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JEE Main 2024
Differentiation
Differentiation
Hard

Question

Let f:R{0}Rf: \mathbb{R}-\{0\} \rightarrow \mathbb{R} be a function satisfying f(xy)=f(x)f(y)f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)} for all x,y,f(y)0x, y, f(y) \neq 0. If f(1)=2024f^{\prime}(1)=2024, then

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Solution

f(xy)=f(x)f(y)f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)} f(1)=2024f(1)=1\begin{aligned} & \mathrm{f}^{\prime}(1)=2024 \\ & \mathrm{f}(1)=1 \end{aligned} Partially differentiating w. r. t. x f(xy)1y=1f(y)f(x)yxf(1)1x=f(x)f(x)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}

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