2025版高考數(shù)學(xué)一輪總復(fù)習(xí)3.2導(dǎo)數(shù)的應(yīng)用習(xí)題

.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(2024屆河北衡水第一中學(xué)調(diào)研一,4)已知定義在R上的函數(shù)f(x),其導(dǎo)函數(shù)f'(x)的大致圖象如圖所示,則下列敘述正確的是()A.f(b)>f(c)>f(d)B.f(b)>f(a)>f(c)C.f(c)>f(b)>f(a)D.f(c)>f(b)>f(d)答案C2.(2024屆贛州十七校期中,9)已知定義在R上的函數(shù)f(x)滿意12f(x)+f'(x)>0且有f(2)=1e,則f(x)>1exA.-∞,12C.(-∞,2)D.(2,+∞)答案D3.(2024甘肅永昌第一高級(jí)中學(xué)期末,8)已知函數(shù)f(x)=ex-2x-1(其中e為自然對(duì)數(shù)的底數(shù)),則y=f(x)圖象大致為()ABCD答案C4.(2024海南第一次聯(lián)考,12)已知函數(shù)f(x)=ax2-4ax-lnx,則f(x)在(1,4)上不單調(diào)的一個(gè)充分不必要條件是()A.a>-12B.0<a<C.a>116或-12<a<0答案D5.(2024呼和浩特一模,10)已知定義域?yàn)镽的連續(xù)函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),且滿意f'(x)m(x-3)<0,當(dāng)m<0時(shí)A.f(1)+f(3)=2f(2)B.f(0)·f(3)=0C.f(4)+f(3)<2f(2)D.f(2)+f(4)>2f(3)答案D6.(2017江蘇,11,5分)已知函數(shù)f(x)=x3-2x+ex-1ex,其中e是自然對(duì)數(shù)的底數(shù).若f(a-1)+f(2a2)≤0,則實(shí)數(shù)a的取值范圍是答案-1,考點(diǎn)二導(dǎo)數(shù)與函數(shù)的極(最)值1.(2024屆河南段考三,6)函數(shù)f(x)=xe2x-x2-x-14的極大值為(A.-12B.-12eC.0答案B2.(2024屆太原期中,10)若x=2是函數(shù)f(x)=12ax2-x-2lnx的極值點(diǎn),則函數(shù)(A.有最小值-2ln2,無(wú)最大值B.有最大值-2ln2,無(wú)最小值C.有最小值-2ln2,最大值2ln2D.無(wú)最大值,無(wú)最小值答案A3.(2024江西重點(diǎn)中學(xué)盟校聯(lián)考,6)若函數(shù)f(x)=x3+ax2+bx+1在x=1處取極值0,則a-b=()A.0B.2C.-2D.1答案A4.(2024屆貴陽(yáng)第一中學(xué)10月月考,16)已知函數(shù)y=f(x)(x>0)滿意xf'(x)-f(x)=x3ex(其中e為自然對(duì)數(shù)的底數(shù)),且f(1)=0.當(dāng)x=時(shí),f(x)取到微小值.

答案55.(2024鄭州二模,14)已知f(x)=(x2+2x+a)·ex,若f(x)存在微小值,則a的取值范圍是.

答案(-∞,2)6.(2024新高考Ⅰ,15,5分)函數(shù)f(x)=|2x-1|-2lnx的最小值為.

