Mathematics 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…

\(g(x) = 3f \left( \frac{x}{3} \right) + f(3 - x)\) and \(f''(x) > 0 \forall x \in (0,3)\)
\(\Rightarrow f'(x)\) is increasing function
\(g'(x) = 3 \times \frac{1}{3} f' \left( \frac{x}{3} \right) - f'(3 - x) = f' \left( \frac{x}{3} \right) - f'(3 - x)\)

If g is decreasing in \((0, \alpha)\)
\(g'(x) < 0\)
\(= f' \left( \frac{x}{3} \right) - f'(3-x) < 0\)
\(\Rightarrow f' \left( \frac{x}{3} \right) < f'(3-x)\)
\(\Rightarrow \frac{x}{3} < 3-x \Rightarrow x < \frac{9}{4}\)
Therefore \(\alpha = \frac{9}{4}\)
Then \(8\alpha = 8 \times \frac{9}{4} = 18\)