電大經(jīng)濟(jì)數(shù)學(xué)基礎(chǔ)1形成性考核冊(cè)答案

1xsinx一、填空題1.lim=.答案：1一x喻0x2k4.設(shè)函數(shù)f(x+1)=x2+2x+5，則f,(x)=.答案2x5.設(shè)f(x)=xsinx，則f,,()=.答案:一二、單項(xiàng)選擇題1.當(dāng)x喻+偽時(shí)，下列變量為無(wú)窮小量的是（D）21x2sin21xD．x2.下列極限計(jì)算正確的是（B）x喻0xx喻0+xx喻0xx喻偽x4.若函數(shù)f(x)在點(diǎn)x0處可導(dǎo)，則(B)是錯(cuò)誤的.A．函數(shù)f(x)在點(diǎn)x0處有定義B．limf(x)=A，但A子f(x0)x喻x0C．函數(shù)f(x)在點(diǎn)x0處連續(xù)D．函數(shù)f(x)在點(diǎn)x0處可微5.若f()=x，則f,(x)=（B）.A.1x2B.1x2 1x1x三、解答題1．計(jì)算極限EQ\*jc3\*hps23\o\al(\s\up0(m),喻1)EQ\*jc3\*hps23\o\al(\s\up8(一),x2)EQ\*jc3\*hps23\o\al(\s\up0(m),喻1)EQ\*jc3\*hps23\o\al(\s\up8(x),x)EQ\*jc3\*hps23\o\al(\s\up8(1),1)EQ\*jc3\*hps23\o\al(\s\up8(x),x)EQ\*jc3\*hps23\o\al(\s\up8(2),1)EQ\*jc3\*hps23\o\al(\s\up0(m),喻1)EQ\*jc3\*hps23\o\al(\s\up8(x),x)EQ\*jc3\*hps23\o\al(\s\up8(2),1)EQ\*jc3\*hps23\o\al(\s\up8(1),1)EQ\*jc3\*hps23\o\al(\s\up8(2),1)EQ\*jc3\*hps23\o\al(\s\up0(im),喻2)EQ\*jc3\*hps23\o\al(\s\up8(x),x)EQ\*jc3\*hps13\o\al(\s\up13(2),2)EQ\*jc3\*hps23\o\al(\s\up8(5),6)EQ\*jc3\*hps23\o\al(\s\up8(x),x)EQ\*jc3\*hps23\o\al(\s\up8(6),8)EQ\*jc3\*hps23\o\al(\s\up0(im),喻2)EQ\*jc3\*hps23\o\al(\s\up8(x),x)EQ\*jc3\*hps23\o\al(\s\up8(2),2)EQ\*jc3\*hps23\o\al(\s\up8(x),x)EQ\*jc3\*hps23\o\al(\s\up8(3),4)EQ\*jc3\*hps23\o\al(\s\up0(im),喻2)EQ\*jc3\*hps23\o\al(\s\up8(x),x)EQ\*jc3\*hps23\o\al(\s\up8(3),4)EQ\*jc3\*hps23\o\al(\s\up8(2),2)EQ\*jc3\*hps23\o\al(\s\up8(3),4)EQ\*jc3\*hps23\o\al(\s\up8(1),2)EQ\*jc3\*hps23\o\al(\s\up0(im),喻0)EQ\*jc3\*hps23\o\al(\s\up6(一x),x)EQ\*jc3\*hps23\o\al(\s\up1(im),喻0)EQ\*jc3\*hps23\o\al(\s\up1(im),喻0)EQ\*jc3\*hps23\o\al(\s\up8(一),1)EQ\*jc3\*hps23\o\al(\s\up1(im),喻0)2EQ\*jc3\*hps23\o\al(\s\up8(2),3)EQ\*jc3\*hps23\o\al(\s\up8(x),x)EQ\*jc3\*hps13\o\al(\s\up13(2),2)EQ\*jc3\*hps23\o\al(\s\up8(3),2)EQ\*jc3\*hps23\o\al(\s\up8(x),x)EQ\*jc3\*hps23\o\al(\s\up8(5),4)EQ\*jc3\*hps23\o\al(\s\up14(2),3)EQ\*jc3\*hps23\o\al(\s\up8(2),3)EQ\*jc3\*hps23\o\al(\s\up8(0),0)EQ\*jc3\*hps23\o\al(\s\up8(0),0)EQ\*jc3\*hps23\o\al(\s\up8(2),3)EQ\*jc3\*hps23\o\al(\s\up0(im),喻0)EQ\*