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<=1
1-a<1-a^2
解得 0<a<1.