Question

[vvvvvvvv]

tìm giá trị lớn nhất và giá trị nhỏ nhất của hàm số sau:

\(y=2cos^2x-2\sqrt{3}sinxcosx+1\)

Answer 1

[Hồng Phúc]

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

\(=2cos^2x-1-2\sqrt{3}sinx.cosx+2\)

\(=cos2x-\sqrt{3}sin2x+2\)

\(=2\left(\dfrac{1}{2}cos2x-\dfrac{\sqrt{3}}{2}sin2x\right)+2\)

\(=2cos\left(2x+\dfrac{\pi}{3}\right)+2\)

Ta có: \(cos\left(2x+\dfrac{\pi}{3}\right)\in\left[-1;1\right]\)

\(\Rightarrow min=0\Leftrightarrow cos\left(2x+\dfrac{\pi}{3}\right)=-1\Leftrightarrow2x+\dfrac{\pi}{3}=\pi+k2\pi\Leftrightarrow x=\dfrac{\pi}{3}+k\pi\)

\(\Rightarrow max=4\Leftrightarrow cos\left(2x+\dfrac{\pi}{3}\right)=1\Leftrightarrow2x+\dfrac{\pi}{3}=k2\pi\Leftrightarrow x=-\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)

Answer 2

[Nguyễn Thị Ngọc Thơ]

\(y=2cos^2x-\sqrt{3}sin2x+1=cos2x-\sqrt{3}sin2x+2\)

\(y=2.cos\left(2x+\dfrac{\pi}{3}\right)+2\)

\(\forall x\in R->-1\le cos\left(2x+\dfrac{\pi}{3}\right)\)

=> \(Min_y=2.\left(-1\right)+2=0\) 

Mặt khác, theo Bunhiacopxki:

\(\left(cos2x+\sqrt{3}sin2x\right)^2\le\left(1^2+\sqrt{3}^2\right)\left(cos^22x+sin^22x\right)=4\)

=>\(Max_y=4\)