歷招收攻讀入學(xué)考試試題及講解數(shù)分

廈門(mén)大考試科目及代碼： 數(shù)學(xué)分x2tarctanI.0xsin2xtan設(shè)0x1c（其中c為常數(shù)xn12xnn(n1,廈門(mén)大考試科目及代碼： 數(shù)學(xué)分x2tarctanI.0xsin2xtan設(shè)0x1c（其中c為常數(shù)xn12xnn(n1,2收斂，并求極限limxnc2 1Idxx0x ..... ln(1x)x .2n sinF(x)1在區(qū)間[0,)連續(xù)，在(0,有連續(xù)導(dǎo)數(shù)6.I(x2y2SSx2y2z2R2z0)x2y2Rx7.I(z2x)dydz(x2y)dzdx(y2其中Szx2y2z0z1之間的部分,且積分沿曲面S下側(cè)12001年數(shù)分答x2tarctanI0xsin2xtanxt2arctanx2arctan0xx21 x0340xncn1nk0xkc(xnk1時(shí),xk12xk c,由于0xk2001年數(shù)分答x2tarctanI0xsin2xtanxt2arctanx2arctan0xx21 x0340xncn1nk0xkc(xnk1時(shí),xk12xk c,由于0xkc,所以(xc)2c2kkkcc即得0xk1c,所以0xnc(nN,x又xn1xnxn n(cxn)0,得xn1xn;故{xn}為單調(diào)有界數(shù)列,必n ca2a斂,我們假設(shè)limxna或者acxn0,所以取limxnacn1xn解:已知ln(1x1,Ix01 0xx1 ...(1)n1 n01(x 0 1 ...2(1 111 1111122n222n1n補(bǔ)充兩點(diǎn):上述能做逐項(xiàng)積分是因?yàn)閘n(1x我 有l(wèi)imln(1x)1,所以我們定義f(x)ln(1x),f(0)1xx(01設(shè)f(x)ln(1x)x ... (2n 1n1n補(bǔ)充兩點(diǎn):上述能做逐項(xiàng)積分是因?yàn)閘n(1x我 有l(wèi)imln(1x)1,所以我們定義f(x)ln(1x),f(0)1xx(01設(shè)f(x)ln(1x)x ... (2n 1從上可知x ... ln(1x) f(x)ln(1x)x...2n2(02 (2n 所以ln(1x)x ... ; (2n g(xtsinsin1sin1dt111t1tsing(xt)F(x)01t2dt在[0,tcosdt0costdtxgx(x,1tg(x,t)dt一致收斂,即x1t0F(x)的導(dǎo)數(shù)存在.又因?yàn)間x(x,t)在[0,(0,)上連續(xù),所以F(x)Sf(x,y,z)ds f(x,y,z(x,D1z2z2 其中z(x,y) R2x2y2,z,zxyR2x2R2x23f(xyz)dS(x2y2RdxdySD令xrcos,yrsin,其f(xyz)dS(x2y2RdxdySD令xrcos,yrsin,其中 x,0rRcos,于223RDr2Rrdr R5(x2y2)Rdxdy227.(GaussPz2x,Qx2yRy2原式3dxdydzy2VS21 (rsin 2200V544編者寄語(yǔ)我登上這列地周?chē)腔靵y，嘈雜，骯臟和4廈門(mén)大考試科目及代碼： 數(shù)學(xué)分證明數(shù)列{x}收斂,其中x1, 1(x3),n1,2...,求極限lim廈門(mén)大考試科目及代碼： 數(shù)學(xué)分證明數(shù)列{x}收斂,其中x1, 1(x3),n1,2...,求極限limxn1nn2xnfI上有定義,滿足xIx(xx,f在(xxI上有界(1證明,Ia,b時(shí),其中(ba0fI上有界(2)當(dāng)I(a,b)時(shí), f在I上一定有界嗎?eyxlnxxxy0(x1,cosF(x)證明dt在區(qū)間(1上連續(xù)可微t(1)(1)n1sin1收斂n(1)n1sin1(2)lim (lnsinn6Ix2dydzy2dzdxz2dxdy,Sx2y2z21,z0S7.計(jì)算積分I Lx2y2z2R2z0)x2y2Rx(R0)的交線,從z軸的正面看去按逆時(shí)針?lè)较?52002年數(shù)分答 1(x3)1233,所以xnn2 n310,所以xn又因?yàn)閤n1xn xn)n22002年數(shù)分答 1(x3)1233,所以xnn2 n310,所以xn又因?yàn)閤n1xn xn)n2可知數(shù)列極限存在,設(shè)limxa,對(duì)原等式兩邊取極限,得a1(a3)n a3,又因?yàn)閤n0,所以limxna 32.