2019年高考數(shù)學(xué)真題分類匯編 3.2 導(dǎo)數(shù)的應(yīng)用 理 .doc

2019年高考數(shù)學(xué)真題分類匯編 3.2 導(dǎo)數(shù)的應(yīng)用 理考點(diǎn)一函數(shù)的單調(diào)性1.(xx課標(biāo),11,5分)已知函數(shù)f(x)=ax3-3x2+1,若f(x)存在唯一的零點(diǎn)x0,且x0>0,則a的取值范圍是()A.(2,+) B.(1,+) C.(-,-2) D.(-,-1)答案C2.(xx課標(biāo),21,12分)已知函數(shù)f(x)=ex-e-x-2x.(1)討論f(x)的單調(diào)性;(2)設(shè)g(x)=f(2x)-4bf(x),當(dāng)x>0時(shí),g(x)>0,求b的最大值;(3)已知1.414 2<<1.414 3,估計(jì)ln 2的近似值(精確到0.001).解析(1)f (x)=ex+e-x-20,等號(hào)僅當(dāng)x=0時(shí)成立.所以f(x)在(-,+)上單調(diào)遞增.(2)g(x)=f(2x)-4bf(x)=e2x-e-2x-4b(ex-e-x)+(8b-4)x,g(x)=2e2x+e-2x-2b(ex+e-x)+(4b-2)=2(ex+e-x-2)(ex+e-x-2b+2).(i)當(dāng)b2時(shí),g(x)0,等號(hào)僅當(dāng)x=0時(shí)成立,所以g(x)在(-,+)上單調(diào)遞增.而g(0)=0,所以對(duì)任意x>0,g(x)>0.(ii)當(dāng)b>2時(shí),若x滿足2<ex+e-x<2b-2,即0<x<ln(b-1+)時(shí),g(x)<0.而g(0)=0,因此當(dāng)0<xln(b-1+)時(shí),g(x)<0.綜上,b的最大值為2.(3)由(2)知,g(ln)=-2b+2(2b-1)ln 2.當(dāng)b=2時(shí),g(ln)=-4+6ln 2>0,ln 2>>0.692 8;當(dāng)b=+1時(shí),ln(b-1+)=ln,g(ln)=-2+(3+2)ln 2<0,ln 2<<0.693 4.所以ln 2的近似值為0.693.3.(xx廣東,21,14分)設(shè)函數(shù)f(x)=,其中k<-2.(1)求函數(shù)f(x)的定義域D(用區(qū)間表示);(2)討論函數(shù)f(x)在D上的單調(diào)性;(3)若k<-6,求D上滿足條件f(x)>f(1)的x的集合(用區(qū)間表示).解析(1)由題意得(x2+2x+k)2+2(x2+2x+k)-3>0,(x2+2x+k)+3(x2+2x+k)-1>0,x2+2x+k<-3或x2+2x+k>1,(x+1)2<-2-k(-2-k>0)或(x+1)2>2-k(2-k>0),|x+1|<或|x+1|>,-1-<x<-1+或x<-1-或x>-1+,函數(shù)f(x)的定義域D為(-,-1-)(-1-,-1+)(-1+,+).(2)f (x)=-=-,由f (x)>0得(x2+2x+k+1)(2x+2)<0,即(x+1+)(x+1-)(x+1)<0,x<-1-或-1<x<-1+,結(jié)合定義域知x<-1-或-1<x<-1+,所以函數(shù)f(x)的單調(diào)遞增區(qū)間為(-,-1-),(-1,-1+),同理,遞減區(qū)間為(-1-,-1),(-1+,+).(3)由f(x)=f(1)得(x2+2x+k)2+2(x2+2x+k)-3=(3+k)2+2(3+k)-3,(x2+2x+k)2-(3+k)2+2(x2+2x+k)-(3+k)=0,(x2+2x+2k+5)(x2+2x-3)=0,(x+1+)(x+1-)(x+3)(x-1)=0,x=-1-或x=-1+或x=-3或x=1,k<-6,1(-1,-1+),-3(-1-,-1),-1-<-1-,-1+>-1+,結(jié)合函數(shù)f(x)的單調(diào)性知f(x)>f(1)的解集為(-1-,-1-)(-1-,-3)(1,-1+)(-1+,-1+).考點(diǎn)二函數(shù)的極值與最值4.(xx課標(biāo),12,5分)設(shè)函數(shù)f(x)=sin.若存在f(x)的極值點(diǎn)x0滿足+f(x0)2<m2,則m的取值范圍是()A.