（通用版）2020版高考數(shù)學(xué)大二輪復(fù)習(xí) 專題突破練9 利用導(dǎo)數(shù)證明問題及討論零點個數(shù) 文

專題突破練9利用導(dǎo)數(shù)證明問題及討論零點個數(shù)1.設(shè)函數(shù)f(x)=e2x-aln x.(1)討論f(x)的導(dǎo)函數(shù)f'(x)零點的個數(shù);(2)證明:當(dāng)a>0時,f(x)2a+aln2a.2.(2019陜西咸陽一模,文21)設(shè)函數(shù)f(x)=x+1-mex,mR.(1)當(dāng)m=1時,求f(x)的單調(diào)區(qū)間;(2)求證:當(dāng)x(0,+)時,lnex-1x>x2.3.(2019河南洛陽三模,理21)已知函數(shù)f(x)=ln x-kx,其中kR為常數(shù).(1)討論函數(shù)f(x)的單調(diào)性;(2)若f(x)有兩個相異零點x1,x2(x1<x2),求證:ln x2>2-ln x1.4.已知函數(shù)f(x)=ln x+ax2+(2a+1)x.(1)討論f(x)的單調(diào)性;(2)當(dāng)a<0時,證明f(x)-34a-2.5.設(shè)函數(shù)f(x)=ln x-x+1.(1)討論f(x)的單調(diào)性;(2)證明當(dāng)x(1,+)時,1<x-1lnx<x;(3)設(shè)c>1,證明當(dāng)x(0,1)時,1+(c-1)x>cx.6.(2019湖南六校聯(lián)考,文21)已知函數(shù)f(x)=ex,g(x)=ax2+x+1(a>0).(1)設(shè)F(x)=g(x)f(x),討論函數(shù)F(x)的單調(diào)性;(2)若0<a12,證明:f(x)>g(x)在(0,+)上恒成立.7.(2019天津卷,文20)設(shè)函數(shù)f(x)=ln x-a(x-1)ex,其中aR.(1)若a0,討論f(x)的單調(diào)性;(2)若0<a<1e,證明f(x)恰有兩個零點;設(shè)x0為f(x)的極值點,x1為f(x)的零點,且x1>x0,證明3x0-x1>2.參考答案專題突破練9利用導(dǎo)數(shù)證明問題及討論零點個數(shù)1.解(1)f(x)的定義域為(0,+),f'(x)=2e2x-ax(x>0).當(dāng)a0時,f'(x)>0,f'(x)沒有零點,當(dāng)a>0時,因為e2x單調(diào)遞增,-ax單調(diào)遞增,所以f'(x)在(0,+)單調(diào)遞增.又f'(a)>0,當(dāng)b滿足0<b<a4且b<14時,f'(b)<0,故當(dāng)a>0時,f'(x)存在唯一零點.(2)由(1),可設(shè)f'(x)在(0,+)的唯一零點為x0,當(dāng)x(0,x0)時,f'(x)<0;當(dāng)x(x0,+)時,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,f(x)取得最小值,最小值為f(x0).因為2e2x0-ax0=0,所以f(x0)=a2x0+2ax0+aln2a2a+aln2a.故當(dāng)a>0時,f(x)2a+aln2a.2.(1)解當(dāng)m=1時,f(x)=x+1-ex,f'(x)=1-ex.令f'(x)=0,則x=0.當(dāng)x<0時,f'(x)>0;當(dāng)x>0時,f'(x)<0,故函數(shù)f(x)的單調(diào)遞增區(qū)間是(-,0);單調(diào)遞減區(qū)間是(0,+).(2)證明由(1)知,當(dāng)m=1時,f(x)max=f(0)=0,當(dāng)x(0,+)時,x+1-ex<0,即ex>x+1.