Question

求解答两题高中函数题

希望给出过程。谢谢

Answer 1

Answer 2

16、解析：∵f(x)=√3sinx+cosx=2sin(x+π/6)
单调增区间为2kπ-π/2<=x+π/6<=2kπ+π/2==>2kπ-2π/3<=x<=2kπ+π/3
∵x∈[-π/2,π/2]
f(-π/2)=2sin(-π/2+π/6)=-√3
f(π/3)=2sin(π/3+π/6)=2
f(π/2)=2sin(π/2+π/6)= √3
∴f(x)值域为[-√3,2]

17、解析：∵f(x)=√3/2sin(x)cos(x)+1/2(cos(x))^2+1
=√3/4sin(2x)+1/4(cos(2x)+1)+1
=1/2sin(2x+π/6)+5/4
∴f(x)最小正周期为π
2x+π/6=2kπ+π/2==>x=kπ+π/6
∴f(x)取得最大的值时，x=kπ+π/6==>{x|x=kπ+π/6，k∈Z}
单调增区间2kπ-π/2<=2x+π/6<=2kπ+π/2==>kπ-π/3<=x<=2kπ+π/6
单调减区间2kπ+π/2<=2x+π/6<=2kπ+3π/2==>kπ+π/6<=x<=kπ+2π/3
由y=sinx图像变换成y=1/2sin(2x+π/6)+5/4图像：
1）将y=sinx图像左平移π/6个单位，得y=sin(x+π/6)图像；
2）将X坐标缩小1/2倍，得y=sin(2x+π/6)图像；
3）将Y坐标缩小1/2倍，得y=1/2sin(2x+π/6)图像；
4）将y=1/2sin(2x+π/6)图像，上移5/4个单位，得到y=1/2sin(2x+π/6)+5/4图像。