(1)證明:設(shè)u{(xxxxxIIBorelI,使得u'(xxxxI xxn iinnI[a,b](xix,xix)I[(xix,xix)I]iiiif(x在(xixxixIf(xIii(2Ia,b時(shí),f(xIf(x1x(0,1xg'(x)lnxygx0xeygx0xeygx0xeyg(e 0,從而 0,即得不等eyxlnxxy0(x1,ycos 1(1)11dt收斂,所以cos tttecostdt tgxt在(1,)(eF(x)te間(1,lntcost,由于tx6g(x,t)xtx11 (x1) x0ttgx(x,t)dt111(1)(1)n1 ,易知當(dāng)n時(shí), 0,且 是關(guān)于n單調(diào)減少的nnn(2)由上分析可知(1)n1sin1收斂,即關(guān)于xtx11 (x1) x0ttgx(x,t)dt111(1)(1)n1 ,易知當(dāng)n時(shí), 0,且 是關(guān)于n單調(diào)減少的nnn(2)由上分析可知(1)n1sin1收斂,即關(guān)于x是一致收斂的,又 關(guān)于nn(ln1單調(diào)遞減(x[)),其中01x(ln1 (ln內(nèi)閉一致收斂,又 n(1)n1sin(1)n1sin1lim nlim n (ln(lnnx0n36.由GaussI(2x2y2z)dxdydzV12 (rsincosrsinsinrcos)r220 127Stoze公式，LPdxQdy(R )dydz )dzdx )dxdy Pyz,QzRyRQ0,PR1,QP dxdydzdxR20R2447廈門(mén)大考試科目及代碼： 數(shù)學(xué)分1.設(shè){xn}xlim(1xn)nexn2f(x為[a,bf(xx0(a,bf廈門(mén)大考試科目及代碼： 數(shù)學(xué)分1.設(shè){xn}xlim(1xn)nexn2f(x為[a,bf(xx0(a,bf(x0)maxf'(x存在，f'(x)f003f(xf(0)0g(xf(x),xg(x) 1x 1dx，其中a0,b0為常數(shù)計(jì)算 ln0計(jì)算(xy)dxdy，其中x2y2xy的內(nèi)部(2n 的和函數(shù)(7.設(shè)yyyFy)x2 dxFy'(8.設(shè)x1y,x2yA2uC 8變換 0，證明,必為C2BA0的兩個(gè)相異的實(shí)根，其 2AB,CACB202003年數(shù).變換 0，證明,必為C2BA0的兩個(gè)相異的實(shí)根，其 2AB,CACB202003年數(shù).{xn}（xnx0f(xnA(n）由{xn}x，我們可知{xn} 0(n)時(shí)，我們先來(lái)求此極限lim(1x)nnn nxnexHeine定理，我們馬上得到lim(1xn)nexnn.f(xf(xx(xxf'(x)f'(x)fx 0 0有f(x)f(x0)xf(x)f(x0)xf(x)f(x0)f'(x)f('x) xx0x0f(x)f(x0)f'(x)f('x) xx0x0f'(x)0 )limg(xlimf(x(0型fxf0，所以g(x為連續(xù)函數(shù) 9xf'(x)fx0g(x'g(x)f(x)xf'(0)(0)f''，x20xf'(x)f(x)(0)f''又limx0g(x'g0gx存在且連續(xù)20111bb14.解：原式 xdy dy )xyyxx0aa011IF(y) )xdxxf'(x)fx0g(x'g(x)f(x)xf'(0)(0)f''，x20xf'(x)f(x)(0)f''又limx0g(x'g0gx存在且連續(xù)20111bb14.解：原式 xdy dy )xyyxx0aa011IF(y) )xdxyx01111x1111對(duì)于F(y) )xydx xy1 )xyxx00011111 )xdx )y )xy2y=yxx0011I，(y(y1)21xb11bdx (y1)21 ln0axrcos25.1yrsin222(xy)dxdy(rcosrsin 22rdrd 2 (2n3)6.解：設(shè)u(x)(2n1)x2n，由于un1x0nn(n1)!(2nun從而收斂域?yàn)?,)2n111 )2n1 xe(xx，(2n =(xe)(12x)e2x2)dxe2(y)y'(y)e2(y)y'((Fy((22(y)2(y)2x(2n =(xe)(12x)e2x2)dxe2(y)y'(y)e2(y)y'((Fy((22(y)2(y)2x(y)y(y)y x e(uuu，2u 2u ，8.，2 ]，11 221 212 12022(A2BC (A2BC22 22CA2，所以必為C22BA0的兩個(gè)相異的實(shí)根11 A2BC222我來(lái)自偶然像一顆塵我看遍這人間坎坷辛廈門(mén)大考試科目及代碼： 數(shù)學(xué)分一1．設(shè){xn}不趨向無(wú)窮大，則{xn}必有收斂的子列對(duì)，無(wú)界必有發(fā)散到無(wú)窮大的子列，非無(wú)窮大2．f(x)在非空開(kāi)區(qū)間(a,b廈門(mén)大考試科目及代碼： 數(shù)學(xué)分一1．設(shè){xn}不趨向無(wú)窮大，則{xn}必有收斂的子列對(duì)，無(wú)界必有發(fā)散到無(wú)窮大的子列，非無(wú)窮大2．f(x)在非空開(kāi)區(qū)間(a,b(a,b),x1x2(a,bx1x2，使f(x1)f(x2)f()x1f(xx3n1k3.