2022新高考總復(fù)習(xí)《數(shù)學(xué)》(人教)第三章　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

1、 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:100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.(2020福建福州模擬,理7)已知函數(shù)f(x)為偶函數(shù),當(dāng)x0時(shí),f(x)=x2-ln(-x),則曲線y=f(x)在x=1處的切線方程為()A.x-y=0B.x-y-2=0C.x+y-2=0D.3x-y-2=02.設(shè)函數(shù)f(x)在R上可導(dǎo),其導(dǎo)函數(shù)為f(x),若函數(shù)f(x)在x=1處取得極大值,則函數(shù)y=-xf(x)的圖象可能是()3.已知函數(shù)f(x)=x+1,g(x)=ln x,若f(x1)=g(x2),則x2-x1的最

2、小值為()A.1B.2+ln 2C.2-ln 2D.24.已知定義在(0,+)上的函數(shù)f(x)滿足xf(x)-f(x)0的解集是()A.(-,ln 2)B.(ln 2,+)C.(0,e2)D.(e2,+)5.(2020北京房山區(qū)二模,5)函數(shù)f(x)=ex-x2的零點(diǎn)個(gè)數(shù)為()A.0B.1C.2D.36.(2020山東青島5月模擬,8)已知函數(shù)f(x)=lnxx2,若f(x)eB.me2C.m1D.me7.已知函數(shù)f(x)=x2+|x-a|,g(x)=(2a-1)x+aln x,若函數(shù)y=f(x)與函數(shù)y=g(x)的圖象恰好有兩個(gè)不同的交點(diǎn),則實(shí)數(shù)a的取值范圍為()A.(1,+)B.(-,1)

3、C.(0,+)D.(-,0)8.(2020河南新鄉(xiāng)三模,理12)已知函數(shù)f(x)=x2-axx1e,e與g(x)=ex的圖象上存在兩對(duì)關(guān)于直線y=x對(duì)稱的點(diǎn),則實(shí)數(shù)a的取值范圍是()A.e-1e,eB.1,e-1eC.1,e-1eD.1,e+1e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.(2020山東濰坊臨朐模擬二,12)已知函數(shù)f(x)=xln x+x2,x0是函數(shù)f(x)的極值點(diǎn),以下結(jié)論中正確的是()A.0 x01eC.f(x0)+2x0010.(2020山東聊城二模,10)下列關(guān)于函

4、數(shù)f(x)=x3-3x2+2x的敘述正確的是()A.函數(shù)f(x)有三個(gè)零點(diǎn)B.點(diǎn)(1,0)是函數(shù)f(x)圖象的對(duì)稱中心C.函數(shù)f(x)的極大值點(diǎn)為x=1-33D.存在實(shí)數(shù)a,使得函數(shù)g(x)=f(x)2+af(x)在R上為增函數(shù)11.(2020海南天一大聯(lián)考第三次模擬,12)已知函數(shù)f(x)=x3+ax+b,其中a,bR,則下列選項(xiàng)中的條件使得f(x)僅有一個(gè)零點(diǎn)的有()A.ab,f(x)為奇函數(shù)B.a=ln(b2+1)C.a=-3,b2-40D.a012.(2020山東師大附中月考,12)設(shè)函數(shù)f(x)=|lnx|,x0,ex(x+1),x0,若方程f(x)2-af(x)+116=0有六個(gè)不

5、等的實(shí)數(shù)根,則實(shí)數(shù)a可能的取值是()A.12B.23C.1D.2三、填空題:本題共4小題,每小題5分,共20分.13.(2020山東、海南兩省4月模擬,13)函數(shù)f(x)=alnxex在點(diǎn)P(1,f(1)處的切線與直線2x+y-3=0垂直,則a=.14.設(shè)f(x)=ex(ln x-a),若函數(shù)f(x)在區(qū)間1e,e上單調(diào)遞減,則實(shí)數(shù)a的取值范圍為.15.已知函數(shù)f(x)=log2x,g(x)=x+a-x(a0),若對(duì)x1x|g(x)=x+a-x,x24,16,使g(x1)=f(x2)成立,則實(shí)數(shù)a的取值范圍是.16.已知函數(shù)f(x)=2ln x,g(x)=ax2-x-12(a0).若直線y=2