jc3\*hps23\o\al(\s\up7(si),si)EQ\*jc3\*hps23\o\al(\s\up7(n),n)EQ\*jc3\*hps23\o\al(\s\up7(3),5)EQ\*jc3\*hps23\o\al(\s\up7(x),x)EQ\*jc3\*hps23\o\al(\s\up0(im),喻0)EQ\*jc3\*hps23\o\al(\s\up21(s),s)EQ\*jc3\*hps23\o\al(\s\up21(i),i)EQ\*jc3\*hps23\o\al(\s\up21(x),x)EQ\*jc3\*hps23\o\al(\s\up7(3),5)EQ\*jc3\*hps23\o\al(\s\up7(3),5)EQ\*jc3\*hps23\o\al(\s\up21(s),s)EQ\*jc3\*hps23\o\al(\s\up21(i),i)EQ\*jc3\*hps23\o\al(\s\up21(x),x)EQ\*jc3\*hps23\o\al(\s\up7(3),5)EQ\*jc3\*hps23\o\al(\s\up7(1),1)EQ\*jc3\*hps23\o\al(\s\up7(3),5)EQ\*jc3\*hps23\o\al(\s\up1(im),喻2)EQ\*jc3\*hps23\o\al(\s\up9(x),n)EQ\*jc3\*hps23\o\al(\s\up1(im),喻2)EQ\*jc3\*hps23\o\al(\s\up9(+),in)EQ\*jc3\*hps23\o\al(\s\up9(一),2)EQ\*jc3\*hps23\o\al(\s\up1(im),喻2)EQ\*jc3\*hps23\o\al(\s\up1(im),喻2)EQ\*jc3\*hps23\o\al(\s\up10(x),x)EQ\*jc3\*hps23\o\al(\s\up10(0),0)問(wèn)1）當(dāng)a,b為何值時(shí)，f(x)在x=0處極限存在？（2）當(dāng)a,b為何值時(shí)，f(x)在x=0處連續(xù).解1）因?yàn)閒(x)在x=0處有極限存在，則有l(wèi)imf(x)=limf(x)x喻0一x喻0+又lim一f(x)=lim一(xsin1+b)=bx喻0x喻0xx喻0x喻0x所以當(dāng)a為實(shí)數(shù)、b=1時(shí)，f(x)在x=0處極限存在.（2）因?yàn)閒(x)在x=0處連續(xù)，則有x喻0x喻0lim一fx喻0x喻0又f(0)=a，結(jié)合（1）可知a=b=1所以當(dāng)a=b=1時(shí)，f(x)在x=0處連續(xù).3．計(jì)算下列函數(shù)的導(dǎo)數(shù)或微分：x3 3x5222], 3x5221x，求y,解：y,=(x2),(xex),=2x2ex1解：y,=(eax),sinbxeax(sinbx),=eax(ax),sinbxeaxcosbx(bx),=aeaxsinbxbeaxcosbxaxsinbxbeaxcosbx)dx 1 121EQ\*jc3\*hps23\o\al(\s\up7(e),x)EQ\*jc3\*hps13\o\al(\s\up11(x),2)EQ\*jc3\*hps23\o\al(\s\up7(3),2)EQ\*jc3\*hps13\o\al(\s\up13(1),2)2222sin+2xex=(cos),(ex),=sin(),ex(x2),=2x2222sin+2xex12)2),)1xx35sin135xln2(cosx)(x),2x2+6x6=x2cosx2x2+6x64.下列各方程中y是x的隱函數(shù)，試求y,或dy4解：方程兩邊同時(shí)對(duì)x求導(dǎo)得：,y2x3y=2yxdy=y,dx=yy2xEQ\*jc3\*hps17\o\al(\s\up5(一),x)3dx解：方程兩邊同時(shí)對(duì)x求導(dǎo)得：,4cos(x+y)yexyy=5．求下列函數(shù)的二階導(dǎo)數(shù)：+x2),,,2x,2(1+x22(11 xx22 xx22經(jīng)濟(jì)數(shù)學(xué)基礎(chǔ)作業(yè)2（一）填空題1.若∫f(x)dx=2x+2x+c，則f(x)=2xln2+2.2.4.設(shè)函數(shù)ln(1+x2)dx=055.若P(x)=dt，則P,(x)=1. .（二）單項(xiàng)選擇題1.下列函數(shù)中D）是xsinx2的原函數(shù).A．cosx2B．2cosx2C．-2cosx2D．-cosx23.