Let $g(x)=3 f\left(\frac{x}{3}\right)+f(3-x)$ and $f^{\prime \prime}(x)>0$ for all $x \in(0,3)$. If $g$ is decreasing in $(0, \alpha)$ and increasing in $(\alpha, 3)$, then $8 \alpha$ is :

<p>$$g(x)=3 f\left(\frac{x}{3}\right)+f(3-x) \text { and } f^{\prime \prime}(x)>0 \forall x \in(0,3)$$</p> <p>$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})$ is increasing function</p> <p>$$\begin{aligned} & g^{\prime}(x)=3 \times \frac{1}{3} \cdot f^{\prime}\left(\frac{x}{3}\right)-f^{\prime}(3-x) \\ & =f^{\prime}\left(\frac{x}{3}\right)-f^{\prime}(3-x) \end{aligned}$$</p> <p>If $\mathrm{g}$ is decreasing in $(0, \alpha)$</p> <p>$$\begin{aligned} & \mathrm{g}^{\prime}(\mathrm{x})<0 \\ & \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)-\mathrm{f}^{\prime}(3-\mathrm{x})<0 \\ & \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)<\mathrm{f}^{\prime}(3-\mathrm{x}) \\ & \Rightarrow \frac{\mathrm{x}}{3}<3-\mathrm{x} \\ & \Rightarrow \mathrm{x}<\frac{9}{4} \end{aligned}$$</p> <p>Therefore $\alpha=\frac{9}{4}$</p> <p>Then $8 \alpha=8 \times \frac{9}{4}=18$</p>