Question

Tìm GTLN GTNN y=\(\sqrt{5-2cos^2xsin^2x}\)

Answer 1

Lời giải:

Vì \(\cos ^2x; \sin ^2x\geq 0, \forall x\Rightarrow 5-2\cos^2x\sin ^2x\leq 5\)

\(\Rightarrow y=\sqrt{5-2\cos ^2x\sin ^2x}\leq \sqrt{5}\)

Vậy \(y_{\max}=\sqrt{5}\Leftrightarrow \sin x=0\) hoặc \(\cos x=0\)

\(y=\sqrt{5-2\cos ^2x\sin ^2x}=\sqrt{5-\frac{(2\sin x\cos x)^2}{2}}\)

\(=\sqrt{5-\frac{\sin ^22x}{2}}\)

Ta thấy: \(\sin ^22x\leq 1\Rightarrow 5-\frac{\sin ^22x}{2}\geq \frac{9}{2}\)

\(\Rightarrow y\geq \frac{3}{\sqrt{2}}\)

Vậy \(y_{\min}=\frac{3}{\sqrt{2}}\Leftrightarrow \sin 2x=\pm 1\)