適用于新高考新教材廣西專版2024屆高考數(shù)學(xué)一輪總復(fù)習(xí)單元質(zhì)檢卷三一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用

單元質(zhì)檢卷三一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用(時(shí)間:120分鐘滿分:150分)一、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)=(2x-a)ex,且f'(1)=3e,則實(shí)數(shù)a的值為()A.-3 B.3 C.-1 D.12.函數(shù)f(x)=x2-2lnx在區(qū)間[1,2]上的最大值是()A.4-2ln2 B.1 C.4+2ln2 D.e2-23.(2022山東日照二模)曲線y=f(x)=lnx-2x在x=1處的切線的傾斜角為α,則cos2α的值為(A.45 B.-C.35 D.-4.冪函數(shù)f(x)的圖象過點(diǎn)22,2,則函數(shù)g(x)=exf(x)A.(0,2)B.(-∞,-2)∪(0,+∞) C.(-2,0)D.(-∞,-2)和(0,+∞)5.曲線y=f(x)=ln(2x-1)上的點(diǎn)到直線2x-y+3=0的最短距離是()A.1 B.2 C.5 D.36.已知函數(shù)f(x)=x+acosx,對(duì)?x1,x2∈R(x1≠x2),f(x1)-f(x2)x1A.[1-2,1+2] B.[1-2,1]C.[-1,1] D.[-1,1-2]7.(2022福建福州模擬)已知a=esin1+1esin1,b=etan2+1etan2,c=ecos3+1eA.a>b>c B.b>c>aC.a>c>b D.c>a>b8.(2022山東泰安模擬預(yù)測(cè))定義在(1,+∞)上的函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),且(x-1)·f'(x)-f(x)>x2-2x對(duì)任意x∈(1,+∞)恒成立.若f(2)=3,則不等式f(x)>x2-x+1的解集為()A.(1,2) B.(2,+∞)C.(1,3) D.(3,+∞)二、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.如果定義域?yàn)镽的函數(shù)f(x)的導(dǎo)函數(shù)的圖象如圖所示,則下列說法正確的是()A.f(x)在區(qū)間-3,-12內(nèi)單調(diào)遞增B.f(x)有且僅有1個(gè)極小值點(diǎn)C.f(x)在區(qū)間(4,5)內(nèi)單調(diào)遞增D.f(x)的極大值為f(2)10.已知函數(shù)f(x)=lnx-ax的圖象在x=1處的切線方程為x+y+b=0,則()A.a=2 B.b=1C.f(x)的極小值為-ln2-1 D.f(x)的極大值為-ln2-111.已知函數(shù)f(x)=lnxx-x,則下列結(jié)論錯(cuò)誤的是(A.f(x)的單調(diào)遞減區(qū)間為(0,1)B.f(x)的極小值點(diǎn)為1C.f(x)的極大值為-1D.f(x)的最小值為-112.已知函數(shù)f(x)=x2-ex+a有兩個(gè)極值點(diǎn)x1與x2,且x1<x2,則下列結(jié)論正確的是()A.a<ln2-1B.0<x1<1C.-1<f(x1)<0D.0<x1ex2三、填空題:本題共4小題,每小題5分,共20分.13.(2022河北張家口三模)函數(shù)f(x)=ex+1ln(1-x)在點(diǎn)(0,f(0))處的切線方程為.

14.若函數(shù)f(x)=-x2+ax在區(qū)間(-1,0)內(nèi)恰有一個(gè)極值點(diǎn),則實(shí)數(shù)a的取值范圍是.

15.(2022廣東一模)已知直線y=t分別與函數(shù)f(x)=2x+1和g(x)=2lnx+x的圖象交于點(diǎn)A,B,則|AB|的最小值為.

