Membrane potential after exposure to glutamate
I believe C is also correct. The excitatory neurotransmitters open $\ce{Na}$ channels which makes the cell membrane more permeable to sodium as compared to potasium and therefore the equilibrium membrane potential would be much closer to the nernst potential for sodium. Now, since the concentration of sodium outside is much greater than inside, the nernst potential for sodium is positive with respect to the inside of the membrane. This makes the overall potential also positive.   

If you want further clarifications regarding the calculation of the equilibrium and the nernst potential across the membrane, feel free to ask.   

ADDENDUM  

The approximate concentrations of sodium and potassium inside and outside a cell are :-
$$\ce {Na+_{in}}=10 \text{mmoll}^{-1}$$
$$\ce {K+_{in}}=105 \text{mmoll}^{-1}$$
$$\ce {Na+_{out}}=140 \text{mmoll}^{-1}$$
$$\ce {K+_{out}}=5 \text{mmoll}^{-1}$$ 

The Nernst potential is given by $E=-\frac{RT}{nF}\ln\frac{\text{in}}{\text{out}}$  

Which for the given concentrations comes out to be $+67.7mV$ for sodium and $-78.1mV$ for potassium. During a sufficiently strong action potential, the relative permeability of sodium with respect to potassium becomes $10$. Using the GHK equation(ignoring the contribution of chlorine) gives us a net potential of $+47.5mV$.

The graph of the action potential clearly indicates that at the peak of the AP, the membrane potential is positive. Source : Graph of action potential