《高等数学基础》作业1参考
答案
八年级地理上册填图题岩土工程勘察试题省略号的作用及举例应急救援安全知识车间5s试题及答案
07春,,07秋水利水电专业,
,08春,任课教师:苏世齐,86206618
1.C 2.C 3.B 4.C 5.D 6.C 7.A
21.eexx, 2.; 3.; 4.; 5.; 6.无穷小量. 3+,;,x=0,,
11.解: f22,,,f00,,fee1.,,,,,,,,
21x,lg2.解:要使 有意义,必须 x
21x,,,0,1,xx,,0,或 解得: x,2,x,0,
211x,,,?,,,,,函数的定义域为ylg,.-,0 ,,,,x2,,
ABDCAB2RDE x,=,高=3.解:如图,梯形ABCD为半圆O的内接梯形,
2222 连接则为直角三角形,ODDEORx,,OD=R,OE=,DCOCRx,,,22,
1
D C 1122?,,,,梯形的面积S=DEDCABxRRx22,,,,22 22A B 即其中S,0,,,,,xRRxxR,,,,O E
sin3233sin323xxxxlimlimlim.,,,,,4.解:原式= A xxx,,,0003sin2223sin22xxxx
A
xx,,11lim1limlim12,,,,,,,xx5.解:原式= ,,,,xxx,,,,,,111sin1sin1xx,,,,,,
sin33sin31xx6.解:原式= ,,,,lim3limlim3.xxx,,,0003cos33cos3xxxx
2221111,,,,xx,,,,x7.解:原式= limlim0.,,xx,,002211sin11sin,,,,,,xxxx,,,,
,4x,3,3,,,4,,41x,,,,,4,,8.解:原式= ,,lim1.e,,,,x,,,,xx,,33,,,,,,
xx,,42,,,,x,229.解:原式= limlim.,,xx,,44xxx,,,4113,,,,
10. 10limlim10,解:ffxx,,,,,,,,,,,,,,xx,,,,11
10limlim110ffxxf,,,,,,,,,,,,,,,xx,,,,11,,
1?,,xfx是的间断点,,,
10limlim1,又ffxx,,,,,,,,xx,,11,,2 10limlim21,11ffxxf,,,,,,且xx,,11,,,,,,,,,,
1?,fxx在处连续.,,
综上可知,函数fx在整个定义域上是不连续的,其间断点为,,
1.x,,
2
07春, ,07秋水利水电专业,
, 08春,1.B 2.D 3.A 4.D 5.C
2ln5x,112x1.0; 2.; 3.; 4.; 5.; 6.. y,,102ln1xx,,,2xx
三、
,1.求下列函数的导数: y
313,,,,3xxx222,(1)3,3解:yxeyxexe,,?,,,,,,,2,,,,
x1, 236.即yexxx,,,,,2
1112, 22ln2ln.解:yxxxxxx,,,,,,,,,,,22sinsinxxx
11x,,2,32ln2ln1.解:yxxxx,,,,, ,,,,22,,lnlnxxx,,
1xx32,,4sin2ln23cos2,解:yxxxx,,,,,,,,,,,6,,x x21ln23sin3cos.,,,,xxxx44,,,,xx
3
11,,,,2,52sinlncos解:yxxxxx,,,,,,,,,2,,,,sinxx,,,, 2212ln,,xxx2,,,cos.xxxxsinsin
sinx3, 64cosln.解:yxxx,,,,,x1xx2,,,73cos23ln3sin解:yxxxx,,,,,,,,,,2x,,3,,
21x,,,,,,cos2sinln3ln3.xxxx3
11xx8tan,解:yexe,,,,,,2cosxx 11x,,tan.,,,ex2,,cosxx,,
,2.求下列函数的导数: y
1,xx,1.解:yexe,,, ,,,,2x1sinx,, 2costan.解:yxx,,,,,,,,,,coscosxx
112,,7112,,,7,,882, 解:,,,?,3,.yxxxxyx,,,,,,8,,,,,,
,, 42sinsin2sincossin2.解:yxxxxx,,,,, ,,,,
22,,5cos2cos.解:yxxxx,,, ,,,,
xxxx,,6sin.解:yeeee=-sin,,,, ,,,,
nn,1,7sincoscossinsin解:ynxxnxxnxn,,,,,,,,,,, nn,,11,,,,,,,,,,nxxnxxnxnxnxsincoscossinsinsincos1.
