函數(shù)的極值人教A版高中數(shù)學(xué)選擇性必修練習(xí)（Word解析版）

第五章一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用導(dǎo)數(shù)在研究函數(shù)中的應(yīng)用函數(shù)的極值與最大(小)值第1課時(shí)函數(shù)的極值課后篇鞏固提升基礎(chǔ)達(dá)標(biāo)練1.(2020福建高二期末)定義在區(qū)間-12,4上的函數(shù)f(x)的導(dǎo)函數(shù)f'(x)圖象如圖所示,則下列結(jié)論不正確的是(A.函數(shù)f(x)在區(qū)間(0,4)單調(diào)遞增B.函數(shù)f(x)在區(qū)間-12,0C.函數(shù)f(x)在x=1處取得極大值D.函數(shù)f(x)在x=0處取得極小值解析根據(jù)導(dǎo)函數(shù)圖象可知,f(x)在區(qū)間(-∞,0)上,f'(x)<0,f(x)單調(diào)遞減,在區(qū)間(0,+∞)上,f'(x)>0,f(x)單調(diào)遞增.所以f(x)在x=0處取得極小值,沒(méi)有極大值.所以選項(xiàng)A,B,D正確,選項(xiàng)C錯(cuò)誤.答案C2.函數(shù)y=2x2-lnx的極值點(diǎn)為(),12,-12 B.12C.12 解析y'=4x-1x,令y'=4x-1x=0,可得x=12x=-12舍去答案C3.函數(shù)f(x)=x3-ax2-bx+a2在x=1時(shí)有極值10,則實(shí)數(shù)a,b的值為()=3,b=-3或a=-4,b=11=-4,b=2或a=-4,b=11=-4,b=11D.以上都不對(duì)解析f'(x)=3x2-2ax-b,f'(1)=0,即2a+b=3①,f(1)=a2-a-b+1=10,即a2-a-b=9②,解由①②組成的方程組,得a=-4,b=11(有極值)或a=3,b=-3(無(wú)極值,舍去).答案C4.(2019廣東高二期末)已知x=1e是函數(shù)f(x)=xln(ax)+1的極值點(diǎn),則a=(A.12 C.1解析f'(x)=ln(ax)+1,由f'1e=0,得a=1又f'(x)=lnx+1,當(dāng)x>1e時(shí),f'(x)>當(dāng)0<x<1e時(shí),f'(x)<0,故x=1e是函數(shù)的極值點(diǎn),故a=1成立.故選答案B5.(多選)(2020江蘇鎮(zhèn)江中學(xué)高二期末)如圖是函數(shù)y=f(x)的導(dǎo)函數(shù)的圖象,對(duì)于下列四個(gè)判斷,其中正確的是()(x)在[-2,-1]上是增函數(shù)B.當(dāng)x=-1時(shí),f(x)取得極小值(x)在(-1,2)上是增函數(shù),在(2,4)上是減函數(shù)D.當(dāng)x=3時(shí),f(x)取得極小值解析根據(jù)圖象知當(dāng)x∈(-2,-1)或x∈(2,4)時(shí),f'(x)<0,函數(shù)單調(diào)遞減;當(dāng)x∈(-1,2)或x∈(4,+∞)時(shí),f'(x)>0,函數(shù)單調(diào)遞增.故A錯(cuò)誤;故當(dāng)x=-1時(shí),f(x)取得極小值,B正確;C正確;當(dāng)x=3時(shí),f(x)不取得極值,D錯(cuò)誤.故選BC.答案BC6.已知曲線f(x)=x3+ax2+bx+1在點(diǎn)(1,f(1))處的切線斜率為3,且x=23是y=f(x)的極值點(diǎn),則a+b=.解析∵f'(x)=3x2+2ax+b,∴f'(1解得a=2,b=-4,∴a+b=2-4=-2.答案-27.設(shè)a∈R,若函數(shù)y=ex+ax(x∈R)有大于零的極值點(diǎn),則a的取值范圍為.

