1+lnt+1=lnt，(2)求f(x(的極值；-ln(x+1(+1，函數(shù)f(x(的定義域?yàn)?-1,+∞(.f/(x(=，x(>0.因此f(x(在(-1,0(單調(diào)遞減，在(0,+∞(單調(diào)遞增，故f(x(的最小值為f(0(=1.(2)f(x(的定義域?yàn)?-1,+∞(,f/(x(=a-.，故f(x(在(-1,+∞(單調(diào)遞減，f(x(無(wú)極值；0得x=-1.當(dāng)-1<x<-1時(shí)，f/(x(<0；因此f(x(在(-1,-1(單調(diào)遞減，在-1,+∞(單調(diào)遞增，故f(x(有極小值f-1(=2-a+lna，無(wú)極大值.(3)解法1：令g(x(=f(x(-ex=ax-ln(x+1(+1-ex(-1<x<0)，/(x(=a--ex，令h(x(=a--ex，則h/(x(=-ex，因此h/(x(>0,h(x(在(-1,0(單調(diào)遞增，h(x(<h(0(=a-2≤0，即g/(x(<0，故g(x(在(-1,0(單調(diào)遞減，g(x(>g(0(=0，即當(dāng)-1<x<0時(shí)，f(x(>ex.解法2：因?yàn)閍≤2,-1<x<0，所以f(x(要證當(dāng)-1<x<0時(shí)，f(x(>ex，即證2x-ln(x+1(+1-er>0，令g(x(=2x-ln(x+1(+1-ex(-1<x<0),g/(x(=2--ex，令h(x(=2--ex，則h/(x(=-ex，因?yàn)?1<x<0，所以>1,ex<1，因此h/(x(>0,h(x(在(-1,0(單調(diào)遞增，h(x(<h(0(=0，解法3：令g(x(=f(x(-ex=ax-ln(x+1(+1-ex(-1<x<0)，g/(x(=a--ex，由(1)可知x-ln(x+1(≥0，當(dāng)且僅當(dāng)x=0時(shí)取等號(hào)，g/(x(=a--ex<a--(x+1(<a-2≤0，即g/(x(<0，故g(x(在(-1,0(單調(diào)遞減，g(x(>g(0(=0，即當(dāng)-1<x<0時(shí)，f(x(>ex.t(=et-t(t<0)，則g/(t(=et-1<0，故g(t(在(-∞,0(單調(diào)遞減.-1<x<0時(shí)，ln(x+1(<x<0，所以g[ln(x+1([>g(x(，即：x+1-ln(x+1(>ex-x，故2x-ln(x+1(+1>ex，而a≤2,-1<x<0，所以f(x(≥2x-ln(x+1(+1>ex.t(=t-lnt(0<t<1)，因?yàn)間/(t(=<0，所以g(t(在(0,1(單調(diào)遞減.由(1)可知：當(dāng)-1<x<0時(shí)，ln(x+1(<x<0，所以0<x+1<er<1,g(x+1(>g(er(，即：x+1-ln(x+1(>ex-x，故2x-ln(x+1(+1>ex，而a≤2,-1<x<0，所以f(x(≥2x-ln(x+1(+1>ex.5.(2024·浙江寧波·一模)已知函數(shù)f(x(=1+2ax2-axsinx.(1)判斷f(x(的奇偶性；(1)f(x(=1+2ax2-axsinx,L-2a-2aLL-2a-2aL且f(-x(=1+2a(-x(2-a(-x(sin(-x(=1+2ax2-axsinx=f(x(，故f(x(為偶函數(shù)，(2)當(dāng)a=-時(shí)，f(x(=1-x2+xsinx為偶函數(shù)，要證f(x(≤1，只需要證1-x2≤1-xsinx，當(dāng)-1≤x≤1時(shí)，1-xsinx>0，只需證明0≤x≤1時(shí)，1-x2≤(1-xsinx(2，即證x2-xsinx+x2sin2x≥0，只需證x-sinx≥0，即證x≥sinx，令g(x(=x-sinx,g/(x)=1-cosx≥0,g(x(在0≤x≤1單調(diào)遞增，故g(x(≥g(0(=0，所以x≥sinx，得證.6.(2024·廣東·二模)已知函數(shù)f(x)=ex-1-xlnx.(1)求曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程；(2)證明：f(x)>0.(1)y=1fl(x(=ex-1-(lnx+1)，則k=fl(1)=0，①當(dāng)0<x<1時(shí)，ex-1>e-1，xlnx<0，則ex-1>xlnx，即f(x)>0；②當(dāng)x≥1時(shí)，fl(x(=ex-1-(lnx+1)=ex-1-lnx-1.x-1-xlnx≥1，要證f(x)>0，只需證ex-1>xlnx，只需證>，x2x2?x-lnx=1-x2x2>≥g(x)，也就是>，即f(x)>0x.x.(3)求證：f(x)≥x-lnx+e-1.(2)結(jié)合函數(shù)f(x)的圖象求解即可；(3)轉(zhuǎn)化為證明-x+lnx≥e-1，構(gòu)造函數(shù)g(x(=-x+lnx，x>0，進(jìn)而結(jié)合導(dǎo)數(shù)證明即可.所以函數(shù)f(x)在(-∞,0(和(0,1(上單調(diào)遞減，在(1,+∞(上單調(diào)遞增，(3)證明：由f(x)≥x-lnx+e-1，x>0，即≥x-lnx+e-1，即-x+lnx≥e-1，設(shè)g(x(=-x+lnx，x>0，令h(x(=ex-x，ex-1>0，所以函數(shù)h(x(在(0,+∞(上單調(diào)遞增，則h(x(=ex-x>h(0(=1>0所以令g/(x(>0所以函數(shù)g(x(在(0,1(上單調(diào)遞減，在(1,+∞(上單調(diào)遞增，所以g(x(=-x+lnx≥g(1(=e-1，所以f(x)≥x-lnx+e-1.①?x∈D,g(a(≤f(x(，則只需要g(a(≤f(x(min=m<f(x(，則只需要g(a(<f(x(min=m②?x∈D,g(a(≥f(x(，則只需要g(a(≥f(x(max=M>f(x(，則只需要g(a(>f(x(max=M<f(x(，則只需要g(a(≤m(注意與(1)中對(duì)應(yīng)情況進(jìn)行對(duì)比)≥f(x(，則只需要g(a(≥M點(diǎn).