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
- \( x f'(x) - 2024 f(x) = 0 \)
- \( x f'(x) + 2024 f(x) = 0 \)
- \( x f'(x) + f(x) = 2024 \)
- \( x f'(x) - 2023 f(x) = 0 \)
The Correct Option is A
Approach Solution - 1
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:
\[ xf'(x) - 2024f(x) = 0. \]
Approach Solution -2
To solve this problem, we need to analyze the given functional equation and the information provided about the derivative of the function.
The function \( f : \mathbb{R} - \{0\} \rightarrow \mathbb{R} \) satisfies:
\(f\left( \frac{x}{y} \right) = \frac{f(x)}{f(y)}\) for all \( x, y \), with condition \( f(y) \neq 0 \).
This is a property of logarithmic functions or power functions. Let's explore the possibility of \( f(x) \) being a power function.
Assume \( f(x) = x^k \), where \( k \) is a constant. Check if the functional equation holds:
For \( f\left(\frac{x}{y}\right) = \left(\frac{x}{y}\right)^k \), and \( \frac{f(x)}{f(y)} = \frac{x^k}{y^k} = \left(\frac{x}{y}\right)^k \),
both expressions are equal, confirming that \( f(x) = x^k \) is indeed a valid assumption.
Next, we use the additional condition \( f'(1) = 2024 \) to find the exponent \( k \).
The derivative of \( f(x) = x^k \) is \( f'(x) = k x^{k-1} \).
Given \( f'(1) = 2024 \), substitute \( x = 1 \):
\(f'(1) = k \cdot 1^{k-1} = k = 2024\)
Hence, \( k = 2024 \), so \( f(x) = x^{2024} \).
Now, let's verify which differential equation matches \( f(x) = x^{2024} \).
Substitute \( f(x) = x^{2024} \) into the given options:
- \(x^{2024} \Rightarrow f'(x) = 2024 x^{2023}\)
Checking each option:
- \( x f'(x) - 2024 f(x) = x (2024 x^{2023}) - 2024 x^{2024} = 2024 x^{2024} - 2024 x^{2024} = 0 \). This satisfies the equation.
- \( x f'(x) + 2024 f(x) = x (2024 x^{2023}) + 2024 x^{2024} = 2024 x^{2024} + 2024 x^{2024} = 4048 x^{2024} \neq 0\). This is incorrect.
- \( x f'(x) + f(x) = 2024 = x (2024 x^{2023}) + x^{2024} = 2025 x^{2024} \neq 2024\). This is incorrect.
- \( x f'(x) - 2023 f(x) = x (2024 x^{2023}) - 2023 x^{2024} = 2024 x^{2024} - 2023 x^{2024} = x^{2024} \neq 0\). This is incorrect.
Thus, the correct option is:
\( x f'(x) - 2024 f(x) = 0 \).
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