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f(x)=√2cosωx(sinωx+cosωx)
=√2(cosωxsinωx+cos²ωx)
=√2(cosωxsinωx+cos²ωx)
=√2/2(sin2ωx+cos2ωx+1)
=sin(2ωx+π/4)+√2/2
(1) 最小正周期的π,则 2π/2ω=π
得 ω=1
(2) f(x)=sin(2x+π/4)+√2/2
x∈[0,π/2]
2x∈[0,π]
故 x∈[0,π/4] 单调增
x∈[π/4,π/2] 单调减