高中三年級(jí)文科數(shù)學(xué)模擬試卷第29章

第三章導(dǎo)數(shù)及其應(yīng)用3.1導(dǎo)數(shù)及其運(yùn)算38導(dǎo)數(shù)的概念與幾何意義1.(2019江西上饒重點(diǎn)中學(xué)一模,文13,導(dǎo)數(shù)的概念與幾何意義,填空題)若曲線:y=ax+1(a>0且a≠1)在點(diǎn)(0,2)處的切線與直線x+2y+1=0垂直,則a=.

解析:y=ax+1的導(dǎo)數(shù)為y'=axlna,即有曲線在點(diǎn)(0,2)處的切線斜率為k=lna,由于切線與直線x+2y+1=0垂直,則lna·-12=-1,解得a=e答案:e22.(2019山西太原一模,文14,導(dǎo)數(shù)的概念與幾何意義,填空題)函數(shù)f(x)=xex在點(diǎn)(1,f(1))處的切線方程是.

解析:函數(shù)f(x)=xex的導(dǎo)數(shù)為f'(x)=ex+xex,在點(diǎn)(1,f(1))處的切線斜率為k=2e,切點(diǎn)為(1,e),則有在點(diǎn)(1,f(1))處的切線方程為y-e=2e(x-1),即為y=2ex-e.答案:y=2ex-e3.(2019黑龍江大慶二模,文9,導(dǎo)數(shù)的概念與幾何意義,選擇題)已知函數(shù)f(x)=13x3-2x2+3x+13,則與f(x)圖象相切的斜率最小的切線方程為(A.2x-y-3=0 B.x+y-3=0C.x-y-3=0 D.2x+y-3=0解析:∵f(x)=13x3-2x2+3x+1∴f'(x)=x2-4x+3=(x-2)2-1.∵當(dāng)x=2時(shí),f'(x)取到最小值為-1,∴f(x)=13x3-2x2+3x+13的切線中,斜率最小的切線方程的斜率為-∵f(2)=1,∴切點(diǎn)坐標(biāo)為(2,1).∴切線方程為y-1=-(x-2),即x+y-3=0.答案:B10.(2019江西上饒二模,文10,導(dǎo)數(shù)的概念與幾何意義,選擇題)已知函數(shù)f(x)=2ex,函數(shù)g(x)=k(x+1),若函數(shù)f(x)圖象恒在函數(shù)g(x)圖象的上方(沒(méi)有交點(diǎn)),則實(shí)數(shù)的取值范圍是()A.k>2 B.k≥2 C.0≤k≤2 D.0≤k<2解析:若函數(shù)f(x)圖象恒在函數(shù)g(x)圖象的上方(沒(méi)有交點(diǎn)),即f(x)-g(x)>0恒成立,即2ex-k(x+1)>0,即2ex>k(x+1),若k=0,滿足條件,若k<0,則不滿足條件.則當(dāng)k>0時(shí),g(x)=k(x+1)過(guò)定點(diǎn)(-1,0),函數(shù)f(x)的導(dǎo)數(shù)為f'(x)=2ex,設(shè)切點(diǎn)為(a,b),則對(duì)應(yīng)的切線斜率k=f'(a)=2ea,則對(duì)應(yīng)的切線方程為y-2ea=2ea(x-a),又直線過(guò)點(diǎn)(-1,0),故-2ea=2ea(-1-a),解得a=0,此時(shí)切線斜率k=f'(0)=2,即此時(shí)k=2,則解得0<k<2.綜上可得0≤k<2.答案:D11.(2019江西紅色六校一模,文11,導(dǎo)數(shù)的概念與幾何意義,填空題)曲線y=-5ex-3x在點(diǎn)(0,-5)處的切線方程為.

解析:∵y=-5ex-3x,∴y'=-5ex-3.∴曲線y=-5ex-3x在點(diǎn)P(0,-5)處的切線的斜率為k=-8.∴曲線y=-5ex-3x在點(diǎn)(0,-5)處的切線的方程為8x+y+5=0.答案:8x+y+5=014.(2019廣西南寧一模,文14,導(dǎo)數(shù)的概念與幾何意義,填空題)若曲線y=x4的一條切線l與直線x+4y-8=0垂直,則l的方程為.

