`f(x/y)=f(x)/f(y)`and `f'(1)=2024`, then

Options

A. `xf'(x)-2024f(x)=0`
B. `2024f'(x)=f(x)`
C. `f'(x)-2024f(x)=0`
D. None of these

Answer

To solve the problem given by the functional equation \( f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)} \) and the condition \( f'(1) = 2024 \), we will follow these steps:

### Step 1: Analyze the functional equation
The functional equation \( f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)} \) suggests that \( f \) is a multiplicative function. A common form for such functions is \( f(x) = x^k \) for some constant \( k \).

### Step 2: Substitute \( y = 1 \)
Substituting \( y = 1 \) into the functional equation gives:
\[ f\left(\frac{x}{1}\right) = \frac{f(x)}{f(1)} \implies f(x) = \frac{f(x)}{f(1)} \]
This implies that \( f(1) = 1 \) (assuming \( f(x) \neq 0 \)).

### Step 3: Assume a form for \( f(x) \)
Given the form \( f(x) = x^k \), we can find \( f(1) \):
\[ f(1) = 1^k = 1 \]
This is consistent with our previous finding.

### Step 4: Differentiate \( f(x) \)
Now we differentiate \( f(x) = x^k \):
\[ f'(x) = kx^{k-1} \]
Evaluating this at \( x = 1 \):
\[ f'(1) = k \cdot 1^{k-1} = k \]
Given that \( f'(1) = 2024 \), we have:
\[ k = 2024 \]

### Step 5: Write the final form of \( f(x) \)
Thus, the function can be expressed as:
\[ f(x) = x^{2024} \]

### Conclusion
The value of \( f(x) \) is \( x^{2024} \).