當(dāng)x(0,+)時,要證lnex-1x>x2,只需證ex-1>xex2.令F(x)=ex-1-xex2=ex-x(e)x-1,F'(x)=ex-(e)x-x(e)xlne=(e)x(e)x-1-x2=ex2ex2-1-x2.由ex>x+1可得,ex2>1+x2,則x(0,+)時,F'(x)>0恒成立,即F(x)在(0,+)內(nèi)單調(diào)遞增,F(x)>F(0)=0.即ex-1>xex2,lnex-1x>x2.3.(1)解f'(x)=1x-k=1-kxx(x>0),當(dāng)k0時,f'(x)>0,f(x)在區(qū)間(0,+)內(nèi)遞增,當(dāng)k>0時,由f'(x)>0,得0<x<1k,故f(x)在區(qū)間0,1k內(nèi)遞增,在區(qū)間1k,+內(nèi)遞減.(2)證明設(shè)f(x)的兩個相異零點為x1,x2,設(shè)x1>x2>0,f(x1)=0,f(x2)=0,lnx1-kx1=0,lnx2-kx2=0,lnx1-lnx2=k(x1-x2),lnx1+lnx2=k(x1+x2),要證明lnx2>2-lnx1,即證明lnx1+lnx2>2,故k(x1+x2)>2,即lnx1-lnx2x1-x2>2x1+x2,即lnx1x2>2(x1-x2)x1+x2,設(shè)t=x1x2>1,上式轉(zhuǎn)化為lnt>2(t-1)t+1(t>1).設(shè)g(t)=lnt-2(t-1)t+1,g'(t)=(t-1)2t(t+1)2>0,g(t)在(1,+)上單調(diào)遞增,g(t)>g(1)=0,lnt>2(t-1)t+1,lnx1+lnx2>2,即lnx2>2-lnx1.4.(1)解f(x)的定義域為(0,+),f'(x)=1x+2ax+2a+1=(x+1)(2ax+1)x.若a0,則當(dāng)x(0,+)時,f'(x)>0,故f(x)在(0,+)單調(diào)遞增.若a<0,則當(dāng)x0,-12a時,f'(x)>0;當(dāng)x-12a,+時,f'(x)<0.故f(x)在0,-12a單調(diào)遞增,在-12a,+單調(diào)遞減.(2)證明由(1)知,當(dāng)a<0時,f(x)在x=-12a取得最大值,最大值為f-12a=ln-12a-1-14a.所以f(x)-34a-2等價于ln-12a-1-14a-34a-2,即ln-12a+12a+10.設(shè)g(x)=lnx-x+1,則g'(x)=1x-1.當(dāng)x(0,1)時,g'(x)>0;當(dāng)x(1,+)時,g'(x)<0.所以g(x)在(0,1)單調(diào)遞增,在(1,+)單調(diào)遞減.故當(dāng)x=1時,g(x)取得最大值,最大值為g(1)=0.所以當(dāng)x>0時,g(x)0.從而當(dāng)a<0時,ln-12a+12a+10,即f(x)-34a-2.5.(1)解由題設(shè),f(x)的定義域為(0,+),f'(x)=1x-1,令f'(x)=0解得x=1.當(dāng)0<x<1時,f'(x)>0,f(x)單調(diào)遞增;當(dāng)x>1時,f'(x)<0,f(x)單調(diào)遞減.(2)證明由(1)知f(x)在x=1處取得最大值,最大值為f(1)=0.所以當(dāng)x1時,lnx<x-1.故當(dāng)x(1,+)時,lnx<x-1,ln1x<1x-1,即1<x-1lnx<x.(3)證明由題設(shè)c>1,設(shè)g(x)=1+(c-1)x-cx,則g'(x)=c-1-cxlnc,令g'(x)=0,解得x0=lnc-1lnclnc.當(dāng)x<x0時,g'(x)>0,g(x)單調(diào)遞增;當(dāng)x>x0時,g'(x)<0,g(x)單調(diào)遞減.