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(2024年整理)电大高等数学基础考试答案完整版(整理).doc

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高等数学基础归类复习

一、单项选择题

1-1下列各函数对中,( C )中的两个函数相等. A.

f(x)?(x)2,g(x)?x B. f(x)?x23,g(x)?x

x2?1 C.f(x)?lnx,g(x)?3lnx D. f(x)?x?1,g(x)?

x?11-⒉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

设函数f(x)的定义域为(??,??),则函数f(x)?f(?x)的图形关于(D )对称.

A. y?x B. x轴 C. y轴 D. 坐标原点 e?x?ex.函数y?2的图形关于( A )对称.

(A) 坐标原点 (B)

x轴 (C) y轴 (D) y?x

1-⒊下列函数中为奇函数是( B ). A.

y?ln(1?x) B. y?xcosx C.

2ax?a?xy?2 D.

y?ln(1?x)

下列函数中为奇函数是(A ). A.

y?x3?x B. y?ex?e?x C. y?ln(x?1) D. y?xsinx

A

下列函数中为偶函数的是( D ).

y?(1?x)sinx B y?x2x C y?xcosx D y?ln(1?x2)

2-1 下列极限存计算不正确的是( D ).

x2?1 B. limln(1?x)?0 A. lim2x??x?2x?0sinx1?0 D. limxsin?0 C. limx??x??xx2-2当x?0时,变量( C )是无穷小量.

1sinx1A. B. C. xsin D. ln(x?2)

xxxx1sinxx当x?0时,变量( C )是无穷小量.A B C e?1 D 2

xxxsinx1x.当x?0时,变量(D )是无穷小量.A B C 2 D ln(x?1)

xx下列变量中,是无穷小量的为( B )

11x?2 Asin?x?0? B ln?x?1??x?0? Cex?x??? D.2?x?2?

xx?4精品 1

学 海 无 涯

f(1?2h)?f(1)3-1设f(x)在点x=1处可导,则lim?( D ).

h?0hA. f?(1) B. ?f?(1) C. 2f?(1) D. ?2f?(1)

f(x0?2h)?f(x0)设f(x)在x0可导,则lim?( D ).

h?0hA f?(x0) B 2f?(x0) C ?f?(x0) D ?2f?(x0)

f(x)在x0可导,则limA.

f(x0?2h)?f(x0)?( D ).

h?02h?2f?(x0) B. f?(x0) C. 2f?(x0) D. ?f?(x0)

?x?0设

f(x)?ex,则limf(1??x)?f(1)11?( A ) A e B. 2e C. e D. e

?x2413-2. 下列等式不成立的是(D ).

1dx?dx D.lnxdx?d()

x2x11dx)?arctanxdxd()??下列等式中正确的是(B ).A.d( B.

x1?x2x2xx C.d(2ln2)?2dx D.d(tanx)?cotxdx

A.exdx?dex B ?sinxdx?d(cosx) C.

f(x)?x2?4x?1的单调增加区间是( D ).

A. (??,2) B. (?1,1) C. (2,??) D. (?2,??)

4-1函数函数.函数

y?x2?4x?5在区间(?6,6)内满足(A ).

A. 先单调下降再单调上升 B. 单调下降 C. 先单调上升再单调下降 D. 单调上升

y?x2?x?6在区间(-5,5)内满足( A )

A 先单调下降再单调上升 B 单调下降 C先单调上升再单调下降 D 单调上升

. 函数

y?x2?2x?6在区间(2,5)内满足(D ).

A. 先单调下降再单调上升 B. 单调下降 C. 先单调上升再单调下降 D. 单调上升

5-1若

f(x)的一个原函数是

1,则f?(x)?(D ). A. lnxx B.

?1x2 C.

21 D. 3xx

.若F(x)是 AC5-2若

f(x) 的一个原函数,则下列等式成立的是( A )。

?xaf(x)dx?F(x)?F(a) B

?babF(x)dx?f(b)?f(a)

f?(x)?F(x) D?f?(x)dx?F(b)?F(a)

af(x)?cosx,则?f?(x)dx?( B ).