設(shè)函數(shù)f(x)在區(qū)間[0,1]上有定義，且極限 f()存在，則此時(shí)f(x)nnk1nnf(k)n1k區(qū)間[0,1]上可積，且0f(x)dxlimDirichle函數(shù)4.設(shè)函數(shù)項(xiàng)級(jí)數(shù)fn(xI,上一致收斂，且fn則(1)n(xnxn1二．（1）liman其中an11...12（2）lim(x2y2 yx,2f(x在[a,f(a0,fa0xafx0f(x)02（2）lim(x2y2 yx,2f(x在[a,f(a0,fa0xafx0f(x)0在[a,bf(xdx0三、設(shè)有界函f(x)在[a,bf(xaf(x01 xln)n(ppnn四、討論 的變換x(uz2z2五、zz(xy滿足y(u,， 220六、若函數(shù)uu(x,yzu2u0,uu(x,yz)為該區(qū)域內(nèi)的調(diào)和函數(shù)。證明若 u(x,y,z)為區(qū)域{(x,y,z)R3:(xx)2(yy)2(z a}內(nèi)的調(diào)和00014a2u(x,y,z)u(x,y, S其中a0S為球2004二．大 1an1，其中a11，我假an12，那么an 1an12an又 1a1，2004二．大 1an1，其中a11，我假an12，那么an 1an12an又 1a1， n1a1n2 a，故{a}為壓縮數(shù)列，所以{a} nnlimaa則可得a1a，解得a5n2x2y2limln(x2y2 (xy)ln(xy yx,y0x2x,x2(x2y2因?yàn)?x2y2x2 x2令x2y2且lim(x2y2)ln(x2y2 limtlnt0，所以原極限e0x,x,.利用Taylorxa處Taylorf((xa2，(ax取區(qū)間為[aaf(a)]f(x)f(a)f'(a)(xa)f'2f(x在區(qū)間的兩個(gè)端點(diǎn)是異號(hào)的，所以在[aaf(a)f'理我們知道(xx 在[a,).0x(xxf(xa0002f(x)dxf(x)dxa2a0bbf(x)dxf(x)dxf(x)dx2aabf(xdx0a設(shè)a1xlnn)n，注意到limlnn0，所以.0x(xxf(xa0002f(x)dxf(x)dxa2a0bbf(x)dxf(x)dxf(x)dx2aabf(xdx0a設(shè)a1xlnn)n，注意到limlnn0，所以N,nN，有a0nnnn項(xiàng)級(jí)數(shù)。利用Taylorln(1x)x1x2o(x2)(x0可2aexp[nln(1xlnn)]exp{n[xlnn1(xlnn)2o((xlnn)2nn n32(xln13xln))]} )] 2 nnn21(1xln px1px1~npnzzx yu5.2(zz)z (zz)z，uuxx ux xy z，2z(zz)z(zz)z zx yxx xy x xy yy y2z2 zx zxxu zxyu zxyu zyyu22 和 ，zzz 結(jié)合22 2z26.f(au(xyz)dSM0x0y0z0xx0asinyyasinsin(0,020zz6.f(au(xyz)dSM0x0y0z0xx0asinyyasinsin(0,020zza f(a)0d0u(xyz)asind2f(a) u(x,y,z)sin'00(usincosusinsinucos)d2I asin2dxyza0(u(n,x)u(n,y)u(n,z))dS（n (a)的單位外法向量xyzd(uxsincosuysinsinuzcos)a2sindI(a00(ux(n,x)uy(n,y)uz(n,z))dSuxdydzuydzdx=(uxxuyyuzz)dxdydza2f'(a)1f(a) u(x,y,z)a1 ，令a0c22a c0d0u(M0)sindu(M0)224u(M0)1 f(a)14a2所以u(píng)(M) f u(xyz)dS，證 0S廈門(mén)大考試科目及代碼： 數(shù)學(xué)分一、判斷數(shù)列f(xn提示：錯(cuò)，數(shù)列廈門(mén)大考試科目及代碼： 數(shù)學(xué)分一、判斷數(shù)列f(xn提示：錯(cuò)，數(shù)列f(xnf(x)x0f(xnf(x0).2.f(x定義在(a,bf(xx0(a,b)limm()limM(其中m(infxxx)f(xM()supxxx)f(x 提示：由上可知lim(m(M(0，infx(x,x)f(x)supx(x,x)f 對(duì)于infx(x,x)f(x)supx(x,xf(x)f(x0)3.fnaxRf '(a)(xf''(a)......(xa)nf(n)f(x)f(a)(xa)fxaTalyor4.設(shè)f(x)在(a,b)[cda,b)(cd(cd)f(df(cf)(dc),則(c,d)是c,d(cdbnf(knkbf(x)在[a,b]Raf(x)dx.nbbnbf(a提示：錯(cuò)，將區(qū)間均分可得af(x)dxk)nbnf(knkbf(x)在[a,b]Raf(x)dx.nbbnbf(a提示：錯(cuò)，將區(qū)間均分可得af(x)dxk)nnk1.