解析:與直線x+4y-8=0垂直的直線l為:4x-y+m=0,即y=x4在某一點(diǎn)的導(dǎo)數(shù)為4,而y'=4x3,所以y=x4在(1,1)處導(dǎo)數(shù)為4,故方程為4x-y-3=0.答案:4x-y-3=012.(2019江西上饒三模,文12,導(dǎo)數(shù)的概念與幾何意義,選擇題)定義:如果函數(shù)f(x)在[a,b]上存在x1,x2(a<x1<x2<b)滿足f'(x1)=f(b)-f(a)b-a,f'(x2)=f(b)-f(a)b-a,則稱函數(shù)f(x)是[a,b]上的“雙中值函數(shù)”.已知函數(shù)A.13,12 B.32,3 解析:由題意可知,f'(x)=3x2-2x,在區(qū)間[0,a]上存在x1,x2(a<x1<x2<b)滿足f'(x1)=f'(x2)=f(a)-f∵f(x)=x3-x2+a,∴f'(x)=3x2-2x.∴方程3x2-2x=a2-a在區(qū)間(0,a)上有兩個(gè)不相等的解.令g(x)=3x2-2x-a2+a(0<x<a),則Δ解得12<a<1∴實(shí)數(shù)a的取值范圍是12答案:C14.(2019江西南昌模擬,文14,導(dǎo)數(shù)的概念與幾何意義,填空題)函數(shù)f(x)=ax+3·ex的圖象存在與直線2x-4y+1=0垂直的切線,則實(shí)數(shù)a的取值范圍是.

解析:由f(x)=ax+3·ex,得f'(x)=a+3·ex,∵函數(shù)f(x)=ax+3·ex的圖象存在與直線2x-4y+1=0垂直的切線,且直線2x-4y+1=0的斜率為12∴方程a+3·ex=-2有實(shí)數(shù)解,即a=3·ex-2有解.∵3·ex<0,∴a=3·ex-2<-2.答案:(-∞,-2)13.(2019江西重點(diǎn)中學(xué)協(xié)作體二模,文13,導(dǎo)數(shù)的概念與幾何意義,填空題)曲線y=x(2lnx-1)在點(diǎn)(1,-1)處的切線方程為.

解析:對(duì)y=x(2lnx-1)求導(dǎo),得y'=2lnx+1,當(dāng)x=1時(shí),y'=1,所以曲線y=x(2lnx-1)在點(diǎn)(1,-1)處的切線斜率為-1.又切點(diǎn)為(1,-1),所以切線方程為y+1=x-1,即x-y-2=0.答案:x-y-2=04.(2019江西紅色六校二模,文4,導(dǎo)數(shù)的概念與幾何意義,選擇題)若冪函數(shù)f(x)=mxα的圖象經(jīng)過(guò)點(diǎn)A14,12,則它在點(diǎn)A.2x-y=0 B.2x+y=0C.4x-4y+1=0 D.4x+4y+1=0解析:因?yàn)閒(x)=mxα為冪函數(shù),故m=1,又圖象經(jīng)過(guò)點(diǎn)A14,12則α=12即有f(x)=x12,則f'(x)=則f(x)在點(diǎn)A處的切線斜率為12·所以切線方程為y-12=x-14,即4x-4y+1=答案:C14.(2019黑龍江哈爾濱九中三模,文14,導(dǎo)數(shù)的概念與幾何意義,填空題)已知f(x)=ex-x,過(guò)原點(diǎn)與f(x)相切的直線方程為.

