2025屆高考數(shù)學(xué)一輪復(fù)習(xí)核心素養(yǎng)測(cè)評(píng)第三章3.3利用導(dǎo)數(shù)研究函數(shù)的極值最值理含解析北師大版

PAGE12-核心素養(yǎng)測(cè)評(píng)十五利用導(dǎo)數(shù)探討函數(shù)的極值、最值(30分鐘60分)一、選擇題(每小題5分,共25分)1.設(shè)函數(shù)f(x)=QUOTE+lnx則 ()A.x=QUOTE為f(x)的極大值點(diǎn)B.x=QUOTE為f(x)的微小值點(diǎn)C.x=2為f(x)的極大值點(diǎn)D.x=2為f(x)的微小值點(diǎn)【解析】選D.f′(x)=-QUOTE+QUOTE=QUOTE,由f′(x)>0,得x>2,所以f(x)的增區(qū)間為QUOTE,f(x)的減區(qū)間為(0,2),所以f(x)只有微小值,微小值點(diǎn)為x=2.2.已知函數(shù)f(x)是R上的可導(dǎo)函數(shù),f(x)的導(dǎo)函數(shù)f′(x)的圖像如圖,則下列結(jié)論正確的是 ()A.a,c分別是極大值點(diǎn)和微小值點(diǎn)B.b,c分別是極大值點(diǎn)和微小值點(diǎn)C.f(x)在區(qū)間(a,c)上是增函數(shù)D.f(x)在區(qū)間(b,c)上是減函數(shù)【解析】選C.由極值點(diǎn)的定義可知,a是微小值點(diǎn),無(wú)極大值點(diǎn);由導(dǎo)函數(shù)的圖像可知,函數(shù)f(x)在區(qū)間(a,+∞)上是增函數(shù).3.(2024·榆林模擬)已知x=2是函數(shù)f(x)=x3-3ax+2的微小值點(diǎn),那么函數(shù)f(x)的極大值為 ()A.15 B.16 C.17 D.18【解析】選D.因?yàn)閤=2是函數(shù)f(x)=x3-3ax+2的微小值點(diǎn),所以f′(2)=12-3a=0,解得a=4,所以函數(shù)f(x)的解析式為f(x)=x3-12x+2,f′(x)=3x2-12,由f′(x)=0,得x=±2,故函數(shù)f(x)在(-2,2)上是削減的,在(-∞,-2),(2,+∞)上是增加的,由此可知當(dāng)x=-2時(shí),函數(shù)f(x)取得極大值f(-2)=18.4.(2024·湘潭模擬)某蓮藕種植塘每年的固定成本是1萬(wàn)元,每年最大規(guī)模的種植是8萬(wàn)斤,每種植一斤藕,成本增加0.5元,銷售額函數(shù)是f(x)=-QUOTEx3+QUOTEax2+QUOTEx,x是蓮藕種植量,單位:萬(wàn)斤;銷售額的單位:萬(wàn)元,a是常數(shù),若種植2萬(wàn)斤,利潤(rùn)是2.5萬(wàn)元,則要使利潤(rùn)最大,每年種植蓮藕 ()A.8萬(wàn)斤 B.6萬(wàn)斤C.3萬(wàn)斤 D.5萬(wàn)斤【解析】選B.設(shè)銷售利潤(rùn)為g(x),得g(x)=-QUOTEx3+QUOTEax2+QUOTEx-1-QUOTEx=-QUOTEx3+QUOTEax2-1,當(dāng)x=2時(shí),g(2)=-QUOTE×23+QUOTEa×22-1=2.5,解得a=2.所以g(x)=-QUOTEx3+QUOTEx2-1,g′(x)=-QUOTEx2+QUOTEx=-QUOTEx(x-6),所以函數(shù)g(x)在(0,6)上單調(diào)遞增,在(6,8)上單調(diào)遞減.所以當(dāng)x=6時(shí),函數(shù)g(x)取得極大值即最大值.5.若函數(shù)f(x)=ax-lnx在區(qū)間(0,e]上的最小值為3,則實(shí)數(shù)a的值為世紀(jì)金榜導(dǎo)學(xué)號(hào)()A.e2B.2eC.QUOTED.QUOTE【解題指南】(1)推斷單調(diào)區(qū)間,把a(bǔ)分為a≤0與a>0兩種狀況來(lái)確定單調(diào)區(qū)間,而a>0時(shí)又要將QUOTE與區(qū)間(0,e]進(jìn)行比較探討;(2)依據(jù)各種狀況的單調(diào)區(qū)間確定各種狀況下的最小值,每計(jì)算一個(gè)a的值都要記得檢驗(yàn)是否滿意前提范圍.【解析】選A.因?yàn)閒(x)=ax-lnx,(x>0),所以f′(x)=a-QUOTE=QUOTE(x>0).①當(dāng)a≤0時(shí),f′(x)<0,則f(x)在(0,e]上為減函數(shù),此時(shí)f(x)min=f(e)=ae-1=3,解得a=QUOTE>0(舍去).②當(dāng)a>0時(shí),當(dāng)0<x<QUOTE時(shí),f′(x)<0,f(x)在QUOTE上為減函數(shù),當(dāng)x≥QUOTE時(shí),f′(x)≥0,f(x)在QUOTE上為增函數(shù).所以當(dāng)0<QUOTE≤e時(shí),即a≥QUOTE時(shí),x=QUOTE為f(x)在(0,e]上的微小值點(diǎn)也是最小值點(diǎn)且最小值為fQUOTE=1-lnQUOTE=3,解得a=e2.當(dāng)QUOTE>e時(shí),即a<QUOTE時(shí),f(x)在(0,e]上為減函數(shù),f(x)min=f(e)=ae-1=3,解得a=QUOTE>QUOTE(舍去),綜上所述:a=e2.二、填空題(每小題5分,共15分)6.(2024·濮陽(yáng)模擬)函數(shù)f(x)=ex-2x的最小值為________________.