下列不定積分中，常用分部積分法計(jì)算的是（C）.x1x2dxC4.下列定積分中積分值為0的是（D）.EQ\*jc3\*hps13\o\al(\s\up11(6),1)5.下列無(wú)窮積分中收斂的是（B）.sinxdx(三)解答題1.計(jì)算下列不定積分EQ\*jc3\*hps13\o\al(\s\up10(3x),ex)dxEQ\*jc3\*hps23\o\al(\s\up8(1),3)dxdx22)dx222EQ\*jc3\*hps13\o\al(\s\up8(x2),x)EQ\*jc3\*hps23\o\al(\s\up8(2),x)EQ\*jc3\*hps23\o\al(\s\up6(x),2)22xx2+x2dx 622232xsindxxdcoscosd()xx(1)dx（3）3dx 3e1e1（5）xlnxdx解：原式=lnxdx22=x22=e2e1lnxe12e4xdx1+42 1 EQ\*jc3\*hps13\o\al(\s\up10(ex),x2)解：原式=ed() x21 2 2=eeπ02xcos2xdx解：原式=xdsin2x=xsin2xsin2xd(2x)402402解：原式=dxxdex=4xexEQ\*jc3\*hps13\o\al(\s\up8(4),0)exd(x)72220經(jīng)濟(jì)數(shù)學(xué)基礎(chǔ)作業(yè)3|（一）填空題1.|（一）填空題1.設(shè)矩陣A=-2322.設(shè)A,B均為3階矩陣，且A=B=-3，則-2ABT=.答案：-723.設(shè)A,B均為n階矩陣，則等式(A-B)2=A2-2AB+B2成立的充分必要條件是.答案：AB=BA4.設(shè)A,B均為n階矩陣，(I-B)可逆，則矩陣A+BX=X的解X=.答案：(I-B)-1A-1（二）單項(xiàng)選擇題1.以下結(jié)論或等式正確的是（C）.A．若A,B均為零矩陣，則有A=BC．對(duì)角矩陣是對(duì)稱(chēng)矩陣2.設(shè)A為3根4矩陣，B為5根2矩陣，且乘積矩陣ACBT有意義，則CT為（A）矩陣.3.設(shè)A,B均為n階可逆矩陣，則下列等式成立的是（C`A．(A+B)-1=A-1+B-1，B．(A.B)-1=A-1.B-1C．AB=BAD．AB=BA4.下列矩陣可逆的是（A）.「1EQ\*jc3\*hps23\o\al(\s\up9(0),0)3]3]3「-1|002--1]1234234EQ\*jc3\*hps23\o\al(\s\up6(3),4)2]EQ\*jc3\*hps23\o\al(\s\up4(3),4)85]0三、解答題5]0=24]5]424]5]41-2|EQ\*jc3\*hps23\o\al(\s\up8(4),3)4]「2EQ\*jc3\*hps23\o\al(\s\up8(0),7)|EQ\*jc3\*hps23\o\al(\s\up8(4),3)4]「2EQ\*jc3\*hps23\o\al(\s\up8(0),7)7]0-7」5]041-7]0-7」5]041-22]041-2-4-2EQ\*jc3\*hps23\o\al(\s\up9(6),3)|EQ\*jc3\*hps23\o\al(\s\up9(7),0)=-EQ\*jc3\*hps23\o\al(\s\up8(2),2)EQ\*jc3\*hps23\o\al(\s\up8(1),2EQ\*jc3\*hps23\o\al(\s\up9(6),3)|EQ\*jc3\*hps23\o\al(\s\up9(7),0)=-「23-1]「123]EQ\*jc3\*hps23\o\al(\s\up7(1),0)EQ\*jc3\*hps23\o\al(\s\up7(1),1)EQ\*jc3\*hps23\o\al(\s\up7(1),0)EQ\*jc3\*hps23\o\al(\s\up7(1),1)EQ\*jc3\*hps23\o\al(\s\up7(2),1)解因?yàn)锳B=AB2222（注意：因?yàn)榉?hào)輸入方面的原因，在題4—題7的矩陣初等行變換中，書(shū)寫(xiě)時(shí)應(yīng)把（1）寫(xiě)成①2）寫(xiě)成②3）寫(xiě)成③；…）4]2λEQ\*jc3\*hps23\o\al(\s\up9(4]2λEQ\*jc3\*hps23\o\al(\s\up9(2),1)EQ\*jc3\*hps23\o\al(\s\up8(1),0)EQ\*jc3\*hps23\o\al(\s\up8(1),0)1921-10] EQ\*jc3\*hps23\o\al(\s\up8(0),0)EQ\*jc3\*hps23\o\al(\s\up8(1),0)EQ\*jc3\*hps23\o\al(\s\up8(0),1)21-10] EQ\*jc3\*hps23\o\al(\s\up8(0),0)EQ\*jc3\*hps23\o\al(\s\up8(1),0)EQ\*jc3\*hps23\o\al(\s\up8(0),1) | 