JEE Main 2020 (Online) 9th January Evening ShiftQuestion 1
Let $[t]$ denote the greatest integer $\le t$ and $\lim_{x\to0}x\left[\frac4x\right]=A$. Then the function, $f(x)=[x^2]\sin(\pi x)$ is discontinuous, when $x$ is equal to :
Sets, Relations and Function
Solution
$\lim_{x\to0}x\left[\frac4x\right]$: $\frac4x-1<[\frac4x]\le\frac4x$, so the limit is $4$; $A=4$. $f(x)=[x^2]\sin\pi x$ is discontinuous at $x=\sqrt n$ ($n$ a positive integer, where $[x^2]$ jumps) only if $\sin(\pi\sqrt n)\ne0$, i.e. $\sqrt n$ not an integer. Options: $\sqrt A=2$, $\sqrt{A+5}=3$, $\sqrt{A+21}=5$ are integers (continuous); $\sqrt{A+1}=\sqrt5$ is discontinuous. Option (B).