已知向量a=(m,cos2x+2sinx-1),向量b=(3-cos2x+4sinx,-1),f(x)=a*b,若对任意的x∈[0,π/2],不等式f(x)>0恒成立,求m的取值范围
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![已知向量a=(m,cos2x+2sinx-1),向量b=(3-cos2x+4sinx,-1),f(x)=a*b,若对任意的x∈[0,π/2],不等式f(x)>0恒成立,求m的取值范围](/uploads/image/z/10458446-14-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%28m%2Ccos2x%2B2sinx-1%29%2C%E5%90%91%E9%87%8Fb%3D%283-cos2x%2B4sinx%2C-1%29%2Cf%28x%29%3Da%2Ab%2C%E8%8B%A5%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84x%E2%88%88%EF%BC%BB0%2C%CF%80%2F2%EF%BC%BD%2C%E4%B8%8D%E7%AD%89%E5%BC%8Ff%EF%BC%88x%EF%BC%89%3E0%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
已知向量a=(m,cos2x+2sinx-1),向量b=(3-cos2x+4sinx,-1),f(x)=a*b,若对任意的x∈[0,π/2],不等式f(x)>0恒成立,求m的取值范围
已知向量a=(m,cos2x+2sinx-1),向量b=(3-cos2x+4sinx,-1),f(x)=a*b,若对任意的x∈[0,π/2],不等式f(x)>0恒成立,求m的取值范围
已知向量a=(m,cos2x+2sinx-1),向量b=(3-cos2x+4sinx,-1),f(x)=a*b,若对任意的x∈[0,π/2],不等式f(x)>0恒成立,求m的取值范围
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