2 câu trả lời
Đáp án:
GTLN $y=\sqrt2$ khi $x=\dfrac{\pi}8+k\pi$
Lời giải:
$y=\cos 2x+\sin 2x=\sqrt2\sin x(2x+\dfrac{\pi}{4})$
Do $-1\le\sin x\le1$ $\forall x$
$\Rightarrow-1\le\sin x(2x+\dfrac{\pi}{4})\le1$
$\Rightarrow-\sqrt2\le\sqrt2\sin x(2x+\dfrac{\pi}{4})\le\sqrt2$
$\Rightarrow-\sqrt2\le y\le \sqrt2$
GTLN $y=\sqrt2$ khi $\sin (2x+\dfrac{\pi}{4})=1\Leftrightarrow 2x+\dfrac{\pi}4=\dfrac{\pi}2+k2\pi$
$\Leftrightarrow x=\dfrac{\pi}8+k\pi$.
Ta có
$\cos(2x) + \sin(2x) = \sqrt{2} (\dfrac{1}{\sqrt{2}} \cos(2x) + \dfrac{1}{\sqrt{2}} \sin(2x)) = \sqrt{2} \sin(2x + \dfrac{\pi}{4})$
Ta luôn có
$-1 \leq \sin(2x + \dfrac{\pi}{4}) \leq 1$
$<-> -\sqrt{2} \leq \sqrt{2} \sin(2x + \dfrac{\pi}{4}) \leq \sqrt{2}$
Vậy GTLN của hso là $\sqrt{2}$, đạt được khi $\sin(2x + \dfrac{\pi}{4}) = 1$ hay $2x + \dfrac{\pi}{4} = \dfrac{\pi}{2} + 2k\pi$ hay $x = \dfrac{\pi}{8} + k\pi$.