答案17.(2024四川南充二模,16)設(shè)函數(shù)f(x)=x+e|x|e|x|的最大值為M,最小值為N,下述四個(gè)結(jié)論:①M(fèi)+N=4;②M-N=2e;③MN=1-1答案②③8.(2024北京,19)已知函數(shù)f(x)=3-2x(1)若a=0,求曲線y=f(x)在(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.解析(1)當(dāng)a=0時(shí),f(x)=3-2xx2,∴f(1)=1,又f'(x)=2x-6x3,故f'(1)=-4,故曲線(2)由題意得f'(x)=2x2-6x-2a(x2+a)2,且f'(-1)=0,故8-2a=0,解得令f'(x)>0,解得x>4或x<-1;令f'(x)<0,解得-1<x<4,故函數(shù)f(x)的單調(diào)增區(qū)間為(-∞,-1)和(4,+∞),單調(diào)減區(qū)間為(-1,4).所以f(x)的極大值為f(-1)=1,f(x)的微小值為f(4)=-14.又當(dāng)x∈(-∞,-1)時(shí),3-2x>0,故f(x)>0;當(dāng)x∈(4,+∞)時(shí),3-2x<0,故f(x)<0,∴f(x)max=f(-1)=1,f(x)min=f(4)=-1考點(diǎn)三導(dǎo)數(shù)的綜合應(yīng)用1.(2024屆太原期中,11)已知f(x)=12ax2-ex+(a+1)x,對(duì)隨意x1,x2∈(0,+∞)都有f(x1)-f(A.(-∞,0]B.(0,1)C.(-∞,1]D.[1,+∞)答案C2.(2024寧夏固原第五中學(xué)期末,12)已知函數(shù)f(x)=lnx-ax(x∈[1,+∞)),若不等式f(x)≤0恒成立,則實(shí)數(shù)a的取值范圍為()A.[1,+∞)B.-∞,C.1e,+∞答案C3.(2024江西新余二模,12)若對(duì)于隨意的0<x1<x2<a,都有x2lnx1-x1lnA.1B.eC.1eD.答案C4.(2024屆成都蓉城名校聯(lián)盟聯(lián)考一,16)已知b>a>1,logab-3logba=m(m為常數(shù)),bea的最大值為27e3,答案25.(2024東北三省三校一模,16)已知函數(shù)f(x)=ax+xlnx,g(x)=x3-x2-5,若對(duì)于隨意的x1,x2∈12,2,都有f(x1)-g(x2)≥2成立,則實(shí)數(shù)a答案[1,+∞)6.(2024新高考Ⅰ,21,12分)已知函數(shù)f(x)=aex-1-lnx+lna.(1)當(dāng)a=e時(shí),求曲線y=f(x)在點(diǎn)(1,f(1))處的切線與兩坐標(biāo)軸圍成的三角形的面積;(2)若f(x)≥1,求a的取值范圍.解析f(x)的定義域?yàn)?0,+∞),f'(x)=aex-1-1x(1)當(dāng)a=e時(shí),f(x)=ex-lnx+1,f'(1)=e-1,曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程為y-(e+1)=(e-1)·(x-1),即y=(e-1)x+2.直線y=(e-1)x+2在x軸,y軸上的截距分別為-2e-1,2.因此所求三角形的面積為(2)當(dāng)0<a<1時(shí),f(1)=a+lna<1.當(dāng)a=1時(shí),f(x)=ex-1-lnx,f'(x)=ex-1-1x.當(dāng)x∈(0,1)時(shí),f'(x)<0;當(dāng)x∈(1,+∞)時(shí),f'(x)>0.所以當(dāng)x=1時(shí),f(x)取得最小值,最小值為f(1)=1,從而f(x)≥1.當(dāng)a>1時(shí),f(x)=aex-1-lnx+lna≥ex-1-lnx≥綜上,a的取值范圍是[1,+∞).綜合篇知能轉(zhuǎn)換考法一利用導(dǎo)數(shù)探討函數(shù)的單調(diào)性1.