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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: -1<=sin x<=1

=>-1+3<=sin x+3<=1+3

=>2<=sinx+3<=4

=>\(\dfrac{1}{2}>=\dfrac{1}{sinx+3}>=\dfrac{1}{4}\)

=>\(2>=\dfrac{4}{sinx+3}>=1\)

=>\(-2< =-\dfrac{4}{sinx+3}< =-1\)

=>-2+3<=y<=-1+3

=>1<=y<=2

y=1 khi \(\dfrac{-4}{sinx+3}+3=1\)

=>\(\dfrac{-4}{sinx+3}=-2\)

=>sinx+3=2

=>sin x=-1

=>x=-pi/2+k2pi

y=3 khi sin x=1

=>x=pi/2+k2pi

b: -1<=cosx<=1

=>4>=-4cosx>=-4

=>9>=-4cosx+5>=1

=>2/9<=2/5-4cosx<=2

=>2/9<=y<=2

\(y_{min}=\dfrac{2}{9}\) khi \(\dfrac{2}{5-4cosx}=\dfrac{2}{9}\)

=>\(5-4\cdot cosx=9\)

=>4*cosx=4

=>cosx=1

=>x=k2pi

y max khi cosx=-1

=>x=pi+k2pi

c: \(0< =cos^2x< =1\)

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

=>\(-1< =y< =2\)

y min=-1 khi cos^2x=0

=>x=pi/2+kpi

y max=2 khi cos^2x=1

=>sin^2x=0

=>x=kpi

 

b: \(-1< =cos4x< =1\)

=>\(-3< =3\cdot cos4x< =3\)

=>\(9< =3\cdot cos4x+12< =15\)

=>\(3< =y< =\sqrt{15}\)

y min=3 khi cos4x=-1

=>4x=pi+k2pi

=>x=pi/4+kpi/2

y max=căn 15 khi cos4x=1

=>4x=k2pi

=>x=kpi/2

c: -1<=sin 9x<=1

=>-1+20<=sin 9x+20<=21

=>19<=y<=21

y min=19 khi sin 9x=-1

=>9x=-pi/2+k2pi

=>x=-pi/18+k2pi/9

y max=21 khi sin 9x=1

=>9x=pi/2+k2pi

=>x=pi/18+k2pi/9

NV
11 tháng 9 2021

1. Không dịch được đề

2.

\(-1\le cos2x\le1\Rightarrow1\le y\le3\)

3.

a. \(-2\le2sinx\le2\Rightarrow-1\le y\le3\)

\(y_{min}=-1\) khi \(sinx=-1\Rightarrow x=-\dfrac{\pi}{2}+k2\pi\)

\(y_{max}=3\) khi \(sinx=1\Rightarrow x=\dfrac{\pi}{2}+k2\pi\)

b.

\(0\le cos^2x\le1\Rightarrow-1\le y\le2\)

\(y_{min}=-1\) khi \(cos^2x=1\Rightarrow x=k\pi\)

\(y_{max}=2\) khi \(cosx=0\Rightarrow x=\dfrac{\pi}{2}+k\pi\)

4.

\(y=\left(tanx-1\right)^2+2\ge2\)

\(y_{min}=2\) khi \(tanx=1\Rightarrow x=\dfrac{\pi}{4}+k\pi\)

\(y=\sqrt{\left(sinx+cosx\right)^2+2\cdot sinx\cdot cosx+2}\)

\(=\sqrt{1+2sinx\cdot cosx+2\cdot sinx\cdot cosx+2}\)

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

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

=>\(-2< =2\cdot sin2x< =2\)

=>\(-2+3< =2\cdot sin2x+3< =5\)

=>\(1< =2\cdot sin2x+3< =5\)

=>\(1< =\sqrt{2\cdot sin2x+3}< =\sqrt{5}\)

=>\(1< =y< =\sqrt{5}\)

\(y_{min}=1\) khi \(sin2x=-1\)

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

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

\(y_{max}=\sqrt{5}\) khi sin 2x=1

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

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

9 tháng 4 2019

Đáp án B

7 tháng 10 2017

1   ≥   2   -   cos x   ≥   2

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