Let g(x)=3f( x 3)+f(3-x) and f^"(x)>0 for all x∈ (0,3). If g is decreasing in (0,α ) and increasing in (α ,3), then 8α is

Given: $g(x)=3f(\frac{x}{3})+f(3-x)$g(x)=3f(3x​)+f(3−x) and ${f}^{"}(x)>0\forall x\in (0,3)$f"(x)>0∀x∈(0,3)

$\Rightarrow {f}^{'}(x)$⇒f′(x) is increasing function

$$\Rightarrow {g}^{'}(x)=3\times \frac{1}{3}\cdot {f}^{'}(\frac{x}{3})-{f}^{'}(3-x)$$⇒g′(x)=3×31​⋅f′(3x​)−f′(3−x)

$$\Rightarrow {g}^{'}(x)={f}^{'}(\frac{x}{3})-{f}^{'}(3-x)$$⇒g′(x)=f′(3x​)−f′(3−x)

It is given that $g$g is decreasing in $(0,\alpha )$(0,α) then ${g}^{'}(x)<0$g′(x)<0.

$$\Rightarrow {f}^{'}(\frac{x}{3})-{f}^{'}(3-x)<0$$⇒f′(3x​)−f′(3−x)<0

$$\Rightarrow {f}^{'}(\frac{x}{3})<{f}^{'}(3-x)$$⇒f′(3x​)<f′(3−x)

$$\Rightarrow \frac{x}{3}<3-x$$⇒3x​<3−x

$$\Rightarrow x<9-3x$$⇒x<9−3x

$$\Rightarrow x<\frac{9}{4}$$⇒x<49​

$$\Rightarrow \alpha =\frac{9}{4}$$⇒α=49​

$$\Rightarrow 8\alpha =8\times \frac{9}{4}$$⇒8α=8×49​

$$\Rightarrow 8\alpha =18$$⇒8α=18