(2024安徽池州期末,10)已知函數(shù)f(x)的定義域?yàn)镽,其導(dǎo)函數(shù)為f'(x),且3f(x)-f'(x)>0在R上恒成立,則下列不等式肯定成立的是()A.f(1)<e3f(0)B.f(1)>e3f(0)C.f(1)<e2f(0)D.f(1)>e2f(0)答案A2.(2024黑龍江大慶二模,12)已知函數(shù)f(x)=ex-e-x,g(x)=cosx+12x2-ax.對(duì)于隨意x1,x2∈-π2,π2且x1≠x2,都有f(A.π2B.π2+1C.1-π答案C3.(2024屆鄭州外國(guó)語(yǔ)中學(xué)調(diào)研二,10)a=3×(2-ln3)e2,b=1e,c=ln33,則a,b,c的大小依次A.a<c<bB.c<a<bC.a<b<cD.b<a<c答案A4.(2024屆贛州十七校期中,19)已知函數(shù)f(x)=13x3-32ax2+(2a2+a-1)x(a∈(1)當(dāng)a=-1時(shí),求f(x)的極值;(2)探討f(x)的單調(diào)性.解析(1)當(dāng)a=-1時(shí),f(x)=13x3+32x2,f'(x)=x2+3x=x(x+3),令f'(x)=0,得x=0或x=-3,令f'(x)<0,得-3<x<0,令f'(x)>0,得x<-3或x>0,所以f(x)在(-3,0)上為減函數(shù),在(-∞,-3)和(0,+∞)上為增函數(shù),所以f(x)的極大值為f(-3)=92,f(x)的(2)f'(x)=x2-3ax+(2a2+a-1)=[x-(2a-1)][x-(a+1)].令f'(x)=0,解得x=2a-1或x=a+1.(i)當(dāng)2a-1=a+1,即a=2時(shí),f'(x)≥0恒成立,則函數(shù)f(x)在R上單調(diào)遞增.(ii)當(dāng)2a-1>a+1,即a>2時(shí),列表得:x(-∞,a+1)a+1(a+1,2a-1)2a-1(2a-1,+∞)f'(x)+0-0+f(x)增函數(shù)極大值減函數(shù)微小值增函數(shù)所以f(x)在(-∞,a+1)和(2a-1,+∞)上單調(diào)遞增,在(a+1,2a-1)上單調(diào)遞減.(iii)當(dāng)2a-1<a+1,即a<2時(shí),列表得:x(-∞,2a-1)2a-1(2a-1,a+1)a+1(a+1,+∞)f'(x)+0-0+f(x)增函數(shù)極大值減函數(shù)微小值增函數(shù)所以f(x)在(-∞,2a-1)和(a+1,+∞)上單調(diào)遞增,在(2a-1,a+1)上單調(diào)遞減.綜上所述,當(dāng)a=2時(shí),f(x)在R上單調(diào)遞增;當(dāng)a>2時(shí),f(x)在(-∞,a+1)和(2a-1,+∞)上單調(diào)遞增,在(a+1,2a-1)上單調(diào)遞減;當(dāng)a<2時(shí),f(x)在(-∞,2a-1)和(a+1,+∞)上單調(diào)遞增,在(2a-1,a+1)上單調(diào)遞減.5.(2024全國(guó)甲,21,12分)已知a>0且a≠1,函數(shù)f(x)=xa(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.解析(1)當(dāng)a=2時(shí),f(x)=x22x=x2·2-x(x>0),則f'(x)=2x·2-x+x2·(-ln2·2-x)=x·2-x(2-xln2).令f'(x)=0,則x=2ln2,f'(x),f(x)x0,22f'(x)+0-f(x)↗極大值↘所以函數(shù)f(x)的單調(diào)遞增區(qū)間是0,2ln2,單調(diào)遞減區(qū)間是(2)令f(x)=1,則xaax=1,所以xa=ax,兩邊同時(shí)取對(duì)數(shù),可得alnx=xlna,即lnxx=lnaa,依據(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)=1-lnxx2,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lnaa∈0,考法二利用導(dǎo)數(shù)探討函數(shù)的極(最)值1.