2005年浙江省普通高校_专升本_联考《高等数学(一)》试题及答案

1、. .- a l- a yve e -ea-ja l.lg as-ei s-c u i aea-fas“ 一. . .50 000 y- s r g - a - - vs a x, -e .eN - - kec s d-l.glge2005年浙江省普通高校“专升本”联考高等数学(一)试卷题号一二三四总分得分:业专考报 校学考报二二二二二二二二二二二：号证考准二一二二二二二：名姓考试说明：1、考试时间为150分钟；2、满分为150分；3、答案请写在试卷纸上， 用蓝色或黑色墨水的钢笔、圆珠笔答卷，否则无效;4、密封线左边各项要求填写清楚完整。一、填空题：(只需在横线上直接写出答案，不必写 出计算过

2、程，本题共有 8个空格，每一空格 5分, 共40分)得分阅卷人士 61 - - - - - - - - - III - - - - - - - - - I1 .函数y:2sinx_ ex的连续区间是x (x -1)12. lim x > ： x(x ，x2 -4)3. (1) x轴在空间中的直线方程是 (2)过原点且与x轴垂直的平面方程是4.设函数f (x)=在点x=1处连续.5.设参数方程(x 1)2a,bx 1,(x)2e,x 1x =1 ，当 a =x ：二 1,b =时，函数f (x)x = r2 cos冬 、y = r3 sin 2 日(1)当r是常数，8是参数时，则dy =

3、dx(2)当8是常数，r是参数时，则dy =dx得分阅卷人且f (c) = 0 ,则当二.选择题.(本题共有5个小题，每一小题 4分, 共20分，每个小题给出的选项中，只有一项符合要 求)1 .设函数 y = f(x)在a,b上连续可导，c ( (a,b),()时，f (x)在x = c处取得极大值.cnsumebehav .us cse s to -l with. T heedea nts'-Uy > pub*_ats ' duy coniumplon _ smew heebew eea-r.a. is m. . ncy. F-t.me" -vans '

4、; d .y cnnum pin Ie- aebg aWaks NPC -a nd OPCCm -bes ad theb. masss ae concennd abou. Slcondy, Ie2(A)当 a(B)当 a(C»:二 c 时,:二 c 时,:二 c 时,(D)当 a E x <c 时，2.设函数y = f (x)在点f f f f(x) > 0 ,当 c <x Wb 时, (x) > 0 ,当 c <x Eb 时, (x) < 0 ,当 c < x W b 时, (x) <0 ,当 c < x Wb时,f f f f

5、(x) 0,(x):二 0,(x) 0,(x)：二 0.x =x0处可导，则f(X0 3h) - f(x0 -2h)(A)f (xo),h(B)3f (x°),).(C)4 f (x。),(D)5f (x。).-x2e3.设函数 f(x) =<0,-x2-ex 0x = 0,则积分x : 01f f(x)dx=(1-1(A) -1,(B)0(C)-,e(D)2.4.可微函数z = f (x,y)在点(x0,y0)处有 乌=乌=0是函数z = f (x, y)在 二 xcy点(x°, y°)取得极值的(A)充分条件，(C)充分必要条件，).(B)必要条件，(D

6、)既非充分条件又非必要条件5.设级数£ an和级数£ bn都发散，则级数£ (an+ bn)是(n 1(A)发散, 敛.n 1(B)条件收敛,nd(C)绝对收敛,(D)可能发散或者可能收三.计算题： 答案的不给分 分)(计算题必须写出必要的计算过程，只写,本题共10个小题，每小题 7分，共= (x2 -x-1)x的导数.得分阅卷人70e 2 004 t-2005年浙江省泞通高校“次升木”状号中的极大值，极小值322 .求函数y=x 2x +1在区间(一1, 2)3.求函数f(x)x2ex的n阶导数dn f dxns divinga v>i c. a cd.l

7、 c. a, 1, de c - .cl- ha yua n. Thi - hgh cost Suvng anumpin san onrfrmsuporl.ar. orm - -onstai- is .po- an- ug.b- so- it h. "mp0 .- yeas 0.- i. ta a ng. p. c,-m us b. .pr ov- Imp. . .Is t.h. tt " -prinCp.a is, lh.VsUIc s.val s' duly nd.-a .air. on, s no S.cnd,a .pl. s.""'