①?x∈D,g(a(≤f(x(，則只需要g(a(≤f(x(max=M<f(x(，則只需要g(a(<f(x(max=M②?x∈D,g(a(≥f(x(，則只需要g(a(≥f(x(min=m>f(x(，則只需要g(a(>f(x(min=m<f(x(，則只需要g(a(<M>f(x(，則只需要g(a(>m點(diǎn).故f/(x)=2ln(1+x)+-1=2ln(1+x)-+1，因?yàn)閥=2ln(1+x),y=-+1在(-1,+∞(上為增函數(shù)，故當(dāng)-1<x<0時(shí)，f/(x)<0，當(dāng)x>0時(shí)，f/(x)>0，故f(x(在x=0處取極小值且極小值為f(0(=0，無(wú)極大值.x(=-aln(1+x(+-1=-aln(1+x(-,x>0，設(shè)s(x(=-aln(1+x(-,x>0，x(=-=-=-，所以f(x(在[0,+∞(上為增函數(shù)，故f(x(≥f(0(=0.當(dāng)-<a<0時(shí)，當(dāng)0<x<-時(shí)，s/(x(<0，x(<0在(0,+∞(上恒成立，綜上，a≤-.=a-=a-=a-則f/(x)=g(t)=a-=當(dāng)a=8,f/(x)=g(t)==f/(x)<0.f/(x)>0.(2)設(shè)g(x)=f(x)-sin2xg/(x)=f/(x)-2cos2x=g(t)-2(2cos2x-1(=-2(2t-1)=a+2-4t+-設(shè)φ(t)=a+2-4t+-t)=-4-+==->0(x)=φ(t)<a-3≤0x0(=0.綜上,a的取值范圍為(-∞,3].x-2elnx+x，要證明f(x(>x，即證g(x(=f(x(-x=ex-2elnx->0；x(<0，即可得h(x(在(0,1(上單調(diào)遞減，x(>0，即可得h(x(在(1,+∞(上單調(diào)遞增，x≥ex；x(<0，即可得p(x(在(0,e(上單調(diào)遞減，x(>0，即可得p(x(在(e,+∞(上單調(diào)遞增，故x≥elnx；x-2elnx->ex-2x-=(e-x>0,故f(x(>x；(2)易知f(x(=ex-2elnx+ax+lna≥令m(x(=ex-2elnx+ex+1，則m/(x(=ex-+e，x(<0，即可得m(x(在(0,1(上單調(diào)遞減，x(>0，即可得m(x(在(1,+∞(上單調(diào)遞增，4.(2024·遼寧·模擬預(yù)測(cè))已知函數(shù)f(x(=(ax-1(ex+1+3(a≠0(.(1)求f(x(的極值；(2)設(shè)a=1，若關(guān)于x的不等式f(x(≤(b-1(ex+1-x在區(qū)間[-1,+∞(內(nèi)有解，求b的取值范圍.f/(x(>0f/(x(<0(2)a=1時(shí)，f(x(≤(b-1(ex+1-x等價(jià)于b≥+x，則b≥+x在區(qū)間[-1,+∞(內(nèi)有解.ex+1-(x+2(令g(x(=+xex+1-(x+2(=ex+1-1在[-1,+∞(上單調(diào)遞增，有h/(x(≥h/(-1=ex+1-1在[-1,+∞(上單調(diào)遞增，有h/(x(≥h/(-1(=0，所以h(x(在區(qū)間[-1,+∞(內(nèi)單調(diào)遞增，即h(x(≥h(-1(=0，所以g/(x(≥0在區(qū)間[-1,+∞(內(nèi)恒成立，(2)若存在x0則f/(x(=1+則f/(x(=1+-2x2+x+1xx-2x==--2x2+x+1xx-2x==-當(dāng)0<x<1時(shí)，f/當(dāng)0<x<1時(shí)，f/(x(=-(2x+1((x-1(xx(x(=-(2x+1((x-1(x8a81-44.28a81-44.2f(x0(>f(1)=0.0>0.而f/(x)=2x-=，(2)因?yàn)椴坏仁絝(x(≤2(lnx(2+x2-2x在區(qū)間(1,+∞)上有解，所以(lnx)2+alnx-x+1≥0在區(qū)間(1,+∞(上有解，此時(shí)lnx>0，即a≥-lnx在區(qū)間(1,+∞(上有解，令g(x(=-lnx(x>1(，則g/(x(=-=.令h(x(=x-lnx-1(x>1(，則h/(x(=1-=>0，當(dāng)1<x<e時(shí)g/(x(<0；當(dāng)x>e時(shí)g/(x(>0，所以g(x(min=g(e(=e-2可.(x)<0，得x<lna；由f/(x)>0，得x>lna，(2)不等式f(x)≥x2+2?ex-ax+1≥x2+2?ax≤ex-x2-1，依題意，當(dāng)x>0時(shí)，a≤-x-恒成立，令g(x)=-x-,x>0，求導(dǎo)得g/(x)=-1+=，令h(x)=ex-1-x,x>0，性即可.遞增.故f(x)min=f(a(=a-1-alna，由f(x(≥0恒成立，得a-1-alna≥0恒成立，令g(a(=a-1-alna(a>0(，求導(dǎo)得g/(a(=-lna，a(<0，所以g(a)max=g(1(=0，故g(a(=a-1-aln(x)=-2a=，max=φ=ln≤0，即f/(x)max≤0，所以f(x)的極值點(diǎn)的個(gè)數(shù)為0.(2)h(x)=xlnx-ax+1，-1(-1,+∞(所以h(x)min=h(ea-1(=1-ea-1，a-1+e-1，a-1+(e-1)(a-3)-1>0成立.(1)討論f(x(的單調(diào)性；(2)(-1,+∞((2)變形得到f(x(min<，在(1)的基礎(chǔ)上得到f(x(min=k-klnk，從而k-klnk<，再令g(k(=k值為h(1(=e1+a【詳解】(1)f(x(=ex-1-k(x-1(的定義域?yàn)镽，則f/(x(=ex-1-k，恒成立，故f(x(=ex-1-k(x-1(在R上單調(diào)遞增，故f(x(=ex-1-k(x-1(在(1+lnk,+∞(上單調(diào)遞增，在(-∞,1+lnk(上單調(diào)遞減，當(dāng)k>0時(shí)，f(x(在(1+lnk,+∞(上單調(diào)遞增，在(-∞,1+lnk(上單調(diào)遞減；f(x(min<，由(1)知，k>0時(shí)，f(x(在(1+lnk,+∞(上單調(diào)遞增，在(-∞,1+lnk(上單調(diào)遞減；故f(x(在x=1+lnk處取得極小值，也是最小值，故f(x(min=e1+lnk-1-k(1+lnk-1(=k-klnk，即證k-klnk<，令g(k(=k-klnk，k>0，則g/(k(=1-lnk-1=-lnk，k(<0，k(>0，1.