第17講導(dǎo)數(shù)與函數(shù)的極值、最值1.A[解析]記y=f'(x)的圖象與x軸的三個(gè)交點(diǎn)的橫坐標(biāo)從小到大依次為x1,x2,x3,則當(dāng)x∈(0,x1)時(shí),f'(x)<0,所以f(x)在(0,x1)上單調(diào)遞減;當(dāng)x∈(x1,x2)時(shí),f'(x)>0,所以f(x)在(x1,x2)上單調(diào)遞增;當(dāng)x∈(x2,x3)時(shí),f'(x)<0,所以f(x)在(x2,x3)上單調(diào)遞減;當(dāng)x∈(x3,+∞)時(shí),f'(x)>0,所以f(x)在(x3,+∞)上單調(diào)遞增.故函數(shù)f(x)有兩個(gè)極小值點(diǎn)x1,x3,有一個(gè)極大值點(diǎn)x2,且函數(shù)f(x)有兩個(gè)單調(diào)遞減區(qū)間,故B,C,D中說法正確.不能確定函數(shù)f(x)的零點(diǎn)個(gè)數(shù),故A中說法不正確.故選A.2.C[解析]f(x)的定義域?yàn)?0,+∞),f'(x)=2xlnx+x=x(2lnx+1),令f'(x)=0,得x=1e,當(dāng)x∈0,1e時(shí),f'(x)<0,f(x)單調(diào)遞減,當(dāng)x∈1e,+∞時(shí),f'(x)>0,f(x)單調(diào)遞增,所以f(x)在x=1e處取得極小值f1e3.C[解析]由題知,函數(shù)f(x)=|x|+|x|12-cosx的定義域?yàn)镽,因?yàn)閒(-x)=|-x|+|-x|12-cos(-x)=|x|+|x|12-cosx=f(x),所以f(x)為偶函數(shù).當(dāng)x>0時(shí),f(x)=x+x12-cosx,可得f'(x)=1+12x+sinx≥1+12x-1>0,所以f(x)在(0,+∞)上單調(diào)遞增,又f(x)為偶函數(shù),所以f(x)在(-∞,0)4.A[解析]f'(x)=ex+k,因?yàn)楹瘮?shù)f(x)=ex+kx在x=0處取得極值,所以f'(0)=e0+k=0,解得k=-1.經(jīng)檢驗(yàn)k=-1滿足題意,故選A.5.A[解析]因?yàn)閒(x)=ax+lnx-a,所以f(x)的定義域?yàn)?0,+∞),f'(x)=-ax2+1x=x-ax2.當(dāng)a≤0時(shí),f'(x)>0恒成立,所以f(x)在(0,+∞)上單調(diào)遞增,f(x)不存在最小值,不滿足題意.當(dāng)a>0時(shí),令f'(x)<0,得0<x<a,此時(shí)f(x)單調(diào)遞減,令f'(x)>0,得x>a,此時(shí)f(x)單調(diào)遞增,所以當(dāng)x=a時(shí),f(x)取得最小值,所以g(a)=f(a)=1+lna-a,所以g'(a)=1a-1=1-aa,令g'(a)>0,得0<a<1,此時(shí)g(a)單調(diào)遞增,令g'(a)<0,得a>1,此時(shí)g(a)單調(diào)遞減,所以當(dāng)a=1時(shí),g(a)取得最大值,所以g6.A[解析]函數(shù)f(x)的定義域?yàn)?0,+∞).因?yàn)楹瘮?shù)f(x)=12x2-x+alnx有兩個(gè)不同的極值點(diǎn),所以f'(x)=x-1+ax=x2-x+ax=0有兩個(gè)不同的正根,即x2-x+a=0有兩個(gè)不同的正根解得0<a<14,故實(shí)數(shù)a的取值范圍為0,147.CD[解析]f'(x)=ex-2,令f'(x)=0,得x=ln2.當(dāng)x∈(-∞,ln2)時(shí),f'(x)<0,f(x)單調(diào)遞減;當(dāng)x∈(ln2,+∞)時(shí),f'(x)>0,f(x)單調(diào)遞增.故函數(shù)f(x)有極小值,也是最小值,函數(shù)f(x)沒有極大值和最大值.故選CD.8.AC[解析]由題意可得,f'(x)=4ax2(x-3),1≤x≤4.當(dāng)a=0時(shí),f(x)=b,顯然不符合題意;當(dāng)a>0時(shí),令f'(x)>0,得3<x≤4,令f'(x)<0,得1≤x<3,故f(x)在[1,3)上單調(diào)遞減,在(3,4]上單調(diào)遞增,又f(3)=b-27a,f(1)=b-3a,f(4)=b,所以f(1)<f(4),所以b-27a=-6,b=3,解得a=13,b=3,則a+b=103;當(dāng)a<0時(shí),令f'(x)>0,得1≤x<3,令f'(x)<0,得3<x≤4,故f(x)在(3,4]上單調(diào)遞減,在[1,3)上單調(diào)遞增,又f(3)=b-27a,f(1)=b-3a,f(4)=b,所以f(1)>f(4),所以9.π3-3[解析]因?yàn)閒(x)=x-2sinx,所以f'(x)=1-2cosx.