2023 年浙江省十校联盟高考数学第三次联考试卷-Kmath科数- 中考、高考、考验数学题库

已知函数 $f(x)=\sin \left(\omega x+\frac{\pi}{4}\right)(\omega>0)$ 在 $\left(0, \frac{\pi}{4}\right)$ 上单调递增, 且 $f\left(\frac{\pi}{2}\right)=f(\pi)$, 则 $\omega=$

$ \text{A.}$ $\frac{5}{3}$ $ \text{B.}$ $\frac{4}{3}$ $ \text{C.}$ $\frac{2}{3}$ $ \text{D.}$ $\frac{1}{3}$

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答案：

解析：

解: $\because x \in\left(0, \frac{\pi}{4}\right)$, 又 $\omega>0, \therefore \omega x+\frac{\pi}{4} \in\left(\frac{\pi}{4}, \frac{\pi \omega}{4}+\frac{\pi}{4}\right)$,
又 $f(x)=\sin \left(\omega x+\frac{\pi}{4}\right)(\omega>0)$ 在 $\left(0, \frac{\pi}{4}\right)$ 上单调递增,
$$
\begin{aligned}
& \therefore \frac{\pi \omega}{4}+\frac{\pi}{4} \leqslant \frac{\pi}{2}, \therefore \omega \in(0,1], \\
& \text { 又 } f\left(\frac{\pi}{2}\right)=f(\pi), \\
& \therefore \sin \left(\frac{\pi \omega}{2}+\frac{\pi}{4}\right)=\sin \left(\pi \omega+\frac{\pi}{4}\right), \\
& \therefore \pi \omega+\frac{\pi}{4}=\frac{\pi \omega}{2}+\frac{\pi}{4}+2 k \pi \text { 或 } \frac{\pi \omega}{2}+\frac{\pi}{4}+\pi \omega+\frac{\pi}{4}=\pi+2 k \pi, k \in Z, \\
& \therefore \omega=4 k \text { 或 } \omega=\frac{1}{3}+\frac{4 k}{3}, k \in Z, \text { 又 } \omega \in(0,1], \\
& \therefore \omega=\frac{1}{3},
\end{aligned}
$$
故选: $D$.

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