(2022·全國(guó)·高考真題)已知函數(shù)=ax-lnx.思路詳解：(1)當(dāng)a=0時(shí)，f(x(=--lnx,x>0，則f/(x(=-=，f/(x(>0，f(x(單調(diào)遞增；，f(x(單調(diào)遞減；所以f(x(max=f(1(=-1；(2)f(x(=ax--(a+1(lnx,x>0，則f/(x(=a+-=，f/(x(>0，f(x(單調(diào)遞增；，f(x(單調(diào)遞減；x(>0，f(x(單調(diào)遞增；x(<0，f(x(單調(diào)遞減；又f(1(=a-1<0，由(1)得+lnx≥1，即ln≥1-x，所以lnx<當(dāng)x>1時(shí)，f(x)=ax--(a+1)lnx>ax--2(a+1)x>ax-(2a+3)x，所以f(x(僅在,+∞(有唯一零點(diǎn)，符合題意；所以f(x(有唯一零點(diǎn)，符合題意；f/(x(>0，f(x(單調(diào)遞增；x(<0，f(x(單調(diào)遞減；此時(shí)f(1(=a-1>0，此時(shí)f(x)=ax--(a+1)lnx<ax--2(a+1)(1-<-+,2.(2022·全國(guó)·高考真題)已知函(2)若f(x(在區(qū)間(-1,0(,(0,+∞(各恰有一個(gè)零點(diǎn)，求a的取值范圍.所以曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程為y=2x(2)f(x)=ln(1+x)+f/(x)=+=設(shè)g(x)=ex+a(1-x2(x+a(1-x2(>0,即f/(x)>0所以f(x)在(-1,0)上單調(diào)遞增,f(x)<f(0)=0(x)=ex-2ax>0所以f(x)在(0,+∞)上單調(diào)遞增,f(x)>f(0)=0若a<-1令h(x)=,x>-1,則h/(x)=,x>-1,又e--1>0，f(e--1(≥-+a?=0，x+a(1-x2(設(shè)h(x)=g/(x)=ex-2axhx)=ex-2a>01)=+2a<0,g/(0)=1>0又g(-1)=>0-1,0(，h(x(>h(-1(=-e，又-1<eae-1<0，f(eae-1(<ae-ae=0即f(x)在(-1,0)上有唯一零點(diǎn)所以a<-1,符合題意所以若f(x)在區(qū)間(-1,0),(0,+∞)各恰有一個(gè)零點(diǎn),求a的取值范圍為(-∞,-1)(1)討論f(x)的單調(diào)性；若x∈(0,+∞(，則f/(x(>0,f(x(單調(diào)遞增；當(dāng)0<a<時(shí)，若x∈(-∞,ln(2a((，則f/(x(>0,f(x(單調(diào)遞增，若x∈(ln(2a(,0(，則f/(x(<0,f(x(單調(diào)遞減，若x∈(0,+∞(，則f/(x(>0,f(x(單調(diào)遞增；x(≥0,f(x(在R上單調(diào)遞增；f/(x(>0,f(x(單調(diào)遞增，若x∈(0,ln(2a((，則f/(x(<0,f(x(單調(diào)遞減，若x∈(ln(2a(,+∞(，則f/(x(>0,f(x(單調(diào)遞增；而f(-b而f(\a a a-be-b+b-b而函數(shù)在區(qū)間(-∞,0(上單調(diào)遞增，故函數(shù)在區(qū)間(-∞,0(上有一個(gè)零點(diǎn).f(ln(2a((=2a[ln(2a(-1[-a[ln(2a([2+b>2a[ln(2a(-1[-a[ln(2a([2+2a=2aln(2a(-a[ln(2a([2=aln(2a([2-ln(2a([，2，故aln(2a([2-ln(2a([≥0，-1≤2a-1<0，當(dāng)b<0時(shí)，構(gòu)造函數(shù)H(x(=ex-x-1，則H/(x(=ex-1，當(dāng)x∈(-∞,0(時(shí)，H/(x(<0,H(x(單調(diào)遞減，x(>0,H(x(單調(diào)遞增，f(x(=(x-1(ex-ax2+b≥(x-1((x+1(-ax2+b=(1-a(x2+(b-1(，當(dāng)x>時(shí)，(1-a(x2+(b-1(>0，取x0=+1，則f(x0(>0，+1>0，f(ln(2a((=2a[ln(2a(-1[-a[ln(2a([2+b≤2a[ln(2a(-1[-a[ln(2a([2+2a=2aln(2a(-a[ln(2a([2=aln(2a([2-ln(2a([，由于0<a<，0<2a<1，故aln(2a([2-ln(2a([<0，(2)若曲線y=f(x(與直線y=1有且僅有兩個(gè)交點(diǎn)，求a的取值范f/(x(==,令f'(x(=0得x=,當(dāng)0<x<時(shí)，f/(x(>0,當(dāng)x>時(shí)，f/(x(<0,f(x(==1?ax=xa?xlna=alnx?=,設(shè)函數(shù)g(x(=,x(>0,g(x(單調(diào)遞增；在(e,+∞(上g/(x(<0,g(x(單調(diào)遞減；∴g(x(max=g(e(=,所以曲線y=f(x(與直線y=1有且僅有兩個(gè)交點(diǎn)，即曲線y=g(x(與直線y=有兩個(gè)交點(diǎn)的充分必要條件是0<<,這即是0<g(a(<g(e(,程alnx=xlna在區(qū)間(0,+∞)內(nèi)有兩個(gè)解.構(gòu)造函數(shù)g(x)=alnx-xlna,x∈(0,+∞)，求導(dǎo)數(shù)得g/(x)=-lna=.(x)<由于0<e-<1<,g(e-=-1-e-<0，僅有兩個(gè)交點(diǎn).lnx0=(x-x0(，即y=x-1+lnx0.(x)=0得x=.(x)>0得0<x<,f(x)在區(qū)間(0,內(nèi)單調(diào)遞增，由f/(x)<0得x>,f(x)在區(qū)間,+∞(內(nèi)單調(diào)遞減.-a,a1->5.(2022·全國(guó)·高考真題)已知函數(shù)f(x)=ex-ax和g(x)=ax-lnx有相同的最小值.g(x)=ax-lnx的定義域?yàn)?0,+∞(，而g/(x)=a-=.