tìm min max của y y= 2sin4x + 3cos4x + 1 y= sinx + 2 căn (2- sin^2 x ) MONG MỌI NGƯỜI GIÚP ĐỠ

$1)\, y = 2\sin4x + 3\cos4x + 1$

Áp dụng bất đẳng thức $Bunyakovsky$ ta được:

$(y – 1)^2 = (2\sin4x + 3\cos4x)^2 \leq (2^2 + 3^2)(\sin^24x + \cos^24x) = 13$

$\Rightarrow -\sqrt{13} \leq y – 1 \leq \sqrt{13}$

$\Rightarrow 1 – \sqrt{13} \leq y \leq 1 + \sqrt{13}$

Vậy $\min y = 1 – \sqrt{13} \Leftrightarrow 2\sin4x + 4\cos4x = – \sqrt{13} \Leftrightarrow x = -\dfrac{\pi}{8} – \dfrac{1}{4}\arccos\dfrac{2}{\sqrt{13}} + k\dfrac{\pi}{2}$

$\max y = 1 + \sqrt{13} \Leftrightarrow 2\sin4x + 4\cos4x = \sqrt{13} \Leftrightarrow x = \dfrac{\pi}{8} – \dfrac{1}{4}\arccos\dfrac{2}{\sqrt{13}} + k\dfrac{\pi}{2} \qquad (k \in \Bbb Z)$

$2)\, y = \sin x + 2\sqrt{2 – \sin^2x}$

Ta có:

$\sin x \geq – 1$

$\sin^2x \geq 0$

$\Leftrightarrow 2 – \sin^2x \geq 1$

$\Leftrightarrow 2\sqrt{2 – \sin^2x} \geq 2$

$\Rightarrow \sin x + 2\sqrt{2 – \sin^2x} \geq -1 + 2 = 1$

Hay $y \geq 1$

Dấu = xảy ra $\Leftrightarrow \sin x = -1 \Leftrightarrow x = – \dfrac{\pi}{2} + k2\pi$

Áp dụng bất đẳng thức $Bunyakovsky$ ta được:

$y^2 = (\sin x+ 2\sqrt{2 – \sin^2x})^2 \leq (1^2 + 2^2)(\sin^2x + 2 – \sin^2x) = 10$

$\Rightarrow y \leq \sqrt{10}$

Dấu = xảy ra $\Leftrightarrow 2\sin x = \sqrt{2 – \sin^2x} \Leftrightarrow \sin x = \pm \sqrt{\dfrac{2}{5}}$