设函数f(x)=ab,其中向量a=（m,cos2x）,b=(1+sin2x,1),x属于R且y=f(x)的图象过（π/4,2）求fx值域

f(x)=m(1+sin2x) + cos2x
f(x)过(π/4,2),所以2= m(1+sinπ/2 ) +cosπ/2 = 2m,所以m=1
所以f(x)= 1 + sin2x +cos2x = 1+ sin(2x+π/4)*根号2
显然sin(2x+π/4)值域为[-1,1],f(x)的值域为[1-根号2,1+根号2]