JEE Main 2022 (Online) 24th June Morning ShiftQuestion 1
The number of points where the function $f(x)=\begin{cases}|2x^{2}-3x-7| & \text{if } x\le-1\\ [4x^{2}-1] & \text{if } -1<x<1\\ |x+1|+|x-2| & \text{if } x\ge1\end{cases}$, $[t]$ denotes the greatest integer $\le t$, is discontinuous is $\_\_\_\_\_\_$ .
Limits & Continuity
Answer$7$
Solution
For $x\le-1$ and $x\ge1$ the pieces are continuous. At $x=-1$: $f(-1)=|2+3-7|=2$ and right limit $[4x^2-1]\to[3^-]=2$: continuous. On $(-1,1)$, $4x^2-1\in[-1,3)$; it equals an integer at $x=0$ (min, continuous) and $x=\pm\frac12,\pm\frac{\sqrt2}2,\pm\frac{\sqrt3}2$: $6$ discontinuities. At $x=1$: $f(1)=3$ but left limit $=2$: discontinuous. Total $7$.