giải giúp mình với tìm GTLN GTNN của hàm số y=sin2x+√2-sin^2.2x 27/09/2021 Bởi Melody giải giúp mình với tìm GTLN GTNN của hàm số y=sin2x+√2-sin^2.2x
\[\begin{array}{l} y = \sin 2x + \sqrt 2 – {\sin ^2}2x\\ = – \left( {{{\sin }^2}2x – \sin 2x} \right) + \sqrt 2 \\ = – \left( {{{\sin }^2}2x – 2.\frac{1}{2}\sin 2x + \frac{1}{4}} \right) + \frac{1}{4} + \sqrt 2 \\ = – {\left( {\sin 2x – \frac{1}{2}} \right)^2} + \frac{{4\sqrt 2 + 1}}{4} \le \frac{{4\sqrt 2 + 1}}{4}\\ \Rightarrow Dau\,\, = \,\,\,xay\,\,ra\, \Leftrightarrow \sin 2x = \frac{1}{2}\\ \Leftrightarrow \left[ \begin{array}{l} 2x = \frac{\pi }{6} + k2\pi \\ 2x = \frac{{5\pi }}{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \frac{\pi }{{12}} + \frac{{k\pi }}{4}\\ x = \frac{{5\pi }}{{12}} + \frac{{k\pi }}{4} \end{array} \right.\,\,\,\left( {k \in Z} \right).\\ Vay\,\,\,Maxy = \frac{{4\sqrt 2 + 1}}{4}. \end{array}\] Bình luận
\[\begin{array}{l}
y = \sin 2x + \sqrt 2 – {\sin ^2}2x\\
= – \left( {{{\sin }^2}2x – \sin 2x} \right) + \sqrt 2 \\
= – \left( {{{\sin }^2}2x – 2.\frac{1}{2}\sin 2x + \frac{1}{4}} \right) + \frac{1}{4} + \sqrt 2 \\
= – {\left( {\sin 2x – \frac{1}{2}} \right)^2} + \frac{{4\sqrt 2 + 1}}{4} \le \frac{{4\sqrt 2 + 1}}{4}\\
\Rightarrow Dau\,\, = \,\,\,xay\,\,ra\, \Leftrightarrow \sin 2x = \frac{1}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \frac{\pi }{6} + k2\pi \\
2x = \frac{{5\pi }}{6} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{{12}} + \frac{{k\pi }}{4}\\
x = \frac{{5\pi }}{{12}} + \frac{{k\pi }}{4}
\end{array} \right.\,\,\,\left( {k \in Z} \right).\\
Vay\,\,\,Maxy = \frac{{4\sqrt 2 + 1}}{4}.
\end{array}\]