解析∵y=ex+ax,∴y'=ex+a.當(dāng)a≥0時(shí),y'>0,函數(shù)y=ex+ax在R上單調(diào)遞增,沒(méi)有極值點(diǎn).當(dāng)a<0時(shí),令y'=ex+a=0,則ex=-a,即x=ln(-a).當(dāng)x∈(-∞,ln(-a))時(shí),y'<0,當(dāng)x∈(ln(-a),+∞)時(shí),y'>0,故x=ln(-a)是函數(shù)的極值點(diǎn),又ln(-a)>0,∴-a>1,即a<-1.答案(-∞,-1)8.已知f(x)=ax3+bx2+cx(a≠0)在x=±1處取得極值,且f(1)=-1.(1)試求常數(shù)a,b,c的值;(2)試判斷x=±1是函數(shù)的極大值點(diǎn)還是極小值點(diǎn),并說(shuō)明理由.解f'(x)=3ax2+2bx+c,(1)方法一:∵x=±1是函數(shù)的極值點(diǎn),∴x=±1是方程3ax2+2bx+c=0的兩根.由根與系數(shù)的關(guān)系知-又f(1)=-1,即a+b+c=-1,③由①②③解得a=12,b=0,c=-3方法二:由f'(1)=f'(-1)=0,得3a+2b+c=0,①3a-2b+c=0,②又f(1)=-1,即a+b+c=-1,③由①②③解得a=12,b=0,c=-3(2)f(x)=12x3-32∴f'(x)=32x2-32=32(當(dāng)x<-1或x>1時(shí),f'(x)>0,當(dāng)-1<x<1時(shí),f'(x)<0.∴函數(shù)f(x)在(-∞,-1)和(1,+∞)上是增函數(shù),在(-1,1)上是減函數(shù).∴當(dāng)x=-1時(shí),函數(shù)取得極大值,-1為極大值點(diǎn);當(dāng)x=1時(shí),函數(shù)取得極小值,1為極小值點(diǎn).能力提升練1.(2019天津高三)已知a>0且a≠1,則函數(shù)f(x)=(x-a)2lnx()A.有極大值,無(wú)極小值B.有極小值,無(wú)極大值C.既有極大值,又有極小值D.既無(wú)極大值,又無(wú)極小值解析∵f'(x)=2(x-a)lnx+(x-a)2·1x=(x-a)2lnx+x-ax,又g(x)=2lnx+x-ax在(0,+∞)上單調(diào)遞增,x→0,g(x)→-∞;x→+∞,g(x)→+∞,所以g(x)=2lnx+x-ax在(0,+∞)上有且僅有一個(gè)零點(diǎn),設(shè)為x0,因?yàn)閍≠1,則x0≠a,所以導(dǎo)函數(shù)f'(x答案C2.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)論中一定成立的是()A.函數(shù)f(x)有極大值f(2)和極小值f(1)B.函數(shù)f(x)有極大值f(-2)和極小值f(1)C.函數(shù)f(x)有極大值f(2)和極小值f(-2)D.函數(shù)f(x)有極大值f(-2)和極小值f(2)解析由題圖可知,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)在x=-2處取得極大值,在x=2處取得極小值.答案D3.已知函數(shù)f(x)=x3+ax2+bx-a2-7a在x=1處取得極大值10,則ab的值為(23或-23 D.解析∵f'(x)=3x2+2ax+b且f(x)在x=1處取得極大值10,∴f'(1)=3+2a+b=0,f(1)=1+a+b-a2-7a=10,∴a2+8a+12=0,∴a=-2,b=1或a=-6,b=9.當(dāng)a=-2,b=1時(shí),f'(x)=3x2-4x+1=(3x-1)(x-1).當(dāng)13<x<1時(shí),f'(x)<0,當(dāng)x>1時(shí),f'(x)>∴f(x)在x=1處取得極小值,與題意不符.當(dāng)a=-6,b=9時(shí),f'(x)=3x2-12x+9=3(x-1)(x-3);當(dāng)x<1時(shí),f'(x)>0,當(dāng)1<x<3時(shí),f'(x)<0,∴f(x)在x=1處取得極大值,符合題意;∴ab=-69=-答案A4.(多選)(2020山東高三期末)已知函數(shù)f(x)=x+sinx-xcosx的定義域?yàn)閇-2π,2π),則()(x)為奇函數(shù)(x)在[0,π)上單調(diào)遞增(x)恰有4個(gè)極大值點(diǎn)(x)有且僅有4個(gè)極值點(diǎn)解析∵f(x)的定義域?yàn)閇-2π,2π),∴f(x)是非奇非偶函數(shù),∵f(x)=x+sinx-xcosx,∴f'(x)=1+cosx-(cosx-xsinx)=1+xsinx,當(dāng)x∈[0,π)時(shí),f'(x)>0,則f(x)在[0,π)上單調(diào)遞增.顯然f'(0)≠0,令f'(x)=0,得sinx=-1x分別作出y=sinx,y=-1x在區(qū)間[-2π,2π)上的圖象由圖可知,這兩個(gè)函數(shù)的圖象在區(qū)間[-2π,2π)上共有4個(gè)公共點(diǎn),且兩圖象在這些公共點(diǎn)上都不相切,故f(x)在區(qū)間[-2π,2π)上的極值點(diǎn)的個(gè)數(shù)為4,且f(x)只有2個(gè)極大值點(diǎn).故選BD.答案BD5.若直線y=a與函數(shù)f(x)=x3-3x的圖象有相異的三個(gè)公共點(diǎn),則a的取值范圍是.