0]-2-3102-111-130-14|4]4]-44]-4-7」2-12λ121λ2-1λ-4 42-5-8-7-135414-15-5-15EQ\*jc3\*hps23\o\al(\s\up2(1),0)3]73]0000233434EQ\*jc3\*hps23\o\al(\s\up8(7),9)|027EQ\*jc3\*hps23\o\al(\s\up8(1),0)|EQ\*jc3\*hps23\o\al(\s\up9(2),1)EQ\*jc3\*hps23\o\al(\s\up9(0),0) EQ\*jc3\*hps23\o\al(\s\up8(1),0)|EQ\*jc3\*hps23\o\al(\s\up9(2),1)EQ\*jc3\*hps23\o\al(\s\up9(0),0) λ-當(dāng)λ=時(shí)，r(EQ\*jc3\*hps23\o\al(\s\up3(3),0)|5EQ\*jc3\*hps23\o\al(\s\up3(3),0)|5「1-75-「1-75-82-50]42-7900EQ\*jc3\*hps23\o\al(\s\up2(4),2)EQ\*jc3\*hps23\o\al(\s\up2(3),0)2-6-2-6EQ\*jc3\*hps23\o\al(\s\up0(3),1EQ\*jc3\*hps23\o\al(\s\up2(4),2)EQ\*jc3\*hps23\o\al(\s\up2(3),0)2-6-2-6EQ\*jc3\*hps23\o\al(\s\up0(3),1)4-500 喻20]20]EQ\*jc3\*hps23\o\al(\s\up0(3),1)喻EQ\*jc3\*hps23\o\al(\s\up0(0),0)→||0→|6．求下列矩陣的逆矩陣：-301--301-30101-834-6-212]EQ\*jc3\*hps23\o\al(\s\up8(2),1)-18]EQ\*jc3\*hps23\o\al(\s\up7(7),9)-3] 喻EQ\*jc3\*hps23\o\al(\s\up9(0),0)「1EQ\*jc3\*hps23\o\al(\s\up9(2),3)|-3「1EQ\*jc3\*hps23\o\al(\s\up9(0),0)「1EQ\*jc3\*hps23\o\al(\s\up9(0),0)-3214-3--30-5|（2）A=|（2）A=-400-21-4|解：|解：1 EQ\*jc3\*hps23\o\al(\s\up9(0),0) EQ\*jc3\*hps23\o\al(\s\up9(0),0)「-「-12||12]2]-31-100-1-1-4-132-60]00]EQ\*jc3\*hps23\o\al(\s\up0(1),0)EQ\*jc3\*hps23\o\al(\s\up0(0),1)30]EQ\*jc3\*hps23\o\al(\s\up0(3),6)EQ\*jc3\*hps23\o\al(\s\up0(0),1)0]「1EQ\*jc3\*hps23\o\al(\s\up3(1),2)EQ\*jc3\*hps23\o\al(\s\up3(0),0)EQ\*jc3\*hps23\o\al(\s\up9(0),0)010-2-10-2-1-402-72]2]-30]EQ\*jc3\*hps23\o\al(\s\up7(1),0)EQ\*jc3\*hps23\o\al(\s\up7(0),1)30]EQ\*jc3\*hps23\o\al(\s\up8(-),1)EQ\*jc3\*hps23\o\al(\s\up8(6),3)EQ\*jc3\*hps23\o\al(\s\up8(1),0)0] 0]2]2502-11-30-53「-52]∴A-1=|「-52]∴X=BA-1=|0]四、證明題1．試證：若B1,B2都與A可交換，則B1+B2，B1B2也與A可交換。22)2也與A可交換。1B2)A2A)(AB21A)B2即B1B2也與A可交換.2．試證：對(duì)于任意方陣A，A+AT，AAT,ATA是對(duì)稱(chēng)矩陣。A+ATTATT∴A+AT是對(duì)稱(chēng)矩陣。∴AAT是對(duì)稱(chēng)矩陣。