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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(diǎn)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(diǎn)的個數(shù);(2)證明:當(dāng)a0時,f(x)2a+aln2a.2.(2019福建漳州質(zhì)檢二,理21)已知函數(shù)f(x)=xln x.(1)若函數(shù)g(x)=f(x)x2-1x,求g(x)的極值;(2)證明:f(x)+1ex-x2.(參考數(shù)據(jù):ln 20.69,ln 31.10,e324.48,e27.39)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(diǎn)x1,x2(x12-ln x1.4
2、.(2019四川成都二模,理21)已知函數(shù)f(x)=ln x+a1x-1,aR.(1)若f(x)0,求實(shí)數(shù)a取值的集合;(2)證明:ex+1x2-ln x+x2+(e-2)x.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.已知函數(shù)f(x)=(x-2)ex+a(x-1)2有兩個零點(diǎn).(1)求a的取值范圍;(2)設(shè)x1,x2是f(x)的兩個零點(diǎn),證明:x1+x22.7.(2019天津卷,理20)設(shè)函數(shù)f(x)=excos x,g(x)為f(x)的導(dǎo)函數(shù).(1)求f(x)的單調(diào)區(qū)間;(2
3、)當(dāng)x4,2時,證明f(x)+g(x)2-x0;(3)設(shè)xn為函數(shù)u(x)=f(x)-1在區(qū)間2n+4,2n+2內(nèi)的零點(diǎn),其中nN,證明2n+2-xn0).當(dāng)a0時,f(x)0,f(x)沒有零點(diǎn),當(dāng)a0時,因?yàn)閑2x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(diǎn).(2)由(1),可設(shè)f(x)在(0,+)的唯一零點(diǎn)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?yàn)?e2x0-ax0=0,所以f(x0
4、)=a2x0+2ax0+aln2a2a+aln2a.故當(dāng)a0時,f(x)2a+aln2a.2.(1)解g(x)=lnx-1x(x0),g(x)=2-lnxx2.令g(x)0,解得0xe2.令g(x)e2,故g(x)在(0,e2)內(nèi)遞增,在(e2,+)內(nèi)遞減,故g(x)的極大值為g(e2)=1e2,沒有極小值.(2)證明要證f(x)+10.先證明lnxx-1,取h(x)=lnx-x+1,則h(x)=1-xx,易知h(x)在(0,1)內(nèi)遞增,在(1,+)內(nèi)遞減,所以h(x)h(1)=0,即lnxx-1,當(dāng)且僅當(dāng)x=1時,取“=”,故xlnxx(x-1),ex-x2-xlnx-1ex-2x2+x-1
5、,故只需證明當(dāng)x0時,ex-2x2+x-10恒成立.令k(x)=ex-2x2+x-1(x0),則k(x)=ex-4x+1.令F(x)=k(x),則F(x)=ex-4,令F(x)=0,解得x=2ln2.F(x)遞增,當(dāng)x(0,2ln2時,F(x)0,F(x)遞減,即k(x)遞減,當(dāng)x(2ln2,+)時,F(x)0,F(x)遞增,即k(x)遞增,且k(2ln2)=5-8ln20,k(2)=e2-8+10,由零點(diǎn)存在定理,可知x1(0,2ln2),x2(2ln2,2),使得k(x1)=k(x2)=0,故0xx2時,k(x)0,k(x)遞增,當(dāng)x1xx2時,k(x)0,故當(dāng)x0時,k(x)0,原不等式
6、成立.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+1(t1).設(shè)g(t)=lnt-2(t-1)t+1,g(t)=(t-1)2t(t+1
7、)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)=1x-ax2=x-ax2(x0).當(dāng)a0時,f(x)0,函數(shù)f(x)在(0,+)上單調(diào)遞增.又f(1)=0,因此當(dāng)0x1時,f(x)0時,可得函數(shù)f(x)在(0,a)上單調(diào)遞減,在(a,+)上單調(diào)遞增,故當(dāng)x=a時,函數(shù)f(x)取得最小值,則f(a)=lna+1-a0.令g(a)=lna+1-a,g(1)=0.由g(a)=1a-1=1-aa,可知當(dāng)a=1時,函數(shù)g(a)取得最大值,而g(1)=0,因此只有當(dāng)a=1時滿足f(a)=lna+1-