(-,-6)(6,+)B.(-,-4)(4,+)C.(-,-2)(2,+)D.(-,-1)(1,+)答案C5.(xx安徽,18,12分)設(shè)函數(shù)f(x)=1+(1+a)x-x2-x3,其中a>0.(1)討論f(x)在其定義域上的單調(diào)性;(2)當(dāng)x0,1時(shí),求f(x)取得最大值和最小值時(shí)的x的值.解析(1)f(x)的定義域?yàn)?-,+), f (x)=1+a-2x-3x2.令f (x)=0,得x1=,x2=,x1<x2,所以f (x)=-3(x-x1)(x-x2).當(dāng)x<x1或x>x2時(shí), f (x)<0;當(dāng)x1<x<x2時(shí), f (x)>0.故f(x)在(-,x1)和(x2,+)內(nèi)單調(diào)遞減,在(x1,x2)內(nèi)單調(diào)遞增.(2)因?yàn)閍>0,所以x1<0,x2>0.當(dāng)a4時(shí),x21.由(1)知, f(x)在0,1上單調(diào)遞增.所以f(x)在x=0和x=1處分別取得最小值和最大值.當(dāng)0<a<4時(shí),x2<1.由(1)知, f(x)在0,x2上單調(diào)遞增,在x2,1上單調(diào)遞減.所以f(x)在x=x2=處取得最大值.又f(0)=1, f(1)=a,所以當(dāng)0<a<1時(shí), f(x)在x=1處取得最小值;當(dāng)a=1時(shí), f(x)在x=0處和x=1處同時(shí)取得最小值;當(dāng)1<a<4時(shí), f(x)在x=0處取得最小值.6.(xx山東,20,13分)設(shè)函數(shù)f(x)=-k(k為常數(shù),e=2.718 28是自然對(duì)數(shù)的底數(shù)).(1)當(dāng)k0時(shí),求函數(shù)f(x)的單調(diào)區(qū)間;(2)若函數(shù)f(x)在(0,2)內(nèi)存在兩個(gè)極值點(diǎn),求k的取值范圍.解析(1)函數(shù)y=f(x)的定義域?yàn)?0,+).f (x)=-k=-=.由k0可得ex-kx>0,所以當(dāng)x(0,2)時(shí), f (x)<0,函數(shù)y=f(x)單調(diào)遞減,當(dāng)x(2,+)時(shí), f (x)>0,函數(shù)y=f(x)單調(diào)遞增.所以f(x)的單調(diào)遞減區(qū)間為(0,2),單調(diào)遞增區(qū)間為(2,+).(2)由(1)知,當(dāng)k0時(shí),函數(shù)f(x)在(0,2)內(nèi)單調(diào)遞減,故f(x)在(0,2)內(nèi)不存在極值點(diǎn);當(dāng)k>0時(shí),設(shè)函數(shù)g(x)=ex-kx,x0,+).因?yàn)間(x)=ex-k=ex-eln k,當(dāng)0<k1時(shí),當(dāng)x(0,2)時(shí),g(x)=ex-k>0,y=g(x)單調(diào)遞增,故f(x)在(0,2)內(nèi)不存在兩個(gè)極值點(diǎn);當(dāng)k>1時(shí),得x(0,ln k)時(shí),g(x)<0,函數(shù)y=g(x)單調(diào)遞減,x(ln k,+)時(shí),g(x)>0,函數(shù)y=g(x)單調(diào)遞增. 所以函數(shù)y=g(x)的最小值為g(ln k)=k(1-ln k).函數(shù)f(x)在(0,2)內(nèi)存在兩個(gè)極值點(diǎn),當(dāng)且僅當(dāng)解得e<k<.綜上所述,函數(shù)f(x)在(0,2)內(nèi)存在兩個(gè)極值點(diǎn)時(shí),k的取值范圍為.7.(xx福建,20,14分)已知函數(shù)f(x)=ex-ax(a為常數(shù))的圖象與y軸交于點(diǎn)A,曲線y=f(x)在點(diǎn)A處的切線斜率為-1.(1)求a的值及函數(shù)f(x)的極值;(2)證明:當(dāng)x>0時(shí),x2<ex;(3)證明:對(duì)任意給定的正數(shù)c,總存在x0,使得當(dāng)x(x0,+)時(shí),恒有x2<cex.