新教材適用2023-2024學(xué)年高中數(shù)學(xué)第二章導(dǎo)數(shù)及其應(yīng)用習(xí)題課用導(dǎo)數(shù)研究函數(shù)的單調(diào)性極值最值課后訓(xùn)練北師大版選擇性必修第二冊(cè)

習(xí)題課——用導(dǎo)數(shù)研究函數(shù)的單調(diào)性、極值、最值課后訓(xùn)練鞏固提升1.函數(shù)y=12x2-lnx的單調(diào)遞減區(qū)間為()A.(-1,1] B.(0,1)C.[1,+∞) D.(0,+∞)解析:函數(shù)y=12x2-lnx,則此函數(shù)的定義域?yàn)?0,+∞),y'=x-1x=x令y'<0,得0<x<1,所以此函數(shù)的單調(diào)遞減區(qū)間為(0,1).答案:B2.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().(第2題)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,f'(-2)=0,f'(2)=0,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.由此可得函數(shù)f(x)在x=-2處取得極大值,在x=2處取得極小值.答案:D3.已知y=f(x)是奇函數(shù),當(dāng)x∈(0,2)時(shí),f(x)=lnx-axa>12,當(dāng)x∈(-2,0)時(shí),f(x)的最小值為1,則a的值等于().A.14 B.13 C.12解析:由題意知,當(dāng)x∈(0,2)時(shí),f(x)的最大值為-1.已知當(dāng)x∈(0,2)時(shí),f(x)=lnx-axa>12.因?yàn)閍>12,所以0<1a<令f'(x)=1x-a=0,得x=1a,當(dāng)0<x<1a時(shí),f'(x)>0,函數(shù)f(x)在區(qū)間當(dāng)1a<x<2時(shí),f'(x)<0,函數(shù)f(x)在區(qū)間1a所以f(x)max=f1a=-lna-1=-1,解得a=1答案:D4.函數(shù)f(x)=x3+ax2+bx-a2-7a(a,b∈R)在x=1處取得極大值10,則ab的值為()A.-23 B.-C.-2或-23 D.2或-解析:f'(x)=3x2+2ax+b.由題意得,f(1)=1+a+b-a2-7a=10,①f'(1)=3+2a+b=0,②聯(lián)立①②,得a當(dāng)a=-2,b=1時(shí),f'(x)=3x2-4x+1.令f'(x)=0,解得x=13或x=1,易知f(x)在x=1處取極小值故a=-2,b=1不符合題意.當(dāng)a=-6,b=9時(shí),f'(x)=3x2-12x+9,令f'(x)=0,解得x=1或x=3,易知f(x)在x=1處取極大值.故a=-6,b=9滿足題意.所以ab=-答案:A5.設(shè)函數(shù)y=f(x)在R上的導(dǎo)函數(shù)為f'(x),且2f(x)+xf'(x)>x2,則下面不等式在R上恒成立的是().A.f(x)>0 B.f(x)<0C.f(x)>x D.f(x)<x解析:∵2f(x)+xf'(x)>x2(x≠0),∴當(dāng)x>0時(shí),2xf(x)+x2f'(x)>x3>0;當(dāng)x<0時(shí),2xf(x)+x2f'(x)<x3<0,即當(dāng)x>0時(shí),[x2f(x)]'>0;當(dāng)x<0時(shí),[x2f(x)]'<0.設(shè)F(x)=x2f(x)(x≠0),則函數(shù)F(x)在區(qū)間(-∞,0)上單調(diào)遞減,在區(qū)間(0,+∞)上單調(diào)遞增,∴F(x)>F(0),即x2f(x)>0(x≠0),∴f(x)>0(x≠0).當(dāng)x=0時(shí),由題意知,2f(0)>0,即f(0)>0.故f(x)>0在R上恒成立.答案:A6.函數(shù)y=2x-1x2+6的極大值是答案:37.已知函數(shù)f(x)=(x+1)ex,x<1,x2-2x,x≥1,則函數(shù)f(x)的值域?yàn)?解析:因?yàn)閒(x)=(x+1)ex,x<1,x2-2x,x≥1,當(dāng)x<1時(shí),f(x)=(x+1)ex,則f'(x)=(x+2)ex,當(dāng)x<-2時(shí),f'(x)<0,當(dāng)-2<x<1時(shí),f'(x)>0,即y=f(x)在(-∞,-2)上單調(diào)遞減,在(-2,1)上單調(diào)遞增,則當(dāng)x=-2時(shí)取得極小值,且f(-2)=-e-2.當(dāng)x≥(第7題)由函數(shù)圖象可知函數(shù)的值域?yàn)閇-1,+∞),函數(shù)g(x)=f(x)-k有3個(gè)零點(diǎn),即y=f(x)的圖象與直線y=k有3個(gè)交點(diǎn),所以-e-2<k<0,即k∈(-e-2,0).答案:[-1,+∞)(-e-2,0)8.設(shè)函數(shù)f(x)=lnx-12ax2-bx,若x=1是f(x)的極大值點(diǎn),則實(shí)數(shù)a的取值范圍為.解析:函數(shù)f(x)的定義域?yàn)?0,+∞),f'(x)=1x-ax-b由f'(1)=0,得b=1-a.所以f'(x)=1x-ax+a-1=-①若a≥0,當(dāng)0<x<1時(shí),f'(x)>0,f(x)單調(diào)遞增;當(dāng)x>1時(shí),f'(x)<0,f(x)單調(diào)遞減,所以x=1是f(x)的極大值點(diǎn).②若a<0,由f'(x)=0,得x=1或x=-1a因?yàn)閤=1是f(x)的極大值點(diǎn),所以-1a>1,解得-1<a<0綜上,實(shí)數(shù)a的取值范圍是(-1,+∞).答案:(-1,+∞)9.已知函數(shù)f(x)=ex-1-x-ax2.(1)當(dāng)a=0時(shí),求證:f(x)≥0.(2)當(dāng)x≥0時(shí),若不等式f(x)≥0恒成立,求實(shí)數(shù)a的取值范圍.(1)證明:當(dāng)a=0時(shí),f(x)=ex-1-x,f'(x)=ex-1.令f'(x)=0,得x=0.當(dāng)x∈(-∞,0)時(shí),f'(x)<0;當(dāng)x∈(0,+∞)時(shí),f'(x)>0.所以函數(shù)f(x)在區(qū)間(-∞,0)上單調(diào)遞減,在區(qū)間(0,+∞)上單調(diào)遞增,所以f(x)min=f(0)=0,所以f(x)≥0.(2)解:f'(x)=ex-1-2ax.設(shè)h(x)=ex-1-2ax,x≥0,則h'(x)=ex-2a.①當(dāng)2a≤1,即a≤12時(shí),h'(x)≥0在區(qū)間[0,+∞)上恒成立,則函數(shù)h(x)在區(qū)間[0,+∞)上單調(diào)遞增,所以h(x)≥h(0)=0,即f'(x)≥0,且只在有限個(gè)點(diǎn)為0,所以f(x)在區(qū)間[0,+∞)上單調(diào)遞增,所以f(x)≥f(0)=0,即當(dāng)a≤12②當(dāng)2a>1,即a>12時(shí),令h'(x)=0,解得x=ln(2a),當(dāng)x∈[0,ln(2a))時(shí),h'(x)<0,函數(shù)h(x)