设函数f(x)=2cos^2+sin2x+a(a€R)
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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)

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