Question

Answer 1

y = \(\sqrt{3}\) sin 2x - cos 2x + 1 . 

Xét y'= \(\sqrt{3}\) sin 2x - cos 2x <=> \(\dfrac{y'}{2}\) = \(\dfrac{\sqrt{3}}{2}\) sin 2x - \(\dfrac{1}{2}\) cos 2x = sin2x. cos \(\dfrac{\pi}{6}\) - cos2x.sin \(\dfrac{\pi}{6}\) = sin ( 2x - \(\dfrac{\pi}{6}\) ) => y'= 2 sin ( 2x - \(\dfrac{\pi}{6}\) ) 

 Suy ra y= 2 sin ( 2x - \(\dfrac{\pi}{6}\) ) + 1 

Ta có  sin ( 2x - \(\dfrac{\pi}{6}\) ) ∈ [ -1,1] => 2 sin ( 2x - \(\dfrac{\pi}{6}\) ) ∈ [ -2,2] => y∈ [ -1,1]