Given that the function $f(x)=a\mathrm{e}^x-\ln x$ is monotonically increasing on the interval $(1,2)$, the minimum value of $a$ is A. $\mathrm{e}^2$ B. $\mathrm{e}$ C. $\mathrm{e}^{-1}$ D. $\mathrm{e}^{-2}$
C From the problem, $f'(x)=a\mathrm{e}^x-\frac{1}{x}\geq 0$ holds for all $x$ in the interval $(1,2)$, so $a\geq\left(\frac{1}{x\mathrm{e}^x}\right)_{\max}$. Let $g(x)=x\mathrm{e}^x$. For $x\in(1,2)$, we have $g'(x)=(x+1)\mathrm{e}^x>0$, so $g(x)_{\min}=g(1)=\mathrm{e}$. Thus $\left(\frac{1}{g(x)}\right)_{\max}=\frac{1}{\mathrm{e}}$, so $a\geq\mathrm{e}^{-1}$. Choose C.
Given that the function $f(x)=a\mathrm{e}^x-\ln x$ is monotonically increasing on the interval $(1,2)$, the minimum value of $a$ is\\
A. $\mathrm{e}^2$\\
B. $\mathrm{e}$\\
C. $\mathrm{e}^{-1}$\\
D. $\mathrm{e}^{-2}$