Tìm GLTN, GTNN của hàm số y = 2 sin x + $\sqrt {3}$ cos x + 7

Đáp án:

$\begin{cases}\min y = 7 – \sqrt7 \Leftrightarrow x = – \dfrac{\pi}{2} – \alpha + k2\pi\\\max y = 7 + \sqrt7  \Leftrightarrow x = \dfrac{\pi}{2} – \alpha + k2\pi\end{cases}\quad \left(\alpha = \arccos\dfrac{2}{\sqrt7}, \, k \in \Bbb Z\right)$

Giải thích các bước giải:

$y = 2\sin x + \sqrt3\cos x + 7$

$y = \sqrt7\left(\dfrac{2}{\sqrt7}\sin x + \dfrac{\sqrt3}{\sqrt7}\cos x\right) + 7$

$y = \sqrt7\sin(x + \alpha) + 7$ Với $\alpha = \arccos\dfrac{2}{\sqrt7}$

Ta có: $-1 \leq \sin(x + \alpha) \leq 1$

$\Leftrightarrow -\sqrt7 \leq \sqrt7\sin(x + \alpha) \leq \sqrt7$

$\Leftrightarrow 7 – \sqrt7 \leq \sqrt7\sin(x + \alpha) + 7 \leq 7 + \sqrt7$

Hay $7 – \sqrt7 \leq y \leq 7+ \sqrt7$

Vậy $\min y = 7 – \sqrt7 \Leftrightarrow \sin(x + \alpha) = – 1 \Leftrightarrow x = – \dfrac{\pi}{2} – \alpha + k2\pi$

$\max y = 7 + \sqrt7 \Leftrightarrow \sin(x + \alpha) = 1 \Leftrightarrow x = \dfrac{\pi}{2} – \alpha + k2\pi \quad (k \in \Bbb Z)$