Question

已知函数f(x)=5sin xcos x-5√3cos^2x+2分之5√3 (x∈R)

Answer 1

f(x)=5sin xcos x-5√3cos²x+5√3 /2

=(5/2)sin2x-(5√3/2)(1+cos2x)+5√3 /2

=√[(5/2)²+(5√3/2)]sin(2x+ψ)

=5sin(2x+ψ)(tanψ=√3,ψ=π/3)

1,∴T(+)min=2π/2=π

2,因为正弦函数在区间（-π/2+2kπ,π/2+2kπ）是增函数

所以由-π/2+2kπ≤2x+π/3≤π/2+2kπ得：

增区间为：kπ-5π/12≤x≤kπ+π/12

又正弦函数在区间（π/2+2kπ,3π/2+2kπ）是减函数

所以由π/2+2kπ≤2x+π/3≤3π/2+2kπ得：

增区间为：kπ+π/12≤x≤kπ+7π/12

3,由2x+π/3=0得：x=-π/6

故对称轴为：x=-π/6

对称中心为：(-π/6,0)