已知函数f(x)=ax^3-3ax,g(x)=bx^2+cInx,且g(x)在点(1,g(1))处的切线方程为2y-1=

f'(x)=3ax?-3a=3a(x+1)(x-1)g'(x)=2bx+c/x由题g(x)在点(1,g(1))处的切线方程为2y-1=0而2y-1=0斜率为0故g(x)在此点达到极值进而g'(1)=2b+c=0,即c=-2b且g(1)=b+cln1=b=1/2所以c=-1F'(x)=3a(x+1)(x-1)+2bx+c/x=3a(x+1)(x-1)+x-1/x=(x-1)(x+1)(3ax+1)/xF'(x)＞0等价于x(x-1)(x+1)(3ax+1)＞0若a=0,则单调增区间为(-1,0),(1,+∞)太麻烦了,%>_