Toán Mn giúp em tìm Max Min của hàm số y= √2Sin2x + Cos2x +1 04/08/2021 By Daisy Mn giúp em tìm Max Min của hàm số y= √2Sin2x + Cos2x +1
Đáp án: $Miny = 1 – \sqrt 3 \Leftrightarrow x = \dfrac{1}{2}\arccos \dfrac{1}{{\sqrt 3 }} + \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)$; $Maxy = 1 + \sqrt 3 \Leftrightarrow x = \dfrac{1}{2}\arccos \dfrac{1}{{\sqrt 3 }} + k\pi \left( {k \in Z} \right)$ Giải thích các bước giải: Ta có: $\begin{array}{l}y = \sqrt 2 \sin 2x + \cos 2x + 1\\ = \sqrt 3 \left( {\dfrac{{\sqrt 2 }}{{\sqrt 3 }}\sin 2x + \dfrac{1}{{\sqrt 3 }}\cos 2x} \right) + 1\\ = \sqrt 3 \left( {\sqrt {\dfrac{2}{3}} \sin 2x + \dfrac{1}{{\sqrt 3 }}\cos 2x} \right) + 1\\ = \sqrt 3 \cos \left( {2x – \alpha } \right) + 1\left( {\alpha :\sin \alpha = \sqrt {\dfrac{2}{3}} ;\cos \alpha = \dfrac{1}{{\sqrt 3 }}} \right)\end{array}$ Khi đó: $\alpha = \arccos \dfrac{1}{{\sqrt 3 }}$ Mặt khác: $\begin{array}{l} – 1 \le \cos \left( {2x – \alpha } \right) \le 1,\forall x\\ \Rightarrow 1 – \sqrt 3 \le \sqrt 3 \cos \left( {2x – \alpha } \right) + 1 \le 1 + \sqrt 3 \end{array}$ $1 – \sqrt 3 \le y \le 1 + \sqrt 3 $ Như vậy: $\begin{array}{l} + )Miny = 1 – \sqrt 3 \\ \Leftrightarrow \cos \left( {2x – \alpha } \right) = – 1\\ \Leftrightarrow 2x – \alpha = \pi + k2\pi \\ \Leftrightarrow x = \dfrac{1}{2}\arccos \dfrac{1}{{\sqrt 3 }} + \dfrac{\pi }{2} + k\pi \\ + )Maxy = 1 + \sqrt 3 \\ \Leftrightarrow \cos \left( {2x – \alpha } \right) = 1\\ \Leftrightarrow 2x – \alpha = k2\pi \\ \Leftrightarrow x = \dfrac{1}{2}\arccos \dfrac{1}{{\sqrt 3 }} + k\pi \end{array}$ Vậy $Miny = 1 – \sqrt 3 \Leftrightarrow x = \dfrac{1}{2}\arccos \dfrac{1}{{\sqrt 3 }} + \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)$; $Maxy = 1 + \sqrt 3 \Leftrightarrow x = \dfrac{1}{2}\arccos \dfrac{1}{{\sqrt 3 }} + k\pi \left( {k \in Z} \right)$ Trả lời
$y=\sqrt2\sin2x+\cos2x+1$ $=\sqrt3(\sqrt{\dfrac{2}{3}}\sin 2x+\dfrac{1}{\sqrt3}\cos 2x)+1$ Đặt $\sin a=\sqrt{\dfrac{2}{3}}$, $\cos a=\dfrac{1}{\sqrt3}$ $\Rightarrow y=\sqrt3\sin(2x+a)+1$ $-1\le \sin(2x+a)\le 1$ $\Rightarrow -\sqrt3+1\le y\le \sqrt3+1$ $\Rightarrow \min y=-\sqrt3+1; max y=\sqrt3+1$ Trả lời
Đáp án:
$Miny = 1 – \sqrt 3 \Leftrightarrow x = \dfrac{1}{2}\arccos \dfrac{1}{{\sqrt 3 }} + \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)$; $Maxy = 1 + \sqrt 3 \Leftrightarrow x = \dfrac{1}{2}\arccos \dfrac{1}{{\sqrt 3 }} + k\pi \left( {k \in Z} \right)$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
y = \sqrt 2 \sin 2x + \cos 2x + 1\\
= \sqrt 3 \left( {\dfrac{{\sqrt 2 }}{{\sqrt 3 }}\sin 2x + \dfrac{1}{{\sqrt 3 }}\cos 2x} \right) + 1\\
= \sqrt 3 \left( {\sqrt {\dfrac{2}{3}} \sin 2x + \dfrac{1}{{\sqrt 3 }}\cos 2x} \right) + 1\\
= \sqrt 3 \cos \left( {2x – \alpha } \right) + 1\left( {\alpha :\sin \alpha = \sqrt {\dfrac{2}{3}} ;\cos \alpha = \dfrac{1}{{\sqrt 3 }}} \right)
\end{array}$
Khi đó: $\alpha = \arccos \dfrac{1}{{\sqrt 3 }}$
Mặt khác:
$\begin{array}{l}
– 1 \le \cos \left( {2x – \alpha } \right) \le 1,\forall x\\
\Rightarrow 1 – \sqrt 3 \le \sqrt 3 \cos \left( {2x – \alpha } \right) + 1 \le 1 + \sqrt 3
\end{array}$
$1 – \sqrt 3 \le y \le 1 + \sqrt 3 $
Như vậy:
$\begin{array}{l}
+ )Miny = 1 – \sqrt 3 \\
\Leftrightarrow \cos \left( {2x – \alpha } \right) = – 1\\
\Leftrightarrow 2x – \alpha = \pi + k2\pi \\
\Leftrightarrow x = \dfrac{1}{2}\arccos \dfrac{1}{{\sqrt 3 }} + \dfrac{\pi }{2} + k\pi \\
+ )Maxy = 1 + \sqrt 3 \\
\Leftrightarrow \cos \left( {2x – \alpha } \right) = 1\\
\Leftrightarrow 2x – \alpha = k2\pi \\
\Leftrightarrow x = \dfrac{1}{2}\arccos \dfrac{1}{{\sqrt 3 }} + k\pi
\end{array}$
Vậy $Miny = 1 – \sqrt 3 \Leftrightarrow x = \dfrac{1}{2}\arccos \dfrac{1}{{\sqrt 3 }} + \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)$; $Maxy = 1 + \sqrt 3 \Leftrightarrow x = \dfrac{1}{2}\arccos \dfrac{1}{{\sqrt 3 }} + k\pi \left( {k \in Z} \right)$
$y=\sqrt2\sin2x+\cos2x+1$
$=\sqrt3(\sqrt{\dfrac{2}{3}}\sin 2x+\dfrac{1}{\sqrt3}\cos 2x)+1$
Đặt $\sin a=\sqrt{\dfrac{2}{3}}$, $\cos a=\dfrac{1}{\sqrt3}$
$\Rightarrow y=\sqrt3\sin(2x+a)+1$
$-1\le \sin(2x+a)\le 1$
$\Rightarrow -\sqrt3+1\le y\le \sqrt3+1$
$\Rightarrow \min y=-\sqrt3+1; max y=\sqrt3+1$