a: -1<=sinx<=1

=>5>=-5sinx>=-5

=>11>=-5sinx+6>=1

=>1<=y<=11

\(y_{min}=1\) khi sin x=1

=>x=pi/2+k2pi

\(y_{max}=11\) khi sin x=-1

=>x=-pi/2+k2pi

b: \(-1< =cosx< =1\)

=>\(1>=-cosx>=-1\)

=>\(-3>=-cosx-4>=-5\)

=>\(-3>=y>=-5\)

\(y_{min}=-5\) khi cosx=1

=>x=k2pi

\(y_{max}=-3\) khi cosx=-1

=>x=pi+k2pi

c: \(-1< =cosx< =1\)

=>\(-\sqrt{3}< \sqrt{3}\cdot cosx< =\sqrt{3}\)

=>\(-\sqrt{3}+8< =y< =\sqrt{3}+8\)

\(y_{min}=-\sqrt{3}+8\) khi cosx=-1

=>x=pi+k2pi

\(y_{max}=\sqrt{3}+8\) khi cosx=1

=>x=k2pi

d: \(-1< =cos3x< =1\)

=>\(1>=-cos3x>=-1\)

=>\(16>=y>=14\)

y min=14 khi cos3x=1

=>3x=k2pi

=>x=k2pi/3

y max=16 khi cos3x=-1

=>3x=pi+k2pi

=>x=pi/3+k2pi/3

e: -1<=sin6x<=1

=>-1+2024<=sin6x+2024<=1+2024

=>2023<=y<=2025

y min=2023 khi sin6x=-1

=>6x=-pi/2+k2pi

=>x=-pi/12+kpi/3

y max=2025 khi sin6x=1

=>6x=pi/2+k2pi

=>x=pi/12+kpi/3

a: \(0< =cos^23x< =1\)

=>\(9< =cos^23x+9< =10\)

=>9<=y<=10

\(y_{min}=9\) khi \(cos^23x=0\)

=>\(cos3x=0\)

=>3x=pi/2+kpi

=>x=pi/6+kpi/3

\(y_{max}=10\) khi \(cos^23x=0\)

=>\(sin^23x=0\)

=>3x=kpi

=>x=kpi/3

b: \(0< =sin^2x< =1\)

=>\(-3< =y< =-2\)

\(y_{min}=-3\) khi \(sin^2x=0\)

=>x=kpi

\(y_{max}=-2\) khi \(sin^2x=1\)

=>\(cos^2x=0\)

=>x=pi/2+kpi

c: \(0< =sin^25x< =1\)

=>12<=y<=13

y min=12 khi sin²5x=0

=>sin 5x=0

=>5x=kpi

=>x=kpi/5

y max=13 khi sin²5x=0

=>cos²5x=0

=>cos5x=0

=>5x=pi/2+kpi

=>x=pi/10+kpi/5

A là đáp án đúng!

Lời giải:

a. $y=\sqrt{x^2+x-2}\geq 0$ (tính chất cbh số học)

Vậy $y_{\min}=0$. Giá trị này đạt tại $x^2+x-2=0\Leftrightarrow x=1$ hoặc $x=-2$
b.

$y^2=6+2\sqrt{(2+x)(4-x)}\geq 6$ do $2\sqrt{(2+x)(4-x)}\geq 0$ theo tính chất căn bậc hai số học

$\Rightarrow y\geq \sqrt{6}$ (do $y$ không âm)

Vậy $y_{\min}=\sqrt{6}$ khi $x=-2$ hoặc $x=4$

$y^2=(\sqrt{2+x}+\sqrt{4-x})^2\leq (2+x+4-x)(1+1)=12$ theo BĐT Bunhiacopxky

$\Rightarrow y\leq \sqrt{12}=2\sqrt{3}$

Vậy $y_{\max}=2\sqrt{3}$ khi $2+x=4-x\Leftrightarrow x=1$

c. ĐKXĐ: $-2\leq x\leq 2$

$y^2=(x+\sqrt{4-x^2})^2\leq (x^2+4-x^2)(1+1)$ theo BĐT Bunhiacopxky

$\Leftrightarrow y^2\leq 8$

$\Leftrightarrow y\leq 2\sqrt{2}$

Vậy $y_{\max}=2\sqrt{2}$ khi $x=\sqrt{2}$

Mặt khác:

$x\geq -2$

$\sqrt{4-x^2}\geq 0$

$\Rightarrow y\geq -2$
Vậy $y_{\min}=-2$ khi $x=-2$

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

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

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

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

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

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

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

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

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

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

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

=> \(Min_y=2.\left(-1\right)+2=0\) 

Mặt khác, theo Bunhiacopxki:

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

=>\(Max_y=4\)