2025版新高考版高考總復(fù)習(xí)數(shù)學(xué)專題四導(dǎo)數(shù)及其應(yīng)用導(dǎo)數(shù)與函數(shù)的單調(diào)性、極值和最值

2025版新高考版高考總復(fù)習(xí)數(shù)學(xué)專題四導(dǎo)數(shù)及其應(yīng)用4.2導(dǎo)數(shù)與函數(shù)的單調(diào)性、極值和最值五年高考考點(diǎn)1導(dǎo)數(shù)與函數(shù)的單調(diào)性1.(2014課標(biāo)Ⅱ文,11,5分,易)若函數(shù)f(x)=kx-lnx在區(qū)間(1,+∞)單調(diào)遞增,則k的取值范圍是()A.(-∞,-2]B.(-∞,-1]C.[2,+∞)D.[1,+∞)答案D2.(2023新課標(biāo)Ⅱ,6,5分,中)已知函數(shù)f(x)=aex-lnx在區(qū)間(1,2)單調(diào)遞增,則a的最小值為()A.e2B.eC.e-1D.e-2答案C3.(2023新課標(biāo)Ⅰ,19,12分,中)已知函數(shù)f(x)=a(ex+a)-x.(1)討論f(x)的單調(diào)性;(2)證明:當(dāng)a>0時(shí),f(x)>2lna+32解析(1)由已知得函數(shù)f(x)的定義域?yàn)镽,f'(x)=aex-1.①當(dāng)a≤0時(shí),f'(x)<0,f(x)在R上單調(diào)遞減;②當(dāng)a>0時(shí),令f'(x)=0,則x=ln1a當(dāng)x<ln1a時(shí),f'(x)<0,f(x)單調(diào)遞減當(dāng)x>ln1a時(shí),f'(x)>0,f(x)單調(diào)遞增綜上所述,當(dāng)a≤0時(shí),f(x)在R上單調(diào)遞減;當(dāng)a>0時(shí),f(x)在?∞,ln1a上單調(diào)遞減,(2)證明:由(1)知,當(dāng)a>0時(shí),f(x)在?∞,ln1a上單調(diào)遞減,在ln1a,+∞上單調(diào)遞增,則f(x)min=要證明f(x)>2lna+32,只需證明1+a2+lna>2lna+3即證a2-lna-12>0令g(x)=x2-lnx-12(x>0),則g'(x)=2x-1當(dāng)0<x<22時(shí),g'(x)<0,g(x)單調(diào)遞減當(dāng)x>22時(shí),g'(x)>0,g(x)單調(diào)遞增∴g(x)min=g22=∴g(x)>0在(0,+∞)上恒成立,即a2-lna-12>0∴f(x)>2lna+324.(2023全國(guó)甲文,20,12分,中)已知函數(shù)f(x)=ax-sinxcos2x(1)當(dāng)a=1時(shí),討論f(x)的單調(diào)性;(2)若f(x)+sinx<0,求a的取值范圍.解析(1)當(dāng)a=1時(shí),f(x)=x-sinxcos2xf'(x)=1-co=cos3所以函數(shù)f(x)在0,π2(2)令g(x)=sin=sinx則g'(x)=3cos因?yàn)閤∈0,π2,所以3cos2xsin2x+2sin4x>0,cos3x則g'(x)>0,所以函數(shù)g(x)在0,π2g(0)=0,當(dāng)x→π2時(shí),g(x)→+∞因?yàn)閒(x)+sinx<0恒成立,所以sinxco即直線y=ax在0<x<π2時(shí)恒在g(x)的圖象下方,如圖所示由圖及g'(0)=0可得a≤0,即a的取值范圍為(-∞,0].5.(2015課標(biāo)Ⅱ文,21,12分,中)已知函數(shù)f(x)=lnx+a(1-x).(1)討論f(x)的單調(diào)性;(2)當(dāng)f(x)有最大值,且最大值大于2a-2時(shí),求a的取值范圍.解析(1)f(x)的定義域?yàn)?0,+∞),f'(x)=1x-若a≤0,則f'(x)>0,所以f(x)在(0,+∞)上單調(diào)遞增.若a>0,則當(dāng)x∈0,1a時(shí),f'(x)>0;當(dāng)x∈1a,+∞時(shí),f'(x)<0.