(2024東北三省四市聯(lián)考,12)已知函數(shù)f(x)=ex-3,g(x)=12+lnx2,若f(m)=g(n)成立,則m-n的最大值為(A.1-ln2B.ln2C.2ln2D.ln2-1答案A2.(2024屆四川遂寧零診,16)已知函數(shù)f(x)=exx2+k(2lnx-x)和g(x)=exx2,若g(x)的微小值點(diǎn)是f(x)的唯一極值點(diǎn)答案e3.(2024四川南充重點(diǎn)中學(xué)月考,16)已知函數(shù)f(x)=4x2-72-x,函數(shù)g(x)=x3-3a2x-2a(a≥1),若對(duì)隨意x1∈[0,1],總存在x0∈[0,1],使得g(x0)=f(x1)成立,答案1,4.(2024課標(biāo)Ⅰ,21,12分)已知函數(shù)f(x)=1x(1)探討f(x)的單調(diào)性;(2)若f(x)存在兩個(gè)極值點(diǎn)x1,x2,證明:f(解析(1)f(x)的定義域?yàn)?0,+∞),f'(x)=-1x2-1+ax(i)若a≤2,則f'(x)≤0,當(dāng)且僅當(dāng)a=2,x=1時(shí),f'(x)=0,所以f(x)在(0,+∞)上單調(diào)遞減.(ii)若a>2,令f'(x)=0,得x=a-a2-4當(dāng)x∈0,a-a2當(dāng)x∈a-a所以f(x)在0,a-a2-42,a(2)證明:由(1)知,f(x)存在兩個(gè)極值點(diǎn)當(dāng)且僅當(dāng)a>2.由于f(x)的兩個(gè)極值點(diǎn)x1,x2滿意x2-ax+1=0,所以x1x2=1,不妨設(shè)x1<x2,則x2>1,由于f(x1)-f(x2)所以f(x1)-f(x2設(shè)函數(shù)g(x)=1x-x+2lnx,由(1)知,g(x)在(0,+∞)上單調(diào)遞減,又g(1)=0,從而當(dāng)x∈(1,+∞)時(shí)所以1x2-x2+2lnx2<0,即5.(2024屆湘豫名校聯(lián)盟11月聯(lián)考,21)已知函數(shù)f(x)=aex-(x+1)lnx+1a,其中a>0,且a∈(1)當(dāng)a=1時(shí),證明:當(dāng)x>0時(shí),f(x)>1;(2)已知函數(shù)f(x)有兩個(gè)極值點(diǎn),求實(shí)數(shù)a的取值范圍.解析(1)證明:當(dāng)a=1時(shí),f(x)=ex-(x+1)ln(x+1)(x>-1).f'(x)=ex-ln(x+1)-1,f″(x)=ex-1x+1,在(-1,+∞)單調(diào)遞增且f″(0)=0.所以當(dāng)x>0時(shí),f″(x)>0,f'(x)單調(diào)遞增,當(dāng)-1<x<0時(shí),f″(x)<0,f'(x)單調(diào)遞減,所以f'(x)≥f'(0)=0,∴函數(shù)f(x)在(-1,+∞)單調(diào)遞增,又f(0)=1,所以當(dāng)x>0(2)f'(x)=aex-ln(x+1)+lna-1,而f″(x)=aex-1x+1,在(-1,+∞)上單調(diào)遞增,且當(dāng)x→-1時(shí),f″(x)→-∞,當(dāng)x→+∞時(shí),f″(x)→+∞,所以函數(shù)f″(x)有唯一零點(diǎn)x0,即aex0-1x0+1當(dāng)-1<x<x0時(shí),f″(x)<0;當(dāng)x>x0時(shí),f″(x)>0,即函數(shù)f'(x)在(-1,x0)上單調(diào)遞減,在(x0,+∞)上單調(diào)遞增,所以函數(shù)f'(x)的微小值為f'(x0),f'(x0)=aex0-ln(x0+1)+lna-1=1x0+1設(shè)函數(shù)h(x)=1x易知函數(shù)h(x)在(-1,+∞)上單調(diào)遞減,且h(0)=0,當(dāng)-1<x0≤0時(shí),f'(x0)≥0,那么函數(shù)f(x)在(-1,+∞)上單調(diào)遞增,此時(shí)函數(shù)f(x)無(wú)極值;當(dāng)x0>0時(shí),f'(x0)<0,當(dāng)x→-1時(shí),f'(x)→+∞;當(dāng)x→+∞時(shí),f'(x)→+∞.