1、2023新高考數(shù)學(xué)第一輪專題練習(xí)4.2導(dǎo)數(shù)的應(yīng)用基礎(chǔ)篇固本夯基考點(diǎn)一導(dǎo)數(shù)與函數(shù)的單調(diào)性1.(2022屆山東師范大學(xué)附中開學(xué)考,4)已知f(x)=14x2+sin52+x, f '(x)為f(x)的導(dǎo)函數(shù),則f '(x)的圖象是()ABCD答案A2.(2022屆山東煙臺萊州一中開學(xué)考,3)函數(shù)f(x)=-2ln x-x-3x的單調(diào)遞增區(qū)間是()A.(0,+)B.(-3,1) C.(1,+)D.(0,1)答案D3.(2022屆河南天一大聯(lián)考,10)若函數(shù)f(x)=sin2+xsin(x-2)-acos(-x)在區(qū)間0,2上單調(diào)遞增,則實(shí)數(shù)a的取值范圍是()A.(-,-1B.(-,2

2、C.(1,2D.1,+)答案A4.(2021湖北開學(xué)考,4)函數(shù)f(x)=13x2-ln x的單調(diào)遞減區(qū)間為()A.62,+B.62,62C.0,62D.,62答案C5.(2021廣東韶關(guān)一模,7)已知函數(shù)f(x)=ln(ex+1)-12x,若a=flog415,b=f(log56),c=f(log64),則a,b,c的大小關(guān)系正確的是()A.b>a>cB.a>b>cC.c>b>aD.c>a>b答案B6.(2021全國甲理,21,12分)已知a>0且a1,函數(shù)f(x)=xaax(x>0).(1)當(dāng)a=2時(shí),求f(x)的單調(diào)區(qū)間;(2)

3、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.解析(1)當(dāng)a=2時(shí), f(x)=x22x=x2·2-x(x>0),則f '(x)=2x·2-x+x2·(-ln 2·2-x)=x·2-x(2-xln 2).令f '(x)=0,則x=2ln2, f '(x), f(x)的變化情況如下:x0,2ln22ln22ln2,+f '(x)+0-f(x)極大值所以函數(shù)f(x)在0,2ln2上單調(diào)遞增,在2ln2,+上單調(diào)遞減.(2)令f(x)=1,則xaax=1,所以xa=ax,兩邊同時(shí)取對數(shù),可得al

4、n x=xln a,即lnxx=lnaa,根據(jù)題意可知,方程lnxx=lnaa有兩個(gè)實(shí)數(shù)解.設(shè)g(x)=lnxx,則g'(x)=1lnxx2,令g'(x)=0,則x=e.當(dāng)x(0,e)時(shí),g'(x)>0,g(x)單調(diào)遞增;當(dāng)x(e,+)時(shí),g'(x)<0,g(x)單調(diào)遞減.又知g(1)=0,limx+g(x)=0,g(x)max=g(e)=1e,所以要使曲線y=f(x)與直線y=1有且僅有兩個(gè)交點(diǎn),則只需lnaa0,1e,即g(a)=lnaa0,1e,所以a(1,e)(e,+).綜上,實(shí)數(shù)a的取值范圍為(1,e)(e,+).考點(diǎn)二導(dǎo)數(shù)與函數(shù)的極(最)

5、值1.(2022屆湖南岳陽一中入學(xué)考,7)若函數(shù)f(x)=2x3-ax2(a<0)在區(qū)間a2,a+63上有最大值,則實(shí)數(shù)a的取值范圍為()A.-2,0)B.(-,-2C.-4,0)D.(-,-4答案D2.(多選)(2021廣東湛江一模,10)已知函數(shù)f(x)=x3-3ln x-1,則()A. f(x)的極大值為0B.曲線y=f(x)在(1, f(1)處的切線為x軸C. f(x)的最小值為0D. f(x)在定義域內(nèi)單調(diào)答案BC3.(2020浙江溫麗聯(lián)盟聯(lián)考一,6)若函數(shù)f(x)=(x-a)3-3x+b的極大值是M,極小值是m,則M-m()A.與a有關(guān),且與b有關(guān)B.與a有關(guān),且與b無關(guān)C.

6、與a無關(guān),且與b無關(guān)D.與a無關(guān),且與b有關(guān)答案C4.(多選)(2021湖南三湘名校教育聯(lián)盟聯(lián)考,11)下列說法正確的是()A. f(x)=x+1ex的最小值為1B. f(x)=exx(x>0)的最小值為1C. f(x)=x-ln x(x>0)的最小值為1D. f(x)=xe1x(x>0)的最小值為1答案AC5.(2021廣東深圳五校聯(lián)考,5)已知函數(shù)f(x)=ax4-4ax3+b(a>0),x1,4, f(x)的最大值為3,最小值為-6,則a+b=()A.53B.73C.103D.113答案C6.(2017課標(biāo)理,11,5分)若x=-2是函數(shù)f(x)=(x2+ax-1

7、)ex-1的極值點(diǎn),則f(x)的極小值為()A.-1B.-2e-3C.5e-3D.1答案A7.(2021北京,19,15分)已知函數(shù)f(x)=32xx2+a.(1)若a=0,求曲線y=f(x)在點(diǎn)(1, f(1)處的切線方程;(2)若f(x)在x=-1處取得極值,求f(x)的單調(diào)區(qū)間,并求其最大值與最小值.解析(1)當(dāng)a=0時(shí), f(x)=3x2-2x, f '(x)=-6x3+2x2.所以f(1)=1, f '(1)=-4.所以曲線y=f(x)在點(diǎn)(1, f(1)處的切線方程為y-1=-4(x-1),即y=-4x+5.(2)由f(x)=32xx2+a得f '(x)=2

