QJEEM24Ma896a: JEE Main 2024 Mathematics - Class 12 JEEM24Ma896a Solution: Let and for all . If is decreasing in and increasing in , then…

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

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

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

If $\mathrm{g}$ is decreasing in $(0, \alpha)$

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

Therefore $\alpha=\frac{9}{4}$

Then $8 \alpha=8 \times \frac{9}{4}=18$