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1、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)a0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-1xx2.3.(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(x12-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)
2、性;(2)當(dāng)a0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,1x-1lnx1,證明當(dāng)x(0,1)時,1+(c-1)xcx.6.(2019湖南六校聯(lián)考,文21)已知函數(shù)f(x)=ex,g(x)=ax2+x+1(a0).(1)設(shè)F(x)=g(x)f(x),討論函數(shù)F(x)的單調(diào)性;(2)若0g(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0ax0,證明3x0-x12.參考答案專題突破練9利用導(dǎo)數(shù)證明問題及討論零
3、點個數(shù)1.解(1)f(x)的定義域為(0,+),f(x)=2e2x-ax(x0).當(dāng)a0時,f(x)0,f(x)沒有零點,當(dāng)a0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0ba4且b14時,f(b)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.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)a0時,f(x)2
4、a+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)x0;當(dāng)x0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-exx+1.當(dāng)x(0,+)時,要證lnex-1xx2,只需證ex-1xex2.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exx+1可得,ex21+x2,則x(0,+)時,F(x)0恒成立,即F
5、(x)在(0,+)內(nèi)單調(diào)遞增,F(x)F(0)=0.即ex-1xex2,lnex-1xx2.3.(1)解f(x)=1x-k=1-kxx(x0),當(dāng)k0時,f(x)0,f(x)在區(qū)間(0,+)內(nèi)遞增,當(dāng)k0時,由f(x)0,得0xx20,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lnx22-lnx1,即證明lnx1+lnx22,故k(x1+x2)2,即lnx1-lnx2x1-x22x1+x2,即lnx1x22(x1-x2)x1+x2,設(shè)t=x1x21,上式轉(zhuǎn)化為lnt2(t-1)t
6、+1(t1).設(shè)g(t)=lnt-2(t-1)t+1,g(t)=(t-1)2t(t+1)20,g(t)在(1,+)上單調(diào)遞增,g(t)g(1)=0,lnt2(t-1)t+1,lnx1+lnx22,即lnx22-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a0;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)a0;當(dāng)x(1,+)時,g(x)0時,g(x)0.從而當(dāng)a0時,ln-12a+12
7、a+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)0x0,f(x)單調(diào)遞增;當(dāng)x1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,lnxx-1.故當(dāng)x(1,+)時,lnxx-1,ln1x1x-1,即1x-1lnx1,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)x0,g(x)單調(diào)遞增;當(dāng)xx0時,g(x)0,g(x)單調(diào)遞減.由(2)知1c-1lncc,故0x01.又g
8、(0)=g(1)=0,故當(dāng)0x0.所以當(dāng)x(0,1)時,1+(c-1)xcx.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a12,則2a-1a0,當(dāng)x2a-1a時,F(x)0,當(dāng)0x0,F(x)在(-,0),2a-1a,+上單調(diào)遞減,在0,2a-1a上單調(diào)遞增.若0a12,則2a-1a0,當(dāng)x0時,F(x)0,當(dāng)2a-1ax0.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00恒成立,h(x)在(0,+)
9、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-10,ex12x2+x+1,ex12x2+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-ax2ex0,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0a0,且gln1a=1-aln1a21a=1-ln1a20,故g(x)=0在(0,
10、+)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1x0g(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)x1時,h(x)=1x-11時,h(x)h(1)=0,所以lnxx-1.從而fln1a=lnln1a-aln1a-1eln1a=lnln1a-ln1a+1=hln1af(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)x1時,lnxx01,故ex1-x0x02(x1-1)x1-1=x02,兩邊取對數(shù),得lnex1-x0lnx02,于是x1-x02lnx02.13
（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
