Đáp án:
$y_{min}=1$ khi $x=\dfrac{7\pi}{6}+k2\pi\,\,(k\in\mathbb Z)$
$y_{max}=5$ khi $x=\dfrac{\pi}{6}+k2\pi\,\,(k\in\mathbb Z)$
Giải thích các bước giải:
$y=\sin x+\sqrt{3}\cos x+3$
$=2.\left(\dfrac{1}{2}\sin x+\dfrac{\sqrt{3}}{2}\cos x\right)+3$
$=2.\left(\sin\dfrac{\pi}{6}.\sin x+\cos\dfrac{\pi}{6}.\cos x\right)+3$
$=2.\cos\left(x-\dfrac{\pi}{6}\right)+3$
Ta có:
$-1\le \cos\left(x-\dfrac{\pi}{6}\right)\le 1$
$⇔-2\le 2.\cos\left(x-\dfrac{\pi}{6}\right)\le 2$
$⇔1\le 2.\cos\left(x-\dfrac{\pi}{6}\right)+3\le 5$
$⇔1\le y\le 5$
$y\ge 1⇒y_{min}=1$
Dấu "=" xảy ra khi:
$2.\cos\left(x-\dfrac{\pi}{6}\right)+3=1$
$⇔\cos\left(x-\dfrac{\pi}{6}\right)=-1$
$⇔x-\dfrac{\pi}{6}=\pi+k2\pi\,\,(k\in\mathbb Z)$
$⇔x=\dfrac{7\pi}{6}+k2\pi\,\,(k\in\mathbb Z)$
$y\le 5⇒y_{max}=5$
Dấu "=" xảy ra khi:
$2.\cos\left(x-\dfrac{\pi}{6}\right)+3=5$
$⇔\cos\left(x-\dfrac{\pi}{6}\right)=1$
$⇔x-\dfrac{\pi}{6}=k2\pi\,\,(k\in\mathbb Z)$
$⇔x=\dfrac{\pi}{6}+k2\pi\,\,(k\in\mathbb Z)$
Vậy $y_{min}=1$ khi $x=\dfrac{7\pi}{6}+k2\pi\,\,(k\in\mathbb Z)$
$y_{max}=5$ khi $x=\dfrac{\pi}{6}+k2\pi\,\,(k\in\mathbb Z)$.
