f(x) = m(1 + sin2x) + cos2x
f(π/4) = m[1 + sin(π/2)] + cos(π/2) = m(1 + 1) + 0 = 2m = 2
m = 1
f(x) = 1 + sin2x + cos2x = 1 + 2[(sin2x)(√2/2) + (cos2x)(√2/2)]
= 1 + 2[sin2xcos(π/4) + cos2xsin(π/4)]
= 1+ 2sin(2x + π/4)
x ∈ [0,π/2], 2x + π/4 ∈ [π/4,5π/4]
π/4 < 2x+ π/4 < π/2时, f(x)为增函数
π/2 < 2x + π/4< 5π/4时, f(x)为减函数
所以只须比较f(0), f(π/2)
f(0) = 1 + 2sin(π/4) = 1 + √2
f(π/2) = 1 + 2sin(5π/4) = 1 - 2sin(π/4) = 1 - √2 < f(0)
fx在区间[0,π/2]上的最小值 = 1 - √2