所以f(x)在0,1(2)由(1)知,當(dāng)a≤0時(shí),f(x)在(0,+∞)上無(wú)最大值;當(dāng)a>0時(shí),f(x)在x=1a處取得最大值,最大值為f1a=ln1a+因此f1a>2a-2等價(jià)于lna+a-1<0令g(a)=lna+a-1,a>0,則g(a)在(0,+∞)上單調(diào)遞增,g(1)=0.于是,當(dāng)0<a<1時(shí),g(a)<0;當(dāng)a>1時(shí),g(a)>0.因此,a的取值范圍是(0,1).考點(diǎn)2導(dǎo)數(shù)與函數(shù)的極(最)值1.(多選)(2023新課標(biāo)Ⅱ,11,5分,中)若函數(shù)f(x)=alnx+bx+cx2(a≠0)既有極大值也有極小值,A.bc>0B.ab>0C.b2+8ac>0D.ac<0答案BCD2.(多選)(2022新高考Ⅰ,10,5分,中)已知函數(shù)f(x)=x3-x+1,則()A.f(x)有兩個(gè)極值點(diǎn)B.f(x)有三個(gè)零點(diǎn)C.點(diǎn)(0,1)是曲線y=f(x)的對(duì)稱中心D.直線y=2x是曲線y=f(x)的切線答案AC3.(2021新高考Ⅰ,15,5分,中)函數(shù)f(x)=|2x-1|-2lnx的最小值為.

答案14.(2022全國(guó)乙理,16,5分,難)已知x=x1和x=x2分別是函數(shù)f(x)=2ax-ex2(a>0且a≠1)的極小值點(diǎn)和極大值點(diǎn).若x1<x2,則a的取值范圍是.

答案15.(2021北京,19,15分,中)已知函數(shù)f(x)=3?2x(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)=3?2x∴f(1)=1,f'(x)=2x?6x3,故f'(1)=-4,故曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程為y=-4(x-1)+1,即4x(2)由題意得f'(x)=2x2?6x?2a(x2+故8-2a=0,解得a=4,故f(x)=3?2xx2+4,則f'(x)=2x令f'(x)>0,得x>4或x<-1;令f'(x)<0,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)的極小值為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)=-146.(2019課標(biāo)Ⅲ文,20,12分,中)已知函數(shù)f(x)=2x3-ax2+2.(1)討論f(x)的單調(diào)性;(2)當(dāng)0<a<3時(shí),記f(x)在區(qū)間[0,1]的最大值為M,最小值為m,求M-m的取值范圍.解析(1)第一步:求函數(shù)的定義域和導(dǎo)函數(shù),并因式分解求出導(dǎo)函數(shù)的零點(diǎn).由題意知x∈R,f'(x)=6x2-2ax=2x(3x-a).令f'(x)=0,得x=0或x=a3第二步:討論a的取值,比較根的大小關(guān)系,寫出單調(diào)區(qū)間.①若a>0,則當(dāng)x∈(-∞,0)∪a3,+∞時(shí),f'(x當(dāng)x∈0,a3時(shí),f'(x)故f(x)在(-∞,0),a3,+∞單調(diào)遞增,在②若a=0,f(x)在(-∞,+∞)單調(diào)遞增;③若a<0,則當(dāng)x∈?∞,a3∪(0,+∞)時(shí),f'(當(dāng)x∈a3,0時(shí),f'(x)故f(x)在?∞,a3,(0,+∞)單調(diào)遞增,(2)當(dāng)0<a<3時(shí),由(1)知,f(x)在0,a3單調(diào)遞減,在a3,1單調(diào)遞增,所以f(x)在[0,1]的最小值為fa3=?a327+2,最大值為f(當(dāng)0<a<2時(shí),f(1)>f(0),最大值為f(1)=4-a.