Question

高中数学题 不等式 求大神解答

设0＜X＜π÷2，函数F（X）=（1+COS2X+8SIN²X）÷SIN2X的最小值为_____解答

Answer 1

解:
cos2x=1-2sin^2x
2sin^2x=1-cos2x
8sin^2x=4-4cos2x

y=f(x)=(1+cos2x+8sin^2x)/sin2x
=(1+cos2x+4-4cos2x)/sin2x
=(5-3cos2x)/sin2x
=(5-3cos2x)/√[1-(cos2x)^2]

已知0<x<π/2
(5-3cos2x)>0,sin2x>0
∴y>0
y*√[1-(cos2x)^2]=5-3cos2x
y^2*[1-(cos2x)^2]=(5-3cos2x)^2
(9+y^2)*(cos2x)^2-30(cos2x)+25-y^2=0
上方程未知数为(cos2x)的判别式△≥0,即
(-30)^2-4*(9+y^2)*(25-y^2)≥0
y^4-16y^2≥0
y^2*(y+4)*(y-4))≥0
y≥4(另一解y≤-4舍去)
y的最小值=4
y=4
(9+y^2)*(cos2x)^2-30(cos2x)+25-y^2=0
(5cos2x-3)^2=0
cos2x=3/5,sin2x=4/5
y=f(x)=(1+cos2x+8sin^2x)/sin2x
=(1+cos2x+4-4cos2x)/sin2x
=(5-3cos2x)/sin2x
=(5-3*3/5)/(4/5)
=4
当0<x<π/2时，f(x)=(1+cos2x+8sin^2x)/sin2x的最小值=4