02-第五章　一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用

第五章一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用全卷滿分150分考試用時(shí)120分鐘一、單項(xiàng)選擇題(本題共8小題,每小題5分,共40分.在每小題給出的四個(gè)選項(xiàng)中,只有一項(xiàng)是符合題目要求的)1.已知函數(shù)f(x)的圖象如圖所示,在區(qū)間[x1,x2],[x2,x3],[x1,x3],[x3,x4]內(nèi),函數(shù)y=f(x)的平均變化率最大的是()A.[x1,x2] B.[x2,x3] C.[x1,x3] D.[x3,x4]2.已知函數(shù)f(x)=sinx+4x,則limΔx→A.12 B.6 C.3 D.33.已知函數(shù)f(x)的導(dǎo)函數(shù)f'(x)在[a,b]上的圖象如圖所示,則下列說法正確的是()A.f(x)的極小值點(diǎn)為x1,x2B.f(x)的極大值點(diǎn)為x2C.f'(x)有唯一的極小值點(diǎn)D.函數(shù)f(x)在(a,b)上的極值點(diǎn)的個(gè)數(shù)為24.已知函數(shù)f(x)=x32x2,x∈[1,3],則下列說法錯(cuò)誤的是()A.函數(shù)f(x)的最大值為9B.函數(shù)f(x)的最小值為3C.函數(shù)f(x)在區(qū)間[1,3]上單調(diào)遞增D.x=0是函數(shù)f(x)的極大值點(diǎn)5.已知函數(shù)f(x)=x3+3ax2+bx+a2在x=1處取得極值0,則a+b=()A.4 B.7 C.11 D.4或116.已知定義在R上的函數(shù)f(x)滿足f(1)=1,且f(x)的導(dǎo)函數(shù)f'(x)在R上恒有f'(x)<12,則不等式f(x)<x2+1A.(1,+∞) B.(∞,1)C.(1,1) D.(∞,1)∪(1,+∞)7.若對(duì)于任意t∈[1,2],函數(shù)f(x)=x3+m2+2x22x在區(qū)間(t,3)上不單調(diào),則實(shí)數(shù)m的取值范圍是(A.-373,-9 B.-3738.已知a=3(2-ln3)e2A.a<c<b B.c<a<b C.a<b<c D.b<a<c二、多項(xiàng)選擇題(本題共4小題,每小題5分,共20分.在每小題給出的選項(xiàng)中,有多項(xiàng)符合題目要求.全部選對(duì)的得5分,部分選對(duì)的得2分,有選錯(cuò)的得0分)9.已知函數(shù)f(x)=exln(x+a),a∈R,則()A.當(dāng)a=0時(shí),f(x)沒有零點(diǎn)B.當(dāng)a=0時(shí),f(x)在(0,+∞)上單調(diào)遞增C.當(dāng)a=2時(shí),直線y=12x+1ln2與曲線y=f(x)D.當(dāng)a=2時(shí),f(x)只有一個(gè)極值點(diǎn)x0,且x0∈(1,0)10.對(duì)于函數(shù)f(x)=2lnxx2,下列說法正確的有A.f(x)在x=e處取得極大值1B.f(x)只有一個(gè)零點(diǎn)C.f(2)>f(π)D.若f(x)<k1x2在(0,+∞)上恒成立,11.已知函數(shù)f(x)=lnx+1,g(x)=ex1,下列說法正確的是(參考數(shù)據(jù):e2≈7.39,e3≈20.09,ln2≈0.69,ln3≈1.10)()A.存在實(shí)數(shù)m,使得直線y=x+m與曲線y=f(x)和曲線y=g(x)都相切B.存在實(shí)數(shù)k,使得直線y=kx1與曲線y=f(x)和曲線y=g(x)都相切C.函數(shù)y=g(x)f(x)在區(qū)間23D.函數(shù)y=g(x)f(x)在區(qū)間23,+12.已知實(shí)數(shù)a,b滿足等式e2aeb=2(2ba),則下列不等式中可能成立的有()A.a<b<0 B.b<a<0 C.0<a<b D.0<b<a三、填空題(本題共4小題,每小題5分,共20分)13.已知函數(shù)f(x)=2f'(1)x+3ex2,f'(x)是f(x)的導(dǎo)函數(shù),則f'(1)=.