設(shè)x1, 1(x5),n1,2...證明x存在，并求xnn0n2xnf(x在區(qū)間[0,l上連續(xù)及當(dāng)0l時(shí)(x)2y2z20l u(x,y,z)dy20(x)2y2n3.設(shè)f(x)anxn的收斂半徑為，令fn(x) k限區(qū)間[a,bf(f(x4.設(shè)函數(shù)f(x)[a,b]上可積，證明存在[a,b]上的多項(xiàng)式函數(shù)列n(x)(n1,2...)使bb (x)dx f(x)dxnna1 XdYYdXI X2YCXaxby,YcxdyC為包圍原點(diǎn)的簡(jiǎn)單封閉曲線(adcb0)2005年數(shù)二．大1.證明： 1(x5)15 5n2 n55 2 x x n n nnnxn10,xn1xn，數(shù)列{xn}是單調(diào)遞減且有下界，所以limnxn接下來(lái)求limnxn，設(shè)limnxnl12005年數(shù)二．大1.證明： 1(x5)15 5n2 n55 2 x x n n nnnxn10,xn1xn，數(shù)列{xn}是單調(diào)遞減且有下界，所以limnxn接下來(lái)求limnxn，設(shè)limnxnl1(l)l f(x)連續(xù)可知， f( 也是連續(xù)函數(shù)，(x)2y22.1f()2(xf()(xl3d3d所以0[(x)2y2z2]0[(x)2y2z2]2 flld，同理我們可 d[(x)y3[(x)y50 0 f()luyd30[(x)2y2z2] f() flld d3252[(x)yz[(x)yz 200f()luzd30[(x)2y2z2]f(f()ll3d5d0[(x)2y2z20[(x)2y2z2] f()[(x)yz flldd所以 [(x)y3[(x)y50 0 ]3.證明：由f(x)的收斂半徑為，可知在任何有限區(qū)間[a,bf(x)（Abelfn(xf(x，所以10,N0,當(dāng)nNfn(xf(x)1對(duì)x[a,b]0,0,x，現(xiàn)在只要讓1f(x)f(x0ffn(xff(x))x[a,bffn（Abelfn(xf(x，所以10,N0,當(dāng)nNfn(xf(x)1對(duì)x[a,b]0,0,x，現(xiàn)在只要讓1f(x)f(x0ffn(xff(x))x[a,bffn(x在任何有限區(qū)間[a,bff(x4.[a,b作n等分，設(shè)分點(diǎn)為a(n)x(n) x(n)x(n)即 1ai(ba),iin(x為過(guò)點(diǎn)在 (n),f(x(n))][x(n),f(x(n)及iiinix(n),x(n) xn(x)f(xi1)[f(xi)f(xi1x(n)i則n(x是[a,b若令m, 及 分別表示函數(shù)f(x)在[x(nx(n上的下確界和 iiii(n),x(n) (x)(x)M(n),m(n)f(x)ifniin x(nnbbbf(x)(x)dxf(x)dx f(x)(x)dx i(n nnn xbn 0時(shí)， nbb所 f(x)dx (x)dxnaaX2Y20PYX5.P，QX2Y ax，行列式adcb0xyX,Y又YcxaxbyR,0YcxdyRxRdcosaxbyR,0YcxdyRxRdcosRbadadbcy R2cos2R2sin2 XdY XdYI d XlX2Y022廈門(mén)大考試科目及代碼： 數(shù)學(xué)分一判斷2.f(x)為可微函數(shù)，則f'(x)dx(f(x)dx)'F廈門(mén)大考試科目及代碼： 數(shù)學(xué)分一判斷2.f(x)為可微函數(shù)，則f'(x)dx(f(x)dx)'F(xFxfx)dxf(x)dxFx(f(x)dx)'f(x)，所以Fxf37（對(duì)1提示錯(cuò)，f(x) x,x(0,1]其導(dǎo)數(shù)為f'(x)x0,fx2xsgn(xr1.我們將有理數(shù)排成一個(gè)數(shù)列r n的連續(xù)性f(x)n2.f(x在(,a(,)x(,)1x [f(th)f(t)]dtf(x)fh0a3.x0(a1a2an是nnf(x)f(x1,x2,...xn)ijxixi,jn在單位球{x(x,x,...x)Rn x21}內(nèi)的極值點(diǎn)，則必存在R使得 nAx0x01n12An221n12An22nnn1f(axbycz)dxdydz u)f(ku)du,其中k a2b2c202x2y2z2四、設(shè)a,bc為常數(shù)。證明由方程axbycz(x2y2z2確zz(xy(cybz)z(azcx)zbx2006二．大1.證明：設(shè)u(x)sgn(xrn，對(duì)于nN，xR1，由1nn斂，所以由M判別法可知u(x)sgn(xrn)n10當(dāng)r時(shí)，u(xx處連續(xù)，取2006二．大1.