JEE MAIN - Mathematics (2024 - 30th January Evening Shift - No. 6)
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 $$f^{\prime}(1)=2024$$, then
$$x f^{\prime}(x)+2024 f(x)=0$$
$$x f^{\prime}(x)-2023 f(x)=0$$
$$x f^{\prime}(x)-2024 f(x)=0$$
$$x f^{\prime}(x)+f(x)=2024$$
Explanation
$$f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$$
$$\begin{aligned} & \mathrm{f}^{\prime}(1)=2024 \\ & \mathrm{f}(1)=1 \end{aligned}$$
Partially differentiating w. r. t. x
$$\begin{aligned} & \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{\mathrm{y}}\right) \cdot \frac{1}{\mathrm{y}}=\frac{1}{\mathrm{f}(\mathrm{y})} \mathrm{f}^{\prime}(\mathrm{x}) \\ & \mathrm{y} \rightarrow \mathrm{x} \\ & \mathrm{f}^{\prime}(1) \cdot \frac{1}{\mathrm{x}}=\frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})} \\ & 2024 \mathrm{f}(\mathrm{x})=\mathrm{xf}^{\prime}(\mathrm{x}) \Rightarrow \mathrm{xf}^{\prime}(\mathrm{x})-2024 \mathrm{~f}(\mathrm{x})=0 \end{aligned}$$
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