2025版高中數(shù)學(xué)一輪復(fù)習(xí)課時(shí)作業(yè)梯級(jí)練十五導(dǎo)數(shù)與函數(shù)的極值最值課時(shí)作業(yè)理含解析新人教A版

PAGE課時(shí)作業(yè)梯級(jí)練十五導(dǎo)數(shù)與函數(shù)的極值、最值一、選擇題(每小題5分,共25分)1.下列結(jié)論錯(cuò)誤的是 ()A.函數(shù)的極大值肯定比微小值大B.導(dǎo)數(shù)等于0的點(diǎn)不肯定是函數(shù)的極值點(diǎn)C.若x0是函數(shù)y=f(x)的極值點(diǎn),則肯定有f′(x0)=0D.函數(shù)的最大值不肯定是極大值,函數(shù)的最小值也不肯定是微小值【解析】選A.對(duì)于A,如圖,在x1處的極大值比在x2處的微小值小.所以A符合題意.對(duì)于B,如y=x3在x=0處,導(dǎo)數(shù)為0,但不是極值點(diǎn),不符合題意.對(duì)于C,由極值點(diǎn)定義知明顯正確,不符合題意.對(duì)于D,如圖知正確,不符合題意.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)y=f′(x)的圖象如圖,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)的圖象可知,函數(shù)f(x)在區(qū)間(a,+∞)上是增函數(shù).3.函數(shù)f(x)=(1-x)ex有 ()A.最大值1 B.最小值1C.最大值e D.最小值e【解析】選A.f′(x)=-ex+(1-x)ex=-xex,當(dāng)x<0時(shí),f′(x)>0,當(dāng)x>0時(shí),f′(x)<0,所以f(x)在(-∞,0)上單調(diào)遞增,在(0,+∞)上單調(diào)遞減,所以f(x)有最大值為f(0)=1.4.(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.(2024·麗江模擬)設(shè)函數(shù)fQUOTE=x2+mlnQUOTE有兩個(gè)極值點(diǎn),則實(shí)數(shù)m的取值范圍是 ()A.QUOTE B.QUOTEC.QUOTE D.QUOTE【解析】選B.fQUOTE的定義域?yàn)镼UOTE.f′QUOTE=QUOTE,令其分子為gQUOTE=2x2+2x+mQUOTE,在區(qū)間QUOTE上有兩個(gè)零點(diǎn),故QUOTE解得m∈QUOTE.二、填空題(每小題5分,共15分)6.函數(shù)f(x)=QUOTEx3-4x+QUOTE的極大值是,微小值是.

【解析】f′(x)=x2-4,令f′(x)=0,解得x1=-2,x2=2.當(dāng)x改變時(shí),f(x),f′(x)的改變狀況如表:x(-∞,-2)-2(-2,2)2(2,+∞)f′(x)+0-0+f(x)單調(diào)遞增極大值單調(diào)遞減微小值單調(diào)遞增因此,當(dāng)x=-2時(shí),f(x)有極大值f(-2)=QUOTE;當(dāng)x=2時(shí),f(x)有微小值f(2)=-5.答案:QUOTE-5【加練備選·拔高】已知函數(shù)f(x)=QUOTEx3+x2-2ax+1,若函數(shù)f(x)在(1,2)上有極值,則實(shí)數(shù)a的取值范圍為.

【解析】f′(x)=x2+2x-2a的圖象是開口向上的拋物線,且對(duì)稱軸為x=-1,則f′(x)在(1,2)上是單調(diào)遞增函數(shù),因此QUOTE解得QUOTE<a<4,故實(shí)數(shù)a的取值范圍為QUOTE.答案:QUOTE7.(2024·鄭州模擬)已知函數(shù)f(x)=axcosx+QUOTE在區(qū)間QUOTE上有最大值QUOTE,則實(shí)數(shù)a=.

