Question

Tìm giá trị lớn nhất, giá trị nhỏ nhất:

a) \(y=3-\dfrac{4}{3+2sinx}\)

b) \(y=\dfrac{2}{5-4cosx}\)

c) \(y=2cos^2x-1\)

d) \(y=3-4sin^22x\)

e) \(y=\sqrt{3-2sinx}\)

f) \(y=\dfrac{5}{\sqrt{5-4sinx}}\)

g) \(y=\dfrac{4}{4-\sqrt{5+4cosx}}\)

h) \(y=sinx-cosx-2\)

i) \(y=\sqrt{3}cosx-sinx+3\)

j) \(y=4cos^2x-4cosx+5\)

Answer 1

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