Q60 / 90 · JEE Main 2024 (06 Apr Shift 2) · Mathematics
Let $[\mathrm{t}]$ denote the greatest integer less than or equal to $\mathrm{t}$. Let $f:[0, \infty) \rightarrow \mathbf{R}$ 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 $\sum_{\mathrm{a} \in \mathrm{S}} \mathrm{a}$ is equal to _______
Correct answer: 17
Solution
$\left[\frac{x}{2}+3\right]$ is discontinuous at $x=2,4,6,8$
$\sqrt{\mathrm{x}}$ is discontinuous at $\mathrm{x}=1,4$ $\mathrm{F}(\mathrm{x})$ is discontinuous at $\mathrm{x}=1,2,6,8$
$\sum \mathrm{a}=1+2+6+8=17$
Hence, the correct answer is 17.