【解析】因?yàn)閒(x)=axcosx+QUOTE?f′(x)=a(cosx-xsinx),因?yàn)閤∈QUOTE?cosx-xsinx<0,所以當(dāng)a<0時(shí),f′(x)>0?f(x)為增函數(shù)?f(x)max=f(π)=QUOTE?a=-QUOTE,當(dāng)a>0時(shí),f′(x)<0?f(x)為減函數(shù)?f(x)max=fQUOTE=QUOTE≠Q(mào)UOTE(舍去),所以a=-QUOTE.答案:-QUOTE8.(2024·樂(lè)山模擬)設(shè)函數(shù)f(x)=[ax2-(4a+1)x+4a+3]ex.若f(x)在x=2處取得微小值,則a的取值范圍為.

【解析】f′(x)=[ax2-(2a+1)x+2]ex=(ax-1)(x-2)ex.若a>QUOTE,則當(dāng)x∈QUOTE時(shí),f′(x)<0;當(dāng)x∈(2,+∞)時(shí),f′(x)>0.所以f(x)在x=2處取得微小值.當(dāng)a≤QUOTE,則當(dāng)x∈(0,2)時(shí),x-2<0,ax-1≤QUOTEx-1<0,所以f′(x)>0.所以2不是f(x)的微小值點(diǎn).綜上可知,a的取值范圍是QUOTE.答案:QUOTE【加練備選·拔高】可導(dǎo)函數(shù)的凹凸性與其導(dǎo)數(shù)的單調(diào)性有關(guān),假如函數(shù)的導(dǎo)函數(shù)在某個(gè)區(qū)間上單調(diào)遞增,那么在這個(gè)區(qū)間上函數(shù)是向下凹的,反之則是向上凸的,曲線上凹凸性的分界點(diǎn)稱為曲線的拐點(diǎn),則函數(shù)f(x)=QUOTE-x2+1的極大值點(diǎn)為,拐點(diǎn)為.

【解析】由題意可知f′(x)=x2-2x=x(x-2),故函數(shù)f(x)在(-∞,0)上單調(diào)遞增,在(0,2)上單調(diào)遞減,在(2,+∞)上單調(diào)遞增,故其極大值在x=0處取到,所以f(x)的極大值點(diǎn)為x=0,極大值為1,又拐點(diǎn)是二階導(dǎo)數(shù)也等于零的點(diǎn),即f″(x)=2x-2=2(x-1)=0,所以x=1,f′(1)=1-2=-1,拐點(diǎn)為(1,-1).答案:x=0(1,-1)三、解答題(每小題10分,共20分)9.某商場(chǎng)銷售某種商品的閱歷表明,該商品每日的銷售量y(單位:千克)與銷售價(jià)格x(單位:元/千克)滿意關(guān)系式y(tǒng)=QUOTE+10(x-5)2,其中2<x<5,a為常數(shù).已知銷售價(jià)格為4元/千克時(shí),每日可售出該商品10.5千克.(1)求a的值;(2)若該商品的成本為2元/千克,試確定銷售價(jià)格x的值,使商場(chǎng)每日銷售該商品所獲得的利潤(rùn)最大.【解析】(1)因?yàn)閤=4時(shí),y=10.5,所以QUOTE+10=10.5,所以a=1.(2)由(1)可知,該商品每日的銷售量y=QUOTE+10(x-5)2,所以商場(chǎng)每日銷售該商品所獲得的利潤(rùn)f(x)=(x-2)QUOTE=1+10(x-2)(x-5)2,2<x<5.從而,f′(x)=10[(x-5)2+2(x-2)(x-5)]=30(x-3)(x-5).于是,當(dāng)x改變時(shí),f′(x),f(x)的改變狀況如表:x(2,3)3(3,5)f′(x)+0-f(x)單調(diào)遞增極大值41單調(diào)遞減由表可得,x=3是函數(shù)f(x)在區(qū)間(2,5)內(nèi)的極大值點(diǎn),也是最大值點(diǎn).所以當(dāng)x=3時(shí),函數(shù)f(x)取得最大值,且最大值等于41.答:當(dāng)銷售價(jià)格為3元/千克時(shí),商場(chǎng)每日銷售該商品所獲得的利潤(rùn)最大.10.