A. A.

sinx?c B. cosx?c C. ?sinx?c D. ?cosx?c

下列等式成立的是(D ).

?f?(x)dx?f(x) B. ?df(x)?f(x)

df(x)dx?f(x) C. d?f(x)dx?f(x) D.

dx?d11233233xf(x)f(x)xf(x)dx?f(x)f(x) ( B ). A. B. C. D. ?dx33d11222xf(x)xf(x)dx xf(x)dx?f(x)dxf(x)( D ) A B C D ?dx221f(x)dx?( B ). ⒌-3若?f(x)dx?F(x)?c,则?x精品

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学 海 无 涯

A.

F(x)?c B. 2F(x)?c C. F(2x)?c D.

?x?x?x ?F(e)?c, 无穷积分收敛的是 ef(e)dx??1xF(x)?c

补充:

???11dx 2x 函数

二、填空题

f(x)?10x?10?x的图形关于 y 轴 对称。

x2?9?ln(1?x)的定义域是 (3,+∞) .

x?3⒈函数f(x)?函数

x?4?x的定义域是 (2,3) ∪ (3,4 ]

ln(x?2)1函数f(x)?ln(x?5)?的定义域是 (-5,2)

2?x?x2?1,x?0若函数f(x)??,则f(0)? 1 .

xx?0?2,1?x?2若函数f(x)??(1?x),x?0,在x?0处连续,则k? e

?x?0?x?k,?sin2x?x?0.函数f(x)??x在x?0处连续,则k? 2

?x?0?k?x?1,x?0函数y??的间断点是 x=0 .

?sinx,x?0x2?2x?3函数y?的间断点是 x=3 。

x?31函数y?的间断点是 x=0 x1?e3-⒈曲线f(x)?x?1在(1,2)处的切线斜率是 1/2 .

y?曲线曲线.曲线

f(x)?x?2在(2,2)处的切线斜率是 1/4 .

f(x)?ex?1在(0,2)处的切线斜率是 1 .

f(x)?x3?1在(1,2)处的切线斜率是 3 .

π3-2 曲线f(x)?sinx在(,1)处的切线方程是 y = 1 .切线斜率是 0 2曲线y = sinx 在点 (0,0)处的切线方程为 y = x 切线斜率是 1

4.函数y?ln(1?x)的单调减少区间是 (-∞,0 ) .

2f(x)?ex的单调增加区间是 (0,+∞) .

2.函数y?(x?1)?1的单调减少区间是 (-∞,-1 ) .

2.函数f(x)?x?1的单调增加区间是 (0,+∞) .

函数 函数5-1d2y?e?x22的单调减少区间是 (0,+∞) .

?x?edx?

e?xdx

2 . .

dsinx2dx? sinx2. ?dx?(tanx)?dx? tan x +C .

若?f(x)dx?sin3x?c,则f?(x)? -9 sin 3x .

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5-2

15(sinx?)dx? 3 . ??323学 海 无 涯 1x3dedx? 0 . ln(x?1)dx? 0 ??1x2?1?1dx1下列积分计算正确的是( B ).

A

?1?1(ex?e?x)dx?0 B?(ex?e?x)dx?0 C?x2dx?0 D

?1?11?1?1|x|dx?0

三、计算题

(一)、计算极限(1小题,11分)

(1)利用极限的四则运算法则,主要是因式分解,消去零因子。 (2)利用连续函数性质:类型1: 利用重要极限 f(x0)有定义,则极限limf(x)?f(x0)

x?x0limsinxsinkxtankx?1 , lim?k, lim?k 计算

x?0x?0x?0xxxsin6xsin6x6x1-1求lim. 解: limsin6x?lim?? x?0sin5xx?0sin5xx?0sin5x5xtanxtanx1tanx11 解: lim?lim??1?

