適用于老高考新教材2023屆高考數(shù)學(xué)二輪總復(fù)習(xí)考點(diǎn)突破練18利用導(dǎo)數(shù)證明不等式含解析

Page6考點(diǎn)突破練18利用導(dǎo)數(shù)證明不等式1.(2022·福建福州質(zhì)檢)已知函數(shù)f(x)=elnx-ax(a∈R).(1)討論f(x)的單調(diào)性;(2)當(dāng)a=e時,證明:xf(x)-ex+2ex≤0.2.(2022·江西鷹潭二模(文))已知函數(shù)f(x)=ax2-(a+2)x+lnx.(1)當(dāng)a>0時,討論函數(shù)f(x)的單調(diào)區(qū)間;(2)當(dāng)a=0時,證明:f(x)<ex-2x-2(其中e為自然對數(shù)的底數(shù)).3.(2022·陜西咸陽二模)已知函數(shù)f(x)=lnx-kx+1.(1)若f(x)≤0恒成立,求實數(shù)k的取值范圍;(2)證明:1+1221+132…1+4.(2021·新高考Ⅰ·22)已知函數(shù)f(x)=x(1-lnx).(1)討論f(x)的單調(diào)性;(2)設(shè)a,b為兩個不相等的正數(shù),且blna-alnb=a-b,證明:2<1a+15.(2022·江蘇泰州一模)已知函數(shù)f(x)=ax+lnx(1)討論f(x)的單調(diào)性;(2)若f(x1)=f(x2)=2(x1≠x2),證明:a2<x1x2<ae.

考點(diǎn)突破練18利用導(dǎo)數(shù)證明不等式1.(1)解f'(x)=ex-a(x>0),①若a≤0,則f'(x)>0,f(x)在(0,+∞)上為增函數(shù);②若a>0,則當(dāng)x<ea時,f'(x)>0;當(dāng)x>ea時,f'(x)<故在0,ea上,f(x)單調(diào)遞增;在ea,+(2)證明因為x>0,所以只需證f(x)≤exx由(1)知,當(dāng)a=e時,f(x)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減,所以f(x)max=f(1)=-e.記g(x)=exx-2e(x>0),則g'(x)=所以,當(dāng)0<x<1時,g'(x)<0,g(x)單調(diào)遞減;當(dāng)x>1時,g'(x)>0,g(x)單調(diào)遞增,所以g(x)min=g(1)=-e.所以當(dāng)x>0時,f(x)≤g(x),即f(x)≤exx即xf(x)-ex+2ex≤0.2.(1)解f(x)的定義域為(0,+∞),f'(x)=2ax-(a+2)+1x=(①當(dāng)0<1a<12,即a>2時,在0,1a,12,在1a,12上,f'(x)<0,f②當(dāng)1a=12,a=2時,f'(x)≥0,f(x)在(0,③當(dāng)1a>12,即0<a<2時,在0,12,1a,在12,1a上,f'(x)<0,f綜上所述,當(dāng)a>2時,f(x)的單調(diào)遞增區(qū)間為0,1a當(dāng)a=2時,f(x)的單調(diào)遞增區(qū)間為(0,+∞).當(dāng)0<a<2時,f(x)的單調(diào)遞增區(qū)間為0,12,1a,+∞,單調(diào)遞減區(qū)間為(2)證明當(dāng)a=0時,由f(x)<ex-2x-2化簡得ex-lnx-2>0,構(gòu)造函數(shù)h(x)=ex-lnx-2(x>0),h'(x)=ex-1x,h″(x)=ex+1x2>0,∴h'(x)在(0,h'12=e-2<0,h'(1)=e-故存在x0∈12,1,使得h'(x0)=當(dāng)x∈(0,x0)時,h'(x)<0,h(x)單調(diào)遞減;當(dāng)x∈(x0,+∞)時,h'(x)>0,h(x)單調(diào)遞增.所以x=x0時,h(x)取得極小值,也即是最小值.h(x0)=ex0-lnx0-2=1x0-ln1ex0-2=1x0+x0所以h(x)=ex-lnx-2>0,故f(x)<ex-2x-2.3.(1)解f(x)≤0?lnx-kx+1≤0?k≥lnx令g(x)=lnx+1x,x>0,則g'(x)當(dāng)0<x<1時,g'(x)>0,g(x)單調(diào)遞增,當(dāng)x>1時,g'(x)<0,g(x)單調(diào)遞減,∴g(x)max=g(1)=1,∴k≥1.(2)證明由(1)知,k=1時,有不等式lnx≤x-1對任意x∈(0,+∞)恒成立,當(dāng)且僅當(dāng)x=1時,取“=”號,∴當(dāng)x∈(1,+∞),lnx<x-1恒成立,令x=1+1n2(n>1,且n∈N*),則ln∴l(xiāng)n1+122+ln1+132即ln1+1221+132…∴1+1221+132…1+4.(1)解由條件知,函數(shù)f(x)的定義域為(0,+∞),f'(x)=-lnx.當(dāng)x∈(0,1)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.