设 $a=0.1 e^{0.1}, b=\frac{1}{9}, c=-\ln 0.9$, 则
$ \text{A.} $ $a < b < c$ $ \text{B.} $ $c < b < a$ $ \text{C.} $ $c < a < b$ $ \text{D.} $ $a < c < b$
【答案】 C

【解析】 $f(x)=(1-x) e^x, 0 < x < 1 . \Rightarrow f^{\prime}(x)=-x e^x < 0, f(x)$ 递减 $\Rightarrow f(0.1) < f(0)=1$.
$$
\begin{gathered}
\Rightarrow 0.9 e^{0.1} < 1 . \Rightarrow a=0.1 e^{0.1} < \frac{0.1}{0.9}=\frac{1}{9}=b . \\
g(x)=x e^x+\ln (1-x), 0 < x < \frac{1}{4} . \Rightarrow g^{\prime}(x)=(x+1) e^x-\frac{1}{1-x} . \\
G(x)=\left(1-x^2\right) e^x, 0 < x < \frac{1}{4} . \Rightarrow G^{\prime}(x)=\left[2-(x+1)^2\right) e^x > 0, G(x) \text { 递增 } \\
\Rightarrow G(x)=\left(1-x^2\right) e^x > G(0)=1 . \Rightarrow g^{\prime}(x)=(x+1) e^x-\frac{1}{1-x} > 0 . \\
\Rightarrow g(x) \text { 递增, } g(0.1) > g(0)=0 . \Rightarrow a=0.1 e^{0.1} > \ln \frac{1}{1-0.1}=-\ln 0.9=c .
\end{gathered}
$$
系统推荐