Question

高一数学题!~已知函数f(x)=msin2x+cos2x+m,若f(π/4)=2.

Answer 1

(1)f(π/4)=msin(2*π/4)+cos(2*π/4)+m=2m=2,所以m=1

(2)f(x)=sin2x+cos2x+1

=√2(sin2x/√2+cos2x/√2)+1

=√2sin(2x+π/4)+1

所以当sin(2x+π/4)=-1时,f(x)取最小值1-√2,

这时x=(3/2+2k)π/2-π/8,k=0、±1、±2、±3...为整数.