2024年高考數(shù)學(xué)二輪復(fù)習(xí)專題突破專題突破練6　利用導(dǎo)數(shù)證明問題（附答案）

專題突破練6利用導(dǎo)數(shù)證明問題1.已知函數(shù)f(x)=x2+2ax(a>0)與g(x)=4a2lnx+b的圖象有公共點P,且在點P處的切線相同.(1)若a=1,求b的值;(2)求證:f(x)≥g(x).2.(2023·廣西南寧一模)已知f(x)=x-aln(1+x).(1)討論f(x)的單調(diào)性;(2)當(dāng)a=1時,證明f(x)≥0;(3)證明對于任意正整數(shù)n,都有1n+1n+1+13.(2023·河北石家莊一模)伯努利不等式,又稱貝努利不等式,由數(shù)學(xué)家伯努利提出:對于實數(shù)x>-1,且x≠0,正整數(shù)n不小于2,那么(1+x)n≥1+nx.研究發(fā)現(xiàn),伯努利不等式可以推廣,請證明以下問題.(1)當(dāng)α∈[1,+∞)時,(1+x)α≥1+αx對任意x>-1恒成立;(2)對任意n∈N*,1n+2n+3n+…+nn<(n+1)n恒成立.4.已知函數(shù)f(x)=aex+sinx+x,x∈[0,π].(1)證明:當(dāng)a=-1時,函數(shù)f(x)有唯一的極大值點;(2)當(dāng)-2<a<0時,證明:f(x)<π.5.已知函數(shù)f(x)=ex,g(x)=2ax+1.(1)若f(x)≥g(x)恒成立,求a的取值集合;(2)若a>0,且方程f(x)-g(x)=0有兩個不同的根x1,x2,證明:x1+x26.(2023·浙江杭州一模)已知函數(shù)f(x)=klnx+1e(1)若函數(shù)y=f(x)為增函數(shù),求實數(shù)k的取值范圍;(2)已知0<x1<x2.①證明:eex2?e②若x1ex1=x2ex2=k,證明:|f(x

專題突破練6利用導(dǎo)數(shù)證明問題1.(1)解設(shè)P(x0,y0)(x0>0),則x02+2ax0=4a2lnx0又f'(x)=2x+2a,g'(x)=4a∴2x0+2a=4∵a=1,∴x02+x0-∴x0=1,則4×1×0+b=1+2=3,解得b=3.(2)證明由(1)得2x0+2a=4a2x0,即x02+ax0-2a2=∴a2+2a2-4a2lna-b=0.令h(x)=f(x)-g(x)=x2+2ax-4a2lnx-b(a>0),則h'(x)=2x+2a-4當(dāng)0<x<a時,h'(x)<0;當(dāng)x>a時,h'(x)>0,故h(x)在區(qū)間(0,a)內(nèi)單調(diào)遞減,在區(qū)間(a,+∞)內(nèi)單調(diào)遞增.∴x=a時,函數(shù)h(x)取得極小值即最小值,且h(a)=a2+2a2-4a2lna-b=0,因此h(x)≥0,故f(x)≥g(x).2.(1)解由題意得,f(x)的定義域為(-1,+∞),f'(x)=1-a①若a≤0,則當(dāng)x∈(-1,+∞)時,f'(x)>0,所以f(x)在區(qū)間(-1,+∞)內(nèi)單調(diào)遞增;②若a>0,則當(dāng)x∈(-1,a-1)時,f'(x)<0;當(dāng)x∈(a-1,+∞)時,f'(x)>0,所以f(x)在區(qū)間(-1,a-1)內(nèi)單調(diào)遞減,在區(qū)間(a-1,+∞)內(nèi)單調(diào)遞增.綜上所述,當(dāng)a≤0時,f(x)在區(qū)間(-1,+∞)內(nèi)單調(diào)遞增;當(dāng)a>0時,f(x)在區(qū)間(-1,a-1)內(nèi)單調(diào)遞減,在區(qū)間(a-1,+∞)內(nèi)單調(diào)遞增.(2)證明當(dāng)a=1時,由(1)知f(x)在區(qū)間(-1,0)內(nèi)單調(diào)遞減,在區(qū)間(0,+∞)內(nèi)單調(diào)遞增,所以f(x)≥f(0)=0.