Question

设函数f(x)=sinax+cosax(0＜a＜1),g(x)=tan(mx+)(0＜m＜1),已知函数f(x)、g(x)的最小正周期相同,且f(1)=2g(1).

(1)试确定f(x)、g(x)的解析式;

(2)求函数f(x)的单调递增区间.

Answer 1

解:(1)f(x)=2sin(ax+),其最小正周期为,g(x)的最小正周期为,

∴=,解得a=2m.                                                    

由f(1)=2g(1)及a=2m,

得2sin[2(m+)]=2tan(m+),

∴2sin(m+)cos(m+)=.                                      

∵0＜m＜1,∴＜m+＜1+＜.

∴sin(m+)≠0.∴cos(m+)=.

∴m+=.解得m=,a=.

∴f(x)=2sin(x+),g(x)=tan(x+).                                     

(2)由2kπ-≤x+≤2kπ+,得12k-5≤x≤12k+1,k∈Z,

∴函数f(x)的单调递增区间是[12k-5,12k+1](k∈Z).