Question

已知函數f(x)=a(2cos²+sinx)+b.

(1)當a=1時,求函數f(x)的單調遞增區間;

(2)當a＜0時,函數f(x)的值域是［3,4］,求a+b的值.

Answer 1

解:f(x)=a(cosx+1+sinx)+b

=asin(x+)+a+b.

(1)當a=1時,f(x)=  sin(x+)+1+b,

∴當2kπ-≤x+≤2kπ+ (k∈Z)時,f(x)是增函數.

∴函數f(x)的單調增區間為［2kπ-,2kπ+］(k∈Z).

(2)當a＜0時,sin(x+)=1,函數f(x)取得最小值3,即a+a+b=3.            ①

當sin(x+)=-1時,函數f(x)取得最大值4,

即-a+a+b=4.                                                                               ②

由①+②,得a+b=.