f(x)=-1/4x^4+2/3x^3+ax^2-2x-2f'(x)=-x^3+2x^2+2ax-2由于在区间[-1,1]上单调递减,在[1,2]上单调递增∴x=1是一个零点即f'(1)=0-1+2+2a-2=0a=1/2f(x)=-1/4x^4+2/3x^3+1/2x^2-2x-2f'(x)=-x^3+2x^2+x-2=-(x-1)(x+1)(x-2...