已知函数f(x)=cosx+5x,x∈[-1,1],若f(1-a)<f(1-a^2),则a的取值范围.解 设x1,x2∈[-1,1],且x1<x2,则(x1+x2)/2∈[-1,1],cos((x1+x2)/2)>0,0<(x2-x1)/2<1,sin((x2-x1)/2)<(x2-x1)/2,于是f(x1)-f(x2)=cosx1+5x1-cosx2-5x2= 展开
已知函数f(x)=cosx+5x,x∈[-1,1],若f(1-a)<f(1-a^2),则a的取值范围.解 设x1,x2∈[-1,1],且x1<x2,则(x1+x2)/2∈[-1,1],cos((x1+x2)/2)>0,0<(x2-x1)/2<1,sin((x2-x1)/2)<(x2-x1)/2,于是f(x1)-f(x2)=cosx1+5x1-cosx2-5x2=5(x1-x2)+2cos((x1+x2)/2)sin((x1-x2)/2)=5(x1-x2)-2cos((x1+x2)/2)sin((x2-x1)/2)=5(x1-x2)-2cos((x1+x2)/2)*((x2-x1)/2)=(x1-x2)[5+cos((x1+x2)/2]<0,所以,f(x)在[-1,1]上单调递增.从而由f(1-a)<f(1-a^2),可得不等式组-1<=1-a<=1,1<=1-a^2<=11-a<1-a^2解得 0<a<1. 收起
