第三講導(dǎo)數(shù)的簡(jiǎn)單應(yīng)用(40分鐘70分)一、選擇題(每小題5分,共30分)1.已知直線y=kx+1與曲線y=x3+mx+n相切于點(diǎn)A(1,3),則n=()A.-1B.1C.3D.4【解析】選C.對(duì)于y=x3+mx+n,y=3x2+m,而直線y=kx+1與曲線y=x3+mx+n相切于點(diǎn)A(1,3),則有3+m=k,k+1=3,1+m+n=3,可解得n=3.2.函數(shù)f(x)=3x2+ln x-2x的極值點(diǎn)的個(gè)數(shù)是()A.0B.1C.2D.無(wú)數(shù)個(gè)【解析】選A.函數(shù)定義域?yàn)?0,+),且f(x)=6x+1x-2=6x2-2x+1x,由于x>0,g(x)=6x2-2x+1中=-20<0,所以g(x)>0恒成立,故f(x)>0恒成立,即f(x)在定義域上單調(diào)遞增,無(wú)極值點(diǎn).3.設(shè)函數(shù)f(x)=2x+ln x,則()A.x=12為f(x)的極大值點(diǎn)B.x=12為f(x)的極小值點(diǎn)C.x=2為f(x)的極大值點(diǎn)D.x=2為f(x)的極小值點(diǎn)【解析】選D.由函數(shù)f(x)=2x+ln x求導(dǎo)數(shù)得f(x)=-2x2+1x=x-2x2,函數(shù)定義域?yàn)?0,+),所以在區(qū)間(0,2)上,f(x)<0,f(x)是減函數(shù),在區(qū)間(2,+)上,f(x)>0,f(x)是增函數(shù),所以x=2為f(x)的極小值點(diǎn).4.(2018菏澤一模)已知函數(shù)f(x)=x3+bx2+cx+d的圖象如圖所示,則函數(shù)y=log2x2+23bx+c3的單調(diào)遞減區(qū)間為()A.12,+B.3,+)C.-2,3D.(-,-2)【解析】選D.因?yàn)閒(x)=x3+bx2+cx+d,所以f(x)=3x2+2bx+c,由圖可知f(-2) =f(3)=0,所以12-4b+c=0,27+6b+c=0,解得b=-32,c=-18.令g(x)=x2+23bx+c3,則g(x)=x2-x-6,g(x)=2x-1,由g(x)=x2-x-6>0,解得x<-2或x>3.當(dāng)x<12時(shí),g(x)<0,所以g(x)=x2-x-6在(-,-2)上為減函數(shù),所以函數(shù)y=log2x2+23bx+c3的單調(diào)遞減區(qū)間為(-,-2).5.設(shè)f(x)和g(x)分別是f(x)和g(x)的導(dǎo)函數(shù),若f(x)g(x)0在區(qū)間上恒成立,則稱f(x)和g(x)在區(qū)間上單調(diào)性相反,若函數(shù)f(x)=13x3-2ax與g(x)=x2+2bx在開區(qū)間(a,b)上單調(diào)性相反(a>0),則b-a的最大值為()A.12B.1C.32D.2【解析】選A.由題意得,b>a>0,所以g(x)=2x+2b>0在(a,b)上恒成立,所以問(wèn)題等價(jià)于f(x)=x2-2a0在(a,b)上恒成立,所以(x2-2a)max=b2-2a0,所以b-ab-12b2=-12(b-1)2+1212,當(dāng)且僅當(dāng)b=1,a=12時(shí),等號(hào)成立,所以b-a的最大值為12.6.(2018唐山一模)設(shè)函數(shù)f(x)與g(x)是定義在同一區(qū)間a,b上的兩個(gè)函數(shù),若對(duì)任意的xa,b,都有|f(x)-g(x)|k(k>0),則稱f(x)和g(x)在a,b上是“k度和諧函數(shù)”,a,b稱為“k度密切區(qū)間”.設(shè)函數(shù)f(x)=ln x與g(x)=mx-1x在1e,e上是“e度和諧函數(shù)”,則m的取值范圍是()A.-e-1,1B.-1,e+1C.1e-e,1+eD.1e+1-e,1+e【解析】選B.設(shè)h(x)=f(x)-g(x)=ln x-mx-1x=-m+1x+ln x,h(x)=-1x2+1x=x-1x2,故當(dāng)x1e,1時(shí),h(x)<0,函數(shù)h(x)單調(diào)遞減;當(dāng)x1,e時(shí),h(x)0,函數(shù)h(x)單調(diào)遞增.