當(dāng)x<lna時(shí)，f/(x)<0，故f(x)在(-∞,lna(上為減函數(shù)，當(dāng)x>lna時(shí)，f/(x)>0，故f(x)在(lna,+∞(上為增函數(shù)，故f(x)min=f(lna(=a-alna.故g(x)min=g=1-ln.因?yàn)閒(x)=ex-ax和g(x)=ax-lnx有相同的最小值，設(shè)g(a(=-lna,a>0，則g/(a(=-=≤0，由(1)可得f(x)=ex-x和g(x)=x-lnx的最小值為1-ln1=1-ln=1.x-x=b的解的個(gè)數(shù)、x-lnx=b的解的個(gè)數(shù).設(shè)S(x(=ex-x-b，S/(x(=ex-1，x(>0，故S(x(在(-∞,0(上為減函數(shù)，在(0,+∞(上為增函數(shù)，所以S(x(min=S(0(=1-b<0，b-2>0，故u(b(在(1,+∞(上為增函數(shù)，故u(b(>u(1(=e-2>0，故S(b(>0，故S(x(=ex-x-b有兩個(gè)不同的零點(diǎn)，即ex-x=b的解的個(gè)數(shù)為2.設(shè)T(x(=x-lnx-b，T/(x(=，當(dāng)0<x<1時(shí)，T/(x(<0，當(dāng)x>1時(shí)，T/(x(>0，故T(x(在(0,1(上為減函數(shù)，在(1,+∞(上為增函數(shù)，所以T(x(min=T(1(=1-b<0，而T(e-b(=e-b>0，T(eb(=eb-2b>0，T(x(=x-lnx-b有兩個(gè)不同的零點(diǎn)即x-lnx=b的解的個(gè)數(shù)為2.故若存在直線y=b與曲線y=f(x(、y=g(x(有三個(gè)不同的交點(diǎn)，設(shè)h(x)=ex+lnx-2x，其中x>0，故h/(x)=ex+-2，設(shè)s(x(=ex-x-1，x>0，則s/(x(=ex-1>0，故s(x(在(0,+∞(上為增函數(shù)，故s(x(>s(0(=0即ex>x+1，所以h/(x)>x+-1≥2-1>0，所以h(x)在(0,+∞(上為增函數(shù)，而h(1)=e-2>0，h=e-3-<e-3-<0，故h(x(在(0,+∞(上有且只有一個(gè)零點(diǎn)x0，<x0<1且：當(dāng)0<x<x0時(shí)，h(x(<0即ex-x<x-lnx即f(x(<g(x(，當(dāng)x>x0(x(>0即ex-x>x-lnx即f(x(>g(x(，因此若存在直線y=b與曲線y=f(x(、y=g(x(有三個(gè)不同的交點(diǎn)，故b=f(x0(=g(x0(>1，此時(shí)ex-x=b有兩個(gè)不同的根x1,x0(x1<0<x0)，此時(shí)x-lnx=b有兩個(gè)不同的根x0,x4(0<x0<1<x4)，x-x1=b，ex-x0=b，x4-lnx4-b=0，x0-lnx0-b=0所以x4-b=lnx4即ex-b=x4即ex-b-(x4-b(-b=0，故x4-b為方程ex-x=b的解，同理x0-b也為方程ex-x=b的解x-x1=b可化為ex=x1+b即x1-ln(x1+b(=0即(x1+b(-ln(x1+b(-b=0，故x1+b為方程x-lnx=b的解，同理x0+b也為方程x-lnx=b的解，所以{x1,x0{={x0-b,x4-b{，而b>1，[方法二]：由(1)知，f(x)=ex-x，g(x)=x-lnx，且f(x)在(-∞,0)上單調(diào)遞減，在(0,+∞)上單調(diào)遞增；g(x)在(0,1)上單調(diào)遞減，在(1,+∞)上單調(diào)遞增，且f(x)min=g(x)min=1.①b<1時(shí)，此時(shí)f(x)min=g(x)min=1>b，顯然y=b與兩條曲線y=f(x)和y=g(x)②b=1時(shí)，此時(shí)f(x)min=g(x)min=1=b，故y=b與兩條曲線y=f(x)和y=g(x)共有2個(gè)交點(diǎn)，交點(diǎn)的橫坐標(biāo)分別為0和1；即證明F(x)=f(x)-b有2個(gè)零點(diǎn)，F(xiàn)′(x)=f′(x)=ex-1，所以F(x)在(-∞,0)上單調(diào)遞減，在(0,+∞)上單調(diào)遞增，又因?yàn)镕(-b)=e-b>0，F(xiàn)(0)=1-b<0，F(xiàn)(b)=eb-2b>0，(令t(b)=eb-2b，則t′(b)=eb-2>0，t(b)>t(1)=e-2>0)所以F(x)=f(x)-b在(-∞,0)上存在且只存在1個(gè)零點(diǎn)，設(shè)為x1，在(0,+∞)上存在且只存在1個(gè)零點(diǎn)，即證明G(x)=g(x)-b有2個(gè)零點(diǎn)，G′(x)=g′(x)=1-，又因?yàn)镚(e-b)=e-b>0，G(1(=1-b<0，G(2b)=b-ln2b>0，(令μ(b)=b-ln2b，則μ′(b)=1->0，μ(b)>μ(1)=1-ln2>0)2=x3:因?yàn)镕(x2)=G(x3)=0，所以b=ex-x2=x3-lnx3，若x2=x3x-x2=x2-lnx2，即ex-2x2+lnx2=0，所以只需證明ex-2x+lnx=0在(0,1)上有解即可，即φ(x)=ex-2x+lnx在(0,1)上有零點(diǎn)，因?yàn)棣?e--3<0，φ(1)=e-2>0，此時(shí)取b=ex-x0則此時(shí)存在直線y=b，其與兩條曲線y=f(x)和y因?yàn)镕(x1)=F(x2)=F(x0)=0=G(x3)=G(x0)=G(x4)所以F(x1)=G(x0)=F(lnx0)，又因?yàn)镕(x)在(-∞,0)上單調(diào)遞減，x1<0，0<x0<1即lnx0<0，所以x1=lnx0，同理，因?yàn)镕(x0)=G(ex)=G(x4)，又因?yàn)閑x-2x0+lnx0=0，所以x1+x4=ex+lnx0=2x0，即直線y=b與兩條曲線y=f(x)和y=g(x)從左到右的三個(gè)交點(diǎn)的橫坐標(biāo)成等差數(shù)列.6.(2022·新Ⅰ卷·高考真題)設(shè)函數(shù)f(x)=+lnx(x>0).