Toán Lớp 11: Tìm giá trị lớn nhất nhỏ nhất của hàm số y = cos ^2 x – cos 2x + 1
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 11: Tìm giá trị lớn nhất nhỏ nhất của hàm số y = cos ^2 x – cos 2x + 1
Comments ( 1 )
{y_{\min }} = 1\\
{y_{\max }} = 2
\end{array} \right.\]
y = {\cos ^2}x – \cos 2x + 1\\
= \left( {{{\cos }^2}x – \dfrac{1}{2}} \right) – \cos 2x + \dfrac{3}{2}\\
= \dfrac{1}{2}.\left( {2{{\cos }^2}x – 1} \right) – \cos 2x + \dfrac{3}{2}\\
= \dfrac{1}{2}.\cos 2x – \cos 2x + \dfrac{3}{2}\\
= \dfrac{3}{2} – \dfrac{1}{2}\cos 2x\\
– 1 \le \cos 2x \le 1\\
\Leftrightarrow – \dfrac{1}{2} \le \dfrac{1}{2}\cos 2x \le \dfrac{1}{2}\\
\Leftrightarrow – \dfrac{1}{2} \le – \dfrac{1}{2}\cos 2x \le \dfrac{1}{2}\\
\Leftrightarrow – \dfrac{1}{2} + \dfrac{3}{2} \le – \dfrac{1}{2}\cos 2x + \dfrac{3}{2} \le \dfrac{1}{2} + \dfrac{3}{2}\\
\Leftrightarrow 1 \le \dfrac{3}{2} – \dfrac{1}{2}\cos 2x \le 2\\
\Leftrightarrow 1 \le y \le 2\\
\Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = 1 \Leftrightarrow \cos 2x = 1 \Leftrightarrow 2x = k2\pi \Leftrightarrow x = k\pi \\
{y_{\max }} = 2 \Leftrightarrow \cos 2x = – 1 \Leftrightarrow 2x = \pi + k2\pi \Leftrightarrow x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
\Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = 1\\
{y_{\max }} = 2
\end{array} \right.
\end{array}\)