Question

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

• A. 24
• B. 0
• C. 18
• D. 20

Answer 1

Answer: (C)

18

Answer 2

Given:
$g(x) = 3f\left(\frac{x}{3}\right) + f(3 - x) \quad \text{and} \quad f''(x) > 0 \quad \text{for } x \in (0, 3).$

Since $f''(x) > 0$, $f'(x)$ is an increasing function.

To find intervals where $g(x)$ is decreasing, we differentiate:
$g'(x) = 3 \times \frac{1}{3} f'\left(\frac{x}{3}\right) - f'(3 - x) = f'\left(\frac{x}{3}\right) - f'(3 - x).$

For $g(x)$ to be decreasing in $(0, \alpha)$:
$g'(x) < 0 \implies f'\left(\frac{x}{3}\right) < f'(3 - x).$

Setting equality for the transition point:
$f'\left(\frac{\alpha}{3}\right) = f'(3 - \alpha).$

From symmetry and the increasing nature of $f'$, we find:
$\alpha = \frac{9}{4}.$

Calculating $8\alpha$:
$8\alpha = 8 \times \frac{9}{4} = 18.$

The Correct answer is: 18