$\Delta S=\frac{Q}{T}$

Now, for given system to be in steady state, heat lost by resorvoir at temperature $T_{1}=$ heat gained by resorvoir at temperature $T_{2}$ (=Q say).

So, change in entropy for heat conduction process is

$\Delta S-\frac{-Q}{}+\frac{(+Q)}{T_{1}}$

$\Rightarrow \quad \Delta S=Q\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)$

So, time rate of change of entropy is

$\frac{d S}{d t}=\frac{d}{d t}\left\{Q\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)\right\}$

$=\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right) \cdot\left(\frac{d Q}{d t}\right)$

$\frac{d S}{d t}=\left(\frac{T_{1}}{T_{2}}-1\right) \cdot T_{1}\left(\frac{d Q}{d t}\right)$

As, $\quad \frac{d Q}{d t}=-k A\left(\frac{d T}{d x}\right)$

So, $\quad \frac{d S}{d t}=-k A \frac{d T}{d x} \cdot T_{1}\left(\frac{T_{1}}{T_{2}}-1\right)$

$\Rightarrow \quad \frac{d S}{d t}=k A T_{1} \cdot\left(1-\frac{T_{1}}{T_{2}}\right) \cdot \frac{d T}{d x}$

$=\frac{k A T_{1}}{x}\left(1-\frac{T_{1}}{T_{2}}\right)\left(T_{1}-T_{2}\right)$

$=\frac{k A T_{1}^{2}}{x}\left(1-\frac{T_{1}}{T_{2}}\right)\left(1-\frac{T_{2}}{T_{1}}\right)$

Clearly, at $\frac{T_{1}}{T_{2}}=1, \frac{d S}{d t}=0$.

Also, graph is asym metrical. So, correct option is $(b)$.