若函数f(x)=(sinx+co
若函数f(x)=(sinx+cosx)^2+2cos^2 x-m在[0,π/2]上有零点，则m的取值范围是
A。[1,2+根号2] B。[-1,2] C。[-1,2+根号2] D。 [1,3]
f(x)=(sinx+cosx)^2+2cos^2 x-m
=sin^2 x+cos^2 x+2sinxcosx+2cos^2 x-m
=1+sin2x+(cos2x+1)-m
=(2-m)+(sin2x+cos2x)
=(2-m)+√2sin[2x+(π/4)]
因为x∈[0,π/2]，所以：2x∈[0,π]
则，2x+(π/4)∈[π/4,5π/4]
那么，sin[2x+(π/4)]∈[-√2...全部
若函数f(x)=(sinx+cosx)^2+2cos^2 x-m在[0,π/2]上有零点，则m的取值范围是
A。[1,2+根号2] B。[-1,2] C。[-1,2+根号2] D。
[1,3]
f(x)=(sinx+cosx)^2+2cos^2 x-m
=sin^2 x+cos^2 x+2sinxcosx+2cos^2 x-m
=1+sin2x+(cos2x+1)-m
=(2-m)+(sin2x+cos2x)
=(2-m)+√2sin[2x+(π/4)]
因为x∈[0,π/2]，所以：2x∈[0,π]
则，2x+(π/4)∈[π/4,5π/4]
那么，sin[2x+(π/4)]∈[-√2/2,1]
所以，√2sin[2x+(π/4)]∈[-1,√2]
所以，f(x)=(2-m)+√2sin[2x+(π/4)]=0时
===> m=2+√2sin[2x+(π/4)]∈[1,2+√2]
——答案：A。收起
