Question

已知函数f（x）＝2sinxcosx.-2cos^2x（x∈R）.1求函数f（x）的最小正周期.2.当x∈【0,兀/2】

Answer 1

f(x)=2sinxcosx-2cos^2x

=sin(2x) -cos(2x)-1

=√2 sin(2x -π/4) -1

最小正周期Tmin=2π/2=π

x∈[0,π/2],则 2x-π/4∈[-π/4,3π/4]

sin(2x-π/4)∈[-√2/2,1]

√2 (-√2/2) -1≤f(x)≤√2 -1

-2≤f(x)≤√2 -1

函数的取值范围为[-2,√2 -1].