f(x)的图像经过点(π/4,1)
∴msin(2×π/4)+cos(2×π/4)=1
∴m=1
f(x)=sin2x+cos2x
=√2(√2/2sin2x+√2/2cos2x)
=√2sin(2x+π/4)
f(x)最小正周期T=2π/2=π
(2)
将f(x)向左平移π/24个单位
得到y=√2sin[2(x+π/24)+π/4]
=√2sin(2x+π/12+π/4)
=√2sin(2x+π/3)图像
∴g(x)=√2sin(2x+π/3)
由2kπ-π/2≤2x+π/3≤2kπ+π/2
得kπ-5π/12≤x≤kπ+π/12
∴g(x)单调递增区间为
[ kπ-5π/12,kπ+π/12],k∈Z