已知函數(shù)f(x)=lnx,g(x)=(1/2)ax+b；(1)若f(x)與g(x)在x=1處相切,試求g(x)的表達(dá)式；
(2)若h(x)=[m(x-1)/(x+1)]-f(x)在[1,+∞)上是減函數(shù),求實數(shù)m的取值范圍.(1).f '(x)=1/x；f '(1)=1；g'(x)=(1/2)a,令g'(1)=(1/2)a=1,即得a=2；
又因為相切,切點處兩函數(shù)的值相等,故得f(1)=ln1=0=g(1)=(1/2)a+b=1+b故b=-1,
于是得g(x)=x-1.
(2).h(x)=[m(x-1)/(x+1)]-lnx；
h'(x)=[m(x+1)-m(x-1)]/(x+1)²-1/x=2m/(x+1)²-1/x
=[2mx-(x+1)²]/[x(x+1)²]=[-x²+2(m-1)x-1]/[x(x+1)²]0,于是得-x²+2(m-1)x-10；故其判別式Δ=4(m-1)²-4=4[(m-1)²-1]=4(m²-2m)=4m(m-2)