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2023年经济数学基础形成性考核册及参考答案完整版.docx

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    • 经济数学基础形成性考核册及参考答案作业(一)(一)填空题x-sinx1. hm .答案:0x-*o x2.设 /(x)=r5 E0,在* k, x = O=0处连续,则* =— .答案:13. 曲线y = G在(1,2)的切线方程是 .答案:y = ^x + |4. 设函数/lx +1)=亍+ 2工+ 5,则f(x) = .答案:2% 5. 设f(x) = xsinx,则广(宣)= .答案:-y(二)单项选择题1.当X- + 8时,下列变量为无穷小量的是( )答案:DA. In(l + x)B.2Xx + \1sin xc. e 了D.X2 .下列极限计算对的的是()答案:B1x1IxlA.lim 旦=1B.lim 旦=1IX.F X1sinxC. hm xsin — = 1D. lim =1I X> X3.设y = \g2x,则dy = ().答案:BA. —dr B. —!—drC.蛀心 D. IcLv2x xlnlOX X4. 若函数/ (x)在点&处可导,则( )是错误的.答案:BA.函数f (x)在点为处有定义B. lim f(x) = A,但Af(x0)C.函数/(x)在点旳)处连续 D.函数f (x)在点xo处可微5.若 f (-)=x,则 f,(x)=( ).答案:BxA . ―— B.— —— C. — D.——(三)解答题1.计算极限⑴ "一2)心i x2-\ 2 (x — ])(x+l) (2)liin4zj£±6=lim(x-2)(x-3)12 x2 -6x + 8 E(x-2)(x-4)lim王纟(x + 1)职(工-4) 2⑶岫正丑=临也寻匝mD Y 5 X(V1-X + 1)-x-1=_1=lim —— = lim , x(Vl-x + l) (Vl-x + 1) 2]_3+_5_..x~ — 3x + 5 xx2 1(4)hm = hm—~~ =-x*3t +2x + 4 2 4 33 1 T5xsin 3x 3 3X X, l r sin 3x r(5)lun = lunsin 5x x^° 3xsin 5x 5 5⑹lim "_4 一蜘“ _2)S + 2)=4 i2 sin(x-2) i2 sin(x-2)2.设函数f(x)=-xsin —+ /?,Q,sinxxx<0x = 0,x> 0问:(1)当a,b为什么值时,f(对在x = 0处有极限存在?(2)当a,b为什么值时,/(X)在x = 0处连续.lim /(x)= lim (xsm-+b) = b 答案:⑴F F . xlim /(x) = lim — = 1W -O' x当b = \,a任意时,f(x)在x = 0处有极限存在;(2)f(0)= a =lim/(O) = ^ = l x->0当a = b = l时,/(x)在x = 0处连续。

