Question

设函数f(x)=2cos²x+2sinxcosx-1(x∈R).

(1)化简函数f(x)的表达式,并求函数f(x)的最小正周期.

(2)若x∈[0,],求函数f(x)的最大值与最小值.

Answer

(1)f(x)=2sin(2x+ T=π

(2)x=时,f(x)min=-1;x=时,f(x)max=2.

【解析】(1)∵f(x)=2cos2x+2sinxcosx-1

=cos2x+sin2x

=2sin(2x+),

∴函数f(x)的最小正周期T=π.

(2)∵0≤x≤,∴≤2x+≤,

∴-≤sin(2x+)≤1,

∴-1≤2sin(2x+)≤2,

∴当2x+=,

即x=时,f(x)min=-1;

当2x+=,

即x=时,f(x)max=2.