2 câu trả lời
\( y=2\sin\sqrt x +1\)
Do \(-1\le \sin x\le 1\) \(\forall x\)
nên \(-1\le\sin \sqrt x\le 1\)
\(\Rightarrow -2\le 2\sin\sqrt x \le 2\)
\(\Rightarrow -2+1\le 2\sin\sqrt x +1\le 2+1\)
\(\Rightarrow -1\le 2\sin\sqrt x +1\le 3\)
\(\Rightarrow -1\le y\le 3\)
Vậy \(y\) đạt GTLN là \(3\) khi \(\sin \sqrt x=1\Leftrightarrow \sqrt x=\dfrac{\pi}{2}+k\pi, k\in\mathbb Z\Leftrightarrow x=(\dfrac{1}{2}+k\pi)^2,k\in\mathbb Z\).
\(y\) đạt GTNN là \(-1\) khi \(\sin \sqrt x=-1\Leftrightarrow \sqrt x=-\dfrac{\pi}{2}+k\pi, k\in\mathbb Z\Leftrightarrow x=(-\dfrac{1}{2}+k\pi)^2,k\in\mathbb Z\).
$$\eqalign{ & y = 2{\mathop{\rm sinx}\nolimits} + 1 \cr & - 1 \le \sin x \le 1 \cr & \Leftrightarrow - 2 \le 2\sin x \le 2 \cr & \Leftrightarrow - 1 \le 2\sin x + 1 \le 3 \cr & \Rightarrow \min y = - 1;\,\,\max y = 3 \cr} $$
