[题目]函数 (x)=dfrac (|x|sin (x-1))(x(x-1)(x-2)) 在下列区间内有界-|||-的是 ()-|||-A.(0,1)-|||-
=(x)^3-3(x)^2 的单减凸区间是()A.(-∞,0)B.(1,2)C.(0,1)D.(2.+∞)A.(-∞,0)B.(1,2)C.(0,1)D.(2.
(A.)(-1,0) (B.)(0,1) (C.)(1,2) (D.)(2,3)【例38】函数$f(x)=\frac{|x|\sin(x-2)}{x(x-
(B.)(0,1). (C.)(1,2). (D.)(2,+∞).2.函数$f(x)=\frac{|x|\sin(x-2)}{x(x-1)(x-2)^{2}
若_(i)sim N(0,1), =1,2,(X)_(1),(X)_(2)独立,则_(i)sim N(0,1), =1,2,(X)_(1),(X)_(2)()A
求由y=ex,x=2,y=1围成的曲边梯形的面积时,若选择x为积分变量,则积分区间为() A.[0,e2] B.[0,2] C.[1,2] D.[0,1
求由曲线y=ex,直线x=2,y=1围成的曲边梯形的面积时,若选择x为积分变量,则积分区间为() A.[0,e2] B.[0,2] C.[1,2] D.
(0,1) , =2s-1,则(0,1) , =2s-1,( )。A.(0,1) , =2s-1,B.(0,1) , =2s-1,C.(0,1) , =2
》[真题6](2018)函数 =arcsin (1-x)+dfrac (1)(2)lg dfrac (1+x)(1-x) 的定义域是 __-|||-A.(0,1
[真题6](2018)函数 =arcsin (1-x)+dfrac (1)(2)lg dfrac (1+x)(1-x) 的定义域是 __-|||-A.(0,1)