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 address this problem, we will analyze the function \( g(x) = 3f\left(\frac{x}{3}\right) + f(3 - x) \) and the provided conditions. The analysis proceeds as follows:

1. The condition \( f''(x) > 0 \) for all \( x \in (0, 3) \) establishes that \( f(x) \) is convex within this interval.
2. The function \( g(x) \) is defined as: \(g(x) = 3f\left(\frac{x}{3}\right) + f(3 - x)\).
3. The information that \( g(x) \) decreases on \( (0, \alpha) \) and increases on \( (\alpha, 3) \) indicates that \( g'(x) = 0 \) at \( x = \alpha \).
4. Differentiating \( g(x) \) with respect to \( x \) yields: \(g'(x) = 3 \cdot \frac{1}{3} \cdot f'\left(\frac{x}{3}\right) + (-1) \cdot f'(3 - x) = f'\left(\frac{x}{3}\right) - f'(3 - x)\).
5. The condition \( g'(x) = 0 \) at \( x = \alpha \) implies: \(f'\left(\frac{\alpha}{3}\right) = f'(3 - \alpha)\).
6. Given that \( f(x) \) is convex on \( (0, 3) \), meaning \( f''(x) > 0 \), its derivative \( f'(x) \) is increasing in this interval.
7. Consequently, the equality of the derivatives \( f'\left(\frac{\alpha}{3}\right) = f'(3 - \alpha) \) occurs when the arguments are equidistant from the midpoint of the interval \( (0, 3) \), which is 1.5.
8. This symmetry leads to the equation: \(\frac{\alpha}{3} + (3 - \alpha) = 3\). Solving this equation: \(\frac{\alpha}{3} = 3 - \alpha \\ \Rightarrow \alpha = 3(3 - \alpha) \\ \Rightarrow \alpha + 3\alpha = 9 \\ \Rightarrow 4\alpha = 9 \\ \Rightarrow \alpha = \frac{9}{4}.\)
9. The final calculation required is \( 8\alpha \): \(8\alpha = 8 \times \frac{9}{4} = 18.\)

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