14.已知函數(shù)f(x)=exmx+1的圖象上存在與直線y=12x垂直的切線,則實(shí)數(shù)m的取值范圍是15.記定義在R上的可導(dǎo)函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),且f'(x)f(x)<0,f(2)=1,則不等式f(x)>ex2的解集為.

16.已知函數(shù)f(x)=x-lnx,x>0,x+4e,x≤0,若存在x1≤0,x2四、解答題(本題共6小題,共70分.解答應(yīng)寫出文字說明、證明過程或演算步驟)17.(10分)已知函數(shù)f(x)=ex+ax+b,曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程為y=ab.(1)求a,b的值;(2)證明:f(x)≥0.18.(12分)某企業(yè)擬建造一個(gè)容器(不計(jì)厚度,長(zhǎng)度單位:m),如圖所示,其中容器的中間為圓柱形,左、右兩端均為半球形,按照設(shè)計(jì)要求,容器的容積為80π3m3,且l≥2r,假設(shè)該容器的建造費(fèi)用僅與其表面積有關(guān),已知圓柱形部分每平方米的建造費(fèi)用為3萬元,半球形部分每平方米的建造費(fèi)用為c(c>3)萬元,該容器的總建造費(fèi)用為y(1)寫出y關(guān)于r的函數(shù)表達(dá)式,并求該函數(shù)的定義域;(2)求該容器的總建造費(fèi)用最少時(shí)r的值.19.(12分)已知函數(shù)f(x)=ex(a+e)x2+ax.(1)當(dāng)a=e時(shí),求f(x)的單調(diào)區(qū)間;(2)若f(x)有三個(gè)零點(diǎn),求a的取值范圍.20.(12分)已知函數(shù)f(x)=12x2alnx+(1a)x+1(a∈R(1)討論函數(shù)f(x)的單調(diào)性;(2)當(dāng)a=1時(shí),求證:f(x)≤x(ex1)+12x221.(12分)已知函數(shù)f(x)=lnx+ax2x.(1)當(dāng)a=1時(shí),求f(x)的單調(diào)區(qū)間;(2)存在x∈[1,+∞),使得f(x)≥12x3+1成立,求a的最小整數(shù)值22.(12分)已知函數(shù)f(x)=x-asinx(x(1)若f(x)≥0恒成立,求實(shí)數(shù)a的取值范圍;(2)若a<14,證明:f(x)在0,π2上有唯一極值點(diǎn)x0,且f(x0)>

答案全解全析1.D2.B3.D4.C5.C6.A7.A8.A9.ACD10.AB11.AB12.ACD1.Df(x)在[x1,x2]上的平均變化率P1=f(f(x)在[x2,x3]上的平均變化率P2=f(f(x)在[x1,x3]上的平均變化率P3=f(f(x)在[x3,x4]上的平均變化率P4=f(結(jié)合題中函數(shù)y=f(x)的圖象,可得P2<P3<0<P1<P4.故選D.2.B易得f'(x)=cosx+4,∴f'(π)=3,∴l(xiāng)imΔx→3.D由題圖知,當(dāng)a<x<x3時(shí),f'(x)>0,f(x)單調(diào)遞增;當(dāng)x3<x<x5時(shí),f'(x)<0,f(x)單調(diào)遞減;當(dāng)x5<x<b時(shí),f'(x)>0,f(x)單調(diào)遞增,故f(x)的極小值點(diǎn)為x5,極大值點(diǎn)為x3,故A,B錯(cuò)誤,D正確.f'(x)的極小值點(diǎn)為x1,x4,共2個(gè),故C錯(cuò)誤.故選D.4.C∵f(x)=x32x2,x∈[1,3],∴f'(x)=3x24x=x(3x4),x∈[1,3].