已知函数f(x)=msin2x+cos2x(x属于r)的图像经过点(π/4,

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

把π/4带入 msin（π/2）+cos（π/2）=m=1
f(x)=sin2x+cos2x
f(x)=sin2x+cos2x=√2sin（4x+π/4）
周期为π/2
2 左移 g(x)=√2sin（4（x+π/24）+π/4）=2sin（4x+5π/12）
3 2kπ-π<=4x+5π/12<=2kπ+π
kπ/2-17π/48<=x<= kπ/2+7π/48

1、
f(π/4)=m=1
所以，f(x)=sin2x+cos2x=√2sin(2x+π/4)

2、
最小正周期T=2π/2=π

3、
g(x)=√2sin[2(x+π/24)+π/4]
=√2sin(2x+π/3)
-π/2+2kπ<2x+π/3<π/2+2kπ
-5π/6+2kπ<2x

f(Pai/4)=m*1+0=1,m=1
f(x)=sin2x+cos2x=根号2sin(2x+Pai/4)
故最小 正周期T=2Pai/2=Pai
(2)向左平移后得到g(x)=根号sin(2(x+Pai/24)+Pai/4)=根号2sin(2x+Pai/3)
单调增区间是：2kPai-Pai/2<=2x+Pai/3<=2kPai+Pai/2
即有[kPai-5Pai/12,kPai+Pai/12]