已知向量a=(2sinx\2,根号下3+1),向量b=(cosx\2-根号下3sinx\2,1),f(x)=向量a•向量b+m1、求f(x)在[0,2派]上的单调区间2、当x属于[0,派\2]时,f(x)的最小值为2,求f(x)大于等于2成立的x的取值集合3、
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![已知向量a=(2sinx\2,根号下3+1),向量b=(cosx\2-根号下3sinx\2,1),f(x)=向量a•向量b+m1、求f(x)在[0,2派]上的单调区间2、当x属于[0,派\2]时,f(x)的最小值为2,求f(x)大于等于2成立的x的取值集合3、](/uploads/image/z/3807580-4-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%282sinx%5C2%2C%E6%A0%B9%E5%8F%B7%E4%B8%8B3%2B1%29%2C%E5%90%91%E9%87%8Fb%3D%28cosx%5C2-%E6%A0%B9%E5%8F%B7%E4%B8%8B3sinx%5C2%2C1%29%2Cf%28x%EF%BC%89%3D%E5%90%91%E9%87%8Fa%26%238226%3B%E5%90%91%E9%87%8Fb%2Bm1%E3%80%81%E6%B1%82f%28x%29%E5%9C%A8%5B0%2C2%E6%B4%BE%5D%E4%B8%8A%E7%9A%84%E5%8D%95%E8%B0%83%E5%8C%BA%E9%97%B42%E3%80%81%E5%BD%93x%E5%B1%9E%E4%BA%8E%5B0%2C%E6%B4%BE%5C2%5D%E6%97%B6%2Cf%EF%BC%88x%EF%BC%89%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E4%B8%BA2%2C%E6%B1%82f%EF%BC%88x%EF%BC%89%E5%A4%A7%E4%BA%8E%E7%AD%89%E4%BA%8E2%E6%88%90%E7%AB%8B%E7%9A%84x%E7%9A%84%E5%8F%96%E5%80%BC%E9%9B%86%E5%90%883%E3%80%81)
已知向量a=(2sinx\2,根号下3+1),向量b=(cosx\2-根号下3sinx\2,1),f(x)=向量a•向量b+m1、求f(x)在[0,2派]上的单调区间2、当x属于[0,派\2]时,f(x)的最小值为2,求f(x)大于等于2成立的x的取值集合3、
已知向量a=(2sinx\2,根号下3+1),向量b=(cosx\2-根号下3sinx\2,1),f(x)=向量a•向量b+m
1、求f(x)在[0,2派]上的单调区间
2、当x属于[0,派\2]时,f(x)的最小值为2,求f(x)大于等于2成立的x的取值集合
3、若存在实数a,b,c,使得a[f(x)-m]+b[f(x-c)-m]=1,对任意x属于R恒成立,求bcosc\a的值
已知向量a=(2sinx\2,根号下3+1),向量b=(cosx\2-根号下3sinx\2,1),f(x)=向量a•向量b+m1、求f(x)在[0,2派]上的单调区间2、当x属于[0,派\2]时,f(x)的最小值为2,求f(x)大于等于2成立的x的取值集合3、
向量a=(2sinx\2,根号下3+1),向量b=(cosx\2-根号下3sinx\2,1)
f(x)=a●b+m
=2sinx/2(cosx/2-√3sinx/2)+√3+1+m
=2sinx/2cox/2-2√3sin²x/2+√3+1+m
=sinx-√3(1-cosx)+√3+1+m
=2(1/2sinx+√3/2cosx)+1+m
=2sin(x+π/3)+1+m
∵x∈[0,2π]
∴x+π/3∈[π/3,7π/3]
∴x+π/3∈[π/3,π/2]或x+π/3∈[3π/2,7π/3],
即x∈[0,π/6],或x∈[7π/6,2π]
函数f(x)递增
当x+π/3∈[π/2,3π/2],即x∈[π/6,7π/6]
函数递减
∴f(x)在[0,2派]上的单调递增区间为
[0,π/6],[7π/6,2π]
单调递增区间为[π/6,7π/6]
2
x属于[0,派\2]时,x+π/3∈[π/3,5π/6]
∴x+π/3=5π/6时,f(x)min=2+m=2
∴m=0
f(x)=2sin(x+π/3)+1
f(x)≥2即 sin(x+π/3)≥1/2
2kπ+π/6≤x+π/3≤ 2kπ+5π/6
2kπ-π/6≤x≤ 2kπ+π/2
∴f(x)≥2成立的x的取值集合
为{x| 2kπ-π/6≤x≤ 2kπ+π/2,k∈Z}
3
a[f(x)-m]+b[f(x-c)-m]=1
即a[2sin(x+π/3)+1]+b[2sin(x-c+π/3)+1]=1
2[asin(x+π/3)+bsin(x+π/3-c)]+a+b=1对任意x属于R恒成立
x=-π/3时,2bsin(-c)+a+b=1
x=2π/3时,2bsinc+a+b=1
x=π/6时,2a+2bcosc+a+b=1
∴a+b=1,sinc=0,cosc=±1
若cosc=1,那么a+b=0矛盾
∴cosc=-1,2a-2b=0,a=b
∴a=b=1/2,c=2kπ+π,k∈Z
此时2[asin(x+π/3)+bsin(x+π/3-c)]+a+b
=sin(x+π/3)+sin(x+π/3-2kπ-π)+a+b
=sin(x+π/3)-sin(x+π/3)+a+b
=a+b=1恒成立
∴ bcos(c/a)=1/2cos(4kπ+2π)=1/2