已知函數(shù)f(x)=sinxcosx-m(sinx+cosx)
(1)求函數(shù)f(x)的最小值g(x)
(2)若函數(shù)f(x)在區(qū)間[π/2,π]上是單調(diào)函數(shù),求實(shí)數(shù)m的取值范圍.
(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)
當(dāng)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處取極小值f(x3)= f(x4)=-(m^2+1)/2
當(dāng)m=0時(shí)
f(x)=sinxcosx=1/2sin2x
2x=2kπ-π/2==>x=kπ-π/4
∴f(x)在x=kπ-π/4處取極小值-1/2
當(dāng)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處取極小值f(x3)= f(x4)=-(m^2+1)/2
當(dāng)m>√2時(shí)
cosx-sinx=0==>√2cos(x+π/4)=0==>x1=2kπ-3π/4,x2=2kπ+π/4
f’’(x1)0
∴f(x)在x2處取極小值1/2-√2m
(2)解析：∵函數(shù)f(x)在區(qū)間[π/2,π]上是單調(diào)函數(shù)
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
∴實(shí)數(shù)m的取值范圍為m=1