所以M-m=2-a+a327,0<a對(duì)于函數(shù)y=a327-a+2,y'=a29-1,當(dāng)0<a<2時(shí),y'<0,從而y=a327-a+2單調(diào)遞減,此時(shí)827<a327-a+2<2,即當(dāng)2≤a<3時(shí),f(1)<f(0),最大值為f(0)=2,所以M-m=a327,而函數(shù)y=a327單調(diào)遞增,所以M-綜上,M-m的取值范圍是827易錯(cuò)警示解題時(shí),易犯以下兩個(gè)錯(cuò)誤:①對(duì)參數(shù)a未討論或?qū)分類討論不全面,尤其易忽略a=0的情形而導(dǎo)致失分;②當(dāng)a>0時(shí),f(x)在(-∞,0),a3,+∞單調(diào)遞增,將這兩個(gè)區(qū)間合并表示為f(x)在(-∞,0)∪a37.(2023新課標(biāo)Ⅱ,22,12分,難)(1)證明:當(dāng)0<x<1時(shí),x-x2<sinx<x;(2)已知函數(shù)f(x)=cosax-ln(1-x2),若x=0是f(x)的極大值點(diǎn),求a的取值范圍.解析(1)證明:令g(x)=x-x2-sinx,0<x<1,則g'(x)=1-2x-cosx,令G(x)=g'(x),得G'(x)=-2+sinx<0在區(qū)間(0,1)上恒成立,所以g'(x)在區(qū)間(0,1)上單調(diào)遞減,因?yàn)間'(0)=0,所以g'(x)<0在區(qū)間(0,1)上恒成立,所以g(x)在區(qū)間(0,1)上單調(diào)遞減,所以g(x)<g(0)=0,即當(dāng)0<x<1時(shí),x-x2<sinx.令h(x)=sinx-x,0<x<1,則h'(x)=cosx-1<0在區(qū)間(0,1)上恒成立,所以h(x)在區(qū)間(0,1)上單調(diào)遞減,所以h(x)<h(0)=0,即當(dāng)0<x<1時(shí),sinx<x.綜上,當(dāng)0<x<1時(shí),x-x2<sinx<x.(2)函數(shù)f(x)的定義域?yàn)?-1,1).當(dāng)a=0時(shí),f(x)=1-ln(1-x2),f(x)在(-1,0)上單調(diào)遞減,在(0,1)上單調(diào)遞增,x=0不是f(x)的極大值點(diǎn),所以a≠0.當(dāng)a>0時(shí),f'(x)=-asinax+2x1?x2,x∈(-1(i)當(dāng)0<a≤2時(shí),取m=min1a,1,x∈(0,m),則ax∈(0,由(1)可得f'(x)=-asinax+2x因?yàn)閍2x2>0,2-a2≥0,1-x2>0,所以f'(x)>0,所以f(x)在(0,m)上單調(diào)遞增,不合題意.(ii)當(dāng)a>2時(shí),取x∈0,1a?(0,1),則ax∈(0,由(1)可得f'(x)=-asinax+2x1?x2<-a(ax-a2x=x1?x2(-a3x3+a2x2+a3x+2-設(shè)h(x)=-a3x3+a2x2+a3x+2-a2,x∈0,1則h'(x)=-3a3x2+2a2x+a3,因?yàn)閔'(0)=a3>0,h'1a=a3-a>0,且h'(x)的圖象是開口向下的拋物線,所以?x∈0,1a,均有h'(x)>0,所以h(x)在因?yàn)閔(0)=2-a2<0,h1a=2>0,所以h(x)在0,1當(dāng)x∈(0,n)時(shí),h(x)<0,又因?yàn)閤>0,1-x2>0.則f'(x)<x1?x2(-a3x3+a2x2+a3x+2-a2即當(dāng)x∈(0,n)?(0,1)時(shí),f'(x)<0,則f(x)在(0,n)上單調(diào)遞減.又因?yàn)閒(x)是偶函數(shù),所以f(x)在(-n,0)上單調(diào)遞增,所以x=0是f(x)的極大值點(diǎn).綜合(i)(ii)知a>2.當(dāng)a<0時(shí),由于將f(x)中的a換為-a所得解析式不變,所以a<-2符合要求.故a的取值范圍為(-∞,-2)∪(2,+∞).三年模擬綜合基礎(chǔ)練1.(2023山東煙臺(tái)開學(xué)考,3)函數(shù)f(x)=-2lnx-x-3x的單調(diào)遞增區(qū)間是()A.(0,+∞)B.(-3,1)C.(1,+∞)D.(0,1)答案D2.