6、x-b與函數(shù)y=f(x),y=g(x)的圖象均相切,則a的值為;若總存在直線與函數(shù)y=f(x),y=g(x)的圖象均相切,則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)(2020河南鄭州質(zhì)量預(yù)測二,理21)已知函數(shù)f(x)=lnxa,g(x)=x+1x(x0).(1)當(dāng)a=1時(shí),求曲線y=f(x)g(x)在x=1處的切線方程;(2)討論函數(shù)F(x)=f(x)-1g(x)在(0,+)上的單調(diào)性.18.(12分)(2020河南開封三模,理20)已知函數(shù)f(x)=axex-ln x+b(a,bR)在x=1處的切線方程為y=(2e-1)x

7、-e.(1)求a,b值;(2)若f(x)mx恒成立,求實(shí)數(shù)m的取值范圍.19.(12分)(2020陜西寶雞三模,文21)已知函數(shù)f(x)=ln x+ax2-(2a+1)x,aR,f(x)為f(x)的導(dǎo)函數(shù).(1)討論f(x)的單調(diào)性;(2)若g(x)=f(x)+a+1,當(dāng)a12時(shí),求證:g(x)有兩個(gè)零點(diǎn).20.(12分)(2020遼寧大連一中6月模擬,文20)已知函數(shù)f(x)=xln x-1,g(x)=(k-1)x-k(kR).(1)若直線y=g(x)是曲線y=f(x)的一條切線,求k的值;(2)當(dāng)x1時(shí),直線y=g(x)與曲線y=f(x)+1無交點(diǎn),求整數(shù)k的最大值.21.(12分)(20

8、20天津,20)已知函數(shù)f(x)=x3+kln x(kR),f(x)為f(x)的導(dǎo)函數(shù).(1)當(dāng)k=6時(shí),()求曲線y=f(x)在點(diǎn)(1,f(1)處的切線方程;()求函數(shù)g(x)=f(x)-f(x)+9x的單調(diào)區(qū)間和極值.(2)當(dāng)k-3時(shí),求證:對(duì)任意的x1,x21,+),且x1x2,有f(x1)+f(x2)2f(x1)-f(x2)x1-x2.22.(12分)(2020浙江,22)已知10時(shí),-x1時(shí),f(x)0;當(dāng)x=1時(shí),f(x)=0;當(dāng)x0.所以當(dāng)x0,當(dāng)0 x1時(shí),y=-xf(x)1時(shí),y=-xf(x)0,可知選項(xiàng)B符合題意.故選B.3.D設(shè)f(x1)=g(x2)=t,所以x1=t-1

9、,x2=et,所以x2-x1=et-t+1,令h(t)=et-t+1,則h(t)=et-1,所以h(t)在(-,0)上單調(diào)遞減,在(0,+)上單調(diào)遞增,所以h(t)min=h(0)=2.4.A令g(x)=f(x)x,g(x)=xf(x)-f(x)x20等價(jià)為f(ex)exf(2)2,即g(ex)g(2),故ex2,即xln 2,則所求的解集為(-,ln 2).故選A.5.B令f(x)=ex-x2=0,得ex=x2,分別畫出y=ex和y=x2的圖象,如圖所示,當(dāng)x0時(shí),f(x)=ex-2x,令g(x)=ex-2x,則g(x)=ex-2,當(dāng)g(x)=0時(shí),可得x=ln 2.當(dāng)x(0,ln 2)時(shí),

10、g(x)0,g(x)單調(diào)遞增.所以g(x)min=g(ln 2)=eln 2-2ln 2=2-ln 40,所以f(x)在(0,+)上單調(diào)遞增.又因?yàn)閒(0)=1,所以當(dāng)x(0,+)時(shí),f(x)0.故f(x)在(0,+)上無零點(diǎn).綜上,函數(shù)f(x)=ex-x2的零點(diǎn)個(gè)數(shù)為1.故選B.6.B若f(x)m-1x2在(0,+)上恒成立,即f(x)+1x2m在(0,+)上恒成立,令g(x)=f(x)+1x2=lnx+1x2,故只需g(x)maxm即可,g(x)=1xx2-(lnx+1)2xx4=-2lnx-1x3,令g(x)=0,得x=e-12,當(dāng)0 x0;當(dāng)xe-12時(shí),g(x)e2.故選B.7.A當(dāng)

