设 $a=\sin \frac{1}{2}, b=\frac{3}{2 \pi}, c=\ln 2$, 则
$ \text{A.} $ $b < a < c$ $ \text{B.} $ $b < c < a$ $ \text{C.} $ $a < b < c$ $ \text{D.} $ $c < b < a$
【答案】 A

【解析】 $\because 4 > \mathrm{e}, \therefore 2 > \sqrt{\mathrm{e}}, \therefore c=\ln 2 > \ln \sqrt{\mathrm{e}}=\frac{1}{2}, \quad \because \frac{1}{2} < \frac{\pi}{6}, \therefore a=\sin \frac{1}{2} < \sin \frac{\pi}{6}=\frac{1}{2}, \quad \because b=\frac{3}{2 \pi} < \frac{3}{6}$ $=\frac{1}{2}, \quad \because a < c, b < c, \quad$ 构造 $f(x)=\frac{\sin x}{x}, x \in\left(0, \frac{\pi}{6}\right], \quad f^{\prime}(x)=\frac{x \cos x-\sin x}{x^2}=\frac{\cos x(x-\tan x)}{x^2}$ $ < 0$, 故 $f(x)=\frac{\sin x}{x}$ 在 $\left(0, \frac{\pi}{6}\right]$ 上单调递减, 所以 $f\left(\frac{1}{2}\right) > f\left(\frac{\pi}{6}\right)=\frac{\sin \frac{\pi}{6}}{\frac{\pi}{6}}=\frac{3}{\pi}$, 即 $\frac{\sin \frac{1}{2}}{\frac{1}{2}} > \frac{3}{\pi}$, 故 $\sin \frac{1}{2} > \frac{3}{2 \pi}$, 从而 $a > b, \therefore b < a < c$, 故选 A.
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