Y=sin^4x+ cos ^4x tính min max giúp em với ạ

Đáp án:

$MinY = \dfrac{1}{2} \Leftrightarrow {\sin ^2}2x = 1 \Leftrightarrow \cos 2x = 0 \Leftrightarrow 2x = \dfrac{\pi }{2} + k\pi  \Leftrightarrow x = \dfrac{\pi }{4} + k\dfrac{\pi }{2}\left( {k \in Z} \right)$

$MaxY = 1 \Leftrightarrow {\sin ^2}2x = 0 \Leftrightarrow \sin 2x = 0 \Leftrightarrow 2x = k\pi  \Leftrightarrow x = k\dfrac{\pi }{2}\left( {k \in Z} \right)$

Giải thích các bước giải:

 Ta có:

$\begin{array}{l}
Y = {\sin ^4}x + {\cos ^4}x\\
 = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} – 2{\sin ^2}x{\cos ^2}x\\
 = {1^2} – \dfrac{{{{\sin }^2}2x}}{2}\\
 = 1 – \dfrac{{{{\sin }^2}2x}}{2}
\end{array}$

Mà $0 \le {\sin ^2}2x \le 1 \Rightarrow \dfrac{{ – 1}}{2} \le \dfrac{{ – {{\sin }^2}2x}}{2} \le 0$

$ \Rightarrow \dfrac{1}{2} \le Y \le 1$

Như vậy:;

$MinY = \dfrac{1}{2} \Leftrightarrow {\sin ^2}2x = 1 \Leftrightarrow \cos 2x = 0 \Leftrightarrow 2x = \dfrac{\pi }{2} + k\pi  \Leftrightarrow x = \dfrac{\pi }{4} + k\dfrac{\pi }{2}\left( {k \in Z} \right)$

$MaxY = 1 \Leftrightarrow {\sin ^2}2x = 0 \Leftrightarrow \sin 2x = 0 \Leftrightarrow 2x = k\pi  \Leftrightarrow x = k\dfrac{\pi }{2}\left( {k \in Z} \right)$