Tìm GTNN, GTLN của: $y=2.cos^2-3cosx+3$ $ y = sin^2x-cosx+1$ với pi/4 $\leq$ x $\leq$ pi/2 10/09/2021 Bởi Parker Tìm GTNN, GTLN của: $y=2.cos^2-3cosx+3$ $ y = sin^2x-cosx+1$ với pi/4 $\leq$ x $\leq$ pi/2
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: \(\begin{array}{l}a.y = 2{\cos ^2}x – 3\cos x + 3\\ = {\left( {\sqrt 2 \cos x} \right)^2} – 2.\sqrt 2 \cos x.\dfrac{3}{{2\sqrt 2 }} + {\left( {\dfrac{3}{{2\sqrt 2 }}} \right)^2} + \dfrac{{15}}{8}\\ = {\left( {\sqrt 2 \cos x – \dfrac{3}{{2\sqrt 2 }}} \right)^2} + \dfrac{{15}}{8}\\Do: – 1 \le \cos x \le 1\\ \to – \sqrt 2 \le \sqrt 2 \cos x \le \sqrt 2 \\ \to – \sqrt 2 – \dfrac{3}{{2\sqrt 2 }} \le \sqrt 2 \cos x – \dfrac{3}{{2\sqrt 2 }} \le \sqrt 2 – \dfrac{3}{{2\sqrt 2 }}\\ \to \dfrac{{49}}{8} \ge {\left( {\sqrt 2 \cos x – \dfrac{3}{{2\sqrt 2 }}} \right)^2} \ge \dfrac{1}{8}\\ \to \dfrac{{49}}{8} + \dfrac{{15}}{8} \ge {\left( {\sqrt 2 \cos x – \dfrac{3}{{2\sqrt 2 }}} \right)^2} \ge \dfrac{1}{8} + \dfrac{{15}}{8}\\ \to 8 \ge y \ge 2\\ \to Min = 2 \Leftrightarrow \cos x = 1 \to x = k2\pi \left( {k \in Z} \right)\\Max = 8 \Leftrightarrow \cos x = – 1 \Leftrightarrow x = \pi + k2\pi \left( {k \in Z} \right)\\b.y = \left( {1 – {{\cos }^2}x} \right) – \cos x + 1\\ = – {\cos ^2}x – \cos x + 2\\ = – \left( {{{\cos }^2}x + 2.\cos x.\dfrac{1}{2} + \dfrac{1}{4} – \dfrac{9}{4}} \right)\\ = – {\left( {\cos x – \dfrac{1}{2}} \right)^2} + \dfrac{9}{4}\\Do: – 1 \le \cos x \le 1\\ \to – 1 – \dfrac{1}{2} \le \cos x – \dfrac{1}{2} \le 1 – \dfrac{1}{2}\\ \to \dfrac{9}{4} \ge {\left( {\cos x – \dfrac{1}{2}} \right)^2} \ge \dfrac{1}{4}\\ \to – \dfrac{9}{4} \le – {\left( {\cos x – \dfrac{1}{2}} \right)^2} \le – \dfrac{1}{4}\\ \to 0 \le – {\left( {\cos x – \dfrac{1}{2}} \right)^2} + \dfrac{9}{4} \le 2\\ \to Min = 0 \Leftrightarrow \cos x = – 1 \to x = \pi + k2\pi \left( {k \in Z} \right)\\Max = 2 \Leftrightarrow \cos x = 1 \to x = k2\pi \left( {k \in Z} \right)\end{array}\) Bình luận
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:
\(\begin{array}{l}
a.y = 2{\cos ^2}x – 3\cos x + 3\\
= {\left( {\sqrt 2 \cos x} \right)^2} – 2.\sqrt 2 \cos x.\dfrac{3}{{2\sqrt 2 }} + {\left( {\dfrac{3}{{2\sqrt 2 }}} \right)^2} + \dfrac{{15}}{8}\\
= {\left( {\sqrt 2 \cos x – \dfrac{3}{{2\sqrt 2 }}} \right)^2} + \dfrac{{15}}{8}\\
Do: – 1 \le \cos x \le 1\\
\to – \sqrt 2 \le \sqrt 2 \cos x \le \sqrt 2 \\
\to – \sqrt 2 – \dfrac{3}{{2\sqrt 2 }} \le \sqrt 2 \cos x – \dfrac{3}{{2\sqrt 2 }} \le \sqrt 2 – \dfrac{3}{{2\sqrt 2 }}\\
\to \dfrac{{49}}{8} \ge {\left( {\sqrt 2 \cos x – \dfrac{3}{{2\sqrt 2 }}} \right)^2} \ge \dfrac{1}{8}\\
\to \dfrac{{49}}{8} + \dfrac{{15}}{8} \ge {\left( {\sqrt 2 \cos x – \dfrac{3}{{2\sqrt 2 }}} \right)^2} \ge \dfrac{1}{8} + \dfrac{{15}}{8}\\
\to 8 \ge y \ge 2\\
\to Min = 2 \Leftrightarrow \cos x = 1 \to x = k2\pi \left( {k \in Z} \right)\\
Max = 8 \Leftrightarrow \cos x = – 1 \Leftrightarrow x = \pi + k2\pi \left( {k \in Z} \right)\\
b.y = \left( {1 – {{\cos }^2}x} \right) – \cos x + 1\\
= – {\cos ^2}x – \cos x + 2\\
= – \left( {{{\cos }^2}x + 2.\cos x.\dfrac{1}{2} + \dfrac{1}{4} – \dfrac{9}{4}} \right)\\
= – {\left( {\cos x – \dfrac{1}{2}} \right)^2} + \dfrac{9}{4}\\
Do: – 1 \le \cos x \le 1\\
\to – 1 – \dfrac{1}{2} \le \cos x – \dfrac{1}{2} \le 1 – \dfrac{1}{2}\\
\to \dfrac{9}{4} \ge {\left( {\cos x – \dfrac{1}{2}} \right)^2} \ge \dfrac{1}{4}\\
\to – \dfrac{9}{4} \le – {\left( {\cos x – \dfrac{1}{2}} \right)^2} \le – \dfrac{1}{4}\\
\to 0 \le – {\left( {\cos x – \dfrac{1}{2}} \right)^2} + \dfrac{9}{4} \le 2\\
\to Min = 0 \Leftrightarrow \cos x = – 1 \to x = \pi + k2\pi \left( {k \in Z} \right)\\
Max = 2 \Leftrightarrow \cos x = 1 \to x = k2\pi \left( {k \in Z} \right)
\end{array}\)