16.已知函數(shù)f(x)=ex-ex+a與g(x)=lnx+1x的圖象上存在關(guān)于x軸對(duì)稱的點(diǎn),則實(shí)數(shù)a的取值范圍是.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)=3-(1)若a=0,求曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程;(2)若函數(shù)f(x)在x=-1處取得極值,求f(x)的單調(diào)區(qū)間以及最大值和最小值.18.(12分)已知函數(shù)f(x)=ln(2x)-ax2.(1)若f(x)在(1,+∞)內(nèi)不單調(diào),求實(shí)數(shù)a的取值范圍;(2)若a=2,求f(x)在區(qū)間12e,e219.(12分)(2022全國甲,文20)已知函數(shù)f(x)=x3-x,g(x)=x2+a,曲線y=f(x)在點(diǎn)(x1,f(x1))處的切線也是曲線y=g(x)的切線.(1)若x1=-1,求a;(2)求a的取值范圍.20.(12分)已知a>0且a≠1,函數(shù)f(x)=xaax((1)當(dāng)a=2時(shí),求f(x)的單調(diào)區(qū)間;(2)若曲線y=f(x)與直線y=1有且僅有兩個(gè)交點(diǎn),求a的取值范圍.21.(12分)(2022廣東廣州三模)已知函數(shù)f(x)=ex-ax(a∈R).(1)若f(x)在(0,+∞)上是增函數(shù),求a的取值范圍;(2)若x1,x2是函數(shù)f(x)的兩個(gè)不同的零點(diǎn),求證:1<x1+x2<2lna-ln2.22.(12分)已知函數(shù)f(x)=x3-kx+k2.(1)討論f(x)的單調(diào)性;(2)若f(x)有三個(gè)零點(diǎn),求k的取值范圍.

單元質(zhì)檢卷三一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用1.D解析:因?yàn)閒'(x)=(2+2x-a)ex,且f'(1)=3e,所以f'(1)=(4-a)e=3e,解得a=1,故選D.2.A解析:因?yàn)閒'(x)=2x-2x=2(x+1)(x-1)x≥0在區(qū)間[1,2]上恒成立,所以函數(shù)f(x)在區(qū)間[1,2]上單調(diào)遞增,所以,當(dāng)x=2時(shí),f(3.B解析:根據(jù)已知條件,f'(x)=1x+2x2,因?yàn)榍€y=lnx-2x在x=1處的切線的傾斜角為α,所以tanα=f'(1)=1+2=3.所以cos24.D解析:設(shè)f(x)=xa,則22a=2,解得a=-2,所以g(x)=exx-2=x2ex,函數(shù)的定義域是{x∈R|x≠0},g'(x)=(x2+2x)ex,令g'(x)>0,得x<-2或x>0,所以g(x)的單調(diào)遞增區(qū)間是(-∞,-2)和(0,5.C解析:因?yàn)橹本€2x-y+3=0的斜率為2,所以令f'(x)=22x-1=2,解得x=1.把x=1代入曲線方程得f(1)=ln(2-1)=0,即曲線f(x)過點(diǎn)(1,0)的切線斜率為2,則點(diǎn)(1,0)到直線2x-y+3=0的距離d=|2-0+3|22+(-1)2=5,即曲線6.B解析:設(shè)x1>x2,由f(x1)-f(x2)x1-x2>a2-a可得f(x1)-f(x2)>(a2-a)(x1-x2),即f(x1)-(a2-a)x1>f(x2)-(a2-a)x2.構(gòu)造函數(shù)g(x)=f(x)-(a2-a)x=acosx+(1-a2+a)x,則函數(shù)g(x)在R上單調(diào)遞增,g'(x)=-asinx+(1-a2+a)≥0對(duì)任意的x∈R恒成立.