9.已知函数f(x)= _{1)-(x)_(2)}lt 0 成-|||-立,则a的取值范围是 ()-|||-A. (0,dfrac (3)(5)] . B. [ dfrac (3)(5),1) C.(0,1) D. [ dfrac (3)(5),2)-|||-10.已知函数 (x)=x+dfrac (4)(x) , (x)=(2)^x+a ,若 forall (x)_(1)in [ dfrac (1)(2),1] ] exists (x)_(2)in [ 1,2] ,使得-|||-((x)_(1))leqslant g((x)_(2)) ,则实数a的取值范围是 ()-|||-A. [ dfrac (1)(2),+infty ) B. (-infty ,dfrac (1)(2)] cup [ 3,+infty )-|||-C. (-infty ,dfrac (1)(2))cup (dfrac (1)(2),+infty ) D. [ dfrac (9)(2),+infty )-|||-11.已知函数 (x)=dfrac ({2)^x-1}({2)^x+1}+3x+1 ,且 ((a)^2)+f(3a-4)gt 2 ,则实数a的取值范围是-|||-()-|||-A. (-4,1) B. (-infty ,-4)cup (1,+infty ) C. (-infty ,-1)cup (4,+infty ) D. (-1,4)-|||-12.已知函数f(x )是定义在R上的奇函数,当 gt 0 时, (x)=dfrac (x-1)({e)^x} ,给出下列命题:-|||-①当 lt 0 时, (x)=(x+1)(e)^x ;-|||-②函数f(x)有2个零点;-|||-③∀x1, _(2)in R ,都有 |f((x)_(1))-f((x)_(2))|leqslant 2 :-|||-④ (x)leqslant 0 的解集为 (-infty ,-1] cup (0,1] .-|||-其中正确的命题是 ()-|||-A.①④ B.②③ C.①③ D.②④