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

The Correct Option is C

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