Đáp án đúng : C

\(y=2\left(\frac{1}{2}-\frac{1}{2}cos2x\right)+cos^22x=cos^22x-cos2x+1\)

\(=\left(cos2x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

\(\Rightarrow y_{min}=\frac{3}{4}\) khi \(cos2x=\frac{1}{2}\)

\(y=cos^22x-2cos2x+cos2x-2+3\)

\(y=\left(cos2x-2\right)\left(cos2x+1\right)+3\)

Do \(-1\le cos2x\le1\Rightarrow\left\{{}\begin{matrix}cos2x-2< 0\\cos2x+1\ge0\end{matrix}\right.\) \(\Rightarrow\left(cos2x-2\right)\left(cos2x+1\right)\le0\)

\(\Rightarrow y\le3\Rightarrow y_{max}=3\) khi \(cos2x=-1\)

\(y=1-cos2x+2sin2x+6=2sin2x-cos2x+7\)

\(y=\sqrt{5}\left(\dfrac{2}{\sqrt{5}}sin2x-\dfrac{1}{\sqrt{5}}cos2x\right)+7\)

Đặt \(\dfrac{2}{\sqrt{5}}=cosa\) với \(a\in\left(0;\dfrac{\pi}{2}\right)\)

\(y=\sqrt{5}sin\left(2x-a\right)+7\)

\(\Rightarrow-\sqrt{5}+7\le y\le\sqrt{5}+7\)

y=(sin2x-3)^2-6

-1<=sin2x<=1

=>-4<=sin2x-3<=-2

=>4<=(sin2x-3)^2<=16

=>-2<=y<=10

y min khi sin2x-3=-2

=>sin 2x=1

=>2x=pi/2+k2pi

=>x=pi/4+kpi

y max khi sin 2x-3=-4

=>sin 2x=-1

=>2x=-pi/2+k2pi

=>x=-pi/4+kpi

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

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

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

ta có \(y=2sin^4x+\left(1-2sin^2x\right)^2\)=\(2sin^4x+4sin^4x-4sinx^2+1=6sin^4x-4sin^2x+1\)

đặt \(t=sin^2x,0\le t\le1\) 

ta đc \(y=6t^2-4t+1\)

ta tính y'=12t-4

giải pt y'=0 suy ra t=1/3

ta có bảng biến thiên 

từ bảng bt ta suy ra hàm số đạt giá trị nhỏ nhất \(y=\frac{1}{3}\) khi \(t=\frac{1}{3}\Rightarrow sin^2x=\frac{1}{3}\)

hàm số đạt giá trị lớn nhất y=3 khi \(t=1\Rightarrow sin^2x=1\)