當(dāng)-π<x<-π3時(shí),f'(x)>0,f(x)單調(diào)遞增;當(dāng)-π3<x<π3時(shí),f'(x)<0,f(x)單調(diào)遞減;當(dāng)π3<x<π時(shí),f'(x)>0,f(x)單調(diào)遞增.故當(dāng)x=π3時(shí),f(x)取得極小值fπ3=π310.2[解析]因?yàn)閒(x)=ae-x+bx,所以f'(x)=-ae-x+b,因?yàn)楹瘮?shù)f(x)=ae-x+bx在x=0處取得極小值1,所以f(0)=a=1,f'(0)=-a+b=0,解得a=1,b=1,所以f'(x)=-e-x+1=ex-1ex.當(dāng)x∈(-∞,0)時(shí),f'(x)<0,函數(shù)f(x)單調(diào)遞減11.[-2,1)[解析]f'(x)=3x2-3,令f'(x)=0,得x=±1.當(dāng)x∈(-∞,-1)∪(1,+∞)時(shí),f'(x)>0,當(dāng)x∈(-1,1)時(shí),f'(x)<0,所以f(x)在(-∞,-1)和(1,+∞)上單調(diào)遞增,在(-1,1)上單調(diào)遞減.若函數(shù)f(x)在(a,8-a2)上有最小值,則其最小值必為f(1),所以有1∈(a,8-a2)且f(a)=a3-3a≥f(1)=-2,解得-2≤a<1,故實(shí)數(shù)a的取值范圍為[-2,1).12.-∞,e2[解析]函數(shù)f(x)的定義域?yàn)?0,+∞),且f'(x)=ex(x-1)x2+2ax-2a=ex(x-1)x2-2a(x-1)x=(x-1)(ex-2ax)x2,依題意可得f'(x)存在唯一的變號(hào)正零點(diǎn).當(dāng)a≤0時(shí),ex-2ax>0,f'(x)有唯一的變號(hào)正零點(diǎn)1,符合題意.當(dāng)a>0,由ex-2ax=0,得exx-2a=0,令g(x)=exx(x>0),則g'(x)=(x-1)exx2,所以當(dāng)0<x<1時(shí),g'(x)<0,當(dāng)x>1時(shí),g'(x)>0,故g(x)在(0,1)上單調(diào)遞減,13.解:(1)由f(x)=13x3-4x+4,得f'(x)=x2-4,則f'(3)=5,所以所求切線方程為y-1=5(x-即5x-y-14=0.(2)f(x)的定義域?yàn)镽,由(1)知,f'(x)=(x-2)(x+2),當(dāng)x<-2或x>2時(shí),f'(x)>0,當(dāng)-2<x<2時(shí),f'(x)<0,故函數(shù)f(x)在(-∞,-2),(2,+∞)上單調(diào)遞增,在(-2,2)上單調(diào)遞減,所以當(dāng)x=-2時(shí),函數(shù)f(x)取得極大值,當(dāng)x=2時(shí),函數(shù)f(x)取得極小值.因?yàn)閒(x)在區(qū)間(a,a+5)上既有最大值又有最小值,所以a<-2<2<a+5,解得-3<a<-2,故實(shí)數(shù)a的取值范圍是(-3,-2).14.解:(1)由題意知f(x)的定義域?yàn)镽,f'(x)=a-(1+2x)e2x.令f'(x)=0,得a=(1+2x)e2x.令g(x)=(1+2x)e2x,則g'(x)=2e2x+2(1+2x)e2x=(4x+4)e2x,當(dāng)x∈(-∞,-1)時(shí),g'(x)<0,當(dāng)x∈(-1,+∞)時(shí),g'(x)>0,所以g(x)在(-∞,-1)上單調(diào)遞減,在(-1,+∞)上單調(diào)遞增,又g(-1)=-e-2,當(dāng)x<-12時(shí),g(x)<0,當(dāng)x>-12時(shí),g(x)>0,所以g(x)的大致圖象如圖所示當(dāng)a≤-e-2時(shí),g(x)≥a恒成立,即f'(x)≤0恒成立,所以f(x)在R上單調(diào)遞減,f(x)無極值點(diǎn);當(dāng)-e-2<a<0時(shí),g(x)的圖象與直線y=a有兩個(gè)不同的交點(diǎn),此時(shí)f'(x)有兩個(gè)變號(hào)零點(diǎn),則f(x)有兩個(gè)極值點(diǎn);當(dāng)a≥0時(shí),g(x)的圖象與直線y=a有且僅有一個(gè)交點(diǎn),此時(shí)f'(x)有且僅有一個(gè)變號(hào)零點(diǎn),則f(x)有且僅有一個(gè)極值點(diǎn).綜上所述,當(dāng)a≤-e-2時(shí),f(x)無極值點(diǎn);當(dāng)-e-2<a<0時(shí),f(x)有兩個(gè)極值點(diǎn);當(dāng)a≥0時(shí),f(x)有且僅有一個(gè)極值點(diǎn).(2)由題意知,當(dāng)x>0時(shí),ax+lnx+1≤xe2x=elnx·e2x=e2x+lnx恒成立.設(shè)h(x)=ex-x-1,則