(2024·貴陽(yáng)模擬)已知函數(shù)f(x)=QUOTE(a>0)的導(dǎo)函數(shù)y=f′(x)的兩個(gè)零點(diǎn)為-3和0.(1)求f(x)的單調(diào)區(qū)間;(2)若f(x)的微小值為-e3,求f(x)在區(qū)間[-5,+∞)上的最大值.【解析】(1)f′(x)=QUOTE=QUOTE.令g(x)=-ax2+(2a-b)x+b-c,因?yàn)閑x>0,所以y=f′(x)的零點(diǎn)就是g(x)=-ax2+(2a-b)x+b-c的零點(diǎn),且f′(x)與g(x)符號(hào)相同.又因?yàn)閍>0.所以當(dāng)-3<x<0時(shí),g(x)>0,即f′(x)>0,當(dāng)x<-3或x>0時(shí),g(x)<0,即f′(x)<0,所以f(x)的單調(diào)遞增區(qū)間是(-3,0),單調(diào)遞減區(qū)間是(-∞,-3),(0,+∞).(2)由(1)知,x=-3是f(x)的微小值點(diǎn),所以QUOTE解得a=1,b=5,c=5,所以f(x)=QUOTE,因?yàn)閒(x)的單調(diào)遞增區(qū)間是(-3,0),單調(diào)遞減區(qū)間是(-∞,-3),(0,+∞),所以f(0)=5為函數(shù)f(x)的極大值,故f(x)在區(qū)間[-5,+∞)上的最大值取f(-5)和f(0)中的最大者,而f(-5)=QUOTE=5e5>5=f(0),所以函數(shù)f(x)在區(qū)間[-5,+∞)上的最大值是5e5.1.(5分)設(shè)函數(shù)f(x)在R上可導(dǎo),其導(dǎo)函數(shù)為f′(x),且函數(shù)y=(1-x)f′(x)的圖象如圖所示,則下列結(jié)論中肯定成立的是 ()A.函數(shù)f(x)有極大值f(2)和微小值f(1)B.函數(shù)f(x)有極大值f(-2)和微小值f(1)C.函數(shù)f(x)有極大值f(2)和微小值f(-2)D.函數(shù)f(x)有極大值f(-2)和微小值f(2)【解析】選D.由題圖可知,當(dāng)x<-2時(shí),f′(x)>0;當(dāng)-2<x<1時(shí),f′(x)<0;當(dāng)1<x<2時(shí),f′(x)<0;當(dāng)x>2時(shí),f′(x)>0.由此可以得到函數(shù)f(x)在x=-2處取得極大值,在x=2處取得微小值.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),A.24m3C.12m3【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15m33.(5分)若函數(shù)f(x)的導(dǎo)數(shù)f′(x)=QUOTE(x-k)k,k≥1,k∈Z,已知x=k是函數(shù)f(x)的極大值點(diǎn),則k=.

【解析】因?yàn)楹瘮?shù)的導(dǎo)數(shù)為f′(x)=QUOTE(x-k)k,k≥1,k∈Z,所以若k是偶數(shù),則x=k不是極值點(diǎn),則k是奇數(shù),若k<QUOTE,由f′(x)>0,解得x>QUOTE或x<k;由f′(x)<0,解得k<x<QUOTE,即當(dāng)x=k時(shí),函數(shù)f(x)取得極大值.因?yàn)閗≥1,k∈Z,所以k=1.若k>QUOTE,由f′(x)>0,解得x>k或x<QUOTE;由f′(x)<0,解得QUOTE<x<k,即當(dāng)x=k時(shí),函數(shù)f(x)取得微小值不滿意條件.答案:14.(10分)設(shè)a>0,函數(shù)f(x)=QUOTEx2-(a+1)x+a(1+lnx).(1)求曲線y=f(x)在(2,f(2))處與直線y=-x+1垂直的切線方程;(2)求函數(shù)f(x)的極值.【解析】(1)由已知得x>0,f′(x)=x-(a+1)+QUOTE,y=f(x)在(2,f(2))處切線的斜率為1,所以f′(2)=1,即2-(a+1)+QUOTE=1,所以a=0,此時(shí)f(2)=2-2=0,故所求的切線方程為y=x-2.