JEE Main 2024 — 27 January, Shift 2 — Question 8

Given the function:

$$g(x) = 3 f\left(\frac{x}{3}\right) + f(3 - x)$$

And the condition that $f''(x) > 0$ for all $x \in (0, 3)$, which implies that $f'(x)$ is a strictly increasing function.

Finding the Derivative Differentiating $g(x)$ with respect to $x$ using the chain rule:

$$g'(x) = 3 \cdot f'\left(\frac{x}{3}\right) \cdot \frac{1}{3} + f'(3 - x) \cdot (-1)$$ $$g'(x) = f'\left(\frac{x}{3}\right) - f'(3 - x)$$

Finding the Critical Point $\alpha$ The function $g$ changes from decreasing to increasing at $x = \alpha$. At this point, $g'(\alpha) = 0$:

$$f'\left(\frac{\alpha}{3}\right) - f'(3 - \alpha) = 0$$ $$f'\left(\frac{\alpha}{3}\right) = f'(3 - \alpha)$$

Since $f'(x)$ is strictly increasing (one-to-one), we can equate the arguments:

$$\frac{\alpha}{3} = 3 - \alpha$$

Solving for $\alpha$ Multiplying by 3:

$$\alpha = 9 - 3\alpha$$ $$4\alpha = 9$$ $$\alpha = \frac{9}{4}$$

Final Calculation The question asks for the value of $8\alpha$:

$$8\alpha = 8 \times \frac{9}{4}$$ $$8\alpha = 2 \times 9$$ $$8\alpha = 18$$