设连续型随机变量X的分布函数F(x),概率密度函数f(x),则()A. F(x)= P(x > x)B. F(x)= P(X = x)C. f(x)= P(X = x)D. P(X = x)= 0

已知α1，α2，α3是齐次方程组Ax=0的基础解系，那么基础解系还可以是（ ）A. k1α1+k2α2+k3α3B. α1+α2，α2+α3，α3+α1C. α1-α2，α2-α3D. α1，α1-α2+α3，α3-α2

2.为描绘函数 =dfrac (1)(1+x)(e)^-x 的图形,须求出:-|||-(1)所给函数的定义域: __ ;-|||-__ __-|||-(2) ^circ = __ __ ,驻点: __-|||-(3) ^11= __ __ ,拐点: __ ;-|||-(4)列表:

a , - . --- -" " ",-, , .-|||-10.某车间靠墙壁要盖一间长方形小屋,现有存砖只够砌20 m长的墙壁.问应围成怎样的长方-|||-形才能使这间小屋的面积最大?

1.用洛必达法则求下列各极限:-|||-(1) lim _(xarrow 0)dfrac (ln (1+x))(x)-|||-(3) lim _(xarrow a)dfrac (cos x-cos a)(x-a);-|||-(5) lim _(xarrow dfrac {pi )(2)}dfrac (ln sin x)({(pi -2x))^2};-|||-(7) lim _(xarrow 0)+dfrac (ln tan 3x)(ln tan 4x);-|||-(9) lim _(xarrow +infty )dfrac (ln (1+dfrac {2)(x))}(arctan x)-|||-(11)lim xcot 3x;-|||-(13) lim _(xarrow 1)(dfrac (2)({x)^2-1}-dfrac (1)(x-1))-|||-1515) limx^(100°;;-|||-(2) lim _(xarrow 0)dfrac ({e)^x-(e)^-x}(sin x)-|||-(4) lim _(xarrow 0)dfrac (sin ax)(tan bx)(bneq 0)-|||-(6) lim _(xarrow a)dfrac ({x)^5-(a)^5}({x)^3-(a)^3}-|||-(8) lim _(xarrow dfrac {pi )(2)}dfrac (tan x)(tan 5x);-|||-(10) lim _(xarrow 0)dfrac (ln (1+{x)^2)}(sec x-cos x)-|||-(12) lim _(xarrow 0)(x)^2(e)^dfrac (1{{x)^2}} =-|||-(14) lim _(xarrow infty )((1+dfrac {3)(x))}^x =-|||-(16) lim _(xarrow 0+{(dfrac {1)(x))}^sin x}

103、设 A、B、C 三个事件两两独立，则 A、B、C 相互独立的充分必要条件是（）。A. A 与 BC 独立B. AB 与 A∪C 独立C. AB 与 AC 独立D. A∪B 与 A∪C 独立

函数f(z)在°0点展开为泰勒级数，其收敛半径是展开点到函数f(z)最近的一个奇点的距离A. 正确B. 错误

int dfrac (dx)(xsqrt {{x)^2-1}}=______________

2.求下列不定积分:-|||-(14) int (dfrac (3)(1+{x)^2}-dfrac (2)(sqrt {1-{x)^2}})dx;

第西意 不定积分-|||-(20)f(f(m(m+x)/√(1+x))dx;-|||-(10) sqrt (tan sqrt {1+{x)^2}}cdot dfrac (x{d)^2}(sqrt {1+{x)^2}}-|||-(22) int dfrac (dx)(sin xcos x)-|||-(21) int dfrac (1+ln x)({(xln x))^2}dx;-|||-(24)∫cos^3xdx-|||-(23) int dfrac (ln tan x)(cos xsin x)dx;-|||-(26)∫sin22xcos3xdx;-|||-(25) int (cos )^2(omega t+varphi )dt;-|||-(28)|sin55xsin7xdx;-|||-(27) int cos xcos dfrac (x)(2)dx;-|||-(29) int (tan )^3xsec xdx;-|||-(30) int dfrac (dx)({e)^x+(e)^-x};-|||-(31) int dfrac (1-x)(sqrt {9-4{x)^2}}dx;-|||-(32) int dfrac ({x)^3}(9+{x)^2}dx;-|||-(33) int dfrac (dx)(2{x)^2-1};-|||-(34) int dfrac (dx)((x+1)(x-2));-|||-(35) int dfrac (x)({x)^2-x-2}dx;-|||-(36) int dfrac ({x)^2dx}(sqrt {{a)^2-(x)^2}}(agt 0);-|||-(37) int dfrac (dx)(xsqrt {{x)^2-1}};-|||-(38) int dfrac (dx)(sqrt {{({x)^2+1)}^3}}-|||-(39) int dfrac (sqrt {{x)^2-9}}(x)dx-|||-(40) int dfrac (dx)(1+sqrt {2x)};-|||-(41) int dfrac (bx)(1+sqrt {1-{x)^2}}-|||-(42) int dfrac (dx)(x+sqrt {1-{x)^2}};-|||-(43) int dfrac (x-1)({x)^2+2x+3}dx;-|||-(44) int dfrac ({x)^3+1}({({x)^2+1)}^2}dx.-|||-前面我们在复合函数求导法则的基础上,得到了换元积分法.-|||-个函数乘积的求导法则,来推得另一个求积分的基本方法-|||-设函数 u=u(x) 及 v=v(x) 具有连续导数,则两个函数乘积的-|||-(uv)'=u'v+uv',-|||-,得