已知函数f(x)=根号3((sinx)^2-(cosx)^2)+2sinxcos+1

(1)f(x)=√3(sin²x-cos²x)+2sinxcosx+1=sin(2x)-√3cos(2x) +1=2[(1/2)sin(2x)-(√3/2)cos(2x)]+1=2sin(2x-π/3) +12kπ+π/2≤2x-π/3≤2kπ+3π/2 (k∈Z)时,函数单调递减,此时kπ+5π/12≤x≤kπ+11π/12 (k∈Z)函数的单调递减区间为[kπ+5π/12,kπ+11π/12] (k∈Z)(2)x∈[0,π/2],则-π/3≤2x-π/3≤2π/3-√3/2≤sin(2x-π/3)≤1-√3≤2sin(2x-π/3)≤21 -√3≤2sin(2x-π/3)+1≤31 -√3≤f(x)≤3函数的值域为[1- √3,3](3)g(x)=af(x/2 +π/6) +cos(2x)=a[2sin[2(x/2 +π/6) -π/3] +1]+1-2sin²x=a(2sinx +1)+1-2sin²x=-2sin²x+2asinx +a+1=-2(sinx -a/2)² +a²/2 +a +1a/2>1时,即a>2时,当sinx=1时,g(x)有最大值g(x)max=-2+2a+a+1=3a-1-1≤a/2≤1时,即-2≤a≤2时,当sinx=a/2时,g(x)有最大值g(x)max=a²/2 +a+1a/2