【解析】f′(x)=ex-2,令f′(x)=ex-2=0,解得x=ln2.可得:函數(shù)f(x)在(-∞,ln2)上單調(diào)遞減,在(ln2,+∞)上單調(diào)遞增.所以x=ln2時(shí),函數(shù)f(x)取得微小值也是最小值,f(ln2)=2-2ln2.答案:2-2ln27.(2024·咸陽(yáng)模擬)已知y=f(x)是奇函數(shù),當(dāng)x∈(0,2)時(shí),f(x)=lnx-axQUOTE,當(dāng)x∈(-2,0)時(shí),f(x)的最小值為1,則a=________________.

【解析】由題意知,當(dāng)x∈(0,2)時(shí),f(x)的最大值為-1.令f′(x)=QUOTE-a=0,得x=QUOTE,當(dāng)0<x<QUOTE時(shí),f′(x)>0;當(dāng)x>QUOTE時(shí),f′(x)<0.所以f(x)max=fQUOTE=-lna-1=-1,解得a=1.答案:18.已知函數(shù)f(x)=QUOTE當(dāng)x∈(-∞,m]時(shí),函數(shù)f(x)的取值范圍為[-16,+∞),則實(shí)數(shù)m的取值范圍是________________. 世紀(jì)金榜導(dǎo)學(xué)號(hào)

【解析】當(dāng)x≤0時(shí),f′(x)=3(2+x)(2-x),所以當(dāng)x<-2時(shí),f′(x)<0,函數(shù)f(x)單調(diào)遞減;當(dāng)-2<x≤0時(shí),f′(x)>0,函數(shù)f(x)單調(diào)遞增,所以函數(shù)f(x)在x=-2處取最小值f(-2)=-16.畫出函數(shù)的圖像,結(jié)合函數(shù)的圖像得-2≤m≤8時(shí),函數(shù)f(x)總能取到最小值-16,故m的取值范圍是[-2,8].答案:[-2,8]三、解答題(每小題10分,共20分)9.若函數(shù)y=f(x)在x=x0處取得極大值或微小值,則稱x0為函數(shù)y=f(x)的極值點(diǎn).已知a,b是實(shí)數(shù),1和-1是函數(shù)f(x)=x3+ax2+bx的兩個(gè)極值點(diǎn).(1)求a,b的值.(2)設(shè)函數(shù)g(x)的導(dǎo)數(shù)g′(x)=f(x)+2,求g(x)的極值點(diǎn).【解析】(1)由題設(shè)知f′(x)=3x2+2ax+b,且f′(-1)=3-2a+b=0,f′(1)=3+2a+b=0,解得a=0,b=-3.(2)由(1)知f(x)=x3-3x,則g′(x)=f(x)+2=(x-1)2(x+2),所以g′(x)=0的根為x1=x2=1,x3=-2,即函數(shù)g(x)的極值點(diǎn)只可能是1或-2.當(dāng)x<-2時(shí),g′(x)<0,當(dāng)-2<x<1時(shí),g′(x)>0,當(dāng)x>1時(shí),g′(x)>0,所以-2是g(x)的極值點(diǎn),1不是g(x)的極值點(diǎn).10.已知函數(shù)f(x)=ax+lnx,其中a為常數(shù). 世紀(jì)金榜導(dǎo)學(xué)號(hào)(1)當(dāng)a=-1時(shí),求f(x)的最大值.(2)若f(x)在區(qū)間(0,e]上的最大值為-3,求a的值.【解析】(1)易知f(x)的定義域?yàn)?0,+∞),當(dāng)a=-1時(shí),f(x)=-x+lnx,f′(x)=-1+QUOTE=QUOTE,令f′(x)=0,得x=1.當(dāng)0<x<1時(shí),f′(x)>0;當(dāng)x>1時(shí),f′(x)<0.所以f(x)在(0,1)上是增函數(shù),在(1,+∞)上是減函數(shù).所以f(x)max=f(1)=-1.所以當(dāng)a=-1時(shí),函數(shù)f(x)在(0,+∞)上的最大值為-1.(2)f′(x)=a+QUOTE,x∈QUOTE,QUOTE∈QUOTE.①若a≥-QUOTE,則f′(x)≥0,從而f(x)在QUOTE上單調(diào)遞增,所以f(x)max=f(e)=ae+1≥0,不符合題意.