Answer
d) is correct.
Work Step by Step
According to the work-energy theorem, $$\sum W=W_1+W_2=\frac{1}{2}m(v_f^2-v_0^2)$$
Since $v_f\gt v_0$, we have $(W_1+W_2)\gt0$. Therefore, it is not possible that the work done by each force $F_1$ and $F_2$ is negative, because its sum is positive.
d) is correct.
