Let \(g(x)=3 f\left(\frac{x}{3}\right)+f(3-x) \forall x \in(0,3)\) and \(f^{\prime\prime}(x)>0 \forall x \in(0,3)\) t…

\(
g(x)=3 f\left(\frac{x}{3}\right)+f(3-x) \text { and } f^{\prime \prime}(x)>0 \forall x \in(0,3)
\)

\(\Rightarrow f^{\prime}(x)\) is increasing function

\(
\begin{aligned}
& g^{\prime}(x)=3 \times \frac{1}{3} \cdot f^{\prime}\left(\frac{x}{3}\right)-f^{\prime}(3-x) \\
& =\mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)-\mathrm{f}^{\prime}(3-\mathrm{x})
\end{aligned}
\)

If g is decreasing in \((0, \alpha)\)

\(
\begin{aligned}
& \mathrm{g}^{\prime}(\mathrm{x})<0 \\
& \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)-\mathrm{f}^{\prime}(3-\mathrm{x})<0 \\
& \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)<\mathrm{f}^{\prime}(3-\mathrm{x}) \\
& \Rightarrow \frac{\mathrm{x}}{3}<3-\mathrm{x} \\
& \Rightarrow \mathrm{x}<\frac{9}{4}
\end{aligned}
\)