Bài tập 3 trang 14 SGK Toán 11 NC

a) Ta có \( - 1 \le \cos \left( {x + \frac{\pi }{3}} \right) \le 1\)

\(\begin{array}{l}
 \Rightarrow  - 2 \le 2\cos \left( {x + \frac{\pi }{3}} \right) \le 2\\
 \Rightarrow 1 \le 2\cos \left( {x + \frac{\pi }{3}} \right) + 3 \le 5\\
 \Rightarrow 1 \le y \le 5
\end{array}\)

Vậy \(\min y = 1\) khi \(x + \frac{\pi }{3} = \pi  + k2\pi  \Leftrightarrow x = \frac{{2\pi }}{3} + k2\pi \) \(k \in Z\)

\(\max y = 5\) khi \(x + \frac{\pi }{3} = k2\pi  \Leftrightarrow x =  - \frac{\pi }{3} + k2\pi \) \(k \in Z\)

b) Ta có  0 ≤ 1−sinx² ≤ 2

\(\begin{array}{l}
 \Rightarrow  - 1 \le \sqrt {1 - \sin \left( {{x^2}} \right)}  - 1 \le \sqrt 2  - 1\\
 \Rightarrow  - 1 \le y \le \sqrt 2  - 1
\end{array}\)

Vậy min y = −1 khi \({x^2} = \frac{\pi }{2} + k2\pi ,k \ge 0,k \in Z\)

\(\max y = \sqrt 2  - 1\) khi \({x^2} =  - \frac{\pi }{2} + k2\pi ,k > 0,k \in Z\)

c) Ta có:

\(\begin{array}{l}
 - 1 \le \sin \sqrt x  \le 1 \Rightarrow  - 4 \le 4\sin \sqrt x  \le 4\\
 \Rightarrow  - 4 \le y \le 4
\end{array}\)

 Vậy min y = −4 khi \(\sqrt x  =  - \frac{\pi }{2} + k2\pi ,k > 0,k \in Z\)

max y = 4 khi \(\sqrt x  = \frac{\pi }{2} + k2\pi ,k \ge 0,k \in Z\)

-- Mod Toán 11 HỌC247