JEE Main 2020 (Online) 4th September Evening ShiftQuestion 1
Let $f:(0,\infty)\to(0,\infty)$ be a differentiable function such that $f(1)=e$ and $\lim_{t\to x}\frac{t^2f^2(x)-x^2f^2(t)}{t-x}=0$. If $f(x)=1$, then $x$ is equal to :
Differentiation and its Application
Solution
Concept. The limit is the derivative of $t^2f^2(x)-x^2f^2(t)$ in $t$ at $t=x$; setting it to $0$ gives a simple equation for $f$. Formulas used. (1) $\lim_{t\to x}\dfrac{F(t)-F(x)}{t-x}=F'(x)$Step 1 - Equation. With $F(t)=t^2f^2(x)-x^2f^2(t)$ and $F(x)=0$, by (1): $$2xf^2(x)-2x^2f(x)f'(x)=0\ \Longrightarrow\ \frac{f'(x)}{f(x)}=\frac1x.$$ Step 2 - Solve. $\ln f=\ln x+C$, so $f=cx$; $f(1)=e$ gives $f(x)=ex$. Step 3 - Answer. $ex=1$ gives $x=\frac1e$. Answer: (B)