Question

设函数f(x)=a*b,其中向量a=(2cosx,1),b=(cosx,√3sin2x),x∈R,求f(x)取得最大值的x的值

Answer 1

f(x)=cos2x+√3sin2x+1
对f(x)求导
f`(x)=2√3cos2x-2sin2x=4cos(2x+π/6)
由余弦函数性质可知最大值为3
f`(x)>0时f(x)单调递增即x∈（kπ-π/3，kπ+π/6).

Answer 2

答案(1)f(x)=ab=2cosx*cosx+√3sin2x=1+cos2x+√3sin2x

=1+2((√3/2)sin2x+(1/2)cos2x)=1+2sin(2x+π/6)
∴取最大值
2x+π/6=2kπ+π/2,
∴x=kπ+π/6(k是任意整数)
2)单调递增区间
2x+π/6∈
[2kπ-π/2,
2kπ+π/2]
∴x∈
[kπ-π/3,
kπ+π/6],
即单调增区间为[kπ-π/3,
kπ+π/6](k是任意整数)

Answer 3

1)f(x)=ab=2cosx*cosx+√3sin2x=1+cos2x+√3sin2x

=1+2((√3/2)sin2x+(1/2)cos2x)=1+2sin(2x+π/6)
∴取最大值
2x+π/6=2kπ+π/2,
∴x=kπ+π/6(k是任意整数)
2)单调递增区间
2x+π/6∈
[2kπ-π/2,
2kπ+π/2]
∴x∈
[kπ-π/3,
kπ+π/6],
即单调增区间为[kπ-π/3,
kπ+π/6](k是任意整数)