高數(shù)一下冊(cè)試卷及答案一、選擇題（每題3分，共30分）1.已知函數(shù)f(x)=x^3-3x+1，求f'(x)。A.3x^2-3B.3x^2+3C.x^2-3xD.3x^2-3x+12.求極限lim(x→0)(1-cosx)/x。A.0B.1C.-1D.23.已知函數(shù)f(x)=x^2+2x+1，求f(-1)。A.0B.1C.2D.34.求不定積分∫x^2dx。A.x^3/3+CB.x^3+CC.3x^3+CD.3x^3/3+C5.求定積分∫_0^1x^2dx。A.1/3B.1/4C.1/2D.16.求極限lim(x→∞)(x^2-3x+2)/(x^2+2x+1)。A.1B.0C.-1D.27.求二重積分?_D(x^2+y^2)dσ，其中D是由x^2+y^2≤1所定義的區(qū)域。A.π/2B.πC.2πD.4π8.求極限lim(x→0)(x^2+x)/(x^2+1)。A.0B.1C.-1D.29.已知函數(shù)f(x)=e^x，求f'(x)。A.e^xB.e^(-x)C.-e^xD.010.求不定積分∫e^xdx。A.e^x+CB.e^(-x)+CC.-e^x+CD.0二、填空題（每題3分，共15分）11.函數(shù)f(x)=x^3-6x^2+11x-6的導(dǎo)數(shù)為_(kāi)________。12.函數(shù)f(x)=ln(x)的不定積分為_(kāi)________。13.函數(shù)f(x)=x^2-4x+4的極值點(diǎn)為_(kāi)________。14.函數(shù)f(x)=x^3-3x的拐點(diǎn)為_(kāi)________。15.函數(shù)f(x)=x^2+2x+1的最小值為_(kāi)________。三、解答題（每題10分，共50分）16.求函數(shù)f(x)=x^3-3x^2+2x的極值點(diǎn)。解：首先求導(dǎo)數(shù)f'(x)=3x^2-6x+2，令f'(x)=0，解得x=1或x=2/3。然后求二階導(dǎo)數(shù)f''(x)=6x-6，代入x=1，f''(1)=0，代入x=2/3，f''(2/3)>0，所以x=2/3為極小值點(diǎn)，x=1為極大值點(diǎn)。17.求函數(shù)f(x)=x^2-4x+4的單調(diào)區(qū)間。解：首先求導(dǎo)數(shù)f'(x)=2x-4，令f'(x)>0，解得x>2，令f'(x)<0，解得x<2。所以函數(shù)在(-∞,2)上單調(diào)遞減，在(2,+∞)上單調(diào)遞增。18.求函數(shù)f(x)=x^3-3x^2+2x的拐點(diǎn)。解：首先求二階導(dǎo)數(shù)f''(x)=6x-6，令f''(x)=0，解得x=1。然后檢查x=1處的凹凸性，f''(1)=0，所以x=1為拐點(diǎn)。19.求極限lim(x→0)(sin2x)/x。解：根據(jù)洛必達(dá)法則，lim(x→0)(sin2x)/x=lim(x→0)(2cos2x)/1=2。20.求定積分∫_0^π/2sinxdx。解：∫_0^π/2sinxdx=[-cosx]_0^π/2=1-(-1)=2。四、證明題（每題5分，共10分）21.證明函數(shù)f(x)=x^3-3x+1在x=1處取得極小值。證明：首先求導(dǎo)數(shù)f'(x)=3x^2-3，令f'(x)=0，解得x=1或x=-1。然后求二階導(dǎo)數(shù)f''(x)=6x-6，代入x=1，f''(1)>0，所以x=1為極小值點(diǎn)。22.證明函數(shù)f(x)=x^2-4x+4在x=2處取得最小值。證明：首先求導(dǎo)數(shù)f'(x)=2x-4，令f'(x)=0，解得x=2。然后求二階導(dǎo)數(shù)f''(x)=2，因?yàn)閒''(x)>0，所以x=2為極小值點(diǎn)，即最小值點(diǎn)。答案：1.A2.D3.A4.A5.