8、;：iiiy systm,，w .Il iVstglton-u.acnsd. to raks a I I . . . L I_oth.r m-iu.s .n- units a nd th. work ta. of - .nc p up. a-su.vs- tana-sou- ahas.asic of a- -y ".in* a- uniu-ting s.ngt. su.一s i mp - .atm sa nnar- ofI-il.n a- is. -ants' -uy cornu-sooalrg htbutsn- -tL .on .to sr.Of sa-a - mus .

9、iacala- a-i,一 cs.vats '-uy cn c：pi-ianpmmoncyFisof al Ci s- its m.pci Gov.la a nts ' -.y cnsum a. big -bac NPC - put.s a I - CPPC m.m b.s an- th. broa- mas a. cn - rnn- a bout.、cn-lk th.4 .计算积分-2一 X- 1- 1dx.3x 2cns-ebeha' .us cse s I dea wI. T heedda nts'duy isde publcle'ats'

10、duyconnumpIon _ smew hee be- eearfrm-ddmonslatd is -po - neandu.e ncy. F-tnmet l- des-'ants ' d uyconnum ptin thee aebg - acks NP. de I utis a nd ，P>"m- be. ad theb.ld ma sss ae conce nnd a bou.Slcondy,Ie4一八15.计算积分J 2x dx.1 ees drv ng a vehiceacc det - nt,1 peple .led ad die ct ecnom

11、c lsse s than 100,000 n. "d,hghf - y ad l ow us.Surveys show that the nsor - ito.ce n ate dine, blbw the cvl - 'at h adpu, sevas ' poSlins but as te naue of the w ok ad the workla d.Ireorm of metod Shang apro-e d Cvl Sevas post ndad itsa -I- musbe so', it Ie seal hee .as bfre a ba-

12、a. fie of I-.n.epac - sm us be -pr o-e-mpr e ns t.hee tt h.ep. . ". the igpol cy s' dUy cornu -" on, a cdhow to u_ pesnal |e Susdies or snal-y 0f.bsidies Tefr*- mus e-bls ad pefct nts' duy cornu onrf-supori "big unl - ,ad .dspered""a be i- mos post soul d acord icyof

13、 i. _r_ of govena nne ss_ s . t he cvi l _ exmialon of resonniblt，m lw aul ivslgalon-uni. d sadad consde I raksaport -i -w|. "e mesurs 1 nd unis a nd Ie work tak of d - encaslex unied of stanad soudp u.Disipie islcton and -pe'ison organs, Inacal ahas ea« c of and dynamic of, maesr.ins

14、ad unis i imp e.I一ng depe nts soud se ngghen -pe's 0fonuiid sa nda d Shi ha s must of flexbe ' dsposl ' ht； bu sine se if sandad mus afe iacia, ad 一onrf.p.ge s, -'at.'y, pulc -'.s ' duy ndaris a d botia nCal ea pa-ugh. T hid,cir audit apro-edHou I imp.me ntaton. C) .I-p.

15、Pulduy meau.sto-pac se- ns ' duy -Icy of ntaton of insect nshiig2 003o "0 y 000 -e - t-2005年浙江省泞通高校“次升木”状号:业专考报 校学考报二二二二二二二二二二二：号证考准二二二二一二二：名姓- - - - - - - - - - - - - - - - -Il - - - - - - 二6 .计算积分7.设函数z(x2 x - 2)exdxcos(xy) +sin(x + y),求偏导数.2二 zx. ys div ng a.hi -a -d-t -ais n ,-ofc a nts

16、 -son.l dp c ecnomc l s -nsm- b-aists '-tf sa.一 .L .一.ow t o Is. 1-snal I- susd-ha 'as or lowt mah ad lwn. Ti d hgh ortos迎3 w.Th-d一OI - s-s,r ow oma8页-ns ptin m on-Zai on r-fm of poli-s tnal- a p-'-t-nar snal -fow-nsm ptin -nom. ad th- -u- of th- w ok ad th- wsowta."or.a d.op-lof-b&

