Đáp án:
$maxy = \sqrt{\dfrac{11}{2}} \Leftrightarrow x = \pm \dfrac{\pi}{6} + k\pi\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$y = \sqrt{1 + \dfrac{1}{2}\cos^2x} + \dfrac{1}{2}\sqrt{5 + 2\sin^2x}$
$= \sqrt{1 + \dfrac{1}{2}\cdot\dfrac{1 + \cos2x}{2}} + \sqrt{\dfrac{1}{4}\cdot\left(5 + 2.\dfrac{1 - \cos2x}{2}\right)}$
$= \sqrt{\dfrac{5 + \cos2x}{4}} + \sqrt{\dfrac{6 - \cos2x}{4}}$
Áp dụng bất đẳng thức Bunhiacopxki, ta được:
$y^2 = \left(1.\sqrt{\dfrac{5 + \cos2x}{4}} + 1.\sqrt{\dfrac{6 - \cos2x}{4}}\right)^2 \leq (1^2+1^2)\left(\dfrac{5 + \cos2x}{4} + \dfrac{6 - \cos2x}{4}\right)$
$\Leftrightarrow y^2 \leq 2.\dfrac{5 + \cos2x + 6 - \cos2x}{4} = \dfrac{11}{2}$
$\Rightarrow y \leq \sqrt{\dfrac{11}{2}}$
Dấu = xảy ra $\Leftrightarrow \sqrt{\dfrac{5 + \cos2x}{4}}= \sqrt{\dfrac{6 - \cos2x}{4}}$
$\Leftrightarrow 5 + \cos2x = 6 - \cos2x$
$\Leftrightarrow \cos2x = \dfrac{1}{2}$
$\Leftrightarrow x = \pm \dfrac{\pi}{6} + k\pi\quad (k \in \Bbb Z)$
Vậy $maxy = \sqrt{\dfrac{11}{2}} \Leftrightarrow x = \pm \dfrac{\pi}{6} + k\pi\quad (k \in \Bbb Z)$
