Đáp án:
\[\left\{ \begin{array}{l}
{y_{\min }} = - 2 \Leftrightarrow x = \pi \\
{y_{\max }} = 1 \Leftrightarrow x = \dfrac{\pi }{3}
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
y = \sqrt 3 \sin x + \cos x - 1\\
= 2.\left( {\dfrac{1}{2}\cos x + \dfrac{{\sqrt 3 }}{2}\sin x} \right) - 1\\
= 2.\left( {\cos x.cos\dfrac{\pi }{3} + \sin x.\sin \dfrac{\pi }{3}} \right) - 1\\
= 2.\cos \left( {x - \dfrac{\pi }{3}} \right) - 1\\
x \in \left[ {0;\pi } \right] \Rightarrow - \dfrac{\pi }{3} \le x - \dfrac{\pi }{3} \le \dfrac{{2\pi }}{3}\\
\Rightarrow - \dfrac{1}{2} \le \cos \left( {x - \dfrac{\pi }{3}} \right) \le 1\\
\Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = 2.\left( { - \dfrac{1}{2}} \right) - 1 = - 2 \Leftrightarrow x - \dfrac{\pi }{3} = \dfrac{{2\pi }}{3} \Leftrightarrow x = \pi \\
{y_{\max }} = 2.1 - 1 = 1 \Leftrightarrow x - \dfrac{\pi }{3} = 0 \Leftrightarrow x = \dfrac{\pi }{3}
\end{array} \right.
\end{array}\)
Vậy \(\left\{ \begin{array}{l}
{y_{\min }} = - 2 \Leftrightarrow x = \pi \\
{y_{\max }} = 1 \Leftrightarrow x = \dfrac{\pi }{3}
\end{array} \right.\)
