Register Now

222-9+11+12:2*14+14 = ? ( )

## Toán Lớp 11: tìm txđ y=căn(3+sinx)/2cos^2-1 tìm GTLN GTNN HÀM SỐ y=4/căn 3sinxcosx

Toán Lớp 11: tìm txđ
y=căn(3+sinx)/2cos^2-1
tìm GTLN GTNN HÀM SỐ
y=4/căn 3sinxcosx

$\begin{array}{l} \left\{ \begin{array}{l} \sin x + 3 \ge 0\\ 2{\cos ^2}x – 1 \ne 0 \end{array} \right. \Leftrightarrow 2{\cos ^2}x – 1 \ne 0\\ \left( { – 1 \le \sin x \le 1 \Rightarrow \sin x + 3 \ge 2 > 0} \right)\\ \Leftrightarrow \cos 2x \ne 0 \Leftrightarrow 2x \ne \dfrac{\pi }{2} + k\pi \\ \Leftrightarrow x \ne \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\left( {k \in \mathbb{Z}} \right)\\ D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{4} + \dfrac{{k\pi }}{2}|k \in \mathbb{Z}} \right\} \end{array}$
$\begin{array}{l} y = \dfrac{4}{{\sqrt 3 \sin x\cos x}} = \dfrac{8}{{2\sqrt 3 \sin x\cos x}} = \dfrac{8}{{\sqrt 3 \sin 2x}}\\ – 1 \le \sin 2x \le 1 \Rightarrow – \sqrt 3 \le \sqrt 3 \sin 2x \le \sqrt 3 \\ \Rightarrow – \dfrac{8}{{\sqrt 3 }} \le y \le \dfrac{8}{{\sqrt 3 }}\\ \Rightarrow \max y = \dfrac{8}{{\sqrt 3 }},\min y = – \dfrac{8}{{\sqrt 3 }} \end{array}$