Đáp án:
b. \(\begin{array}{l}
Min = 0 \Leftrightarrow \cos x = - 1 \to x = \pi + k2\pi \left( {k \in Z} \right)\\
Max = 2 \Leftrightarrow \cos x = 1 \to x = k2\pi \left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.y = 2{\cos ^2}x - 3\cos x + 3\\
= {\left( {\sqrt 2 \cos x} \right)^2} - 2.\sqrt 2 \cos x.\dfrac{3}{{2\sqrt 2 }} + {\left( {\dfrac{3}{{2\sqrt 2 }}} \right)^2} + \dfrac{{15}}{8}\\
= {\left( {\sqrt 2 \cos x - \dfrac{3}{{2\sqrt 2 }}} \right)^2} + \dfrac{{15}}{8}\\
Do: - 1 \le \cos x \le 1\\
\to - \sqrt 2 \le \sqrt 2 \cos x \le \sqrt 2 \\
\to - \sqrt 2 - \dfrac{3}{{2\sqrt 2 }} \le \sqrt 2 \cos x - \dfrac{3}{{2\sqrt 2 }} \le \sqrt 2 - \dfrac{3}{{2\sqrt 2 }}\\
\to \dfrac{{49}}{8} \ge {\left( {\sqrt 2 \cos x - \dfrac{3}{{2\sqrt 2 }}} \right)^2} \ge \dfrac{1}{8}\\
\to \dfrac{{49}}{8} + \dfrac{{15}}{8} \ge {\left( {\sqrt 2 \cos x - \dfrac{3}{{2\sqrt 2 }}} \right)^2} \ge \dfrac{1}{8} + \dfrac{{15}}{8}\\
\to 8 \ge y \ge 2\\
\to Min = 2 \Leftrightarrow \cos x = 1 \to x = k2\pi \left( {k \in Z} \right)\\
Max = 8 \Leftrightarrow \cos x = - 1 \Leftrightarrow x = \pi + k2\pi \left( {k \in Z} \right)\\
b.y = \left( {1 - {{\cos }^2}x} \right) - \cos x + 1\\
= - {\cos ^2}x - \cos x + 2\\
= - \left( {{{\cos }^2}x + 2.\cos x.\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{9}{4}} \right)\\
= - {\left( {\cos x - \dfrac{1}{2}} \right)^2} + \dfrac{9}{4}\\
Do: - 1 \le \cos x \le 1\\
\to - 1 - \dfrac{1}{2} \le \cos x - \dfrac{1}{2} \le 1 - \dfrac{1}{2}\\
\to \dfrac{9}{4} \ge {\left( {\cos x - \dfrac{1}{2}} \right)^2} \ge \dfrac{1}{4}\\
\to - \dfrac{9}{4} \le - {\left( {\cos x - \dfrac{1}{2}} \right)^2} \le - \dfrac{1}{4}\\
\to 0 \le - {\left( {\cos x - \dfrac{1}{2}} \right)^2} + \dfrac{9}{4} \le 2\\
\to Min = 0 \Leftrightarrow \cos x = - 1 \to x = \pi + k2\pi \left( {k \in Z} \right)\\
Max = 2 \Leftrightarrow \cos x = 1 \to x = k2\pi \left( {k \in Z} \right)
\end{array}\)
