JEE Main 2024 (Online) 6th April Evening ShiftQuestion 1
Let $[t]$ denote the greatest integer less than or equal to $t$. Let $f:[0,\infty)\to\mathbf{R}$ be a function defined by $f(x)=\left[\frac{x}{2}+3\right]-[\sqrt{x}]$. Let $S$ be the set of all points in the interval $[0,8]$ at which $f$ is not continuous. Then $\sum_{a\in S}a$ is equal to $\_\_\_\_\_\_$ .
Limits & Continuity
Answer$17$
Solution
Concept. $[\frac x2+3]=3+[\frac x2]$. List the jumps of each GIF and check where they cancel. Formulas used. (1) $[t+n]=[t]+n$ for integer $n$Step 1 - Jumps. On $[0,8]$: $[\frac x2]$ jumps at $x=2,4,6,8$; $[\sqrt x]$ jumps at $x=1,4$. Step 2 - Values. $f=3$ on $[0,1)$, $2$ on $[1,2)$, $3$ on $[2,6)$, $4$ on $[6,8)$, and $f(8)=3+4-2=5$.Step 3 - $x=4$. Both terms jump by $1$ and cancel: $f=3$ on both sides. Continuous. Step 4 - $x=8$. Left limit $4$, $f(8)=5$: discontinuous. Step 5. $S=\{1,2,6,8\}$, sum $=17$. Answer: 17