Question

設函數f(x)=2cos²x+2sinxcosx-1(x∈R).
(1)化簡函數f(x)的表達式,并求函數f(x)的最小正周期.
(2)若x∈[0,],求函數f(x)的最大值與最小值.

Answer 1

(1)f(x)=2sin(2x+  T=π
(2)x=時,f(x)_min=-1;x=時,f(x)_max=2.

(1)∵f(x)=2cos²x+2sinxcosx-1
=cos2x+sin2x
=2sin(2x+),
∴函數f(x)的最小正周期T=π.
(2)∵0≤x≤,∴≤2x+≤,
∴-≤sin(2x+)≤1,
∴-1≤2sin(2x+)≤2,
∴當2x+=,
即x=時,f(x)_min=-1;
當2x+=,
即x=時,f(x)_max=2.