Question

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

• A. \(\frac{31}{11x}-\frac{9}{11}x^{10}\)
• B. \(\frac{9}{11x}+\frac{13}{11}x^{10}\)
• C. \(\frac{-9}{11x}+\frac{31}{11}x^{10}\)
• D. \(\frac{13}{11x}+\frac{9}{11}x^{10}\)

Answer 1

The Correct Option is B

Given:

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

By simplifying the numerator and denominator:

\[ \lim_{{t \to x}} \frac{10t^9 f(x) - 10x^9 f(t)}{9x^8} = 1 \]

As \( t \to x \), we get:

\[ 10x^9 f(x) - x^{10} f'(x) = 9x^8 \]

Rearranging terms:

\[ xf'(x) - 10f(x) = -\frac{9}{x^2} \]

This is a linear differential equation of the form:

\[ f'(x) + \frac{10}{x} f(x) = -\frac{9}{x^2} \]

Integrating factor (I.F.):

\[ e^{\int \frac{10}{x}dx} = e^{-10 \ln x} = \frac{1}{x^{10}} \]

Multiplying through by I.F.:

\[ \frac{1}{x^{10}} f(x) = \int \frac{-9}{x^{12}} dx \] \[ \frac{1}{x^{10}} f(x) = \frac{9}{11x^{11}} + C \]

where \( C \) is the constant of integration.

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

Using \( f(1) = 2 \):

\[ 2 = \frac{9}{11} + C \] \[ C = 2 - \frac{9}{11} = \frac{13}{11} \]

Thus, the final function is:

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

Final Answer:

\[ \frac{9}{11x} + \frac{13}{11x^{10}} \]