Google
Câu hỏi: Tìm giá trị lớn nhất của hàm số $f\left( x \right)=\sin 3x+\cos 3x$.
A. $3\sqrt{2}$.
B. $2$.
C. $\sqrt{2}$.
D. $1$.
A. $3\sqrt{2}$.
B. $2$.
C. $\sqrt{2}$.
D. $1$.
Ta có $f\left( x \right)=\sin 3x+\cos 3x=\sqrt{2}\sin \left( 3x+\dfrac{\pi }{4} \right)$.
Ta có $-1\le \sin \left( 3x+\dfrac{\pi }{4} \right)\le 1\Rightarrow -\sqrt{2}\le \sqrt{2}\sin \left( 3x+\dfrac{\pi }{4} \right)\le \sqrt{2}\Rightarrow -\sqrt{2}\le f\left( x \right)\le \sqrt{2}$.
$\max f\left( x \right)=\sqrt{2}\Leftrightarrow \sin \left( 3x+\dfrac{\pi }{4} \right)=1\Leftrightarrow 3x+\dfrac{\pi }{4}=\dfrac{\pi }{2}+k2\pi \Leftrightarrow x=\dfrac{\pi }{12}+\dfrac{k2\pi }{3}$.
Ta có $-1\le \sin \left( 3x+\dfrac{\pi }{4} \right)\le 1\Rightarrow -\sqrt{2}\le \sqrt{2}\sin \left( 3x+\dfrac{\pi }{4} \right)\le \sqrt{2}\Rightarrow -\sqrt{2}\le f\left( x \right)\le \sqrt{2}$.
$\max f\left( x \right)=\sqrt{2}\Leftrightarrow \sin \left( 3x+\dfrac{\pi }{4} \right)=1\Leftrightarrow 3x+\dfrac{\pi }{4}=\dfrac{\pi }{2}+k2\pi \Leftrightarrow x=\dfrac{\pi }{12}+\dfrac{k2\pi }{3}$.
Đáp án C.