證明：設(shè)u(x)sgn(xrn，對(duì)于nN，xR1，由1nn斂，所以由M判別法可知u(x)sgn(xrn)n10當(dāng)r時(shí)，u(xx處連續(xù)，取0充分小時(shí)，u(x在(,nnn仍連續(xù)，又因?yàn)閡n(x在(,f(xx20rk1,2...時(shí)，f(xsgn(xrn在，ksgn(xrxrxr處間斷，f(xxr處間斷kkkk1x2.證明：左邊= [f(th)fh0a1x= f(t)dt f halimf(xhf(ah(L'Hospital法則=f(xf(af(xx2x2 x2a(0a1 nnL(x1x2xni,jxx(xxxaxaa,...a應(yīng)滿 2iji n 0，即xxxx0(ii1 i2in ix1x1x1 a1x xx aA22其中22... ...x ax xn nn n4.axbycz0lk1{abcuk1(axbycz),va1xb1yuk1(axbycz),va1xb1yc1z,wa2xb2ycJ1D(uvw)1則由正交變換性質(zhì)可知：{(uvwu2v2w21}D(x,y,1f(ku)Jdudvdw f f(axbycz)dxdydzx2y2z2u2v2w2v2w21= u)f25.證明：對(duì)方程axbycz(x2y2z2a2x'x球偏導(dǎo)acz(2x2zzzx2z'xb2y''(2y2zy求偏導(dǎo)by，解zy2z'代入(cybzzazcx(cybz)a2x'(azcx)b2y'(bxay)2z'(bx2z'bx2z'2z'廈門(mén)大考試科目及代碼： 數(shù)學(xué)分一判斷若{xn}無(wú)界，則limnxn無(wú)界廈門(mén)大考試科目及代碼： 數(shù)學(xué)分一判斷若{xn}無(wú)界，則limnxn無(wú)界至少存在一個(gè)發(fā)散的子列；非無(wú)窮大若{xn}無(wú)界，則{xn}提示：對(duì)，從上我們可知，若{xn}是無(wú)界的，設(shè)其為是非無(wú)窮大的，明顯個(gè)子列收斂情況是不一樣的，可知其是發(fā)散的。也可以考慮其逆否命題：若{xn是收斂的，則{xn}有界，顯然逆否命題是從3.若{xn}單調(diào)有下界，則{xn}提示：錯(cuò)，一個(gè)數(shù)列單調(diào)上升到無(wú)窮大，但是有下界，明顯{xn}是發(fā)散的4.若{xn}收斂，則{xn}有界設(shè)函數(shù)fn(x為閉區(qū)間[abf(xf(xf(xf(c)3.設(shè)f在c處右可微， f'RfRx在0使得對(duì)所有t(ccf(tf(c0f(ak(ba))]n[14.設(shè)f(x)在區(qū)間[a,b上連續(xù),求極限nnkns1s1時(shí)發(fā)散，其中[a表示ans1s1時(shí)發(fā)散，其中[a表示a226.設(shè)uf(r)，r x2y2z2，f為兩次可微函數(shù)。證明uF(r)，其u2uF(r) 2007年數(shù)分答二．大設(shè)數(shù)列{an}單調(diào)上升且有上確界，則nNan又對(duì)0，存在aN，使得aN，又因?yàn)閍n單調(diào)上升，所以當(dāng)nN時(shí)，有an，即an2007年數(shù)分答二．大設(shè)數(shù)列{an}單調(diào)上升且有上確界，則nNan又對(duì)0，存在aN，使得aN，又因?yàn)閍n單調(diào)上升，所以當(dāng)nN時(shí)，有an，即an所以an2.fn(xf(x30N當(dāng)nNf(x)f，對(duì)x[a,b]都成立n0,xf(x)f(x0n 3f(x)f(x0f(x)fn(x)fn(x)fn(x0)fn(x0)f(x0 f(x)f(x)f(x)f(x)f(x)f(xnn 0f(x3.證明： A0，由極限的保號(hào)性知,存在0,使得當(dāng)0xc時(shí)fRf(x)f(c)A0x(cc)fxc0Rx2f(x)f(c)0證明:[a,b上的連續(xù)函數(shù)在[a,b上必可積.參見(jiàn)課本把[a,b作n等分,分點(diǎn)為abakk1,2,...nn[aba(k1abak中取點(diǎn)abakff(abak)kknnnn是f(abanbabf(x)dxnk1所以limn[abbf(a))]f(x)dxnknan{111}(I(n2[(n1)2111設(shè)u ，級(jí)數(shù)(I中共有2n1項(xiàng)，前nn(n2[(n1)1所以limn[abbf(a))]f(x)dxnknan{111}(I(n2[(n1)2111設(shè)u ，級(jí)數(shù)(I中共有2n1項(xiàng)，前nn(n2[(n1)21n21后n1和(ns1222u(x)u0，又因?yàn)閧unn(nn以(I即設(shè)原來(lái)級(jí)數(shù)的部分和數(shù)列為{Sn，而級(jí)數(shù)(I部分和數(shù)列為{n，那么nNmmSnm1或者m1Snm，從Snm10s1u0，n2y2x6.