（浙江專用）2020版高考數(shù)學(xué)大一輪復(fù)習(xí) 專項強化練四 導(dǎo)函數(shù)不可求零點的導(dǎo)數(shù)綜合問題.docx

專項強化練四導(dǎo)函數(shù)不可求零點的導(dǎo)數(shù)綜合問題1.已知函數(shù)f(x)=x2ex-ln x,求證:當x>0時,不等式f(x)>1.證明f (x)=x(x+2)ex-1x,x>0.則f (x)=(x2+4x+2)ex+1x2>0,故f (x)在(0,+)上單調(diào)遞增.又f 14=916e14-4<0, f 12=54e12-2>0,根據(jù)零點存在性定理可知,存在x014,12,使得f (x0)=0.當x(0,x0)時, f (x)<0,故f(x)在(0,x0)上單調(diào)遞減;當x(x0,+)時, f (x)>0,故f(x)在(x0,+)上單調(diào)遞增.故f(x)min=f(x0)=x02ex0-ln x0.由f (x0)=0,得x0(x0+2)ex0-1x0=0,即x0(x0+2)ex0=1x0,ex0=1x02(x0+2).故f(x0)=x02ex0-ln x0=1x0+2-ln x0,其中x014,12.令g(x)=1x+2-ln x,x14,12.則g(x)=-1(x+2)2-1x<0,故g(x)在x14,12上單調(diào)遞減.故g(x)>g12=25-ln12>1,即f(x0)>1.綜上,有f(x)min>1,則當x>0時,不等式f(x)>1.2.已知函數(shù)f(x)=e2x-aln x,求證:當a>0時, f(x)2a+aln2a.證明f (x)=2e2x-ax,x>0.f (x)有零點,等價于方程2e2x-ax=0有實根,等價于方程2e2x=ax有實根,等價于函數(shù)y=2e2x與函數(shù)y=ax的圖象有交點.顯然,當a<0時,兩個函數(shù)圖象無交點;當a>0時,兩個函數(shù)圖象有一個交點.因此,當a<0時, f (x)無零點,當a>0時, f (x)只有一個零點.當a>0時, f (x)在(0,+)上單調(diào)遞增,且只有一個零點,設(shè)此零點為x0,則f (x0)=0.當x(0,x0)時, f (x)<0, f(x)在(0,x0)上單調(diào)遞減;當x(x0,+)時, f (x)>0, f(x)在(x0,+)上單調(diào)遞增.故f(x)min=f(x0)=e2x0-aln x0.由f (x0)=0,得2e2x0-ax0=0,即e2x0=a2x0,即ln e2x0=ln a-ln 2x0,化簡得ln x0=ln a-ln 2-2x0.故f(x0)=a2x0-a(ln a-ln 2-2x0)=a2x0+2ax0+aln 2a2a+aln2a.故f(x)min2a+aln2a,即當a>0時, f(x)2a+aln2a.3.已知函數(shù)f(x)=1+ln(x+1)x,當x>0時, f(x)>kx+1恒成立,求正整數(shù)k的最大值.解析由已知得k<(x+1)1+ln(x+1)x在x>0上恒成立.令h(x)=(x+1)1+ln(x+1)x,x>0,只需k<h(x)min.h(x)=x-1-ln(x+1)x2.令(x)=x-1-ln(x+1),由(x)=xx+1>0,得(x)在(0,+)上單調(diào)遞增.又(2)=1-ln 3<0,(3)=2-ln 4>0,根據(jù)零點存在性定理可知,存在x0(2,3),使得(x0)=0.當x(0,x0)時,(x)<0,h(x)<0,h(x)在(0,x0)上單調(diào)遞減;當x(x0,+)時,(x)>0,h(x)>0,h(x)在(x0,+)上單調(diào)遞增.故h(x)min=h(x0)=(x0+1)1+ln(x0+1)x0.由(x0)=0,得x0-1-ln(x0+1)=0,即x0=1+ln(x0+1).則h(x0)=x0+1(3,4).故正整數(shù)k的最大值為3.4.已知函數(shù)f(x)=aex+a+1x-2(a+1)0對任意的x(0,+)恒成立,其中a>0.求a的取值范圍.解析f (x)=aex-a+1x2=ax2ex-(a+1)x2,令g(x)=ax2ex-(a+1),其中x>0,a>0.則g(x)=a(2x+x2)ex>0,故g(x)在(0,+)上單調(diào)遞增.又g(0)=-(a+1)<0,當x+時,g(x)+,故存在x0(0,+),使得g(x0)=0.當x(0,x0)時,g(x)<0, f (x)<0, f(x)在(0,x0)上單調(diào)遞減;當x(x0,+)時,g(x)>0, f (x)>0, f(x)在(x0,+)上單調(diào)遞增.