12、已知$1 < m < n$，过点$(m, \log_2^m)$和$(n, \log_2^n)$的直线为$l_1$，过点$(m, \log_5^m)$和$(n, \log_5^n)$的直线为$l_2$，$l_1$与$l_2$在$y$轴上的截距相等，设函数$f(x) = m^{nx} + n^{-mx}$，则（ ）
$A.f(x)$在R上单调递增 $B.$若$m = 2$，则$f(1) = 32$
$C.$若$f(2) = 6$，则$f(4) = 34$ $D.m, n$均不为$e$（e为自然对数的底数）
解：依题可知
$l_1$方程为$y - \log_2^m = \frac{\log_2^n - \log_2^m}{n - m}(x - m)$
$l_2$方程为$y - \log_5^m = \frac{\log_5^n - \log_5^m}{n - m}(x - m)$
$\therefore \frac{\log_2^n - \log_2^m}{n - m}(-m) + \log_2^m = \frac{\log_5^n - \log_5^m}{n - m}(-m) + \log_5^m$
$\therefore m\left( \frac{\log_5^n - \log_5^m}{n - m} - \frac{\log_2^n - \log_2^m}{n - m} \right) = \log_5^m - \log_2^m$
$\therefore \log_5^{\frac{n^m}{m^n}} = \log_2^{\frac{n^m}{m^n}}$
$\therefore n^m = m^n$
设$m^n = t$（$t > 1$）
$\therefore f(x) = m^{nx} + n^{-mx} = (m^n)^x + (n^m)^{-x} = t^x + \left( \frac{1}{t} \right)^x$
对于$A$选项
$f(x) = t^x + \left( \frac{1}{t} \right)^x$为偶函数，在$y$轴两侧单调性相反，故A错误
对于$B$选项
$\because 1 < m < n, 2^n = n^2$
$\therefore n = 4$
$\therefore f(1) = 16 + \frac{1}{16} \neq 32$，故B错误
对于$C$选项
$\because f(2) = t^2 + \frac{1}{t^2} = 6$
$\therefore f(4) = t^4 + \frac{1}{t^4} = \left( t^2 + \frac{1}{t^2} \right)^2 - 2 = 36 - 2 = 34$故C正确
对$D$选项
$\because m^n = n^m$
$\therefore \ln m^n = \ln n^m$
$\therefore n\ln m = m\ln n$
$\because \frac{\ln m^n}{mn} = \frac{\ln m}{m}$，$\frac{\ln n^m}{mn} = \frac{\ln n}{n}$
$\therefore \frac{\ln m}{m} = \frac{\ln n}{n}$
设$g(x) = \frac{\ln x}{x}$，则$g'(x) = \frac{1 - \ln x}{x^2}$
$\therefore x > e$时，$g(x)$单减，$0 < x < e$时$g(x)$单增
$\therefore g(x)_{极大值} = g(e) = \frac{1}{e}$
$\therefore m < e < n$，故D正确