證明：uxf' ，uxx 'uyyuzz可以類似求得rrry2所以u(píng)xxuyyuzz 'rrx2+f f'rx2+f f'rf'rf''=2f所以u(píng)F(r)=f r廈門(mén)大考試科目及代碼： 數(shù)學(xué)分一．判斷f(xIR設(shè)廈門(mén)大考試科目及代碼： 數(shù)學(xué)分一．判斷f(xIR設(shè))x（4）f(xI,上有界且limxf(xf(xI.1.設(shè){x}為有界正實(shí)數(shù)列，求nxx... nlimxg(xu0f在uu0處連續(xù)。證明：2.設(shè)limxf(g(x))f(u03.f(x在(,1(nxCex其中Cf(n)(x)f(n1)(x)f4.設(shè)DR2若f(xy))2dxdyD5.設(shè){nan}收斂，級(jí)數(shù)n(5.設(shè){nan}收斂，級(jí)數(shù)n(anan1收斂，證明anIcos(axbycz)dxdydz，其6.計(jì)算在單位球Vx2y2z21Va,bc2008年數(shù)分答二．大1.解：設(shè)Snx1x2 S存在時(shí)，那么 x0，所以當(dāng)nnxx... n當(dāng)limnSn時(shí)，因?yàn)閧xn}xnM（2008年數(shù)分答二．大1.解：設(shè)Snx1x2 S存在時(shí)，那么 x0，所以當(dāng)nnxx... n當(dāng)limnSn時(shí)，因?yàn)閧xn}xnM（M0數(shù)，所以xx... n2.limxg(xu010M0,Mg(x)f(u在uu0f(u)f(u0)00,u我們?nèi)?，所以f(g(xf(u0Mg(x)x3.1f(np)(x)f(n)(x)f(n1)(x)由f(np)(x)f(n)f(np1)(x)f(np1)(x)....f(n1)(x)f(n)11f(np)(x)f(np1)(x) f(n1)(x)f(n)p(n(n考慮級(jí)數(shù)1，該級(jí)數(shù)是收斂的，所以由Cauchy收斂準(zhǔn)則0,NnN11，對(duì)pN(n(n對(duì)pN都成立由Cauchy收斂準(zhǔn)則可知 f(n)f(np)(x)f(n)()nny'y111yedx(o() dxCCedxAedxo()edxC'exBo(A'nnn為常數(shù),所以當(dāng)n時(shí),有l(wèi)imnny'y111yedx(o() dxCCedxAedxo()edxC'exBo(A'nnn為常數(shù),所以當(dāng)n時(shí),有l(wèi)imnf(n)(xCe'x其中C'4.證明：反證法，fD中的連續(xù)點(diǎn)(x0y0上的取值不為0，當(dāng)x(x0x0f(x0,y0)A，Ay(y,D，(xy)Df(xy0002因此f(xy))2dxdyf(xy))2dxdyf(xy))2dxdyf(xy))2DA20，這與f(xy))dxdy0dxdy244Dnnn(i1)(ai1ai)nannan(i1)(ai(iinn又(i1)(aiai1i(aiai1)，n(anan1)收斂又 }單調(diào)有界ii{nan 1Abel判別法可知(i即ian6.解：作坐標(biāo)系的旋轉(zhuǎn)變換，將oxy旋轉(zhuǎn)到平面axbycz0axbya2b2這時(shí)x軸和y軸被旋轉(zhuǎn)到0的平面內(nèi)，把它們分別記為,軸，根據(jù)解析1,{(,,)222J記k a2b2c2，則Icos(axbycz)dxdydzcos(kVsin41 sin41 )cos(k)d 2coskk0廈門(mén)大考試科目及代碼： 數(shù)學(xué)分一選擇1.設(shè){an}為單調(diào)數(shù)列，若存在一收斂子列{an}，這時(shí)（AjB.{an}C.{an}j廈門(mén)大考試科目及代碼： 數(shù)學(xué)分一選擇1.設(shè){an}為單調(diào)數(shù)列，若存在一收斂子列{an}，這時(shí)（AjB.{an}C.{an}j2.f(xR上為一連續(xù)函數(shù)，則有（CIAIf(IBf(I為閉區(qū)間時(shí)DA.B.C都不一3.設(shè)在某去心鄰域U0x0 ]'(x) f'(x)=；Ｂ．若在連續(xù)，則A0flim004.f(x在[a,b]R.]f(x)在[a,bf(xf(xf(x在[a,b5.設(shè)un．[DＡ．若limu0,則收斂，則limＢ． nnn收斂，則 nn1.設(shè)a0,0a1(a1(a),n1n2a2an1.設(shè)a0,0a1(a1(a),n1n2a2an2.f(x在區(qū)間[a)g(x在區(qū)間[a)limx(f(x)g(x))f(x)在區(qū)間[a)3.f(x在(,f(0)f0)0.0,xg(x)f,xxg(x在(,)sinn（1）級(jí)數(shù)在(,)（2）存在，使得 22009年數(shù)分答二．