解析解法一:(1)由f(x)=ex-ax,得f (x)=ex-a.又f (0)=1-a=-1,得a=2.所以f(x)=ex-2x,f (x)=ex-2.令f (x)=0,得x=ln 2.當(dāng)x<ln 2時(shí), f (x)<0,f(x)單調(diào)遞減;當(dāng)x>ln 2時(shí), f (x)>0,f(x)單調(diào)遞增.所以當(dāng)x=ln 2時(shí),f(x)取得極小值,且極小值為f(ln 2)=eln 2-2ln 2=2-ln 4,f(x)無極大值.(2)令g(x)=ex-x2,則g(x)=ex-2x.由(1)得g(x)=f(x)f(ln 2)>0,故g(x)在R上單調(diào)遞增,又g(0)=1>0,因此,當(dāng)x>0時(shí),g(x)>g(0)>0,即x2<ex.(3)若c1,則excex.又由(2)知,當(dāng)x>0時(shí),x2<ex.所以當(dāng)x>0時(shí),x2<cex.取x0=0,當(dāng)x(x0,+)時(shí),恒有x2<cex.若0<c<1,令k=>1,要使不等式x2<cex成立,只要ex>kx2成立.而要使ex>kx2成立,則只要x>ln(kx2),只要x>2ln x+ln k成立.令h(x)=x-2ln x-ln k,則h(x)=1-=,所以當(dāng)x>2時(shí),h(x)>0,h(x)在(2,+)內(nèi)單調(diào)遞增.取x0=16k>16,所以h(x)在(x0,+)內(nèi)單調(diào)遞增,又h(x0)=16k-2ln(16k)-ln k=8(k-ln 2)+3(k-ln k)+5k,易知k>ln k,k>ln 2,5k>0,所以h(x0)>0.即存在x0=,當(dāng)x(x0,+)時(shí),恒有x2<cex.綜上,對(duì)任意給定的正數(shù)c,總存在x0,當(dāng)x(x0,+)時(shí),恒有x2<cex.解法二:(1)同解法一.(2)同解法一.(3)對(duì)任意給定的正數(shù)c,取x0=,由(2)知,當(dāng)x>0時(shí),ex>x2,所以ex=>,當(dāng)x>x0時(shí),ex>>=x2,因此,對(duì)任意給定的正數(shù)c,總存在x0,當(dāng)x(x0,+)時(shí),恒有x2<cex.解法三:(1)同解法一.(2)同解法一.(3)首先證明當(dāng)x(0,+)時(shí),恒有x3<ex.證明如下:令h(x)=x3-ex,則h(x)=x2-ex.由(2)知,當(dāng)x>0時(shí),x2<ex,從而h(x)<0,h(x)在(0,+)內(nèi)單調(diào)遞減,所以h(x)<h(0)=-1<0,即x3<ex.取x0=,當(dāng)x>x0時(shí),有x2<x3<ex.因此,對(duì)任意給定的正數(shù)c,總存在x0,當(dāng)x(x0,+)時(shí),恒有x2<cex.注:對(duì)c的分類可有不同的方式,只要解法正確,均相應(yīng)給分.考點(diǎn)三導(dǎo)數(shù)的綜合應(yīng)用8.(xx陜西,10,5分)如圖,某飛行器在4千米高空水平飛行,從距著陸點(diǎn)A的水平距離10千米處開始下降,已知下降飛行軌跡為某三次函數(shù)圖象的一部分,則該函數(shù)的解析式為()A.y=x3-xB.y=x3-xC.y=x3-xD.y=-x3+x答案A9.(xx遼寧,11,5分)當(dāng)x-2,1時(shí),不等式ax3-x2+4x+30恒成立,則實(shí)數(shù)a的取值范圍是()A.-5,-3B.C.-6,-2D.-4,-3答案C10.(xx課標(biāo),21,12分)設(shè)函數(shù)f(x)=aexln x+,曲線y=f(x)在點(diǎn)(1, f(1)處的切線方程為y=e(x-1)+2.(1)求a,b;(2)證明:f(x)>1.解析(1)函數(shù)f(x)的定義域?yàn)?0,+), f (x)=aexln x+ex-ex-1+ex-1.由題意可得f(1)=2, f (1)=e.故a=1,b=2.(2)由(1)知, f(x)=exln x+ex-1,從而f(x)>1等價(jià)于xln x>xe-x-.設(shè)函數(shù)g(x)=xln x,則g(x)=1+ln x.