Can initial and final state of isothermal and adiabatic process be same? The following question was in my mid-sem test:

If $W_1$ and $W_2$ be the work done in an isothermal process and adiabatic process between same initial state and final state respectively, then
(a) $W_1$=$W_2$
(b) $W_1$>$W_2$
(c) $W_1$<$W_2$
(d) $W_1\le W_2$

I was thinking that the statement in the problem is wrong. Let the initial state be $(P_1,V_1)$ and final state be $(P_2,V_2)$. Let the isothermal curve passes through the initial state and final state. Also there is a relation between the slope of adiabatic curve and the isothermal curve.
slope of adiabatic curve = $\gamma \frac{dP}{dV}$ where $\frac{dP}{dV}$ is the slope of the isothermal curve.
Now if the adiabatic curve passes through the initial state $(P_1,V_1)$ then it won't pass through the final state.
Am I right or wrong? Is the statement in the problem means something else?
 A: If both processes are reversible, then, as you said, they can't pass through the same initial and final states.  If they are irreversible, I'm not so sure.  
In any event, the intent of the problem seems to have been to get you to use the first law of thermodynamics.  Both processes have the same $\Delta U$ because the initial and final states are the same.  So, $W_I-W_A=Q_I-Q_A$.  But, $Q_A=0$.  Therefore, $$W_I=W_A+Q_I$$$Q_I$ can be either positive of negative.  So, it seems to me, none of the answers are right.
A: My previous answer (below) only showed that an ideal gas can undergo an irreversible process that is both isothermal and adiabatic. However OP's question seems to be about the feasibility of two distinct processes, one isothermal but not adiabatic and another adiabatic but not isothermal, between the same initial and final states. This is possible as shown in T-S diagram below. 
----------------Previous answer---------------------
In an irreversible process it is possible for two distinct states to be connected by isothermal and adiabatic processes. Let the two states have pressure and volume $(p_1,V_1)$ and $(p_2,V_2)$ respectively. The for an isothermal process:
\begin{align}
p_1V_1=p_2V_2=\alpha
\end{align}
where $\alpha$ is a constant that depends on the magnitude of the temperature being held constant. For a reversible adiabatic process, which is also an isentropic process, we have:
\begin{align}
p_1V_1^\gamma=p_2V_2^\gamma=\beta
\end{align}
where $\beta$ is a constant which depends on the magnitude of the entropy being held constant. However an irreversible adiabatic process is not isentropic, and for this case:
\begin{align}
p_1V_1^\gamma & =\beta_1 \\
p_2V_2^\gamma & =\beta_2
\end{align}
where $\beta_1\neq \beta_2$, because entropy of the two states is different. Thus we have:
\begin{align}
\frac{V_2}{V_1} & =\left( \frac{\beta_2}{\beta_1} \right)^{1/(\gamma-1)} \\
\frac{p_2}{p_1} & =\left( \frac{\beta_2}{\beta_1} \right)^{1/(1-\gamma)}
\end{align}
Therefore for the set of initial and final states satisfying above relation, an isothermal and (irreversible) adiabatic curve can pass through the same set of initial and final states. This presumes that the same system can be run in reversible and irreversible modes between a given set of initial and final states.
As regards work done, Chester Miller has already answered your question.
