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

Answer 2

Step 1: Interpret the given limit as a differential relation
For each \( x > 0 \), \[ \lim_{t \to x} \frac{t^2 f(x) - x^2 f(t)}{t - x} = 1. \] Rewrite the numerator by adding and subtracting \(x^2 f(x)\): \[ t^2 f(x) - x^2 f(t) = \big(t^2 f(x) - x^2 f(x)\big) + \big(x^2 f(x) - x^2 f(t)\big) = f(x)(t^2 - x^2) - x^2\big(f(t) - f(x)\big). \] Divide by \(t - x\) and take \(t \to x\): \[ \lim_{t \to x} \frac{t^2 f(x) - x^2 f(t)}{t - x} = \lim_{t \to x} \left[ f(x)\frac{t^2 - x^2}{t - x} - x^2 \frac{f(t) - f(x)}{t - x} \right] = f(x)\cdot(2x) - x^2 f'(x). \] Thus, for all \( x > 0 \), \[ 2x\,f(x) - x^2 f'(x) = 1. \]

Step 2: Convert to a linear first-order ODE and solve
Rearrange: \[ x^2 f'(x) - 2x f(x) = -1. \] Divide by \(x^2\) \((x > 0)\): \[ f'(x) - \frac{2}{x} f(x) = -\frac{1}{x^2}. \] This is linear with integrating factor \[ \mu(x) = e^{\int -\tfrac{2}{x}\,dx} = e^{-2\ln x} = x^{-2}. \] Then \[ \frac{d}{dx}\big(\mu f\big) = \mu \left(f' - \frac{2}{x} f\right) = x^{-2}\left(-\frac{1}{x^2}\right) = -x^{-4}. \] Integrate: \[ \mu f = \int -x^{-4}\,dx = \frac{1}{3}x^{-3} + C \quad \Longrightarrow \quad x^{-2} f(x) = \frac{1}{3}x^{-3} + C. \] Hence \[ f(x) = \frac{1}{3}x^{-1} + Cx^2. \]

Step 3: Use the initial condition \( f(1) = 1 \)
\[ 1 = f(1) = \frac{1}{3}\cdot 1 + C \cdot 1^2 \quad \Longrightarrow \quad C = \frac{2}{3}. \] Therefore \[ f(x) = \frac{1}{3x} + \frac{2}{3}x^2. \]

Step 4: Evaluate \( 2f(2) + 3f(3) \)
\[ f(2) = \frac{1}{3\cdot 2} + \frac{2}{3}\cdot 4 = \frac{1}{6} + \frac{8}{3} = \frac{17}{6},\qquad f(3) = \frac{1}{3\cdot 3} + \frac{2}{3}\cdot 9 = \frac{1}{9} + 6 = \frac{55}{9}. \] Thus \[ 2f(2) + 3f(3) = 2\cdot \frac{17}{6} + 3\cdot \frac{55}{9} = \frac{17}{3} + \frac{165}{9} = \frac{51}{9} + \frac{165}{9} = \frac{216}{9} = 24. \]

Final answer

24