ATA)TAT(AT)T∴ATA是對(duì)稱(chēng)矩陣.3．設(shè)A,B均為n階對(duì)稱(chēng)矩陣，則AB對(duì)稱(chēng)的充分必要條件是：AB=BA。證：必要性：若AB是對(duì)稱(chēng)矩陣，即(AB)T=AB而(AB)=BTAT=BA因此A充分性：∴AB是對(duì)稱(chēng)矩陣.4．設(shè)A為n階對(duì)稱(chēng)矩陣，B為n階可逆矩陣，且B一1=BT，證明B一1AB是對(duì)稱(chēng)矩陣。=BTB1AB)TT.T=BT.AT.BTT1AB∴B1AB是對(duì)稱(chēng)矩陣.證畢.經(jīng)濟(jì)數(shù)學(xué)基礎(chǔ)作業(yè)4ln(x1)______________2.函數(shù)y=3(x1)2的駐點(diǎn)是，極值點(diǎn)是，它是極值點(diǎn)。答案：x=11，0小。p3.設(shè)某商品的需求函數(shù)為q(p)=10e一2，則需求彈性Ep=.答案：Ep=一111111111=__________.答案：4.__ A喻5.設(shè)線性方程組AX=b，且 A喻5.設(shè)線性方程組AX=b，且|EQ\*jc3\*hps18\o\al(\s\up7(2),0)（二）單項(xiàng)選擇題1.下列函數(shù)在指定區(qū)間(-偽,+偽)上單調(diào)增加的是（B）.A．sinxB．exC．x22.設(shè)f(x)=，則f(f(x))=（C）.23.下列積分計(jì)算正確的是（A）.EQ\*jc3\*hps13\o\al(\s\up8(1),1)EQ\*jc3\*hps13\o\al(\s\up8(1),1)EQ\*jc3\*hps13\o\al(\s\up7(1),1)-EQ\*jc3\*hps13\o\al(\s\up7(1),1)(x2+x3)dx=04.設(shè)線性方程組AmxnX=b有無(wú)窮多解的充分必要條件是（D）.(x1215.設(shè)線性方程組〈|EQ\*jc3\*hps23\o\al(\s\up10(|),l)x122EQ\*jc3\*hps23\o\al(\s\up3(+),x)2xEQ\*jc3\*hps23\o\al(\s\up3(=),x)3a3，則方程組有解(x121231三、解答題1．求解下列可分離變量的微分方程：(1)y,=ex+yEQ\*jc3\*hps23\o\al(\s\up8(dy),dx)dx3y2xdexy3=xex-∫exdxy3=xex-ex+c2.求解下列一階線性微分方程：（1）y,-y=(x+1)3解:y=e-（2)2(2（2)（2）y,-=2xsin2x3.求解下列微分方程的初值問(wèn)題：(1)y,=e2x-y,y(0)=0dydxeye2xdxey=e2xe00解：y,+y=exy=e-=e-lnxx)（注意：因?yàn)榉?hào)輸入方面的原因，在題4—題7的矩陣初等行變換中，書(shū)寫(xiě)時(shí)應(yīng)把（1）寫(xiě)成①2）寫(xiě)成②3）寫(xiě)成③；…）4.求解下列線性方程組的一般解：2]EQ\*jc3\*hps23\o\al(\s\up7(1),5)4750453 +52]2]EQ\*jc3\*hps23\o\al(\s\up7(1),5)4750453 +52]-32-552-14-13-34-77 ||EQ\*jc3\*hps23\o\al(\s\up9(0),0) 喻-1]1-1]12-100 ||| 解：A=0-11-32L2-15-所以一般解為(x1=-2x3+x4(x1=-2x3+x4lx2=x3-x4EQ\*jc3\*hps13\o\al(\s\up0(3),x)2-1422-50-13042-50-1304-7 ||2]-30210-1350xx3-x4x3-x4因?yàn)橹華(A)=2，所以方程組有解，一般解為〈|x1=|x=l2其中x3,x4是自由未知量。5.當(dāng)λ為何值時(shí)，線性方程組(x1-x2-5x3+4x4=2|2x-x+EQ\*jc3\*hps23\o\al(\s\up11(3),λ)有解，并求一般解。2]-EQ\*jc3\*hps23\o\al(\s\up0(3),3) λ-14」-EQ\*jc3\*hps23\o\al(\s\up0(3),3) λ-14」2]-1]-11004-900-2]-1]-11004-900-500EQ\*jc3\*hps23\o\al(\s\up0(0),0) 喻λ-8」λ-8」可見(jiàn)當(dāng)λ=8時(shí)，方程組有解，其一般解為lx2