令f'(x)>0,可得1≤x<0或43<x≤3;令f'(x)<0,可得0<x<4∴函數(shù)f(x)在區(qū)間[1,0),43,3上單調(diào)遞增,在區(qū)間∴x=0是函數(shù)f(x)的極大值點(diǎn),x=43是函數(shù)f(x)的極小值點(diǎn),故C中說法錯(cuò)誤,D中說法正確∵f(0)=0,f(3)=272×9=9,f(1)=12×1=3,f43∴函數(shù)y=f(x)在區(qū)間[1,3]上的最大值為9,最小值為3,故A,B中說法正確.故選C.5.C易得f'(x)=3x2+6ax+b.因?yàn)閒(x)在x=1處取得極值0,所以f(-1)=0,f當(dāng)a=1,b=3時(shí),f'(x)=3x2+6x+3=3(x+1)2≥0,f(x)在R上單調(diào)遞增,無極值,不符合題意,舍去.當(dāng)a=2,b=9時(shí),f'(x)=3x2+12x+9,令f'(x)=0,得x=1或x=3,經(jīng)檢驗(yàn),x=1和x=3都為函數(shù)f(x)的極值點(diǎn),符合題意.所以a=2,b=9,故a+b=2+9=11.故選C.6.Af(x)<x2+12令g(x)=f(x)x2-12,因?yàn)閒'(x)<12,所以g'(x)<0,所以g(x)在R上單調(diào)遞減因?yàn)閒(1)=1,所以g(1)=f(1)12所以當(dāng)x>1時(shí),g(x)<0,當(dāng)x<1時(shí),g(x)>0.所以不等式f(x)<x2+127.Af'(x)=3x2+(m+4)x2,則f'(0)=2,∵對(duì)于任意t∈[1,2],f(x)在區(qū)間(t,3)上不單調(diào),∴f'(t∴m+4<2t∴實(shí)數(shù)m的取值范圍是-373,-8.A由題知,a=3(b=1e=lnee,c=ln33,易得f'(x)=1-lnxx2,x>0,令f'(x)=0,得x=e,則當(dāng)0<x<e時(shí),f'(x)>0,f(x)單調(diào)遞增,當(dāng)易知f(1)=0,當(dāng)x>1時(shí),f(x)>0,畫出y=f(x)的圖象如圖所示,∵1<e23<e<3,若t=lnxx有兩個(gè)解x1,x2,不妨設(shè)x1<x2,則1<x1<e<x2,t∈0,1e,lnx1=tx1,lnx2=tx2∵e23×3=e2,∴只要比較x1x2與e2令g(x)=lnx2(x-1)x+1(x>1),則g'(x)=(所以g(x)>g(1)=0,即在(1,+∞)上,lnx>2(若x=x2x1,則lnx2-lnx1x2-x所以當(dāng)x2=3時(shí),e>x1>e23,所以fe23<f(x綜上,b>c>a.故選A.9.ACD當(dāng)a=0時(shí),f(x)=exlnx,則f'(x)=xex-1x,x>0,設(shè)g(x)=xex1,則g'(x)=(x+1)ex,當(dāng)x>0時(shí),g'(x)>0,所以g(x)在(0,+∞)上單調(diào)遞增,又g(0)=1<0,當(dāng)x→+∞時(shí),g(x)→+∞,所以g(x)在(0,+∞)上存在唯一的零點(diǎn),設(shè)為m,則f(x)在(0,m)上單調(diào)遞減,在(m,+∞)上單調(diào)遞增,由em=1m,得m=ln1m=lnm,所以f(x)min=f(m)=emlnm=em+m>0,從而f(x)沒有零點(diǎn)當(dāng)a=2時(shí),f(x)=exln(x+2),則f'(x)=ex1x+2,x>2,因?yàn)閒(0)=1ln2且f'(0)=12,所以曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程為y=12x+1ln2,由C選項(xiàng)易知f'(x)在(2,+∞)上單調(diào)遞增,且f'(1)<0,f'(0)>0,所以存在唯一的x=x0,使f'(x0)=0,即f(x)只有一個(gè)極值點(diǎn)x0,且x0∈(1,0),故D正確.