解析:設(shè)切點(diǎn)坐標(biāo)為(a,ea-a),又切線過(guò)(0,0),得到切線的斜率k=ea又f'(x)=ex-1,把x=a代入得:斜率k=f'(a)=ea-1,則ea-1=ea由于ea>0,則得到a=1,即切點(diǎn)坐標(biāo)為(1,e-1),所以切線方程為y-e+1=(e-1)(x-1),即y=(e-1)x.答案:y=(e-1)x15.(2019山西太原外國(guó)語(yǔ)學(xué)校4月模擬,文15,導(dǎo)數(shù)的概念與幾何意義,填空題)已知函數(shù)f(x)=x3對(duì)應(yīng)的曲線在點(diǎn)(ak,f(ak))(k∈N*)處的切線與x軸的交點(diǎn)為(ak+1,0),若a1=1,則f(3a1解析:由f'(x)=3x2得曲線的切線的斜率k=3ak故切線方程為y-ak3=3ak2令y=0得ak+1=23ak?a故數(shù)列{an}是首項(xiàng)a1=1,公比q=23的等比數(shù)列又f(3a1)+f(3a2)+…=a1+a2+…+a10=a1(1-q所以f(3a答案:310.(2019甘肅蘭州一模,文10,導(dǎo)數(shù)的概念與幾何意義,選擇題)在直角坐標(biāo)系xOy中,設(shè)P是曲線C:xy=1(x>0)上任意一點(diǎn),l是曲線C在點(diǎn)P處的切線,且l交坐標(biāo)軸于A,B兩點(diǎn),則下列結(jié)論正確的是()A.△OAB的面積為定值2B.△OAB的面積有最小值為3C.△OAB的面積有最大值為4D.△OAB的面積的取值范圍是[3,4]解析:由題意,y=1x(x>0),則y'=-1設(shè)Pa,1a,則曲線C在點(diǎn)P處的切線方程為y-1a=-x=0可得y=2a,y=0可得x=2a故△OAB的面積為12×2a×2a=答案:A39導(dǎo)數(shù)的運(yùn)算1.(2019山西太原一模,文12,導(dǎo)數(shù)的運(yùn)算,選擇題)已知函數(shù)f(x)=lnx+tanα,α∈0,π2的導(dǎo)函數(shù)為f'(x),若使得f'(x0)=f(x0)成立的x0<1,則實(shí)數(shù)αA.π4,π2 B.0,π3 解析:∵f'(x)=1x,f'(x0)=1x0,f'(x0)=f(x∴1x0=lnx0+tanα.∴tanα=1x0-ln又0<x0<1,可得1x0-lnx0>1,即tanα>∴α∈π4答案:A3.2導(dǎo)數(shù)與函數(shù)的單調(diào)性、極值、最值40導(dǎo)數(shù)與函數(shù)的單調(diào)性1.(2019吉林省實(shí)驗(yàn)中學(xué)二模,文9,導(dǎo)數(shù)與函數(shù)的單調(diào)性,選擇題)已知函數(shù)f(x)的導(dǎo)函數(shù)圖象如圖所示,若△ABC為銳角三角形,則一定成立的是()A.f(cosA)<f(cosB)B.f(sinA)<f(cosB)C.f(sinA)>f(sinB)D.f(sinA)>f(cosB)解析:根據(jù)導(dǎo)數(shù)函數(shù)圖象可判斷:f(x)在(0,1)單調(diào)遞增,(1,+∞)單調(diào)遞減,∵△ABC為銳角三角形,∴A+B>π2,0<π2-B<A<∴0<sinπ2-B<sinA<1,0<cosB<sinA<1,f(sinA)即f(sinA)>f(cosB).答案:D2.(2019江西上饒重點(diǎn)中學(xué)一模,文10,導(dǎo)數(shù)與函數(shù)的單調(diào)性,選擇題)定義在R上的函數(shù)y=f(x),滿足f(2-x)=f(x),(x-1)f'(x)<0,若f(3a+1)<f(3),則實(shí)數(shù)a的取值范圍是()A.-∞B.2C.-2D.-解析:當(dāng)x>1時(shí),f'(x)<0,此時(shí)函數(shù)單調(diào)遞減,當(dāng)x<1時(shí),f'(x)>0,此時(shí)函數(shù)單調(diào)遞增,∵f(2-x)=f(x),∴函數(shù)關(guān)于x=1對(duì)稱.若f(3a+1)<f(3),則滿足①3a+1≥1,3②3a+1≤1,3綜上,實(shí)數(shù)a的取值范圍為-∞答案:D3.(2019廣西玉林、貴港4月模擬,文11,導(dǎo)數(shù)與函數(shù)的單調(diào)性,選擇題)若函數(shù)f(x)=ex+4x-kx在區(qū)間12,+∞上是增函數(shù),則實(shí)數(shù)A.2+e B.2+e C.4+e D.4ln2+e解析:函數(shù)f(x)=ex+4x-kx的導(dǎo)數(shù)為f'(x)=ex+4xln4-k,由題意可得f'(x)≥0在區(qū)間12,即有k≤ex+4xln4在區(qū)間12,令g(x)=ex+4xln4,則g(x)為12,即有g(shù)(x)>e+2ln4=4ln2+e.