(ⅱ)若0<a<e,x1<x2<x3，則+<+<-.x(=-+=，故f(x(的減區(qū)間為(0,f(x(的增區(qū)間為,+∞(.i故f(xi(-b=f/(xi((xi-a(，故方程f(x(-b=f/(x((x-a(有3個(gè)不同的根，該方程可整理為-(x-a(--lnx+b=0，設(shè)g(x(=-(x-a(--lnx+b，則g/(x(=-+(-+(x-a(-+=-(x-e((x-a(，故-(e-a(--lne+b<0且-(a-a(--lna+b>0，整理得到：b<+1且b>+lna=f(a(，此時(shí)b-f(a(--1(<+lna(-+=--lna，設(shè)u(a(=--lna，則u/(a(=<0，故u(a(為(e,+∞(上的減函數(shù)，故u(a(<--lne=0，故0<b-f(a(<-1(.不妨設(shè)x1<x2<x3，則0<x1<a<x2<e<x3，故-(e-a(--lne+b>0且-(a-a(--lna+b<0，因?yàn)閤1<x2<x3<a<x2<e<x3，又g(x(=1-+-lnx+b，-+-lnx+b=0即為：-t+t2+lnt+b=0即為-(m+1(t+t2+lnt+b=0，記t1=,t2=,t3=,則t1,t2,t3為-(m+1(t+t2+lnt+b=0有三個(gè)不同的根，+t3<-，即證：<t1+t3<-，t3-t1+t3-+<0，t3-2-<，而-(m+1(t1+t+lnt1+b=0且-(m+1(t3+t+lnt3+b=0，故lnt1-lnt3+(t-t(-(m+1((t1-t3(=0，故t1+t3-2-=-×,mt1-t336m(t1+t3(，故即證：-2×lnt1-lnt3mt1-t336m(t1+t3(，k(=-k--2lnk(，φ/(k(>0，72(m+1(記ω(m(=lnm+(m-1((m-13((m2-m+12(,0<72(m+1( 7.(2024·河北邯鄲·模擬預(yù)測(cè))已知函數(shù)f(x(=(lnx+x((ex-(a∈R(.【答案】(1)y=(3e-1(x-2e；則f/(x(=+1((ex-+(lnx+x((ex+，所以f(1(=e-1,f/(1(=3e-1，所以y=f(x(在點(diǎn)(1,f(1((處的切線方程為y-(e-1(=(3e-1((x-1(，即y=(3e-1(x-2e.所以g(x(在(0,+∞(單調(diào)遞增，且g=-ln2+=ln<0,g(1(=1>0，x，設(shè)h(x(=xex(x>0)，可得h/(x(=(x+1(ex>0，8.(2024·四川·一模)設(shè)f(x(=ex-x-ax(2)討論f(x(的零點(diǎn)數(shù)量.x-x,f/(x(=(3x2-1(ex-x.注意到ex-x>0，從而f/(x(的正負(fù)只和(3x2-1(有關(guān)，從而可作出下表：(-∞,--33(-3,+∞(f/(x(+0-0+f(x(↗e↘-e↗從而f(x(的單調(diào)遞增區(qū)間是(-∞,-,+∞(，單調(diào)遞減區(qū)間是(-,(.當(dāng)a≠0時(shí)，注意到所求可以化為x3-x=ln(ax((ax>0(的解的數(shù)量.(-∞,-1(-1(-1,-√3-30+++0--g(x(↗0↗9↘00、31--0+++g(x(0↘-9↗0↗則(g(x(-h(x((/=3x2-1->0，從而g(x(-h(x(單調(diào)遞增，零點(diǎn)若有則至多有一個(gè)，1≥-從而g(x1(>0>-1≥h(x1(，設(shè)x2=min{((2≤-2,，從而g(x2(<0<1≤h(x2(，從而在(x2,x1(上必然有一個(gè)零點(diǎn).從而總是有一個(gè)零點(diǎn).(-∞,-1-3(-, 31k+0-0+++k(x(↗7-9↘-9↗1↗其在(0,+∞(上的正負(fù)性和k(x(一樣，從而(g(x(-h(x((先單調(diào)減少后單調(diào)遞增，注意到α3=(α+1(，從而此處g(α(-h(α(=-α-ln(α(-ln(a(，當(dāng)a>b時(shí)，g(x(-h(x(在x=α處小于0，在x1=min{x1(<-1<g(x1(，從而g(x(-h(x(在(x1,α(上有一個(gè)零點(diǎn).設(shè)u(x(=lnx-x+1,x>1，則u/(x(=-1<0，所以函數(shù)u(x(在(1,+∞(上單調(diào)遞減，則u(x(=lnx-x+1<u(1(=0，即lnx<x-1，當(dāng)x≥2時(shí)，g(x(-h(x(≥3x-x+1-ln(a(=2x+(1-ln(a((，從而在x2=max{x(-h(x(>0，=，綜上a=1,b=-1.所以f(x)=有兩個(gè)不同的實(shí)數(shù)根可轉(zhuǎn)化為：關(guān)于x的方程=m(x≠1,x≠有兩個(gè)不同的實(shí)數(shù)根.設(shè)g(x)=x≠1,x≠,令g/(x)=0得，x=0或x=.x(-∞,0)0 32g/(x(+0---0+g(x)12×-1(1-110.(2023·江蘇南通·一模)已知函數(shù)f(x(=和g(x(=在同一處取得相同的最大值.(1)求實(shí)數(shù)a；(2)設(shè)直線y=b與兩條曲線y=f(x(和y=g(x(共有四個(gè)不同的交點(diǎn)，其橫坐標(biāo)分別為x1,x2,x3,x4(x1<x2<x3<x4)，證明：x1x4=x2x3.(1)a=1(1)利用導(dǎo)數(shù)分別求f(x(,g(x(的最值點(diǎn)，列式求解即可；(2)構(gòu)建F(x(=f(x(-g(x(，利用同構(gòu)思想分析f(x(,g(x(的大小關(guān)系，進(jìn)而可得直線y=b與曲線y=f(x(和y=g(x(的交點(diǎn)，再結(jié)合G(x(=的單調(diào)性分析即可證出.