若存在实数$x$使得关于$x$的不等式$\left( e^{x} - a \right)^{2} + x^{2} - 2ax + a^{2} \leq \frac{1}{2}$成立，则实数$a$的取值范围()
A．$\left\{ \frac{1}{2} \right\}$ B．$\left\{ \frac{1}{4} \right\}$ C．$\left\lbrack \frac{1}{2}, + \infty \right)$ D．$\left\lbrack \frac{1}{4}, + \infty \right)$
解：原式可化为$\left( e^{x} - a \right)^{2} + (x - a)^{2} \leq \frac{1}{2}$
可看成点$(x,e^{x})$到点$(a,a)$的距离小于等于$\frac{\sqrt{2}}{2}$
设$f(x) = e^{x}$，则$f'(x) = e^{x}$
$\therefore f'(0) = e^{0} = 1$
$\therefore$函数$f(x) = e^{x}$在点$(0,1)$处的切线方程为$y = x + 1$
画出$f(x) = e^{x}$，$y = x$，$y = x + 1$的图像，如右图所示
注意到直线$y = x$和$y = x + 1$的距离为$\frac{\sqrt{2}}{2}$
$\therefore$符合题意的点为直线$y = - x + 1$与$y = x$的交点$\left( \frac{1}{2},\frac{1}{2} \right)$
$\therefore a = \frac{1}{2}$ 选 A