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电大经济数学基础形成性考核册及参考答案
( 一) 填空题 1.limx?02.设
x?sinx?___________________.答案: 0 x?x2?1,x?0, 在x?0处连续, 则k?________.答案: 1 f(x)???k,x?0?121 23.曲线y?x在(1,1)的切线方程是 .答案: y?x?4.设函数f(x?1)?x2?2x?5, 则f?(x)?____________.答案: 2x 5.设f(x)?xsinx, 则f??()?__________.答案: ?( 二) 单项选择题 1. 函数y?x?1的连续区间是( D ) x2?x?2π2π 2A.(??,1)?(1,??) B.(??,?2)?(?2,??) C.(??,?2)?(?2,1)?(1,??) D.(??,?2)?(?2,??)或(??,1)?(1,??) 2. 下列极限计算正确的是( B )
A.limx?0xx?1 B.lim?x?0xx?1
xsin?1 D.limC.limx?0x??1xsinx?1 x3. 设y?lg2x, 则dy?( B ) .
A.
1dx2x B.
1ln10dx C.dxxln10x D.dx1x
4. 若函数f (x)在点x0处可导, 则( B )是错误的.
f(x)?A, 但 A.函数f (x)在点x0处有定义 B.xlim?x0A?f(x0)
C.函数f (x)在点x0处连续 D.函数f (x)在点x0处可微
5.当x?0时, 下列变量是无穷小量的是( C ) .
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A.2x B.(三)解答题 1.计算极限
x2?3x?21??( 1) lim x?12x2?1sinx C.ln(1?x) D.cosx x原式?lim(x?1)(x?2)x?1(x?1)(x?1)x?2 ?limx?1x?11??2x2?5x?61? ( 2) limx?2x2?6x?82原式=limx?2(x-2)(x-3)
(x-2)(x-4)x?3?limx?2x?4 1?2( 3) limx?0原式=limx?0=limx?01?x?11?? x2(1?x?1)(1?x?1)
x(1?x?1)?1 1?x?1
=?1 2x2?3x?51? ( 4) limx??3x2?2x?4335?21原式=xx= 3433??2xx1?资料内容仅供您学习参考,如有不当或者侵权,请联系改正或者删除。
( 5) limx?0sin3x3? sin5x5sin3x33原式=lim3x =
55x?0sin5x5xx2?4( 6) lim?4 x?2sin(x?2)原式=limx?2x?2 sin(x?2)x?2=
lim(x?2)x?2sin(x?2)limx?2x?2 = 4
2.设函数
1?xsin?b,x?0?x?f(x)??a,x?0,
?sinxx?0?x?问: ( 1) 当a,b为何值时, f(x)在x?0处有极限存在? ( 2) 当a,b为何值时, f(x)在x?0处连续. 解: (1)limf(x)?b,limf(x)?1
x?0?x?0?当 a?b?1时,有(2). 当limf(x)?f(0)?1
x?0a?b?1时,有limf(x)?f(0)?1
x?0 函数f(x)在x=0处连续. 3.计算下列函数的导数或微分: ( 1) y?x2?2x?log2x?22, 求y?
答案: y??2x?2xln2?( 2) y?ax?b, 求y? cx?d1 xln2答案: y??( 3) y?13x?5a(cx?d)?c(ax?b)ad?bc?2(cx?d)(cx?d)2
, 求y?
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3?3答案: y???(3x?5)2
2( 4) y?x?xex, 求y?
答案: y??12x?(e?xe)=
xx12x?ex?xex
( 5) y?eaxsinbx, 求dy
y??(eax)?(sinbx?eax(sinbx)?答案: ∵
?aeaxsinbx?beaxcosbx?eax(sinbx?bcosbx)
∴dy?eax(asinbx?bcosbx)dx( 6) y?e?xx, 求dy
113答案: ∵y???2ex?x
x2311 ∴dy?(x?2ex)dx
2x1x( 7) y?cosx?e?x, 求dy
2答案: ∵y???sinx?(x)??e?x?(?x2)?
2 =? ∴dy?(?sinx?2xe?x2x22
sinx?2xe?x)dx 2x( 8) y?sinnx?sinnx, 求y? 答案: y??nsinn?1x?cosx?ncosnx ( 9) y?ln(x?1?x2), 求y? 答案: y?? =( 10) y?21x?1?x1x?1?xcot1x22?(x?1?x2)? =1?x2?x1?x21x?1?x2?(1?x1?x2)
? =
11?x2
?1?3x2?2xx, 求y?
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11?1y??2?ln2?(cos)??(x2?x6?2)?x答案:
1cos1111??2?2xln2?sin??xx2x36x5cos1x
4.下列各方程中y是x的隐函数, 试求y?或dy
(1) 方程两边对x求导: 2x?2y?y??y?xy??3?0 (2y?x)y??y?2x?3 因此 dy?y?2x?3dx 2y?x (2) 方程两边对x求导: cos(x?y)(1?y?)?exy?(y?xy?)?4 [cos(x?y)?xexy]y??4?cos(x?y)?yexy
4?cos(x?y)?yexy 因此 y??cos(x?y)?xexy
5.求下列函数的二阶导数: ( 1) y?ln(1?x2), 求y?? 答案: (1) y??2x1?x2
2(1?x2)?2x?2x2?2x2 y????(1?x2)2(1?x2)2 (2) y??(x?121?1?31?x)???x2?x2
22123?51?32 y???x?x2
44 y?(1)???1
3414
作业( 二)