Do \(\left( { - 2\pi ; - \pi } \right) = \left( { - 2\pi ;\pi  - 2\pi } \right)\) nên hàm số \(y = \cos x\) nghịch biến trên khoảng \(\left( { - 2\pi ; - \pi } \right)\)

Đáp án A

Hàm số chẵn là: y = cos x

a.

\(y'=\dfrac{2-x}{2x^2\sqrt{x-1}}=0\Rightarrow x=2\)

\(y\left(1\right)=0\) ; \(y\left(2\right)=\dfrac{1}{2}\) ; \(y\left(5\right)=\dfrac{2}{5}\)

\(\Rightarrow y_{min}=y\left(1\right)=0\)

\(y_{max}=y\left(2\right)=\dfrac{1}{2}\)

b.

\(y'=\dfrac{1-3x}{\sqrt{\left(x^2+1\right)^3}}< 0\) ; \(\forall x\in\left[1;3\right]\Rightarrow\) hàm nghịch biến trên [1;3]

\(\Rightarrow y_{max}=y\left(1\right)=\dfrac{4}{\sqrt{2}}=2\sqrt{2}\)

\(y_{min}=y\left(3\right)=\dfrac{6}{\sqrt{10}}=\dfrac{3\sqrt{10}}{5}\)

c.

\(y=1-cos^2x-cosx+1=-cos^2x-cosx+2\)

Đặt \(cosx=t\Rightarrow t\in\left[-1;1\right]\)

\(y=f\left(t\right)=-t^2-t+2\)

\(f'\left(t\right)=-2t-1=0\Rightarrow t=-\dfrac{1}{2}\)

\(f\left(-1\right)=2\) ; \(f\left(1\right)=0\) ; \(f\left(-\dfrac{1}{2}\right)=\dfrac{9}{4}\)

\(\Rightarrow y_{min}=0\) ; \(y_{max}=\dfrac{9}{4}\)

d.

Đặt \(sinx=t\Rightarrow t\in\left[-1;1\right]\)

\(y=f\left(t\right)=t^3-3t^2+2\Rightarrow f'\left(t\right)=3t^2-6t=0\Rightarrow\left[{}\begin{matrix}t=0\\t=2\notin\left[-1;1\right]\end{matrix}\right.\)

\(f\left(-1\right)=-2\) ; \(f\left(1\right)=0\) ; \(f\left(0\right)=2\)

\(\Rightarrow y_{min}=-2\) ; \(y_{max}=2\)

1: \(y=x+\dfrac{4}{\left(x-2\right)^2}\)

\(\Leftrightarrow y'=1+\left(\dfrac{4}{\left(x-2\right)^2}\right)'\)

=>\(y'=1+\dfrac{4'\left(x-2\right)^2-4\left[\left(x-2\right)^2\right]'}{\left(x-2\right)^4}\)

=>\(y'=1+\dfrac{-4\cdot2\cdot\left(x-2\right)'\left(x-2\right)}{\left(x-2\right)^4}\)

=>\(y'=1-\dfrac{8}{\left(x-2\right)^3}\)

Đặt y'=0

=>\(\dfrac{8}{\left(x-2\right)^3}=1\)

=>\(\left(x-2\right)^3=8\)

=>x-2=2

=>x=4

Đặt \(f\left(x\right)=x+\dfrac{4}{\left(x-2\right)^2}\)

\(f\left(4\right)=4+\dfrac{4}{\left(4-2\right)^2}=4+1=5\)

\(f\left(0\right)=0+\dfrac{4}{\left(0-2\right)^2}=0+\dfrac{4}{4}=1\)

\(f\left(5\right)=5+\dfrac{4}{\left(5-2\right)^2}=5+\dfrac{4}{9}=\dfrac{49}{9}\)

Vì f(0)<f(4)<f(5)

nên \(f\left(x\right)_{max\left[0;5\right]\backslash\left\{2\right\}}=f\left(5\right)=\dfrac{49}{9}\) và \(f\left(x\right)_{min\left[0;5\right]\backslash\left\{2\right\}}=1\)

2: \(y=cos^22x-sinx\cdot cosx+4\)

\(=1-sin^22x-\dfrac{1}{2}\cdot sin2x+4\)

\(=-sin^22x-\dfrac{1}{2}\cdot sin2x+5\)

\(=-\left(sin^22x+\dfrac{1}{2}\cdot sin2x-5\right)\)

\(=-\left(sin^22x+2\cdot sin2x\cdot\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{81}{16}\right)\)

\(=-\left(sin2x+\dfrac{1}{4}\right)^2+\dfrac{81}{16}\)

\(-1< =sin2x< =1\)

=>\(-\dfrac{3}{4}< =sin2x+\dfrac{1}{4}< =\dfrac{5}{4}\)

=>\(0< =\left(sin2x+\dfrac{1}{4}\right)^2< =\dfrac{25}{16}\)

=>\(0>=-\left(sin2x+\dfrac{1}{4}\right)^2>=-\dfrac{25}{16}\)

=>\(\dfrac{81}{16}>=-sin\left(2x+\dfrac{1}{4}\right)^2+\dfrac{81}{16}>=-\dfrac{25}{16}+\dfrac{81}{16}=\dfrac{7}{2}\)

=>\(\dfrac{81}{16}>=y>=\dfrac{7}{2}\) 

\(y_{min}=\dfrac{7}{2}\) khi \(sin2x+\dfrac{1}{4}=\dfrac{5}{4}\)

=>\(sin2x=1\)

=>\(2x=\dfrac{\Omega}{2}+k2\Omega\)

=>\(x=\dfrac{\Omega}{4}+k\Omega\)

\(y_{max}=\dfrac{81}{16}\) khi sin 2x=-1

=>\(2x=-\dfrac{\Omega}{2}+k2\Omega\)

=>\(x=-\dfrac{\Omega}{4}+k\Omega\)

\(cos^2t=cos^2\left(2\pi-x-y-z\right)=cos^2\left(x+y+z\right)\)

\(VT=cos^2x+cos^2y-cos^2z-cos^2\left(x+y+z\right)\)

\(=\frac{1}{2}+\frac{1}{2}cos2x+\frac{1}{2}+\frac{1}{2}cos2y-\frac{1}{2}-\frac{1}{2}cos2z-\frac{1}{2}-\frac{1}{2}cos\left(2x+2y+2z\right)\)

\(=\frac{1}{2}\left(cos2x-cos2z\right)+\frac{1}{2}\left(cos2y-cos\left(2x+2y+2z\right)\right)\)

\(=-sin\left(x+z\right)sin\left(x-z\right)-sin\left(x+2y+z\right)sin\left(-x-z\right)\)

\(=sin\left(x+z\right)\left[sin\left(x+2y+z\right)-sin\left(x-z\right)\right]\)

\(=2sin\left(x+z\right).sin\left(y+z\right).cos\left(x+y\right)\)

Không giống :D

Hàm số nghịch biến trên khoảng \(\left( {\pi ;2\pi } \right)\) là:\(y = \cos x\)

Chọn B