(2023吉林長(zhǎng)春六中月考,9)函數(shù)f(x)=cosx+(x+1)sinx+1在區(qū)間[0,2π]上的最小值、最大值分別為()A.-π2,π2C.-π2,π2+2D.?答案D3.(2024屆江蘇無(wú)錫期中,5)當(dāng)x=2時(shí),函數(shù)f(x)=x3+bx2-12x取得極值,則f(x)在區(qū)間[-4,4]上的最大值為()A.8B.12C.16D.32答案C4.(2024屆湖南師大附中第4次月考,6)已知x=0是函數(shù)f(x)=x2ex-2xex+2ex-a3x3的一個(gè)極值點(diǎn),則a的取值集合為()A.{a|a≥-1}B.{0}C.{1}D.R答案C5.(2024屆河北石家莊二中月考,5)已知函數(shù)f(x)=x3-3mx2+9mx+1在(1,+∞)上為單調(diào)遞增函數(shù),則實(shí)數(shù)m的取值范圍為()A.(-∞,-1)B.[-1,1]C.[1,3]D.[-1,3]答案D6.(2024屆重慶長(zhǎng)壽中學(xué)期中,7)已知函數(shù)f(x)=2x-2x-alnx,則“a>5”是“函數(shù)f(x)在(1,2)上單調(diào)遞減”的()A.充分不必要條件B.必要不充分條件C.充要條件D.既不充分也不必要條件答案A7.(多選)(2024屆福建福州聯(lián)考,10)設(shè)函數(shù)f(x)=x3-12x+b,則下列結(jié)論錯(cuò)誤的是()A.函數(shù)f(x)在(-∞,-1)上單調(diào)遞增B.函數(shù)f(x)在(-∞,-1)上單調(diào)遞減C.若b=-6,則函數(shù)f(x)的圖象在點(diǎn)(-2,f(-2))處的切線方程為y=10D.若b=0,則函數(shù)f(x)的圖象與直線y=10只有一個(gè)公共點(diǎn)答案ABD8.(2024屆江蘇蘇州中學(xué)模擬,14)已知函數(shù)g(x)=2x+lnx-ax在區(qū)間[1,2]上不單調(diào),則實(shí)數(shù)a的取值范圍是答案(-10,-3)9.(2024屆河南省實(shí)驗(yàn)中學(xué)月考,15)若函數(shù)f(x)=x3-12x在區(qū)間(a,a+4)上存在最大值,則實(shí)數(shù)a的取值范圍是.

答案(-6,-2)10.(2024屆湖北武漢二中測(cè)試,15)已知函數(shù)f(x)=ax4-4ax3+b,x∈[1,4],f(x)的最大值為3,最小值為-6,則a+b的值是.

答案1011.(2023重慶八中入學(xué)考,18)已知函數(shù)f(x)=ax+b+cosx(a,b∈R),若曲線f(x)在點(diǎn)(0,f(0))處的切線方程為y=12x+2(1)求f(x)的解析式;(2)求函數(shù)f(x)在[0,2π]上的值域.解析(1)因?yàn)閒(x)=ax+b+cosx(a,b∈R),所以f'(x)=a-sinx,由題意得f(0)=b+cos0=2,f'(0)=a?sin0=12,即b(2)由(1)得f(x)=12x+1+cosx,f'(x)=12-sin由f'(x)≥0且x∈[0,2π]可得0≤x≤π6或5π6≤x≤2π,函數(shù)f(x)由f'(x)<0且x∈[0,2π]可得π6<x<5π6,函數(shù)f(因此當(dāng)x=π6時(shí),函數(shù)取得極大值fπ6=12×π6+1+cosπ6=1+又f(0)=2,f(2π)=12×2π+1+cos2π=1+π+1=2+π1+5π12?32<2<1+π12+32<2+π,所以函數(shù)f(x)在[0,2π]上的最大值為2+π,最小值為1+5π12?32綜合拔高練11.(2024屆湖南長(zhǎng)沙長(zhǎng)郡中學(xué)月考,4)若0<x1<x2<1,則()A.ex2?ex1C.x2e答案C2.(多選)(2024屆廣東東莞月考,11)已知函數(shù)f(x)=ax2-2x+lnx存在極值點(diǎn),則實(shí)數(shù)a的值可以是()A.0B.-eC.1答案ABD3.(2024屆山東泰安月考,15)設(shè)a∈R,若函數(shù)y=ex+ax,x∈R有大于零的極值點(diǎn),則a的取值范圍是.