令t=sinx,則t∈[-1,1],所以-at+(1-a2+a)≥0在t∈[-1,1]時(shí)恒成立,所以7.B解析:設(shè)函數(shù)f(x)=ex+1ex,則f(x)為偶函數(shù),且當(dāng)x≥0時(shí),f'(x)=ex-1所以f(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增.因?yàn)閟in1<32,tan2<-1<cos3<-3所以-tan2>1>-cos3>32>sin1>又a=f(sin1),b=f(tan2)=f(-tan2),c=f(cos3)=f(-cos3),所以b>c>a.8.B解析:由(x-1)f'(x)-f(x)>x2-2x,得(x-1)·f'(x)-f(x)+1>(x-1)2,即(x-1)即f(x)-1x-1-x'>0對(duì)x∈(1,令g(x)=f(x)-1x-1-x,則g(x)因?yàn)閒(2)=3,所以g(2)=0.所以f(x)>x2-x+1,即f(x即g(x)>g(2),不等式的解集為x>2.9.CD解析:由f'(x)的圖象知,在區(qū)間(-∞,-2)和(2,4)內(nèi)f'(x)<0,則f(x)單調(diào)遞減;在區(qū)間(-2,2)和(4,+∞)上f'(x)>0,則f(x)單調(diào)遞增,所以f(x)的極大值為f(2),極小值為f(-2)和f(4).所以f(x)在區(qū)間-3,-12內(nèi)不單調(diào),故A項(xiàng)錯(cuò)誤;f(x)有2個(gè)極小值點(diǎn),故B項(xiàng)錯(cuò)誤;f(x)在區(qū)間(4,5)內(nèi)單調(diào)遞增,故C項(xiàng)正確;f(x)的極大值為f(2),故D項(xiàng)正確.故選CD.10.ABD解析:因?yàn)閒(x)=lnx-ax,所以f'(x)=1x-a.又因?yàn)楹瘮?shù)f(x)的圖象在x=1處的切線方程為x+y+b=0,所以f(1)=-a=-b-1,f'(1)=1-a=-1,解得a=2,b=1.所以A,B正確f(x)的定義域?yàn)?0,+∞),f'(x)=1x-2=1-2xx.令f'(x)=0,得x=12.令f'(x)>0,得0<x<12,則f(x)在區(qū)間0,12內(nèi)單調(diào)遞增;令f'(x)<0,得x>12,則f(x)在區(qū)間12,+∞上單調(diào)遞減,所以f(x)在x=12處取得極大值,且f12=ln12-1=-ln211.ABD解析:f(x)的定義域?yàn)?0,+∞),f'(x)=1-lnxx2-1=1-lnx-x2x2.令φ(x)=1-lnx-x2,則φ'(x)=-1x-2x<0,所以φ(x)=1-lnx-x2在(0,+∞)上單調(diào)遞減.因?yàn)棣?1)=0,所以當(dāng)0<x<1時(shí),φ(x)>0;當(dāng)x>1,φ(x)<0.所以f(x)的單調(diào)遞增區(qū)間為(0,1),單調(diào)遞減區(qū)間為(1,+∞),故f(x)12.ABC解析:f(x)=x2-ex+a,則該函數(shù)的定義域?yàn)镽,f'(x)=2x-ex+a.由已知可得2所以函數(shù)f'(x)有兩個(gè)正零點(diǎn).由2x=ex+a,其中x>0,可得x+a=ln2+lnx,可得a=lnx-x+ln2.構(gòu)造函數(shù)g(x)=lnx-x+ln2,x>0,則g'(x)=1x-1=1令g'(x)=0,得x=1.當(dāng)0<x<1時(shí),g'(x)>0,此時(shí)函數(shù)g(x)單調(diào)遞增;當(dāng)x>1時(shí),g'(x)<0,此時(shí)函數(shù)g(x)單調(diào)遞減,所以,函數(shù)g(x)的極大值為g(1)=ln2-1,作出函數(shù)g(x)的圖象,如圖所示.