2)\(y=1-2.sin\dfrac{x}{2}.cos\dfrac{x}{2}+cos2x=1-sinx+\left(1-2sin^2x\right)\)
\(=-2sin^2x-sinx+2\)
Đặt \(t=sinx,t\in\left[-1;1\right]\)
Xét \(f\left(t\right)=-2t^2+2-t\)
Vẽ BBT(dạng như này, lười quá)
\(t\) \(-1\) \(-\dfrac{1}{4}\) \(1\)
\(f\left(t\right)\) \(1\) \(\dfrac{17}{8}\) \(-1\)
\(miny=minf\left(t\right)=-1\Leftrightarrow sinx=-1\)\(\Leftrightarrow x=-\dfrac{\pi}{2}+k2\pi\)
\(maxy=maxf\left(t\right)=\dfrac{17}{8}\Leftrightarrow sinx=-\dfrac{1}{4}\)\(\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.\)
1)\(y=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x-4+cos2x\)
\(=-3-\dfrac{1}{2}sin^22x+cos2x\)
\(=-\dfrac{1}{2}\left(1-cos^22x\right)-3+cos2x\)
\(=\dfrac{1}{2}cos^22x+cos2x-\dfrac{7}{2}\)
Đặt \(t=cos2x,t\in\left[-1;1\right]\)
Xét \(f\left(t\right)=\dfrac{1}{2}t^2+t-\dfrac{7}{2}\), \(I\left(-1;-4\right)\)
Vẽ BBT, khoảng từ \(\left(-1;+vc\right)\) hàm \(f\left(t\right)\) đồng biến
\(miny=\min\limits_{\left[-1;1\right]}f\left(t\right)=-4\Leftrightarrow t=-1\Leftrightarrow cos2x=-1\) \(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)
\(maxy=\max\limits_{\left[-1;1\right]}f\left(t\right)=-2\Leftrightarrow t=1\Leftrightarrow cos2x=1\)\(\Leftrightarrow x=k\pi\)
Vậy...
4)\(y=2\left(\dfrac{1}{2}sin2x-\dfrac{\sqrt{3}}{2}cos2x\right)+1\)\(=2\left(sin2x.cos\dfrac{\pi}{3}-cos2x.sin\dfrac{\pi}{3}\right)+1\)
\(=2sin\left(2x-\dfrac{\pi}{3}\right)+1\)
mà \(-2\le2sin\left(2x-\dfrac{\pi}{3}\right)\le2\)
\(\Leftrightarrow-1\le y\le3\)
\(miny=-1\Leftrightarrow sin\left(2x-\dfrac{\pi}{3}\right)=-1\)\(\Leftrightarrow x=-\dfrac{\pi}{12}+k\pi\)
\(maxy=3\Leftrightarrow sin\left(2x-\dfrac{\pi}{3}\right)=1\)\(\Leftrightarrow x=\dfrac{5\pi}{12}+k\pi\)
Vậy
.Nhưng khoảng từ pi/2 đến -pi/4 thì vẫn bao gồm đáp án A mà sao mình lại chọn ạ