則k≤4ln2+e.故k的最大值為4ln2+e.答案:D4.(2019廣西桂林、防城港聯(lián)合調(diào)研,文21,導(dǎo)數(shù)與函數(shù)的單調(diào)性,解答題)設(shè)函數(shù)f(x)=ex-3e-x-ax.(1)當(dāng)a=4時(shí),求函數(shù)f(x)的單調(diào)遞增區(qū)間;(2)若函數(shù)f(x)在[-2,2]上為單調(diào)函數(shù),求實(shí)數(shù)a的取值范圍.解:∵f(x)=ex-3e-x-ax,∴f'(x)=ex+3e-x-a.(1)當(dāng)a=4時(shí),f'(x)=e2由f'(x)≥0,解得x≥ln3或x≤0,由f'(x)≤0,解得0≤x≤ln3,∴函數(shù)f(x)在(-∞,0]和[ln3,+∞)上遞增,在[0,ln3]上遞減.(2)∵函數(shù)f(x)在[-2,2]上是單調(diào)函數(shù),∴f'(x)在[-2,2]上不變號(hào),即a≤ex+3exmin,x∈[-2,2],求得a≤23,或a≥ex+3exmax=∴實(shí)數(shù)a的取值范圍是a≤23或a≥e-2+3e2.5.(2019廣西柳州一模,文20,導(dǎo)數(shù)與函數(shù)的單調(diào)性,解答題)已知函數(shù)f(x)=ax2-blnx在點(diǎn)(1,f(1))處的切線為y=2.(1)求實(shí)數(shù)a,b的值;(2)是否存在實(shí)數(shù)m,當(dāng)x∈(0,1]時(shí),函數(shù)g(x)=f(x)-2x2+m(x-1)的最小值為0?若存在,求出m的取值范圍;若不存在,說(shuō)明理由.解:(1)f'(x)=2ax-bx(x>依題意可得f(1)=2ax=2(2)∵g(x)=f(x)-2x2+m(x-1)=m(x-1)-4lnx,x∈(0,1],∴g'(x)=m-4x①當(dāng)m≤0時(shí),g'(x)<0,∴g(x)在(0,1]上單調(diào)遞減.∴g(x)min=g(1)=0.②當(dāng)0<m≤4時(shí),g'(x)=mx-∴g(x)在(0,1]上單調(diào)遞減.∴g(x)min=g(1)=0.③當(dāng)m>4時(shí),g'(x)<0在0,4m上恒成立,g'(x)>0在∴g(x)在0,4m上單調(diào)遞減,在∴g4m<g(1)=0.∴g(x)min≠0綜上所述,存在m滿足題意,其范圍為(-∞,4].9.(2019山西太原二模,文9,導(dǎo)數(shù)與函數(shù)的單調(diào)性,選擇題)已知函數(shù)f(x)的導(dǎo)函數(shù)在(a,b)上的圖象關(guān)于直線x=a+b2對(duì)稱,則函數(shù)y=f(x)在[a,b]上的圖象可能是(解析:∵函數(shù)y=f(x)的導(dǎo)函數(shù)在區(qū)間(a,b)上的圖象關(guān)于直線x=a+b∴導(dǎo)函數(shù)的圖象無(wú)增減性,或在直線x=a+b對(duì)于A,由圖知,在a處切線斜率最小,在b處切線斜率最大,∴導(dǎo)函數(shù)圖象不關(guān)于直線x=a+b2對(duì)稱對(duì)于B,由圖知,在a處切線斜率最大,在b處切線斜率最小,∴導(dǎo)函數(shù)圖象不關(guān)于直線x=a+b2對(duì)稱對(duì)于C,由圖知,在a處切線的斜率最小,在b處切線的斜率最大,其導(dǎo)函數(shù)圖象不關(guān)于直線x=a+b2對(duì)稱對(duì)于D,由圖知,原函數(shù)是中心對(duì)稱函數(shù),對(duì)稱中心在直線x=a+b2與原函數(shù)圖象的交點(diǎn)處,∴導(dǎo)函數(shù)圖象關(guān)于直線x=a+答案:D11.(2019山西太原二模,文11,導(dǎo)數(shù)與函數(shù)的單調(diào)性,選擇題)下列不等式正確的是()A.sin1<2sin12<3sinB.3sin13<2sin12C.sin1<3sin13<2sinD.2sin12<sin1<3sin解析:設(shè)y=xsinx,x∈0,π2,則y'=sinx+xcos∵x∈0,π2,y'>∴函數(shù)y=xsinx,x∈0,π∴sin1<2sin12<3sin1答案:A21.(2019山西太原二模,文21,導(dǎo)數(shù)與函數(shù)的單調(diào)性,解答題)已知函數(shù)f(x)=lnx-ax(a∈R)有兩個(gè)不相等的零點(diǎn)x1,x2(x1<x2).(1)求a的取值范圍;(2)判斷2x1+x解:(1)由題意得x1,x2