The number of points of discontinuity of the function…
Solution
$\{0,1, \sqrt{2}, 2, \sqrt{6}, \sqrt{8}, \sqrt{10}, \sqrt{12}, \sqrt{14}, 4\}$
Continuous at $0^{+}$, continuous at $4^{-}$ $\left[\frac{x^2}{2}\right]=[\sqrt{x}]$, occurs at $x=\sqrt{2}$
$\Rightarrow$ Not continuous
Asked in: JEE Main 2025 (07 Apr Shift 1)
Practice more Continuity and Differentiability questions on Aicharya