JEE Main 2024 (Online) 1st February Morning ShiftQuestion 1
Let $\{x\}$ denote the fractional part of $x$ and $f(x)=\frac{\cos^{-1}\left(1-\{x\}^{2}\right)\sin^{-1}(1-\{x\})}{\{x\}-\{x\}^{3}}$, $x\neq0$. If $L$ and $R$ respectively denotes the left hand limit and the right hand limit of $f(x)$ at $x=0$, then $\frac{32}{\pi^{2}}\left(L^{2}+R^{2}\right)$ is equal to $\_\_\_\_\_\_$ .
Limits & Continuity
Answer$18$
Solution
Concept. $\{x\}\to0^+$ as $x\to0^+$ and $\{x\}\to1^-$ as $x\to0^-$. Formulas used. (1) $\cos^{-1}(1-u)\sim\sqrt{2u}$ as $u\to0^+$(2) $\sin^{-1}s\sim s$, $\sin^{-1}1=\frac\pi2$, $\cos^{-1}0=\frac\pi2$Step 1 - Right limit. $t=\{x\}=x\to0^+$. By (1), $\cos^{-1}(1-t^2)\sim\sqrt2\,t$; $\sin^{-1}(1-t)\to\frac\pi2$; $t-t^3\sim t$. $$R=\frac{\sqrt2\,t\cdot\frac\pi2}{t}=\frac{\pi}{\sqrt2},\quad R^2=\frac{\pi^2}2$$ Step 2 - Left limit. $t=\{x\}=1+x$; write $t=1-s$, $s\to0^+$. Then $\cos^{-1}(1-t^2)\to\cos^{-1}0=\frac\pi2$; $\sin^{-1}(1-t)=\sin^{-1}s\sim s$; $t-t^3=t(1-t)(1+t)\sim2s$. $$L=\frac{\frac\pi2\cdot s}{2s}=\frac\pi4,\quad L^2=\frac{\pi^2}{16}$$ Step 3. $L^2+R^2=\frac{9\pi^2}{16}$, so $\frac{32}{\pi^2}\cdot\frac{9\pi^2}{16}=18$. Answer: 18