JEE Main 2024 (Online) 30th January Evening ShiftQuestion 1
Let $f:\mathbb{R}-\{0\}\to\mathbb{R}$ be a function satisfying $f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$ for all $x,y$, $f(y)\neq0$. If $f'(1)=2024$, then
Differentiation and its Application
Solution
Concept. Differentiate $f\left(\frac xy\right)=\frac{f(x)}{f(y)}$ with respect to $x$, then put $x=y$. Formulas used. (1) Chain rule: $\dfrac{d}{dx}f\left(\frac xy\right)=\dfrac1y\,f'\left(\frac xy\right)$Step 1 - $f(1)$. Put $x=y$: $f(1)=1$. Step 2 - Differentiate in $x$. By (1), $$\frac1y\,f'\left(\frac xy\right)=\frac{f'(x)}{f(y)}.$$ Step 3 - Put $x=y$. $\frac{f'(1)}{y}=\frac{f'(y)}{f(y)}$, i.e. $y\,f'(y)=2024\,f(y)$ for every $y\ne0$. Step 4 - Conclude. $xf'(x)-2024f(x)=0$. Answer: (A)