(3)證明由(2)知當(dāng)a=1,且x∈(0,+∞)時,x-ln(1+x)>0,對于任意正整數(shù)n,令x=1n,得1n>ln(1+所以1n+1n+1+1n+2+…+14n>ln(1+1n)+ln(1+1n=lnn+1n+lnn+2n+1+lnn+3n+2+…+>ln(n+1n·n+2n+1·n+3n3.證明(1)令f(x)=(1+x)α-1-αx,若α=1,則f(x)=0,原不等式成立;若α>1,則f'(x)=α(1+x)α-1-α=α[(1+x)α-1-1],當(dāng)x∈(-1,0)時,(1+x)α-1<1,f'(x)<0,f(x)單調(diào)遞減;當(dāng)x∈(0,+∞)時,f'(x)>0,f(x)單調(diào)遞增.所以f(x)≥f(0)=0,即(1+x)α≥1+αx.綜上所述,當(dāng)α∈[1,+∞)時,(1+x)α≥1+αx對任意x>-1恒成立.(2)要證對任意n∈N*,1n+2n+3n+…+nn<(n+1)n恒成立,只需證(1n+1)n+(2n+1)n+(3n+1)n+…+即證(1-nn+1)n+(1-n-1n+1)n+(1-n-2n+1)n由(1)知對于任意正整數(shù)i∈{1,2,3,…,n},1-in+1≤(1-1所以(1-in+1)n≤(1-1n+1)ni=[(1-1n那么(1-nn+1)n+(1-n-1n+1)n+(1-n-2n+1≤(1-1n+1)nn+(1-1n+1)n(n-1)+(1-1n+1)n(n-2)+…=[(1-1n+1)n]n+[(1-1n+1)n]n-1+[(1-1n+1)n]n-2+…+[(1-下面證明(1-1n+1)n≤要證(1-1n+1)n≤12成立,只需證令1n=x∈(0,1],即證明2x≤x+1成立令g(x)=2x-1-x,則g'(x)=2xln2-1.當(dāng)0<x<log2(1ln2)時,g'(x)<0,g(x)單調(diào)遞減當(dāng)log2(1ln2)<x≤1時,g'(x)>0,g(x)單調(diào)遞增又g(0)=0,g(1)=0,所以當(dāng)x∈(0,1]時,g(x)≤0,即2x≤x+1.所以(1-1n+1)所以[(1-1n+1)n]n+[(1-1n+1)n]n-1+[(1-1n+1)n]n-2+…+[(1≤(12)n+)12)n-1+(12)n-2+…+(12)2+(12)1=1-(1所以命題得證.4.證明(1)當(dāng)a=-1時,f(x)=x+sinx-ex,f'(x)=1+cosx-ex.因為x∈[0,π],所以1+cosx≥0.令g(x)=1+cosx-ex,x∈[0,π],則g'(x)=-ex-sinx<0,所以g(x)在區(qū)間[0,π]上單調(diào)遞減.又因為g(0)=2-1=1>0,g(π)=-eπ<0,所以存在x0∈(0,π),使得f'(x0)=0,且當(dāng)0<x<x0時,f'(x)>0;當(dāng)x0<x<π時,f'(x)<0.所以函數(shù)f(x)的單調(diào)遞增區(qū)間是[0,x0],單調(diào)遞減區(qū)間是[x0,π].所以函數(shù)f(x)存在唯一的極大值點x0.(2)當(dāng)-2<a<0,0≤x≤π時,令h(x)=f(x)-π=aex+sinx+x-π,則h'(x)=aex+cosx+1,令p(x)=aex+cosx+1,則p'(x)=aex-sinx<0,所以函數(shù)h'(x)在區(qū)間[0,π]上單調(diào)遞減.因為h'(0)=a+2>0,h'(π)=aeπ<0,所以存在t∈(0,π),使得h'(t)=0,即aet+cost+1=0,且當(dāng)0<x<t時,h'(x)>0;當(dāng)t<x<π時,h'(x)<0.所以函數(shù)h(x)在區(qū)間[0,t]上單調(diào)遞增,在區(qū)間[t,π]上單調(diào)遞減.