所以函數(shù)h(x)的最小值為h(1)=-m+1,而h1e=-m+e-1.h(e)=-m+1e+1,顯然e-1>1e+1,所以h1e>h(e),故函數(shù)h(x)的最大值為h1e=-m+e-1.故函數(shù)h(x)在1e,e上的值域?yàn)?m+1,-m+e-1.由題意,|h(x)|e,即-eh(x)e,所以-m+1-e,-m+e-1e,解得-1m1+e.二、填空題(每小題5分,共10分)7.曲線y=xex+2x+1在點(diǎn)(0,1)處的切線方程為_. 【解析】y=ex+xex+2,斜率k=e0+0+2=3,所以,y-1=3x,即y=3x+1.答案:y=3x+18.已知函數(shù)f(x)=x(ln x-ax)有兩個(gè)極值點(diǎn),則實(shí)數(shù)a的取值范圍是_. 【解析】函數(shù)f(x)=x(ln x-ax),則f(x)=ln x-ax+x1x-a=ln x-2ax+1, 令f(x)=ln x-2ax+1=0,得ln x=2ax-1,因?yàn)楹瘮?shù)f(x)=x(ln x-ax)有兩個(gè)極值點(diǎn),所以f(x)=ln x-2ax+1有兩個(gè)零點(diǎn),等價(jià)于函數(shù)y=ln x與y=2ax-1的圖象有兩個(gè)交點(diǎn),在同一個(gè)坐標(biāo)系中作出它們的圖象,過(guò)點(diǎn)(0,-1)作y=ln x的切線,設(shè)切點(diǎn)為(x0,y0),則切線的斜率k=1x0,切線方程為y=1x0x-1. 切點(diǎn)在切線上,則y0=x0x0-1=0,又切點(diǎn)在曲線y=ln x上,則ln x0=0x0=1,即切點(diǎn)為(1,0).切線方程為y=x-1.再由直線y=2ax-1與曲線y=ln x有兩個(gè)交點(diǎn),知直線y=2ax-1位于兩直線y=-1和y=x-1之間,其斜率2a滿足0<2a<1,解得實(shí)數(shù)a的取值范圍是0,12.答案:0,12三、解答題(每小題10分,共30分)9.已知函數(shù)f(x)=x-eax(a>0).(1)求函數(shù)f(x)的單調(diào)區(qū)間.(2)求函數(shù)f(x)在1a,2a上的最大值.【解析】(1)f(x)=x-eax(a>0),則f(x)=1-aeax,令f(x)=1-aeax=0,則x=1aln 1a.當(dāng)x變化時(shí),f(x),f(x)的變化情況如下表:x-,1aln1a1aln 1a1aln1a,+f(x)+0-f(x)極大值故函數(shù)f(x)的單調(diào)遞增區(qū)間為-,1aln 1a;單調(diào)遞減區(qū)間為1aln 1a,+.(2)當(dāng)1aln 1a2a,即0<a1e2時(shí),f(x)max=f2a=2a-e2;當(dāng)1a<1aln 1a<2a,即1e2<a<1e時(shí),f(x)max=f1aln 1a=1aln 1a-1a;當(dāng)1aln 1a1a,即a1e時(shí),f(x)max=f1a=1a-e.綜上,當(dāng)0<a1e2時(shí),f(x)max=2a-e2;當(dāng)1e2<a<1e時(shí),f(x)max=1aln 1a-1a;當(dāng)a1e時(shí),f(x)max=1a-e.10.已知函數(shù)f(x)=ln x-x2+x.(1)求函數(shù)f(x)的單調(diào)遞減區(qū)間.(2)若關(guān)于x的不等式f(x)a2-1x2+ax-1恒成立,求整數(shù)a的最小值.(3)若正實(shí)數(shù)x1,x2滿足f(x1)+f(x2)+2(x12+x22)+x1x2=0,證明x1+x25-12.【解題指南】(1)求f(x),而使f(x)0的x所在區(qū)間便為f(x)的單調(diào)遞減區(qū)間.(2)構(gòu)造函數(shù)g(x)=f(x)-a2-1x2+ax-1=ln x-12ax2+(1-a)x+1,求g(x) =-ax2+(1-a)x+1x,容易判斷當(dāng)a0時(shí)不合題意;而a>0時(shí),能夠求出f(x)的最大值為g1a=12a-ln a,可設(shè)h(a)=12a-ln a,該函數(shù)在(0,+)上為減函數(shù),并且h(1)>0,h(2)<0,從而得到整數(shù)a最小為2.