所以f'(x)在(-1,x0)和(x0+∞)內(nèi)各有一個(gè)零點(diǎn),即函數(shù)f(x)在(-1,x0)和(x0+∞)內(nèi)各有一個(gè)極值,合計(jì)兩個(gè)極值,此時(shí),易知1a=(x0+1)ex0綜上所述,當(dāng)且僅當(dāng)0<a<1時(shí),函數(shù)f(x)有兩個(gè)極值點(diǎn).考法三利用導(dǎo)數(shù)探討不等式1.(2024屆安徽淮南一中月考三,12)若a>0,b>0,且ln(2a)+lnb≥a2+b2-1,則a+b=()A.2B.3C.322答案A2.(2024屆湘豫名校聯(lián)盟11月聯(lián)考,16)若關(guān)于x的不等式ex+1+a>aln(ax)對(duì)于隨意x∈(0,+∞)恒成立,則實(shí)數(shù)a的取值范圍是.

答案(0,e2)3.(2024陜西榆林一模,16)若0<x1<x2<1,則下面不等式正確的是.

①x1lnx1<x2lnx2;②x2lnx1<x1lnx2;③x1ex1>x2ex2;④x2ex1>x1ex2;⑤答案②④4.(2024屆山東省試驗(yàn)中學(xué)期中,22)已知函數(shù)f(x)=lnx-ax+1(a∈R).(1)函數(shù)f(x)≤0在定義域內(nèi)恒成立,求實(shí)數(shù)a的取值范圍;(2)求證:當(dāng)n∈N*,n≥2時(shí),1+1221+1(3)已知f(x)有兩個(gè)不同的零點(diǎn)x1,x2,求證:x1x2<1a解析(1)函數(shù)定義域?yàn)?0,+∞),f'(x)=1x-a=1-axx.當(dāng)a≤0時(shí),f(1)=1-a>0,不滿意題設(shè);當(dāng)a>0時(shí),令f'(x)=0,得x=1a,所以在0,1a上,f'(x)>0,f(x)單調(diào)遞增,在1a,+∞上,f'(x)<0,f(x)單調(diào)遞減,所以f(x)max=f1a=ln1a(2)證明:由(1)得,當(dāng)a=1時(shí),lnx≤x-1,當(dāng)且僅當(dāng)x=1時(shí)等號(hào)成立,所以ln1+1n2<1n2,結(jié)合對(duì)數(shù)的運(yùn)算法則可得ln1+1221+132…1+1n2=ln1+122+ln1+132+…+ln1+1n2<122+所以1+1221+1(3)證明:由題意得lnx1-ax1+1=0①,lnx2-ax2+1=0②,②-①得lnx2x1-a(x2-x1)=0,即a=lnx2x1x2-x1,故要證明x1x2<1a2,即證明x1x2<(x2-x1)2lnx2x12,即證明lnx2x12<(x2-x1)2x1x2=x2x1-2+x1x2,不妨設(shè)0<x1<x2,令g(t)=(lnt)2-t-1t+2(t>1),則g'(t)=2lntt-1+1t2=15.(2024全國(guó)乙,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(解析(1)由題意得y=xf(x)=xln(a-x),x∈(-∞,a),∴y'=ln(a-x)+x·1a-x·(-1)=ln(a-x)-xa-x,x∈(-∞,a),∵x=0是函數(shù)y=xf(x)的極值點(diǎn),∴l(xiāng)n(a-0)-0a-0=0,可得a=1.當(dāng)a=1時(shí),y'=ln(1-x)-x1-x,x∈(-∞,1),令p(x)=ln(1-x)-x1-x,則p'(x)=1x-1-1(1-x)2=x-2(x-1)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;(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=-ln(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)6.