可得f(x(在x=1處取到最大值f(1(=；可得f(x(在x=1處取到最小值f(1(=，不合題意；綜上所述：a>0，f(x(在x=1處取到最大值f(1(=.0<x<e1-a；令g/(x(<0，解得x>e1-a；可得g(x(在x=e1-a處取到最大值g(e1-a(=ea-1；>-a可得直線y=b與曲線y=f(x(至多有兩個(gè)交點(diǎn)；若直線y=b與兩條曲線y=f(x(和y=g(x(共有四個(gè)不同的交點(diǎn)，則0<b<1，此時(shí)直線y=b與曲線y=f(x(、y=g(x(均有兩個(gè)交點(diǎn)，構(gòu)建F(x(=f(x(-g(x(=---(x>0(，構(gòu)建G(x(==f(x(，且e>0，則F(x(=e[G(x(-G(lnex([(x>0(，構(gòu)建φ(x(=x-lnex=x-lnx-1(x>0(，則φ/(x(=1-=，則φ(x(在(1,+∞(上單調(diào)遞增，在(0,1(上單調(diào)遞減，可得φ(x(=x-lnex≥φ(1(=0，-G(lnex([=0，所以f(1(=g(1(；當(dāng)0<x<1時(shí)，lnex<x<1，且G(x(在(-∞,1(上單調(diào)遞增，則G(lnex(<G(x(，可得F(x(=e[G(x(-G(lnex([>0，所以f(x(>g(x(；當(dāng)x>1時(shí)，1<lnex<x，且G(x(在(1,+∞(上單調(diào)遞減，則G(lnex(>G(x(，可得F(x(=e[G(x(-G(lnex([<0，所以f(x(<g(x(；當(dāng)0<x<1時(shí)，f(x(>g(x(；當(dāng)x>1時(shí)，f(x(<g(x(.為x2,x4，且0<x1<x2<1<x3<x4，(=G(lnex2(，因?yàn)镚(x(在(-∞,1(上單調(diào)遞增，且x1<1,lnex2<1，則x1=lnex2，可得ex-1=x2所以==f(x1(=b；(=G(lnex4(，因?yàn)镚(x(在(1,+∞(上單調(diào)遞增，且x3>1,lnex4>1，則x3=lnex4，可得ex-1=x4，所以==f(x3(=b；=x2x3.a≤blnb，a≤lnbelnb，構(gòu)建f(x(=xex；alnea≤blnb，構(gòu)建f(x(=xlnx；③a+lna≤lnb+ln(lnb(，構(gòu)建f(x(=x+lnx.①≤,構(gòu)建f(x(=；②≤,構(gòu)建f(x(=；③a-lna≤lnb-ln(lnb(，構(gòu)建f(x(=x-lnx.a±lnea≤b±lnb，構(gòu)建f(x(=x±l1.(2024·廣東佛山·二模)已知f(x(=-e2x+4ex-ax-5.x-3x-5，f/(x(=-e2x+4ex-3=-(ex-1((ex-3(，f/(x(<0，xf/(x(>0，故f(x(的單調(diào)遞減區(qū)間為(-∞,0(、(ln3,+∞(，單調(diào)遞增區(qū)間為(0,ln3(；(2)f/(x(=-e2x+4ex-a，令t=ex，即f/(x(=-t2+4t-a，則f(x1(+f(x2(+x1+x2=-e2x1+4ex1-ax1-5-e2x2+4ex2-ax2-5+x1+x2=-(t+t(+4(t1+t2(-(a-1((lnt1+lnt2(-10=-[(t1+t2(2-2t1t2[+4(t1+t2(-(a-1(lnt1t2-10=-(16-2a(+16-(a-1(lna-10=a-(a-1(lna-2，要證f(x1(+f(x2(+x1+x2<0，即證a-(a-1(lna-2<0(0<a<4(，令g(x(=x-(x-1(lnx-2(0<x<4(，則g(x(=1-(lnx+=-lnx，令h(x(=-lnx(0<x<4(，則h(x(=--<0，則g(x(在(0,4(上單調(diào)遞減，又g(1(=-ln1=1，g(2(=-ln2<0，=-lnx0=0，即=lnx0，<0，則g(x(≤g(x0(=x0-(x0-1(lnx0-2=x0-(x0-1(×-2=x0+-3，又x0=x0+-3<0，即g(x(<0，即f(x1(+f(x2(+x1+x2<0.2.已知f(x(=(x+1(ekx,k≠0.f(m+n(+1>f(m(+f(n(.故f(x(在(0,f(0((處的切線斜率為f(0(=(0+2(e0=2，而f(0(=(0+1(e0=1，所以f(x(在(0,f(0((處的切線方程為y-1=2(x-0)，即2x-y+1=0.(2)由題意得g(x(=f(x(=(kx+k+1(ekx，則g(x(=(k2x+k2+2k(ekx，令g(x(=(k2x+k2+2k(ekx<0，即k2x+k2+2k<0,∴x<-1-，令g(x(=(k2x+k2+2k(ekx>0，即k2x+k2+2k>0,∴x>-1-，>0時(shí)，g(x(=f(x(在(0,+∞(上單調(diào)遞增，而g(0(=f(0(=k+1>0，即f(x(>0在(0,+∞(上恒成立，故f(x(在(0,+∞(上單調(diào)遞增，設(shè)h(x)=f(x+m)-f(x)，則h(x)=f(x+m)-f(x)，所以h(x)=f(x+m)-f(x)在(0,+∞(上單調(diào)遞增，而n∈(0,+∞(，則h(n)>h(0)，即f(n+m)-f(n)>f(m)-f(0)，而f(0(=1，故f(n+m)-f(n)>f(m)-1，即f(m+n(+1>f(m(+f(n(.(2)若函數(shù)f(x(有兩個(gè)極值點(diǎn)x1,x2(x1<x2(，求2f(x1(-f(x2(的最小值.思路詳解：(1)因?yàn)閒(x(=lnx+x2-2ax,x>0，所以f/(x)=+2x-2a=,3=,x4=3<x4，當(dāng)0<x<x3或x>x4時(shí)，g(x)>0，即f/(x)>0；其中x3=,x4=.3此時(shí)a>2，x1+x2=a>0，x1?x2=1、1=2x+1，2ax2=2x+1、則2f(x1(-f(x2(=2(lnx1+x-2ax1(-(lnx2+x-2ax2(=2(lnx1+x-2x-1(-(lnx2+x-2x-1(=-2x+2lnx1-lnx2+x-1=x-2(222+2ln22-lnx2-1=x-2-lnx-2ln2-1t(=t-t-lnt-2ln2-1，則g/(t(=1+22-=(2t-2(t-1(，t(>0，則g(t(單調(diào)遞增，所以g(t(min=g(1(=-1+ln2，所以2f(x1(-f(x2(的最小值為-1+ln2.，使得f(x1(=f(x2(x1x2(>0.(2)不等式轉(zhuǎn)化為證明x1<<ea-1，即證明f(x2(<f，構(gòu)造函數(shù)g(x(=f(x(-fx∈【詳解】(1)f/(x(=a-1-lnx，當(dāng)0<x<ea-1時(shí)，f/(x(>0,f(x(單調(diào)遞增；當(dāng)x>ea-1x(<0,f(x(單調(diào)遞減．