基礎(chǔ)數(shù)學(xué)微積分試題2含答案

f(x)

1

,g(x)

ex1,f[g1(ex1

.1ln(x2f(x)ex2，則

f(12x)f(1)x

4e解f'(x)2xex2limf(12xf(1)2f'(1) e2xf(x)x(x1x0

；補充定義f(x0) -e2x e2x 解x0

x0

2,x1x(x

x1x(xf(x)1sin3xacosxx3

a f解

為極 f'(x)cos3xasinx0 當(dāng)x時,f'(x0,當(dāng)x時,f'(x)0,f當(dāng)x x3

3 3 f(t)dtx，則f(7) 1 f(x31)3x21,當(dāng)x2時

二、單項選擇（220分2xx2xxf(x)222

ln(2x

的定義域區(qū)間是（ 22

（D）(1,函數(shù)f(x) ，則f(x)（B（A）單調(diào) （B）有界 1f(x)ex2

x2x

有（B） 4在同一變化過程中，結(jié)論（ 無窮大與有界變量之積為無窮大（D）有限個無窮大之當(dāng)x 時，下列函數(shù)那個是其它三個的高階無窮?。―

f(x為定義在(的可導(dǎo)的偶函數(shù)，則函數(shù)（A）為奇f(sin（C）f(cos

f(x)sin（D）[f(x)sinf(x)f(x)

f(xf(n)

（

[f

f(xx=af(a)limnf(a1)f

limf(ah)f(an

limf(ah)f （D）limf(a2h)f 若f(x)的導(dǎo)函數(shù)是 ，則f(x)的一個原函數(shù)是（

設(shè)fx在1，2上可積,且f(11f(242f(x)dx2,則2xf'(x)dx（A (A) (B) (C) (D)求極限lim[x 解lim[xx2ln(11lim11ln(1t)limtln(1

t

t1lim 1

t02t(1 x2axf(x)

x

,x

f(2)5

x2axbx2

xli（x2x2ax(4

42ab0b(4 x2 a

xbexysin(x

x，求

解exyydxxdycos(x2y)(2xydxx2dy)2當(dāng)x0時，y1,則dx2dy

12f(xf(xef(xf(nx解f'(x)ef(xf"(x)f'(x)ef(x)e2ff"'(x)2f'(x)e2f(x)2!e3ff(n)(x)(1)n1(n1)!enf設(shè)曲線f(x)x3 1，yxa，b，c f'(x)3x22axb,f"(x)6x2a,f"(1)6f1）＝1又因為曲線平行與yx,則f'(0)b1c曲線方程為f(x)x33x2計算不定積分cos(lnx)dx解

x)dxxcos(lnx) xsin(lnx)1xxcos(lnx)sin(lnx)dxxcos(lnx)xsin(lnx)xcos(lnx)1xcos(lnx)dx1[xcos(lnx)xsin(lnx)]2 1解令xsintdxcos

t

t

11 11

1cos

sin2t

sin2

dsinttcott sin 11

sinarcsinx x2xf(x)

在(,

解f(xf(x在[0，）內(nèi)的最值222令f'(x)2x2(2x2)ex20，則得駐點為x 222且當(dāng)0x

時，f'(x)0 當(dāng)x

f'(x)0故x 為f(x)在[0，+]的極大值點，也是最大值點， f()limf(x) (2t)etdt(2t)et etdt f(0) minf(x)f(0)x=125-5pC(x)=100x+x2,若生產(chǎn)的商品都能全部售出。求：(1)使利潤最大時的產(chǎn)量；(2最51.2x224xL'(x)2.4x240x

（

sinxx1x3證明

f(x)sinxx1x3,f(0)f'