8、(x2+a)2x(32x)(x2+a)2=2(x23xa)(x2+a)2.由題意知f '(-1)=0,所以(-1)2-3×(-1)-a=0.故a=4.當(dāng)a=4時(shí), f(x)=32xx2+4, f '(x)=2(x+1)(x4)(x2+4)2. f '(x)與f(x)的變化情況如下:x(-,-1)-1(-1,4)4(4,+)f '(x)+0-0+f(x)1-14因此, f(x)的單調(diào)遞增區(qū)間是(-,-1)和(4,+),單調(diào)遞減區(qū)間是(-1,4).所以f(x)在區(qū)間(-,4上的最大值是f(-1)=1.又因?yàn)楫?dāng)x(4,+)時(shí), f(x)<0,所以f(-

9、1)=1是f(x)的最大值.同理可知, f(4)=-14是f(x)的最小值.8.(2021山東泰安一模,21)已知函數(shù)f(x)=xln x-12x2+(2a-1)x(aR),討論函數(shù)f(x)的極值點(diǎn)的個(gè)數(shù).解析函數(shù)f(x)的定義域?yàn)?0,+), f '(x)=ln x-x+2a,令h(x)=ln x-x+2a,x>0,則h'(x)=1x-1=1xx,當(dāng)x(0,1)時(shí),h'(x)>0,h(x)單調(diào)遞增;當(dāng)x(1,+)時(shí),h'(x)<0,h(x)單調(diào)遞減,h(x)max=h(1)=2a-1,當(dāng)a12時(shí),h(1)=2a-10,f '(x)0,

10、f(x)在(0,+)上單調(diào)遞減,此時(shí), f(x)無極值點(diǎn);當(dāng)a>12時(shí),h(1)=2a-1>0,0<e-2a<1,h(e-2a)=-2a-e-2a+2a=-e-2a<0,h(x)在(0,1)上有且只有一個(gè)零點(diǎn).f(x)在(0,1)上有且只有一個(gè)極值點(diǎn).又e4a>e2>1,h(e5a)=5a-e5a+2a<7a-e4aa=a(7-e4a)<a(7-e2)<0,h(x)在(1,+)上有且只有一個(gè)零點(diǎn),f(x)在(1,+)上有且只有一個(gè)極值點(diǎn).綜上所述,當(dāng)a12時(shí), f(x)無極值點(diǎn);當(dāng)a>12時(shí), f(x)有2個(gè)極值點(diǎn).綜合篇知能轉(zhuǎn)

11、換考法一利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性1.(2022屆山東平邑一中開學(xué)考,8)若函數(shù)f(x)=x2-ax+ln x在區(qū)間(2,e)上單調(diào)遞增,則a的取值范圍是()A.92,+B.,92C.92,e2+1D.e2+1,+)答案B2.(2022屆重慶八中8月入學(xué)摸底,6)已知定義在R上的奇函數(shù)f(x)滿足f(2)=0,當(dāng)x>0時(shí),2xf(x)+x2f '(x)>0,則使得f(x)>0成立的x的取值范圍是()A.(-,-2)B.(2,+)C.(-2,0)(2,+)D.(-,-2)(0,2)答案C3.(多選)(2022屆山東濟(jì)寧實(shí)驗(yàn)中學(xué)開學(xué)考,9)下列選項(xiàng)中,在(-,+)上單調(diào)遞增

12、的函數(shù)有()A.f(x)=x4B.f(x)=x-sin xC.f(x)=xexD.f(x)=ex-e-x-2x答案BD4.(2021山東青島三模,7)定義在R上的奇函數(shù)f(x)的圖象連續(xù)不斷,其導(dǎo)函數(shù)為f '(x),對任意正實(shí)數(shù)x恒有xf '(x)>2f(-x),若g(x)=x2f(x),則不等式g(log3(x2-1)+g(-1)<0的解集是()A.(0,2)B.(-2,2)C.(-3,2)D.(-2,-1)(1,2)答案D5.(多選)(2021八省聯(lián)考,9)已知函數(shù)f(x)=xln(1+x),則()A. f(x)在(0,+)上單調(diào)遞增B. f(x)有兩個(gè)零點(diǎn)C.

13、曲線y=f(x)在點(diǎn)12, f12處切線的斜率為-1-ln 2D. f(x)是偶函數(shù)答案AC6.(2022廣東深圳實(shí)驗(yàn)學(xué)校10月月考改編,22)已知函數(shù)g(x)=ln x+2x+ax(aR),討論g(x)的單調(diào)性.解析函數(shù)g(x)=ln x+2x+ax的定義域?yàn)?0,+),g'(x)=1x+2-ax2=2x2+xax2(aR),方程2x2+x-a=0的判別式=1+8a.當(dāng)a-18時(shí),0,g'(x)0,g(x)在(0,+)上單調(diào)遞增;當(dāng)a>-18時(shí),>0,方程2x2+x-a=0的兩根為x1=11+8a4,x2=1+1+8a4.(i)當(dāng)-18<a0時(shí),x1<

14、x20,對任意的x>0,g'(x)>0,g(x)在(0,+)上單調(diào)遞增;(ii)當(dāng)a>0時(shí),x1<0<x2,令g'(x)<0,可得0<x<x2,令g'(x)>0,可得x>x2.所以,g(x)在1+8a14,+上為單調(diào)遞增,在0,1+8a14上單調(diào)遞減.7.(2021山東青島二模,21)已知函數(shù)f(x)=aln x-x+1(x>0),aR.(1)討論f(x)的單調(diào)性;(2)若對任意x(0,+),均有f(x)0,求a的值;(3)假設(shè)某籃球運(yùn)動員每次投籃命中的概率均為0.81,若其10次投籃全部命中的概率為p,

