第20練導(dǎo)數(shù)與函數(shù)的極值、最值(分值：40分)一、單項(xiàng)選擇題(每小題5分,共20分)1.(2024·成都模擬)若函數(shù)f(x)=x(x+a)2在x=1處有極大值,則實(shí)數(shù)a的值為()A.1 B.-1或-3C.-1 D.-3答案D解析函數(shù)f(x)=x(x+a)2,f'(x)=(x+a)2+2x(x+a)=(x+a)(3x+a),由函數(shù)f(x)=x(x+a)2在x=1處有極大值,可得f'(1)=(1+a)(3+a)=0,解得a=-1或a=-3,當(dāng)a=-1時(shí),f'(x)=(x-1)(3x-1),當(dāng)x∈13,1時(shí),f'(x)<0,當(dāng)x∈(1,+∞)時(shí),f'(x)所以f(x)在13,1上單調(diào)遞減,在(1,+∞)上單調(diào)遞增,f(x)在x=1處有極小值,當(dāng)a=-3時(shí),f'(x)=(x-3)(3x-3),當(dāng)x∈(-∞,1)時(shí),f'(x)>0,當(dāng)x∈(1,3)時(shí),f'(x)<0,所以f(x)在(-∞,1)上單調(diào)遞增,在(1,3)上單調(diào)遞減,f(x)在x=1處有極大值,符合題意.綜上可得,a=-3.2.函數(shù)f(x)=cosx+(x+1)sinx+1在區(qū)間[0,2π]的最小值、最大值分別為()A.-π2,π2C.-π2,π2+2 D答案D解析f(x)=cosx+(x+1)sinx+1,x∈[0,2π],則f'(x)=-sinx+sinx+(x+1)·cosx=(x+1)cosx,x∈[0,2π].令f'(x)=0,解得x=-1(舍去)或x=π2或x=3π因?yàn)閒

π2=cosπ2+π2=2+πf

3π2=cos3π2+3π2+1又f(0)=cos0+(0+1)sin0+1=2,f(2π)=cos2π+(2π+1)sin2π+1=2,所以f(x)max=f

π2f(x)min=f

3π2=-3.已知函數(shù)f(x)=13x3-axlna在(0,+∞)內(nèi)既有極大值也有極小值,則實(shí)數(shù)aA.(0,1)∪1,e2e B.(0C.e2e,+答案D解析f(x)=13x3-axlna,則f'(x)=x要使函數(shù)f(x)=13x3-axlna在(0,+只需方程x2-ax=0在(0,+∞)內(nèi)有兩個(gè)不相等的實(shí)根.即lna=2lnxx,令g(x)=2lnxx則g'(x)=2(1-ln當(dāng)x∈(0,e)時(shí),g'(x)>0,當(dāng)x∈(e,+∞)時(shí),g'(x)<0,則g(x)在(0,e)上單調(diào)遞增,在(e,+∞)上單調(diào)遞減,當(dāng)x>e時(shí),g(x)>0恒成立,g(e)=2g(x)的大致圖象如圖所示.由圖可知lna∈0,∴1<a<e24.(2025·咸陽(yáng)模擬)已知函數(shù)f(x)=cosx+a2x2,若x=0是函數(shù)f(x)的唯一極小值點(diǎn),則a的取值范圍為(A.[1,+∞) B.(-1,1)C.[-1,+∞) D.(-∞,1]答案A解析f'(x)=-sinx+ax,令g(x)=f'(x)=-sinx+ax,則g'(x)=-cosx+a,當(dāng)a≥1時(shí),g'(x)=-cosx+a≥0,故g(x)在R上單調(diào)遞增,又g(0)=-sin0+0=0,故當(dāng)x>0時(shí),g(x)>0,當(dāng)x<0時(shí),g(x)<0,故f(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增,故x=0是函數(shù)f(x)的唯一極小值點(diǎn),符合題意；當(dāng)a<1時(shí),g'(0)=-cos0+a=-1+a<0,故一定存在m>0,使g(x)在(0,m)上單調(diào)遞減,此時(shí)x=0不是函數(shù)f(x)的極小值點(diǎn),故a<1時(shí)不符合題意.綜上所述,a的取值范圍為[1,+∞).二、填空題(共5分)5.(2024·浙江省金麗衢十二校聯(lián)考)已知函數(shù)f(x)=-1x,x<0,ex-2,x≥0,若x1<x2,且f(x1)=f(x2)答案2ln2+1解析設(shè)f(x1)=f(x2)=t,如圖所示,由f(x)的圖象知,t>0,且x1<0<x2,則-1x1=ex所以x1=-1t,x2=ln(t+2則x2-x1=ln(t+2)+1t令g(t)=ln(t+2)+1t(t>0)則g'(t)=1t+2-1t2所以當(dāng)t∈(0,2)時(shí),g'(t)<0,函數(shù)g(t)單調(diào)遞減；當(dāng)t∈(2,+∞)時(shí),g'(t)>0,函數(shù)g(t)單調(diào)遞增,所以當(dāng)t=2時(shí),g(t)取得最小值,為g(2)=2ln2+12,即x2-x1的最小值為2ln2+三、解答題(共15分)6.(15分)(2024·黃山模擬)已知函數(shù)f(x)=32x2-4ax+a2lnx在x=1處取得極大值(1)求實(shí)數(shù)a的值；(9分)(2)求f(x)在區(qū)間1e,e上的最大值.(6解(1)由已知f(x)的定義域?yàn)?0,+∞),f'(x)=3x-4a+a2x=3令f'(x)=0得x=a或x=a當(dāng)a=1時(shí),令f'(x)>0得,0<x<13或x>1令f'(x)<0,得13<x<1故函數(shù)f(x)在0,13上單調(diào)遞增,在13,1上單調(diào)遞減,在(1,+此時(shí)函數(shù)f(x)在x=13處取得極大值,在x=1處取得極小值,與函數(shù)f(x)在x=1當(dāng)a3=1,即a=3時(shí),令f'(x)>0,得0<x<1或x>3,令f'(x)<0,得1<x<3故函數(shù)f(x)在(0,1)上單調(diào)遞增,在(1,3)上單調(diào)遞減,在(3