答案(-∞,-1)4.(2024屆遼寧遼東教學(xué)共同體期中,19)已知函數(shù)f(x)=ex,g(x)=ex(1)直接寫出曲線y=f(x)與曲線y=g(x)的公共點(diǎn)坐標(biāo),并求曲線y=f(x)在公共點(diǎn)處的切線方程;(2)已知直線y=a分別交曲線y=f(x)和y=g(x)于點(diǎn)A,B,當(dāng)a∈(0,e)時(shí),設(shè)△OAB的面積為S(a),其中O是坐標(biāo)原點(diǎn),求S(a)的最大值.解析(1)易得曲線y=f(x)與曲線y=g(x)的公共點(diǎn)坐標(biāo)為(1,e).因?yàn)閒'(x)=ex,所以f'(1)=e,所以曲線y=f(x)在公共點(diǎn)處的切線方程為y-e=e(x-1),即y=ex.(2)因?yàn)橹本€y=a分別交曲線y=f(x)和y=g(x)于點(diǎn)A,B,所以A(lna,a),BeaS(a)=12a·|AB|=12ae因?yàn)閍∈(0,e)時(shí),ea>1,lna<1,所以ea>ln所以S(a)=e2?12alna,a∈(求導(dǎo)得S'(a)=-12(1+lna令S'(a)=0,得a=1e,所以S'(a),S(a)的變化情況如表a0,11S'(a)+0-S(a)↗極大值↘因此,S(a)的極大值也是最大值,為S1e5.(2024屆湖南長(zhǎng)沙南雅中學(xué)開學(xué)考,21)已知函數(shù)f(x)=ax-1x-(a+1)lnx(a≠0)(1)討論函數(shù)f(x)的單調(diào)性;(2)若f(x)既有極大值又有極小值,且極大值和極小值的和為g(a),解不等式g(a)<2a-2.解析(1)函數(shù)f(x)的定義域?yàn)?0,+∞),對(duì)f(x)求導(dǎo)得f'(x)=a+1x令f'(x)=0,則x1=1,x2=1a當(dāng)a<0時(shí),ax-1<0,令f'(x)>0,解得0<x<1,令f'(x)<0,解得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)a>0時(shí),①當(dāng)1a=1,即a=1時(shí),f'(x)≥0恒成立所以f(x)在(0,+∞)上單調(diào)遞增;②當(dāng)1a>1,即0<a<1時(shí)令f'(x)>0,解得0<x<1或x>1a,令f'(x)<0,解得1<x<1所以f(x)在(0,1)上單調(diào)遞增,在1,1a上單調(diào)遞減,在1③當(dāng)1a<1,即a>1時(shí)令f'(x)>0,解得0<x<1a或x>1,令f'(x)<0,解得1a<x所以f(x)在0,1a上單調(diào)遞增,在1a,1上單調(diào)遞減,在(1,綜上所述:當(dāng)a<0時(shí),f(x)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減;當(dāng)0<a<1時(shí),f(x)在(0,1)上單調(diào)遞增,在1,1a上單調(diào)遞減,在1當(dāng)a=1時(shí),f(x)在(0,+∞)上單調(diào)遞增;當(dāng)a>1時(shí),f(x)在0,1a上單調(diào)遞增,在1a,1上單調(diào)遞減,在(1,(2)由(1)知:a>0且a≠1,且g(a)=f1a+f(1)=1-a+(a+1)lna+a-1=(a+1)lng(a)<2a-2等價(jià)于(a+1)lna<2a-2(a>0且a≠1),等價(jià)于解不等式lna-2(a?1)令m(a)=lna-2(a?1)a+1(a>0),(構(gòu)造函數(shù)m(a),結(jié)合函數(shù)的單調(diào)性以及特殊值m(1)m'(a)=1a?