Verified Solution • 2025–26 Session moderate QJEEM24Mac71a: JEE Main 2024 Mathematics - Class 12 JEEM24Mac71a Solution: Let be a function satisfying for all . If , then Let f:R−{0}→Rf: \mathbb{R}-\{0\} \rightarrow \mathbb{R}f:R−{0}→R be a function satisfying f(xy)=f(x)f(y)f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}f(yx)=f(y)f(x) for all x,y,f(y)≠0x, y, f(y) \neq 0x,y,f(y)=0 . If f′(1)=2024f^{\prime}(1)=2024f′(1)=2024 , then Multiple Choice Options A xf′(x)+2024f(x)=0x f^{\prime}(x)+2024 f(x)=0xf′(x)+2024f(x)=0 B xf′(x)−2023f(x)=0x f^{\prime}(x)-2023 f(x)=0xf′(x)−2023f(x)=0 C xf′(x)−2024f(x)=0x f^{\prime}(x)-2024 f(x)=0xf′(x)−2024f(x)=0 D xf′(x)+f(x)=2024x f^{\prime}(x)+f(x)=2024xf′(x)+f(x)=2024 Step-by-Step Verified Solution f(xy)=f(x)f(y)f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}f(yx)=f(y)f(x) amp;f′(1)=2024amp;f(1)=1\begin{aligned}& \mathrm{f}^{\prime}(1)=2024 \\& \mathrm{f}(1)=1\end{aligned}amp;f′(1)=2024amp;f(1)=1 Partially differentiating w. r. t. x amp;f′(xy)⋅1y=1f(y)f′(x)amp;y→xamp;f′(1)⋅1x=f′(x)f(x)amp;2024f(x)=xf′(x)⇒xf′(x)−2024 f(x)=0\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}amp;f′(yx)⋅y1=f(y)1f′(x)amp;y→xamp;f′(1)⋅x1=f(x)f′(x)amp;2024f(x)=xf′(x)⇒xf′(x)−2024 f(x)=0 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring