Let f: R-{0} rightarrow R be a function satis | vedclass

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(A) Given the functional equation $f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$.Setting $x=y=1$,we get $f(1) = \frac{f(1)}{f(1)} = 1$.Now,differentiat

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$xf^{\prime}(x)-2024 f(x)=0$

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(A) Given the functional equation $f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$.Setting $x=y=1$,we get $f(1) = \frac{f(1)}{f(1)} = 1$.Now,differentiate the given equation partially with respect to $x$:$f^{\prime}\left(\frac{x}{y}\right) \cdot \frac{1}{y} = \frac{f^{\prime}(x)}{f(y)}$.Substitute $y=x$ into the equation:$f^{\prime}(1) \cdot \frac{1}{x} = \frac{f^{\prime}(x)}{f(x)}$.Since $f^{\prime}(1) = 2024$,we have:$2024 \cdot \frac{1}{x} = \frac{f^{\prime}(x)}{f(x)}$.Rearranging the terms,we get:$x f^{\prime}(x) = 2024 f(x)$,which implies $x f^{\prime}(x) - 2024 f(x) = 0$.

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

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