已知函数f(x)=1/3x^3-(2a+1)x^2+3a(a+2)x+1,a为R,当函数y=f'(x)在(0,4)上有唯一的零点时,求实数a的取值范围.

f(x)=1/3x³-(2a+1)x²+3a(a+2)x+1f(x)′=3*1/3x²-2(2a+1)x+3a(a+2)f(x)′=x²-2(2a+1)x+3a(a+2)当函数y=f'(x)在(0,4)上有唯一的零点时4=3a(a+2)4=3(a²+2a)3(a²+2a+1)=4+3(a+1)²=7/...