Question

Let \( f \) be a differentiable function in the interval \( (0, \infty) \) such that \( f(1) = 1 \) and \(\lim_{t \to x} \frac{t^2 f(x) - x^2 f(t)}{t - x} = 1\) for each \( x > 0 \).
Then \( 2f(2) + 3f(3) \) is equal to _________.

Answer 1

Correct Answer: 24

Applying L'Hôpital's rule, we differentiate the numerator and the denominator:

\[ \lim_{t \to x} \frac{t^2 f(x) - x^2 f(t)}{t - x} = 1 \] \[ \lim_{t \to x} \frac{2t f(x) - x^2 f'(x)}{1} = 1 \]

Simplifying, we get:

\[ 2x f(x) - x^2 f'(x) = 1 \]

Rearranging terms:

\[ \frac{dy}{dx} = \frac{2}{x} \quad \text{where } y = -\frac{1}{x^2} \]

The differential equation becomes:

\[ \frac{dy}{dx} + \frac{y}{x} = \frac{2}{x^2} \]

Solving this differential equation, we assume:

\[ y = \frac{1}{3x} + \frac{2x^2}{3}, \quad \text{leading to: } y = \frac{2x^3 + 1}{3x} \]

Calculating specific values:

\[ f(2) = \frac{17}{6}, \quad f(3) = \frac{55}{9} \]

Thus: \[ 2f(2) + 3f(3) = \frac{17}{3} + \frac{55}{3} = \frac{72}{3} = 24 \]