Question:

Step 1: Define an auxiliary function to apply Rolle's theorem. We define $g(x) = \frac{f(x)}{x}$. Since $a > 0$, the function $g$ is well-defined on the interval $[a, b]$. Step 2: Verify that the auxiliary function satisfies the conditions of Rolle's theorem. Since $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, and since $x > 0$ on this interval, the function $g(x) = \frac{f(x)}{x}$ is also continuous on $[a, b]$ and differentiable on $(a, b)$. Step 3: Check the boundary condition for Rolle's theorem. We are given that $\dfrac{f(a)}{a} = \dfrac{f(b)}{b}$. This means: $$g(a) = g(b)$$ Step 4: Apply Rolle's theorem. Since $g$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $g(a) = g(b)$, by Rolle's theorem, there exists at least one point $x_0 \in (a, b)$ such that: $$g'(x_0) = 0$$ Step 5: Compute the derivative of $g(x)$ and evaluate at $x_0$. Using the quotient rule: $$g'(x) = \frac{d}{dx}\left(\frac{f(x)}{x}\right) = \frac{x \cdot f'(x) - f(x) \cdot 1}{x^2} = \frac{xf'(x) - f(x)}{x^2}$$ Setting $g'(x_0) = 0$: $$\frac{x_0 f'(x_0) - f(x_0)}{x_0^2} = 0$$ Since $x_0 > 0$, we have $x_0^2 \neq 0$, so the numerator must equal zero: $$x_0 f'(x_0) - f(x_0) = 0$$ Therefore: $$x_0 f'(x_0) = f(x_0)$$ **Final Answer:** There exists $x_0 \in (a, b)$ such that $x_0 f'(x_0) = f(x_0)$, which corresponds to **Option 1**.