Đáp án:
$\min y = 0 \Leftrightarrow x =-\dfrac{\pi}{2} + k2\pi$
$\max y = 2 \Leftrightarrow x =\dfrac{\pi}{2} + k2\pi\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$y =\left(\sin\dfrac x2 +\cos \dfrac x2\right)^2$
$\to y = \left(\sin\dfrac x2\right)^2 + 2\sin\dfrac x2\cos\dfrac x2 + \left(\cos\dfrac x2\right)^2$
$\to y = 1+ \sin x$
Ta có:
$\quad -1\leq \sin x \leq 1$
$\to 0 \leq 1 +\sin x \leq 2$
$\to 0\leq y \leq 2$
Vậy $\min y = 0 \Leftrightarrow \sin x =-1\Leftrightarrow x =-\dfrac{\pi}{2} + k2\pi$
$\max y = 2 \Leftrightarrow \sin x = 1 \Leftrightarrow x =\dfrac{\pi}{2} + k2\pi\quad (k\in\Bbb Z)$