11、a0時(shí)函數(shù)g(x)的定義域?yàn)?0,+),所以只研究這兩個(gè)函數(shù)在x(0,+)上的圖象,當(dāng)a0時(shí),f(x)單調(diào)遞增,又g(x)單調(diào)遞減,兩者的圖象最多只有一個(gè)交點(diǎn),不符合題意.當(dāng)a0時(shí),設(shè)(x)=f(x)-g(x),即(x)=x2-2ax-alnx+a,0 xa,x2+(2-2a)x-alnx-a,xa,因?yàn)?x)=2(x-a)-ax0,0 x0,xa,所以(x)在(0,a)上單調(diào)遞減,(a,+)上單調(diào)遞增,所以(x)min=-a2-aln a+a,因?yàn)閤0,x+時(shí),(x)+,所以(x)有兩個(gè)零點(diǎn),當(dāng)且僅當(dāng)(x)min=-a2-aln a+a1,即a的取值范圍為(1,+).8.Bf(x)與g(x)

12、的圖象在x1e,e上存在兩對(duì)關(guān)于直線y=x對(duì)稱的點(diǎn),則函數(shù)f(x)與函數(shù)(x)=ln x的圖象在x1e,e上有兩個(gè)交點(diǎn),ln x=x2-ax在x1e,e上有兩個(gè)實(shí)數(shù)解,即a=x-lnxx在x1e,e上有兩個(gè)實(shí)數(shù)解,令h(x)=x-lnxx,則h(x)=x2+lnx-1x2.令k(x)=x2+ln x-1,k(x)在x1e,e上單調(diào)遞增,且k(1)=0,當(dāng)x1e,1時(shí),h(x)0,h(x)單調(diào)遞增.h(x)min=h(1)=1.對(duì)g1e=e+1e,g(e)=e-1e,a的取值范圍是1,e-1e.9.AD函數(shù)f(x)=xln x+x2(x0),f(x)=ln x+1+2x.x0是函數(shù)f(x)的極值

13、點(diǎn),f(x0)=0,即ln x0+1+2x0=0,f(x)在(0,+)上單調(diào)遞增,且f1e=2e0,又x0,f(x)-,0 x00,即選項(xiàng)D正確,選項(xiàng)C不正確.故選AD.10.ABC令f(x)=0,即x(x-1)(x-2)=0,解得x=0或x=1或x=2,故函數(shù)f(x)有三個(gè)零點(diǎn),故選項(xiàng)A正確;因?yàn)閒(1+x)+f(1-x)=0,所以點(diǎn)(1,0)是函數(shù)f(x)圖象的對(duì)稱中心,故選項(xiàng)B正確;令f(x)=3x2-6x+2=0,解得x=333,故f(x)在-,3-33上單調(diào)遞增,在3-33,3+33上單調(diào)遞減,在3+33,+上單調(diào)遞增,函數(shù)f(x)的極大值點(diǎn)為x=1-33,故選項(xiàng)C正確;因?yàn)閒(x)

14、在R上不單調(diào),所以不存在實(shí)數(shù)a,使得函數(shù)g(x)=f(x)2+af(x)在R上為增函數(shù),故D錯(cuò)誤.故選ABC.11.BD由題知f(x)=3x2+a.對(duì)于A,由f(x)是奇函數(shù),知b=0,因?yàn)閍0,所以f(x)存在兩個(gè)極值點(diǎn),易知f(x)有三個(gè)零點(diǎn),故A錯(cuò)誤;對(duì)于B,因?yàn)閎2+11,所以a0,f(x)0,所以f(x)單調(diào)遞增,則f(x)僅有一個(gè)零點(diǎn),故B正確;對(duì)于C,若取b=2,則f(x)的極大值為f(-1)=4,極小值為f(1)=0,此時(shí)f(x)有兩個(gè)零點(diǎn),故C錯(cuò)誤;對(duì)于D,f(x)的極大值為f-a3=b-2a3-a3,極小值為f-a3=b+2a3-a3.因?yàn)閍b2+a360,所以b2-4a3