故選ACD.10.AB對(duì)于A,f'(x)=2x令f'(x)=0,得4lnx=2,解得x=e,當(dāng)0<x<e時(shí),f'(x)>0,f(x)單調(diào)遞增,當(dāng)x>e時(shí),f'(x)<0,f(x)單調(diào)遞減,∴f(x)在x=e處取得極大值,為f(e)=2lne(e對(duì)于B,∵f(x)在(0,e)上單調(diào)遞增,f(1)=2ln112=0,∴函數(shù)f(x)在(0,e)當(dāng)x≥e時(shí),f(x)=2lnxx2>0恒成立,即函數(shù)f(x)在[e,+∞)上沒有零點(diǎn),故f(x)有唯一零點(diǎn),故對(duì)于C,∵f(x)在(e,+∞)上單調(diào)遞減,2>π>e,∴f(2)<f(π對(duì)于D,由題意得k>f(x)+1x2=2ln設(shè)g(x)=2lnx+1x2,令g'(x)=0,解得x=1,當(dāng)0<x<1時(shí),g'(x)>0,g(x)單調(diào)遞增,當(dāng)x>1時(shí),g'(x)<0,g(x)單調(diào)遞減,∴當(dāng)x=1時(shí),函數(shù)g(x)取得極大值,也為最大值,最大值為g(1)=1,所以k>1,故D錯(cuò)誤.故選AB.11.AB易得f'(x)=1x,g'(x)=ex設(shè)直線l與曲線y=f(x)、y=g(x)分別相切于點(diǎn)P(x1,y1),Q(x2,y2),因?yàn)閒'(x1)=1x1,f(x1)=lnx1+1,所以直線l的方程為y(lnx1+1)=1x1·(xx1),即y=因?yàn)間'(x2)=ex2,g(x2)=ex21,所以直線l的方程為y(e故1x1=ex2,lnx1=-x2當(dāng)x2=0時(shí),直線l的方程為y=x;當(dāng)x2=1時(shí),直線l的方程為y=ex1.對(duì)于A,存在m=0滿足條件,故A正確.對(duì)于B,存在k=e滿足條件,故B正確.令h(x)=g(x)f(x)=exlnx2,x∈(0,+∞),則h'(x)=ex1x,x∈令φ(x)=h'(x)=ex1x,x∈(0,+∞),則φ'(x)=ex+1所以h'(x)在(0,+∞)上單調(diào)遞增,又h'23所以當(dāng)x∈23,+∞時(shí),h'(x)>0,h(x)單調(diào)遞增,故C,D12.ACDe2aeb=2(2ba)=4b2a,則e2a+2a=eb+4b,令f(b)=e2b+2beb4b=e2beb2b,則f'(b)=2e2beb2,當(dāng)b<0時(shí),f'(b)<0,f(b)在(∞,0)上單調(diào)遞減,f(b)>f(0)=0,此時(shí)e2b+2b>eb+4b,∴e2b+2b>e2a+2a,令g(x)=e2x+2x,則g(x)在R上單調(diào)遞增,∴g(b)>g(a)?a<b<0,A正確,B錯(cuò)誤.當(dāng)b>0時(shí),取b=1,則e2a+2a=eb+4b=e+4,此時(shí)g(1)=e2+2>e+4=g(a),又g(x)在R上單調(diào)遞增,∴a<1=b,∴0<a<b可能成立,C正確.取b=14,則e2a+2a=eb+4b=e又g(x)在R上單調(diào)遞增,∴a>14=b,∴a>b>0可能成立,D正確.故選13.答案3e解析易得f'(x)=2f'(1)+3ex,故f'(1)=2f'(1)+3e,即f'(1)=3e.14.