所以f(x)max=f(ea-1(=ea-1≤1，<x2<ea-1<x2<ea，要證f/(所以只需證f(x1(<f，即f(x2(<f.令g(x(=f(x(-fx∈(ea-1,ea(，則g(ea-1(=0，g/(x(=.當(dāng)x>ea-1時(shí)，a-1-lnx<0,x2-e2a-2>0，則g/(x(<0，所以g(x(在(ea-1,ea(上單調(diào)遞減，則g(x(<g(ea-1(=0．所以f(x1(<f.5.(2024·山西·模擬預(yù)測(cè))已知函數(shù)f(x)=2(x1<x2(是函數(shù)f(x)的兩個(gè)極值點(diǎn)，求證：f(x1(-f(x2(<(a-(x1-x2(.(3)化簡(jiǎn)f(x1(-f(x2(=ln-，將f(x1(-f(x2(<(a-(x1-x2(轉(zhuǎn)化成ln<，再構(gòu)造函數(shù)h(t)=lnt-，通過討論其單調(diào)性即可求證.x(<0，x(==，x(>0，所以f(x)<.因?yàn)閒(x)有兩個(gè)極值點(diǎn)x1，x2(x1<x2(，,所以f(x1(-f(x2(=(lnx1+x-x1+2(-(lnx2+x-x2+2(=ln+(x-x(-(x1-x2(=ln+-(x1-x2(=ln-，所以要證f(x1(-f(x2(<(a-(x1-x2(，即證ln-<(a-(x1-x2(，令h(t)=lnt-，則h/(t(=>0，所以原不等式f(x1(-f(x2(<(a-(x1-x2(成立.6.已知函數(shù)f(x(=ln(x+1(-x2-ax-1(a∈R(.f(x1(-f(x2(≥M，求M的最大值；f/<0.f(x1(-f(x2([max=f(x(max-f(x(min=【詳解】(1)當(dāng)a=-2時(shí)，f(x(=ln(x+1(-x2+2x-1，則f(x(的定義域?yàn)?-1,+∞(，且f/(x(=-2x+2=，f/(x(>0，所以f(x(在[0,1[上單調(diào)遞增，所以f(x(在[0,1[上的最大值為f(1(=ln2，最小值為f(0(=-1，由題意知M≤[f(x1(-f(x2([max=f(x(max-f(x(min=ln2+1，(2)證明：由題意知f(m(=ln(m+1(-m2-am-1=0，f(n(=ln(n+1(-n2-an-1=0，所以f(m(-f(n(=ln-(m+n((m-n(-a(m-n(=0，所以a=-ln-(m+n(.因?yàn)閒/(x(=-2x-a，所以f/=-(m+n(-a=-(m+n(--ln+(m+n(令g(t(=+lnt(0<t<1(，則g/(t(=+=，所以g(t(在(0,1(上單調(diào)遞增，零點(diǎn)問題是高考的熱點(diǎn)問題，隱零點(diǎn)的代換與估計(jì)問題是函數(shù)f(x(=xex?f(-lnx(=-?x2ex+lnx=0所以在解決形如?x+lnx=0,這些常見的代換都是隱零點(diǎn)中常見的操作.∵f(x)=aex-1-lnx+lna,∴f/(x)=aex-1-，且a>0.x-1+>0,f(x(min=f(1(=1,∴f(x(≥1成立.-1<1，1-1-1((a-1)<0,)時(shí)f/(x)<0x0-1=，∴l(xiāng)na+x0-1=-lnx0，因此f(x)min=f(x0)=aex-1-lnx0+lna=+lna+x0-1+lna≥2lna-1+2?x0=2lna+1>1,當(dāng)0<a<1時(shí)，f(1)=a+lna<a<1,∴f(1)<1,f(x)≥1不是恒成立.由f(x)≥1得aex-1-lnx+lna≥1，即elna+x-1+lna+x-1≥lnx+x，而lnx+x=elnx+lnx，所以elna+x-1+lna+x-1≥elnx+lnx.令h(m)=em+m，則h/(m)=em+1>0，所以h(m)在R上單調(diào)遞增.lna+x-1+lna+x-1≥elnx+lnx，可知h(lna+x-1)≥h(lnx)，所以lna+x-1≥lnx，所以lna≥(lnx-x+1)max.令F(x)=lnx-x+1，則F(x)=-1=.所以a的取值范圍為a≥1.由題意知a>0,x>0，令aex-1=t，所以lna+x-1=lnt，所以lna=lnt-x+1.于是f(x)=aex-1-lnx+lna=t-lnx+lnt-x+1.由于f(x)≥1,t-lnx+lnt-x+1≥1?t+lnt≥x+lnx，而y=x+lnx在x∈(0,+∞)時(shí)為增函數(shù)，故t≥x，即aex-1≥x，分離參數(shù)后有a≥.下面證明當(dāng)a≥1時(shí)，f(x)≥1恒成立.令T(a)=aex-1-lnx+lna，只需證當(dāng)a≥1時(shí)，T(a)≥1[T(a)]min=T(1)=ex-1-lnx.x≥x+1,lnx≤x-1，得ex-1≥x,-lnx≥1-x.上面兩個(gè)不等式兩邊相加可得ex-1-lnx≥1，故a≥1時(shí)，f(x)≥1恒成立.當(dāng)0<a<1時(shí)，因?yàn)閒(1)=a+lna<1，顯然不滿足f(x)≥1恒成立.所以a的取值范圍為a≥1.則f(x(=-a=，x-lnx-x-1,x>0，gI(x(=(x+1(ex--1，令h(x(=(x+1(ex--1，則hI(x(=(x+2(ex+>0,(x>0(，即h(x(在(0,+∞)上單調(diào)遞增，h=e,h(e(=(e+1(ee--1>0，故g(x(min=g(x0(=x0ex0-ln-x0-1=0即g(x(≥0，即xex≥lnx+x+1，則lnx+x+1≤xex.3.已知函數(shù)f(x)=ln(ax),a>0，若f(x)≤(x-1)ex-a,求a的取值范圍.x-a-f(x)=(x-1)ex-a-lnx-lna,x>0,依題意,h(x)≥0恒成立,求導(dǎo)得hI(x)=xex-a-,x>0,令y=hI(x)=xex-a-,yI=(x+1)ex-a+>0,又hI=e-2<0,hI(a+1)=(a=0,即x0ex0-a=成立,h(x)min=h(x0(=(x0-1(ex-a-lnx0-lna,ex0-a=,得ex0-a=,a=x0+2lnx0,于是得h(x0(=-lnx0-ln(x0+2lnx0(,于是得函數(shù)φ(x)=-lnx-ln(x+2lnx)0+2lnx0>0時(shí),由(1)中l(wèi)nx≤x-1知,-lnx0≥1-x0,有-ln(x0+2lnx0(≥1-(x0+2lnx0(,h(x0(=-lnx0-ln(x0+2lnx0(≥-lnx0+1-(x0+2lnx0(=-3lnx0-x0+1≥-3(x0-1(-x0+1=,≥0,因此滿足f(x)≤(x-1)ex-a,又a=x0+2lnx0,y=x+2lnx(x)=a-=(x>0)，(2)證明:由(1)知g(x)=x2-x-xlnx,g/(x)=2x-2-lnx，令h(x)=2x-2-lnx,h/(x)=2-=(x>0)，g(x0(=x-x0-x0lnx0，2(x0-1(，所以g(x0(=x-x0-2x0(x0-1)=-(x0-2+，<.