Question

已知O为坐标原点,向量OA=(2sin^2x,1),向量OB=(1,-2√3sinxcosx+1),f(x)=向量OA×向量OB+m

1、求f（x）的单调递增区间
2、若当x∈[π\2,π]时，f（x）的取值范围是[2,5]，求m的值

Answer 1

f(x)=向量OA×向量OB+m
=2sin^2x-2√3sinxcosx+1+m
=1-cos2x-√3sin2x+1+m
=-2sin(2x+π/6)+2+m
1. 单增区间为2x+π/6∈[2kπ+π/2, 2kπ+3π/2]
x∈[kπ+π/6, kπ+2π/3]
2. 当x∈[π/2,π] 2x+π/6∈[7π/6, 13π/6]
f(x)max=2+m+2=m+4 f(x)min=2*(-1/2)+2+m=m+1
所以m+4=5 m+1=2
解得m=1