理科高數(shù)第二套試卷及答案一、單項(xiàng)選擇題（每題3分，共15分）1.函數(shù)f(x)=x^2-4x+3的零點(diǎn)個(gè)數(shù)為（）A.0B.1C.2D.32.極限lim(x→0)(1-cosx)/x^2的值為（）A.0B.1/2C.1D.-13.函數(shù)y=x^3+1的導(dǎo)數(shù)為（）A.3x^2+1B.3x^2C.x^2+1D.3x^34.函數(shù)f(x)=x^2-2x+1的極值點(diǎn)為（）A.0B.1C.2D.35.函數(shù)y=x^2的不定積分為（）A.x^3/3+CB.x^2+CC.2x^2+CD.x^3+C二、填空題（每題4分，共20分）6.函數(shù)f(x)=x^3-3x的導(dǎo)數(shù)為_(kāi)________。7.極限lim(x→∞)(x^2-1)/(x^3+1)的值為_(kāi)________。8.函數(shù)f(x)=x^2-4x+4的極值點(diǎn)為_(kāi)________。9.函數(shù)y=x^3的不定積分為_(kāi)________。10.函數(shù)f(x)=x^2-4x+3的零點(diǎn)為_(kāi)________。三、計(jì)算題（每題10分，共30分）11.求極限lim(x→0)(sin2x)/x。12.求函數(shù)f(x)=x^3-3x的極值點(diǎn)。13.求函數(shù)y=x^2-4x+4的不定積分。四、證明題（每題15分，共30分）14.證明：對(duì)于任意x∈R，有e^x>1+x。15.證明：函數(shù)f(x)=x^3-3x在x=1處取得極小值。答案一、單項(xiàng)選擇題1.C2.B3.B4.B5.A二、填空題6.3x^2-37.08.29.(1/4)x^4+C10.1和3三、計(jì)算題11.極限lim(x→0)(sin2x)/x=2。12.函數(shù)f(x)=x^3-3x的導(dǎo)數(shù)為f'(x)=3x^2-3，令f'(x)=0，解得x=±1。當(dāng)x<-1或x>1時(shí)，f'(x)>0，函數(shù)單調(diào)遞增；當(dāng)-1<x<1時(shí)，f'(x)<0，函數(shù)單調(diào)遞減。因此，x=-1為極大值點(diǎn)，x=1為極小值點(diǎn)。13.函數(shù)y=x^2-4x+4的不定積分為∫(x^2-4x+4)dx=(1/3)x^3-2x^2+4x+C。四、證明題14.證明：令g(x)=e^x-1-x，則g'(x)=e^x-1。當(dāng)x<0時(shí)，g'(x)<0，函數(shù)g(x)單調(diào)遞減；當(dāng)x>0時(shí)，g'(x)>0，函數(shù)g(x)單調(diào)遞增。因此，g(x)≥g(0)=0，即e^x≥1+x。15.證明：函數(shù)f(x)=x^3-3x的導(dǎo)數(shù)為f'(x)=3x^2-3，令f'(x)=0，解得x=±1。當(dāng)x<-1或x>1時(shí)，f'(x)>0，函數(shù)單調(diào)遞增；當(dāng)-1<x<1時(shí)，f'(x)<0，函數(shù)單調(diào)遞減。因此