2022年高考數(shù)學(xué)二輪復(fù)習(xí) 專(zhuān)題突破練8 利用導(dǎo)數(shù)證明問(wèn)題及討論零點(diǎn)個(gè)數(shù) 理

2022年高考數(shù)學(xué)二輪復(fù)習(xí) 專(zhuān)題突破練8 利用導(dǎo)數(shù)證明問(wèn)題及討論零點(diǎn)個(gè)數(shù) 理1.(2018河南鄭州二模,理21)已知函數(shù)f(x)=ex-x2.(1)求曲線f(x)在x=1處的切線方程;(2)求證:當(dāng)x>0時(shí),ln x+1.2.(2018河南鄭州一模,理21)已知函數(shù)f(x)=ln x+,aR且a0.(1)討論函數(shù)f(x)的單調(diào)性;(2)當(dāng)x時(shí),試判斷函數(shù)g(x)=(ln x-1)ex+x-m的零點(diǎn)個(gè)數(shù).3.設(shè)函數(shù)f(x)=x2-aln x,g(x)=(a-2)x.(1)略;(2)若函數(shù)F(x)=f(x)-g(x)有兩個(gè)零點(diǎn)x1,x2,求滿足條件的最小正整數(shù)a的值;求證:F'>0.4.(2018河北保定一模,理21節(jié)選)已知函數(shù)f(x)=ln x-(aR).(1)略;(2)若f(x)有兩個(gè)極值點(diǎn)x1,x2,證明:f.5.已知函數(shù)f(x)=(x-2)ex+a(x-1)2有兩個(gè)零點(diǎn).(1)求a的取值范圍;(2)設(shè)x1,x2是f(x)的兩個(gè)零點(diǎn),證明:x1+x2<2.6.(2018山西名校二模,理21)已知函數(shù)f(x)=mln x.(1)討論函數(shù)F(x)=f(x)+-1的單調(diào)性;(2)定義:“對(duì)于在區(qū)域D上有定義的函數(shù)y=f(x)和y=g(x),若滿足f(x)g(x)恒成立,則稱(chēng)曲線y=g(x)為曲線y=f(x)在區(qū)域D上的緊鄰曲線”.試問(wèn)曲線y=f(x+1)與曲線y=是否存在相同的緊鄰直線,若存在,請(qǐng)求出實(shí)數(shù)m的值;若不存在,請(qǐng)說(shuō)明理由.參考答案專(zhuān)題突破練8利用導(dǎo)數(shù)證明問(wèn)題及討論零點(diǎn)個(gè)數(shù)1.解 (1)f'(x)=ex-2x,由題設(shè)得f'(1)=e-2,f(1)=e-1,f(x)在x=1處的切線方程為y=(e-2)x+1.(2)f'(x)=ex-2x,f(x)=ex-2,f'(x)在(0,ln 2)上單調(diào)遞減,在(ln 2,+)上單調(diào)遞增,所以f'(x)f'(ln 2)=2-2ln 2>0,所以f(x)在0,1上單調(diào)遞增,所以f(x)max=f(1)=e-1,x0,1.f(x)過(guò)點(diǎn)(1,e-1),且y=f(x)在x=1處的切線方程為y=(e-2)x+1,故可猜測(cè):當(dāng)x>0,x1時(shí),f(x)的圖象恒在切線y=(e-2)x+1的上方.下證:當(dāng)x>0時(shí),f(x)(e-2)x+1,設(shè)g(x)=f(x)-(e-2)x-1,x>0,則g'(x)=ex-2x-(e-2),g(x)=ex-2,g'(x)在(0,ln 2)上單調(diào)遞減,在(ln 2,+)上單調(diào)遞增,又g'(0)=3-e>0,g'(1)=0,0<ln 2<1,g'(ln 2)<0,所以,存在x0(0,ln 2),使得g'(x0)=0,所以,當(dāng)x(0,x0)(1,+)時(shí),g'(x)>0;當(dāng)x(x0,1)時(shí),g'(x)<0,故g(x)在(0,x0)上單調(diào)遞增,在(x0,1)上單調(diào)遞減,在(1,+)上單調(diào)遞增,又g(0)=g(1)=0,g(x)=ex-x2-(e-2)x-10,當(dāng)且僅當(dāng)x=1時(shí)取等號(hào),故x,x>0.