电大经济数学基础形成性考核册答案材料

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一、填空题1.limx?02.设

x?sinx?___________________.答案: 1 x?x2?1,x?0, 在x?0处连续, 则k?________.答案1 f(x)???k,x?0?3.曲线y?x+1在(1,1)的切线方程是 . 答案:y=1/2X+3/2 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: ?π2? 2二、单项选择题1. 当x???时, 下列变量为无穷小量的是( D )

?2x2A．ln(1?x) B． C．exx?11 D．

sinx x2. 下列极限计算正确的是( B ) A.limx?0xx?1 B.lim?x?0xx?1 C.limxsinx?01sinx?1 D.lim?1

x??xx3. 设y?lg2x, 则dy?( B ) ． A．

1dx2x B．

1ln101dx C．dx D．dx xln10xx4. 若函数f (x)在点x0处可导, 则( B )是错误的．

A．函数f (x)在点x0处有定义 B．xlimf(x)?A, 但A?f(x0)

?x0 C．函数f (x)在点x0处连续 D．函数f (x)在点x0处可微 5.若f()?x, 则f?(x)?( B ) . A．

1x21x B．?1x2 C． D．?1x1 x三、解答题1．计算极限

x2?3x?2(x?1)(x?2)x?21?21limlim??( 1) lim 解: 原式=== 2x?1x?1x?1(x?1)(x?1)x?1x?11?12x2?5x?6(x?2)(x?3)lim( 2) lim 解: 原式=x?2x2?6x?8x?2(x?2)(x?4)=limx?2x?32?31?? x?42?42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。

( 3) limx?01?x?11(1?x?1)(1?x?1)11?x?1lim? 解: 原式=lim===? limx?0x?0x2x(1?x?1)x(1?x?1)x?01?x?12352??22x?3x?5xx?2?0?0?2 ( 4) lim 解: 原式=limx??3x2?2x?4x??243??23?0?03xxsin3xsin3xlim33x?03x313sin3xlim3x?????? ( 5) lim 解: 原式=x?0sin5xx?0sin5xsin5x51555limx?05x5x(x?2)(x?2)x?2x2?4?lim(x?2)?lim?4?1?4 ( 6) lim 解: 原式=limx?2x?2x?2sin(x?2)x?2sin(x?2)sin(x?2)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处极限存在?

( 2) 当a,b为何值时, f(x)在x?0处连续. 解: ( 1) 因为f(x)在x?0处有极限存在, 则有

x?0?limf(x)?lim?f(x)

x?0x?0?x?0?又 limf(x)?lim(xsin?b)?b limf(x)?limx?0?1xsinx?1

x?0?x即 b?1

因此当a为实数、 b?1时, f(x)在x?0处极限存在. ( 2) 因为f(x)在x?0处连续, 则有 limf(x)?limf(x)?f(0)

x?0?x?0?又 f(0)?a, 结合( 1) 可知a?b?1 因此当a?b?1时, f(x)在x?0处连续. 3．计算下列函数的导数或微分:

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( 1) y?x2?2x?log2x?22, 求y? 解: y??2x?2xln2?( 2) 1 xln2ax?b(ax?b)?(cx?d)?(ax?b)(cx?d)?a(cx?d)?(ax?b)c

y?cx?d, 求y? 解: y??(cx?d)2=

(cx?d)2=

ad?bc(cx?d)2 ( 3) y?1, 求y? 解: y??[(3x?5)?1132x?5]???1(3x?5)?2?1(3x?5)???3(3x?5)?2322

11( 4) y?x?xex, 求y? 解: y??(x2)??(xex)??1x?2?ex2?xex

( 5) y?eaxsinbx, 求dy

解: y??(eax)?sinbx?eax(sinbx)??eax(ax)?sinbx?eaxcosbx(bx)?=aeaxsinbx?beaxcosbx

dy?y?dx?(aeaxsinbx?beaxcosbx)dx

1( 6) y?ex?xx, 求dy

11313解: y??(e)??(x)??e(13x1x2x2?1x)??2x??e32x2?2x

1 dy?y?dx?(?ex31x2?2x2)dx

( 7)

y?cosx?e?x2,

求

dy 解y??(cosx)??(e?x2)???sinx(x)??e?x2(?x2)???sinx?2xe?x22x

( 8) y?sinnx?sinnx, 求y?

解: y??[(sinx)n]??(sinnx)??n(sinx)n?1(sinx)??cosnx(nx)??n(sinx)n?1cosx?ncosnx ( 9) y?ln(x?1?x2), 求y? 1解: y??1(112x?1?x2(x?1?x2)??1x?1?x2?((?x2))?)

=11x?1?x2(1?122?11x?1?x212(1?x)?2x)?x?1?x2?1?x2?1?x2 :

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( 10) y?2解: y??(2sin1xcot1x?1?3x2?2xx?1216, 求y?

sin1x)??(x)??(x)??(2)??23?211?31?ln2(sin)??x2?x6?0

x265 ?21sinx111ln2()()??xcosxx21?x65?62ln21?31??2?x2?x6 xcosx26sin1x54.下列各方程中y是x的隐函数, 试求y?或dy ( 1) x2?y2?xy?3x?1, 求dy 解: 方程两边同时对x求导得: (x2)??(y2)??(xy)??(3x)??(1)? 2x?2yy??y?xy??3?0 y??y?2x?3

2y?xy?2x?3dx

2y?x dy?y?dx?( 2) sin(x?y)?exy?4x, 求y? 解: 方程两边同时对x求导得:

cos(x?y)?(x?y)??exy?(xy)??4 cos(x?y)?(1?y?)?exy?(y?xy?)?4 y?(cos(x?y)?xexy)?4?cos(x?y)?yexy

4?cos(x?y)?yexy y??xycos(x?y)?xe5．求下列函数的二阶导数: ( 1) y?ln(1?x2), 求y?? 解: y??12x2?(1?x)? 221?x1?x2x2(1?x2)?2x(0?2x)2?2x2 y???( )???222221?x(1?x)(1?x)( 2) y?1?x?1?x1?21?222, 求y??及y??(1) 解: y??()??(x)??(x)???x?x

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1?21?213?211?23?21?2??? y?(?x?x)???(?x)??(?)x?x?x=1

22222244315353经济数学基础作业2

( 一) 填空题 1.若?f(x)dx?2x?2x?c, 则f(x)?2xln2?2. 2. ?(sinx)?dx?sinx?c. 3. 若?f(x)dx?F(x)?c, 则?xf(1?x2)dx??F(1?x2)?c

de4.设函数?1ln(1?x2)dx?0

dx125. 若P(x)??x011?t2dt, 则P?(x)??11?x2.

( 二) 单项选择题 1. 下列函数中, ( D ) 是xsinx的原函数． A．cosx B．2cosx C．-2cosx D．-cosx 2. 下列等式成立的是( C ) ．

A．sinxdx?d(cosx) B．lnxdx?d() C．2xdx?D．

1xdx?dx

12222

122

1x1d(2x) ln2