(2024課標(biāo)Ⅰ,21,12分)已知函數(shù)f(x)=ex+ax2-x.(1)當(dāng)a=1時(shí),探討f(x)的單調(diào)性;(2)當(dāng)x≥0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)<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12x3-ax2+x+1e-x≤則g'(x)=-12x=-12x[x2-(2a+3)x+4a+2]e=-12x(x-2a-1)(x-2)e-x(i)若2a+1≤0,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+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-2≤1,即a≥7-e24.所以當(dāng)7-e2(iii)若2a+1≥2,即a≥12,則g(x)≤12x由于0∈7-e故由(ii)可得12x3+x+1故當(dāng)a≥12時(shí),g(x)≤綜上,a的取值范圍是7-e7.(2024天津,20,16分)已知函數(shù)f(x)=x3+klnx(k∈R),f'(x)為f(x)的導(dǎo)函數(shù).(1)當(dāng)k=6時(shí),(i)求曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程;(ii)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)隨意的x1,x2∈[1,+∞),且x1>x2,有f'(x1解析(1)(i)當(dāng)k=6時(shí),f(x)=x3+6lnx,故f'(x)=3x2+6x.可得f(1)=1,f'(1)=9,所以曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程為y-1=9(x-1),即(ii)依題意,g(x)=x3-3x2+6lnx+3x,x∈(0,+∞).從而可得g'(x)=3x2-6x+6x-3x2,整理可得g'(x)=3(x-1)3(x+1)x2.令g'(x)=0,x(0,1)1(1,+∞)g'(x)-0+g(x)↘微小值↗所以,函數(shù)g(x)的單調(diào)遞減區(qū)間為(0,1),單調(diào)遞增區(qū)間為(1,+∞);g(x)的微小值為g(1)=1,無(wú)極大值.(2)證明:由f(x)=x3+klnx,得f'(x)=3x2+kx.對(duì)隨意的x1,x2∈[1,+∞),且x1>x2,令x1x2=t(t>1),則(x1-x2)[f'(x1)+f'(x2)]-2[f(x1)-f(x2)]=(x1-x2)3x12+kx1+3x22+kx2-2x13-x23+klnx1x2=x1令h(x)=x-1x-2lnx,x∈[1,+∞).當(dāng)x>1時(shí),h'(x)=1+1x2-2x=1-1x2>0,由此可得h(x)在[1,+∞)單調(diào)遞增,所以當(dāng)因?yàn)閤2≥1,t3-3t2+3t-1=(t-1)3>0,k≥-3,所以x23(t3-3t2+3t-1)+kt-1t-2lnt>(t3-3t2+3t-1)-3t由(1)(ii)可知,當(dāng)t>1時(shí),g(t)>g(1),即t3-3t2+6lnt+3t>1,故t3-3t2+6lnt+3由①②③可得(x1-x2)[f'(x1)+f'(x2)]-2[f(x1)-f(x2)]>0.所以,當(dāng)k≥-3時(shí),對(duì)隨意的x1,x2∈[1,+∞),且x1>x2,有f'(x1考法四利用導(dǎo)數(shù)探討函數(shù)的零點(diǎn)或方程的根1.