Question

设函数f（x）=sinxcosx-sin^2 x+1 （1）求f（x）的最小正周期（2）当x属于【0，π/2】时，求函数最大值

Answer 1

f(x)=(1/2)sin2x-(1/2)(1-cos2x)+1
=(1/2)sin2x+(1/2)cox2x+1/2
=√2/2sin(2x+π/4)+1/2
(1)
T=2π/2=π
(2)
0≤x≤π/2
π/2≤2x+π/2≤3π/2
当2x+π/2=π/2
即x=0时，f(x)取最大值，
f(MAX)=f(0)=1

Answer 2

f(x)=sinxcosx-sin^2 x+1=0.5*2*sinxcosx+0.5*(1-2sin^2 x)+1-0.5
=0.5sin2x+0.5cos2x+0.5=0.5*(sin2x+cos2x+1)=0.5*2^2/2sin(2x+π/4)+0.5
=2^2/4sin(2x+π/4)+0.5
(1)Tmin=2π/W=π
(2)当x属于【0，π/2】时，2x+π/4属于【π/4，5π/4】，sin(2x+π/4)属于【-2^2/2，1】
2^2/4sin(2x+π/4)属于【-1/4，2^2/4】,2^2/4sin(2x+π/4)+0.5属于【1/4，（2+2^2）/4】,