大證（單調(diào)有界必有極限從題中a0所以 1(a)12nnn2 ann即數(shù)列{an}有下2 a1a a 1a0，即 n nnn2 所以數(shù)列{an}是單調(diào)遞減的，所以{an}必有由 2009年數(shù)分答二．大證（單調(diào)有界必有極限從題中a0所以 1(a)12nnn2 ann即數(shù)列{an}有下2 a1a a 1a0，即 n nnn2 所以數(shù)列{an}是單調(diào)遞減的，所以{an}必有由 1(a)，兩邊令n，設(shè)limal，則l1(l)，解得lnn2a n2.f(xg(x00,ax時(shí)，f(x因。3g(x)在區(qū)間[a上一致收斂，故對(duì)0,10,1x''g(x'')g(x')xx',x'31f(xg(xf(x'')f(x'f(x'')g(x''g(x'')g(x'利用Cantorf(x在[a1]020,xx'[a1]xx2時(shí)，f(x'')f(x'取min{1,2}xx'[axxf(xf(x，首先 g(x) f(x)(0型)fxf00，所以g(x) x0g(x在(,g(x)f' f''f(x)(0)g'(0)x20g(x)(f(x)'xf(xf'xf'(x)f(x)(0)f''(0)g'(x)x02g(x在(,)4.解：（Guass公式的運(yùn)用Pxy2,Qyz2Rzx2，由GuassPQRVxydydzyzdzdxzxdxdy222 dxdydz (xyz sVxrcosyrsin,5V11(xyg(x在(,)4.解：（Guass公式的運(yùn)用Pxy2,Qyz2Rzx2，由GuassPQRVxydydzyzdzdxzxdxdy222 dxdydz (xyz sVxrcosyrsin,5V11(xyz)dxdydz (rz)rdr 30nsinsinn2（1）證：設(shè)un(x ，級(jí)數(shù)nsinn拉斯判別法可知級(jí)數(shù)在(,)（2）證：設(shè)a(x)ncosx 1n斯判別法知an(x)是一致收斂的，又剛好an(x)un(x)的導(dǎo)數(shù)形式，所以sinnn2sinnsin2=f(x)，問(wèn)題現(xiàn)在轉(zhuǎn)化為：存在 )，使得f() '= 2sinn222cosf'(x)(2sin (2sin110xcos1]x0時(shí)g(0)4g(x)(2sin621g() )0，由介值定理可知存在 ) )，使 662 .廈門(mén)大考試科目及代碼： 數(shù)學(xué)分一選擇fx3[fx)]22exf(x0x1f(x)0f(xf(x00f(x廈門(mén)大考試科目及代碼： 數(shù)學(xué)分一選擇fx3[fx)]22exf(x0x1f(x)0f(xf(x00f(x（BB.取極小值；C.不取極值；D2f(x)lnxxk,(k0)在區(qū)間(0,（CeBCA.D3.已知當(dāng)x0時(shí)，函數(shù)etanxex與xn為同階無(wú)窮小，則n （CABCD（AAf(x在[abRiemannbf(x)dxF(b)FaxBf(x在[abRiemann可積，則f(t)dt在[abaf2x在[abRiemannCf不一定在[abRiemannfD若在[abRiemannf(x在[abRiemanna5.設(shè)a0f(x在(aa（Bfx0x，I f(x)dx2fBI0CI0IAD1．設(shè)f(x)a1sinxa2sin(2x) ansin(nx)，f(xsinxai(i1,2...n實(shí)常數(shù)，證明a12a2 2.f在(,上一致連續(xù)，0，在(,Igf(z):yz(xx)}，證明g在(,g(x)sup{f(f(z):yz(xx)}，證明g在(,g(x)sup{f(y)xIf''(x)011aaaf a00x4.設(shè)f0(x)在區(qū)間[0,a]上連續(xù)，令f(x) fn1(t)dt,nn05.f(xyax22bxycy2x2y21其中（b2ac0,a,bc06.zz(xyF(xzyz0Fyxxzyzxy 2010年數(shù)學(xué)答二．大1.f(xsinxf(00f'f(xcosnx，令x0并取其絕對(duì)nf'(x)acosx2acos2x 12 f'2.f在(,上一致連續(xù)，所以0,0x''xxI2010年數(shù)學(xué)答二．大1.f(xsinxf(00f'f(xcosnx，令x0并取其絕對(duì)nf'(x)acosx2acos2x 12 f'2.f在(,上一致連續(xù)，所以0,0x''xxIf(x'')f(x'g(x'')g(x'f(y'')f(z'')f(y')f(z'現(xiàn),x'')，y',z'(x',x')y'',z''supf(y'')f(z'')supf(y')f(z'f(y'')f(z'')f(y')f(z'f(y'')f(z'')f(y')f(z'y''y'f(xg(x'')g(x'f(z')f(z'')}f(y'')f(z'')f(y')f(z'sup{f(y'')f(y')g在(,3.證明：（Jessen不等式nn是下凸函數(shù)，所以f(xf(xfx0我們可知函數(shù)i nnaina1，上述可以用歸納法證明。