8、a0.故a=1.故實(shí)數(shù)a取值的集合是1.(2)證明由(1)可知,當(dāng)a=1時,f(x)0,即lnx1-1x在(0,+)內(nèi)恒成立.要證明ex+1x2-lnx+x2+(e-2)x,即證明ex1+x2+(e-2)x,即ex-1-x2-(e-2)x0.令h(x)=ex-1-x2-(e-2)x,x0.h(x)=ex-2x-(e-2),令u(x)=ex-2x-(e-2),u(x)=ex-2,令u(x)=ex-2=0,解得x=ln2.則函數(shù)u(x)在(0,ln2)內(nèi)單調(diào)遞減,在(ln2,+)內(nèi)單調(diào)遞增.即函數(shù)h(x)在(0,ln2)內(nèi)單調(diào)遞減,在(ln2,+)內(nèi)單調(diào)遞增.而h(0)=1-(e-2)=3-e0,
9、h(ln2)0,h(x)單調(diào)遞增;當(dāng)x(x0,1)時,h(x)0,h(x)單調(diào)遞增.又h(0)=1-1=0,h(1)=e-1-1-(e-2)=0,故對x0,h(x)0恒成立,即ex-1-x2-(e-2)x0.綜上可知,ex+1x2-lnx+x2+(e-2)x成立.5.(1)解由題設(shè),f(x)的定義域?yàn)?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,
10、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(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)=(x-1)ex+2a(x-1)=(x-1)(ex+2a).設(shè)a=0,則f(x)=(x-2)ex,f(x)只有一個零點(diǎn).設(shè)a0,則當(dāng)x(-,1)時,f(x)0,所以f(x)在(-,1)單調(diào)遞減,在(1,+)單調(diào)遞增.又f(1)=-e,f(2)=a,取b滿足b0且ba2(b
11、-2)+a(b-1)2=ab2-32b0,故f(x)存在兩個零點(diǎn).設(shè)a0,因此f(x)在(1,+)單調(diào)遞增.又當(dāng)x1時,f(x)0,所以f(x)不存在兩個零點(diǎn).若a1,故當(dāng)x(1,ln(-2a)時,f(x)0.因此f(x)在(1,ln(-2a)單調(diào)遞減,在(ln(-2a),+)單調(diào)遞增.又當(dāng)x1時f(x)0,所以f(x)不存在兩個零點(diǎn).綜上,a的取值范圍為(0,+).(2)證明不妨設(shè)x1x2,由(1)知,x1(-,1),x2(1,+),2-x2(-,1),f(x)在(-,1)單調(diào)遞減,所以x1+x2f(2-x2),即f(2-x2)1時,g(x)1時,g(x)0.從而g(x2)=f(2-x2)0
12、,故x1+x2cosx,得f(x)0,則f(x)單調(diào)遞減;當(dāng)x2k-34,2k+4(kZ)時,有sinx0,則f(x)單調(diào)遞增.所以,f(x)的單調(diào)遞增區(qū)間為2k-34,2k+4(kZ),f(x)的單調(diào)遞減區(qū)間為2k+4,2k+54(kZ).(2)證明記h(x)=f(x)+g(x)2-x.依題意及(1),有g(shù)(x)=ex(cosx-sinx),從而g(x)=-2exsinx.當(dāng)x4,2時,g(x)0,故h(x)=f(x)+g(x)2-x+g(x)(-1)=g(x)2-x0.因此,h(x)在區(qū)間4,2上單調(diào)遞減,進(jìn)而h(x)h2=f2=0.所以,當(dāng)x4,2時,f(x)+g(x)2-x0.(3)證明依題意,u(xn)=f(xn)-1=0,即exncosxn=1.記yn=xn-2n,則yn4,2,且f(yn)=eyncosyn=exn-2ncos(xn-2n)=e-2n(nN).由f(yn)=e-2n1=f(y0)及(1),得yny0.由(2)知,當(dāng)x4,2時,g(x)0,所以g(x)在4,2上為減函數(shù),因此g(yn)g(y0)g4=0.又由(2)知,f(yn)+g(yn)2-yn0,故2-yn-f(yn)g(yn)=-e-2ng(yn)-e-2ng(y0)=e-2ney0(siny0-cosy0)e-2nsinx0-cosx0.所以,2n+2-xne-2nsinx0-cosx0.13
（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(diǎn)個數(shù) 理
