Let f: R - 0 R be a function satisfying f ( xy) = f(x)f(y) for all x, y with f(y) 0. If f'(1) = 2024, then:

Let f: R - 0 R be a function satisfying f ( xy) = f(x)f(y) for al… — Solution | JEE Mathematics

MCQMediumJEE 2024Linear Differential Equations

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

A
$xf'(x) - 2024f(x) = 0$
B
$xf(x) + 2024f(x) = 0$
C
$f(x) + xf'(x) = 2024$
D
$xf(x) - 2023f(x) = 0$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: $f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)}$ and $f'(1) = 2024$ .

Find: Which differential equation is satisfied by $f(x)$ .

From the functional equation, take the standard power-form assumption used in the solution:

f(x) = x^k

Then

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

so this form satisfies the given relation.

Differentiate:

f'(x) = kx^{k-1}

Using $f'(1) = 2024$ ,

f'(1) = k = 2024

Hence,

f(x) = x^{2024}

and

f'(x) = 2024x^{2023}

Now test option A:

xf'(x) - 2024f(x) = x\big(2024x^{2023}\big) - 2024x^{2024} = 0

Therefore, the correct option is A.

Option Verification

Given: $f(x) = x^{2024}$ obtained from the solution working.

Find: Which of the four options is true.

Substitute $f(x) = x^{2024}$ and $f'(x) = 2024x^{2023}$ into each option:

\begin{aligned} \text{A: } xf'(x) - 2024f(x) &= x(2024x^{2023}) - 2024x^{2024} = 0 \\ \text{B: } xf(x) + 2024f(x) &= xx^{2024} + 2024x^{2024} = (x+2024)x^{2024} \neq 0 \\ \text{C: } f(x) + xf'(x) &= x^{2024} + x(2024x^{2023}) = 2025x^{2024} \neq 2024 \\ \text{D: } xf(x) - 2023f(x) &= xx^{2024} - 2023x^{2024} = (x-2023)x^{2024} \neq 0 \end{aligned}

Only option A matches.

Pattern Recognition

Given: the solution identifies the function as a power form.

Find: The matching differential equation quickly.

If $f(x) = x^k$ , then

f'(x) = kx^{k-1}

so multiplying by $x$ gives

xf'(x) = kx^k = kf(x)

Hence every such function satisfies

xf'(x) - kf(x) = 0

Since $f'(1) = k = 2024$ , we get

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

So the correct option is A.

Common mistakes

Assuming the condition directly gives a logarithmic function. That is incorrect because the given ratio form $f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$ matches multiplicative power-type behavior here. Use the form $f(x)=x^k$ as done in the solution.
Using $f'(1)=2024$ incorrectly as $f(1)=2024$ . The condition is about the derivative at $x=1$ , so after differentiating $f(x)=x^k$ , substitute into $f'(1)=k=2024$ .
Checking the options with the wrong derivative. For $f(x)=x^{2024}$ , the derivative is $f'(x)=2024x^{2023}$ , not $2024x^{2024}$ . Differentiate first, then substitute carefully.

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Answer

Step-by-step solution

Common mistakes

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