所以當(dāng)x時(shí),g(x)<0;當(dāng)x時(shí),g(x)>0.故g(x)在上單調(diào)遞減,在上單調(diào)遞增,從而g(x)在(0,+)上的最小值為g=-.設(shè)函數(shù)h(x)=xe-x-,則h(x)=e-x(1-x).所以當(dāng)x(0,1)時(shí),h(x)>0;當(dāng)x(1,+)時(shí),h(x)<0.故h(x)在(0,1)上單調(diào)遞增,在(1,+)上單調(diào)遞減,從而h(x)在(0,+)上的最大值為h(1)=-.綜上,當(dāng)x>0時(shí),g(x)>h(x),即f(x)>1.11.(xx北京,18,13分)已知函數(shù)f(x)=xcos x-sin x,x.(1)求證:f(x)0;(2)若a<<b對(duì)x恒成立,求a的最大值與b的最小值.解析(1)由f(x)=xcos x-sin x得f (x)=cos x-xsin x-cos x=-xsin x.因?yàn)樵趨^(qū)間上f (x)=-xsin x<0,所以f(x)在區(qū)間上單調(diào)遞減.從而f(x)f(0)=0.(2)當(dāng)x>0時(shí),“>a”等價(jià)于“sin x-ax>0”;“<b”等價(jià)于“sin x-bx<0”.令g(x)=sin x-cx,則g(x)=cos x-c.當(dāng)c0時(shí),g(x)>0對(duì)任意x恒成立.當(dāng)c1時(shí),因?yàn)閷?duì)任意x,g(x)=cos x-c<0,所以g(x)在區(qū)間上單調(diào)遞減.從而g(x)<g(0)=0對(duì)任意x恒成立.當(dāng)0<c<1時(shí),存在唯一的x0使得g(x0)=cos x0-c=0.g(x)與g(x)在區(qū)間上的情況如下:x(0,x0)x0g(x)+0-g(x)因?yàn)間(x)在區(qū)間0,x0上是增函數(shù),所以g(x0)>g(0)=0.進(jìn)一步,“g(x)>0對(duì)任意x恒成立”當(dāng)且僅當(dāng)g=1-c0,即0<c.綜上所述,當(dāng)且僅當(dāng)c時(shí),g(x)>0對(duì)任意x恒成立;當(dāng)且僅當(dāng)c1時(shí),g(x)<0對(duì)任意x恒成立.所以,若a<<b對(duì)任意x恒成立,則a的最大值為,b的最小值為1.12.(xx江西,18,12分)已知函數(shù)f(x)=(x2+bx+b)(bR).(1)當(dāng)b=4時(shí),求f(x)的極值;(2)若f(x)在區(qū)間上單調(diào)遞增,求b的取值范圍.解析(1)當(dāng)b=4時(shí), f (x)=,由f (x)=0得x=-2或x=0.當(dāng)x(-,-2)時(shí), f (x)<0, f(x)單調(diào)遞減;當(dāng)x(-2,0)時(shí), f (x)>0, f(x)單調(diào)遞增;當(dāng)x時(shí), f (x)<0, f(x)單調(diào)遞減,故f(x)在x=-2處取極小值f(-2)=0,在x=0處取極大值f(0)=4.(2)f (x)=,因?yàn)楫?dāng)x時(shí),<0,依題意,當(dāng)x時(shí),有5x+(3b-2)0,從而+(3b-2)0.所以b的取值范圍為.13.(xx天津,20,14分)設(shè)f(x)=x-aex(aR),xR.已知函數(shù)y=f(x)有兩個(gè)零點(diǎn)x1,x2,且x1<x2.(1)求a的取值范圍;(2)證明隨著a的減小而增大;(3)證明x1+x2隨著a的減小而增大.解析(1)由f(x)=x-aex,可得f (x)=1-aex,下面分兩種情況討論:a0時(shí),f (x)>0在R上恒成立,可得f(x)在R上單調(diào)遞增,不合題意.a>0時(shí),由f (x)=0,得x=-ln a.當(dāng)x變化時(shí), f (x), f(x)的變化情況如下表:x(-,-ln a)-ln a(-ln a,+)f (x)+0-f(x)-ln a-1這時(shí), f(x)的單調(diào)遞增區(qū)間是(-,-ln a);單調(diào)遞減區(qū)間是(-ln a,+).