17、#187;- s. Th-or- w- must -In r-orm of m-hod Shasals ad p-I-I wi ats - duty -on-mpl0ns connump.n sandad . i . . > >. and 0 -P l . 一 .“of t- s-is on r-frm - porton as sm-w h- b-w - ar- orm d-onsratd is mporan- and ug-e reform pro Cea-tsho 丁-Il 工ifiiif sfstm.，w aul ivstgaionn-fFisof al-vi sa .s .

18、pp-iGog-l-d-sa ns'd"-ns-“一，s bead ie-a- bg d - ba tss ng- pa d ma-n Io raks a I I . . . L I_oth-r m-o-s to _-.-NPC- pu.sa .d "cm-mus b- mpr ov-d mpr - -nd unis a nd t- work tak of d - -n- -p up.I-in ad suu-vb-s and t- broad mas- soudlr- prl nc P-s is. .nerthesin as-sa- -nn-rnnd a bout.

19、 -ndfth-has-asi - of ad df.ILI . csvatudiig s- m- rigins ad unl. I . . L . I . . . . .ngth- s- -s i ' 0 -at.L . IL.s' -t. sanda ds -air. on. s not -ntaf toon uniid sa ndar d Si ha s mus of il-n ad is- -ants' duf -on-mpton mon-iatdsooalrg htbutsnn- -tL .po> -"-. a on -s, t ocal In

20、a nC a siuitlof sada d mus f- ianala. a-i.-I I II. Ll. I I. I II- sd. -n，-18.把函数y =展开成x-1的哥级数，并求出它的收敛区间x 1cnsumebehav .us cse s to -l with. T heedea nts'-Uy > pub*_ats ' duy coniumplon _ smew heebew eea-r.a. is m. . ncy. F-tnmet l_ des 'd uy connum pin Ie- ae bg aWa.sNPCa nd CPCC m_ b

21、es a masss aeconce nnd aboU. Slcondy, Ie69.求二阶微分方程d-ydx2- 2s dxy = x的通解.-2005年浙江省泞通高校“次升木”状号2 003u y to ship e ate 59 csd a tt 5 veis as ed ieig trF iex eses c rus ,5H0 uc evecS u _ 1 000 yu一一_ 2一. 210.设a,b是两个向量，且 a =2, b =3,求a+2b +|a-2b的值，其中a表 示向量a的模.四.综合题：分，共20分）（本题共2个小题，每小题得分阅卷人1 .计算积分二 2n 12m 1，

22、sin1xsin 1 xdx，其中 n, m 是整数.o 22-H ng - -hi c” , kl de c - no一 h-.1 a n. Thi - hgh How"-" -e 3.n-l” !-s or low8页nsum ptinm onez-i on rfm of pollc* the en.-c X n.-d onr.r.portee-re orm - d-onia-mpor-ne -nd .eb3'. it he -mp0 . be 0- ie .f rs s nge .-m . be mpr . .,. r.h” r t “一，”.，” r re

23、c -t .' -y -* . , s not -.pl- d - iny s tm, Iw -ul i's'-ton " te recnsde to r-ks -I I . . . L I_other m-l>e. t leend units - nd te work t-k of dieenccs sopup. sino.-s-h-s.-sic of -d dy -d uniudtig s.s i mp - e.t.L . IL.on uni- s- nn-r d Sih- s m.s of -ilengthe sue-n-dise dutyconl

24、umption mone"，dsoo-lr. htbutsnne -t.L .of s-d- d m.s .e (n-d-l-d -i.ieI I II. Ll. I I. I IIon re s,t st.-s' d. cnncyFis of-、, sec . ts -ppci Gove des nts' dd. ccnsum -ebig d - b-c- NPCd. putes- .d ""membes -nd thebro-d m-ses -e cnner nnd -bout、cnd. the322 .已知函数 f(x)=4ax +3bx +

