2021年高考数学新高考Ⅰ-4 |
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2021-06-14 21:34:10 |
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(5分)下列区间中,函数$f(x)=7\sin (x-\dfrac{\pi }{6})$单调递增的区间是( ) A.$(0,\dfrac{\pi }{2})$ B.$(\dfrac{\pi }{2},\pi )$ C.$(\pi ,\dfrac{3\pi }{2})$ D.$(\dfrac{3\pi }{2},2\pi )$ 分析:本题需要借助正弦函数单调增区间的相关知识点求解. 解:令$-\dfrac{\pi }{2}+2k\pi \leqslant x-\dfrac{\pi }{6}\leqslant \dfrac{\pi }{2}+2k\pi$,$k\in Z$. 则$-\dfrac{\pi }{3}+2k\pi \leqslant x\leqslant \dfrac{2\pi }{3}+2k\pi$,$k\in Z$. 当$k=0$时,$k\in [-\dfrac{\pi }{3}$,$\dfrac{2\pi }{3}]$, $(0,\dfrac{\pi }{2})\subseteq [-\dfrac{\pi }{3}$,$\dfrac{2\pi }{3}]$, 故选:A. 点评:本题考查正弦函数单调性,是简单题.
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