對(duì)于A選項(xiàng),當(dāng)a<ln2-1時(shí),直線y=a與函數(shù)g(x)的圖象有兩個(gè)交點(diǎn),即函數(shù)f(x)有兩個(gè)極值點(diǎn),A正確;對(duì)于B選項(xiàng),x1,x2為直線y=a與函數(shù)g(x)圖象兩個(gè)交點(diǎn)的橫坐標(biāo),因?yàn)楹瘮?shù)g(x)在區(qū)間(0,1)內(nèi)單調(diào)遞增,在(1,+∞)上單調(diào)遞減,且x1<x2,所以0<x1<1,x2>1,B正確;對(duì)于C選項(xiàng),0<x1<1,則f(x1)=x12-ex1+a=x12-2x1=(x對(duì)于D選項(xiàng),由2x所以x1ex2=x2又0<x1<1,x2>1,所以x2ex1>1,從而x1ex2故選ABC.13.ex+y=0解析:∵f(0)=0,∴切點(diǎn)為(0,0).∵f'(x)=ex+1ln(1-x)+1x-1·e∴f'(0)=-e,即切線的斜率為-e.故該切線方程為y=-ex,即ex+y=0.14.(-2,0)解析:二次函數(shù)f(x)=-x2+ax圖象的對(duì)稱軸為直線x=-a-2,即x=a2.因?yàn)楹瘮?shù)f(x)=-x2+ax在區(qū)間(-1,0)內(nèi)恰有一個(gè)極值點(diǎn),所以-1<a2<0,可得-2<a<0.故實(shí)數(shù)a的取值范圍是15.32-ln2解析:如圖,作出函數(shù)y=g(x)=2lnx+x的圖象,作直線y=2x+1,平移到與函數(shù)y=g(x)的圖象相切由圖象知直線y=t與這兩條平行直線的交點(diǎn)的橫坐標(biāo)之差為所求最小值.由g(x)=x+2lnx,得g'(x)=1+2x令g'(x)=1+2x=2,得x=2,此時(shí)g(2)=2+即切點(diǎn)坐標(biāo)為(2,2+2ln2).由2x+1=2+2ln2,得x=12+ln2故|AB|min=2-12+ln2=32-ln2.16.(-∞,-1]解析:由題意知,方程ex-ex+a=-lnx-1x在(0,+∞)上有解,即a=ex-ex-lnx-1x在(0,+∞)上有解.令h(x)=ex-ex-lnx-1x,x>0,則h'(x)=e-ex-1x+1x2=e-ex+1-xx2,顯然h'(1)=0,且當(dāng)0<x<1時(shí),h'(x)>0,函數(shù)h(x)單調(diào)遞增;當(dāng)x>1時(shí),h'(x)<0,函數(shù)h(x)單調(diào)遞減,因此h(x)在x=1處取得極大值亦即最大值h(1)=-1,所以h(x)的值域?yàn)?-∞,17.解(1)當(dāng)a=0時(shí),f(x)=3-2xx2,則f'(x)=2(x-3)此時(shí)曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程為y-1=-4(x-1),即4x+y-5=0.(2)因?yàn)閒(x)=3-所以f'(x)=-2由題意可得f'(-1)=2(4-a)故f(x)=3-2xx2+4,f'(xx(-∞,-1)-1(-1,4)4(4,+∞)f'(x)+0-0+f(x)單調(diào)遞增極大值單調(diào)遞減極小值單調(diào)遞增所以,函數(shù)f(x)的單調(diào)遞增區(qū)間為(-∞,-1),(4,+∞),單調(diào)遞減區(qū)間為(-1,4).當(dāng)x<32時(shí),f(x)>0;當(dāng)x>32時(shí),f(x)<0.所以,f(x)max=f(-1)=1,f(x)min=f(4)=-18.