所以h(x)max=h(t)=aet+sint+t-π,t∈(0,π).因為aet+cost+1=0,所以只需證φ(t)=sint-cost+t-1-π<0即可,而φ'(t)=cost+sint+1=sint+(1+cost)>0,所以函數(shù)φ(t)在區(qū)間(0,π)內(nèi)單調(diào)遞增,所以φ(t)<φ(π)=0,故f(x)<π.5.(1)解令u(x)=f(x)-g(x)=ex-2ax-1,則u'(x)=ex-2a.若a≤0,則u'(x)>0,所以u(x)在R上單調(diào)遞增,而u(0)=0,所以當(dāng)x<0時,u(x)<0,不符合題意;若a>0,由u'(x)=0,得x=ln(2a),當(dāng)x<ln(2a)時,u'(x)<0,u(x)單調(diào)遞減;當(dāng)x>ln(2a)時,u'(x)>0,u(x)單調(diào)遞增,故u(x)min=u(ln(2a))=2a-2aln(2a)-1≥0.令h(x)=x-xlnx-1,則h'(x)=-lnx.令h'(x)=0,得x=1,當(dāng)0<x<1時,h'(x)>0,當(dāng)x>1時,h'(x)<0,故h(x)在區(qū)間(0,1)內(nèi)單調(diào)遞增,在區(qū)間(1,+∞)內(nèi)單調(diào)遞減,故h(x)≤h(1)=0,即x-xlnx-1≤0,所以2a-2aln(2a)-1≤0,故2a-2aln(2a)-1=0,所以2a=1,即a=12,故a的取值集合為(2)證明方程f(x)-g(x)=0有兩個不同的根x1,x2,不妨令x1<x2,則e∴2a=e要證x1+x22<ln2a,即證ex1+x22<ex2-ex1x2-x1?(x2-x1)ex1+x22<ex2?ex1?(x2-x1)ex2-x12<ex2易證et>t+1,故G'(t)>0,故G(t)在區(qū)間(0,+∞)內(nèi)單調(diào)遞增,所以G(t)>G(0)=0.故原不等式成立.6.(1)解∵f(x)=klnx+1e∴f'(x)=kx?1若f(x)是增函數(shù),則f'(x)=kx又x>0,可得k≥x故k≥xex對?x>構(gòu)建φ(x)=xex(x>0),則φ'(x)=1-令φ'(x)>0,解得0<x<1;令φ'(x)<0,解得x>1.則φ(x)在區(qū)間(0,1)內(nèi)單調(diào)遞增,在區(qū)間(1,+∞)內(nèi)單調(diào)遞減,故φ(x)≤φ(1)=1e故實數(shù)k的取值范圍為[1e,+∞)(2)證明①由(1)可知,當(dāng)k=1e時,f(x)=lnx∵0<x1<x2,則f(x2)>f(x1),即1elnx2+1ex2>1e整理得eex2?eex1>lnx1-構(gòu)建g(x)=x-lnx-1(x>0),則g'(x)=1-1x=x令g'(x)<0,解得0<x<1;令g'(x)>0,解得x>1.則g(x)在區(qū)間(0,1)內(nèi)單調(diào)遞減,在區(qū)間(1,+∞)內(nèi)單調(diào)遞增,故g(x)=x-lnx-1≥g(1)=0,即-lnx≥1-x,當(dāng)且僅當(dāng)x=1時等號成立,令x=x2x1>1,可得-lnx2x故eex2?②∵x∴kx可知f'(x)=kx?1ex=0有兩個不相等的實數(shù)根x1,x2,由(1)知0<x1可得f(x1)=klnx1+1ex1=x1e同理可得f(x2)=x2構(gòu)建p(x)=xlnx+1ex(x>0),則p'(x)當(dāng)0<x<1時,(1-x)lnx<0;當(dāng)x>1時,(1-x)lnx<0;當(dāng)x=1時,(1-x)lnx=0;且ex>0,故p'(x)≤0對?x∈(0,+∞)恒成立,故p(x)在區(qū)間(0,+∞)內(nèi)單調(diào)遞減.又0<x1<1<x2,∴p(x2)<p(1)<p(