于是,“函數(shù)y=f(x)有兩個(gè)零點(diǎn)”等價(jià)于如下條件同時(shí)成立:(i)f(-ln a)>0;(ii)存在s1(-,-ln a),滿足f(s1)<0;(iii)存在s2(-ln a,+),滿足f(s2)<0.由f(-ln a)>0,即-ln a-1>0,解得0<a<e-1.而此時(shí),取s1=0,滿足s1(-,-ln a),且f(s1)=-a<0;取s2=+ln,滿足s2(-ln a,+),且f(s2)=+<0.所以a的取值范圍是(0,e-1).(2)證明:由f(x)=x-aex=0,有a=.設(shè)g(x)=,由g(x)=,知g(x)在(-,1)上單調(diào)遞增,在(1,+)上單調(diào)遞減.并且,當(dāng)x(-,0時(shí),g(x)0;當(dāng)x(0,+)時(shí),g(x)>0.由已知,x1,x2滿足a=g(x1),a=g(x2).由a(0,e-1),及g(x)的單調(diào)性,可得x1(0,1),x2(1,+).對(duì)于任意的a1,a2(0,e-1),設(shè)a1>a2,g(1)=g(2)=a1,其中0<1<1<2;g(1)=g(2)=a2,其中0<1<1<2.因?yàn)間(x)在(0,1)上單調(diào)遞增,故由a1>a2,即g(1)>g(1),可得1>1;類似可得2<2.又由1,1>0,得<<.所以隨著a的減小而增大.(3)證明:由x1=a,x2=a,可得ln x1=ln a+x1,ln x2=ln a+x2.故x2-x1=ln x2-ln x1=ln.設(shè)=t,則t>1,且解得x1=,x2=.所以x1+x2=.(*)令h(x)=,x(1,+),則h(x)=.令u(x)=-2ln x+x-,得u(x)=.當(dāng)x(1,+)時(shí),u(x)>0.因此,u(x)在(1,+)上單調(diào)遞增,故對(duì)于任意的x(1,+),u(x)>u(1)=0,由此可得h(x)>0,故h(x)在(1,+)上單調(diào)遞增.因此,由(*)可得x1+x2隨著t的增大而增大.而由(2),知t隨著a的減小而增大,所以x1+x2隨著a的減小而增大.14.(xx遼寧,21,12分)已知函數(shù)f(x)=(cos x-x)(+2x)-(sin x+1),g(x)=3(x-)cos x-4(1+sin x)ln.證明:(1)存在唯一x0,使f(x0)=0;(2)存在唯一x1,使g(x1)=0,且對(duì)(1)中的x0,有x0+x1<.證明(1)當(dāng)x時(shí), f (x)=-(1+sin x)(+2x)-2x-cos x<0,函數(shù)f(x)在上為減函數(shù),又f(0)=->0, f=-2-<0,所以存在唯一x0,使f(x0)=0.(2)考慮函數(shù)h(x)=-4ln,x.令t=-x,則x時(shí),t.記u(t)=h(-t)=-4ln,則u(t)=.由(1)得,當(dāng)t(0,x0)時(shí),u(t)>0,當(dāng)t時(shí),u(t)<0.在(0,x0)上u(t)是增函數(shù),又u(0)=0,從而當(dāng)t(0,x0時(shí),u(t)>0,所以u(píng)(t)在(0,x0上無零點(diǎn).在上u(t)為減函數(shù),由u(x0)>0,u=-4ln 2<0,知存在唯一t1,使u(t1)=0.所以存在唯一的t1,使u(t1)=0.因此存在唯一的x1=-t1,使h(x1)=h(-t1)=u(t1)=0.因?yàn)楫?dāng)x時(shí),1+sin x>0,故g(x)=(1+sin x)h(x)與h(x)有相同的零點(diǎn),所以存在唯一的x1,使g(x1)=0.因x1=-t1,t1>x0,所以x0+x1<.15.(xx湖南,22,13分)已知常數(shù)a>0,函數(shù)f(x)=ln(1+ax)-.(1)討論f(x)在區(qū)間(0,+)上的單調(diào)性;(2)若f(x)存在兩個(gè)極值點(diǎn)x1,x2,且f(x1)+f(x2)>0,求a的取值范圍.解析(1)f (x)=-=.(*)當(dāng)a1時(shí), f (x)>0,此時(shí), f(x)在區(qū)間(0,+)上單調(diào)遞增.