Question

Answer 1

Ta có : \(y=2cos^2x-2\sqrt{3}sinx.cosx+1\)  \(=cos2x-\sqrt{3}sin2x+2\) 

\(=2\left[cos2x.cos\dfrac{\pi}{3}-sin\dfrac{\pi}{3}.sin2x\right]+2\)  \(=2cos\left(2x+\dfrac{\pi}{3}\right)+2\)

\(\Rightarrow y'=-2.2sin\left(2x+\dfrac{\pi}{3}\right)=-4sin\left(2x+\dfrac{\pi}{3}\right)\)  

( Đạo hàm : \(\left(cosu\right)'=-sinu.u'\)  ; \(c'=0\) ( c là hằng số ) ) 

y ' = 0 \(\Leftrightarrow2x+\dfrac{\pi}{3}=k\pi\left(k\in Z\right)\)

Có : \(x\in\left[0;\dfrac{7\pi}{12}\right]\) . Suy ra : k = 1 \(\Rightarrow x=\dfrac{\pi}{3}\)

Ta có : \(y\left(\dfrac{\pi}{3}\right)=0;y\left(0\right)=3;y\left(\dfrac{7}{12}\pi\right)=2\)

=> Chọn C