4.20若已知 (t)arrow F(jomega ) ,试求下列函数的频谱。-|||-(1)tf(2t)-|||-(2) (t-2)f(t)-|||-(3) dfrac (df(t))(dt)-|||-(4) f(1-t)-|||-(5) (1-t)f(1-t)-|||-(6) f(2t-5)-|||-(7)(int )_(-infty )^1-dfrac (1{2)t}f(t)dt-|||-(8) ^itf(3-2t)-|||-(9) dfrac (df(t))(dt)neq dfrac (1)(pi t)

用来衡量一个城市两点间的距离，可以使用()A. 欧式距离B. 曼哈顿距离C. 余弦距离D. 海明距离

7.求下列函数的导数:-|||-(1) =arcsin (1-2x) ;-|||-(2) =dfrac (1)(sqrt {1-{x)^2}} ;-|||-(3) =(e)^-dfrac (x{2)}cos 3x ;-|||-(4) =arccos dfrac (1)(x) =-|||-(5) =dfrac (1-ln x)(1+ln x) =-|||-(6) =dfrac (sin 2x)(x) =-|||-(7) =arcsin sqrt (x) ;-|||-(8) =ln (x+sqrt ({a)^2+(x)^2}) =-|||-(9) =ln (sec x+tan x)-|||-(10) =ln (csc x-cot x)

3.验证极lim _(xarrow 0)dfrac ({x)^2sin dfrac (1)(x)}(sin x)存在，但不能用洛必达法则得出.

(6) arctan dfrac (y)(x)=ln sqrt ({x)^2+(y)^2}

3.对于函数y=sin(tan x)-tan(sin x)(0le xle pi)，x=(pi)/(2)是____. (A.)连续点 (B.)第一类间断点 (C.)可去间断点 (D.)第二类间断点

17. (10.0分) 已知矩阵A,B,X满足AX=B，其中 A=}1&-2&11&-2&20&-1&1,试求X.

求微分方程sqrt (x)y'=(y)^2+1的通解。

4.8 将下列各函数在指定圆环内展开为洛朗级数:-|||-(1) dfrac (z+1)({z)^2(z-1)} ,lt 1=1lt 1,1lt 1=1lt +infty ;-|||-(2) ^2(e)^dfrac (1{x)},0lt |z|lt +infty ;-|||-(3) dfrac ({z)^2-2z+5}((z-2)({z)^2+1)},1lt |z|lt 2 ;-|||-(4) cos dfrac (1)(1-z),0lt |z-1|lt +infty .

设fx在[0,1]上有连续的二阶导数,且f(0)=0,f(1)=0.5,f(1/2)=0, 证明在(0,1)内至少有一点ξ,使f(ξ)≥2 错了，应该证明在(0,1)内至少有一点ξ，使f’’(ξ)≥2

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