(3)由f(x1)+f(x2)+2(x12+x22)+x1x2=0便得到(x1+x2)2+(x1+x2)=x1x2-ln(x1x2),這樣令t=x1x2,t>0,容易求得函數(shù)t-ln t的最小值為1,從而得到(x1+x2)2+(x1+x2)1,解這個(gè)關(guān)于x1+x2的一元二次不等式即可得出要證的結(jié)論.【解析】(1)f(x)=-2x2+x+1x(x>0),所以x1時(shí),f(x)0;所以f(x)的單調(diào)減區(qū)間為1,+).(2)令g(x)=f(x)-a2-1x2+ax-1=ln x-12ax2+(1-a)x+1,所以g(x)=-ax2+(1-a)x+1x=(-ax+1)(x+1)x,當(dāng)a0時(shí),因?yàn)閤>0,所以g(x)>0;所以此時(shí)g(x)在(0,+)上是遞增函數(shù);又g(1)=-32a+2>0;所以g(x)0不能恒成立,即關(guān)于x的不等式f(x)a2-1x2+ax-1不能恒成立;所以這種情況不存在;當(dāng)a>0時(shí),g(x)=ax-1a(x+1)x;所以當(dāng)x0,1a時(shí),g(x)>0;當(dāng)x1a,+時(shí),g(x)<0;所以函數(shù)g(x)的最大值為g1a=ln 1a-12a1a2+(1-a)1a+1=12a-ln a;令h(a)=12a-ln a;因?yàn)閔(1)=12>0,h(2)=14-ln 2<0,又h(a)在a(0,+)上是減函數(shù);所以當(dāng)a2時(shí),h(a)<0;所以整數(shù)a的最小值為2.(3)由f(x1)+f(x2)+2(x12+x22)+x1x2=0,即ln x1+x12+x1+ln x2+x22+x2+x1x2=0,從而(x1+x2)2+(x1+x2)=x1x2-ln(x1x2).令t=x1x2,則由h(t)=t-ln t得,h(t)=t-1t;可知,h(t)在區(qū)間(0,1)上單調(diào)遞減,在區(qū)間(1,+)上單調(diào)遞增;所以h(t)h(1)=1;所以(x1+x2)2+(x1+x2)1,又x1+x2>0,因此x1+x25-12成立.11.設(shè)函數(shù)f(x)=xlnx-ax.(1)若函數(shù)f(x)在(1,+)上為減函數(shù),求實(shí)數(shù)a的最小值.(2)若存在x1,x2e,e2,使f(x1)f(x2)+a成立,求實(shí)數(shù)a的取值范圍.【解析】(1)由已知得x>0,x1.因?yàn)閒(x)在(1,+)上為減函數(shù),故f(x)= lnx-1(lnx)2-a0在(1,+)上恒成立.所以當(dāng)x(1,+)時(shí),f(x)max0.又f(x)=lnx-1(lnx)2-a=-1lnx2+1lnx-a=-1lnx-122+14-a,故當(dāng)1lnx=12,即x=e2時(shí),f(x)max=14-a.所以14-a0,于是a14,故a的最小值為14. (2)命題“若存在x1,x2e,e2,使f(x1)f(x2)+a成立”等價(jià)于“當(dāng)xe,e2時(shí),有f(x)minf(x)max+a”.由(1),當(dāng)xe,e2時(shí),f(x)max=14-a,所以f(x)max+a=14.問(wèn)題等價(jià)于“當(dāng)xe,e2時(shí),有f(x)min14”.當(dāng)a14時(shí),由(1),f(x)在e,e2上為減函數(shù).則f(x)min=f(e2)= e22-ae214,故a12-14e2.當(dāng)-14<-a<0,即0<a<14時(shí),因?yàn)閤e,e2,所以ln x1,2.因?yàn)閒(x)=-a+lnx-1(lnx)2,由復(fù)合函數(shù)的單調(diào)性知f(x)在e,e2上為增函數(shù),所以存在唯一x0(e,e2),使f(x0)=0且滿足:f(x)min=f(x0)=-ax0+x0ln x0,要使f(x)min14,所以-a14x0-1ln x0<14-12=-14,與-14<-a<0矛盾,所以-14<-a<0不合題意.綜上,實(shí)數(shù)a的取值范圍為12-14e2,+.(20分鐘20分)1.(10分)(2018黃岡一模)已知函數(shù)f(x)滿足f(x)=f(1)ex-1-f(0)x+12x2.(1)求f(x)的解析式及單調(diào)區(qū)間.(2)若f(x)12x2+ax+b,求(a+1)b的最大值.