sinsinxx,,85ln5sin5cosln5.解:yxx,,,, ,,,,
coscosxx,,9cossin.解:yexxe,,,, ,,,,
, 3. 在下列方程中,是由方程确定的函数,求: yyyx,,,
4
2y,,1cossin2,解:yxyxey,,,,,,,,
2y, yxeyxcossin,,,,,
yxsin,?,y.2ycosxe,
cosy,,2sinln,解:yyyx,,,,,,x
cosy,,, 1sinln,,,yxyx
cosy,?,y.,,xyx1sinln,
x3sin,解:两边求导,得yy,,,2
1,, yyyyysincos,,,,2
1,?,y.,,2sincosyyy,
,1y,,41.解:yy=1+,,, ,,yy
1y52,,,解:,,,,eyyy,,x
1y2,,yey,, ,,x
1,.?,yy2xye,,,
xx,,62sincos,解:yyeyeyy,,,,,,,
xx, 2cossin,yeyyey,,,,x
eysinx,?,y.2cosyey,
yx2,,73,解:eyeyy,,,,,,
yx2,eyye,,3, ,,xe,?,y.y2ey,3
xy,,85ln52ln2,解:yy,,,,,
yx,12ln25ln5,,,y ,,
x5ln5,?,y.y12ln2,
5
4.求下列函数的微分: dy
22,1csccotcsccsccotcsc,解:yxxxxxx,,,,,,,,,, ,,,?,,,,dyydxxxxdxcsccotcsc.
1sincoslnxxx,sincoslnxxxx,x,,,2,解:y,,22 sinsinxxx
sincoslnxxxx,?,dydx.2
xxsin
,32sincossin2,解:yxxx,,,,
?,dyxdxsin2.
22xxx,4secsec,解:yeeex,,,,, x2?,dyexdxsec.
5.求下列函数的二阶导数:
1,112,1,yx解:,,,,22x 33,,,,11122,,yxx.?,,,,,,,224,,
x,23ln3,解:y,,, 2x,,?,y3ln3.
1,3,解:y,,,x 1,,?,,y.2x
,4sincos,解:yxxx,,,, ,,,,?,,,,,yxxxxxxxcoscossin2cossin.
证:由题设,有fxfx,,,,,,,,
,,,,,,?,,,fxfx,,,,,,,,,,,,,,,
,,即fxfx,,,,,1,,,,,
,,fxfx,,,,
,?fx是偶函数. 6
07春,,07秋水利水电专业, ,08春,1.D 2.D 3.A 4.C 5.C 6.A
1.极小值; 2.0; 3.; 4.; 5.; 6.. ,,,00,,,fa0,2,,,,,,,,
2,1.解:令,yxxxxx,,,,,,,,,52153510,,,,,,,,,,
得:xx,,1,5. 12
列表如下
x (-?,1) 1 (1,5) 5 (5,+?)
y' + 0 - 0 +
极大值 极小值 y ? ? ? 32 0
?,,,函数的单调增区间为 ,5,,5.y-,1单调减区间为1,,,,,,, 当x=1时,函数取得极大值;当3250.x=时,函数取得极小值
,2.22210, 1.解:令得yxxx,,,,,,,,
,,当时,当时, 0,10; 1,30.xyxy,,,,,,,,
2?,,,,函数在区间上的极值点为yxxx230,31. ,,
又yyy03,12,36,,,,,,,,,,2
?,函数yxx-2+3在,上的最大值为,最小值为0360.,,
7
23. ,,, 2,(0)解:设所求的点则PxyPAdyxx,,,,,
2222,,,,dxyxxxxx=,,,,,,,,,,2044224
221xx,,22,令d,,,0,
22424xxxx,,,,
得x,1
2易知,是函数x,1d的极小值点,也是最小值点.