(2024江西宜春中學(xué)等三校聯(lián)考一,12)已知函數(shù)f(x)=ex-x+1x-1,對(duì)于函數(shù)①函數(shù)f(x)在其定義域上為增函數(shù);②對(duì)于隨意的a<1,都有f(a)>-1成立;③f(x)有且僅有兩個(gè)零點(diǎn);④若曲線y=ex在點(diǎn)(x0,ex0)(x0≠1)處的切線也是曲線y=lnx的切線,則x0必是f(x)其中全部正確的結(jié)論序號(hào)是()A.①②③B.②③C.②④D.②③④答案D2.(2024屆云南師大附中月考五,12)已知函數(shù)f(x)=ax-x2+lnx-xex-e,若函數(shù)f(x)有兩個(gè)不同的零點(diǎn),則實(shí)數(shù)a的取值范圍是()A.(1,+∞)B.(1+e,+∞)C.(1+2e,+∞)D.(1+3e,+∞)答案C3.(2024屆山西長(zhǎng)治其次中學(xué)月考,22)已知函數(shù)f(x)=lnex2-ax,g(x)=(1)求函數(shù)f(x)的極值點(diǎn);(2)當(dāng)a>0時(shí),函數(shù)h(x)=f(x)-g(x)恰有三個(gè)不同的零點(diǎn),求實(shí)數(shù)a的取值范圍.解析(1)因?yàn)閒(x)=lnex2-ax=ln所以f'(x)=1x-a=1-當(dāng)a≤0時(shí),f'(x)>0,所以函數(shù)f(x)無(wú)極值點(diǎn);當(dāng)a>0時(shí),令f'(x)=0,解得x=1a當(dāng)0<x<1a時(shí),f'(x)>0,f(x)單調(diào)遞增當(dāng)x>1a時(shí),f'(x)<0,f(x)單調(diào)遞減,故函數(shù)f(x)有極大值點(diǎn)1a,無(wú)微小綜上,當(dāng)a≤0時(shí),函數(shù)f(x)無(wú)極值點(diǎn),當(dāng)a>0時(shí),函數(shù)f(x)有極大值點(diǎn)1a,無(wú)微小值點(diǎn)(2)當(dāng)a>0時(shí),h(x)=f(x)-g(x)=lnx2-ax+4a所以h'(x)=1x-a-4ax設(shè)k(x)=-ax2+x-4a,則Δ=1-16a2,①當(dāng)Δ≤0,a>0,即a≥14時(shí),h'(x)≤0,h(x)在(0,+∞)單調(diào)遞減②當(dāng)Δ>0,a>0,x1=1-1-16a22a,x2=1+1-16a又因?yàn)閗(x)=-ax2+x-4a的圖象開口向下,所以當(dāng)0<x<x1時(shí),k(x)<0,即h'(x)<0,所以h(x)在(0,x1)上單調(diào)遞減;當(dāng)x1<x<x2時(shí),k(x)>0,即h'(x)>0,所以h(x)在(x1,x2)上單調(diào)遞增;當(dāng)x>x2時(shí),k(x)<0,即h'(x)<0,所以h(x)在(x2,+∞)上單調(diào)遞減.因?yàn)閔(2)=ln1-2a+4a2=0,又x1x2=4,所以x1<2<x∴h(x1)<h(2)=0<h(x2).∵h(yuǎn)1a2=ln12a2-a·1a2令m(a)=-ln2-2lna-1a+4a30則m'(a)=-2a+1a2+12a2=12a4-2a+1a2>1-2aa2>0,所以m(a)在0,14單調(diào)遞增,所以m(a)<m14=-ln2-2ln14-4+4×143=3ln2-4+116<0,即h1a2<0,由零點(diǎn)存在性定理知,h(x)在區(qū)間x2,1a2上有唯一的零點(diǎn)x0,∵h(yuǎn)(x0)+h4x0=lnx02-ax0+4ax0+ln12·4x0-a·4x0+4.(2024課標(biāo)Ⅰ,20,12分)已知函數(shù)f(x)=sinx-ln(1+x),f'(x)為f(x)的導(dǎo)數(shù).證明:(1)f'(x)在區(qū)間-1,π2(2)f(x)有且僅有2個(gè)零點(diǎn).證明(1)設(shè)g(x)=f'(x),則g(x)=cosx-11+x,g'(x)=-sinx+當(dāng)x∈-1,π2時(shí),g'(x)單調(diào)遞減,而g'(0)>0,g'π2<0,可得g'(x)在-1,π2有唯一零點(diǎn),設(shè)為α.