Question

函数y=sin2xcos2x的最小正周期是[π/2][π/2]，最大值是[1/2][1/2]．

Answer 1

y=sin2xcos2x=[1/2]sin4x

∴T=[2π/w]=[2π/4＝

π

2]

∵-1≤sin4x≤1

y_max=[1/2]

故答案为：[π/2]；[1/2]．