函数f(x)=2acos²x+bsinxcosx,且f(0)=2,f(π/3)=1/2+√3/2,(1)求f(x)的最大值,(2)求f(x)的单调增区间
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![函数f(x)=2acos²x+bsinxcosx,且f(0)=2,f(π/3)=1/2+√3/2,(1)求f(x)的最大值,(2)求f(x)的单调增区间](/uploads/image/z/14588106-42-6.jpg?t=%E5%87%BD%E6%95%B0f%28x%29%3D2acos%26sup2%3Bx%2Bbsinxcosx%2C%E4%B8%94f%280%29%3D2%2Cf%28%CF%80%2F3%29%3D1%2F2%2B%E2%88%9A3%2F2%2C%EF%BC%881%EF%BC%89%E6%B1%82f%28x%29%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%2C%EF%BC%882%EF%BC%89%E6%B1%82f%28x%29%E7%9A%84%E5%8D%95%E8%B0%83%E5%A2%9E%E5%8C%BA%E9%97%B4)
函数f(x)=2acos²x+bsinxcosx,且f(0)=2,f(π/3)=1/2+√3/2,(1)求f(x)的最大值,(2)求f(x)的单调增区间
函数f(x)=2acos²x+bsinxcosx,且f(0)=2,f(π/3)=1/2+√3/2,(1)求f(x)的最大值,(2)求f(x)的单调增区间
函数f(x)=2acos²x+bsinxcosx,且f(0)=2,f(π/3)=1/2+√3/2,(1)求f(x)的最大值,(2)求f(x)的单调增区间
(1)因f(0)=2,所以2a=2,a=1
又f(π/3)=2cos²π/3+bsinπ/3 cosπ/3=1/2+√3/2,所以b=2,
f(x)=2cos²x+2sinx cosx=1+cos2x+sin2x=√2sin(2x+π/4)+1
当sin(2x+π/4)=1时取最大值,最大值为1+√2
(2)f(x)=√2sin(2x+π/4)+1,单调增区间为:
2kπ-π/2≤2x+π/4≤2kπ+π/2
kπ-3π/8≤x≤kπ+π/8(k∈z)
f(x)=2acos²x+bsinxcosx=cosx(2acosx+bsinx)=(4a^2+b^)^(1/2)sin(x+c),tanc=2a/b;
f(0)=2,f(π/3)=1/2+√3/2代入得a=1;b=1/2,最大值=(4a^2+b^)^(1/2)=5^(1/2)/2
(1)f(x)=0,所以2a=2,a=1
f(π/3)=1/2+根号3/2,所以b=2,所以原式=1+sin2x+cos2x=根号2sin(2x+π/4)+1
所以f(x)最大值为3
(2)2kπ-π/2<=2x+π/4<=2kπ+π/2
kπ-3π/8<=x<=kπ+π/8(k∈z)