设 $a=e^{0.1}-1, b=\tan 0.1, c=\frac{0.4}{\pi}$, 则
$ \text{A.} $ $a < b < c$ $ \text{B.} $ $b < a < c$ $ \text{C.} $ $a < c < b$ $ \text{D.} $ $b < c < a$
【答案】 B

【解析】 $f(x)=\frac{e^x-1}{x}, g(x)=e^x \cos x-\cos x-\sin x, 0 < x < 1$.
$$
\therefore f^{\prime}(x)=\frac{e^x(x-1)+1}{x^2} > 0, g^{\prime}(x)=\left(e^x-1\right)(\cos x-\sin x) > 0 \text {. }
$$
(因为: $F(x)=e^x(x-1)+1, F^{\prime}(x)=e^x x > 0, F(x) > F(0)=0$.)
$$
\therefore f(0.1) < f(0.25)=4(\sqrt[4]{e}-1) < 4 \times 0.3 < \frac{4}{\pi}, g(0.1) > g(0)=0 \text {. }
$$
即: $a=e^{0.1}-1 < \frac{0.4}{\pi}=c, a=e^{0.1}-1 > \frac{\sin 0.1}{\cos 0.1}=\tan 0.1=b$.
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