设函数f(x)=2cos^2+sin2x+a(a€R)

f(x)=2(cosx)^2+sin(2x)+a
=1+cos(2x)+sin(2x)+a
=√2sin(2x+π/4)+a+1
=√2sin[2(x+π/8)]+a+1
函数最小正周期为2π/2=π
2kπ-π/2≤2(x+π/8)≤2kπ+π/2(k∈Z)时,函数单调递增,此时
kπ-3π/8≤x≤kπ+π/8
函数的单调递增区间为[kπ-3π/8,kπ+π/8] (k∈Z)
x∈[0,π/4]
2x+π/4∈[π/4,3π/4]
2x+π/4=π/2时,函数有最大值f(x)max=a+1=2
a=1
函数为f(x)= √2sin[2(x+π/8)]+2
令2(x+π/8)=kπ x=kπ/2-π/8
函数对称轴方程为x=kπ/2-π/8 (k∈Z)