Find Function from Limit Condition Involving Differentiable Function

To find the function $f(x)$, we start by evaluating the given limit for any $x > 0$:

$\lim_{t \to x} \frac{t^{10} f(x) - x^{10} f(t)}{t^9 - x^9} = 1$

As $t \to x$, the numerator approaches $x^{10}f(x) - x^{10}f(x) = 0$, and the denominator approaches $x^9 - x^9 = 0$. Since this is an indeterminacy of the form $\left[\frac{0}{0}\right]$ and $f(x)$ is continuously differentiable, we can apply L'Hôpital's Rule with respect to $t$:

$\lim_{t \to x} \frac{\frac{d}{dt}\left(t^{10} f(x) - x^{10} f(t)\right)}{\frac{d}{dt}\left(t^9 - x^9\right)} = 1$

$\lim_{t \to x} \frac{10t^9 f(x) - x^{10} f'(t)}{9t^8} = 1$

Substituting $t = x$ into the simplified expression gives:

$\frac{10x^9 f(x) - x^{10} f'(x)}{9x^8} = 1$

Multiply both sides by $9x^8$:

$10x^9 f(x) - x^{10} f'(x) = 9x^8$

Divide through by $x^9$ (since $x > 0$):

$10 f(x) - x f'(x) = \frac{9}{x}$

Rearranging into the standard form of a first-order linear differential equation:

$f'(x) - \frac{10}{x} f(x) = -\frac{9}{x^2}$

To solve this linear differential equation, we find the Integrating Factor ($\text{I.F.}$):

$\text{I.F.} = e^{\int -\frac{10}{x} \, dx} = e^{-10 \ln x} = x^{-10}$

Multiplying the differential equation by the integrating factor gives:

$\frac{d}{dx} \left[ f(x) \cdot x^{-10} \right] = -\frac{9}{x^2} \cdot x^{-10} = -9x^{-12}$

Integrating both sides with respect to $x$:

$f(x) \cdot x^{-10} = \int -9x^{-12} \, dx + C$

$f(x) \cdot x^{-10} = -9 \left( \frac{x^{-11}}{-11} \right) + C$

$f(x) \cdot x^{-10} = \frac{9}{11} x^{-11} + C$

Multiplying through by $x^{10}$:

$f(x) = \frac{9}{11x} + C x^{10}$

Using the given initial condition $f(1) = 2$:

$f(1) = \frac{9}{11(1)} + C(1)^{10} = 2$

$\frac{9}{11} + C = 2 \implies C = 2 - \frac{9}{11} = \frac{13}{11}$

Thus, the function $f(x)$ is:

$f(x) = \frac{9}{11x} + \frac{13}{11}x^{10}$

This matches Option (B).