（5分）已知函数$f(x)={e^{-{{(x-1)}^2}}}$．记$a=f(\dfrac{\sqrt{2}}{2})$，$b=f(\dfrac{\sqrt{3}}{2})$，$c=f(\dfrac{\sqrt{6}}{2})$，则$($　　$)$
A．$b > c > a$              B．$b > a > c$              C．$c > b > a$              D．$c > a > b$
答案：$A$
分析：令$g(x)=-(x-1)^{2}$，先利用作差比较法及一元二次函数的性质，可得$g(\dfrac{\sqrt{2}}{2}) < g(\dfrac{\sqrt{6}}{2}) < g(\dfrac{\sqrt{3}}{2})$，再根据$y=e^{x}$的单调性，即可求解．
解：令$g(x)=-(x-1)^{2}$，则$g(x)$的开口向下，对称轴为$x=1$，
$\because$$\dfrac{\sqrt{6}}{2}-1-(1-\dfrac{\sqrt{3}}{2})=\dfrac{\sqrt{6}+\sqrt{3}}{2}-\dfrac{4}{2}$，
而$(\sqrt{6}+\sqrt{3})^{2}-{4}^{2}=9+6\sqrt{2}-16=6\sqrt{2}-7 > 0$，
$\therefore$$\dfrac{\sqrt{6}}{2}-1-(1-\dfrac{\sqrt{3}}{2})=\dfrac{\sqrt{6}+\sqrt{3}-4}{2} > 0$，
$\therefore$$\dfrac{\sqrt{6}}{2}-1 > 1-\dfrac{\sqrt{3}}{2}$，
$\therefore$由一元二次函数的性质可知$g(\dfrac{\sqrt{6}}{2}) < g(\dfrac{\sqrt{3}}{2})$，
$\because$$\dfrac{\sqrt{6}}{2}-1-(1-\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{6}+\sqrt{2}-4}{2}$，
而$(\sqrt{6}+\sqrt{2})^{2}-{4}^{2}=4\sqrt{3}-8 < 0$，
$\therefore$$\dfrac{\sqrt{6}}{2}-1 < 1-\dfrac{\sqrt{2}}{2}$，$\therefore$$g(\dfrac{\sqrt{6}}{2}) > g(\dfrac{\sqrt{2}}{2})$，
综合可得$g(\dfrac{\sqrt{2}}{2}) < g(\dfrac{\sqrt{6}}{2}) < g(\dfrac{\sqrt{3}}{2})$，又$y=e^{x}$为增函数，
$\therefore a < c < b$，即$b > c > a$．
故选：$A$．
点评：本题考查利用函数的单调性比较大小，作差比较法的应用，化归转化思想，属中档题．