JEE Main 202430 Jan, Evening shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002Let 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, thenAOption A. xf′(x)+2024f(x)=0x f^{\prime}(x)+2024 f(x)=0xf′(x)+2024f(x)=0BOption B. xf′(x)−2023f(x)=0x f^{\prime}(x)-2023 f(x)=0xf′(x)−2023f(x)=0COption C. xf′(x)−2024f(x)=0x f^{\prime}(x)-2024 f(x)=0xf′(x)−2024f(x)=0DOption D. xf′(x)+f(x)=2024x f^{\prime}(x)+f(x)=2024xf′(x)+f(x)=2024CheckSkip · show solution