15、27,則b-2a3-a3或b0,f-a30或f-a30,f-a30,可知f(x)僅有一個(gè)零點(diǎn),故D正確.12.BC當(dāng)x0時(shí),f(x)=ex(x+1),則f(x)=ex(x+1)+ex=ex(x+2).由f(x)0得,x+20,即x0得,x+20,即-2x0,此時(shí)f(x)單調(diào)遞增,即當(dāng)x=-2時(shí),f(x)取得極小值f(-2)=-1e2,作出f(x)的圖象如圖:由圖象可知當(dāng)00,g(1)0,0,0a20,1-a+1160,a2-41160,0a21,解得120,所以只需ln x+1x-a0,即aln x+1x在1e,e上恒成立.令g(x)=ln x+1x.因?yàn)間(x)=1x-1x2=x-1x2.由

16、g(x)=0,得x=1.則g(x)在1e,1上單調(diào)遞減,在(1,e)上單調(diào)遞增,g1e=ln1e+e=e-1,g(e)=1+1e,因?yàn)閑-11+1e,所以g(x)max=g1e=e-1.故a的取值范圍為e-1,+).15.4,8結(jié)合題意可得log24=2f(x)log216=4,要使得對(duì)x1x|g(x)=x+a-x,x24,16,使g(x1)=f(x2)成立,則要求g(x)的值域在2,4上,對(duì)g(x)求導(dǎo)得g(x)=a-x-x2xa-x,令g(x)0,解得x0,則切線方程為y-2ln x1=2x1(x-x1),代入g(x)=ax2-x-12(a0),得2x1x-2+2ln x1=ax2-x-1

17、2,即ax2-1+2x1x+32-2ln x1=0.所以=1+2x12-4a32-2ln x1=0.所以a=(x1+2)22x12(3-4ln x1)(x10).令y=(x1+2)22x12(3-4ln x1)(x10),則y=2(x1+2)(4ln x1+x1-1)x13(3-4ln x1)2.令y=0,解得x1=1.當(dāng)x11時(shí),y0,y單調(diào)遞增,當(dāng)0 x11時(shí),y0,y單調(diào)遞減,因此y(1+2)2212(3-4ln1)=32,即a32.17.解 (1)當(dāng)a=1時(shí),y=f(x)g(x)=xlnxx+1,y=(1+lnx)(x+1)-xlnx(x+1)2=lnx+x+1(x+1)2,所以y|x

18、=1=ln1+1+1(1+1)2=12,即當(dāng)x=1時(shí),切線的斜率為12,又切線過點(diǎn)(1,0),所以切線方程為x-2y-1=0.(2)f(x)=1ax,1g(x)=1(x+1)2,F(x)=f(x)-1g(x)=1ax-1(x+1)2=(x+1)2-axax(x+1)2,當(dāng)a0時(shí),F(x)0時(shí),令h(x)=1ax2+2a-1x+1a,=1-4a,當(dāng)0,即00,即a4時(shí),方程1ax2+2a-1x+1a=0有兩個(gè)不等實(shí)數(shù)根x1,x2,設(shè)x1x2,則x1=a-2-a2-4a2,x2=a-2+a2-4a2,所以0 x11x2,此時(shí),函數(shù)F(x)在(0,x1),(x2,+)上單調(diào)遞增,在(x1,x2)上單

19、調(diào)遞減.綜上所述,當(dāng)a4時(shí),F(x)的單調(diào)遞減區(qū)間是a-2-a2-4a2,a-2+a2-4a2,單調(diào)遞增區(qū)間是0,a-2-a2-4a2,a-2+a2-4a2,+.當(dāng)00),即mxex-lnx-1x.令(x)=xex-lnx-1x,則(x)=x2ex+lnxx2.令h(x)=x2ex+ln x,h(x)在(0,+)上單調(diào)遞增,則h1e=1e2e1e-10.所以h(x)在1e,1上存在零點(diǎn)x0,即h(x0)=x02ex0+ln x0=0,即x0ex0=-ln x0 x0=ln1x0(eln1x0).由于y=xex在(0,+)上單調(diào)遞增,故x0=ln1x0=-ln x0,即ex0=1x0.因?yàn)?x)

