已知$\frac{a}{\ln a} = \frac{b}{\ln b} = 2.86$，$c\ln c = d\ln d = - 0.35$，$a < b$，$c < d$，则有（ ）
A．$a + b < 2\text{e}$ B．$c + d > \frac{2}{\text{e}}$
C．$ad < 1$ D．$bc > 1$
解 令$f(x) = \frac{x}{\ln x}(x > 1)$，$g(x) = x\ln x$，
$\because f'(x) = \frac{\ln x - 1}{\ln^{2}x}$，$\therefore$当$x \in \left( 1,\text{e} \right)$时，$f'(x) < 0$；当$x \in \left( \text{e}, + \infty \right)$时，$f'(x) > 0$；
$\therefore f(x)$在$\left( 1,\text{e} \right)$上单调递减，在$\left( \text{e}, + \infty \right)$上单调递增，且$f(x)_{\min} = f\left( \text{e} \right) = \text{e}$；
若$f(a) = f(b) = 2.86$，则$1 < a < \text{e} < b$，
令$F(x) = f(x) - f\left( 2\text{e} - x \right)$，$1 < x < \text{e}$
则$F'(x) = \frac{\ln x - 1}{\ln^{2}x} + \frac{\ln\left( 2\text{e} - x \right) - 1}{\ln^{2}\left( 2\text{e} - x \right)} = \frac{\ln x \cdot \ln^{2}\left( 2\text{e} - x \right) + \ln^{2}x \cdot \ln\left( 2\text{e} - x \right) - \ln^{2}x - \ln^{2}\left( 2\text{e} - x \right)}{\ln^{2}x \cdot \ln^{2}\left( 2\text{e} - x \right)} = \frac{\ln x\ln\left( 2\text{e} - x \right)\left\lbrack \ln\left( 2\text{e} - x \right) + \ln x \right\rbrack - \ln^{2}x - \ln^{2}\left( 2\text{e} - x \right)}{\ln^{2}x \cdot \ln^{2}\left( 2\text{e} - x \right)} = \frac{\ln x\ln\left( 2\text{e} - x \right) \cdot \ln\left( - x^{2} + 2\text{e}x \right) - \ln^{2}x - \ln^{2}\left( 2\text{e} - x \right)}{\ln^{2}x \cdot \ln^{2}\left( 2\text{e} - x \right)}$，
$\because$当$x \in \left( 1,\text{e} \right)$时，$- x^{2} + 2\text{e}x < \text{e}^{2}$，
$\therefore\ln x\ln\left( 2\text{e} - x \right) \cdot \ln\left( - x^{2} + 2\text{e}x \right) - \ln^{2}x - \ln^{2}\left( 2\text{e} - x \right) < 2\ln x\ln\left( 2\text{e} - x \right) - \ln^{2}x - \ln^{2}\left( 2\text{e} - x \right) = - \left\lbrack \ln x - \ln\left( 2\text{e} - x \right) \right\rbrack^{2}$，
$\therefore F'(x) < 0$在$\left( 1,\text{e} \right)$上恒成立，$\therefore F(x)$在$\left( 1,\text{e} \right)$上单调递减，$\therefore F(x) > F\left( \text{e} \right) = 0$，
即$f(x) > f\left( 2\text{e} - x \right)$，又$1 < a < \text{e}$，$\therefore f(a) > f\left( 2\text{e} - a \right)$，
$\because f(a) = f(b)$，$\therefore f(b) > f\left( 2\text{e} - a \right)$，
$\because b > \text{e}$，$2\text{e} - a > \text{e}$，$f(x)$在$\left( \text{e}, + \infty \right)$上单调递增，
$\therefore b > 2\text{e} - a$，即$a + b > 2\text{e}$，A错误；
$\because g'(x) = \ln x + 1$，$\therefore$当$x \in \left( 0,\frac{1}{\text{e}} \right)$时，$g'(x) < 0$；当$x \in \left( \frac{1}{\text{e}}, + \infty \right)$时，$g'(x) > 0$；
$\therefore g(x)$在$\left( 0,\frac{1}{\text{e}} \right)$上单调递减，在$\left( \frac{1}{\text{e}}, + \infty \right)$上单调递增，且$g(x)_{\min} = g\left( \frac{1}{\text{e}} \right) = - \frac{1}{\text{e}}$；
由$c\ln c = d\ln d = - 0.35$得：$0 < c < \frac{1}{\text{e}} < d < 1$；
设$G(x) = g(x) - g\left( \frac{2}{\text{e}} - x \right)$，$0 < x < \frac{1}{\text{e}}$，
则$G'(x) = \ln x + 1 + \ln\left( \frac{2}{\text{e}} - x \right) + 1 = \ln\left( \frac{2}{\text{e}}x - x^{2} \right) + 2$；
当$0 < x < \frac{1}{\text{e}}$时，$- x^{2} + \frac{2}{\text{e}}x \in \left( 0,\frac{1}{\text{e}^{2}} \right)$，$\therefore G'(x) < 0$，
$\therefore G(x)$在$\left( 0,\frac{1}{\text{e}} \right)$上单调递减，$\therefore G(x) > G\left( \frac{1}{\text{e}} \right) = 0$，即$g(x) > g\left( \frac{2}{\text{e}} - x \right)$，
又$0 < c < \frac{1}{\text{e}}$，$\therefore g(c) > g\left( \frac{2}{\text{e}} - c \right)$，又$g(c) = g(d)$，$\therefore g(d) > g\left( \frac{2}{\text{e}} - c \right)$，
$\because d > \frac{1}{\text{e}}$，$\frac{2}{\text{e}} - c > \frac{1}{\text{e}}$，$g(x)$在$\left( \frac{1}{\text{e}}, + \infty \right)$上单调递增，
$\therefore d > \frac{2}{\text{e}} - c$，即$c + d > \frac{2}{\text{e}}$，B正确；
$\because 1 < a < \text{e} < b$，$0 < c < \frac{1}{\text{e}} < d < 1$，$f\left( \frac{1}{x} \right) = \frac{\frac{1}{x}}{\ln\frac{1}{x}} = - \frac{1}{x\ln x} = - \frac{1}{g(x)}$，
$\therefore f\left( \frac{1}{d} \right) = - \frac{1}{g(d)} = \frac{1}{0.35} \approx 2.857 < 2.86 = f(a)$，又$1 < \frac{1}{d} < \text{e}$，$1 < a < \text{e}$，$f(x)$在$\left( 1,\text{e} \right)$上单调递减，
$\therefore\frac{1}{d} > a$，则$ad < 1$，C正确；
$\because f\left( \frac{1}{c} \right) = - \frac{1}{g(c)} \approx 2.857 < 2.86 = f(b)$，又$\frac{1}{c} > \text{e}$，$b > \text{e}$，$f(x)$在$\left( \text{e}, + \infty \right)$上单调递增，
$\therefore\frac{1}{c} < b$，则$bc > 1$，D正确.故选：BCD.