xiaozengqz
幼苗
共回答了19个问题采纳率:89.5% 举报
这题其实与向量关系都不大了,是函数和不等式的题:
f(x)=a·b=(m,cos2x+2sinx-1)·(3-cos2x+4sinx,-1)
=3m-mcos2x+4msinx-cos2x-2sinx+1
=3m+1-(m+1)(1-2sinx^2)+(4m-2)sinx
=2(m+1)sinx^2+(4m-2)sinx+2m
f(x)>0对任意x∈[0,π/2]恒成
即:(m+1)sinx^2+(2m-1)sinx+m>0恒成立
二次函数的对称轴:(1-2m)/(2(m+1))
1
当0≤(1-2m)/(2(m+1))≤1时,须Δ=(2m-1)^2-4m(m+1)
=1-8m1/8
(1-2m)/(2(m+1))≤1,即:(1-2m-2m-2)/(m+1)≤0
即:(4m+1)/(m+1)≥0
即:m≥-1/4,或m>-1
即:m≥-1/4
(1-2m)/(2(m+1))≥0,即:(2m-1)/(m+1)≤0
即:-10
即:(m+1)+(2m-1)+m>0
即:m>0
故:m>1/2
3
(1-2m)/(2(m+1))>1,即:(1-2m-2m-2)/(m+1)>0
即:(4m+1)/(m+1)1/8
1年前
6