25、2cx + d ,其中常数a, b, c, d满足a+b+c+d = 0,(1)证明函数f (x)在(0, 1)内至少有一个根，2(2)当3b <8ac时，证明函数f (x)在(0, 1)内只有一个根cns-ebeha- . uscse s I-a wt. Theed- nts'.uy . is- pubic-ats ' duy cornump.on _ smew heebew ee a rfrm ._ d_on_a.d i. imp.raneanduge nny.Fistnmetiia desan. ' . uyc.nnum pin tee aebg .awac

26、ks NP. - I uis a nd CPPCCm- be. ad te b.ld ma sss ae conce nnd ab.u. Sicondy, te8es dr. ng a vehiceacc det ca us nt,1 peple .led a. die ct ecnomc lsse s than 100,000 n. "d,hghf y a.i ow us.Suveys show that the op-aing css of sa it sa that musbe solved it Ie ".al t hee yeas bfre - a ba- a.

27、lie of I-singe pac - sm ust be mpr o- mpro - e ns tadhee tot h.epri . 一 、-the tigpol cy s' .Uy -“ on, a cd fiancai.“a. peoples - y, pul c servants ' .* Sa .”a .lea pa -ugh. T hid, ord ns oktcen ate d-ef.wthe cvi at wth a'how to u- p.al L susdies or low tma . po.in* butals te nas.1 -y -ok

28、 ad the wof .usidie s Tefr* - mus eI reorm of metod Shad cvl sevas post san.a.- bls a. nts' duy cornu onrfm supori "big uniid,ad - .spere.""a be ta mos post0fte -r- nne sstms .cis t sould acord -uni. d sa.ad co.sde traksaport -i .Wie '.he mesurs to kend unis a nd the work tak

29、of dieenccsle* unied of stadad sou.p u.Disipie is.cton and iupe-isinor1ans, ianalahas eax c of and dynamc of, maesrrgins ad unis i imp - eat-iig -pame nts sou. se nghen -pe cton ofonu - sa nda d Shi ha s must of fleXbe ' dispo - rg ht； bu snne se if sanda d mu. ale iaca, a.a-it -onrfmp.ge s, at.

30、', r audt apro- d Hutmpeme ntaton. (C) efmpak.e Pul-" meauest.一-pac - n. -"y of .tain of in.em .hiig2 003o "0 y 000 -e - t-2005年浙江省泞通高校“次升木”状号2005年高数（一）答案（A）卷.填空题:（每空格5分，共40分）连续区间是（O0,0)U(0,1)U(1,)，2.3.'y =0或者4. a=0, b =一,或者x=t, y = 0,z = 0 （其中t是参数）， 0(2)(2)5. (1).选择题.（每一小题4分，

31、共20分）题号12345答案BDBBD计算题。2ln y = xl x _nx 1)(y =x(2x-1)2dx - x 1(3分)22ln(x - x 1)( x - x 1)(7分)y'= 3x2 - 4x = x(3x -4)驻点为x1 二 0, x2(2分)y (0)y(0) =1(5分)y q) =4 04y(R 一327(7分)（法二）x1(T, 0)0(0, %)43(% , 2)2'y正0负0正y-2递增1递减-%7递增5分)当x =0时，y =1 （极大值），当=%时，y = - 7 （极小值）(7分）3 .解：利用莱布尼兹公式dx= x2 2nx n(n -

32、 1)ex(7分)-es d.nga .hiceacdet. nts , kl dec nomc l s ha . ,a n. Ti _ high eiceow sowta."or.a sme beaiopeati- ofc- se.ns d pu. sevas '-. had low ueof te w ok ad the w .unigm-sitoktce n ate dine. folowthe cvi - at wt ahow to-e,esnal tte susdes or lowIeapesnal cnsm pin u_ of susde s. Theeo

33、re we must e页，共8页or to stp. sei I us cass t .« wit. T hee id-s. pul c se. ns ' . cns- ption m oneizai onrf. theecnomy ad the ee nig of te reorm. public-vats ' du. connumptop-iIn reorm of mehod Shasals adpe - twing a“.I c se-a-n.a. i . . > >. onrfrmsuportof the seison as smew hee b

34、ew ee are orm - d-onstatd is imporane and ugebeso it he ef. proe- Cd- sula IncyFisof aI" se-a . ts <sppciGo “一，s bead fie d uieiii. s . lw aul ivs'Kon lysem. te reportla a ns ' .u. cnslrn ae big d - baif this s nge pa Ltds macn to raks a I I . . . L I_other m-lues to lee.NPC putesa .

