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Đề là:
\(y=\sqrt{4-3cos^23x}+1\) đúng không nhỉ?
Ta có:
\(0\le cos^23x\le1\Rightarrow1\le\sqrt{4-3cos^23x}\le2\)
\(\Rightarrow2\le y\le3\)
\(y_{min}=2\) khi \(cos^23x=1\)
\(y_{max}=3\) khi \(cos3x=0\)
\(-1\le cos\left(\sqrt{x}+\dfrac{\pi}{4}\right)\le1\Rightarrow-5\le y\le5\)
\(y_{max}=5\) khi \(cos\left(\sqrt{x}+\dfrac{\pi}{4}\right)=1\)
\(y_{min}=-5\) khi \(cos\left(\sqrt{x}+\dfrac{\pi}{4}\right)=-1\)
\(x\in\left[\dfrac{1}{4};\dfrac{3}{2}\right]\Rightarrow\pi x\in\left[\dfrac{\pi}{4};\dfrac{3\pi}{2}\right]\)
\(\Rightarrow cos\left(\pi x\right)\in\left[-1;\dfrac{\sqrt{2}}{2}\right]\)
\(y_{max}=\dfrac{\sqrt{2}}{2}\) khi \(x=\dfrac{1}{4}\)
\(y_{min}=-1\) khi \(x=1\)
Đặt \(sin^24x=t\left(t\in\left[0;1\right]\right)\)
\(y=1-8sin^22x.cos^22x+2sin^42x\)
\(=1-2sin^24x+2sin^42x\)
\(\Rightarrow y=f\left(t\right)=1-2t+2t^2\)
\(y_{min}=min\left\{f\left(0\right);f\left(1\right);f\left(\dfrac{1}{2}\right)\right\}=\dfrac{1}{2}\)
\(y_{max}=max\left\{f\left(0\right);f\left(1\right);f\left(\dfrac{1}{2}\right)\right\}=1\)
\(y=-1-cos^2\left(2x+\dfrac{\pi}{3}\right)\)
\(=-\dfrac{3}{2}+\dfrac{1}{2}-cos^2\left(2x+\dfrac{\pi}{3}\right)\)
\(=-\dfrac{3}{2}-\dfrac{1}{2}\left[2cos^2\left(2x+\dfrac{\pi}{3}\right)-1\right]\)
\(=-\dfrac{3}{2}-\dfrac{1}{2}cos\left(4x+\dfrac{2\pi}{3}\right)\)
Vì \(cos\left(4x+\dfrac{2\pi}{3}\right)\in\left[-1;1\right]\)
\(\Rightarrow min=-\dfrac{3}{2}-\dfrac{1}{2}=-2\Leftrightarrow cos\left(4x+\dfrac{2\pi}{3}\right)=1\)
\(\Rightarrow max=-\dfrac{3}{2}+\dfrac{1}{2}=-1\Leftrightarrow cos\left(4x+\dfrac{2\pi}{3}\right)=-1\)