此时, 212,2,yy,,,,,
?,所求的点为PP1,2或1,2.,,,,
r如图所示,圆柱体高与底半径满足 4.解:h
222hrL,,
圆柱体的体积MATCH_
word
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_1714283720851_0为 L h2 V,,rh
r 222rLh,,将代入得
22VLhh,,,()
求导得
22222,VhLhLh,,(2())(3),,,,,,
3663hL,rL,rL,hL,,令得,并由此解出.即当底半径,高时,V,03333
圆柱体的体积最大.
5.解:设圆柱体半径为R,高为h,则 R VV222,,,,,,,, hSRhRR,222表面积2,RRh
2VV3,SRR40令得,,,,,2R2,
,,,,VV33,,当R0,,,,,,,时,当SS0,,0R时, ,,,,,,,,,,22,,,,
V3?,R是函数的极小值点,也是最小值点S.2,
4V3此时h=.,
8
V4V33R,h,答:当 时表面积最大. 2,,
x6.解:设长方体的底边长为米,高为h米. 则
62.52 由得62.5,,xhh2x
25022h 用料的面积为: Sxxhxx,,,,,40,,,x
2503,令 Sxxx,,,,,201255得,x 2x
x,5S是函数的极小值点,也是最小值点.易知, 答:当该长方体的底边长为5米,高为2.5米时用料最省。
1.ln1,证:令则fxxx,,,,,,,
1x,,,当时,xfx,,,,,010,11,,xx
,,?,函数在区间,为增函数fx0+ .,,
又f00,,,,
,,,,?,,,当时,xfxf000,
,,xx,,,ln10,
,,于是 ln1xx,,.
x2.1, 证:令则fxex,,,,,,,
x,,,当时,xfxe,,,,010,x,,,,?,,,函数fxex=10+ .在区间,为增函数,,
,,又f00,,
,,,,?,,,当时,xfxf000,x
,,xex,,,10,
于是 1ex,,.
9
07春, ,07秋水利水电专业,
, 08春,
1.D 2.D 3.B 4.B 5.B 6.D
2x1.edx; 2.; 3.; 4.; FxGxc,,fxdxtanxc,,,,,,,,
5.; 6.. 7.. ,9cos3x3,1
111,,1.解:原式=,,,,cossin.dc ,,,xxx,,
xx2. 解:原式=22.edxec,,,
13.解:原式= dxxclnlnln.,,,,,,,lnx
114.解:原式=sin22cos2.xdxxc,,, ,,,22
10
e5.解:原式=3lnln,xdx,,,,,1e
21,,,,3lnlnxx,,12,, 1,,,,,30,,2,,
7,.2
112,x6.解:原式=,xde,02
111212,,xx0,,,xeedx0,22122,,x11,,,,,,ee,, 024,,
11122,,,,244ee
132.,,44e
e127.ln解:原式=xdx,12ee22111,,xxxdxlnln1,22e2111,,exdx,22e22 111,,ex2224
111,,,ee2224
11.,,e,,4
e1,,8.ln解:原式=,xd,,,1x,,
ee11lnln,,,xdx1,,1,xxe11,,,dx21,exe 111,,,ex
111,,,,ee
21.,,e
11
1.,,证:由题设,fxfxxaa,,,,,,,,,,,
aa0
,,,,,,,,,?,,fxdxfxdxfxdx,,aa0
,aa,,,,,, ,,,,,,,,,fxdxfxdx00,,aa,,,,
,,00,,,fxdxfxdx
,0.
2.,,证:由题设,fxfxxaa,,,,,,,,,,
aa0,,,,,,?,,fxdxfxdxfxdx,,,,,aa0
,aa,,,,,, ,,,,,,fxdxfxdx,,00,,
,,,,aa
,,fxdxfxdx,,00,,a
,2.fxdx0,
12