Question

已知函数f(x)=2sin的平方（π/4+x)减去根号下3cos2x-1,x∈R,若在x∈[π/4,π/2]上恒成立，不等式f(x)＞m的

已知函数f(x)=2sin^2(π/4+x)-根号3cos2x-1,x∈R,若在x∈[π/4,π/2]上恒成立,不等式f(x)>m的平方减去3m+3恒成立,求实数m的取值范围？请帮帮忙明天就要用了谢谢！

Answer 1

f(x)=2sin²(π/4+x)- √3cos2x-1
=1-cos(π/2+2x)-√3cos2x-1
=sin2x-√3cos2x
=2sin(2x-π/3),
π/4≤x≤π/2, π/6 ≤2x-π/3≤2π/3, 1/2 ≤sin(2x-π/3)≤1
1≤ f(x)≤2,
f(x)＞m²-3m+3恒成立，就是1>m²-3m+3恒成立，
1<m<2

Answer 2

f(x)=2sin²（π/4+x)-√3cos2x-1,
=[1-cos(π/2+2x)]-√3cos2x-1
=sin2x-√3cos2x
=2(1/2*sin2x-√3/2*cos2x)
=2sin(2x-π/3)
∵x∈[π/4,π/2]
∴2x-π/3∈[π/6,2π/3]
∴sin(2x-π/3)∈[1/2,1]
∴f(x)∈[1,2]
∵不等式f(x)>m²-3m+3恒成立
∴f(x)min>m²-3m+3
∴1>m²-3m+3
∴m²-3m+2<0
∴1<m<2

Answer 3

你的题目没有说清楚。。