两条曲线$y = a \cdot 2^x - \ln 2$与$y = \frac{\ln x}{x}$存在两个公共点，则实数$a$的取值范围（ ）
$A.(-\infty, \frac{1}{e\ln 2})$ $B.(0, \frac{1}{e\ln 2})$ $C.(0, \frac{1}{e})$ $D.(-\infty, \frac{1}{e})$
解：$y = a \cdot 2^x - \ln 2$与$y = \frac{\ln x}{x}$存在两个公共点
$\therefore a \cdot 2^x - \ln 2 = \frac{\ln x}{x}$有两个不同的正解
$\therefore a \cdot x \cdot 2^x - x\ln 2 = \ln x$有两个不同的正解
$\therefore a = \frac{\ln(x \cdot 2^x)}{x \cdot 2^x}$有两个不同的正解
设$f(t) = \frac{\ln t}{t}$，则$f'(t) = \frac{1 - \ln t}{t^2}$
$\therefore 0 < t < e$时$f(t)$单增，$t > e$时$f(t)$单减
$\therefore f(t)_{极大值} = f(e) = \frac{1}{e}$
$\therefore 0 < a < \frac{1}{e}$，选C