數(shù)學(xué)分析配套題目及答案一、選擇題（共20分，每題4分）1.已知函數(shù)f(x)在區(qū)間[a,b]上連續(xù)，則以下哪個(gè)選項(xiàng)是正確的？A.f(x)在[a,b]上一定有最大值和最小值B.f(x)在[a,b]上一定有界C.f(x)在[a,b]上一定單調(diào)D.f(x)在[a,b]上一定有無窮多個(gè)零點(diǎn)答案：B2.極限lim(x→0)(sinx)/x的值是多少？A.0B.1C.π/2D.-1答案：B3.以下哪個(gè)函數(shù)是偶函數(shù)？A.f(x)=x^2+xB.f(x)=x^3-xC.f(x)=x^2-1D.f(x)=x^3+1答案：C4.以下哪個(gè)級(jí)數(shù)是收斂的？A.1+1/2+1/4+1/8+...B.1-1/2+1/3-1/4+...C.1+1/2^2+1/3^2+1/4^2+...D.1+2+3+4+...答案：C5.以下哪個(gè)積分是發(fā)散的？A.∫(0to1)1/xdxB.∫(0to1)x^2dxC.∫(0to1)e^xdxD.∫(0to1)sinxdx答案：A二、填空題（共20分，每題4分）6.函數(shù)f(x)=x^3-3x+2的導(dǎo)數(shù)是________。答案：3x^2-37.函數(shù)f(x)=e^x的不定積分是________。答案：e^x+C8.函數(shù)f(x)=lnx的反導(dǎo)數(shù)是________。答案：xlnx-x+C9.函數(shù)f(x)=x^2在區(qū)間[0,1]上的定積分是________。答案：1/310.函數(shù)f(x)=sinx的周期是________。答案：2π三、計(jì)算題（共30分，每題10分）11.計(jì)算極限lim(x→∞)(x^2+3x+2)/(2x^2+x-3)。解：lim(x→∞)(x^2+3x+2)/(2x^2+x-3)=lim(x→∞)(1+3/x+2/x^2)/(2+1/x-3/x^2)=1/2。12.計(jì)算不定積分∫(1/(x^2-1))dx。解：∫(1/(x^2-1))dx=∫(1/((x-1)(x+1)))dx=1/2∫(1/(x-1)-1/(x+1))dx=1/2(ln|x-1|-ln|x+1|)+C=1/2ln|(x-1)/(x+1)|+C。13.計(jì)算定積分∫(0to1)(x^2-2x+1)dx。解：∫(0to1)(x^2-2x+1)dx=[x^3/3-x^2+x](0to1)=(1/3-1+1)-(0)=1/3。14.計(jì)算二重積分?(0≤x≤1,0≤y≤1)(x^2+y^2)dxdy。解：?(0≤x≤1,0≤y≤1)(x^2+y^2)dxdy=∫(0to1)∫(0to1)(x^2+y^2)dydx=∫(0to1)[x^2y+y^3/3](0to1)dx=∫(0to1)(x^2+1/3)dx=[x^3/3+x/3](0to1)=4/9。四、證明題（共30分，每題10分）15.證明：若函數(shù)f(x)在區(qū)間[a,b]上連續(xù)，則f(x)在[a,b]上一定有界。證明：由函數(shù)f(x)在區(qū)間[a,b]上連續(xù)，根據(jù)極值定理，f(x)在[a,b]上一定有最大值M和最小值m。因此，對(duì)于任意x∈[a,b]，都有m≤f(x)≤M，即f(x)在[a,b]上有界。16.證明：若函數(shù)f(x)在區(qū)間[a,b]上可導(dǎo)，則f(x)在[a,b]上連續(xù)。證明：由函數(shù)f(x)在區(qū)間[a,b]上可導(dǎo)，根據(jù)導(dǎo)數(shù)的定義，對(duì)于任意x∈(a,b)，都有f'(x)=lim(h→0)(f(x+h)-f(x))/h。由于f'(x)存在，根據(jù)極限的性質(zhì)，f(x)在x處連續(xù)。又因?yàn)閒(x)在[a,b]上可導(dǎo)，所以f(x)在[a,b]上連續(xù)。17.證明：若函數(shù)f(x)在區(qū)間[a,b]上單調(diào)遞增，則f(x)在[a,b]上連續(xù)。證明：由函數(shù)f(x)在區(qū)間[a,b]上單調(diào)遞增，對(duì)于任意x1,x2∈[a,b]，若x1<x2，則f(x1)≤f(x2)。由于f(x)單調(diào)遞增，根據(jù)單調(diào)性的定義，f(x)在[a,b]上連續(xù)。18.證明：若函數(shù)f(x)在區(qū)間[a,b]上可積，則f(x)在[a,b]上連續(xù)。證明：由函數(shù)f(x)在區(qū)間[a,b]上可積，根據(jù)可積性的定義，對(duì)于任意ε>0，存在δ>0，使得對(duì)于任意的區(qū)間[a,b]的劃分P，只要∑(Δxi)^2<δ，都有∑f(ξi)Δxi-∫(atob)f(x)dx<ε。由于f(x)可積，根據(jù)可積性的性質(zhì)，f(x)在[a,b]上連續(xù)。五、綜合題（共30分，每題10分）19.已知函數(shù)f(x)=x^2-2x+3，求f(x)的單調(diào)區(qū)間和極值點(diǎn)。解：首先求f(x)的導(dǎo)數(shù)f'(x)=2x-2。令f'(x)=0，解得x=1。當(dāng)x<1時(shí)，f'(x)<0，f(x)單調(diào)遞減；當(dāng)x>1時(shí)，f'(x)>0，f(x)單調(diào)遞增。因此，f(x)的單調(diào)遞減區(qū)間為(-∞,1)，單調(diào)遞增區(qū)間為(1,+∞)。又因?yàn)閒(x)在x=1處由減變?cè)?，所以x=1是f(x)的極小值點(diǎn)，極小值為f(1)=2。20.已知函數(shù)f(x)=e^x-x^2，求f(x)的零點(diǎn)個(gè)數(shù)和零點(diǎn)。解：首先求f(x)的導(dǎo)數(shù)f'(x)=e^x-2x。令f'(x)=0，解得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。因此，f(x)在x=ln2處取得極小值f(ln2)=2-2ln2。又