又xln x+1,即ln x+1,當(dāng)x=1時(shí),等號(hào)成立.2.解 (1)f'(x)=(x>0),當(dāng)a<0時(shí),f'(x)>0恒成立,函數(shù)f(x)在(0,+)上遞增;當(dāng)a>0時(shí),由f'(x)>0,得x>,由f'(x)<0,得0<x<,函數(shù)單調(diào)遞增區(qū)間為,單調(diào)遞減區(qū)間為綜上所述,當(dāng)a<0時(shí),函數(shù)f(x)的單調(diào)遞增區(qū)間為(0,+).當(dāng)a>0時(shí),函數(shù)f(x)的單調(diào)遞增區(qū)間為,單調(diào)遞減區(qū)間為(2)x時(shí),函數(shù)g(x)=(ln x-1)ex+x-m的零點(diǎn),即方程(ln x-1)ex+x=m的根.令h(x)=(ln x-1)ex+x,h'(x)=ex+1.由(1)知當(dāng)a=1時(shí),f(x)=ln x+-1在遞減,在1,e上遞增,f(x)f(1)=0.+ln x-10在x上恒成立.h'(x)=ex+10+1>0,h(x)=(ln x-1)ex+x在x上單調(diào)遞增.h(x)min=h=-2,h(x)max=e.所以當(dāng)m<-2或m>e時(shí),沒(méi)有零點(diǎn),當(dāng)-2me時(shí)有一個(gè)零點(diǎn).3.解 (1)略;(2)F(x)=x2-aln x-(a-2)x,F'(x)=2x-(a-2)-(x>0).因?yàn)楹瘮?shù)F(x)有兩個(gè)零點(diǎn),所以a>0,此時(shí)函數(shù)F(x)在單調(diào)遞減,在單調(diào)遞增.所以F(x)的最小值F<0,即-a2+4a-4aln<0.a>0,a+4ln-4>0.令h(a)=a+4ln-4,顯然h(a)在(0,+)上為增函數(shù),且h(2)=-2<0,h(3)=4ln-1=ln-1>0,所以存在a0(2,3),h(a0)=0.當(dāng)a>a0時(shí),h(a)>0,所以滿足條件的最小正整數(shù)a=3.證明:不妨設(shè)0<x1<x2,于是-(a-2)x1-aln x1=-(a-2)x2-aln x2,即+2x1-2x2=ax1+aln x1-ax2-aln x2=a(x1+ln x1-x2-ln x2).所以a=F'=0,當(dāng)x時(shí),F'(x)<0,當(dāng)x時(shí),F'(x)>0,故只要證即可,即證x1+x2>,即證+(x1+x2)(ln x1-ln x2)<+2x1-2x2,也就是證ln設(shè)t=(0<t<1).令m(t)=ln t-,則m'(t)=因?yàn)閠>0,所以m'(t)0,當(dāng)且僅當(dāng)t=1時(shí),m'(t)=0,所以m(t)在(0,+)上是增函數(shù).又m(1)=0,所以當(dāng)t(0,1),m(t)<0總成立,所以原題得證.4.解 (1)略;(2)f'(x)=(x>0),令p(x)=x2+(2-a)x+1,由f(x)在(0,+)有兩個(gè)極值點(diǎn)x1,x2,則方程p(x)=0在(0,+)有兩個(gè)實(shí)根x1,x2,得a>4.f(x1)+f(x2)=ln x1-+ln x2-=ln x1x2-=-a,f=f=ln=ln-(a-2).f=ln-a-2+=ln+2.設(shè)h(a)=ln+2(a>4),則h'(a)=<0,h(a)在(4,+)上為減函數(shù),又h(4)=0,h(a)<0,f5.(1)解 f'(x)=(x-1)ex+2a·(x-1)=(x-1)(ex+2a).()若a=0,則f(x)=(x-2)ex,f(x)只有一個(gè)零點(diǎn).