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

To solve this problem, we need to analyze the function \( g(x) = 3f\left(\frac{x}{3}\right) + f(3 - x) \) and the given conditions. Let's break this down step by step: 

1. We know that \( f''(x) > 0 \) for all \( x \in (0, 3) \). This implies that \( f(x) \) is a convex function in this interval.
2. The function \( g(x) \) is given by: \(g(x) = 3f\left(\frac{x}{3}\right) + f(3 - x)\).
3. We are told that \( g(x) \) is decreasing in \( (0, \alpha) \) and increasing in \( (\alpha, 3) \). This implies that \( g'(x) = 0 \) at \( x = \alpha \).
4. Let's differentiate \( g(x) \) with respect to \( x \): \(g'(x) = 3 \cdot \frac{1}{3} \cdot f'\left(\frac{x}{3}\right) + (-1) \cdot f'(3 - x) = f'\left(\frac{x}{3}\right) - f'(3 - x).\)
5. From the condition that \( g'(x) = 0 \) at \( x = \alpha \), we get: \(f'\left(\frac{\alpha}{3}\right) = f'(3 - \alpha).\)
6. Given that \( f(x) \) is convex in the interval (0, 3) (i.e., \( f''(x) > 0 \)), it means that \( f'(x) \) is increasing in this interval.
7. Thus, the point where the derivatives are equal occurs when \(\frac{\alpha}{3}\) and \(3 - \alpha\) are symmetrical with respect to the midpoint of the interval (0, 3), which is 1.5.
8. So, we can set: \(\frac{\alpha}{3} + (3 - \alpha) = 3\)(since it's symmetric about 1.5), solving this equation gives us: \(\frac{\alpha}{3} = 3 - \alpha \\ \Rightarrow \alpha = 3(3 - \alpha) \\ \Rightarrow \alpha + 3\alpha = 9 \\ \Rightarrow 4\alpha = 9 \\ \Rightarrow \alpha = \frac{9}{4}.\)
9. Finally, we are asked for \( 8\alpha \): \(8\alpha = 8 \times \frac{9}{4} = 18.\)

Therefore, the value of \( 8\alpha \) is 18. The correct answer is 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