Câu 3 trang 14 SGK Đại số và Giải tích 11 Nâng cao

Hướng dẫn giải

a. Ta có: $$-1 ≤ \cos \left( {x + {\pi \over 3}} \right) ≤ 1$$

\eqalign{ & \Rightarrow - 2 \le 2\cos \left( {x + {\pi \over 3}} \right) \le 2\cr& \Rightarrow 1 \le 2\cos \left( {x + {\pi \over 3}} \right) + 3 \le 5 \Rightarrow 1 \le y \le 5 \cr &\text{ Vậy }\cr&\min \,y = 1\,khi\,x + {\pi \over 3} = \pi + k2\pi \,\cr&\,\,\,\,\,\,\,\text{ khi} \,x = {{2\pi } \over 3} + k2\pi \cr &\max \,y = 5\,khi\,x + {\pi \over 3} = k2\pi \,\text{ khi} \,x = - {\pi \over 3} + k2\pi \cr&\left( {k \in \mathbb Z} \right) \cr}

b. Ta có:  $$0 \le 1 - \sin {x^2} \le 2$$

$$\Rightarrow - 1 \le \sqrt {1 - \sin {x^2}} - 1 \le \sqrt 2 - 1$$

$$\Rightarrow - 1 \le y \le \sqrt 2 - 1$$

\eqalign{ & \text{ Vậy }\,\min \,y = - 1\,\text{ khi} \,{x^2} = {\pi \over 2} + k2\pi ,k \ge 0,k \in\mathbb Z \cr &\max\,y = \sqrt 2 - 1\text{ khi}\,{x^2} = - {\pi \over 2} + k2\pi ,k > 0,k \in \mathbb Z \cr}

c. Ta có:  $$- 1 \le \sin \sqrt x \le 1 \Rightarrow - 4 \le 4\sin \sqrt x \le 4$$

$$⇒ -4 ≤ y ≤ 4$$

\eqalign{ & \text{ Vậy }\cr&\min \,y = - 4\,\text{ khi}\,\sqrt x = - {\pi \over 2} + k2\pi ,k > 0,k \in\mathbb Z \cr & \max \,y = 4\,\text{ khi}\,\sqrt x = {\pi \over 2} + k2\pi ,k \ge 0,k \in\mathbb Z \cr}