(2)f′(x)=x-(a+1)+QUOTE=QUOTE=QUOTE.①當(dāng)0<a<1時(shí),若x∈(0,a),f′(x)>0,函數(shù)f(x)單調(diào)遞增;若x∈(a,1),f′(x)<0,函數(shù)f(x)單調(diào)遞減;若x∈(1,+∞),f′(x)>0,函數(shù)f(x)單調(diào)遞增.此時(shí)x=a是f(x)的極大值點(diǎn),x=1是f(x)的微小值點(diǎn),函數(shù)f(x)的極大值是f(a)=-QUOTEa2+alna,微小值是f(1)=-QUOTE.②當(dāng)a=1時(shí),f′(x)=QUOTE>0,所以函數(shù)f(x)在定義域(0,+∞)內(nèi)單調(diào)遞增,此時(shí)f(x)沒(méi)有極值點(diǎn),故無(wú)極值.③當(dāng)a>1時(shí),若x∈(0,1),f′(x)>0,函數(shù)f(x)單調(diào)遞增;若x∈(1,a),f′(x)<0,函數(shù)f(x)單調(diào)遞減;若x∈(a,+∞),f′(x)>0,函數(shù)f(x)單調(diào)遞增.此時(shí)x=1是f(x)的極大值點(diǎn),x=a是f(x)的微小值點(diǎn),函數(shù)f(x)的極大值是f(1)=-QUOTE,微小值是f(a)=-QUOTEa2+alna.綜上,當(dāng)0<a<1時(shí),f(x)的極大值是-QUOTEa2+alna,微小值是-QUOTE;當(dāng)a=1時(shí),f(x)沒(méi)有極值;當(dāng)a>1時(shí),f(x)的極大值是-QUOTE,微小值是-QUOTEa2+alna.5.(10分)已知函數(shù)f(x)=ex-ax,a>0.(1)記f(x)的微小值為g(a),求g(a)的最大值;(2)若對(duì)隨意實(shí)數(shù)x,恒有f(x)≥0,求f(a)的取值范圍.【解析】(1)函數(shù)f(x)的定義域是(-∞,+∞),f′(x)=ex-a.令f′(x)=0,得x=lna,易知當(dāng)x∈(lna,+∞)時(shí),f′(x)>0,當(dāng)x∈(-∞,lna)時(shí),f′(x)<0,所以函數(shù)f(x)在x=lna處取微小值,g(a)=f(x)微小值=f(lna)=elna-alna=a-alna.g′(a)=1-(1+lna)=-lna,當(dāng)0<a<1時(shí),g′(a)>0,g(a)在(0,1)上單調(diào)遞增;當(dāng)a>1時(shí),g′(a)<0,g(a)在(1,+∞)上單調(diào)遞減.所以a=1是函數(shù)g(a)在(0,+∞)上的極大值點(diǎn),也是最大值點(diǎn),所以g(a)max=g(1)=1.(2)明顯,當(dāng)x≤0時(shí),ex-ax≥0(a>0)恒成立.當(dāng)x>0時(shí),由f(x)≥0,即ex-ax≥0,得a≤QUOTE.令h(x)=QUOTE,x∈(0,+∞),則h′(x)=QUOTE=QUOTE,當(dāng)0<x<1時(shí),h′(x)<0,當(dāng)x>1時(shí),h′(x)>0,故h(x)的最小值為h(1)=e,所以a≤e,故實(shí)數(shù)a的取值范圍是(0,e].f(a)=ea-a2,a∈(0,e],f′(a)=ea-2a,易知ea-2a≥0對(duì)a∈(0,e]恒成立,故f(a)在(0,e]上單調(diào)遞增,所以f(0)=1<f(a)≤f(e)=ee-e2,即f(a)的取值范圍是(1,ee-e2].【加練備選·拔高】(2024·廈門模擬)已知函數(shù)f(x)=QUOTE-x+alnx.(1)若f(x)在(0,+∞)上為單調(diào)函數(shù),求實(shí)數(shù)a的取值范圍;(2)若QUOTE≤a≤QUOTE,記f(x)的兩個(gè)極值點(diǎn)為x1,x2,記QUOTE的最大值與最小值分別為M,m,求M-m的值.