Question Id : 17975 |
Context : JEE Main 2026 (22 January Evening Shift)
Let $[\,\cdot\,]$ denote the greatest integer function, and let $f(x)=\min\{\sqrt{2}\,x,x^2\}$. Let $S=\{x\in(-2,2):\text{ the function } g(x)=|x|[x^2] \text{ is discontinuous at }x\}$. Then $\sum_{x\in S} f(x)$ equals :
🎥 Video solution / Text Solution of this question is given below:
The function $[x^2]$ changes value when $x^2$ crosses an integer. In $(-2,2)$, discontinuities occur at $x=\pm1$. Hence $S=\{-1,1\}$. Now $f(-1)=\min\{-\sqrt2,1\}=-\sqrt2$ and $f(1)=\min\{\sqrt2,1\}=1$. Therefore $\sum_{x\in S}f(x)=1-\sqrt2$.