則當(dāng)x∈(-1,α)時(shí),g'(x)>0;當(dāng)所以g(x)在(-1,α)單調(diào)遞增,在α,π2單調(diào)遞減,故g(x)在-1,π2存在唯一極大值點(diǎn),即(2)f(x)的定義域?yàn)?-1,+∞).(i)當(dāng)x∈(-1,0]時(shí),由(1)知,f'(x)在(-1,0)單調(diào)遞增,而f'(0)=0,所以當(dāng)x∈(-1,0)時(shí),f'(x)<0,故f(x)在(-1,0)單調(diào)遞減.又f(0)=0,從而x=0是f(x)在(-1,0]的唯一零點(diǎn).(ii)當(dāng)x∈0,π2時(shí),由(1)知,f'(x)在(0,α)單調(diào)遞增,在α,π2單調(diào)遞減,而f'(0)=0,f'π2<0,所以存在β∈α,π2,使得f'(β)=0,且當(dāng)x∈(0,β)時(shí),f'(x)>0;當(dāng)x∈β,π2時(shí),f'(x)<0.故f(x)在(0,β)單調(diào)遞增,在β,π2單調(diào)遞減.又f(0)=0,fπ(iii)當(dāng)x∈π2,π時(shí),f'(x)<0,所以f(x)在π2,π單調(diào)遞減.而fπ2(iv)當(dāng)x∈(π,+∞)時(shí),ln(x+1)>1,所以f(x)<0,從而f(x)在(π,+∞)沒有零點(diǎn).綜上,f(x)有且僅有2個(gè)零點(diǎn).5.(2024新高考Ⅱ,22)已知函數(shù)f(x)=(x-1)·ex-ax2+b.(1)探討函數(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).①12<a≤e22,b>2a;②0<a<1解析(1)f(x)=(x-1)ex-ax2+b.f'(x)=xex-2ax=x(ex-2a).①當(dāng)a≤0時(shí),ex-2a>0對(duì)隨意x∈R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.②當(dāng)a>0時(shí),令ex-2a=0?x=ln(2a).圖1當(dāng)0<a<12時(shí),ln(2a)<0.y=f'(x)的大致圖象如圖1所示.因此當(dāng)x∈(-∞,ln(2a))∪(0,+∞)時(shí)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)a=12時(shí),ln(2a)=0,此時(shí)f'(x)≥0對(duì)隨意x∈R恒成立,故f(x)在R上單調(diào)遞增.當(dāng)a>12圖2y=f'(x)的大致圖象如圖2所示,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①.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,f-ba=-ba-1e-ba<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)-1]2a-a[ln(2a)]2+b=aln(2a)·[2-ln(2a)]+b-2a>aln(2a)·[2-ln(2a)].∵12<a≤e綜上,f(x)在R上有唯一零點(diǎn).選②.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,f(x)=(x-1)ex-ax2+b?f(0)=b-1<0,又f2-ba=2-ba=2-ba-1e2-∴f(x)在[0,+∞)上有唯一零點(diǎn).在x∈(-∞,0)上,f(x)≤f(ln(2a))=aln(2a)·[2-ln(2a)]+b-2a<0,即f(x)<0對(duì)隨意x<0恒成立.綜上,f(x)在R上有唯一零點(diǎn).6.(2024課標(biāo)Ⅲ,21,12分)設(shè)函數(shù)f(x)=x3+bx+c,曲線y=f(x)在點(diǎn)12,f1(1)求b;(2)若f(x)有一個(gè)肯定值不大于1的零點(diǎn),證明:f(x)全