答案(2,+∞)解析f'(x)=exm,∵f(x)的圖象上存在與直線y=12x垂直的切線,∴f'(x)=exm=2有解,即m=2+ex有解,又2+ex>2,∴m>2.故實(shí)數(shù)m的取值范圍是15.答案(∞,2)解析構(gòu)造函數(shù)g(x)=f(x)ex因?yàn)閒'(x)f(x)<0,所以g'(x)=f'(所以g(x)=f(x)e不等式f(x)>ex2可化為f(x)ex>1e2,因?yàn)間(2)=f(2)16.答案4e2解析當(dāng)x>0時(shí),f(x)=xlnx,f'(x)=11x當(dāng)x>1時(shí),f'(x)>0,f(x)單調(diào)遞增,當(dāng)0<x<1時(shí),f'(x)<0,f(x)單調(diào)遞減.故當(dāng)x=1時(shí),f(x)取得極小值,為f(1)=1.當(dāng)x≤0時(shí),f(x)=x+4e單調(diào)遞增,且f(x)≤4e.設(shè)f(x1)=f(x2)=t,則1≤t≤4e,由f(x1)=t得x1+4e=t,則x1=t4e,則x1f(x2)=t(t4e)=(t2e)24e2,因?yàn)?≤t≤4e,所以當(dāng)t=2e時(shí),x1f(x2)取得最小值,為4e2.17.解析(1)∵f(x)=ex+ax+b,∴f'(x)=ex+a.(2分)∵曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程為y=ab,∴f'(0)=1+a(2)證明:由(1)知f(x)=exx1,f'(x)=ex1,(7分)∴當(dāng)x<0時(shí),f'(x)<0,f(x)單調(diào)遞減,當(dāng)x>0時(shí),f'(x)>0,f(x)單調(diào)遞增,(9分)∴f(x)的最小值為f(0)=0,∴f(x)≥0.(10分)18.解析(1)設(shè)該容器的容積為Vm3,則V=πr2l+43πr3因?yàn)閂=80π3,所以l=因?yàn)閘≥2r,所以4320r2-r≥2r,所以y=2πrl×3+4πr2c=2πr×4320r2-(2)由(1)得y'=8π(c2)r160πr2因?yàn)閏>3,所以c2>0,故20c-2令y'=0,即r320c-2=0,則r=320則y'=8π(c-2)①若0<m<2,即c>92,則當(dāng)r∈(0,m)時(shí)當(dāng)r∈(m,2)時(shí),y'>0,所以r=m是極小值點(diǎn),也是最小值點(diǎn);(9分)②若m≥2,即3<c≤92,則當(dāng)r∈(0,2)時(shí),y'<0,函數(shù)單調(diào)遞減所以r=2是函數(shù)的最小值點(diǎn).(11分)綜上,若3<c≤92,則當(dāng)r=2時(shí)總建造費(fèi)用最少;若c>92,則當(dāng)r=320c19.解析(1)當(dāng)a=e時(shí),f(x)=exex,則f'(x)=exe,(2分)令f'(x)<0,得x<1;令f'(x)>0,得x>1,∴f(x)的單調(diào)遞減區(qū)間為(∞,1),單調(diào)遞增區(qū)間為(1,+∞).(4分)(2)∵f(x)有三個(gè)零點(diǎn),∴ex(a+e)x2+ax=0有三個(gè)不同的根,又x=0不是方程的根,∴exx=(a+e)xa令g(x)=exx,h(x)=(a+e)xa,則y=g(x)與y=h(x)的圖象有三個(gè)不同的交點(diǎn),(7易得g'(x)=ex(x-1)x2,令g'(x)>0,∴g(x)在(∞,0)和(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增,∴g(x)在x=1處取得極小值,為g(1)=e,易知當(dāng)x<0時(shí),g(x)<0,當(dāng)x>0時(shí),g(x)>0.