Tìm GTLN, GTNN của hàm số \(y = {\sin ^2}x - {\cos ^2}x + 2\sin x\cos x + 5\).

Lời giải:

\(y = {\sin ^2}x - {\cos ^2}x + 2\sin x\cos x + 5 = \sin 2x - \cos 2x + 5 = \sqrt 2 \sin \left( {2x - \frac{\pi }{4}} \right) + 5\)

\(\begin{array}{l} - 1 \le \sin \left( {2x - \frac{\pi }{4}} \right) \le 1\\ \Rightarrow - \sqrt 2 \le \sqrt 2 \sin \left( {2x - \frac{\pi }{4}} \right) \le \sqrt 2 \\ \Rightarrow 5 - \sqrt 2 \le y \le 5 + \sqrt 2 \end{array}\)

\(\begin{array}{l} \Rightarrow \left\{ {\begin{array}{*{20}{c}}{{{\max }_y} = 5 + \sqrt 2 \Rightarrow 2x - \frac{\pi }{4} = \frac{\pi }{2} + k2\pi }\\{{{\min }_y} = 5 - \sqrt 2 \Rightarrow 2x - \frac{\pi }{4} = - \frac{\pi }{2} + k2\pi }\end{array}} \right.\\ \Rightarrow \left\{ {\begin{array}{*{20}{c}}{{{\max }_y} = 5 + \sqrt 2 \Rightarrow x = \frac{{3\pi }}{8} + k\pi }\\{{{\min }_y} = 5 - \sqrt 2 \Rightarrow x = - \frac{\pi }{8} + k\pi }\end{array},k \in \mathbb{Z}} \right.\end{array}\)