please see problem attached to this post
The global minimum of $\displaystyle f(x)$ occurs at an end point of the interval if the calculus like minimum (the solution of $\displaystyle f'(x)=0$) does not lie in the interval $\displaystyle (0,1).$
Now find the solution of $\displaystyle f'(u)=0$ (in terms of $\displaystyle a$) and use the fact that $\displaystyle f(u)=3$ (which gives a quadratic in $\displaystyle a$) and solve.
Now how many values of $\displaystyle a$ do you have and do the corresponding values of $\displaystyle u$ lie in $\displaystyle (0,1)$ ?
CB