(1) ∵f(x)=√3sin²x+sinxcosx
=(√3/2)*(2sin²x-1+1)+(1/2)*2sinxcosx
=(√3/2)*(-cos2x+1)+(1/2)*(sin2x)
=(1/2)*sin2x/-(√3/2)*cos2x+√3/2
=sin(2x-π/3)+√3/2
∴ f(24π/6)=f(4π)=sin(2*4π-π/3)+√3/2
=sin(8π-π/3)+√3/2
=sinπ/3+√3/2=√3
(2) g(x)=f(x)+√3/2
=sin(2x-π/3)+√3/2+√3/2
=sin(2x-π/3)+√3
当 2x-π/3=2kπ-π/2,即x=kπ-π/12(k∈z)时,g(x)取得最小值
此时,最小值为-1+√3
(3) f(x)单调递增区间为
2kπ≤2x-π/3≤2kπ+π/2
或 2kπ+3π/2≤2x-π/3≤2kπ+2π (k∈z)
解,得
x∈[kπ+π/6,kπ+5π/12]或x∈[kπ+11π/12,kπ+7π/6] (k∈z)