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Mathematics · Limits, Continuity and Differentiability

JEE Main 2026 — 22 January, Evening Shift — Question 15

Let [∙][\bullet] denote the greatest integer function, and let f(x)=min⁡{2x,x2}f(x)=\min \left\{\sqrt{2} x, x^{2}\right\}. Let S={x∈(−2,2):S=\{x \in(-2,2): the function g(x)=∣x∣[x2]\mathrm{g}(\mathrm{x})=|\mathrm{x}|\left[\mathrm{x}^{2}\right] is discontinuous at x}\}.Then ∑x∈Sf(x)\sum_{x \in S} f(x) equals :

  1. Option A:

    2−22-\sqrt{2}

  2. Option B:

    26−322 \sqrt{6}-3 \sqrt{2}

  3. Option C:

    1−21-\sqrt{2}

    Correct
  4. Option D:

    6−22\sqrt{6}-2 \sqrt{2}

Answer: C

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 \mathrm{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}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
22 January, Evening Shift
Subject
Mathematics
Chapter
Limits, Continuity and Differentiability
Topic
Continuity

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