大一上學(xué)期高數(shù)期末考試試題五套詳解答案

1、20102010級(jí)高等數(shù)學(xué)級(jí)高等數(shù)學(xué)(上)A)A解答解答一、填空題：一、填空題：(每題每題3分，共分，共1818分)()(請(qǐng)將正確答案填入下表，否則不給分請(qǐng)將正確答案填入下表，否則不給分)1已知極限已知極限，則常數(shù)，則常數(shù)的值分別是的值分別是(空1)1)。01lim2??????????????baxxxxba解：解：0xba1xxlimbax1xxx1limx2x???????????????????????????1a=01a=0?

2、a=1a=1??????????????????????????x1xxlimax1xxlimb2x2x1x111lim1xxlim1xxxxlimxx22x??????????????????????????或：01xbx)ba(x)a1(limbax1xxlim2x2x??????????????????????????????所以所以1a=0ab=01a=0ab=0?a=1b=1a=1b=1?；颍?????????????????

3、?????????????1x1bax1x1xlimbax1xxlim2x2x01x1)b1(x)a1(lim1x1bax1xlimxx?????????????????????????????所以所以1a=01b=01a=01b=0?a=1b=1a=1b=1。2函數(shù)函數(shù)的第一類間斷點(diǎn)是的第一類間斷點(diǎn)是(空2)2)。xxxxxf323)(23????解：解：f(x)f(x)在x=301x=301處無定義，是間斷點(diǎn)。處無定義，是間斷點(diǎn)。，

4、121)3x)(1x(x3xlimx3x2x3xlim)x(flim3x233x3x????????????x=3x=3是第一類間斷點(diǎn)。是第一類間斷點(diǎn)。??????????x3x2x3xlim)x(flim231x1xx=1x=1是第二類間斷點(diǎn)。是第二類間斷點(diǎn)。????????x3x2x3xlim)x(flim230x0xx=0x=0是第二類間斷點(diǎn)。是第二類間斷點(diǎn)。3設(shè)函數(shù)設(shè)函數(shù)可導(dǎo)，可導(dǎo)，，則，則＝(空3)3)。)(xf)(1)(2x

5、fxg??)(xg解：解：)x(f1)x(f)x(f)x(f)x(f2)x(f121)x(g22????????4設(shè)函數(shù)設(shè)函數(shù)在處取得極值，則處取得極值，則(空4)4)。3223???axxy1?x?a解：解：ax2x6y2???在x=1x=1處取極值，則處取極值，則，即，即62a=062a=0，解得，解得0y1x???3a??5設(shè)設(shè)是函數(shù)是函數(shù)的一個(gè)原函數(shù)，則不定積分的一個(gè)原函數(shù)，則不定積分(空5)5)。2xe)(xf???dxxf)

6、(解：解：A.A.???1dxxxln????1xlnxdln????????12xln21B.B.???e2dx)x(lnx1????e2xlnd)x(ln11xln1e???????????C.C.???edxxlnx1????exlndxln1????????exlnlnD.D.???e21dx)x(lnx1????e21xlnd)x(ln1????????????e21)x(ln2三、三、(6(6分)求極限求極限??xxxex1

7、302lim??解：解：....................................3分??x1x30xex2lim??x1)1ex2(1ex21x30xx3x3)]1ex2(1[lim???????????由于由于................................................2分5xx3lim2x1elim2x1ex2lim0xx30xx30x???????????所以所以=.......

8、.......................1分??x1x30xex2lim??x1)1ex2(1ex21x30xx3x3)]1ex2(1[lim???????????5e或設(shè)，則，則..........................................................................................................1分??x1x3ex2y??.......

9、.....................................................................................................1分??x3ex2lnx1yln??..............................................................2分51ex2e32limx)ex2ln(limylnlimx3x30xx30

10、x0x?????????所以所以......................................................................................2分??5ylnlimx1x30xeeex2lim0x?????四、四、(6(6分)22202)(cossindxydxyytyduuxt、、、、、、?????????解：解：..............................

11、..........................................................2分22tsint2dtdytsindtdx???..........................................................................................2分t2tsintsint2dtdxdtdydxdy22?????..............