      3. 计算卜.列函数的导数或微分: (l)y = x2+2x+log2x-22,^/答案:HF2F容案 y _ a(cx+ - c(ax+b) _ ad-cb(cx+d)2(cx+d)2y- <32y/(3x-5)3答案:j = -?J= = (3x-5)'7V3x-5⑷ y = 4x-xex,求 y'答案:y = —^-(x + De、2y/x (5) j = sin/?%,求dy答案:y' = (ex + eav(sinZ?x)z=“e* sin bx + eat cosbx • b=eax (asin bx+b cosbx)dy = e av (s in Z>x + Z? cos bx)dx(6) j = ev +x4x,求 dy(7) y = cosVx-e r » 求 dy答案:dy = (2xe-『・(8) y = sinH x + sinnx,求 y'答案:y' = nsmni xcos x+cos/irn= z?(sina_l x cos x +cos fix) (9)y = ]n(x+yli+xi),求y'案:y' = (x + Vl + X2y = (l + -(l + x2) 22x) = 0+ /X『X + Vl + X X + yjl + X 乙 X + Vl + X" VI+X(10) "2呻+】+气J屈,求VCO(—2 』ln2 1y/X答案:y,= _ --/7+yX^2 • 1 2 6x sm —x4. 下列各方程中y是x的隐函数,试求),'或dy (1 )x2 +y2 -xy + 3x = \,^.dy答案:解:方程两边关于X求导:2x +2yyf -y-xy, + 3 = 0(2y-x)/ = y-2x-3 ,dy= J~3~2xdx2y-x(2 )sin(x+y) + e" =4x,求 y'答案:解:方程两边关于X求导cos(x + y)(l + y1) + exy\y + xy') = 4 (cos(x + y) + exy^y' = 4 - yexy - cos(x + y),_ 4 -yexy - cos(r + y) W +cos(r+ y)5. 求下列函数的二阶导数:⑴ j = ln(l + x2),求dx答案:y"=2-2x2(l + /)2(2) y = ^,^y"及 y"l)答案顼弓方戸"E作业(二)(一)填空题.答案:2rln2 + 21.若J/3dx = 2”+2x + c,则/(x) =2. J(sinx),dr =答案:sinx + c3. 若 J f(x)dx =F(x) + c 测 J e ' /(g ")dr =4. 设函数 ^pln(l + x2)dr =,答案:— F(D + c.答案:05.若 PW = ^-j==dt,则 P'(x)=•答案:一志(二)单项选择题1.下列函数中,()是A-s i IU-?的原函数.A. —cosx22B ,2cos.vC. — 2 co s .v2答案:D2.卜列等式成立的是(A. sinidr = d(cosv)B.ln xdx = d(—)C.2xdx = —d(2x)«In 2答案:C3 .下列不定积分中,常用分部积分法计算的是().A. jcos(2r + l)dr,B. |xvl-x2dxC. xsin2'VdxD.X1 + x2答案:c4.下列定积分计算对的的是(A.jl2Adx = 2C. f 7 (x2+x3)dr = O* -71答案:D5,下列无穷积分中收敛的是(r +8 1 r +8 1A・[~dx B・[歹心答案:BB.>16i dx = 15D. I sin Adx = 0J -Ae'dxCsinAdx(三)解答题1.计算下列不定积分(l)f—dxJ evy答案:f^-dx=((—)'dx=-^- + c% e 启e⑵J罕h答案:j"顼 dx = j("2;')心=r(x4 +2x2 +x&(kJ Jx J >Jx 」+ -x^+cdr答案:5J—― dr = j(x- 2)dx = -^x2 - 2x+c答钊为蚌侦商d(f)Tm|fw(5) jxv2 + x2dx答案:Jxj2 +》dx = Sjj2 + Fd(2 + ;)=!(2 + x2)^+c(6 )J#位答案:= 2jsinVxdVx =-2cos7x+ c(7) jxsin^dx答案:Jxsin;dx = _2jxJcas|dx= -2xcos—+ 2fc<>s—d^=-2xcos- + 4sin- + c2 J 2 2 2(8) jln(x+1)ck答案:j ln(x+1 )dv = j ln(x +1 )d(x 4-1)= (x + l)ln(x+l)- j(x +1 )dln(x +1) = (x +1)ln(x + \)-x+c2 .计算下列定积分(1仆栖答案:j2Jl-.¥|dx=Jli(l-x)dx +Ji2(x-l)dA- = (x--ix2)|l.l +(^x2-x)⑵萨I答案:《事〜广力卜-屛⑶-扁 答案:, I dr =. 1 d(l + 1ILV)=2(l + ln xy f =2(4 ) Rxcos2«ukJo[色 I 1 — 1答案:Rxcos2Adx = — f 2xt/sin 2x = —xsin 2x J ——[2sin2xdx=—— Jo 2 2 2」° 2(5)j\lnAdx-答案:Adx = ;j:ln 血/ = !尤2 批 x|e _ j^^in x = l(e2 +1)(6 )匚(1 + 官*答案:j*:(l + Ae-")dx=* - J*xde~x =3-xe~x\^ + J*e_tdx=5 + 5e^作业三(一)填空题1.设矩阵厶=,则A的元素。

      23 = ■答案:3-12.设均为3阶矩阵,且舛=固=—3测2人码= .答案:一 72.答案:3.设A, 8均为〃阶矩阵,则等式(A - B)2 =A2-2AB + B2成立的充足必要条件是. AB=BA4.设A,B均为〃阶矩阵,(1-8)可逆,则矩阵A+BX = X的解X = 答案05.设矩阵A =0-3,则 A-1 =.答案:A0\_2(-)单项选择题以下结论或等式对的的是().A.若A, B均为零矩阵,则有A = BB. 若AB=AC,且A#O,则B = CC. 对角矩阵是对称矩阵D. 若A「O,B女O,则AB^O答案C)矩阵2.设A为3x4矩阵,B为5x2矩阵,且乘积矩阵ACBT故意义,则为(A. 2x4B. 4x2C.3x5D.5x3答案A3.设A, B均为〃阶可逆矩阵,则下列等式成立的是(A.(A+B)-' = A-'+B-',C. |A&=|明D. AB = BA答案C4.下列矩阵可逆的是(0 -1B.A.0 0D.答案A5.矩阵A =-10-3-14的秩是(A. 0B. 1 C. 2D.3 答案D三、解答題1 .计算I 0-2(1)(2)(3) [-1=[0]00-120 0D 02-3 0。

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