Question

(理)已知函数f(x)=2sin²x+2sinxcosx+1.

(1)求f(x)的单调递增区间;

(2)若不等式f(x)≥m对x∈［0,］都成立,求实数m的最大值.

(文)已知函数f(x)=2sin²x+2sinxcosx+1.

求:(1)f(x)的最小正周期;

(2)f(x)的单调递增区间;

(3)f(x)在［0,］上的最值.

Answer 1

(理)解：(1)因为f(x)=2sin²x+2sinxcosx+1

=1-cos2x+sinxcosx+1                                                   

=2sin(2x)+2,                                                           

由2kπ≤2x≤2kπ+(k∈Z),

得kπ≤x≤kπ+(k∈Z).                                                

所以f(x)的单调增区间是［kπ,kπ+］(k∈Z).                              

(2)因为0≤x≤,

所以≤2x≤.

所以≤sin(2x)≤1.                                                  

所以f(x)=2sin(2x)+2∈［1,4］.                                          

所以m≤1,即m的最大值为1.                                               

(文)解:(1)因为f(x)=2sin²x+23sinxcosx+1

=1-cos2x+sinxcosx+1                                                   

=sin2x-cos2x+2                                                          

=2sin(2x)+2,                                                           

所以f(x)的最小正周期T==π.                                             

(2)因为f(x)=2sin(2x)+2,

所以由2kπ≤2x≤2kπ+(k∈Z),                                      

得kπ≤x≤kπ+(k∈Z).

所以f(x)的单调增区间是［kπ,kπ+］(k∈Z).                              

(3)因为0≤x≤,

所以≤2x≤.

所以≤sin(2x)≤1.                                                    

所以f(x)=2sin(2x)+2∈［1,4］,

即f(x)的最小值为1,最大值为4.