f(x)=sin2x+cos2x+2=√2*sin(2x+π/4)+2
所以f(x)的最小正周期为2π/2=π
最大值为√2+2,最小值为-√2+2
而当2x+π/4=2kπ+π/2时得到最大值
而当2x+π/4=2kπ+3π/2时得到最小值
求解如下。祝你好运!学习进步!~~
f(x) = ab = 2sinxcosx + cos2x + 2
= sin2x + cos2x + 2
= 根号2 sin(2x+pi/4) + 2
T = pi
2 + 根号2 2x+pi/4 = pi/2 + 2 k pi
2 - 根号2 2x+pi/4 = -pi/2 + 2 k pi
.......