JEE Main 2025 (Online) 7th April Morning ShiftQuestion 1
The number of points of discontinuity of the function $f(x)=\left[\frac{x^{2}}{2}\right]-\left[\sqrt{x}\right]$, $x\in[0,4]$, where $[\cdot]$ denotes the greatest integer function, is $\_\_\_\_\_\_$ .
Limits & Continuity
Answer$8$
Solution
Concept. List the jump points of each GIF term; a point is a discontinuity unless the jumps cancel. Formulas used. (1) $[u(x)]$ jumps where $u(x)$ crosses an integer ($u$ increasing)Step 1 - Jumps of $[\frac{x^2}2]$. On $[0,4]$, $\frac{x^2}2$ runs from $0$ to $8$. It equals $m=1,\dots,8$ at $x=\sqrt{2m}$: $$\sqrt2,\,2,\,\sqrt6,\,2\sqrt2,\,\sqrt{10},\,2\sqrt3,\,\sqrt{14},\,4$$ Step 2 - Jumps of $[\sqrt x]$. $\sqrt x\in[0,2]$ equals $1,2$ at $x=1,4$. Step 3 - Common point $x=4$. It is the right end, so only the left limit matters: $\lim_{x\to4^-}f=7-1=6$ and $f(4)=8-2=6$. Continuous. Step 4 - $x=0$. Both terms are $0$ near $0^+$: continuous. Step 5 - Count. The $7$ points from Step 1 (other than $4$) plus $x=1$: $7+1=8$. Answer: 8