Đáp án D

Chọn A.

Ta có:  . Đặt , do 0 ≤ cos²x ≤ 1 nên ta có 

Xét hàm số  có 

Lại có 

Vậy 

\(y=\sqrt{3}sin2x-cos2x=2\left(\dfrac{\sqrt{3}}{2}sin2x-\dfrac{1}{2}cos2x\right)=2sin\left(2x-\dfrac{\pi}{6}\right)\)

Do \(-1\le sin\left(2x-\dfrac{\pi}{6}\right)\le1\Rightarrow-2\le y\le2\)

\(y_{max}=2\) khi \(sin\left(2x-\dfrac{\pi}{6}\right)=1\)

\(y_{min}=-2\) khi \(sin\left(2x-\dfrac{\pi}{6}\right)=-1\)

Đáp án B

Ta có:

y = s   i n 2 x − sin 2 x + 11 = − sin 2 x − c os 2 x + 12 = − 2 sin 2 x − π 4 + 12. − 1 ≤ sin 2 x − π 4 ≤ 1 ⇒ − 2 ≤ − 2 sin 2 x − π 4 ≤ 12 ⇒ − 2 sin 2 x − π 4 + 12 ≤ 12 + 2 ⇒ M = 12 + 2

1: \(-1<=cosx\le1\)

=>\(-3\le-3\cdot cosx\le3\)

=>\(-3+5\le-3\cdot cosx+5\le3+5\)

=>2<=y<=8

y min=2 khi cosx=1

=>\(x=k2\pi\)

y min=8 khi cosx=-1

=>\(x=\pi+k2\pi\)

3: \(y=cos^2x+2\cdot cos2x\)

\(=\frac{1+cos2x}{2}+2\cdot cos2x=2,5\cdot cos2x+0,5\)

Ta có: \(-1\le cos2x\le1\)

=>\(-2,5\le2,5cos2x\le2,5\)

=>\(-2,5+0,5\le2,5cos2x+0,5\le2,5+0,5\)

=>-2<=y<=3

y min=-2 khi cos2x=-1

=>\(2x=\pi+k2\pi\)

=>\(x=\frac{\pi}{2}+k\pi\)

y max=3 khi cos2x=1

=>\(2x=k2\pi\)

=>\(x=k\pi\)

6: \(y=\sqrt3\cdot\sin x-cosx-2\)

\(=2\left(\frac{\sqrt3}{2}\cdot\sin x-\frac12\cdot cosx\right)-2=2\cdot\sin\left(x-\frac{\pi}{6}\right)-2\)

Ta có: \(-1\le\sin\left(x-\frac{\pi}{6}\right)\le1\)

=>\(-2\le2\sin\left(x-\frac{\pi}{6}\right)\le2\)

=>\(-2-2\le2\sin\left(x-\frac{\pi}{6}\right)-2\le2-2\)

=>-4<=y<=0

y min=-4 khi \(\sin\left(x-\frac{\pi}{6}\right)=-1\)

=>\(x-\frac{\pi}{6}=-\frac{\pi}{2}+k2\pi\)

=>\(x=-\frac{\pi}{2}+\frac{\pi}{6}+k2\pi=-\frac26\pi+k2\pi=-\frac13\pi+k2\pi\)

y max=0 khi \(\sin\left(x-\frac{\pi}{6}\right)=1\)

=>\(x-\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)

=>\(x=\frac23\pi+k2\pi\)

Đáp án D

Đặt

Chọn D