所以m(a)在(0,+∞)上單調(diào)遞增,且m(1)=0,所以m(a)<0=m(1),即不等式的解集為{a|0<a<1}.6.(2024屆北京一零一中學(xué)測(cè)試,18)已知函數(shù)f(x)=ax3+bx+2在x=2處取得極值-14.(1)求a,b的值;(2)求曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程;(3)求函數(shù)f(x)在[-3,3]上的最值.解析(1)因?yàn)閒(x)=ax3+bx+2,所以f'(x)=3ax2+b,又函數(shù)f(x)在x=2處取得極值-14,所以f經(jīng)檢驗(yàn),a=1,b=-12符合題意,故a=1,b=-12.(2)由(1)知:f(x)=x3-12x+2,f'(x)=3x2-12,故f(1)=-9,f'(1)=-9.所以曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程為y-(-9)=-9(x-1),即9x+y=0.(3)由(1)知:f(x)=x3-12x+2,f'(x)=3x2-12,令f'(x)=0,解得x1=-2,x2=2,x∈[-3,3]時(shí),隨x的變化f'(x),f(x)的變化情況如表:x-3(-3,-2)-2(-2,2)2(2,3)3f'(x)+0-0+f(x)11↗18↘-14↗-7由表可知:當(dāng)x=-2時(shí),函數(shù)f(x)有極大值f(-2)=18;當(dāng)x=2時(shí),函數(shù)f(x)有極小值f(2)=-14;因?yàn)閒(-2)=18>f(3)=-7,f(2)=-14<f(-3)=11,故函數(shù)f(x)在[-3,3]上的最小值為f(2)=-14,最大值為f(-2)=18.綜合拔高練21.(多選)(2024屆湖北宜昌中學(xué)階段練,12)已知函數(shù)f(x)=ax+exx+aln1x在xA.-eB.-2e答案BD2.(多選)(2024屆安徽池州一中階段練,10)已知函數(shù)f(x)=x3-2x2+ax,則下列說(shuō)法正確的是()A.函數(shù)f(x)的極值點(diǎn)個(gè)數(shù)可能為0,1,2B.若函數(shù)f(x)有兩個(gè)極值點(diǎn),則a<4C.若a=1,則函數(shù)f(x)在1D.若a=1,則函數(shù)f(x)在12答案BD3.(2024屆湖北黃岡中學(xué)月考,14)定義在R上的函數(shù)f(x)=13x3-x+3①f(x)在(0,1)上是減函數(shù),在(1,+∞)上是增函數(shù).②y=f'(x)x在(③f(x)的圖象在x=0處的切線與直線y=2x+2垂直.④設(shè)g(x)=4lnx-m,若存在x∈[1,e],使得g(x)<f'(x),則m>5-e2.以上描述中正確的是.(填序號(hào))

答案①④4.(2024屆北京海淀北大附中?？?20)已知函數(shù)f(x)=eax(x-1)2.(1)若a=1,求曲線f(x)在(0,f(0))處的切線方程;(2)求f(x)的極大值與極小值.解析(1)當(dāng)a=1時(shí),f(x)=ex(x-1)2,f'(x)=ex(x2-1),所以f'(0)=e0(02-1)=-1,又f(0)=e0(0-1)2=1,所以切線方程為y-1=-(x-0),即x+y-1=0.(2)f'(x)=aeax(x-1)2+2eax(x-1)=eax(x-1)(ax-a+2),當(dāng)a=0時(shí),令f'(x)=2(x-1)=0,解得x=1,故x<1時(shí),