x?03xx?03x3x?0x33tan3xtan3xtan3x.3?1?3?3 1-3 求lim 解:lim=limx?0x?0x?0xx3xsin(x?a)x?a?1, lim?1 化简计算。 类型2: 因式分解并利用重要极限 limx?a(x?a)x?asin(x?a)1-2 求 lim(x?1)x2?1x2?1.(x?1)?1?(?1?1)??2 2-1求lim. 解: lim=limx??1sin(x?1)x??1sin(x?1)x??1sin(x?1)sin?x?1?sin(x?1)sin(x?1)111lim?lim.?1??2-2lim 解: 22x?1x?1x?1(x?1)(x?1)1?12x?1x?1x2?4x?3x2?4x?3(x?3)(x?1)?lim?lim(x?1)?2 2-3lim 解: limx?3sin(x?3)x?3x?3x?3sin(x?3)sin(x?3)类型3:因式分解并消去零因子,再计算极限

x2?6x?8x2?6x?8(x?4)(x?2)x?22?lim? 3-1 lim2 解: lim2=limx?4x?5x?4x?4x?5x?4x?4(x?4)(x?1)x?4x?13x2?x?6?x?3??x?2??limx?2?5 x2?x?63-2 lim2 lim2?limx??3x?x?12x??3x?x?12x??3?x?3??x?4?x??3x?47x2?3x?2x2?3x?2(x?2)(x?1)x?113-3 lim 解 lim?lim?lim? 22x?2x?2x?2x?2x?4(x?2)(x?2)x?24x?412x2sinxsin1?x?12其他: lim?lim?2 ?lim?0, limx?0x?0x?0x?01sinxsinxx?1?1x22222x?6xx?6x?5x2x22lim2?lim2?1, lim2?lim2?

x??3x?4x?5x??x?4x?5x??xx??3x3tan8xtan8xx.?8?2 (0807考题)计算lim. 解: lim=limx?0sin4xx?0sin4xx?0sin4x4x精品

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tan8x学 海 无 涯

sinxsinx1sinx1(0801考题. )计算lim. 解 lim?lim?

x?02xx?02xx?02x2(x?1).(x?3)x2?2x?3?1?(?1?3)??4 (0707考题.)lim=limx??1sin(x?1)x??1sin(x?1)(二) 求函数的导数和微分(1小题,11分)

(1)利用导数的四则运算法则

(u?v)??u??v? (uv)??u?v?uv?

(2)利用导数基本公式和复合函数求导公式

1aa?1 (x)??ax xuuxx (e)??e (e)??e.u? (sinx)??cosx222(ex)??ex.(x2)??2xex(cosx)???sinxsinxsinxsinx?? (e)?e.(sinx)?ecosx 2?(tanx)?secx(ecosx)??ecosx.(cosx)???ecosxsinx2(cotx)???cscx(sinu)??cosu.u?(cosu)???sinu.u?(lnx)??(sinx2)??cosx2.(x2)??2xcosx2(sinex)??cosex.(ex)??excosex1-1 (cosx2)???sinx2(x2)???2xsinx2(cose)???sinex.(ex)???exsinex 类型1:加减法与乘法混合运算的求导,先加减求导,后乘法求导;括号求导最后计算。 y?(xx?3)ex 13?133??x???3???33?xxxx 解:y?=?x2?3?e??x2?3??e??x2e??x2?3?e??x2?x2?3?e

2?2???????21-2 y?cotx?xlnx

22222 解:y??(cotx)??(xlnx)???cscx?(x)?lnx?x(lnx)???cscx?2xlnx?x

x1-3 设y?etanx?lnx,求y?.

11xxxxx2解: y??(etanx)??(lnx)??(e)?tanx?e(tanx)???etanx?esecx?

xx类型2:加减法与复合函数混合运算的求导,先加减求导,后复合求导 2-1

y?sinx2?lnx,求y? 解:y??(sinx2)??(lnx)??2xcosx2?1 x2-2 y?cosex?sinx2,求

x2xx22xx2解:y??(cose)??(sinx)???sine.(e)??cosx.(x)???esine?2xcosx

2-3

y?ln5x?e?5x,求

2, 解:

y??(ln5x)??.(e?5x)??2254lnx?5e?5x x2类型3: 乘积与复合函数混合运算的求导,先乘积求导,后复合求导 y?excosx,求y? 。 解:y??(ex)?cosx?ex(cosx)??2xexcosx?exsinx

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(2024年整理)电大高等数学基础考试答案完整版(整理).doc

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