Question

f(x)=|sin2x|+|cos2x|

Answer 1

(1)
f(-7π/12)=|sin(-7π/6)|+|cos(-7π/6)|
=1/2+√3/2
=(1+√3)/2
(2)
f(x)=｜sin2x｜+｜cos2x｜
那么f²(x)=1+2|sin2x||cos2x|=1+|sin4x|
当0≤x≤π/4时，0≤4x≤π，0≤sin4x≤1
∴1≤1+|sin4x|≤2，即1≤f²(x)≤2
∴1≤f(x)≤√2，即f(x)的取值范围为[1,√2]
(3)f(x)是偶函数，也是周期函数，最小正周期T=π/4
f(x)max=√2，f(x)min=1
单调递增区间：[kπ/4，kπ/4+π/8]
(k∈Z)
单调递减区间：[kπ/4+π/8，(k+1)π/4]
(k∈Z)