20、在(0,x0)上單調(diào)遞減,在(x0,+)上單調(diào)遞增,所以(x)min=x0ex0-ln x0-1x0=1+x0-1x0=1.所以m1.實(shí)數(shù)m的取值范圍為(-,1.19.(1)解 f(x)=1x+2ax-(2a+1)=(x-1)(2ax-1)x(x0).當(dāng)a0時(shí),令f(x)0,得0 x1;令f(x)1.所以f(x)在(0,1)上單調(diào)遞增,在(1,+)上單調(diào)遞減.當(dāng)a0時(shí),令f(x)=0,得x1=1,x2=12a.()當(dāng)a=12時(shí),f(x)=(x-1)2x0,所以f(x)在(0,+)上單調(diào)遞增.()當(dāng)a12時(shí),令f(x)0,得0 x1;令f(x)0,得12ax1.所以f(x)在0,12a和(1,+

21、)上單調(diào)遞增,在12a,1上單調(diào)遞減.()當(dāng)0a0,得0 x12a;令f(x)0,得1x12時(shí),f(x)在0,12a和(1,+)上單調(diào)遞增,在12a,1上單調(diào)遞減;當(dāng)0a12時(shí),f(x)在0,12a和(1,+)上單調(diào)遞增,在12a,1上單調(diào)遞減.則g(x)在0,12a和(1,+)上單調(diào)遞增,在12a,1上單調(diào)遞減.因?yàn)間(1)=0,所以1是函數(shù)g(x)的一個(gè)零點(diǎn),且g12a0.當(dāng)x0,12a時(shí),取0 x0e-a-1且x012a,則ax02-(2a+1)x0+a+1=ax02-x0-2ax0+a+1a+1,g(x0)-a-1+a+1=0.所以g12ag(x0)0),設(shè)切點(diǎn)為P(x0,x0ln x

22、0-1),在點(diǎn)P處的切線方程為y-(x0ln x0-1)=(1+ln x0)(x-x0).整理得y=(1+ln x0)x-(x0+1).由1+ln x0=k-1,k=x0+1,即ln x0=k-2,x0=k-1,得ln x0=x0-1.令h(x)=ln x-x+1,則h(x)=1x-1=1-xx.當(dāng)0 x0,h(x)在(0,1)上單調(diào)遞增;當(dāng)x1時(shí),h(x)1).當(dāng)k-20時(shí),F(x)0,所以f(x)在(1,+)上單調(diào)遞增.所以F(x)F(1)=1,即F(x)在(1,+)上無零點(diǎn).當(dāng)k-20時(shí),由F(x)=0,得x=ek-2.當(dāng)1xek-2時(shí),F(x)ek-2時(shí),F(x)0,所以F(x)在(e

23、k-2,+)上單調(diào)遞增.F(x)的最小值為F(ek-2)=(k-1)ek-2-k(ek-2-1)=k-ek-2.令m(k)=k-ek-2,則m(k)=1-ek-20,m(4)=4-e2x2,令x1x2=t(t1),則(x1-x2)f(x1)+f(x2)-2f(x1)-f(x2)=(x1-x2)3x12+kx1+3x22+kx2-2x13-x23+klnx1x2=x13-x23-3x12x2+3x1x22+kx1x2-x2x1-2klnx1x2=x23(t3-3t2+3t-1)+kt-1t-2ln t.令h(x)=x-1x-2ln x,x(1,+).當(dāng)x1時(shí),h(x)=1+1x2-2x=1-1x20,由此可得h(x)在(1,+)上單調(diào)遞增,所以當(dāng)t1時(shí),h(t)h(1),即t-1t-2ln t0.因?yàn)閤21,t3-3t2+3t-1=(t-1)30,k-3,所以,x23(t3-