Question

一道高一数学题，急！！！

求函数f（x）=2sinxcosx+2cos²x的周期，最小值，递增区间

Answer 1

f（x）=2sinxcosx+2cos²x=sin2x+cos2x+1=√2sin(2x+π/4)+1
当sin(2x+π/4)=-1时，f(x)有最小值f(x)min=1-√2

最小正周期为T=2π/2=π

递增区间为
-π/2+2Kπ<2x+π/4<π/2+2Kπ
=>-3π/4+2Kπ<2x<π/4+2Kπ
=>-3π/8+Kπ<x<π/8+Kπ (K为整数)

Answer 2

f(x)=sin(2x)+cos(2x)+1
=(√2)sin[2x+(π/4)]+1.
[[[1]]
T=π
[[[2]]
min=1-√2
[[[3]]]
2kπ-(π/2)≤2x+(π/4)≤2kπ+(π/2)
2kπ-(3π/4)≤2x≤2kπ+(π/4)
kπ-(3π/8)≤x≤kπ+(π/8)

Answer 3

f(x)=sin2x+2*(1+cos2x)/2
=sin2x+cos2x+1
=√2sin(2x+π/4)+1
周期T=2π/2=π
最小值为1-√2
递增区间的计算如下：
-π/2+2kπ<=2x+π/4<=π/2+2kπ
解得-3π/8+kπ<x<π/8+kπ (k为整数)

Answer 4

f（x）=2sinxcosx+2cos²x
=sin2x+cos2x +1
=（根号2）(（根号2）/2 sin2x+ (根号2）/2 cos2x)+1
=(根号2）(sin(2x +45度）+1

Answer 5