()若a>0,則當(dāng)x(-,1)時(shí),f'(x)<0;當(dāng)x(1,+)時(shí),f'(x)>0,所以f(x)在(-,1)內(nèi)單調(diào)遞減,在(1,+)內(nèi)單調(diào)遞增.又f(1)=-e,f(2)=a,取b滿足b<0且b<ln,則f(b)>(b-2)+a(b-1)2=a>0,故f(x)存在兩個(gè)零點(diǎn).()若a<0,由f'(x)=0得x=1或x=ln(-2a).若a-,則ln(-2a)1,故當(dāng)x(1,+)時(shí),f'(x)>0,因此f(x)在(1,+)內(nèi)單調(diào)遞增.又當(dāng)x1時(shí),f(x)<0,所以f(x)不存在兩個(gè)零點(diǎn).若a<-,則ln(-2a)>1,故當(dāng)x(1,ln(-2a)時(shí),f'(x)<0;當(dāng)x(ln(-2a),+)時(shí),f'(x)>0.因此f(x)在(1,ln(-2a)內(nèi)單調(diào)遞減,在(ln(-2a),+)內(nèi)單調(diào)遞增.又當(dāng)x1時(shí)f(x)<0,所以f(x)不存在兩個(gè)零點(diǎn).綜上,a的取值范圍為(0,+).(2)證明 不妨設(shè)x1<x2,由(1)知,x1(-,1),x2(1,+),2-x2(-,1),f(x)在(-,1)內(nèi)單調(diào)遞減,所以x1+x2<2等價(jià)于f(x1)>f(2-x2),即f(2-x2)<0.由于f(2-x2)=-x2+a(x2-1)2,而f(x2)=(x2-2)+a(x2-1)2=0,所以f(2-x2)=-x2-(x2-2)設(shè)g(x)=-xe2-x-(x-2)ex,則g'(x)=(x-1)(e2-x-ex).所以當(dāng)x>1時(shí),g'(x)<0,而g(1)=0,故當(dāng)x>1時(shí),g(x)<0.從而g(x2)=f(2-x2)<0,故x1+x2<2.6.解 (1)F'(x)=(x>0).當(dāng)m0時(shí),F'(x)<0,函數(shù)F(x)在(0,+)上單調(diào)遞減;當(dāng)m>0時(shí),令F'(x)<0,得x<,函數(shù)F(x)在上單調(diào)遞減;令F'(x)>0,得x>,函數(shù)F(x)在上單調(diào)遞增;綜上所述,當(dāng)m0時(shí),函數(shù)F(x)在(0,+)上單調(diào)遞減;當(dāng)m>0時(shí),函數(shù)F(x)在上單調(diào)遞減,在上單調(diào)遞增.(2)原命題等價(jià)于曲線y=f(x+1)與曲線y=是否相同的外公切線.函數(shù)f(x+1)=mln(x+1)在點(diǎn)(x1,mln(x1+1)處的切線方程為y-mln(x1+1)=(x-x1),即y=x+mln(x1+1)-,曲線y=在點(diǎn)處的切線方程為y-(x-x2),即y=x+曲線y=f(x+1)與y=的圖象有且僅有一條外公切線,所以有唯一一對(duì)(x1,x2)滿足這個(gè)方程組,且m>0,由得x1+1=m(x2+1)2,代入消去x1,整理得2mln(x2+1)+mln m-m-1=0,關(guān)于x2(x2>-1)的方程有唯一解.令g(x)=2mln(x+1)+mln m-m-1(x>-1),g'(x)=當(dāng)m>0時(shí),g(x)在上單調(diào)遞減,在上單調(diào)遞增;g(x)min=g=m-mln m-1.因?yàn)閤+,g(x)+;x-1,g(x)+,只需m-mln m-1=0.令h(m)=m-mln m-1,h'(m)=-ln m在(0,+)上為單調(diào)遞減函數(shù),且m=1時(shí),h'(m)=0,即h(m)max=h(1)=0,所以m=1時(shí),關(guān)于x2的方程2mln(x2+1)+mln m-m-1=0有唯一解,此時(shí)x1=x2=0,外公切線的方程為y=x.這兩條曲線存在相同的緊鄰直線,此時(shí)m=1.