（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) 理

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)+1<ex-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x1,x2(x1<x2),求證:ln x2>2-ln x1.4.(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,1<x-1lnx<x;(3)設(shè)c>1,證明當(dāng)x(0,1)時,1+(c-1)x>cx.6.已知函數(shù)f(x)=(x-2)ex+a(x-1)2有兩個零點.(1)求a的取值范圍;(2)設(shè)x1,x2是f(x)的兩個零點,證明:x1+x2<2.7.(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)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,其中nN,證明2n+2-xn<e-2nsinx0-cosx0.參考答案專題突破練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g(x)=lnx-1x(x>0),g'(x)=2-lnxx2.令g'(x)>0,解得0<x<e2.令g'(x)<0,解得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)+1<ex-x2.即證ex-x2-xlnx-1>0.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,故只需證明當(dāng)x>0時,ex-2x2+x-1>0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)=5-8ln2<0,k'(0)=2>0,k'(2)=e2-8+1>0,由零點存在定理,可知x1(0,2ln2),x2(2ln2,2),使得k'(x1)=k'(x2)=0,故0<x<x1或x>x2時,k'(x)>0,k(x)遞增,當(dāng)x1<x<x2時,k'(x)<0,k(x)遞減,故k(x)的最小值是k(0)=0或k(x2),由k'(x2)=0,得ex2=4x2-1,k(x2)=ex2-2x22+x2-1=-(x2-2)(2x2-1).x2(2ln2,2),k(x2)>0,故當(dāng)x>0時,k(x)>0,原不等式成立.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)=1x-ax2=x-ax2(x>0).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)0<x<1時,f(x)<0,不合題意.當(dāng)a>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-a0.故a=1.故實數(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,x>0.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-e>0,h'(ln2)<h'(1)=0.故存在x0(0,ln2),使得h'(x0)=0,當(dāng)x(0,x0)時,h'(x)>0,h(x)單調(diào)遞增;當(dāng)x(x0,1)時,h'(x)<0,h(x)單調(diào)遞減.當(dāng)x(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x>0,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)=(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.設(shè)a>0,則當(dāng)x(-,1)時,f'(x)<0;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b<0且b<lna2,則f(b)>a2(b-2)+a(b-1)2=ab2-32b>0,故f(x)存在兩個零點.設(shè)a<0,由f'(x)=0得x=1或x=ln(-2a).若a-e2,則ln(-2a)1,故當(dāng)x(1,+)時,f'(x)>0,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.若a<-e2,則ln(-2a)>1,故當(dāng)x(1,ln(-2a)時,f'(x)<0;當(dāng)x(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.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)單調(diào)遞減,所以x1+x2<2等價于f(x1)>f(2-x2),即f(2-x2)<0.由于f(2-x2)=-x2e2-x2+a(x2-1)2,而f(x2)=(x2-2)ex2+a(x2-1)2=0,所以f(2-x2)=-x2e2-x2-(x2-2)ex2.設(shè)g(x)=-xe2-x-(x-2)ex,則g'(x)=(x-1)(e2-x-ex).所以當(dāng)x>1時,g'(x)<0,而g(1)=0,故當(dāng)x>1時,g(x)<0.從而g(x2)=f(2-x2)<0,故x1+x2<2.7.(1)解由已知,有f'(x)=ex(cosx-sinx).因此,當(dāng)x2k+4,2k+54(kZ)時,有sinx>cosx,得f'(x)<0,則f(x)單調(diào)遞減;當(dāng)x2k-34,2k+4(kZ)時,有sinx<cosx,得f'(x)>0,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-x<0.因此,h(x)在區(qū)間4,2上單調(diào)遞減,進而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-xn<e-2nsinx0-cosx0.13