(9分)又y=h(x)的圖象為恒過點(diǎn)(1,e)的直線,斜率為a+e,g'(1)=0,∴a+e>0,即a>e,∴a的取值范圍為(e,+∞).(12分)20.解析(1)易得f'(x)=xax+1-a=x2+①當(dāng)a≤0時(shí),f'(x)>0在(0,+∞)上恒成立,所以f(x)在(0,+∞)上單調(diào)遞增.(3分)②當(dāng)a>0時(shí),若x>a,則f'(x)>0,若0<x<a,則f'(x)<0,所以f(x)在(0,a)上單調(diào)遞減,在(a,+∞)上單調(diào)遞增.(5分)綜上,當(dāng)a≤0時(shí),f(x)在(0,+∞)上單調(diào)遞增;當(dāng)a>0時(shí),f(x)在(0,a)上單調(diào)遞減,在(a,+∞)上單調(diào)遞增.(6分)(2)證明:當(dāng)a=1時(shí),f(x)=12x2lnx+1,要證f(x)≤x(ex1)+12x22lnx,即證xexlnxx1≥0.(7令g(x)=xexlnxx1(x>0),則g'(x)=ex+xex1x令u(x)=ex1x,則u'(x)=ex+1x2>0,故u(x)在又u12=e2<0,u(1)=e1>0,所以由函數(shù)零點(diǎn)存在定理知,存在x0,使u(x0所以當(dāng)x∈(0,x0)時(shí),u(x)<0,即g'(x)<0,當(dāng)x∈(x0,+∞)時(shí),u(x)>0,即g'(x)>0,所以g(x)在(0,x0)上單調(diào)遞減,在(x0,+∞)上單調(diào)遞增,所以g(x)min=g(x0)=x0ex0lnx0x由u(x0)=0,得ex0-1x0=0,即x0ex0=1,所以ln(x0所以g(x)min=g(x0)=x0ex0lnx0x01=0,即xexlnxx1≥0所以當(dāng)a=0時(shí),f(x)≤x(ex1)+12x22lnx.(12分21.解析(1)當(dāng)a=1時(shí),f(x)=lnx+x2x,則f'(x)=1+2x當(dāng)x>0時(shí),f'(x)>0恒成立,(2分)所以f(x)的單調(diào)遞增區(qū)間為(0,+∞),無單調(diào)遞減區(qū)間.(4分)(2)由f(x)≥12x3+1得a≥1令g(x)=12x3因?yàn)榇嬖趚∈[1,+∞),使得f(x)≥12x3+1成立,所以a≥g(x)min.(5分g'(x)=12x3令h(x)=12x33x+2lnx,x≥1,則h'(x)=3x3令n(x)=3x32x+4(x≥1),則n'(x)=9x22(x≥1),因?yàn)楫?dāng)x≥1時(shí),n'(x)>0恒成立,所以n(x)單調(diào)遞增,所以n(x)≥n(1)=5>0,所以h'(x)>0恒成立,所以h(x)單調(diào)遞增.因?yàn)閔32=-4516所以?m∈32,2,使得h(m)=0,且當(dāng)x∈[1,m)時(shí),h(x)<0,此時(shí)g'(x)<0,g(x)單調(diào)遞減,當(dāng)x∈(m,+∞)時(shí),h(x)>0,此時(shí)g'(x)>0,g(x)單調(diào)遞增,且則g(x)min=g(m)=12m因?yàn)間'(m)=0,即12m33m+2lnm=0,所以lnm=1所以g(x)min=g(m)=3m3+2令φ(x)=3x3+2x-24令t(x)=3x32x+4,32<x<2,易得t(x)在32所以t(x)>t32=898>0,所以當(dāng)x∈3所以φ32<φ(x)