(x)=lnx+2(1)=2f(1)=-2=令h(x)=3lnx-x+2，則h/(x)=<0恒成立.00-x0+2=0，則lnx0=，00(x)<0值范圍.(2)(-∞,2-ln2[①當(dāng)a≤0時(shí)，x2-a>0，則f/(x(>0，故f(x(單調(diào)遞增.調(diào)遞增.min=f(a(=a-alna=，即a-alna=1.令m(x(=x-xlnx，則m/(x(=-lnx，令m/(x(=-lnx>0，解得0<x<1，由題可得x2-lnx-(x-1(2ex+e2≥m在,2上恒成立.令h(x(=x2-lnx-(x-1(2ex+e2,x∈,2，則h/(x(=(x+1((x-1(-ex(，令t(x(=-ex，則t/(x(=--ex<0，可得t(x(在(0,+∞(上單調(diào)遞減，x0,-lnx0=x0，易知h(2(=2-ln2,h(x0(=x-lnx0-(x0-1(2ex0+e2=-+2+e2，由于-<-<-，故2-ln2<+e2<h(x0(，因此h(x)min=2-ln2，故m≤2-ln2，即m的取值范圍為(-∞,2-ln2[.(1)求f(x(的極值；(2)證明：xg(x(+2>exf(x(-.(2)構(gòu)造h(x(=xg(x(+2-exf(x(+=xlnx+2-x+(x>0(，求出其單調(diào)性進(jìn)而求得最小值為f/(x(<0，所以f(x(在(-∞,1(單調(diào)遞增，在(1,+∞(單調(diào)遞減，(2)解：令h(x(=xg(x(+2-exf(x(+=xlnx+2-x+(x>0(，則r/(x(=+>0在x>0上恒成立，所以r(x(在(0,+∞(上單調(diào)遞增，又r(1(=ln1-=-2<0，r(e(=lne-=1->0，=0，h(x(min=h(x0(=x0lnx0+2-x0+=x0?+2-x0+=2-x0+(x0∈(1,e((，則m/(x(=-1-<0在(1,e(上恒成立，所以m(x(在(1,e(上單調(diào)遞減，所以m(x(>m(e(=2-e+>0，所以h(x(min=h(x0(=2-x0+>0，所以xg(x(+2>exf(x(-.(2)構(gòu)造一元差函數(shù)F(x)=f(x0+x)-f(x0-x)；(3)確定函數(shù)F(x)的單調(diào)性；1.(2022·全國(guó)·統(tǒng)考高考真題)已知函數(shù)f(x(=-lnx+x-a.f/(x)=令f/(x(=0,得x=1(x)<0,f(x)單調(diào)遞減(x)>0,f(x)單調(diào)遞增f(x)≥f(1)=e+1-a，所以a的取值范圍為(-∞,e+1]由f(x(≥0得：e-lnx+x+x-lnx-a≥0令t=x-lnx,t≥1，則f(t(=et+t-a≥0即a≤et+t故g(t(=et+t在區(qū)間[1,+∞(上是增函數(shù)所以a的取值范圍為(-∞,e+1]∈(0,1),即證f(x1(>f又因?yàn)閒(x1(=f(x2(,故只需證f(x2(>f即證-lnx+x-即證--2lnx-x->0下面證明x>1時(shí)，->0,lnx-x-<0設(shè)g(x)=-,x>1，則g/(x)=-x+-=1-x-e1-=(1--e=-e設(shè)φ(x(=(x>1(,φ/(x(=-x=ex>01所以-e>0,所以g/(x)>0即g(x)>g(1)=0,所以->0令h(x)=lnx-x-x>1h/(x)=-1+==<0即h(x)<h(1)=0,所以lnx-x-<0；綜上,--2lnx-x->0,所以x1x2<1.由題意得：f(x(=+ln-a令t=>1，則f(t(=t+lnt-a，f'(t(=1+>0所以g(t(=t+lnt-a在(1,+∞(上單調(diào)遞增，故g(t(=0只有1個(gè)解又因?yàn)閒(x(=+ln-a有兩個(gè)零點(diǎn)x1,x2，故t==兩邊取對(duì)數(shù)得：x1-lnx1=x2下證x1x2<(*(因?yàn)閤1x2<?lnx1-lnx2<?ln<-不妨設(shè)t=>1，則只需證2lnt<t-構(gòu)造h(t(=2lnt-t+,t>1，則h'(t(=-1-1-2<0故h(t(=2lnt-t+在(1,+∞(上單調(diào)遞減故h(t(<h(1(=0，即2lnt<t-得證2.已知函數(shù)f(x(=lnx-ax+1,a∈R.由f(x(=lnx-ax+1≤0，得a≥.x(=.從而g(x)max=g(e(=.ln(ex(+ln(ex2(=eax+ax.設(shè)F(x(=x+ex，則eln(ex(+ln(ex2(=eax+ax等價(jià)于F(ln(ex2((=F(ax(.由h/(x(>0則h(x(在(0,、e(上單調(diào)遞增，在(、e,+∞(上單調(diào)遞減，方程f(x2(=eax-ex2有兩個(gè)不同的正實(shí)根x1,x2，不妨設(shè)x1<x2，<x<x1<e<x2,0<a<.e(t-1)2t(t+1)2>0,(t-1)2t(t+1)2>0,則G(t(在(1,+∞(上單調(diào)遞增.2(x2-x1(x1+x2x1+x22(x2-x1(x1+x2x1+x2<lnx2-lnx1，則<.2因?yàn)榉匠蘤(x2(=eax-ex2有兩個(gè)不同的正實(shí)根x1,x2，x2-x12=lnx2x2-x12=lnx2-lnx1a.x2-x1lnx2-lnx1x2-x1lnx2-lnx1則x1x2>e3.(2024·全國(guó)·模擬預(yù)測(cè))已知函數(shù)f(x(=1-lnx-(a∈R(.(1)求f(x(的單調(diào)區(qū)間；x(=，因?yàn)閒(