Question

Let \( f : \mathbb{R} - \{0\} \rightarrow \mathbb{R} \) be a function satisfying\[f\left( \frac{x}{y} \right) = \frac{f(x)}{f(y)}\]for all \( x, y \), \( f(y) \neq 0 \). If \( f'(1) = 2024 \),then

• A. \( x f'(x) - 2024 f(x) = 0 \)
• B. \( x f'(x) + 2024 f(x) = 0 \)
• C. \( x f'(x) + f(x) = 2024 \)
• D. \( x f'(x) - 2023 f(x) = 0 \)

Answer 1

The Correct Option is A

The functional equation given is:

\[ f\left( \frac{x}{y} \right) = \frac{f(x)}{f(y)}. \]

This suggests an exponential-like function. Assume \( f(x) = x^k \) for some constant \( k \).

Verify that \( f(x) = x^k \) satisfies the functional equation:

\[ f\left( \frac{x}{y} \right) = \left( \frac{x}{y} \right)^k = \frac{x^k}{y^k} = \frac{f(x)}{f(y)}. \]

So, \( f(x) = x^k \) is a valid solution.

Differentiating \( f(x) = x^k \):

\[ f'(x) = kx^{k-1}. \]

Given \( f'(1) = 2024 \):

\[ f'(1) = k = 2024. \]

Therefore, \( f(x) = x^{2024} \).

Check which option matches: Substitute \( f(x) = x^{2024} \) and \( f'(x) = 2024x^{2023} \) in each option.

Option (1):

\[ xf'(x) - 2024f(x) = x \times 2024x^{2023} - 2024x^{2024} = 0. \]

Therefore, the correct answer is:

\[ xf'(x) - 2024f(x) = 0. \]