故f(x)min=f(x0)=aex0+a+1x0-2(a+1).由g(x0)=0,得ax02ex0-(a+1)=0,即aex0=a+1x02.則f(x0)=aex0+a+1x0-2(a+1)=a+1x02+a+1x0-2(a+1),令a+1x02+a+1x0-2(a+1)0,由x0>0,a>0,得0<x01.因為g(x)=ax2ex-(a+1)在(0,+)上單調(diào)遞增,0<x01,所以ae-(a+1)0,解得a1e-1.5.已知函數(shù)f(x)=ex-ln(x+m).(1)設(shè)x=0是f(x)的極值點,求m,并討論f(x)的單調(diào)性;(2)當m2時,求證: f(x)>0.解析(1)f (x) =ex-1x+m.由x=0是f(x)的極值點,得f (0)=0,所以m=1.于是f(x)=ex-ln(x+1),定義域為(-1,+),f (x)=ex-1x+1.函數(shù)f (x)=ex-1x+1在(-1,+)上單調(diào)遞增,且f (0)=0,因此當x(-1,0)時, f (x)<0;當x(0,+)時, f (x)>0.所以f(x)在(-1,0)上單調(diào)遞減,在(0,+)上單調(diào)遞增.(2)證明:當m2,x(-m,+)時,ln(x+m)ln(x+2),故只需證明當m=2時, f(x)>0.當m=2時,函數(shù)f (x)=ex-1x+2在(-2,+)上單調(diào)遞增.又f (-1)<0, f (0)>0,故f (x)=0在(-2,+)上有唯一實根x0,且x0(-1,0).當x(-2,x0)時, f (x)<0;當x(x0,+)時, f (x)>0,從而當x=x0時, f(x)取得極小值,也是最小值.f (x)、 f(x)的大致圖象如圖.由f (x0)=0得ex0=1x0+2,即ln(x0+2)=-x0,故f(x)f(x0)=1x0+2+x0=(x0+1)2x0+2>0.所以當m2時, f(x)>0.6.已知函數(shù)f(x)=x2-ax+ln x(aR).(1)當a=1時,求曲線f(x)在點P(1,0)處的切線方程;(2)若函數(shù)f(x)有兩個極值點x1,x2,求f(x1+x2)的取值范圍.解析(1)當a=1時, f(x)=x2-x+ln x,則f (x)=2x-1+1x.所以f (1)=2.因此曲線f(x)在點P(1,0)處的切線方程為2x-y-2=0.(2)由題意得f (x)=2x-a+1x,令f (x)=0,則2x-a+1x=0,即2x2-ax+1=0(x>0).由題意知2x2-ax+1=0有兩個不等的實根,為x1,x2.由根與系數(shù)的關(guān)系得=a2-8>0,x1+x2=a2>0,x1x2=12>0,解得a>22.故f(x1+x2)=(x1+x2)2-a(x1+x2)+ln(x1+x2)=-a24+lna2.設(shè)g(a)=-a24+lna2(a>22),則g(a)=-a2+1a=2-a22a<0.故g(a)在(22,+)上單調(diào)遞減,所以g(a)<g(22)=-2+ln 2.因此f(x1+x2)的取值范圍是(-,-2+ln 2).7.已知函數(shù)f(x)=ex-56x2-2x,g(x)=-13x2-x+1.(1)設(shè)h(x)=f(x)-g(x),求h(x)在-1,0上的最大值;(2)當-1x1時,求證:e-176f(x)<53.解析(1)h(x)=f(x)-g(x)=ex-12x2-x-1,則h(x)=ex-x-1,令k(x)=ex-x-1,則k(x)=ex-1,所以當x0時,k(x)0,即h(x)在-1,0上單調(diào)遞減,又h(0)=0,故h(x)0,所以h(x)在-1,0上單調(diào)遞增.即當x=0時,h(x)取得最大值,為h(0)=0.(2)證明: f (x)=ex-53x-2,令m(x)=ex-53x-2,則m(x)=ex-53,由m(x)=ex-53>0,得x>ln53,由m(x)=ex-53<0,得x<ln53.m(x)在-1,ln53上單調(diào)遞減,在ln53,1上單調(diào)遞增,又mln53=53-53ln53-2<0,m(-1)=1e-13>0,m(1)=e-53-2<0,所以m(x)在-1,ln53上必有一零點x0,在(-1,x0)上f(x)單調(diào)遞增,在(x0,1)上 f(x)單調(diào)遞減.所以f(x)min=minf(-1), f(1)=e-176;f(x)max=f(x0)g(x0),又因為g(x)在(-1,x0)上單調(diào)遞減,所以f(x)maxg(x0)<g(-1)=53.綜上所述,當-1x1時,e-176f(x)<53.