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

Answer: (A)

$x f'(x) - 2024 f(x) = 0$

Answer 2

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:

$xf'(x) - 2024f(x) = 0.$