Cho y= (√sinx) + ( √cosx). Tim a) TXD cua no b) GTLN va GTNN cua no 03/10/2021 Bởi Kennedy Cho y= (√sinx) + ( √cosx). Tim a) TXD cua no b) GTLN va GTNN cua no
a) ĐKXĐ: \(\left\{ \begin{array}{l} \sin x\ge 0 \\ \cos x\ge 0\end{array} \right .\) \(\Leftrightarrow \left\{ \begin{array}{l} x\in [k2\pi;\pi+k2\pi] \\ x\in [-\dfrac{\pi}{2}+k2\pi;\dfrac{\pi}{2}+k2\pi] \end{array} \right .\) \(\Leftrightarrow x\in [k2\pi;\dfrac{\pi}{2}+k2\pi]\) b) Ta có: \(y= \sqrt{\sin x} + \sqrt{\cos x}\ge 2\sqrt{\sqrt{\sin x} . \sqrt{\cos x}}=2\sqrt{\sqrt{\sin x\cos x}}=2\sqrt{\sqrt{\dfrac{\sin 2x}{2}}}=2\sqrt[4]{\dfrac{\sin 2x}{2}}\) Ta có: \(0\le \sin 2x\le 1\) \(\Rightarrow 0\le \dfrac{\sin 2x}{2}\le \dfrac{1}{2}\) \(\Rightarrow 0\le \sqrt[4]{\dfrac{\sin 2x}{2}}\le \dfrac{1}{\sqrt[4] 2}\) \(\Rightarrow 0\le 2\sqrt[4]{\dfrac{\sin 2x}{2}}\le 2\dfrac{1}{\sqrt[4] 2}\) \(\Rightarrow y\ge[0;2\dfrac{1}{\sqrt[4] 2}]\) Vậy \(y\) đạt GTNN bằng \(2\dfrac{1}{\sqrt[4] 2}\) tại \(\sin 2x= 1\). Bình luận
a) ĐKXĐ: \(\left\{ \begin{array}{l} \sin x\ge 0 \\ \cos x\ge 0\end{array} \right .\) \(\Leftrightarrow \left\{ \begin{array}{l} x\in [k2\pi;\pi+k2\pi] \\ x\in [-\dfrac{\pi}{2}+k2\pi;\dfrac{\pi}{2}+k2\pi] \end{array} \right .\) \(\Leftrightarrow x\in [k2\pi;\dfrac{\pi}{2}+k2\pi]\)
b) Ta có: \(y= \sqrt{\sin x} + \sqrt{\cos x}\ge 2\sqrt{\sqrt{\sin x} . \sqrt{\cos x}}=2\sqrt{\sqrt{\sin x\cos x}}=2\sqrt{\sqrt{\dfrac{\sin 2x}{2}}}=2\sqrt[4]{\dfrac{\sin 2x}{2}}\)
Ta có: \(0\le \sin 2x\le 1\)
\(\Rightarrow 0\le \dfrac{\sin 2x}{2}\le \dfrac{1}{2}\)
\(\Rightarrow 0\le \sqrt[4]{\dfrac{\sin 2x}{2}}\le \dfrac{1}{\sqrt[4] 2}\)
\(\Rightarrow 0\le 2\sqrt[4]{\dfrac{\sin 2x}{2}}\le 2\dfrac{1}{\sqrt[4] 2}\)
\(\Rightarrow y\ge[0;2\dfrac{1}{\sqrt[4] 2}]\)
Vậy \(y\) đạt GTNN bằng \(2\dfrac{1}{\sqrt[4] 2}\) tại \(\sin 2x= 1\).