15、證明:p<e-2.解析(1)函數(shù)f(x)的定義域?yàn)?0,+), f '(x)=ax-12x=2ax2x.若a0,則f '(x)<0對任意的x>0恒成立,此時(shí)函數(shù)f(x)在(0,+)上為減函數(shù);若a>0,由f '(x)>0可得0<x<4a2,由f '(x)<0可得x>4a2,此時(shí),函數(shù)f(x)的單調(diào)遞增區(qū)間為(0,4a2),單調(diào)遞減區(qū)間為(4a2,+).綜上所述,當(dāng)a0時(shí),函數(shù)f(x)在(0,+)上為減函數(shù);當(dāng)a>0時(shí),函數(shù)f(x)在(0,4a2)上單調(diào)遞增,在(4a2,+)上單調(diào)遞減.(2)當(dāng)a0時(shí),函

16、數(shù)f(x)在(0,+)上為減函數(shù),且f(1)=0,當(dāng)0<x<1時(shí), f(x)>f(1)=0,不合題意.當(dāng)a>0時(shí),由(1)知f(x)max=f(4a2)=aln(4a2)-2a+1=2aln(2a)-2a+10,令t=2a,t>0,可得tln t-t+10,即ln t-1+1t0,令g(t)=ln t+1t-1,其中t>0,則g'(t)=1t-1t2=t1t2.當(dāng)0<t<1時(shí),g'(t)<0,此時(shí)函數(shù)g(t)單調(diào)遞減;當(dāng)t>1時(shí),g'(t)>0,此時(shí)函數(shù)g(t)單調(diào)遞增.所以,g(t)min=g(1)=0

17、,則g(t)g(1)=0,又g(t)0,所以g(t)=0,所以2a=t=1,解得a=12.(3)由題意可得p=0.8110,由(2)可知,當(dāng)a=12時(shí),lnx2-x+10,即ln x2(x-1),所以ln 0.8110=10ln 0.81<20(0.81-1)=-2,因此,p<e-2.8.(2021全國甲文,20,12分)設(shè)函數(shù)f(x)=a2x2+ax-3ln x+1,其中a>0.(1)討論f(x)的單調(diào)性;(2)若y=f(x)的圖象與x軸沒有公共點(diǎn),求a的取值范圍.解析(1)f(x)=a2x2+ax-3ln x+1,x(0,+),f '(x)=2a2x+a-3x=2

18、a2x2+ax3x=(2ax+3)(ax1)x.a>0,x>0,2ax+3x>0,當(dāng)x0,1a時(shí), f '(x)<0;當(dāng)x1a,+時(shí), f '(x)>0,函數(shù)f(x)在0,1a上單調(diào)遞減,在1a,+上單調(diào)遞增.(2)y=f(x)的圖象與x軸沒有公共點(diǎn),函數(shù)f(x)在(0,+)上沒有零點(diǎn),由(1)可得函數(shù)f(x)在0,1a上單調(diào)遞減,在1a,+上單調(diào)遞增,f(x)min=f1a=3-3ln 1a=3+3ln a>0,ln a>-1,解得a>1e,故實(shí)數(shù)a的取值范圍是1e,+.考法二利用導(dǎo)數(shù)研究函數(shù)的極(最)值1.(2022屆山東煙臺

19、萊州一中開學(xué)考,8)已知直線y=a分別與直線y=2x-2和曲線y=2ex+x相交于點(diǎn)A,B,則線段AB長度的最小值為()A.12(3+ln 2)B.3-ln 2C.2e-1D.3答案A2.(2022屆重慶巴蜀中學(xué)月考(一),8)已知函數(shù)f(x)的定義域是(0,+),且滿足2x2f(x)+x3f '(x)=ln x, f(e)=14e(其中e為自然對數(shù)的底數(shù)),則下列說法正確的是()A.f(x)在(0,+)上單調(diào)遞增B.f(x)在(0,+)上單調(diào)遞減C.f(x)在(0,+)上有極大值D.f(x)在(0,+)上有極小值答案B3.(2021山東日照模擬,8)關(guān)于函數(shù)f(x)=sinxx,x(

20、0,+)的性質(zhì),以下說法正確的是()A.函數(shù)f(x)的周期是2B.函數(shù)f(x)在(0,)上有極值C.函數(shù)f(x)在(0,+)上單調(diào)遞減D.函數(shù)f(x)在(0,+)上有最小值答案D4.(2020河北衡水中學(xué)七調(diào),12)已知函數(shù)f(x)=xex-ln x-x-2,g(x)=ex2x+ln x-x的最小值分別為a,b,則()A.a=bB.a<bC.a>bD.a,b的大小關(guān)系不確定答案A5.(多選)(2022屆山東濟(jì)寧實(shí)驗(yàn)中學(xué)開學(xué)考,12)已知函數(shù)f(x)=x2+x1ex,則下列結(jié)論正確的是()A.函數(shù)f(x)存在兩個(gè)不同的零點(diǎn)B.函數(shù)f(x)既存在極大值又存在極小值C.當(dāng)-e<k&

21、lt;0時(shí),方程f(x)=k有且只有兩個(gè)實(shí)根D.當(dāng)xt,+)時(shí), f(x)max=5e2,t的最小值為2答案ABC6.(2021新高考,15,5分)函數(shù)f(x)=|2x-1|-2ln x的最小值為. 答案17.(2018江蘇,11,5分)若函數(shù)f(x)=2x3-ax2+1(aR)在(0,+)內(nèi)有且只有一個(gè)零點(diǎn),則f(x)在-1,1上的最大值與最小值的和為. 答案-38.(2021山東日照模擬,21改編)已知函數(shù)f(x)=ex-ax2-bx-1,其中a,bR,e=2.718 28為自然對數(shù)的底數(shù),設(shè)g(x)是函數(shù)f(x)的導(dǎo)函數(shù),求函數(shù)g(x)在區(qū)間0,1上的最小值.解析g(

