Let $$f(x)$$ be a continuously differentiable function on the interval $$(0, \infty)$$ such that $$f(1) = 2$$ and $$\lim_{t \to x} \frac{t^{10}f(x) - x^{10}f(t)}{t^9 - x^9} = 1$$ for each $$x \gt 0$$. Then, for all $$x \gt 0$$, $$f(x)$$ is equal to

The limit condition is
$$\lim_{t \to x}\;\frac{t^{10}f(x)-x^{10}f(t)}{t^{9}-x^{9}} = 1 \qquad\forall\,x\gt0$$

For fixed $$x$$, treat the numerator and denominator as functions of the variable $$t$$:
  • $$g(t)=t^{10}f(x)-x^{10}f(t)$$ with $$g(x)=0$$
  • $$h(t)=t^{9}-x^{9}$$ with $$h(x)=0$$

Because both approach $$0$$ as $$t\to x$$, apply L’Hospital’s Rule (differentiate with respect to $$t$$):
$$\lim_{t\to x}\frac{g(t)}{h(t)} =\frac{g'(x)}{h'(x)}$$

Compute the derivatives:
$$g'(t)=10t^{9}f(x)-x^{10}f'(t)\;\;\Longrightarrow\;\;g'(x)=10x^{9}f(x)-x^{10}f'(x)$$
$$h'(t)=9t^{8}\;\;\Longrightarrow\;\;h'(x)=9x^{8}$$

The limit equals 1, so
$$\frac{10x^{9}f(x)-x^{10}f'(x)}{9x^{8}} = 1$$

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

Divide by $$x^{8}$$:
$$10x\,f(x)-x^{2}f'(x)=9$$

Re-arrange to a first-order linear differential equation:
$$f'(x)-\frac{10}{x}f(x)=-\frac{9}{x^{2}} \quad -(1)$$

Let integrating factor $$I(x)=e^{\int -10/x\,dx}=e^{-10\ln x}=x^{-10}$$.

Multiply $$-(1)$$ by $$x^{-10}$$:
$$x^{-10}f'(x)-\frac{10}{x}x^{-10}f(x)=-9x^{-12}$$

The left side is the derivative of $$x^{-10}f(x)$$:
$$\frac{d}{dx}\bigl(x^{-10}f(x)\bigr)=-9x^{-12}$$

Integrate:
$$x^{-10}f(x)=\int -9x^{-12}\,dx=-9\cdot\frac{x^{-11}}{-11}+C=\frac{9}{11}x^{-11}+C$$

Multiply by $$x^{10}$$ to get $$f(x)$$:
$$f(x)=x^{10}\Bigl(\frac{9}{11}x^{-11}+C\Bigr) =\frac{9}{11x}+Cx^{10}$$

Use the given value $$f(1)=2$$:
$$2=\frac{9}{11}+C\;\;\Longrightarrow\;\;C=2-\frac{9}{11}=\frac{13}{11}$$

Hence, for all $$x\gt0$$,
$$f(x)=\frac{9}{11x}+\frac{13}{11}x^{10}$$

Option B which is: $$\frac{9}{11x} + \frac{13}{11}x^{10}$$