高考數(shù)學(xué)二輪復(fù)習(xí)專題一函數(shù)與導(dǎo)數(shù)專題突破練6利用導(dǎo)數(shù)證明問題含解析

PAGEPAGE9專題突破練6利用導(dǎo)數(shù)證明問題1.(2021·海南海口月考)已知函數(shù)f(x)=x2+2ax(a>0)與g(x)=4a2lnx+b的圖象有公共點(diǎn)P,且在點(diǎn)P處的切線相同.(1)若a=1,求b的值;(2)求證:f(x)≥g(x).2.(2021·遼寧朝陽一模)已知函數(shù)f(x)=ex-asinx-x,曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程為x+y-1=0.(1)求實(shí)數(shù)a的值;(2)證明:對?x∈R,f(x)>0恒成立.3.(2021·河北石家莊三模)已知函數(shù)f(x)=alnx-x2+x+3a.(1)討論函數(shù)f(x)的單調(diào)性;(2)若0<a<14,求證:f(x)<exx-x4.(2021·江蘇百校聯(lián)盟3月聯(lián)考)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(diǎn);(2)當(dāng)-2<a<0時,證明:f(x)<π.5.(2021·廣東湛江一模)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.(2021·廣州一模)已知函數(shù)f(x)=xlnx-ax2+x(a∈R).(1)證明:曲線y=f(x)在點(diǎn)(1,f(1))處的切線l恒過定點(diǎn);(2)若f(x)有兩個零點(diǎn)x1,x2,且x2>2x1,證明:x1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)=4a2x,∴2x0+∵a=1,∴x02+x0-2=0,∴x0=1,則4×1×0+b=1+2=3,解得(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)上單調(diào)遞減,在區(qū)間(a,+∞)上單調(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f'(x)=ex-acosx-1.∵曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程為x+y-1=0,∴f'(0)=-1,∴1-a-1=-1,得a=1.(2)證明由于f(x)=ex-sinx-x,要證明對?x∈R,f(x)>0恒成立,需證明對?x∈R,ex-x>sinx.令g(x)=ex-x,∴g'(x)=ex-1.令g'(x)=0,得x=0.∴當(dāng)x∈(-∞,0)時,g'(x)<0,當(dāng)x∈(0,+∞)時,g'(x)>0.∴函數(shù)g(x)在區(qū)間(-∞,0)上單調(diào)遞減,在區(qū)間(0,+∞)上單調(diào)遞增.故g(x)min=g(0)=1,即對?x∈R,ex-x≥1都成立,∴ex-x-sinx≥1-sinx≥0,兩個等號不同時成立,∴ex-x>sinx,∴對?x∈R,f(x)>0恒成立.3.(1)解f'(x)=ax-2x+1=-2x2+x+ax,x>0,令f'(x)=0,即-2x2當(dāng)Δ≤0,即a≤-18時,f'(x∴f(x)在區(qū)間(0,+∞)上單調(diào)遞減.當(dāng)Δ>0,即a>-18時,由f'(x)=0,得x1=1+1+8a4,x2=1-1+8①當(dāng)a≥0時,x1>0,x2≤0,x∈(0,x1)時,f'(x)>0,x∈(x1,+∞)時,f'(x)<0,∴f(x)在區(qū)間(0,x1)上單調(diào)遞增,在區(qū)間(x1,+∞)上單調(diào)遞減.②當(dāng)-18<a<0時,x1>0,x2>0,x∈(0,x2)∪(x1,+∞)時,f'(x)<0,x∈(x2,x1)時,f'(x)>∴f(x)在區(qū)間(0,x2)和(x1,+∞)上單調(diào)遞減,在區(qū)間(x2,x1)上單調(diào)遞增.綜上所述,當(dāng)a≤-18時,f(x)在區(qū)間(0,+∞當(dāng)a≥0時,f(x)在區(qū)間0,1+1+8a4上單調(diào)遞增,在區(qū)間1+1+8a4,+當(dāng)-18<a<0時,f(x)在區(qū)間0,1-1+8a4和1+1+8a4,+∞上單調(diào)遞減,在區(qū)間1(2)證明由已知得需證a(lnx+3)<e∵a>0,x>0,∴exx>0,當(dāng)lnx+3當(dāng)lnx+3>0時,由于0<a<14,∴a(lnx+3)<14(lnx+3),因此只需證14(lnx+3)<令g(x)=lnx+34x(x>0),∴g'(令g'(x)=0,得x=e-2.當(dāng)x∈(0,e-2)時,g'(x)>0,當(dāng)x∈(e-2,+∞)時,g'(x)<0,即g(x)在區(qū)間(0,e-2)上單調(diào)遞增,在區(qū)間(e-2,+∞)上單調(diào)遞減.∴g(x)max=g(e-2)=e令h(x)=exx2(x>0),則h'(x當(dāng)x∈(0,2)時,h'(x)<0,當(dāng)x∈(2,+∞)時,h'(x)>0,∴h(x)在區(qū)間(0,2)上單調(diào)遞減,在區(qū)間(2,+∞)上單調(diào)遞增,∴h(x)min=h(2)=e∴g(x)max≤h(x)min,但兩邊取最值的條件不一樣,∴l(xiāng)nx+34x<exx2.4.證明(1)當(dāng)a=-1時,f(x)=x+sinx-ex,f'(x)=1+cosx-ex.因?yàn)閤∈[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?yàn)間(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(diǎn)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?yàn)閔'(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?yàn)閍et+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)上單調(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)上單調(diào)遞增,在區(qū)間(1,+∞)上單調(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ex1=2要證x1+x22<ln2a,即證ex1+x22<ex2-ex1x2-x1?(x2-x1)ex1+x22<ex2-ex1?(x2-x1)ex2-x易證et>t+1,故G'(t)>0,故G(t)在區(qū)間(0,+∞)上單調(diào)遞增,所以G(t)>G(0)=0.故原不等式成立.6.證明(1)f'(x)=lnx-2ax+2,則f'(1)=2-2a,所以切線l的斜率為2-2a.又f(1)=1-a,所以切線l的方程為y-(1-a)=(2-2a)(x-1),即y=(2-2a)x-12,可得當(dāng)x=12時,y=0,故切線l恒過定點(diǎn)12,0.(2)∵x1,x2是f(x)的零點(diǎn),x2>2x1,且x1>0,x2>0,∴∴a=lnx即ln(x1x2)+2=(令t=x2x1,則t>