22、x)=ex-2ax-b,g'(x)=ex-2a.當(dāng)a0時(shí),g'(x)=ex-2a>0,所以g(x)g(0)=1-b.當(dāng)a>0時(shí),由g'(x)=ex-2a>0得ex>2a,即x>ln(2a).若a>12,則ln(2a)>0;若a>e2,則ln(2a)>1.所以當(dāng)0<a12時(shí),g(x)在0,1上單調(diào)遞增,所以g(x)g(0)=1-b.當(dāng)12<ae2時(shí),g(x)在0,ln(2a)上單調(diào)遞減,在ln(2a),1上單調(diào)遞增,所以g(x)g(ln(2a)=2a-2aln(2a)-b.當(dāng)a>e2時(shí),g(x)在0,

23、1上單調(diào)遞減,所以g(x)g(1)=e-2a-b.綜上所述:當(dāng)a12時(shí),g(x)在0,1上的最小值為1-b;當(dāng)12<ae2時(shí),g(x)在0,1上的最小值為2a-2·aln(2a)-b;當(dāng)a>e2時(shí),g(x)在0,1上的最小值為e-2a-b.9.(2020北京,19,15分)已知函數(shù)f(x)=12-x2.(1)求曲線y=f(x)的斜率等于-2的切線方程;(2)設(shè)曲線y=f(x)在點(diǎn)(t, f(t)處的切線與坐標(biāo)軸圍成的三角形的面積為S(t),求S(t)的最小值.解析(1)因?yàn)閒(x)=12-x2,所以f '(x)=-2x,令-2x=-2,解得x=1,又f(1)=11

24、,所以所求切線方程為y-11=-2(x-1),整理得2x+y-13=0.(2)由(1)可知f '(x)=-2x,所以曲線y=f(x)在點(diǎn)(t, f(t)處的切線斜率k=-2t,又f(t)=12-t2,所以切線方程為y-(12-t2)=-2t(x-t),整理得2tx+y-(t2+12)=0,當(dāng)x=0時(shí),y=t2+12,所以切線與y軸的交點(diǎn)為(0,t2+12),當(dāng)y=0時(shí),x=t2+122t,所以切線與x軸的交點(diǎn)為t2+122t,0.當(dāng)t>0時(shí),S(t)=12·t2+122t·(t2+12)=(t2+12)24t,則S'(t)=3(t24)(t2+12)4

25、t2,當(dāng)0<t<2時(shí),S'(t)<0,此時(shí)S(t)在(0,2)上單調(diào)遞減;當(dāng)t>2時(shí),S'(t)>0,此時(shí)S(t)在(2,+)上單調(diào)遞增,所以S(t)min=S(2)=32.當(dāng)t<0時(shí),S(t)=-(t2+12)24t,則S'(t)=-3(t24)(t2+12)4t2,當(dāng)t<-2時(shí),S'(t)<0,此時(shí)S(t)在(-,-2)上單調(diào)遞減;當(dāng)-2<t<0時(shí),S'(t)>0,此時(shí)S(t)在(-2,0)上單調(diào)遞增,所以S(t)min=S(-2)=32.綜上所述,當(dāng)t=±2時(shí),S(t)取最

26、小值,為32.10.(2018天津文,20,14分)設(shè)函數(shù)f(x)=(x-t1)(x-t2)(x-t3),其中t1,t2,t3R,且t1,t2,t3是公差為d的等差數(shù)列.(1)若t2=0,d=1,求曲線y=f(x)在點(diǎn)(0, f(0)處的切線方程;(2)若d=3,求f(x)的極值;(3)若曲線y=f(x)與直線y=-(x-t2)-63有三個(gè)互異的公共點(diǎn),求d的取值范圍.解析(1)由已知,可得f(x)=x(x-1)(x+1)=x3-x,故f '(x)=3x2-1.因此f(0)=0, f '(0)=-1,又因?yàn)榍€y=f(x)在點(diǎn)(0, f(0)處的切線方程為y-f(0)=f &#

27、39;(0)·(x-0),故所求切線方程為x+y=0.(2)由已知可得f(x)=(x-t2+3)(x-t2)(x-t2-3)=(x-t2)3-9(x-t2)=x3-3t2x2+(3t22-9)x-t23+9t2.故f '(x)=3x2-6t2x+3t22-9.令f '(x)=0,解得x1=t2-3,x2=t2+3.當(dāng)x變化時(shí), f '(x), f(x)的變化情況如下表:x(-,x1)x1(x1,x2)x2(x2,+)f '(x)+0-0+f(x)極大值極小值所以函數(shù)f(x)的極大值為f(t2-3)=(-3)3-9×(-3)=63;函數(shù)f(x)

28、的極小值為f(t2+3)=(3)3-9×3=-63.(3)曲線y=f(x)與直線y=-(x-t2)-63有三個(gè)互異的公共點(diǎn)等價(jià)于關(guān)于x的方程(x-t2+d)(x-t2)(x-t2-d)+(x-t2)+63=0有三個(gè)互異的實(shí)數(shù)解.令u=x-t2,可得u3+(1-d2)u+63=0.設(shè)函數(shù)g(x)=x3+(1-d2)x+63,則曲線y=f(x)與直線y=-(x-t2)-63有三個(gè)互異的公共點(diǎn)等價(jià)于函數(shù)y=g(x)有三個(gè)零點(diǎn).g'(x)=3x2+(1-d2).當(dāng)d21時(shí),g'(x)0,這時(shí)g(x)在R上單調(diào)遞增,不合題意.當(dāng)d2>1時(shí),令g'(x)=0,解得x

29、1=-d213,x2=d213.易得,g(x)在(-,x1)上單調(diào)遞增,在x1,x2上單調(diào)遞減,在(x2,+)上單調(diào)遞增.g(x)的極大值g(x1)=gd213=23(d21)329+63>0.g(x)的極小值g(x2)=gd213=-23(d21)329+63.若g(x2)0,由g(x)的單調(diào)性可知函數(shù)y=g(x)至多有兩個(gè)零點(diǎn),不合題意.若g(x2)<0,即(d2-1)32>27,也就是|d|>10,此時(shí)|d|>x2,g(|d|)=|d|+63>0,且-2|d|<x1,g(-2|d|)=-6|d|3-2|d|+63<-6210+63<0

