Toán Tìm Max, Min y=2cosx^2 x giúp mk nha mk cần gấp 06/08/2021 By Josie Tìm Max, Min y=2cosx^2 x giúp mk nha mk cần gấp
Đáp án: $\begin{array}{l}y = 2{\cos ^2}x\\Do: – 1 \le \cos x \le 1\\ \Rightarrow 0 \le {\cos ^2}x \le 1\\ \Rightarrow 0 \le 2{\cos ^2}x \le 2\\ \Rightarrow 0 \le y \le 2\\ \Rightarrow \left\{ \begin{array}{l}GTNN:y = 0\,\\khi:\cos x = 0 \Rightarrow x = \frac{\pi }{2} + k\pi \\GTLN:y = 2\\khi:\cos x = \pm 1 \Rightarrow x = k\pi \end{array} \right.\end{array}$ Trả lời
$y=2\cos^2x$ $=2.\dfrac{1+\cos 2x}{2}$ $=1+\cos 2x$ $-1\le \cos 2x\le 1$ $\Rightarrow 0\le 1+\cos 2x\le 2$ Vậy: $\min y=0\Leftrightarrow \cos 2x=-1\Leftrightarrow x=\dfrac{\pi}{2}+k\pi$ $\max y=2\Leftrightarrow \cos 2x=1\Leftrightarrow x=k\pi$ Trả lời
Đáp án:
$\begin{array}{l}
y = 2{\cos ^2}x\\
Do: – 1 \le \cos x \le 1\\
\Rightarrow 0 \le {\cos ^2}x \le 1\\
\Rightarrow 0 \le 2{\cos ^2}x \le 2\\
\Rightarrow 0 \le y \le 2\\
\Rightarrow \left\{ \begin{array}{l}
GTNN:y = 0\,\\
khi:\cos x = 0 \Rightarrow x = \frac{\pi }{2} + k\pi \\
GTLN:y = 2\\
khi:\cos x = \pm 1 \Rightarrow x = k\pi
\end{array} \right.
\end{array}$
$y=2\cos^2x$
$=2.\dfrac{1+\cos 2x}{2}$
$=1+\cos 2x$
$-1\le \cos 2x\le 1$
$\Rightarrow 0\le 1+\cos 2x\le 2$
Vậy:
$\min y=0\Leftrightarrow \cos 2x=-1\Leftrightarrow x=\dfrac{\pi}{2}+k\pi$
$\max y=2\Leftrightarrow \cos 2x=1\Leftrightarrow x=k\pi$