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解 设x1,x2∈[-1,1],且x10, 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-... |
