题目

已知函数f(x)=sinxcosx-m(sinx+cosx)
(1)求函数f(x)的最小值g(x)
(2)若函数f(x)在区间[π/2,π]上是单调函数,求实数m的取值范围.

提问时间：2021-03-23

答案

已知函数f(x)=sinxcosx-m(sinx+cosx)
(1)求函数f(x)的最小值g(x)
(2)若函数f(x)在区间[π/2,π]上是单调函数,求实数m的取值范围.
(1)解析：∵f(x)=sinxcosx-m(sinx+cosx)
f’(x)=cos^2x-sin^2x-m(cosx-sinx)=
(cosx-sinx) (cosx+sinx-m)
f’’(x)=-2sin2x+m(cosx+sinx)
当m√2cos(x+π/4)=0==>x1=2kπ-3π/4,x2=2kπ+π/4
f’’(x1)>0,f’’(x2)0
令θ=arcos(-m√2/2)
f(x3)= sinx3cosx3-m(sinx3+cosx3)=1/2sin2x3-m^2=1/2cos2θ-m^2
=(2cos^2θ-1)/2-m^2=-(m^2+1)/2
∴f(x)在x3,x4处取极小值f(x3)= f(x4)=-(m^2+1)/2
当m=0时
f(x)=sinxcosx=1/2sin2x
2x=2kπ-π/2==>x=kπ-π/4
∴f(x)在x=kπ-π/4处取极小值-1/2
当0x1=2kπ-3π/4,x2=2kπ+π/4
f’’(x1)0,f’’(x4)>0
令θ=arcos(m√2/2)
f(x3)= sinx3cosx3-m(sinx3+cosx3)=1/2sin2x3-m^2=1/2cos2θ-m^2
=(2cos^2θ-1)/2-m^2=-(m^2+1)/2
∴f(x)在x3,x4处取极小值f(x3)= f(x4)=-(m^2+1)/2
当m>√2时
cosx-sinx=0==>√2cos(x+π/4)=0==>x1=2kπ-3π/4,x2=2kπ+π/4
f’’(x1)0
∴f(x)在x2处取极小值1/2-√2m
(2)解析：∵函数f(x)在区间[π/2,π]上是单调函数
f’(x)=cos^2x-sin^2x-m(cosx-sinx)=cos2x-m(cosx-sinx)
令f’(π/2)=-1-m>=0==>m=0==>m=1
取二者交m>=1
∴实数m的取值范围为m=1