30、,從而由g(x)的單調(diào)性,可知函數(shù)y=g(x)在區(qū)間(-2|d|,x1),(x1,x2),(x2,|d|)內(nèi)各有一個(gè)零點(diǎn),符合題意.所以,d的取值范圍是(-,-10)(10,+).考法三利用導(dǎo)數(shù)研究不等式1.(2021八省聯(lián)考,8)已知a<5且ae5=5ea,b<4且be4=4eb,c<3且ce3=3ec,則()A.c<b<aB.b<c<aC.a<c<bD.a<b<c答案D2.(2021山東濱州二模,8)已知a=ln22,b=1e(e=2.718 28為自然對數(shù)的底數(shù)),c=2ln39,則a,b,c的大小關(guān)系為()A.a>

31、b>cB.a>c>bC.b>a>cD.b>c>a答案C3.(2022屆江蘇淮安車橋中學(xué)入學(xué)調(diào)研,10)已知函數(shù)y=f(x)在R上可導(dǎo)且f(x)<f '(x)恒成立,則下列不等式中一定成立的是()A.f(3)>e3f(0), f(2 018)>e2 018f(0)B.f(3)<e3f(0), f(2 018)>e2 018f(0)C.f(3)>e3f(0), f(2 018)<e2 018f(0)D.f(3)<e3f(0), f(2 018)<e2 018f(0)答案A4.(2022屆長沙雅禮

32、中學(xué)月考一,7)已知函數(shù)f(x)=x22x2,x0,ln(x+1),x>0,若關(guān)于x的不等式f(x)ax+a-12在R上恒成立,則實(shí)數(shù)a的取值范圍是()A.e12,2B.2,e12C.e12,6D.e12,6答案A5.(2021濟(jì)南一模,8)設(shè)a=2 022ln 2 020,b=2 021ln 2 021,c=2 020ln 2 022,則()A.a>c>bB.c>b>aC.b>a>cD.a>b>c答案D6.(多選)(2022屆福州八縣(市、區(qū))一中期中,12)已知函數(shù)f(x)的定義域、值域都是(0,+),且滿足12f(x)<f &#

33、39;(x),則下列結(jié)論一定正確的是()A.若f(1)=e,則f(2)>e32B.f(2)<f(3)C.3f(2)>2f(4)D.7f14<6f13e18答案ABD7.(2022屆遼寧六校期初聯(lián)考,8)已知奇函數(shù)f(x)的定義域?yàn)?,2,其導(dǎo)函數(shù)為f '(x),當(dāng)0<x<2時(shí),有f '(x)cos x+f(x)sin x>0成立,則關(guān)于x的不等式|f(x)|<2f4cos x的解集為()A.4,4B.2,44,2C.4,00,4D.4,04,2答案A8.(多選)(2021山東淄博二模,12)已知e是自然對數(shù)的底數(shù),則下列不等關(guān)系中

34、不正確的是()A.ln 2>2eB.ln 3<3eC.ln >eD.ln3ln <3答案ACD9.(2021新高考,22,12分)已知函數(shù)f(x)=x(1-ln x).(1)討論f(x)的單調(diào)性;(2)設(shè)a,b為兩個(gè)不相等的正數(shù),且bln a-aln b=a-b,證明:2<1a+1b<e.解析(1)函數(shù)f(x)的定義域?yàn)?0,+), f '(x)=-ln x,令f '(x)>0,解得0<x<1,令f '(x)<0,解得x>1,所以函數(shù)f(x)在(0,1)上單調(diào)遞增,在(1,+)上單調(diào)遞減.(2)證明:由b

35、ln a-aln b=a-b得1a(1+ln a)=1b(1+ln b),即1a1ln 1a=1b1ln 1b,令x1=1a,x2=1b,則x1,x2為f(x)=k的兩個(gè)實(shí)根,當(dāng)x0+時(shí),f(x)0+,當(dāng)x+時(shí), f(x)-,且f(1)=1,故k(0,1),不妨令x1(0,1),x2(1,e),則2-x1>1,e-x1>1,先證明x1+x2>2,即證x2>2-x1,即證f(x2)=f(x1)<f(2-x1).令h(x)=f(x)-f(2-x),x(0,1),則h'(x)=f '(x)+f '(2-x)=-ln x-ln(2-x)=-lnx(

36、2-x).x(0,1),x(2-x)(0,1),h'(x)>0恒成立,h(x)為增函數(shù),h(x)<h(1)=0.f(x2)<f(2-x1),x2>2-x1,x1+x2>2.再證x1+x2<e,即證x2<e-x1,即證f(x2)=f(x1)>f(e-x1).令(x)=f(x)-f(e-x),x(0,1),則'(x)=-lnx(e-x),x0+時(shí),'(x)+,'(1)=-ln(e-1)<0,'(x)在(0,1)上單調(diào)遞減,在(0,1)上必存在唯一x0,使'(x0)=0,且當(dāng)x(0,x0)時(shí),

37、9;(x)>0,(x)單調(diào)遞增,當(dāng)x(x0,1)時(shí),'(x)<0,(x)單調(diào)遞減,又x0+時(shí), f(x)0+,且f(e)=0,x0+時(shí),(x)0+,又(1)=f(1)-f(e-1)>0,(x)>0恒成立,f(x2)>f(e-x1),x2<e-x1,x1+x2<e.綜上,2<1a+1b<e成立.10.(2021全國乙理,20,12分)設(shè)函數(shù)f(x)=ln(a-x),已知x=0是函數(shù)y=xf(x)的極值點(diǎn).(1)求a;(2)設(shè)函數(shù)g(x)=x+f(x)xf(x).證明:g(x)<1.解析(1)由題意得y=xf(x)=xln(a-x

