数学高一急需答案

Question

如题所述

Answer 1

(1) f(x)=2cos^2x+2√3sinxcosx+m
=cos2x+1+√3sin2x+m
=2sin(2x+π/6)+m+1
T=2π/2=π
(2) 0=<x<=π/2
π/6=<2x+π/6<=7π/6
-1/2=<sin(2x+π/6)<=1
-1=<2sin(2x+π/6)<=2
-1+m+1=<2sin(2x+π/6)<=2+m+1
m=<2sin(2x+π/6)<=m+3
1/2=<2sin(2x+π/6)+m+1<=7/2
∴m=1/2
2x+π/6=2kπ
2x=-π/6+2kπ
x=-π/12+kπ (k∈Z)
f(x)=0+1/2+1=3/2
对称中心：(-π/12+kπ,0) (k∈Z)

Answer 2

希望能帮助到你~望采纳~

Answer 3

追答