Question

Let $f(x)$ be a differentiable function satisfying the equations $\lim_{t \to x} \dfrac{t^2 f(x)-x^2 f(t)}{t-x} = 3$ and $f(1)=2$. Find the value of $2f(2)$.

Hint:

Limits resembling derivative definitions often convert directly into differential equations.

Answer 1

Correct Answer: 23

Step 1: Simplify the given limit.
\[ \lim_{t \to x} \frac{t^2 f(x)-x^2 f(t)}{t-x} = 3 \] Rewriting, \[ \lim_{t \to x} \frac{x^2 f(t)-t^2 f(x)}{x-t} = 3 \] Step 2: Apply derivative definition.
\[ x^2 f'(x) - 2x f(x) = 3 \] Step 3: Form differential equation.
\[ f'(x) - \frac{2}{x}f(x) = \frac{3}{x^2} \] Step 4: Solve using integrating factor.
Integrating factor, \[ \text{I.F.} = e^{\int -\frac{2}{x} dx} = \frac{1}{x^2} \] \[ \frac{d}{dx}\left(\frac{f(x)}{x^2}\right) = \frac{3}{x^4} \] Step 5: Integrate.
\[ \frac{f(x)}{x^2} = \int \frac{3}{x^4} dx = -\frac{1}{x^3} + c \] \[ f(x) = cx^2 - \frac{1}{x} \] Step 6: Use given condition.
\[ f(1) = c - 1 = 2 \Rightarrow c = 3 \] \[ f(x) = 3x^2 - \frac{1}{x} \] Step 7: Find $2f(2)$.
\[ f(2) = 12 - \frac{1}{2} = \frac{23}{2} \] \[ 2f(2) = 23 \] Final conclusion.
The value of $2f(2)$ is 23.