`~rai~`
\(y=\sin2x+2\cos^2x\\\quad=\sin2x+2.\dfrac{\cos2x+1}{2}\\\quad=\sin2x+\cos2x+1\\\quad=\sqrt{2}\left(\dfrac{\sqrt{2}}{2}\sin2x+\dfrac{\sqrt{2}}{2}\cos2x\right)+1\\\quad=\sqrt{2}\sin\left(2x+\dfrac{\pi}{4}\right)+1\\\text{Ta có:}-1\le \sin\left(2x+\dfrac{\pi}{4}\right)\le 1\\\Leftrightarrow -\sqrt{2}\le\sqrt{2}\sin\left(2x+\dfrac{\pi}{4}\right)\le\sqrt{2}\\\Leftrightarrow 1-\sqrt{2}\le\sqrt{2}\sin\left(2x+\dfrac{\pi}{4}\right)\le 1+\sqrt{2}\\\Leftrightarrow 1-\sqrt{2}\le y\le1+\sqrt{2}\\\Rightarrow\text{Max}_y=1+\sqrt{2}.\\\to\text{Chọn ý D.}\)
$y=sin2x+cos^2x$
$=sin2x+cos2x+1$
$=\sqrt[]{2}sin(2x+\dfrac{\pi}{4})+1$
$-1≤sin(2x+\dfrac{\pi}{4})≤1$
$⇔-\sqrt[]{2}≤\sqrt[]{2}sin(2x+\dfrac{\pi}{4})≤\sqrt[]{2}$
$⇔1-\sqrt[]{2}≤\sqrt[]{2}sin(2x+\dfrac{\pi}{4})+1≤1+\sqrt[]{2}$
Vậy $GTLN:y=1+\sqrt[]{2}⇒D$
nhanh lên các bạn ơi gấp gấp gấp gấp
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Gấp gấp gấp gấp gấp gấp gấp gấp gấp gấp gấp ạ
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jup nhá........................
cho ba điểm H, I , K thẳng hàng :
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