已知$a\in \bf{R}$，函数${f(x)=a x^{3}-2 x+1}$有两个极值点${{x}_{1}},{{x}_{2}}$，则$(\qquad)$．

$a$可能为负值

$f\\left( {{x}_{1}} \\right)+f\\left( {{x}_{2}} \\right)$为定值

若$a=1$，则过点$(1,0)$作曲线$y={f(x)}$的切线，切线方程为$y=x-1$或$y=-2x+2$

若存在${{x}_{0}}\\in \\bf{R}$，使得$\\left| f\\left( {{x}_{0}}+2 \\right)-f\\left( {{x}_{0}} \\right) \\right|\\le \\dfrac{1}{2}$，则$0\\lt a\\le \\dfrac{9}{4}$

$\because {f(x)=a x^{3}-2 x+1}$，则${f}'(x)=3a{{x}^{2}}-2$，

对于$\rm A$：当$a\le 0$时，${f}'\left( x \right)\lt 0$恒成立，

$\therefore f\left( x \right)$单调递减，故$f\left( x \right)$没有极值，故$\rm A$错误；

对于$\rm B$：当$a\gt 0$时，由$3a{{x}^{2}}-2=0$解得${{x}_{1}}=-\sqrt{\dfrac{2}{3a}}$，${{x}_{2}}=\sqrt{\dfrac{2}{3a}}$，

$\therefore f\left( x \right)$在区间$\left( -\infty ,{{x}_{1}} \right)$，$\left( {{x}_{2}},+\infty \right)$上${f}'\left( x \right)\gt 0,f\left( x \right)$单调递增，

在区间$\left( {{x}_{1}},{{x}_{2}} \right)$上${f}'\left( x \right)\lt 0,f\left( x \right)$单调递减，

$\therefore {{x}_{1}}$是$f\left( x \right)$的极大值点，${{x}_{2}}$是$f\left( x \right)$的极小值点，

而$3ax_{1}^{2}-2=0,3ax_{2}^{2}-2=0,ax_{1}^{2}=\dfrac{2}{3},ax_{2}^{2}=\dfrac{2}{3}$，

$\therefore f\left( {{x}_{1}} \right)+f\left( {{x}_{2}} \right)=ax_{1}^{3}-2{{x}_{1}}+1+ax_{2}^{3}-2{{x}_{2}}+1={{x}_{1}}\left( ax_{1}^{2}-2 \right)+{{x}_{2}}\left( ax_{2}^{2}-2 \right)+2$

$=-\dfrac{4}{3}\left( {{x}_{1}}+{{x}_{2}} \right)+2=2$为定值，故$\rm B$正确$.$

对于$\rm C$：若$a=1$，$f\left( x \right)={{x}^{3}}-2x+1$，${f}'\left( x \right)=3{{x}^{2}}-2$，

设切点为$\left( {{x}_{0}},x_{0}^{3}-2{{x}_{0}}+1 \right)$，则${f}'\left( {{x}_{0}} \right)=3x_{0}^{2}-2$，

$\therefore $ 切线方程为$y-\left( x_{0}^{3}-2{{x}_{0}}+1 \right)=\left( 3x_{0}^{2}-2 \right)\left( x-{{x}_{0}} \right)$，

又切线过点$\left( 1,0 \right)$，则$-\left( x_{0}^{3}-2{{x}_{0}}+1 \right)=\left( 3x_{0}^{2}-2 \right)\left( 1-{{x}_{0}} \right)$，

整理得$2x_{0}^{3}-3x_{0}^{2}+1=0$，

令$m\left( x \right)=2x_{{}}^{3}-3x_{{}}^{2}+1$，则${m}'\left( x \right)=6x_{{}}^{2}-6x=6x\left( x-1 \right)$，

$\therefore $ 当$x\lt 0$或$x\gt 1$时${{m}^{\prime }}\left( x \right)\gt 0$，当$0\lt x\lt 1$时${m}'\left( x \right)\lt 0$，

$\therefore m\left( x \right)$在$\left( -\infty ,0 \right)$，$\left( 1,+\infty \right)$上单调递增，在$\left( 0,1 \right)$上单调递减，

又$m\left( 0 \right)=1$，$m\left( 1 \right)=0$，$m\left( -\dfrac{1}{2} \right)=0$，

$\therefore $ 方程$2x_{0}^{3}-3x_{0}^{2}+1=0$的解为${{x}_{0}}=-\dfrac{1}{2}$或$x_0=1$，

$\therefore $ 切线方程为$y=-\dfrac{5}{4}x+\dfrac{5}{4}$或$y=x-1$，

$\therefore $ 函数$f\left( x \right)$过点$\left( 1,0 \right)$的切线方程为$y=-\dfrac{5}{4}x+\dfrac{5}{4}$或$y=x-1$，故$\rm C$错误；

对于$\rm D$：若存在${{x}_{0}}\in \bf{R}$，使得$\left| f\left( {{x}_{0}}+2 \right)-f\left( {{x}_{0}} \right) \right|\le \dfrac{1}{2}$，

即$\left| a{{\left( {{x}_{0}}+2 \right)}^{3}}-2\left( {{x}_{0}}+2 \right)+1-\left( ax_{0}^{3}-2{{x}_{0}}+1 \right) \right|\le \dfrac{1}{2}$，

即$\left| 3ax_{0}^{2}+6a{{x}_{0}}+4a-2 \right|\le \dfrac{1}{4}$，

即$-\dfrac{1}{4}\le 3ax_{0}^{2}+6a{{x}_{0}}+4a-2\le \dfrac{1}{4}$，

即$\begin{cases} 3ax_{0}^{2}+6a{{x}_{0}}+4a-2\ge -\dfrac{1}{4} \\ 3ax_{0}^{2}+6a{{x}_{0}}+4a-2\le \dfrac{1}{4} \\ \end{cases}$，即$\begin{cases} 3ax_{0}^{2}+6a{{x}_{0}}+4a-\dfrac{7}{4}\ge 0 \\ 3ax_{0}^{2}+6a{{x}_{0}}+4a-\dfrac{9}{4}\le 0 \\ \end{cases}$，

由于$a\gt 0$，

$\therefore 3ax_{0}^{2}+6a{{x}_{0}}+4a-\dfrac{7}{4}\ge 0$必存在，

对于$3ax_{0}^{2}+6a{{x}_{0}}+4a-\dfrac{9}{4}\le 0$，则有$\varDelta =36{{a}^{2}}-12a\left( 4a-\dfrac{9}{4} \right)=-12{{a}^{2}}+27a\ge 0$，

即$\begin{cases} a\left( 4a-9 \right)\le 0 \\ a\gt 0 \\ \end{cases}$，解得$0\lt a\le \dfrac{9}{4}$，故$\rm D$正确$.$

故选：$\rm BD$