Q38 / 75 · JEE Main 2026 (22 January Shift 2) · Mathematics
Let $[\cdot]$ denote the greatest integer function, and let $f(x)=\min \left\{\sqrt{2} x, x^{2}\right\}$. Let $\mathrm{S}=\left\{x \in(-2,2)\right.$ : the function $\mathrm{g}(x)=|x|\left[x^{2}\right]$ is discontinuous at $\left.x\right\}$. Then $\sum_{x \in S} f(x)$ equals
- A$\sqrt{6}-2 \sqrt{2}$
- Correct answerB$1-\sqrt{2}$
- C$2-\sqrt{2}$
- D$2 \sqrt{6}-3 \sqrt{2}$
Solution
$\mathrm{g}(\mathrm{x})=|\mathrm{x}|\left[\mathrm{x}^2\right]$
Points of discontinuity of $\mathrm{g}(\mathrm{x})$ in $(-2,2)$ are
$( \pm 1, \pm \sqrt{2}, \pm \sqrt{3})$
$\therefore S=\{-1,1,-\sqrt{2}, \sqrt{2},-\sqrt{3}, \sqrt{3}\}$
$\because f(x)=\min \left\{\sqrt{2} x, x^2\right\}$
$\therefore \sum_{x \in S} f(x)=-\sqrt{2}+1-2+2-\sqrt{6}+\sqrt{6}$
$=1-\sqrt{2}$
Points of discontinuity of $\mathrm{g}(\mathrm{x})$ in $(-2,2)$ are
$( \pm 1, \pm \sqrt{2}, \pm \sqrt{3})$
$\therefore S=\{-1,1,-\sqrt{2}, \sqrt{2},-\sqrt{3}, \sqrt{3}\}$
$\because f(x)=\min \left\{\sqrt{2} x, x^2\right\}$
$\therefore \sum_{x \in S} f(x)=-\sqrt{2}+1-2+2-\sqrt{6}+\sqrt{6}$
$=1-\sqrt{2}$
Hence, the correct answer is option: 'B'.