Đáp án: $B$
Giải thích các bước giải:
Ta có:
$y=\sqrt{1+\dfrac12\cos^2x}+\dfrac12\sqrt{5+2\sin^2x}$
$\to y=\sqrt{1+\dfrac12\cos^2x}+\sqrt{\dfrac54+\dfrac12\sin^2x}$
$\to y\le \sqrt{2(1+\dfrac12\cos^2x+\dfrac54+\dfrac12\sin^2x)}$
$\to y\le \sqrt{2(\dfrac94+\dfrac12(\cos^2x+\sin^2x))}$
$\to y\le \sqrt{\dfrac{11}{2}}$
Dấu = xảy ra khi
$1+\dfrac12\cos^2x=\dfrac54+\dfrac12\sin^2x$
$\to \dfrac12(\cos^2x-\sin^2x)=\dfrac14$
$\to \dfrac12\cos 2x=\dfrac14$
$\to \cos 2x=\dfrac12$
$\to x\in\{\dfrac{\pi}{6}+k\pi, \dfrac56\pi+k\pi\}$