35、d ""memus be impr ove. mpr - e . ts t.hee to tnd unis a nd te work of dieenc -pup.conadsuevse. soudlr- pri nc p” . unerthe ahaseasic of ad dy ".in* ad univs csevat s' _. san.a.s clirl on. s not Secnd.a udtig ses i mp eat.L . IL.on uniid sa ndar d Si ha s m.s of -ilengte sueadise -

36、ts' duyconnumpton monetzatdsooalrg htbutsnne -IL .of sa.a d mus .e iacala. adi.0I I II. Ll. I I. I II asiuate pies son re s,to sr UId. cnn lap；"：9-3x 2dx =(x-1)(x.2)dx =(3分)(7分)dx(3分)11n(1 e2x) c2(7分)(x2x -2)exdx2_(x x-2)e(3分)J(2x+1)exdx =2- (3e-1)+2ex3 -3e 2e 2 =1 e(7分)(3分).x .:yi nxy _xyc

37、o sxy) sin/ y)(7分)8:解:(2分)(5分)(7分)9.解：特征方程为九2齐次方程(3分)eaders driving a-hi-ea-idet -a ns,1 pepe all 1d ad dre -t e-nom- lse-ns-ptids ol -isea ns on, o.-e n . dine, "w”. -iil _ at -h .d p-l- seas ' d-y-nn-mpts -thow”.- pesnal L sU-on sand han 100,000 dxn. Ti d, hg h f-d ls or low tma postions

38、but as te naow S-2snal-y -ns-e ol the w ok ad the woS"" s efre - mus eorkla d.x -1x -1x-2x -1dxInx-2x -12x1 e2x)2 -(x -1+1-1)n=edx 二是任意常数）(2x 1)exdx =;z二-ysin(xy) cos(x y).x(-1)n(x-1)-1x -1+,2九+1=0,特征值为九=1 （二重根）一2业dxy = 0 的通解是 = (c+ c2x)ex其中Ci,C2是任I reorm ol metod Sha- b's a. pel-3.ng

39、aprove d spo.i itsad - "musbe sod it I- -se- nts'd-y-nn-mptsan - onrfm s-porol te-ris“a be tamos -nsumptionav-rg t h-e .sbfre a ba-soul d a-ord -<e ad linetund on the-uni. d sadad -onsde t raksabass ol ts s nge pa -ie. m us be m.r o.dm.rme ns dhee t.o3. dWie “e mes-rs kend unis a nd te w

40、ork tak ol dieen-lsp -.D-»e isl-in and -e - Xu.h-e pr n- - l-t unde the igd oistanad soudsinorgans, iana”has eax-iibpo' -y pr so - at s'd-y-nn-mpt一.m. nt. so.d s.s nmp - eatnghen -pis se-in o.on -ar -, 一“on- - sa nda d Shi ha s must ol lleXbeon, -ayto ' d-y -on-dspo ' h3>t si

41、ce seonrfmp.gepo'-y. Se -y, p-'- - at s ' d-y -ns-ptil Sanda d mus ale Ina-a, ad - 1 一s, - r audl apro-dHou t mpeme- a.'d-y mea-e.ontain. C) elm p.eP"n- ndar ds a dbot.an-al ea - pals T一-p a-ns ' _y -n-mpti on e 2y ol-nnumption t“ntaon o. inse-I -ns-hid,n shitng-nsume beha I us -se s t -a wt. Theedda "a nts'd-y >is, te e pub* ats'd-y-nn-mpton _ smew heebew ee a rfrm d-onnlatd -