38、),x(-,a),y'=ln(a-x)+x·1ax·(-1)=ln(a-x)-xax,x(-,a),x=0是函數(shù)y=xf(x)的極值點(diǎn),ln(a-0)-0a0=0,可得a=1.當(dāng)a=1時(shí),y'=ln(1-x)-x1x,x(-,1),令p(x)=ln(1-x)-x1x,x(-,1),則p'(x)=1x1-1(1x)2=x2(x1)2,易知當(dāng)x(-,1)時(shí),p'(x)<0恒成立.p(x)在(-,1)上為減函數(shù),又p(0)=0,當(dāng)x(-,0)時(shí),p(x)>0;當(dāng)x(0,1)時(shí),p(x)<0,函數(shù)y=xf(x)=xln(1-x)在(

39、-,0)上為增函數(shù),在(0,1)上為減函數(shù).當(dāng)a=1時(shí),x=0是函數(shù)y=xf(x)的極值點(diǎn).a=1.(2)證明:由(1)知a=1,f(x)=ln(1-x),x(-,1),當(dāng)x(0,1)時(shí), f(x)=ln(1-x)<0,xf(x)<0,當(dāng)x(-,0)時(shí), f(x)=ln(1-x)>0,xf(x)<0,要證g(x)=x+f(x)xf(x)<1,只需證x+f(x)>xf(x).只需證x+ln(1-x)>xln(1-x),只需證x+(1-x)ln(1-x)>0,令h(x)=x+(1-x)ln(1-x),則h'(x)=1-ln(1-x)-1=-l

40、n(1-x),當(dāng)x(0,1)時(shí),h'(x)>0,h(x)單調(diào)遞增,當(dāng)x(-,0)時(shí),h'(x)<0,h(x)單調(diào)遞減,當(dāng)x(-,0)(0,1)時(shí),h(x)>h(0)=0,x+(1-x)ln(1-x)>0在(-,0)(0,1)上恒成立.g(x)<1.11.(2020課標(biāo)理,21,12分)已知函數(shù)f(x)=ex+ax2-x.(1)當(dāng)a=1時(shí),討論f(x)的單調(diào)性;(2)當(dāng)x0時(shí), f(x)12x3+1,求a的取值范圍.解析(1)當(dāng)a=1時(shí), f(x)=ex+x2-x, f '(x)=ex+2x-1.故當(dāng)x(-,0)時(shí), f '(x)&l

41、t;0;當(dāng)x(0,+)時(shí),f '(x)>0.所以f(x)在(-,0)單調(diào)遞減,在(0,+)單調(diào)遞增.(2)f(x)12x3+1等價(jià)于12x3ax2+x+1e-x1.設(shè)函數(shù)g(x)=12x3ax2+x+1e-x(x0),則g'(x)=-12x3ax2+x+132x2+2ax1e-x=-12xx2-(2a+3)x+4a+2e-x=-12x(x-2a-1)(x-2)e-x.(i)若2a+10,即a-12,則當(dāng)x(0,2)時(shí),g'(x)>0.所以g(x)在(0,2)單調(diào)遞增,而g(0)=1,故當(dāng)x(0,2)時(shí),g(x)>1,不合題意.(ii)若0<2a+

42、1<2,即-12<a<12,則當(dāng)x(0,2a+1)(2,+)時(shí),g'(x)<0;當(dāng)x(2a+1,2)時(shí),g'(x)>0.所以g(x)在(0,2a+1),(2,+)單調(diào)遞減,在(2a+1,2)單調(diào)遞增.由于g(0)=1,所以g(x)1當(dāng)且僅當(dāng)g(2)=(7-4a)e-21,即a7e24.所以當(dāng)7e24a<12時(shí),g(x)1.(iii)若2a+12,即a12,則g(x)12x3+x+1e-x.由于07e24,12,故由(ii)可得12x3+x+1e-x1.故當(dāng)a12時(shí),g(x)1.綜上,a的取值范圍是7e24,+.12.(2021山東泰安三模,2

43、1改編)已知函數(shù)f(x)=x22+ax(a>0),g(x)=(x+1)ln(x+1),且曲線y=f(x)和y=g(x)在原點(diǎn)處有相同的切線,求實(shí)數(shù)a的值,并證明:當(dāng)x>0時(shí),f(x)>g(x).解析由條件可得f(0)=g(0)=0,且f '(x)=x+a,g'(x)=1+ln(x+1).因?yàn)榍€y=f(x)和y=g(x)在原點(diǎn)處有相同的切線,所以f '(0)=g'(0),則a=1.要證f(x)>g(x),即證(x+1)ln(x+1)<x22+x.令h(x)=(x+1)ln(x+1)-x22-x,x>0,則h'(x)=l

44、n(x+1)-x,令(x)=h'(x)=ln(x+1)-x,x>0,則'(x)=1x+1-1=-xx+1,易知'(x)<0,所以h'(x)在(0,+)上單調(diào)遞減,所以h'(x)<h'(0)=0,所以h(x)在(0,+)上單調(diào)遞減,所以h(x)<h(0)=0.所以當(dāng)x>0時(shí),(x+1)ln(x+1)<x22+x,即f(x)>g(x).考法四利用導(dǎo)數(shù)研究函數(shù)的零點(diǎn)或方程的根1.(2022屆廣東深圳六校聯(lián)考二,8)已知函數(shù)f(x)=xlnx,x>0,1x2,x0,若函數(shù)g(x)=f(x)-k有三個(gè)零點(diǎn),則

