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JEE Main 202622 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2026 · 22 January · Evening Shift · Mathematics · Continuity and Differentiability

Let [·] denote the greatest integer function, and let f(x)= \ 2 x,...

Let [⋅][\cdot] denote the greatest integer function, and let f(x)=min⁡{2x,x2}.LetSf(x)=\min \left\{\sqrt{2} x, x^{2}\right\}. Let \mathrm{S}=\left\{x \in(-2,2)\right.:thefunction: the function\mathrm{g}(x)=∣x∣[x2](x)=|x|\left[x^{2}\right] is discontinuous at \left.x\right\}.Then. Then\sum_{x \in S} f(x)$ equals

Options

  1. A6−22\sqrt{6}-2 \sqrt{2}
  2. B1−21-\sqrt{2}
  3. C2−22-\sqrt{2}
  4. D26−322 \sqrt{6}-3 \sqrt{2}

Correct answer

B. 1−21-\sqrt{2}

Step-by-step solution

g(x)=∣x∣[x2]\mathrm{g}(\mathrm{x})=|\mathrm{x}|\left[\mathrm{x}^2\right]
Points of discontinuity of g(x)\mathrm{g}(\mathrm{x}) in (−2,2)(-2,2) are
(±1,±2,±3)( \pm 1, \pm \sqrt{2}, \pm \sqrt{3})
∴S={−1,1,−2,2,−3,3}\therefore S=\{-1,1,-\sqrt{2}, \sqrt{2},-\sqrt{3}, \sqrt{3}\}
∵f(x)=min⁡{2x,x2}\because f(x)=\min \left\{\sqrt{2} x, x^2\right\}
∴∑x∈Sf(x)=−2+1−2+2−6+6\therefore \sum_{x \in S} f(x)=-\sqrt{2}+1-2+2-\sqrt{6}+\sqrt{6}
=1−2=1-\sqrt{2}
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