解析令f'(x)=3x2-3=0,得x=±1,則極大值為f(-1)=2,極小值為f(1)=-2.如圖,觀察圖象可知當(dāng)-2<a<2時(shí),直線y=a與f(x)=x3-3x的圖象恰有三個(gè)不同的公共點(diǎn).答案(-2,2)6.若函數(shù)f(x)=x3+x2-ax-4在區(qū)間(-1,1)上恰有一個(gè)極值點(diǎn),則實(shí)數(shù)a的取值范圍為,若恰有兩個(gè)極值點(diǎn),則實(shí)數(shù)a的取值范圍是.

解析∵f'(x)=3x2+2x-a,函數(shù)f(x)在區(qū)間(-1,1)上恰有一個(gè)極值點(diǎn),即f'(x)=0在(-1,1)內(nèi)恰有一個(gè)根.又函數(shù)f'(x)=3x2+2x-a的對(duì)稱(chēng)軸為x=-13∴應(yīng)滿(mǎn)足f'(-1)≤0,f若在(-1,1)內(nèi)恰有兩個(gè)極值點(diǎn),則應(yīng)滿(mǎn)足f∴3-2-a>0,答案[1,5)-137.設(shè)f(x)=2x3+ax2+bx+1的導(dǎo)數(shù)為f'(x),若函數(shù)y=f'(x)的圖象關(guān)于直線x=-12對(duì)稱(chēng),且f'(1)=0(1)求實(shí)數(shù)a,b的值;(2)求函數(shù)f(x)的極值.解(1)因?yàn)閒(x)=2x3+ax2+bx+1,所以f'(x)=6x2+2ax+b.從而f'(x)=6x+a6即y=f'(x)的圖象關(guān)于直線x=-a6對(duì)稱(chēng),從而由題設(shè)條件知-a6=-12,解得又因?yàn)閒'(1)=0,即6+2a+b=0,解得b=-12.所以,實(shí)數(shù)a,b的值分別為3,-12.(2)由(1)知f(x)=2x3+3x2-12x+1,f'(x)=6x2+6x-12=6(x-1)(x+2).令f'(x)=0,即6(x-1)(x+2)=0.解得x1=-2,x2=1.當(dāng)x∈(-∞,-2)時(shí),f'(x)>0,故f(x)在(-∞,-2)內(nèi)為增函數(shù);當(dāng)x∈(-2,1)時(shí),f'(x)<0,故f(x)在(-2,1)內(nèi)為減函數(shù);當(dāng)x∈(1,+∞)時(shí),f'(x)>0,故f(x)在(1,+∞)內(nèi)為增函數(shù);從而函數(shù)f(x)在x1=-2處取得極大值f(-2)=21,在x2=1處取得極小值f(1)=-6.素養(yǎng)培優(yōu)練(2020湖南高二期末)對(duì)于函數(shù)f(x)=2sinx-x,x∈[0,π],下列說(shuō)法正確的有()①f(x)在x=π3處取得極大值3②f(x)有兩個(gè)不同的零點(diǎn);③f(π)<fπ2<fπ④f(x)在[0,π]上是單調(diào)函數(shù).個(gè) 個(gè) 個(gè) 個(gè)解析∵f'(x)=2cosx-1,∴當(dāng)x∈0,π3時(shí),f'(x)>0;當(dāng)x∈π3,π時(shí),f'(x)<0,∴f(x)在0,π3上單調(diào)遞增,在π3,π上單調(diào)遞減∴f(x)在x=π3