2022年高考數(shù)學(xué)二輪復(fù)習(xí) 專(zhuān)題二 函數(shù)與導(dǎo)數(shù) 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) 文

2022年高考數(shù)學(xué)二輪復(fù)習(xí) 專(zhuān)題二 函數(shù)與導(dǎo)數(shù) 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) 文1.(2018全國(guó)卷3,文21)已知函數(shù)f(x)=.(1)求曲線y=f(x)在點(diǎn)(0,-1)處的切線方程;(2)證明:當(dāng)a1時(shí),f(x)+e0.2.設(shè)函數(shù)f(x)=e2x-aln x.(1)討論f(x)的導(dǎo)函數(shù)f'(x)零點(diǎn)的個(gè)數(shù);(2)證明當(dāng)a>0時(shí),f(x)2a+aln.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(mǎn)足條件的最小正整數(shù)a的值;求證:F'>0.4.(2018福建龍巖4月質(zhì)檢,文21節(jié)選)已知函數(shù)f(x)=-2ln x,mR.(1)略;(2)若函數(shù)f(x)有兩個(gè)極值點(diǎn)x1,x2,且x1<x2,證明:f(x2)<x2-1.5.已知函數(shù)f(x)=aln x,g(x)=x+f'(x).(1)討論h(x)=g(x)-f(x)的單調(diào)性;(2)若h(x)的極值點(diǎn)為3,設(shè)方程f(x)+mx=0的兩個(gè)根為x1,x2,且ea,求證:.6.(2018河南安陽(yáng)一模,文21)已知函數(shù)f(x)=,g(x)=3eln x,其中e為自然對(duì)數(shù)的底數(shù).(1)討論函數(shù)f(x)的單調(diào)性.(2)試判斷曲線y=f(x)與y=g(x)是否存在公共點(diǎn)并且在公共點(diǎn)處有公切線.若存在,求出公切線l的方程;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)=,f'(0)=2.因此曲線y=f(x)在(0,-1)處的切線方程是2x-y-1=0.(2)當(dāng)a1時(shí),f(x)+e(x2+x-1+ex+1)e-x.令g(x)=x2+x-1+ex+1,則g'(x)=2x+1+ex+1.當(dāng)x<-1時(shí),g'(x)<0,g(x)單調(diào)遞減;當(dāng)x>-1時(shí),g'(x)>0,g(x)單調(diào)遞增;所以g(x)g(-1)=0.因此f(x)+e0.2.(1)解 f(x)的定義域?yàn)?0,+),f'(x)=2e2x-(x>0).當(dāng)a0時(shí),f'(x)>0,f'(x)沒(méi)有零點(diǎn),當(dāng)a>0時(shí),因?yàn)閑2x單調(diào)遞增,-單調(diào)遞增,所以f'(x)在(0,+)單調(diào)遞增.又f'(a)>0,當(dāng)b滿(mǎn)足0<b<且b<時(shí),f'(b)<0,故當(dāng)a>0時(shí),f'(x)存在唯一零點(diǎn).(2)證明 由(1),可設(shè)f'(x)在(0,+)的唯一零點(diǎn)為x0,當(dāng)x(0,x0)時(shí),f'(x)<0;當(dāng)x(x0,+)時(shí),f'(x)>0.故f(x)在(0,x0)單調(diào)遞減,在(x0,+)單調(diào)遞增,所以當(dāng)x=x0時(shí),f(x)取得最小值,最小值為f(x0).由于2=0,所以f(x0)=+2ax0+aln2a+aln.故當(dāng)a>0時(shí),f(x)2a+aln.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(mǎn)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.證明 (1)略;(2)函數(shù)f(x)的定義域?yàn)閤>0,f'(x)=,f(x)有兩個(gè)極值點(diǎn)x1,x2,且x1<x2,g(x)=x2-2x-m=0有兩個(gè)不同的正根x1=1-,x2=1+,-1<m<0.欲證明f(x2)=x2+-2ln x2<x2-1,即證明2ln x2->1,m=-2x2,證明2ln x2->1成立,等價(jià)于證明2ln x2-x2>-1成立.m=x2(x2-2)(-1,0),x2=1+(1,2).設(shè)函數(shù)h(x)=2ln x-x,x(1,2),求導(dǎo)可得h'(x)=-1.易知h'(x)>0在x(1,2)上恒成立,即h(x)在x(1,2)上單調(diào)遞增,h(x)>h(1)=-1,即2ln x2-x2>-1在x2(1,2)上恒成立,函數(shù)f(x)有兩個(gè)極值點(diǎn)x1,x2,且x1<x2時(shí),f(x2)<x2-1.5.(1)解 h(x)=g(x)-f(x)=x-aln x+,其定義域是(0,+),h'(x)=.1+a0即a-1時(shí),x(0,+)時(shí),h'(x)>0,h(x)在(0,+)遞增;a+1>0即a>-1時(shí),x(0,1+a)時(shí),h'(x)<0,x(1+a,+)時(shí),h'(x)>0,h(x)在(0,1+a)遞減,在(1+a,+)遞增,綜上,a>-1時(shí),h(x)在(0,1+a)遞減,在(1+a,+)遞增,a-1時(shí),h(x)在(0,+)遞增.(2)證明 由(1)得x=1+a是函數(shù)h(x)的唯一極值點(diǎn),故a=2.2ln x1+mx1=0,2ln x2+mx2=0,2(ln x2-ln x1)=m(x1-x2),又f(x)=2ln x,f'(x)=,=+m+ln.令=te2,(t)=+ln t,則'(t)=>0,(t)在e2,+)上遞增,(t)(e2)=1+>1+,故.6.解 (1)f'(x)=,令f'(x)=0得x=.當(dāng)x=且x0時(shí),f'(x)<0;當(dāng)x>時(shí),f'(x)>0.所以f(x)在(-,0)上單調(diào)遞減,在上單調(diào)遞減,在上單調(diào)遞增.(2)假設(shè)曲線y=f(x)與y=g(x)存在公共點(diǎn)且在公共點(diǎn)處有公切線,且切點(diǎn)橫坐標(biāo)為x0>0,則即其中式即4-3e2x0-e3=0.記h(x)=4x3-3e2x-e3,x(0,+),則h'(x)=3(2x+e)(2x-e),得h(x)在上單調(diào)遞減,在上單調(diào)遞增,又h(0)=-e3,h=-2e3,h(e)=0,故方程h(x0)=0在(0,+)上有唯一實(shí)數(shù)根x0=e,經(jīng)驗(yàn)證也滿(mǎn)足式.于是,f(x0)=g(x0)=3e,f'(x0)=g'(x0)=3,曲線y=g(x)與y=g(x)的公切線l的方程為y-3e=3(x-e),即y=3x.