當(dāng)0<a<1時(shí),由f (x)=0得x1=2x2=-2舍去.當(dāng)x(0,x1)時(shí), f (x)<0;當(dāng)x(x1,+)時(shí), f (x)>0,故f(x)在區(qū)間(0,x1)上單調(diào)遞減,在區(qū)間(x1,+)上單調(diào)遞增.綜上所述,當(dāng)a1時(shí), f(x)在區(qū)間(0,+)上單調(diào)遞增;當(dāng)0<a<1時(shí), f(x)在區(qū)間上單調(diào)遞減,在區(qū)間上單調(diào)遞增.(2)由(*)式知,當(dāng)a1時(shí), f (x)0,此時(shí)f(x)不存在極值點(diǎn).因而要使得f(x)有兩個(gè)極值點(diǎn),必有0<a<1,又f(x)的極值點(diǎn)只可能是x1=2和x2=-2,且由f(x)的定義可知,x>-且x-2,所以-2>-,-2-2,解得a.此時(shí),由(*)式易知,x1,x2分別是f(x)的極小值點(diǎn)和極大值點(diǎn).而f(x1)+f(x2)=ln(1+ax1)-+ln(1+ax2)-=ln1+a(x1+x2)+a2x1x2-=ln(2a-1)2-=ln(2a-1)2+-2,令2a-1=x,由0<a<1且a知,當(dāng)0<a<時(shí),-1<x<0;當(dāng)<a<1時(shí),0<x<1,記g(x)=ln x2+-2.(i)當(dāng)-1<x<0時(shí),g(x)=2ln(-x)+-2,所以g(x)=-=<0,因此,g(x)在區(qū)間(-1,0)上單調(diào)遞減,從而g(x)<g(-1)=-4<0,故當(dāng)0<a<時(shí), f(x1)+f(x2)<0.(ii)當(dāng)0<x<1時(shí),g(x)=2ln x+-2,所以g(x)=-=<0,因此,g(x)在區(qū)間(0,1)上單調(diào)遞減,從而g(x)>g(1)=0,故當(dāng)<a<1時(shí), f(x1)+f(x2)>0.綜上所述,滿足條件的a的取值范圍為.16.(xx湖北,22,14分)為圓周率,e=2.718 28為自然對(duì)數(shù)的底數(shù).(1)求函數(shù)f(x)=的單調(diào)區(qū)間;(2)求e3,3e,e,e,3,3這6個(gè)數(shù)中的最大數(shù)與最小數(shù);(3)將e3,3e,e,e,3,3這6個(gè)數(shù)按從小到大的順序排列,并證明你的結(jié)論.解析(1)函數(shù)f(x)的定義域?yàn)?0,+).因?yàn)閒(x)=,所以f (x)=.當(dāng)f (x)>0,即0<x<e時(shí),函數(shù)f(x)單調(diào)遞增;當(dāng)f (x)<0,即x>e時(shí),函數(shù)f(x)單調(diào)遞減.故函數(shù)f(x)的單調(diào)遞增區(qū)間為(0,e),單調(diào)遞減區(qū)間為(e,+).(2)因?yàn)閑<3<,所以eln 3<eln ,ln e<ln 3,即ln 3e<ln e,ln e<ln 3.于是根據(jù)函數(shù)y=ln x,y=ex,y=x在定義域上單調(diào)遞增,可得3e<e<3,e3<e<3.故這6個(gè)數(shù)的最大數(shù)在3與3之中,最小數(shù)在3e與e3之中.由e<3<及(1)的結(jié)論,得f()<f(3)<f(e),即<<.由<,得ln 3<ln 3,所以3>3;由<,得ln 3e<ln e3,所以3e<e3.綜上,6個(gè)數(shù)中的最大數(shù)是3,最小數(shù)是3e.(3)由(2)知,3e<e<3<3,3e<e3.又由(2)知,<,得e<e.故只需比較e3與e和e與3的大小.由(1)知,當(dāng)0<x<e時(shí), f(x)<f(e)=,即<.在上式中,令x=,又<e,則ln<,從而2-ln <,即得ln >2-.由得,eln >e>2.7>2.7(2-0.88)=3.024>3,即eln >3,亦即ln e>ln e3,所以e3<e.又由得,3ln >6->6-e>,即3ln >,所以e<3.綜上可得,3e<e3<e<e<3<3,即6個(gè)數(shù)從小到大的順序?yàn)?e,e3,e,e,3,3.17.