As given that slope of the graph $=f(V)$, where $V$ is volume. Slope of $p=f(V)$ curve at $\left(V \cdot p_0\right)=f\left(V_0\right)$
$$
\begin{aligned}
&p V_0^\gamma=K \text { (constant) } \\
&\left.p=\frac{K}{V_0^\gamma} \text { or } \frac{d p}{d V}=K(-\gamma) \cdot V_0^{-\gamma-1} \quad \text { (Put } K=p_0 V_0^\gamma\right) \\
&\left(\frac{d p}{d V}\right)^{\text {slope of adiabatic at }\left(V_0 \cdot p_0\right)} \\
&=p_0(-\gamma) V_0^{-1-\gamma+\gamma} \\
&\left(\frac{d p}{d V}\right)_{\left(V_0, p_0\right)}=\frac{-\gamma p_0}{V_0}
\end{aligned}
$$
Heat absorbed by in the process $p=f(V)$
$$
\begin{aligned}
&d Q=d U+d W=n C_V d T+p d V \\
&\because p V=n R T \\
&\Rightarrow T=\left(\frac{p V}{n R}\right)=\frac{V}{n R} f(V) \quad(\because p=f(V) \text { given }) \\
&\text { or } T=\left(\frac{1}{n R}\right) V \cdot f(V)
\end{aligned}
$$
Differentiated both sides w.r.t. $V$,
$$
\begin{aligned}
&d T=\left(\frac{1}{n R}\right)\left[f(V)+V f^{\prime}(V)\right] d V \\
&\frac{d T}{d V}=\frac{1}{n R}\left[f(V)+V f^{\prime}(V)\right]
\end{aligned}
$$
From Eq. (i)
$$
\begin{aligned}
\frac{d Q}{d V} &=n C_V \frac{d T}{d V}+p \frac{d V}{d V}=n C \frac{d T}{d V}+p \\
&=\frac{n C_V}{n R} \times\left[f(V)+V f^{\prime}(V)\right]+p \text { [from Eq. (ii)] }
\end{aligned}
$$
$$
\begin{aligned}
=\frac{C_V}{R} \times & {\left[f(V)+V f^{\prime}(V)\right]+f(V) \quad[\because p=f(V)] \text { given } } \\
{\left[\frac{d Q}{d V}\right]_{V=V_0} } &=\frac{C_V}{R}\left[f\left(V_0\right)+V_0 f^{\prime}\left(V_0\right)\right]+f\left(V_0\right) \\
&=f\left(V_0\right)\left[\frac{C_V}{R}+1\right]+V_0 f^{\prime}\left(V_0\right) \frac{C_V}{R}
\end{aligned}
$$
As we know that
$$
\begin{aligned}
&C_P-C_V=R \\
&\frac{C_P}{C_V}-1=\frac{R}{C_V} \Rightarrow \gamma-1=\frac{R}{C_V} \quad \text { So, } \frac{C_V}{R}=\frac{1}{\gamma-1} \\
&{\left[\frac{d Q}{d V}\right]_{V=V_0}=\left[\frac{1}{\gamma-1}+1\right] f\left(V_0\right)+\frac{V_0 f^{\prime}\left(V_0\right)}{\gamma-1}\left(\because f\left(V_0\right)=p_0\right)} \\
&\quad=\frac{\gamma}{\gamma-1} p_0+\frac{V_0}{\gamma-1} f^{\prime}\left(V_0\right)=\frac{1}{(\gamma-1)}\left[\gamma p_0+V_0 f^{\prime}\left(V_0\right)\right]
\end{aligned}
$$
If $\gamma>1$, so $\left(\frac{1}{\gamma-1}\right)$ is positive.
Then, heat is absorbed where $\left[\frac{d Q}{d V}\right]_{V=V_0}>0$ when gas expands.
Hence, $\gamma p_0+V_0 f^{\prime}\left(V_0\right)>0$
$$
\begin{gathered}
V_0 f^{\prime}\left(V_0\right)>\left[-\gamma p_0\right] \\
f^{\prime}\left(V_0\right)>\left(-\gamma \frac{p_0}{V_0}\right)
\end{gathered}
$$