Question

函数f（x）＝2√3cosx²x＋sin2x-√3＋1

求解析

Answer 1

答：
f(x)=2√3cos²x+sin2x-√3+1
f(x)=√3(cos2x+1)+sin2x-√3+1
f(x)=√3cos2x+sin2x+1
f(x)=2*[(1/2)sin2x+(√3/2)cos2x]+1
f(x)=2sin(2x+π/3)+1
1）
f(x)的最小正周期T=2π/2=π
单调递增区间满足：
2kπ-π/2<=2x+π/3<=2kπ+π/2
2kπ-5π/6<=2x<=2kπ+π/6
kπ-5π/12<=x<=kπ+π/12
单调递增区间为 [kπ-5π/12，kπ+π/12] ，k为任意整数
2）
-π/4<=x<=π/4
-π/2<=2x<=π/2
-π/6<=2x+π/3<=5π/6
所以：-1/2<=sin(2x+π/3)<=1
所以：-1+1<=f(x)<=2+1
所以：值域为 [0，3]

Answer 2

1)f(x)=√3(1+cos2x)+sin2x-√3+1
=√3cos2x+sin2x+1
=2sin(2x+π/3)+1
最小正周期T=2π/2=π
单调增区间为： 2kπ-π/2=<2x+π/3<=2kπ+π/2
即： 2kπ-5π/12=<x<=kπ+π/12
这里k是任意整数

2）若x在[-π/4,π/4],
则2x+π/3在[-π/6, 5π/6]
sin(2x+π/3)的值域为[-1/2, 1]
因此f(x)的值域为 [0,3]