Đáp án:
\(Max = \dfrac{9}{8}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = {\sin ^4}x + {\cos ^4}x + \sin x.\cos x\\
= {\sin ^4}x + {\cos ^4}x + 2{\sin ^2}x.{\cos ^2}x - 2{\sin ^2}x.{\cos ^2}x + \sin x.\cos x\\
= \left( {{{\sin }^2}x + {{\cos }^2}x} \right) - 2{\left( {\sin x.\cos x} \right)^2} - \sin x.\cos x\\
= 1 - 2{\left( {\sin x.\cos x} \right)^2} - \sin x.\cos x\\
= - 2{\left( {\sin x.\cos x} \right)^2} - \sin x.\cos x + 1\\
= - \left[ {2{{\left( {\sin x.\cos x} \right)}^2} + 2.\sqrt 2 .\sin x.\cos x.\dfrac{1}{{2\sqrt 2 }} + \dfrac{1}{8} - \dfrac{9}{8}} \right]\\
= - {\left( {\sqrt 2 .\sin x.\cos x + \dfrac{1}{{2\sqrt 2 }}} \right)^2} + \dfrac{9}{8}\\
Do:{\left( {\sqrt 2 .\sin x.\cos x + \dfrac{1}{{2\sqrt 2 }}} \right)^2} \ge 0\forall x\\
\to - {\left( {\sqrt 2 .\sin x.\cos x + \dfrac{1}{{2\sqrt 2 }}} \right)^2} \le 0\\
\to - {\left( {\sqrt 2 .\sin x.\cos x + \dfrac{1}{{2\sqrt 2 }}} \right)^2} + \dfrac{9}{8} \le \dfrac{9}{8}\\
\to Max = \dfrac{9}{8}\\
\Leftrightarrow \sqrt 2 .\sin x.\cos x + \dfrac{1}{{2\sqrt 2 }} = 0\\
\Leftrightarrow \sin x.\cos x = - \dfrac{1}{4}\\
\to \dfrac{{\sin 2x}}{2} = - \dfrac{1}{4}\\
\to \sin 2x = - \dfrac{1}{2}\\
\to \left[ \begin{array}{l}
2x = - \dfrac{\pi }{6} + k2\pi \\
2x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\to \left[ \begin{array}{l}
x = - \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{{7\pi }}{{12}} + k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