②若a<-QUOTE,令f′(x)>0得a+QUOTE>0,結(jié)合x∈QUOTE,解得0<x<-QUOTE;令f′(x)<0得a+QUOTE<0,結(jié)合x∈QUOTE,解得-QUOTE<x≤e.從而f(x)在QUOTE上單調(diào)遞增,在QUOTE上單調(diào)遞減,所以f(x)max=fQUOTE=-1+lnQUOTE,令-1+lnQUOTE=-3,得lnQUOTE=-2,所以a=-e2,因?yàn)?e2<-QUOTE,所以a=-e2為所求,故實(shí)數(shù)a的值為-e2.(15分鐘35分)1.(5分)設(shè)函數(shù)f(x)=(x+1)ex+1,則 ()A.x=2為f(x)的極大值點(diǎn)B.x=2為f(x)的微小值點(diǎn)C.x=-2為f(x)的極大值點(diǎn)D.x=-2為f(x)的微小值點(diǎn)【解析】選D.函數(shù)f(x)=(x+1)ex+1,所以f′(x)=(x+2)ex,令(x+2)ex=0,可得x=-2,當(dāng)x<-2時(shí),f′(x)<0,函數(shù)是減函數(shù);當(dāng)x>-2時(shí),f′(x)>0,函數(shù)是增函數(shù),所以x=-2是函數(shù)的微小值點(diǎn).2.(5分)用長(zhǎng)為30m的鋼條圍成一個(gè)長(zhǎng)方體形態(tài)的框架(即12條棱長(zhǎng)總和為30m),要求長(zhǎng)方體的長(zhǎng)與寬之比為3∶2,則該長(zhǎng)方體最大體積是 ()A.24m3B.15m3C.12m3D.6m3【解析】選B.設(shè)該長(zhǎng)方體的寬是xm,由題意知,其長(zhǎng)是QUOTEm,高是QUOTE=QUOTEm(0<x<3),則該長(zhǎng)方體的體積V(x)=x·QUOTE·QUOTE=-QUOTEx3+QUOTEx2,V′(x)=-QUOTEx2+QUOTEx,由V′(x)=0,得到x=2(x=0舍去),且當(dāng)0<x<2時(shí),V′(x)>0;當(dāng)2<x<3時(shí),V′(x)<0,即體積函數(shù)V(x)在x=2處取得極大值V(2)=15,也是函數(shù)V(x)在定義域上的最大值.所以該長(zhǎng)方體體積的最大值是15m3.【變式備選】用邊長(zhǎng)為120cm的正方形鐵皮做一個(gè)無(wú)蓋水箱,先在四周分別截去一個(gè)小正方形,然后把四邊翻轉(zhuǎn)90°角,再焊接成水箱,則水箱的最大容積為 ()A.120000cm3 B.128000cm3C.150000cm3 D.158000cm3【解析】選B.設(shè)水箱底長(zhǎng)為xcm,則高為QUOTEcm.由QUOTE得0<x<120.設(shè)水箱的容積為ycm3,則有y=-QUOTEx3+60x2.求導(dǎo)數(shù),有y′=-QUOTEx2+120x.令y′=0,解得x=80(x=0舍去).當(dāng)x∈(0,80)時(shí),y′>0;當(dāng)x∈(80,120)時(shí),y′<0.因此,x=80是函數(shù)y=-QUOTEx3+60x2的極大值點(diǎn),也是最大值點(diǎn),此時(shí)y=128000.3.(5分)(2024·昆明模擬)已知函數(shù)f(x)=ax2+bx+clnx(a>0)在x=1和x=2處取得極值,且極大值為-QUOTE,則函數(shù)f(x)在區(qū)間(0,4]上的最大值為 世紀(jì)金榜導(dǎo)學(xué)號(hào)()A.0B.-QUOTEC.2ln2-4D.4ln2-4【解析】選D.函數(shù)的導(dǎo)數(shù)為f′(x)=2ax+b+QUOTE=QUOTE.