函数 f(t) = 8u(t-4) + e^-2tsin(10t) 的拉氏变换为：A. F(s) = (8)/(s) e^-4s + (10)/((s+2)^2 + 100)B. F(s) = (8)/(s) e^-4s + (10)/((s-2)^2 + 100)C. F(s) = (8)/(s) e^4s + (s+2)/((s+2)^2 + 100)D. F(s) = (8)/(s) e^4s + (s-2)/((s-2)^2 + 100)

5.【证明题】设函数f(x)在[a,b]上连续，p,q为正数，证明：至少存在一点xiin[a,b]使得pf(a)+qf(b)=(p+q)f(xi).

9.单选题函数f(x)=x^4在[1,2]上满足拉格郎日中值定理，则定理中的ξ=（）.A. xi=sqrt[3]((4)/(15))B. xi=sqrt[3]((15)/(4))C. xi=sqrt[3]((6)/(15))D. xi=sqrt[3]((15)/(6))

3.若 =ln (x(y)^2(z)^3), 则 du=

18. (2.5分)设f(2x)=x²-2x，那么f(x)=（）。A. (1)/(4)x+1B. (1)/(4)x^2-1C. (1)/(4)x^2-xD. (1)/(4)x^2+1

10.12个乒乓球中有9个新球,3个旧球.第一次比赛,取出3个球,用完以后放回去,第-|||-二次比赛又从中取出3个球.-|||-(2)若第二次取出的3个球中有2个新球,求第一次取到的3个球中恰有一个新球的概率.

设由 F(x, y, z) = 0 确定了隐函数 z = f(x, y)，且 (partial F)/(partial x) bigg|_((1,1,1)) = -1，(partial F)/(partial y) bigg|_((1,1,1)) = 2，(partial z)/(partial y) bigg|_((1,1)) = 1，则 (partial z)/(partial x) bigg|_((1,1)) = ( )。A. -(1)/(2)B. (1)/(2)C. -2D. 2

设方程 xy - e^x + e^y = 0 确定了隐函数 y = y(x)，求 y'(0) = ( )A. 2B. 0C. -1D. 1

曲线 y = x + sin^2 x 在点 ((pi)/(2), 1 + (pi)/(2)) 处的切线方程的斜率是____

2.在同一个平面直角坐标系中作出 =2(x)^2,-|||-=-2(x)^2 =dfrac (1)(2)(x)^2, =-dfrac (1)(2)(x)^2 的图象.-|||-(1)列表:-|||-x -2 -1.5 -1 -0.5 0 0.5 1 1.5 2-|||-=2(x)^2 8 4.5 2 0.5 0 0.5 2 4.5 8-|||-=-2(x)^2 -8 -4.5 -2 -0.5 0 -0.5 -2 -4.5 -8-|||-x -4 -3 -2 -1 0 1 2 3 4-|||-=dfrac (1)(2)(x)^2 8 4.5 2 0.5 0.5 2 4.5 8-|||-=-dfrac (1)(2)(x)^2 ... -8 -4.5 -2 -0.5 -0.5 -2 -4.5 -8-|||-(2)在平面直角坐标系中画出 =2(x)^2 =-|||-dfrac (1)(2)(x)^2 和 =-2(x)^2, =-dfrac (1)(2)(x)^2 的图象.2.在同一个平面直角坐标系中作出 =2(x)^2,-|||-=-2(x)^2 =dfrac (1)(2)(x)^2, =-dfrac (1)(2)(x)^2 的图象.-|||-(1)列表:-|||-x -2 -1.5 -1 -0.5 0 0.5 1 1.5 2-|||-=2(x)^2 8 4.5 2 0.5 0 0.5 2 4.5 8-|||-=-2(x)^2 -8 -4.5 -2 -0.5 0 -0.5 -2 -4.5 -8-|||-x -4 -3 -2 -1 0 1 2 3 4-|||-=dfrac (1)(2)(x)^2 8 4.5 2 0.5 0.5 2 4.5 8-|||-=-dfrac (1)(2)(x)^2 ... -8 -4.5 -2 -0.5 -0.5 -2 -4.5 -8-|||-(2)在平面直角坐标系中画出 =2(x)^2 =-|||-dfrac (1)(2)(x)^2 和 =-2(x)^2, =-dfrac (1)(2)(x)^2 的图象.