由(2)知1<c-1lnc<c,故0<x0<1.又g(0)=g(1)=0,故當(dāng)0<x<1時,g(x)>0.所以當(dāng)x(0,1)時,1+(c-1)x>cx.6.(1)解F(x)=g(x)f(x)=ax2+x+1ex,F'(x)=-ax2+(2a-1)xex=-ax(x-2a-1a)ex,若a=12,則F'(x)=-ax2ex0,F(x)在R上單調(diào)遞減.若a>12,則2a-1a>0,當(dāng)x<0或x>2a-1a時,F'(x)<0,當(dāng)0<x<2a-1a時,F'(x)>0,F(x)在(-,0),2a-1a,+上單調(diào)遞減,在0,2a-1a上單調(diào)遞增.若0<a<12,則2a-1a<0,當(dāng)x<2a-1a或x>0時,F'(x)<0,當(dāng)2a-1a<x<0時,F'(x)>0.F(x)在-,2a-1a,(0,+)上單調(diào)遞減,在2a-1a,0上單調(diào)遞增.(2)證明0<a12,ax2+x+112x2+x+1.設(shè)h(x)=ex-12x2-x-1,則h'(x)=ex-x-1.設(shè)p(x)=h'(x)=ex-x-1,則p'(x)=ex-1,在(0,+)上,p'(x)>0恒成立,h'(x)在(0,+)上單調(diào)遞增.又h'(0)=0,當(dāng)x(0,+)時,h'(x)>0,h(x)在(0,+)上單調(diào)遞增,h(x)>h(0)=0,ex-12x2-x-1>0,ex>12x2+x+1,ex>12x2+x+1ax2+x+1,綜上,f(x)>g(x)在(0,+)上恒成立.7.(1)解由已知,f(x)的定義域為(0,+),且f'(x)=1x-aex+a(x-1)ex=1-ax2exx.因此當(dāng)a0時,1-ax2ex>0,從而f'(x)>0,所以f(x)在(0,+)內(nèi)單調(diào)遞增.(2)證明由(1)知,f'(x)=1-ax2exx.令g(x)=1-ax2ex,由0<a<1e,可知g(x)在(0,+)內(nèi)單調(diào)遞減,又g(1)=1-ae>0,且gln1a=1-aln1a21a=1-ln1a2<0,故g(x)=0在(0,+)內(nèi)有唯一解,從而f'(x)=0在(0,+)內(nèi)有唯一解,不妨設(shè)為x0,則1<x0<ln1a.當(dāng)x(0,x0)時,f'(x)=g(x)x>g(x0)x=0,所以f(x)在(0,x0)內(nèi)單調(diào)遞增;當(dāng)x(x0,+)時,f'(x)=g(x)x<g(x0)x=0,所以f(x)在(x0,+)內(nèi)單調(diào)遞減,因此x0是f(x)的唯一極值點.令h(x)=lnx-x+1,則當(dāng)x>1時,h'(x)=1x-1<0,故h(x)在(1,+)內(nèi)單調(diào)遞減,從而當(dāng)x>1時,h(x)<h(1)=0,所以lnx<x-1.從而fln1a=lnln1a-aln1a-1eln1a=lnln1a-ln1a+1=hln1a<0,又因為f(x0)>f(1)=0,所以f(x)在(x0,+)內(nèi)有唯一零點.又f(x)在(0,x0)內(nèi)有唯一零點1,從而,f(x)在(0,+)內(nèi)恰有兩個零點.由題意,f'(x0)=0,f(x1)=0,即ax02ex0=1,lnx1=a(x1-1)ex1,從而lnx1=x1-1x02ex1-x0,即ex1-x0=x02lnx1x1-1.因為當(dāng)x>1時,lnx<x-1,又x1>x0>1,故ex1-x0<x02(x1-1)x1-1=x02,兩邊取對數(shù),得lnex1-x0<lnx02,于是x1-x0<2lnx0<2(x0-1),整理得3x0-x1>2.13