(xx重慶,20,12分)已知函數(shù)f(x)=ae2x-be-2x-cx(a,b,cR)的導(dǎo)函數(shù)f (x)為偶函數(shù),且曲線y=f(x)在點(diǎn)(0, f(0)處的切線的斜率為4-c.(1)確定a,b的值;(2)若c=3,判斷f(x)的單調(diào)性;(3)若f(x)有極值,求c的取值范圍.解析(1)對(duì)f(x)求導(dǎo)得f (x)=2ae2x+2be-2x-c,由f (x)為偶函數(shù),知f (-x)=f (x),即2(a-b)(e2x+e-2x)=0,因?yàn)閑2x+e-2x>0,所以a=b.又f (0)=2a+2b-c=4-c,故a=1,b=1.(2)當(dāng)c=3時(shí), f(x)=e2x-e-2x-3x,那么f (x)=2e2x+2e-2x-32-3=1>0,故f(x)在R上為增函數(shù).(3)由(1)知f (x)=2e2x+2e-2x-c,而2e2x+2e-2x2=4,當(dāng)x=0時(shí)等號(hào)成立.下面分三種情況進(jìn)行討論.當(dāng)c<4時(shí),對(duì)任意xR, f (x)=2e2x+2e-2x-c>0,此時(shí)f(x)無極值;當(dāng)c=4時(shí),對(duì)任意x0, f (x)=2e2x+2e-2x-4>0,此時(shí)f(x)無極值;當(dāng)c>4時(shí),令e2x=t,注意到方程2t+-c=0有兩根t1,2=>0,即f (x)=0有兩個(gè)根x1=ln t1,x2=ln t2.當(dāng)x1<x<x2時(shí), f (x)<0;又當(dāng)x>x2時(shí), f (x)>0,從而f(x)在x=x2處取得極小值.綜上,若f(x)有極值,則c的取值范圍為(4,+).18.(xx浙江,22,14分)已知函數(shù)f(x)=x3+3|x-a|(aR).(1)若f(x)在-1,1上的最大值和最小值分別記為M(a),m(a),求M(a)-m(a);(2)設(shè)bR.若f(x)+b24對(duì)x-1,1恒成立,求3a+b的取值范圍.解析(1)因?yàn)閒(x)=所以f (x)=由于-1x1,(i)當(dāng)a-1時(shí),有xa,故f(x)=x3+3x-3a.此時(shí)f(x)在(-1,1)上是增函數(shù),因此,M(a)=f(1)=4-3a,m(a)=f(-1)=-4-3a,故M(a)-m(a)=(4-3a)-(-4-3a)=8.(ii)當(dāng)-1<a<1時(shí),若x(a,1),則f(x)=x3+3x-3a,在(a,1)上是增函數(shù);若x(-1,a),則f(x)=x3-3x+3a,在(-1,a)上是減函數(shù),所以,M(a)=maxf(1), f(-1),m(a)=f(a)=a3,由于f(1)-f(-1)=-6a+2,因此,當(dāng)-1<a時(shí),M(a)-m(a)=-a3-3a+4;當(dāng)<a<1時(shí),M(a)-m(a)=-a3+3a+2.(iii)當(dāng)a1時(shí),有xa,故f(x)=x3-3x+3a,此時(shí)f(x)在(-1,1)上是減函數(shù),因此,M(a)=f(-1)=2+3a,m(a)=f(1)=-2+3a,故M(a)-m(a)=(2+3a)-(-2+3a)=4.綜上,M(a)-m(a)=(2)令h(x)=f(x)+b,則h(x)=h(x)=因?yàn)閒(x)+b24對(duì)x-1,1恒成立,即-2h(x)2對(duì)x-1,1恒成立,所以由(1)知,(i)當(dāng)a-1時(shí),h(x)在(-1,1)上是增函數(shù),h(x)在-1,1上的最大值是h(1)=4-3a+b,最小值是h(-1)=-4-3a+b,則-4-3a+b-2且4-3a+b2,矛盾.(ii)當(dāng)-1<a時(shí),h(x)在-1,1上的最小值是h(a)=a3+b,最大值是h(1)=4-3a+b,所以a3+b-2且4-3a+b2,從而-2-a3+3a3a+b6a-2且0a.令t(a)=-2-a3+3a,則t(a)=3-3a2>0,t(a)在上是增函數(shù),故t(a)t(0)=-2,因此-23a+b0.