Đáp án:
$\min y =0$
Giải thích các bước giải:
$y = \cos^2x + \sin x + 1$
$\to y = 1 -\sin^2x +\sin x +1$
$\to y = -\left(\sin^2x -2.\dfrac{1}{2}\sin x +\dfrac{1}{4}\right) +\dfrac{9}{4}$
$\to y = -\left(\sin x -\dfrac{1}{2}\right)^2 +\dfrac{9}{4}$
Ta có:
$-1 \leq \sin x \leq 1$
$\to -\dfrac{3}{2} \leq \sin x -\dfrac{1}{2}\leq \dfrac{1}{2}$
$\to 0 \leq \left(\sin x -\dfrac{1}{2}\right)^2 \leq \dfrac{9}{4}$
$\to -\dfrac{9}{4} \leq -\left(\sin x -\dfrac{1}{2}\right)^2 \leq 0$
$\to 0 \leq -\left(\sin x -\dfrac{1}{2}\right)^2 +\dfrac 94 \leq \dfrac 94$
$\to 0 \leq y \leq \dfrac 94$
Vậy $\min y = 0$
