乐山师范学院数学与应用数学

出自:乐山师范学院数学与应用数学

f (x) 在 x0 点连续,则下列命题不成立的是( )。 A. f (x0 +0) 、f (x0 - 0) 存在 B. f (x) 在 x0 点的极限存在 C. f (x) 在 x0 点的某邻域内有界 D. f (x) 在 x0 点的某空心邻域内连续

下面广义积分发散的一个是 A.∫dx/1-x B.∫1-cosx/xdx C.∫x/1+x2sin2xdx D.∫lnxdx

级数∑ln(n+1-ln n)/(ln n)2为( )级数。 A. 收敛 B. 绝对收敛 C. 条件收敛 D. 发散

f (x) 在 x0 点连续的充分条件是( )。 A. f (x0 +0) 、f (x0 - 0) 存在 B. f (x) 在 x0 点的极限存在 C. f-. (x0 ) 、f+. (x0 ) 存在 D. f (x) 在 x0 点的某空心邻域内连续

已知∫f(x)dx=cosx+C, 则f(x)=( )； A.sinx B.cosx C.-sinx D.-cosx

下列广义积分中,收敛的是( ) A.∫1/x2dx B.∫1/x dx C.∫1/x dx D.∫1/x dx

设的一个原函数为 ,则 ( )。 A.0 B.π/2 C.π/2+1 D.π/2-1

lima=0是级数∑a收敛的( )条件； A. 充分但不必要 B. 必要但不充分 C. 充要 D. 既非充分也非必要

f (x) 在 x0 点可导是 f (x) 在( x0 , f (x0)) 点有切线的( ) 条件。 A. 充分 B. 必要 C. 充分必要 D. 非充分亦非必要

幂级数的收敛半径为 A. 1/2； B. 1； C. 2； D.2/2

若为的一个原函数,则 ( )。 A.lnx/x+C B.1+lnx/x2+C C.1/x+C D.1/x-2lnx/x+C

幂级数∑X/n!的收敛域为( )； A. (-1,1) B.(-∞,+∞) C.[-1,1) D.(-1,1]

∫df(x)=() A.f(X) B.f.(x) C.f(x)+C D.f.(x)+C

∫1/x2dx=() A. -2 B. 2 C. 0 D. 发散

下列广义积分中,收敛的是( )。 A.∫1/x dx B.∫1/x dx C.∫1/x dx D.∫1/x dx

幂级数∑(x-1)/2n的收敛域为( )。 A.(-2,2) B.[-2,2) C.[-1,3) D.(-1,3)

设 f (x) = x (x + 1)(x + 2) … (x +2004) , 则 f . (0) = ( ) A. 0 B. 2003! C. 2004! D. 2005!

已知F.(x)=f(x), 则∫f(x)dx=( )； A.F(X)+sinC B.F(X)+e C.F(X)+C2 D.F(X)+lnC

下面级数发散的一个是 A. ∑√N+1/N； B. ∑2/3； C. ∑(√n3+1-√n3-1)； D. ∑2+(-1)/2

lim(√n+2-2√n+1+√n)=（） A. 0 B. 1 C. 2 D. 3

函数 f (x, y) =x/y 的全微分为( ) 。 A.1/xdx*x/y2dy B.1/xdx+x/y2dy C.1/ydx*x/y2dy D.1/xdx+x/y2dy

使函数序列f(x)=xⁿ/1+xⁿ一致收敛的区域为 A.[0,1] ； B.[0,1-e] ； C. [1-e,1+e]； D. (0,+∞)。其中0<e<1 。

幂级数∑x/2n的收敛域为( )。 A.(-2,2) B.[-2,2) C.(-2,2] D.[-2,2]

φ (x) 在 a 点连续, f (x) = | x - a |φ (x), f.(a)存在的条件是 ( ) 。 A. φ(a) = 0 B. φ(a) = 1 C. φ(a) = -1 D. φ(a) = a

函数f(x)=|x|在(-π ,π )的富里埃级数展开式为π/2-4/π∑cos(2k+1)x/(2k+1)2 ,记此级数的和函数为s(x),则使s(x)=f(x)成立的范围是 A. [-π,π)； B.(-π,π) ； C.[-π,π] D.(-π,π]

下列反常积分收敛的是( )。 A.∫1/√x dx B.∫lnx/x dx C.∫cosxdx D.∫xedx

函数在 [a,b] 上可积的充要条件是( ) A. "e>0,$ s>0和d>0使得对任一分法D，当l（D）<d时，对应于wi3e的那些区间Dxi长度之和∑Dxi< s B. "e>0,s>0, d>0使得对某一分法D，当l（D）<d时，对应于wi3e的那些区间Dxi长度之和∑Dxi< s C. "e>0,$d>0使得对任一分法D，当l（D） D. "e>0, s>0，$ d>0使得对任一分法D，当l（D）

积分∫xe dx = A. 1； B.2e+1 ； C. -2e+e； D. -2e+1

已知x2y-3x22y2-1=0 ,则dy/dx=( )。 A.2xy+12x3y2/x2+6x22y B.12x2y2-2y/x-6x3y C.12x2y2+2y/x-6x3y D.6x2y2+y/x-3x3y

、 Passage B Nowadays, a standard for measuring power has changed. These changes foretell a new standard for measuring power. No longer will a nation’s political influence be based solely on the strength of its military forces. Of course, military effectiveness will remain a primary measure of power. But political influence is also closely tied to industrial competitiveness. It’s often said that without its military the Soviet Union would really be a third-world nation. The new standard of power and influence that is evolving now places more emphasis on the ability of a country to compete effectively in the economic markets of the world. America must recognize this new course of events. Our success in shaping world events over the past 40 years has been the direct result of our ability to adapt technology and to take advantage of the capabilities of our people for the purpose of maintaining peace. Our industrial prowess over most of this period was unchallenged. It is ironic that it is just this prowess that has enabled other countries to prosper and in turn to threaten our industrial leadership. The competitiveness of America’s industrial base is an issue bigger than the Department of Defense and is going to require the efforts of the major institutional forces in our society---government, industry, and education. That is not to say that the Defense Department will not be a strong force in the process because we will. But we simply cannot be, nor should we be, looked upon by others as the savior of American industry. 61. Now a nation’s political influence depends on（） . A. the strength of its military forces B. its ability to compete in industry C. economic markets D. both A and B