45、()A.-e<k1B.-1e<k<1C.-e<k<0D.-1e<k<0答案D2.(2022屆廣東珠海二中10月月考,7)已知函數(shù)f(x)=eax-x存在兩個(gè)零點(diǎn),則正數(shù)a的取值范圍是()A.0,e2B.e2,+C.0,12eD.12e,+答案C3.(2022屆重慶八中8月入學(xué)摸底,8)設(shè)實(shí)數(shù)>0,若對任意的x(1,+),不等式e3x-lnx30恒成立,則實(shí)數(shù)的取值范圍是()A.1e,+B.13e,+C.e,+)D.3e,+)答案B4.(多選)(2022屆湖湘教育三新探索協(xié)作體11月期中,12)已知函數(shù)f(x)=ex,g(x)=ln x2+12,下

46、列說法正確的是()A.mR,h(x)=f(x)-g(x)+m都存在零點(diǎn)B.若x>1, f(ax)-axx-g(2x)+12恒成立,則正實(shí)數(shù)a的最小值為1eC.若f(x),g(x)的圖象與直線y=m分別交于A,B兩點(diǎn),則|AB|的最小值為2+ln 2D.存在直線y=m與f(x),g(x)的圖象分別交于A,B兩點(diǎn),使得曲線f(x)在A處的切線與曲線g(x)在B處的切線平行答案BCD5.(多選)(2021山東新高考聯(lián)盟聯(lián)考,12)已知函數(shù)f(x)=3xx3,x<0,2x1,x0,若關(guān)于x的方程4f 2(x)-4a·f(x)+2a+3=0有5個(gè)不同的實(shí)根,則實(shí)數(shù)a可能的取值有()

47、A.-32B.-43C.-54D.-76答案BCD6.(2021山東泰安模擬,8)已知0<a<b且滿足ea-b=ab,則下列說法正確的是()A.ab<a-b+1B.ln a+2a=ln b+2bC.a>12D.不存在a,b滿足a+b=1答案D7.(2021山東青島模擬,14)設(shè)函數(shù)f(x)=ex(x+1)的圖象在點(diǎn)(0,1)處的切線為y=ax+b,若方程|ax-b|=m有兩個(gè)不等實(shí)根,則實(shí)數(shù)m的取值范圍是. 答案(0,1)8.(2021山東青島三模,13)定義方程f(x)=f '(x)的實(shí)數(shù)根x0叫做函數(shù)f(x)的“新駐點(diǎn)”,若函數(shù)g(x)=12x,h

48、(x)=ln(2x),(x)=sin x(0<x<)的“新駐點(diǎn)”分別為,則,的大小關(guān)系為. 答案<<9.(2021山東泰安三模,15)已知函數(shù)f(x)=x(sin x+1)+acos x,當(dāng)a>2時(shí),函數(shù)g(x)=f(x)-3在區(qū)間0,2上有唯一零點(diǎn),則實(shí)數(shù)a的取值范圍是. 答案(2,310.(2021山東聊城三模,16)已知函數(shù)f(x)=xex2+(a-2)xex+2-a有三個(gè)不同的零點(diǎn)x1,x2,x3,其中x1<x2<x3,則1-x1ex121-x2ex2·1x3ex3的值為. 答案111.(2022屆T8聯(lián)

49、考,22)已知函數(shù)f(x)=aln x-sin x+x,其中a為非零常數(shù).(1)若函數(shù)f(x)在(0,+)上單調(diào)遞增,求a的取值范圍;(2)設(shè),32,且cos =1+sin ,證明:當(dāng)2sin <a<0時(shí),函數(shù)f(x)在(0,2)上恰有兩個(gè)極值點(diǎn).解析(1)f '(x)=ax-cos x+1(x>0).a>0,因?yàn)閤>0,所以1-cos x0,則f '(x)>0,f(x)在(0,+)上單調(diào)遞增,符合要求.a<0,當(dāng)x0,a2時(shí),ax<-2,從而f '(x)<-2-cos x+1=-(1+cos x)0,所以f(x)在

50、0,a2上單調(diào)遞減,不符合要求.綜上,a的取值范圍是(0,+).(2)證明:令f '(x)=0,則ax-cos x+1=0,即a=xcos x-x.設(shè)g(x)=xcos x-x(x>0),則g'(x)=cos x-xsin x-1.當(dāng)x(0,)時(shí),cos x<1,sin x>0,則cos x-1<0,-xsin x<0,從而g'(x)<0,g(x)單調(diào)遞減.當(dāng)x,32時(shí),g(x)=-sin x-(sin x+xcos x)=-(2sin x+xcos x).因?yàn)閟in x<0,cos x<0,所以g(x)>0,從而g

51、'(x)單調(diào)遞增.因?yàn)間'()=-2<0,g'32=32-1>0,所以g'(x)在,32上有唯一零點(diǎn),記為x0,且當(dāng)x(,x0)時(shí),g'(x)<0,g(x)單調(diào)遞減;當(dāng)xx0,32時(shí),g'(x)>0,g(x)單調(diào)遞增.當(dāng)x32,2時(shí),g(x)=-(2cos x+cos x-xsin x)=xsin x-3cos x.因?yàn)閟in x<0,cos x>0,所以g(x)<0,從而g(x)單調(diào)遞減.因?yàn)間32=2>0,g(2)=-2<0,所以g(x)在32,2內(nèi)有唯一零點(diǎn),記為x1,且當(dāng)x32,x1時(shí)

52、,g(x)>0,g'(x)單調(diào)遞增;當(dāng)x(x1,2)時(shí),g(x)<0,g'(x)單調(diào)遞減.因?yàn)間'32=32-1>0,g'(2)=0,所以當(dāng)x32,2時(shí),g'(x)>0,g(x)單調(diào)遞增.綜上,g(x)在(0,x0)上單調(diào)遞減,在(x0,2)上單調(diào)遞增.因?yàn)間(0)=g(2)=0,所以當(dāng)g(x0)<a<0時(shí),直線y=a與函數(shù)g(x)的圖象在(0,2)上有兩個(gè)交點(diǎn),從而f '(x)有兩個(gè)變號零點(diǎn),即f(x)在(0,2)上恰有兩個(gè)極值點(diǎn).因?yàn)間'(x0)=0,所以cos x0-x0sin x0-1=0,即c

