JEE Main 2021 (Online) 16th March Evening ShiftQuestion 1
Let $\alpha\in\mathbf{R}$ be such that the function $f(x)=\begin{cases}\frac{\cos^{-1}\left(1-\{x\}^{2}\right)\sin^{-1}\left(1-\{x\}\right)}{\{x\}-\{x\}^{3}}, & x\neq0\\ \alpha, & x=0\end{cases}$ is continuous at $x=0$, where $\{x\}=x-[x]$, $[x]$ is the greatest integer less than or equal to $x$. Then :
Limits & Continuity
Solution
Right limit ($\{x\}=t\to0^+$): $\cos^{-1}(1-t^2)\sim\sqrt2t$, $\sin^{-1}1=\frac\pi2$, denominator $\sim t$: limit $\frac{\pi}{\sqrt2}$. Left limit ($x\to0^-$, $t\to1^-$): $\cos^{-1}0=\frac\pi2$, $\sin^{-1}(1-t)\sim1-t$, denominator $t(1-t)(1+t)\to2(1-t)$: limit $\frac\pi4$. The two differ, so no $\alpha$ makes $f$ continuous: option B.