(B)当 ''(x)lt 0 时, (dfrac (1)(2))lt 0.-|||-(C)当 '(x)gt 0 时, (dfrac (1)(2))lt 0. (D)当 ''(x)gt 0 时, (dfrac (1)(2))lt 0.A、AB、BC、CD、D

12、单选 设函数 -|||-A (x)lt 0 ,(x)lt 0-|||-B (x)lt 0 ,(x)gt 0-|||-C (x)gt 0 ,(x)gt 0-

(5)设 (x,y)=ln (x+dfrac (y)(2x)) ,则 _(y)(1,0)= () .-|||-(A)1 (B) dfrac (1)(2) (C)

当x→0时,变量 dfrac (1)({x)^2}sin dfrac (1)(x) 是-|||-__当x→0时,变量 dfrac (1)({x)^2}sin d

[题目]当 gt 0 时,证明: -dfrac ({x)^2}(2)lt ln (1+x)lt x.

(int )_(0)^1dfrac (1)(2)xdx= A.1B.0C.(int )_(0)^1dfrac (1)(2)xdx=D.2A.1B.0C.D.2

(B)对任意x,f(-x)<0.(C)对任意x,f(-x)>0. (D)对任意x,f(-x)≥0.1.设f(x)在(-∞,+∞)内可导，且对任意$x_{1},x

(2) (x)^n-((x))^2+((x))^3=0-|||-(3) ^n+dfrac (2)(1-x)((x))^2=0 ;-|||-(4) +sqrt (

设((x)^2)=dfrac (1)(x) ( x > 0 ) ,则 f ( x ) = _____.设(x>0),则f(x)=_____.

5.设函数 (x)=cos x, 则 (int )_(0)^dfrac (pi {2)}xf(x)dx= ()-|||-【】-|||-A.0 B 1 C -1

设f(x)在[0,1]上可导, (0)=0, 且当 in (0,1) , lt f(x)lt 1 试证当-|||-in (0,1),-|||-(({int )_