解(1)f'(x)=1-2ax2x,因?yàn)閒(x)在所以關(guān)于x的方程1-2ax2=0在(1,+∞)內(nèi)有解,所以1-2a>0,a>(2)因?yàn)閍=2,所以f'(x)=(1令f'(x)>0,得12e≤x<12;令f'(x)<0,得12<x所以f(x)在區(qū)間12e,12上單調(diào)遞增,在區(qū)間12,e2上單調(diào)遞減,所以f(x)max=f12因?yàn)閒12e=-1-12e2,fe2=1-e22,且f12e-fe2=e22-2-所以f(x)在12e,e2上的值域?yàn)?-e22,-19.解(1)∵f'(x)=3x2-1,∴f'(-1)=2.當(dāng)x1=-1時(shí),f(-1)=0,故曲線y=f(x)在點(diǎn)(-1,0)處的切線方程為y=2x+2.又直線y=2x+2與曲線y=g(x)相切,將y=2x+2代入g(x)=x2+a,得x2-2x+a-2=0.由Δ=4-4(a-2)=0,得a=3.(2)∵f'(x)=3x2-1,∴f'(x1)=3x12-1,則曲線y=f(x)在點(diǎn)x1,fx1處的切線為y-x1由g(x)=x2+a,得g'(x)=2x.設(shè)曲線y=g(x)在點(diǎn)(x2,g(x2))處的切線為y-(x22+a)=2x2(x-x整理得y=2x2x-x22由題可得3∴a=x22-2令h(x)=9x4-8x3-6x2+1,則h'(x)=36x3-24x2-12x=12x(x-1)(3x+1).當(dāng)x<-13或0<x<1時(shí),h'x<0,函數(shù)hx單調(diào)遞減當(dāng)-13<x<0或x>1時(shí),h'x>0,函數(shù)hx單調(diào)遞增又h-13=2027,h(0)=∴hxmin=h(1)=-∴a≥-44=-1,即a的取值范圍為[-1,+∞20.解(1)當(dāng)a=2時(shí),f(x)=x22x(f'(x)=2x當(dāng)x∈0,2ln2時(shí),f'(x)>0,f(x)單調(diào)遞增;當(dāng)x∈2ln2,+∞時(shí),f'(x)<所以函數(shù)f(x)的單調(diào)遞增區(qū)間為0,2ln2,單調(diào)遞減區(qū)間為2ln2,+∞.(2)曲線y=f(x)與直線y=1有且僅有兩個(gè)交點(diǎn),則轉(zhuǎn)化為方程xaax=1(x>即方程lnxx令g(x)=lnxx(x>0),即函數(shù)g(x)=lnxx的圖象與直線g'(x)=1-lnxx令g'(x)=1-lnxx2=當(dāng)0<x<e時(shí),g'(x)>0,函數(shù)g(x)單調(diào)遞增;當(dāng)x>e時(shí),g'(x)<0,函數(shù)g(x)單調(diào)遞減,故g(x)max=g(e)=1e因?yàn)楫?dāng)0<x<1時(shí),g(x)∈(-∞,0);當(dāng)x>1時(shí),g(x)∈0,1e,g(1)=0,所以要使函數(shù)g(x)的圖象與直線y=lnaa有兩個(gè)交點(diǎn),則0<ln所以a>1且a≠e.故a的取值范圍為(1,e)∪(e,+∞).21.(1)解函數(shù)f(x)=ex-ax,則f'(x)=ex-a2①若a≤0,則?x>0都有f'(x)>0,所以f(x)在(0,+∞)為增函數(shù),符合題意.②若a>0,因?yàn)閒(x)在(0,+∞)為增函數(shù),所以?x>0,f'(x)≥0恒成立,即?x>0,a≤2x·ex恒成立,令φ(x)=2x·ex,則φ'(x)=2exx+12x所以函數(shù)φ(x)在(0,+∞)上單調(diào)遞增,φ(x)>φ(0)=0,所以a≤0,這與a>0矛盾,所以舍去.綜上,a的取值范圍是(-∞,0].(2)證明因?yàn)閤1,x2是函數(shù)f(x)的兩個(gè)不同的零點(diǎn),所以ex1=ax1顯然x1>0,x2>0,則