Let g ( x ) = 3f((x)/(3)) + f ( 3 - x ) and f ^ " ( x ) > 0 for all...

Given: $g ( x ) = 3f\left(\dfrac{x}{3}\right) + f ( 3 - x )$ and {f}^{"} ( x ) > 0 \forall x \in ( 0, 3 )$</p><p>$⇒ {f}^{'} ( x )$is increasing function</p><p>$⇒ {g}^{'} ( x ) = 3 \times \dfrac{1}{3} \cdot {f}^{'}\left(\dfrac{x}{3}\right) - {f}^{'} ( 3 - x )$</p><p>$⇒ {g}^{'} ( x ) = {f}^{'}\left(\dfrac{x}{3}\right) - {f}^{'} ( 3 - x )$</p><p>It is given that$g$is decreasing in \left(0, \alpha\right)$ then ${g}^{'} ( x ) < 0$.

$⇒ {f}^{'}\left(\dfrac{x}{3}\right) - {f}^{'} ( 3 - x ) < 0$

$⇒ {f}^{'}\left(\dfrac{x}{3}\right) < {f}^{'} ( 3 - x )$

$⇒ \dfrac{x}{3} < 3 - x$

$⇒ x < 9 - 3x$

$⇒ x < \dfrac{9}{4}$

$⇒ \alpha = \dfrac{9}{4}$

$⇒ 8\alpha = 8 \times \dfrac{9}{4}$

$⇒ 8\alpha = 18$