Question

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

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
求过程，O(∩_∩)O谢谢！

Answer 1

（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)

Answer 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)

Answer 3

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