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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 11: Tìm GTLN,GTNN của hàm số: 1) y= 3sin^2(2x)-4 2) y= 1- sinx.cosx

Toán Lớp 11: Tìm GTLN,GTNN của hàm số:
1) y= 3sin^2(2x)-4
2) y= 1- sinx.cosx

Comments ( 1 )

  1. ~rai~
    \(1.y=3\sin^22x-4\\\text{Ta có:}0\le\sin^22x\le 1\\\Leftrightarrow 0\le3\sin^22x\le 3\\\Leftrightarrow -4\le3\sin^22x-4\le -1\\\Leftrightarrow -4\le y\le1.\\+)Min_y=-4\Leftrightarrow \sin^22x=0\\\Leftrightarrow \sin2x=0\\\Leftrightarrow 2x=k\pi\\\Leftrightarrow x=k\dfrac{\pi}{2}.(k\in\mathbb{Z})\\+)Max_y=-1\Leftrightarrow \sin^22x=1\\\Leftrightarrow \cos^22x=0\\\Leftrightarrow \cos2x=0\\\Leftrightarrow 2x=\dfrac{\pi}{2}+k\pi\\\Leftrightarrow x=\dfrac{\pi}{4}+k\dfrac{\pi}{2}.(k\in\mathbb{Z})\\\text{Vậy Min}_y=-4\text{ khi x=}k\dfrac{\pi}{2};\\\text{Max}_y=-1\text{ khi x=}\dfrac{\pi}{4}+k\dfrac{\pi}{2}.(k\in\mathbb{Z})\\2.y=1-\sin x\cos x\\\quad=\dfrac{1}{2}(2-2\sin x\cos x)\\\quad=\dfrac{1}{2}(2-\sin2x).\\\text{Ta có:}-1\le \sin2x\le 1\\\Leftrightarrow -1\le -\sin2x\le 1\\\Leftrightarrow 1\le2-\sin2x\le 3\\\Leftrightarrow \dfrac{1}{2}\le\dfrac{1}{2}(1-\sin2x)\le\dfrac{3}{2}\\\Leftrightarrow \dfrac{1}{2}\le y\le \dfrac{3}{2}.\\+)Min_y=\dfrac{1}{2}\Leftrightarrow \sin2x=1\\\Leftrightarrow 2x=\dfrac{\pi}{2}+k2\pi\\\Leftrightarrow x=\dfrac{\pi}{4}+k\pi.\\+)Max_y=\dfrac{3}{2}\Leftrightarrow \sin2x=-1\\\Leftrightarrow 2x=-\dfrac{\pi}{2}+k2\pi\\\Leftrightarrow x=-\dfrac{\pi}{4}+k\pi.(k\in\mathbb{Z})\\\text{Vậy Min}_y=\dfrac{1}{2}\text{ khi x=}\dfrac{\pi}{4}+k\pi;\\\text{Max}_y=\dfrac{3}{2}\text{ khi x=}-\dfrac{\pi}{4}+k\pi.(k\in\mathbb{Z})\)

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222-9+11+12:2*14+14 = ? ( )