因?yàn)閒(x)在x=1和x=2處取得極值,所以f′(1)=2a+b+c=0①,f′(2)=4a+b+QUOTE=0②,因?yàn)閒(x)極大值為-QUOTE,a>0,所以由函數(shù)性質(zhì)知當(dāng)x=1時(shí),函數(shù)取得極大值為-QUOTE,則f(1)=a+b+cln1=a+b=-QUOTE③,由①②③得a=QUOTE,b=-3,c=2,即f(x)=QUOTEx2-3x+2lnx,f′(x)=x-3+QUOTE=QUOTE=QUOTE,由f′(x)>0得2<x≤4或0<x<1,此時(shí)為增函數(shù),由f′(x)<0得1<x<2,此時(shí)f(x)為減函數(shù),則當(dāng)x=1時(shí),f(x)取得極大值,極大值為-QUOTE,又f(4)=8-12+2ln4=4ln2-4>-QUOTE,即函數(shù)在區(qū)間(0,4]上的最大值為4ln2-4.4.(10分)(2024·成都模擬)已知函數(shù)f(x)=alnx-x2+QUOTEx-QUOTE. 世紀(jì)金榜導(dǎo)學(xué)號(hào)(1)當(dāng)曲線f(x)在x=3時(shí)的切線與直線y=-4x+1平行,求曲線f(x)在QUOTE處的切線方程.(2)求函數(shù)f(x)的極值,并求當(dāng)f(x)有極大值且極大值為正數(shù)時(shí),實(shí)數(shù)a的取值范圍.【解析】(1)f′(x)=QUOTE-2x+a-2.由題意得f′(3)=QUOTE-2×3+a-2=-4,得a=3.當(dāng)x=1時(shí),f(1)=-12+QUOTE×1-QUOTE=-QUOTE,f′(1)=QUOTE-2×1+3-2=2,故曲線f(x)在QUOTE處的切線方程為y+QUOTE=2QUOTE,即8x-4y-17=0.(2)f′(x)=QUOTE-2x+a-2=QUOTE(x>0),①當(dāng)a≤0時(shí),f′(x)≤0,所以f(x)在QUOTE上單調(diào)遞減,f(x)無(wú)極值.②當(dāng)a>0時(shí),由f′(x)=0得x=QUOTE,隨x的改變,f′(x)、f(x)的改變狀況如下:xf′(x)+0-f(x)↗極大值↘故f(x)有極大值,無(wú)微小值,極大值為fQUOTE=alnQUOTE-QUOTE+QUOTE×QUOTE-QUOTE=alnQUOTE-a,由alnQUOTE-a>0,結(jié)合a>0可得a>2e,所以當(dāng)f(x)有極大值且極大值為正數(shù)時(shí),實(shí)數(shù)a的取值范圍是QUOTE.5.(10分)(2024·濟(jì)寧模擬)已知函數(shù)f(x)=lnx-xex+ax(a∈R). 世紀(jì)金榜導(dǎo)學(xué)號(hào)(1)若函數(shù)f(x)在[1,+∞)上單調(diào)遞減,求實(shí)數(shù)a的取值范圍.(2)若a=1,求f(x)的最大值.【解題指南】(1)由題意分別參數(shù),將原問題轉(zhuǎn)化為函數(shù)求最值的問題,然后利用導(dǎo)函數(shù)即可確定實(shí)數(shù)a的取值范圍.(2)結(jié)合函數(shù)的解析式求導(dǎo)函數(shù),將其分解因式,利用導(dǎo)函數(shù)探討函數(shù)的單調(diào)性,最終利用函數(shù)的單調(diào)性結(jié)合函數(shù)的解析式即可確定函數(shù)的最大值.【解析】(1)由題意知,f′(x)=QUOTE-(ex+xex)+a=QUOTE-(x+1)ex+a≤0在[1,+∞)上恒成立,所以a≤(x+1)ex-QUOTE在[1,+∞)上恒成立.令g(x)=-QUOTE+(x+1)ex,則g′(x)=(x+2)ex+QUOTE>0,所以g(x)在[1,+∞)上單調(diào)遞增,所以g(x)min=g(1)=2e-1,所以a≤2e-1.(2)當(dāng)a=1時(shí),f(x)=lnx-xex+x(x>0),則f′(x)=QUOTE-(x+1)ex+1=(x+1)QUOTE,令m(x)=QUOTE-ex,則m′(x)=-QUOTE-ex<0,所以m(x)在(0,+∞)上單調(diào)遞減.由于mQUOTE>0,m(1)<0,所以存在x0>0滿意m(x0)=0,即QUOTE=QUOTE.當(dāng)x∈(0,x0),m(x)>0,f′(x)>0;當(dāng)x∈(x0,+∞)時(shí),m(x)<0,f′(x)<0.所以f(x)在(0,x0)上單調(diào)遞增,在(x0,+∞)上單調(diào)遞減.所以f(x)max=f(x0)=lnx0-x0QUOTE+x0,因?yàn)镼UOTE