又(xixi a(t)dtnn0n1n1n n11af( aa(x))，所以f f[(t)]dt))))(t)dt)ffiin a00sinxfxf(x)fxfx0我們可f是下凸函數(shù)，所f(xf(xfx)(xx0001f 11aaaa(t)dt，則f(x)f (t)dt)a(t)dt)(x0aaa0000x(t111aaaf((t))f (t)dt)f (t)dt)((t) 'fx0我們可f是下凸函數(shù)，所f(xf(xfx)(xx0001f 11aaaa(t)dt，則f(x)f (t)dt)a(t)dt)(x0aaa0000x(t111aaaf((t))f (t)dt)f (t)dt)((t) 'aaa000兩邊都在區(qū)間[0,a111aaaaaa(t))af (t)dt)f (t)dt (t)dt)af 'faaa00000011aaa則f (t)dt)a00f0(x在區(qū)間[0,a上連續(xù)，故在[0,a0，使得4.證明：因Mf0xxxf0(t)dtMxf(t)dt tdt；f21000xdt0對(duì)x[0,a]；fn(n(n05.axby0， f'fbxcy bc0，故只有唯一解(0,0)f(0,0)x2y21Lax22bxycy2(x2y20，得到方 (a)xbybx(c)yx2y21上(x,y)(0,0)a cac (ac)2得1,22f(xyax2bxycy22x2y21aa cac (ac)2得1,22f(xyax2bxycy22x2y21ac(ac)2maxf(x,y)max{0,1,2}2ac(ac)2minf(x,y)min{0,1,2}26.F(xzyz0xyyxz)0，F(xiàn) zz)F(1zy)01yF(11z(11 2yxxFF12 1，zxxxyyF F 1 1 xzyzxy 廈門(mén)大考試科目及代碼： 數(shù)學(xué)分一 選擇1yf(xx2'(x)00''(x)0，則（Af B廈門(mén)大考試科目及代碼： 數(shù)學(xué)分一 選擇1yf(xx2'(x)00''(x)0，則（Af BCD2.f(x)lnxx在區(qū)間(0,內(nèi)的零點(diǎn)個(gè)數(shù)為（ABCAD3.x0時(shí)，函數(shù)esinxexxn為同階無(wú)窮小量，則n（CABCDAf(x在[a,bRiemannF(xx f(x)dx]'fafBf(x在[a,bRiemann在[a,bf2x在[a,bRiemannCf一定在[a,bRiemann若f在[a,bRiemannf(x在[a,bRiemannD1x，I 45.f(x在(1,1)f I0 I0I0CD1.f(x在[0,1]f(0)f(1)0minf(x1存在(0,1)，使二、fC[0,1 f()1x2dx f2三、f(x在[0,1上連續(xù)，證明t01f(x)dx二、fC[0,1 f()1x2dx f2三、f(x在[0,1上連續(xù)，證明t01f(x)dx]2f210t2x[dxt 0四、設(shè)01limanaa1lim(na a a).01 五、設(shè)fn(x) cosx,x[1,1],n為正整數(shù)，證明（1）fn(xx[1,111f(x)dx limf(x)dx（2）nnnf(a17.設(shè)f(x)在a點(diǎn)可微，且f(a)0，求極限 nf8.計(jì)算[0,][0,1]y1kk.9.計(jì)算2011年數(shù)學(xué)答二．大f(x10可知函數(shù)的最小值在內(nèi)部達(dá)到，所以x0(0,1)1.證明：由f(x)minf(x1f(xf')0000,(0,1)使得xx0處按Taylor0f(0)f(x)1f''()(0x)2；0f(1)f(x)1f''()(1x)22011年數(shù)學(xué)答二．大f(x10可知函數(shù)的最小值在內(nèi)部達(dá)到，所以x0(0,1)1.證明：由f(x)minf(x1f(xf')0000,(0,1)使得xx0處按Taylor0f(0)f(x)1f''()(0x)2；0f(1)f(x)1f''()(1x)20000222212f，f''()所以當(dāng)x(0,]時(shí)，f''() 80(1200012x[,10(12f2.x dx f(0) ] [f()f(0)] lim f(1110 f()fdx10該反常積分是收斂的，由Cauchy1對(duì)0,AAAA 1x1x1x1A0[f()f1x 1xdx [f()f dx [f()f 120x0x[0,A]，又，所以0，當(dāng)取的適當(dāng)大時(shí)fxf(0)，所以當(dāng)fC[0,10x1x1AA0[f()f(0)]1x2[f()f(0)]1x2dxM現(xiàn)在考慮第二個(gè)積分：當(dāng)x1x1A[f()f(0)]1x2A[f()f(0)]1x2dxMx1所以 [f()fdx0101f(x) fdx]21f21111 dx]2dx[dxtttt2 t20000d(x)1arctan(1)1 dx11又xtx1x