Question

Tìm max, min (nếu có) của:

a, y = sin²x + 3sin2x + 3cos²x

b, y = \(\dfrac{3}{1+\sqrt{2+sin^2x}}\)

 

Answer 1

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

=1/2-1/2cos2x+3sin2x+3/2+3/2cos2x

=3sin2x+cos2x+2

=>3sin2x+cos2x+2-y=0

Để PT có nghiệm thì 3^2+1^2>=(y-2)^2

=>(y-2)^2<=10

=>-căn 10+2<=y<=căn 10+2

y min khi 3sin2x+cos2x+2+căn 10-2=0

=>3sin2x+cos2x=-căn 10

=>3/căn 10*sin2x+1/căn 10*cos2x=-1

=>sin(2x+a)=-1

=>2x+a=-pi/2+k2pi

=>x=-pi/4+kpi-a/2

y max khi 3sin2x+cos2x+2-căn 10-2=0

=>sin(2x+a)=1

=>2x+a=pi/2+k2pi

=>x=pi/4-a/2+kpi

b: 0<=sin^2x<=1

=>2<=sin^2x+2<=3

=>căn 2<căn (sin^2x+2)<=căn 3

=>căn 2+1<=căn (sin^2x+2)+1<=căn 3+1

=>-3+3căn 2>=y>=1/2(-3+3*căn 3)

y max khi sin2x=0

=>2x=kpi

=>x=kpi/2

y min khi cos2x=0

=>2x=pi/2+kpi

=>x=pi/4+kpi/2