8．已知函数$f(x) = x^{2} + \ln|x - 1| - 2x$，$a = f(e^{0.1})$（$e$为自然对数的底数），$b = f(1 - \ln 1.1)$，$c = f\left( \frac{8}{9} \right)$，则（ ）
A．$b < c < a$ B．$b < a < c$ C．$a < b < c$ D．$c < b < a$
解 依题可知$f(x) = (x - 1)^{2} + \ln|x - 1| - 1$，$f(x)$图像关于$x = 1$对称
显然$x > 1$时$f(x)$单增，$x < 1$时$f(x)$单减，图像如下图所示（在$x = 1$处趋于$-\infty$）：
$\because e^{0.1} - 1 - \left( 1 - \frac{8}{9} \right) = e^{0.1} - \frac{10}{9} < \frac{1}{1 - 0.1} - \frac{10}{9} = 0$
$\therefore f(e^{0.1}) < f\left( \frac{8}{9} \right)$
$\therefore a < c$
$\because e^{0.1} - 1 - (1 - 1 + \ln 1.1) = e^{0.1} - \ln 1.1 - 1 > 1 + 0.1 - 0.1 - 1 = 0$
$\therefore f(e^{0.1}) > f(1 - \ln 1.1)$
$\therefore a > b$
$\therefore b < a < c$
故选：B.