即在區(qū)間(0,1)內(nèi),函數(shù)f(x)單調(diào)遞增;在區(qū)間(1,+∞)內(nèi),函數(shù)f(x)單調(diào)遞減.(2)證明(方法一)由blna-alnb=a-b得1a1-ln1a由a≠b,得1a由(1)不妨設(shè)1a∈(0,1),1b∈(1,+∞),則f1a>0,從而f1b>①令g(x)=f(2-x)-f(x),則g'(x)=-f'(2-x)-f'(x)=ln[1-(x-1)2],當(dāng)x∈(0,1)時,g'(x)<0,g(x)在區(qū)間(0,1)內(nèi)單調(diào)遞減,g(x)>g(1)=0,從而f(2-x)>f(x),所以f2-1a>f1由(1)得2-1a<1b即2<令h(x)=x+f(x),則h'(x)=1+f'(x)=1-lnx,當(dāng)x∈(1,e)時,h'(x)>0,h(x)在區(qū)間(1,e)內(nèi)單調(diào)遞增,h(x)<h(e)=e,從而x+f(x)<e,所以1b+f1b<又由1a∈(0,1),可得1a<1a所以1a+1b<f1b由①②得2<1a+1(方法二)blna-alnb=a-b變形為lnaa-令1a=m,1b=n.則上式變?yōu)閙(1-lnm)=n(1-ln于是命題轉(zhuǎn)換為證明2<m+n<e.令f(x)=x(1-lnx),則有f(m)=f(n),不妨設(shè)m<n.令f'(x)=0,得x=1,且f(e)=0.結(jié)合(1)知0<m<1,1<n<e,先證m+n>2.要證m+n>2?n>2-m?f(n)<f(2-m)?f(m)<f(2-m)?f(m)-f(2-m)<0.令g(x)=f(x)-f(2-x),x∈(0,1),則g'(x)=f'(x)+f'(2-x)=-ln[x(2-x)]>0,∴g(x)在區(qū)間(0,1)內(nèi)單調(diào)遞增,所以g(x)<g(1)=0,即m+n>2.再證m+n<e.因為m(1-lnm)=n(1-lnn)>m,所以n(1-lnn)+n<e?m+n<e.令h(x)=x(1-lnx)+x,x∈(1,e),所以h'(x)=1-lnx>0,故h(x)在區(qū)間(1,e)內(nèi)單調(diào)遞增.所以h(x)<h(e)=e.故h(n)<e,即m+n<e.綜合可知2<1a+1(方法三)證明1a+1b>2同方法二.以下證明x1+x不妨設(shè)x2=tx1,則t=x2x由x1(1-lnx1)=x2(1-lnx2)得x1(1-lnx1)=tx1[1-ln(tx1)],lnx1=1-tln要證x1+x2<e,只需證(1+t)x1<e,兩邊取對數(shù)得ln(1+t)+lnx1<1,即ln(1+t)+1-tlntt-記g(s)=ln(1+s)s,s則g'(s)=s1+記h(s)=s1+s-ln(1+s),則h'(s)=1所以h(s)在區(qū)間(0,+∞)內(nèi)單調(diào)遞減.h(s)<h(0)=0,則g'(s)<0,所以g(s)在區(qū)間(0,+∞)內(nèi)單調(diào)遞減.由t∈(1,+∞)得t-1∈(0,+∞),所以g(t)<g(t-1),即ln((方法四)由已知得lnaa-lnbb=1b-1a,令1a=x1,1b=x2,不妨設(shè)x1<x2,所以f(x1)=f(x2).令f'(x)=0,得x=1,且f(e)=0.由(1)及前面分析知,0<x證明x1+x2>2同方法二.再證明x1+x2<e.令h(x)=1-lnxx-e(0<x<e),h'令φ(x)=lnx+ex-2(0<x<e),則φ'(x)=1x-所以φ(x)>φ(e)=0,h'(x)>0,h(x)在區(qū)間(0,e)內(nèi)單調(diào)遞增.因為0<x1<x2<e,所以1-即1-又因為f(x1)=f(x2),所以1-即x22-ex2<x12-ex1,(x1-x2)(x1+x2-因為x1<x2,所以x1+x2<e,即1a+1綜上,有2<1a+15.(1)解函數(shù)f(x)=ax+lnx的定義域為(0,+∞),求導(dǎo)得f'(x)=x當(dāng)a≤0時,f'(x)>0恒成立,則f(x)在(0,+∞)上單調(diào)遞增,當(dāng)a>0時,f'(x)<0的解集為(0,a),f'(x)>0的解集為(a,+∞),即f(x)的單調(diào)遞增區(qū)間為(a,+∞),單調(diào)遞減區(qū)間為(0,a),所以當(dāng)a≤0時,f(x)在(0,+∞)上單調(diào)遞增,當(dāng)a>0時,f(x)在(a,+∞)上單調(diào)遞增,在(0,a)上單調(diào)遞減.(2)證明因為f(x1)=f(x2)=2(x1≠x2),由(1)知,a>0,且f(x)min=f(a)=lna+1<2,解得a∈(0,e),設(shè)x1<x2,則0<x1<a<x2,要證x1x2>a2,即證x2>a2x1>a,即證f(x2)>fa2x1,即證f(x1)>fa2x1,設(shè)g(x)=f(x)-fa2x=2ln則g'(x)=2x-ax2-1a=-(x-a)2ax2<0,即g(x即f(x)>fa2x(x∈(0,a)),則f(x1)>fa2x1成立,因此x1要證x1x2<ae,即證a<x2<aex1,即證