Đáp án:
\(\begin{array}{l}
B1:\\
c)Max = \dfrac{3}{2}\\
Min = \dfrac{1}{2}\\
B2:\\
c)Max = 12\\
Min = \dfrac{{23}}{2}\\
d)Max = 9\\
Min = \dfrac{{17}}{2}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
c)y = - \sin x.\cos x + 1\\
= - \dfrac{{\sin 2x}}{2} + 1\\
Do: - 1 \le \sin 2x \le 1\\
\to - \dfrac{1}{2} \le \dfrac{{\sin 2x}}{2} \le \dfrac{1}{2}\\
\to \dfrac{1}{2} \ge - \dfrac{{\sin 2x}}{2} \ge - \dfrac{1}{2}\\
\to \dfrac{3}{2} \ge 1 - \dfrac{{\sin 2x}}{2} \ge \dfrac{1}{2}\\
\to Max = \dfrac{3}{2} \Leftrightarrow \sin 2x = - 1 \to 2x = - \dfrac{\pi }{2} + k2\pi \to x = - \dfrac{\pi }{4} + k\pi \left( {k \in Z} \right)\\
Min = \dfrac{1}{2} \Leftrightarrow \sin 2x = 1 \to 2x = \dfrac{\pi }{2} + k2\pi \to x = \dfrac{\pi }{4} + k\pi \left( {k \in Z} \right)\\
B2:\\
c)y = 12 - 2{\left( {\sin x\cos x} \right)^2} = 12 - 2.{\left( {\dfrac{{\sin 2x}}{2}} \right)^2}\\
= 12 - \dfrac{{{{\sin }^2}2x}}{2}\\
Do:0 \le {\sin ^2}2x \le 1\\
\to 0 \le \dfrac{{{{\sin }^2}2x}}{2} \le \dfrac{1}{2}\\
\to 0 \ge - \dfrac{{{{\sin }^2}2x}}{2} \ge - \dfrac{1}{2}\\
\to 12 \ge 12 - \dfrac{{{{\sin }^2}2x}}{2} \ge \dfrac{{23}}{2}\\
\to Max = 12 \Leftrightarrow {\sin ^2}2x = 0 \to 2x = k\pi \to x = \dfrac{{k\pi }}{2}\left( {k \in Z} \right)\\
Min = \dfrac{{23}}{2} \Leftrightarrow {\sin ^2}2x = 1 \to 2x = \pm \dfrac{\pi }{2} + k2\pi \to x = \pm \dfrac{\pi }{4} + k\pi \left( {k \in Z} \right)\\
d)y = {\sin ^4}x + {\cos ^4}x + 2{\left( {\sin x\cos x} \right)^2} - 2{\left( {\sin x\cos x} \right)^2} + 8\\
= {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\left( {\sin x\cos x} \right)^2} + 8\\
= 1 - 2.{\left( {\dfrac{{\sin 2x}}{2}} \right)^2} + 8\\
= 9 - \dfrac{{{{\sin }^2}2x}}{2}\\
Do:0 \le {\sin ^2}2x \le 1\\
\to 0 \le \dfrac{{{{\sin }^2}2x}}{2} \le \dfrac{1}{2}\\
\to 0 \ge - \dfrac{{{{\sin }^2}2x}}{2} \ge - \dfrac{1}{2}\\
\to 9 \ge 9 - \dfrac{{{{\sin }^2}2x}}{2} \ge \dfrac{{17}}{2}\\
\to Max = 9 \Leftrightarrow {\sin ^2}2x = 0 \to 2x = k\pi \to x = \dfrac{{k\pi }}{2}\left( {k \in Z} \right)\\
Min = \dfrac{{17}}{2} \Leftrightarrow {\sin ^2}2x = 1 \to 2x = \pm \dfrac{\pi }{2} + k2\pi \to x = \pm \dfrac{\pi }{4} + k\pi \left( {k \in Z} \right)
\end{array}\)