【解析】(1)f(x)的定義域?yàn)?0,+∞),f′(x)=-QUOTE-1+QUOTE=-QUOTE.因?yàn)閒(x)為單調(diào)函數(shù),所以x2-ax+1≥0對(duì)x>0恒成立,即a≤x+QUOTE對(duì)x∈(0,+∞)恒成立,所以x+QUOTE≥2,當(dāng)且僅當(dāng)x=1時(shí)取等號(hào),所以a≤2.(2)由(1)知x1,x2是x2-ax+1=0的兩個(gè)根,從而x1+x2=a,x1x2=1,不妨設(shè)x1<x2,則t=QUOTE=QUOTE=QUOTE,0<t<1,由已知QUOTE≤a≤QUOTE,所以t為關(guān)于a的減函數(shù),所以QUOTE≤t≤QUOTE,QUOTE=-QUOTE-1+aQUOTE=-2+(x1+x2)QUOTE=-2+QUOTElnt.令h(t)=-2+QUOTElnt,則h′(t)=QUOTE.因?yàn)楫?dāng)a=2時(shí),f(x)=QUOTE-x+2lnx在(0,+∞)上為減函數(shù),所以當(dāng)t<1時(shí),f(t)=QUOTE-t+2lnt>f(1)=0,從而h′(t)<0,所以h(t)在(0,1)上為減函數(shù),所以當(dāng)QUOTE≤a≤QUOTE時(shí),M-m=hQUOTE-hQUOTE=QUOTE.(2024·全國(guó)Ⅲ卷)已知函數(shù)f(x)=2x3-ax2+b.(1)探討f(x)的單調(diào)性.(2)是否存在a,b,使得f(x)在區(qū)間[0,1]的最小值為-1且最大值為1?若存在,求出a,b的全部值;若不存在,說(shuō)明理由.【解析】(1)f′(x)=6x2-2ax=2x(3x-a).令f′(x)=0,得x=0或x=QUOTE.若a>0,則當(dāng)x∈(-∞,0)∪QUOTE時(shí),f′(x)>0;當(dāng)x∈QUOTE時(shí),f′(x)<0.故f(x)在(-∞,0),QUOTE上單調(diào)遞增,在QUOTE上單調(diào)遞減;若a=0,f(x)在(-∞,+∞)上單調(diào)遞增;若a<0,則當(dāng)x∈QUOTE∪(0,+∞)時(shí),f′(x)>0;當(dāng)x∈QUOTE時(shí),f′(x)<0.故f(x)在QUOTE,(0,+∞)上單調(diào)遞增,在QUOTE上單調(diào)遞減.(2)滿意題設(shè)條件的a,b存在.①當(dāng)a≤0時(shí),由(1)知,f(x)在[0,1]上單調(diào)遞增,所以f(x)在區(qū)間[0,1]上的最小值為f(0)=b,最大值為f(1)=2-a+b.此時(shí)a,b滿意題設(shè)條件,當(dāng)且僅當(dāng)b=-1,2-a+b=1,即a=0,b=-1.②當(dāng)a≥3時(shí),由(1)知,f(x)在[0,1]上單調(diào)遞減,所以f(x)在區(qū)間[0,1]上的最大值為f(0)=b,最小值為f(1)=2-a+b.此時(shí)a,b滿意題設(shè)條件當(dāng)且僅當(dāng)2-a+b=-1,b=1,即a=4,b=1.③當(dāng)0<a<3時(shí),由(1)知,f(x)在[0,1]上的最小值為fQUOTE=-QUOTE+b,最大值為b或2-a+b.若-QUOTE+b=-1,b=1,則a=3QUOTE,與0<a<3沖突.若-QUOTE+b=-1,2-a+b=1,則a=3QUOTE或a=-3QUOTE或a=0,與0<a<3沖突.綜上,當(dāng)且僅當(dāng)a=0,b=-1或a=4,b=1時(shí),f(x)在區(qū)間[0,1]上的最小值為-1,最大值為1.【加練備選·拔高】設(shè)函數(shù)f(x)=(x-a)(x-b)(x-c),a,b,c∈R,f′(x)為f(x)的導(dǎo)函數(shù).(1)若a≠b,b=c,且f(x)和f′(x)的零點(diǎn)均在集合{-3,1,3}中,求f(x)的微小值;(2)若a=0,0<b≤1,c=1,且f(x)的極大值