(iii)當(dāng)<a<1時(shí),h(x)在-1,1上的最小值是h(a)=a3+b,最大值是h(-1)=3a+b+2,所以a3+b-2且3a+b+22,解得-<3a+b0.(iv)當(dāng)a1時(shí),h(x)在-1,1上的最大值是h(-1)=2+3a+b,最小值是h(1)=-2+3a+b,所以3a+b+22且3a+b-2-2,解得3a+b=0.綜上,得3a+b的取值范圍是-23a+b0.19.(xx四川,21,14分)已知函數(shù)f(x)=ex-ax2-bx-1,其中a,bR,e=2.718 28為自然對(duì)數(shù)的底數(shù).(1)設(shè)g(x)是函數(shù)f(x)的導(dǎo)函數(shù),求函數(shù)g(x)在區(qū)間0,1上的最小值;(2)若f(1)=0,函數(shù)f(x)在區(qū)間(0,1)內(nèi)有零點(diǎn),求a的取值范圍.解析(1)由f(x)=ex-ax2-bx-1,有g(shù)(x)=f (x)=ex-2ax-b.所以g(x)=ex-2a.因此,當(dāng)x0,1時(shí),g(x)1-2a,e-2a.當(dāng)a時(shí),g(x)0,所以g(x)在0,1上單調(diào)遞增.因此g(x)在0,1上的最小值是g(0)=1-b;當(dāng)a時(shí),g(x)0,所以g(x)在0,1上單調(diào)遞減,因此g(x)在0,1上的最小值是g(1)=e-2a-b;當(dāng)<a<時(shí),令g(x)=0,得x=ln(2a)(0,1).所以函數(shù)g(x)在區(qū)間0,ln(2a)上單調(diào)遞減,在區(qū)間(ln(2a),1上單調(diào)遞增.于是,g(x)在0,1上的最小值是g(ln(2a)=2a-2aln(2a)-b.綜上所述,當(dāng)a時(shí),g(x)在0,1上的最小值是g(0)=1-b;當(dāng)<a<時(shí),g(x)在0,1上的最小值是g(ln(2a)=2a-2aln(2a)-b;當(dāng)a時(shí),g(x)在0,1上的最小值是g(1)=e-2a-b.(2)設(shè)x0為f(x)在區(qū)間(0,1)內(nèi)的一個(gè)零點(diǎn),則由f(0)=f(x0)=0可知, f(x)在區(qū)間(0,x0)上不可能單調(diào)遞增,也不可能單調(diào)遞減.則g(x)不可能恒為正,也不可能恒為負(fù).故g(x)在區(qū)間(0,x0)內(nèi)存在零點(diǎn)x1.同理g(x)在區(qū)間(x0,1)內(nèi)存在零點(diǎn)x2.所以g(x)在區(qū)間(0,1)內(nèi)至少有兩個(gè)零點(diǎn).由(1)知,當(dāng)a時(shí),g(x)在0,1上單調(diào)遞增,故g(x)在(0,1)內(nèi)至多有一個(gè)零點(diǎn).當(dāng)a時(shí),g(x)在0,1上單調(diào)遞減,故g(x)在(0,1)內(nèi)至多有一個(gè)零點(diǎn).所以<a<.此時(shí)g(x)在區(qū)間0,ln(2a)上單調(diào)遞減,在區(qū)間(ln(2a),1上單調(diào)遞增.因此x1(0,ln(2a),x2(ln(2a),1),必有g(shù)(0)=1-b>0,g(1)=e-2a-b>0.由f(1)=0有a+b=e-1<2,有g(shù)(0)=1-b=a-e+2>0,g(1)=e-2a-b=1-a>0.解得e-2<a<1.當(dāng)e-2<a<1時(shí),g(x)在區(qū)間0,1內(nèi)有最小值g(ln(2a).若g(ln(2a)0,則g(x)0(x0,1),從而f(x)在區(qū)間0,1上單調(diào)遞增,這與f(0)=f(1)=0矛盾,所以g(ln(2a)<0.又g(0)=a-e+2>0,g(1)=1-a>0,故此時(shí)g(x)在(0,ln(2a)和(ln(2a),1)內(nèi)各只有一個(gè)零點(diǎn)x1和x2.由此可知f(x)在0,x1上單調(diào)遞增,在(x1,x2)上單調(diào)遞減,在x2,1上單調(diào)遞增.所以f(x1)>f(0)=0, f(x2)<f(1)=0,故f(x)在(x1,x2)內(nèi)有零點(diǎn).綜上可知,a的取值范圍是(e-2,1).