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2024年电大高数基础形考1-4答案
《高等数学基础》作业一
第1章 函数 第2章 极限与连续
(一) 单项选择题
⒈下列各函数对中,(C )中的两个函数相等.
2 A. f(x)?(x),g(x)?x B. f(x)?x2,g(x)?x
x2?13 C. f(x)?lnx,g(x)?3lnx D. f(x)?x?1,g(x)?
x?1⒉设函数f(x)的定义域为(??,??),则函数f(x)?f(?x)的图形关于(C)对称.
A. 坐标原点 B. x轴 C. y轴 D. y?x ⒊下列函数中为奇函数是(B).
A. y?ln(1?x) B. y?xcosx
2ax?a?x C. y? D. y?ln(1?x)
2⒋下列函数中为基本初等函数是(C). A. y?x?1 B. y??x C. y?x
⒌下列极限存计算不正确的是(D).
2 D. y????1,x?0 x?0?1,x2?1 B. limln(1?x)?0 A. lim2x??x?2x?0sinx1?0 D. limxsin?0 C. limx??x??xx⒍当x?0时,变量(C)是无穷小量.
sinx1 A. B.
xx1 C. xsin D. ln(x?2)
x⒎若函数f(x)在点x0满足(A),则f(x)在点x0连续。
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A. limf(x)?f(x0) B. f(x)在点x0的某个邻域内有定义
x?x0 C. lim?f(x)?f(x0) D. lim?f(x)?lim?f(x)
x?x0x?x0x?x0(二)填空题 ⒈函数f(x)?x2?9?ln(1?x)的定义域是
x?32?x|x?3?
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⒉已知函数f(x?1)?x?x,则f(x)? x2-x .
1x)? .
x??2x11x12x?12lim(1?)?lim(1?)?e2 x??x??2x2x1?x?⒋若函数f(x)??(1?x),x?0,在x?0处连续,则k? e .
?x?0?x?k,?x?1,x?0⒌函数y??的间断点是 x?0 .
?sinx,x?0⒍若limf(x)?A,则当x?x0时,f(x)?A称为 x?x0时的无穷小量 .
⒊lim(1?x?x0(二) 计算题 ⒈设函数
?ex,x?0 f(x)???x,x?0求:f(?2),f(0),f(1).
解:f??2???2,f?0??0,f?1??e?e
12x?1的定义域. x?2x?1??x?0??2x?11?解:y?lg有意义,要求?解得?x?或x?0
x2??x?0???x?0?1?? 则定义域为?x|x?0或x??
2??⒊在半径为R的半圆内内接一梯形,梯形的一个底边与半圆的直径重合,另一底边的两个
⒉求函数y?lg端点在半圆上,试将梯形的面积表示成其高的函数. 解: D A R O h E
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B C
设梯形ABCD即为题中要求的梯形,设高为h,即OE=h,下底CD=2R 直角三角形AOE中,利用勾股定理得
AE?OA2?OE2?R2?h2
则上底=2AE?2R2?h2 h2R?2R2?h2?hR?R2?h2 2sin3x⒋求lim.
x?0sin2xsin3xsin3x?3xsin3x3133解:lim?lim3x?lim3x?=??
x?0sin2xx?0sin2xx?0sin2x2122?2x2x2xx2?1⒌求lim.
x??1sin(x?1)故S?????x2?1(x?1)(x?1)x?1?1?1?lim?lim???2 解:limx??1sin(x?1)x??1sin(x?1)x??1sin(x?1)1x?1tan3x⒍求lim.
x?0xtan3xsin3x1sin3x11解:lim?lim?lim??3?1??3?3
x?0x?0xxcos3xx?03xcos3x11?x2?1⒎求lim.
x?0sinx1?x2?1(1?x2?1)(1?x2?1)x2?lim?lim解:lim2x?0x?0x?0sinx(1?x?1)sinx(1?x2?1)sinx ?limx?0
x(1?x2?1)sinxx?0?0
?1?1??1⒏求lim(x??x?1x). x?3111(1?)x[(1?)?x]?1?1x?1xex?4x?lim?x解:lim( )?lim(x)?lim??ex3x??x?3x??x??33xx??e11?(1?)[(1?)3]3xxx3x2?6x?8⒐求lim2.
x?4x?5x?4x2?6x?8?x?4??x?2??limx?2?4?2?2
解:lim2?limx?4x?5x?4x?4?x?4??x?1?x?4x?14?131?. .
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⒑设函数
?(x?2)2,x?1?f(x)??x,?1?x?1
?x?1,x??1?讨论f(x)的连续性,并写出其连续区间. 解:分别对分段点x??1,x?1处讨论连续性 (1)
x??1?x??1?limf?x??limx??1x??1?x??1?limf?x??lim?x?1???1?1?0x??1?x??1?
所以limf?x??limf?x?,即f?x?在x??1处不连续 (2)
x?1?x?1?limf?x??lim?x?2???1?2??1x?1?x?1?22limf?x??limx?1f?1??1
所以limf?x??limf?x??f?1?即f?x?在x?1处连续
x?1?x?1?由(1)(2)得f?x?在除点x??1外均连续 故f?x?的连续区间为???,?1?
??1,???
《高等数学基础》作业二
第3章 导数与微分
(一)单项选择题
f(x)f(x)存在,则lim?(C ).
x?0x?0xx A. f(0) B. f?(0) C. f?(x) D. 0cvx
f(x0?2h)?f(x0)?(D ). ⒉设f(x)在x0可导,则limh?02h A. ?2f?(x0) B. f?(x0)
⒈设f(0)?0且极限lim C. 2f?(x0) D. ?f?(x0)
f(1??x)?f(1)?(A ).
?x?0?x A. e B. 2e
11 C. e D. e
24 ⒊设f(x)?e,则limx. .
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⒋设f(x)?x(x?1)(x?2)?(x?99),则f?(0)?(D ). A. 99 B. ?99 C. 99! D. ?99! ⒌下列结论中正确的是( C ).
A. 若f(x)在点x0有极限,则在点x0可导. B. 若f(x)在点x0连续,则在点x0可导. C. 若f(x)在点x0可导,则在点x0有极限. D. 若f(x)在点x0有极限,则在点x0连续.
(二)填空题
1?2xsin,x?0? ⒈设函数f(x)??,则f?(0)? 0 . x?x?0?0,df(lnx)2lnxx2xx5. ? ⒉设f(e)?e?5e,则?dxxx1 ⒊曲线f(x)?x?1在(1,2)处的切线斜率是k?
222?π
x?(1?) ⒋曲线f(x)?sinx在(,1)处的切线方程是y?2244
2x2x ⒌设y?x,则y??2x(1?lnx)
1 ⒍设y?xlnx,则y???
x(三)计算题
⒈求下列函数的导数y?:
3x⑴y?(xx?3)e y??(x?3)e?x2e
222⑵y?cotx?xlnx y???cscx?x?2xlnx
x321xx22xlnx?x⑶y? y??
lnxln2xcosx?2xx(?sinx?2xln2)?3(cosx?2x)⑷y? y?? 34xx1sinx(?2x)?(lnx?x2)cosx2lnx?xx⑸y? y?? 2sinxsinxsinx43?cosxlnx ⑹y?x?sinxlnx y??4x?x3x(cosx?2x)?(sinx?x2)3xln3sinx?x2⑺y? y?? x32x3ex1xx?⑻y?etanx?lnx y??etanx?
cos2xx. .