【解析】(1)由已知得f(x)=f(1)ex-1-f(0)+x.所以f(1)=f(1)-f(0)+1,即f(0)=1.又f(0)=f(1)e-1,所以f(1)=e.從而f(x)=ex-x+12x2.由于f(x)=ex-1+x,故當(dāng)x(-,0)時(shí),f(x)<0;當(dāng)x(0,+)時(shí),f(x)>0.從而,f(x)的單調(diào)減區(qū)間是(-,0),單調(diào)增區(qū)間是(0,+).(2)由已知條件得ex-(a+1)xb.()若a+1<0,則對(duì)任意常數(shù)b,當(dāng)x<0,且x<1-ba+1時(shí),可得ex-(a+1)x<b,因此式不成立.()若a+1=0,則(a+1)b=0.()若a+1>0,設(shè)g(x)=ex-(a+1)x,則g(x)=ex-(a+1).當(dāng)x(-,ln(a+1)時(shí),g(x)<0;當(dāng)x(ln(a+1),+)時(shí),g(x)>0.從而g(x)在(-,ln(a+1)上單調(diào)遞減,在(ln(a+1),+)上單調(diào)遞增.故g(x)有最小值g(ln(a+1)=a+1-(a+1)ln(a+1).所以f(x)12x2+ax+b等價(jià)于ba+1-(a+1)ln(a+1).因此(a+1)b(a+1)2-(a+1)2ln(a+1).設(shè)h(a)=(a+1)2-(a+1)2ln(a+1),則h(a)=(a+1)1-2ln(a+1).所以h(a)在(-1,e12-1)上單調(diào)遞增,在(e12-1,+)上單調(diào)遞減,故h(a)在a=e12-1處取得最大值.從而h(a)e2,即(a+1)be2.當(dāng)a=e12-1,b=e122時(shí),式成立,故f(x)12x2+ax+b.綜合得,(a+1)b的最大值為e2.2.(10分)(2018新鄉(xiāng)一模)已知函數(shù)f(x)=(x+a)ln x,g(x)=x2ex,曲線y=f(x)在點(diǎn)(1,f(1)處的切線與直線2x-y-3=0平行.(1)求證:方程f(x)=g(x)在(1,2)內(nèi)存在唯一的實(shí)根.(2)設(shè)函數(shù)m(x)=minf(x),g(x)(minp,q表示p,q中的較小者),求m(x)的最大值.【解析】(1)由題意知,曲線y=f(x)在點(diǎn)(1,f(1)處的切線斜率為2,所以f(1)=2,又f(x)=ln x+ax+1,所以a=1.所以f(x)=(x+1)ln x.設(shè)h(x)=f(x)-g(x)=(x+1)ln x-x2ex,當(dāng)x(0,1時(shí),h(x)<0,又h(2)=3ln 2-4e2=ln 8-4e2>1-1=0,所以存在x0(1,2),使h(x0)=0.因?yàn)閔(x)=ln x+1x+1-x(2-x)ex,當(dāng)x(1,2)時(shí),0<x(2-x)=-(x-1)2+1<1,ex>e,所以0<1ex<1e,所以x(2-x)ex<1e,所以h(x)>1-1e>0,所以當(dāng)x(1,2)時(shí),h(x)單調(diào)遞增,所以方程f(x)=g(x)在(1,2)內(nèi)存在唯一的實(shí)根.(2)由(1)知,方程f(x)=g(x)在(1,2)內(nèi)存在唯一的實(shí)根x0,且x(0,x0)時(shí),f(x)<g(x),又當(dāng)x(x0,2)時(shí),h(x)>0,當(dāng)x(2,+)時(shí),h(x)>0,所以當(dāng)x(x0,+)時(shí),h(x)>0,所以當(dāng)x(x0,+)時(shí),f(x)>g(x),所以m(x)=(x+1)lnx,x(0,x0,x2ex,x(x0,+).當(dāng)x(0,x0)時(shí),若x(0,1,則m(x)0;若x(1,x0,由m(x)=ln x+1x+1>0,可知0<m(x)m(x0),故當(dāng)x(0,x0時(shí),m(x)m(x0).當(dāng)x(x0,+)時(shí),由m(x)=x(2-x)ex可得當(dāng)x(x0,2)時(shí),m(x)>0,m(x)單調(diào)遞增;x(2,+)時(shí),m(x)<0,m(x)單調(diào)遞減.可知m(x)m(2)=4e2,且m(x0)<m(2).綜上可得,函數(shù)m(x)的最大值為4e2.