53、os x0=1+x0sin x0.從而g(x0)=x0cos x0-x0=x0(1+x0sin x0)-x0=x02sin x0.取=x0,則cos =1+sin ,且當(dāng)2sin <a<0時(shí),函數(shù)f(x)在(0,2)上恰有兩個(gè)極值點(diǎn).12.(2021山東淄博二模,21)已知函數(shù)f(x)=|ln x|+ax(a<0).(1)討論函數(shù)f(x)的單調(diào)性;(2)若函數(shù)f(x)恰好有三個(gè)零點(diǎn),求a的取值范圍.解析(1)函數(shù)f(x)的定義域?yàn)?0,+),由f(x)=|ln x|+ax=lnx+ax,x1,lnx+ax,0<x<1,得f '(x)=1x+a,x1,1x+

54、a,0<x<1.因?yàn)閍<0,所以-1x+a<0,即在區(qū)間(0,1)上, f '(x)<0,故函數(shù)f(x)在(0,1)上單調(diào)遞減.當(dāng)-1<a<0時(shí), f '(x), f(x)的變化如下表:x1,1a-1a1a,+f '(x)+0-f(x)增減當(dāng)a-1時(shí),1x+a0,即在區(qū)間1,+)上, f '(x)0,故函數(shù)f(x)在1,+)上單調(diào)遞減.綜上,當(dāng)-1<a<0時(shí),函數(shù)f(x)在區(qū)間(0,1)上單調(diào)遞減,在區(qū)間1,1a上單調(diào)遞增,在區(qū)間1a,+上單調(diào)遞減;當(dāng)a-1時(shí),函數(shù)f(x)在區(qū)間(0,+)上單調(diào)遞減.(2)結(jié)

55、合(1)知,只有當(dāng)-1<a<0時(shí)函數(shù)f(x)才可能存在三個(gè)零點(diǎn).當(dāng)-1<a<0時(shí), f(1)=a<0, f(ea)=-ln ea+a·ea=a(ea-1)>0(0<ea<1), f(x)在區(qū)間(0,1)上恰好存在一個(gè)零點(diǎn);在區(qū)間1,+)上存在兩個(gè)零點(diǎn),需要保證f1a=ln1a-1>0,即-1e<a<0,且此時(shí)f(1)=a<0, f1a>0,在區(qū)間1,1a上存在一個(gè)零點(diǎn),同時(shí)1a2>-1a, f1a2=2ln1a+1a,設(shè)t=-1a>e,對于函數(shù)y=2ln t-t,y'=2tt<0,

56、所以y<2ln e-e=2-e<0,故f1a2<0,又f1a>0,所以f(x)在區(qū)間1a,+上存在一個(gè)零點(diǎn).綜上,當(dāng)-1e<a<0時(shí),在區(qū)間(0,1),1,1a,1a,+上各存在一個(gè)零點(diǎn).13.(2021新高考,22,12分)已知函數(shù)f(x)=(x-1)·ex-ax2+b.(1)討論函數(shù)f(x)的單調(diào)性;(2)從下面兩個(gè)條件中選一個(gè),證明: f(x)有一個(gè)零點(diǎn).12<ae22,b>2a;0<a<12,b2a.解析(1)f(x)=(x-1)ex-ax2+b,f '(x)=xex-2ax=x(ex-2a).當(dāng)a0時(shí),ex

57、-2a>0對任意xR恒成立,當(dāng)x(-,0)時(shí), f '(x)<0,當(dāng)x(0,+)時(shí), f '(x)>0.因此y=f(x)在(-,0)上單調(diào)遞減,在(0,+)上單調(diào)遞增.當(dāng)a>0 時(shí),令ex-2a=0x=ln(2a).當(dāng)0<a<12時(shí),ln(2a)<0.y=f '(x)的大致圖象如圖1所示.圖1因此當(dāng)x(-,ln(2a)(0,+)時(shí), f '(x)>0,當(dāng)x(ln(2a),0)時(shí), f '(x)<0,因此f(x)在(-,ln(2a)上單調(diào)遞增,在(ln(2a),0)上單調(diào)遞減,在(0,+)上單調(diào)遞增.當(dāng)

58、a=12時(shí),ln(2a)=0,此時(shí)f '(x)0對任意xR恒成立,故f(x)在R上單調(diào)遞增.當(dāng)a>12時(shí),ln(2a)>0,y=f '(x)的大致圖象如圖2所示,圖2因此,當(dāng)x(-,0)(ln(2a),+)時(shí), f '(x)>0,當(dāng)x(0,ln(2a)時(shí),f '(x)<0,因此f(x)在(-,0)上單調(diào)遞增,在(0,ln(2a)上單調(diào)遞減,在(ln(2a),+)上單調(diào)遞增.(2)證明:選:由(1)知, f(x)在(-,0)上單調(diào)遞增,在(0,ln(2a)上單調(diào)遞減,在(ln(2a),+)上單調(diào)遞增,又f(0)=b-1>0,fba=ba1eba<0,故f(x)在(-,0上有唯一零點(diǎn).當(dāng)x(0,+)時(shí), f(x)f(ln(2a)=ln(2a)-12a-aln(2a)2+b=aln(2a)·2-ln(2a)+b-2a>aln(2a)·2-ln(2a).12<ae22,0<ln(2a)2,即f(x)>0對任意x>0恒成立.綜上, f(x)在R上有唯一零點(diǎn).選:由(1)知f(x)在(-,ln(2a)上單調(diào)遞增,在(ln(2a),0)上單調(diào)遞減,在(0,+)上單調(diào)遞增, 又f(0)=b-1<0,f2ba=2ba1e2ba-a·2ba+b=2ba1e2ba+2